Hamming numbers

Hamming numbers
You are encouraged to solve this task according to the task description, using any language you may know.

Hamming numbers are numbers of the form  

    H = 2i × 3j × 5k
           where 
     i,  j,  k  ≥  0 

Hamming numbers   are also known as   ugly numbers   and also   5-smooth numbers   (numbers whose prime divisors are less or equal to 5).

Task

Generate the sequence of Hamming numbers, in increasing order.   In particular:

1. Show the   first twenty   Hamming numbers.
2. Show the   1691^st   Hamming number (the last one below   2³¹).
3. Show the   one million^th   Hamming number (if the language – or a convenient library – supports arbitrary-precision integers).

Related tasks

• Humble numbers
• N-smooth numbers

References

• Wikipedia entry:   Hamming numbers     (this link is re-directed to   Regular number).
• Wikipedia entry:   Smooth number
• OEIS entry:   A051037   5-smooth   or   Hamming numbers
• Hamming problem from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread here and here).

11l

Translation of: Python

<lang 11l>F hamming(limit)

  V h = [1] * limit
  V (x2, x3, x5) = (2, 3, 5)
  V i = 0
  V j = 0
  V k = 0

  L(n) 1 .< limit
     h[n] = min(x2, x3, x5)
     I x2 == h[n]
        i++
        x2 = 2 * h[i]
     I x3 == h[n]
        j++
        x3 = 3 * h[j]
     I x5 == h[n]
        k++
        x5 = 5 * h[k]
  R h.last

print((1..20).map(i -> hamming(i))) print(hamming(1691))</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000

360 Assembly

<lang 360asm>* Hamming numbers 12/03/2017 HAM CSECT

        USING  HAM,R13            base register
        B      72(R15)            skip savearea
        DC     17F'0'             savearea
        STM    R14,R12,12(R13)    save previous context
        ST     R13,4(R15)         link backward
        ST     R15,8(R13)         link forward
        LR     R13,R15            set addressability
        LA     R6,1               ii=1
      DO WHILE=(C,R6,LE,=F'20')   do ii=1 to 20
        BAL    R14,PRTHAM           call prtham
        LA     R6,1(R6)             ii++
      ENDDO    ,                  enddo ii
        LA     R6,1691            ii=1691
        BAL    R14,PRTHAM         call prtham
        L      R13,4(0,R13)       restore previous savearea pointer
        LM     R14,R12,12(R13)    restore previous context
        XR     R15,R15            rc=0
        BR     R14                exit

PRTHAM EQU * ---- prtham

        ST     R14,R14PRT         save return addr
        LR     R1,R6              ii
        XDECO  R1,XDEC            edit
        MVC    PG+2(4),XDEC+8     output ii
        LR     R1,R6              ii
        BAL    R14,HAMMING        call hamming(ii)
        XDECO  R0,XDEC            edit
        MVC    PG+8(10),XDEC+2    output hamming(ii)
        XPRNT  PG,L'PG            print buffer
        L      R14,R14PRT         restore return addr
        BR     R14           ---- return

HAMMING EQU * ---- hamming(ll)

        ST     R14,R14HAM         save return addr
        ST     R1,LL              ll
        MVC    HH,=F'1'           h(1)=1
        SR     R0,R0              0
        ST     R0,I               i=0
        ST     R0,J               j=0
        ST     R0,K               k=0
        MVC    X2,=F'2'           x2=2
        MVC    X3,=F'3'           x3=3
        MVC    X5,=F'5'           x5=5
        LA     R7,1               n=1
        L      R2,LL              ll
        BCTR   R2,0               -1
        ST     R2,LLM1            ll-1
      DO WHILE=(C,R7,LE,LLM1)     do n=1 to ll-1
        L      R4,X2                m=x2
      IF C,R4,GT,X3 THEN            if m>x3 then
        L      R4,X3                  m=x3
      ENDIF    ,                    endif
      IF C,R4,GT,X5 THEN            if m>x5 then
        L      R4,X5                  m=x5
      ENDIF    ,                    endif
        LR     R1,R7                n
        SLA    R1,2                 *4
        ST     R4,HH(R1)            h(n+1)=m
      IF C,R4,EQ,X2 THEN            if m=x2 then
        L      R1,I                   i
        LA     R1,1(R1)               i+1
        ST     R1,I                   i=i+1
        SLA    R1,2                   *4
        L      R2,HH(R1)              h(i+1)
        MH     R2,=H'2'               *2
        ST     R2,X2                  x2=2*h(i+1)
      ENDIF    ,                    endif
      IF C,R4,EQ,X3 THEN            if m=x3 then
        L      R1,J                   j
        LA     R1,1(R1)               j+1
        ST     R1,J                   j=j+1
        SLA    R1,2                   *4
        L      R2,HH(R1)              h(j+1)
        MH     R2,=H'3'               *3
        ST     R2,X3                  x3=3*h(j+1)
      ENDIF    ,                    endif
      IF C,R4,EQ,X5 THEN            if m=x5 then
        L      R1,K                   k
        LA     R1,1(R1)               k+1
        ST     R1,K                   k=k+1
        SLA    R1,2                   *4
        L      R2,HH(R1)              h(k+1)
        MH     R2,=H'5'               *5
        ST     R2,X5                  x5=5*h(k+1)
      ENDIF    ,                    endif
        LA     R7,1(R7)             n++
      ENDDO    ,                  enddo n
        L      R1,LL              ll
        SLA    R1,2               *4
        L      R0,HH-4(R1)        return h(ll) 
        L      R14,R14HAM         restore return addr
        BR     R14           ---- return

R14HAM DS A return addr of hamming R14PRT DS A return addr of print LL DS F ll LLM1 DS F ll-1 I DS F i J DS F j K DS F k X2 DS F x2 X3 DS F x3 X5 DS F x5 PG DC CL80'H(xxxx)=xxxxxxxxxx' XDEC DS CL12 temp

        LTORG                     positioning literal pool

HH DS 1691F array h(1691)

        YREGS
        END    HAM</lang>

Output:

H(   1)=         1
H(   2)=         2
H(   3)=         3
H(   4)=         4
H(   5)=         5
H(   6)=         6
H(   7)=         8
H(   8)=         9
H(   9)=        10
H(  10)=        12
H(  11)=        15
H(  12)=        16
H(  13)=        18
H(  14)=        20
H(  15)=        24
H(  16)=        25
H(  17)=        27
H(  18)=        30
H(  19)=        32
H(  20)=        36
H(1691)=2125764000

Ada

Works with: GNAT

GNAT provides the datatypes Integer, Long_Integer and Long_Long_Integer, which are not large enough to store hamming numbers. In this program, we represent them as the factors for each of the prime numbers 2, 3 and 5, and only convert them to a base-10 numbers for display. We use the gmp library binding part of GNATCOLL, though a simple 'pragma import' would be enough.

This version is very fast (20ms for the million-th hamming number), thanks to a good algorithm. We also do not manipulate large numbers directly (gmp lib), but only the factors of the prime.

It will fail to compute the billion'th number because we use an array of the stack to store all numbers. It is possible to get rid of this array though it will make the code slightly less readable.

<lang Ada> with Ada.Numerics.Generic_Elementary_Functions; with Ada.Text_IO; use Ada.Text_IO; with GNATCOLL.GMP.Integers; with GNATCOLL.GMP.Lib;

procedure Hamming is

  type Log_Type is new Long_Long_Float;
  package Funcs is new Ada.Numerics.Generic_Elementary_Functions (Log_Type);

  type Factors_Array is array (Positive range <>) of Positive;

  generic
     Factors : Factors_Array := (2, 3, 5);
     --  The factors for smooth numbers. Hamming numbers are 5-smooth.
  package Smooth_Numbers is
     type Number is private;
     function Compute (Nth : Positive) return Number;
     function Image (N : Number) return String;

  private
     type Exponent_Type is new Natural;
     type Exponents_Array is array (Factors'Range) of Exponent_Type;
     --  Numbers are stored as the exponents of the prime factors.

     type Number is record
        Exponents : Exponents_Array;
        Log       : Log_Type;
        --  The log of the value, used to ease sorting.
     end record;

     function "=" (N1, N2 : Number) return Boolean
       is (for all F in Factors'Range => N1.Exponents (F) = N2.Exponents (F));
  end Smooth_Numbers;

  package body Smooth_Numbers is
     One : constant Number := (Exponents => (others => 0), Log => 0.0);
     Factors_Log : array (Factors'Range) of Log_Type;

     function Image (N : Number) return String is
        use GNATCOLL.GMP.Integers, GNATCOLL.GMP.Lib;
        R, Tmp : Big_Integer;
     begin
        Set (R, "1");
        for F in Factors'Range loop
           Set (Tmp, Factors (F)'Image);
           Raise_To_N (Tmp, GNATCOLL.GMP.Unsigned_Long (N.Exponents (F)));
           Multiply (R, Tmp);
        end loop;
        return Image (R);
     end Image;

     function Compute (Nth : Positive) return Number is
        Candidates : array (Factors'Range) of Number;

        Values     : array (1 .. Nth) of Number;
        --  Will result in Storage_Error for very large values of Nth

        Indices    : array (Factors'Range) of Natural :=
           (others => Values'First);
        Current    : Number;
        Tmp        : Number;
     begin
        for F in Factors'Range loop
           Factors_Log (F) := Funcs.Log (Log_Type (Factors (F)));
           Candidates (F) := One;
           Candidates (F).Exponents (F) := 1;
           Candidates (F).Log := Factors_Log (F);
        end loop;

        Values (1) := One;

        for Count in 2 .. Nth loop
           --  Find next value (the lowest of the candidates)
           Current := Candidates (Factors'First);
           for F in Factors'First + 1 .. Factors'Last loop
              if Candidates (F).Log < Current.Log then
                 Current := Candidates (F);
              end if;
           end loop;

           Values (Count) := Current;

           --  Update the candidates. There might be several candidates with
           --  the same value
           for F in Factors'Range loop
              if Candidates (F) = Current then
                 Indices (F) := Indices (F) + 1;

                 Tmp := Values (Indices (F));
                 Tmp.Exponents (F) := Tmp.Exponents (F) + 1;
                 Tmp.Log := Tmp.Log + Factors_Log (F);

                 Candidates (F) := Tmp;
              end if;
           end loop;
        end loop;

        return Values (Nth);
     end Compute;
  end Smooth_Numbers;

  package Hamming is new Smooth_Numbers ((2, 3, 5));

begin

  for N in 1 .. 20 loop
     Put (" " & Hamming.Image (Hamming.Compute (N)));
  end loop;
  New_Line;

  Put_Line (Hamming.Image (Hamming.Compute (1691)));
  Put_Line (Hamming.Image (Hamming.Compute (1_000_000)));

end Hamming; </lang>

Output:

 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ALGOL 68

Works with: Algol 68 Genie 1.19.0

Hamming numbers are generated in a trivial iterative way as in the Python version below. This program keeps the series needed to generate the numbers as short as possible using flexible rows; on the downside, it spends considerable time on garbage collection. <lang algol68>PR precision=100 PR

MODE SERIES = FLEX [1 : 0] UNT, # Initially, no elements #

    UNT = LONG LONG INT; # A 100-digit unsigned integer #

PROC hamming number = (INT n) UNT: # The n-th Hamming number #

    CASE n
    IN 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 # First 10 in a table #
    OUT # Additional operators #
        OP MIN = (INT i, j) INT: (i < j | i | j), MIN = (UNT i, j) UNT: (i < j | i | j);
        PRIO MIN = 9;
        OP LAST = (SERIES h) UNT: h[UPB h]; # Last element of a series #
        OP +:= = (REF SERIES s, UNT elem) VOID:
           # Extend a series by one element, only keep the elements you need #
           (INT lwb = (i MIN j) MIN k, upb = UPB s; 
            REF SERIES new s = HEAP FLEX [lwb : upb + 1] UNT;
            (new s[lwb : upb] := s[lwb : upb], new s[upb + 1] := elem);
            s := new s
           );
        # Determine the n-th hamming number iteratively #
        SERIES h := 1, # Series, initially one element #
        UNT m2 := 2, m3 := 3, m5 := 5, # Multipliers #
        INT i := 1, j := 1, k := 1; # Counters #
        TO n - 1
        DO h +:= (m2 MIN m3) MIN m5;
           (LAST h = m2 | m2 := 2 * h[i +:= 1]);
           (LAST h = m3 | m3 := 3 * h[j +:= 1]);
           (LAST h = m5 | m5 := 5 * h[k +:= 1])
        OD;
        LAST h
    ESAC;

FOR k TO 20 DO print ((whole (hamming number (k), 0), blank)) OD; print ((newline, whole (hamming number (1 691), 0))); print ((newline, whole (hamming number (1 000 000), 0)))</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ALGOL W

Algol W only has 32 bit integers, so we just show the first 20 Hamming Numbers and Hamming number 1691.
This uses the algorithm from Dr Dobbs (as in the Python version). The Coffee Script solution has some notes on how it works. <lang algolw>begin

   % returns the minimum of a and b                    %
   integer procedure min ( integer value a, b ) ; if a < b then a else b;
   % find and print Hamming Numbers                        %
   % Algol W only supports 32-bit integers so we just find %
   % the 1691 32-bit Hamming Numbers                       %
   integer MAX_HAMMING;
   MAX_HAMMING := 1691;
   begin
       integer array H( 1 :: MAX_HAMMING );
       integer p2, p3, p5, last2, last3, last5;
       H( 1 ) := 1;
       last2  := last3 := last5 := 1;
       p2     := 2;
       p3     := 3;
       p5     := 5;
       for hPos := 2 until MAX_HAMMING do begin
           integer m;
           % the next Hamming number is the lowest of the next multiple of 2, 3, and 5 %
           m := min( min( p2, p3 ), p5 );
           H( hPos ) := m;
           if m = p2 then begin
               last2 := last2 + 1;
               p2    := 2 * H( last2 )
           end if_used_power_of_2 ;
           if m = p3 then begin
               last3 := last3 + 1;
               p3    := 3 * H( last3 )
           end if_used_power_of_3 ;
           if m = p5 then begin
               last5 := last5 + 1;
               p5    := 5 * H( last5 )
           end if_used_power_of_5 ;
       end for_hPos ;
       i_w := 1;
       s_w := 1;
       write( H( 1 ) );
       for i := 2 until 20 do writeon( H( i ) );
       write( H( MAX_HAMMING ) )
   end

end.</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

Arturo

<lang rebol>hamming: function [limit][

   if limit=1 -> return 1
   h: map 0..limit-1 'z -> 1
   x2: 2, x3: 3, x5: 5
   i: 0, j: 0, k: 0
   
   loop 1..limit-1 'n [
       set h n min @[x2 x3 x5]
       if x2 = h\[n] [
           i: i + 1
           x2: 2 * h\[i]
       ]
       if x3 = h\[n] [
           j: j + 1
           x3: 3 * h\[j]
       ]
       if x5 = h\[n] [
           k: k + 1
           x5: 5 * h\[k]
       ]
   ]
   last h

] print map 1..20 => hamming print hamming 1691 print hamming 1000000</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

ATS

<lang ATS> // // How to compile: // patscc -DATS_MEMALLOC_LIBC -o hamming hamming.dats //

1. include

"share/atspre_staload.hats"

fun min3 (

 A: arrayref(int, 3)

) : natLt(3) = i where {

 var x: int = A[0]
 var i: natLt(3) = 0
 val () = if A[1] < x then (x := A[1]; i := 1)
 val () = if A[2] < x then (x := A[2]; i := 2)

} (* end of [min3] *)

fun hamming {n:pos} (

 n: int(n)

) : int = let // var A = @[int](2, 3, 5) val A = $UNSAFE.cast{arrayref(int, 3)}(addr@A) var I = @[int](1, 1, 1) val I = $UNSAFE.cast{arrayref(int, 3)}(addr@I) val H = arrayref_make_elt<int> (i2sz(succ(n)), 0) val () = H[0] := 1 // fun loop{k:pos}

 (k: int(k)) : void =

( // if k < n then let

 val i = min3(A)
 val k =
 (
   if A[i] > H[k-1] then (H[k] := A[i]; k+1) else k
 ) : intBtwe(k, k+1)
 val ii = I[i]
 val () = I[i] := ii+1
 val ii = $UNSAFE.cast{natLte(n)}(ii)
 val () = if i = 0 then A[i] := 2*H[ii]
 val () = if i = 1 then A[i] := 3*H[ii]
 val () = if i = 2 then A[i] := 5*H[ii]

in

 loop(k)

end // end of [then] else () // end of [else] // ) (* end of [loop] *) // in

 loop (1); H[n-1]

end (* end of [hamming] *)

implement main0 () = { val () = loop(1) where { fun loop {n:pos} (

 n: int(n)

) : void = if n <= 20 then let

 val () =
 println! ("hamming(",n,") = ", hamming(n))

in

 loop(n+1)

end // end of [then] // end of [if] } (* end of [val] *) val n = 1691 val () = println! ("hamming(",n,") = ", hamming(n)) // } (* end of [main0] *) </lang>

Output:

hamming(1) = 1
hamming(2) = 2
hamming(3) = 3
hamming(4) = 4
hamming(5) = 5
hamming(6) = 6
hamming(7) = 8
hamming(8) = 9
hamming(9) = 10
hamming(10) = 12
hamming(11) = 15
hamming(12) = 16
hamming(13) = 18
hamming(14) = 20
hamming(15) = 24
hamming(16) = 25
hamming(17) = 27
hamming(18) = 30
hamming(19) = 32
hamming(20) = 36
hamming(1691) = 2125764000

AutoHotkey

<lang AutoHotKey>SetBatchLines, -1 Msgbox % hamming(1,20) Msgbox % hamming(1690) return

hamming(first,last=0) { if (first < 1) ans=ERROR

if (last = 0) last := first

i:=0, j:=0, k:=0

num1 := ceil((last * 20)**(1/3)) num2 := ceil(num1 * ln(2)/ln(3)) num3 := ceil(num1 * ln(2)/ln(5))

loop { H := (2**i) * (3**j) * (5**k) if (H > 0) ans = %H%`n%ans% i++ if (i > num1) { i=0 j++ if (j > num2) { j=0 k++ } } if (k > num3) break } Sort ans, N

Loop, parse, ans, `n, `r { if (A_index > last) break if (A_index < first) continue Output = %Output%`n%A_LoopField% }

return Output }</lang>

AWK

<lang AWK>

1. syntax: gawk -M -f hamming_numbers.awk

BEGIN {

   for (i=1; i<=20; i++) {
     printf("%d ",hamming(i))
   }
   printf("\n1691: %d\n",hamming(1691))
   printf("\n1000000: %d\n",hamming(1000000))
   exit(0)

} function hamming(limit, h,i,j,k,n,x2,x3,x5) {

   h[0] = 1
   x2 = 2
   x3 = 3
   x5 = 5
   for (n=1; n<=limit; n++) {
     h[n] = min(x2,min(x3,x5))
     if (h[n] == x2) { x2 = 2 * h[++i] }
     if (h[n] == x3) { x3 = 3 * h[++j] }
     if (h[n] == x5) { x5 = 5 * h[++k] }
   }
   return(h[limit-1])

} function min(x,y) {

   return((x < y) ? x : y)

} </lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

BASIC256

Translation of: FreeBASIC

<lang freebasic>print "The first 20 Hamming numbers are :" for i = 1 to 20

   print Hamming(i);" ";

next i

print print "H( 1691) = "; Hamming(1691) end

function min(a, b)

   if a < b then return a else return b

end function

function Hamming(limit)

   dim h(1000000)

   h[0] = 1
   x2 = 2 : x3 = 3 : x5 = 5
   i  = 0 : j  = 0 : k  = 0
   for n = 1 to limit
       h[n]  = min(x2, min(x3, x5))
       if x2 = h[n] then i += 1: x2 = 2 *h[i]
       if x3 = h[n] then j += 1: x3 = 3 *h[j]
       if x5 = h[n] then k += 1: x5 = 5 *h[k]
   next n
   return h[limit -1]

end function</lang>

BBC BASIC

<lang bbcbasic> @% = &1010

     FOR h% = 1 TO 20
       PRINT "H("; h% ") = "; FNhamming(h%)
     NEXT
     PRINT "H(1691) = "; FNhamming(1691)
     END
     
     DEF FNhamming(l%)
     LOCAL i%, j%, k%, n%, m, x2, x3, x5, h%()
     DIM h%(l%) : h%(0) = 1
     x2 = 2 : x3 = 3 : x5 = 5
     FOR n% = 1 TO l%-1
       m = x2
       IF m > x3 m = x3
       IF m > x5 m = x5
       h%(n%) = m
       IF m = x2 i% += 1 : x2 = 2 * h%(i%)
       IF m = x3 j% += 1 : x3 = 3 * h%(j%)
       IF m = x5 k% += 1 : x5 = 5 * h%(k%)
     NEXT
     = h%(l%-1)</lang>

Output:

H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36
H(1691) = 2125764000

Bc

<lang bc>cat hamming_numbers.bc define min(x,y) {

if (x < y) {

return x

} else {

return y

}

} define hamming(limit) {

i = 0
j = 0
k = 0
h[0] = 1
x2 = 2
x3 = 3
x5 = 5
for (n=1; n<=limit; n++) {
 h[n] = min(x2,min(x3,x5))
 if (h[n] == x2) { x2 = 2 * h[++i] }
 if (h[n] == x3) { x3 = 3 * h[++j] }
 if (h[n] == x5) { x5 = 5 * h[++k] }
}
return (h[limit-1])

} for (lab=1; lab<=20; lab++) {

hamming(lab)

} hamming(1691) hamming(1000000) quit </lang>

Output:

$ bc hamming_numbers.bc 
bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'. 
1
2
3
4
5
6
8
9
10
12
15
16
18
20
24
25
27
30
32
36
2125764000
51931278044838873608958984375000000000000000000000000000000000000000\
0000000000000000

Bracmat

Translation of: D

<lang bracmat>( ( hamming

 =   x2 x3 x5 n i j k min
   .   tbl$(h,!arg)        { This creates an array. Arrays are always global in Bracmat. }
     & 1:?(0$h)
     & 2:?x2
     & 3:?x3
     & 5:?x5
     & 0:?n:?i:?j:?k
     &   whl
       ' ( !n+1:<!arg:?n
         & !x2:?min
         & (!x3:<!min:?min|)
         & (!x5:<!min:?min|)
         & !min:?(!n$h)               { !n is index into array h }
         & (   !x2:!min
             & 2*!((1+!i:?i)$h):?x2
           |
           )
         & (   !x3:!min
             & 3*!((1+!j:?j)$h):?x3
           |
           )
         & (   !x5:!min
             & 5*!((1+!k:?k)$h):?x5
           |
           )
         )
     & !((!arg+-1)$h) (tbl$(h,0)&)    { We delete the array by setting its size to 0 }
 )

& 0:?I & whl'(!I+1:~>20:?I&put$(hamming$!I " ")) & out$ & out$(hamming$1691) & out$(hamming$1000000) );</lang>

Output:

1  2  3  4  5  6  8  9  10  12  15  16  18  20  24  25  27  30  32  36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

C

Using a min-heap to keep track of numbers. Does not handle big integers. <lang c>#include <stdio.h>

1. include <stdlib.h>

typedef unsigned long long ham;

size_t alloc = 0, n = 1; ham *q = 0;

void qpush(ham h) { int i, j; if (alloc <= n) { alloc = alloc ? alloc * 2 : 16; q = realloc(q, sizeof(ham) * alloc); }

for (i = n++; (j = i/2) && q[j] > h; q[i] = q[j], i = j); q[i] = h; }

ham qpop() { int i, j; ham r, t; /* outer loop for skipping duplicates */ for (r = q[1]; n > 1 && r == q[1]; q[i] = t) { /* inner loop is the normal down heap routine */ for (i = 1, t = q[--n]; (j = i * 2) < n;) { if (j + 1 < n && q[j] > q[j+1]) j++; if (t <= q[j]) break; q[i] = q[j], i = j; } }

return r; }

int main() { int i; ham h;

for (qpush(i = 1); i <= 1691; i++) { /* takes smallest value, and queue its multiples */ h = qpop(); qpush(h * 2); qpush(h * 3); qpush(h * 5);

if (i <= 20 || i == 1691) printf("%6d: %llu\n", i, h); }

/* free(q); */ return 0; }</lang>

Alternative

Standard algorithm. Numbers are stored as exponents of factors instead of big integers, while GMP is only used for display. It's much more efficient this way. <lang c>#include <stdio.h>

1. include <stdlib.h>
2. include <string.h>
3. include <math.h>
4. include <gmp.h>

/* number of factors. best be mutually prime -- duh. */

1. define NK 3
2. define MAX_HAM (1 << 24)
3. define MAX_POW 1024

int n_hams = 0, idx[NK] = {0}, fac[] = { 2, 3, 5, 7, 11};

/* k-smooth numbers are stored as their exponents of each factor;

   v is the log of the number, for convenience. */

typedef struct { int e[NK]; double v; } ham_t, *ham;

ham_t *hams, values[NK] = {{{0}, 0}}; double inc[NK][MAX_POW];

/* most of the time v can be just incremented, but eventually

* floating point precision will bite us, so better recalculate */

inline void _setv(ham x) { int i; for (x->v = 0, i = 0; i < NK; i++) x->v += inc[i][x->e[i]]; }

inline int _eq(ham a, ham b) { int i; for (i = 0; i < NK && a->e[i] == b->e[i]; i++);

return i == NK; }

ham get_ham(int n) { int i, ni; ham h;

n--; while (n_hams < n) { for (ni = 0, i = 1; i < NK; i++) if (values[i].v < values[ni].v) ni = i;

*(h = hams + ++n_hams) = values[ni];

for (ni = 0; ni < NK; ni++) { if (! _eq(values + ni, h)) continue; values[ni] = hams[++idx[ni]]; values[ni].e[ni]++; _setv(values + ni); } }

return hams + n; }

void show_ham(ham h) { static mpz_t das_ham, tmp; int i;

	mpz_init_set_ui(das_ham, 1);

mpz_init_set_ui(tmp, 1); for (i = 0; i < NK; i++) { mpz_ui_pow_ui(tmp, fac[i], h->e[i]); mpz_mul(das_ham, das_ham, tmp); } gmp_printf("%Zu\n", das_ham); }

int main() { int i, j; hams = malloc(sizeof(ham_t) * MAX_HAM);

for (i = 0; i < NK; i++) { values[i].e[i] = 1; inc[i][1] = log(fac[i]); _setv(values + i);

for (j = 2; j < MAX_POW; j++) inc[i][j] = j * inc[i][1]; }

printf(" 1,691: "); show_ham(get_ham(1691)); printf(" 1,000,000: "); show_ham(get_ham(1e6)); printf("10,000,000: "); show_ham(get_ham(1e7)); return 0; }</lang>

Output:

     1,691: 2125764000
 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
10,000,000: 16244105063830431823239 ..<a gadzillion digits>.. 000000000000000000000

C#

Translation of: D

<lang csharp>using System; using System.Numerics; using System.Linq;

namespace Hamming {

   class MainClass {

       public static BigInteger Hamming(int n) {
           BigInteger two = 2, three = 3, five = 5;
           var h = new BigInteger[n];
           h[0] = 1;
           BigInteger x2 = 2, x3 = 3, x5 = 5;
           int i = 0, j = 0, k = 0;
           
           for (int index = 1; index < n; index++) {
               h[index] = BigInteger.Min(x2, BigInteger.Min(x3, x5));
               if (h[index] == x2) x2 = two * h[++i];
               if (h[index] == x3) x3 = three * h[++j];
               if (h[index] == x5) x5 = five * h[++k];
           }
           return h[n - 1];
       }

       public static void Main(string[] args) {
           Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).ToList().Select(x => Hamming(x))));
           Console.WriteLine(Hamming(1691));
           Console.WriteLine(Hamming(1000000));
       }
   }

}</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Generic version for any set of numbers

The algorithm is similar to the one above. <lang csharp>using System; using System.Numerics; using System.Linq;

namespace Hamming {

   class MainClass {

       public static BigInteger[] Hamming(int n, int[] a) {
           var primes = a.Select(x => (BigInteger)x).ToArray();
           var values = a.Select(x => (BigInteger)x).ToArray();
           var indexes = new int[a.Length];
           var results = new BigInteger[n];
           results[0] = 1;
           for (int iter = 1; iter < n; iter++) {
               results[iter] = values[0];
               for (int p = 1; p < primes.Length; p++)
                   if (results[iter] > values[p])
                       results[iter] = values[p];
               for (int p = 0; p < primes.Length; p++)
                   if (results[iter] == values[p])
                       values[p] = primes[p] * results[++indexes[p]];
           }
           return results;
       }
       
       public static void Main(string[] args) {
           foreach (int[] primes in new int[][] { new int[] {2,3,5}, new int[] {2,3,5,7} }) {
               Console.WriteLine("{0}-Smooth:", primes.Last());
               Console.WriteLine(string.Join(" ", Hamming(20, primes)));
               Console.WriteLine(Hamming(1691, primes).Last());
               Console.WriteLine(Hamming(1000000, primes).Last());
               Console.WriteLine();
           }
       }
   }

}</lang>

Output:

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

Fast version

Like some of the other implementations on this page, this version represents each number as a list of exponents which would be applied to each prime number. So the number 60 would be represented as int[3] { 2, 1, 1 } which is interpreted as 2^2 * 3^1 * 5^1.

As often happens, optimizing for speed caused a marked increase in code size and complexity. Clearly the versions I wrote above are easier to read & understand. They were also much quicker to write. But the generic version above runs in 3+ seconds for the 1000000th 5-smooth number whereas this version does it in 0.35 seconds, 8-10 times faster.

I've tried to comment it as best I could, without bloating the code too much.

--Mike Lorenz

<lang csharp>using System; using System.Linq; using System.Numerics;

namespace HammingFast {

   class MainClass {

       private static int[] _primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 };

       public static BigInteger Big(int[] exponents) {
           BigInteger val = 1;
           for (int i = 0; i < exponents.Length; i++)
               for (int e = 0; e < exponents[i]; e++)
                   val = val * _primes[i];
           return val;
       }

       public static int[] Hamming(int n, int nprimes) {
           var hammings  = new int[n, nprimes];                    // array of hamming #s we generate
           var hammlogs  = new double[n];                          // log values for above
           var primelogs = new double[nprimes];                    // pre-calculated prime log values
           var indexes   = new int[nprimes];                       // intermediate hamming values as indexes into hammings
           var listheads = new int[nprimes, nprimes];              // intermediate hamming list heads
           var listlogs  = new double[nprimes];                    // log values of list heads
           for (int p = 0; p < nprimes; p++) {
               listheads[p, p] = 1;                                // init list heads to prime values
               primelogs[p]    = Math.Log(_primes[p]);             // pre-calc prime log values
               listlogs[p]     = Math.Log(_primes[p]);             // init list head log values
           }
           for (int iter = 1; iter < n; iter++) {
               int min = 0;                                        // find index of min item in list heads
               for (int p = 1; p < nprimes; p++)
                   if (listlogs[p] < listlogs[min])
                       min = p;
               hammlogs[iter] = listlogs[min];                     // that's the next hamming number
               for (int i = 0; i < nprimes; i++)
                   hammings[iter, i] = listheads[min, i];
               for (int p = 0; p < nprimes; p++) {                 // update each list head if it matches new value
                   bool equal = true;                              // test each exponent to see if number matches
                   for (int i = 0; i < nprimes; i++) {
                       if (hammings[iter, i] != listheads[p, i]) {
                           equal = false;
                           break;
                       }
                   }
                   if (equal) {                                    // if it matches...
                       int x = ++indexes[p];                       // set index to next hamming number
                       for (int i = 0; i < nprimes; i++)           // copy each hamming exponent
                           listheads[p, i] = hammings[x, i];
                       listheads[p, p] += 1;                       // increment exponent = mult by prime
                       listlogs[p] = hammlogs[x] + primelogs[p];   // add log(prime) to log(value) = mult by prime
                   }
               }
           }

           var result = new int[nprimes];
           for (int i = 0; i < nprimes; i++)
               result[i] = hammings[n - 1, i];
           return result;
       }

       public static void Main(string[] args) {
           foreach (int np in new int[] { 3, 4, 5 }) {
               Console.WriteLine("{0}-Smooth:", _primes[np - 1]);
               Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).Select(x => Big(Hamming(x, np)))));
               Console.WriteLine(Big(Hamming(1691, np)));
               Console.WriteLine(Big(Hamming(1000000, np)));
               Console.WriteLine();
           }
       }
   }

}</lang>

Output:

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

11-Smooth:
1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 20 21 22 24
296352
561912530929780078125000

C# Enumerator Version

I wanted to fix the enumerator (old) version, as it wasn't working. It became a bit of an obsession... after a few iterations I came up with the following, which is the fastest C# version on my computer - your mileage may vary. It combines the speed of the Log method; Log(2)+Log(3)=Log(2*3) to help determine which is the next one to use. Then I have added some logic (using the series property) to ensure that exponent sets are never duplicated - which speeds the calculations up a bit.... Adding this trick to the Fast Version will probably result in the fastest version, but I'll leave that to someone else to implement. Finally it's all enumerated through a crazy one-way-linked-list-type-structure that only exists as long as the enumerator and is left up to the garbage collector to remove the bits no longer needed... I hope it's commented enough... follow it if you dare!

<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Numerics;

namespace HammingTest {

   class HammingNode
   {
       public double log;
       public int[] exponents;
       public HammingNode next;
       public int series;
   }
   
   class HammingListEnumerator : IEnumerable<BigInteger>
   {
       private int[] primes;
       private double[] primelogs;
       private HammingNode next;
       private HammingNode[] values;
       private HammingNode[] indexes;

       public HammingListEnumerator(IEnumerable<int> seeds)
       {
           // Ensure our seeds are properly ordered, and generate their log values
           primes = seeds.OrderBy(x => x).ToArray();
           primelogs = primes.Select(x => Math.Log10(x)).ToArray();
           // Start at 1 (log(1)=0, exponents are all 0, series = none)
           next = new HammingNode { log = 0, exponents = new int[primes.Length], series = primes.Length };
           // Set all exponent sequences to the start, and calculate the first value for each exponent
           indexes = new HammingNode[primes.Length];
           values = new HammingNode[primes.Length];
           for(int i = 0; i < primes.Length; ++i)
           {
               indexes[i] = next;
               values[i] = AddExponent(next, i);
           }
       }

       // Make a copy of a node, and increment the specified exponent value
       private HammingNode AddExponent(HammingNode node, int i)
       {
           HammingNode ret = new HammingNode { log = node.log + primelogs[i], exponents = (int[])node.exponents.Clone(), series = i };
           ++ret.exponents[i];
           return ret;
       }

       private void GetNext()
       {
           // Find which exponent value is the lowest
           int min = 0;
           for(int i = 1; i < values.Length; ++i)
               if(values[i].log < values[min].log)
                   min = i;
           
           // Add it to the end of the 'list', and move to it
           next.next = values[min];
           next = values[min];

           // Find the next node in an allowed sequence (skip those that would be duplicates) 
           HammingNode val = indexes[min].next;
           while(val.series < min)
               val = val.next;

           // Keep the current index, and calculate the next value in the series for that exponent
           indexes[min] = val;
           values[min] = AddExponent(val, min);
       }

       // Skip values without having to calculate the BigInteger value from the exponents
       public HammingListEnumerator Skip(int count)
       {
           for(int i = count; i > 0; --i)
               GetNext();

           return this;
       }

       // Calculate the BigInteger value from the exponents
       internal BigInteger ValueOf(HammingNode n)
       {
           BigInteger val = 1;
           for(int i = 0; i < n.exponents.Length; ++i)
               for(int e = 0; e < n.exponents[i]; e++)
                   val = val * primes[i];
           return val;
       }

       public IEnumerator<BigInteger> GetEnumerator()
       {
           while(true)
           {
               yield return ValueOf(next);
               GetNext();
           }
       }

       System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
       {
           return this.GetEnumerator();
       }
   }

   class Program
   {
       static void Main(string[] args)
       {
           foreach(int[] primes in new int[][] { 
               new int[] { 2, 3, 5 },
               new int[] { 2, 3, 5, 7 },
               new int[] { 2, 3, 5, 7, 9}})
           {
               HammingListEnumerator hammings = new HammingListEnumerator(primes);
               System.Diagnostics.Debug.WriteLine("{0}-Smooth:", primes.Last());
               System.Diagnostics.Debug.WriteLine(String.Join(" ", hammings.Take(20).ToArray()));
               System.Diagnostics.Debug.WriteLine(hammings.Skip(1691 - 20).First());
               System.Diagnostics.Debug.WriteLine(hammings.Skip(1000000 - 1691).First());
               System.Diagnostics.Debug.WriteLine("");
           }
       }
   }

} </lang>

Output:

5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

7-Smooth:
1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27
3317760
4157409948433216829957008507500000000

11-Smooth:
1 2 3 4 5 6 7 8 9 10 11 12 14 15 16 18 20 21 22 24
296352
561912530929780078125000

Alternate Generic Enumerating version

YMMV, but unlike the author of the above code, I found the above version to be much slower on my machine than the "Generic version". The following version is actually just a little slower than the Generic version but uses much less memory due to avoiding duplicates and only keeping in memory those "lazy list" streams necessary for calculation from 1/5 of the current range to 1/2 (for Smooth-5 numbers), and not successive values in those ranges but only the values the are the multiples of previous ranges. Like the Haskell code from which it is translated, the head of the streams is not retained so can be garbage collected when no longer necessary and it is recommended that the primes be processed in reverse order so that the least dense streams are processed first for slightly less memory use and operations.

It also shows that one can use somewhat functional programming techniques in C#.

The class implements its own partial version of a lazy list using the Lazy class and uses lambda closures for the recursive use of the successive streams to avoid stack use. It uses Aggregate to implement the Haskell "foldl" function. The code demonstrates that even though C# is primarily imperative in paradigm, with its ability to implement closures using delegates/lambdas, it can express some algorithms such as this mostly functionally.

It isn't nearly as fast as a Haskell, Scala or even Clojure and Scheme (GambitC) versions of this algorithm, being about five times slower is primarily due to its use of many small heap based instances of classes, both for the LazyList's and for closures (implemented using at least one class to hold the captured free variables) and the inefficiency of DotNet's allocation and garbage collection of many small instance objects (although about twice as fast as F#'s implementation, whose closures must require even more small object instances); it seems Haskell and the (Java) JVM are much more efficient at doing these allocations/garbage collections for many small objects. The slower speed to a relatively minor extent is also due to less efficient BigInteger operations:

Translation of: Haskell

<lang csharp>using System; using System.Collections; using System.Collections.Generic; using System.Linq; using System.Numerics;

namespace Hamming {

 class Hammings : IEnumerable<BigInteger> {
   private class LazyList<T> {
     public T v; public Lazy<LazyList<T>> cont;
     public LazyList(T v, Lazy<LazyList<T>> cont) {
       this.v = v; this.cont = cont;
     }
   }
   private uint[] primes;
   private Hammings() { } // must have an argument!!!
   public Hammings(uint[] prms) { this.primes = prms; }
   private LazyList<BigInteger> merge(LazyList<BigInteger> xs,
                                      LazyList<BigInteger> ys) {
     if (xs == null) return ys; else {
       var x = xs.v; var y = ys.v;
       if (BigInteger.Compare(x, y) < 0) {
         var cont = new Lazy<LazyList<BigInteger>>(() =>
                      merge(xs.cont.Value, ys));
         return new LazyList<BigInteger>(x, cont);
       }
       else {
         var cont = new Lazy<LazyList<BigInteger>>(() =>
                      merge(xs, ys.cont.Value));
         return new LazyList<BigInteger>(y, cont);
       }
     }
   }
   private LazyList<BigInteger> llmult(uint mltplr,
                                       LazyList<BigInteger> ll) {      
     return new LazyList<BigInteger>(mltplr * ll.v,
                                     new Lazy<LazyList<BigInteger>>(() =>
                                       llmult(mltplr, ll.cont.Value)));
   }
   public IEnumerator<BigInteger> GetEnumerator() {
     Func<LazyList<BigInteger>,uint,LazyList<BigInteger>> u =
       (acc, p) => { LazyList<BigInteger> r = null;
                     var cont = new Lazy<LazyList<BigInteger>>(() => r);
                     r = new LazyList<BigInteger>(1, cont);
                     r = this.merge(acc, llmult(p, r));
                     return r; };
     yield return 1;
     for (var stt = primes.Aggregate(null, u); ; stt = stt.cont.Value)
       yield return stt.v;
   }
   IEnumerator IEnumerable.GetEnumerator() {
     return this.GetEnumerator();
   }
 }

 class Program {
   static void Main(string[] args) {
     Console.WriteLine("Calculates the Hamming sequence of numbers.\r\n");

     var primes = new uint[] { 5, 3, 2 };
     Console.WriteLine(String.Join(" ", (new Hammings(primes)).Take(20).ToArray()));
     Console.WriteLine((new Hammings(primes)).ElementAt(1691 - 1));

     var n = 1000000;

     var elpsd = -DateTime.Now.Ticks;

     var num = (new Hammings(primes)).ElementAt(n - 1);

     elpsd += DateTime.Now.Ticks;

     Console.WriteLine(num);
     Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

     Console.Write("\r\nPress any key to exit:");
     Console.ReadKey(true);
     Console.WriteLine();
   }
 }

}</lang>

Fast enumerating logarithmic version

The so-called "fast" generic version above isn't really all that fast due to all the extra array accesses required by the generic implementation and that it doesn't avoid duplicates as the above functional code does avoid. It also uses a lot of memory as it has arrays that are the size of the range for which the Hamming numbers are calculated.

The following code eliminates or reduces all of those problems by being non-generic, eliminating duplicate calculations, saving memory by "draining" the growable List's used in blocks as back pointer indexes are used (thus using memory at the same rate as the functional version), thus avoiding excessive allocations/garbage collections; it also is enumerates through the Hamming numbers although that comes at a slight cost in overhead function calls:

Translation of: Nim

<lang csharp>using System; using System.Collections; using System.Collections.Generic; using System.Linq; using System.Numerics;

class HammingsLogArr : IEnumerable<Tuple<uint, uint, uint>> {

 public static BigInteger trival(Tuple<uint, uint, uint> tpl) {
   BigInteger rslt = 1;
   for (var i = 0; i < tpl.Item1; ++i) rslt *= 2;
   for (var i = 0; i < tpl.Item2; ++i) rslt *= 3;
   for (var i = 0; i < tpl.Item3; ++i) rslt *= 5;
   return rslt;
 }
 private const double lb3 = 1.5849625007211561814537389439478; // Math.Log(3) / Math.Log(2);
 private const double lb5 = 2.3219280948873623478703194294894; // Math.Log(5) / Math.Log(2);
 private struct logrep {
   public uint x2, x3, x5;
   public double lg;
   public logrep(uint x, uint y, uint z, double lg) {
     this.x2 = x; this.x3 = y; this.x5 = z; this.lg = lg;
   }
   public static bool operator <(logrep x, logrep y) {
     return x.lg < y.lg;
   }
   public static bool operator >(logrep x, logrep y) {
     return x.lg > y.lg;
   }
   public logrep mul2() {
     return new logrep(this.x2 + 1, this.x3, this.x5, this.lg + 1.0);
   }
   public logrep mul3() {
     return new logrep(this.x2, this.x3 + 1, this.x5, this.lg + lb3);
   }
   public logrep mul5() {
     return new logrep(this.x2, this.x3, this.x5 + 1, this.lg + lb5);
   }
 }
 public IEnumerator<Tuple<uint, uint, uint>> GetEnumerator() {
   var one = new logrep();
   var m = new List<logrep>(); var h = new List<logrep>();
   var x5 = one.mul5();
   var mrg = one.mul3();
   var x53 = one.mul3().mul3(); // already advanced one step
   var x532 = one.mul2();
   var i = 0; var j = 0;
   yield return Tuple.Create(0u, 0u, 0u); // trivial case for one representation
   while (true) {
     if (i >= h.Capacity >> 1) { h.RemoveRange(0, i); i = 0; } // assume capacity stays the same...
     if (x532 < mrg) { h.Add(x532); x532 = h[i].mul2(); i++; }
     else {
       h.Add(mrg);
       if (j >= m.Capacity) { m.RemoveRange(0, j); j = 0; }
       if (x53 < x5) { mrg = x53; x53 = m[j].mul3(); j++; }
       else { mrg = x5; x5 = x5.mul5(); }
       m.Add(mrg);
     }
     var rslt = h[h.Count - 1];
     yield return Tuple.Create(rslt.x2, rslt.x3, rslt.x5);
   }
 }
 IEnumerator IEnumerable.GetEnumerator() {
   return this.GetEnumerator();
 }

}

class Program {

 static void Main(string[] args) {
   Console.WriteLine(String.Join(" ", (new HammingsLogArr()).Take(20)
                                       .Select(t => HammingsLogArr.trival(t))
                                       .ToArray()));
   Console.WriteLine(HammingsLogArr.trival(NthHamming.findNth(1691)));

   var n = 1000000UL;
   var elpsd = -DateTime.Now.Ticks;

   var rslt = (new HammingsLogArr()).ElementAt((int)n - 1);

   elpsd += DateTime.Now.Ticks;

   Console.WriteLine("2^{0} times 3^{1} times 5^{2}", rslt.Item1, rslt.Item2, rslt.Item3);
   var lgrthm = Math.Log10(2.0) * ((double)rslt.Item1 +
                 ((double)rslt.Item2 * Math.Log(3.0) + (double)rslt.Item3 * Math.Log(5.0)) / Math.Log(2.0));
   var pwr = Math.Floor(lgrthm); var mntsa = Math.Pow(10.0, lgrthm - pwr);
   Console.WriteLine("Approximately:  {0}E+{1}", mntsa, pwr);
   var s = HammingsLogArr.trival(rslt).ToString();
   var lngth = s.Length;
   Console.WriteLine("Decimal digits:  {0}", lngth);
   if (lngth <= 10000) {
     var i = 0;
     for (; i < lngth - 100; i += 100) Console.WriteLine(s.Substring(i, 100));
     Console.WriteLine(s.Substring(i));
   }
   Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

   Console.Write("\r\nPress any key to exit:");
   Console.ReadKey(true);
   Console.WriteLine();
 }

}</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
2^55 times 3^47 times 5^64
Approximately:  5.19312780448414E+83
Decimal digits:  84
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
The 1000000th hamming number took 55 milliseconds

The above code is about 30 times faster than the functional code due to both eliminating the lambda closures that were the main problem with that code as well as eliminating the BigInteger operations. It has about O(n) empirical performance and can find the billionth Hamming number in about 60 seconds.

Extremely fast non-enumerating version calculating the error band

The above code is about as fast as one can go generating sequences; however, if one is willing to forego sequences and just calculate the nth Hamming number (again), then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: regular number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows:

Translation of: Nim

<lang csharp>using System; using System.Collections; using System.Collections.Generic; using System.Linq; using System.Numerics;

static class NthHamming {

 public static BigInteger trival(Tuple<uint, uint, uint> tpl) {
   BigInteger rslt = 1;
   for (var i = 0; i < tpl.Item1; ++i) rslt *= 2;
   for (var i = 0; i < tpl.Item2; ++i) rslt *= 3;
   for (var i = 0; i < tpl.Item3; ++i) rslt *= 5;
   return rslt;
 }

 private struct logrep {
   public uint x2, x3, x5;
   public double lg;
   public logrep(uint x, uint y, uint z, double lg) {
     this.x2 = x; this.x3 = y; this.x5 = z; this.lg = lg;
   }
 }

 private const double lb3 = 1.5849625007211561814537389439478; // Math.Log(3) / Math.Log(2);
 private const double lb5 = 2.3219280948873623478703194294894; // Math.Log(5) / Math.Log(2);
 private const double fctr = 6.0 * lb3 * lb5;
 private const double crctn = 2.4534452978042592646620291867186; // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)

 public static Tuple<uint, uint, uint> findNth(UInt64 n) {
   if (n < 1) throw new Exception("NthHamming.findNth:  argument must be > 0!");
   if (n < 2) return Tuple.Create(0u, 0u, 0u); // trivial case for argument of one
   var lgest = Math.Pow(fctr * (double)n, 1.0/3.0) - crctn; // from WP formula
   var frctn = (n < 1000000000) ? 0.509 : 0.105;
   var lghi = Math.Pow(fctr * ((double)n + frctn * lgest), 1.0/3.0) - crctn;
   var lglo = 2.0 * lgest - lghi; // upper and lower bound of upper "band"
   var count = 0UL; // need 64 bit precision in case...
   var bnd = new List<logrep>();
   for (uint k = 0, klmt = (uint)(lghi / lb5) + 1; k < klmt; ++k) {
     var p = (double)k * lb5;
     for (uint j = 0, jlmt = (uint)((lghi - p) / lb3) + 1; j < jlmt; ++j) {
       var q = p + (double)j * lb3;
       var ir = lghi - q;
       var lg = q + Math.Floor(ir); // current log2 value (estimated)
       count += (ulong)ir + 1;
       if (lg >= lglo) bnd.Add(new logrep((UInt32)ir, j, k, lg));
     }
   }
   if (n > count) throw new Exception("NthHamming.findNth:  band high estimate is too low!");
   var ndx = (int)(count - n);
   if (ndx >= bnd.Count) throw new Exception("NthHamming.findNth:  band low estimate is too high!");
   bnd.Sort((a, b) => (b.lg < a.lg) ? -1 : 1); // sort in decending order

   var rslt = bnd[ndx];
   return Tuple.Create(rslt.x2, rslt.x3, rslt.x5);
 }

}

class Program {

 static void Main(string[] args) {
   Console.WriteLine(String.Join(" ", Enumerable.Range(1,20).Select(i =>
                                         NthHamming.trival(NthHamming.findNth((ulong)i))).ToArray()));
   Console.WriteLine(NthHamming.trival((new HammingsLogArr()).ElementAt(1691 - 1)));

   var n = 1000000000000UL;
   var elpsd = -DateTime.Now.Ticks;

   var rslt = NthHamming.findNth(n);

   elpsd += DateTime.Now.Ticks;

   Console.WriteLine("2^{0} times 3^{1} times 5^{2}", rslt.Item1, rslt.Item2, rslt.Item3);
   var lgrthm = Math.Log10(2.0) * ((double)rslt.Item1 +
                 ((double)rslt.Item2 * Math.Log(3.0) + (double)rslt.Item3 * Math.Log(5.0)) / Math.Log(2.0));
   var pwr = Math.Floor(lgrthm); var mntsa = Math.Pow(10.0, lgrthm - pwr);
   Console.WriteLine("Approximately:  {0}E+{1}", mntsa, pwr);
   var s = HammingsLogArr.trival(rslt).ToString();
   var lngth = s.Length;
   Console.WriteLine("Decimal digits:  {0}", lngth);
   if (lngth <= 10000) {
     var i = 0;
     for (; i < lngth - 100; i += 100) Console.WriteLine(s.Substring(i, 100));
     Console.WriteLine(s.Substring(i));
   }
   Console.WriteLine("The {0}th hamming number took {1} milliseconds", n, elpsd / 10000);

   Console.Write("\r\nPress any key to exit:");
   Console.ReadKey(true);
   Console.WriteLine();
 }

}</lang>

The output is the same as above except that the time is too small to be measured. The billionth number in the sequence can be calculated in just about 10 milliseconds, the trillionth in about one second, the thousand trillionth in about a hundred seconds, and it should be possible to calculate the 10^19th value in less than a day (untested) on common personal computers. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit.

C++

C++11 For Each Generator

<lang cpp>

1. include <iostream>
2. include <vector>

// Hamming like sequences Generator // // Nigel Galloway. August 13th., 2012 // class Ham { private: std::vector<unsigned int> _H, _hp, _hv, _x; public: bool operator!=(const Ham& other) const {return true;} Ham begin() const {return *this;}

       Ham end() const {return *this;}

unsigned int operator*() const {return _x.back();} Ham(const std::vector<unsigned int> &pfs):_H(pfs),_hp(pfs.size(),0),_hv({pfs}),_x({1}){} const Ham& operator++() { for (int i=0; i<_H.size(); i++) for (;_hv[i]<=_x.back();_hv[i]=_x[++_hp[i]]*_H[i]); _x.push_back(_hv[0]); for (int i=1; i<_H.size(); i++) if (_hv[i]<_x.back()) _x.back()=_hv[i]; return *this; } }; </lang>

5-Smooth

<lang cpp> int main() {

 int count = 1;
 for (unsigned int i : Ham({2,3,5})) {
   if (count <= 62) std::cout << i << ' ';
   if (count++ == 1691) {
     std::cout << "\nThe one thousand six hundred and ninety first Hamming Number is " << i << std::endl;
     break;
   }
 }
 return 0;

} </lang> Produces:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 45 48 50 54 60 64 72 75 80 81 90 96 100 108 120 125 128 135 144 150 160 162 180 192 200 216 225 240 243 250 256 270 288 300 320 324 360 375 384 400 405
The one thousand six hundred and ninety first Hamming Number is 2125764000

7-Smooth

<lang cpp> int main() {

 int count = 1;
 for (unsigned int i : Ham({2,3,5,7})) {
   std::cout << i << ' ';
   if (count++ == 64) break;
 }
 std::cout << std::endl;
 return 0;

} </lang> Produces:

1 2 3 4 5 6 7 8 9 10 12 14 15 16 18 20 21 24 25 27 28 30 32 35 36 40 42 45 48 49 50 54 56 60 63 64 70 72 75 80 81 84 90 96 98 100 105 108 112 120 125 126 128 135 140 144 147 150 160 162 168 175 180 189

Avoiding Duplicates with Functional Code

If converted to use multi-precision integers (via GMP, as in this code), the above code is slower than it needs to be due to several reasons: 1) It uses the an adaptation of the original Dijkstra's algorithm, which is somewhat slower due to repeated calculations (2 time 3, 3 times 2, etc.), 2) the generator is written generally to handle any series of multiples, and 3) for finding the nth Hamming number, the code as `for (auto hmg : Ham({5, 3, 2})` means that there is a copy of the sometimes very large multi-precision number which can consume more time than the actual computation. The following code isn't particularly fast due to other reasons that will be discussed, but avoids duplication of calculations by a modification of the algorithm; it is written functionally as a LazyList (which could easily also have iteration abilities added, with the a basic LazyList type defined here since there is no library available:

Translation of: Haskell

Works with: C++11

<lang cpp>#include <chrono>

1. include <iostream>
2. include <gmpxx.h>
3. include <functional>
4. include <memory>

template<class T> class Lazy { public: T _v; private: std::function<T()> _f;

public: explicit Lazy(std::function<T()> thnk) : _v(T()), _f(thnk) {}; T value() { // not thread safe! if (this->_f != nullptr) { this->_v = this->_f(); this->_f = nullptr; } return this->_v; } };

template<class T> class LazyList { public: T head; std::shared_ptr<Lazy<LazyList<T>>> tail; LazyList(): head(T()) {} // only used in initializing Lazy... LazyList(T head, std::function<LazyList<T>()> thnk) : head(head), tail(std::make_shared<Lazy<LazyList<T>>>(thnk)) {} // default Copy/Move constructors and assignment operators seem to work well enough bool isEmpty() { return this->tail == nullptr; } };

typedef std::shared_ptr<mpz_class> PBI; typedef LazyList<PBI> LL; typedef std::function<LL(LL)> FLL2LL;

LL merge(LL a, LL b) { auto ha = a.head; auto hb = b.head; if (*ha < *hb) { return LL(ha, [=]() { return merge(a.tail->value(), b); }); } else { return LL(hb, [=]() { return merge(a, b.tail->value()); }); } }

LL smult(int m, LL s) { const auto im = mpz_class(m); const auto psmlt = std::make_shared<FLL2LL>([](LL ss) { return ss; }); *psmlt = [=](LL ss) { return LL(std::make_shared<mpz_class>(*ss.head * im), [=]() { return (*psmlt)(ss.tail->value()); }); }; return (*psmlt)(s); // worker wrapper pattern with recursive closure as worker... }

LL u(LL s, int n) { const auto r = std::make_shared<LL>(LL()); // interior mutable... *r = smult(n, LL(std::make_shared<mpz_class>(1), [=]() { return *r; })); if (!s.isEmpty()) { *r = merge(s, *r); } return *r; }

LL hammings() { auto r = LL(); for (auto pn : std::vector<int>({5, 3, 2})) { r = u(r, pn); } return LL(std::make_shared<mpz_class>(1), [=]() { return r; }); }

int main() { auto hmgs = hammings(); for (auto i = 0; i < 20; ++i) { std::cout << *hmgs.head << " "; hmgs = hmgs.tail->value(); } std::cout << "\n";

hmgs = hammings(); for (auto i = 1; i < 1691; ++i) hmgs = hmgs.tail->value(); std::cout << *hmgs.head << "\n";

auto start = std::chrono::steady_clock::now(); hmgs = hammings(); for (auto i = 1; i < 1000000; ++i) hmgs = hmgs.tail->value(); auto stop = std::chrono::steady_clock::now();

auto ms = std::chrono::duration_cast<std::chrono::milliseconds>(stop - start); std::cout << *hmgs.head << " in " << ms.count() << "milliseconds.\n"; }</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000 in 1079 milliseconds.

Note that the repeat loop to increment to the desired value is written so as not to oopy unnecessary Hamming values, unlike the use of the first Generator class.

This shows that one can program functionally in C++, but the performance is many times slower than a language more suitable for functional paradigms such as Haskell or even Kotlin; this is likely due to the cost of memory allocation with (multi-threading-safe) reference counting and that the memory system isn't tuned to many small allocations/de-allocations such as are generally necessary with functional programming. One can easily see how to adapt this algorithm to make it work for the general case by just having an argument which is an vector of the required multipliers used in the `hammings` function.

There is another problem in using languages such as this that do not have cyclic reference breaking capbilities: the code will leak memory due to the cyclic memory reference cycles and it is perhaps impossible to change the function algorithm to manually free, even though the code uses "shared"/reference counting facilities.

Avoiding Duplicates with Imperative Code

To show that it is the execution time for the functional LazyList that is taking the time, here is the same algorithm implemented imperatively using vectors, also avoiding duplicate calculations; it is not written as a general function for any set of multipliers as the extra vector addressing takes some extra time; again, it minimizes copying of values:

Translation of: Rust

Works with: C++11

<lang cpp>#include <chrono>

1. include <iostream>
2. include <vector>
3. include <gmpxx.h>

class Hammings { private: const mpz_class _two = 2, _three = 3, _five = 5; std::vector<mpz_class> _m = {}, _h = {1}; mpz_class _x5 = 5, _x53 = 9, _mrg = 3, _x532 = 2; int _i = 1, _j = 0; public: Hammings() {_m.reserve(65536); _h.reserve(65536); }; bool operator!=(const Hammings& other) const { return true; } Hammings begin() const { return *this; } Hammings end() const { return *this; } mpz_class operator*() { return _h.back(); } const Hammings& operator++() { if (_i > _h.capacity() / 2) { _h.erase(_h.begin(), _h.begin() + _i); _i = 0; } if (_x532 < _mrg) { _h.push_back(_x532); _x532 = _h[_i++] * _two; } else { _h.push_back(_mrg); if (_x53 < _x5) { _mrg = _x53; _x53 = _m[_j++] * _three; } else { _mrg = _x5; _x5 = _x5 * _five; } if (_j > _m.capacity() / 2) { _m.erase(_m.begin(), _m.begin() + _j); _j = 0; } _m.push_back(_mrg); } return *this; } };

int main() { auto cnt = 1; for (auto hmg : Hammings()) { if (cnt <= 20) std::cout << hmg << " "; if (cnt == 20) std::cout << "\n"; if (cnt++ >= 1691) { std::cout << hmg << "\n"; break; } }

auto start = std::chrono::steady_clock::now(); hmgs = hammings(); auto&& hmgitr = Hammings(); for (auto i = 1; i < 1000000; ++i) ++hmgitr; auto stop = std::chrono::steady_clock::now();

auto ms = std::chrono::duration_cast<std::chrono::milliseconds>(stop - start); std::cout << *hmgitr << " in " << ms.count() << "milliseconds.\n"; }</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000 in 79 milliseconds.

This code takes about the same amount of time as Haskell for the functional algorithm calculating multi-precision values (also uses GMP; not including the list processing time), but greatly reduces the C++ processing time compared to the functional version due to the use of imperative code and vectors.

Chapel

Chapel is by no means a functional language although it has some Higher Order Functional (HOF) concepts such as zippering, scanning, and reducing of iterations, it lacks closures (functions that can capture variable bindings from the enclosing scope(s)) even though it has first class functions that can be passed as values and lambdas (anonymous functions), nor is tail-call-optimization of recursive functions and iterators guarantied. However, now that Chapel supports class fields that can be of any type including references to other classes of any storage type, we can emulate closures using shared classes (shared classes are automatically de-allocated when they have no more references, currently using reference counting). The following code does that for the non-duplicating version of the sequence of Hamming numbers:

Translation of: Haskell

Hamming_numbers#Avoiding_generation_of_duplicates

Works with: 1.24.1 version or greater, maybe lesser

<lang chapel>use BigInteger; use Time;

// Chapel doesn't have closure functions that can capture variables from // outside scope, so we use a class to emulate them for this special case; // the member fields mult, mrglst, and mltlst, emulate "captured" variables // that would normally be captured by the `next` continuation closure... class HammingsList {

   const head: bigint;
   const mult: uint(8);
   var mrglst: shared HammingsList?;
   var mltlst: shared HammingsList?;
   var tail: shared HammingsList? = nil;
   proc init(hd: bigint, mlt: uint(8), mrgl: shared HammingsList?,
                                       mltl: shared HammingsList?) {
       head = hd; mult = mlt; mrglst = mrgl; mltlst = mltl; }
   proc next(): shared HammingsList {
       if tail != nil then return tail: shared HammingsList;
       const nhd: bigint = mltlst!.head * mult;
       if mrglst == nil then {
           tail = new shared HammingsList(nhd, mult,
                                          nil: shared HammingsList?,
                                          nil: shared HammingsList?);
           mltlst = mltlst!.next();
           tail!.mltlst <=> mltlst;
       }
       else {
           if mrglst!.head < nhd then {
               tail = new shared HammingsList(mrglst!.head, mult,
                                              nil: shared HammingsList?,
                                              nil: shared HammingsList?);
               mrglst = mrglst!.next(); mrglst <=> tail!.mrglst;
               mltlst <=> tail!.mltlst;
           }
           else {
               tail = new shared HammingsList(nhd, mult,
                                              nil: shared HammingsList?,
                                              nil: shared HammingsList?);
               mltlst = mltlst!.next(); mltlst <=> tail!.mltlst;
               mrglst <=> tail!.mrglst;
           }
       }
       return tail: shared HammingsList;
   }

}

proc u(n: uint(8), s: shared HammingsList?): shared HammingsList {

   var r = new shared HammingsList(1: bigint, n, s,
                                   nil: shared HammingsList?);
   r.mltlst = r; // lazy recursion!
   return r.next();

}

iter hammings(): bigint {

   var nxt: shared HammingsList? = nil: shared HammingsList?;
   const mlts: [ 0 .. 2 ] int = [ 5, 3, 2 ];
   for m in mlts do nxt = u(m: uint(8), nxt);
   yield 1 : bigint;
   while true { yield nxt!.head; nxt = nxt!.next(); }

}

write("The first 20 Hamming numbers are: "); var cnt: int = 0; for h in hammings() { write(" ", h); cnt += 1; if cnt >= 20 then break; } write(".\nThe 1691st Hamming number is "); cnt = 0; for h in hammings() { cnt += 1; if cnt < 1691 then continue; write(h); break; } writeln(".\nThe millionth Hamming number is "); var timer: Timer; timer.start(); cnt = 0; for h in hammings() { cnt += 1; if cnt < 1000000 then continue; write(h); break; } timer.stop(); writeln(".\nThis last took ",

                     timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36.
The 1691st Hamming number is 2125764000.
The millionth Hamming number is 
519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 224.652 milliseconds.

The above time is as run on an Intel Skylake i5-6500 at 3.6 GHz (turbo, single-threaded).

It isn't as fast as the following versions due to the many memory allocations and de-allocations as for typically functional forms of code, but it is in the order of speed of many actual functional languages and faster than many, depending on how well their memory management is adapted for this programming paradigm.

This shows that the functional forms of most algorithms can be translated into Chapel, although some concessions need to be made for the functional facilities that Chapel doesn't have. However, there is one major problem in using languages such as this for functional algorithms when such languages do not have cyclic reference breaking capabilities: the code will leak memory due to the cyclic memory reference cycles and it is perhaps impossible to change the functional algorithm to manually free, even though the code uses "shared"/reference counting facilities.

Alternate Imperative Version Using "Growable" Arrays

In general, we can convert functional algorithms into imperative style algorithms using Array's to emulate memoizing lazy lists and simple mutable variables to express the recursion within a while loop, as in the following codes (as also used when necessary in the above code). Rather than implement the true Dijkstra merge algorithm which is slower and uses more memory, the following codes implement the better non-duplicating algorithm:

Translation of: Nim

<lang chapel>use BigInteger; use Time;

iter nodupsHamming(): bigint {

 var mdom = { 0 .. 0 }; var m: [mdom] bigint; // init so can double!
 var hdom = { 0 .. 0 }; var h: [hdom] bigint; // init so can double!
 var x5 = 5: bigint; var mrg = 3: bigint;
 var x53 = 9: bigint; // already advanced one step!
 var x532 = 2: bigint;
 var ih, jm, i, j: int;
 yield 1: bigint; // first trivial case
 while true {
   const cph = h.size; // move in place to avoid allocation!
   if i + i >= cph {
     h[0 .. ih - i - 1] = h[i .. ih - 1]; ih -= i; i = 0; }
   if ih >= cph then hdom = { 0 .. cph + cph - 1 };
   if x532 < mrg { h[ih] = x532; x532 = h[i] * 2; i += 1; }
   else {
     h[ih] = mrg; const cpm = m.size; // move in place to avoid allocation!
     if j + j >= cpm {
       m[0 .. jm - j - 1] = m[j .. jm - 1]; jm -= j; j = 0; }
     if jm >= cpm then mdom = { 0 .. cpm + cpm - 1 };
     if x53 < x5 { mrg = x53; x53 = m[j] * 3; j += 1; }
     else { mrg = x5; x5 = x5 * 5; }
     m[jm] = mrg; jm += 1;
   }
   yield h[ih]; ih += 1;
 }

}

// test it... write("The first 20 hamming numbers are: "); var cnt = 0; for h in nodupsHamming() { if cnt >= 20 then break; cnt += 1; write(" ", h); }

write("\nThe 1691st hamming number is "); cnt = 1; for h in nodupsHamming() { if cnt >= 1691 then { writeln(h); break; } cnt += 1; }

write("The millionth hamming number is "); var timer: Timer; cnt = 1; timer.start(); for h in nodupsHamming() { if cnt >= 1000000 { write(h); break; } cnt += 1; } timer.stop(); writeln(".\nThis last took ", timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</lang>

Output:

The first 20 hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
The 1691st hamming number is 2125764000
The millionth hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 125.51 milliseconds.

The above time is as run on an Intel Skylake i5-6500 at 3.6 GHz (turbo, single-threaded).

As you can see, this algorithm is quite fast, as it minimizes the memory allocations/de-allocations, but it still takes considerable time for the many multi-precision (BigInteger) calculations even though the GMP library is being used under-the-covers.

Alternate version using logarithm approximations for sorting order

To greatly reduce the time used for BigInteger calculations, the following algorithm uses logarithmic approximations (real(64)) for internal calculations and only output the final answer(s) as BigInteger(s): <lang chapel>use BigInteger; use Math; use Time;

config const nth: uint(64);

type Trival = 3*uint(32);

proc trival2bigint(x: Trival): bigint {

 proc xpnd(bs: uint, v: uint(32)): bigint {
   var rslt = 1: bigint; var bsm = bs: bigint; var vm = v: uint;
   while vm > 0 { if vm & 1 then rslt *= bsm; bsm *= bsm; vm >>= 1; }
   return rslt;
 }
 const (x2, x3, x5) = x;
 return xpnd(2: uint, x2) * xpnd(3: uint, x3) * xpnd(5: uint, x5);

}

iter nodupsHammingL(): Trival {

 const lb2 = 1: real(64); // log base 2 of 2!
 const lb3 = log2(3: real(64)); const lb5 = log2(5: real(64));
 record LogRep { var lg: real(64); var x2: uint(32);
                 var x3: uint(32); var x5: uint(32); }
 proc <(a: LogRep, b: LogRep): bool { return a.lg < b.lg; }
 proc mul2(x: LogRep): LogRep {
   return new LogRep(x.lg + lb2, x.x2 + 1, x.x3, x.x5); }
 proc mul3(x: LogRep): LogRep {
   return new LogRep(x.lg + lb3, x.x2, x.x3 + 1, x.x5); }
 proc mul5(x: LogRep): LogRep {
   return new LogRep(x.lg + lb5, x.x2, x.x3, x.x5 + 1); }
 const one = new LogRep(0, 0, 0, 0);
 var mdom = { 0 .. 0 }; var m: [mdom] LogRep; // init so can double!
 var hdom = { 0 .. 0 }; var h: [hdom] LogRep; // init so can double!
 var x5 = mul5(one); var mrg = mul3(one); var x532 = mul2(one);
 var x53 = mul3(mrg); // 9 -> already advanced one step!
 var ih, jm, i, j: int;
 yield (0: uint(32), 0: uint(32), 0: uint(32)); // first trivial case of 1
 while true {
   const cph = h.size; // move in place to avoid allocation!
   if i + i >= cph {
     h[0 .. ih - i - 1] = h[i .. ih - 1]; ih -= i; i = 0; }
   if ih >= cph then hdom = { 0 .. cph + cph - 1 };
   if x532 < mrg { h[ih] = x532; x532 = mul2(h[i]); i += 1; }
   else {
     h[ih] = mrg; const cpm = m.size; // move in place to avoid allocation!
     if j + j >= cpm {
       m[0 .. jm - j - 1] = m[j .. jm - 1]; jm -= j; j = 0; }
     if jm >= cpm then mdom = { 0 .. cpm + cpm - 1 };
     if x53 < x5 { mrg = x53; x53 = mul3(m[j]); j += 1; }
     else { mrg = x5; x5 = mul5(x5); }
     m[jm] = mrg; jm += 1;
   }
   const rslt = h[ih]; yield (rslt.x2, rslt.x3, rslt.x5); ih += 1;
 }

}

// test it... write("The first 20 hamming numbers are: "); var cnt = 0: uint(64); for h in nodupsHammingL() {

 if cnt >= 20 then break; cnt += 1; write(" ", trival2bigint(h)); }

write("\nThe 1691st hamming number is "); cnt = 1; for h in nodupsHammingL() {

 if cnt >= 1691 { writeln(trival2bigint(h)); break; } cnt += 1; }

write("The ", nth, "th hamming number is "); var timer: Timer; cnt = 1; timer.start(); for h in nodupsHammingL() {

 if cnt >= nth { write(trival2bigint(h)); break; } cnt += 1; }

timer.stop(); writeln(".\nThis last took ", timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</lang>

Output:

The first 20 hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
The 1691st hamming number is 2125764000
The 1000000th hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 8.926 milliseconds.

The above time is as run on an Intel Skylake i5-6500 at 3.6 GHz (turbo, single-threaded).

As you can see, the time expended for the required task is almost too fast to measure, meaning that much of the time expended in previous versions was just the time doing multi-precision arithmetic; the program takes about 13.4 seconds to find the billionth Hamming number.

Very Fast Algorithm Using a Sorted Error Band

The above code is about as fast as one can go generating sequences; however, if one is willing to forego sequences and just calculate the nth Hamming number (repeatedly), then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: Regular Number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows:

Translation of: Nim

Works with: 1.22 version for zero based tuple indices

<lang chapel>use BigInteger; use Math; use Sort; use Time;

config const nth = 1000000: uint(64);

type TriVal = 3*uint(32);

proc trival2bigint(x: TriVal): bigint {

 proc xpnd(bs: uint, v: uint(32)): bigint {
   var rslt = 1: bigint; var bsm = bs: bigint; var vm = v: uint;
   while vm > 0 { if vm & 1 then rslt *= bsm; bsm *= bsm; vm >>= 1; }
   return rslt;
 }
 const (x2, x3, x5) = x;
 return xpnd(2: uint, x2) * xpnd(3: uint, x3) * xpnd(5: uint, x5);

}

proc nthHamming(n: uint(64)): TriVal {

 if n < 1 {
   writeln("nthHamming - argument must be at least one!"); exit(1); }
 if n < 2 then return (0: uint(32), 0: uint(32), 0: uint(32)); // TriVal for 1

 type LogRep = (real(64), uint(32), uint(32), uint(32));
 record Comparator {} // used for sorting in reverse order!
 proc Comparator.compare(a: LogRep, b: LogRep): real(64) {
   return b[0] - a[0]; }
 var logrepComp: Comparator;

 const lb3 = log2(3.0: real(64)); const lb5 = log2(5.0: real(64));
 const fctr = 6.0: real(64) * lb3 * lb5;
 const crctn = log2(sqrt(30.0: real(64))); // log base 2 of sqrt 30
 // from Wikipedia Regular Numbers formula...
 const lgest = (fctr * n: real(64))**(1.0: real(64) / 3.0: real(64)) - crctn;
 const frctn = if n < 1000000000 then 0.509: real(64) else 0.105: real(64);
 const lghi = (fctr * (n: real(64) + frctn * lgest))**
                (1.0: real(64) / 3.0: real(64)) - crctn;
 const lglo = 2.0: real(64) * lgest - lghi; // lower limit of the upper "band"
 var count = 0: uint(64); // need to use extended precision, might go over
 var bndi = 0; var dombnd = { 0 .. bndi }; // one value so doubling size works!
 var bnd: [dombnd] LogRep; const klmt = (lghi / lb5): uint(32);
 for k in 0 .. klmt { // i, j, k values can be just uint(32) values!
   const p = k: real(64) * lb5; const jlmt = ((lghi - p) / lb3): uint(32);
   for j in 0 .. jlmt {
     const q = p + j: real(64) * lb3;
     const ir = lghi - q; const lg = q + floor(ir); // current log value (est)
     count += ir: uint(64) + 1;
     if lg >= lglo {
       const sz = dombnd.size; if bndi >= sz then dombnd = { 0..sz + sz - 1 };
       bnd[bndi] = (lg, ir: uint(32), j, k); bndi += 1;
     }
   }
 }
 if n > count {
   writeln("nth_hamming: band high estimate is too low!"); exit(1); }
 dombnd = { 0 .. bndi - 1 }; const ndx = (count - n): int;
 if ndx >= dombnd.size {
   writeln("nth_hamming: band low estimate is too high!"); exit(1); }
 sort(bnd, comparator = logrepComp); // descending order leaves zeros at end!

 const rslt = bnd[ndx]; return (rslt[1], rslt[2], rslt[3]);

}

// test it... write("The first 20 Hamming numbers are: "); for i in 1 .. 20 do write(" ", trival2bigint(nthHamming(i: uint(64))));

writeln("\nThe 1691st hamming number is ",

       trival2bigint(nthHamming(1691: uint(64))));

var timer: Timer; timer.start(); const answr = nthHamming(nth); timer.stop(); write("The ", nth, "th Hamming number is 2**",

     answr[0], " * 3**", answr[1], " * 5**", answr[2]);

const lgrslt = (answr[0]: real(64) + answr[1]: real(64) * log2(3: real(64)) +

               answr[2]: real(64) * log2(5: real(64))) * log10(2: real(64));

const whl = lgrslt: uint(64); const frac = lgrslt - whl: real(64); write(",\nwhich is approximately ", 10: real(64)**frac, "E+", whl); const bganswr = trival2bigint(answr); const answrstr = bganswr: string; const asz = answrstr.size; writeln(" and has ", asz, " digits."); if asz <= 2000 then write("Can be printed as: ", answrstr); else write("It's too long to print"); writeln("!\nThis last took ",

       timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
The 1691st hamming number is 2125764000
The 1000000th Hamming number is 2**55 * 3**47 * 5**64,
which is approximately 5.19313E+83 and has 84 digits.
Can be printed as:  519312780448388736089589843750000000000000000000000000000000000000000000000000000000!
This last took 0.0 milliseconds.

As you can see, the execution time is much too small to be measured. The billionth number in the sequence can be calculated in about 15 milliseconds and the trillionth in about 0.359 seconds. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit. However, this version gives inaccurate results much about the 1e13th Hamming number due to the log base two (double) approximate representation not having enough precision to accurately sort the values put into the error band array.

Alternate version with a greatly increased range without error

To solve the problem of inadequate precision in the double log base two representation, the following code uses a BigInt representation of the log value with about twice the significant bits, which is then sufficient to extend the usable range well beyond any reasonable requirement:

Translation of: Nim

Works with: 1.22 version for zero based tuple indices

<lang chapel>use BigInteger; use Math; use Sort; use Time;

config const nth = 1000000: uint(64);

type TriVal = 3*uint(32);

proc trival2bigint(x: TriVal): bigint {

 proc xpnd(bs: uint, v: uint(32)): bigint {
   var rslt = 1: bigint; var bsm = bs: bigint; var vm = v: uint;
   while vm > 0 { if vm & 1 then rslt *= bsm; bsm *= bsm; vm >>= 1; }
   return rslt;
 }
 const (x2, x3, x5) = x;
 return xpnd(2: uint, x2) * xpnd(3: uint, x3) * xpnd(5: uint, x5);

}

proc nthHamming(n: uint(64)): TriVal {

 if n < 1 {
   writeln("nthHamming - argument must be at least one!"); exit(1); }
 if n < 2 then return (0: uint(32), 0: uint(32), 0: uint(32)); // TriVal for 1

 type LogRep = (bigint, uint(32), uint(32), uint(32));
 record Comparator {} // used for sorting in reverse order!
 proc Comparator.compare(a: LogRep, b: LogRep): int {
   return (b[0] - a[0]): int; }
 var logrepComp: Comparator;

 const lb3 = log2(3.0: real(64)); const lb5 = log2(5.0: real(64));
 const bglb2 = "1267650600228229401496703205376": bigint;
 const bglb3 = "2009178665378409109047848542368": bigint;
 const bglb5 = "2943393543170754072109742145491": bigint;
 const fctr = 6.0: real(64) * lb3 * lb5;
 const crctn = log2(sqrt(30.0: real(64))); // log base 2 of sqrt 30
 // from Wikipedia Regular Numbers formula...
 const lgest = (fctr * n: real(64))**(1.0: real(64) / 3.0: real(64)) - crctn;
 const frctn = if n < 1000000000 then 0.509: real(64) else 0.105: real(64);
 const lghi = (fctr * (n: real(64) + frctn * lgest))**
                (1.0: real(64) / 3.0: real(64)) - crctn;
 const lglo = 2.0: real(64) * lgest - lghi; // lower limit of the upper "band"
 var count = 0: uint(64); // need to use extended precision, might go over
 var bndi = 0; var dombnd = { 0 .. bndi }; // one value so doubling size works!
 var bnd: [dombnd] LogRep; const klmt = (lghi / lb5): uint(32);
 for k in 0 .. klmt { // i, j, k values can be just uint(32) values!
   const p = k: real(64) * lb5; const jlmt = ((lghi - p) / lb3): uint(32);
   for j in 0 .. jlmt {
     const q = p + j: real(64) * lb3;
     const ir = lghi - q; const lg = q + floor(ir); // current log value (est)
     count += ir: uint(64) + 1;
     if lg >= lglo {
       const sz = dombnd.size; if bndi >= sz then dombnd = { 0..sz + sz - 1 };
       const bglg =
         bglb2 * ir: int(64) + bglb3 * j: int(64) + bglb5 * k: int(64);
       bnd[bndi] = (bglg, ir: uint(32), j, k); bndi += 1;
     }
   }
 }
 if n > count {
   writeln("nth_hamming: band high estimate is too low!"); exit(1); }
 dombnd = { 0 .. bndi - 1 }; const ndx = (count - n): int;
 if ndx >= dombnd.size {
   writeln("nth_hamming: band low estimate is too high!"); exit(1); }
 sort(bnd, comparator = logrepComp); // descending order leaves zeros at end!

 const rslt = bnd[ndx]; return (rslt[1], rslt[2], rslt[3]);

}

// test it... write("The first 20 Hamming numbers are: "); for i in 1 .. 20 do write(" ", trival2bigint(nthHamming(i: uint(64))));

writeln("\nThe 1691st hamming number is ",

       trival2bigint(nthHamming(1691: uint(64))));

var timer: Timer; timer.start(); const answr = nthHamming(nth); timer.stop(); write("The ", nth, "th Hamming number is 2**",

     answr[0], " * 3**", answr[1], " * 5**", answr[2]);

const lgrslt = (answr[0]: real(64) + answr[1]: real(64) * log2(3: real(64)) +

               answr[2]: real(64) * log2(5: real(64))) * log10(2: real(64));

const whl = lgrslt: uint(64); const frac = lgrslt - whl: real(64); write(",\nwhich is approximately ", 10: real(64)**frac, "E+", whl); const bganswr = trival2bigint(answr); const answrstr = bganswr: string; const asz = answrstr.size; writeln(" and has ", asz, " digits."); if asz <= 2000 then write("Can be printed as: ", answrstr); else write("It's too long to print"); writeln("!\nThis last took ",

       timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</lang>

The above code has the same output as before and doesn't take an appreciably different amount time to execute; it can produce the billionth Hamming number in about 31 milliseconds, the trillionth in about 0.546 seconds and the thousand trillionth (which is now possible without error) in about 39.36 seconds. Thus, it successfully extends the usable range of the algorithm to near the maximum expressible 64 bit number in a few hours of execution time on a modern desktop computer although the (2^64 - 1)th Hamming number can't be found due to the restrictions of the expressible range limit in sizing of the required error band.

That said, if one actually needed a sequence of Hamming numbers for fairly large ranges, one would likely be better off to make this last adjustment to the final logarithmic sequence version above as although this error-band version is extremely fast for single values, the accumulative cost for repeating use will be more than the incremental cost of the sequence version at some range limit.

Clojure

This version implements Dijkstra's merge solution, so is closely related to the Haskell version. <lang clojure>(defn smerge [xs ys]

 (lazy-seq
   (let [x (first xs),
         y (first ys),
         [z xs* ys*]
         (cond
           (< x y) [x (rest xs) ys]
           (> x y) [y xs (rest ys)]
           :else   [x (rest xs) (rest ys)])]
     (cons z (smerge xs* ys*)))))

(def hamming

 (lazy-seq
   (->> (map #(*' 5 %) hamming)
        (smerge (map #(*' 3 %) hamming))
        (smerge (map #(*' 2 %) hamming))
        (cons 1))))</lang>

Note that the above version uses a lot of space and time after calculating a few hundred thousand elements of the sequence. This is no doubt due to not avoiding the generation of duplicates in the sequences as well as its "holding on to the head": it maintains the entire generated sequences in memory.

Avoiding duplicates and reducing memory use

In order to fix the problems with the above program as to memory use and extra time expended, the following code implements the Haskell idea as a function so that it does not retain the pointers to the streams used so that they can be garbage collected from the beginning as they are consumed. it avoids duplicate number generation by using intermediate streams for each of the multiples and building each on the results of the last; also, it orders the streams from least dense to most so that the intermediate streams retained are as short as possible, with the "s5" stream only from one fifth to a third of the current value, the "s35" stream only between a third and a half of the current output value, and the s235 stream only between a half and the current output - as the sequence is not very dense with increasing range, mot many values need be retained:

Translation of: Haskell

<lang clojure>(defn hamming

 "Computes the unbounded sequence of Hamming 235 numbers."
 []
 (letfn [(merge [xs ys]
           (if (nil? xs) ys
             (let [xv (first xs), yv (first ys)]
               (if (< xv yv) (cons xv (lazy-seq (merge (next xs) ys)))
                             (cons yv (lazy-seq (merge xs (next ys)))))))),
         (smult [m s] ;; equiv to map (* m) s -- faster
           (cons (*' m (first s)) (lazy-seq (smult m (next s))))),
         (u [s n] (let [r (atom nil)]
                     (reset! r (merge s (smult n (cons 1 (lazy-seq @r)))))))]
   (cons 1 (lazy-seq (reduce u nil (list 5 3 2))))))</lang>

Much of the time expended for larger ranges (say 10 million or more) is due to the time doing extended precision arithmetic, with also a significant percentage spent in garbage collection. Following is the output from the REPL after compiling the program:

After compiling code in REPL:

Output:

(take 20 (hamming))
(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
(->> (hamming) (drop 1690) (first) (time))
"Elapsed time: 1.105582 msecs"
2125764000
(->> (hamming) (drop 999999) (first) (time))
"Elapsed time: 447.561128 msecs"
519312780448388736089589843750000000000000000000000000000000000000000000000000000000N

So that generated '.class' files in a folder or a generated '.jar' file (possibly standalone, containing the run time library) run at about the same speed as inside the IDE (after compilation), the Leiningen "project.clj" file needs to be modified to contain the following line so as to eliminate JVM options that slow the performance:

  :jvm-opts ^:replace []

CoffeeScript

<lang coffeescript># Generate hamming numbers in order. Hamming numbers have the

1. property that they don't evenly divide any prime numbers outside
2. a given set, such as [2, 3, 5].

generate_hamming_sequence = (primes, max_n) ->

 # We use a lazy algorithm, only ever keeping N candidates
 # in play, one for each of our seed primes.  Let's say
 # primes is [2,3,5].  Our virtual streams are these:
 #
 # hammings:    1,2,3,4,5,6,8,10,12,15,16,18,20,...
 # hammings*2:  2,4,6,9.10,12,16,20,24,30,32,36,40...
 # hammings*3:  3,6,9,12,15,18,24,30,36,45,...
 # hammings*5:  5,10,15,20,25,30,40,50,...
 #
 # After encountering 40 for the last time, our candidates 
 # will be
 #   50 = 2 * 25
 #   45 = 3 * 15
 #   50 = 5 * 10
 # Then, after 45
 #   50 = 2 * 25
 #   48 = 3 * 16 <= new
 #   50 = 5 * 10 
 hamming_numbers = [1]
 candidates = ([p, p, 1] for p in primes)
 last_number = 1
 while hamming_numbers.length < max_n
   # Get the next candidate Hamming Number tuple.
   i = min_idx(candidates)
   candidate = candidates[i]
   [n, p, seq_idx] = candidate
   
   # Add to sequence unless it's a duplicate.
   if n > last_number
     hamming_numbers.push n
     last_number = n

   # Replace the candidate with its successor (based on
   # p = 2, 3, or 5).
   #
   # This is the heart of the algorithm.  Let's say, over the 
   # primes [2,3,5], we encounter the hamming number 32 based on it being 
   # 2 * 16, where 16 is the 12th number in the sequence.
   # We'll be passed in [32, 2, 12] as candidate, and
   # hamming_numbers will be [1,2,3,4,5,6,8,9,10,12,16,18,...]
   # by now.  The next candidate we need to enqueue is
   # [36, 2, 13], where the numbers mean this:
   #
   #    36 - next multiple of 2 of a Hamming number
   #     2 - prime number
   #    13 - 1-based index of 18 in the sequence
   # 
   # When we encounter [36, 2, 13], we will then enqueue
   # [40, 2, 14], based on 20 being the 14th hamming number.
   q = hamming_numbers[seq_idx]
   candidates[i] = [p*q, p, seq_idx+1]
   
 hamming_numbers

min_idx = (arr) ->

 # Don't waste your time reading this--it just returns
 # the index of the smallest tuple in an array, respecting that
 # the tuples may contain integers. (CS compiles to JS, which is
 # kind of stupid about sorting.  There are libraries to work around
 # the limitation, but I wanted this code to be standalone.)
 less_than = (tup1, tup2) ->
   i = 0
   while i < tup2.length
     return true if tup1[i] <= tup2[i]
     return false if tup1[i] > tup2[i]
     i += 1

 min_i = 0
 for i in [1...arr.length]
   if less_than arr[i], arr[min_i]
     min_i = i
 return min_i

primes = [2, 3, 5] numbers = generate_hamming_sequence(primes, 10000) console.log numbers[1690] console.log numbers[9999]</lang>

Common Lisp

Maintaining three queues, popping the smallest value every time. <lang lisp>(defun next-hamm (factors seqs)

 (let ((x (apply #'min (map 'list #'first seqs))))
   (loop for s in seqs

for f in factors for i from 0 with add = t do (if (= x (first s)) (pop s))  ;; prevent a value from being added to multiple lists (when add (setf (elt seqs i) (nconc s (list (* x f)))) (if (zerop (mod x f)) (setf add nil)))

   finally (return x))))

(loop with factors = '(2 3 5)

     with seqs    = (loop for i in factors collect '(1))
     for n from 1 to 1000001 do
     (let ((x (next-hamm factors seqs)))

(if (or (< n 21) (= n 1691) (= n 1000000)) (format t "~d: ~d~%" n x))))</lang> A much faster method: <lang lisp>(defun hamming (n)

 (let ((fac '(2 3 5))

(idx (make-array 3 :initial-element 0)) (h (make-array (1+ n) :initial-element 1 :element-type 'integer)))

   (loop for i from 1 to n

with e with x = '(1 1 1) do (setf e (setf (aref h i) (apply #'min x)) x (loop for y in x for f in fac for j from 0 collect (if (= e y) (* f (aref h (incf (aref idx j)))) y))))

   (aref h n)))

(loop for i from 1 to 20 do

     (format t "~2d: ~d~%" i (hamming i)))

(loop for i in '(1691 1000000) do

     (format t "~d: ~d~%" i (hamming i)))</lang>

Output:

 1: 1
 2: 2
 3: 3
 4: 4
 5: 5
 6: 6
 7: 8
 8: 9
 9: 10
10: 12
11: 15
12: 16
13: 18
14: 20
15: 24
16: 25
17: 27
18: 30
19: 32
20: 36
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Crystal

Translation of: Bc

<lang ruby>require "big"

def hamming(limit)

 h = Array.new(limit, 1.to_big_i)     # h = Array.new(limit+1, 1.to_big_i)
 x2, x3, x5 = 2.to_big_i, 3.to_big_i, 5.to_big_i
 i, j, k = 0, 0, 0
 (1...limit).each do |n|              # (1..limit).each do |n|
   h[n] = Math.min(x2, Math.min(x3, x5))
   x2 = 2 * h[i += 1] if x2 == h[n]
   x3 = 3 * h[j += 1] if x3 == h[n]
   x5 = 5 * h[k += 1] if x5 == h[n]
 end
 h[limit - 1]

end

start = Time.monotonic print "Hamming Number (1..20): "; (1..20).each { |i| print "#{hamming(i)} " } puts puts "Hamming Number 1691: #{hamming 1691}" puts "Hamming Number 1,000,000: #{hamming 1_000_000}" puts "Elasped Time: #{(Time.monotonic - start).total_seconds} secs" </lang>

System: I7-6700HQ, 3.5 GHz, Linux Kernel 5.6.17, Crystal 0.35
Run as: $ crystal run hammingnumbers.cr --release

Output:

Hamming Number (1..20): 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
Hamming Number 1691: 2125764000
Hamming Number 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Elasped Time: 0.21420532 secs

Functional Non-Duplicates Version

The above implementation is true to the original Dijkstra algorithm but it's one of the few times where Dijkstra's analysis wasn't complete; there has been developed a later algorithm that is at least twice as fast due to only processing non-duplicate Hamming numbers and keeping only the numbers as necessary for further extensions of the sequence (the tails of the lists). Although Crystal isn't really a functional language, it is capable of enough functional forms of code to be able to implement this new algorithm. The algorithm requires lazy lists, for which currently Crystal has no library module, but as Crystal does have full first class functions including the ability to capture environment variables as closures, the `LazyList` type is easy enough to implement, as in the following code:

Translation of: Kotlin

<lang ruby>require "big"

1. Unlike some languages like Kotlin, Crystal doesn't have a Lazy module,
2. but it has closures, so it is easy to implement a LazyList class;
3. Memoizes the results of the thunk so only executed once...

class LazyList(T)

 getter head
 @tail : LazyList(T)? = nil

 def initialize(@head : T, @thnk : Proc(LazyList(T)))
 end
 def initialize(@head : T, @thnk : Proc(Nil))
 end
 def initialize(@head : T, @thnk : Nil)
 end

 def tail # not thread safe without a lock/mutex...
   if thnk = @thnk
     @tail = thnk.call; @thnk = nil
   end
   @tail
 end

end

class Hammings

 include Iterator(BigInt)
 private BASES = [ 5, 3, 2 ] of Int32
 private EMPTY = nil.as(LazyList(BigInt)?)
 @ll : LazyList(BigInt)

 def initialize
   rst = uninitialized LazyList(BigInt)
   BASES.each.accumulate(EMPTY) { |u, n| Hammings.unify(u, n) }
            .skip(1).each { |ll| rst = ll.not_nil! }
   @ll = LazyList.new(BigInt.new(1), ->{ rst } )
 end

 protected def self.unify(s : LazyList(BigInt)?, n : Int32)
   r = uninitialized LazyList(BigInt)?
   if ss = s
     r = merge(ss, mults(n, LazyList.new(BigInt.new(1), -> { r.not_nil! })))
   else
     r = mults(n, LazyList.new(BigInt.new(1), -> { r.not_nil! }))
   end
   r
 end

 private def self.mults(m : Int32, lls : LazyList(BigInt))
   mlts = uninitialized Proc(LazyList(BigInt), LazyList(BigInt))
   mlts = -> (ill : LazyList(BigInt)) {
     LazyList.new(ill.head * m, -> { mlts.call(ill.tail.not_nil!) }) }
   mlts.call(lls)
 end

 private def self.merge(x : LazyList(BigInt), y : LazyList(BigInt))
   xhd = x.head; yhd = y.head
   if xhd < yhd
     LazyList.new(xhd, -> { merge(x.tail.not_nil!, y) })
   else
     LazyList.new(yhd, -> { merge(x, y.tail.not_nil!) })
   end
 end

 def next
   rslt = @ll.head; @ll = @ll.tail.not_nil!; rslt
 end

end

print "The first 20 Hamming numbers are: " Hammings.new.first(20).each { |h| print(" ", h) } print ".\r\nThe 1691st Hamming number is " Hammings.new.skip(1690).first(1).each { |h| print h } print ".\r\nThe millionth Hamming number is " start_time = Time.monotonic Hammings.new.skip(999_999).first(1).each { |h| print h } elpsd = (Time.monotonic - start_time).total_milliseconds printf(".\r\nThis last took %f milliseconds.\r\n", elpsd)</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36.
The 1691st Hamming number is 2125764000.
The millionth Hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 162.713293 milliseconds.

The time is as run on an Intel SkyLake i5-6500 CPU running at 3.6 GHz single threaded as here. The code is a little slower than the fastest functional languages, such as Haskell or Kotlin due to that the speed of the Boehm Garbage Collector used by Crystal isn't as tuned for the many small allocations as necessary for functional forms of code such as the `LazyList` as those other languages which use memory pools to reduce the allocation/deallocation time from many small blocks of memory; that said, many common languages are much slower than this for functional algorithms due to their memory allocators being even slower and less tuned for this use.

About a quarter of the time is spent doing extended precision calculations (which time will increase disproportional to range as the numbers get larger) but over two thirds of the the is spent just handling memory allocations/deallocations.

Functional Non-Duplicates Version Using Log Estimations

In order to show the time expended in multi-precision integer calculations, the following code implements the same algorithm as above but uses logarithmic estimations rather than multi-precision integer arithmetic to compute each instance of the Hamming number sequence, only converting to `BigInt` for the results:

<lang ruby>require "big"

1. Unlike some languages like Kotlin, Crystal doesn't have a Lazy module,
2. but it has closures, so it is easy to implement a LazyList class;
3. Memoizes the results of the thunk so only executed once...

class LazyList(T)

 getter head
 @tail : LazyList(T)? = nil

 def initialize(@head : T, @thnk : Proc(LazyList(T)))
 end
 def initialize(@head : T, @thnk : Proc(Nil))
 end
 def initialize(@head : T, @thnk : Nil)
 end

 def tail # not thread safe without a lock/mutex...
   if thnk = @thnk
     @tail = thnk.call; @thnk = nil
   end
   @tail
 end

end

class LogRep

 private LOG2_2 = 1.0_f64
 private LOG2_3 = Math.log2 3.0_f64
 private LOG2_5 = Math.log2 5.0_f64

 def initialize(@logrep : Float64, @x2 : Int32, @x3 : Int32, @x5 : Int32)
 end

 def self.mult2(x : LogRep)
   LogRep.new(x.@logrep + LOG2_2, x.@x2 + 1, x.@x3, x.@x5)
 end

 def self.mult3(x : LogRep)
   LogRep.new(x.@logrep + LOG2_3, x.@x2, x.@x3 + 1, x.@x5)
 end

 def self.mult5(x : LogRep)
   LogRep.new(x.@logrep + LOG2_5, x.@x2, x.@x3, x.@x5 + 1)
 end

 def <(other : LogRep)
   self.@logrep < other.@logrep
 end

 def toBigInt
   expnd = -> (x : Int32, mlt : Int32) do
     rslt = BigInt.new(1); m = BigInt.new(mlt)
     while x > 0
       rslt *= m if (x & 1) > 0; m *= m; x >>= 1
     end
     rslt
   end
   expnd.call(@x2, 2) * expnd.call(@x3, 3) * expnd.call(@x5, 5)
 end

end

class HammingsLogRep

 include Iterator(LogRep)
 private BASES = [ -> (x : LogRep) { LogRep.mult5 x },
                   -> (x : LogRep) { LogRep.mult3 x },
                   -> (x : LogRep) { LogRep.mult2 x } ]
 private EMPTY = nil.as(LazyList(LogRep)?)
 private ONE = LogRep.new(0.0, 0, 0, 0)
 @ll : LazyList(LogRep)

 def initialize
   rst = uninitialized LazyList(LogRep)
   BASES.each.accumulate(EMPTY) { |u, n| HammingsLogRep.unify(u, n) }
            .skip(1).each { |ll| rst = ll.not_nil! }
   @ll = LazyList.new(ONE, ->{ rst } )
 end

 protected def self.unify(s : LazyList(LogRep)?, n : LogRep -> LogRep)
   r = uninitialized LazyList(LogRep)?
   if ss = s
     r = merge(ss, mults(n, LazyList.new(ONE, -> { r.not_nil! })))
   else
     r = mults(n, LazyList.new(ONE, -> { r.not_nil! }))
   end
   r
 end

 private def self.mults(m : LogRep -> LogRep, lls : LazyList(LogRep))
   mlts = uninitialized Proc(LazyList(LogRep), LazyList(LogRep))
   mlts = -> (ill : LazyList(LogRep)) {
     LazyList.new(m.call(ill.head), -> { mlts.call(ill.tail.not_nil!) }) }
   mlts.call(lls)
 end

 private def self.merge(x : LazyList(LogRep), y : LazyList(LogRep))
   xhd = x.head; yhd = y.head
   if xhd < yhd
     LazyList.new(xhd, -> { merge(x.tail.not_nil!, y) })
   else
     LazyList.new(yhd, -> { merge(x, y.tail.not_nil!) })
   end
 end

 def next
   rslt = @ll.head; @ll = @ll.tail.not_nil!; rslt
 end

end

print "The first 20 Hamming numbers are: " HammingsLogRep.new.first(20).each { |h| print(" ", h.toBigInt) } print ".\r\nThe 1691st Hamming number is " HammingsLogRep.new.skip(1690).first(1).each { |h| print h.toBigInt } print ".\r\nThe millionth Hamming number is " start_time = Time.monotonic HammingsLogRep.new.skip(999_999).first(1).each { |h| print h.toBigInt } elpsd = (Time.monotonic - start_time).total_milliseconds printf(".\r\nThis last took %f milliseconds.\r\n", elpsd)</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36.
The 1691st Hamming number is 2125764000.
The millionth Hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 131.661941 milliseconds.

As can be seen by comparing with the above results using the same Intel Skylake i5-6500 CPU, this is about 20 percent faster due to less time spent doing the increasingly long multi-precision `BigInt`'s. Note that using a `struct` rather than a `class` would make this code about twice as slow due to the larger memory copies required in copying "value's" rather than "reference" pointers.

Functional Non-Duplicates Version Using Log Estimations and Imperative Code

To show that the majority of the time for the above implementations is used in memory allocations/deallocations for the functional lazy list form of code, the following code implements this imperatively by using home-grown "growable" arrays; these "growable" arrays were hand implemented using pointer allocations to avoid the automatic bounds checking done for conventional Array's; note that the `LogRep` is now a `struct` rather than a `class` as now there aren't many value copies and to save the quite large amount of time required to allocation/deallocate memory as if `class`'s were used:

Translation of: Kotlin

<lang ruby>require "big"

struct LogRep

 private LOG2_2 = 1.0_f64
 private LOG2_3 = Math.log2 3.0_f64
 private LOG2_5 = Math.log2 5.0_f64

 def initialize(@logrep : Float64, @x2 : Int32, @x3 : Int32, @x5 : Int32)
 end

 def self.mult2(x : LogRep)
   LogRep.new(x.@logrep + LOG2_2, x.@x2 + 1, x.@x3, x.@x5)
 end

 def self.mult3(x : LogRep)
   LogRep.new(x.@logrep + LOG2_3, x.@x2, x.@x3 + 1, x.@x5)
 end

 def self.mult5(x : LogRep)
   LogRep.new(x.@logrep + LOG2_5, x.@x2, x.@x3, x.@x5 + 1)
 end

 def <(other : LogRep)
   self.@logrep < other.@logrep
 end

 def toBigInt
   expnd = -> (x : Int32, mlt : Int32) do
     rslt = BigInt.new(1); m = BigInt.new(mlt)
     while x > 0
       rslt *= m if (x & 1) > 0; m *= m; x >>= 1
     end
     rslt
   end
   expnd.call(@x2, 2) * expnd.call(@x3, 3) * expnd.call(@x5, 5)
 end

end

class HammingsImpLogRep

 include Iterator(LogRep)
 private ONE = LogRep.new(0.0, 0, 0, 0)
 @x5 = LogRep.mult5 ONE; @mrg = LogRep.mult3 ONE; @x2 = LogRep.mult2 ONE
 @hh = 0; @mh = 0; @hl = 0; @ml = 0; @rslt = ONE
 @ha = Pointer(LogRep).malloc 1; @ma = Pointer(LogRep).malloc 1
 @hasz = 1; @masz = 1

 def initialize
   @x3 = LogRep.mult3 @mrg # already advance one step to 9
 end

 def next   
   if @hl + @hl >= @hasz # @ha.size
     @ha.move_from(@ha + @hl, @hh - @hl); @hh -= @hl; @hl = 0
   end
   if @hh >= @hasz
     @hasz += @hasz; nha = Pointer(LogRep).malloc @hasz
     nha.move_from(@ha, @hh); @ha = nha
   end
   if @x2 < @mrg
     @ha[@hh] = @x2; @x2 = LogRep.mult2(@ha[@hl]); @hl += 1
   else
     @ha[@hh] = @mrg
     if @ml + @ml >= @masz # @ma.size
       @ma.move_from(@ma + @ml, @mh - @ml); @mh -= @ml; @ml = 0
     end
     if @mh >= @masz
       @masz += @masz; nma = Pointer(LogRep).malloc @masz
       nma.move_from(@ma, @mh); @ma = nma
     end
     if @x3 < @x5
       @mrg = @x3; @x3 = LogRep.mult3(@ma[@ml]); @ml += 1
     else
       @mrg = @x5; @x5 = LogRep.mult5(@x5)
     end      
     @ma[@mh] = @mrg; @mh += 1
   end
   rslt = @rslt; @rslt = @ha[@hh]; @hh += 1; rslt
 end

end

print "The first 20 Hamming numbers are: " HammingsImpLogRep.new.first(20).each { |h| print(" ", h.toBigInt) } print ".\r\nThe 1691st Hamming number is " HammingsImpLogRep.new.skip(1690).first(1).each { |h| print h.toBigInt } print ".\r\nThe millionth Hamming number is " start_time = Time.monotonic HammingsImpLogRep.new.skip(999_999).first(1).each { |h| print h.toBigInt } elpsd = (Time.monotonic - start_time).total_milliseconds printf(".\r\nThis last took %f milliseconds.\r\n", elpsd)</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36.
The 1691st Hamming number is 2125764000.
The millionth Hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 10.722201 milliseconds.

As can be seen by comparing with the above results using the same Intel Skylake i5-6500 CPU, this is about thirteen times faster than the version also using logarithmic representations due to less time spent doing memory allocations/deallocations for the imperative form of code.

D

Basic Version

This version keeps all numbers in memory, computing all the Hamming numbers up to the needed one. Performs constant number of operations per Hamming number produced. <lang d>import std.stdio, std.bigint, std.algorithm, std.range, core.memory;

auto hamming(in uint n) pure nothrow /*@safe*/ {

   immutable BigInt two = 2, three = 3, five = 5;
   auto h = new BigInt[n];
   h[0] = 1;
   BigInt x2 = 2, x3 = 3, x5 = 5;
   size_t i, j, k;

   foreach (ref el; h.dropOne) {
       el = min(x2, x3, x5);
       if (el == x2) x2 = two   * h[++i];
       if (el == x3) x3 = three * h[++j];
       if (el == x5) x5 = five  * h[++k];
   }
   return h.back;

}

void main() {

   GC.disable;
   iota(1, 21).map!hamming.writeln;
   1_691.hamming.writeln;
   1_000_000.hamming.writeln;

}</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Runtime is about 1.6 seconds with LDC2.

Alternative Version 1

This keeps $O(n^{2/3})$ numbers in memory, but over-computes a sequence by a factor of about $\Theta (n^{2/3})$, calculating extra multiples past that as well. Incurs an extra $O(\log n)$ factor of operations per each number produced (reinserting its multiples into a tree). Doesn't stop when the target number is reached, instead continuing until it is no longer needed:

Translation of: Java

<lang d>import std.stdio, std.bigint, std.container, std.algorithm, std.range,

      core.memory;

BigInt hamming(in int n) in {

  assert(n > 0);

} body {

   auto frontier = redBlackTree(2.BigInt, 3.BigInt, 5.BigInt);
   auto lowest = 1.BigInt;
   foreach (immutable _; 1 .. n) {
       lowest = frontier.front;
       frontier.removeFront;
       frontier.insert(lowest * 2);
       frontier.insert(lowest * 3);
       frontier.insert(lowest * 5);
   }
   return lowest;

}

void main() {

   GC.disable;
   writeln("First 20 Hamming numbers: ", iota(1, 21).map!hamming);
   writeln("hamming(1691) = ", 1691.hamming);
   writeln("hamming(1_000_000) = ", 1_000_000.hamming);

}</lang>

Output:

First 20 Hamming numbers: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
hamming(1691) = 2125764000
hamming(1_000_000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

About 3.2 seconds run time with LDC2.

Alternative Version 2

Does exactly what the first version does, creating an array and filling it with Hamming numbers, keeping the three back pointers into the sequence for next multiples calculations, except that it represents the numbers as their coefficients triples and their logarithm values (for comparisons), thus saving on BigInt calculations.

Translation of: C

<lang d>import std.stdio: writefln; import std.bigint: BigInt; import std.conv: text; import std.numeric: gcd; import std.algorithm: copy, map; import std.array: array; import core.stdc.stdlib: calloc; import std.math: log; // ^^

// Number of factors. enum NK = 3;

enum MAX_HAM = 10_000_000; static assert(gcd(NK, MAX_HAM) == 1);

enum int[NK] factors = [2, 3, 5];

/// K-smooth numbers (stored as their exponents of each factor). struct Hamming {

   double v; // Log of the number, for convenience.
   ushort[NK] e; // Exponents of each factor.

   public static __gshared immutable double[factors.length] inc =
       factors[].map!log.array;

   bool opEquals(in ref Hamming y) const pure nothrow @nogc {
       //return this.e == y.e; // Too much slow.
       foreach (immutable i; 0 .. this.e.length)
           if (this.e[i] != y.e[i])
               return false;
       return true;
   }

   void update() pure nothrow @nogc {
       //this.v = dotProduct(inc, this.e); // Too much slow.
       this.v = 0.0;
       foreach (immutable i; 0 .. this.e.length)
           this.v += inc[i] * this.e[i];
   }

   string toString() const {
       BigInt result = 1;
       foreach (immutable i, immutable f; factors)
           result *= f.BigInt ^^ this.e[i];
       return result.text;
   }

}

// Global variables. __gshared Hamming[] hams; __gshared Hamming[NK] values;

nothrow @nogc static this() {

   // Slower than calloc if you don't use all the MAX_HAM items.
   //hams = new Hamming[MAX_HAM];

   auto ptr = cast(Hamming*)calloc(MAX_HAM, Hamming.sizeof);
   static const err = new Error("Not enough memory.");
   if (!ptr)
       throw err;
   hams = ptr[0 .. MAX_HAM];

   foreach (immutable i, ref v; values) {
       v.e[i] = 1;
       v.v = Hamming.inc[i];
   }

}

ref Hamming getHam(in size_t n) nothrow @nogc in {

   assert(n <= MAX_HAM);

} body {

   // Most of the time v can be just incremented, but eventually
   // floating point precision will bite us, so better recalculate.
   __gshared static size_t[NK] idx;
   __gshared static int n_hams;

   for (; n_hams < n; n_hams++) {
       {
           // Find the index of the minimum v.
           size_t ni = 0;
           foreach (immutable i; 1 .. NK)
               if (values[i].v < values[ni].v)
                   ni = i;

           hams[n_hams] = values[ni];
           hams[n_hams].update;
       }

       foreach (immutable i; 0 .. NK)
           if (values[i] == hams[n_hams]) {
               values[i] = hams[idx[i]];
               idx[i]++;
               values[i].e[i]++;
               values[i].update;
           }
   }

   return hams[n - 2];

}

void main() {

   foreach (immutable n; [1691, 10 ^^ 6, MAX_HAM])
       writefln("%8d: %s", n, n.getHam);

}</lang> The output is similar to the second C version. Runtime is about 0.11 seconds if MAX_HAM = 1_000_000 (as the task requires), and 0.90 seconds if MAX_HAM = 10_000_000.

Alternative Version 3

This version is similar to the precedent, but frees unused values. It's a little slower than the precedent version, but it uses much less RAM, so it allows to compute the result for larger n. <lang d>import std.stdio: writefln; import std.bigint: BigInt; import std.conv: text; import std.algorithm: map; import std.array: array; import core.stdc.stdlib: malloc, calloc, free; import std.math: log; // ^^

// Number of factors. enum NK = 3;

__gshared immutable int[NK] primes = [2, 3, 5]; __gshared immutable double[NK] lnPrimes = primes[].map!log.array;

/// K-smooth numbers (stored as their exponents of each factor).

struct Hamming {

   double ln; // Log of the number.
   ushort[NK] e; // Exponents of each factor.
   Hamming* next;
   size_t n;

   // Recompute the logarithm from the exponents.
   void recalculate() pure nothrow @safe @nogc {
       this.ln = 0.0;
       foreach (immutable i, immutable ei; this.e)
           this.ln += lnPrimes[i] * ei;
   }

   string toString() const {
       BigInt result = 1;
       foreach (immutable i, immutable f; primes)
           result *= f.BigInt ^^ this.e[i];
       return result.text;
   }

}

Hamming getHam(in size_t n) nothrow @nogc in {

   assert(n && n != size_t.max);

} body {

   static struct Candidate {
       typeof(Hamming.ln) ln;
       typeof(Hamming.e) e;

       void increment(in size_t n) pure nothrow @safe @nogc {
           e[n] += 1;
           ln += lnPrimes[n];
       }

       bool opEquals(T)(in ref T y) const pure nothrow @safe @nogc {
           // return this.e == y.e; // Slow.
           return !((this.e[0] ^ y.e[0]) |
                    (this.e[1] ^ y.e[1]) |
                    (this.e[2] ^ y.e[2]));
       }

       int opCmp(T)(in ref T y) const pure nothrow @safe @nogc {
           return (ln > y.ln) ? 1 : (ln < y.ln ? -1 : 0);
       }
   }

   static struct HammingIterator { // Not a Range.
       Candidate cand;
       Hamming* base;
       size_t primeIdx;

       this(in size_t i, Hamming* b) pure nothrow @safe @nogc {
           primeIdx = i;
           base = b;
           cand.e = base.e;
           cand.ln = base.ln;
           cand.increment(primeIdx);
       }

       void next() pure nothrow @safe @nogc {
           base = base.next;
           cand.e = base.e;
           cand.ln = base.ln;
           cand.increment(primeIdx);
       }
   }

   HammingIterator[NK] its;
   Hamming* head = cast(Hamming*)calloc(Hamming.sizeof, 1);
   Hamming* freeList, cur = head;
   Candidate next;

   foreach (immutable i, ref it; its)
       it = HammingIterator(i, cur);

   for (size_t i = cur.n = 1; i < n; ) {
       auto leastReferenced = size_t.max;
       next.ln = double.max;
       foreach (ref it; its) {
           if (it.cand == *cur)
               it.next;
           if (it.base.n < leastReferenced)
               leastReferenced = it.base.n;
           if (it.cand < next)
               next = it.cand;
       }

       // Collect unferenced numbers.
       while (head.n < leastReferenced) {
           auto tmp = head;
           head = head.next;
           tmp.next = freeList;
           freeList = tmp;
       }

       if (!freeList) {
           cur.next = cast(Hamming*)malloc(Hamming.sizeof);
       } else {
           cur.next = freeList;
           freeList = freeList.next;
       }

       cur = cur.next;
       version (fastmath) {
           cur.ln = next.ln;
           cur.e = next.e;
       } else {
           cur.e = next.e;
           cur.recalculate; // Prevent FP error accumulation.
       }

       cur.n = i++;
       cur.next = null;
   }

   auto result = *cur;
   version (leak) {}
   else {
       while (head) {
           auto tmp = head;
           head = head.next;
           tmp.free;
       }

       while (freeList) {
           auto tmp = freeList;
           freeList = freeList.next;
           tmp.free;
       }
   }

   return result;

}

void main() {

   foreach (immutable n; [1691, 10 ^^ 6, 10_000_000])
       writefln("%8d: %s", n, n.getHam);

}</lang> The output is the same as the second alternative version.

Dart

In order to produce reasonable ranges of Hamming numbers, one needs the BigInt type, but processing of many BigInt's in generating a sequence slows the code; for that reason the following code records the determined values as a combination of an approximation of the log base two value and the triple of the powers of two, three and five, only generating the final output values as BigInt's as required: <lang dart>import 'dart:math';

final lb2of2 = 1.0; final lb2of3 = log(3.0) / log(2.0); final lb2of5 = log(5.0) / log(2.0);

class Trival {

 final double log2;
 final int twos;
 final int threes;
 final int fives;
 Trival mul2() {
   return Trival(this.log2 + lb2of2, this.twos + 1, this.threes, this.fives);
 }
 Trival mul3() {
   return Trival(this.log2 + lb2of3, this.twos, this.threes + 1, this.fives);
 }
 Trival mul5() {
   return Trival(this.log2 + lb2of5, this.twos, this.threes, this.fives + 1);
 }
 @override String toString() {
   return this.log2.toString() + " "
     + this.twos.toString() + " "
     + this.threes.toString() + " "
     + this.fives.toString();
 }
 const Trival(this.log2, this.twos, this.threes, this.fives);

}

Iterable<Trival> makeHammings() sync* {

 var one = Trival(0.0, 0, 0, 0);
 yield(one);
 var s532 = one.mul2();
 var mrg = one.mul3();
 var s53 = one.mul3().mul3(); // equivalent to 9 for advance step
 var s5 = one.mul5();
 var i = -1; var j = -1;
 List<Trival> h = [];
 List<Trival> m = [];
 Trival rslt;
 while (true) {
   if (s532.log2 < mrg.log2) {
     rslt = s532; h.add(s532); ++i; s532 = h[i].mul2(); 
   } else {
     rslt = mrg; h.add(mrg); 
     if (s53.log2 < s5.log2) {
       mrg = s53; m.add(s53); ++j; s53 = m[j].mul3();
     } else {
       mrg = s5; m.add(s5); s5 = s5.mul5();
     }
     if (j > (m.length >> 1)) {m.removeRange(0, j); j = 0; }
   }
   if (i > (h.length >> 1)) {h.removeRange(0, i); i = 0; }
   yield(rslt);
 }

}

BigInt trival2Int(Trival tv) {

 return BigInt.from(2).pow(tv.twos)
          * BigInt.from(3).pow(tv.threes)
          * BigInt.from(5).pow(tv.fives);

}

void main() {

 final numhams = 1000000000000;
 var hamseqstr = "The first 20 Hamming numbers are:  ( ";
 makeHammings().take(20)
     .forEach((h) => hamseqstr += trival2BigInt(h).toString() + " ");
 print(hamseqstr + ")");
 var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
 for (var i = 1; i <= 20; ++i) {
   nthhamseqstr += trival2BigInt(nthHamming(i)).toString() + " ";
 }
 print(nthhamseqstr + ")");
 final strt = DateTime.now().millisecondsSinceEpoch;
 final answr = makeHammings().skip(999999).first;
 final elpsd = DateTime.now().millisecondsSinceEpoch - strt;
 print("The ${numhams}th Hamming number is:  $answr");
 print("in full as:  ${trival2BigInt(answr)}");
 print("This test took $elpsd milliseconds.");

}</lang>

Output:

The first 20 Hamming numbers are:  ( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
The 1000000th Hamming number is:  278.096635606686 55 47 64
in full as:  519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This test took 311 milliseconds.

Due to using a mutable extendable List (Array) and mutation, the above generator is reasonably fast, and as well has the feature that List memory is recovered as it is no longer required, with a considerable saving in both execution speed and memory requirement.

Alternate extremely fast version using an "error band"

Although not a Hamming sequence generator, the following code uses the known characteristics of the distribution of Hamming numbers to just scan through to find all possibilities in a relatively narrow "error band" which then can be sorted based on the log base two approximation and the nth element determined inside that band; it has a huge advantage that memory requirements drop to O(n^(1/3)) and asymptotic execution complexity drops from O(n) to O(n^(2/3)) for an extremely fast execution speed (thanks to WillNess for the start of this algorithm as referenced in the Haskell section): Template:Translated from <lang dart>import 'dart:math';

final lb2of2 = 1.0; final lb2of3 = log(3.0) / log(2.0); final lb2of5 = log(5.0) / log(2.0);

class Trival {

 final double log2;
 final int twos;
 final int threes;
 final int fives;
 Trival mul2() {
   return Trival(this.log2 + lb2of2, this.twos + 1, this.threes, this.fives);
 }
 Trival mul3() {
   return Trival(this.log2 + lb2of3, this.twos, this.threes + 1, this.fives);
 }
 Trival mul5() {
   return Trival(this.log2 + lb2of5, this.twos, this.threes, this.fives + 1);
 }
 @override String toString() {
   return this.log2.toString() + " "
     + this.twos.toString() + " "
     + this.threes.toString() + " "
     + this.fives.toString();
 }
 const Trival(this.log2, this.twos, this.threes, this.fives);

}

BigInt trival2BigInt(Trival tv) {

 return BigInt.from(2).pow(tv.twos)
          * BigInt.from(3).pow(tv.threes)
          * BigInt.from(5).pow(tv.fives);

}

Trival nthHamming(int n) {

 if (n < 1) throw Exception("nthHamming:  argument must be higher than 0!!!");
 if (n < 7) {
   if (n & (n - 1) == 0) {
     final bts = n.bitLength - 1;
     return Trival(bts.toDouble(), bts, 0, 0);
   }
   switch (n) {
     case 3: return Trival(lb2of3, 0, 1, 0);
     case 5: return Trival(lb2of5, 0, 0, 1);
     case 6: return Trival(lb2of2 + lb2of3, 1, 1, 0);
   }
 }
 final fctr = 6.0 * lb2of3 * lb2of5;
 final crctn = log(sqrt(30.0)) / log(2.0);
 final lb2est = pow(fctr * n.toDouble(), 1.0/3.0) - crctn;
 final lb2rng = 2.0/lb2est;
 final lb2hi = lb2est + 1.0/lb2est;
 List<Trival> ebnd = [];
 var cnt = 0;
 for (var k = 0; k < (lb2hi / lb2of5).ceil(); ++k) {
   final lb2p = lb2hi - k * lb2of5;
   for (var j = 0; j < (lb2p / lb2of3).ceil(); ++j) {
     final lb2q = lb2p - j * lb2of3;
     final i = lb2q.floor(); final lb2frac = lb2q - i;
     cnt += i + 1;
     if (lb2frac <= lb2rng) {
       final lb2v = i * lb2of2 + j * lb2of3 + k * lb2of5;
       ebnd.add(Trival(lb2v, i, j, k));
     }
   }
 }
 ebnd.sort((a, b) => b.log2.compareTo(a.log2)); // descending order
 final ndx = cnt - n;
 if (ndx < 0) throw Exception("nthHamming:  not enough triples generated!!!");
 if (ndx >= ebnd.length) throw Exception("nthHamming:  error band is too narrow!!!");
 return ebnd[ndx];

}

void main() {

 final numhams = 1000000;
 var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
 for (var i = 1; i <= 20; ++i) {
   nthhamseqstr += trival2BigInt(nthHamming(i)).toString() + " ";
 }
 print(nthhamseqstr + ")");
 final strt = DateTime.now().millisecondsSinceEpoch;
 final answr = nthHamming(numhams);
 final elpsd = DateTime.now().millisecondsSinceEpoch - strt0;
 print("The ${numhams}th Hamming number is:  $answr");
 print("in full as:  ${trival2BigInt(answr)}");
 print("This test took $elpsd milliseconds.");

}</lang>

The output from the above code is the same as the above version but it is so fast that the time to find the millionth Hamming number is too small to be measured other than the Dart VM JIT time. It can find the billionth prime in a fraction of a second and the trillionth prime in seconds.

Increasing the range above 1e13 by using a BigInt log base two representation

For arguments higher than about 1e13, the precision of the Double log base two approximations used above is not adequate to do an accurate sort, but the algorithm continues to work (although perhaps slightly slower) by changing the code to use BigInt log base two representations as follows: <lang dart>import 'dart:math';

final biglb2of2 = BigInt.from(1) << 100; // 100 bit representations... final biglb2of3 = (BigInt.from(1784509131911002) << 50) + BigInt.from(134114660393120); final biglb2of5 = (BigInt.from(2614258625728952) << 50) + BigInt.from(773584997695443);

class BigTrival {

 final BigInt log2;
 final int twos;
 final int threes;
 final int fives;
 @override String toString() {
   return this.log2.toString() + " "
     + this.twos.toString() + " "
     + this.threes.toString() + " "
     + this.fives.toString();
 }
 const BigTrival(this.log2, this.twos, this.threes, this.fives);

}

BigInt bigtrival2BigInt(BigTrival tv) {

 return BigInt.from(2).pow(tv.twos)
          * BigInt.from(3).pow(tv.threes)
          * BigInt.from(5).pow(tv.fives);

}

BigTrival nthHamming(int n) {

 if (n < 1) throw Exception("nthHamming:  argument must be higher than 0!!!");
 if (n < 7) {
   if (n & (n - 1) == 0) {
     final bts = n.bitLength - 1;
     return BigTrival(BigInt.from(bts) << 100, bts, 0, 0);
   }
   switch (n) {
     case 3: return BigTrival(biglb2of3, 0, 1, 0);
     case 5: return BigTrival(biglb2of5, 0, 0, 1);
     case 6: return BigTrival(biglb2of2 + biglb2of3, 1, 1, 0);
   }
 }
 final fctr = lb2of3 * lb2of5 * 6;
 final crctn = log(sqrt(30.0)) / log(2.0);
 final lb2est = pow(fctr * n.toDouble(), 1.0/3.0) - crctn;
 final lb2rng = 2.0/lb2est;
 final lb2hi = lb2est + 1.0/lb2est;
 List<BigTrival> ebnd = [];
 var cnt = 0;
 for (var k = 0; k < (lb2hi / lb2of5).ceil(); ++k) {
   final lb2p = lb2hi - k * lb2of5;
   for (var j = 0; j < (lb2p / lb2of3).ceil(); ++j) {
     final lb2q = lb2p - j * lb2of3;
     final i = lb2q.floor(); final lb2frac = lb2q - i;
     cnt += i + 1;
     if (lb2frac <= lb2rng) {

// final lb2v = i * lb2of2 + j * lb2of3 + k * lb2of5; // ebnd.add(Trival(lb2v, i, j, k));

       final lb2v = BigInt.from(i) * biglb2of2
                       + BigInt.from(j) * biglb2of3
                       + BigInt.from(k) * biglb2of5;
       ebnd.add(BigTrival(lb2v, i, j, k));
     }
   }
 }
 ebnd.sort((a, b) => b.log2.compareTo(a.log2)); // descending order
 final ndx = cnt - n;
 if (ndx < 0) throw Exception("nthHamming:  not enough triples generated!!!");
 if (ndx >= ebnd.length) throw Exception("nthHamming:  error band is too narrow!!!");
 return ebnd[ndx];

}

void main() {

 final numhams = 1000000000;
 var nthhamseqstr = "The first 20 Hamming numbers are:  ( ";
 for (var i = 1; i <= 20; ++i) {
   nthhamseqstr += bigtrival2BigInt(nthHamming(i)).toString() + " ";
 }
 print(nthhamseqstr + ")");
 final strt = DateTime.now().millisecondsSinceEpoch;
 final answr = nthHamming(numhams);
 final elpsd = DateTime.now().millisecondsSinceEpoch - strt;
 print("The ${numhams}th Hamming number is:  $answr");
 print("in full as:  ${bigtrival2BigInt(answr)}");
 print("This test took $elpsd milliseconds.");

}</lang>

With these changes, the algorithm can find the 1e19'th prime in the order af days depending on the CPU used.

DCL

<lang DCL>$ limit = p1 $ $ n = 0 $ h_'n = 1 $ x2 = 2 $ x3 = 3 $ x5 = 5 $ i = 0 $ j = 0 $ k = 0 $ $ n = 1 $ loop: $ x = x2 $ if x3 .lt. x then $ x = x3 $ if x5 .lt. x then $ x = x5 $ h_'n = x $ if x2 .eq. h_'n $ then $ i = i + 1 $ x2 = 2 * h_'i $ endif $ if x3 .eq. h_'n $ then $ j = j + 1 $ x3 = 3 * h_'j $ endif $ if x5 .eq. h_'n $ then $ k = k + 1 $ x5 = 5 * h_'k $ endif $ n = n + 1 $ if n .le. limit then $ goto loop $ $ i = 0 $ loop2: $ write sys$output h_'i $ i = i + 1 $ if i .lt. 20 then $ goto loop2 $ $ n = limit - 1 $ write sys$output h_'n</lang>

Output:

Here's the output;

$ @hamming 1691
1
2
3
4
5
6
8
9
10
12
15
16
18
20
24
25
27
30
32
36
2125764000

Delphi

See Pascal.

Eiffel

<lang Eiffel> note description  : "Initial part, in order, of the sequence of Hamming numbers" math  : "[ Hamming numbers, also known as regular numbers and 5-smooth numbers, are natural integers that have 2, 3 and 5 as their only prime factors. ]" computer_arithmetic : "[ This version avoids integer overflow and stops at the last representable number in the sequence. ]" output  : "[

   			   Per requirements of the RosettaCode example, execution will produce items of indexes 1 to 20 and 1691.
   			   The algorithm (procedure `hamming') is more general and will produce the first `n' Hamming numbers
   			   for any `n'.
   			  ]"

source  : "This problem was posed in Edsger W. Dijkstra, A Discipline of Programming, Prentice Hall, 1978" date  : "8 August 2012" authors  : "Bertrand Meyer", "Emmanuel Stapf" revision  : "1.0" libraries  : "Relies on SORTED_TWO_WAY_LIST from EiffelBase" implementation : "[ Using SORTED_TWO_WAY_LIST provides an elegant illustration of how to implement a lazy scheme in Eiffel through the use of object-oriented data structures. ]" warning  : "[ The formatting (<lang>) specifications for Eiffel in RosettaCode are slightly obsolete: `note' and other newer keywords not supported, red color for manifest strings. This should be fixed soon. ]"

class APPLICATION

create make

feature {NONE} -- Initialization

make -- Print first 20 Hamming numbers, in order, and the 1691-st one. local Hammings: like hamming -- List of Hamming numbers, up to 1691-st one. do Hammings := hamming (1691) across 1 |..| 20 as i loop io.put_natural (Hammings.i_th (i.item)); io.put_string (" ") end io.put_new_line; io.put_natural (Hammings.i_th (1691)); io.put_new_line end

feature -- Basic operations

hamming (n: INTEGER): ARRAYED_LIST [NATURAL] -- First `n' elements (in order) of the Hamming sequence, -- or as many of them as will not produce overflow. local sl: SORTED_TWO_WAY_LIST [NATURAL] overflow: BOOLEAN first, next: NATURAL do create Result.make (n); create sl.make sl.extend (1); sl.start across 1 |..| n as i invariant -- "The numbers output so far are the first `i' - 1 Hamming numbers, in order". -- "Result.first is the `i'-th Hamming number." until sl.is_empty loop first := sl.first; sl.start Result.extend (first); sl.remove across << 2, 3, 5 >> as multiplier loop next := multiplier.item * first overflow := overflow or next <= first if not overflow and then not sl.has (next) then sl.extend (next) end end end end end </lang>

Output:

1
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

Elixir

<lang elixir>defmodule Hamming do

 def generater do
   queues = [{2, queue}, {3, queue}, {5, queue}]
   Stream.unfold({1, queues}, fn {n, q} -> next(n, q) end)
 end
 
 defp next(n, queues) do
   queues = Enum.map(queues, fn {m, queue} -> {m, push(queue, m*n)} end)
   min    = Enum.map(queues, fn {_, queue} -> top(queue) end) |> Enum.min
   queues = Enum.map(queues, fn {m, queue} ->
              {m, (if min==top(queue), do: erase_top(queue), else: queue)}
            end)
   {n, {min, queues}}
 end
 
 defp queue, do: {[], []}
 
 defp push({input, output}, term), do: {[term | input], output}
 
 defp top({input, []}), do: List.last(input)
 defp top({_, [h|_]}), do: h
 
 defp erase_top({input, []}), do: erase_top({[], Enum.reverse(input)})
 defp erase_top({input, [_|t]}), do: {input, t}

end

IO.puts "first twenty Hamming numbers:" IO.inspect Hamming.generater |> Enum.take(20) IO.puts "1691st Hamming number:" IO.puts Hamming.generater |> Enum.take(1691) |> List.last IO.puts "one millionth Hamming number:" IO.puts Hamming.generater |> Enum.take(1_000_000) |> List.last</lang>

Output:

first twenty Hamming numbers:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
1691st Hamming number:
2125764000
one millionth Hamming number:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Elm

The Elm language has many restrictions that make the implementation of the Hamming Number sequence algorithms difficult, as the classic Edsger Dijkstra algorithm as written in Haskell Hamming_numbers#The_classic_version cannot be written in Elm as current Elm forbids cyclic value references (the value "hamming" is back referenced three times), and the implementation wouldn't be efficient even if it could as the current Elm version 0.19.x has removed the "Lazy" package the would defer the memoization of the result of a computation as necessary in implementing Haskell's lazy lists. Thus, one has to implement memoization using a different data structure than a lazy list; however, all current Elm data structures forbid mutation and can only implement some sort of Copy On Write (COW), thus there is no implementation of a linear array and the "Array" module is a tree based structure (with some concessions to data blocks for slightly better performance) that will have a logarithmic execution complexity when the size increases above a minimum. In fact, all Elm data structures that could be used for this also have a logarithmic response (Dict, Set, Array). The implementation of List is not lazy so new elements can't be added to the "tail" but need to be added to the "head" for efficiency, which means if one wants to add higher elements to a list in increasing order, one needs to (COW) reverse the List (twice) in order to do it!

The solution here uses a pure functional implementation of a Min Heap (Binary Heap) Priority Queue so that the minimum element can be viewed in O(1) time although inserting new elements/replacing elements still takes O(log n) time where "n" is the number of elements in the queue. As written, no queue needs to be maintained for the multiples of five, but two ques are maintained, one for the merge of the multiples of five and three, and the larger one for the merge of all the multiples of five, three, and two. In order to minimize redundant computation time, the implementation maintains the "next" comparison values as part of the recursive function loop states that can change with every loop.

To express the sequence, a Co-Inductive Stream (CIS) is used as a deferred execution (lazy) stream; it does not memoize computations (as discussed above) but that isn't necessary for this application where the sequence is only traversed once and consumed as being traversed.

In addition, in order to reduce the "BigInt" computation time, the calculations are done on the basis of a "Float" logarithmic approxitmation while maintaining "Trival" triple representation of the number of times two, three, and five, are multiplied in order to obtain the current value represented by the logarithmic approximation. The working code is as follows:

<lang elm>module Main exposing ( main )

import Bitwise exposing (..) import BigInt import Task exposing ( Task, succeed, perform, andThen ) import Html exposing (Html, div, text) import Browser exposing ( element ) import Time exposing ( now, posixToMillis )

cLIMIT : Int cLIMIT = 1000000

-- an infinite non-empty non-memoizing Co-Inductive Stream (CIS)... type CIS a = CIS a (() -> CIS a)

takeCIS2List : Int -> CIS a -> List a takeCIS2List n cis =

 let loop i (CIS hd tl) lst =
       if i < 1 then List.reverse lst
       else loop (i - 1) (tl()) (hd :: lst)
 in loop n cis []

nthCIS : Int -> CIS a -> a nthCIS n (CIS hd tl) =

 if n <= 1 then hd else nthCIS (n - 1) (tl())

type PriorityQ comparable v =

 Mt
 | Br comparable v (PriorityQ comparable v)
                   (PriorityQ comparable v)

emptyPQ : PriorityQ comparable v emptyPQ = Mt

peekMinPQ : PriorityQ comparable v -> Maybe (comparable, v) peekMinPQ pq = case pq of

                 (Br k v _ _) -> Just (k, v)
                 Mt -> Nothing

pushPQ : comparable -> v -> PriorityQ comparable v

          -> PriorityQ comparable v

pushPQ wk wv pq =

 case pq of
   Mt -> Br wk wv Mt Mt
   (Br vk vv pl pr) -> 
     if wk <= vk then Br wk wv (pushPQ vk vv pr) pl
     else Br vk vv (pushPQ wk wv pr) pl

siftdown : comparable -> v -> PriorityQ comparable v

            -> PriorityQ comparable v -> PriorityQ comparable v

siftdown wk wv pql pqr =

 case pql of
   Mt -> Br wk wv Mt Mt
   (Br vkl vvl pll prl) ->
     case pqr of
       Mt -> if wk <= vkl then Br wk wv pql Mt
             else Br vkl vvl (Br wk wv Mt Mt) Mt
       (Br vkr vvr plr prr) ->
         if wk <= vkl && wk <= vkr then Br wk wv pql pqr
         else if vkl <= vkr then Br vkl vvl (siftdown wk wv pll prl) pqr
              else Br vkr vvr pql (siftdown wk wv plr prr)

replaceMinPQ : comparable -> v -> PriorityQ comparable v

                -> PriorityQ comparable v

replaceMinPQ wk wv pq = case pq of

                         Mt -> Mt
                         (Br _ _ pl pr) -> siftdown wk wv pl pr

type alias Trival = (Int, Int, Int) showTrival : Trival -> String showTrival tv =

 let (x2, x3, x5) = tv
     xpnd x m r =
       if x <= 0 then r
       else xpnd (shiftRightBy 1 x) (BigInt.mul m m)
                 (if (and 1 x) /= 0 then BigInt.mul m r else r)
 in BigInt.fromInt 1 |> xpnd x2 (BigInt.fromInt 2)
      |> xpnd x3 (BigInt.fromInt 3) |> xpnd x5 (BigInt.fromInt 5)
      |> BigInt.toString

type alias LogRep = { lr: Float, trv: Trival } ltLogRep : LogRep -> LogRep -> Bool ltLogRep lra lrb = lra.lr < lrb.lr oneLogRep : LogRep oneLogRep = { lr = 0.0, trv = (0, 0, 0) } lg2_2 : Float lg2_2 = 1.0 -- log base two of two lg2_3 : Float lg2_3 = logBase 2.0 3.0 lg2_5 : Float lg2_5 = logBase 2.0 5.0 multLR2 : LogRep -> LogRep multLR2 lr = let (x2, x3, x5) = lr.trv

            in LogRep (lr.lr + lg2_2) (x2 + 1, x3, x5)

multLR3 : LogRep -> LogRep multLR3 lr = let (x2, x3, x5) = lr.trv

            in LogRep (lr.lr + lg2_3) (x2, x3 + 1, x5)

multLR5 : LogRep -> LogRep multLR5 lr = let (x2, x3, x5) = lr.trv

            in LogRep (lr.lr + lg2_5) (x2, x3, x5 + 1)

hammingsLog : () -> CIS Trival hammingsLog() =

 let im235 = multLR2 oneLogRep
     im35 = multLR3 oneLogRep
     imrg = im35
     im5 = multLR5 oneLogRep
     next bpq mpq m235 mrg m35 m5 =
       if ltLogRep m235 mrg then
         let omin = case peekMinPQ bpq of
                      Just (lr, trv) -> LogRep lr trv
                      otherwise -> m235 -- at the beginning!
             nm235 = multLR2 omin
             nbpq = replaceMinPQ m235.lr m235.trv bpq
         in CIS m235.trv <| \ () ->
              next nbpq mpq nm235 mrg m35 m5
       else
         if ltLogRep mrg m5 then
           let omin = case peekMinPQ mpq of
                        Just (lr, trv) -> LogRep lr trv
                        otherwise -> mrg -- at the beginning!
               nm35 = multLR3 omin
               nmrg = if ltLogRep nm35 m5 then nm35 else m5
               nmpq = replaceMinPQ mrg.lr mrg.trv mpq
               nbpq = pushPQ mrg.lr mrg.trv bpq
           in CIS mrg.trv <| \ () ->
                next nbpq nmpq m235 nmrg nm35 m5
         else
           let nm5 = multLR5 m5
               nmrg = if ltLogRep m35 nm5 then m35 else nm5
               nmpq = pushPQ m5.lr m5.trv mpq
               nbpq = pushPQ m5.lr m5.trv bpq
           in CIS m5.trv <| \ () ->
                next nbpq nmpq m235 nmrg m35 nm5
 in CIS (0, 0, 0) <| \ () ->
      next emptyPQ emptyPQ im235 imrg im35 im5

-- following code has to do with outputting to a web page using MUV/TEA...

type alias Model = { time: Int

                  , str1: String, str2: String, str3: String }

type Msg = Start Int | Done ( Int, Trival )

timemillis : () -> Task Never Int -- a side effect function timemillis() = now |> andThen (\ t -> succeed (posixToMillis t))

test : Int -> Cmd Msg -- side effect function chain (includes "perform")... test lmt =

 timemillis()
   |> andThen (\ t -> succeed ( t, nthCIS lmt (hammingsLog()) ))
   |> perform Done

myupdate : Msg -> Model -> (Model, Cmd Msg) myupdate msg mdl =

 case msg of
   Start tm -> ( Model tm mdl.str1 mdl.str2 "Running...", test cLIMIT )
   Done (stptm, answr) ->
     ( ( Model stptm mdl.str1 mdl.str2
                 <| "The " ++ String.fromInt cLIMIT ++
                       "th Hamming number is " ++
                       showTrival answr ++ " in " ++
                       String.fromInt (stptm - mdl.time) ++
                       " milliseconds." )
     , Cmd.none ) -- terminates computation; allows UI update...

myview : Model -> Html msg myview mdl =

 div [] [ div [] [text mdl.str1]
        , div [] [text mdl.str2]
        , div [] [text mdl.str3] ]

main : Program () Model Msg main =

 element { init = \ _ -> ( Model 0
                             ("The first 20 Hamming numbers are:  " ++
                               (hammingsLog() |> takeCIS2List 20
                                 |> List.map showTrival
                                 |> String.join ", ") ++ ".")
                             ("The 1691st Hamming number is " ++
                                (hammingsLog() |> nthCIS 1691
                                               |> showTrival) ++ ".")
                             "", perform Start (timemillis()) )
         , update = myupdate
         , subscriptions = \ _ -> Sub.none
         , view = myview }</lang>

Output:

The first 20 Hamming numbers are: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36.
The 1691st Hamming number is 2125764000.
The 1000000th Hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000 in 756 milliseconds.

Do note that, due to the logarithmic response of the Min Heap Priority Queue, the execution time is logarithmic with number of elements evaluation and not linear as it would otherwise be, so if it takes 0.7 seconds to find the millionth Hamming number, it takes something about 10 seconds to find the ten millionth value instead of about 7 seconds. Considering that the generated "native" code is just JavaScript, it is reasonably fast and competitive with easier implementations in other languages such as F#.

ERRE

For bigger numbers, you have to use an external program, like MULPREC.R <lang ERRE>PROGRAM HAMMING

!$DOUBLE

DIM H[2000]

PROCEDURE HAMMING(L%->RES)

     LOCAL I%,J%,K%,N%,M,X2,X3,X5
     H[0]=1
     X2=2  X3=3  X5=5
     FOR N%=1 TO L%-1 DO
       M=X2
       IF M>X3 THEN M=X3 END IF
       IF M>X5 THEN M=X5 END IF
       H[N%]=M
       IF M=X2 THEN I%+=1  X2=2*H[I%]  END IF
       IF M=X3 THEN J%+=1  X3=3*H[J%]  END IF
       IF M=X5 THEN K%+=1  X5=5*H[K%]  END IF
     END FOR
     RES=H[L%-1]

END PROCEDURE

BEGIN

     FOR H%=1 TO 20 DO
       HAMMING(H%->RES)
       PRINT("H(";H%;")=";RES)
     END FOR
     HAMMING(1691->RES)
     PRINT("H(1691)=";RES)

END PROGRAM</lang>

Output:

H( 1 )= 1
H( 2 )= 2
H( 3 )= 3
H( 4 )= 4
H( 5 )= 5
H( 6 )= 6
H( 7 )= 8
H( 8 )= 9
H( 9 )= 10
H( 10 )= 12
H( 11 )= 15
H( 12 )= 16
H( 13 )= 18
H( 14 )= 20
H( 15 )= 24
H( 16 )= 25
H( 17 )= 27
H( 18 )= 30
H( 19 )= 32
H( 20 )= 36
H(1691)= 2125764000

F#

This version implements Dijkstra's merge solution, so is closely related to the Haskell classic version. <lang fsharp>type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>> // ': fix colouring

let rec hammings() =

 let rec (-|-) (Cons(x, nxf) as xs) (Cons(y, nyf) as ys) =
   if x < y then Cons(x, lazy(nxf.Value -|- ys))
   elif x > y then Cons(y, lazy(xs -|- nyf.Value))
   else Cons(x, lazy(nxf.Value -|- nyf.Value))
 let rec inf_map f (Cons(x, nxf)) =
   Cons(f x, lazy(inf_map f nxf.Value))
 Cons(1I, lazy(let x = inf_map ((*) 2I) hamming
               let y = inf_map ((*) 3I) hamming
               let z = inf_map ((*) 5I) hamming
               x -|- y -|- z))

// testing... [<EntryPoint>] let main args =

 let rec iterLazyListFor f n (Cons(v, rf)) = 
   if n > 0 then f v; iterLazyListFor f (n - 1) rf.Value
 let rec nthLazyList n ((Cons(v, rf)) as ll) =
   if n <= 1 then v else nthLazyList (n - 1) rf.Value
 printf "( "; iterLazyListFor (printf "%A ") 20 (hammings()); printfn ")"
 printfn "%A" (hammings() |> nthLazyList 1691)
 printfn "%A" (hammings() |> nthLazyList 1000000)
 0</lang>

The above code memory residency is quite high as it holds the entire lazy sequence in memory due to the reference preventing garbage collection as the sequence is consumed,

The following code reduces that high memory residency by making the routine a function and using internal local stream references for the intermediate streams so that they can be collected as the stream is consumed as long as no reference is held to the main results stream (which is not in the sample test functions); it also avoids duplication of factors by successively building up streams and further reduces memory use by ordering of the streams so that the least dense are determined first:

Translation of: Haskell

<lang fsharp>type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>> // ': fix colouring

let cNUMVALS = 1000000

let hammings() =

 let rec merge (Cons(x, f) as xs) (Cons(y, g) as ys) =
   if x < y then Cons(x, lazy(merge (f.Force()) ys))
   else Cons(y, lazy(merge xs (g.Force())))
 let rec smult m (Cons(x, rxs)) =
   Cons(m * x, lazy(smult m (rxs.Force())))
 let rec first = smult 5I (Cons(1I, lazy first))
 let u s n = 
   let rec r = merge s (smult n (Cons(1I, lazy r))) in r
 Seq.unfold (fun (Cons(hd, rst)) -> Some (hd, rst.Value))
            (Cons(1I, lazy(Seq.fold u first [| 3I; 2I |])))

[<EntryPoint>] let main argv =

 printf "( "; hammings() |> Seq.take 20 |> Seq.iter (printf "%A "); printfn ")"
 printfn "%A" (hammings() |> Seq.item (1691 - 1))
 let strt = System.DateTime.Now.Ticks

 let rslt = (hammings()) |> Seq.item (cNUMVALS - 1)

 let stop = System.DateTime.Now.Ticks

 printfn "%A" rslt  
 printfn "Found this last up to %d in %d milliseconds." cNUMVALS ((stop - strt) / 10000L)

 0 // return an integer exit code</lang>

Both codes output the same results as follows but the second is over three times faster:

Output:

( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Found this last up to 1000000 in 1302 milliseconds.

Both codes are over 10 times slower as compared to Haskell (or Kotlin or Scala or Clojure) when all are written in exactly the same style, perhaps due in some small degree to the BigInteger implementation being much slower for these operations than GMP and the JVM's implementation of BigInteger. Much of this is due to that the DotNet runtime does not allocate from a memory pool as the Haskell and JVM runtime's do, which is much slower when allocating for these functional algorithms where many small allocations/de-allocations are necessary.

Fast somewhat imperative sequence version using logarithms

Since the above pure functional approach isn't very efficient, a more imperative approach using "growable" arrays which are "drained" of unnecessary older values in blocks once the back pointer indices are advanced is used in the following code. The code also implements an algorithm to avoid duplicate calculations and thus does the same number of operations as the above code but faster due to using integer and floating point operations rather an BigInteger ones. Due to the "draining" the memory use is the same as the above by a constant factor. However, note that other than the contents of these arrays, pure functional code using immutable values is used. Note that the implementation of IEnumerable using sequences in F# is also not very efficient and a "roll-your-own" IEnumerable implementation would likely be somewhat faster:

<lang fsharp>let cCOUNT = 1000000

let hammingsLog() = // imperative arrays, eliminates the BigInteger operations...

 let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
 let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
 let inline mul2 (lg, x2, x3, x5) = (lg + 1.0, x2 + 1u, x3, x5)
 let inline mul3 (lg, x2, x3, x5) = (lg + lb3, x2, x3 + 1u, x5)
 let inline mul5 (lg, x2, x3, x5) = (lg + lb5, x2, x3, x5 + 1u)
 let one = (0.0, 0u, 0u, 0u)
 let s532, mrg = one |> mul2, one |> mul3
 let s53 = one |> mul3 |> mul3 // equivalent to 9 for advance step
 let s5 = one |> mul5
 let h = ResizeArray<_>()
 let m = ResizeArray<_>()
 let inline drplg (_, x2, x3, x5) = (x2, x3, x5)
 let inline nontriv() = Seq.unfold (fun (i, j, s532, mrg, s53, s5) -> // THIS STILL IS PATTERN MATCHING!!!!!
   let inline (<) (lga, _, _, _) (lgb, _, _, _) = lga < lgb
   let nv, ni, nj, ns532, nmrg, ns53, ns5 =
     if s532 < mrg then h.Add(s532)
                        s532, i + 1, j, h.[i] |> mul2, mrg, s53, s5
     else if s53 < s5 then h.Add(mrg); m.Add(s53)
                           mrg, i, j + 1, s532, s53, m.[j] |> mul3, s5
          else h.Add(mrg); m.Add(s5)
               mrg, i, j, s532, s5, s53, s5 |> mul5
   let nj = if nj >= m.Capacity / 2 then m.RemoveRange(0, nj); 0 else nj
   let ni = if ni >= h.Capacity / 2 then h.RemoveRange(0, ni); 0 else ni
   Some (drplg nv, (ni, nj, ns532, nmrg, ns53, ns5))) (0,0,s532,mrg,s53,s5)
 seq { yield drplg one; yield! nontriv() } // this is very slow

let trival (x2, x3, x5) = // convert trival to BigInteger

 let rec loop n mlt rslt =
   if n <= 0u then rslt
   else loop (n - 1u) mlt (mlt * rslt)
 loop x2 2I 1I |> loop x3 3I |> loop x5 5I

[<EntryPoint>] let main argv =

 printf "( "; hammingsLog() |> Seq.take 20 |> Seq.iter (printf "%A " << trival); printfn ")"
 printfn "%A" (hammingsLog() |> Seq.item (1691 - 1))
 let strt = System.DateTime.Now.Ticks

 let rslt = (hammingsLog()) |> Seq.item (1000000 - 1)

 let stop = System.DateTime.Now.Ticks

 printfn "%A" (rslt |> trival)  
 printfn "Found this last up to %d in %d milliseconds." cCOUNT ((stop - strt) / 10000L)
 
 printfn ""
 0 // return an integer exit code</lang>

Output:

( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
(5u, 12u, 3u)
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Found this last up to 1000000 in 253 milliseconds.

The above code produces the same outputs as above, but takes only about 253 milliseconds rather than over a second.

Extremely fast non-enumerating version sorting values in error band

If one is willing to forego sequences and just calculate the nth Hamming number, then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: regular number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows:

Translation of: Haskell

<lang fsharp>let nthHamming n =

 if n < 1UL then failwith "nthHamming; argument must be > 0!"
 if n < 2UL then 0u, 0u, 0u else // trivial case for first value of one
 let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
 let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
 let fctr = 6.0 * lb3 * lb5
 let crctn = 2.4534452978042592646620291867186 // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)
 let lbest = (fctr * double n) ** (1.0/3.0) - crctn // from WP formula
 let lbhi = lbest + 1.0 / lbest
 let lblo = 2.0 * lbest - lbhi // upper and lower bound of upper "band"
 let klmt = uint32 (lbhi / lb5)
 let rec loopk k kcnt kbnd =
   if k > klmt then kcnt, kbnd else
     let p = lbhi - double k * lb5
     let jlmt = uint32 (p / lb3)
     let rec loopj j jcnt jbnd =
       if j > jlmt then loopk (k + 1u) jcnt jbnd else
         let q = p - double j * lb3
         let i = uint32 q
         let lg = lbhi - q + double i // current log 2 value (estimated)
         let nbnd = if lg >= lblo then (lg, (uint32 i, j, k)) :: jbnd else jbnd
         loopj (j + 1u) (jcnt + uint64 i + 1UL) nbnd in loopj 0u kcnt kbnd
 let count, bnd = loopk 0u 0UL [] // 64-bit value so doesn't overflow
 if n > count then failwith "nthHamming:  band high estimate is too low!"
 let ndx = int (count - n)
 if ndx >= bnd.Length then failwith "NthHamming.findNth:  band low estimate is too high!"
 let sbnd = bnd |> List.sortBy (fun (lg, _) -> -lg) // sort in decending order
 let _, rslt = sbnd.[ndx]
 rslt

[<EntryPoint>] let main argv =

 let topNum = 1000000UL
 printf "( "; {1..20} |> Seq.iter (printf "%A " << trival << nthHamming << uint64); printfn ")"
 printfn "%A" (nthHamming 1691UL |> trival)
 let rslt = nthHammingx topNum
 let strt = System.DateTime.Now.Ticks

 let rslt = nthHamming topNum

 let stop = System.DateTime.Now.Ticks

 let x2, x3, x5 = rslt
 printfn "2**%A times 3**%A times 5**%A" x2 x3 x5
 let lgrthm = log10 2.0 * (double x2 + (double x3 * log 3.0 + double x5 * log 5.0) / log 2.0)
 let exp = floor lgrthm |> int
 let mntsa = 10.0 ** (lgrthm - double exp)
 printfn "Approximately %AE+%A" mntsa exp
 let s = trival rslt |> string
 let lngth = s.Length
 printfn "Digits:  %A" lngth
 if lngth <= 10000 then
   {0..100..lngth-1}
     |> Seq.iter (fun i ->
       printfn "%s" (s.Substring(i, if i + 100 < lngth then 100 else lngth - i)))
 
 printfn "\r\nFound this last up to %A in %A milliseconds." topNum ((stop - strt) / 10000L)
 
 printf "\r\nPress any key to exit:"
 System.Console.ReadKey(true) |> ignore
 printfn ""
 0 // return an integer exit code</lang>

Output:

( 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 )
2125764000
2**55u times 3**47u times 5**64u
Approximately 5.193127804E+83
Digits:  84
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Found this last up to 1000000UL in 0L milliseconds.

Even though the above code is implemented in a completely functional style using immutable bindings and (non-lazy) lists (without closures), it is about as fast as implementations in the fastest of languages. It is faster than the Haskell version due to that version using lazy lists with the overhead of creating the requisite "thunks".

It takes too short a time to be measured to calculate the millionth Hamming number, the billionth number in the sequence can be calculated in just about 15 milliseconds, the trillionth in about one second, the thousand trillionth in about a hundred seconds, and it should be possible to calculate the 10^19th value in less than a day (untested) on common personal computers. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit.

Enhancement to by able to find Hamming numbers beyond the ten trillionth one

Due to the limited 53-bit mantissa of 64-bit double floating piint numbers, the above code can't properly sort the error band for input arguments somewhere above 10**13; the following code makes the sort accurate by using a multi-precision logarithm representation of sufficient precision so that the sort is accurate for arguments well beyond the uint64 input argument range, at about a doubling in cost in execution speed:

Translation of: Haskell

<lang fsharp>let nthHamming n =

 if n < 1UL then failwith "nthHamming:  argument must be > 0!"
 if n < 2UL then 0u, 0u, 0u else // trivial case for first value of one
 let lb3 = 1.5849625007211561814537389439478 // Math.Log(3) / Math.Log(2);
 let lb5 = 2.3219280948873623478703194294894 // Math.Log(5) / Math.Log(2);
 let fctr = 6.0 * lb3 * lb5
 let crctn = 2.4534452978042592646620291867186 // Math.Log(Math.sqrt(30.0)) / Math.Log(2.0)
 let lbest = (fctr * double n) ** (1.0/3.0) - crctn // from WP formula
 let lbhi = lbest + 1.0/lbest
 let lblo = 2.0 * lbest - lbhi // upper and lower bound of upper "band"
 let bglb2 = 1267650600228229401496703205376I
 let bglb3 = 2009178665378409109047848542368I
 let bglb5 = 2943393543170754072109742145491I
 let klmt = uint32 (lbhi / lb5)
 let rec loopk k kcnt kbnd =
   if k > klmt then kcnt, kbnd else
     let p = lbhi - double k * lb5
     let jlmt = uint32 (p / lb3)
     let rec loopj j jcnt jbnd =
       if j > jlmt then loopk (k + 1u) jcnt jbnd else
         let q = p - double j * lb3
         let i = uint32 q
         let lg = lbhi - q + double i // current log 2 value (estimated)
         let nbnd = if lg < lblo then jbnd else
                      let bglg = bglb2 * bigint i + bglb3 * bigint j + bglb5 * bigint k in
                      (bglg, (uint32 i, j, k)) :: jbnd
         loopj (j + 1u) (jcnt + uint64 i + 1UL) nbnd in loopj 0u kcnt kbnd
 let count, bnd = loopk 0u 0UL [] // 64-bit value so doesn't overflow
 if n > count then failwith "nthHamming:  band high estimate is too low!"
 let ndx = int (count - n)
 if ndx >= bnd.Length then failwith "NthHamming.findNth:  band low estimate is too high!"
 let sbnd = bnd |> List.sortBy (fun (lg, _) -> -lg) // sort in decending order
 let _, rslt = sbnd.[ndx]
 rslt</lang>

Factor

Translation of: Scala

<lang factor>USING: accessors deques dlists fry kernel make math math.order

IN: rosetta.hamming

TUPLE: hamming-iterator 2s 3s 5s ;

<hamming-iterator> ( -- hamming-iterator )

   hamming-iterator new
       1 1dlist >>2s
       1 1dlist >>3s
       1 1dlist >>5s ;

enqueue ( n hamming-iterator -- )

   [ [ 2 * ] [ 2s>> ] bi* push-back ]
   [ [ 3 * ] [ 3s>> ] bi* push-back ]
   [ [ 5 * ] [ 5s>> ] bi* push-back ] 2tri ;

next ( hamming-iterator -- n )

   dup [ 2s>> ] [ 3s>> ] [ 5s>> ] tri
   3dup [ peek-front ] tri@ min min
   [
       '[
           dup peek-front _ =
           [ pop-front* ] [ drop ] if
       ] tri@
   ] [ swap enqueue ] [ ] tri ;

next-n ( hamming-iterator n -- seq )

   swap '[ _ [ _ next , ] times ] { } make ;

nth-from-now ( hamming-iterator n -- m )

   1 - over '[ _ next drop ] times next ;</lang>

 <hamming-iterator> 20 next-n .
 <hamming-iterator> 1691 nth-from-now .
 <hamming-iterator> 1000000 nth-from-now .

Translation of: Haskell

Lazy lists are quite slow in Factor, but still. <lang factor>USING: combinators fry kernel lists lists.lazy locals math ; IN: rosetta.hamming-lazy

sort-merge ( xs ys -- result )

   xs car :> x
   ys car :> y
   {
       { [ x y < ] [ [ x ] [ xs cdr ys sort-merge ] lazy-cons ] }
       { [ x y > ] [ [ y ] [ ys cdr xs sort-merge ] lazy-cons ] }
       [ [ x ] [ xs cdr ys cdr sort-merge ] lazy-cons ]
   } cond ;

hamming ( -- hamming )

   f :> h!
   [ 1 ] [
       h 2 3 5 [ '[ _ * ] lazy-map ] tri-curry@ tri
       sort-merge sort-merge
   ] lazy-cons h! h ;</lang>

 20 hamming ltake list>array .
 1690 hamming lnth .
 999999 hamming lnth .

Forth

Works with: Gforth version 0.7.0

This version uses a compact representation of Hamming numbers: each 64-bit cell represents a number 2^l*3^m*5^n, where l, n, and m are bitfields in the cell (20 bits for now). It also uses a fixed-point logarithm to compare the Hamming numbers and prints them in factored form. This code has been tested up to the 10^9th Hamming number. <lang Forth>\ manipulating and computing with Hamming numbers:

extract2 ( h -- l )

   40 rshift ;

extract3 ( h -- m )

   20 rshift $fffff and ;

extract5 ( h -- n )

   $fffff and ;

' + alias h* ( h1 h2 -- h )

h. { h -- }

   ." 2^"  h extract2 0 .r
   ." *3^" h extract3 0 .r
   ." *5^" h extract5 . ;

\ the following numbers have been produced with bc -l as follows 1 62 lshift constant ldscale2

7309349404307464679 constant ldscale3 \ 2^62*l(3)/l(2) (rounded up)

10708003330985790206 constant ldscale5 \ 2^62*l(5)/l(2) (rounded down)

hld { h -- ud }

   \ ud is a scaled fixed-point representation of the logarithm dualis of h
   h extract2 ldscale2 um*
   h extract3 ldscale3 um* d+
   h extract5 ldscale5 um* d+ ;

h<= ( h1 h2 -- f )

   2dup = if
       2drop true exit
   then
   hld rot hld assert( 2over 2over d<> )
   du>= ;

hmin ( h1 h2 -- h )

   2dup h<= if
       drop
   else
       nip
   then ;

\ actual algorithm

0 value seq variable seqlast 0 seqlast !

lastseq ( -- u )

   \ last stored number in the sequence 
   seq seqlast @ th @ ;

genseq ( h1 "name" -- )

   \ h1 is the factor for the sequence
   create , 0 , \ factor and index of element used for last return
 does> ( -- u2 )
   \ u2 is the next number resulting from multiplying h1 with numbers
   \ in the sequence that is larger than the last number in the
   \ sequence
   dup @ lastseq { h1 l } cell+ dup @ begin ( index-addr index )
       seq over th @ h1 h* dup l h<= while
           drop 1+ repeat
   >r swap ! r> ;

$10000000000 genseq s2 $00000100000 genseq s3 $00000000001 genseq s5

nextseq ( -- )

   s2 s3 hmin s5 hmin , 1 seqlast +! ;

nthseq ( u1 -- h )

   \ the u1 th element in the sequence
   dup seqlast @ u+do
       nextseq
   loop
   1- 0 max cells seq + @ ;

.nseq ( u1 -- )

   dup seqlast @ u+do
       nextseq
   loop
   0 u+do
       seq i th @ h.
   loop ;

here to seq 0 , \ that's 1

20 .nseq cr 1691 nthseq h. cr 1000000 nthseq h.</lang>

Output:

2^0*3^0*5^0 2^1*3^0*5^0 2^0*3^1*5^0 2^2*3^0*5^0 2^0*3^0*5^1 2^1*3^1*5^0 2^3*3^0*5^0 2^0*3^2*5^0 2^1*3^0*5^1 2^2*3^1*5^0 2^0*3^1*5^1 2^4*3^0*5^0 2^1*3^2*5^0 2^2*3^0*5^1 2^3*3^1*5^0 2^0*3^0*5^2 2^0*3^3*5^0 2^1*3^1*5^1 2^5*3^0*5^0 2^2*3^2*5^0 
2^5*3^12*5^3 
2^55*3^47*5^64

A smaller, less capable solution is presented here. It solves two out of three requirements and is ANS-Forth compliant. <lang Forth>2000 cells constant /hamming create hamming /hamming allot

                  ( n1 n2 n3 n4 n5 n6 n7 -- n3 n4 n5 n6 n1 n2 n8)

min? >r dup r> min >r 2rot r> ;

hit? ( n1 n2 n3 n4 n5 n6 n7 n8 -- n3 n4 n9 n10 n1 n2 n7)

 >r 2dup =        \ compare number with found minimum
 if -rot drop 1+ hamming over cells + @ r@ * rot then
 r> drop >r 2rot r>

\ if so, increment and rotate

hamming# ( n1 -- n2)

 1 hamming ! >r   \ set first cell and initialize parms
 0 5 over 3 over 2
 r@ 1 ?do         \ determine minimum and set cell
    dup min? min? min? dup hamming i cells + !
    2 hit? 5 hit? 3 hit? drop
 loop             \ find if minimum equals value
 2drop 2drop 2drop hamming r> 1- cells + @

\ clean up stack and fetch hamming number

test

 cr 21 1 ?do i . i hamming# . cr loop
 1691 hamming# . cr

</lang>

Fortran

Works with: Fortran version 90 and later

Using big_integer_module from here [1] <lang fortran>program Hamming_Test

 use big_integer_module
 implicit none
 
 call Hamming(1,20)
 write(*,*)
 call Hamming(1691)
 write(*,*)
 call Hamming(1000000)
  

contains

subroutine Hamming(first, last)

 integer, intent(in) :: first
 integer, intent(in), optional :: last
 integer :: i, n, i2, i3, i5, lim
 type(big_integer), allocatable :: hnums(:)

 if(present(last)) then
   lim = last
 else
   lim = first
 end if

 if(first < 1 .or. lim > 2500000 ) then
   write(*,*) "Invalid input"
   return
 end if
 
 allocate(hnums(lim))
 
 i2 = 1 ;  i3 = 1 ; i5 = 1  
 hnums(1) = 1
 n = 1
 do while(n < lim)
   n = n + 1
   hnums(n) = mini(2*hnums(i2), 3*hnums(i3), 5*hnums(i5))
   if(2*hnums(i2) == hnums(n)) i2 = i2 + 1
   if(3*hnums(i3) == hnums(n)) i3 = i3 + 1
   if(5*hnums(i5) == hnums(n)) i5 = i5 + 1
 end do
 
 if(present(last)) then
   do i = first, last
     call print_big(hnums(i))
     write(*, "(a)", advance="no") " "
   end do
 else
   call print_big(hnums(first))
 end if
 
 deallocate(hnums)

end subroutine

function mini(a, b, c)

 type(big_integer) :: mini
 type(big_integer), intent(in) :: a, b, c
  
 if(a < b ) then 
   if(a < c) then
     mini = a
   else
     mini = c
   end if
 else if(b < c) then
   mini = b
 else
   mini = c
 end if 

end function mini end program</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

FreeBASIC

<lang freebasic>' FB 1.05.0 Win64

' The biggest integer which FB supports natively is 8 bytes so unable ' to calculate 1 millionth Hamming number without using an external ' "bigint" library such as GMP

Function min(x As Integer, y As Integer) As Integer

 Return IIf(x < y, x, y)

End Function

Function hamming(n As Integer) As Integer

 Dim h(1 To n) As Integer
 h(1) = 1
 Dim As Integer i = 1, j = 1, k = 1   
 Dim As Integer x2 = 2, x3 = 3, x5 = 5
 
 For m As Integer = 2 To n
   h(m) = min(x2, min(x3, x5))
   If h(m) = x2 Then 
     i += 1
     x2 = 2 * h(i)
   End If
   If h(m) = x3 Then
     j += 1
     x3 = 3 * h(j)
   End if
   If h(m) = x5 Then
     k += 1
     x5 = 5 * h(k)
   End If
 Next     

 Return h(n)

End Function

Print "The first 20 Hamming numbers are :" For i As Integer = 1 To 20

 Print hamming(i); " ";

Next Print : Print Print "The 1691st hamming number is :" Print hamming(1691) Print Print "Press any key to quit" Sleep</lang>

Output:

The first 20 Hamming numbers are :
 1  2  3  4  5  6  8  9  10  12  15  16  18  20  24  25  27  30  32  36

The 1691st Hamming number is :
 2125764000

FunL

Translation of: Scala

<lang funl>native scala.collection.mutable.Queue

val hamming =

 q2 = Queue()
 q3 = Queue()
 q5 = Queue()
 
 def enqueue( n ) =
   q2.enqueue( n*2 )
   q3.enqueue( n*3 )
   q5.enqueue( n*5 )

 def stream =
   val n = min( min(q2.head(), q3.head()), q5.head() )
   
   if q2.head() == n then q2.dequeue()
   if q3.head() == n then q3.dequeue()
   if q5.head() == n then q5.dequeue()
   
   enqueue( n )
   n # stream()
   
 for q <- [q2, q3, q5] do q.enqueue( 1 )
 
 stream()</lang>

Translation of: Haskell

<lang funl>val hamming = 1 # merge( map((2*), hamming), merge(map((3*), hamming), map((5*), hamming)) )

def

 merge( inx@x:_, iny@y:_ )
   | x < y     = x # merge( inx.tail(), iny )
   | x > y     = y # merge( inx, iny.tail() )
   | otherwise = merge( inx, iny.tail() )

println( hamming.take(20) ) println( hamming(1690) ) println( hamming(2000) )</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
8100000000

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

Go

Concise version using dynamic-programming

<lang go>package main

import (

   "fmt"
   "math/big"

)

func min(a, b *big.Int) *big.Int {

   if a.Cmp(b) < 0 {
       return a
   }
   return b

}

func hamming(n int) []*big.Int {

   h := make([]*big.Int, n)
   h[0] = big.NewInt(1)
   two, three, five    := big.NewInt(2), big.NewInt(3), big.NewInt(5)
   next2, next3, next5 := big.NewInt(2), big.NewInt(3), big.NewInt(5)
   i, j, k := 0, 0, 0
   for m := 1; m < len(h); m++ {
       h[m] = new(big.Int).Set(min(next2, min(next3, next5)))
       if h[m].Cmp(next2) == 0 { i++; next2.Mul(  two, h[i]) } 
       if h[m].Cmp(next3) == 0 { j++; next3.Mul(three, h[j]) } 
       if h[m].Cmp(next5) == 0 { k++; next5.Mul( five, h[k]) } 
   }
   return h

}

func main() {

   h := hamming(1e6)
   fmt.Println(h[:20])
   fmt.Println(h[1691-1])
   fmt.Println(h[len(h)-1])

}</lang>

Output:

[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Longer version using dynamic-programming and logarithms

More than 10 times faster. <lang go>package main

import ( "flag" "fmt" "log" "math" "math/big" "os" )

var ( // print the whole sequence or just one element?

seqMode = flag.Bool("s", false, "sequence mode") // precomputed base-2 logarithms for 3 and 5 lg3, lg5 float64 = math.Log2(3), math.Log2(5)

// state of the three multiplied sequences front = [3]cursor{ {0, 0, 1}, // 2 {1, 0, lg3}, // 3 {2, 0, lg5}, // 5 }

// table for dynamic-programming stored results table [][3]int16 )

type cursor struct { f int // index (0, 1, 2) corresponding to factor (2, 3, 5) i int // index into table for the entry being multiplied lg float64 // base-2 logarithm of the multiple (for ordering) }

func (c *cursor) val() [3]int16 { x := table[c.i] x[c.f]++ // multiply by incrementing the exponent return x }

func (c *cursor) advance() { c.i++ // skip entries that would produce duplicates for (c.f < 2 && table[c.i][2] > 0) || (c.f < 1 && table[c.i][1] > 0) { c.i++ } x := c.val() c.lg = float64(x[0]) + lg3*float64(x[1]) + lg5*float64(x[2]) }

func step() { table = append(table, front[0].val()) front[0].advance() // re-establish sorted order if front[0].lg > front[1].lg { front[0], front[1] = front[1], front[0] if front[1].lg > front[2].lg { front[1], front[2] = front[2], front[1] } } } func show(elem [3]int16) { z := big.NewInt(1) for i, base := range []int64{2, 3, 5} { b := big.NewInt(base) x := big.NewInt(int64(elem[i])) z.Mul(z, b.Exp(b, x, nil)) } fmt.Println(z) }

func main() { log.SetPrefix(os.Args[0] + ": ") log.SetOutput(os.Stderr) flag.Parse() if flag.NArg() != 1 { log.Fatalln("need one positive integer argument") } var ordinal int // ordinal of last sequence element to compute _, err := fmt.Sscan(flag.Arg(0), &ordinal) if err != nil || ordinal <= 0 { log.Fatalln("argument must be a positive integer") } table = make([][3]int16, 1, ordinal) for i, n := 1, ordinal; i < n; i++ { if *seqMode { show(table[i-1]) } step() } show(table[ordinal-1]) }</lang>

Output:

$ ./hamming -s 20 | xargs
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
$ time ./hamming 1000000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

real	0m0.110s
user	0m0.090s
sys	0m0.020s
$ uname -a
Linux lance 3.0-ARCH #1 SMP PREEMPT Sat Aug 6 16:18:35 CEST 2011 x86_64 Intel(R) Core(TM)2 Duo CPU P8400 @ 2.26GHz GenuineIntel GNU/Linux

Low Memory Use Enumerating Version Eliminating Duplicates

While the above code is fast due to avoiding big.Int operations and being tuned to avoid duplication of work, it has two problems: It uses memory at about six times the value of "n", the nth value and has a practical upper range where the logarithm estimate used to compare currently processed value round off error will become too big so that values will become not in proper order for some values for large ranges. This latter problem could be fixed by using double precision of two 64-bit uint's for the accumulated estimate, but the algorithm would still consume quite a lot of memory.

The following algorithm implements a continuously increasing enumeration of the Hamming numbers at about the same speed as the first solution by eliminating duplicate calculations, by organizing the streams/lazylists so that the least dense ones are processed first, and by using local variables that don't retain the heads of the streams/lazylists so that they can be garbage collected as consumed. In this way, the billionth value can be calculated using only about a billion bytes of memory (one sixth of the above), with most of that used for storage of the necessary big.Int's. If a tweaked logarithm algorithm were used, it would reduce the memory use almost to zero and would speed it up, although not to the same extent as the immediately above code as much of the remaining time would be spent in allocation of new stream/lazylist values and garbage collection.

The program implements the memoized streams/lazylists with a "roll-your-own" implementation and only the necessary methods as required by this algorithm as Go does not have a library to supply such, and uses a function closure to implement a simple form of enumeration of the Hamming values. It used "llmult to perform the function of the "map" function used in the Haskell code, which is to produce a new stream which has each value of the input stream multiplied by a constant. Instead of the Haskell "foldl" function, this program uses a simple Go "for" comprehension of the input primes array.

Translation of: Haskell

<lang go>// Hamming project main.go package main

import ( "fmt" "math/big" "time" )

type lazyList struct { head *big.Int tail *lazyList contf func() *lazyList }

func (oll *lazyList) next() *lazyList { if oll.contf != nil { // not thread-safe oll.tail = oll.contf() oll.contf = nil } return oll.tail }

func merge(a *lazyList, b *lazyList) *lazyList { rslt := new(lazyList) x := a.head y := b.head if x.Cmp(y) < 0 { rslt.head = x rslt.contf = func() *lazyList { return merge(a.next(), b) } } else { rslt.head = y rslt.contf = func() *lazyList { return merge(a, b.next()) } } return rslt }

func llmult(m *big.Int, ll *lazyList) *lazyList { rslt := new(lazyList) rslt.head = new(big.Int).Set(big.NewInt(0)).Mul(m, ll.head) rslt.contf = func() *lazyList { return llmult(m, ll.next()) } return rslt }

func u(s *lazyList, n *big.Int) *lazyList { rslt := new(lazyList) cr := new(lazyList) cr.head = big.NewInt(1) cr.contf = func() *lazyList { return rslt } if s == nil { rslt = llmult(n, cr) } else { rslt = merge(s, llmult(n, cr)) } return rslt }

func Hamming() func() *big.Int { prms := []int64{5, 3, 2} curr := new(lazyList) curr.head = big.NewInt(1) curr.contf = func() *lazyList { var r *lazyList = nil for _, v := range prms { r = u(r, big.NewInt(v)) } return r } return func() *big.Int { temp := curr curr = curr.next() return temp.head } }

func main() { n := 1000000

hamiter := Hamming() rarr := make([]*big.Int, 20) for i, _ := range rarr { rarr[i] = hamiter() } fmt.Println(rarr)

hamiter = Hamming() for i := 1; i < 1691; i++ { hamiter() } fmt.Println(hamiter())

strt := time.Now()

hamiter = Hamming() for i := 1; i < n; i++ { hamiter() } rslt := hamiter()

end := time.Now() fmt.Printf("Found the %vth Hamming number as %v in %v.\r\n", n, rslt.String(), end.Sub(strt)) } </lang>

The outputs are about the same as the above versions. In order to perform this algorithm, one can see how much more verbose Go is than more functional languages such as Haskell or F# for this primarily functional algorithm.

Fast imperative version avoiding duplicates, reducing memory, and using logarithmic representation

While the above version can calculate to larger ranges due to somewhat reduced memory use, it is still somewhat limited as to range by memory limits due to the increasing size of the big integers used, limited in speed due to those big integer calculations, and also limited in speed due to Go's slow memory allocations and de-allocations. The following code uses combined techniques to overcome all three limitations: 1) as for other solutions on this page, it uses a representation using integer exponents of 2, 3, and 5 and a scaled integer logarithm only for value comparisons (scaled such that round-off errors aren't a factor over the applicable range); thus memory use per element is constant rather than growing with range for big integers, and operations are simple integer comparisons and additions and are thus very fast. 2) memory reductions are by draining the used arrays by batches (rather than one by one as above) in place to reduce the time required for constant allocations and de-allocations. The code is as follows:

Translation of: Rust

<lang go>package main

import ( "fmt" "math/big" "time" )

// constants as expanded integers to minimize round-off errors, and // reduce execution time using integer operations not float... const cLAA2 uint64 = 35184372088832 // 2.0f64.ln() * 2.0f64.powi(45)).round() as u64; const cLBA2 uint64 = 55765910372219 // 3.0f64.ln() / 2.0f64.ln() * 2.0f64.powi(45)).round() as u64; const cLCA2 uint64 = 81695582054030 // 5.0f64.ln() / 2.0f64.ln() * 2.0f64.powi(45)).round() as u64;

type logelm struct { // log representation of an element with only allowable powers exp2 uint16 exp3 uint16 exp5 uint16 logr uint64 // log representation used for comparison only - not exact }

func (self *logelm) lte(othr *logelm) bool { if self.logr <= othr.logr { return true } else { return false } } func (self *logelm) mul2() logelm { return logelm{ exp2: self.exp2 + 1, exp3: self.exp3, exp5: self.exp5, logr: self.logr + cLAA2, } } func (self *logelm) mul3() logelm { return logelm{ exp2: self.exp2, exp3: self.exp3 + 1, exp5: self.exp5, logr: self.logr + cLBA2, } } func (self *logelm) mul5() logelm { return logelm{ exp2: self.exp2, exp3: self.exp3, exp5: self.exp5 + 1, logr: self.logr + cLCA2, } }

func log_nodups_hamming(n uint) *big.Int { if n < 1 { panic("log_nodups_hamming: argument < 1!") } if n < 2 { // trivial case of first in sequence return big.NewInt(1) } if n > 1.2e15 { panic("log_nodups_hamming: argument too large!") }

one := logelm{} next5, merge := one.mul5(), one.mul3() next53, next532 := merge.mul3(), one.mul2()

g := make([]logelm, 1, 65536) g[0] = one // never used, just so append works h := make([]logelm, 1, 65536) h[0] = one // never used, just so append works

i, j := 1, 1 for m := uint(1); m < n; m++ { cph := cap(h) if i >= cph/2 { nm := copy(h[0:i], h[i:]) h = h[0:nm:cph] i = 0 } if next532.lte(&merge) { h = append(h, next532) next532 = h[i].mul2() i++ } else { h = append(h, merge) if next53.lte(&next5) { merge = next53 next53 = g[j].mul3() j++ } else { merge = next5 next5 = next5.mul5() } cpg := cap(g) if j >= cpg/2 { nm := copy(g[0:j], g[j:]) g = g[0:nm:cpg] j = 0 } g = append(g, merge) } }

two, three, five := big.NewInt(2), big.NewInt(3), big.NewInt(5) o := h[len(h)-1] // convert last element to big integer... ob := big.NewInt(1) for i := uint16(0); i < o.exp2; i++ { ob.Mul(two, ob) } for i := uint16(0); i < o.exp3; i++ { ob.Mul(three, ob) } for i := uint16(0); i < o.exp5; i++ { ob.Mul(five, ob) } return ob }

func main() { n := uint(1e6)

rarr := make([]*big.Int, 20) for i, _ := range rarr { rarr[i] = log_nodumps_hamming(i) } fmt.Println(rarr)

fmt.Println(log_nodups_hamming(1691))

strt := time.Now()

rslt := log_nodups_hamming(n)

end := time.Now()

rs := rslt.String() lrs := len(rs) fmt.Printf("%v digits:\r\n", lrs) ndx := 0 for ; ndx < lrs-100; ndx += 100 { fmt.Println(rs[ndx : ndx+100]) } fmt.Println(rs[ndx:])

fmt.Printf("This last found the %vth hamming number in %v.\r\n", n, end.Sub(strt)) }</lang>

Output:

[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]
2125764000
84 digits:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last found the 1000000th hamming number in 10.0006ms.

The above code can produce the billionth hamming number (844 digits) in about 14 seconds and given a machine with over 9 Gigabytes, it can calculate to the limit of 1.2e13 (about 19,335 digits) in about a day or so. Functional enumerating versions as the immediately precedent code, even if adapted to the logarithm algorithm, will take longer because of the time required for enumeration, but much worse is the time required for allocations/de-allocations (garbage collection) of individual elements rather than as here in batches and for the majority of times done in place not requiring allocation/de-allocation at all.

Extremely fast version inserting logarithms into the top error band

The above code is not as fast as one can go as it is limited by the need to calculate all Hamming numbers in the sequence up to the required one: some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: regular number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows:

Translation of: Nim

<lang go>package main

import ( "fmt" "math" "math/big" "sort" "time" )

type logrep struct { lg float64 x2, x3, x5 uint32 } type logreps []logrep

func (s logreps) Len() int { // necessary methods for sorting return len(s) } func (s logreps) Swap(i, j int) { s[i], s[j] = s[j], s[i] } func (s logreps) Less(i, j int) bool { return s[j].lg < s[i].lg // sort in decreasing order (reverse order compare) }

func nthHamming(n uint64) (uint32, uint32, uint32) { if n < 2 { if n < 1 { panic("nthHamming: argument is zero!") } return 0, 0, 0 } const lb3 = 1.5849625007211561814537389439478 // math.Log2(3.0) const lb5 = 2.3219280948873623478703194294894 // math.Log2(5.0) fctr := 6.0 * lb3 * lb5 crctn := math.Log2(math.Sqrt(30.0)) // from WP formula lgest := math.Pow(fctr*float64(n), 1.0/3.0) - crctn var frctn float64 if n < 1000000000 { frctn = 0.509 } else { frctn = 0.106 } lghi := math.Pow(fctr*(float64(n)+frctn*lgest), 1.0/3.0) - crctn lglo := 2.0*lgest - lghi // and a lower limit of the upper "band" var count uint64 = 0 bnd := make(logreps, 0) // give it one value so doubling size works klmt := uint32(lghi/lb5) + 1 for k := uint32(0); k < klmt; k++ { p := float64(k) * lb5 jlmt := uint32((lghi-p)/lb3) + 1 for j := uint32(0); j < jlmt; j++ { q := p + float64(j)*lb3 ir := lghi - q lg := q + math.Floor(ir) // current log value estimated count += uint64(ir) + 1 if lg >= lglo { bnd = append(bnd, logrep{lg, uint32(ir), j, k}) } } } if n > count { panic("nthHamming: band high estimate is too low!") } ndx := int(count - n) if ndx >= bnd.Len() { panic("nthHamming: band low estimate is too high!") } sort.Sort(bnd) // sort decreasing order due definition of Less above

rslt := bnd[ndx] return rslt.x2, rslt.x3, rslt.x5 }

func convertTpl2BigInt(x2, x3, x5 uint32) *big.Int { result := big.NewInt(1) two := big.NewInt(2) three := big.NewInt(3) five := big.NewInt(5) for i := uint32(0); i < x2; i++ { result.Mul(result, two) } for i := uint32(0); i < x3; i++ { result.Mul(result, three) } for i := uint32(0); i < x5; i++ { result.Mul(result, five) } return result }

func main() { for i := 1; i <= 20; i++ { fmt.Printf("%v ", convertTpl2BigInt(nthHamming(uint64(i)))) } fmt.Println() fmt.Println(convertTpl2BigInt(nthHamming(1691)))

strt := time.Now() x2, x3, x5 := nthHamming(uint64(1e6)) end := time.Now()

fmt.Printf("2^%v times 3^%v times 5^%v\r\n", x2, x3, x5) lrslt := convertTpl2BigInt(x2, x3, x5) lgrslt := (float64(x2) + math.Log2(3.0)*float64(x3) + math.Log2(5.0)*float64(x5)) * math.Log10(2.0) exp := math.Floor(lgrslt) mant := math.Pow(10.0, lgrslt-exp) fmt.Printf("Approximately: %vE+%v\r\n", mant, exp) rs := lrslt.String() lrs := len(rs) fmt.Printf("%v digits:\r\n", lrs) if lrs <= 10000 { ndx := 0 for ; ndx < lrs-100; ndx += 100 { fmt.Println(rs[ndx : ndx+100]) } fmt.Println(rs[ndx:]) }

fmt.Printf("This last found the %vth hamming number in %v.\r\n", uint64(1e6), end.Sub(strt)) }</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
2^55 times 3^47 times 5^64
Approximately:  5.193127804483804E+83
84 digits:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last found the 1000000th hamming number in 0s.

As can be seen above, the time to calculate the millionth Hamming number is now too small to be measured. The billionth number in the sequence can be calculated in just about 15 milliseconds, the trillionth in about 1.5 seconds, the thousand trillionth in about 150 seconds, and it should be possible to calculate the 10^19th value in less than a day (untested) on common personal computers. The (2^64 - 1)th value (18446744073709551615th value) cannot be calculated due to a slight overflow problem as it approaches that limit.

Groovy

<lang groovy>class Hamming {

   static final ONE = BigInteger.ONE
   static final THREE = BigInteger.valueOf(3)
   static final FIVE = BigInteger.valueOf(5)

   static void main(args) {
       print 'Hamming(1 .. 20) ='
       (1..20).each {
           print " ${hamming it}"
       }
       println "\nHamming(1691) = ${hamming 1691}"
       println "Hamming(1000000) = ${hamming 1000000}"
   }

   static hamming(n) {
       def priorityQueue = new PriorityQueue<BigInteger>()
       priorityQueue.add ONE

       def lowest

       n.times {
           lowest = priorityQueue.poll()
           while (priorityQueue.peek() == lowest) {
               priorityQueue.poll()
           }
           updateQueue(priorityQueue, lowest)
       }

       lowest
   }

   static updateQueue(priorityQueue, lowest) {
       priorityQueue.add(lowest.shiftLeft 1)
       priorityQueue.add(lowest.multiply THREE)
       priorityQueue.add(lowest.multiply FIVE)
   }

}</lang>

Haskell

The classic version

<lang haskell>hamming = 1 : map (2*) hamming `union` map (3*) hamming `union` map (5*) hamming

union a@(x:xs) b@(y:ys) = case compare x y of

           LT -> x : union  xs  b
           EQ -> x : union  xs  ys
           GT -> y : union  a   ys

main = do

   print $ take 20 hamming
   print  (hamming !! (1691-1), hamming !! (1692-1))
   print $ hamming !! (1000000-1)

-- Output: -- [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36] -- (2125764000,2147483648) -- 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</lang> Runs in about a second on Ideone.com. The nested unions' effect is to produce the minimal value at each step, with duplicates removed. As Haskell evaluation model is on-demand, the three map expressions are in effect iterators, maintaining hidden pointers back into the shared named storage with which they were each created (a name is a pointer/handle in Haskell; to name is to point at, to refer to, to take a handle on).

The amount of operations is constant for each number produced, so the time complexity should be $O(n)$. Empirically, it is slightly above that and worsening, suggestive of extra cost of bignum arithmetics. Using triples representation with logarithm values for comparisons amends this problem, but runs ~ 1.2x slower for the 1,000,000.

This is what that DDJ blog post's "pseudo-C" code was transcribing, mentioned at the Python entry that started this task ( curiously, it is in almost word-for-word correspondence with Edsger Dijkstra's code from his book A Discipline of Programming, p. 132 ). D, Go, PARI/GP, Prolog all implement the same idea of back-pointers into shared storage. A Haskell run-time system can actually free up the storage automatically at the start of the shared list and only keep the needed portion of it, from the (5*) back-pointer, – which is about $O(n^{2/3})$ in length – behind the scenes, as long as there's no re-use evident in the code.

Avoiding generation of duplicates

The classic version can be sped up quite a bit (about twice, with roughly the same empirical orders of growth) by avoiding generation of duplicate values in the first place: <lang haskell>hammings :: () -> [Integer] hammings() = 1 : foldr u [] [2,3,5] where

 u n s = -- fix (merge s . map (n*) . (1:))
         r where
   r = merge s (map (n*) (1:r))
   merge [] b = b
   merge a@(x:xs) b@(y:ys) | x < y     = x : merge xs b
                           | otherwise = y : merge a ys

main :: IO () main = do

 print $ take 20 (hammings ())
 print $ (hammings ()) !! 1690
 print $ (hammings ()) !! (1000000-1)</lang>

Explicit multiples reinserting

This is a common approach which explicitly maintains an internal buffer of $O(n^{2/3})$ elements, removing the numbers from its front and reinserting their 2- 3- and 5-multiples in order. It overproduces the sequence, stopping when the n-th number is no longer needed instead of when it's first found. Also overworks by maintaining this buffer in total order where just heap would be sufficient. Worse, this particular version uses a sequential list for its buffer. That means $O(n^{2/3})$ operations for each number, instead of $O(1)$ of the above version (and thus $O(n^{5/3})$ overall). Translation of Java (which does use priority queue though, so should have O‍ ‍(n‍ ‍logn) operations overall). Uses union from the "classic" version above: <lang Haskell>hammFrom n = drop n $

  iterate (\(_ , (a:t)) -> (a, union t [2*a,3*a,5*a])) (0, [1])</lang> 

Output:

<lang Haskell>> take 20 $ map fst $ hammFrom 1 [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]

> take 2 $ map fst $ hammFrom 1691 [2125764000,2147483648]

> mapM_ print $ take 10 $ hammFrom 1 (1,[2,3,5]) (2,[3,4,5,6,10]) (3,[4,5,6,9,10,15]) (4,[5,6,8,9,10,12,15,20]) (5,[6,8,9,10,12,15,20,25]) (6,[8,9,10,12,15,18,20,25,30]) (8,[9,10,12,15,16,18,20,24,25,30,40]) (9,[10,12,15,16,18,20,24,25,27,30,40,45]) (10,[12,15,16,18,20,24,25,27,30,40,45,50]) (12,[15,16,18,20,24,25,27,30,36,40,45,50,60])

> map (length . snd . head . hammFrom) [2000,4000,8000,16000] [402,638,1007,1596]

> map (logBase 2) $ zipWith (/) =<< tail $ [402,638,1007,1596] [0.67,0.66,0.66]</lang> Runs too slowly to reach 1,000,000, with empirical orders of growth above ~‍ ‍(n‍ ‍^1.7‍ ‍) and worsening. Last two lines show the internal buffer's length for several sample n‍ ‍s, and its empirical orders of growth which strongly support the $O(n^{2/3})$ claim.

Enumeration by a chain of folded merges

<lang Haskell> hamm = foldr merge1 [] . iterate (map (5*)) .

        foldr merge1 [] . iterate (map (3*))
                               $ iterate (2*) 1 
   where
   merge1 (x:xs) ys = x : merge xs ys

{- 1, 2, 4, 8, 16, 32, ...

  3,  6, 12, 24,  48,  96, ...
  9, 18, 36, 72, 144, 288, ...
  27, ... -}</lang>

Uses merge, as there's no need for duplicates-removing union because each number is produced only once here, too.

The merges are arranged in a chain of folds. Might be suitable for parallel execution, because of their large number.

Twice slower than the classic version at producing 1,000,000th Hamming number, and worsening, running at ~n^1.14..1.16 empirically (vs. the classic version's linear operations). This is surprisingly efficient considering the large number of merges going on (about 300 for the 1Mth number, and ~3n^1/3 in general).

Can be significantly improved, both in time complexity and absolute run time, by replacing the linear fold with the tree-shaped mergeAll from the Data.List.Ordered module of data-ordlist package.

Direct calculation through triples enumeration

It is also possible to more or less directly calculate the n-th Hamming number by enumerating (and counting) all the (i,j,k) triples below its estimated value – with ordering according to their exponents, i*ln2 + j*ln3 + k*ln5 – while storing only the "band" of topmost triples close enough to the target value (more in the original post on DDJ). The savings come from enumerating only pairs of indices, and finding the corresponding third index by a direct calculation, thus achieving the O(n^(2/3)) time complexity. Space complexity, with constant empirical estimation correction, is ~n^(2/3); but is further tweaked to ~n^(1/3) (following the idea from the entry below).

The total count of thus produced triples is then the band's topmost value's index in the Hamming sequence, 1-based. The nth number in the sequence is then found by a simple lookup in the sorted band, provided it was wide enough. This produces the 1,000,000-th value instantaneously. Following the 2017-10 IDEOne update to a faster 64-bit system, the 1 trillionth number is found in 0.7s on Ideone.com: <lang haskell>-- directly find n-th Hamming number, in ~ O(n^{2/3}) time. -- based on "top band" idea by Louis Klauder, from the DDJ discussion. -- by Will Ness, original post: drdobbs.com/blogs/architecture-and-design/228700538

import Data.List (sortBy, foldl') -- ' import Data.Function (on)

main = let (r,t) = nthHam 1000000 in print t >> print (trival t)

trival (i,j,k) = 2^i * 3^j * 5^k

nthHam :: Int -> (Double, (Int, Int, Int)) -- ( 64bit: use Int!!! NB! ) nthHam n -- n: 1-based: 1,2,3...

 | n <= 0    = error $ "n is 1--based: must be n > 0: " ++ show n
 | n < 2     = ( 0.0, (0, 0, 0) ) -- trivial case so estimation works for rest
 | w >= 1    = error $ "Breach of contract: (w < 1):  " ++ show w
 | m <  0    = error $ "Not enough triples generated: " ++ show ((c,n) :: (Int, Int))
 | m >= nb   = error $ "Generated band is too narrow: " ++ show (m,nb)
 | otherwise = sortBy (flip compare `on` fst) b !! m     -- m-th from top in sorted band
where  
 lb3 = logBase 2 3;  lb5 = logBase 2 5;  lb30_2 = logBase 2 30 / 2
 v = (6*lb3*lb5* fromIntegral n)**(1/3) - lb30_2         -- estimated logval, base 2
 estval n = (v + (1/v), 2/v)                             -- the space tweak! (thx, GBG!)
 (hi,w) = estval n                                       --   hi > logval > hi-w
 m      = fromIntegral (c - n)                           -- target index, from top
 nb     = length (b :: [(Double, (Int, Int, Int))])      -- length of the band
 (c,b) = foldl_ (\(c,b) (i,t)-> let c2=c+i in c2 `seq`   -- ( total count, the band )
                    case t of []-> (c2,b);[v]->(c2,v:b) ) (0,[])  -- ( =~= mconcat )
           [ ( fromIntegral i+1,                         -- total triples w/ this (j,k)
               [ (r,(i,j,k)) | frac < w ] )              -- store it, if inside band
             | k <- [ 0 .. floor ( hi   /lb5) ],  let p = fromIntegral k*lb5,    
               j <- [ 0 .. floor ((hi-p)/lb3) ],  let q = fromIntegral j*lb3 + p,
               let (i,frac) = pr  (hi-q) ;            r = hi - frac  -- r = i + q 
           ] where  pr      = properFraction             -- pr 1.24 => (1,0.24)
                    foldl_  = foldl'</lang>

Output:

-- time: 0.00s    memory: 4.2MB
(55,47,64)
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Using loops for a faster code, and a narrower band to save space

The DDJ blog post by Will Ness doesn't use the fact mentioned by the Wikipedia article that the error term in the estimation of the log of the resulting value for the nth Hamming number is directly proportional to this same log result. Using this fact, we are able to reduce the span of the "band" to only a constant fraction of the estimated log result for large n, and thus reduce memory space requirements to O(n^(1/3)) from O(n^(2/3)) for a considerable space saving for larger ranges.

As well, although it isn't quite as elegant in a Haskell style sense, one can get an additional constant factor in execution time by replacing the "loops" based on list comprehensions to tail-recursive function call "loops", as in the following code:

<lang haskell>{-# OPTIONS_GHC -O3 -XStrict #-}

import Data.Word import Data.List (sortBy) import Data.Function (on)

nthHam :: Word64 -> (Int, Int, Int) nthHam n -- n: 1-based 1,2,3...

 | n < 2     = case n of
                 0 -> error "nthHam:  Argument is zero!"
                 _ -> (0, 0, 0)                       -- trivial case for 1
 | m <  0    = error $ "Not enough triples generated: " ++ show (c,n)
 | m >= nb   = error $ "Generated band is too narrow: " ++ show (m,nb)
 | otherwise = case res of (_, tv) -> tv -- 2^i * 3^j * 5^k
where
 lb3     = logBase 2 3; lb5 = logBase 2 5.0
 lbrt30  = logBase 2 $ sqrt 30 :: Double -- estimate adjustment as per WP
 lg2est  = (6 * lb3 * lb5 * fromIntegral n)**(1/3) - lbrt30 -- estimated logval, base 2
 (hi,lo) = (lg2est + 1/lg2est, 2 * lg2est - hi) -- hi > log2est > lo
 (c, b)  = let klmt = floor (hi / lb5)
               loopk k ck bndk =
                 if k > klmt then (ck, bndk) else
                 let p = hi - fromIntegral k * lb5; jlmt = floor (p / lb3)
                     loopj j cj bndj =
                       if j > jlmt then loopk (k + 1) cj bndj else
                       let q = p - fromIntegral j * lb3
                           (i, frac) = properFraction q
                           nj = j + 1; ncj = cj + fromIntegral i + 1
                           r = hi - frac
                           nbndj = i `seq` bndj `seq`
                                   if r < lo then bndj
                                   else case (r, (i, j, k)) of
                                          nhd -> nhd `seq` nhd : bndj
                       in ncj `seq` nbndj `seq` loopj nj ncj nbndj
                 in loopj 0 ck bndk
           in loopk 0 0 []
 (m,nb)  = ( fromIntegral $ c - n, length b )         -- m 0-based from top, |band|
 (s,res) = ( sortBy (flip compare `on` fst) b, s!!m ) -- sorted decreasing, result<

main = putStrLn $ show $ nthHam 1000000000000</lang>

This implementation can likely calculate the 10^19th Hamming number in less than a day and can't quite reach the (2^64-1)th (18446744073709551615th) Hamming due to a slight range overflow as it approaches this limit. Maximum memory used to these limits is less than a few hundred Megabytes, so possible on typical personal computers given the required day or two of computing time.

On IdeOne (64-bit), this takes 0.03 seconds for the 10 billionth and 0.70 seconds for the trillionth number (October 2017 update to a faster 64-bit system).

Using "roll-your-own" extended precision logarithm values in the error band to extend range

All of these codes using algorithms can't do an accurate sort of the error band for arguments somewhere above 10^13 due to the limited precision of the Double logarithm values, but this is easily fixed by using "roll-your-own" Integer logarithm values as follows with very little cost in execution time as it only applies to the relatively very small error band: <lang haskell>{-# OPTIONS_GHC -O3 -XStrict #-}

import Data.Word import Data.List (sortBy) import Data.Function (on)

nthHam :: Word64 -> (Int, Int, Int) nthHam n -- n: 1-based 1,2,3...

 | n < 2     = case n of
                 0 -> error "nthHam:  Argument is zero!"
                 _ -> (0, 0, 0)                       -- trivial case for 1
 | m <  0    = error $ "Not enough triples generated: " ++ show (c,n)
 | m >= nb   = error $ "Generated band is too narrow: " ++ show (m,nb)
 | otherwise = case res of (_, tv) -> tv -- 2^i * 3^j * 5^k
where
 lb3     = logBase 2 3; lb5 = logBase 2 5.0
 lbrt30  = logBase 2 $ sqrt 30 :: Double -- estimate adjustment as per WP
 lg2est  = (6 * lb3 * lb5 * fromIntegral n)**(1/3) - lbrt30 -- estimated logval, base 2
 (hi,lo) = (lg2est + 1/lg2est, 2 * lg2est - hi) -- hi > log2est > lo
 bglb2 = 1267650600228229401496703205376 :: Integer
 bglb3 = 2009178665378409109047848542368 :: Integer
 bglb5 = 2943393543170754072109742145491 :: Integer
 (c, b)  = let klmt = floor (hi / lb5)
               loopk k ck bndk =
                 if k > klmt then (ck, bndk) else
                 let p = hi - fromIntegral k * lb5; jlmt = floor (p / lb3)
                     loopj j cj bndj =
                       if j > jlmt then loopk (k + 1) cj bndj else
                       let q = p - fromIntegral j * lb3
                           (i, frac) = properFraction q
                           nj = j + 1; ncj = cj + fromIntegral i + 1
                           r = hi - frac
                           nbndj = i `seq` bndj `seq`
                                   if r < lo then bndj
                                   else 
                                     let bglg = bglb2 * fromIntegral i +
                                                  bglb3 * fromIntegral j +
                                                  bglb5 * fromIntegral k in
                                     bglg `seq` case (bglg, (i, j, k)) of
                                                  nhd -> nhd `seq` nhd : bndj
                       in ncj `seq` nbndj `seq` loopj nj ncj nbndj
                 in loopj 0 ck bndk
           in loopk 0 0 []
 (m,nb)  = ( fromIntegral $ c - n, length b )         -- m 0-based from top, |band|

-- (s,res) = (b, s!!m)

 (s,res) = ( sortBy (flip compare `on` fst) b, s!!m ) -- sorted decreasing, result<

main = putStrLn $ show $ nthHam 1000000000000</lang>

All of these codes run a constant factor faster using the forced "Strict" mode, which shows that it is very difficult to anticipate the Haskell strictness analyser, especially in the case of the first code using List comprehensions.

Icon and Unicon

This solution uses Unicon's object oriented extensions. An Icon only version has not been provided.

Lazy evaluation is used to improve performance. <lang Unicon># Lazily generate the three Hamming numbers that can be derived directly

1. from a known Hamming number h

class Triplet : Class (cv, ce)

   method nextVal()
       suspend cv := @ce
   end

   initially (baseNum)
       cv := 2*baseNum
       ce := create (3|5)*baseNum

end

1. Generate Hamming numbers, in order. Default is first 30
2. But an optional argument can be used to generate more (or less)
3. e.g. hamming 5000 generates the first 5000.

procedure main(args)

   limit := integer(args[1]) | 30
   every write("\t", generateHamming() \ limit)

end

1. Do the work. Start with known Hamming number 1 and maintain
2. a set of triplet Hamming numbers as they get derived from that
3. one. Most of the code here is to figure out which Hamming
4. number is next in sequence (while removing duplicates)

procedure generateHamming()

   triplers := set()
   insert(triplers, Triplet(1))

   suspend 1
   repeat {
       # Pick a Hamming triplet that *may* have the next smallest number
       t1 := !triplers         # any will do to start

       every t1 ~=== (t2 := !triplers) do {
           if t1.cv > t2.cv then {
              # oops we were wrong, switch assumption
              t1 := t2
              }
           else if t1.cv = t2.cv then {
              # t2's value is a duplicate, so
              # advance triplet t2, if none left in t2, remove it
              t2.nextVal() | delete(triplers, t2)
              }
          }

       # Ok, t1 has the next Hamming number, grab it
       suspend t1.cv
       insert(triplers, Triplet(t1.cv)) 
       # Advance triplet t1, if none left in t1, remove it
       t1.nextVal() | delete(triplers, t1)
       }

end</lang>

J

Solution:
A concise tacit expression using a (right) fold: <lang j>hamming=: {. (/:~@~.@] , 2 3 5 * {)/@(1x ,~ i.@-)</lang> Example usage: <lang j> hamming 20 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36

  {: hamming 1691

2125764000</lang> For the millionth (and billionth (1e9)) Hamming number see the nh verb from Hamming Number essay on the J wiki.

Explanation:
I'll explain this J-sentence by dividing it in three parts from left to right omitting the leftmost {.:

• sort and remove duplicates

<lang j> /:~@~.@]</lang>

• produce 3 elements by selection and multiplication (we have already produced smaller values, this will overproduce slightly larger values, but the extra values overlap, and we handle that by discarding duplicates):

<lang j> 2 3 5 * {</lang> note that LHA (2 3 5 * {) RHA is equivalent to <lang j> 2 3 5 * LHA { RHA</lang>

• the RH part forms an array of descending indices and the initial Hamming number 1

<lang j> (1x ,~ i.@-)</lang> e.g. if we want the first 5 Hamming numbers, it produces the array:

4 3 2 1 0 1

in other words, we compute a sequence which begins with the desired hamming sequence and then take the first n elements (which will be our desired hamming sequence) <lang j> ({. (/:~@~.@] , 2 3 5 * {)/@(1x ,~ i.@-)) 7 1 2 3 4 5 6 8</lang> This starts using a descending sequence with 1 appended: <lang j> (1x ,~ i.@-) 7 6 5 4 3 2 1 0 1</lang> and then the fold expression is inserted between these list elements and the result computed: <lang j> 6(/:~@~.@] , 2 3 5 * {) 5(/:~@~.@] , 2 3 5 * {) 4(/:~@~.@] , 2 3 5 * {) 3(/:~@~.@] , 2 3 5 * {) 2(/:~@~.@] , 2 3 5 * {) 1(/:~@~.@] , 2 3 5 * {) 0(/:~@~.@] , 2 3 5 * {) 1 1 2 3 4 5 6 8 9 10 12 15 18 20 25 30 16 24 40</lang> (Note: A train of verbs in J is evaluated by supplying arguments to the every other verb (counting from the right) and the combining these results with the remaining verbs. Also: { has been implemented so that an index of 0 will select the only item from an array with no dimensions.)

Java

Works with: Java version 1.5+

Has a common shortcoming of overproducing the sequence by about $O(n^{2/3})$ entries, until the n-th number is no longer needed, instead of stopping as soon as it is reached. See Haskell for an illustration.

Inserting the top number's three multiples deep into the priority queue as it does, incurs extra cost for each number produced. To not worsen the expected algorithm complexity, the priority queue should have (amortized) $O(1)$ implementation for both insertion and deletion operations but it looks like it's $O(\log n)$ in Java. <lang java5>import java.math.BigInteger; import java.util.PriorityQueue;

final class Hamming {

   private static BigInteger THREE = BigInteger.valueOf(3);
   private static BigInteger FIVE = BigInteger.valueOf(5);

   private static void updateFrontier(BigInteger x,
                                      PriorityQueue<BigInteger> pq) {
       pq.offer(x.shiftLeft(1));
       pq.offer(x.multiply(THREE));
       pq.offer(x.multiply(FIVE));
   }

   public static BigInteger hamming(int n) {
       if (n <= 0)
           throw new IllegalArgumentException("Invalid parameter");
       PriorityQueue<BigInteger> frontier = new PriorityQueue<BigInteger>();
       updateFrontier(BigInteger.ONE, frontier);
       BigInteger lowest = BigInteger.ONE;
       for (int i = 1; i < n; i++) {
           lowest = frontier.poll();
           while (frontier.peek().equals(lowest))
               frontier.poll();
           updateFrontier(lowest, frontier);
       }
       return lowest;
   }

   public static void main(String[] args) {
       System.out.print("Hamming(1 .. 20) =");
       for (int i = 1; i < 21; i++)
            System.out.print(" " + hamming(i));
       System.out.println("\nHamming(1691) = " + hamming(1691));
       System.out.println("Hamming(1000000) = " + hamming(1000000));
   }

}</lang>

Output:

Hamming(1 .. 20) = 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Hamming(1691) = 2125764000
Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Another possibility is to realize that Hamming numbers can be represented and stored as triples of nonnegative integers which are the powers of 2, 3 and 5, and do all operations needed by the algorithms directly on these triples without converting to $BigInteger$, which saves tremendous memory and time. Also, the search frontier through this three-dimensional grid can be generated in an order that never reaches the same state twice, so we don't need to keep track which states have already been encountered, saving even more memory. The objects of $HammingTriple$ encode Hamming numbers in three fields $a$, $b$ and $c$. Multiplying by 2, 3 and 5 can now be done just by incrementing that field. The order comparison of triples needed by the priority queue is implemented with simple logarithm formulas of multiplication and addition, resorting to conversion to $BigInteger$ only in the rare cases that the floating point arithmetic produces too close results.

<lang java5> import java.math.BigInteger; import java.util.*;

public class HammingTriple implements Comparable<HammingTriple> {

   // Precompute a couple of constants that we need all the time
   private static final BigInteger two = BigInteger.valueOf(2);
   private static final BigInteger three = BigInteger.valueOf(3);
   private static final BigInteger five = BigInteger.valueOf(5);
   private static final double logOf2 = Math.log(2);
   private static final double logOf3 = Math.log(3);
   private static final double logOf5 = Math.log(5);
       
   // The powers of this triple
   private int a, b, c;
     
   public HammingTriple(int a, int b, int c) {
       this.a = a; this.b = b; this.c = c;
   }
   
   public String toString() {
       return "[" + a + ", " + b + ", " + c + "]";
   }
   
   public BigInteger getValue() {
       return two.pow(a).multiply(three.pow(b)).multiply(five.pow(c));
   }
   
   public boolean equals(Object other) {
       if(other instanceof HammingTriple) {
           HammingTriple h = (HammingTriple) other;
           return this.a == h.a && this.b == h.b && this.c == h.c;
       }
       else { return false; }
   }
   
   // Return 0 if this == other, +1 if this > other, and -1 if this < other
   public int compareTo(HammingTriple other) {
       // equality
       if(this.a == other.a && this.b == other.b && this.c == other.c) {
           return 0;
       }
       // this dominates
       if(this.a >= other.a && this.b >= other.b && this.c >= other.c) {
           return +1;
       }
       // other dominates
       if(this.a <= other.a && this.b <= other.b && this.c <= other.c) {
           return -1;
       }
      
       // take the logarithms for comparison
       double log1 = this.a * logOf2 + this.b * logOf3 + this.c * logOf5;
       double log2 = other.a * logOf2 + other.b * logOf3 + other.c * logOf5;
       
       // are these different enough to be reliable?
       if(Math.abs(log1 - log2) > 0.0000001) {
           return (log1 < log2) ? -1: +1;
       }
       
       // oh well, looks like we have to do this the hard way
       return this.getValue().compareTo(other.getValue());
       // (getting this far should be pretty rare, though)
   }
   
   public static BigInteger computeHamming(int n, boolean verbose) {
       if(verbose) {
           System.out.println("Hamming number #" + n);
       }
       long startTime = System.currentTimeMillis();
       
       // The elements of the search frontier
       PriorityQueue<HammingTriple> frontierQ = new PriorityQueue<HammingTriple>();
       int maxFrontierSize = 1;
       
       // Initialize the frontier
       frontierQ.offer(new HammingTriple(0, 0, 0)); // 1
       
       while(true) {
           if(frontierQ.size() > maxFrontierSize) {
               maxFrontierSize = frontierQ.size();
           }
           // Pop out the next Hamming number from the frontier
           HammingTriple curr = frontierQ.poll();
           
           if(--n == 0) {
               if(verbose) {
                   System.out.println("Time: " + (System.currentTimeMillis() - startTime) + " ms");
                   System.out.println("Frontier max size: " + maxFrontierSize);
                   System.out.println("As powers: " + curr.toString());
                   System.out.println("As value: " + curr.getValue());
               }
               return curr.getValue();
           }
           
           // Current times five, if at origin in (a,b) plane
           if(curr.a == 0 && curr.b == 0) {
               frontierQ.offer(new HammingTriple(curr.a, curr.b, curr.c + 1));
           }
           // Current times three, if at line a == 0
           if(curr.a == 0) {
               frontierQ.offer(new HammingTriple(curr.a, curr.b + 1, curr.c));
           }
           // Current times two, unconditionally
           curr.a++;
           frontierQ.offer(curr); // reuse the current HammingTriple object
       }
   }

} </lang>

Hamming number #1000000
Time: 650 ms
Frontier max size: 10777
As powers: [55, 47, 64]
As value: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Hamming number #1000000000
Time: 1763306 ms
Frontier max size: 1070167
As powers: [1334, 335, 404]
As value: 62160757555652448616308163328720720039470565190896527065916324096423370220027531418244175407
772567327803701726166152919355404186200255249167295000868314547113136940786355040041603128729517887
0364794838245609107270160079056207179759030665476588225699039176388785014115448224991592743918456282
8227449023750262318234797192076792208033475638322151983772515798004125909334741121595323950448656375
1044570269974247729669174417794061727369755885568000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000

Alternative

This uses memoized streams - similar in principle to the classic lazy-evaluated sequence, but with a java flavor. Hope you like this recipe!

<lang java> import java.math.BigInteger;

public class Hamming {

   public static void main(String args[])
   {
       Stream hamming = makeHamming();
       System.out.print("H[1..20] ");
       for (int i=0; i<20; i++) {
           System.out.print(hamming.value());
           System.out.print(" ");
           hamming = hamming.advance();
       }
       System.out.println();

       System.out.print("H[1691] ");
       hamming = makeHamming();
       for (int i=1; i<1691; i++) {
           hamming = hamming.advance();
       }
       System.out.println(hamming.value());

       hamming = makeHamming();
       System.out.print("H[10^6] ");
       for (int i=1; i<1000000; i++) {
           hamming = hamming.advance();
       }
       System.out.println(hamming.value());
   }

   public interface Stream
   {
       BigInteger value();
       Stream advance();
   }

   public static class MultStream implements Stream
   {
       MultStream(int mult)
       { m_mult = BigInteger.valueOf(mult); }
       MultStream setBase(Stream s)
       { m_base = s; return this; }
       public BigInteger value()
       { return m_mult.multiply(m_base.value()); }
       public Stream advance()
       { return setBase(m_base.advance()); }

       private final BigInteger m_mult;
       private Stream m_base;
   }

   private final static class RegularStream implements Stream
   {
       RegularStream(Stream[] streams, BigInteger val)
       {
           m_streams = streams;
           m_val = val;
       }
       public BigInteger value()
       { return m_val; }

       public Stream advance()
       {
           // memoized value for the next stream instance.
           if (m_advance != null) { return m_advance; }

           int minidx = 0 ;
           BigInteger next = nextStreamValue(0);
           for (int i=1; i<m_streams.length; i++) {
               BigInteger v = nextStreamValue(i);
               if (v.compareTo(next) < 0) {
                   next = v;
                   minidx = i;
               }
           }
           RegularStream ret = new RegularStream(m_streams, next);
           // memoize the value!
           m_advance = ret;
           m_streams[minidx].advance();
           return ret;
       }
       private BigInteger nextStreamValue(int streamidx)
       {
           // skip past duplicates in the streams we're merging.
           BigInteger ret = m_streams[streamidx].value();
           while (ret.equals(m_val)) {
               m_streams[streamidx] = m_streams[streamidx].advance();
               ret = m_streams[streamidx].value();
           }
           return ret;
       }
       private final Stream[] m_streams;
       private final BigInteger m_val;
       private RegularStream m_advance = null;
   }

   private final static Stream makeHamming()
   {
       MultStream nums[] = new MultStream[] {
           new MultStream(2),
           new MultStream(3),
           new MultStream(5)
       };
       Stream ret = new RegularStream(nums, BigInteger.ONE);
       for (int i=0; i<nums.length; i++) {
           nums[i].setBase(ret);
       }
       return ret;
   }

} </lang>

$ java Hamming 
H[1..20] 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
H[1691] 2125764000
H[10^6] 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
$ 

JavaScript

Works with: JavaScript version 1.7

Works with: Firefox version 2

Translation of: Ruby

This does not calculate the 1,000,000th Hamming number.

Note the use of for (x in obj) to iterate over the properties of an object, versus for each (y in obj) to iterate over the values of the properties of an object. <lang javascript>function hamming() {

   var queues = {2: [], 3: [], 5: []};
   var base;
   var next_ham = 1;
   while (true) {
       yield next_ham;

       for (base in queues) {queues[base].push(next_ham * base)}

       next_ham = [ queue[0] for each (queue in queues) ].reduce(function(min, val) {
           return Math.min(min,val)
       });

       for (base in queues) {if (queues[base][0] == next_ham) queues[base].shift()}
   }

}

var ham = hamming(); var first20=[], i=1;

for (; i <= 20; i++)

   first20.push(ham.next());

print(first20.join(', ')); print('...'); for (; i <= 1690; i++)

   ham.next();

print(i + " => " + ham.next());</lang>

Output:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36
...
1691 => 2125764000 

Fast & complete version

Translation of: C#

A translation of my fast C# version. I was curious to see how much slower JavaScript is. The result: it runs about 5x times slower than C#, though YMMV. You can try it yourself here: http://jsfiddle.net/N7AFN/

--Mike Lorenz

<lang javascript><html> <head></head> <body>

</body> <script src="http://code.jquery.com/jquery-latest.min.js"></script> <script src="http://peterolson.github.com/BigInteger.js/BigInteger.min.js"></script> <script type="text/javascript">

   var _primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37];

   function log(text) {
       $('#main').append(text + "\n");
   }

   function big(exponents) {
       var i, e, val = bigInt.one;
       for (i = 0; i < exponents.length; i++)
           for (e = 0; e < exponents[i]; e++)
               val = val.times(_primes[i]);
       return val.toString();
   }

   function hamming(n, nprimes) {
       var i, iter, p, q, min, equal, x;

       var hammings = new Array(n);                            // array of hamming #s we generate
       hammings[0] = new Array(nprimes);
       for (p = 0; p < nprimes; p++) {
           hammings[0][p] = 0;
       }

       var hammlogs = new Array(n);                            // log values for above
       hammlogs[0] = 0;

       var primelogs = new Array(nprimes);                     // pre-calculated prime log values
       var listlogs  = new Array(nprimes);                     // log values of list heads
       for (p = 0; p < nprimes; p++) {
           primelogs[p] = listlogs[p] = Math.log(_primes[p]);
       }

       var indexes = new Array(nprimes);                       // intermediate hamming values as indexes into hammings
       for (p = 0; p < nprimes; p++) {
           indexes[p] = 0;
       }

       var listheads = new Array(nprimes);                     // intermediate hamming list heads
       for (p = 0; p < nprimes; p++) {
           listheads[p] = new Array(nprimes);
           for (q = 0; q < nprimes; q++) {
               listheads[p][q] = 0;
           }
           listheads[p][p] = 1;
       }

       for (iter = 1; iter < n; iter++) {
           min = 0;
           for (p = 1; p < nprimes; p++)
               if (listlogs[p] < listlogs[min])
                   min = p;
           hammlogs[iter] = listlogs[min];                     // that's the next hamming number
           hammings[iter] = listheads[min].slice();
           for (p = 0; p < nprimes; p++) {                     // update each list head if it matches new value
               equal = true;                                   // test each exponent to see if number matches
               for (i = 0; i < nprimes; i++) {
                   if (hammings[iter][i] != listheads[p][i]) {
                       equal = false;
                       break;
                   }
               }
               if (equal) {                                    // if it matches...
                   x = ++indexes[p];                           // set index to next hamming number
                   listheads[p] = hammings[x].slice();         // copy hamming number
                   listheads[p][p] += 1;                       // increment exponent = mult by prime
                   listlogs[p] = hammlogs[x] + primelogs[p];   // add log(prime) to log(value) = mult by prime
               }
           }
       }

       return hammings[n - 1];
   }

   $(document).ready(function() {
       var i, nprimes;
       var t = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1691,1000000];

       for (nprimes = 3; nprimes <= 4; nprimes++) {
           var start = new Date();

log('

' + _primes[nprimes - 1] + '-Smooth:' + '

'); log('

'); for (i = 0; i < t.length; i++) log('' + '');

           var end = new Date();

log('' + ''); log('

' + t[i] + ':' + '	' + big(hamming(t[i], nprimes)) + '
' + 'Elapsed time:' + '	' + (end-start)/1000 + ' seconds' + '

');

       }
   });

</script> </html></lang>

Output:

5-Smooth:

1:		1
2:		2
3:		3
4:		4
5:		5
6:		6
7:		8
8:		9
9:		10
10:		12
11:		15
12:		16
13:		18
14:		20
15:		24
16:		25
17:		27
18:		30
19:		32
20:		36
1691:		2125764000
1000000:	519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Elapsed time:	1.73 seconds

7-Smooth:

1:		1
2:		2
3:		3
4:		4
5:		5
6:		6
7:		7
8:		8
9:		9
10:		10
11:		12
12:		14
13:		15
14:		16
15:		18
16:		20
17:		21
18:		24
19:		25
20:		27
1691:		3317760
1000000:	4157409948433216829957008507500000000
Elapsed time:	1.989 seconds

jq

Works with: jq version 1.4

We take the primary challenge here to be to write a Hamming number generator that can generate a given number of Hamming numbers, or the n-th Hamming number, without storing previously generated numbers.

To motivate a more complex version, in Part 1 of this section hamming(n) is defined as a generator of Hamming numbers, as numbers. This function uses an efficient algorithm and can run indefinitely, but it has one disadvantage: currently, jq converts large integers to floating point approximations, and thus precision is lost. For example, it reports the millionth Hamming number as 1.926511252902403e+44.

In Part 2, the algorithm in the first part is modified to use the [p,q,r] representation of Hamming numbers, where p, q, and r are the relevant exponents respectively of 2, 3, and 5.

The task description focuses on finding the n-th element of an infinite sequence and so it should be mentioned that using jq versions greater than 1.4, it would be possible to simply the generator so that is always unbounded, and then harness it with new builtins such as "limit" and "nth".

Hamming number generator

<lang jq># Return the index in the input array of the min_by(f) value def index_min_by(f):

 . as $in
 | if length == 0 then null
   else .[0] as $first
   | reduce range(0; length) as $i 
       ([0, $first, ($first|f)];   # state: [ix; min; f|min]
        ($in[$i]|f) as $v
        | if $v < .[2] then [ $i, $in[$i], $v ] else . end) 
   | .[0]
   end;

1. Emit n Hamming numbers if n>0; the nth if n<0

def hamming(n):

 # input: [twos, threes, fives] of which at least one is assumed to be non-empty
 # output: the index of the array holding the min of the firsts
 def next: map( .[0] ) | index_min_by(.);

 # input: [value, [twos, threes, fives] ....]
 # ix is the index in [twos, threes, fives] of the array to be popped
 # output: [popped, updated_arrays ...]
 def pop(ix):
   .[1] as $triple
   | setpath([0];    $triple[ix][0])
   | setpath([1,ix]; $triple[ix][1:]);

 # input: [x, [twos, threes, fives], count]
 # push value*2 to twos, value*3 to threes, value*5 to fives and increment count
 def push(v):
   [.[0], [.[1][0] + [2*v], .[1][1] + [3*v], .[1][2] + [5*v]], .[2] + 1];

 # _hamming is the workhorse
 # input: [previous, [twos, threes, fives], count]
 def _hamming:
   .[0] as $previous
   | if (n > 0 and .[2] == n) or (n<0 and .[2] == -n) then $previous
     else (.[1]|next) as $ix     # $ix cannot be null
     | pop($ix)
     | .[0] as $next
     | (if $next == $previous then empty elif n>=0 then $previous else empty end), 
       (if $next == $previous then . else push($next) end | _hamming) 
     end;
 [1, [[2],[3],[5]], 1] | _hamming;

. as $n | hamming($n)</lang> Examples: <lang jq># First twenty: hamming(20)

1. See elsewhere for output

1. 1691st Hamming number:

hamming(-1691)

1. => 2125764000

1. Millionth:

hamming(-1000000)

1. => 1.926511252902403e+44

</lang>

Hamming numbers as triples

In this section, Hamming numbers are represented as triples, [p,q,r], where p, q and r are the relevant powers of 2, 3, and 5 respectively. We therefore begin with some functions for managing Hamming numbers represented in this manner: <lang jq># The log (base e) of a Hamming triple: def ln_hamming:

 if length != 3 then error("ln_hamming: \(.)") else . end
 | (.[0] * (2|log)) + (.[1] * (3|log)) + (.[2] * (5|log));

1. The numeric value of a Hamming triple:

def hamming_tof: ln_hamming | exp;

def hamming_toi:

 def pow(n): . as $in | reduce range(0;n) as $i (1; . * $in);
 . as $in | (2|pow($in[0])) * (3|pow($in[1])) * (5|pow($in[2]));

1. Return the index in the input array of the min_by(f) value

def index_min_by(f):

 . as $in
 | if length == 0 then null
   else .[0] as $first
   | reduce range(0; length) as $i 
       ([0, $first, ($first|f)];   # state: [ix; min; f|min]
        ($in[$i]|f) as $v
        | if $v < .[2] then [ $i, $in[$i], $v ] else . end) 
   | .[0]
   end;

1. Emit n Hamming numbers (as triples) if n>0; the nth if n<0; otherwise indefinitely.

def hamming(n):

 # n must be 2, 3 or 5
 def hamming_times(n): n as $n
   | if $n==2 then .[0] += 1 elif $n==3 then .[1] += 1 else .[2] += 1 end;

 # input: [twos, threes, fives] of which at least one is assumed to be non-empty
 # output: the index of the array holding the min of the firsts
 def next: map( .[0] ) | index_min_by( ln_hamming );

 # input: [value, [twos, threes, fives] ....]
 # ix is the index in [twos, threes, fives] of the array to be popped
 # output: [popped, updated_arrays ...]
 def pop(ix):
   .[1] as $triple
   | setpath([0];    $triple[ix][0])
   | setpath([1,ix]; $triple[ix][1:]);

 # input: [x, [twos, threes, fives], count]
 # push value*2 to twos, value*3 to threes, value*5 to fives and increment count
 def push(v):
   [.[0], [.[1][0] + [v|hamming_times(2)], .[1][1] + [v|hamming_times(3)],
           .[1][2] + [v|hamming_times(5)]], .[2] + 1];

 # _hamming is the workhorse
 # input: [previous, [twos, threes, fives], count]
 def _hamming:
   .[0] as $previous
   | if (n > 0 and .[2] == n) or (n<0 and .[2] == -n) then $previous
     else (.[1]|next) as $ix     # $ix cannot be null
     | pop($ix)
     | .[0] as $next
     | (if $next == $previous then empty elif n>=0 then $previous else empty end), 
       (if $next == $previous then . else push($next) end | _hamming) 
     end;
 [[0,0,0], [ 1,0,0 ,0,1,0, 0,0,1 ], 1] | _hamming;

</lang> Examples <lang jq># The first twenty Hamming numbers as integers: hamming(-20) | hamming_toi

1. => (see elsewhere)

1. 1691st as a Hamming triple:

hamming(-1691)

1. => [5,12,3]

1. The millionth:

hamming(-1000000)

1. => [55,47,64]</lang>

Julia

Simple brute force algorithm, derived from the discussion at ProgrammingPraxis.com. <lang julia>function hammingsequence(N)

   if N < 1
       throw("Hamming sequence exponent must be a positive integer")
   end
   ham = N > 4000 ? Vector{BigInt}([1]) : Vector{Int}([1])
   base2, base3, base5 = (1, 1, 1)
   for i in 1:N-1
       x = min(2ham[base2], 3ham[base3], 5ham[base5])
       push!(ham, x)
       if 2ham[base2] <= x
           base2 += 1
       end
       if 3ham[base3] <= x
           base3 += 1
       end
       if 5ham[base5] <= x
           base5 += 1
       end
   end
   ham

end

println(hammingsequence(20)) println(hammingsequence(1691)[end])

println(hammingsequence(1000000)[end])</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

The above code is terribly inefficient, just as said, but can be improved by about a factor of two by using intermediate variables (next2, next3, and next5) to avoid recalculating the long multi-precision integers for each comparison, as it seems that the Julia compiler (version 1.0.2) is not doing common sub expression elimination: <lang julia>function hammingsequence(N::Int)

   if N < 1
       throw("Hamming sequence index must be a positive integer")
   end
   ham = Vector{BigInt}([1])
   base2, base3, base5 = 1, 1, 1
   next2, next3, next5 = BigInt(2), BigInt(3), BigInt(5)
   for _ in 1:N-1
       x = min(next2, next3, next5)
       push!(ham, x)
       next2 <= x && (base2 += 1; next2 = 2ham[base2])
       next3 <= x && (base3 += 1; next3 = 3ham[base3])
       next5 <= x && (base5 += 1; next5 = 5ham[base5])
   end
   ham

end</lang>

Infinite generator, avoiding duplicates, using logarithms for faster processing

The above code is slow for several reasons, partly because it is doing many multi-precision integer multiplications requiring much memory allocation and garbage collection for which the current Julia version 1.0.2 is quite slow, but also because there are many repeated calculations (3 time 2 equals 2 times three, etc.). The following code is about 30 times faster by using floating point logarithms for multiplication and comparison; it also is an infinite generator (an iterator), which means that memory consumption can be greatly reduced by eliminating values which are no longer of any use: <lang julia>function trival((twos, threes, fives))

   BigInt(2)^twos * BigInt(3)^threes * BigInt(5)^fives

end

mutable struct Hammings

   ham532 :: Vector{Tuple{Float64,Tuple{Int,Int,Int}}}
   ham53 :: Vector{Tuple{Float64,Tuple{Int,Int,Int}}}
   ndx532 :: Int
   ndx53 :: Int
   next2 :: Tuple{Float64,Tuple{Int,Int,Int}}
   next3 :: Tuple{Float64,Tuple{Int,Int,Int}}
   next5 :: Tuple{Float64,Tuple{Int,Int,Int}}
   next53 :: Tuple{Float64,Tuple{Int,Int,Int}}
   Hammings() = new(
       Vector{Tuple{Float64,Tuple{Int,Int,Int}}}(),
       Vector{Tuple{Float64,Tuple{Int,Int,Int}}}(),
       1, 1,
       (1.0, (1, 0, 0)), (log(2, 3), (0, 1, 0)),
       (log(2, 5), (0, 0, 1)), (0.0, (0, 0, 0))
   )

end Base.eltype(::Type{Hammings}) = Tuple{Int,Int,Int} function Base.iterate(HM::Hammings, st = HM) # :: Union{Nothing,Tuple{Tuple{Float64,Tuple{Int,Int,Int}},Hammings}}

   log2of2, log2of3, log2of5 = 1.0, log(2,3), log(2,5)
   if st.next2[1] < st.next53[1]
       push!(st.ham532, st.next2); st.ndx532 += 1
       last, (twos, threes, fives) = st.ham532[st.ndx532]
       st.next2 = (log2of2 + last, (twos + 1, threes, fives))
   else
       push!(st.ham532, st.next53)
       if st.next3[1] < st.next5[1]
           st.next53 = st.next3; push!(st.ham53, st.next3)
           last, (_, threes, fives) = st.ham53[st.ndx53]; st.ndx53 += 1
           st.next3 = (log2of3 + last, (0, threes + 1, fives))
       else
           st.next53 = st.next5; push!(st.ham53, st.next5)
           last, (_, _, fives) = st.next5
           st.next5 = (log2of5 + last, (0, 0, fives + 1))
       end
   end
   len53 = length(st.ham53)
   if st.ndx53 > (len53 >>> 1)
       nlen53 = len53 - st.ndx53 + 1
       copyto!(st.ham53, 1, st.ham53, st.ndx53, nlen53)
       resize!(st.ham53, nlen53); st.ndx53 = 1
   end
   len532 = length(st.ham532)
   if st.ndx532 > (len532 >>> 1)
       nlen532 = len532 - st.ndx532 + 1
       copyto!(st.ham532, 1, st.ham532, st.ndx532, nlen532)
       resize!(st.ham532, nlen532); st.ndx532 = 1
   end
   _, tri = st.ham532[end]
   tri, st

1. convert(Union{Nothing,Tuple{Tuple{Float64,Tuple{Int,Int,Int}},Hammings}},(st.ham532[end], st))
2. (length(st.ham532), length(st.ham53)), st

end

foreach(x -> print(trival(x)," "), (Iterators.take(Hammings(), 20))); println() let count = 1691; for t in Hammings() count <= 1 && (println(trival(t)); break); count -= 1 end end let count = 1000000; for t in Hammings() count <= 1 && (println(trival(t)); break); count -= 1 end end</lang>

Determination of the nth Hamming number by processing of error band

For some phenomenal speed in determining the nth Hamming/regular number, one doesn't need to find all the values up to that limit but rather only the values within an error band which is a factor of two either way from the correct value; this has the advantage that the number of processing loops are reduced from O(n^3) to O(n^(2/3)) for a considerable saving for larger ranges and has the further advantage that memory consumption is reduced to O(n^(1/3)) meaning that huge ranges can be computed on a common desktop computer. The folwingcode can compute the trillionth (10^12th) Hamming number is a couple of seconds: <lang julia>function nthhamming(n :: UInt64) # :: Tuple{UInt32, UInt32, UInt32}

   # take care of trivial cases too small for band size estimation to work...
   n < 1 && throw("nthhamming:  argument must be greater than zero!!!")
   n < 2 && return (0, 0, 0)
   n < 3 && return (1, 0, 0)

   # some constants...
   log2of2, log2of3, log2of5 = 1.0, log(2, 3), log(2, 5)
   fctr, crctn = 6.0 * log2of3 * log2of5, log(2, sqrt(30))
   log2est = (fctr * Float64(n))^(1.0 / 3.0) - crctn # log2 answer from WP formula
   log2hi = log2est + 1.0 / log2est; width = 2.0 / log2est # up to 2X higher/lower

   # loop to find the count of regular numbers and band of possible candidates...
   count :: UInt64 = 0; band = Vector{Tuple{Float64,Tuple{UInt32,UInt32,UInt32}}}()
   fiveslmt = UInt32(ceil(log2hi / log2of5)); fives :: UInt32 = 0
   while fives < fiveslmt
       log2p = log2hi - fives * log2of5
       threeslmt = UInt32(ceil(log2p / log2of3)); threes :: UInt32 = 0
       while threes < threeslmt
           log2q = log2p - threes * log2of3
           twos = UInt32(floor(log2q)); frac = log2q - twos; count += twos + 1
           frac <= width && push!(band, (log2hi - frac, (twos, threes, fives)))
           threes += 1
       end
       fives += 1
   end

   # process the band found including checks for validity and range...
   n > count && throw("nthhamming:  band high estimate is too low!!!")
   ndx = count - n + 1
   ndx > length(band) && throw("nthhamming:  band width estimate is too narrow!!!")
   sort!(band, by=(tpl -> let (lg,_) = tpl; -lg end)) # sort in decending order

   # get and return the answer...
   _, tri = band[ndx]
   tri

end

foreach(x-> print(trival(nthhamming(UInt(x))), " "), 1:20); println() println(trival(nthhamming(UInt64(1691)))) println(trival(nthhamming(UInt64(1000000))))</lang>

Above about a range of 10^13, a Float64 logarithm doesn't have enough precision to be able to sort the error band properly, so a refinement of using a "roll-your-own" extended precision logarithm must be used, as follows: <lang julia>function nthhamming(n :: UInt64) # :: Tuple{UInt32, UInt32, UInt32}

   # take care of trivial cases too small for band size estimation to work...
   n < 1 && throw("nthhamming:  argument must be greater than zero!!!")
   n < 2 && return (0, 0, 0)
   n < 3 && return (1, 0, 0)

   # some constants...
   log2of2, log2of3, log2of5 = 1.0, log(2, 3), log(2, 5)
   fctr, crctn = 6.0 * log2of3 * log2of5, log(2, sqrt(30))
   log2est = (fctr * Float64(n))^(1.0 / 3.0) - crctn # log2 answer from WP formula
   log2hi = log2est + 1.0 / log2est; width = 2.0 / log2est # up to 2X higher/lower

   # some really really big constants representing the "roll-your-own" big logs...
   biglog2of2 = BigInt(1267650600228229401496703205376)
   biglog2of3 = BigInt(2009178665378409109047848542368)
   biglog2of5 = BigInt(2943393543170754072109742145491)

   # loop to find the count of regular numbers and band of possible candidates...
   count :: UInt64 = 0; band = Vector{Tuple{BigInt,Tuple{UInt32,UInt32,UInt32}}}()
   fiveslmt = UInt32(ceil(log2hi / log2of5)); fives :: UInt32 = 0
   while fives < fiveslmt
       log2p = log2hi - fives * log2of5
       threeslmt = UInt32(ceil(log2p / log2of3)); threes :: UInt32 = 0
       while threes < threeslmt
           log2q = log2p - threes * log2of3
           twos = UInt32(floor(log2q)); frac = log2q - twos; count += twos + 1
           if frac <= width
               biglog = biglog2of2 * twos + biglog2of3 * threes + biglog2of5 * fives
               push!(band, (biglog, (twos, threes, fives)))
           end
           threes += 1
       end
       fives += 1
   end

   # process the band found including checks for validity and range...
   n > count && throw("nthhamming:  band high estimate is too low!!!")
   ndx = count - n + 1
   ndx > length(band) && throw("nthhamming:  band width estimate is too narrow!!!")
   sort!(band, by=(tpl -> let (lg,_) = tpl; -lg end)) # sort in decending order

   # get and return the answer...
   _, tri = band[ndx]
   tri

end</lang>

The above code can find the trillionth Hamming number in about two seconds (very little slower) and the thousand trillionth value in about 192 seconds. This routine would be able to find the million trillionth Hamming number in about 20,000 seconds or about five and a half hours.

Kotlin

Translation of: Java

<lang scala>import java.math.BigInteger import java.util.*

val Three = BigInteger.valueOf(3)!! val Five = BigInteger.valueOf(5)!!

fun updateFrontier(x : BigInteger, pq : PriorityQueue<BigInteger>) {

   pq.add(x.shiftLeft(1))
   pq.add(x.multiply(Three))
   pq.add(x.multiply(Five))

}

fun hamming(n : Int) : BigInteger {

   val frontier = PriorityQueue<BigInteger>()
   updateFrontier(BigInteger.ONE, frontier)
   var lowest = BigInteger.ONE
   for (i in 1 .. n-1) {
       lowest = frontier.poll() ?: lowest
       while (frontier.peek() == lowest)
           frontier.poll()
       updateFrontier(lowest, frontier)
   }
   return lowest

}

fun main(args : Array<String>) {

   System.out.print("Hamming(1 .. 20) =")
   for (i in 1 .. 20)
       System.out.print(" ${hamming(i)}")
   System.out.println("\nHamming(1691) = ${hamming(1691)}")
   System.out.println("Hamming(1000000) = ${hamming(1000000)}")

}</lang>

Output:

Hamming(1 .. 20) = 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Hamming(1691) = 2125764000
Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Overloaded function:

<lang scala>import java.math.BigInteger import java.util.*

val One = BigInteger.ONE!! val Three = BigInteger.valueOf(3)!! val Five = BigInteger.valueOf(5)!!

fun PriorityQueue<BigInteger>.update(x: BigInteger) : PriorityQueue<BigInteger> {

   add(x.shiftLeft(1))
   add(x.multiply(Three))
   add(x.multiply(Five))
   return this

}

fun hamming(n: Int): BigInteger {

   val frontier = PriorityQueue<BigInteger>().update(One)
   var lowest = One
   repeat(n - 1) {
       lowest = frontier.poll() ?: lowest
       while (frontier.peek() == lowest)
           frontier.poll()
       frontier.update(lowest)
   }
   return lowest

}

fun hamming(i : Iterable<Int>) : Iterable<BigInteger> = i.map { hamming(it) }

fun main(args: Array<String>) {

   val r = 1..20
   println("Hamming($r) = " + hamming(r))
   arrayOf(1691, 1000000).forEach { println("Hamming($it) = " + hamming(it)) }

}</lang>

Recursive function:

<lang scala>import java.math.BigInteger import java.util.*

val One = BigInteger.ONE!! val Three = BigInteger.valueOf(3)!! val Five = BigInteger.valueOf(5)!!

infix fun PriorityQueue<BigInteger>.update(x: BigInteger) : PriorityQueue<BigInteger> {

   add(x.shiftLeft(1))
   add(x.multiply(Three))
   add(x.multiply(Five))
   return this

}

fun hamming(a: Any?): Any = when (a) {

   is Number -> {
       val pq = PriorityQueue<BigInteger>() update One
       var lowest = One
       repeat(a.toInt() - 1) {
           lowest = pq.poll() ?: lowest
           while (pq.peek() == lowest) pq.poll()
           pq update lowest
       }
       lowest
   }
   is Iterable<*> -> a.map { hamming(it) }
   else -> throw IllegalArgumentException("cannot parse argument")

}

fun main(args: Array<String>) {

   arrayOf(1..20, 1691, 1000000).forEach { println("Hamming($it) = " + hamming(it)) }

}</lang>

Output:

Hamming(1..20) = [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
Hamming(1691) = 2125764000
Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Functional Style Eliminating Duplicates, Optional Sequence Output

The following code implements a functional version, with the only mutable state that required to implement a recursive binding as commented in the code. It is fast because it uses non-genereric functions so that much of the boxing/unboxing can be optimized away, and it takes very little memory because of the avoiding duplicates, the order that the primes are processed with least dense first, and because it is implemented in such a way so as to use only local bindings for the heads of the lazy lists so that they can be consumed as used and garbage collected away. Kotlin does not have a lazy list like Haskell or a memoized lazy Stream like Scala, so the code implements a basic version of LazyList to be used by the algorithm (Java 8 Streams are not memoized as required here):

Translation of: scala

<lang scala>import java.math.BigInteger as BI

data class LazyList<T>(val head: T, val lztail: Lazy<LazyList<T>?>) {

   fun toSequence() = generateSequence(this) { it.lztail.value }
           .map { it.head }

}

fun hamming(): LazyList<BI> {

   fun merge(s1: LazyList<BI>, s2: LazyList<BI>): LazyList<BI> {
       val s1v = s1.head; val s2v = s2.head
       if (s1v < s2v) {
           return LazyList(s1v, lazy({->merge(s1.lztail.value!!, s2)}))
       } else {
           return LazyList(s2v, lazy({->merge(s1, s2.lztail.value!!)}))
       }
   }
   fun llmult(m: BI, s: LazyList<BI>): LazyList<BI> {
       fun llmlt(ss: LazyList<BI>): LazyList<BI> {
           return LazyList(m * ss.head, lazy({->llmlt(ss.lztail.value!!)}))
       }
       return llmlt(s)
   }
   fun u(s: LazyList<BI>?, n: Long): LazyList<BI> {
       var r: LazyList<BI>? = null // mutable nullable so can do the below
       if (s == null) { // recursively referenced variables are ugly!!!
           r = llmult(BI.valueOf(n), LazyList(BI.valueOf(1), lazy{ -> r }))
       } else { // recursively referenced variables only work with lazy
           r = merge(s, llmult(BI.valueOf(n), // or a loop race limit
                               LazyList(BI.valueOf(1), lazy{ -> r })))
       }
       return r
   }
   val prms = arrayOf(5L, 3L, 2L)
   val thunk = {->prms.fold<Long,LazyList<BI>?>(null, {s, n -> u(s,n)})!!}
   return LazyList(BI.valueOf(1), lazy(thunk))

}

fun main(args: Array<String>) { tailrec fun nth(n: Int, h: LazyList<BI>): BI = if (n > 1) { nth(n - 1, h.lztail.value!!) } else { h.head } // non-generic faster: boxing optimized away println(hamming().toSequence().take(20).toList()) println(nth(1691, hamming())) val strt = System.currentTimeMillis() println(nth(1000000, hamming())) val stop = System.currentTimeMillis() println("Took ${stop - strt} milliseconds for the last.") }</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Took 381 milliseconds for the last.

Run on a AMD Bulldozer FX8120 3.1 GHz which is about half the speed as an equivalent Intel (but also half the price).

Liberty BASIC

LB has unlimited precision integers. <lang lb> dim h( 1000000)

for i =1 to 20

   print hamming( i); " ";

next i

print print "H( 1691)", hamming( 1691) print "H( 1000000)", hamming( 1000000)

end

function hamming( limit)

   h( 0) =1
   x2 =2: x3 =3: x5 =5
   i  =0: j  =0: k  =0
   for n =1 to limit
       h( n) = min( x2, min( x3, x5))
       if x2 = h( n) then i = i +1: x2 =2 *h( i)
       if x3 = h( n) then j = j +1: x3 =3 *h( j)
       if x5 = h( n) then k = k +1: x5 =5 *h( k)
   next n
   hamming =h( limit -1)

end function</lang>

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
H( 1691)
2125764000
H( 1000000) 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Logo

<lang logo>to init.ham

 ; queues
 make "twos   [1]
 make "threes [1]
 make "fives  [1]

end to next.ham

                         localmake "ham first :twos
 if less? first :threes :ham [make "ham first :threes]
 if less? first :fives  :ham [make "ham first :fives]

 if equal? :ham first :twos   [ignore dequeue "twos]
 if equal? :ham first :threes [ignore dequeue "threes]
 if equal? :ham first :fives  [ignore dequeue "fives]

 queue "twos   :ham * 2
 queue "threes :ham * 3
 queue "fives  :ham * 5

 output :ham

end

init.ham repeat 20 [print next.ham] repeat 1690-20 [ignore next.ham] print next.ham</lang>

Lua

<lang lua>function hiter()

 hammings = {1}
 prev, vals = {1, 1, 1}
 index = 1
 local function nextv()
   local n, v = 1, hammings[prev[1]]*2

if hammings[prev[2]]*3 < v then n, v = 2, hammings[prev[2]]*3 end if hammings[prev[3]]*5 < v then n, v = 3, hammings[prev[3]]*5 end prev[n] = prev[n] + 1 if hammings[index] == v then return nextv() end index = index + 1 hammings[index] = v return v

 end
 return nextv

end

j = hiter() for i = 1, 20 do

 print(j())

end n, l = 0, 0 while n < 2^31 do n, l = j(), n end print(l)</lang>

M2000 Interpreter

For Long Only

We have to exit loop (and function) before calculating new X2 or X3 or X4 and get overflow error <lang M2000 Interpreter> Module hamming_long { function hamming(l as long, &h(),&last()) { l=if(l<1->1&, l) long oldlen=len(h()) if oldlen<l then dim h(l) else =h(l-1): exit def long i, j, k, n, m, x2, x3, x5, ll stock last(0) out x2,x3,x5,i,j,k n=oldlen : ll=l-1 { m=x2 if m>x3 then m=x3 if m>x5 then m=x5 h(n)=m if n>=1690 then =h(n):break if m=x2 then i++:x2=2&*h(i) if m=x3 then j++:x3=3&*h(j) if m=x5 then k++:x5=5&*h(k) if n<ll then n++: loop } stock last(0) in x2,x3,x5,i,j,k =h(ll) } dim h(1)=1&, last() def long i const nl$={ } document doc$ last()=(2&,3&,5&,0&,0&,0&) for i=1 to 20 Doc$=format$("{0::-10} {1::-10}", i, hamming(i,&h(), &last()))+nl$ next i i=1691 Doc$=format$("{0::-10} {1::-10}", i, hamming(i,&h(), &last()))+nl$ print #-2,Doc$ clipboard Doc$ }

hamming_long

</lang>

Output:

         1          1
         2          2
         3          3
         4          4
         5          5
         6          6
         7          8
         8          9
         9         10
        10         12
        11         15
        12         16
        13         18
        14         20
        15         24
        16         25
        17         27
        18         30
        19         32
        20         36
      1691 2125764000

Using Decimal type

Max hamming number is the 43208th

We have to exit loop (and function) before calculating new X2 or X3 or X4 and get overflow error

<lang M2000 Interpreter> Module hamming { function hamming(l as long, &h(),&last()) { l=if(l<1->1&, l) oldlen=len(h()) if oldlen<l then dim h(l) else =h(l-1): exit def decimal i, j, k, m, x2, x3, x5 stock last(0) out x2,x3,x5,i,j,k n=oldlen : ll=l-1& { m=x2 if m>x3 then m=x3 if m>x5 then m=x5 h(n)=m if n>=43207& then =h(n):break if m=x2 then i++:x2=2@*h(i) if m=x3 then j++:x3=3@*h(j) if m=x5 then k++:x5=5@*h(k) if n<ll then n++: loop } stock last(0) in x2,x3,x5,i,j,k =h(ll) } dim h(1)=1@, last() last()=(2@,3@,5@,0@,0@,0@) Document doc$ const nl$={ } for i=1 to 20 Doc$=format$("{0::-10} {1::-28}", i, hamming(i,&h(), &last()))+nl$ next i i=1691 Doc$=format$("{0::-10} {1::-28}", i, hamming(i,&h(), &last()))+nl$ i=9999 Doc$=format$("{0::-10} {1::-28}", i, hamming(i,&h(), &last()))+nl$ i=43208 Doc$=format$("{0::-10} {1::-28}", i, hamming(i,&h(), &last()))+nl$ print #-2, Doc$ clipboard Doc$ } hamming </lang>

Output:

         1                            1
         2                            2
         3                            3
         4                            4
         5                            5
         6                            6
         7                            8
         8                            9
         9                           10
        10                           12
        11                           15
        12                           16
        13                           18
        14                           20
        15                           24
        16                           25
        17                           27
        18                           30
        19                           32
        20                           36
      1691                   2125764000
      9999           288230376151711744
     43208 9164837199872000000000000000

Mathematica / Wolfram Language

<lang mathematica>HammingList[N_] := Module[{A, B, C}, {A, B, C} = (N^(1/3))*{2.8054745679851933, 1.7700573778298891, 1.2082521307023026} - {1, 1, 1};

Take[ Sort@Flatten@Table[ 2^x * 3^y * 5^z , 

{x, 0, A}, {y, 0, (-B/A)*x + B}, {z, 0, C - (C/A)*x - (C/B)*y}], N]];</lang>

HammingList[20]
-> {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36}
HammingList[1691] // Last
-> 2125764000
HammingList[1000000] // Last
->519312780448388736089589843750000000000000000000000000000000000000000000000000000000

MATLAB / Octave

Translation of: Julia

The n parameter was chosen by trial and error. You have to pick an n large enough that the powers of 2, 3 and 5 will all be greater than n at the 1691st Hamming number.

<lang Matlab>n = 40;

powers_2 = 2.^[0:n-1]; powers_3 = 3.^[0:n-1]; powers_5 = 5.^[0:n-1];

matrix = powers_2' * powers_3; powers_23 = sort(reshape(matrix,n*n,1));

matrix = powers_23 * powers_5; powers_235 = sort(reshape(matrix,n*n*n,1));

% % Remove the integer overflow values. % powers_235 = powers_235(powers_235 > 0);

disp(powers_235(1:20)) disp(powers_235(1691))</lang>

MUMPS

<lang MUMPS>Hamming(n) New count,ok,next,number,which For which=2,3,5 Set number=1 For count=1:1:n Do . Set ok=0 Set:count<21 ok=1 Set:count=1691 ok=1 Set:count=n ok=1 . Write:ok !,$Justify(count,5),": ",number . For which=2,3,5 Set next(number*which)=which . Set number=$Order(next("")) . Kill next(number) . Quit Quit Do Hamming(2000)

   1: 1
   2: 2
   3: 3
   4: 4
   5: 5
   6: 6
   7: 8
   8: 9
   9: 10
  10: 12
  11: 15
  12: 16
  13: 18
  14: 20
  15: 24
  16: 25
  17: 27
  18: 30
  19: 32
  20: 36
1691: 2125764000
2000: 8062156800</lang>

Nim

Library: bigints

Classic Dijkstra algorithm

<lang nim>import bigints

proc min(a: varargs[BigInt]): BigInt =

 result = a[0]
 for i in 1..a.high:
   if a[i] < result: result = a[i]

proc hamming(limit: int): BigInt =

 var
   h = newSeq[BigInt](limit)
   x2 = initBigInt(2)
   x3 = initBigInt(3)
   x5 = initBigInt(5)
   i, j, k = 0
 for i in 0..h.high: h[i] = initBigInt(1)

 for n in 1 ..< limit:
   h[n] = min(x2, x3, x5)
   if x2 == h[n]:
     inc i
     x2 = h[i] * 2
   if x3 == h[n]:
     inc j
     x3 = h[j] * 3
   if x5 == h[n]:
     inc k
     x5 = h[k] * 5

 result = h[h.high]

for i in 1 .. 20:

 stdout.write hamming(i), " "

echo "" echo hamming(1691) echo hamming(1_000_000)</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

The above takes over a second to produce the millionth Hamming number on many machines.

Slightly more efficient version

The following code improves on the above by reducing the number of computationally-time-expensive BigInt comparisons slightly: <lang nim>import bigints, times

proc hamming(limit: int): BigInt =

 doAssert limit > 0
 var
   h = newSeq[BigInt](limit)
   x2 = initBigInt(2)
   x3 = initBigInt(3)
   x5 = initBigInt(5)
   i, j, k = 0
 h[0] = initBigInt(1)

 # BigInt comparisons are expensive, reduce them...
 proc min3(x, y, z: BigInt): (int, BigInt) =
   let (cs, r1) = if y == z: (6, y)
                  elif y < z: (2, y) else: (4, z)
   if x == r1: (cs or 1, x)
   elif x < r1: (1, x) else: (cs, r1)

 for n in 1 ..< limit:
   let (cs, e1) = min3(x2, x3, x5)
   h[n] = e1
   if (cs and 1) != 0: i += 1; x2 = h[i] * 2
   if (cs and 2) != 0: j += 1; x3 = h[j] * 3
   if (cs and 4) != 0: k += 1; x5 = h[k] * 5
 
 h[h.high]

for i in 1 .. 20:

 stdout.write hamming(i), " "

echo "" echo hamming(1691)

let strt = epochTime() let rslt = hamming(1_000_000) let stop = epochTime()

echo rslt echo "This last took ", (stop - strt)*1000, " milliseconds."</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 566.3743019104004 milliseconds.

It can be shown that the above reduces the execution time by about 20 per cent. But note that compiling with --gc:arc allows to lower execution time to 380-390 ms.

Functional iterator sequence, eliminating duplicate calculations and reducing memory use

The above code still wastes quite a lot of time doing redundant BigInt calculations (ie. 2 times 3, 3 times 2, etc.) and as well consumes a huge amount of memory for larger Hamming number determination as it uses an array as large as the range. The below code eliminates duplicate calculations and reduces memory use by using a Nim version of a lazy list internally so that unused back calculated values can be eliminated by the garbage collector. Thus, execution time for BigInt calculations is reduced by a constant factor of about two and a half and memory use is reduced from O(n) to O(n^(2/3)) in the following code:

Translation of: Haskell

Works with: Nim 1.4.0

Note, the following code uses the "bigints" library that doesn't ship with the Nim compiler; install it with "nimble install bigints". <lang nim>import bigints, times

iterator func_hamming() : BigInt =

 type Thunk[T] = proc(): T {.closure.}
 type Lazy[T] = ref object of RootObj # tuple[val: T, thnk: Thunk[T]]
   val: T
   thnk: Thunk[T]
 proc force[T](me: var Lazy[T]): T = # not thread-safe; needs lock on thunk
   if me.thnk != nil: me.val = me.thnk(); me.thnk = nil
   me.val
 type LazyList[T] = ref object of RootObj # tuple[hd: T, tl: Lazy[LazyList[T]]]
   hd: T
   tl: Lazy[LazyList[T]]
 type Mytype = LazyList[BigInt]
 proc merge(x, y: Mytype): Mytype =
   let xh = x.hd; let yh = y.hd
   if xh < yh:
     let mthnk = proc(): Mytype = merge x.tl.force, y
     let mlzy = Lazy[Mytype](thnk: mthnk)
     Mytype(hd: xh, tl: mlzy)
   else:
     let mthnk = proc(): Mytype = merge x, y.tl.force
     let mlzy = Lazy[Mytype](thnk: mthnk)
     Mytype(hd: yh, tl: mlzy)
 proc smult(m: int32, s: Mytype): Mytype =
   proc smults(ss: Mytype): Mytype =
     let mthnk = proc(): Mytype = ss.tl.force.smults
     let mlzy = Lazy[Mytype](thnk: mthnk)
     Mytype(hd: ss.hd * m, tl: mlzy)
   s.smults
 proc u(s: Mytype, n: int32): Mytype =
   var r: Mytype
   let mthnk = proc(): Mytype = r
   let mlzy = Lazy[Mytype](thnk: mthnk)
   let frst = Mytype(hd: initBigInt 1, tl: mlzy)
   if s == nil: r = smult(n, frst) else: r = merge(s, smult(n, frst))
   r
 var hmg: Mytype = nil
 for p in [5i32, 3i32, 2i32]: hmg = u(hmg, p)
 yield initBigInt 1
 while true: # loop almost forever
   yield initBigInt hmg.hd
   hmg = hmg.tl.force

var cnt = 1 for h in func_hamming():

 if cnt > 20: break
 write stdout, h, " "; cnt += 1

echo "" cnt = 1 for h in func_hamming():

 if cnt < 1691: cnt += 1; continue
 else: echo h; break

let strt = epochTime() var rslt: BigInt cnt = 1 for h in func_hamming():

 if cnt < 1000000: cnt += 1; continue
 else: rslt = h; break

let stop = epochTime()

echo rslt echo "This last took ", (stop - strt)*1000, " milliseconds."</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 464.9641513824463 milliseconds.

The above result was obtained by compiling with the default "mark-and-sweep" Garbage collector with -d:release -d:danger (all checking including bounds checks turned off); One should not use the new --gc:arc compilation argument (automatic reference counting) with this implementation as the lazy lists are cyclic but compiling with --gc:orc gives an execution time of about 80% of the execution time as compared to the conventional garbage collection, and is slower than the --gc:arc garbage collection by about half again the time (but correct as to not causing memory leaks) due to the extra time spent tracing cycles.

The beauty of Nim inline iterators as used here is that they are zero overhead (tested) so there is no run time penalty for using them.

Functional iterator sequence, eliminating duplicate calculations and using log approximations

Much of the time for above algorithm is spent doing big integer calculations using the extended precision bit integer library; the following code eliminates most of the big integer calculations by using logarithmic aproximations and just converting to big integers for the display of the results:

Works with: Nim 1.4.0

Note, the following code uses the "bigints" library that doesn't ship with the Nim compiler; install it with "nimble install bigints". <lang nim>from times import inMilliseconds import std/monotimes, bigints from math import log2

type TriVal = (uint32, uint32, uint32) type LogRep = (float64, TriVal) type LogRepf = proc(x: LogRep): LogRep const one: LogRep = (0.0f64, (0'u32, 0'u32, 0'u32)) proc `<`(me: LogRep, othr: LogRep): bool = me[0] < othr[0]

proc convertTrival2BigInt(tv: TriVal): BigInt =

 proc xpnd(bs: uint, v: uint32): BigInt =
   result = initBigInt 1;
   var bsm = initBigInt bs;
   var vm = v.uint
   while vm > 0:
     if (vm and 1) != 0: result *= bsm
     bsm = bsm * bsm   # bsm *= bsm crashes.
     vm = vm shr 1 
 result = (2.xpnd  tv[0]) * (3.xpnd tv[1]) * (5.xpnd tv[2])

const lb2 = 1.0'f64 const lb3 = 3.0'f64.log2 const lb5 = 5.0'f64.log2

proc mul2(me: LogRep): LogRep =

 let (lr, tpl) = me; let (x2, x3, x5) = tpl
 (lr + lb2, (x2 + 1, x3, x5))

proc mul3(me: LogRep): LogRep =

 let (lr, tpl) = me; let (x2, x3, x5) = tpl
 (lr + lb3, (x2, x3 + 1, x5))

proc mul5(me: LogRep): LogRep =

 let (lr, tpl) = me; let (x2, x3, x5) = tpl
 (lr + lb5, (x2, x3, x5 + 1))

type

 LazyList = ref object
   hd: LogRep
   tlf: proc(): LazyList {.closure.}
   tl: LazyList

proc rest(ll: LazyList): LazyList = # not thread-safe; needs lock on thunk

 if ll.tlf != nil: ll.tl = ll.tlf(); ll.tlf = nil
 ll.tl

iterator log_func_hammings(until: int): TriVal =

 proc merge(x, y: LazyList): LazyList =
   let xh = x.hd
   let yh = y.hd
   if xh < yh: LazyList(hd: xh, tlf: proc(): auto = merge x.rest, y)
   else: LazyList(hd: yh, tlf: proc(): auto = merge x, y.rest)
 proc smult(mltf: LogRepf; s: LazyList): LazyList =
   proc smults(ss: LazyList): LazyList =
     LazyList(hd: ss.hd.mltf, tlf: proc(): auto = ss.rest.smults)
   s.smults
 proc unnsm(s: LazyList, mltf: LogRepf): LazyList =
   var r: LazyList = nil
   let frst = LazyList(hd: one, tlf: proc(): LazyList = r)
   r = if s == nil: smult mltf, frst else: s.merge smult(mltf, frst)
   r
 yield one[1]
 var hmpll: LazyList = ((nil.unnsm mul5).unnsm mul3).unnsm mul2
 for _ in 2 .. until:
   yield hmpll.hd[1]; hmpll = hmpll.rest # almost forever

proc main =

 stdout.write "The first 20 hammings are:  "
 for h in log_func_hammings(20): stdout.write h.convertTrival2BigInt, " "
 
 var lsth: TriVal
 for h in log_func_hammings(1691): lsth = h
 echo "\r\nThe 1691st Hamming number is:  ", lsth.convertTriVal2BigInt

 let strt = getMonotime()
 for h in log_func_hammings(1000000): lsth = h
 let elpsd = (getMonotime() - strt).inMilliseconds
 echo "The millionth Hamming number is:  ", lsth.convertTriVal2BigInt
 echo "This last took ", elpsd, " milliseconds."

main()</lang>

Output:

The first 20 hammings are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
The 1691st Hamming number is:  2125764000
The millionth Hamming number is:  519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 157 milliseconds.

As you can see, this new version is over twice as fast as the version using many big integer calculations, both due to much less computation and also due to not having to allocate and de-allocate the memory required for many big integer representations. Again, it is about 80% faster if the new --gc:orc memory management is used, which is slower than using the --gc:arc memory management that is yet another 25% faster but incorrect as it has a memory leak due to the cyclic lazy lists that it can't properly handle.

Most of the remaining time is spent in the many allocations and de-allocations of small structures in heap memory as is typical of functional algorithms. Further speed could be gained for the same algorithm as above by making allocations and de-allocations (now all the same size) from an implemented memory pool, which is what Haskell actually does inside its memory management system.

Imperative iterator implementation of the above functional version

The following code uses imperative techniques to implement the same algorithm, using sequences for storage, indexes for back pointers to the results of previous calculations, and custom deleting unused values in chunks in place (using constantly growing capacity) so that the same size of sequence can be longer used and many less new memory allocations need be made:

<lang nim>import bigints, times

iterator nodups_hamming(): BigInt =

 var
   m = newSeq[BigInt](1) # give it two values so doubling size works
   h = newSeq[BigInt](1) # reasonably size
   x5 = initBigInt 5
   mrg = initBigInt 3
   x53 = initBigInt 9 # already advanced one step
   x532 = initBigInt 2
   ih, jm, i, j = 0

 yield initBigInt 1 # trivial case of 1
 while true:
   let cph = h.len # move in-place to avoid allocation
   if i >= cph div 2: # move in-place to avoid allocation
     var s = i; var d = 0
     while s < ih: shallowCopy(h[d], h[s]); s += 1; d += 1
     ih -= i; i = 0
   if ih >= cph: h.setLen(2 * cph)
   if x532 < mrg: h[ih] = x532; x532 = h[i] * 2; i += 1
   else:
     h[ih] = mrg
     let cpm = m.len
     if j >= cpm div 2: # move in-place to avoid allocation
       var s = j; var d = 0
       while s < jm: shallowCopy(m[d], m[s]); s += 1; d += 1
       jm -= j; j = 0
     if jm >= cpm: m.setLen(2 * cpm)
     if x53 < x5: mrg = x53; x53 = m[j] * 3; j += 1
     else: mrg = x5; x5 = x5 * 5
     m[jm] = mrg
     jm += 1
   ih += 1

   yield h[ih - 1]

var cnt = 1 for h in nodups_hamming():

 if cnt > 20: break
 write stdout, h, " "; cnt += 1

echo "" cnt = 1 for h in nodups_hamming():

 if cnt < 1691: cnt += 1; continue
 else: echo h; break

let strt = epochTime() var rslt: BigInt cnt = 1 for h in nodups_hamming():

 if cnt < 1000000: cnt += 1; continue
 else: rslt = h; break

let stop = epochTime()

echo rslt echo "This last took ", (stop - strt)*1000, " milliseconds."</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 307.5404167175293 milliseconds.

Compiling with --gc:arc gives an execution time of 220-230 ms.

So, in both cases, the execution time is reduced which shows that a high percentage of the previous time was not used by BigInt calculations (as this code does exactly the same number of calculations) but rather by the memory allocatons/deallocations required for pure functional lazy algorithms. This may show that the current Nim version (1.4.2) is not so suitable for pure lazy functional algorithms, nor is it as terse as many modern functional languages (Haskell, OcaML, F#, Scala, etc.).

Much faster iterating version using logarithmic calculations

Still, much of the above time is used by BigInt calculations and still many heap allocations/deallocations, as BigInt's have an internal sequence to contain the infinite precision binary digits. The following code uses an internal logarithmic representation of the values rather than BigInt and thus all mathematic operations required are just integer and floating point additions and comparison; as well, since these don't require heap space there is almost no allocation/deallocation at all for greatly increased speed:

<lang nim>import bigints, math, times

type TriVal = (uint32, uint32, uint32)

proc convertTrival2BigInt(tv: TriVal): BigInt =

 proc xpnd(bs: uint, v: uint32): BigInt =
   result = initBigInt 1;
   var bsm = initBigInt bs;
   var vm = v.uint
   while vm > 0:
     if (vm and 1) != 0: result *= bsm
     bsm = bsm * bsm   # bsm *= bsm crashes.
     vm = vm shr 1

 result = (2.xpnd  tv[0]) * (3.xpnd tv[1]) * (5.xpnd tv[2])

iterator log_nodups_hamming(): TriVal =

 type Logrep = (float64, uint32, uint32, uint32)
 proc `<`(me: Logrep, othr: Logrep): bool = me[0] < othr[0]
 proc mul2(me: Logrep): Logrep = (me[0] + 1.0f64, me[1]+1, me[2], me[3])
 proc mul3(me: Logrep): Logrep = (me[0] + 3.0f64.log2, me[1], me[2]+1, me[3])
 proc mul5(me: Logrep): Logrep = (me[0] + 5.0f64.log2, me[1], me[2], me[3]+1)

 let one: Logrep = (0.0f64, 0u32, 0u32, 0u32)
 var
   m = newSeq[Logrep](1) # give it two values so doubling size works
   h = newSeq[Logrep](1) # reasonably size
   x5 = one.mul5 # initBigInt 5
   mrg = one.mul3 # initBigInt 3
   x53 = one.mul3().mul3 # initBigInt 9 # already advanced one step
   x532 = one.mul2 # initBigInt 2
   ih, jm, i, j = 0

 yield (0u32, 0u32, 0u32)
 while true:
   let cph = h.len # move in-place to avoid allocation
   if i >= cph div 2: # move in-place to avoid allocation
     var s = i; var d = 0
     while s < ih: shallowCopy(h[d], h[s]); s += 1; d += 1
     ih -= i; i = 0
   if ih >= cph: h.setLen(2 * cph)
   if x532 < mrg: h[ih] = x532; x532 = h[i].mul2; i += 1
   else:
     h[ih] = mrg
     let cpm = m.len
     if j >= cpm div 2: # move in-place to avoid allocation
       var s = j; var d = 0
       while s < jm: shallowCopy(m[d], m[s]); s += 1; d += 1
       jm -= j; j = 0
     if jm >= cpm: m.setLen(2 * cpm)
     if x53 < x5: mrg = x53; x53 = m[j].mul3; j += 1
     else: mrg = x5; x5 = x5.mul5
     m[jm] = mrg
     jm += 1

   let lr = h[ih]; yield (lr[1], lr[2], lr[3]); ih += 1

var cnt = 1 for h in log_nodups_hamming():

 if cnt > 20: break
 write stdout, h.convertTrival2BigInt, " "; cnt += 1

echo "" cnt = 1 for h in log_nodups_hamming():

 if cnt < 1691: cnt += 1; continue
 else: echo h.convertTrival2BigInt; break

let strt = epochTime() var rslt: (uint32, uint32, uint32) cnt = 1 for h in log_nodups_hamming():

 if cnt < 1000000: cnt += 1; continue
 else: rslt = h; break # """

let stop = epochTime()

let (x2, x3, x5) = rslt writeLine stdout, "2^", x2, " + 3^", x3, " + 5^", x5 let lgrslt = (x2.float64 + x3.float64 * 3.0f64.log2 +

              x5.float64 * 5.0f64.log2) * 2.0f64.log10

let (whl, frac) = lgrslt.splitDecimal echo "Approximately: ", 10.0f64.pow(frac), "E+", whl.uint64 let brslt = rslt.convertTrival2BigInt() let s = brslt.to_string let ls = s.len echo "Number of digits: ", ls if ls <= 2000:

 for i in countup(0, ls - 1, 100):
   if i + 100 < ls: echo s[i .. i + 99]
   else: echo s[i .. ls - 1]

echo "This last took ", (stop - strt)*1000, " milliseconds."</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
2^55 + 3^47 + 5^64
Approximately:  5.193127804483804E+83
Number of digits:  84
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 13.85593414306641 milliseconds.

The time is done for compilation with option -d:release. With -d:danger (no checks), the time drops to 8 ms. And adding --gc:arc to 4 ms.

With these options, the billionth can be calculated in about 9-10 seconds.

Extremely fast version inserting logarithms into the top error band

The above code is about as fast as one can go generating sequences; however, if one is willing to forego sequences and just calculate the nth Hamming number (repeatedly), then some reading on the relationship between the size of numbers to the sequence numbers is helpful (Wikipedia: Regular Number). One finds that there is a very distinct relationship and that it quite quickly reduces to quite a small error band proportional to the log of the output value for larger ranges. Thus, the following code just scans for logarithmic representations to insert into a sequence for this top error band and extracts the correct nth representation from that band. It reduces time complexity to O(n^(2/3)) from O(n) for the sequence versions, but even more amazingly, reduces memory requirements to O(n^(1/3)) from O(n^(2/3)) and thus makes it possible to calculate very large values in the sequence on common personal computers. The code is as follows:

Translation of: Rust

<lang nim>import bigints, math, algorithm, times

type TriVal = (uint32, uint32, uint32)

proc convertTrival2BigInt(tv: TriVal): BigInt =

 proc xpnd(bs: uint, v: uint32): BigInt =
   result = initBigInt 1
   var bsm = initBigInt bs
   var vm = v.uint
   while vm > 0:
     if (vm and 1) != 0: result *= bsm
     bsm = bsm * bsm     # bsm *= bsm causes a crash.
     vm = vm shr 1

 result = (2.xpnd  tv[0]) * (3.xpnd tv[1]) * (5.xpnd tv[2])

proc nth_hamming(n: uint64): TriVal =

 doAssert n > 0u64
 if n < 2: return (0'u32, 0'u32, 0'u32) # trivial case for 1

 type LogRep = (float64, uint32, uint32, uint32)

 let lb3 = 3.0'f64.log2; let lb5 = 5.0'f64.log2; let fctr = 6.0'f64*lb3*lb5
 let
   crctn = 30.0'f64.sqrt().log2 # log base 2 of sqrt 30
   lgest = (fctr * n.float64).pow(1.0'f64/3.0'f64) - crctn # from WP formula
   frctn = if n < 1000000000: 0.509'f64 else: 0.105'f64
   lghi = (fctr * (n.float64 + frctn * lgest)).pow(1.0'f64/3.0'f64) - crctn
   lglo = 2.0'f64 * lgest - lghi # and a lower limit of the upper "band"
 var count = 0'u64 # need to use extended precision, might go over
 var bnd = newSeq[LogRep](1) # give itone value so doubling size works
 let klmt = (lghi / lb5).uint32 + 1
 for k in 0 ..< klmt: # i, j, k values can be just u32 values
   let p = k.float64 * lb5; let jlmt = ((lghi - p) / lb3).uint32 + 1
   for j in 0 ..< jlmt:
     let q = p + j.float64 * lb3
     let ir = lghi - q; let lg = q + ir.floor # current log value (estimated)
     count += ir.uint64 + 1;
     if lg >= lglo: bnd.add((lg, ir.uint32, j, k))
 if n > count: raise newException(Exception, "nth_hamming: band high estimate is too low!")
 let ndx = (count - n).int
 if ndx >= bnd.len: raise newException(Exception, "nth_hamming: band low estimate is too high!")
 bnd.sort((proc (a, b: LogRep): int = a[0].cmp b[0]), SortOrder.Descending)

 let rslt = bnd[ndx]; (rslt[1], rslt[2], rslt[3])

for i in 1 .. 20:

 write stdout, nth_hamming(i.uint64).convertTrival2BigInt, " "

echo "" echo nth_hamming(1691).convertTrival2BigInt

let strt = epochTime() let rslt = nth_hamming(1_000_000'u64) let stop = epochTime()

let (x2, x3, x5) = rslt writeLine stdout, "2^", x2, " + 3^", x3, " + 5^", x5 let lgrslt = (x2.float64 + x3.float64 * 3.0f64.log2 +

              x5.float64 * 5.0f64.log2) * 2.0f64.log10

let (whl, frac) = lgrslt.splitDecimal echo "Approximately: ", 10.0f64.pow(frac), "E+", whl.uint64 let brslt = rslt.convertTrival2BigInt() let s = brslt.to_string let ls = s.len echo "Number of digits: ", ls if ls <= 2000:

 for i in countup(0, ls - 1, 100):
   if i + 100 < ls: echo s[i .. i + 99]
   else: echo s[i .. ls - 1]

echo "This last took ", (stop - strt) * 1000, " milliseconds."</lang>

The output is the same as above except that the execution time is much too small to be measured. The billionth number in the sequence can be calculated in under 5 milliseconds, the trillionth in about 0.38 seconds. The (2^64 - 1)th value (18446744073709551615) cannot be calculated due to a slight overflow problem as it approaches that limit. However, this version gives inaccurate results much about the 1e13th Hamming number due to the log base two (double) approximate representation not having enough precision to accurately sort the values put into the error band array.

Alternate version with a greatly increased range without error

To solve the problem of inadequate precision in the double log base two representation, the following code uses a BigInt representation of the log value with about twice the significant bits, which is then sufficient to extend the usable range well beyond any reasonable requirement: <lang nim>import bigints, math, algorithm, times

type TriVal = (uint32, uint32, uint32)

proc convertTrival2BigInt(tv: TriVal): BigInt =

 proc xpnd(bs: uint, v: uint32): BigInt =
   result = initBigInt 1
   var bsm = initBigInt bs
   var vm = v.uint
   while vm > 0:
     if (vm and 1) != 0: result *= bsm
     bsm = bsm * bsm   # bsm *= bsm causes a crash.
     vm = vm shr 1

 result = (2.xpnd  tv[0]) * (3.xpnd tv[1]) * (5.xpnd tv[2])

proc nth_hamming(n: uint64): TriVal =

 doAssert n > 0u64
 if n < 2: return (0'u32, 0'u32, 0'u32) # trivial case for 1

 type LogRep = (BigInt, uint32, uint32, uint32)

 let lb3 = 3.0'f64.log2; let lb5 = 5.0'f64.log2; let fctr = 6.0'f64*lb3*lb5
 let # manually produce the BigInt "limb's"!
   bglb2 = initBigInt @[0'u32, 0, 0, 16] # 1267650600228229401496703205376
   # 2009178665378409109047848542368
   bglb3 = initBigInt @[11608224'u32, 3177740794'u32, 1543611295, 25]
   # 2943393543170754072109742145491
   bglb5 = initBigInt @[1258143699'u32, 1189265298, 647893747, 37]
   crctn = 30.0'f64.sqrt().log2 # log base 2 of sqrt 30
   lgest = (fctr * n.float64).pow(1.0'f64/3.0'f64) - crctn # from WP formula
   frctn = if n < 1000000000: 0.509'f64 else: 0.105'f64
   lghi = (fctr * (n.float64 + frctn * lgest)).pow(1.0'f64/3.0'f64) - crctn
   lglo = 2.0'f64 * lgest - lghi # and a lower limit of the upper "band"
 var count = 0'u64 # need to use extended precision, might go over
 var bnd = newSeq[LogRep](1) # give it one value so doubling size works
 let klmt = (lghi / lb5).uint32 + 1
 for k in 0 ..< klmt: # i, j, k values can be just u32 values
   let p = k.float64 * lb5; let jlmt = ((lghi - p) / lb3).uint32 + 1
   for j in 0 ..< jlmt:
     let q = p + j.float64 * lb3
     let ir = lghi - q; let lg = q + ir.floor # current log value (estimated)
     count += ir.uint64 + 1;
     if lg >= lglo:
       let bglg = bglb2 * ir.int32 + bglb3 * j.int32 + bglb5 * k.int32
       bnd.add((bglg, ir.uint32, j, k))
 if n > count: raise newException(Exception, "nth_hamming: band high estimate is too low!")
 let ndx = (count - n).int
 if ndx >= bnd.len: raise newException(Exception, "nth_hamming: band low estimate is too high!")
 bnd.sort((proc (a, b: LogRep): int = (a[0].cmp b[0]).int), SortOrder.Descending)

 let rslt = bnd[ndx]; (rslt[1], rslt[2], rslt[3])

for i in 1 .. 20:

 write stdout, nth_hamming(i.uint64).convertTrival2BigInt, " "

echo "" echo nth_hamming(1691).convertTrival2BigInt

let strt = epochTime() let rslt = nth_hamming(1_000_000'u64) let stop = epochTime()

let (x2, x3, x5) = rslt writeLine stdout, "2^", x2, " + 3^", x3, " + 5^", x5 let lgrslt = (x2.float64 + x3.float64 * 3.0f64.log2 +

              x5.float64 * 5.0f64.log2) * 2.0f64.log10

let (whl, frac) = lgrslt.splitDecimal echo "Approximately: ", 10.0f64.pow(frac), "E+", whl.uint64 let brslt = rslt.convertTrival2BigInt() let s = brslt.to_string let ls = s.len echo "Number of digits: ", ls if ls <= 2000:

 for i in countup(0, ls - 1, 100):
   if i + 100 < ls: echo s[i .. i + 99]
   else: echo s[i .. ls - 1]

echo "This last took ", (stop - strt) * 1000, " milliseconds."</lang>

The above code has the same output as before and doesn't take an appreciable amount time different to execute; it can produce the trillionth Hamming number in about 0.35 seconds and the thousand trillionth (which is now possible without error) in about 34.8 seconds. Thus, it successfully extends the usable range of the algorithm to near the maximum expressible 64 bit number in a few hours of execution time on a modern desktop computer although the (2^64 - 1)th Hamming number can't be found due to the restrictions of the expressible range limit in sizing of the required error band.

OCaml

A simple implementation using an integer Set as a priority queue. The semantics of the standard library Set provide a minimum element and prevent duplicate entries. min_elt and add are O(log N).

<lang OCaml>module ISet = Set.Make(struct type t = int let compare=compare end)

let pq = ref (ISet.singleton 1)

let next () =

 let m = ISet.min_elt !pq in
 pq := ISet.(remove m !pq  |> add (2*m) |> add (3*m) |> add (5*m));
 m

let () =

 print_string "The first 20 are: ";

 for i = 1 to 20
 do
   Printf.printf "%d " (next ())
 done;

 for i = 21 to 1690
 do
   ignore (next ())
 done;

 Printf.printf "\nThe 1691st is %d\n" (next ());</lang>

Output:

The first 20 are: 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36

The 1691st is 2125764000

Arbitrary precision

An arbitrary precision version for the one millionth number. Compile with eg: ocamlopt -o hamming.exe nums.cmxa hamming.ml <lang OCaml>open Big_int

module APSet = Set.Make(

 struct
   type t = big_int
   let compare = compare_big_int
 end)

let pq = ref (APSet.singleton (big_int_of_int 1))

let next () =

 let m = APSet.min_elt !pq in
 let ( * ) = mult_int_big_int in
 pq := APSet.(remove m !pq  |> add (2*m) |> add (3*m) |> add (5*m));
 m

let () =

 let n = 1_000_000 in

 for i = 1 to (n-1)
 do
   ignore (next ())
 done;

 Printf.printf "\nThe %dth is %s\n" n (string_of_big_int (next ()));</lang>

Output:

The 1000000th is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Oz

Lazy Version

Translation of: Haskell

<lang oz>declare

 fun lazy {HammingFun}
    1|{FoldL1 [{MultHamming 2} {MultHamming 3} {MultHamming 5}] LMerge}
 end

 Hamming = {HammingFun}

 fun {MultHamming N}
    {LMap Hamming fun {$ X} N*X end}
 end

 fun lazy {LMap Xs F}
    case Xs
    of nil  then nil
    [] X|Xr then {F X}|{LMap Xr F}
    end
 end

 fun lazy {LMerge Xs=X|Xr Ys=Y|Yr}
    if     X < Y then X|{LMerge Xr Ys}
    elseif X > Y then Y|{LMerge Xs Yr}
    else              X|{LMerge Xr Yr}
    end
 end

 fun {FoldL1 X|Xr F}
    {FoldL Xr F X}
 end

in

 {ForAll {List.take Hamming 20} System.showInfo}
 {System.showInfo {Nth Hamming 1690}}
 {System.showInfo {Nth Hamming 1000000}}</lang>

Strict Version

The strict version uses iterators and a priority queue. Note that it can calculate other variations of the hamming numbers too. By changing K, it will calculate the p(K)-smooth numbers. (E.g. K = 3, it will use the first three primes 2,3 and 5, thus resulting in the 5-smooth numbers, see [2]) <lang oz> functor import Application System define

class Multiplier attr lst factor current

meth init(Factor Lst) lst := Lst factor := Factor {self next} end meth next local A AS in A|AS = @lst current := A*@factor lst := AS end end meth peek(?X) X = @current end

meth dump {System.showInfo "DUMP"} {System.showInfo "Factor: "#@factor} {System.showInfo "current: "#@current} end end

% a priority queue of multipliers. The one which currently holds the smallest value is put on front class PriorityQueue attr mults current % for duplicate detection

meth init(Mults) mults := Mults current := 0 end

meth insert(Mult) local fun {Insert M Lst} local Av Mv in case Lst of nil then M|Lst [] A|AS then {A peek(Av)} {M peek(Mv)} if Av < Mv then A|{Insert M AS} else M|A|AS end end end end in mults := {Insert Mult @mults} end end

meth next(Tail NextTail) local M Ms X Curr in M|Ms = @mults {M peek(X)} % gets value of lowest iterator Curr = @current if Curr == X then skip else Tail = X|NextTail % if we found a new value: append end {M next} mults := Ms {self insert(M)} if Curr == X then {self next(Tail NextTail)} else current := X end end end

end

local

% Sieve of erasthothenes, adapted from http://rosettacode.org/wiki/Sieve_of_Eratosthenes#Oz fun {Sieve N} S = {Array.new 2 N true} M = {Float.toInt {Sqrt {Int.toFloat N}}} in for I in 2..M do if S.I then for J in I*I..N;I do S.J := false end end end S end

fun {Primes N} S = {Sieve N} in for I in 2..N collect:C do if S.I then {C I} end end end

% help method to extract args proc {GetNK ArgList N K} case ArgList of A|B|_ then N={StringToInt A} K={StringToInt B} end end

proc {Generate N PriorQ Tail} local NewTail in if N == 0 then Tail = nil else {PriorQ next(Tail NewTail)} {Generate (N-1) PriorQ NewTail} end end end

K = 3 PrimeFactors Lst Tail in ArgList = {Application.getArgs plain} Lst = 1|Tail PrimeFactors = {List.take {Primes K*K} K} Mults = {List.map PrimeFactors fun {$ A} {New Multiplier init(A Lst) } end} PriorQ = {New PriorityQueue init(Mults)} {Generate 20 PriorQ Tail} {ForAll Lst System.showInfo} {Application.exit 0} end end </lang> Strict version made by pietervdvn; do what you want with the code.

PARI/GP

This is a basic implementation; finding the millionth term requires 1 second and 54 MB. Much better algorithms exist. <lang parigp>Hupto(n)={

 my(v=vector(n),x2=2,x3=3,x5=5,i=1,j=1,k=1,t);
 v[1]=1;
 for(m=2,n,
   v[m]=t=min(x2,min(x3,x5));
   if(x2 == t, x2 = v[i++] << 1);
   if(x3 == t, x3 = 3 * v[j++]);
   if(x5 == t, x5 = 5 * v[k++]);
 );
 v

}; H(n)=Hupto(n)[n];

Hupto(20) H(1691) H(10^6)</lang>

Output:

%1 = [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
%2 = 2125764000
%3 = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Pascal

Simple brute force til 2^32-1.I was astonished by the speed.The inner loop is taken 2^32 -1 times.DIV by constant is optimized to Mul and shift. Using FPC_64 3.1.1, i4330 3.5 Ghz <lang pascal> program HammNumb; {$IFDEF FPC}

 {$MODE DELPHI}
 {$OPTIMIZATION ON}

{$ELSE}

 {$APPTYPE CONSOLE}

{$ENDIF} { type

 NativeUInt = longWord;

} var

 pot   : array[0..2] of NativeUInt;

function NextHammNumb(n:NativeUInt):NativeUInt; var

 q,p,nr : NativeUInt;

begin

 repeat
   nr := n+1;
   n := nr;

   p := 0;
   while NOT(ODD(nr)) do
   begin
     inc(p);
     nr := nr div 2;
   end;
   Pot[0]:= p;

   p := 0;
   q := nr div 3;
   while q*3=nr do
   Begin
     inc(P);
     nr := q;
     q := nr div 3;
   end;
   Pot[1] := p;

   p := 0;
   q := nr div 5;
   while q*5=nr do
   Begin
     inc(P);
     nr := q;
     q := nr div 5;
   end;
   Pot[2] := p;

 until nr = 1;
 result:= n;

end;

procedure Check; var

 i,n: NativeUint;

begin

 n := 1;
 for i := 1 to 20 do
 begin
   n := NextHammNumb(n);
   write(n,' ');
 end;
 writeln;
 writeln;
 n := 1;
 for i := 1 to 1690 do
   n := NextHammNumb(n);
 writeln('No ',i:4,' | ',n,' = 2^',Pot[0],' 3^',Pot[1],' 5^',Pot[2]);

end;

Begin

 Check;

End.</lang> Output

2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 

No 1690 | 2125764000 = 2^5 3^12 5^3

real    0m17.328s
user    0m17.310s

Alternate Using Non-Duplicates Logarithmic Estimation Ordering

The above is not a true sequence of Hamming numbers as it doesn't generate an iteration or enumeration of the numbers where each new value is generated from the accumulated state of all the generated numbers up to that point, but rather regenerates all the previous values very inefficiently for each new value, and thus does not have a linear execution complexity with number of generated values. Much more elegant solutions are those using functional programming paradigms, but as Pascal is by no means a functional language, lacking many of the requirements of functional programming such as closure functions to be functional and being difficult (although not impossible) to emulate those functions using classes/objects, the following code implements an imperative version of the non-duplicating Hamming sequence which also saves both time and space in not processing the duplicates (for instance, with two times three already accounted for, there is no need to process three times two); as well, since there is no standard "infinite" precision integer library for Pascal so that numbers larger than 64-bit can't easily be handled, the following code uses the "triplet" method and does the sorting based on a logarithmic estimation of the multiples: <lang pascal>{$OPTIMIZATION LEVEL4} program Hammings(output);

{$mode objfpc} uses Math, SysUtils;

const

 lb22 : Double = 1.0; (* log base 2 of 2 *)
 lb23 : Double = 1.58496250072115618147; (* log base 2 of 3 *)
 lb25 : Double = 2.32192809488736234781; (* log base 2 of 5 *)

type

 TLogRep = record
   lr : Double;
   x2, x3, x5 : Word;
 end;

const oneLogRep : TLogRep = (lr:0.0; x2:0; x3:0; x5:0);

function LogRepMult2(lr : TLogRep) : TLogRep; begin

 Result := lr;
 Result.lr := lr.lr + lb22;
 Result.x2 := lr.x2 + 1

end;

function LogRepMult3(lr : TLogRep) : TLogRep; begin

 Result := lr;
 Result.lr := lr.lr + lb23;
 Result.x3 := lr.x3 + 1

end;

function LogRepMult5(lr : TLogRep) : TLogRep; begin

 Result := lr;
 Result.lr := lr.lr + lb25;
 Result.x5 := lr.x5 + 1

end;

function LogRep2QWord(lr : TLogRep) : QWord; function xpnd(x : Word; m : QWord) : QWord; var mlt : QWord; begin

 mlt := m;
 Result := 1;
 while x > 0 do
 begin
   if x and 1 > 0 then Result := Result * mlt;
   mlt := mlt * mlt; x := x shr 1
 end

end; begin

 Result := xpnd(lr.x2, 2) * xpnd(lr.x3, 3) * xpnd(lr.x5, 5)

end;

function LogRep2String(lr : TLogRep) : AnsiString; type TBI = array of LongWord;

    TDigitStr = String[1];

function mul2(bi : TBI) : TBI; var cry : QWord;

   i : Integer;

begin

 cry := 0;
 for i := 0 to High(bi) do
 begin
   cry := (QWord(bi[i]) shl 1) + cry; bi[i] := cry; cry := cry shr 32
 end;
 if cry <> 0 then
 begin
   SetLength(bi, Length(bi) + 1); bi[High(bi)] := cry
 end;
 Result := bi

end; function add(bia : TBI; bib : TBI) : TBI; var cry : QWord;

   i : Integer;

begin

 cry := 0;
 for i := 0 to High(bia) do
 begin
   cry := QWord(bia[i]) + QWord(bib[i]) + cry;
   bia[i] := cry; cry := cry shr 32
 end;
 if cry <> 0 then
 begin
   SetLength(bia, Length(bia) + 1); bia[High(bia)] := cry
 end;
 Result := bia

end; function div10(bi : TBI) : TDigitStr; var brw : QWord;

   i : Integer;

begin

 brw := 0;
 for i := High(bi) downto 0 do
 begin
   brw := (brw shl 32) + QWord(bi[i]);
   bi[i] := brw div 10; brw := brw - QWord(bi[i]) * 10
 end;
 Result := IntToStr(brw)

end; var v : Word;

   xpnd, xpndt : TBI;

begin

 Result := ;
 SetLength(xpnd, 1); xpnd[0] := 1;
 for v := lr.x2 downto 1 do xpnd := mul2(xpnd);
 for v := lr.x3 downto 1 do
 begin
   xpndt := Copy(xpnd, 0, Length(xpnd));
   xpnd := mul2(xpnd); xpnd := add(xpnd, xpndt)
 end;
 for v := lr.x5 downto 1 do
 begin
   xpndt := Copy(xpnd, 0, Length(xpnd)); xpnd := mul2(xpnd);
   xpnd := mul2(xpnd); xpnd := add(xpnd, xpndt)
 end;
 while Length(xpnd) > 0 do
 begin
   Result := div10(xpnd) + Result;
   if xpnd[High(xpnd)] <= 0 then SetLength(xpnd, Length(xpnd) - 1)
 end

end;

type

 TLogReps = array of TLogRep;
 THammings = class
 private
   FCurrent : TLogRep;
   FBA, FMA : TLogReps;
   Fnxt2, Fnxt3, Fnxt5, Fmrg35 : TLogRep;
   FBb, FBe, FMb, FMe : Integer;
 public
   constructor Create;
   function GetEnumerator : THammings;
   function MoveNext : Boolean;
   property Current : TLogRep read FCurrent;
 end;

constructor THammings.Create; begin

 inherited Create;
 FCurrent := oneLogRep; FCurrent.lr := -1.0;
 SetLength(FBA, 4); SetLength(FMA, 4);
 Fnxt5 := LogRepMult5(oneLogRep);
 Fmrg35 := LogRepMult3(oneLogRep);
 Fnxt3 := LogRepMult3(Fmrg35);
 Fnxt2 := LogRepMult2(oneLogRep);
 FBb := 0; FBe := 0; FMb := 0; FMe := 0

end;

function THammings.GetEnumerator : THammings; begin

 Result := Self

end;

function THammings.MoveNext : Boolean; var blen, mlen, i, j : Integer; begin

 if FCurrent.lr < 0.0 then FCurrent.lr := 0.0 else
 begin
   blen := Length(FBA);
   if FBb >= blen shr 1 then
   begin
     i := 0;
     for j := FBb to FBe - 1 do
     begin
       FBA[i] := FBA[j]; Inc(i)
     end;
     FBe := FBe - FBb; FBb := 0
   end;
   if FBe >= blen then SetLength(FBA, blen shl 1);
   if Fnxt2.lr < Fmrg35.lr then
   begin
     FCurrent := Fnxt2; FBA[FBe] := FCurrent;
     Fnxt2 := LogRepMult2(FBA[FBb]); Inc(FBb)
   end
   else
   begin
     mlen := Length(FMA);
     if FMb >= mlen shr 1 then
     begin
       i := 0;
       for j := FMb to FMe - 1 do
       begin
         FMA[i] := FMA[j]; Inc(i)
       end;
       FMe := FMe - FMb; FMb := 0
     end;
     if FMe >= mlen then SetLength(FMA, mlen shl 1);
     if Fmrg35.lr < Fnxt5.lr then
     begin
       FCurrent := Fmrg35; FMA[FMe] := FCurrent;
       Fnxt3 := LogRepMult3(FMA[FMb]); Inc(FMb)
     end
     else
     begin
       FCurrent := Fnxt5; FMA[FMe] := FCurrent;
       Fnxt5 := LogRepMult5(Fnxt5)
     end;
     if Fnxt3.lr < Fnxt5.lr then Fmrg35 := Fnxt3 else Fmrg35 := Fnxt5;
     FBA[FBe] := FCurrent; Inc(FMe)
   end;
   Inc(FBe)
 end;
 Result := True

end;

var elpsd : QWord;

   count : Integer;
   h : TLogRep;

begin

 write('The first 20 Hamming numbers are: ');
 count := 0;
 for h in THammings.Create do
 begin
   Inc(count);
   if count > 20 then break;
   write(' ', LogRep2QWord(h));
 end;
 writeln('.');
 count := 1;
 for h in THammings.Create do
 begin
   Inc(count);
   if count > 1691 then break;
 end;
 writeln('The 1691st Hamming number is ', LogRep2QWord(h), '.');
 elpsd := GetTickCount64;
 count := 1;
 for h in THammings.Create do
 begin
   Inc(count);
   if count > 1000000 then break;
 end;
 elpsd := GetTickCount64 - elpsd;
 writeln('The millionth Hamming number is approximately ', 2.0**h.lr, '.');
 write('The millionth Hamming triplet is ');
 writeln('2^', h.x2, ' * 3^', h.x3, ' * 5^', h.x5, '.');
 writeln('The millionth Hamming number is ', LogRep2String(h), '.');
 writeln('This last took ', elpsd, ' milliseconds.')

end.</lang>

Output:

The first 20 Hamming numbers are:  1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36.
The 1691st Hamming number is 2125764000.
The millionth Hamming number is approximately  5.19312780448555124533E+0083.
The millionth Hamming triplet is 2^55 * 3^47 * 5^64.
The millionth Hamming number is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000.
This last took 13 milliseconds.

The above was as run on a modern Intel CPU at 4 GHz.

Note that as the millionth Hamming number has 84 decimal digits and the largest standard 64-bit value that is easily expressed in standard Pascal is only about 19 decimal digits, enough of an "infinite" precision integer library has been implemented to be able to convert the produced "triplet" into the resulting millionth value; this does not need to be of maximum efficiency as it is used only for the final answer.

a fast alternative

The first Pascal code is much slower.

The following is easy to use for smooth-3 .. smooth-37.

Big(O) is nearly linear to sub-linear . 1E7-> 0.028s => x10 =>1e8 ->0.273s => x1000 => 100'200'300'400 ~ 1e11 35.907s // estimated 270 s! This depends extreme on sorting speed.

http://rosettacode.org/wiki/Hamming_numbers#Direct_calculation_through_triples_enumeration is head to head, but still faster for very big numbers >1e8 (10^8: 4 MB 0.27 sec)

100'200'300'400 calculates in 8.33 s

For fpc 3.1.1_64 linux on 3.5 Ghz i4330, depends on 64-Bit by a factor of 4 slower on 32-Bit

/* For 12 primes "smooth-37" 1e8 it takes 02.807 s */

I collect only the factors between p^n and p^(n+1), in a recursive way in different lists

5 is a list consisting only 5^? = 1 factor

3 is a sorted list 3^?..3^?+1 and inserted values of 5

2 is a sorted list 2^?..2^?+1 and inserted values of list 3

Changing sizeOf(tElem) to 32 {maxPrimFakCnt = 3+8} instead of 16 ( x2) {maxPrimFakCnt = 3} results in increasing the runtime by x4 ( 2^2 )

<lang pascal>program hammNumb; {$IFDEF FPC}

  {$MODE DELPHI}
  {$OPTIMIZATION ON,ALL}
  {$ALIGN 16}

{$ELSE}

 {$APPTYPE CONSOLE}

{$ENDIF} uses

 sysutils;

const

 maxPrimFakCnt = 3;//3 or 3+8 if tNumber= double, else -1 for extended to keep data aligned
 minElemCnt = 10;

type

 tPrimList = array of NativeUint;
 tnumber = double;
 tpNumber= ^tnumber;
 tElem = record
            n   : tnumber;//ln(prime[0]^Pots[0]*...
            Pots: array[0..maxPrimFakCnt] of word;
          end;
 tpElem  = ^tElem;
 tElems  = array of tElem;
 tElemArr  = array [0..0] of tElem;
 tpElemArr  = ^tElemArr;

 tpFaktorRec = ^tFaktorRec;
 tFaktorRec = record
                frElems  : tElems;
                frInsElems: tElems;
                frAktIdx : NativeUint;
                frMaxIdx : NativeUint;
                frPotNo  : NativeUint;
                frActPot : NativeUint;
                frNextFr : tpFaktorRec;
                frActNumb: tElem;
                frLnPrime: tnumber;
              end;
 tArrFR = array of tFaktorRec;

var

 Pl : tPrimList;
 ActIndex  : NativeUint;
 ArrInsert : tElems;

procedure PlInit(n: integer); const

cPl : array[0..11] of byte=(2,3,5,7,11,13,17,19,23,29,31,37);

var

 i : integer;

Begin

 IF n>High(cPl)+1 then
    n := High(cPl)
 else
     IF n < 0 then
        n := 1;
 setlength(Pl,n);
 dec(n);
 For i := 0 to n do
   Pl[i] := cPl[i];

end;

procedure AusgabeElem(pElem: tElem); var

 i : integer;

Begin

 with pElem do
 Begin
   IF n < 23 then
   begin
     write(round(exp(n)),' ');
     if n < ln(100)then
        EXIT;
   end
   else
     write('ln ',n:13:7);
   For i := 0 to maxPrimFakCnt-1 do
     write(' ',PL[i]:2,'^',Pots[i]);
 end;
 writeln

end;

//LoE == List of Elements function LoEGetNextNumber(pFR :tpFaktorRec):tElem;forward;

procedure LoECreate(const Pl: tPrimList;var FA:tArrFR); var

 i : integer;

Begin

 setlength(ArrInsert,100);
 setlength(FA,Length(PL));
 For i := 0 to High(PL) do
   with FA[i] do
   Begin
     //automatic zeroing
     IF i < High(PL) then
     Begin
       setlength(frElems,minElemCnt);
       setlength(frInsElems,minElemCnt);
       frNextFr := @FA[i+1]
     end
     else
     Begin
       setlength(frElems,2);
       setlength(frInsElems,0);
       frNextFr := NIL;
     end;
     frPotNo  := i;
     frLnPrime:= ln(PL[i]);
     frMaxIdx := 0;
     frAktIdx := 0;
     frActPot := 1;
     With frElems[0] do
     Begin
       n := frLnPrime;
       Pots[i]:= 1;
     end;
     frActNumb := frElems[0];
 end;

end;

procedure LoEFree(var FA:tArrFR); var

 i : integer;

Begin

 For i := High(FA) downto Low(FA) do
   setlength(FA[i].frElems,0);
 setLength(FA,0);

end;

function LoEGetActElem(pFr:tpFaktorRec):tElem; Begin

 with pFr^ do
   result := frElems[frAktIdx];

end;

function LoEGetActLstNumber(pFr:tpFaktorRec):tpNumber; Begin

 with pFr^ do
   result := @frElems[frAktIdx].n;

end;

procedure LoEIncInsArr(var a:tElems); Begin

 setlength(a,Length(a)*8 div 5);

end;

procedure LoEIncreaseElems(pFr:tpFaktorRec;minCnt:NativeUint); var

 newLen: NativeUint;

Begin

 with pFR^ do
 begin
   newLen := Length(frElems);
   minCnt := minCnt+frMaxIdx;
   repeat
     newLen := newLen*8 div 5 +1;
   until newLen > minCnt;
   setlength(frElems,newLen);
 end;

end;

procedure LoEInsertNext(pFr:tpFaktorRec;Limit:tnumber); var

 pNum : tpNumber;
 pElems : tpElemArr;
 cnt,i,u : NativeInt;

begin

 with pFr^ do
 Begin
   //collect numbers of heigher primes
   cnt := 0;
   pNum := LoEGetActLstNumber(frNextFr);
   while Limit > pNum^ do
   Begin
     frInsElems[cnt] := LoEGetNextNumber(frNextFr);

// writeln( 'Ins ',frInsElems[cnt].n:10:8,' < ',pNum^:10:8);

     inc(cnt);
     IF cnt > High(frInsElems) then
       LoEIncInsArr(frInsElems);
     pNum := LoEGetActLstNumber(frNextFr);
   end;

   if cnt = 0 then
    EXIT;

   i := frMaxIdx;
   u := frMaxIdx+cnt+1;

   IF u > High(frElems) then
     LoEIncreaseElems(pFr,cnt);

   IF frPotNo =  0 then
     inc(ActIndex,u);
   //Merge
   pElems := @frElems[0];
   dec(cnt);
   dec(u);
   frMaxIdx:= u;
   repeat

// writeln(i:10,cnt:10,u:10); writeln( pElems^[i].n:10:8,' < ',frInsElems[cnt].n:10:8);

     IF pElems^[i].n < frInsElems[cnt].n then
     Begin
       pElems^[u] := frInsElems[cnt];
       dec(cnt);
     end
     else
     Begin
       pElems^[u] := pElems^[i];
       dec(i);
     end;
     dec(u);
   until (i<0) or (cnt<0);
   IF i < 0 then
     For u := cnt downto 0 do
       pElems^[u] := frInsElems[u];

 end;

end;

procedure LoEAppendNext(pFr:tpFaktorRec;Limit:tnumber); var

 pNum : tpNumber;
 pElems : tpElemArr;
 i : NativeInt;

begin

 with pFr^ do
 Begin
   i := frMaxIdx+1;
   pElems := @frElems[0];
   pNum := LoEGetActLstNumber(frNextFr);
   while Limit > pNum^ do
   Begin
     IF i > High(frElems) then
     Begin
       LoEIncreaseElems(pFr,10);
       pElems := @frElems[0];
     end;
     pElems^[i] := LoEGetNextNumber(frNextFr);
     inc(i);
     pNum := LoEGetActLstNumber(frNextFr);
   end;
   inc(ActIndex,i);
   frMaxIdx:= i-1;
 end;

end;

procedure LoENextList(pFr:tpFaktorRec); var

 pElems : tpElemArr;
 j : NativeUint;

begin

 with pFR^ do
 Begin
   //increase Elements by factor
   pElems := @frElems[0];
   for j := frMaxIdx Downto 0 do
     with pElems^[j] do
     Begin
       n := n+frLnPrime;
       inc(Pots[frPotNo]);
     end;
   //x^j -> x^(j+1)
   j := frActPot+1;
   with frActNumb do
   begin
     n:= j*frLnPrime;
     Pots[frPotNo]:= j;
   end;
   frActPot := j;
   //if something follows
   IF frNextFr <> NIL then
     LoEInsertNext(pFR,frActNumb.n);
   frAktIdx := 0;
 end;

end;

function LoEGetNextNumber(pFR :tpFaktorRec):tElem; Begin

 with pFr^ do
 Begin
   result := frElems[frAktIdx];
   inc(frAktIdx);
   IF frMaxIdx < frAktIdx then
     LoENextList(pFr);
 end;

end;

procedure LoEGetNumber(pFR :tpFaktorRec;no:NativeUint); Begin

 dec(no);
 while ActIndex < no do
   LoENextList(pFR);
 with pFr^ do
   frAktIdx := (no-(ActIndex-frMaxIdx)-1);

end;

var

 T1,T0: tDateTime;
 FA: tArrFR;
 i : integer;

Begin

 PlInit(3);// 3 -> 2,3,5
 LoECreate(Pl,FA);
 i := 1;
 i := 1;
 T0 := time;
 write('First 20 :');
 For i := 1 to 20 do
   AusgabeElem(LoEGetNextNumber(@FA[0]));
 writeln;
 write(' 1691.th :');
 LoEGetNumber(@FA[0],1691);
 AusgabeElem(LoEGetNextNumber(@FA[0]));

 LoEGetNumber(@FA[0],1000*1000);
 AusgabeElem(LoEGetNextNumber(@FA[0]));
 T1 := time;
 Writeln('Timed 1,000,000 in ',FormatDateTime('HH:NN:SS.ZZZ',T1-T0));

 LoEGetNumber(@FA[0],1000*1000*1000);  
 AusgabeElem(LoEGetNextNumber(@FA[0]));
 Writeln('Timed 1,000,000,000 in ',FormatDateTime('HH:NN:SS.ZZZ',time-T1));

 Writeln('Actual Index ',ActIndex );
 AusgabeElem(LoEGetNextNumber(@FA[0]));
 For i := 0 to High(FA) do
   writeln(pL[i]:2,
    ' elemcount  ',FA[i].frMaxIdx+1:7,' out of',length(FA[i].frElems):7);
 LoEFree(FA);

End.</lang>

@ TIO.RUN:

First 20 :2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 
 1691.th :2125764000   2^5  3^12  5^3
ln   192.7618989  2^55  3^47  5^64
Timed 1,000,000 in 00:00:00.003
ln  1942.9063722  2^1334  3^335  5^404
Timed 1,000,000,000 in 00:00:04.456
Actual Index 1001046828
ln  1942.9063727  2^761  3^572  5^489
 2 elemcount  1069703 out of1426063
 3 elemcount     1209 out of   1236
 5 elemcount        1 out of      2
...
change zu use 12 primes [2..37] ( 32 bit ) -> 2.2x runtime  over using 3 primes 
Begin
  PlInit(12)

ln    40.8834947  2^14  3^0  5^6  7^4 11^2 13^1 17^0 19^1 23^0 29^0 31^1 37^0
Actual Index  100269652
Timed 100000000 in 00:00:02.807

 2 elemcount   14322779 out of 14953361
 3 elemcount    3387290 out of  3650722
 5 elemcount     891236 out of   891289
 7 elemcount     289599 out of   348159
11 elemcount      92240 out of   135999
13 elemcount      28272 out of    33202
17 elemcount       9394 out of    12969
19 elemcount       2639 out of     3165
23 elemcount        676 out of      772
29 elemcount        119 out of      188
31 elemcount         15 out of       17
37 elemcount          1 out of        2

Perl

<lang perl>use strict; use warnings; use List::Util 'min';

1. If you want the large output, uncomment either the one line
2. marked (1) or the two lines marked (2)
3. use Math::GMP qw/:constant/; # (1) uncomment this to use Math::GMP
4. use Math::GMPz; # (2) uncomment this plus later line for Math::GMPz

sub ham_gen {

   my @s = ([1], [1], [1]);
   my @m = (2, 3, 5);
   #@m = map { Math::GMPz->new($_) } @m;     # (2) uncomment for Math::GMPz

   return sub {

my $n = min($s[0][0], $s[1][0], $s[2][0]); for (0 .. 2) { shift @{$s[$_]} if $s[$_][0] == $n; push @{$s[$_]}, $n * $m[$_] } return $n

   }

}

my $h = ham_gen; my $i = 0; ++$i, print $h->(), " " until $i > 20; print "...\n";

++$i, $h->() until $i == 1690; print ++$i, "-th: ", $h->(), "\n";

1. You will need to pick one of the bigint choices
2. ++$i, $h->() until $i == 999999;
3. print ++$i, "-th: ", $h->(), "\n";

</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 ...
1691-th: 2125764000
1000000-th: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

The core module bigint (Math::BigInt) is very slow, even with the GMP backend, and not supported here. Alternatives shown are Math::GMP and Math::GMPz (about 4x faster).

Phix

Translation of: AWK

Library: Phix/mpfr

standard and gmp versions

with javascript_semantics 
function hamming(integer N)
sequence h = repeat(1,N)
atom x2 = 2, x3 = 3, x5 = 5, hn
integer i = 1, j = 1, k = 1
    for n=2 to N do
        hn = min(x2,min(x3,x5))
        h[n] = hn
        if hn==x2 then i += 1 x2 = 2*h[i] end if
        if hn==x3 then j += 1 x3 = 3*h[j] end if
        if hn==x5 then k += 1 x5 = 5*h[k] end if
    end for
    return h[N]
end function
 
include builtins\mpfr.e
 
function mpz_hamming(integer N)
sequence h = mpz_inits(N,1)
mpz x2 = mpz_init(2),
    x3 = mpz_init(3),
    x5 = mpz_init(5),
    hn = mpz_init()
integer i = 1, j = 1, k = 1
    for n=2 to N do
        mpz_set(hn,mpz_min({x2,x3,x5}))
        mpz_set(h[n],hn)
        if mpz_cmp(hn,x2)=0 then i += 1 mpz_mul_si(x2,h[i],2) end if
        if mpz_cmp(hn,x3)=0 then j += 1 mpz_mul_si(x3,h[j],3) end if
        if mpz_cmp(hn,x5)=0 then k += 1 mpz_mul_si(x5,h[k],5) end if
    end for
    return h[N]
end function
 
sequence s = {}
for i=1 to 20 do
    s = append(s,hamming(i))
end for
?s
printf(1,"%d\n",hamming(1691))
printf(1,"%d (wrong!)\n",hamming(1000000)) --(the hn==x2 etc fail, so multiplies are all wrong)
 
printf(1,"%s\n",{mpz_get_str(mpz_hamming(1691))})
printf(1,"%s\n",{mpz_get_str(mpz_hamming(1000000))})

Output:

{1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36}
2125764000
246192725545902804828662268200 (wrong!)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

A much faster logarithmic version

This proved much easier to implement than scanning the other entries suggested [not copied, they all frighten me].
At some point, comparing logs will no doubt get it wrong, but I have no idea when that might happen.

with javascript_semantics 
-- numbers kept as {log,{pow2,pow3,pow5}},
--  value is ~= exp(log), == (2^pow2)*(3^pow3)*(5^pow5)
enum LOG, POWS
enum POW2, POW3, POW5

function lnmin(sequence a, sequence b)
    return iff(a[LOG]<b[LOG]?a:b)
end function

constant ln1 = log(1), ln2 = log(2), ln3 = log(3), ln5 = log(5)

function hamming(integer N)
sequence h = repeat(0,N)
sequence x2 = {ln2,{1,0,0}}, 
         x3 = {ln3,{0,1,0}},
         x5 = {ln5,{0,0,1}}
integer i = 1, j = 1, k = 1
    h[1] = {ln1,{0,0,0}}
    for n=2 to N do
        h[n] = deep_copy(lnmin(x2,lnmin(x3,x5)))
        sequence p = h[n][POWS]
        if p=x2[POWS] then i += 1 x2 = deep_copy(h[i]) x2[LOG] += ln2 x2[POWS][POW2] += 1 end if
        if p=x3[POWS] then j += 1 x3 = deep_copy(h[j]) x3[LOG] += ln3 x3[POWS][POW3] += 1 end if
        if p=x5[POWS] then k += 1 x5 = deep_copy(h[k]) x5[LOG] += ln5 x5[POWS][POW5] += 1 end if
    end for
    return h[N]
end function

function hint(sequence hm)
-- (obviously not accurate above 53 bits on a 32-bit system, or 64 bits on a 64 bit system)
    sequence p = hm[POWS]
    return power(2,p[POW2])*power(3,p[POW3])*power(5,p[POW5])
end function

sequence s = {}
for i=1 to 20 do
    s = append(s,hint(hamming(i)))
end for
printf(1,"hamming[1..20]: %v\n",{s})
?hint(hamming(1691))
?hint(hamming(1000000))
printf(1," %d (approx)\n",hint(hamming(1000000)))

include builtins\mpfr.e

function mpz_hint(sequence hm)
-- (as accurate as you like)
    integer {p2,p3,p5} = hm[POWS]
    mpz {tmp2,tmp3,tmp5} = mpz_inits(3)
    mpz_ui_pow_ui(tmp2,2,p2)
    mpz_ui_pow_ui(tmp3,3,p3)
    mpz_ui_pow_ui(tmp5,5,p5)
    mpz_mul(tmp3,tmp3,tmp5)
    mpz_mul(tmp2,tmp2,tmp3)
    return mpz_get_str(tmp2)
end function

?mpz_hint(hamming(1000000))

Output:

hamming[1..20]: {1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36}
2125764000.0
5.193127804e+83
 519312780448389068266824288284848486280402222226888608420684482660084484246042460000 (approx)
"519312780448388736089589843750000000000000000000000000000000000000000000000000000000"

Under pwa/p2js, no real idea or any fretting over why, we instead get:

 519312780448388740000000000000000000000000000000000000000000000000000000000000000000 (approx)

Picat

<lang Picat>go =>

  println([hamming(I) : I in 1..20]),
  time(println(hamming_1691=hamming(1691))),
  time(println(hamming_1000000=hamming(1000000))),
  nl.

hamming(1) = 1. hamming(2) = 2. hamming(3) = 3. hamming(N) = Hamming =>

  A = new_array(N),
  [Next2, Next3, Next5] = [2,3,5],
  A[1] := Next2, A[2] := Next3, A[3] := Next5,
  I = 0, J = 0, K = 0, M = 1,  
  while (M < N) 
     A[M] := min([Next2,Next3,Next5]),
     if A[M] == Next2 then I := I+1, Next2 := 2*A[I] end,
     if A[M] == Next3 then J := J+1, Next3 := 3*A[J] end,
     if A[M] == Next5 then K := K+1, Next5 := 5*A[K] end,
     M := M + 1
  end,
  Hamming = A[N-1].</lang>

Output:

[1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
hamming_1691 = 2125764000
CPU time 0.0 seconds.

hamming_1000000 = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
CPU time 2.721 seconds.

PicoLisp

<lang PicoLisp>(de hamming (N)

  (let (L (1)  H)
     (do N
        (for (X L X (cadr X))      # Find smallest result
           (setq H (car X)) )
        (idx 'L H NIL)             # Remove it
        (for I (2 3 5)             # Generate next results
           (idx 'L (* I H) T) ) )
     H ) )

(println (make (for N 20 (link (hamming N))))) (println (hamming 1691)) # very fast (println (hamming 1000000)) # runtime about 13 minutes on i5-3570S</lang>

Output:

(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

PL/I

<lang pli>(subscriptrange): Hamming: procedure options (main); /* 14 November 2013 with fixes 2021 */

  declare (H(2000), p2, p3, p5, twoTo31, Hm, tenP(11)) decimal(12)fixed;
  declare (i, j, k, m, d, w) fixed binary;

  /* Quicksorts in-place the array of integers H, from lb to ub */
  quicksortH: procedure( lb, ub ) recursive;
     declare ( lb,   ub    )binary(15)fixed;
     declare ( left, right )binary(15)fixed;
     declare ( pivot, swap )decimal(12)fixed;
     declare sorting        bit(1);
     if ub > lb then do
        /* more than one element, so must sort */
        left    = lb;
        right   = ub;
        /* choosing the middle element of the array as the pivot */
        pivot   = H( left + ( ( right + 1 ) - left ) / 2 );
        sorting = '1'b;
        do while( sorting );
           do while( left  <= ub & H( left  ) < pivot ); left  = left  + 1; end;
           do while( right >= lb & H( right ) > pivot ); right = right - 1; end;
           sorting = ( left <= right );
           if sorting then do;
              swap       = H( left  );
              H( left  ) = H( right );
              H( right ) = swap;
              left       = left  + 1;
              right      = right - 1;
           end;
        end;
        call quicksortH( lb,   right );
        call quicksortH( left, ub    );
     end;
  end quicksortH ;

  /* find 2^31 - the limit for Hamming numbers we need to find */
  twoTo31 = 2;
  do i = 2 to 31;
     twoTo31 = twoTo31 * 2;
  end;
  /* calculate powers of 10 so we can check the number of digits */
  /* the numbers will have */
  tenP( 1 ) = 10;
  do i = 2 to 11;
     tenP( i ) = 10 * tenP( i - 1 );
  end;

  /* find the numbers */
  m = 0;
  p5 = 1;
  do k = 0 to 13;
     p3 = 1;
     do j = 0 to 19;
        Hm = 0;
        p2 = 1;
        do i = 0 to 31 while( Hm < twoTo31 );
           /* count the number of digits p2 * p3 * p5 will have */
           d = 0;
           do w = 1 to 11 while( tenP(w) < p2 ); d = d + 1; end;
           do w = 1 to 11 while( tenP(w) < p3 ); d = d + 1; end;
           do w = 1 to 11 while( tenP(w) < p5 ); d = d + 1; end;
           if d < 11 then do;
              /* the product will be small enough */
              Hm = p2 * p3 * p5;
              if Hm < twoTo31 then do;
                 m = m + 1;
                 H(m) = Hm;
              end;
           end;
           p2 = p2 * 2;
        end;
        p3 = p3 * 3;
     end;
     p5 = p5 * 5;
  end;

  /* sort the numbers */
  call quicksortH( 1, m );

  put skip list( 'The first 20 Hamming numbers:' );
  do i = 1 to 20;
     put skip list (H(i));
  end;
  put skip list( 'Hamming number 1691:' );
  put skip list (H(1691));

end Hamming;</lang> Results:

The first 20 Hamming numbers:
              1
              2
              3
              4
              5
              6
              8
              9
             10
             12
             15
             16
             18
             20
             24
             25
             27
             30
             32
             36
Hamming number 1691:
     2125764000

Prolog

Generator idiom

<lang Prolog>%% collect N elements produced by a generator in a row

take( 0, Next, Z-Z, Next). take( N, Next, [A|B]-Z, NZ):- N>0, !, next(Next,A,Next1),

 N1 is N-1,
 take(N1,Next1,B-Z,NZ).

%% a generator provides specific {next} implementation

next( hamm( A2,B,C3,D,E5,F,[H|G] ), H, hamm(X,U,Y,V,Z,W,G) ):-

 H is min(A2, min(C3,E5)),
 (   A2 =:= H -> B=[N2|U],X is N2*2 ; (X,U)=(A2,B) ),
 (   C3 =:= H -> D=[N3|V],Y is N3*3 ; (Y,V)=(C3,D) ),
 (   E5 =:= H -> F=[N5|W],Z is N5*5 ; (Z,W)=(E5,F) ).

mkHamm( hamm(1,X,1,X,1,X,X) ).  % Hamming numbers generator init state

main(N) :-

   mkHamm(G),take(20,G,A-[],_),           write(A), nl, 
   take(1691-1,G,_,G2),take(2,G2,B-[],_),     write(B), nl,  
   take(  N  -1,G,_,G3),take(2,G3,[C1|_]-_,_),   write(C1), nl.</lang>

SWI Prolog 6.2.6 produces (in about 7 ideone seconds):

 ?- time( main(1000000) ).
 [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
 [2125764000,2147483648]
 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
 % 10,017,142 inferences

Laziness flavor

Works with SWI-Prolog. Laziness is simulate with freeze/2 and ground/2.
Took inspiration from this code : http://chr.informatik.uni-ulm.de/~webchr (click on hamming.pl: Solves Hamming Problem). <lang Prolog>hamming(N) :-

    % to stop cleanly
    nb_setval(go, 1),

    % display list
    (	 N = 20 -> watch_20(20, L); watch(1,N,L)),

    % go
    L=[1|L235],
    multlist(L,2,L2),
    multlist(L,3,L3),
    multlist(L,5,L5),
    merge_(L2,L3,L23),
    merge_(L5,L23,L235).

%% multlist(L,N,LN) %% multiply each element of list L with N, resulting in list LN %% here only do multiplication for 1st element, then use multlist recursively multlist([X|L],N,XLN) :- % the trick to stop nb_getval(go, 1) ->

% laziness flavor when(ground(X), ( XN is X*N, XLN=[XN|LN], multlist(L,N,LN)));

true.

merge_([X|In1],[Y|In2],XYOut) :- % the trick to stop nb_getval(go, 1) ->

% laziness flavor ( X < Y -> XYOut = [X|Out], In11 = In1, In12 = [Y|In2] ; X = Y -> XYOut = [X|Out], In11 = In1, In12 = In2 ; XYOut = [Y|Out], In11 = [X | In1], In12 = In2), freeze(In11,freeze(In12, merge_(In11,In12,Out)));

true.

%% display nth element watch(Max, Max, [X|_]) :- % laziness flavor when(ground(X), (format('~w~n', [X]),

 % the trick to stop nb_linkval(go, 0))).

watch(N, Max, [_X|L]):- N1 is N + 1, watch(N1, Max, L).

%% display nth element watch_20(1, [X|_]) :- % laziness flavor when(ground(X), (format('~w~n', [X]),

 % the trick to stop nb_linkval(go, 0))).

watch_20(N, [X|L]):- % laziness flavor when(ground(X), (format('~w ', [X]), N1 is N - 1, watch_20(N1, L))).</lang>

Output:

?- hamming(20).
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
true .

?- hamming(1691).
2125764000
true .

?- hamming(1000000).
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
true .

PureBasic

<lang PureBasic>#X2 = 2

1. X3 = 3
2. X5 = 5

Macro Ham(w)

 PrintN("H("+Str(w)+") = "+Str(Hamming(w)))

EndMacro

Procedure.i Hamming(l.i)

 Define.i i,j,k,n,m,x=#X2,y=#X3,z=#X5
 Dim h.i(l) : h(0)=1
 For n=1 To l-1
   m=x
   If m>y : m=y : EndIf
   If m>z : m=z : EndIf
   h(n)=m
   If m=x : i+1 : x=#X2*h(i) : EndIf
   If m=y : j+1 : y=#X3*h(j) : EndIf
   If m=z : k+1 : z=#X5*h(k) : EndIf    
 Next
 ProcedureReturn h(l-1)

EndProcedure

OpenConsole("Hamming numbers") For h.i=1 To 20

 Ham(h)  

Next Ham(1691) Input()</lang>

Output:

H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36

H(1691) = 2125764000

Python

Version based on example from Dr. Dobb's CodeTalk

<lang python>from itertools import islice

def hamming2():

   \
   This version is based on a snippet from:
       https://web.archive.org/web/20081219014725/http://dobbscodetalk.com:80
                        /index.php?option=com_content&task=view&id=913&Itemid=85
       http://www.drdobbs.com/architecture-and-design/hamming-problem/228700538
       Hamming problem
       Written by Will Ness
       December 07, 2008

       When expressed in some imaginary pseudo-C with automatic
       unlimited storage allocation and BIGNUM arithmetics, it can be
       expressed as:
           hamming = h where
             array h;
             n=0; h[0]=1; i=0; j=0; k=0;
             x2=2*h[ i ]; x3=3*h[j]; x5=5*h[k];
             repeat:
               h[++n] = min(x2,x3,x5);
               if (x2==h[n]) { x2=2*h[++i]; }
               if (x3==h[n]) { x3=3*h[++j]; }
               if (x5==h[n]) { x5=5*h[++k]; } 
   
   h = 1
   _h=[h]    # memoized
   multipliers  = (2, 3, 5)
   multindeces  = [0 for i in multipliers] # index into _h for multipliers
   multvalues   = [x * _h[i] for x,i in zip(multipliers, multindeces)]
   yield h
   while True:
       h = min(multvalues)
       _h.append(h)
       for (n,(v,x,i)) in enumerate(zip(multvalues, multipliers, multindeces)):
           if v == h:
               i += 1
               multindeces[n] = i
               multvalues[n]  = x * _h[i]
       # cap the memoization
       mini = min(multindeces)
       if mini >= 1000:
           del _h[:mini]
           multindeces = [i - mini for i in multindeces]
       #
       yield h</lang>

Output:

>>> list(islice(hamming2(), 20))
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
>>> list(islice(hamming2(), 1690, 1691))
[2125764000]
>>> list(islice(hamming2(), 999999, 1000000))
[519312780448388736089589843750000000000000000000000000000000000000000000000000000000]

Another implementation of same approach

This version uses a lot of memory, it doesn't try to limit memory usage. <lang python>import psyco

def hamming(limit):

   h = [1] * limit
   x2, x3, x5 = 2, 3, 5
   i = j = k = 0

   for n in xrange(1, limit):
       h[n] = min(x2, x3, x5)
       if x2 == h[n]:
           i += 1
           x2 = 2 * h[i]
       if x3 == h[n]:
           j += 1
           x3 = 3 * h[j]
       if x5 == h[n]:
           k += 1
           x5 = 5 * h[k]

   return h[-1]

psyco.bind(hamming) print [hamming(i) for i in xrange(1, 21)] print hamming(1691) print hamming(1000000)</lang>

Implementation based on priority queue

This is inspired by the Picolisp implementation further down, but uses a priority queue instead of a search tree. Computes 3x more numbers than necessary, but discards them quickly so memory usage is not too bad.

<lang python>from heapq import heappush, heappop from itertools import islice

def h():

   heap = [1]
   while True:
       h = heappop(heap)
       while heap and h==heap[0]:
           heappop(heap)
       for m in [2,3,5]:
           heappush(heap, m*h)
       yield h

print list(islice(h(), 20)) print list(islice(h(), 1690, 1691)) print list(islice(h(), 999999, 1000000)) # runtime 9.5 sec on i5-3570S </lang>

"Cyclical Iterators"

The original author is Raymond Hettinger and the code was first published here under the MIT license. Uses iterators dubbed "cyclical" in a sense that they are referring back (explicitly, with p2, p3, p5 iterators) to the previously produced values, same as the above versions (through indices into shared storage) and the classic Haskell version (implicitly timed by lazy evaluation).

Memory is efficiently maintained automatically by the tee function for each of the three generator expressions, i.e. only that much is maintained as needed to produce the next value (although it looks like the storage is not shared so three copies are maintained implicitly there). <lang python>from itertools import tee, chain, groupby, islice from heapq import merge

def raymonds_hamming():

   # Generate "5-smooth" numbers, also called "Hamming numbers"
   # or "Regular numbers".  See: http://en.wikipedia.org/wiki/Regular_number
   # Finds solutions to 2**i * 3**j * 5**k  for some integers i, j, and k.

   def deferred_output():
       for i in output:
           yield i

   result, p2, p3, p5 = tee(deferred_output(), 4)
   m2 = (2*x for x in p2)                          # multiples of 2
   m3 = (3*x for x in p3)                          # multiples of 3
   m5 = (5*x for x in p5)                          # multiples of 5
   merged = merge(m2, m3, m5)
   combined = chain([1], merged)                   # prepend a starting point
   output = (k for k,g in groupby(combined))       # eliminate duplicates

   return result

print list(islice(raymonds_hamming(), 20)) print islice(raymonds_hamming(), 1689, 1690).next() print islice(raymonds_hamming(), 999999, 1000000).next()</lang> Results are the same as before.

Non-sharing recursive generator

Another formulation along the same lines, but greatly simplified, found here. Lacks data sharing, i.e. calls self recursively thus creating a separate copy of the data stream fed to the tee() call, again and again, instead of using its own output. This gravely impacts the efficiency. Not to be used.

<lang python>from heapq import merge from itertools import tee

def hamming_numbers():

   last = 1
   yield last

   a,b,c = tee(hamming_numbers(), 3)

   for n in merge((2*i for i in a), (3*i for i in b), (5*i for i in c)):
       if n != last:
           yield n
           last = n</lang>

Cyclic generator method #2.

Cyclic generator method #2. Considerably faster due to early elimination (before merge) of duplicates. Currently the faster Python version. Direct copy of Haskell code. <lang python>from itertools import islice, chain, tee

def merge(r, s):

   # This is faster than heapq.merge.
   rr = r.next()
   ss = s.next()
   while True:
       if rr < ss:
           yield rr
           rr = r.next()
       else:
           yield ss
           ss = s.next()

def p(n):

   def gen():
       x = n
       while True:
           yield x
           x *= n
   return gen()

def pp(n, s):

   def gen():
       for x in (merge(s, chain([n], (n * y for y in fb)))):
           yield x
   r, fb = tee(gen())
   return r

def hamming(a, b = None):

   if not b:
       b = a + 1
   seq = (chain([1], pp(5, pp(3, p(2)))))
   return list(islice(seq, a - 1, b - 1))

print hamming(1, 21) print hamming(1691)[0] print hamming(1000000)[0]</lang>

QBasic

Works with: QBasic version 1.1

Works with: QuickBasic version 4.5

<lang qbasic>FUNCTION min (a, b)

   IF a < b THEN min = a ELSE min = b

END FUNCTION

FUNCTION Hamming (limit)

   DIM h(limit)
   
   h(0) = 1
   x2 = 2
   x3 = 3
   x5 = 5
   i = 0
   j = 0
   k = 0
   FOR n = 1 TO limit
       h(n) = min(x2, min(x3, x5))
       IF x2 = h(n) THEN
           i = i + 1
           x2 = 2 * h(i)
       END IF
       IF x3 = h(n) THEN
           j = j + 1
           x3 = 3 * h(j)
       END IF
       IF x5 = h(n) THEN
           k = k + 1
           x5 = 5 * h(k)
       END IF
   NEXT n
   Hamming = h(limit - 1)

END FUNCTION

PRINT "The first 20 Hamming numbers are :" FOR i = 1 TO 20

   PRINT Hamming(i); " ";

NEXT i

PRINT PRINT "H( 1691) = "; Hamming(1691)</lang>

Qi

This example is incomplete. Parts 2 & 3 of task missing. Please ensure that it meets all task requirements and remove this message.

Translation of: Clojure

<lang qi>(define smerge

 [X|Xs] [Y|Ys] -> [X | (freeze (smerge (thaw Xs) [Y|Ys]))] where (< X Y)
 [X|Xs] [Y|Ys] -> [Y | (freeze (smerge [X|Xs] (thaw Ys)))] where (> X Y)
 [X|Xs] [_|Ys] -> [X | (freeze (smerge (thaw Xs) (thaw Ys)))])

(define smerge3

 Xs Ys Zs -> (smerge Xs (smerge Ys Zs)))

(define smap

 F [S|Ss] -> [(F S)|(freeze (smap F (thaw Ss)))])

(set hamming [1 | (freeze (smerge3 (smap (* 2) (value hamming))

                                  (smap (* 3) (value hamming))
                                  (smap (* 5) (value hamming))))])

(define stake

 _      0 -> []
 [S|Ss] N -> [S|(stake (thaw Ss) (1- N))])

(stake (value hamming) 20)</lang>

Output:

[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]

Quackery

Uses smoothwith from N-smooth numbers#Quackery.

<lang Quackery> ' [ 2 3 5 ] smoothwith [ size 1000000 = ]

 dup 20 split drop echo cr
 dup 1690 peek echo cr
 -1 peek echo 

</lang>

Output:

[ 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 ]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

R

Recursively find the Hamming numbers below $2^{31}$. Shown are results for tasks 1 and 2. Arbitrary precision integers are not supported natively. <lang R>hamming=function(hamms,limit) {

 tmp=hamms
 for(h in c(2,3,5)) {
   tmp=c(tmp,h*hamms)
 }
 tmp=unique(tmp[tmp<=limit])
 if(length(tmp)>length(hamms)) {
   hamms=hamming(tmp,limit)
 }
 hamms

} h <- sort(hamming(1,limit=2^31-1)) print(h[1:20]) print(h[length(h)])</lang>

Output:

[1]  1  2  3  4  5  6  8  9 10 12 15 16 18 20 24 25 27 30 32 36
[1] 2125764000

Alternate version

The nextn R function provides the needed functionality:

<lang r>hamming <- function(n) {

 a <- numeric(n)
 a[1] <- 1
 for (i in 2:n) {
   a[i] <- nextn(a[i-1]+1)
 }
 a

}</lang>

Output

 hamming(20)
 [1]  1  2  3  4  5  6  8  9 10 12 15 16 18 20 24 25 27 30 32 36

Racket

<lang scheme>#lang racket (require racket/stream) (define first stream-first) (define rest stream-rest)

(define (merge s1 s2)

 (define x1 (first s1))
 (define x2 (first s2))
 (cond [(= x1 x2) (merge s1 (rest s2))]
       [(< x1 x2) (stream-cons x1 (merge (rest s1) s2))]
       [else      (stream-cons x2 (merge s1 (rest s2)))]))

(define (mult k) (λ(x) (* x k)))

(define hamming

 (stream-cons 
  1 (merge (stream-map (mult 2) hamming)
           (merge (stream-map (mult 3) hamming)
                  (stream-map (mult 5) hamming)))))

(for/list ([i 20] [x hamming]) x) (stream-ref hamming 1690) (stream-ref hamming 999999)</lang>

Output:

'(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Translation of Haskell code avoiding duplicates

The above version consumes quite a lot of memory as streams are retained since the head of the stream is a global defined binding "hamming". The following code implements (hamming) as a function and all heads of streams are locally defined so that they can be garbage collected as they are consumed; as well it is formulated so that no duplicate values are generated so as to simplify the calculation and minimize the number of values in the streams; to further the latter it also evaluates the least dense stream first. The following code is about three times faster than the above code:

Translation of: Haskell

<lang scheme>#lang racket (require racket/stream) (define first stream-first) (define rest stream-rest)

(define (hamming)

 (define (merge s1 s2)
   (let ([x1 (first s1)]
         [x2 (first s2)])
     (if (< x1 x2) ; don't have to handle duplicate case
         (stream-cons x1 (merge (rest s1) s2))
         (stream-cons x2 (merge s1 (rest s2))))))
 (define (smult m s) ; faster than using map (* m)
   (define (smlt ss)
     (stream-cons (* m (first ss)) (smlt (rest ss))))
   (smlt s))
 (define (u n s)
   (if (stream-empty? s) ; checking here more efficient than in merge
       (letrec ([r          (smult n (stream-cons 1 r)) ])
         r)
       (letrec ([r (merge s (smult n (stream-cons 1 r)))])
         r)))
 ;; (stream-cons 1 (u 2 (u 3 (u 5 empty-stream))))
 (stream-cons 1 (foldr u empty-stream '(2 3 5))))

(for/list ([i 20] [x (hamming)]) x) (newline) (stream-ref (hamming) 1690) (newline) (stream-ref (hamming) 999999) (newline)</lang>

The output of the above code is the same as that of the earlier code.

Raku

(formerly Perl 6)

Merge version

Works with: rakudo version 2015-11-04

The limit scaling is not required, but it cuts down on a bunch of unnecessary calculation. <lang perl6>my $limit = 32;

sub powers_of ($radix) { 1, |[\*] $radix xx * }

my @hammings =

 (   powers_of(2)[^ $limit ]       X*
     powers_of(3)[^($limit * 2/3)] X* 
     powers_of(5)[^($limit * 1/2)]
  ).sort;

say @hammings[^20]; say @hammings[1690]; # zero indexed</lang>

Output:

(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000

Iterative version

Works with: rakudo version 6.c

This version uses a lazy list, storing a maximum of two extra value from the highest index requested

<lang perl6>my \Hammings := gather {

 my %i = 2, 3, 5 Z=> (Hammings.iterator for ^3);
 my %n = 2, 3, 5 Z=> 1 xx 3;

 loop {
   take my $n := %n{2, 3, 5}.min;

   for 2, 3, 5 -> \k {
     %n{k} = %i{k}.pull-one * k if %n{k} == $n;
   }
 }

}

say Hammings.[^20]; say Hammings.[1691 - 1]; say Hammings.[1000000 - 1]; </lang>

Output:

(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Raven

Translation of: Liberty Basic

<lang raven>define hamming use $limit

   [ ] as $h
   1 $h 0 set
   2 as $x2   3 as $x3    5 as $x5
   0 as $i    0 as $j     0 as $k
   1 $limit 1 + 1 range each as $n
       $x3 $x5 min $x2 min    $h $n   set
       $h $n get   $x2 =  if
           $i  1 +   as $i
           $h $i get    2 *     as $x2
       $h $n get   $x3 =  if
           $j  1 +   as $j
           $h $j get    3 *     as $x3
       $h $n get   $x5 =  if
           $k  1 +   as $k
           $h $k get    5 *     as $x5
   $h   $limit 1 -   get

1 21 1 range each as $lim

   $lim hamming print " " print

"\n" print

"Hamming(1691) is: " print 1691 hamming print "\n" print

1. Raven can't handle > 2^31 using integers
3. "Hamming(1000000) is: " print 1000000 hamming print "\n" print</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
Hamming(1691) is: 2125764000

REXX

The three REXX versions compute and present the Hamming numbers in numerical order.

idiomatic

This REXX program was a direct copy from my old REXX subroutine to compute   UGLY   numbers,
it computes   just enough   Hamming numbers   (two Hamming numbers after the current number). <lang rexx>/*REXX program computes Hamming numbers: 1 ──► 20, # 1691, and the one millionth. */ numeric digits 100 /*ensure enough decimal digits. */ call hamming 1, 20 /*show the 1st ──► twentieth Hamming #s*/ call hamming 1691 /*show the 1,691st Hamming number. */ call hamming 1000000 /*show the 1 millionth Hamming number.*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hamming: procedure; parse arg x,y; if y== then y= x; w= length(y)

                                  #2= 1;     #3= 1;      #5= 1;      @.= 0;        @.1= 1
           do n=2  for y-1
           @.n= min(2*@.#2,  3*@.#3,  5*@.#5)   /*pick the minimum of 3 Hamming numbers*/
           if 2*@.#2 == @.n   then #2= #2 + 1   /*number already defined?  Use next #. */
           if 3*@.#3 == @.n   then #3= #3 + 1   /*   "      "       "       "    "  "  */
           if 5*@.#5 == @.n   then #5= #5 + 1   /*   "      "       "       "    "  "  */
           end   /*n*/                          /* [↑]  maybe assign next 3 Hamming#s. */
                        do j=x  to y;     say  'Hamming('right(j, w)") ="    @.j
                        end   /*j*/

        say right( 'length of last Hamming number ='     length(@.y), 70);        say
        return</lang>

output   when using the default inputs:

Hamming( 1) = 1
Hamming( 2) = 2
Hamming( 3) = 3
Hamming( 4) = 4
Hamming( 5) = 5
Hamming( 6) = 6
Hamming( 7) = 8
Hamming( 8) = 9
Hamming( 9) = 10
Hamming(10) = 12
Hamming(11) = 15
Hamming(12) = 16
Hamming(13) = 18
Hamming(14) = 20
Hamming(15) = 24
Hamming(16) = 25
Hamming(17) = 27
Hamming(18) = 30
Hamming(19) = 32
Hamming(20) = 36
                                     length of last Hamming number = 2

Hamming(1691) = 2125764000
                                    length of last Hamming number = 10

Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
                                    length of last Hamming number = 84

optimized

This REXX version is roughly twice as fast as the 1^st REXX version. <lang rexx>/*REXX program computes Hamming numbers: 1 ──► 20, # 1691, and the one millionth.*/ call hamming 1, 20 /*show the 1st ──► twentieth Hamming #s*/ call hamming 1691 /*show the 1,691st Hamming number. */ call hamming 1000000 /*show the 1 millionth Hamming number.*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hamming: procedure; arg x,y; if y== then y= x; w= length(y); L= length(y-1); p= 2**L

        numeric digits max(9, p + p%4 + p%16)   /*ensure enough decimal digits.        */
                                   #2= 1;    #3= 1;     #5= 1;     @.= 0;       @.1= 1
           do n=2  for y-1
           _2= @.#2 + @.#2                      /*this is faster than:      @.#2 * 2   */
           _3= @.#3 + @.#3 + @.#3               /*  "   "    "     "        @,#3 * 3   */
           _5= @.#5 * 5
                             m= _2              /*assume a minimum (of the 3 Hammings).*/
           if _3  < m   then m= _3              /*is this number less than the minimum?*/
           if _5  < m   then m= _5              /* "   "     "     "    "   "     "    */
               @.n= m                           /*now,  assign the next Hamming number.*/
           if _2 == m   then #2= #2 + 1         /*number already defined?   Use next #.*/
           if _3 == m   then #3= #3 + 1         /*   "      "       "        "    "  " */
           if _5 == m   then #5= #5 + 1         /*   "      "       "        "    "  " */
           end   /*n*/                          /* [↑]  maybe assign next Hamming #'s. */
                        do j=x  to y;      say 'Hamming('right(j, w)") ="     @.j
                        end   /*j*/

        say right( 'length of last Hamming number ='     length(@.y), 70);        say
        return</lang>

output   is identical to the 1^st REXX version.

for huge numbers

This REXX version is slightly slower than the 2^nd REXX version.

It can, however, computer much larger Hamming numbers   (by storing the larger numbers in exponential format).
This is possible because larger Hamming numbers have a significant number of trailing zeros. <lang rexx>/*REXX program computes Hamming numbers: 1 ──► 20, # 1691, and the one millionth.*/ call hamming 1, 20 /*show the 1st ──► twentieth Hamming #s*/ call hamming 1691 /*show the 1,691st Hamming number. */ call hamming 1000000 /*show the 1 millionth Hamming number.*/ exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ hamming: procedure; arg x,y; if y== then y= x; w= length(y); L= length(y-1); p= 2**L

        numeric digits max(9, p + p%4 + p%16)   /*ensure enough decimal digits.        */
                                   #2= 1;    #3= 1;     #5= 1;     @.= 0;       @.1= 1
           do n=2  for y-1
           _2= @.#2 + @.#2                      /*this is faster than:      @.#2 * 2   */
           _3= @.#3 + @.#3 + @.#3               /*  "   "    "     "        @,#3 * 3   */
           _5= @.#5 * 5
                             m= _2              /*assume a minimum (of the 3 Hammings).*/
           if _3  < m   then m= _3              /*is this number less than the minimum?*/
           if _5  < m   then m= _5              /* "   "     "     "    "   "     "    */
                   @.n= format(m,,,,0)          /*now,  assign the next Hamming number.*/
           if _2 == m   then #2= #2 + 1         /*number already defined?   Use next #.*/
           if _3 == m   then #3= #3 + 1         /*   "      "       "        "    "  " */
           if _5 == m   then #5= #5 + 1         /*   "      "       "        "    "  " */
           end   /*n*/                          /* [↑]  maybe assign next Hamming #'s. */
                        do j=x  to y;      say 'Hamming('right(j, w)") ="     @.j / 1
                        end   /*j*/

        say right( 'length of last Hamming number ='     length(@.y / 1), 70);        say
        return</lang>

output   is identical to the 1^st REXX version.

Ring

<lang ring> see "h(1) = 1" + nl for nr = 1 to 19

    see "h(" + (nr+1) + ") = " + hamming(nr) + nl

next see "h(1691) = " + hamming(1690) + nl see nl

func hamming limit

    h = list(1690)
    h[1] =1
    x2 = 2 x3 = 3 x5 =5
    i  = 0 j  = 0 k  =0   
    for n =1 to limit
        h[n]  = min(x2, min(x3, x5))
        if x2 = h[n]  i = i +1  x2 =2 *h[i] ok
        if x3 = h[n]  j = j +1  x3 =3 *h[j] ok
        if x5 = h[n]  k = k +1  x5 =5 *h[k] ok
    next
    hamming = h[limit]
    return hamming

</lang> Output:

h(1) = 1
h(2) = 2
h(3) = 3
h(4) = 4
h(5) = 5
h(6) = 6
h(7) = 8
h(8) = 9
h(9) = 10
h(10) = 12
h(11) = 15
h(12) = 16
h(13) = 18
h(14) = 20
h(15) = 24
h(16) = 25
h(17) = 27
h(18) = 30
h(19) = 32
h(20) = 36
h(1691) = 2125764000

Ruby

Translation of: Scala

Works with: Ruby version 1.9.3

<lang ruby>hamming = Enumerator.new do |yielder|

 next_ham = 1
 queues = [[ 2, []], [3, []], [5, []] ]
 
 loop do
   yielder << next_ham   # or: yielder.yield(next_ham)
   
   queues.each {|m,queue| queue << next_ham * m}
   next_ham = queues.collect{|m,queue| queue.first}.min
   queues.each {|m,queue| queue.shift if queue.first==next_ham}
 end

end</lang> And the "main" part of the task <lang ruby>start = Time.now

hamming.each.with_index(1) do |ham, idx|

 case idx
 when (1..20), 1691
   puts "#{idx} => #{ham}"
 when 1_000_000
   puts "#{idx} => #{ham}"
   break
 end

end

puts "elapsed: #{Time.now - start} seconds"</lang>

Output:

1 => 1
2 => 2
3 => 3
4 => 4
5 => 5
6 => 6
7 => 8
8 => 9
9 => 10
10 => 12
11 => 15
12 => 16
13 => 18
14 => 20
15 => 24
16 => 25
17 => 27
18 => 30
19 => 32
20 => 36
1691 => 2125764000
1000000 => 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
elapsed: 6.522811 seconds

System: I7-6700HQ, 3.5 GHz, Linux Kernel 5.6.17
Run as: $ ruby hammingnumbers.rb
elapsed: 2.589248076 seconds  # Ruby 2.7.1
elapsed: 2.067365 seconds     # JRuby 9.2.11.1
elapsed: N/A - too long       # Truffleruby 20.0.0

Alternative version:

Translation of: Crystal

<lang ruby>def hamming(limit)

 h = Array.new(limit, 1)
 x2, x3, x5 = 2, 3, 5
 i, j, k = 0, 0, 0
 (1...limit).each do |n|
   # h[n] = [x2, [x3, x5].min].min    # not as fast on all VMs
   h[n] = (x3 < x5 ? (x2 < x3 ? x2 : x3) : (x2 < x5 ? x2 : x5))
   x2 = 2 * h[i += 1] if x2 == h[n]
   x3 = 3 * h[j += 1] if x3 == h[n]
   x5 = 5 * h[k += 1] if x5 == h[n]
 end
 h[limit - 1]

end

start = Time.new print "Hamming Number (1..20): "; (1..20).each { |i| print "#{hamming(i)} " } puts puts "Hamming Number 1691: #{hamming 1691}" puts "Hamming Number 1,000,000: #{hamming 1_000_000}" puts "Elasped Time: #{Time.new - start} secs"</lang>

System: I7-6700HQ, 3.5 GHz, Linux Kernel 5.6.17
Run as: $ ruby hammingnumbers.rb

Output:

Hamming Number (1..20): 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
Hamming Number 1691: 2125764000
Hamming Number 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Elasped Time: 1.566937062 secs # Ruby 2.7.1
Elasped Time: 1.3442580 secs   # JRuby 9.2.11.1
Elasped Time: 1.627 secs       # Truffleruby 20.1.0

Run BASIC

<lang runbasic> dim h(1000000) for i =1 to 20

   print hamming(i);" ";

next i

print print "Hamming List First(1691) =";chr$(9);hamming(1691) print "Hamming List Last(1000000) =";chr$(9);hamming(1000000)

end

function hamming(limit)

   h(0) =1
   x2 = 2: x3 = 3: x5 =5
   i  = 0: j  = 0: k  =0
   for n =1 to limit
       h(n)  = min(x2, min(x3, x5))
       if x2 = h(n) then i = i +1: x2 =2 *h(i)
       if x3 = h(n) then j = j +1: x3 =3 *h(j)
       if x5 = h(n) then k = k +1: x5 =5 *h(k)
   next n
   hamming = h(limit -1)

end function</lang>

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
Hamming List First(1691)   =	2125764000
Hamming List Last(1000000) =	519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Rust

Library: num

Basic version

Translation of: D

Improved by minimizing the number of BigUint comparisons: <lang rust>extern crate num; num::bigint::BigUint;

use std::time::Instant;

fn basic_hamming(n: usize) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   let mut h = vec![BigUint::from(0u8); n];
   h[0] = BigUint::from(1u8);
   let mut x2 = BigUint::from(2u8);
   let mut x3 = BigUint::from(3u8);
   let mut x5 = BigUint::from(5u8);
   let mut i = 0usize; let mut j = 0usize; let mut k = 0usize;

   // BigUint comparisons are expensive, so do it only as necessary...
   fn min3(x: &BigUint, y: &BigUint, z: &BigUint) -> (usize, BigUint) {
       let (cs, r1) = if y == z { (0x6, y) } 
                       else if y < z { (2, y) } else { (4, z) };
       if x == r1 { (cs | 1, x.clone()) }
       else if x < r1 { (1, x.clone()) } else { (cs, r1.clone()) }
   }

   let mut c = 1;
   while c < n { // satisfy borrow checker with extra blocks: {  }
       let (cs, e1) = { min3(&x2, &x3, &x5) };
       h[c] = e1; // vector now owns the generated value
       if (cs & 1) != 0 { i += 1; x2 = &two * &h[i] }
       if (cs & 2) != 0 { j += 1; x3 = &three * &h[j] }	
       if (cs & 4) != 0 { k += 1; x5 = &five * &h[k] }
       c += 1;
   }

   match h.pop() {
       Some(v) => v,
       _ => panic!("basic_hamming: arg is zero; no elements")
   }

}

fn main() {

   print!("[");
   for (i, h) in (1..21).map(basic_hamming).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", h)
   }
   println!(" ]");
   println!("{}", basic_hamming(1691));

   let strt = Instant::now();

   let rslt = basic_hamming(1000000);

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;

   let rs = rslt.to_str_radix(10);
   let mut s = rs.as_str();
   println!("{} digits:", s.len());
       while s.len() > 100 {
           let (f, r) = s.split_at(100);
           s = r;
           println!("{}", f);
       }
       println!("{}", s);

   println!("This last took {} milliseconds", dur);

}</lang>

Output:

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
84 digits:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 677 milliseconds.

Eliminating duplicate calculations

Much of the time above is wasted doing big integer multiplications that are duplicated elsewhere as in 2 times 3 and 3 times 2, etc. The following code eliminates such duplicate multiplications and reduces the number of comparisons, as follows: <lang rust>fn nodups_hamming(n: usize) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   let mut m = vec![BigUint::from(0u8); 1];
   m[0] = BigUint::from(1u8);
   let mut h = vec![BigUint::from(0u8); n];
   h[0] = BigUint::from(1u8);
   if n > 1 {
       m.push(BigUint::from(3u8)); // for initial x53 advance
       h[1] = BigUint::from(2u8); // for initial x532 advance
   }
   let mut x5 = BigUint::from(5u8);
   let mut x53 = BigUint::from(9u8); // 3 times 3 because already merged one step
   let mut mrg = BigUint::from(3u8);
   let mut x532 = BigUint::from(2u8);

   let mut i = 0usize; let mut j = 1usize;
   let mut c = 1usize;
   while c < n { // satisfy borrow checker with extra blocks: {  }
       if &x532 < &mrg { h[c] = x532; i += 1; x532 = &two * &h[i]; }
       else {	h[c] = mrg;
               if &x53 < &x5 { mrg = x53; j += 1; x53 = &three * &m[j]; }
               else { mrg = x5.clone(); x5 = &five * &x5; };
               m.push(mrg.clone()); };
       c += 1;
   }
   match h.pop() {
       Some(v) => v,
       _ => panic!("nodups_hamming: arg is zero; no elements")
   }

}</lang>

Substitute the calls to the above code for the calls to "basic_hamming" (three places) in the "main" function above. The output is the same except that the time expended is less (249 milliseconds), for over two and a half times faster.

Much faster logarithmic version with low memory use

The above versions spend much of their time doing BigUint calculations. The below version eliminates much of that time by using integer powers of 2, 3, and 5 representations and all normal integer calculations except for the final conversion to a BitUint for the final result for about a 30 times speed-up.

Another problem is that the above versions use so much memory that they can't compute even the billionth hamming number without running out of memory on a 16 Gigabyte machine. This version greatly reduces the memory use to about O(n^(2/3)) by eliminating no longer required back values in batches so that with about 9 Gigabytes it will calculate the hamming numbers to 1.2e13 (it's limit due to the ranges of the exponents) in a day or so. The code is as follows: <lang rust>fn log_nodups_hamming(n: u64) -> BigUint {

   if n <= 0 { panic!("nodups_hamming: arg is zero; no elements") }
   if n < 2 { return BigUint::from(1u8) } // trivial case for n == 1
   if n > 1.2e13 as u64 { panic!("log_nodups_hamming: argument too large to guarantee results!") }

   // constants as expanded integers to minimize round-off errors, and
   // reduce execution time using integer operations not float...
   const LAA2: u64 = 35184372088832; // 2.0f64.powi(45)).round() as u64;
   const LBA2: u64 = 55765910372219; // 3.0f64.log2() * 2.0f64.powi(45)).round() as u64;
   const LCA2: u64 = 81695582054030; // 5.0f64.log2() * 2.0f64.powi(45)).round() as u64;

   #[derive(Clone, Copy)]
   struct Logelm { // log representation of an element with only allowable powers
       exp2: u16,
       exp3: u16,
       exp5: u16,
       logr: u64 // log representation used for comparison only - not exact
   }

   impl Logelm {
       fn lte(&self, othr: &Logelm) -> bool {
           if self.logr <= othr.logr { true } else { false }
       }
       fn mul2(&self) -> Logelm {
           Logelm { exp2: self.exp2 + 1, logr: self.logr + LAA2, .. *self }
       }
       fn mul3(&self) -> Logelm {
           Logelm { exp3: self.exp3 + 1, logr: self.logr + LBA2, .. *self }
       }
       fn mul5(&self) -> Logelm {
           Logelm { exp5: self.exp5 + 1, logr: self.logr + LCA2, .. *self }
       }
   }

   let one = Logelm { exp2: 0, exp3: 0, exp5: 0, logr: 0 };
   let mut x532 = one.mul2();
   let mut mrg = one.mul3();
   let mut x53 = one.mul3().mul3(); // advance as mrg has the former value...
   let mut x5 = one.mul5();

   let mut h = Vec::with_capacity(65536); // vec!(one.clone(); 0);
   let mut m = Vec::<Logelm>::with_capacity(65536); // vec!(one.clone(); 0);

   let mut i = 0usize; let mut j = 0usize;
   for _ in 1 .. n {
       let cph = h.capacity();
       if i > cph / 2 { // drain extra unneeded values...
           h.drain(0 .. i);
           i = 0;
       }
       if x532.lte(&mrg) {
           h.push(x532);
           x532 = h[i].mul2();
           i += 1;
       } else {
           h.push(mrg);
           if x53.lte(&x5) {
               mrg = x53;
               x53 = m[j].mul3();
               j += 1;
           } else {
               mrg = x5;
               x5 = x5.mul5();
           }
           let cpm = m.capacity();
           if j > cpm / 2 { // drain extra unneeded values...
               m.drain(0 .. j);
               j = 0;
           }
           m.push(mrg);
       }
   }

   let o = &h[&h.len() - 1];
   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   let mut ob = BigUint::from(1u8); // convert to BigUint at the end
   for _ in 0 .. o.exp2 { ob = ob * &two }
   for _ in 0 .. o.exp3 { ob = ob * &three }
   for _ in 0 .. o.exp5 { ob = ob * &five }
   ob

}</lang>

Again, this function can be used with the same "main" as above and the outputs are the same except that the execution time is only 7 milliseconds. It calculates the hamming number to a billion and just over a second and to one hundred billion in just over 100 seconds - O(n) time complexity. As well as eliminating duplicate calculations and calculating using exponents rather than BitUint operations, it also reduces the time required as compared to other similar algorithms by scaling the logarithms of two, three, and five into 64-bit integers so no floating point operations are required. The scaling is such that round-off errors will not affect the order of results for well past the usable range.

Memory used is greatly reduced to O(n^(2/3)) by draining the arrays of back values no longer required in batches (rather than one by one) so that less time is used. It also saves time by not requiring as many allocations and de-allocations as the draining is done in place, thus making the current capacity of arrays longer usable.

Sequence version

As the task actually asks for a sequence of Hamming numbers, any of the above three solutions can easily be adapted to output an Iterator sequence; in this case the last fastest one is converted as follows:

<lang rust>extern crate num; // requires dependency on the num library use num::bigint::BigUint;

use std::time::Instant;

fn log_nodups_hamming_iter() -> Box<Iterator<Item = (u16, u16, u16)>> {

   // constants as expanded integers to minimize round-off errors, and
   // reduce execution time using integer operations not float...
   const LAA2: u64 = 35184372088832; // 2.0f64.powi(45)).round() as u64;
   const LBA2: u64 = 55765910372219; // 3.0f64.log2() * 2.0f64.powi(45)).round() as u64;
   const LCA2: u64 = 81695582054030; // 5.0f64.log2() * 2.0f64.powi(45)).round() as u64;

   #[derive(Clone, Copy)]
   struct Logelm { // log representation of an element with only allowable powers
       exp2: u16,
       exp3: u16,
       exp5: u16,
       logr: u64 // log representation used for comparison only - not exact
   }
   impl Logelm {
       fn lte(&self, othr: &Logelm) -> bool {
           if self.logr <= othr.logr { true } else { false }
       }
       fn mul2(&self) -> Logelm {
           Logelm { exp2: self.exp2 + 1, logr: self.logr + LAA2, .. *self }
       }
       fn mul3(&self) -> Logelm {
           Logelm { exp3: self.exp3 + 1, logr: self.logr + LBA2, .. *self }
       }
       fn mul5(&self) -> Logelm {
           Logelm { exp5: self.exp5 + 1, logr: self.logr + LCA2, .. *self }
       }
   }

   let one = Logelm { exp2: 0, exp3: 0, exp5: 0, logr: 0 };
   let mut x532 = one.mul2();
   let mut mrg = one.mul3();
   let mut x53 = one.mul3().mul3(); // advance as mrg has the former value...
   let mut x5 = one.mul5();

   let mut h = Vec::with_capacity(65536);
   let mut m = Vec::<Logelm>::with_capacity(65536);

   let mut i = 0usize; let mut j = 0usize;
   Box::new((0u64 .. ).map(move |it| if it < 1 { (0, 0, 0) } else {
       let cph = h.capacity();
       if i > cph / 2 {
           h.drain(0 .. i);
           i = 0;
       }
       if x532.lte(&mrg) {
           h.push(x532);
           x532 = h[i].mul2();
           i += 1;
       } else {
           h.push(mrg);
           if x53.lte(&x5) {
               mrg = x53;
               x53 = m[j].mul3();
               j += 1;
           } else {
               mrg = x5;
               x5 = x5.mul5();
           }
           let cpm = m.capacity();
           if j > cpm / 2 {
               m.drain(0 .. j);
               j = 0;
           }
           m.push(mrg);
       }
       let o = &h[&h.len() - 1];
       (o.exp2, o.exp3, o.exp5)
   }))

}

fn convert_log2big(o: (u16, u16, u16)) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   let (x2, x3, x5) = o;
   let mut ob = BigUint::from(1u8); // convert to BigUint at the end
   for _ in 0 .. x2 { ob = ob * &two }
   for _ in 0 .. x3 { ob = ob * &three }
   for _ in 0 .. x5 { ob = ob * &five }
   ob

}

fn main() {

   print!("[");
   for (i, h) in log_nodups_hamming_iter().take(20).map(convert_log2big).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", h)
   }
   println!(" ]");
   println!("{}", convert_log2big(log_nodups_hamming_iter().take(1691).last().unwrap()));

   let strt = Instant::now();

// let rslt = convert_log2big(log_nodups_hamming_iter().take(1000000000).last().unwrap());

   let mut it = log_nodups_hamming_iter().into_iter();
   for _ in 0 .. 100-1 { // a little faster; less one level of iteration
       let _ = it.next();
   }
   let rslt = convert_log2big(it.next().unwrap());

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;

   println!("2^{} times 3^{} times 5^{}", rslt.0, rslt.1, rslt.2);
   let rs = convert_log2big(rslt).to_str_radix(10);
   let mut s = rs.as_str();
   println!("{} digits:", s.len());
   let lg3 = 3.0f64.log2();
   let lg5 = 5.0f64.log2();
   let lg = (rslt.0 as f64 + rslt.1 as f64 * lg3

+ rslt.2 as f64 * lg5) * 2.0f64.log10();

   println!("Approximately {}E+{}", 10.0f64.powf(lg.fract()), lg.trunc());
   if s.len() <= 10000 {
       while s.len() > 100 {
           let (f, r) = s.split_at(100);
           s = r;
           println!("{}", f);
       }
       println!("{}", s);
   }

   println!("This last took {} milliseconds.", dur);

}</lang>

Output:

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
2^55 times 3^47 times 5^64
84 digits:
Approximately 5.193127804483804E+83
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 17 milliseconds.

The above final output is the same as the last one, but the function is called differently; also note that it is somewhat slower than the last version due to the extra function calls required to enumerate over an Iterator. It can enumerate the Hamming numbers up to a billion in about 20 seconds instead of the about ten seconds for the last version - about O(n) time complexity, and has the same O(n^(2/3)) space complexity as the last version.

Functional sequence version avoiding duplicates

It has been said by some that Rust is basically a functional language; however that isn't quite true in several respects, at least as per the following:

1. It does not guarantee tail call optimization for functions, thus sometimes requiring imperative forms of code to produce that effect.
2. It does not have currying or partial application of function arguments without using kludges of nested function/closure calls.
3. move closures cannot use recursive shared values without using interior mutability inside a reference counted value (required here)
4. Closures are not recursive without using a trick involving shared state reference counted values (demonstrated here).
5. It currently does not have a standard library implementation of a lazily computed non-static value (required to implement a Lazy List, and
6. It accordingly is not as easy as in most other languages to implement Co-Inductive Streams or (also memoized) Lazy Lists (a form of Lazy List as is required here).

Many of these come about due to the Rust memory model where pieces of programs "own" data and its disposal but can assign references to other pieces of code (with limits if mutability as required), instead of the Garbage Collected model used by most other functional languages where variables are owned by the system and program code just uses references to that data other than for primitives which are owned by whoever uses them.

The lack of the Lazy type and thus the Lazy List type is partly due to Rust's still being relatively unstable, as Lazy requires a "thunk" (a zero argument move closure acting on owned data - FnOnce in Rust), and in Rust these must be boxed (allocated on the heap) to be usable. However, the new versions of Rust allow boxing of the FnOnce closure so it can be used as a Thunk.

Jeremy Reems had implemented Lazy and also LazyList, but they haven't been maintained for many years and don't compile. According, I have implemented enough of this functionality as required by this algorithm, as per the following code (tested on Rust version 1.53.0, run in --release mode):

Translation of: Haskell

Works with: Rust 1.53.0

<lang rust>extern crate num; use num::bigint::BigUint;

use std::rc::Rc; use std::cell::{UnsafeCell, RefCell}; use std::mem;

use std::time::Instant;

// implementation of Thunk closure here...

pub struct Thunk<'a, R>(Box<dyn FnOnce() -> R + 'a>);

impl<'a, R: 'a> Thunk<'a, R> {

   #[inline(always)]
   fn new<F: 'a + FnOnce() -> R>(func: F) -> Thunk<'a, R> {
       Thunk(Box::new(func))
   }
   #[inline(always)]
   fn invoke(self) -> R { self.0() }

}

// actual Lazy implementation starts here...

use self::LazyState::*;

pub struct Lazy<'a, T: 'a>(UnsafeCell<LazyState<'a, T>>);

enum LazyState<'a, T: 'a> {

   Unevaluated(Thunk<'a, T>),
   EvaluationInProgress,
   Evaluated(T)

}

impl<'a, T: 'a> Lazy<'a, T>{

   #[inline]
   pub fn new<'b, F>(thunk: F) -> Lazy<'b, T>
           where F: 'b + FnOnce() -> T {
       Lazy(UnsafeCell::new(Unevaluated(Thunk::new(thunk))))
   }
   #[inline]
   pub fn evaluated(val: T) -> Lazy<'a, T> {
       Lazy(UnsafeCell::new(Evaluated(val)))
   }
   #[inline]
   fn force<'b>(&'b self) { // not thread-safe
       unsafe {
           match *self.0.get() {
               Evaluated(_) => return, // nothing required; already Evaluated
               EvaluationInProgress =>
                   panic!("Lazy::force called recursively!!!"),
               _ => () // need to do following something else if Unevaluated...
           } // following eliminates recursive race; drops neither on replace:
           match mem::replace(&mut *self.0.get(), EvaluationInProgress) {
               Unevaluated(thnk) => { // Thunk can't call force on same Lazy
                   *self.0.get() = Evaluated(thnk.invoke());
               },
               _ => unreachable!() // already took care of other cases above.
           }
       }
   }
   #[inline]
   pub fn value<'b>(&'b self) -> &'b T {
       self.force(); // evaluatate if not evealutated
       match unsafe { &*self.0.get() } {
           &Evaluated(ref v) => v, // return value
           _ => { unreachable!() } // previous force guarantees Evaluated
       }
   }
   #[inline] // consumes the object to produce the value
   pub fn unwrap<'b>(self) -> T where T: 'b {
       self.force(); // evaluatate if not evealutated
       match { self.0.into_inner() } {
           Evaluated(v) => v,
           _ => unreachable!() // previous code guarantees Evaluated
       }
   }

}

// now for immutable persistent shareable (memoized) LazyList via Lazy above...

type RcLazyListNode<'a, T> = Rc<Lazy<'a, LazyList<'a, T>>>;

use self::LazyList::*;

1. [derive(Clone)]

enum LazyList<'a, T: 'a + Clone> {

   /// The Empty List
   Empty,
   /// A list with one member and possibly another list.
   Cons(T, RcLazyListNode<'a, T>)

}

impl<'a, T: 'a + Clone> LazyList<'a, T> {

   #[inline]
   pub fn cons<F>(v: T, cntf: F) -> LazyList<'a, T>
           where F: 'a + FnOnce() -> LazyList<'a, T> {
       Cons(v, Rc::new(Lazy::new(cntf)))
   }
   #[inline]
   pub fn head<'b>(&'b self) -> &'b T {
       if let Cons(ref hd, _) = *self { return hd }
       panic!("LazyList::head called on an Empty LazyList!!!")
   }

/* // not used

   #[inline]
   pub fn tail<'b>(&'b self) -> &'b Lazy<'a, LazyList<'a, T>> {
       if let Cons(_, ref rlln) = *self { return &*rlln }
       panic!("LazyList::tail called on an Empty LazyList!!!")
   }

• /

   #[inline]
   pub fn unwrap(self) -> (T, RcLazyListNode<'a, T>) { // consumes the object
       if let Cons(hd, rlln) = self { return (hd, rlln) }
       panic!("LazyList::unwrap called on an Empty LazyList!!!")
   }

}

impl<'a, T: 'a + Clone> Iterator for LazyList<'a, T> {

   type Item = T;
   #[inline]
   fn next(&mut self) -> Option<Self::Item> {
       if let Empty = *self { return None }
       let oldll = mem::replace(self, Empty);
       let (hd, rlln) = oldll.unwrap();
       let mut newll = rlln.value().clone();
       // self now contains tail, newll contains the Empty
       mem::swap(self, &mut newll);
       Some(hd)
   }

}

// implements worker wrapper recursion closures using shared RcMFn variable...

type RcMFn<'a, T> = Rc<UnsafeCell<Box<dyn FnMut(T) -> T + 'a>>>;

// #[derive(Clone)] // struct RcMFn<'a, T: 'a>(Rc<UnsafeCell<Box<FnMut() -> T + 'a>>>);

trait RcMFnMethods<'a, T> {

   fn create<F: FnMut(T) -> T + 'a>(v: F) -> RcMFn<'a, T>;
   fn invoke(&self, v: T) -> T;
   fn set<F: FnMut(T) -> T + 'a>(&self, v: F);

}

impl<'a, T: 'a> RcMFnMethods<'a, T> for RcMFn<'a, T> {

   // creates new value wrapper...
   fn create<F: FnMut(T) -> T + 'a>(v: F) -> RcMFn<'a, T> {
       Rc::new(UnsafeCell::new(Box::new(v)))
   }
   #[inline(always)] // needs to be faster to be worth it
   fn invoke(&self, v: T) -> T {
       unsafe { (*(*(*self).get()))(v) }
   }
   fn set<F: FnMut(T) -> T + 'a>(&self, v: F) {
       unsafe { *self.get() = Box::new(v); }
   }

}

type RcMVar<T> = Rc<RefCell<T>>;

trait RcMVarMethods<T> {

   fn create(v: T) -> Self;
   fn get(self: &Self) -> T;
   fn set(self: &Self, v: T);

}

impl<T: Clone> RcMVarMethods<T> for RcMVar<T> {

   fn create(v: T) -> RcMVar<T> { // creates new value wrapped in RcMVar
       Rc::new(RefCell::new(v))
   }
   #[inline]
   fn get(&self) -> T {
       self.borrow().clone()
   }
   fn set(&self, v: T) {
       *self.borrow_mut() = v;
   }

}

// finally what the task objective requires...

fn hammings() -> Box<dyn Iterator<Item = Rc<BigUint>>> {

   type LL<'a> = LazyList<'a, Rc<BigUint>>;
   fn merge<'a>(x: LL<'a>, y: LL<'a>) -> LL<'a> {
       let lte = { x.head() <= y.head() }; // private context for borrow
       if lte {
           let (hdx, tlx) = x.unwrap();
           LL::cons(hdx, move || merge(tlx.value().clone(), y))
       } else {
           let (hdy, tly) = y.unwrap();
           LL::cons(hdy, move || merge(x, tly.value().clone()))
       }
   }
   fn smult<'a>(m: BigUint, s: LL<'a>) -> LL<'a> { // like map m * but faster
       let smlt = RcMFn::create(move |ss: LL<'a>| ss);
       let csmlt = smlt.clone();
       smlt.set(move |ss: LL<'a>| {
           let (hd, tl) = ss.unwrap();
           let ccsmlt = csmlt.clone();
           LL::cons(Rc::new(&m * &*hd),
                            move || ccsmlt.invoke(tl.value().clone()))
       });
       smlt.invoke(s)
   }
   fn u<'a>(s: LL<'a>, n: usize) -> LL<'a> {
       let nb = BigUint::from(n);
       let rslt = RcMVar::create(Empty);
       let crslt = rslt.clone(); // same interior data...
       let cll = LL::cons(Rc::new(BigUint::from(1u8)),
                          move || crslt.get()); // gets future value
       // below sets future value for above closure...
       rslt.set(if let Empty =
           s { smult(nb, cll) } else { merge(s, smult(nb, cll)) });
       rslt.get()
   }
   fn rll<'a>() -> LL<'a> { [5, 3, 2].iter()
                               .fold(Empty, |ll, n| u(ll, *n) ) }
   let hmng = LL::cons(Rc::new(BigUint::from(1u8)), move || rll());
   Box::new(hmng.into_iter())

}

// and the required test outputs...

fn main() {

   print!("[");
   for (i, h) in hammings().take(20).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", h)
   }
   println!(" ]");

   println!("{}", hammings().take(1691).last().unwrap());

   let strt = Instant::now();

   let rslt = hammings().take(1000000).last().unwrap();

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;

   println!("{}", rslt);

   println!("This last took {} milliseconds.", dur);

}</lang> As can be seen, there is little code necessary for the "hammings" and "main" functions if the rest were available in libraries, as they really should be.

Output:

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 172 milliseconds.

In order to run this fast. the BigUint LazyList values are wrapped in a reference counted heap wrapper to make it more efficient for cloning operations as necessary to extract interior values from the nested RcLazyListNode structure.

This is reasonably fast, with it a little slower than some languages, but a fairly high percentage of the time is spent on LazyList processing. This is likely due to the many small heap allocations and de-allocations required as well as the time required to process all of the reference counting. At that, on the same machine (Intel Sky Lake i5-6500 @ 3.6 Gigahertz - turbo when single-threaded as here), it is still about eight times faster than F# running the same functional algorithm, although much more "wordy" as also much more "wordy" than [the Haskell code from which it was translated](https://rosettacode.org/wiki/Hamming_numbers#Avoiding_generation_of_duplicates). However, it is just a little slower than Java JVM based languages (Scala, Kotlin, Clojure, etc.) and about twice slower than Haskell, likely due to those languages having very efficient memory management using memory pools for frequent small-byte-sizes per allocation/collection as for such functional algorithms, and as well not requiring reference counting due to garbage collection (although sometimes this is about a wash, as garbage collection adds its own overheads).

So Rust can be used to implement purely functional algorithms, but it isn't the best at it especially as to conciseness of code.

The other (and likely biggest) wart with implementing such functional algorithms in Rust as here is that when there are cyclic references as here, then the reference counting memory management can't handle automatic reclaiming of memory so as to produce a memory leak, which the above code has; As there is no easy way (or perhaps no way) to demote/downgrade those references to being "weak" references for this algorithm, one likely wouldn't be able to use the above method in "production" code and would have to revert to a more imperative algorithm. The memory leaks don't matter for the above code, which runs and exits taking the leaks away on program termination, it would be a problem if using in a library that would be called from a running applications many many times.

Functional sequence version avoiding duplicates, increasing speed using logarithms

Although we can't eliminate the memory leak of the ahove code, we can increase the speed by eliminating the many BigUint calculations and also reduce the memory used (and thus leaked) by using a LogRep structure instead of the variable length container where the contained BigUint gets constantly bigger with increasing range as per the following code:

Works with: Rust 1.53.0

<lang rust>extern crate num; use num::bigint::BigUint;

use core::cmp::Ordering;

use std::rc::Rc; use std::cell::{UnsafeCell, RefCell}; use std::mem;

use std::time::Instant;

// implementation of Thunk closure here...

pub struct Thunk<'a, R>(Box<dyn FnOnce() -> R + 'a>);

impl<'a, R: 'a> Thunk<'a, R> {

   #[inline(always)]
   fn new<F: 'a + FnOnce() -> R>(func: F) -> Thunk<'a, R> {
       Thunk(Box::new(func))
   }
   #[inline(always)]
   fn invoke(self) -> R { self.0() }

}

// actual Lazy implementation starts here...

use self::LazyState::*;

pub struct Lazy<'a, T: 'a>(UnsafeCell<LazyState<'a, T>>);

enum LazyState<'a, T: 'a> {

   Unevaluated(Thunk<'a, T>),
   EvaluationInProgress,
   Evaluated(T)

}

impl<'a, T: 'a> Lazy<'a, T>{

   #[inline]
   pub fn new<'b, F>(thunk: F) -> Lazy<'b, T>
           where F: 'b + FnOnce() -> T {
       Lazy(UnsafeCell::new(Unevaluated(Thunk::new(thunk))))
   }
   #[inline]
   pub fn evaluated(val: T) -> Lazy<'a, T> {
       Lazy(UnsafeCell::new(Evaluated(val)))
   }
   #[inline]
   fn force<'b>(&'b self) { // not thread-safe
       unsafe {
           match *self.0.get() {
               Evaluated(_) => return, // nothing required; already Evaluated
               EvaluationInProgress =>
                   panic!("Lazy::force called recursively!!!"),
               _ => () // need to do following something else if Unevaluated...
           } // following eliminates recursive race; drops neither on replace:
           match mem::replace(&mut *self.0.get(), EvaluationInProgress) {
               Unevaluated(thnk) => { // Thunk can't call force on same Lazy
                   *self.0.get() = Evaluated(thnk.invoke());
               },
               _ => unreachable!() // already took care of other cases above.
           }
       }
   }
   #[inline]
   pub fn value<'b>(&'b self) -> &'b T {
       self.force(); // evaluatate if not evealutated
       match unsafe { &*self.0.get() } {
           &Evaluated(ref v) => v, // return value
           _ => { unreachable!() } // previous force guarantees Evaluated
       }
   }
   #[inline] // consumes the object to produce the value
   pub fn unwrap<'b>(self) -> T where T: 'b {
       self.force(); // evaluatate if not evealutated
       match { self.0.into_inner() } {
           Evaluated(v) => v,
           _ => unreachable!() // previous code guarantees Evaluated
       }
   }

}

// now for immutable persistent shareable (memoized) LazyList via Lazy above...

type RcLazyListNode<'a, T> = Rc<Lazy<'a, LazyList<'a, T>>>;

use self::LazyList::*;

1. [derive(Clone)]

enum LazyList<'a, T: 'a + Clone> {

   /// The Empty List
   Empty,
   /// A list with one member and possibly another list.
   Cons(T, RcLazyListNode<'a, T>)

}

impl<'a, T: 'a + Clone> LazyList<'a, T> {

   #[inline]
   pub fn cons<F>(v: T, cntf: F) -> LazyList<'a, T>
           where F: 'a + FnOnce() -> LazyList<'a, T> {
       Cons(v, Rc::new(Lazy::new(cntf)))
   }
   #[inline]
   pub fn head<'b>(&'b self) -> &'b T {
       if let Cons(ref hd, _) = *self { return hd }
       panic!("LazyList::head called on an Empty LazyList!!!")
   }
   #[inline]
   pub fn unwrap(self) -> (T, RcLazyListNode<'a, T>) { // consumes the object
       if let Cons(hd, rlln) = self { return (hd, rlln) }
       panic!("LazyList::unwrap called on an Empty LazyList!!!")
   }

}

impl<'a, T: 'a + Clone> Iterator for LazyList<'a, T> {

   type Item = T;
   #[inline]
   fn next(&mut self) -> Option<Self::Item> {
       if let Empty = *self { return None }
       let oldll = mem::replace(self, Empty);
       let (hd, rlln) = oldll.unwrap();
       let mut newll = rlln.value().clone();
       // self now contains tail, newll contains the Empty
       mem::swap(self, &mut newll);
       Some(hd)
   }

}

// implements worker wrapper recursion closures using shared RcMFn variable...

type RcMFn<'a, T> = Rc<UnsafeCell<Box<dyn FnMut(T) -> T + 'a>>>;

trait RcMFnMethods<'a, T> {

   fn create<F: FnMut(T) -> T + 'a>(v: F) -> RcMFn<'a, T>;
   fn invoke(&self, v: T) -> T;
   fn set<F: FnMut(T) -> T + 'a>(&self, v: F);

}

impl<'a, T: 'a> RcMFnMethods<'a, T> for RcMFn<'a, T> {

   // creates new value wrapper...
   fn create<F: FnMut(T) -> T + 'a>(v: F) -> RcMFn<'a, T> {
       Rc::new(UnsafeCell::new(Box::new(v)))
   }
   #[inline(always)] // needs to be faster to be worth it
   fn invoke(&self, v: T) -> T {
       unsafe { (*(*(*self).get()))(v) }
   }
   fn set<F: FnMut(T) -> T + 'a>(&self, v: F) {
       unsafe { *self.get() = Box::new(v); }
   }

}

type RcMVar<T> = Rc<RefCell<T>>;

trait RcMVarMethods<T> {

   fn create(v: T) -> Self;
   fn get(self: &Self) -> T;
   fn set(self: &Self, v: T);

}

impl<T: Clone> RcMVarMethods<T> for RcMVar<T> {

   fn create(v: T) -> RcMVar<T> { // creates new value wrapped in RcMVar
       Rc::new(RefCell::new(v))
   }
   #[inline]
   fn get(&self) -> T {
       self.borrow().clone()
   }
   fn set(&self, v: T) {
       *self.borrow_mut() = v;
   }

}

// finally what the task objective requires...

1. [derive(Clone)]

struct LogRep {lg: f64, x2: u32, x3: u32, x5: u32}

const ONE: LogRep = LogRep { lg: 0f64, x2: 0u32, x3: 0u32, x5: 0u32 }; const LB3: f64 = 1.5849625007211563f64; // log base two of 3f64 const LB5: f64 = 2.321928094887362f64; // log base two of 5f64

impl PartialEq for LogRep {

   #[inline]
   fn eq(&self, other: &Self) -> bool {
       self.lg == other.lg
   }

}

impl Eq for LogRep {}

impl PartialOrd for LogRep {

   #[inline]
   fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
       self.lg.partial_cmp(&other.lg)
   }

}

trait LogRepMults {

   fn mult2(lr: LogRep) -> LogRep;
   fn mult3(lr: LogRep) -> LogRep;
   fn mult5(lr: LogRep) -> LogRep;

}

impl LogRepMults for LogRep {

   #[inline]
   fn mult2(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + 1f64, x2: lr.x2 + 1, x3: lr.x3, x5: lr.x5 }
   }
   #[inline]
   fn mult3(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + LB3, x2: lr.x2, x3: lr.x3 + 1, x5: lr.x5 }
   }
   #[inline]
   fn mult5(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + LB5, x2: lr.x2, x3: lr.x3, x5: lr.x5 + 1 }
   }

}

fn logrep2biguint(lr: LogRep) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   fn xpnd(vm: u32, n: BigUint) -> BigUint {
       let mut rslt = BigUint::from(1u8);
       let mut v = vm; let mut bsm = n;
       while v > 0u32 {
           if v & 1u32 != 0u32 { rslt = rslt * &bsm }
           bsm = &bsm.clone() * bsm; v = v >> 1;
       }
       rslt
   }
   xpnd(lr.x2, two) * xpnd(lr.x3, three) * xpnd(lr.x5, five)

}

fn hammings() -> Box<dyn Iterator<Item = LogRep>> {

   type LR = LogRep;
   type LL<'a> = LazyList<'a, LR>;
   fn merge<'a>(x: LL<'a>, y: LL<'a>) -> LL<'a> {
       let lte = { x.head() <= y.head() }; // private context for borrow
       if lte {
           let (hdx, tlx) = x.unwrap();
           LL::cons(hdx, move || merge(tlx.value().clone(), y))
       } else {
           let (hdy, tly) = y.unwrap();
           LL::cons(hdy, move || merge(x, tly.value().clone()))
       }
   }
   fn smult<'a>(m: fn(LogRep) -> LogRep, s: LL<'a>) -> LL<'a> { // like map m * but faster
       let smlt = RcMFn::create(move |ss: LL<'a>| ss);
       let csmlt = smlt.clone();
       smlt.set(move |ss: LL<'a>| {
           let (hd, tl) = ss.unwrap();
           let ccsmlt = csmlt.clone();
           LL::cons(m(hd), move || ccsmlt.invoke(tl.value().clone()))
       });
       smlt.invoke(s)
   }
   fn u<'a>(s: LL<'a>, f: fn(LogRep) -> LogRep) -> LL<'a> {
       let rslt = RcMVar::create(Empty);
       let crslt = rslt.clone(); // same interior data...
       let cll = LL::cons(ONE, move || crslt.get()); // gets future value
       // below sets future value for above closure...
       rslt.set(if let Empty =
           s { smult(f, cll) } else { merge(s, smult(f, cll)) });
       rslt.get()
   }
   fn rll<'a>() -> LL<'a> { [LR::mult5, LR::mult3, LR::mult2].iter()
                               .fold(Empty, |ll, mf| u(ll, *mf) ) }
   let hmng = LL::cons(ONE, move || rll());
   Box::new(hmng.into_iter())

}

// and the required test outputs...

fn main() {

   print!("[");
   for (i, h) in hammings().take(20).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", logrep2biguint(h))
   }
   println!(" ]");

   println!("{}", logrep2biguint(hammings().take(1691).last().unwrap()));

   let strt = Instant::now();

   let rslt = hammings().take(1000000).last().unwrap();

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;

   println!("{}", logrep2biguint(rslt));

   println!("This last took {} milliseconds.", dur);

}</lang>

Output:

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 122 milliseconds.

As can be seen, the above version takes about two thirds of the time as the previous version running on the same Intel Skylake i5-6500 - although it still has a memory leak, the size of the leak for a given range will be many times smaller. It still isn't as fast as Haskell running the same algorithm, but it is only about 30% slower and about as fast as most other languages that compile their code to a running executable.

Very fast sequence version using imperative code (mutable vectors) and logarithmic approximations for sorting

Most of the remaining execution time for the above version is due to the many allocations/deallocations used in implementing the functional lazy list sequence; the following code avoids that overhead by memoizing the pst values using linear vectors with the head and tail values marked by tracking indices:

Translation of: Nim

<lang rust> extern crate num; use num::bigint::BigUint;

use core::cmp::Ordering;

use std::time::Instant;

1. [derive(Clone, Copy)]

struct LogRep { lg: f64, x2: u32, x3: u32, x5: u32 }

const ONE: LogRep = LogRep { lg: 0f64, x2: 0u32, x3: 0u32, x5: 0u32 }; const LB3: f64 = 1.5849625007211563f64; // log base two of 3f64 const LB5: f64 = 2.321928094887362f64; // log base two of 5f64

impl PartialEq for LogRep {

   #[inline]
   fn eq(&self, other: &Self) -> bool {
       self.lg == other.lg
   }

}

impl Eq for LogRep {}

impl PartialOrd for LogRep {

   #[inline]
   fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
       self.lg.partial_cmp(&other.lg)
   }

}

trait LogRepMults {

   fn mult2(lr: LogRep) -> LogRep;
   fn mult3(lr: LogRep) -> LogRep;
   fn mult5(lr: LogRep) -> LogRep;

}

impl LogRepMults for LogRep {

   #[inline]
   fn mult2(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + 1f64, x2: lr.x2 + 1, x3: lr.x3, x5: lr.x5 }
   }
   #[inline]
   fn mult3(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + LB3, x2: lr.x2, x3: lr.x3 + 1, x5: lr.x5 }
   }
   #[inline]
   fn mult5(lr: LogRep) -> LogRep {
       LogRep { lg: lr.lg + LB5, x2: lr.x2, x3: lr.x3, x5: lr.x5 + 1 }
   }

}

fn logrep2biguint(lr: LogRep) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   fn xpnd(vm: u32, n: BigUint) -> BigUint {
       let mut rslt = BigUint::from(1u8);
       let mut v = vm; let mut bsm = n;
       while v > 0u32 {
           if v & 1u32 != 0u32 { rslt = rslt * &bsm }
           bsm = &bsm.clone() * bsm; v = v >> 1;
       }
       rslt
   }
   xpnd(lr.x2, two) * xpnd(lr.x3, three) * xpnd(lr.x5, five)

}

fn hammings_log() -> Box<dyn Iterator<Item = LogRep>> {

   type LR = LogRep;
   struct State {
       i: usize, j: usize, ih: usize, jh: usize,
       s532: LR, s53: LR, s5: LR, mrg: LR,
       h: Vec<LR>,
       m: Vec<LR>,
   }
   impl Iterator for State {
       type Item = LR;
       #[inline]
       fn next(&mut self) -> Option<Self::Item> {
           let cph = self.h.len();
           let cpm = self.m.len();
           if cph == 0 && cpm == 0 {
               self.h.resize(1, ONE);
               self.m.resize(1, ONE);
               Some(ONE)
           } else {
               if self.i > cph >> 1 {
                   self.h.copy_within(self.i .. self.ih, 0);
                   self.ih -= self.i; self.i = 0;
               }
               if self.ih >= cph {
                   self.h.reserve(cph);
                   unsafe { self.h.set_len(cph + cph); }
               }
               if self.s532 < self.mrg {
                   let nv = self.s532;
                   unsafe { *self.h.get_unchecked_mut(self.ih) = nv; }

                   self.s532 = unsafe {
                       LR::mult2(self.h.get_unchecked(self.i).clone()) };
                   self.ih += 1; self.i += 1; Some(nv)
               } else {
                   if self.j > cpm >> 1 {
                       self.m.copy_within(self.j .. self.jh, 0);
                       self.jh -= self.j; self.j = 0;
                   }
                   if self.jh >= cpm {
                       self.m.reserve(cpm);
                       unsafe { self.m.set_len(cpm + cpm); }
                   }
                   let nv = self.mrg;
                   unsafe { *self.h.get_unchecked_mut(self.ih) = nv; }
                   if self.s53 < self.s5 {
                       self.mrg = self.s53;
                       self.s53 = unsafe {
                           LR::mult3(self.m.get_unchecked(self.j).clone()) };
                       self.j += 1;
                   } else {
                       self.mrg = self.s5;
                       self.s5 = LR::mult5(self.s5);
                   }
                   unsafe { *self.m.get_unchecked_mut(self.jh) = self.mrg; }
                   self.ih += 1; self.jh += 1; Some(nv)
               }
           }
       }
   }
   let mut stt = State {
       i: 0, j: 0, ih: 0, jh: 0,
       s532: LR::mult2(ONE),
       s53: LR::mult3(LR::mult3(ONE)),
       s5: LR::mult5(ONE),
       mrg: LR::mult3(ONE),
       h: Vec::with_capacity(0), m: Vec::with_capacity(0) };
   Box::new(stt.into_iter())

}

// and the required test outputs...

fn main() {

   print!("[");
   for (i, h) in hammings_log().take(20).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", logrep2biguint(h))
   }
   println!(" ]");

   println!("{}", logrep2biguint(hammings_log().take(1691).last().unwrap()));

   let strt = Instant::now();

   let rslt = hammings_log().take(1000000).last().unwrap();

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;

   println!("{}", logrep2biguint(rslt));

   println!("This last took {} milliseconds.", dur);

} </lang>

Output:

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 10 milliseconds.

The code above is over ten times faster than the previous functional lazy list sequence code due to not losing the time for the many small allocations/deallocations of small heap (reference counted) objects and not having recursive references, also does not leak memory.

Extremely fast non-sequence version by calculation of top band of Hamming numbers

One might ask "What could possibly be done to further speed up finding Hamming numbers?": the answer is quite a lot, but one has to dump the ability to iterate a sequence as that depends on being able to refer to past calculated values by back pointers to the memorized O(n^(2/3)) arrays or lists and thus quite large amounts of memory. If one just wants to find very large Hamming numbers individually, one looks to the [mathematical analysis of Hamming/regular numbers on Wikipedia](https://en.wikipedia.org/wiki/Regular_number) and finds there is quite an exact relationship between 'n', the sequence number, and the logarithmic magnitude of the resulting Hamming number, and that the error term is directly proportional to the logarithm of that output number. This means that only the band of Hamming values as wide of this error and including the estimated value need to be generated, and that we need only iterate over two of the three prime exponents, thus O(n^(2/3)) time complexity and O(n^(1/3)) space complexity. The following code was adapted from [an article in DDJ](http://www.drdobbs.com/architecture-and-design/hamming-problem/228700538) and the Haskell code with the further refinements to decrease the memory requirements as described above:

Translation of: Haskell

<lang rust>extern crate num; // requires dependency on the num library use num::bigint::BigUint;

use std::time::Instant;

fn nth_hamming(n: u64) -> (u32, u32, u32) {

   if n < 2 {
       if n <= 0 { panic!("nth_hamming: argument is zero; no elements") }
       return (0, 0, 0) // trivial case for n == 1
   }

   let lg3 = 3.0f64.ln() / 2.0f64.ln(); // log base 2 of 3
   let lg5 = 5.0f64.ln() / 2.0f64.ln(); // log base 2 of 5
   let fctr = 6.0f64 * lg3 * lg5;
   let crctn = 30.0f64.sqrt().ln() / 2.0f64.ln(); // log base 2 of sqrt 30
   let lgest = (fctr * n as f64).powf(1.0f64/3.0f64)
                   - crctn; // from WP formula
   let frctn = if n < 1000000000 { 0.509f64 } else { 0.105f64 };
   let lghi = (fctr * (n as f64 + frctn * lgest)).powf(1.0f64/3.0f64)
                   - crctn; // calculate hi log limit based on log(N) - WP article
   let lglo = 2.0f64 * lgest - lghi; // and a lower limit of the upper "band"
   let mut count = 0; // need to use extended precision, might go over
   let mut bnd = Vec::with_capacity(0);
   let klmt = (lghi / lg5) as u32 + 1;
   for k in 0 .. klmt { // i, j, k values can be just u32 values
       let p = k as f64 * lg5;
       let jlmt = ((lghi - p) / lg3) as u32 + 1;
       for j in 0 .. jlmt {
           let q = p + j as f64 * lg3;
           let ir = lghi - q;
           let lg = q + (ir as u32) as f64; // current log value (estimated)
           count += ir as u64 + 1;
           if lg >= lglo {
               bnd.push((lg, (ir as u32, j, k)))
           }
       }
   }
   if n > count { panic!("nth_hamming: band high estimate is too low!") };
   let ndx = (count - n) as usize;
   if ndx >= bnd.len() { panic!("nth_hamming: band low estimate is too high!") };
   bnd.sort_by(|a, b| b.0.partial_cmp(&a.0).unwrap()); // sort decreasing order

   bnd[ndx].1

}

fn convert_log2big(o: (u32, u32, u32)) -> BigUint {

   let two = BigUint::from(2u8);
   let three = BigUint::from(3u8);
   let five = BigUint::from(5u8);
   let (x2, x3, x5) = o;
   let mut ob = BigUint::from(1u8); // convert to BigUint at the end
   for _ in 0 .. x2 { ob = ob * &two }
   for _ in 0 .. x3 { ob = ob * &three }
   for _ in 0 .. x5 { ob = ob * &five }
   ob

}

fn main() {

   print!("[");
   for (i, h) in (1 .. 21).map(nth_hamming).enumerate() {
       if i != 0 { print!(",") }
       print!(" {}", convert_log2big(h))
   }
   println!(" ]");
   println!("{}", convert_log2big(nth_hamming(1691)));

   let strt = Instant::now();

   let rslt = nth_hamming(1000000);

   let elpsd = strt.elapsed();
   let secs = elpsd.as_secs();
   let millis = (elpsd.subsec_nanos() / 1000000)as u64;
   let dur = secs * 1000 + millis;
   
   println!("2^{} times 3^{} times 5^{}", rslt.0, rslt.1, rslt.2);
   let rs = convert_log2big(rslt).to_str_radix(10);
   let mut s = rs.as_str();
   println!("{} digits:", s.len());
   let lg3 = 3.0f64.log2();
   let lg5 = 5.0f64.log2();
   let lg = (rslt.0 as f64 + rslt.1 as f64 * lg3
           + rslt.2 as f64 * lg5) * 2.0f64.log10();
   println!("Approximately {}E+{}", 10.0f64.powf(lg.fract()), lg.trunc());
   if s.len() <= 10000 {
       while s.len() > 100 {
           let (f, r) = s.split_at(100);
           s = r;
           println!("{}", f);
       }
       println!("{}", s);
   }

   println!("This last took {} milliseconds.", dur);

}</lang>

[ 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ]
2125764000
2^55 times 3^47 times 5^64
84 digits:
Approximately 5.193127804483804E+83
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This last took 0 milliseconds.

The above code takes too little time to calculate the millionth Hamming numbers to be measured (as seen above), calculates the billionth number in under 10 milliseconds, calculates the trillionth in less than a second, and the thousand trillionth (10^15) in just over a minute (72 seconds). However, the program needs to be tuned for correctness for ranges of about the 100 trillionth value and above as the precision of the log approximation is not sufficient above about that level to maintain the proper sort order, and thus the answers will start to be out by one value or more. The answers are likely correct up to that point as they are the same to a trillion as the equivalent Haskell program, although this version is much faster due to no garbage collection (the Haskell version spends about half its time garbage collecting) and doing the calculations using loops and array/vector accesses rather than the lazy list processing used in the Haskell version. The program should be able to determine the 10^19th hamming number in a few hours and can't quite find the 2^64th (18446744073709551615th) Hamming number due to a slight overflow near the limit.

The above code uses the library vector sort capabilities; custom sorting versions could be written but with the reduced array size, sorting is a very small percentage of the execution time and maximum space requirements are only a few 10's of Megabytes so that neither the time nor the space used for sorting are a concern.

Note that I'm not knocking Haskell, just that (as here) many Haskell programmers like to use lazy list processing which has its costs; the Haskell version could be re-written to use arrays and functional loops and likely be about the same speed although perhaps not as concise. By simply converting the Haskell program to force strictness and to use this same method of determining the width of the upper band, the Haskell program would have the same time and space complexity as here, but would still be a constant factor of almost eight times slower due to the list processing (with a constant factor for extra space as well). Use of a mutable array or vector would solve that, but unfortunately not as easily as here as there would be the question of "unboxed" versus "boxed" arrays/vectors, and the complexities of implementing the (faster) unboxed type in which to sort the band - in short, not as easy as here in Rust.

Scala

<lang scala>class Hamming extends Iterator[BigInt] {

 import scala.collection.mutable.Queue
 val qs = Seq.fill(3)(new Queue[BigInt])
 def enqueue(n: BigInt) = qs zip Seq(2, 3, 5) foreach { case (q, m) => q enqueue n * m }
 def next = {
   val n = qs map (_.head) min;
   qs foreach { q => if (q.head == n) q.dequeue }
   enqueue(n)
   n
 }
 def hasNext = true
 qs foreach (_ enqueue 1)

}</lang> However, the usage of closures adds a significant amount of time. The code below, though a bit uglier because of the repetitions, is twice as fast: <lang scala>class Hamming extends Iterator[BigInt] {

 import scala.collection.mutable.Queue
 val q2 = new Queue[BigInt]
 val q3 = new Queue[BigInt]
 val q5 = new Queue[BigInt]
 def enqueue(n: BigInt) = {
   q2 enqueue n * 2
   q3 enqueue n * 3
   q5 enqueue n * 5
 }
 def next = {
   val n = q2.head min q3.head min q5.head
   if (q2.head == n) q2.dequeue
   if (q3.head == n) q3.dequeue
   if (q5.head == n) q5.dequeue
   enqueue(n)
   n
 }
 def hasNext = true
 List(q2, q3, q5) foreach (_ enqueue 1)

}</lang> Usage:

scala> new Hamming take 20 toList
res87: List[BigInt] = List(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36)

scala> new Hamming drop 1690 next
res88: BigInt = 2125764000

scala> new Hamming drop 999999 next
res89: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

There's also a fairly mechanical translation from Haskell using purely functional lazy streams

Translation of: Haskell

<lang scala>val hamming : Stream[BigInt] = {

  def merge(inx : Stream[BigInt], iny : Stream[BigInt]) : Stream[BigInt] = {
     if (inx.head < iny.head) inx.head #:: merge(inx.tail, iny) else 
     if (iny.head < inx.head) iny.head #:: merge(inx, iny.tail) else
     merge(inx, iny.tail)
  }

  1 #:: merge(hamming map (_ * 2), merge(hamming map (_ * 3), hamming map (_ * 5)))

}</lang> Use of "force" ensures that the stream is computed before being printed, otherwise it would just be left suspended and you'd see "Stream(1, ?)"

scala> (hamming take 20).force
res0: scala.collection.immutable.Stream[BigInt] = Stream(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36)

To get the nth code find the n-1th element because indexes are 0 based

scala> hamming(1690)
res1: BigInt = 2125764000

To calculate the 1000000th code I had to increase the JVM heap from the default

scala> hamming(999999)
res2: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Translation of Haskell code avoiding duplicates

One can fix the problems of the memory use of the above code resulting from the entire stream being held in memory due to the use a "val hamming: Stream[BigInt]" by using a function "def hamming(): Stream[BigInt]" and making temporary local variables for intermediate streams so that the beginnings of the streams are garbage collected as the output stream is consumed; one can also implement the other Haskell algorithm to avoid factor duplication by building each stream on successive streams, again with memory conserved by building the least dense first: <lang scala> def hamming(): Stream[BigInt] = {

   def merge(a: Stream[BigInt], b: Stream[BigInt]): Stream[BigInt] = {
     if (a.isEmpty) b else {
       val av = a.head; val bv = b.head
       if (av < bv) av #:: merge(a.tail, b)
       else bv #:: merge(a, b.tail) } }
   def smult(m:Int, s: Stream[BigInt]): Stream[BigInt] =
     (m * s.head) #:: smult(m, s.tail) // equiv to map (m *) s; faster
   def u(s: Stream[BigInt], n: Int): Stream[BigInt] = {
     lazy val r: Stream[BigInt] = merge(s, smult(n, 1 #:: r))
     r }
   1 #:: List(5, 3, 2).foldLeft(Stream.empty[BigInt]) { u } }</lang>

Usage:

scala> hamming() take 20 force
res0: scala.collection.immutable.Stream[BigInt] = Stream(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36)

scala> hamming() drop 1690 head
res1: BigInt = 2125764000

scala> hamming() drop 999999 head
res2: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

It only takes under a half second to find the millionth number in the sequence in the last output.

Scheme

<lang scheme>(define-syntax lons

 (syntax-rules ()
   ((_ lar ldr) (delay (cons lar (delay ldr))))))

(define (lar lons)

 (car (force lons)))

(define (ldr lons)

 (force (cdr (force lons))))

(define (lap proc . llists)

 (lons (apply proc (map lar llists)) (apply lap proc (map ldr llists))))

(define (take n llist)

 (if (zero? n)
     (list)
     (cons (lar llist) (take (- n 1) (ldr llist)))))

(define (llist-ref n llist)

 (if (= n 1)
     (lar llist)
     (llist-ref (- n 1) (ldr llist))))

(define (merge llist-1 . llists)

 (define (merge-2 llist-1 llist-2)
   (cond ((null? llist-1) llist-2)
         ((null? llist-2) llist-1)
         ((< (lar llist-1) (lar llist-2))
          (lons (lar llist-1) (merge-2 (ldr llist-1) llist-2)))
         ((> (lar llist-1) (lar llist-2))
          (lons (lar llist-2) (merge-2 llist-1 (ldr llist-2))))
         (else (lons (lar llist-1) (merge-2 (ldr llist-1) (ldr llist-2))))))
 (if (null? llists)
     llist-1
     (apply merge (cons (merge-2 llist-1 (car llists)) (cdr llists)))))

(define hamming

 (lons 1
       (merge (lap (lambda (x) (* x 2)) hamming)
              (lap (lambda (x) (* x 3)) hamming)
              (lap (lambda (x) (* x 5)) hamming))))

(display (take 20 hamming)) (newline) (display (llist-ref 1691 hamming)) (newline) (display (llist-ref 1000000 hamming)) (newline)</lang>

Output:

(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
out of memory

Avoiding Generation of Duplicates, including reduced memory use

Translation of: Haskell

Although the algorithm above is true to the classic Dijkstra version and although the algorithm does require a form of lazy list/stream processing in order to utilize memoization and avoid repeated recalculations/comparisons, the stream implementation can be simplified, and the modified algorithm as per the Haskell code avoids duplicate generations of factors. As well, the following code implements the algorithm as a procedure/function so that it restarts the calculation from the beginning on every new call and so that internal stream variables are not top level so that the garbage collector can collect the beginning of all intermediate and final streams when they are no longer referenced; in this way total memory used (after interspersed garbage collections) is almost zero for a sequence of the first million numbers. Note that Scheme R5RS does not define "map" or "foldl" functions, so these are provided (a simplified "smult" which is faster than using map for this one purpose): <lang scheme>(define (hamming)

 (define (foldl f z l)
   (define (foldls zs ls)
     (if (null? ls) zs (foldls (f zs (car ls)) (cdr ls))))
   (foldls z l))
 (define (merge a b)
   (if (null? a) b
     (let ((x (car a)) (y (car b)))
       (if (< x y) (cons x (delay (merge (force (cdr a)) b)))
                   (cons y (delay (merge a (force (cdr b)))))))))
 (define (smult m s) (cons (* m (car s)) ;; equiv to map (* m) s; faster
            (delay (smult m (force (cdr s))))))
 (define (u s n) (letrec ((a (merge s (smult n (cons 1 (delay a)))))) a))
 (cons 1 (delay (foldl u '() '(5 3 2)))))

test...

(define (stream-take->list n strm)

 (if (= n 0) (list) (cons (car strm) 
   (stream-take->list (- n 1) (force (cdr strm))))))

(define (stream-ref strm nth)

 (do ((nxt strm (force (cdr nxt))) (cnt 0 (+ cnt 1)))
     ((>= cnt nth) (car nxt))))

(display (stream-take->list 20 (hamming))) (newline) (display (stream-ref (hamming) (- 1691 1))) (newline) (display (stream-ref (hamming) (- 1000000 1))) (newline)</lang>

Output:

{1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36}
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

The "stream-ref" procedure is zero based as is the Scheme standard for array indices, thus the subtraction of one from the desired nth number in the sequence.

Seed7

<lang seed7>$ include "seed7_05.s7i";

 include "bigint.s7i";

const func bigInteger: min (in bigInteger: a, in bigInteger: b, in bigInteger: c) is func

 result
   var bigInteger: min is 0_;
 begin
   if a < b then
     min := a;
   else
     min := b;
   end if;
   if c < min then
     min := c;
   end if;
 end func;

const func bigInteger: hamming (in integer: n) is func

 result
   var bigInteger: hammingNum is 1_;
 local
   var array bigInteger: hammingNums is 0 times 0_;
   var integer: index is 0;
   var bigInteger: x2 is 2_;
   var bigInteger: x3 is 3_;
   var bigInteger: x5 is 5_;
   var integer: i is 1;
   var integer: j is 1;
   var integer: k is 1;
 begin
   hammingNums := n times 1_;
   for index range 2 to n do
     hammingNum := min(x2, x3, x5);
     hammingNums[index] := hammingNum;
     if x2 = hammingNum then
       incr(i);
       x2 := 2_ * hammingNums[i];
     end if;
     if x3 = hammingNum then
       incr(j);
       x3 := 3_ * hammingNums[j];
     end if;
     if x5 = hammingNum then
       incr(k);
       x5 := 5_ * hammingNums[k];
     end if;
   end for;
 end func;

const proc: main is func

 local
   var integer: n is 0;
 begin
   for n range 1 to 20 do
     write(hamming(n) <& " ");
   end for;
   writeln;
   writeln(hamming(1691));
   writeln(hamming(1000000));
 end func;</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Sidef

<lang ruby>func ham_gen {

   var s = [[1], [1], [1]]
   var m = [2, 3, 5]

 

   func {
       var n = [s[0][0], s[1][0], s[2][0]].min
       { |i|
           s[i].shift if (s[i][0] == n)
           s[i].append(n * m[i])
       } << ^3
       return n
   }

}

var h = ham_gen()

var i = 20; say i.of { h() }.join(' ')

{ h() } << (i+1 ..^ 1691) say h()</lang>

Output:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

Smalltalk

Works with: GNU Smalltalk

This is a straightforward implementation of the pseudocode snippet found in the Python section. Smalltalk supports arbitrary-precision integers, but the implementation is too slow to try it with 1 million. <lang smalltalk>Object subclass: Hammer [

 Hammer class >> hammingNumbers: howMany [
   |h i j k x2 x3 x5| 
     h := OrderedCollection new.
     i := 0. j := 0. k := 0.
     h add: 1.
     x2 := 2. x3 := 2. x5 := 5.
     [ ( h size) < howMany ] whileTrue: [
       |m|
       m := { x2. x3. x5 } sort first.
       (( h indexOf: m ) = 0) ifTrue: [ h add: m ].
       ( x2 = (h last) ) ifTrue: [ i := i + 1. x2 := 2 * (h at: i) ].
       ( x3 = (h last) ) ifTrue: [ j := j + 1. x3 := 3 * (h at: j) ].
       ( x5 = (h last) ) ifTrue: [ k := k + 1. x5 := 5 * (h at: k) ]. 
     ].
     ^ h sort
 ]

].

(Hammer hammingNumbers: 20) displayNl. (Hammer hammingNumbers: 1690) last displayNl.</lang>

Works with: Pharo Smalltalk

<lang smalltalk> limit := 10 raisedToInteger: 84. tape := Set new.

hammingProcess := [:newHamming| (newHamming <= limit) ifTrue: [| index | index := tape scanFor: newHamming. (tape array at: index) ifNil: [tape atNewIndex: index put: newHamming asSetElement. hammingProcess value: newHamming * 2. hammingProcess value: newHamming * 3. hammingProcess value: newHamming * 5]]].

hammingProcess value: 1.

sc := tape asSortedCollection. sc first: 20. "a SortedCollection(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)" sc at: 1691. "2125764000" sc at: 1000000. "519312780448388736089589843750000000000000000000000000000000000000000000000000000000" </lang>

Works with: Squeak Smalltalk with Xtream package (and probably on Pharo too)

This is using the Xtreams package (see http://www.squeaksource.com/Xtreams.html) The tape is a Heap of associations, the key is a hamming number, the value is its greatest prime factor. Associations responds to <, so can be used in Heap, and are sorted by key. The stream can only move forward, for economy, we don't bother buffering past values. The counterpart is that we have no direct indexing and must keep the position counter by ourself. <lang smalltalk>tape := Heap with: 1 -> 1. hammingStream := [| next | next := tape removeFirst. next value <= 2 ifTrue: [tape add: next key * 2 -> 2]. next value <= 3 ifTrue: [tape add: next key * 3 -> 3]. next value <= 5 ifTrue: [tape add: next key * 5 -> 5]. next key] reading.

hammingStream read: 20. "get first 20 values => #(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)" hammingStream ++ 1670. "skip the next 1670 values" hammingStream get. "and the 1691th value is => 2125764000". hammingStream ++ (999999 - 1691). "now skip more to position at 999,999". hammingStream get. "and the 1,000,000th value is => 519312780448388736089589843750000000000000000000000000000000000000000000000000000000". tape size. "See how many we have buffered => 24904"

</lang>

SQL

This uses SQL99's "WITH RECURSIVE" (more like co-recursion) to build a table of Hamming numbers, then selects out the desired ones. With sqlite it is very fast. It doesn't try to get the millionth number because sqlite doesn't have bignums.

<lang sql> CREATE TEMPORARY TABLE factors(n INT); INSERT INTO factors VALUES(2); INSERT INTO factors VALUES(3); INSERT INTO factors VALUES(5);

CREATE TEMPORARY TABLE hamming AS

   WITH RECURSIVE ham AS (
         SELECT 1 as h
         UNION
         SELECT h*n x FROM ham JOIN factors ORDER BY x
         LIMIT 1700
       )
   SELECT h FROM ham;

sqlite> SELECT h FROM hamming ORDER BY h LIMIT 20; 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 sqlite> SELECT h FROM hamming ORDER BY h LIMIT 1 OFFSET 1690; 2125764000 </lang>

Tcl

This uses coroutines to simplify the description of what's going on.

Works with: Tcl version 8.6

<lang tcl>package require Tcl 8.6

1. Simple helper: Tcl-style list "map"

proc map {varName list script} {

   set l {}
   upvar 1 $varName v
   foreach v $list {lappend l [uplevel 1 $script]}
   return $l

}

1. The core of a coroutine to compute the product of a hamming sequence.
3. Tricky bit: we don't automatically advance to the next value, and instead
4. wait to be told that the value has been consumed (i.e., is the result of
5. the [yield] operation).

proc ham {key multiplier} {

   global hammingCache
   set i 0
   yield [info coroutine]
   # Cannot use [foreach]; that would take a snapshot of the list in
   # the hammingCache variable, so missing updates.
   while 1 {

set n [expr {[lindex $hammingCache($key) $i] * $multiplier}] # If the number selected was ours, we advance to compute the next if {[yield $n] == $n} { incr i }

   }

}

1. This coroutine computes the hamming sequence given a list of multipliers.
2. It uses the [ham] helper from above to generate indivdual multiplied
3. sequences. The key into the cache is the list of multipliers.
5. Note that it is advisable for the values to be all co-prime wrt each other.

proc hammingCore args {

   global hammingCache
   set hammingCache($args) 1
   set hammers [map x $args {coroutine ham$x,$args ham $args $x}]
   yield
   while 1 {

set n [lindex $hammingCache($args) [incr i]-1] lappend hammingCache($args) \ [tcl::mathfunc::min {*}[map h $hammers {$h $n}]] yield $n

   }

}

1. Assemble the pieces so as to compute the classic hamming sequence.

coroutine hamming hammingCore 2 3 5

1. Print the first 20 values of the sequence

for {set i 1} {$i <= 20} {incr i} {

   puts [format "hamming\[%d\] = %d" $i [hamming]]

} for {} {$i <= 1690} {incr i} {set h [hamming]} puts "hamming{1690} = $h" for {} {$i <= 1000000} {incr i} {set h [hamming]} puts "hamming{1000000} = $h"</lang>

Output:

hamming{1} = 1
hamming{2} = 2
hamming{3} = 3
hamming{4} = 4
hamming{5} = 5
hamming{6} = 6
hamming{7} = 8
hamming{8} = 9
hamming{9} = 10
hamming{10} = 12
hamming{11} = 15
hamming{12} = 16
hamming{13} = 18
hamming{14} = 20
hamming{15} = 24
hamming{16} = 25
hamming{17} = 27
hamming{18} = 30
hamming{19} = 32
hamming{20} = 36
hamming{1690} = 2123366400
hamming{1000000} = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

A faster version can be built that also works on Tcl 8.5 (or earlier, if only small hamming numbers are being computed): <lang tcl>variable hamming 1 hi2 0 hi3 0 hi5 0 proc hamming {n} {

   global hamming hi2 hi3 hi5
   set h2 [expr {[lindex $hamming $hi2]*2}]
   set h3 [expr {[lindex $hamming $hi3]*3}]
   set h5 [expr {[lindex $hamming $hi5]*5}]
   while {[llength $hamming] < $n} {

lappend hamming [set h [expr { $h2<$h3  ? $h2<$h5 ? $h2 : $h5  : $h3<$h5 ? $h3 : $h5 }]] if {$h==$h2} { set h2 [expr {[lindex $hamming [incr hi2]]*2}] } if {$h==$h3} { set h3 [expr {[lindex $hamming [incr hi3]]*3}] } if {$h==$h5} { set h5 [expr {[lindex $hamming [incr hi5]]*5}] }

   }
   return [lindex $hamming [expr {$n - 1}]]

}

1. Print the first 20 values of the sequence

for {set i 1} {$i <= 20} {incr i} {

   puts [format "hamming\[%d\] = %d" $i [hamming $i]]

} puts "hamming{1690} = [hamming 1690]" puts "hamming{1691} = [hamming 1691]" puts "hamming{1692} = [hamming 1692]" puts "hamming{1693} = [hamming 1693]" puts "hamming{1000000} = [hamming 1000000]"</lang>

uBasic/4tH

uBasic's single array does not have the required size to calculate the 1691st number, let alone the millionth. <lang>For H = 1 To 20

 Print "H("; H; ") = "; Func (_FnHamming(H))

Next

End

_FnHamming Param (1)

 @(0) = 1

 X = 2 : Y = 3 : Z = 5
 I = 0 : J = 0 : K = 0

 For N = 1 To a@ - 1
   M = X
   If M > Y Then M = Y
   If M > Z Then M = Z
   @(N) = M

   If M = X Then I = I + 1 : X = 2 * @(I)
   If M = Y Then J = J + 1 : Y = 3 * @(J)
   If M = Z Then K = K + 1 : Z = 5 * @(K)
 Next

Return (@(a@-1))</lang>

Output:

H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36

0 OK, 0:379

UNIX Shell

Works with: ksh93

Large numbers are not supported. <lang bash>typeset -a hamming=(1) function nextHamming {

   typeset -Sa q2 q3 q5
   integer h=${hamming[${#hamming[@]}-1]}
   q2+=( $(( h*2 )) )
   q3+=( $(( h*3 )) )
   q5+=( $(( h*5 )) )
   h=$( min3 ${q2[0]} ${q3[0]} ${q5[0]} )
   (( ${q2[0]} == h )) && ashift q2 >/dev/null
   (( ${q3[0]} == h )) && ashift q3 >/dev/null
   (( ${q5[0]} == h )) && ashift q5 >/dev/null
   hamming+=($h)

}

function ashift {

   nameref ary=$1
   print -- "${ary[0]}"
   ary=( "${ary[@]:1}" )

}

function min3 {

   if (( $1 < $2 )); then
       (( $1 < $3 )) && print -- $1 || print -- $3
   else
       (( $2 < $3 )) && print -- $2 || print -- $3
   fi

}

for ((i=1; i<=20; i++)); do

   nextHamming
   printf "%d\t%d\n" $i ${hamming[i-1]}

done for ((; i<=1690; i++)); do nextHamming; done nextHamming printf "%d\t%d\n" $i ${hamming[i-1]} print "elapsed: $SECONDS"</lang>

Output:

1	1
2	2
3	3
4	4
5	5
6	6
7	8
8	9
9	10
10	12
11	15
12	16
13	18
14	20
15	24
16	25
17	27
18	30
19	32
20	36
1690	2125764000
elapsed: 0.568

Ursala

Smooth is defined as a second order function taking a list of primes and returning a function that takes a natural number $n$ to the $n$-th smooth number with respect to them. An elegant but inefficient formulation based on the J solution is the following. <lang Ursala>#import std

1. import nat

smooth"p" "n" = ~&z take/"n" nleq-< (rep(length "n") ^Ts/~& product*K0/"p") <1></lang> This test program <lang Ursala>main = smooth<2,3,5>* nrange(1,20)</lang> yields this list of the first 20 Hamming numbers.

<1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36>

Although all calculations are performed using unlimited precision, the version above is impractical for large numbers. A more hardcore approach is the following. <lang Ursala>#import std

1. import nat

smooth"p" "n" =

~&H\"p" *-<1>; @NiXS ~&/(1,1); ~&ll~="n"->lr -+

  ^\~&rlPrrn2rrm2Zlrrmz3EZYrrm2lNCTrrm2QAX*rhlPNhrnmtPA2XtCD ~&lrPrhl2E?/~&l ^|/successor@l ~&hl,
  ^|/~& nleq-<&l+ * ^\~&r ~&l|| product@rnmhPX+-

1. cast %nL

main = smooth<2,3,5>* nrange(1,20)--<1691,1000000></lang>

Output:

The great majority of time is spent calculating the millionth Hamming number.

<
   1,
   2,
   3,
   4,
   5,
   6,
   8,
   9,
   10,
   12,
   15,
   16,
   18,
   20,
   24,
   25,
   27,
   30,
   32,
   36,
   2125764000,
   519312780448388736089589843750000000000000000000000000000000000000000000000000000000>

VBA

<lang vb>'RosettaCode Hamming numbers 'This is a well known hard problem in number theory: 'counting the number of lattice points in a 'n-dimensional tetrahedron, here n=3. Public a As Double, b As Double, c As Double, d As Double Public p As Double, q As Double, r As Double Public cnt() As Integer 'stores the number of lattice points indexed on the exponents of 3 and 5 Public hn(2) As Integer 'stores the exponents of 2, 3 and 5 Public Declare Function GetTickCount Lib "kernel32.dll" () As Long Private Function log10(x As Double) As Double

   log10 = WorksheetFunction.log10(x)

End Function Private Function pow(x As Variant, y As Variant) As Double

   pow = WorksheetFunction.Power(x, y)

End Function Private Sub init(N As Long)

   'Computes a, b and c as the vertices
   '(a,0,0), (0,b,0), (0,0,c) of a tetrahedron
   'with apex (0,0,0) and volume N
   'volume N=a*b*c/6
   Dim k As Double
   k = log10(2) * log10(3) * log10(5) * 6 * N
   k = pow(k, 1 / 3)
   a = k / log10(2)
   b = k / log10(3)
   c = k / log10(5)
   p = -b * c
   q = -a * c
   r = -a * b

End Sub Private Function x_given_y_z(y As Integer, z As Integer) As Double

   x_given_y_z = -(q * y + r * z + a * b * c) / p

End Function Private Function cmp(i As Integer, j As Integer, k As Integer, gn() As Integer) As Boolean

   cmp = (i * log10(2) + j * log10(3) + k * log10(5)) > (gn(0) * log10(2) + gn(1) * log10(3) + gn(2) * log10(5))

End Function Private Function count(N As Long, step As Integer) As Long

   'Loop over y and z, compute x and
   'count number of lattice points within tetrahedron.
   'Step 1 is indirectly called by find_seed to calibrate the plane through A, B and C
   'Step 2 fills the matrix cnt with the number of lattice points given the exponents of 3 and 5
   'Step 3 the plane is lowered marginally so one or two candidates stick out
   Dim M As Long, j As Integer, k As Integer
   If step = 2 Then ReDim cnt(0 To Int(b) + 1, 0 To Int(c) + 1)
   M = 0: j = 0: k = 0
   Do While -c * j - b * k + b * c > 0
       Do While -c * j - b * k + b * c > 0
           Select Case step
               Case 1: M = M + Int(x_given_y_z(j, k))
               Case 2
                   cnt(j, k) = Int(x_given_y_z(j, k))
               Case 3
                   If Int(x_given_y_z(j, k)) < cnt(j, k) Then
                       'This is a candidate, and ...
                       If cmp(cnt(j, k), j, k, hn) Then
                           'it is bigger dan what is already in hn
                           hn(0) = cnt(j, k)
                           hn(1) = j
                           hn(2) = k
                       End If
                   End If
           End Select
           k = k + 1
       Loop
       k = 0
       j = j + 1
   Loop
   count = M

End Function Private Sub list_upto(ByVal N As Integer)

   Dim count As Integer
   count = 1
   Dim hn As Integer
   hn = 1
   Do While count < N
       k = hn
       Do While k Mod 2 = 0
           k = k / 2
       Loop
       Do While k Mod 3 = 0
           k = k / 3
       Loop
       Do While k Mod 5 = 0
           k = k / 5
       Loop
       If k = 1 Then
           Debug.Print hn; " ";
           count = count + 1
       End If
       hn = hn + 1
   Loop
   Debug.Print

End Sub Private Function find_seed(N As Long, step As Integer) As Long

   Dim initial As Long, total As Long
   initial = N
   Do 'a simple iterative goal search, takes a handful iterations only
       init initial
       total = count(initial, step)
       initial = initial + N - total
   Loop Until total = N
   find_seed = initial

End Function Private Sub find_hn(N As Long)

   Dim fs As Long, err As Long
   'Step 1: find fs such that the number of lattice points is exactly N
   fs = find_seed(N, 1)
   'Step 2: fill the matrix cnt
   init fs
   err = count(fs, 2)
   'Step 3: lower the plane by diminishing fs, the candidates for
   'the Nth Hamming number will stick out and be recorded in hn
   init fs - 1
   err = count(fs - 1, 3)
   Debug.Print "2^" & hn(0) - 1; " * 3^" & hn(1); " * 5^" & hn(2); "=";
   If N < 1692 Then
       'The task set a limit on the number size
       Debug.Print pow(2, hn(0) - 1) * pow(3, hn(1)) * pow(5, hn(2))
   Else
       Debug.Print
       If N <= 1000000 Then
           'The big Hamming Number will end in a lot of zeroes. The common exponents of 2 and 5
           'are split off to be printed separately.
           If hn(0) - 1 < hn(2) Then
               'Conversion to Decimal datatype with CDec allows to print numbers upto 10^28
               Debug.Print CDec(pow(3, hn(1))) * CDec(pow(5, hn(2) - hn(0) + 1)) & String$(hn(0) - 1, "0")
           Else
               Debug.Print CDec(pow(2, hn(0) - 1 - hn(2))) * CDec(pow(3, hn(1))) & String$(hn(2), "0")
           End If
       End If
   End If

End Sub Public Sub main()

   Dim start_time As Long, finis_time As Long
   start_time = GetTickCount
   Debug.Print "The first twenty Hamming numbers are:"
   list_upto 20
   Debug.Print "Hamming number 1691 is: ";
   find_hn 1691
   Debug.Print "Hamming number 1000000 is: ";
   find_hn 1000000
   finis_time = GetTickCount
   Debug.Print "Execution time"; (finis_time - start_time); " milliseconds"

End Sub</lang>

Output:

The first twenty Hamming numbers are:
 1   2   3   4   5   6   8   9   10   12   15   16   18   20   24   25   27   30   32  
Hamming number 1691 is: 2^5 * 3^12 * 5^3= 2125764000 
Hamming number 1000000 is: 2^55 * 3^47 * 5^64=
519312780448388671875000000000000000000000000000000000000000000000000000000000000000
Execution time 79  milliseconds

VBScript

Translation of: BBC BASIC

<lang vb> For h = 1 To 20 WScript.StdOut.Write "H(" & h & ") = " & Hamming(h) WScript.StdOut.WriteLine Next WScript.StdOut.Write "H(" & 1691 & ") = " & Hamming(1691) WScript.StdOut.WriteLine

Function Hamming(l) Dim h() : Redim h(l) : h(0) = 1 i = 0 : j = 0 : k = 0 x2 = 2 : x3 = 3 : x5 = 5 For n = 1 To l-1 m = x2 If m > x3 Then m = x3 End If If m > x5 Then m = x5 End If h(n) = m If m = x2 Then i = i + 1 : x2 = 2 * h(i) End If If m = x3 Then j = j + 1 : x3 = 3 * h(j) End If If m = x5 Then k = k + 1 : x5 = 5 * h(k) End If Next Hamming = h(l-1) End Function </lang>

Output:

H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36
H(1691) = 2125764000

Vlang

Translation of: go

Concise version using dynamic-programming

<lang vlang>import math.big

fn min(a big.Integer, b big.Integer) big.Integer {

   if a < b {
       return a
   }
   return b

}

fn hamming(n int) []big.Integer {

   mut h := []big.Integer{len: n}
   h[0] = big.one_int
   two, three, five    := big.two_int, big.integer_from_int(3), big.integer_from_int(5)
   mut next2, mut next3, mut next5 := big.two_int, big.integer_from_int(3), big.integer_from_int(5)
   mut i, mut j, mut k := 0, 0, 0
   for m in 1..h.len {
       h[m] = min(next2, min(next3, next5))
       if h[m] == next2 {
           i++
           next2 = two * h[i]
       } 
       if h[m] == next3 {
           j++
           next3 = three * h[j]
       } 
       if h[m] == next5 {
           k++
           next5 = five * h[k]
       } 
   }
   return h

}

fn main() {

   h := hamming(int(1e6))
   println(h[..20])
   println(h[1691-1])
   println(h[h.len-1])

}</lang>

Output:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Wren

Library: Wren-big

Simple but slow. <lang ecmascript>import "/big" for BigInt, BigInts

var primes = [2, 3, 5].map { |p| BigInt.new(p) }.toList

var hamming = Fn.new { |size|

   if (size < 1) Fiber.abort("size must be at least 1")
   var ns = List.filled(size, null)
   ns[0] = BigInt.one
   var next = primes.toList
   var indices = List.filled(3, 0)
   for (m in 1...size) {
       ns[m] = BigInts.min(next)
       for (i in 0..2) {
           if (ns[m] == next[i]) {
               indices[i] = indices[i] + 1
               next[i] = primes[i] * ns[indices[i]]
           }
       }
   }
   return ns

}

var h = hamming.call(1e6) System.print("The first 20 Hamming numbers are:") System.print(h[0..19]) System.print() System.print("The 1,691st Hamming number is:") System.print(h[1690]) System.print() System.print("The 1,000,000th Hamming number is:") System.print(h[999999])</lang>

Output:

The first 20 Hamming numbers are:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]

The 1,691st Hamming number is:
2125764000

The 1,000,000th Hamming number is:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Yabasic

Translation of: Run BASIC

<lang Yabasic>dim h(1000000) for i =1 to 20

   print hamming(i)," ";

next i

print print "Hamming List First(1691) = ",hamming(1691) end

sub hamming(limit)

   local x2,x3,x5,i,j,k,n

   h(0) =1
   x2 = 2: x3 = 3: x5 =5
   i  = 0: j  = 0: k  =0
   for n =1 to limit
       h(n)  = min(x2, min(x3, x5))
       if x2 = h(n) then i = i +1: x2 =2 *h(i):end if
       if x3 = h(n) then j = j +1: x3 =3 *h(j):end if
       if x5 = h(n) then k = k +1: x5 =5 *h(k):end if
   next n
   return h(limit -1)

end sub</lang>

zkl

<lang zkl>var BN=Import("zklBigNum"); // only needed for large N fcn hamming(N){

  h:=List.createLong(N+1); (0).pump(N+1,h.write,Void); // fill list with stuff
  h[0]=1;

1. if 1 // regular (64 bit) ints

  x2:=2; x3:=3; x5:=5; i:=j:=k:=0;

1. else // big ints

  x2:=BN(2); x3:=BN(3); x5:=BN(5); i:=j:=k:=0;

1. endif

  foreach n in ([1..N]){
     z:=(x2<x3) and x2 or x3; z=(z<x5) and z or x5; h[n]=z;
     if (h[n] == x2) { x2 = h[i+=1]*2 }
     if (h[n] == x3) { x3 = h[j+=1]*3 }
     if (h[n] == x5) { x5 = h[k+=1]*5 }
  }
  return(h[N-1])

} [1..20].apply(hamming).println(); hamming(1691).println();</lang>

Output:

L(1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36)
2125764000

While the other algorithms save [lots of] space, run time still sucks when n > 100,000 so memory usage might as well too. Changing the #if 0 to 1 will use Big Int and lots of space.

Output:

hamming(0d1_000_000).println();
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Direct calculation through triples enumeration

OK, I was wrong, calculating the nth Hamming number can be fast and efficient.

Translation of: Haskell

as direct a translation as I can, except using a nested for loop instead of list comprehension (which makes it easier to keep the count).

<lang zkl>#-- directly find n-th Hamming number, in ~ O(n^{2/3}) time

1. -- by Will Ness, based on "top band" idea by Louis Klauder, from DDJ discussion
2. -- http://drdobbs.com/blogs/architecture-and-design/228700538

var BN=Import("zklBigNum"); var lg3 = (3.0).log()/(2.0).log(), lg5 = (5.0).log()/(2.0).log(); fcn logval(i,j,k){ lg5*k + lg3*j + i } fcn trival(i,j,k){ BN(2).pow(i) * BN(3).pow(j) * BN(5).pow(k) } fcn estval(n){ (6.0*lg3*lg5*n).pow(1.0/3) } #-- estimated logval, base 2 fcn rngval(n){

  if(n > 500000) return(2.4496 , 0.0076);	#-- empirical estimation 
  if(n > 50000)  return(2.4424 , 0.0146);	#--   correction, base 2
  if(n > 500)	  return(2.3948 , 0.0723);	#--     (dist,width)
  if(n > 1)	  return(2.2506 , 0.2887);	#-- around (log $ sqrt 30), 

return(2.2506 , 0.5771); #-- says WP }

fcn nthHam(n){ // -> (Double, (Int, Int, Int)) #-- n: 1-based: 1,2,3...

 d,w := rngval(n);				#-- correction dist, width
 hi  := estval(n.toFloat()) - d;		#--   hi > logval > hi-w
 c,b := band(hi,w);				#-- total count, the band
 s   := b.sort(fcn(a,b){ a[0]>b[0] });		#-- sorted decreasing, result
 m   := c - n;					#-- m 0-based from top
 nb  := b.len();				#-- |band|
 res := s[m];					#-- result

 if(w >= 1) throw(Exception.Generic("Breach of contract: (w < 1):  " + w));
 if(m <  0) throw(Exception.Generic("Not enough triples generated: " +c+n));
 if(m >= nb)throw(Exception.Generic("Generated band is too narrow: " +m+nb));
 return(res);

}

fcn band(hi,w){ //--> #-- total count, the band

  b := Sink(List); cnt := 0;
  foreach k in ([0 .. (hi/lg5).floor()]){        p := lg5*k;
     foreach j in ([0 .. ((hi-p)/lg3).floor()]){ q := lg3*j + p;
        i,frac := (hi-q).modf(); r := hi-frac;		#-- r = i + q

cnt+=(i+1); #-- total count if(frac<w) b.write(T(r,T(i,j,k))); #-- store it, if inside band

     }
  }
  return(cnt,b.close());

}</lang> <lang zkl>fcn printHam(n){

  r,t:=nthHam(n); i,j,k:=t; h:=trival(i,j,k);
  println("Hamming(%,d)-->2^%d * 3^%d * 5^%d-->\n%s".fmt(n,i,j,k,h));

}

printHam(1691); //(5,12,3), 10 digits printHam(0d1_000_000); //(55,47,64), 84 digits printHam(0d10_000_000); //(80,92,162), 182 digits, 80 zeros at end printHam(0d1_000_000_000); //(1334,335,404), 845 digits</lang>

Output:

Hamming(1,691)-->2^5 * 3^12 * 5^3-->
2125764000
Hamming(1,000,000)-->2^55 * 3^47 * 5^64-->
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Hamming(10,000,000)-->2^80 * 3^92 * 5^162-->
162441050638304318232392153117595750351085388205966408633356724833252116013682098127901554107666015625 <80 zeros>
Hamming(1,000,000,000)-->2^1334 * 3^335 * 5^404-->
621607575556524486163081633287207200394705651908965270659163240.......

ZX Spectrum Basic

Translation of: BBC_BASIC

<lang zxbasic>10 FOR h=1 TO 20: GO SUB 1000: NEXT h 20 LET h=1691: GO SUB 1000 30 STOP 1000 REM Hamming 1010 DIM a(h) 1030 LET a(1)=1: LET x2=2: LET x3=3: LET x5=5: LET i=1: LET j=1: LET k=1 1040 FOR n=2 TO h 1050 LET m=x2 1060 IF m>x3 THEN LET m=x3 1070 IF m>x5 THEN LET m=x5 1080 LET a(n)=m 1090 IF m=x2 THEN LET i=i+1: LET x2=2*a(i) 1100 IF m=x3 THEN LET j=j+1: LET x3=3*a(j) 1110 IF m=x5 THEN LET k=k+1: LET x5=5*a(k) 1120 NEXT n 1130 PRINT "H(";h;")= ";a(h) 1140 RETURN </lang>