Hamming numbers: Difference between revisions
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{{task|Prime Numbers}}
'''[[wp:Hamming numbers|Hamming numbers]]''' are numbers of the form
<big><big> H = 2<sup>i</sup> × 3<sup>j</sup> × 5<sup>k</sup></big></big>
where
<big> i, j, k ≥ 0 </big>
''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:
# Show the first twenty Hamming numbers.
# Show the 1691<sup>st</sup> Hamming number (the last one below 2<sup>31</sup>).
# Show the one million<sup>th</sup> Hamming number (if the language – or a convenient library – supports arbitrary-precision integers).
;Related tasks:
* [[Humble numbers]]
* [[N-smooth numbers]]
;References:
* Wikipedia entry: [[wp:Hamming numbers|Hamming numbers]] (this link is re-directed to '''Regular number''').
* Wikipedia entry: [[wp:Smooth number|Smooth number]]
* OEIS entry: [[oeis:A051037|A051037 5-smooth or Hamming numbers]]
* [http://dobbscodetalk.com/index.php?option=com_content&task=view&id=913&Itemid=85 Hamming problem] from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread [http://drdobbs.com/blogs/architecture-and-design/228700538 here] and [http://www.jsoftware.com/jwiki/Essays/Hamming%20Number here]).
<br><br>
=={{header|11l}}==
{{trans|Python}}
<syntaxhighlight 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))</syntaxhighlight>
{{out}}
<pre>
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
</pre>
=={{header|360 Assembly}}==
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|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.
<syntaxhighlight 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
function Compute (Nth : Positive) return Number;
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
for F in
Set (Tmp, Factors (F)'Image);
Raise_To_N (Tmp, GNATCOLL.GMP.Unsigned_Long (N.Exponents (F)));
Multiply (R, Tmp);
end loop;
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
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;
Put_Line
Put_Line (Hamming.Image (Hamming.Compute (1_000_000)));
end Hamming;
</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
=={{header|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.
<syntaxhighlight 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);
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)))</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
=={{header|ALGOL W}}==
Algol W only has 32 bit integers, so we just show the first 20 Hamming Numbers and Hamming number 1691.
<br>
This uses the algorithm from Dr Dobbs (as in the Python version). The Coffee Script solution has some notes on how it works.
<syntaxhighlight 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.</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
</pre>
=={{header|Arturo}}==
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|ATS}}==
<syntaxhighlight lang="ats">
//
// How to compile:
// patscc -DATS_MEMALLOC_LIBC -o hamming hamming.dats
//
#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] *)
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|AutoHotkey}}==
<
Msgbox % hamming(1,20)
Msgbox % hamming(1690)
Line 175 ⟶ 625:
return Output
}</
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax:
BEGIN {
for (i=1; i<=20; i++) {
Line 232 ⟶ 635:
}
printf("\n1691: %d\n",hamming(1691))
printf("\n1000000: %d\n",hamming(1000000))
exit(0)
}
Line 250 ⟶ 654:
return((x < y) ? x : y)
}
</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
=={{header|BASIC256}}==
{{trans|FreeBASIC}}
<syntaxhighlight 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</syntaxhighlight>
=={{header|BBC BASIC}}==
<
FOR h% = 1 TO 20
PRINT "H("; h% ") = "; FNhamming(h%)
Line 277 ⟶ 714:
IF m = x5 k% += 1 : x5 = 5 * h%(k%)
NEXT
= h%(l%-1)</
{{out}}
<pre>
H(1) = 1
Line 301 ⟶ 738:
H(20) = 36
H(1691) = 2125764000
</pre>
=={{header|Bc}}==
<syntaxhighlight 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
</syntaxhighlight>
{{out}}
<pre>
$ 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
</pre>
=={{header|Bracmat}}==
{{trans|D}}
<
= x2 x3 x5 n i j k min
. tbl$(h,!arg) { This creates an array. Arrays are always global in Bracmat. }
Line 339 ⟶ 841:
& out$(hamming$1691)
& out$(hamming$1000000)
);</
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|Bruijn}}==
{{trans|Haskell}}
<code>n1000000</code> takes a very long time but eventually reduces to the correct result.
<syntaxhighlight lang="bruijn">
:import std/Combinator .
:import std/Number .
:import std/List .
merge y [[[∅?1 0 (1 [[2 [[go]]]])]]]
go 3 <? 1 (3 : (6 2 4)) (1 : (6 5 0))
# classic version while avoiding duplicate generation
hammings-classic (+1) : (foldr u empty ((+2) : ((+3) : {}(+5))))
u [[y [merge 1 ((mul 2) <$> ((+1) : 0))]]]
:test ((hammings-classic !! (+42)) =? (+162)) ([[1]])
# enumeration by a chain of folded merges (faster)
hammings-folded ([(0 ∘ a) ∘ (0 ∘ b)] (foldr merge1 empty)) $ c
merge1 [[1 [[1 : (merge 0 2)]]]]
a iterate (map (mul (+5)))
b iterate (map (mul (+3)))
c iterate (mul (+2)) (+1)
:test ((hammings-folded !! (+42)) =? (+162)) ([[1]])
# --- output ---
main [first-twenty : (n1691 : {}n1000000)]
first-twenty take (+20) hammings-folded
n1691 hammings-folded !! (+1690)
n1000000 hammings-folded !! (+999999)
</syntaxhighlight>
=={{header|C}}==
Using a min-heap to keep track of numbers. Does not handle big integers.
<
#include <stdlib.h>
Line 402 ⟶ 938:
/* free(q); */
return 0;
}</
===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.
<
#include <stdlib.h>
#include <string.h>
Line 500 ⟶ 1,036:
printf("10,000,000: "); show_ham(get_ham(1e7));
return 0;
}</
{{out}}
<pre> 1,691: 2125764000
1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
10,000,000: 16244105063830431823239 ..<a gadzillion digits>.. 000000000000000000000</pre>
=={{header|C sharp|C#}}==
{{trans|D}}
<
using System.Numerics;
using System.Linq;
Line 602 ⟶ 1,074:
}
}
}</
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Line 610 ⟶ 1,082:
===Generic version for any set of numbers===
The algorithm is similar to the one above.
<
using System.Numerics;
using System.Linq;
Line 646 ⟶ 1,118:
}
}
}</
{{out}}
Line 674 ⟶ 1,146:
--Mike Lorenz
<
using System.Linq;
using System.Numerics;
Line 746 ⟶ 1,218:
}
}
}</
{{out}}
Line 767 ⟶ 1,239:
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!
<
using System.Collections.Generic;
using System.Linq;
Line 890 ⟶ 1,362:
}
}
</syntaxhighlight>
{{out}}
<pre>5-Smooth:
Line 906 ⟶ 1,378:
296352
561912530929780078125000</pre>
'''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:
{{trans|Haskell}}
<syntaxhighlight 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();
}
}
}</syntaxhighlight>
===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:
{{trans|Nim}}
<syntaxhighlight 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 double lg;
public uint x2, x3, x5;
public logrep(double lg, uint x, uint y, uint z) {
this.lg = lg; this.x2 = x; this.x3 = y; this.x5 = z;
}
public logrep mul2() {
return new logrep (this.lg + 1.0, this.x2 + 1, this.x3, this.x5);
}
public logrep mul3() {
return new logrep(this.lg + lb3, this.x2, this.x3 + 1, this.x5);
}
public logrep mul5() {
return new logrep(this.lg + lb5, this.x2, this.x3, this.x5 + 1);
}
}
public IEnumerator<Tuple<uint, uint, uint>> GetEnumerator() {
var one = new logrep();
var s2 = new List<logrep>(); var s3 = new List<logrep>();
s2.Add(one); s3.Add(one.mul3());
var s5 = one.mul5(); var mrg = one.mul3();
var s2hdi = 0; var s3hdi = 0;
while (true) {
if (s2hdi >= s2.Count) { s2.RemoveRange(0, s2hdi); s2hdi = 0; } // assume capacity stays the same...
var v = s2[s2hdi];
if ( v.lg < mrg.lg) { s2.Add(v.mul2()); s2hdi++; }
else {
if (s3hdi >= s3.Count) { s3.RemoveRange(0, s3hdi); s3hdi = 0; }
v = mrg; s2.Add(v.mul2()); s3.Add(v.mul3());
s3hdi++; var chkv = s3[s3hdi];
if (chkv.lg < s5.lg) { mrg = chkv; }
else { mrg = s5; s5 = s5.mul5(); s3hdi--; }
}
yield return Tuple.Create(v.x2, v.x3, v.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((new HammingsLogArr()).ElementAt((int)1691 - 1)));
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();
}
}</syntaxhighlight>
{{output}}
<pre>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</pre>
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:
{{trans|Nim}}
<syntaxhighlight 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();
}
}</syntaxhighlight>
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.
=={{header|C++}}==
===C++11 For Each Generator===
<syntaxhighlight lang="cpp">
#include <iostream>
#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;
}
};
</syntaxhighlight>
===5-Smooth===
<syntaxhighlight 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;
}
</syntaxhighlight>
Produces:
<pre>
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
</pre>
===7-Smooth===
<syntaxhighlight 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;
}
</syntaxhighlight>
Produces:
<pre>
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
</pre>
===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:
{{trans|Haskell}}
{{works with|C++11}}
<syntaxhighlight lang="cpp">#include <chrono>
#include <iostream>
#include <gmpxx.h>
#include <functional>
#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";
}</syntaxhighlight>
{{output}}
<pre>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.</pre>
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:
{{trans|Rust}}
{{works with|C++11}}
<syntaxhighlight lang="cpp">#include <chrono>
#include <iostream>
#include <vector>
#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";
}</syntaxhighlight>
{{output}}
<pre>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.</pre>
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.
=={{header|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:
{{trans|Haskell}}[[Hamming_numbers#Avoiding_generation_of_duplicates]]
{{works with|Chapel|1.24.1|or greater, maybe lesser}}
<syntaxhighlight 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.");</syntaxhighlight>
{{out}}
<pre>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.</pre>
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 and also because the "bigint" implementation isn't as fast as the "gmp" package used by many languages for multi-precision integer caluclations.
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:
{{trans|Nim}}
<syntaxhighlight lang="chapel">use BigInteger; use Time;
iter nodupsHamming(): bigint {
var s2dom = { 0 .. 1023 }; var s2: [s2dom] bigint; // init so can double!
var s3dom = { 0 .. 1023 }; var s3: [s3dom] bigint; // init so can double!
s2[0] = 1: bigint; s3[0] = 3: bigint;
var x5 = 5: bigint; var mrg = 3: bigint;
var s2hdi, s2tli, s3hdi, s3tli: int;
while true {
s2tli += 1;
if s2hdi + s2hdi >= s2tli { // move in place to avoid allocation!
s2[0 .. s2tli - s2hdi - 1] = s2[s2hdi .. s2tli - 1];
s2tli -= s2hdi; s2hdi = 0; }
const s2sz = s2.size;
if s2tli >= s2sz then s2dom = { 0 .. s2sz + s2sz - 1 };
var rslt: bigint; const s2hd = s2[s2hdi];
if s2hd < mrg { rslt = s2hd; s2hdi += 1; }
else {
s3tli += 1;
if s3hdi + s3hdi >= s2tli { // move in place to avoid allocation!
s3[0 .. s3tli - s3hdi - 1] = s3[s3hdi .. s3tli - 1];
s3tli -= s3hdi; s3hdi = 0; }
const s3sz = s3.size;
if s3tli >= s3sz then s3dom = { 0 .. s3sz + s3sz - 1 };
rslt = mrg; s3[s3tli] = rslt * 3;
s3hdi += 1; const s3hd = s3[s3hdi];
if s3hd < x5 { mrg = s3hd; }
else { mrg = x5; x5 = x5 * 5; s3hdi -= 1; }
}
s2[s2tli] = rslt * 2;
yield rslt;
}
}
// test it...
write("The first 20 hamming numbers are: ");
var cnt = 0: uint(64);
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 { writeln(h); break; } cnt += 1; }
write("The millionth hamming number is ");
var timer: Timer; cnt = 1;
timer.start(); var rslt: bigint;
for h in nodupsHamming() {
if cnt >= 1000000 { rslt = h; break; } cnt += 1; }
timer.stop();
write(rslt);
writeln(".\nThis last took ",
timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</syntaxhighlight>
{{out}}
<pre>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 114.867 milliseconds.</pre>
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 for sorting and only outputs the final answer(s) as BigInteger(s):
{{trans|Nim}}
<syntaxhighlight lang="chapel">use BigInteger; use Math; use Time;
config const nth: uint(64) = 1000000;
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);
inline proc mul2(): LogRep {
return new LogRep(this.lg + lb2, this.x2 + 1, this.x3, this.x5); }
inline proc mul3(): LogRep {
return new LogRep(this.lg + lb3, this.x2, this.x3 + 1, this.x5); }
inline proc mul5(): LogRep {
return new LogRep(this.lg + lb5, this.x2, this.x3, this.x5 + 1); }
proc lr2bigint(): 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;
}
return xpnd(2: uint, this.x2) *
xpnd(3: uint, this.x3) * xpnd(5: uint, this.x5);
}
proc writeThis(lr) throws {
lr <~> this.lr2bigint();
}
}
operator <(const ref a: LogRep, const ref b: LogRep): bool { return a.lg < b.lg; }
const one = new LogRep(0, 0, 0, 0);
iter nodupsHammingLog(): LogRep {
var s2dom = { 0 .. 1023 }; var s2: [s2dom] LogRep; // init so can double!
var s3dom = { 0 .. 1023 }; var s3: [s3dom] LogRep; // init so can double!
s2[0] = one; s3[0] = one.mul3();
var x5 = one.mul5(); var mrg = one.mul3();
var s2hdi, s2tli, s3hdi, s3tli: int;
while true {
s2tli += 1;
if s2hdi + s2hdi >= s2tli { // move in place to avoid allocation!
s2[0 .. s2tli - s2hdi - 1] = s2[s2hdi .. s2tli - 1];
s2tli -= s2hdi; s2hdi = 0; }
const s2sz = s2.size;
if s2tli >= s2sz then s2dom = { 0 .. s2sz + s2sz - 1 };
var rslt: LogRep; const s2hd = s2[s2hdi];
if s2hd.lg < mrg.lg { rslt = s2hd; s2hdi += 1; }
else {
s3tli += 1;
if s3hdi + s3hdi >= s2tli { // move in place to avoid allocation!
s3[0 .. s3tli - s3hdi - 1] = s3[s3hdi .. s3tli - 1];
s3tli -= s3hdi; s3hdi = 0; }
const s3sz = s3.size;
if s3tli >= s3sz then s3dom = { 0 .. s3sz + s3sz - 1 };
rslt = mrg; s3[s3tli] = mrg.mul3(); s3hdi += 1;
const s3hd = s3[s3hdi];
if s3hd.lg < x5.lg { mrg = s3hd; }
else { mrg = x5; x5 = x5.mul5(); s3hdi -= 1; }
}
s2[s2tli] = rslt.mul2();
yield rslt;
}
}
// test it...
write("The first 20 hamming numbers are: ");
var cnt = 0: uint(64);
for h in nodupsHammingLog() {
if cnt >= 20 then break; cnt += 1; write(" ", h); }
write("\nThe 1691st hamming number is "); cnt = 1;
for h in nodupsHammingLog() {
if cnt >= 1691 { writeln(h); break; } cnt += 1; }
write("The ", nth, "th hamming number is ");
var timer: Timer; cnt = 1;
timer.start(); var rslt: LogRep;
for h in nodupsHammingLog() {
if cnt >= nth { rslt = h; break; } cnt += 1; }
timer.stop();
write(rslt);
writeln(".\nThis last took ",
timer.elapsed(TimeUnits.milliseconds), " milliseconds.");</syntaxhighlight>
{{out}}
<pre>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 6.372 milliseconds.</pre>
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 8.1 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:
{{trans|Nim}}
{{works with|Chapel|1.22|for zero based tuple indices}}
<syntaxhighlight 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.");</syntaxhighlight>
{{out}}
<pre>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.</pre>
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:
{{trans|Nim}}
{{works with|Chapel|1.22|for zero based tuple indices}}
<syntaxhighlight 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.");</syntaxhighlight>
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.
=={{header|Clojure}}==
This version implements Dijkstra's merge solution, so is closely related to the Haskell version.
<
(lazy-seq
(let [x (first xs),
Line 925 ⟶ 2,402:
(smerge (map #(*' 3 %) hamming))
(smerge (map #(*' 2 %) hamming))
(cons 1))))</
Note that
'''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:
{{trans|Haskell}}
<syntaxhighlight 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))))))</syntaxhighlight>
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}}
<pre>
(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
</pre>
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:
<pre> :jvm-opts ^:replace []</pre>
=={{header|CoffeeScript}}==
<
# property that they don't evenly divide any prime numbers outside
# a given set, such as [2, 3, 5].
Line 1,010 ⟶ 2,523:
numbers = generate_hamming_sequence(primes, 10000)
console.log numbers[1690]
console.log numbers[9999]</
=={{header|Common Lisp}}==
Maintaining three queues, popping the smallest value every time.
<
(let ((x (apply #'min (map 'list #'first seqs))))
(loop for s in seqs
Line 1,033 ⟶ 2,546:
(if (or (< n 21)
(= n 1691)
(= n 1000000)) (format t "~d: ~d~%" n x))))</
A much faster method:
<
(let ((fac '(2 3 5))
(idx (make-array 3 :initial-element 0))
Line 1,054 ⟶ 2,567:
(loop for i in '(1691 1000000) do
(format t "~d: ~d~%" i (hamming i)))</
{{out}}
<pre> 1: 1
Line 1,078 ⟶ 2,591:
1691: 2125764000
1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|Crystal}}==
{{trans|Bc}}
<syntaxhighlight 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"
</syntaxhighlight>
System: I7-6700HQ, 3.5 GHz, Linux Kernel 5.6.17, Crystal 0.35
Run as: $ crystal run hammingnumbers.cr --release
{{out}}
<pre>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</pre>
'''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:
{{trans|Kotlin}}
<syntaxhighlight lang="ruby">require "big"
# Unlike some languages like Kotlin, Crystal doesn't have a Lazy module,
# but it has closures, so it is easy to implement a LazyList class;
# 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)</syntaxhighlight>
{{out}}
<pre>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.</pre>
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:
<syntaxhighlight lang="ruby">require "big"
# Unlike some languages like Kotlin, Crystal doesn't have a Lazy module,
# but it has closures, so it is easy to implement a LazyList class;
# 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)</syntaxhighlight>
{{out}}
<pre>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.</pre>
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:
{{trans|Nim}}
<syntaxhighlight 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 mult2
LogRep.new(@logrep + LOG2_2, @x2 + 1, @x3, @x5)
end
def mult3
LogRep.new(@logrep + LOG2_3, @x2, @x3 + 1, @x5)
end
def mult5
LogRep.new(@logrep + LOG2_5, @x2, @x3, @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)
# use pointers to avoid bounds checking...
@s2 = Pointer(LogRep).malloc 1024; @s3 = Pointer(LogRep).malloc 1024
@s5 : LogRep = ONE.mult5; @mrg : LogRep = ONE.mult3
@s2sz = 1024; @s3sz = 1024
@s2hdi = 0; @s2tli = 0; @s3hdi = 0; @s3tli = 0
def initialize
@s2[0] = ONE; @s3[0] = ONE.mult3
end
def next
@s2tli += 1
if @s2hdi + @s2hdi >= @s2sz # unused is half of used
@s2.move_from(@s2 + @s2hdi, @s2tli - @s2hdi)
@s2tli -= @s2hdi; @s2hdi = 0
end
if @s2tli >= @s2sz # grow array, copying former contents
@s2sz += @s2sz; ns2 = Pointer(LogRep).malloc @s2sz
ns2.move_from(@s2, @s2tli); @s2 = ns2
end
rsltp = @s2 + @s2hdi;
if rsltp.value < @mrg
@s2[@s2tli] = rsltp.value.mult2; @s2hdi += 1
else
@s3tli += 1
if @s3hdi + @s3hdi >= @s3sz # unused is half of used
@s3.move_from(@s3 + @s3hdi, @s3tli - @s3hdi)
@s3tli -= @s3hdi; @s3hdi = 0
end
if @s3tli >= @s3sz # grow array, copying former contents
@s3sz += @s3sz; ns3 = Pointer(LogRep).malloc @s3sz
ns3.move_from(@s3, @s3tli); @s3 = ns3
end
@s2[@s2tli] = @mrg.mult2; @s3[@s3tli] = @mrg.mult3
@s3hdi += 1; ns3hdp = @s3 + @s3hdi
rslt = @mrg; rsltp = pointerof(rslt)
if ns3hdp.value < @s5
@mrg = ns3hdp.value
else
@mrg = @s5; @s5 = @s5.mult5; @s3hdi -= 1
end
end
rsltp.value
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)</syntaxhighlight>
{{out}}
<pre>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 7.330211 milliseconds.</pre>
As can be seen by comparing with the above results using the same Intel Skylake i5-6500 CPU, this is about eighteen times faster than the functional version also using logarithmic representations due to less time spent doing memory allocations/deallocations by using the imperative form of code. This version can find the billionth Hamming number in about 7.6 seconds on this machine.
=={{header|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.
<
auto hamming(in uint n) pure nothrow /*@safe*/ {
Line 1,105 ⟶ 2,983:
1_691.hamming.writeln;
1_000_000.hamming.writeln;
}</
{{out}}
<pre>[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
Line 1,115 ⟶ 2,993:
This keeps <math>O(n^{2/3})</math> numbers in memory, but over-computes a sequence by a factor of about <math>\Theta(n^{2/3})</math>, calculating extra multiples past that as well. Incurs an extra <math>O(\log n)</math> 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:
{{trans|Java}}
<
core.memory;
Line 1,139 ⟶ 3,017:
writeln("hamming(1691) = ", 1691.hamming);
writeln("hamming(1_000_000) = ", 1_000_000.hamming);
}</
{{out}}
<pre>First 20 Hamming numbers: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
Line 1,149 ⟶ 3,027:
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.
{{trans|C}}
<
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; // ^^
Line 1,171 ⟶ 3,050:
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 {
Line 1,263 ⟶ 3,134:
foreach (immutable n; [1691, 10 ^^ 6, MAX_HAM])
writefln("%8d: %s", n, n.getHam);
}</
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.
<syntaxhighlight 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; // ^^
Line 1,278 ⟶ 3,155:
__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.
Line 1,424 ⟶ 3,296:
foreach (immutable n; [1691, 10 ^^ 6, 10_000_000])
writefln("%8d: %s", n, n.getHam);
}</
The output is the same as the second alternative version.
=={{header|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:
<syntaxhighlight 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.");
}</syntaxhighlight>
{{Output}}
<pre>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.</pre>
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):
{{translated from|Haskell}}
<syntaxhighlight 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.");
}</syntaxhighlight>
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:
<syntaxhighlight 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 $elpsd milliseconds.");
}</syntaxhighlight>
With these changes, the algorithm can find the 1e19'th prime in the order af days depending on the CPU used.
=={{header|DCL}}==
<syntaxhighlight lang="dcl">$ 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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Delphi}}==
See [https://rosettacode.org/wiki/Hamming_numbers#Pascal Pascal].
=={{header|EasyLang}}==
{{trans|11l}}
<syntaxhighlight>
func hamming lim .
len h[] lim
h[1] = 1
x2 = 2 ; x3 = 3 ; x5 = 5
i = 1 ; j = 1 ; k = 1
for n = 2 to lim
h[n] = lower x2 lower 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[lim]
.
for nr = 1 to 20
write hamming nr & " "
.
print ""
print hamming 1691
</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
</pre>
=={{header|Eiffel}}==
<syntaxhighlight lang="eiffel">
note
description : "Initial part, in order, of the sequence of Hamming numbers"
Line 1,454 ⟶ 3,712:
]"
warning : "[
The formatting (<syntaxhighlight lang="text">) 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.
Line 1,506 ⟶ 3,764:
end
end
</syntaxhighlight>
{{out}}
Line 1,512 ⟶ 3,770:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000</pre>
=={{header|Elixir}}==
<syntaxhighlight 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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|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 are persistent/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 queues 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 approximation while maintaining "Trival" triple representation of the number of powers of two, three, and five, are multiplied in order to obtain the current value represented by the logarithmic approximation. The working code is as follows:
<syntaxhighlight lang="elm">module Main exposing ( main )
import Bitwise exposing (..)
import BigInt
import Task exposing ( Task, succeed, perform, andThen )
import Html exposing ( 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
Nothing -> 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
Nothing -> 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
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 =
let msg1 = "The first 20 Hamming numbers are: " ++
(hammingsLog() |> takeCIS2List 20
|> List.map showTrival
|> String.join ", ") ++ "."
msg2 = "The 1691st Hamming number is " ++
(hammingsLog() |> nthCIS 1691
|> showTrival) ++ "."
msg3 = "The " ++ String.fromInt cLIMIT ++ "th Hamming number is:"
in timemillis()
|> andThen (\ strt ->
let rsltstr = hammingsLog() |> nthCIS lmt
|> showTrival in
timemillis()
|> andThen (\ stop ->
succeed [msg1, msg2, msg3, rsltstr ++ " in "
++ String.fromInt (stop - strt)
++ " milliseconds."]))
|> perform Done
-- following code has to do with outputting to a web page using MUV/TEA...
type alias Model = List String
type Msg = Done Model
main : Program () Model Msg
main = -- starts with empty list of strings; views model of filled list...
element { init = \ _ -> ( [], test cLIMIT )
, update = \ (Done mdl) _ -> ( mdl , Cmd.none )
, subscriptions = \ _ -> Sub.none
, view = div [] << List.map (div [] << List.singleton << text) }</syntaxhighlight>
{{out}}
<pre>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 767 milliseconds.</pre>
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 somewhat competitive with easier implementations in other languages such as F#.
=={{header|Erlang}}==
For relatively small values of n we can use an elegant code:
<syntaxhighlight lang="erlang">
list(N) -> array:to_list(element(1, array(N, [2, 3, 5]))).
nth(N) -> array:get(N-1, element(1, array(N, [2, 3, 5]))).
array(N, Primes) -> array(array:new(), N, 1, [{P, 1, P} || P <- Primes]).
array(Array, Max, Max, Candidates) -> {Array, Candidates};
array(Array, Max, I, Candidates) ->
Smallest = smallest(Candidates),
N_array = array:set(I, Smallest, Array),
array(N_array, Max, I+1, update(Smallest, N_array, Candidates)).
update(Val, Array, Candidates) -> [update_(Val, C, Array) || C <- Candidates].
update_(Val, {Val, Ind, Mul}, Array) ->
{Mul*array:get(Ind, Array), Ind+1, Mul};
update_(_, X, _) -> X.
smallest(L) -> lists:min([element(1, V) || V <- L]).
</syntaxhighlight>
However, when n become large (let say above 5e7) the memory needed grew very large as I store all the values. Fortunately, the algorithm uses only a small fraction of the end of the array. So I can drop the beginning of the array when it is no longer needed.
<syntaxhighlight lang="erlang">
nth(N, Batch) ->
array:get(N-1, element(1, compact_array(N, Batch, [2, 3, 5]))).
compact_array(Goal, Lim, Primes) ->
{Array, Candidates} = array(Lim, Primes),
compact_array(Goal, Lim, Lim, Array, Candidates).
compact_array(Goal, _, Index, Array, Candidates) when Index > Goal ->
{Array, Candidates};
compact_array(Goal, Lim, Index, Array, Candidates) ->
{N_array, N_candidates} =
array(compact(Array, Candidates), Index + Lim, Index, Candidates),
compact_array(Goal, Lim, Index+Lim, N_array, N_candidates).
compact(Array, L) ->
Index = lists:min([element(2, V) || V <- L]),
Keep = [E || E <- array:sparse_to_orddict(Array), element(1, E) >= Index],
array:from_orddict(Keep).
</syntaxhighlight>
With this approach memory is no longer an issue:
{{out}}
<pre>
timer:tc(task_hamming_numbers, nth, [100_000_000, 1_000_000]).
{232894309,
18140143309611363532953342430693354584669635033709097929462505366714035156593135818380467866054222964635144914854949550271375442721368122191972041094311075107507067573147191502194201568268202614781694681859513649083616294200541611489469967999559505365172812095568020073934100699850397033005903158113691518456912149989919601385875227049401605594538145621585911726469930727034807205200195312500}
</pre>
So a bit less than 4 minutes to get the 100 000 000th regular number. The complexity is slightly worse than linear which is not a surprise given than all the regular numbers are computed.
=={{header|ERRE}}==
For bigger numbers, you have to use an external program, like MULPREC.R
<syntaxhighlight 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</syntaxhighlight>
{{out}}<pre>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
</pre>
=={{header|F_Sharp|F#}}==
This version implements Dijkstra's merge solution, so is closely related to the Haskell classic version.
<syntaxhighlight lang="fsharp">type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>>
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</syntaxhighlight>
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:
{{trans|Haskell}}
<syntaxhighlight lang="fsharp">let cNUMVALS = 1000000
type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>>
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</syntaxhighlight>
Both codes output the same results as follows but the second is over three times faster:
{{out}}
<pre>( 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.</pre>
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. 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:
F# has a particularly slow enumeration ability in the use of the `Seq` type (although easy to use) so in order to be able to bypass that, the following code still uses the imperative `ResizeArray`'s but outputs a closure "next" function that can be used directly to avoid the generation of a `Seq` sequence where maximum speed is desired:
{{tran|Nim}}
<syntaxhighlight lang="fsharp">let cCOUNT = 1000000
type LogRep = struct val lr: double; val x2: uint32; val x3: uint32; val x5: uint32
new(lr, x2, x3, x5) = {lr = lr; x2 = x2; x3 = x3; x5 = x5 } end
let one: LogRep = LogRep(0.0, 0u, 0u, 0u)
let lg2_2: double = 1.0
let lg3_2: double = log 3.0 / log 2.0
let lg5_2: double = log 5.0 / log 2.0
let inline mul2 (lr: LogRep): LogRep = LogRep(lr.lr + lg2_2, lr.x2 + 1u, lr.x3, lr.x5)
let inline mul3 (lr: LogRep): LogRep = LogRep(lr.lr + lg3_2, lr.x2, lr.x3 + 1u, lr.x5)
let inline mul5 (lr: LogRep): LogRep = LogRep(lr.lr + lg5_2, lr.x2, lr.x3, lr.x5 + 1u)
let hammingsLog() = // imperative arrays, eliminates the BigInteger operations...
let s2 = ResizeArray<_>() in let s3 = ResizeArray<_>()
s2.Add(one); s3.Add(mul3 one)
let mutable s5 = mul5 one in let mutable mrg = mul3 one
let mutable s2hdi = 0 in let mutable s3hdi = 0
let next() = // imperative next function to advance value
if s2hdi + s2hdi >= s2.Count then s2.RemoveRange(0, s2hdi); s2hdi <- 0
let mutable rslt: LogRep = s2.[s2hdi]
if rslt.lr < mrg.lr then s2.Add(mul2 rslt); s2hdi <- s2hdi + 1
else
if s3hdi + s3hdi >= s3.Count then s3.RemoveRange(0, s3hdi); s3hdi <- 0
rslt <- mrg; s2.Add(mul2 rslt); s3.Add(mul3 rslt); s3hdi <- s3hdi + 1
let chkv: LogRep = s3.[s3hdi]
if chkv.lr < s5.lr then mrg <- chkv
else mrg <- s5; s5 <- mul5 s5; s3hdi <- s3hdi - 1
rslt
next
let hl2Seq f = Seq.unfold (fun v -> Some(v, f())) (f())
let nthLogHamming n f =
let rec nxt i = if i >= n then f() else f() |> ignore; nxt (i + 1) in nxt 0
let lr2BigInt (lr: LogRep) = // convert trival to BigInteger
let rec xpnd n mlt rslt =
if n <= 0u then rslt
else xpnd (n - 1u) mlt (mlt * rslt)
xpnd lr.x2 2I 1I |> xpnd lr.x3 3I |> xpnd lr.x5 5I
[<EntryPoint>]
let main argv =
printf "( "; hammingsLog() |> hl2Seq |> Seq.take 20
|> Seq.iter (printf "%A " << lr2BigInt); printfn ")"
printfn "%A" (hammingsLog() |> hl2Seq |> Seq.item (1691 - 1) |> lr2BigInt)
let strt = System.DateTime.Now.Ticks
// slow way using Seq:
// let rslt = (hammingsLog()) |> hl2Seq |> Seq.item (1000000 - 1)
// fast way using closure directly:
let rslt = (hammingsLog()) |> nthLogHamming (1000000 - 1)
let stop = System.DateTime.Now.Ticks
printfn "%A" (rslt |> lr2BigInt)
printfn "Found this last up to %d in %d milliseconds." cCOUNT ((stop - strt) / 10000L)
printfn ""
0 // return an integer exit code</syntaxhighlight>
{{out}}
<pre>( 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 57 milliseconds.</pre>
The above code can find the billionth Hamming number in about 60 seconds on the same Intel i5-6500 at 3.6 GHz (single threaded boosted). If the "fast way" is commented out and the commenting out removed from the "slow way", the code is about twice as slow.
===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:
{{trans|Haskell}}
<syntaxhighlight 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</syntaxhighlight>
{{output}}
<pre>( 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.</pre>
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:
{{trans|Haskell}}
<syntaxhighlight 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</syntaxhighlight>
=={{header|Factor}}==
{{trans|Scala}}
<
;
IN: rosetta.hamming
Line 1,546 ⟶ 4,418:
: nth-from-now ( hamming-iterator n -- m )
1 - over '[ _ next drop ] times next ;</
<hamming-iterator> 20 next-n .
Line 1,553 ⟶ 4,425:
{{trans|Haskell}}
Lazy lists
<
IN: rosetta.hamming-lazy
Line 1,571 ⟶ 4,443:
h 2 3 5 [ '[ _ * ] lazy-map ] tri-curry@ tri
sort-merge sort-merge
] lazy-cons h! h ;</
20 hamming ltake list>array .
Line 1,582 ⟶ 4,454:
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.
<
: extract2 ( h -- l )
Line 1,673 ⟶ 4,545:
20 .nseq
cr 1691 nthseq h.
cr 1000000 nthseq h.</
{{out}}
<pre>
Line 1,681 ⟶ 4,553:
</pre>
A smaller, less capable solution is presented here. It solves two out of three requirements and is ANS-Forth compliant.
<
create hamming /hamming allot
( n1 n2 n3 n4 n5 n6 n7 -- n3 n4 n5 n6 n1 n2 n8)
Line 1,705 ⟶ 4,577:
cr 21 1 ?do i . i hamming# . cr loop
1691 hamming# . cr
;</
=={{header|Fortran}}==
{{works with|Fortran|90 and later}}
Using big_integer_module from here [http://fortran.com/big_integer_module.f95]
<
use big_integer_module
implicit none
Line 1,780 ⟶ 4,653:
end if
end function mini
end program</
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Line 1,786 ⟶ 4,659:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|
<syntaxhighlight 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
End Function
Dim h(1 To n) As Integer
h(1) = 1
Dim As Integer i = 1, j = 1,
Dim As Integer x2 = 2, x3 = 3, x5 =
For m As Integer = 2 To n
h(m) = min(x2, min(x3, x5))
End
If h(m) = x3
j +=
x3 = 3 *
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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|FunL}}==
{{trans|Scala}}
<
val hamming =
Line 1,844 ⟶ 4,741:
for q <- [q2, q3, q5] do q.enqueue( 1 )
stream()</
{{trans|Haskell}}
<
def
Line 1,857 ⟶ 4,754:
println( hamming.take(20) )
println( hamming(1690) )
println( hamming(2000) )</
{{out}}
Line 1,866 ⟶ 4,763:
8100000000
</pre>
=={{header|Fōrmulæ}}==
{{FormulaeEntry|page=https://formulae.org/?script=examples/Hamming_numbers}}
'''Solution'''
[[File:Fōrmulæ - Hamming numbers 01.png]]
'''Case 1.''' First twenty Hamming numbers
[[File:Fōrmulæ - Hamming numbers 02.png]]
[[File:Fōrmulæ - Hamming numbers 03.png]]
'''Case 2.''' 1691-st Hamming number
[[File:Fōrmulæ - Hamming numbers 04.png]]
[[File:Fōrmulæ - Hamming numbers 05.png]]
'''Case 3.''' One million-th Hamming number
[[File:Fōrmulæ - Hamming numbers 06.png]]
[[File:Fōrmulæ - Hamming numbers 07.png]]
=={{header|Go}}==
===Concise version using dynamic-programming===
<
import (
Line 1,903 ⟶ 4,826:
fmt.Println(h[1691-1])
fmt.Println(h[len(h)-1])
}</
{{out}}
<pre>
Line 1,912 ⟶ 4,835:
===Longer version using dynamic-programming and logarithms===
More than 10 times faster.
<
import (
Line 1,964 ⟶ 4,887:
func step() {
table = append(table, front[0].
front[0].advance()
// re-establish sorted order
Line 1,989 ⟶ 4,912:
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 {
}
table = make([][3]int16, 1, ordinal)
Line 2,004 ⟶ 4,927:
}
show(table[ordinal-1])
}</
{{out}}
<pre>
Line 2,018 ⟶ 4,941:
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
</pre>
===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.
{{trans|Haskell}}
<syntaxhighlight 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))
}
</syntaxhighlight>
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:
{{trans|Rust}}
<syntaxhighlight 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))
}</syntaxhighlight>
{{output}}
<pre>[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.</pre>
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:
{{trans|Nim}}
<syntaxhighlight 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))
}</syntaxhighlight>
{{output}}
<pre>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.</pre>
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.
=={{header|Groovy}}==
<syntaxhighlight 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)
}
}</syntaxhighlight>
=={{header|Haskell}}==
===The classic version===
<
union a@(x:xs) b@(y:ys) = case compare x y of
Line 2,036 ⟶ 5,420:
-- [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
-- (2125764000,2147483648)
-- 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</
Runs in about a second on [http://ideone.com/q3fma Ideone.com].
The nested <code>union</code>s' effect is to produce the minimal value at each step,
with duplicates removed. As Haskell evaluation model is ''on-demand'',
the three <code>map</code> 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 <math>O(n)</math>. [http://ideone.com/k8PU3 Empirically], it is slightly above that and worsening, suggestive of extra cost of bignum arithmetics. [http://ideone.com/k8PU3 Using triples representation] with logarithm values for comparisons amends
This is what that [http://drdobbs.com/blogs/architecture-and-design/228700538 DDJ blog post's] "pseudo-C" code was transcribing, mentioned at [[#Python|the Python entry]] that started this task (<small> curiously, it is in almost word-for-word correspondence with Edsger Dijkstra's code from his book A Discipline of Programming, [http://web.cecs.pdx.edu/~cs410aph/Lectures/Smalltalk%20II/Dijkstra%20on%20Hamming's%20Problem.pdf p. 132] </small>). [[#D|D]], [[#Go|Go]], [[#PARI/GP|PARI/GP]], [[#Prolog|Prolog]] all implement the same idea of back-pointers into shared storage. A Haskell run-time system can actually [http://ideone.com/q3fma free up the storage] automatically at the start of the shared list and only keep the needed portion of it, from the <code>(5*)</code> back-pointer, – which is about <math>O(n^{2/3})</math> 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 (
<syntaxhighlight lang
r where
merge a@(x:xs) b@(y:ys) | x < y = x : merge xs b
| otherwise = y : merge a ys
main :: IO ()
main = do
===Explicit multiples reinserting===
This is a common approach which explicitly maintains an internal buffer of <math>O(n^{2/3})</math> 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 <math>O(n^{2/3})</math> operations for each number, instead of <math>O(1)</math> of the above version (and thus <math>O(n^{5/3})</math> overall). Translation of [[#Java|Java]] (which does use priority queue though, so should have ''O''‍ ‍(''n''‍ ‍log''n'') operations overall). Uses <code>union</code> from the "classic" version above:
<syntaxhighlight lang="haskell">hammFrom n = drop n $
iterate (\(_ , (a:t)) -> (a, union t [2*a,3*a,5*a])) (0, [1])</syntaxhighlight>
{{out}}
<
[1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
[2125764000,2147483648]
(1,[2,3,5])
(2,[3,4,5,6,10])
Line 2,081 ⟶ 5,471:
(12,[15,16,18,20,24,25,27,30,36,40,45,50,60])
[402,638,1007,1596]
> map (logBase 2) $ zipWith (/) =<< tail $ [402,638,1007,1596]
[0.67,0.66,0.66]</syntaxhighlight>
Runs too slowly to reach 1,000,000, with empirical orders of growth above ''~''‍ ‍(''n''‍ ‍<sup>1.7</sup>‍ ‍) and worsening. Last two lines show the internal buffer's length for several sample ''n''‍ ‍s, and its [http://en.wikipedia.org/wiki/Analysis_of_algorithms#Empirical_orders_of_growth ''empirical'' orders of growth] which strongly support the <math>O(n^{2/3})</math> claim.
===Enumeration by a chain of folded merges===
<syntaxhighlight 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, ... -}</syntaxhighlight>
Uses <code>merge</code>, as there's no need for duplicates-removing <code>union</code> 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<sup>1.14..1.16</sup> 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<sup>1/3</sup> in general).
Can be ''significantly'' improved, both in time complexity and absolute run time, by replacing the linear fold with the tree-shaped <code>mergeAll</code> from the <code>Data.List.Ordered</code> module of <code>data-ordlist</code> 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 <code>(i,j,k)</code> triples below its [http://en.wikipedia.org/wiki/Regular_number#Number_theory estimated value] – with ordering according to their exponents, <code>i*ln2 + j*ln3 + k*ln5</code> – while storing only the "band" of topmost triples close enough to the target value (more in the [http://drdobbs.com/blogs/architecture-and-design/228700538 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
<
--
-- by Will
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'</syntaxhighlight>
{{out}}
<pre>-- time: 0.00s memory: 4.2MB
(55,47,64)
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
'''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:
<syntaxhighlight lang="haskell">{-# OPTIONS_GHC -O3 -XStrict #-}
import Data.Word
import Data.List (sortBy)
import Data.Function (on)
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</syntaxhighlight>
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.
[https://ideone.com/RnAh5X 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:
<syntaxhighlight 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
(
bglb2 = 1267650600228229401496703205376 :: Integer
bglb3 = 2009178665378409109047848542368 :: Integer
bglb5 = 2943393543170754072109742145491 :: Integer
(c, b) = let klmt
loopk k
loopj j cj bndj =
(i,
nj = j + 1; ncj = cj + fromIntegral i +
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</syntaxhighlight>
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.
=={{header|Icon}} and {{header|Unicon}}==
Line 2,142 ⟶ 5,653:
Lazy evaluation is used to improve performance.
<
# from a known Hamming number h
class Triplet : Class (cv, ce)
Line 2,194 ⟶ 5,705:
t1.nextVal() | delete(triplers, t1)
}
end</
=={{header|J}}==
'''Solution:'''<br>
A concise tacit expression using a (right) fold:
<
'''Example usage:'''
<
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
{: hamming 1691
2125764000</
For the millionth (and billionth (1e9)) Hamming number see the <code>nh</code> verb from [[j:Essays/Hamming Number|Hamming Number essay]] on the J wiki.
Line 2,211 ⟶ 5,722:
I'll explain this J-sentence by dividing it in three parts from left to right omitting the leftmost <code>{.</code>:
* sort and remove duplicates
<syntaxhighlight lang
* produce
<syntaxhighlight lang
note that LHA (2 3 5 * {) RHA [http://www.jsoftware.com/help/dictionary/intro05.htm is equivalent] to
<
* the RH part forms an array of descending indices and the initial Hamming number 1
<
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)
<
1 2 3 4 5 6 8</
This starts using a descending sequence with 1 appended:
<
6 5 4 3 2 1 0 1</
and then the fold expression is inserted between these list elements and the result computed:
<
1 2 3 4 5 6 8 9 10 12 15 18 20 25 30 16 24 40</
(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: <code>{</code> has been implemented so that an index of 0 will select the only item from an array with no dimensions.)
Line 2,236 ⟶ 5,747:
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) <math>O(1)</math> implementation for both insertion and deletion operations but it looks like it's <math>O(\log n)</math> in Java.
<
import java.util.PriorityQueue;
Line 2,272 ⟶ 5,783:
System.out.println("Hamming(1000000) = " + hamming(1000000));
}
}</
{{out}}
<pre>Hamming(1 .. 20) = 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Line 2,280 ⟶ 5,791:
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 <math>BigInteger</math>, 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 <math>HammingTriple</math> encode Hamming numbers in three fields <math>a</math>, <math>b</math> and <math>c</math>. 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 <math>BigInteger</math> only in the rare cases that the floating point arithmetic produces too close results.
<
import java.math.BigInteger;
import java.util.*;
public class HammingTriple implements Comparable<HammingTriple> {
Line 2,392 ⟶ 5,902:
}
}
</syntaxhighlight>
<pre>
Hamming number #1000000
Line 2,413 ⟶ 5,923:
0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
00000000000000000000000000000000000000000000000000000
</pre>
===Alternative===
This uses memoized streams - similar in principle to the classic lazy-evaluated sequence, but with a java flavor. Hope you like this recipe!
<syntaxhighlight 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;
}
}
</syntaxhighlight>
<pre>
$ 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
$
</pre>
Line 2,421 ⟶ 6,056:
Note the use of <code>'''for''' (x in obj)</code> to iterate over the ''properties'' of an object, versus <code>'''for each''' (y in obj)</code> to iterate over the ''values'' of the properties of an object.
<
var queues = {2: [], 3: [], 5: []};
var base;
Line 2,447 ⟶ 6,082:
for (; i <= 1690; i++)
ham.next();
print(i + " => " + ham.next());</
{{out}}
<pre>1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36
Line 2,461 ⟶ 6,096:
--Mike Lorenz
<
<head></head>
<body>
Line 2,558 ⟶ 6,193:
});
</script>
</html></
{{out}}
Line 2,612 ⟶ 6,247:
1000000: 4157409948433216829957008507500000000
Elapsed time: 1.989 seconds</pre>
=={{header|jq}}==
{{works with| jq | 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 ====
<syntaxhighlight 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;
# 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)</syntaxhighlight>
'''Examples''':
<syntaxhighlight lang="jq"># First twenty:
hamming(20)
# See elsewhere for output
# 1691st Hamming number:
hamming(-1691)
# => 2125764000
# Millionth:
hamming(-1000000)
# => 1.926511252902403e+44
</syntaxhighlight>
==== 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:
<syntaxhighlight 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));
# 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]));
# 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;
# 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;
</syntaxhighlight>
'''Examples'''
<syntaxhighlight lang="jq"># The first twenty Hamming numbers as integers:
hamming(-20) | hamming_toi
# => (see elsewhere)
# 1691st as a Hamming triple:
hamming(-1691)
# => [5,12,3]
# The millionth:
hamming(-1000000)
# => [55,47,64]</syntaxhighlight>
=={{header|Julia}}==
Simple brute force algorithm, derived from the discussion at ProgrammingPraxis.com.
<syntaxhighlight 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])</syntaxhighlight>{{output}}<pre>
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
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:
<syntaxhighlight 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</syntaxhighlight>
===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 Julia is quite slow, but also because there are many repeated calculations (3 times 2 equals 2 times three, etc.). The following code is about 60 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:
{{trans|Nim}}
<syntaxhighlight lang="julia">struct LogRep
lg :: Float64
x2 :: UInt32
x3 :: UInt32
x5 :: UInt32
end
const ONE = LogRep(0.0, 0, 0, 0)
const LB2_2 = 1.0
const LB2_3 = log(2,3)
const LB2_5 = log(2,5)
function mult2(lr :: LogRep) # :: LogRep
LogRep(lr.lg + LB2_2, lr.x2 + 1, lr.x3, lr.x5)
end
function mult3(lr :: LogRep) # :: LogRep
LogRep(lr.lg + LB2_3, lr.x2, lr.x3 + 1, lr.x5)
end
function mult5(lr :: LogRep) # :: LogRep
LogRep(lr.lg + LB2_5, lr.x2, lr.x3, lr.x5 + 1)
end
function lr2BigInt(lr :: LogRep) # :: BigInt
BigInt(2)^lr.x2 * BigInt(3)^lr.x3 * BigInt(5)^lr.x5
end
mutable struct HammingsLog
s2 :: Vector{LogRep}
s3 :: Vector{LogRep}
s5 :: LogRep
mrg :: LogRep
s2hdi :: Int
s3hdi :: Int
HammingsLog() = new(
[ONE],
[mult3(ONE)],
mult5(ONE),
mult3(ONE),
1, 1
)
end
Base.eltype(::Type{HammingsLog}) = LogRep
function Base.iterate(HM::HammingsLog, st = HM) # :: Union{Nothing,Tuple{LogRep,HammingsLog}}
s2sz = length(st.s2)
if st.s2hdi + st.s2hdi - 2 >= s2sz
ns2sz = s2sz - st.s2hdi + 1
copyto!(st.s2, 1, st.s2, st.s2hdi, ns2sz)
resize!(st.s2, ns2sz); st.s2hdi = 1
end
rslt = @inbounds(st.s2[st.s2hdi])
if rslt.lg < st.mrg.lg
st.s2hdi += 1
else
s3sz = length(st.s3)
if st.s3hdi + st.s3hdi - 2 >= s3sz
ns3sz = s3sz - st.s3hdi + 1
copyto!(st.s3, 1, st.s3, st.s3hdi, ns3sz)
resize!(st.s3, ns3sz); st.s3hdi = 1
end
rslt = st.mrg; push!(st.s3, mult3(rslt))
st.s3hdi += 1; chkv = @inbounds(st.s3[st.s3hdi])
if chkv.lg < st.s5.lg
st.mrg = chkv
else
st.mrg = st.s5; st.s5 = mult5(st.s5); st.s3hdi -= 1
end
end
push!(st.s2, mult2(rslt)); rslt, st
end
function test(n :: Int) :: Tuple{LogRep, Float64}
start = time(); rslt :: LogRep = ONE
count = n; for t in HammingsLog() count <= 1 && (rslt = t; break); count -= 1 end
elpsd = (time() - start) * 1000
rslt, elpsd
end
foreach(x -> print(lr2BigInt(x)," "), (Iterators.take(HammingsLog(), 20))); println()
let count = 1691; for t in HammingsLog() count <= 1 && (println(lr2BigInt(t)); break); count -= 1 end end
rslt, elpsd = test(1000000)
println(lr2BigInt(rslt))
println("This last took $elpsd milliseconds.")</syntaxhighlight>
{{out}}
<pre>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 16.8759822845459 milliseconds.</pre>
The above execution time is as run on an Intel i5-6500 at 3.6 GHz (single threaded boost), and the program can find the billionth Hamming number in about 17 seconds.
===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:
<syntaxhighlight 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))))</syntaxhighlight>
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:
<syntaxhighlight 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</syntaxhighlight>
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.
=={{header|Kotlin}}==
{{trans|Java}}
<
import java.util.
val Three = BigInteger.valueOf(3)!!
val Five = BigInteger.valueOf(5)!!
fun updateFrontier(x : BigInteger, pq : PriorityQueue<BigInteger>) {
pq
pq
pq
}
Line 2,634 ⟶ 6,667:
for (i in 1 .. n-1) {
lowest = frontier.poll() ?: lowest
while (frontier.peek()
frontier.poll()
updateFrontier(lowest, frontier)
Line 2,642 ⟶ 6,675:
fun main(args : Array<String>) {
System.out
for (i in 1 .. 20)
System.out
System.out
System.out
}</
{{out}}
Line 2,653 ⟶ 6,686:
Hamming(1691) = 2125764000
Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
===Overloaded function:===
<syntaxhighlight 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($it) = " + hamming(it)) }
}</syntaxhighlight>
===Recursive function:===
<syntaxhighlight lang="scala">import java.math.BigInteger
import java.util.*
val One = BigInteger.ONE!!
val Three = BigInteger.valueOf(3)!!
val Five = BigInteger.valueOf(5)!!
infix fun PriorityQueue<BigInteger>.update(x: BigInteger) : PriorityQueue<BigInteger> {
add(x.shiftLeft(1))
add(x.multiply(Three))
add(x.multiply(Five))
return this
}
fun hamming(a: Any?): Any = when (a) {
is Number -> {
val pq = PriorityQueue<BigInteger>() update One
var lowest = One
repeat(a.toInt() - 1) {
lowest = pq.poll() ?: lowest
while (pq.peek() == lowest) pq.poll()
pq update lowest
}
lowest
}
is Iterable<*> -> a.map { hamming(it) }
else -> throw IllegalArgumentException("cannot parse argument")
}
fun main(args: Array<String>) {
arrayOf(1..20, 1691, 1000000).forEach { println("Hamming($it) = " + hamming(it)) }
}</syntaxhighlight>
{{out}}
<pre>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</pre>
===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):
{{trans|scala}}
<syntaxhighlight 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.")
}</syntaxhighlight>
{{output}}
<pre>[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.</pre>
Run on a AMD Bulldozer FX8120 3.1 GHz which is about half the speed as an equivalent Intel (but also half the price).
=={{header|Lambdatalk}}==
===1) recursive version===
<syntaxhighlight lang="scheme">
{def hamming
{def hamming.loop
{lambda {:h :a :i :b :j :c :k :m :n}
{if {>= :n :m}
then {A.last :h}
else {let { {:h {A.set! :n {min :a :b :c} :h}}
{:a :a} {:i :i}
{:b :b} {:j :j}
{:c :c} {:k :k}
{:m :m} {:n :n}
} {hamming.loop :h
{if {= :a {A.get :n :h}}
then {* 2 {A.get {+ :i 1} :h}} {+ :i 1}
else :a :i}
{if {= :b {A.get :n :h}}
then {* 3 {A.get {+ :j 1} :h}} {+ :j 1}
else :b :j}
{if {= :c {A.get :n :h}}
then {* 5 {A.get {+ :k 1} :h}} {+ :k 1}
else :c :k}
:m
{+ :n 1} }
}}}}
{lambda {:n}
{hamming.loop {A.new {S.serie 1 :n}} 2 0 3 0 5 0 :n 1}
}}
-> hamming
{S.map hamming {S.serie 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 // < 200ms
Currently limited to javascript's integers and by stackoverflow on some computers.
</syntaxhighlight>
===2) iterative version===
Build a table of 2^i•3^j•5^k from i,j,k = 0 to n and sort it.
===2.1) compute===
<syntaxhighlight lang="scheme">
{def ham
{lambda {:n}
{S.sort <
{S.map {{lambda {:n :i}
{S.map {{lambda {:n :i :j}
{S.map {{lambda {:i :j :k}
{* {pow 2 :i} {pow 3 :j} {pow 5 :k}}} :i :j}
{S.serie 0 :n} } } :n :i}
{S.serie 0 :n} } } :n}
{S.serie 0 :n} }
}}}
-> ham
{def H {ham 30}}
-> H
{S.slice 0 19 {H}}
-> 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
{S.get 1690 {H}}
-> 2125764000 // on my macbook pro
</syntaxhighlight>
===2.2) display===
Display a hamming number as 2<sup>a</sup>•3<sup>b</sup>•5<sup>c</sup>
<syntaxhighlight lang="scheme">
{def factor
{def factor.r
{lambda {:n :i}
{if {> :i :n}
then
else {if {= {% :n :i} 0}
then :i {factor.r {/ :n :i} :i}
else {factor.r :n {+ :i 1}} }}}}
{lambda {:n}
:n is the product of 1 {factor.r :n 2} }}
-> factor
{def asproductofpowers
{def asproductofpowers.loop
{lambda {:a :b :c :n}
{if {= {S.first :n} 1}
then 2{sup :a}•3{sup :b}•5{sup :c}
else {asproductofpowers.loop
{if {= {S.first :n} 2} then {+ :a 1} else :a}
{if {= {S.first :n} 3} then {+ :b 1} else :b}
{if {= {S.first :n} 5} then {+ :c 1} else :c}
{W.rest :n} }
}}}
{lambda {:n}
{asproductofpowers.loop 0 0 0 {S.reverse :n}}}}
-> asproductofpowers
{factor 2125764000}
-> 2125764000 is the product of 1 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5
{asproductofpowers {factor 2125764000}}
-> 2^5•3^12•5^3
{S.map {lambda {:i} {div}:i: {S.get :i {H}} =
{asproductofpowers {factor {S.get :i {H}}}}}
{S.serie 0 19}}
->
0: 1 = 2^0•3^0•5^0
1: 2 = 2^1•3^0•5^0
2: 3 = 2^0•3^1•5^0
3: 4 = 2^2•3^0•5^0
4: 5 = 2^0•3^0•5^1
5: 6 = 2^1•3^1•5^0
6: 8 = 2^3•3^0•5^0
7: 9 = 2^0•3^2•5^0
8: 10 = 2^1•3^0•5^1
9: 12 = 2^2•3^1•5^0
10: 15 = 2^0•3^1•5^1
11: 16 = 2^4•3^0•5^0
12: 18 = 2^1•3^2•5^0
13: 20 = 2^2•3^0•5^1
14: 24 = 2^3•3^1•5^0
15: 25 = 2^0•3^0•5^2
16: 27 = 2^0•3^3•5^0
17: 30 = 2^1•3^1•5^1
18: 32 = 2^5•3^0•5^0
19: 36 = 2^2•3^2•5^0
</syntaxhighlight>
See http://lambdaway.free.fr/lambdawalks/?view=hamming_numbers3 for a better display as 2<sup>a</sup>•3<sup>b</sup>•5<sup>c</sup>.
=={{header|Liberty BASIC}}==
LB has unlimited precision integers.
<syntaxhighlight lang="lb">
dim h( 1000000)
Line 2,680 ⟶ 6,990:
next n
hamming =h( limit -1)
end function</
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Line 2,687 ⟶ 6,997:
H( 1000000) 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
=={{header|Logo}}==
<
; queues
make "twos [1]
Line 2,714 ⟶ 7,024:
repeat 20 [print next.ham]
repeat 1690-20 [ignore next.ham]
print next.ham</
=={{header|Lua}}==
<
hammings = {1}
prev, vals = {1, 1, 1}
Line 2,740 ⟶ 7,050:
n, l = 0, 0
while n < 2^31 do n, l = j(), n end
print(l)</
=={{header|
===For Long Only===
We have to exit loop (and function) before calculating new X2 or X3 or X4 and get overflow error
<syntaxhighlight 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
</syntaxhighlight>
{{out}}
<pre style="height:30ex;overflow:scroll">
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
</pre >
===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
<syntaxhighlight 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
</syntaxhighlight>
{{out}}
<pre style="height:30ex;overflow:scroll">
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
</pre >
=={{header|Mathematica}} / {{header|Wolfram Language}}==
<syntaxhighlight 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]];</
<pre>HammingList[20]
-> {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36}
Line 2,752 ⟶ 7,203:
HammingList[1000000] // Last
->519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|MATLAB}} / {{header|Octave}}==
{{trans|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.
<syntaxhighlight 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))</syntaxhighlight>
=={{header|Mojo}}==
Since current Mojo (version 0.7) does not have many forms of recursive expression, the below is an imperative version of the First In Last Out (FILO) Queue version of the fastest iterative Nim version using logarithmic approximations for the comparison and final conversion of the power tuples to a big integer output. Since Mojo does not currently have a big integer library, enough of the required functionality of one (multiplication and conversion to string) is implemented in the following code:
{{trans|Nim}}
<syntaxhighlight lang="mojo">from collections.vector import (DynamicVector, CollectionElement)
from math import (log2, trunc, pow)
from memory import memset_zero #, memcpy)
from time import now
alias cCOUNT: Int = 1_000_000
struct BigNat(Stringable): # enough just to support conversion and printing
''' Enough "infinite" precision to support as required here - multiply and
divide by 10 conversion to string...
'''
var contents: DynamicVector[UInt32]
fn __init__(inout self):
self.contents = DynamicVector[UInt32]()
fn __init__(inout self, val: UInt32):
self.contents = DynamicVector[UInt32](4)
self.contents.resize(1, val)
fn __copyinit__(inout self, existing: Self):
self.contents = existing.contents
fn __moveinit__(inout self, owned existing: Self):
self.contents = existing.contents^
fn __str__(self) -> String:
var rslt: String = ""
var v = self.contents
while len(v) > 0:
var t: UInt64 = 0
for i in range(len(v) - 1, -1, -1):
t = ((t << 32) + v[i].to_int())
v[i] = (t // 10).to_int(); t -= v[i].to_int() * 10
var sz = len(v) - 1
while sz >= 0 and v[sz] == 0: sz -= 1
v.resize(sz + 1, 0)
rslt = str(t) + rslt
return rslt
fn mult(inout self, mltplr: Self):
var rslt = DynamicVector[UInt32]()
rslt.resize(len(self.contents) + len(mltplr.contents), 0)
for i in range(len(mltplr.contents)):
var t: UInt64 = 0
for j in range(len(self.contents)):
t += self.contents[j].to_int() * mltplr.contents[i].to_int() + rslt[i + j].to_int()
rslt[i + j] = (t & 0xFFFFFFFF).to_int(); t >>= 32
rslt[i + len(self.contents)] += t.to_int()
var sz = len(rslt) - 1
while sz >= 0 and rslt[sz] == 0: sz -= 1
rslt.resize(sz + 1, 0); self.contents = rslt
alias lb2: Float64 = 1.0
alias lb3: Float64 = log2[DType.float64, 1](3.0)
alias lb5: Float64 = log2[DType.float64, 1](5.0)
@value
struct LogRep(CollectionElement, Stringable):
var logrep: Float64
var x2: UInt32
var x3: UInt32
var x5: UInt32
fn __del__(owned self): return
@always_inline
fn mul2(self) -> Self:
return LogRep(self.logrep + lb2, self.x2 + 1, self.x3, self.x5)
@always_inline
fn mul3(self) -> Self:
return LogRep(self.logrep + lb3, self.x2, self.x3 + 1, self.x5)
@always_inline
fn mul5(self) -> Self:
return LogRep(self.logrep + lb5, self.x2, self.x3, self.x5 + 1)
fn __str__(self) -> String:
var rslt = BigNat(1)
fn expnd(inout rslt: BigNat, bs: UInt32, n: UInt32):
var bsm = BigNat(bs); var nm = n
while nm > 0:
if (nm & 1) != 0: rslt.mult(bsm)
bsm.mult(bsm); nm >>= 1
expnd(rslt, 2, self.x2); expnd(rslt, 3, self.x3); expnd(rslt, 5, self.x5)
return str(rslt)
alias oneLR: LogRep = LogRep(0.0, 0, 0, 0)
alias LogRepThunk = fn() escaping -> LogRep
fn hammingsLogImp() -> LogRepThunk:
var s2 = DynamicVector[LogRep](); var s3 = DynamicVector[LogRep](); var s5 = oneLR; var mrg = oneLR
s2.resize(512, oneLR); s2[0] = oneLR.mul2(); s3.resize(1, oneLR); s3[0] = oneLR.mul3()
# var s2p = s2.steal_data(); var s3p = s3.steal_data()
var s2hdi = 0; var s2tli = -1; var s3hdi = 0; var s3tli = -1
@always_inline
fn next() escaping -> LogRep:
var rslt = s2[s2hdi]
var s2len = len(s2)
s2tli += 1;
if s2tli >= s2len:
s2tli = 0
if s2hdi == s2tli:
if s2len < 1024:
s2.resize(1024, oneLR)
else:
s2.resize(s2len + s2len, oneLR) # ; s2p = s2.steal_data()
for i in range(s2hdi):
s2[s2len + i] = s2[i]
# memcpy[UInt8, 0](s2p + s2len, s2p, sizeof[LogRep]() * s2hdi)
s2tli += s2len; s2len += s2len
if rslt.logrep < mrg.logrep:
s2hdi += 1
if s2hdi >= s2len:
s2hdi = 0
else:
rslt = mrg
var s3len = len(s3)
s3tli += 1;
if s3tli >= s3len:
s3tli = 0
if s3hdi == s3tli:
if s3len < 1024:
s3.resize(1024, oneLR)
else:
s3.resize(s3len + s3len, oneLR) # ; s3p = s3.steal_data()
for i in range(s3hdi):
s3[s3len + i] = s3[i]
# memcpy[UInt8, 0](s3p + s3len, s3p, sizeof[LogRep]() * s3hdi)
s3tli += s3len; s3len += s3len
if mrg.logrep < s5.logrep:
s3hdi += 1
if s3hdi >= s3len:
s3hdi = 0
else:
s5 = s5.mul5()
s3[s3tli] = rslt.mul3(); let t = s3[s3hdi];
mrg = t if t.logrep < s5.logrep else s5
s2[s2tli] = rslt.mul2(); return rslt
return next
fn main():
print("The first 20 Hamming numbers are:")
var f = hammingsLogImp();
for i in range(20): print_no_newline(f(), " ")
print()
f = hammingsLogImp(); var h: LogRep = oneLR
for i in range(1691): h = f()
print("The 1691st Hamming number is", h)
let strt: Int = now()
f = hammingsLogImp()
for i in range(cCOUNT): h = f()
let elpsd = (now() - strt) / 1000
print("The " + str(cCOUNT) + "th Hamming number is:")
print("2**" + str(h.x2) + " * 3**" + str(h.x3) + " * 5**" + str(h.x5))
let lg2 = lb2 * Float64(h.x2.to_int()) + lb3 * Float64(h.x3.to_int()) + lb5 * Float64(h.x5.to_int())
let lg10 = lg2 / log2(Float64(10))
let expnt = trunc(lg10); let num = pow(Float64(10.0), lg10 - expnt)
let apprxstr = str(num) + "E+" + str(expnt.to_int())
print("Approximately: ", apprxstr)
let answrstr = str(h)
print("The result has", len(answrstr), "digits.")
print(answrstr)
print("This took " + str(elpsd) + " microseconds.")</syntaxhighlight>
{{out}}
<pre>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
Approximately: 5.1931278110620553E+83
The result has 84 digits.
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This took 3626.192 microseconds.</pre>
The above was as run on an AMD 7840HS CPU single-thread boosted to 5.1 GHz. It is about the same speed as the Nim version from which it was translated.
=={{header|MUMPS}}==
<
For which=2,3,5 Set number=1
For count=1:1:n Do
Line 2,786 ⟶ 7,441:
20: 36
1691: 2125764000
2000: 8062156800</
=={{header|Nim}}==
{{libheader|bigints}}
===Classic Dijkstra algorithm===
<syntaxhighlight 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)
Line 2,802 ⟶ 7,463:
for i in 0..h.high: h[i] = initBigInt(1)
for n in 1 ..
h[n] = min(x2, x3, x5)
if x2 == h[n]:
Line 2,817 ⟶ 7,478:
for i in 1 .. 20:
echo ""
echo hamming(1691)
echo hamming(1_000_000)</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
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:
<syntaxhighlight 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."</syntaxhighlight>
{{output}}
<pre>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.</pre>
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:
{{trans|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".
<syntaxhighlight 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."</syntaxhighlight>
{{output}}
<pre>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.</pre>
The above result was obtained by compiling with the default "mark-and-sweep" Garbage collector with <code>-d:release -d:danger</code> (all checking including bounds checks turned off); One should not use the new <code>--gc:arc</code> compilation argument (automatic reference counting) with this implementation as the lazy lists are cyclic but compiling with <code>--gc:orc</code> gives an execution time of about 80% of the execution time as compared to the conventional garbage collection, and is slower than the <code>--gc:arc</code> 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".
<syntaxhighlight 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()</syntaxhighlight>
{{out}}
<pre>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.</pre>
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 <code>--gc:orc</code> memory management is used, which is slower than using the <code>--gc:arc</code> 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:
<syntaxhighlight 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."</syntaxhighlight>
{{out}}
<pre>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.</pre>
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 for the sorting comparisons 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:
<syntaxhighlight lang="nim"># HammingsLogImp.nim
# compile with: nim c -d:danger -t:-march=native -d:LTO --gc:arc HammingsLogImp
import bigints, std/math
from std/times import inMicroseconds
from std/monotimes import getMonoTime, `-`
type LogRep = (float64, uint32, uint32, uint32)
let one: LogRep = (0.0, 0'u32, 0'u32, 0'u32)
let lb2 = 1.0'f64; let lb3 = 3.0.log2; let lb5 = 5.0.log2
proc mul2(me: Logrep): Logrep {.inline.} =
(me[0] + lb2, me[1] + 1, me[2], me[3])
proc mul3(me: Logrep): Logrep {.inline.} =
(me[0] + lb3, me[1], me[2] + 1, me[3])
proc mul5(me: Logrep): Logrep {.inline.} =
(me[0] + lb5, me[1], me[2], me[3] + 1)
proc lr2BigInt(lr: Logrep): 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; vm = vm shr 1
xpnd(2, lr[1]) * xpnd(3, lr[2]) * xpnd(5, lr[3])
iterator hammingsLogImp(): LogRep =
var
s2 = newSeq[Logrep](1024) # give it size one so doubling size works
s3 = newSeq[Logrep](1024) # reasonably sized
s5 = one.mul5 # initBigInt 5
mrg = one.mul3 # initBigInt 3
s2hdi, s2tli, s3hdi, s3tli = 0
yield one
s2[0] = one.mul2; s3[0] = one.mul3
while true:
s2tli += 1
if s2hdi + s2hdi >= s2tli: # move in-place to avoid allocation
copyMem(addr(s2[0]), addr(s2[s2hdi]), sizeof(LogRep) * (s2tli - s2hdi))
s2tli -= s2hdi; s2hdi = 0
let cps2 = s2.len # move in-place to avoid allocation
if s2tli >= cps2: s2.setLen(cps2 + cps2)
var rsltp = addr(s2[s2hdi])
if rsltp[][0] < mrg[0]: s2[s2tli] = rsltp[].mul2; s2hdi += 1; yield rsltp[]
else:
s3tli += 1
if s3hdi + s3hdi >= s3tli: # move in-place to avoid allocation
copyMem(addr(s3[0]), addr(s3[s3hdi]), sizeof(LogRep) * (s3tli - s3hdi))
s3tli -= s3hdi; s3hdi = 0
let cps3 = s3.len
if s3tli >= cps3: s3.setLen(cps3 + cps3)
s2[s2tli] = mrg.mul2; s3[s3tli] = mrg.mul3; s3hdi += 1
let arsltp = addr(s3[s3hdi])
let rslt = mrg
if arsltp[][0] < s5[0]: mrg = arsltp[]
else: mrg = s5; s5 = s5.mul5; s3hdi -= 1
yield rslt
var cnt = 0
for h in hammingsLogImp():
write stdout, h.lr2BigInt, " "; cnt += 1
if cnt >= 20: break
echo ""
cnt = 0
for h in hammingsLogImp():
cnt += 1
if cnt >= 1691: echo h.lr2BigInt; break
let strt = getMonoTime()
var rslt: LogRep
cnt = 0
for h in hammingsLogImp():
cnt += 1
if cnt >= 1_000_000: rslt = h; break # """
let elpsd = (getMonoTime() - strt).inMicroseconds
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.lr2BigInt()
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 ", elpsd, " microseconds."</syntaxhighlight>
{{output}}
<pre>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 6004 microseconds.</pre>
The time as shown is for for compilation as in the second line of code; with these options, the billionth Hamming number can be calculated in about 7 seconds.
'''Faster alternate to the above using a ring buffer'''
As other language contributions refer to it, the above code is left in place; however, it seems that the amount of time spent "draining" the buffers by already-used values using copying as used in the above code can be eliminated by using the buffers as "ring buffers" by making the indices wrap around from the end of the buffers to the beginning and detecting when the buffer needs to be "grown" by when the next/last/tail index runs into the first/head index, and changing the "grow" logic a little so as to open up a hole between the next and first indexes by the size of the expansion once the buffer size has "grown". The code is as follows:
<syntaxhighlight lang=nim># HammingsLogDQ.nim
# compile with: nim c -d:danger -t:-march=native -d:LTO --gc:arc HammingsImpLogQ
import bigints, std/math
from std/times import inMicroseconds
from std/monotimes import getMonoTime, `-`
type LogRep = (float64, uint32, uint32, uint32)
let one: LogRep = (0.0, 0'u32, 0'u32, 0'u32)
let lb2 = 1.0'f64; let lb3 = 3.0.log2; let lb5 = 5.0.log2
proc mul2(me: Logrep): Logrep {.inline.} =
(me[0] + lb2, me[1] + 1, me[2], me[3])
proc mul3(me: Logrep): Logrep {.inline.} =
(me[0] + lb3, me[1], me[2] + 1, me[3])
proc mul5(me: Logrep): Logrep {.inline.} =
(me[0] + lb5, me[1], me[2], me[3] + 1)
proc lr2BigInt(lr: Logrep): 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; vm = vm shr 1
xpnd(2, lr[1]) * xpnd(3, lr[2]) * xpnd(5, lr[3])
proc `$`(lr: LogRep): string {.inline.} = $lr2BigInt(lr)
iterator hammingsLogQ(): LogRep =
var s2msk, s3msk = 1024
var s2 = newSeq[LogRep] s2msk; var s3 = newSeq[LogRep] s3msk
s2msk -= 1; s3msk -= 1; s2[0] = one; var s2nxti = 1
var s2hdi, s3hdi, s3nxti = 0
var s5 = one.mul5; var mrg = one.mul3
while true:
let s2hdp = addr(s2[s2hdi])
if s2hdp[][0] < mrg[0]:
s2[s2nxti] = s2hdp[].mul2; s2hdi += 1; s2hdi = s2hdi and s2msk
yield s2hdp[]
else:
s2[s2nxti] = mrg.mul2; s3[s3nxti] = mrg.mul3; yield mrg
let s3hdp = addr(s3[s3hdi])
if s3hdp[0] < s5[0]:
mrg = s3hdp[]; s3hdi += 1; s3hdi = s3hdi and s3msk
else: mrg = s5; s5 = s5.mul5
s3nxti += 1; s3nxti = s3nxti and s3msk
if s3nxti == s3hdi: # buffer full - expand...
let sz = s3msk + 1; s3msk = sz + sz; s3.setLen(s3msk); s3msk -= 1
if s3hdi == 0: s3nxti = sz
else: # put extra space between next and head...
copyMem(addr(s3[s3hdi + sz]), addr(s3[s3hdi]),
sizeof(LogRep) * (sz - s3hdi)); s3hdi += sz
s2nxti += 1; s2nxti = s2nxti and s2msk
if s2nxti == s2hdi: # buffer full - expand...
let sz = s2msk + 1; s2msk = sz + sz; s2.setLen s2msk; s2msk -= 1
if s2hdi == 0: s2nxti = sz # copy all in a single block...
else: # make extra space between next and head...
copyMem(addr(s2[s2hdi + sz]), addr(s2[s2hdi]),
sizeof(LogRep) * (sz - s2hdi)); s2hdi += sz
# testing it...
var cnt = 0
for h in hammingsLogQ():
write stdout, h, " "; cnt += 1
if cnt >= 20: break
echo ""
cnt = 0
for h in hammingsLogQ():
cnt += 1
if cnt >= 1691: echo h; break
let strt = getMonoTime()
var rslt: LogRep
cnt = 0
for h in hammingsLogQ():
cnt += 1
if cnt >= 1_000_000: rslt = h; break # """
let elpsd = (getMonoTime() - strt).inMicroseconds
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 s = $rslt
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 ", elpsd, " microseconds."</syntaxhighlight>
{{out}}
<pre>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 5044 microseconds.</pre>
As tested on an Intel i5-6500 (3.6 GHz single-threaded boosted), this is about a millisecond or about twenty percent faster than the version above, and can find the billionth Hamming number in about 4.5 seconds on this machine. The reason this is faster is mostly due to the elimination of the majority of the copy operations.
===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:
{{trans|Rust}}
<syntaxhighlight 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."</syntaxhighlight>
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:
<syntaxhighlight 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."</syntaxhighlight>
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.
=={{header|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. <i>min_elt</i> and <i>add</i> are <em>O</em>(log N).
<syntaxhighlight 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 ());</syntaxhighlight>
Output:
<
The 1691st is 2125764000
</blockquote>
===Arbitrary precision===
An arbitrary precision version for the one millionth number. Compile with eg: <i>ocamlopt -o hamming.exe nums.cmxa hamming.ml</i>
<syntaxhighlight 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 ()));</syntaxhighlight>
Output:
<blockquote>The 1000000th is 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</blockquote>
=={{header|Oz}}==
===Lazy Version===
{{trans|Haskell}}
<
fun lazy {HammingFun}
1|{FoldL1 [{MultHamming 2} {MultHamming 3} {MultHamming 5}] LMerge}
Line 2,858 ⟶ 8,274:
{ForAll {List.take Hamming 20} System.showInfo}
{System.showInfo {Nth Hamming 1690}}
{System.showInfo {Nth Hamming 1000000}}</
===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 [https://en.wikipedia.org/wiki/Hamming_numbers])
<syntaxhighlight 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
</syntaxhighlight>
Strict version made by pietervdvn; do what you want with the code.
=={{header|PARI/GP}}==
This is a basic implementation; finding the millionth term requires
<
my(
for(m=2,n,
if(
if(
if(
);
};
H(n)=Hupto(n)[n];
Line 2,877 ⟶ 8,464:
Hupto(20)
H(1691)
H(10^6)</
{{out}}
<pre>%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</pre>
=={{header|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
<syntaxhighlight 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.</syntaxhighlight>
Output
<pre>
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</pre>
===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:
<syntaxhighlight 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.</syntaxhighlight>
{{out}}
<pre>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.</pre>
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 )
<syntaxhighlight 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.</syntaxhighlight>
{{out|@ TIO.RUN}}
<pre>
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
@home:
//tested til 1E12 with 4.4 Ghz 5600G Free Pascal Compiler version 3.2.2-[2022/11/22] for x86_64
Timed 1,000,000,000,000 in 57:53.015
ln 19444.3672890 2^1126 3^16930 5^40 -> see Haskell-Version [https://ideone.com/RnAh5X]
Actual Index 1000075683108
ln 19444.3672890 2^8295 3^2426 5^6853
2 elemcount 106935365 out of 156797362
3 elemcount 12083 out of 12969
5 elemcount 1 out of 2
user 57m51.015s <<
sys 0m1.616s
</pre>
=={{header|Perl}}==
<syntaxhighlight lang
use warnings;
use List::Util 'min';
# If you want the large output, uncomment either the one line
# marked (1) or the two lines marked (2)
#use Math::GMP qw/:constant/; # (1) uncomment this to use Math::GMP
#use Math::GMPz; # (2) uncomment this plus later line for Math::GMPz
sub ham_gen {
#@m = map { Math::GMPz->new($_) } @m; # (2) uncomment for Math::GMPz
my $n = min($s[0][0], $s[1][0], $s[2][0]);
for (0 .. 2) {
shift @{$s[$_]} if $s[$_][0] == $n;
}
return $n
}
}
my
my $i = 0;
++$i, print $h->(), " " until $i > 20;
print "...\n";
++$i, $h->() until $i == 1690;
print ++$i, "-th: ", $h->(), "\n";
# You will need to pick one of the bigint choices
#++$i, $h->() until $i == 999999;
#print ++$i, "-th: ", $h->(), "\n";
</syntaxhighlight>
{{out}}
<pre>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</pre>
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).
=={{header|Phix}}==
{{trans|AWK}}
{{libheader|Phix/mpfr}}
standard and gmp versions
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">N</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">h</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">N</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">atom</span> <span style="color: #000000;">x2</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">2</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">x3</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">x5</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">5</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">hn</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">N</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">hn</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">min</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">min</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">))</span>
<span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">hn</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">hn</span><span style="color: #0000FF;">==</span><span style="color: #000000;">x2</span> <span style="color: #008080;">then</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x2</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">2</span><span style="color: #0000FF;">*</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">hn</span><span style="color: #0000FF;">==</span><span style="color: #000000;">x3</span> <span style="color: #008080;">then</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x3</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">*</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">j</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">hn</span><span style="color: #0000FF;">==</span><span style="color: #000000;">x5</span> <span style="color: #008080;">then</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x5</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">5</span><span style="color: #0000FF;">*</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">k</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">N</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">include</span> <span style="color: #000000;">builtins</span><span style="color: #0000FF;">\</span><span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">mpz_hamming</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">N</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">h</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_inits</span><span style="color: #0000FF;">(</span><span style="color: #000000;">N</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">mpz</span> <span style="color: #000000;">x2</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">2</span><span style="color: #0000FF;">),</span>
<span style="color: #000000;">x3</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">3</span><span style="color: #0000FF;">),</span>
<span style="color: #000000;">x5</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">(</span><span style="color: #000000;">5</span><span style="color: #0000FF;">),</span>
<span style="color: #000000;">hn</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_init</span><span style="color: #0000FF;">()</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">N</span> <span style="color: #008080;">do</span>
<span style="color: #7060A8;">mpz_set</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hn</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">mpz_min</span><span style="color: #0000FF;">({</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">}))</span>
<span style="color: #7060A8;">mpz_set</span><span style="color: #0000FF;">(</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">],</span><span style="color: #000000;">hn</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">mpz_cmp</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hn</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #7060A8;">mpz_mul_si</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">],</span><span style="color: #000000;">2</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">mpz_cmp</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hn</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #7060A8;">mpz_mul_si</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">j</span><span style="color: #0000FF;">],</span><span style="color: #000000;">3</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #7060A8;">mpz_cmp</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hn</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">)=</span><span style="color: #000000;">0</span> <span style="color: #008080;">then</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #7060A8;">mpz_mul_si</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">k</span><span style="color: #0000FF;">],</span><span style="color: #000000;">5</span><span style="color: #0000FF;">)</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">N</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">20</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">append</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">))</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #0000FF;">?</span><span style="color: #000000;">s</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%d\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1691</span><span style="color: #0000FF;">))</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%d (wrong!)\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1000000</span><span style="color: #0000FF;">))</span> <span style="color: #000080;font-style:italic;">--(the hn==x2 etc fail, so multiplies are all wrong)</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">mpz_get_str</span><span style="color: #0000FF;">(</span><span style="color: #000000;">mpz_hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1691</span><span style="color: #0000FF;">))})</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"%s\n"</span><span style="color: #0000FF;">,{</span><span style="color: #7060A8;">mpz_get_str</span><span style="color: #0000FF;">(</span><span style="color: #000000;">mpz_hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1000000</span><span style="color: #0000FF;">))})</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
{1
2125764000
246192725545902804828662268200 (wrong!)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
===A much faster logarithmic version===
This proved much easier to implement than scanning the other entries suggested [not copied, they all frighten me].<br>
At some point, comparing logs will no doubt get it wrong, but I have no idea when that might happen.
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>
<span style="color: #000080;font-style:italic;">-- numbers kept as {log,{pow2,pow3,pow5}},
-- value is ~= exp(log), == (2^pow2)*(3^pow3)*(5^pow5)</span>
<span style="color: #008080;">enum</span> <span style="color: #000000;">LOG</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">POWS</span>
<span style="color: #008080;">enum</span> <span style="color: #000000;">POW2</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">POW3</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">POW5</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">lnmin</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">a</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">sequence</span> <span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">return</span> <span style="color: #008080;">iff</span><span style="color: #0000FF;">(</span><span style="color: #000000;">a</span><span style="color: #0000FF;">[</span><span style="color: #000000;">LOG</span><span style="color: #0000FF;">]<</span><span style="color: #000000;">b</span><span style="color: #0000FF;">[</span><span style="color: #000000;">LOG</span><span style="color: #0000FF;">]?</span><span style="color: #000000;">a</span><span style="color: #0000FF;">:</span><span style="color: #000000;">b</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">constant</span> <span style="color: #000000;">ln1</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">log</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">),</span> <span style="color: #000000;">ln2</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">log</span><span style="color: #0000FF;">(</span><span style="color: #000000;">2</span><span style="color: #0000FF;">),</span> <span style="color: #000000;">ln3</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">log</span><span style="color: #0000FF;">(</span><span style="color: #000000;">3</span><span style="color: #0000FF;">),</span> <span style="color: #000000;">ln5</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">log</span><span style="color: #0000FF;">(</span><span style="color: #000000;">5</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #004080;">integer</span> <span style="color: #000000;">N</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">h</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">repeat</span><span style="color: #0000FF;">(</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">N</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">x2</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">ln2</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">}},</span>
<span style="color: #000000;">x3</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">ln3</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">}},</span>
<span style="color: #000000;">x5</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">ln5</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">1</span><span style="color: #0000FF;">}}</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1</span>
<span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">1</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">ln1</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">,</span><span style="color: #000000;">0</span><span style="color: #0000FF;">}}</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">N</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">deep_copy</span><span style="color: #0000FF;">(</span><span style="color: #000000;">lnmin</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">lnmin</span><span style="color: #0000FF;">(</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">)))</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">p</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">n</span><span style="color: #0000FF;">][</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">p</span><span style="color: #0000FF;">=</span><span style="color: #000000;">x2</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">then</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x2</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">deep_copy</span><span style="color: #0000FF;">(</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">i</span><span style="color: #0000FF;">])</span> <span style="color: #000000;">x2</span><span style="color: #0000FF;">[</span><span style="color: #000000;">LOG</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">ln2</span> <span style="color: #000000;">x2</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">][</span><span style="color: #000000;">POW2</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">p</span><span style="color: #0000FF;">=</span><span style="color: #000000;">x3</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">then</span> <span style="color: #000000;">j</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x3</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">deep_copy</span><span style="color: #0000FF;">(</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">j</span><span style="color: #0000FF;">])</span> <span style="color: #000000;">x3</span><span style="color: #0000FF;">[</span><span style="color: #000000;">LOG</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">ln3</span> <span style="color: #000000;">x3</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">][</span><span style="color: #000000;">POW3</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">p</span><span style="color: #0000FF;">=</span><span style="color: #000000;">x5</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span> <span style="color: #008080;">then</span> <span style="color: #000000;">k</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #000000;">x5</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">deep_copy</span><span style="color: #0000FF;">(</span><span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">k</span><span style="color: #0000FF;">])</span> <span style="color: #000000;">x5</span><span style="color: #0000FF;">[</span><span style="color: #000000;">LOG</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">ln5</span> <span style="color: #000000;">x5</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">][</span><span style="color: #000000;">POW5</span><span style="color: #0000FF;">]</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span> <span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">return</span> <span style="color: #000000;">h</span><span style="color: #0000FF;">[</span><span style="color: #000000;">N</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">hint</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">hm</span><span style="color: #0000FF;">)</span>
<span style="color: #000080;font-style:italic;">-- (obviously not accurate above 53 bits on a 32-bit system, or 64 bits on a 64 bit system)</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">p</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">hm</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span>
<span style="color: #008080;">return</span> <span style="color: #7060A8;">power</span><span style="color: #0000FF;">(</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POW2</span><span style="color: #0000FF;">])*</span><span style="color: #7060A8;">power</span><span style="color: #0000FF;">(</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POW3</span><span style="color: #0000FF;">])*</span><span style="color: #7060A8;">power</span><span style="color: #0000FF;">(</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POW5</span><span style="color: #0000FF;">])</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #004080;">sequence</span> <span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">20</span> <span style="color: #008080;">do</span>
<span style="color: #000000;">s</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">append</span><span style="color: #0000FF;">(</span><span style="color: #000000;">s</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">i</span><span style="color: #0000FF;">)))</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"hamming[1..20]: %v\n"</span><span style="color: #0000FF;">,{</span><span style="color: #000000;">s</span><span style="color: #0000FF;">})</span>
<span style="color: #0000FF;">?</span><span style="color: #000000;">hint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1691</span><span style="color: #0000FF;">))</span>
<span style="color: #0000FF;">?</span><span style="color: #000000;">hint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1000000</span><span style="color: #0000FF;">))</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">" %d (approx)\n"</span><span style="color: #0000FF;">,</span><span style="color: #000000;">hint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1000000</span><span style="color: #0000FF;">)))</span>
<span style="color: #008080;">include</span> <span style="color: #000000;">builtins</span><span style="color: #0000FF;">\</span><span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>
<span style="color: #008080;">function</span> <span style="color: #000000;">mpz_hint</span><span style="color: #0000FF;">(</span><span style="color: #004080;">sequence</span> <span style="color: #000000;">hm</span><span style="color: #0000FF;">)</span>
<span style="color: #000080;font-style:italic;">-- (as accurate as you like)</span>
<span style="color: #004080;">integer</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">p2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p5</span><span style="color: #0000FF;">}</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">hm</span><span style="color: #0000FF;">[</span><span style="color: #000000;">POWS</span><span style="color: #0000FF;">]</span>
<span style="color: #004080;">mpz</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">tmp2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp5</span><span style="color: #0000FF;">}</span> <span style="color: #0000FF;">=</span> <span style="color: #7060A8;">mpz_inits</span><span style="color: #0000FF;">(</span><span style="color: #000000;">3</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_ui_pow_ui</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p2</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_ui_pow_ui</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p3</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_ui_pow_ui</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">5</span><span style="color: #0000FF;">,</span><span style="color: #000000;">p5</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_mul</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp3</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp5</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">mpz_mul</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp2</span><span style="color: #0000FF;">,</span><span style="color: #000000;">tmp3</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">return</span> <span style="color: #7060A8;">mpz_get_str</span><span style="color: #0000FF;">(</span><span style="color: #000000;">tmp2</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">function</span>
<span style="color: #0000FF;">?</span><span style="color: #000000;">mpz_hint</span><span style="color: #0000FF;">(</span><span style="color: #000000;">hamming</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1000000</span><span style="color: #0000FF;">))</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
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"
</pre>
Under pwa/p2js, no real idea or any fretting over why, we instead get:
<pre>
519312780448388740000000000000000000000000000000000000000000000000000000000000000000 (approx)
</pre>
=={{header|Picat}}==
<syntaxhighlight 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].</syntaxhighlight>
{{out}}
<pre>[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.</pre>
=={{header|PicoLisp}}==
<
(let (L (1) H)
(do N
Line 2,949 ⟶ 9,447:
(println (make (for N 20 (link (hamming N)))))
(println (hamming 1691)) # very fast
(println (hamming 1000000)) # runtime about 13 minutes on i5-3570S</
{{out}}
<pre>(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
Line 2,957 ⟶ 9,455:
=={{header|PL/I}}==
<
Hamming: procedure options (main); /* 14 November 2013 with fixes 2021 */
declare (H(
declare (i, j, k, m,
/* 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 sorting
if ub > lb then
/* more than
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 =
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
put skip list (H(1691));
end Hamming;</syntaxhighlight>
Results:
<pre>
The first 20 Hamming numbers:
36
Hamming number 1691:
2125764000
</pre>
=={{header|Prolog}}==
===Generator idiom===
<
take( 0, Next, Z-Z, Next).
Line 3,033 ⟶ 9,591:
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.</
SWI Prolog 6.2.6 produces (in about 7 ideone seconds):
?- time( main(1000000) ).
Line 3,044 ⟶ 9,602:
Works with SWI-Prolog. Laziness is simulate with '''freeze/2''' and '''ground/2'''.<br>
Took inspiration from this code : http://chr.informatik.uni-ulm.de/~webchr (click on ''hamming.pl: Solves Hamming Problem'').
<
% to stop cleanly
nb_setval(go, 1),
Line 3,117 ⟶ 9,675:
(format('~w ', [X]),
N1 is N - 1,
watch_20(N1, L))).</
{{out}}
<pre>?- hamming(20).
Line 3,131 ⟶ 9,689:
true .
</pre>
=={{header|PureBasic}}==
<syntaxhighlight lang="purebasic">#X2 = 2
#X3 = 3
#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()</syntaxhighlight>
{{out}}<pre>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</pre>
=={{header|Python}}==
===Version based on example from Dr. Dobb's CodeTalk===
<
def hamming2():
'''\
This version is based on a snippet from:
https://web.archive.org/web/20081219014725/http://dobbscodetalk.com
/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
Line 3,174 ⟶ 9,789:
multindeces = [i - mini for i in multindeces]
#
yield h</
{{out}}
<pre>>>> list(islice(hamming2(), 20))
Line 3,186 ⟶ 9,801:
===Another implementation of same approach===
This version uses a lot of memory, it doesn't try to limit memory usage.
<
def hamming(limit):
Line 3,210 ⟶ 9,825:
print [hamming(i) for i in xrange(1, 21)]
print hamming(1691)
print hamming(1000000)</
===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.
<syntaxhighlight 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
</syntaxhighlight>
==="Cyclical Iterators"===
The original author is Raymond Hettinger and the code was first published [http://code.activestate.com/recipes/576961/ here] under the MIT license. Uses iterators dubbed "cyclical" in a sense that they are referring back (explicitly, with <code>p2, p3, p5</code> iterators) to the previously produced values, same as the above versions (through
Memory is efficiently maintained automatically by the <code>tee</code> function for each of the three generator expressions, i.e. only that much is maintained as needed to produce the next value (although, for Python versions older than 3.6 it looks like the storage is not shared so three copies are maintained implicitly there -- whereas for 3.6 and up the storage <i>is</i> shared between the returned iterators, so only a single underlying FIFO queue is maintained, according to the [https://docs.python.org/3.6/library/itertools.html#itertools.tee documentation]).
<
from heapq import merge
Line 3,239 ⟶ 9,876:
print list(islice(raymonds_hamming(), 20))
print islice(raymonds_hamming(), 1689, 1690).next()
print islice(raymonds_hamming(), 999999, 1000000).next()</
Results are the same as before.
====Non-sharing recursive generator====
Another formulation along the same lines, but greatly simplified, found [http://programmingpraxis.com/2011/08/30/hamming-numbers/#comment-3486 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 [http://ideone.com/PIkWEN gravely impacts the efficiency]. Not to be used.
<
from itertools import tee
Line 3,257 ⟶ 9,896:
if n != last:
yield n
last = n</
====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 [[Hamming_numbers#Avoiding_generation_of_duplicates | Haskell code]].
<
def merge(r, s):
Line 3,298 ⟶ 9,937:
print hamming(1, 21)
print hamming(1691)[0]
print hamming(1000000)[0]</
=={{header|QBasic}}==
{{works with|QBasic|1.1}}
{{works with|QuickBasic|4.5}}
<syntaxhighlight 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)</syntaxhighlight>
=={{header|Qi}}==
{{incomplete|Qi|Parts 2 & 3 of task missing.}}
{{trans|Clojure}}
<
[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)
Line 3,322 ⟶ 10,004:
[S|Ss] N -> [S|(stake (thaw Ss) (1- N))])
(stake (value hamming) 20)</
{{out}}
<pre>
[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]
</pre>
=={{header|Quackery}}==
Uses <code>smoothwith</code> from [[N-smooth numbers#Quackery]].
<syntaxhighlight lang="quackery"> ' [ 2 3 5 ] smoothwith [ size 1000000 = ]
dup 20 split drop echo cr
dup 1690 peek echo cr
-1 peek echo
</syntaxhighlight>
{{out}}
<pre>[ 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 ]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
=={{header|R}}==
Recursively find the Hamming numbers below <math>2^{31}</math>.
<
tmp=hamms
for(h in c(2,3,5)) {
Line 3,341 ⟶ 10,039:
hamms
}
h <- sort(hamming(1,limit=2^31
print(h[1:20])
print(h[length(h)])</syntaxhighlight>
{{out}}
<pre>
[1] 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
[1] 2125764000
</pre>
=== Alternate version ===
The '''nextn''' R function provides the needed functionality:
<syntaxhighlight lang="r">hamming <- function(n) {
a <- numeric(n)
a[1] <- 1
for (i in 2:n) {
a[i] <- nextn(a[i-1]+1)
}
a
}</syntaxhighlight>
'''Output'''
<pre> hamming(20)
[1] 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36</pre>
=={{header|Racket}}==
<syntaxhighlight lang="scheme">#lang racket
(require racket/stream)
(define first stream-first)
Line 3,367 ⟶ 10,088:
(for/list ([i 20] [x hamming]) x)
(stream-ref hamming 1690)
(stream-ref hamming 999999)</syntaxhighlight>
{{out}}
<pre>'(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
'''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:
{{trans|Haskell}}
<syntaxhighlight 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)</syntaxhighlight>
The output of the above code is the same as that of the earlier code.
=={{header|Raku}}==
(formerly Perl 6)
=== Merge version ===
{{Works with|rakudo|2015-11-04}}
The limit scaling is not <em>required</em>, but it cuts down on a bunch of unnecessary calculation.
<syntaxhighlight lang="raku" line>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</syntaxhighlight>
{{out}}
<pre>(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000</pre>
=== Iterative version ===
{{Works with|rakudo|6.c}}
This version uses a lazy list, storing a maximum of two extra value from the highest index requested
<syntaxhighlight lang="raku" line>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];
</syntaxhighlight>
{{out}}
<pre>(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
Line 3,377 ⟶ 10,181:
=={{header|Raven}}==
{{trans|Liberty Basic}}
<
[ ] as $h
1 $h 0 set
Line 3,403 ⟶ 10,207:
# Raven can't handle > 2^31 using integers
#
#"Hamming(1000000) is: " print 1000000 hamming print "\n" print</
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
Line 3,409 ⟶ 10,213:
=={{header|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,
<
<syntaxhighlight lang="rexx">/*REXX program computes Hamming numbers: 1 ──► 20, # 1691, and the one millionth. */
call hamming
call hamming
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
hamming: procedure; parse arg x,y;
#2= 1; #3= 1; #5= 1; @.= 0; @.1= 1
do n=2 for y-1
@.n
if 2*@.#2 == @.n then #2
if 3*@.#3 == @.n then #3
if 5*@.#5 == @.n then #5
end /*n*/ /* [↑] maybe assign next 3 Hamming#s. */
do j=x to y; say 'Hamming('right(j, w)")
say right( 'length of last Hamming number =' length(@.y),
return</
<pre>
Hamming( 1) = 1
Hamming( 2) = 2
Line 3,465 ⟶ 10,268:
</pre>
===
This REXX version is roughly twice as fast as the
<
call hamming
call hamming
/*──────────────────────────────────────────────────────────────────────────────────────*/
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;
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</syntaxhighlight>
{{out|output|text= is identical to the 1<sup>st</sup> REXX version.}}
===for huge numbers===
This REXX version is slightly slower than the 2<sup>nd</sup> REXX version.
It can, however, computer much larger Hamming numbers (by storing the larger numbers in exponential format).
<br>This is possible because larger Hamming numbers have a significant number of trailing zeros.
<syntaxhighlight 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</syntaxhighlight>
{{out|output|text= is identical to the 1<sup>st</sup> REXX version.}} <br><br>
=={{header|Ring}}==
<syntaxhighlight 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
</syntaxhighlight>
Output:
<pre>
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
</pre>
=={{header|RPL}}==
RPL does not provide any multi-precision capability, so only parts 1 and 2 of the task can be implemented.
Using global variables <code>In</code> and <code>Xn</code> avoids stack acrobatics that would have made the code slower and unintelligible, despite the ugly <code> 'var_name' STO</code> syntax inherited from vintage HP calculators.
≪ 1 ‘I2’ STO 1 ‘I3’ STO 1 ‘I5’ STO 2 ‘X2’ STO 3 ‘X3’ STO 5 ‘X5’ STO
{ 1 } 1 ROT 1 - '''FOR''' n
X2 X3 MIN X5 MIN
SWAP OVER + SWAP
'''IF''' X2 OVER == '''THEN''' 1 ‘I2’ STO+ OVER I2 GET 2 * ‘X2’ STO '''END'''
'''IF''' X3 OVER == '''THEN''' 1 ‘I3’ STO+ OVER I3 GET 3 * ‘X3’ STO '''END'''
'''IF''' X5 == '''THEN''' 1 ‘I5’ STO+ DUP I5 GET 5 * ‘X5’ STO '''END'''
'''NEXT'''
≫ 'HAMM' STO
20 HAMM
1691 HAMM DUP SIZE GET
{{out}}
<pre>
2: { 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 }
1: 2125764000
</pre>
=={{header|Ruby}}==
{{trans|Scala}}
{{works with|Ruby|1.
<syntaxhighlight lang="ruby">hamming = Enumerator.new do |yielder|
next_ham = 1
queues =
loop do
next_ham =
end
end</
And the "main" part of the task
<
hamming.each.with_index(1) do |ham, idx|
case idx
when (1..20), 1691
when 1_000_000
break
end
end
puts "elapsed: #{Time.now - start} seconds"</
{{out}}
<pre style='height: 30ex; overflow: scroll'>
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
</pre>
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:
{{trans|Crystal}}
<syntaxhighlight 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"</syntaxhighlight>
System: I7-6700HQ, 3.5 GHz, Linux Kernel 5.6.17
Run as: $ ruby hammingnumbers.rb
{{out}}
<pre>
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
</pre>
=={{header|Run BASIC}}==
<
dim h(1000000)
for i =1 to 20
Line 3,594 ⟶ 10,522:
next n
hamming = h(limit -1)
end function</
<pre>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</pre>
=={{header|Rust}}==
{{libheader|num}}
===Basic version===
{{trans|D}}
Improved by minimizing the number of BigUint comparisons:
<syntaxhighlight 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);
}</syntaxhighlight>
{{output}}
<pre>[ 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.</pre>
===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:
<syntaxhighlight 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")
}
}</syntaxhighlight>
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:
<syntaxhighlight 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
}</syntaxhighlight>
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:
<syntaxhighlight 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);
}</syntaxhighlight>
{{output}}
<pre>[ 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.</pre>
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:
# It does not guarantee tail call optimization for functions, thus sometimes requiring imperative forms of code to produce that effect.
# It does not have currying or partial application of function arguments without using kludges of nested function/closure calls.
# move closures cannot use recursive shared values without using interior mutability inside a reference counted value (required here)
# Closures are not recursive without using a trick involving shared state reference counted values (demonstrated here).
# It currently does not have a standard library implementation of a lazily computed non-static value (required to implement a Lazy List, and
# 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.
[http://https://github.com/reem/rust-lazy/blob/master/src/single.rs Jeremy Reems had implemented Lazy] and [http://https://github.com/reem/rust-lazylist/blob/master/src/lib.rs 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):
{{trans|Haskell}}
{{works with|Rust 1.53.0}}
<syntaxhighlight 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::*;
#[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);
}</syntaxhighlight>
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}}
<pre>[ 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.</pre>
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}}
<syntaxhighlight 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::*;
#[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...
#[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);
}</syntaxhighlight>
{{out}}
<pre>[ 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.</pre>
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:
{{trans|Nim}}
<syntaxhighlight lang="rust">extern crate num;
use num::bigint::BigInt;
use core::fmt::Display;
use std::time::Instant;
use std::iter;
const NUM_ELEMENTS: usize = 1000000;
const LB2_2: f64 = 1.0_f64; // log2(2.0)
const LB2_3: f64 = 1.5849625007211563_f64; // log2(3.0)
const LB2_5: f64 = 2.321928094887362_f64; // log2(5.0)
#[derive (Clone)]
struct LogRep {
lr: f64,
x2: u32,
x3: u32,
x5: u32,
}
impl LogRep {
fn int_value(&self) -> BigInt {
BigInt::from(2).pow(self.x2) * BigInt::from(3).pow(self.x3) * BigInt::from(5).pow(self.x5)
}
#[inline(always)]
fn mul2(&self) -> Self {
LogRep {
lr: self.lr + LB2_2,
x2: self.x2 + 1,
x3: self.x3,
x5: self.x5,
}
}
#[inline(always)]
fn mul3(&self) -> Self {
LogRep {
lr: self.lr + LB2_3,
x2: self.x2,
x3: self.x3 + 1,
x5: self.x5,
}
}
#[inline(always)]
fn mul5(&self) -> Self {
LogRep {
lr: self.lr + LB2_5,
x2: self.x2,
x3: self.x3,
x5: self.x5 + 1,
}
}
}
impl Display for LogRep {
fn fmt(&self, f: &mut std::fmt::Formatter<'_>) -> std::fmt::Result {
let val = self.int_value();
let x2 = self.x2;
let x3 = self.x3;
let x5 = self.x5;
write!(f, "[{x2} {x3} {x5}]=>{val}")
}
}
const ONE: LogRep = LogRep { lr: 0.0, x2: 0, x3: 0, x5: 0 };
struct LogRepImperativeIterator {
s2: Vec<LogRep>,
s3: Vec<LogRep>,
s5: LogRep,
mrg: LogRep,
s2i: usize,
s3i: usize,
}
impl LogRepImperativeIterator {
pub fn new() -> Self {
LogRepImperativeIterator {
s2: vec![ONE.mul2()],
s3: vec![ONE.mul3()],
s5: ONE.mul5(),
mrg: ONE.mul3(),
s2i: 0,
s3i: 0,
}
}
fn iter(&self) -> impl Iterator<Item = LogRep> {
iter::once(ONE).chain(LogRepImperativeIterator::new())
}
}
impl Iterator for LogRepImperativeIterator {
type Item = LogRep;
#[inline(always)]
fn next(&mut self) -> Option<Self::Item> {
if self.s2i + self.s2i >= self.s2.len() {
self.s2.drain(0..self.s2i);
self.s2i = 0;
}
let result: LogRep;
if self.s2[self.s2i].lr < self.mrg.lr {
self.s2.push(self.s2[self.s2i].mul2());
result = self.s2[self.s2i].clone(); self.s2i += 1;
} else {
if self.s3i + self.s3i >= self.s3.len() {
self.s3.drain(0..self.s3i);
self.s3i = 0;
}
result = self.mrg.clone();
self.s2.push(self.mrg.mul2());
self.s3.push(self.mrg.mul3());
self.s3i += 1;
if self.s3[self.s3i].lr < self.s5.lr {
self.mrg = self.s3[self.s3i].clone();
} else {
self.mrg = self.s5.clone();
self.s5 = self.s5.mul5();
self.s3i -= 1;
}
};
Some(result)
}
}
fn main() {
LogRepImperativeIterator::new().iter().take(20)
.for_each(&|h: LogRep| print!("{} ", h.int_value()));
println!();
println!("{} ", LogRepImperativeIterator::new().iter()
.take(1691).last().unwrap().int_value());
let t0 = Instant::now();
let rslt = LogRepImperativeIterator::new().iter()
.take(NUM_ELEMENTS).last().unwrap();
let elpsd = t0.elapsed().as_micros() as f64;
println!("{}", rslt.int_value());
println!("This took {} microseconds for {} elements!", elpsd, NUM_ELEMENTS)
}</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
This took 6517 microseconds for 1000000 elements!</pre>
The code above is almost twenty 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 it does not leak memory. This version can calculate the billionth Hamming number in about 8.1 seconds.
===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:
{{trans|Haskell}}
<syntaxhighlight 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);
}</syntaxhighlight>
<pre>[ 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.</pre>
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.
=={{header|Scala}}==
<
import scala.collection.mutable.Queue
val qs = Seq.fill(3)(new Queue[BigInt])
Line 3,612 ⟶ 11,761:
def hasNext = true
qs foreach (_ enqueue 1)
}</
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:
<
import scala.collection.mutable.Queue
val q2 = new Queue[BigInt]
Line 3,634 ⟶ 11,783:
def hasNext = true
List(q2, q3, q5) foreach (_ enqueue 1)
}</
Usage:
<pre>
Line 3,648 ⟶ 11,797:
There's also a fairly mechanical translation from Haskell using purely functional lazy streams
{{trans|Haskell}}
<
def merge(inx : Stream[BigInt], iny : Stream[BigInt]) : Stream[BigInt] = {
if (inx.head < iny.head) inx.head #:: merge(inx.tail, iny) else
Line 3,656 ⟶ 11,805:
1 #:: merge(hamming map (_ * 2), merge(hamming map (_ * 3), hamming map (_ * 5)))
}</
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, ?)"
<pre>
Line 3,672 ⟶ 11,821:
res2: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
'''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:
<syntaxhighlight 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 } }</syntaxhighlight>
Usage:
<pre>
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
</pre>
It only takes under a half second to find the millionth number in the sequence in the last output.
=={{header|Scheme}}==
<
(syntax-rules ()
((_ lar ldr) (delay (cons lar (delay ldr))))))
Line 3,721 ⟶ 11,898:
(newline)
(display (llist-ref 1000000 hamming))
(newline)</
{{out}}
<pre>(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36)
2125764000
out of memory</pre>
===Avoiding Generation of Duplicates, including reduced memory use===
{{trans|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):
<syntaxhighlight 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)</syntaxhighlight>
{{output}}
<pre>{1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36}
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
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.
=={{header|Seed7}}==
<
include "bigint.s7i";
Line 3,787 ⟶ 12,002:
writeln(hamming(1691));
writeln(hamming(1000000));
end func;</
{{out}}
<pre>
Line 3,793 ⟶ 12,008:
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
</pre>
=={{header|Sidef}}==
<syntaxhighlight 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()</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
</pre>
Line 3,798 ⟶ 12,042:
{{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.
<
Hammer class >> hammingNumbers: howMany [
|h i j k x2 x3 x5|
Line 3,818 ⟶ 12,062:
(Hammer hammingNumbers: 20) displayNl.
(Hammer hammingNumbers: 1690) last displayNl.</
{{works with|Pharo Smalltalk}}
<syntaxhighlight 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"
</syntaxhighlight>
{{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.
<syntaxhighlight 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"
</syntaxhighlight>
=={{header|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.
<syntaxhighlight 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
</syntaxhighlight>
=={{header|Tcl}}==
This uses coroutines to simplify the description of what's going on.
{{works with|Tcl|8.6}}
<
# Simple helper: Tcl-style list "map"
Line 3,880 ⟶ 12,218:
puts "hamming{1690} = $h"
for {} {$i <= 1000000} {incr i} {set h [hamming]}
puts "hamming{1000000} = $h"</
{{out}}
<pre>
Line 3,907 ⟶ 12,245:
</pre>
A faster version can be built that also works on Tcl 8.5 (or earlier, if only small hamming numbers are being computed):
<
proc hamming {n} {
global hamming hi2 hi3 hi5
Line 3,940 ⟶ 12,278:
puts "hamming{1692} = [hamming 1692]"
puts "hamming{1693} = [hamming 1693]"
puts "hamming{1000000} = [hamming 1000000]"</
=={{header|uBasic/4tH}}==
uBasic's single array does not have the required size to calculate the 1691st number, let alone the millionth.
<syntaxhighlight lang="text">For H = 1 To 20
Print "H("; H; ") = "; Func (_FnHamming(H))
Next
End
_FnHamming Param (1)
@(0) = 1
X = 2 : Y = 3 : Z = 5
I = 0 : J = 0 : K = 0
For N = 1 To a@ - 1
M = X
If M > Y Then M = Y
If M > Z Then M = Z
@(N) = M
If M = X Then I = I + 1 : X = 2 * @(I)
If M = Y Then J = J + 1 : Y = 3 * @(J)
If M = Z Then K = K + 1 : Z = 5 * @(K)
Next
Return (@(a@-1))</syntaxhighlight>
{{out}}
<pre>H(1) = 1
H(2) = 2
H(3) = 3
H(4) = 4
H(5) = 5
H(6) = 6
H(7) = 8
H(8) = 9
H(9) = 10
H(10) = 12
H(11) = 15
H(12) = 16
H(13) = 18
H(14) = 20
H(15) = 24
H(16) = 25
H(17) = 27
H(18) = 30
H(19) = 32
H(20) = 36
0 OK, 0:379</pre>
=={{header|UNIX Shell}}==
{{works with|ksh93}}
{{works with|Bourne Again SHell|4+}}
Large numbers are not supported.
<syntaxhighlight lang="bash">
typeset -a hamming=(1) q2 q3 q5
function nextHamming {
typeset -i h=${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 {
typeset -n ary=$1
printf '%s\n' "${ary[0]}"
ary=( "${ary[@]:1}" )
}
function min3 {
if (( $1 < $2 )); then
(( $1 < $3 )) && printf '%s\n'$1 || printf '%s\n'$3
else
(( $2 < $3 )) && printf '%s\n'$2 || printf '%s\n'$3
fi
}
for ((i=1; i<=20; i++)); do
nextHamming
printf '%d\t%d\n' "$i" "${hamming[i-1]}"
done
for ((; i<=1690; i++)); do nextHamming; done
nextHamming
printf '%d\t%d\n' "$i" "${hamming[i-1]}"
printf 'elapsed: %s\n' "$SECONDS"
</syntaxhighlight>
{{out}}
<pre>
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</pre>
=={{header|Ursala}}==
Line 3,947 ⟶ 12,402:
number with respect to them. An elegant but inefficient formulation based on the J solution is the
following.
<
#import nat
smooth"p" "n" = ~&z take/"n" nleq-< (rep(length "n") ^Ts/~& product*K0/"p") <1></
This test program
<
yields this list of the first 20 Hamming numbers.
<pre>
Line 3,959 ⟶ 12,414:
Although all calculations are performed using unlimited precision, the version
above is impractical for large numbers. A more hardcore approach is the following.
<
#import nat
Line 3,970 ⟶ 12,425:
#cast %nL
main = smooth<2,3,5>* nrange(1,20)--<1691,1000000></
{{out}}
The great majority of time is spent calculating the millionth Hamming number.
Line 3,996 ⟶ 12,451:
2125764000,
519312780448388736089589843750000000000000000000000000000000000000000000000000000000></pre>
=={{header|VBA}}==
<syntaxhighlight 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</syntaxhighlight>{{out}}
<pre>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
</pre>
=={{header|VBScript}}==
{{trans|BBC BASIC}}
<syntaxhighlight 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
</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
=={{header|V (Vlang)}}==
{{trans|go}}
===Concise version using dynamic-programming===
<syntaxhighlight lang="v (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])
}</syntaxhighlight>
{{out}}
<pre>[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000</pre>
===Fast version with no duplicates algorithm using arrays for memoization and logarithmic approximations===
The V (Vlang) language isn't yet stable enough (version 0.30) to support a fully functional version using generic lazy lists as per the Haskell language versions and in truth is mostly an imperative language anyway; however, it already can do the page task very quickly with a more imperative algorithm using arrays for memoization storage and logarithmic approximations for sorting comparisons to avoid "infinite" precision integer calculations except for the final result values, as per the following code, which is Nim's "ring buffer" version as that is faster due to less copying required:
{{trans|Nim}}
<syntaxhighlight lang="v (vlang)">// compile with: v -cflags -march=native -cflags -O3 -prod HammingsLogQ.v
import time
import math.big
import math { log2 }
import arrays { copy }
const num_elements = 1_000_000
struct LogRep {
lg f64
x2 u32
x3 u32
x5 u32
}
const (
one = LogRep { 0.0, 0, 0, 0 }
lb2_2 = 1.0
lb2_3 = log2(3.0)
lb2_5 = log2(5.0)
)
[inline]
fn (lr &LogRep) mul2() LogRep {
return LogRep { lg: lr.lg + lb2_2,
x2: lr.x2 + 1,
x3: lr.x3,
x5: lr.x5 }
}
[inline]
fn (lr &LogRep) mul3() LogRep {
return LogRep { lg: lr.lg + lb2_3,
x2: lr.x2,
x3: lr.x3 + 1,
x5: lr.x5 }
}
[inline]
fn (lr &LogRep) mul5() LogRep {
return LogRep { lg: lr.lg + lb2_5,
x2: lr.x2,
x3: lr.x3,
x5: lr.x5 + 1 }
}
[inline]
fn xpnd(x u32, mlt u32) big.Integer {
mut r := big.integer_from_int(1)
mut m := big.integer_from_u32(mlt)
mut v := x
for {
if v <= 0 { break }
else {
if v & 1 != 0 { r = r * m }
m = m * m
v >>= 1
}
}
return r
}
fn (lr &LogRep) to_integer() big.Integer {
return xpnd(lr.x2, 2) * xpnd(lr.x3, 3) * xpnd(lr.x5, 5)
}
fn (lr LogRep) str() string {
return (&lr).to_integer().str()
}
struct HammingsLog {
mut:
// automatically initialized with LogRep = one (defult)...
s2 []LogRep = []LogRep { len: 1024, cap: 1024 }
s3 []LogRep = []LogRep { len: 1024, cap: 1024 }
s5 LogRep = one.mul5()
mrg LogRep = one.mul3()
s2msk int = 1023
s2hdi int
s2nxti int = 1
s3msk int = 1023
s3hdi int
s3nxti int
}
[direct_array_access][inline]
fn (mut hl HammingsLog) next() ?LogRep {
mut rsltp := &hl.s2[hl.s2hdi]
if rsltp.lg < hl.mrg.lg {
hl.s2[hl.s2nxti] = rsltp.mul2()
hl.s2hdi++
hl.s2hdi &= hl.s2msk
} else {
mut rslt := hl.mrg
rsltp = &rslt
hl.s2[hl.s2nxti] = hl.mrg.mul2()
hl.s3[hl.s3nxti] = hl.mrg.mul3()
s3hdp := &hl.s3[hl.s3hdi]
if unsafe { s3hdp.lg < hl.s5.lg } {
hl.mrg = *s3hdp
hl.s3hdi++
hl.s3hdi &= hl.s3msk
} else {
hl.mrg = hl.s5
hl.s5 = hl.s5.mul5()
}
hl.s3nxti++
hl.s3nxti &= hl.s3msk
if hl.s3nxti == hl.s3hdi { // buffer full: grow it
sz := hl.s3msk + 1
hl.s3msk = sz + sz
unsafe { hl.s3.grow_len(sz) }
hl.s3msk--
if hl.s3hdi == 0 {
hl.s3nxti = sz
} else {
unsafe { vmemcpy(&hl.s3[hl.s3hdi + sz], &hl.s3[hl.s3hdi],
int(sizeof(LogRep)) * (sz - hl.s3hdi)) }
hl.s3hdi += sz
}
}
}
hl.s2nxti++
hl.s2nxti &= hl.s2msk
if hl.s2nxti == hl.s2hdi { // buffer full: grow it
sz := hl.s2msk + 1
hl.s2msk = sz + sz
unsafe { hl.s2.grow_len(sz) }
hl.s2msk--
if hl.s2hdi == 0 {
hl.s2nxti = sz
} else {
unsafe { vmemcpy(&hl.s2[hl.s2hdi + sz], &hl.s2[hl.s2hdi],
int(sizeof(LogRep)) * (sz - hl.s2hdi)) }
hl.s2hdi += sz
}
}
return *rsltp
}
fn (hmgs HammingsLog) nth_hammings_log(n int) LogRep {
mut cnt := 0
if n > 0 { for h in hmgs {
cnt++
if cnt >= n { return h } }
}
panic("argument less than 1 for nth!")
}
{
hs := HammingsLog {}
mut cnt := 0
for h in hs {
print("$h ")
cnt++
if cnt >= 20 { break }
}
println("")
}
println("${(HammingsLog{}).nth_hammings_log(1691)}")
start_time := time.now()
rslt := (HammingsLog{}).nth_hammings_log(num_elements)
duration := (time.now() - start_time).microseconds()
println("$rslt")
println("Above result for $num_elements elements in $duration microseconds.")</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Above result for 1000000 elements in 4881 microseconds.</pre>
The above result is as computed on an Intel i5-6500 at 3.6 GHz (single-threaded, boosted); the execution time is somewhat variable due to V currently using Garbage Collection by default, but the intention is to eventually use automatic reference counting by default which should make it slightly faster and more consistent other than for any variations caused by the memory allocator. The above version can calculate the billionth Hamming number in about 5.3 seconds.
===Extremely fast version inserting values into the error band and using logarithmic approximations for sorting===
The above code is about as fast as one can go generating sequences/iterations; however, if one is willing to forego sequences/iterations and just calculate the nth Hamming number (repeatedly when a sequence is desired, but that is only for the first required task of three and then only for a trivial range), 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. This version uses a multi-precision integer as the representation of the logarithmic approximation of the value for sorting of the error band to extend the precision for accurate results up to almost the 64-bit number range (in about a day on common desktop computers). The code is as follows:
{{trans|Nim}}
<syntaxhighlight lang="v (vlang)">// compile with: v -cflags -march=native -cflags -O3 -prod HammingsLog.v
import time
import math.big
import math { log2, sqrt, pow, floor }
const num_elements = 1_000_000
struct LogRep {
lg big.Integer
x2 u32
x3 u32
x5 u32
}
const (
one = LogRep { big.zero_int, 0, 0, 0 }
// 1267650600228229401496703205376
lb2_2 = big.Integer { digits: [u32(0), 0, 0, 16],
signum: 1, is_const: true }
// 2009178665378409109047848542368
lb2_3 = big.Integer { digits: [u32(11608224), 3177740794, 1543611295, 25]
signum: 1, is_const: true }
// 2943393543170754072109742145491
lb2_5 = big.Integer { digits: [u32(1258143699), 1189265298, 647893747, 37],
signum: 1, is_const: true }
smlb2_2 = f64(1.0)
smlb2_3 = log2(3.0)
smlb2_5 = log2(5.0)
fctr = f64(6.0) * smlb2_3 * smlb2_5
crctn = log2(sqrt(30.0))
)
fn xpnd(x u32, mlt u32) big.Integer {
mut r := big.integer_from_int(1)
mut m := big.integer_from_u32(mlt)
mut v := x
for {
if v <= 0 { break }
else {
if v & 1 != 0 { r = r * m }
m = m * m
v >>= 1
}
}
return r
}
fn (lr LogRep) to_integer() big.Integer {
return xpnd(lr.x2, 2) * xpnd(lr.x3, 3) * xpnd(lr.x5, 5)
}
fn (lr LogRep) str() string {
return lr.to_integer().str()
}
fn nth_hamming_log(n u64) LogRep {
if n < 2 { return one }
lgest := pow(fctr * f64(n), f64(1.0)/f64(3.0)) - crctn // from WP formula
frctn := if n < 1_000_000_000 { f64(0.509) } else { f64(0.105) }
lghi := pow(fctr * (f64(n) + frctn * lgest), f64(1.0)/f64(3.0)) - crctn
lglo := f64(2.0) * lgest - lghi // and a lower limit of the upper "band"
mut count := u64(0) // need to use extended precision, might go over
mut band := []LogRep { len: 1, cap: 1 } // give it one value so doubling size works
mut ih := 0 // band array insertion index
klmt := u32(lghi / smlb2_5) + 1
for k in u32(0) .. klmt {
p := f64(k) * smlb2_5
jlmt := u32((lghi - p) / smlb2_3) + 1
for j in u32(0) .. jlmt {
q := p + f64(j) * smlb2_3
ir := lghi - q
lg := q + floor(ir) // current log value (estimated)
count += u64(ir) + 1
if lg >= lglo {
len := band.len
if ih >= len { unsafe { band.grow_len(len) } }
bglg := lb2_2 * big.integer_from_u32(u32(ir)) +
lb2_3 * big.integer_from_u32(j) +
lb2_5 * big.integer_from_u32(k)
band[ih] = LogRep { lg: bglg, x2: u32(ir), x3: j, x5: k }
ih++
}
}
}
band.sort_with_compare(fn(a &LogRep, b &LogRep) int {
return b.lg.abs_cmp(a.lg)
}
)
if n > count { panic("nth_hamming_log: band high estimate is too low!") }
ndx := int(count - n)
if ndx >= band.len { panic("nth_hamming_log: band low estimate is too high!") }
return band[ndx]
}
for i in 1 .. 21 { print("${nth_hamming_log(i)} ") }
println("")
println("${nth_hamming_log(1691)}")
start_time := time.now()
rslt := nth_hamming_log(num_elements)
duration := (time.now() - start_time).microseconds()
println("$rslt")
println("Above result for $num_elements elements in $duration microseconds.")</syntaxhighlight>
{{out}}
<pre>1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Above result for 1000000 elements in 277 microseconds.</pre>
The output is the same as above except that the execution time is almost too small to be measured; it can produce the billionth Hamming number in about five milliseconds, the trillionth Hamming number in about 440 milliseconds, and the thousand trillionth (which is now possible without error) in about 42.4 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. This is in spite of the current Vlang standard library using its own implementation of multi-precision integers rather than the highly optimized "gmp" library used by some languages which could be somewhat faster.
=={{header|Wren}}==
===Simple but slow===
{{libheader|Wren-big}}
<syntaxhighlight lang="wren">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])</syntaxhighlight>
{{out}}
<pre>
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
</pre>
===Much faster logarithmic version===
{{trans|Go}}
{{libheader|Wren-dynamic}}
{{libheader|Wren-long}}
{{libheader|Wren-math}}
A translation of Go's 'extremely fast version inserting logarithms into the top error band'.
Not as fast as the statically typed languages but fast enough for me :)
<syntaxhighlight lang="wren">import "./dynamic" for Struct
import "./long" for ULong
import "./big" for BigInt
import "./math" for Math
var Logrep = Struct.create("LogRep", ["lg", "x2", "x3", "x5"])
var nthHamming = Fn.new { |n|
if (n < 2) {
if (n < 1) Fiber.abort("nthHamming: argument is zero!")
return [0, 0, 0]
}
var lb3 = 1.5849625007211561814537389439478
var lb5 = 2.3219280948873623478703194294894
var fctr = 6 * lb3 * lb5
var crctn = 2.4534452978042592646620291867186
var lgest = (n.toNum * fctr).cbrt - crctn
var frctn = (n < 1000000000) ? 0.509 : 0.106
var lghi = ((n.toNum + lgest * frctn) * fctr).cbrt - crctn
var lglo = lgest * 2 - lghi
var count = ULong.zero
var bnd = []
var klmt = (lghi/lb5).truncate.abs + 1
for (k in 0...klmt) {
var p = k * lb5
var jlmt = ((lghi - p)/lb3).truncate.abs + 1
for (j in 0...jlmt) {
var q = p + j * lb3
var ir = lghi - q
var lg = q + ir.floor
count = count + ir.truncate.abs + 1
if (lg >= lglo) bnd.add(Logrep.new(lg, ir.truncate.abs, j, k))
}
}
if (n > count) Fiber.abort("nthHamming: band high estimate is too low!")
var ndx = (count - n).toSmall
if (ndx >= bnd.count) Fiber.abort("nthHamming: band low estimate is too high!")
bnd.sort { |a, b| b.lg < a.lg }
var rslt = bnd[ndx]
return [rslt.x2, rslt.x3, rslt.x5]
}
var convertTpl2BigInt = Fn.new { |tpl|
var result = BigInt.one
for (i in 0...tpl[0]) result = result * 2
for (i in 0...tpl[1]) result = result * 3
for (i in 0...tpl[2]) result = result * 5
return result
}
System.print("The first 20 Hamming numbers are:")
for (i in 1..20) {
System.write("%(convertTpl2BigInt.call(nthHamming.call(ULong.new(i)))) ")
}
System.print("\n\nThe 1,691st Hamming number is:")
System.print(convertTpl2BigInt.call(nthHamming.call(ULong.new(1691))))
var start = System.clock
var res = nthHamming.call(ULong.new(1e6))
var end = System.clock
System.print("\nThe 1,000,000 Hamming number is:")
System.print(convertTpl2BigInt.call(res))
var duration = ((end-start) * 1000).round
System.print("The last of these found in %(duration) milliseconds.")</syntaxhighlight>
{{out}}
<pre>
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 1,000,000 Hamming number is:
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
The last of these found in 16 milliseconds.
</pre>
=={{header|XPL0}}==
<syntaxhighlight lang "XPL0">func Hamming(N); \Return 'true' if N is a Hamming number
int N;
[if N = 1 then return true;
if rem(N/2) = 0 then return Hamming(N/2);
if rem(N/3) = 0 then return Hamming(N/3);
if rem(N/5) = 0 then return Hamming(N/5);
return false;
];
int N, C;
[N:= 1; C:= 0;
loop [if Hamming(N) then
[C:= C+1;
IntOut(0, N); ChOut(0, ^ );
if C >= 20 then quit;
];
N:= N+1;
];
CrLf(0);
N:= 1<<31; \ 8-)
repeat N:= N-1 until Hamming(N);
IntOut(0, N);
]</syntaxhighlight>
{{out}}
<pre>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000</pre>
=={{header|Yabasic}}==
{{trans|Run BASIC}}
<syntaxhighlight 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</syntaxhighlight>
=={{header|zkl}}==
<
fcn hamming(N){
h:=List.createLong(N+1); (0).pump(N+1,h.write,Void); // fill list with stuff
Line 4,016 ⟶ 13,196:
}
[1..20].apply(hamming).println();
hamming(1691).println();</
{{out}}
<pre>
Line 4,022 ⟶ 13,202:
2125764000
</pre>
While the other algorithms save [lots of] space, run time still sucks when n > 100,000 so memory usage might as well too.
{{out}}
<pre>
Line 4,031 ⟶ 13,211:
OK, I was wrong, calculating the nth Hamming number can be fast and efficient.
{{trans|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).
<syntaxhighlight lang="zkl">#-- directly find n-th Hamming number, in ~ O(n^{2/3}) time
#-- by Will Ness, based on "top band" idea by Louis Klauder, from DDJ discussion
#-- http://drdobbs.com/blogs/architecture-and-design/228700538
Line 4,074 ⟶ 13,253:
}
return(cnt,b.close());
}</
<syntaxhighlight 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));
Line 4,083 ⟶ 13,262:
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</
{{out}}
<pre>
Line 4,095 ⟶ 13,274:
621607575556524486163081633287207200394705651908965270659163240.......
</pre>
=={{header|ZX Spectrum Basic}}==
{{trans|BBC_BASIC}}
<syntaxhighlight 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 </syntaxhighlight>
| |||