Hamming numbers: Difference between revisions
m (Fixed misplaced quote (in comment)) |
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<lang Eiffel> |
<lang Eiffel> |
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note |
note |
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description : " |
description : "Initial part, in order, of the sequence of Hamming numbers, also known as regular numbers" |
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arithmetic : "[ |
arithmetic : "[ |
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This version avoids integer overflow and stops at the last representable number in the sequence. |
This version avoids integer overflow and stops at the last representable number in the sequence. |
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make |
make |
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-- Print first 20 Hamming numbers and the 1691-st one. |
-- Print first 20 Hamming numbers, in order, and the 1691-st one. |
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local |
local |
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Hammings: like hamming |
Hammings: like hamming |
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hamming (n: INTEGER): ARRAYED_LIST [NATURAL] |
hamming (n: INTEGER): ARRAYED_LIST [NATURAL] |
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-- First `n' elements of the Hamming sequence, |
-- First `n' elements (in order) of the Hamming sequence, |
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-- or as many as will not produce overflow. |
-- or as many of them as will not produce overflow. |
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local |
local |
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sl: SORTED_TWO_WAY_LIST [NATURAL] |
sl: SORTED_TWO_WAY_LIST [NATURAL] |
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sl.extend (1); sl.start |
sl.extend (1); sl.start |
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across 1 |..| n as i invariant |
across 1 |..| n as i invariant |
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-- "The numbers output so far are the first `i' - 1 Hamming numbers". |
-- "The numbers output so far are the first `i' - 1 Hamming numbers, in order". |
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-- "Result.first is the `i'-th Hamming number." |
-- "Result.first is the `i'-th Hamming number." |
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until sl.is_empty loop |
until sl.is_empty loop |
Revision as of 02:09, 9 August 2012
You are encouraged to solve this task according to the task description, using any language you may know.
Hamming numbers are numbers of the form
- .
Hamming numbers are also known as ugly numbers and also 5-smooth numbers (numbers whose prime divisors are less or equal to 5).
Generate the sequence of Hamming numbers, in increasing order. In particular:
- Show the first twenty Hamming numbers.
- Show the 1691st Hamming number (the last one below ).
- Show the one millionth Hamming number (if the language – or a convenient library – supports arbitrary-precision integers).
References
- wp:Hamming_numbers
- wp:Smooth_number
- Hamming problem from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread here and here).
Ada
GNAT provides the datatypes Integer, Long_Integer and Long_Long_Integer.
Values for GNAT Pro 6.3.1, 64 bit Linux version:
- Integer covers the range -2**31 .. 2**31-1 (-2147483648 .. 2147483647).
- Long_Integer and Long_Long_Integer cover the range -2**63 .. 2**63-1 (-9223372036854775808 .. 9223372036854775807).
Using your own modular integer type (for example type My_Unsigned_Integer is mod 2**64;), you can expand the range to 0 .. 18446744073709551615, but this still is not enough for the millionth Hamming number.
For bigger numbers, you have to use an external library, for example Big_Number.
The code for calculating the Hamming numbers is kept generic, to easily expand the range by changing the concrete type. <lang Ada>with Ada.Text_IO; procedure Hamming is
generic type Int_Type is private; Zero : Int_Type; One : Int_Type; Two : Int_Type; Three : Int_Type; Five : Int_Type; with function "mod" (Left, Right : Int_Type) return Int_Type is <>; with function "/" (Left, Right : Int_Type) return Int_Type is <>; with function "+" (Left, Right : Int_Type) return Int_Type is <>; function Get_Hamming (Position : Positive) return Int_Type;
function Get_Hamming (Position : Positive) return Int_Type is function Is_Hamming (Number : Int_Type) return Boolean is Temporary : Int_Type := Number; begin while Temporary mod Two = Zero loop Temporary := Temporary / Two; end loop; while Temporary mod Three = Zero loop Temporary := Temporary / Three; end loop; while Temporary mod Five = Zero loop Temporary := Temporary / Five; end loop; return Temporary = One; end Is_Hamming; Result : Int_Type := One; Previous : Positive := 1; begin while Previous /= Position loop Result := Result + One; if Is_Hamming (Result) then Previous := Previous + 1; end if; end loop; return Result; end Get_Hamming;
-- up to 2**32 - 1 function Integer_Get_Hamming is new Get_Hamming (Int_Type => Integer, Zero => 0, One => 1, Two => 2, Three => 3, Five => 5);
-- up to 2**64 - 1 function Long_Long_Integer_Get_Hamming is new Get_Hamming (Int_Type => Long_Long_Integer, Zero => 0, One => 1, Two => 2, Three => 3, Five => 5);
begin
Ada.Text_IO.Put ("1) First 20 Hamming numbers: "); for I in 1 .. 20 loop Ada.Text_IO.Put (Integer'Image (Integer_Get_Hamming (I))); end loop; Ada.Text_IO.New_Line; Ada.Text_IO.Put_Line ("2) 1_691st Hamming number: " & Integer'Image (Integer_Get_Hamming (1_691))); -- even Long_Long_Integer overflows here Ada.Text_IO.Put_Line ("3) 1_000_000st Hamming number: " & Long_Long_Integer'Image (Long_Long_Integer_Get_Hamming (1_000_000)));
end Hamming;</lang>
- Output:
1) First 20 Hamming numbers: 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2) 1_691 st Hamming number: 2125764000 Execution terminated by unhandled exception Exception name: CONSTRAINT_ERROR Message: hamming.adb:34 overflow check failed Call stack traceback locations: 0x403212 0x402fd7 0x402a87 0x7f8b99517584 0x4026d7
For using Big_Number, you just have to add this to the code (additional to with Big_Number; and with Ada.Strings.Unbounded; in context clause): <lang Ada> type My_Index is mod 2**8;
package My_Big_Numbers is new Big_Number (Index_type => My_Index, Nb_Item => 64); function Int2Big is new My_Big_Numbers.Generic_Conversion.Int_Number2Big_Unsigned (Integer);
function Big_Get_Hamming is new Get_Hamming (Int_Type => My_Big_Numbers.Big_Unsigned, Zero => My_Big_Numbers.Big_Unsigned_Zero, One => My_Big_Numbers.Big_Unsigned_One, Two => My_Big_Numbers.Big_Unsigned_Two, Three => Int2Big(3), Five => Int2Big(5), "mod" => My_Big_Numbers.Unsigned_Number."mod", "+" => My_Big_Numbers.Unsigned_Number."+", "/" => My_Big_Numbers.Unsigned_Number."/");</lang>
and then use it like this: <lang Ada> Ada.Text_IO.Put_Line ("3) 1_000_000st Hamming number: " &
Ada.Strings.Unbounded.To_String (My_Big_Numbers.String_Conversion.Big_Unsigned2UString (Big_Get_Hamming (1_000_000))));</lang>
AutoHotkey
<lang AutoHotKey>SetBatchLines, -1 Msgbox % hamming(1,20) Msgbox % hamming(1690) return
hamming(first,last=0) { if (first < 1) ans=ERROR
if (last = 0) last := first
i:=0, j:=0, k:=0
num1 := ceil((last * 20)**(1/3)) num2 := ceil(num1 * ln(2)/ln(3)) num3 := ceil(num1 * ln(2)/ln(5))
loop { H := (2**i) * (3**j) * (5**k) if (H > 0) ans = %H%`n%ans% i++ if (i > num1) { i=0 j++ if (j > num2) { j=0 k++ } } if (k > num3) break } Sort ans, N
Loop, parse, ans, `n, `r { if (A_index > last) break if (A_index < first) continue Output = %Output%`n%A_LoopField% }
return Output }</lang>
ALGOL 68
Hamming numbers are generated in a trivial iterative way as in the Python version below. This program keeps the series needed to generate the numbers as short as possible using flexible rows; on the downside, it spends considerable time on garbage collection. <lang algol68>PR precision=100 PR
MODE SERIES = FLEX [1 : 0] UNT, # Initially, no elements #
UNT = LONG LONG INT; # A 100-digit unsigned integer #
PROC hamming number = (INT n) UNT: # The n-th Hamming number #
CASE n IN 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 # First 10 in a table # OUT # Additional operators # OP MIN = (INT i, j) INT: (i < j | i | j), MIN = (UNT i, j) UNT: (i < j | i | j); PRIO MIN = 9; OP LAST = (SERIES h) UNT: h[UPB h]; # Last element of a series # OP +:= = (REF SERIES s, UNT elem) VOID: # Extend a series by one element, only keep the elements you need # (INT lwb = (i MIN j) MIN k, upb = UPB s; REF SERIES new s = HEAP FLEX [lwb : upb + 1] UNT; (new s[lwb : upb] := s[lwb : upb], new s[upb + 1] := elem); s := new s ); # Determine the n-th hamming number iteratively # SERIES h := 1, # Series, initially one element # UNT m2 := 2, m3 := 3, m5 := 5, # Multipliers # INT i := 1, j := 1, k := 1; # Counters # TO n - 1 DO h +:= (m2 MIN m3) MIN m5; (LAST h = m2 | m2 := 2 * h[i +:= 1]); (LAST h = m3 | m3 := 3 * h[j +:= 1]); (LAST h = m5 | m5 := 5 * h[k +:= 1]) OD; LAST h ESAC;
FOR k TO 20 DO print ((whole (hamming number (k), 0), blank)) OD; print ((newline, whole (hamming number (1 691), 0))); print ((newline, whole (hamming number (1 000 000), 0)))</lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
C
Using a min-heap to keep track of numbers. Does not handle big integers. <lang c>#include <stdio.h>
- include <stdlib.h>
typedef unsigned long long ham;
size_t alloc = 0, n = 1; ham *q = 0;
void qpush(ham h) { int i, j; if (alloc <= n) { alloc = alloc ? alloc * 2 : 16; q = realloc(q, sizeof(ham) * alloc); }
for (i = n++; (j = i/2) && q[j] > h; q[i] = q[j], i = j); q[i] = h; }
ham qpop() { int i, j; ham r, t; /* outer loop for skipping duplicates */ for (r = q[1]; n > 1 && r == q[1]; q[i] = t) { /* inner loop is the normal down heap routine */ for (i = 1, t = q[--n]; (j = i * 2) < n;) { if (j + 1 < n && q[j] > q[j+1]) j++; if (t <= q[j]) break; q[i] = q[j], i = j; } }
return r; }
int main() { int i; ham h;
for (qpush(i = 1); i <= 1691; i++) { /* takes smallest value, and queue its multiples */ h = qpop(); qpush(h * 2); qpush(h * 3); qpush(h * 5);
if (i <= 20 || i == 1691) printf("%6d: %llu\n", i, h); }
/* free(q); */ return 0; }</lang>
Alternative
Standard algorithm. Numbers are stored as exponents of factors instead of big integers, while GMP is only used for display. It's much more efficient this way. <lang c>#include <stdio.h>
- include <stdlib.h>
- include <string.h>
- include <math.h>
- include <gmp.h>
/* number of factors. best be mutually prime -- duh. */
- define NK 3
- define MAX_HAM 10000000
int n_hams = 0, idx[NK] = {0}, fac[] = { 2, 3, 5, 7, 11};
/* k-smooth numbers are stored as their exponents of each factor;
v is the log of the number, for convenience. */
typedef struct { unsigned short e[NK]; double v; } ham_t, *ham;
ham_t *hams, values[NK] = {{{0}, 0}}; double inc[NK];
/* most of the time v can be just incremented, but eventually
* floating point precision will bite us, so better recalculate */
inline void _setv(ham x) { int i; for (x->v = 0, i = 0; i < NK; i++) x->v += inc[i] * x->e[i]; }
inline int _eq(ham x, ham y) { int i; for (i = 0; i < NK; i++) if (x->e[i] != y->e[i]) return 0; return 1; }
ham get_ham(int n) { int i, ni; ham h;
n--; while (n_hams < n) { for (ni = 0, i = 1; i < NK; i++) if (values[i].v < values[ni].v) ni = i;
*(h = hams + ++n_hams) = values[ni]; _setv(h);
for (ni = 0; ni < NK; ni++) if ( _eq(values + ni, h) ) { values[ni] = hams[++idx[ni]]; values[ni].e[ni]++; _setv(values + ni); } }
return hams + n; }
void show_ham(ham h) { static mpz_t das_ham, tmp; int i;
mpz_init_set_ui(das_ham, 1);
mpz_init_set_ui(tmp, 1); for (i = 0; i < NK; i++) { mpz_ui_pow_ui(tmp, fac[i], h->e[i]); mpz_mul(das_ham, das_ham, tmp); } gmp_printf("%Zu\n", das_ham); }
int main() { int i; hams = malloc(sizeof(ham_t) * MAX_HAM);
for (i = 0; i < NK; i++) { values[i].e[i] = 1; values[i].v = inc[i] = log(fac[i]); }
printf(" 1,691: "); show_ham(get_ham(1691)); printf(" 1,000,000: "); show_ham(get_ham(1e6)); printf("10,000,000: "); show_ham(get_ham(1e7)); return 0; }</lang>
- Output:
1,691: 2125764000 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000 10,000,000: 16244105063830431823239 ..<a gadzillion digits>.. 000000000000000000000
C#
<lang csharp>using System; using System.Collections; using System.Collections.Generic; using System.Linq;
public class LazyList : IEnumerable<int> {
int First; Func<LazyList> Rest;
public LazyList(int first, Func<LazyList> rest) { First = first; Rest = rest; }
public LazyList Map(Func<int, int> function) { return new LazyList(function(First), () => Rest().Map(function)); }
public LazyList Merge(LazyList list) { if (this == null) { return list; } if (list == null) { return this; } if (First < list.First) { return new LazyList(First, () => Rest().Merge(list)); } if (First > list.First) { return new LazyList(list.First, () => Merge(list.Rest())); } return new LazyList(First, () => Rest().Merge(list.Rest())); }
public IEnumerator<int> GetEnumerator() { yield return First; if (Rest() != null) { foreach (var first in Rest()) { yield return first; } } }
IEnumerator IEnumerable.GetEnumerator() { return (IEnumerator)GetEnumerator(); }
}
class Program {
static void Main() { var hamming = default(LazyList); hamming = new LazyList(1, () => hamming.Map(x => x * 2).Merge(hamming.Map(x => x * 3).Merge(hamming.Map(x => x * 5)))); foreach (var number in hamming.Take(20)) { Console.WriteLine(number); } }
}</lang>
Clojure
This version implements Dijkstra's merge solution, so is closely related to the Haskell version. <lang clojure>(defn smerge [xs ys]
(lazy-seq (let [x (first xs), y (first ys), [z xs* ys*] (cond (< x y) [x (rest xs) ys] (> x y) [y xs (rest ys)] :else [x (rest xs) (rest ys)])] (cons z (smerge xs* ys*)))))
(defn smerge3 [xs ys zs]
(smerge xs (smerge ys zs)))
(defn map*n [n ks] (map #(* n %) ks))
(def hamming
(lazy-seq (cons 1 (smerge3 (map*n 2 hamming) (map*n 3 hamming) (map*n 5 hamming)))))</lang>
Note that this version uses a lot of space and time after calculating a few hundred thousand elements of the sequence. This is no doubt due to its "holding on to the head": it maintains the entire generated sequence in memory.
CoffeeScript
<lang coffeescript># Generate hamming numbers in order. Hamming numbers have the
- property that they don't evenly divide any prime numbers outside
- a given set, such as [2, 3, 5].
generate_hamming_sequence = (primes, max_n) ->
# We use a lazy algorithm, only ever keeping N candidates # in play, one for each of our seed primes. Let's say # primes is [2,3,5]. Our virtual streams are these: # # hammings: 1,2,3,4,5,6,8,10,12,15,16,18,20,... # hammings*2: 2,4,6,9.10,12,16,20,24,30,32,36,40... # hammings*3: 3,6,9,12,15,18,24,30,36,45,... # hammings*5: 5,10,15,20,25,30,40,50,... # # After encountering 40 for the last time, our candidates # will be # 50 = 2 * 25 # 45 = 3 * 15 # 50 = 5 * 10 # Then, after 45 # 50 = 2 * 25 # 48 = 3 * 16 <= new # 50 = 5 * 10 hamming_numbers = [1] candidates = ([p, p, 1] for p in primes) last_number = 1 while hamming_numbers.length < max_n # Get the next candidate Hamming Number tuple. i = min_idx(candidates) candidate = candidates[i] [n, p, seq_idx] = candidate # Add to sequence unless it's a duplicate. if n > last_number hamming_numbers.push n last_number = n
# Replace the candidate with its successor (based on # p = 2, 3, or 5). # # This is the heart of the algorithm. Let's say, over the # primes [2,3,5], we encounter the hamming number 32 based on it being # 2 * 16, where 16 is the 12th number in the sequence. # We'll be passed in [32, 2, 12] as candidate, and # hamming_numbers will be [1,2,3,4,5,6,8,9,10,12,16,18,...] # by now. The next candidate we need to enqueue is # [36, 2, 13], where the numbers mean this: # # 36 - next multiple of 2 of a Hamming number # 2 - prime number # 13 - 1-based index of 18 in the sequence # # When we encounter [36, 2, 13], we will then enqueue # [40, 2, 14], based on 20 being the 14th hamming number. q = hamming_numbers[seq_idx] candidates[i] = [p*q, p, seq_idx+1] hamming_numbers
min_idx = (arr) ->
# Don't waste your time reading this--it just returns # the index of the smallest tuple in an array, respecting that # the tuples may contain integers. (CS compiles to JS, which is # kind of stupid about sorting. There are libraries to work around # the limitation, but I wanted this code to be standalone.) less_than = (tup1, tup2) -> i = 0 while i < tup2.length return true if tup1[i] <= tup2[i] return false if tup1[i] > tup2[i] i += 1
min_i = 0 for i in [1...arr.length] if less_than arr[i], arr[min_i] min_i = i return min_i
primes = [2, 3, 5] numbers = generate_hamming_sequence(primes, 10000) console.log numbers[1690] console.log numbers[9999]</lang>
Common Lisp
Maintaining three queues, popping the smallest value every time. <lang lisp>(defun next-hamm (factors seqs)
(let ((x (apply #'min (map 'list #'first seqs)))) (loop for s in seqs
for f in factors for i from 0 with add = t do (if (= x (first s)) (pop s)) ;; prevent a value from being added to multiple lists (when add (setf (elt seqs i) (nconc s (list (* x f)))) (if (zerop (mod x f)) (setf add nil)))
finally (return x))))
(loop with factors = '(2 3 5)
with seqs = (loop for i in factors collect '(1)) for n from 1 to 1000001 do (let ((x (next-hamm factors seqs)))
(if (or (< n 21) (= n 1691) (= n 1000000)) (format t "~d: ~d~%" n x))))</lang> A much faster method: <lang lisp>(defun hamming (n)
(let ((fac '(2 3 5))
(idx (make-array 3 :initial-element 0)) (h (make-array (1+ n) :initial-element 1 :element-type 'integer)))
(loop for i from 1 to n
with e with x = '(1 1 1) do (setf e (setf (aref h i) (apply #'min x)) x (loop for y in x for f in fac for j from 0 collect (if (= e y) (* f (aref h (incf (aref idx j)))) y))))
(aref h n)))
(loop for i from 1 to 20 do
(format t "~2d: ~d~%" i (hamming i)))
(loop for i in '(1691 1000000) do
(format t "~d: ~d~%" i (hamming i)))</lang>
- Output:
1: 1 2: 2 3: 3 4: 4 5: 5 6: 6 7: 8 8: 9 9: 10 10: 12 11: 15 12: 16 13: 18 14: 20 15: 24 16: 25 17: 27 18: 30 19: 32 20: 36 1691: 2125764000 1000000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
D
This version keeps all numbers in memory and runs in a couple of seconds, computing all the Hamming numbers up to the needed one. Performs constant number of operations per Hamming number produced. <lang d>import std.stdio, std.bigint, std.algorithm, std.range;
auto hamming(int n) {
BigInt two = 2, three = 3, five = 5; auto h = new BigInt[n]; h[0] = 1; BigInt x2 = 2, x3 = 3, x5 = 5; int i, j, k;
foreach (ref el; h[1 .. $]) { el = min(x2, x3, x5); if (x2 == el) x2 = two * h[++i]; if (x3 == el) x3 = three * h[++j]; if (x5 == el) x5 = five * h[++k]; } return h[$ - 1];
}
void main() {
writeln(map!hamming(iota(1, 21))); writeln(hamming(1691)); writeln(hamming(1_000_000));
}</lang>
- Output:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Alternative version 1
This keeps numbers in memory (5.4 seconds run time) but overcomputes a sequence by a factor of about , calculating extra multiples past that as well. Incurs an extra 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:
<lang d>import std.stdio,std.bigint,std.container,std.algorithm,std.range;
BigInt hamming(int n) in {
assert(n > 0);
} body {
auto frontier = redBlackTree(BigInt(2), BigInt(3), BigInt(5)); auto lowest = BigInt(1); foreach (_; 1 .. n) { lowest = frontier.front(); frontier.removeFront(); frontier.insert(lowest * 2); frontier.insert(lowest * 3); frontier.insert(lowest * 5); } return lowest;
}
void main() {
writeln("First 20 Hamming numbers: ", map!hamming(iota(1, 21))); writeln("hamming(1691) = ", hamming(1691)); writeln("hamming(1_000_000) = ", hamming(1_000_000));
}</lang>
- Output:
First 20 Hamming numbers: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] hamming(1691) = 2125764000 hamming(1_000_000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Alternative version 2
Does exactly what the first version does, creating an array and filling it with Hamming numbers, keeping the three back pointers into the sequence for next multiples calculations, except that it represents the numbers as their coefficients triples and their logarithm values (for comparisons), thus saving on BigInt calculations.
<lang d>import std.stdio: writefln; import std.bigint: BigInt, toDecimalString; import std.numeric: gcd; import std.algorithm: copy, map; import std.math; // log, ^^
// number of factors enum NK = 3;
enum MAX_HAM = 10_000_000; static assert(gcd(NK, MAX_HAM) == 1);
enum int[NK] fac = [2, 3, 5];
/// k-smooth numbers (stored as their exponents of each factor).
struct Hamming {
double v; // log of the number, for convenience. ushort[NK] e; // exponents of each factor.
// Compile-time constant, map!log(fac) // log can't be used in CTFE yet public static __gshared /*const*/ double[fac.length] inc;
/*nothrow*/ static this() { // not nothrow because map isn't nothrow copy(map!log(fac[]), inc[]); }
bool opEquals(in ref Hamming y) const pure nothrow { //return this.e == y.e; // too much slow foreach (size_t i; 0 .. this.e.length) if (this.e[i] != y.e[i]) return false; return true; }
void update() /*pure*/ nothrow { // can't be pure because inc is not enum/const //this.v = dotProduct(inc, this.e); // too much slow this.v = 0.0; foreach (size_t i; 0 .. this.e.length) this.v += inc[i] * this.e[i]; }
string toString() const { BigInt result = 1; foreach (size_t i, f; fac) result *= BigInt(f) ^^ this.e[i]; return toDecimalString(result); }
}
// global variables __gshared Hamming[] hams; __gshared Hamming[NK] values;
nothrow static this() {
// Slower than malloc if you don't use all the MAX_HAM items hams = new Hamming[MAX_HAM];
foreach (i, ref v; values) { v.e[i] = 1; v.v = Hamming.inc[i]; }
}
ref Hamming getHam(in size_t n) nothrow
in {
assert(n <= MAX_HAM);
} body {
// most of the time v can be just incremented, but eventually // floating point precision will bite us, so better recalculate __gshared static size_t[NK] idx; __gshared static int n_hams;
for (; n_hams < n; n_hams++) { { // find the index of the minimum v size_t ni = 0; foreach (size_t i; 1 .. NK) if (values[i].v < values[ni].v) ni = i;
hams[n_hams] = values[ni]; hams[n_hams].update(); }
foreach (size_t i; 0 .. NK) if (values[i] == hams[n_hams]) { values[i] = hams[idx[i]]; idx[i]++; values[i].e[i]++; values[i].update(); } }
return hams[n - 2];
}
void main() {
foreach (n; [1691, 10 ^^ 6, MAX_HAM]) writefln("%8d: %s", n, getHam(n));
}</lang> The output is similar to the second C version. Runtime is about 0.2 seconds if MAX_HAM = 1_000_000 (as the task requires), and 1.7 seconds if MAX_HAM = 10_000_000.
Eiffel
<lang Eiffel> note description : "Initial part, in order, of the sequence of Hamming numbers, also known as regular numbers" arithmetic : "[ This version avoids integer overflow and stops at the last representable number in the sequence. ]" output : "[
Per requirements of the RosettaCode example, execution will produce items of indexes 1 to 20 and 1691. The algorithm (procedure `hamming') is more general and will produce the first `n' Hamming numbers for any `n'. ]"
source : "This problem was posed in Edsger W. Dijkstra, A Discipline of Programming, Prentice Hall, 1978" date : "8 August 2012" authors : "Bertrand Meyer", "Emmanuel Stapf" revision : "1.0" libraries : "Relies on SORTED_TWO_WAY_LIST from EiffelBase" implementation : "[ Using SORTED_TWO_WAY_LIST provides an elegant illustration of how to implement a lazy scheme in Eiffel through the use of object-oriented data structures. ]" warning : "[ The formatting (<lang>) specifications for Eiffel in RosettaCode are slightly obsolete: `note' and other newer keywords not supported, red color for manifest strings. This should be fixed soon. ]"
class APPLICATION
create make
feature {NONE} -- Initialization
make -- Print first 20 Hamming numbers, in order, and the 1691-st one. local Hammings: like hamming -- List of Hamming numbers, up to 1691-st one. do Hammings := hamming (1691) across 1 |..| 20 as i loop io.put_natural (Hammings.i_th (i.item)); io.put_new_line end io.put_natural (Hammings.i_th (1691)); io.put_new_line end
feature -- Basic operations
hamming (n: INTEGER): ARRAYED_LIST [NATURAL] -- First `n' elements (in order) of the Hamming sequence, -- or as many of them as will not produce overflow. local sl: SORTED_TWO_WAY_LIST [NATURAL] overflow: BOOLEAN first, next: NATURAL do create Result.make (n); create sl.make sl.extend (1); sl.start across 1 |..| n as i invariant -- "The numbers output so far are the first `i' - 1 Hamming numbers, in order". -- "Result.first is the `i'-th Hamming number." until sl.is_empty loop first := sl.first; sl.start Result.extend (first); sl.remove across << 2, 3, 5 >> as multiplier loop next := multiplier.item * first overflow := overflow or next <= first if not overflow and then not sl.has (next) then sl.extend (next) end end end end end </lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2125764000
Factor
<lang factor>USING: accessors deques dlists fry kernel make math math.order
IN: rosetta.hamming
TUPLE: hamming-iterator 2s 3s 5s ;
- <hamming-iterator> ( -- hamming-iterator )
hamming-iterator new 1 1dlist >>2s 1 1dlist >>3s 1 1dlist >>5s ;
- enqueue ( n hamming-iterator -- )
[ [ 2 * ] [ 2s>> ] bi* push-back ] [ [ 3 * ] [ 3s>> ] bi* push-back ] [ [ 5 * ] [ 5s>> ] bi* push-back ] 2tri ;
- next ( hamming-iterator -- n )
dup [ 2s>> ] [ 3s>> ] [ 5s>> ] tri 3dup [ peek-front ] tri@ min min [ '[ dup peek-front _ = [ pop-front* ] [ drop ] if ] tri@ ] [ swap enqueue ] [ ] tri ;
- next-n ( hamming-iterator n -- seq )
swap '[ _ [ _ next , ] times ] { } make ;
- nth-from-now ( hamming-iterator n -- m )
1 - over '[ _ next drop ] times next ;</lang>
<hamming-iterator> 20 next-n . <hamming-iterator> 1691 nth-from-now . <hamming-iterator> 1000000 nth-from-now .
Lazy lists aren quite slow in Factor, but still. <lang factor>USING: combinators fry kernel lists lists.lazy locals math ; IN: rosetta.hamming-lazy
- sort-merge ( xs ys -- result )
xs car :> x ys car :> y { { [ x y < ] [ [ x ] [ xs cdr ys sort-merge ] lazy-cons ] } { [ x y > ] [ [ y ] [ ys cdr xs sort-merge ] lazy-cons ] } [ [ x ] [ xs cdr ys cdr sort-merge ] lazy-cons ] } cond ;
- hamming ( -- hamming )
f :> h! [ 1 ] [ h 2 3 5 [ '[ _ * ] lazy-map ] tri-curry@ tri sort-merge sort-merge ] lazy-cons h! h ;</lang>
20 hamming ltake list>array . 1690 hamming lnth . 999999 hamming lnth .
Forth
This version uses a compact representation of Hamming numbers: each 64-bit cell represents a number 2^l*3^m*5^n, where l, n, and m are bitfields in the cell (20 bits for now). It also uses a fixed-point logarithm to compare the Hamming numbers and prints them in factored form. This code has been tested up to the 10^9th Hamming number. <lang Forth>\ manipulating and computing with Hamming numbers:
- extract2 ( h -- l )
40 rshift ;
- extract3 ( h -- m )
20 rshift $fffff and ;
- extract5 ( h -- n )
$fffff and ;
' + alias h* ( h1 h2 -- h )
- h. { h -- }
." 2^" h extract2 0 .r ." *3^" h extract3 0 .r ." *5^" h extract5 . ;
\ the following numbers have been produced with bc -l as follows 1 62 lshift constant ldscale2
7309349404307464679 constant ldscale3 \ 2^62*l(3)/l(2) (rounded up)
10708003330985790206 constant ldscale5 \ 2^62*l(5)/l(2) (rounded down)
- hld { h -- ud }
\ ud is a scaled fixed-point representation of the logarithm dualis of h h extract2 ldscale2 um* h extract3 ldscale3 um* d+ h extract5 ldscale5 um* d+ ;
- h<= ( h1 h2 -- f )
2dup = if 2drop true exit then hld rot hld assert( 2over 2over d<> ) du>= ;
- hmin ( h1 h2 -- h )
2dup h<= if drop else nip then ;
\ actual algorithm
0 value seq variable seqlast 0 seqlast !
- lastseq ( -- u )
\ last stored number in the sequence seq seqlast @ th @ ;
- genseq ( h1 "name" -- )
\ h1 is the factor for the sequence create , 0 , \ factor and index of element used for last return does> ( -- u2 ) \ u2 is the next number resulting from multiplying h1 with numbers \ in the sequence that is larger than the last number in the \ sequence dup @ lastseq { h1 l } cell+ dup @ begin ( index-addr index ) seq over th @ h1 h* dup l h<= while drop 1+ repeat >r swap ! r> ;
$10000000000 genseq s2 $00000100000 genseq s3 $00000000001 genseq s5
- nextseq ( -- )
s2 s3 hmin s5 hmin , 1 seqlast +! ;
- nthseq ( u1 -- h )
\ the u1 th element in the sequence dup seqlast @ u+do nextseq loop 1- 0 max cells seq + @ ;
- .nseq ( u1 -- )
dup seqlast @ u+do nextseq loop 0 u+do seq i th @ h. loop ;
here to seq 0 , \ that's 1
20 .nseq cr 1691 nthseq h. cr 1000000 nthseq h.</lang>
- Output:
2^0*3^0*5^0 2^1*3^0*5^0 2^0*3^1*5^0 2^2*3^0*5^0 2^0*3^0*5^1 2^1*3^1*5^0 2^3*3^0*5^0 2^0*3^2*5^0 2^1*3^0*5^1 2^2*3^1*5^0 2^0*3^1*5^1 2^4*3^0*5^0 2^1*3^2*5^0 2^2*3^0*5^1 2^3*3^1*5^0 2^0*3^0*5^2 2^0*3^3*5^0 2^1*3^1*5^1 2^5*3^0*5^0 2^2*3^2*5^0 2^5*3^12*5^3 2^55*3^47*5^64
Fortran
Using big_integer_module from here[1] <lang fortran>program Hamming_Test
use big_integer_module implicit none call Hamming(1,20) write(*,*) call Hamming(1691) write(*,*) call Hamming(1000000)
contains
subroutine Hamming(first, last)
integer, intent(in) :: first integer, intent(in), optional :: last integer :: i, n, i2, i3, i5, lim type(big_integer), allocatable :: hnums(:)
if(present(last)) then lim = last else lim = first end if
if(first < 1 .or. lim > 2500000 ) then write(*,*) "Invalid input" return end if allocate(hnums(lim)) i2 = 1 ; i3 = 1 ; i5 = 1 hnums(1) = 1 n = 1 do while(n < lim) n = n + 1 hnums(n) = mini(2*hnums(i2), 3*hnums(i3), 5*hnums(i5)) if(2*hnums(i2) == hnums(n)) i2 = i2 + 1 if(3*hnums(i3) == hnums(n)) i3 = i3 + 1 if(5*hnums(i5) == hnums(n)) i5 = i5 + 1 end do if(present(last)) then do i = first, last call print_big(hnums(i)) write(*, "(a)", advance="no") " " end do else call print_big(hnums(first)) end if deallocate(hnums)
end subroutine
function mini(a, b, c)
type(big_integer) :: mini type(big_integer), intent(in) :: a, b, c if(a < b ) then if(a < c) then mini = a else mini = c end if else if(b < c) then mini = b else mini = c end if
end function mini end program</lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Go
Concise version using dynamic-programming
<lang go>package main
import (
"fmt" "math/big"
)
func min(a, b *big.Int) *big.Int {
if a.Cmp(b) < 0 { return a } return b
}
func hamming(n int) []*big.Int {
h := make([]*big.Int, n) h[0] = big.NewInt(1) two, three, five := big.NewInt(2), big.NewInt(3), big.NewInt(5) next2, next3, next5 := big.NewInt(2), big.NewInt(3), big.NewInt(5) i, j, k := 0, 0, 0 for m := 1; m < len(h); m++ { h[m] = new(big.Int).Set(min(next2, min(next3, next5))) if h[m].Cmp(next2) == 0 { i++; next2.Mul( two, h[i]) } if h[m].Cmp(next3) == 0 { j++; next3.Mul(three, h[j]) } if h[m].Cmp(next5) == 0 { k++; next5.Mul( five, h[k]) } } return h
}
func main() {
h := hamming(1e6) fmt.Println(h[:20]) fmt.Println(h[1691-1]) fmt.Println(h[len(h)-1])
}</lang>
- Output:
[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36] 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Longer version using dynamic-programming and logarithms
More than 10 times faster. <lang go>package main
import (
"flag" "fmt" "math" "math/big" "os"
)
var ordinal int // ordinal of last sequence element to compute var sequenceMode bool // print the whole sequence or just one element?
var lg3, lg5 float64 // precomputed base-2 logarithms for 3 and 5
var table [][3]int16 // table for dynamic-programming stored results var front [3]cursor // state of the three multiplied sequences
type cursor struct {
f int // index (0, 1, 2) corresponding to factor (2, 3, 5) i int // index into table for the entry being multiplied lg float64 // base-2 logarithm of the multiple (for ordering)
}
func (c *cursor) value() [3]int16 {
x := table[c.i] x[c.f]++ // multiply by incrementing the exponent return x
}
func (c *cursor) advance() {
c.i++ // skip entries that would produce duplicates for (c.f < 2 && table[c.i][2] > 0) || (c.f < 1 && table[c.i][1] > 0) { c.i++ } x := c.value() c.lg = float64(x[0]) + lg3*float64(x[1]) + lg5*float64(x[2])
}
func step() {
table = append(table, front[0].value()) front[0].advance() // re-establish sorted order if front[0].lg > front[1].lg { front[0], front[1] = front[1], front[0] if front[1].lg > front[2].lg { front[1], front[2] = front[2], front[1] } }
} func show(elem [3]int16) {
z := big.NewInt(1) for i, base := range []int64{2, 3, 5} { b := big.NewInt(base) x := big.NewInt(int64(elem[i])) z.Mul(z, b.Exp(b, x, nil)) } fmt.Println(z)
}
func fail(msg string) {
fmt.Fprintf(os.Stderr, "%s: %s\n", os.Args[0], msg) os.Exit(1)
}
func parse() {
flag.Parse() if flag.NArg() != 1 { fail("need one argument") } _, err := fmt.Sscan(flag.Arg(0), &ordinal) if err != nil || ordinal <= 0 { fail("argument must be a positive integer") }
}
func init() {
flag.BoolVar(&sequenceMode, "s", false, "sequence mode") lg3 = math.Log2(3) lg5 = math.Log2(5) front = [3]cursor{ {0, 0, 1}, // 2 {1, 0, lg3}, // 3 {2, 0, lg5}, // 5 }
}
func main() {
parse() table = make([][3]int16, 1, ordinal) for i, n := 1, ordinal; i < n; i++ { if sequenceMode { show(table[i-1]) } step() } show(table[ordinal-1])
}</lang>
- Output:
$ ./hamming -s 20 | xargs 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 $ time ./hamming 1000000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000 real 0m0.110s user 0m0.090s sys 0m0.020s $ uname -a Linux lance 3.0-ARCH #1 SMP PREEMPT Sat Aug 6 16:18:35 CEST 2011 x86_64 Intel(R) Core(TM)2 Duo CPU P8400 @ 2.26GHz GenuineIntel GNU/Linux
Haskell
The classic version
<lang haskell>hamming = 1 : map (2*) hamming `union` map (3*) hamming `union` map (5*) hamming
union a@(x:xs) b@(y:ys) = case compare x y of
LT -> x : union xs b EQ -> x : union xs ys GT -> y : union a ys
main = do
print $ take 20 hamming print (hamming !! (1691-1), hamming !! (1692-1)) print $ hamming !! (1000000-1)
-- Output:
-- [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
-- (2125764000,2147483648)
-- 519312780448388736089589843750000000000000000000000000000000000000000000000000000000</lang>
Runs in about a second on Ideone.com. The nested union
s' effect is to produce the minimal value at each step, without duplicates. As Haskell evaluation model is on-demand, the three map
expressions are in effect iterators, maintaining hidden pointers back into the shared named storage with which they were each created (a name is a pointer/handle in Haskell; to name is to point at, to refer to, to take a handle on).
The amount of operations is constant for each number produced, so the time complexity should be . Empirically, it is slightly above that and worsening, suggestive of extra cost of bignum arithmetics. Using triples representation with logarithm values for comparisons amends that problem, but runs ~ 1.5x slower for 1,000,000.
This is what that DDJ blog "pseudo-C" code was transcribing, mentioned at the Python entry that started this task ( curiously, it is in almost word-for-word correspondence with Edsger Dijkstra's code from his book A Discipline of Programming, p. 132 ). D, Go, PARI/GP, Prolog all implement the same idea of back-pointers into shared storage. A Haskell run-time system can actually free up the storage automatically at the start of the shared list and only keep the needed portion of it, from the (5*)
back-pointer, – which is about in length – behind the scenes, as long as there's no re-use evident in the code.
Explicit multiples reinserting
This is a common approach which explicitly maintains an internal buffer of 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 operations for each number, instead of of the above version. Translation of Java (which does use priority queue though, so should have O (n logn) operations overall). Uses union
from the "classic" version above:
<lang Haskell>hamm n = drop n $ iterate (\(_,(a:t))-> (a,union t [2*a,3*a,5*a])) (0,[1])</lang>
- Output:
<lang Haskell>*Main> map fst $ take 20 $ hamm 1 [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
- Main> map fst $ take 2 $ hamm 1691
[2125764000,2147483648]
- Main> mapM_ print $ take 10 $ hamm 1
(1,[2,3,5]) (2,[3,4,5,6,10]) (3,[4,5,6,9,10,15]) (4,[5,6,8,9,10,12,15,20]) (5,[6,8,9,10,12,15,20,25]) (6,[8,9,10,12,15,18,20,25,30]) (8,[9,10,12,15,16,18,20,24,25,30,40]) (9,[10,12,15,16,18,20,24,25,27,30,40,45]) (10,[12,15,16,18,20,24,25,27,30,40,45,50]) (12,[15,16,18,20,24,25,27,30,36,40,45,50,60])
- Main> map (length.snd.head.hamm) [2000,4000,8000,16000]
[402,638,1007,1596]</lang> Runs too slowly to reach 1,000,000, with empirical complexity above O (n 1.7 ) and worsening. Last two lines show the internal buffer's length for several sample n s.
Direct calculation through triples enumeration
It is also possible to more or less directly calculate the n-th Hamming number by enumerating (and counting) all the (i,j,k)
triples below its estimated value – with ordering according to their exponents, i*ln2 + j*ln3 + k*ln5
– while storing only the "band" of topmost triples close enough to the target value (more in the original post on DDJ).
The total count of the enumerated triples is then the band's topmost value's position in the Hamming sequence, 1-based. The nth number is then found by a simple lookup in the sorted band, if it was wide enough. This produces the 1,000,000-th value in a few hundredths of a second on Ideone.com, running at about empirical time and space complexity: <lang haskell>-- 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
{-# OPTIONS -O2 -XBangPatterns #-} import Data.List (sortBy) import Data.Function (on)
main = do let (r,t) = nthHam 1000000
sequence_ [print t, print $ trival t]
lb3 = logBase 2 3; lb5 = logBase 2 5 logval (i,j,k) = fromIntegral i + fromIntegral j*lb3 + fromIntegral k*lb5 trival (i,j,k) = 2^i * 3^j * 5^k estval n = (6*lb3*lb5* fromIntegral n)**(1/3) -- estimated logval, base 2 rngval n
| n > 500000 = (2.4496 , 0.0076 ) -- empirical estimation | n > 50000 = (2.4424 , 0.0146 ) -- correction, base 2 | n > 500 = (2.3948 , 0.0723 ) -- (dist,width) | n > 1 = (2.2506 , 0.2887 ) -- around (log $ sqrt 30), | otherwise = (2.2506 , 0.5771 ) -- says WP
nthHam n -- n: 1-based 1,2,3...
| w >= 1 = error $ "Breach of contract: (w < 1): " ++ show w | m < 0 = error $ "Not enough triples generated: " ++ show (c,n) | m >= nb = error $ "Generated band is too narrow: " ++ show (m,nb) | True = res where (d,w) = rngval n -- correction dist, width hi = estval n - d -- hi > logval > hi-w (m,nb) = ( fromInteger $ c - n, length b ) -- m 0-based from top, |band| (s,res) = ( sortBy (flip compare `on` fst) b, s!!m ) -- sorted decreasing, result (c,b) = -- f 0 == prod (sum,concat) . unzip ; prod(f,g)(x,y) = (f x,g y) f 0 -- total count, the band [ ( 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 f !c [] = (c,[]) f !c ((c1,b1):r) = let (cr,br) = f (c+c1) r in case b1 of { [v] -> (cr,v:br) ; _ -> (cr, br) }</lang>
- Output:
-- time: 0.01s memory: 3688 kB (55,47,64) 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Icon and Unicon
This solution uses Unicon's object oriented extensions. An Icon only version has not been provided.
Lazy evaluation is used to improve performance. <lang Unicon># Lazily generate the three Hamming numbers that can be derived directly
- from a known Hamming number h
class Triplet : Class (cv, ce)
method nextVal() suspend cv := @ce end
initially (baseNum) cv := 2*baseNum ce := create (3|5)*baseNum
end
- Generate Hamming numbers, in order. Default is first 30
- But an optional argument can be used to generate more (or less)
- e.g. hamming 5000 generates the first 5000.
procedure main(args)
limit := integer(args[1]) | 30 every write("\t", generateHamming() \ limit)
end
- Do the work. Start with known Hamming number 1 and maintain
- a set of triplet Hamming numbers as they get derived from that
- one. Most of the code here is to figure out which Hamming
- number is next in sequence (while removing duplicates)
procedure generateHamming()
triplers := set() insert(triplers, Triplet(1))
suspend 1 repeat { # Pick a Hamming triplet that *may* have the next smallest number t1 := !triplers # any will do to start
every t1 ~=== (t2 := !triplers) do { if t1.cv > t2.cv then { # oops we were wrong, switch assumption t1 := t2 } else if t1.cv = t2.cv then { # t2's value is a duplicate, so # advance triplet t2, if none left in t2, remove it t2.nextVal() | delete(triplers, t2) } }
# Ok, t1 has the next Hamming number, grab it suspend t1.cv insert(triplers, Triplet(t1.cv)) # Advance triplet t1, if none left in t1, remove it t1.nextVal() | delete(triplers, t1) }
end</lang>
J
Solution:
A concise tacit expression using a (right) fold:
<lang j>hamming=: {. (/:~@~.@], 2 3 5 * {)/ @ (1x,~i.@-)</lang>
Example usage:
<lang j> hamming 20
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
{: hamming 1691
2125764000</lang>
For the millionth (and billionth (1e9)) Hamming number see the hn
verb from Hamming Number essay on the J wiki.
Explanation:
I'll explain this J-sentence by dividing it in three parts from left to right omitting the leftmost {.
:
- sort and remove duplicates
<lang j> /:~@~.@]</lang>
- produce (the next) 3 elements by selection and multiplication:
<lang j> 2 3 5 * {</lang> or <lang j> 2 3 5 * LHA { RHA</lang>
- the RH part forms an array of descending indices and the initial Hamming number 1
<lang j> (1x,~i.@-)</lang> e.g. if we want the first 5 Hamming numbers, it produces the array:
4 3 2 1 0 1
in other words, we compute a sequence which begins with the desired hamming sequence and then take the first n elements (which will be our desired hamming sequence)
<lang j> ({. (/:~@~.@], 2 3 5 * {)/ @ (1x,~i.@-)) 7
1 2 3 4 5 6 8</lang>
This starts using a descending sequence with 1 appended:
<lang j> (1x,~i.@-) 7
6 5 4 3 2 1 0 1</lang>
and then the fold expression is inserted between these list elements and the result computed:
<lang j> 6(/:~@~.@], 2 3 5 * {) 5(/:~@~.@], 2 3 5 * {) 4(/:~@~.@], 2 3 5 * {) 3(/:~@~.@], 2 3 5 * {) 2(/:~@~.@], 2 3 5 * {) 1(/:~@~.@], 2 3 5 * {) 0(/:~@~.@], 2 3 5 * {) 1
1 2 3 4 5 6 8 9 10 12 15 18 20 25 30 16 24 40</lang>
(Note: A train of verbs in J is evaluated by supplying arguments to the every other verb (counting from the right) and the combining these results with the remaining verbs. Also: {
has been implemented so that an index of 0 will select the only item from an array with no dimensions.)
Java
Has a common shortcoming of overproducing the sequence by about entries, until the n-th number is no longer needed, instead of stopping as soon as it is reached. See Haskell for an illustration.
Inserting the top number's three multiples deep into the priority queue as it does, incurs extra cost for each number produced. To not worsen the expected algorithm complexity, the priority queue should have (amortized) implementation for both insertion and deletion operations but it looks like it's in Java. <lang java5>import java.math.BigInteger; import java.util.PriorityQueue;
final class Hamming {
private static BigInteger THREE = BigInteger.valueOf(3); private static BigInteger FIVE = BigInteger.valueOf(5);
private static void updateFrontier(BigInteger x, PriorityQueue<BigInteger> pq) { pq.offer(x.shiftLeft(1)); pq.offer(x.multiply(THREE)); pq.offer(x.multiply(FIVE)); }
public static BigInteger hamming(int n) { if (n <= 0) throw new IllegalArgumentException("Invalid parameter"); PriorityQueue<BigInteger> frontier = new PriorityQueue<BigInteger>(); updateFrontier(BigInteger.ONE, frontier); BigInteger lowest = BigInteger.ONE; for (int i = 1; i < n; i++) { lowest = frontier.poll(); while (frontier.peek().equals(lowest)) frontier.poll(); updateFrontier(lowest, frontier); } return lowest; }
public static void main(String[] args) { System.out.print("Hamming(1 .. 20) ="); for (int i = 1; i < 21; i++) System.out.print(" " + hamming(i)); System.out.println("\nHamming(1691) = " + hamming(1691)); System.out.println("Hamming(1000000) = " + hamming(1000000)); }
}</lang>
- Output:
Hamming(1 .. 20) = 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 Hamming(1691) = 2125764000 Hamming(1000000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
JavaScript
This does not calculate the 1,000,000th Hamming number.
Note the use of for (x in obj)
to iterate over the properties of an object, versus for each (y in obj)
to iterate over the values of the properties of an object.
<lang javascript>function hamming() {
var queues = {2: [], 3: [], 5: []}; var base; var next_ham = 1; while (true) { yield next_ham;
for (base in queues) {queues[base].push(next_ham * base)}
next_ham = [ queue[0] for each (queue in queues) ].reduce(function(min, val) { return Math.min(min,val) });
for (base in queues) {if (queues[base][0] == next_ham) queues[base].shift()} }
}
var ham = hamming(); var first20=[], i=1;
for (; i <= 20; i++)
first20.push(ham.next());
print(first20.join(', ')); print('...'); for (; i <= 1690; i++)
ham.next();
print(i + " => " + ham.next());</lang>
- Output:
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36 ... 1691 => 2125764000
Liberty BASIC
LB has unlimited precision integers. <lang lb> dim h( 1000000)
for i =1 to 20
print hamming( i); " ";
next i
print print "H( 1691)", hamming( 1691) print "H( 1000000)", hamming( 1000000)
end
function hamming( limit)
h( 0) =1 x2 =2: x3 =3: x5 =5 i =0: j =0: k =0 for n =1 to limit h( n) = min( x2, min( x3, x5)) if x2 = h( n) then i = i +1: x2 =2 *h( i) if x3 = h( n) then j = j +1: x3 =3 *h( j) if x5 = h( n) then k = k +1: x5 =5 *h( k) next n hamming =h( limit -1)
end function</lang>
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 H( 1691) 2125764000 H( 1000000) 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Logo
<lang logo>to init.ham
; queues make "twos [1] make "threes [1] make "fives [1]
end to next.ham
localmake "ham first :twos if less? first :threes :ham [make "ham first :threes] if less? first :fives :ham [make "ham first :fives]
if equal? :ham first :twos [ignore dequeue "twos] if equal? :ham first :threes [ignore dequeue "threes] if equal? :ham first :fives [ignore dequeue "fives]
queue "twos :ham * 2 queue "threes :ham * 3 queue "fives :ham * 5
output :ham
end
init.ham repeat 20 [print next.ham] repeat 1690-20 [ignore next.ham] print next.ham</lang>
Lua
<lang lua>function hiter()
hammings = {1} prev, vals = {1, 1, 1} index = 1 local function nextv() local n, v = 1, hammings[prev[1]]*2
if hammings[prev[2]]*3 < v then n, v = 2, hammings[prev[2]]*3 end if hammings[prev[3]]*5 < v then n, v = 3, hammings[prev[3]]*5 end prev[n] = prev[n] + 1 if hammings[index] == v then return nextv() end index = index + 1 hammings[index] = v return v
end return nextv
end
j = hiter() for i = 1, 20 do
print(j())
end n, l = 0, 0 while n < 2^31 do n, l = j(), n end print(l)</lang>
MUMPS
<lang MUMPS>Hamming(n) New count,ok,next,number,which For which=2,3,5 Set number=1 For count=1:1:n Do . Set ok=0 Set:count<21 ok=1 Set:count=1691 ok=1 Set:count=n ok=1 . Write:ok !,$Justify(count,5),": ",number . For which=2,3,5 Set next(number*which)=which . Set number=$Order(next("")) . Kill next(number) . Quit Quit Do Hamming(2000)
1: 1 2: 2 3: 3 4: 4 5: 5 6: 6 7: 8 8: 9 9: 10 10: 12 11: 15 12: 16 13: 18 14: 20 15: 24 16: 25 17: 27 18: 30 19: 32 20: 36 1691: 2125764000 2000: 8062156800</lang>
Oz
<lang oz>declare
fun lazy {HammingFun} 1|{FoldL1 [{MultHamming 2} {MultHamming 3} {MultHamming 5}] LMerge} end
Hamming = {HammingFun}
fun {MultHamming N} {LMap Hamming fun {$ X} N*X end} end
fun lazy {LMap Xs F} case Xs of nil then nil [] X|Xr then {F X}|{LMap Xr F} end end
fun lazy {LMerge Xs=X|Xr Ys=Y|Yr} if X < Y then X|{LMerge Xr Ys} elseif X > Y then Y|{LMerge Xs Yr} else X|{LMerge Xr Yr} end end
fun {FoldL1 X|Xr F} {FoldL Xr F X} end
in
{ForAll {List.take Hamming 20} System.showInfo} {System.showInfo {Nth Hamming 1690}} {System.showInfo {Nth Hamming 1000000}}</lang>
PARI/GP
This is a basic implementation; finding the millionth term requires 3 seconds and 54 MB. Much better algorithms exist. <lang parigp>Hupto(n)={
my(v=vector(n),x2=2,x3=3,x5=5,i=1,j=1,k=1,t); v[1]=1; for(m=2,n, v[m]=t=min(x2,min(x3,x5)); if(x2 == t, x2 = v[i++] << 1); if(x3 == t, x3 = 3 * v[j++]); if(x5 == t, x5 = 5 * v[k++]); ); v
}; H(n)=Hupto(n)[n];
Hupto(20) H(1691) H(10^6)</lang>
- Output:
%1 = [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] %2 = 2125764000 %3 = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Perl
<lang perl>use List::Util 'min';
sub ham_gen { my @s = ([1], [1], [1]); my @m = (2, 3, 5);
return sub { # use bigint; my $n = min($s[0][0], $s[1][0], $s[2][0]); for (0 .. 2) { shift @{$s[$_]} if $s[$_][0] == $n; push @{$s[$_]}, $n * $m[$_] }
return $n } }
my ($h, $i) = ham_gen;
++$i, print $h->(), " " until $i > 20; print "...\n";
++$i, $h->() until $i == 1690; print ++$i, "-th: ", $h->(), "\n";</lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 40 ... 1691-th: 2125764000 <perl's bigint is so horribly slow that I didn't have the patience for the 1000000th output> <some big number presumably>
Perl 6
The limit scaling is not required, but it cuts down on a bunch of unnecessary calculation. <lang perl6>my $limit = 32;
sub powers_of ($radix) { 1, [\*] $radix xx * }
my @hammings =
( powers_of(2)[^ $limit ] X* ( powers_of(3)[^($limit * 2/3)] X* powers_of(5)[^($limit * 1/2)] ) ).sort;
say ~@hammings[^20]; say @hammings[1690]; # zero indexed</lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2125764000
PicoLisp
<lang PicoLisp>(de hamming (N)
(let (L (1) H) (do N (for (X L X (cadr X)) # Find smallest result (setq H (car X)) ) (idx 'L H NIL) # Remove it (for I (2 3 5) # Generate next results (idx 'L (* I H) T) ) ) H ) )
(println (make (for N 20 (link (hamming N))))) (println (hamming 1691)) (println (hamming 1000000))</lang>
- Output:
(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36) 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Prolog
Generator idiom
<lang Prolog>%% collect N elements produced by a generator in a row
take( 0, Next, Z-Z, Next). take( N, Next, [A|B]-Z, NZ):- N>0, !, next(Next,A,Next1),
N1 is N-1, take(N1,Next1,B-Z,NZ).
%% a generator provides specific {next} implementation
next( hamm( A2,B,C3,D,E5,F,[H|G] ), H, hamm(X,U,Y,V,Z,W,G) ):-
H is min(A2, min(C3,E5)), ( A2 =:= H -> B=[N2|U],X is N2*2 ; (X,U)=(A2,B) ), ( C3 =:= H -> D=[N3|V],Y is N3*3 ; (Y,V)=(C3,D) ), ( E5 =:= H -> F=[N5|W],Z is N5*5 ; (Z,W)=(E5,F) ).
mkHamm( hamm(1,X,1,X,1,X,X) ). % Hamming numbers generator init state
main(N) :-
mkHamm(G),take(20,G,A-[],_), write(A), nl, take(1691-1,G,_,G2),take(2,G2,B-[],_), write(B), nl, take( N -1,G,_,G3),take(2,G3,[C1|_]-_,_), write(C1), nl.</lang>
Run on Ideone.com, which uses Prolog (swi) (swipl 5.6.64), it produces:
?- main(1000000). [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] [2125764000, 2147483648] ERROR: */2: Arithmetic: evaluation error: `int_overflow'
Laziness flavor
Works with SWI-Prolog. Laziness is simulate with freeze/2 and ground/2.
Took inspiration from this code : http://chr.informatik.uni-ulm.de/~webchr (click on hamming.pl: Solves Hamming Problem).
<lang Prolog>hamming(N) :-
% to stop cleanly nb_setval(go, 1),
% display list ( N = 20 -> watch_20(20, L); watch(1,N,L)),
% go L=[1|L235], multlist(L,2,L2), multlist(L,3,L3), multlist(L,5,L5), merge_(L2,L3,L23), merge_(L5,L23,L235).
%% multlist(L,N,LN)
%% multiply each element of list L with N, resulting in list LN
%% here only do multiplication for 1st element, then use multlist recursively
multlist([X|L],N,XLN) :-
% the trick to stop
nb_getval(go, 1) ->
% laziness flavor when(ground(X), ( XN is X*N, XLN=[XN|LN], multlist(L,N,LN)));
true.
merge_([X|In1],[Y|In2],XYOut) :- % the trick to stop nb_getval(go, 1) ->
% laziness flavor ( X < Y -> XYOut = [X|Out], In11 = In1, In12 = [Y|In2] ; X = Y -> XYOut = [X|Out], In11 = In1, In12 = In2 ; XYOut = [Y|Out], In11 = [X | In1], In12 = In2), freeze(In11,freeze(In12, merge_(In11,In12,Out)));
true.
%% display nth element watch(Max, Max, [X|_]) :- % laziness flavor when(ground(X), (format('~w~n', [X]),
% the trick to stop nb_linkval(go, 0))).
watch(N, Max, [_X|L]):-
N1 is N + 1,
watch(N1, Max, L).
%% display nth element
watch_20(1, [X|_]) :-
% laziness flavor
when(ground(X),
(format('~w~n', [X]),
% the trick to stop nb_linkval(go, 0))).
watch_20(N, [X|L]):-
% laziness flavor
when(ground(X),
(format('~w ', [X]),
N1 is N - 1,
watch_20(N1, L))).</lang>
- Output:
?- hamming(20). 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 true . ?- hamming(1691). 2125764000 true . ?- hamming(1000000). 519312780448388736089589843750000000000000000000000000000000000000000000000000000000 true .
Python
Version based on example from Dr. Dobb's CodeTalk
<lang python>from itertools import islice
def hamming2():
\ This version is based on a snippet from: http://dobbscodetalk.com/index.php?option=com_content&task=view&id=913&Itemid=85
When expressed in some imaginary pseudo-C with automatic unlimited storage allocation and BIGNUM arithmetics, it can be expressed as: hamming = h where array h; n=0; h[0]=1; i=0; j=0; k=0; x2=2*h[ i ]; x3=3*h[j]; x5=5*h[k]; repeat: h[++n] = min(x2,x3,x5); if (x2==h[n]) { x2=2*h[++i]; } if (x3==h[n]) { x3=3*h[++j]; } if (x5==h[n]) { x5=5*h[++k]; } h = 1 _h=[h] # memoized multipliers = (2, 3, 5) multindeces = [0 for i in multipliers] # index into _h for multipliers multvalues = [x * _h[i] for x,i in zip(multipliers, multindeces)] yield h while True: h = min(multvalues) _h.append(h) for (n,(v,x,i)) in enumerate(zip(multvalues, multipliers, multindeces)): if v == h: i += 1 multindeces[n] = i multvalues[n] = x * _h[i] # cap the memoization mini = min(multindeces) if mini >= 1000: del _h[:mini] multindeces = [i - mini for i in multindeces] # yield h</lang>
- Output:
>>> list(islice(hamming2(), 20)) [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36] >>> list(islice(hamming2(), 1689, 1690)) [2123366400] >>> list(islice(hamming2(), 999999, 1000000)) [519312780448388736089589843750000000000000000000000000000000000000000000000000000000]
Another implementation of same approach
This is the fastest of the Python implementations, it uses a lot of memory. Does not try to limit memory usage. <lang python>import psyco
def hamming(limit):
h = [1] * limit x2, x3, x5 = 2, 3, 5 i = j = k = 0
for n in xrange(1, limit): h[n] = min(x2, x3, x5) if x2 == h[n]: i += 1 x2 = 2 * h[i] if x3 == h[n]: j += 1 x3 = 3 * h[j] if x5 == h[n]: k += 1 x5 = 5 * h[k]
return h[-1]
psyco.bind(hamming) print [hamming(i) for i in xrange(1, 21)] print hamming(1691) print hamming(1000000)</lang>
"Cyclical Iterators"
The original author is Raymond Hettinger and the code was first published here under the MIT license. Uses iterators dubbed "cyclical" in a sense that they are referring back (explicitly, with p2, p3, p5
iterators) to the previously produced values, same as the above versions (through indecies into shared storage) and the classic Haskell version (implicitly timed by lazy evaluation).
Memory is efficiently maintained automatically by the tee
function for each of the three generator expressions, i.e. only that much is maintained as needed to produce the next value (although it looks like the storage is not shared so three copies are maintained implicitly there).
<lang python>from itertools import tee, chain, groupby, islice
from heapq import merge
def raymonds_hamming():
# Generate "5-smooth" numbers, also called "Hamming numbers" # or "Regular numbers". See: http://en.wikipedia.org/wiki/Regular_number # Finds solutions to 2**i * 3**j * 5**k for some integers i, j, and k.
def deferred_output(): for i in output: yield i
result, p2, p3, p5 = tee(deferred_output(), 4) m2 = (2*x for x in p2) # multiples of 2 m3 = (3*x for x in p3) # multiples of 3 m5 = (5*x for x in p5) # multiples of 5 merged = merge(m2, m3, m5) combined = chain([1], merged) # prepend a starting point output = (k for k,g in groupby(combined)) # eliminate duplicates
return result
print list(islice(raymonds_hamming(), 20)) print islice(raymonds_hamming(), 1689, 1690).next() print islice(raymonds_hamming(), 999999, 1000000).next()</lang> Results are the same as before. Another formulation along the same lines, but greatly simplified, can be found here.
Qi
<lang qi>(define smerge
[X|Xs] [Y|Ys] -> [X | (freeze (smerge (thaw Xs) [Y|Ys]))] where (< X Y) [X|Xs] [Y|Ys] -> [Y | (freeze (smerge [X|Xs] (thaw Ys)))] where (> X Y) [X|Xs] [_|Ys] -> [X | (freeze (smerge (thaw Xs) (thaw Ys)))])
(define smerge3
Xs Ys Zs -> (smerge Xs (smerge Ys Zs)))
(define smap
F [S|Ss] -> [(F S)|(freeze (smap F (thaw Ss)))])
(set hamming [1 | (freeze (smerge3 (smap (* 2) (value hamming))
(smap (* 3) (value hamming)) (smap (* 5) (value hamming))))])
(define stake
_ 0 -> [] [S|Ss] N -> [S|(stake (thaw Ss) (1- N))])
(stake (value hamming) 20)</lang>
- Output:
[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]
R
Recursively find the Hamming numbers below . Works for larger limits, however to find the value, one needs guess the correct limit <lang R>hamming=function(hamms,limit) {
tmp=hamms for(h in c(2,3,5)) { tmp=c(tmp,h*hamms) } tmp=unique(tmp[tmp<=limit]) if(length(tmp)>length(hamms)) { hamms=hamming(tmp,limit) } hamms
} sort(hamming(1,limit=2^31)[-1])</lang>
REXX
<lang rexx>/*REXX program computes Hamming numbers: 1──>20, #1691, #one millionth.*/ numeric digits 100 /*ensure we have enough precision*/ call hamming 1,20 /*show the first ──> twentyth #s.*/ call hamming 1691 /*show the 1,691st Hamming number*/ call hamming 1000000 /*show the one millioneth number.*/ exit /*stick a fork in it, we're done.*/ /*───────────────────────────────────HAMMING subroutine─────────────────*/ hamming: procedure; parse arg x,y; if y== then y=x p2=1; p3=1; p5=1; a.=0; a.1=1
do n=2 for y─1 a.n=min(2*a.p2, 3*a.p3, 5*a.p5) /*pick the minimum of three pigs.*/ if 2*a.p2==a.n then p2=p2+1 if 3*a.p3==a.n then p3=p3+1 if 5*a.p5==a.n then p5=p5+1 end /*n*/
say
do j=x to y say 'hamming('j")=" a.j /*list 'em, Dano. */ end /*j*/
say say right('length of last hamming number=' length(a.y), 70) return</lang> output
hamming(1)= 1 hamming(2)= 2 hamming(3)= 3 hamming(4)= 4 hamming(5)= 5 hamming(6)= 6 hamming(7)= 8 hamming(8)= 9 hamming(9)= 10 hamming(10)= 12 hamming(11)= 15 hamming(12)= 16 hamming(13)= 18 hamming(14)= 20 hamming(15)= 24 hamming(16)= 25 hamming(17)= 27 hamming(18)= 30 hamming(19)= 32 hamming(20)= 36 length of last hamming number=2 hamming(1691)= 2125764000 length of last hamming number=10 hamming(1000000)= 519312780448388736089589843750000000000000000000000000000000000000000000000000000000 length of last hamming number=84
Ruby
<lang ruby>require 'generator'
- the Hamming number generator
hamming = Generator.new do |generator|
next_ham = 1 queues = { 2 => [], 3 => [], 5 => [] } loop do generator.yield next_ham [2,3,5].each {|m| queues[m] << (next_ham * m)} next_ham = [2,3,5].collect {|m| queues[m][0]}.min [2,3,5].each {|m| queues[m].shift if queues[m][0] == next_ham} end
end</lang>
This method does not require a library module. <lang ruby>hamming = Enumerator.new do |yielder|
next_ham = 1 queues = { 2 => [], 3 => [], 5 => [] }
loop do yielder << next_ham # or: yielder.yield(next_ham)
[2,3,5].each {|m| queues[m]<< (next_ham * m)} next_ham = [2,3,5].collect {|m| queues[m][0]}.min [2,3,5].each {|m| queues[m].shift if queues[m][0]== next_ham} end
end</lang> And the "main" part of the task <lang ruby>start = Time.now
idx = 1 hamming.each do |ham|
case idx when (1..20), 1691 p [idx, ham] when 1_000_000 p [idx, ham] break end idx += 1
end
puts "elapsed: #{Time.now - start} seconds"</lang>
- Output:
[1, 1] [2, 2] [3, 3] [4, 4] [5, 5] [6, 6] [7, 8] [8, 9] [9, 10] [10, 12] [11, 15] [12, 16] [13, 18] [14, 20] [15, 24] [16, 25] [17, 27] [18, 30] [19, 32] [20, 36] [1691, 2125764000] [1000000, 519312780448388736089589843750000000000000000000000000000000000000000000000000000000] elapsed: 143.96875 seconds
Scala
<lang scala>class Hamming extends Iterator[BigInt] {
import scala.collection.mutable.Queue val qs = Seq.fill(3)(new Queue[BigInt]) def enqueue(n: BigInt) = qs zip Seq(2, 3, 5) foreach { case (q, m) => q enqueue n * m } def next = { val n = qs map (_.head) min; qs foreach { q => if (q.head == n) q.dequeue } enqueue(n) n } def hasNext = true qs foreach (_ enqueue 1)
}</lang> However, the usage of closures adds a significant amount of time. The code below, though a bit uglier because of the repetitions, is twice as fast: <lang scala>class Hamming extends Iterator[BigInt] {
import scala.collection.mutable.Queue val q2 = new Queue[BigInt] val q3 = new Queue[BigInt] val q5 = new Queue[BigInt] def enqueue(n: BigInt) = { q2 enqueue n * 2 q3 enqueue n * 3 q5 enqueue n * 5 } def next = { val n = q2.head min q3.head min q5.head if (q2.head == n) q2.dequeue if (q3.head == n) q3.dequeue if (q5.head == n) q5.dequeue enqueue(n) n } def hasNext = true List(q2, q3, q5) foreach (_ enqueue 1)
}</lang> Usage:
scala> new Hamming take 20 toList res87: List[BigInt] = List(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36) scala> new Hamming drop 1690 next res88: BigInt = 2125764000 scala> new Hamming drop 999999 next res89: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
There's also a fairly mechanical translation from Haskell using purely functional lazy streams
<lang scala>val hamming : Stream[BigInt] = {
def merge(inx : Stream[BigInt], iny : Stream[BigInt]) : Stream[BigInt] = { if (inx.head < iny.head) inx.head #:: merge(inx.tail, iny) else if (iny.head < inx.head) iny.head #:: merge(inx, iny.tail) else merge(inx, iny.tail) }
1 #:: merge(hamming map (_ * 2), merge(hamming map (_ * 3), hamming map (_ * 5)))
}</lang> Use of "force" ensures that the stream is computed before being printed, otherwise it would just be left suspended and you'd see "Stream(1, ?)"
scala> (hamming take 20).force res0: scala.collection.immutable.Stream[BigInt] = Stream(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36)
To get the nth code find the n-1th element because indexes are 0 based
scala> hamming(1690) res1: BigInt = 2125764000
To calculate the 1000000th code I had to increase the JVM heap from the default
scala> hamming(999999) res2: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Scheme
<lang scheme>(define-syntax lons
(syntax-rules () ((_ lar ldr) (delay (cons lar (delay ldr))))))
(define (lar lons)
(car (force lons)))
(define (ldr lons)
(force (cdr (force lons))))
(define (lap proc . llists)
(lons (apply proc (map lar llists)) (apply lap proc (map ldr llists))))
(define (take n llist)
(if (zero? n) (list) (cons (lar llist) (take (- n 1) (ldr llist)))))
(define (llist-ref n llist)
(if (= n 1) (lar llist) (llist-ref (- n 1) (ldr llist))))
(define (merge llist-1 . llists)
(define (merge-2 llist-1 llist-2) (cond ((null? llist-1) llist-2) ((null? llist-2) llist-1) ((< (lar llist-1) (lar llist-2)) (lons (lar llist-1) (merge-2 (ldr llist-1) llist-2))) ((> (lar llist-1) (lar llist-2)) (lons (lar llist-2) (merge-2 llist-1 (ldr llist-2)))) (else (lons (lar llist-1) (merge-2 (ldr llist-1) (ldr llist-2)))))) (if (null? llists) llist-1 (apply merge (cons (merge-2 llist-1 (car llists)) (cdr llists)))))
(define hamming
(lons 1 (merge (lap (lambda (x) (* x 2)) hamming) (lap (lambda (x) (* x 3)) hamming) (lap (lambda (x) (* x 5)) hamming))))
(display (take 20 hamming)) (newline) (display (llist-ref 1691 hamming)) (newline) (display (llist-ref 1000000 hamming)) (newline)</lang>
- Output:
(1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36) 2125764000 out of memory
Seed7
<lang seed7>$ include "seed7_05.s7i";
include "bigint.s7i";
const func bigInteger: min (in bigInteger: a, in bigInteger: b, in bigInteger: c) is func
result var bigInteger: min is 0_; begin if a < b then min := a; else min := b; end if; if c < min then min := c; end if; end func;
const func bigInteger: hamming (in integer: n) is func
result var bigInteger: hammingNum is 1_; local var array bigInteger: hammingNums is 0 times 0_; var integer: index is 0; var bigInteger: x2 is 2_; var bigInteger: x3 is 3_; var bigInteger: x5 is 5_; var integer: i is 1; var integer: j is 1; var integer: k is 1; begin hammingNums := n times 1_; for index range 2 to n do hammingNum := min(x2, x3, x5); hammingNums[index] := hammingNum; if x2 = hammingNum then incr(i); x2 := 2_ * hammingNums[i]; end if; if x3 = hammingNum then incr(j); x3 := 3_ * hammingNums[j]; end if; if x5 = hammingNum then incr(k); x5 := 5_ * hammingNums[k]; end if; end for; end func;
const proc: main is func
local var integer: n is 0; begin for n range 1 to 20 do write(hamming(n) <& " "); end for; writeln; writeln(hamming(1691)); writeln(hamming(1000000)); end func;</lang>
- Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 2125764000 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Smalltalk
This is a straightforward implementation of the pseudocode snippet found in the Python section. Smalltalk supports arbitrary-precision integers, but the implementation is too slow to try it with 1 million. <lang smalltalk>Object subclass: Hammer [
Hammer class >> hammingNumbers: howMany [ |h i j k x2 x3 x5| h := OrderedCollection new. i := 0. j := 0. k := 0. h add: 1. x2 := 2. x3 := 2. x5 := 5. [ ( h size) < howMany ] whileTrue: [ |m| m := { x2. x3. x5 } sort first. (( h indexOf: m ) = 0) ifTrue: [ h add: m ]. ( x2 = (h last) ) ifTrue: [ i := i + 1. x2 := 2 * (h at: i) ]. ( x3 = (h last) ) ifTrue: [ j := j + 1. x3 := 3 * (h at: j) ]. ( x5 = (h last) ) ifTrue: [ k := k + 1. x5 := 5 * (h at: k) ]. ]. ^ h sort ]
].
(Hammer hammingNumbers: 20) displayNl. (Hammer hammingNumbers: 1690) last displayNl.</lang>
Tcl
This uses coroutines to simplify the description of what's going on.
<lang tcl>package require Tcl 8.6
- Simple helper: Tcl-style list "map"
proc map {varName list script} {
set l {} upvar 1 $varName v foreach v $list {lappend l [uplevel 1 $script]} return $l
}
- The core of a coroutine to compute the product of a hamming sequence.
- Tricky bit: we don't automatically advance to the next value, and instead
- wait to be told that the value has been consumed (i.e., is the result of
- the [yield] operation).
proc ham {key multiplier} {
global hammingCache set i 0 yield [info coroutine] # Cannot use [foreach]; that would take a snapshot of the list in # the hammingCache variable, so missing updates. while 1 {
set n [expr {[lindex $hammingCache($key) $i] * $multiplier}] # If the number selected was ours, we advance to compute the next if {[yield $n] == $n} { incr i }
}
}
- This coroutine computes the hamming sequence given a list of multipliers.
- It uses the [ham] helper from above to generate indivdual multiplied
- sequences. The key into the cache is the list of multipliers.
- Note that it is advisable for the values to be all co-prime wrt each other.
proc hammingCore args {
global hammingCache set hammingCache($args) 1 set hammers [map x $args {coroutine ham$x,$args ham $args $x}] yield while 1 {
set n [lindex $hammingCache($args) [incr i]-1] lappend hammingCache($args) \ [tcl::mathfunc::min {*}[map h $hammers {$h $n}]] yield $n
}
}
- Assemble the pieces so as to compute the classic hamming sequence.
coroutine hamming hammingCore 2 3 5
- Print the first 20 values of the sequence
for {set i 1} {$i <= 20} {incr i} {
puts [format "hamming\[%d\] = %d" $i [hamming]]
} for {} {$i <= 1690} {incr i} {set h [hamming]} puts "hamming{1690} = $h" for {} {$i <= 1000000} {incr i} {set h [hamming]} puts "hamming{1000000} = $h"</lang>
- Output:
hamming{1} = 1 hamming{2} = 2 hamming{3} = 3 hamming{4} = 4 hamming{5} = 5 hamming{6} = 6 hamming{7} = 8 hamming{8} = 9 hamming{9} = 10 hamming{10} = 12 hamming{11} = 15 hamming{12} = 16 hamming{13} = 18 hamming{14} = 20 hamming{15} = 24 hamming{16} = 25 hamming{17} = 27 hamming{18} = 30 hamming{19} = 32 hamming{20} = 36 hamming{1690} = 2123366400 hamming{1000000} = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
A faster version can be built that also works on Tcl 8.5 (or earlier, if only small hamming numbers are being computed): <lang tcl>variable hamming 1 hi2 0 hi3 0 hi5 0 proc hamming {n} {
global hamming hi2 hi3 hi5 set h2 [expr {[lindex $hamming $hi2]*2}] set h3 [expr {[lindex $hamming $hi3]*3}] set h5 [expr {[lindex $hamming $hi5]*5}] while {[llength $hamming] < $n} {
lappend hamming [set h [expr { $h2<$h3 ? $h2<$h5 ? $h2 : $h5 : $h3<$h5 ? $h3 : $h5 }]] if {$h==$h2} { set h2 [expr {[lindex $hamming [incr hi2]]*2}] } if {$h==$h3} { set h3 [expr {[lindex $hamming [incr hi3]]*3}] } if {$h==$h5} { set h5 [expr {[lindex $hamming [incr hi5]]*5}] }
} return [lindex $hamming [expr {$n - 1}]]
}
- Print the first 20 values of the sequence
for {set i 1} {$i <= 20} {incr i} {
puts [format "hamming\[%d\] = %d" $i [hamming $i]]
} puts "hamming{1690} = [hamming 1690]" puts "hamming{1691} = [hamming 1691]" puts "hamming{1692} = [hamming 1692]" puts "hamming{1693} = [hamming 1693]" puts "hamming{1000000} = [hamming 1000000]"</lang>
Ursala
Smooth is defined as a second order function taking a list of primes and returning a function that takes a natural number to the -th smooth number with respect to them. An elegant but inefficient formulation based on the J solution is the following. <lang Ursala>#import std
- import nat
smooth"p" "n" = ~&z take/"n" nleq-< (rep(length "n") ^Ts/~& product*K0/"p") <1></lang> This test program <lang Ursala>main = smooth<2,3,5>* nrange(1,20)</lang> yields this list of the first 20 Hamming numbers.
<1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36>
Although all calculations are performed using unlimited precision, the version above is impractical for large numbers. A more hardcore approach is the following. <lang Ursala>#import std
- import nat
smooth"p" "n" =
~&H\"p" *-<1>; @NiXS ~&/(1,1); ~&ll~="n"->lr -+
^\~&rlPrrn2rrm2Zlrrmz3EZYrrm2lNCTrrm2QAX*rhlPNhrnmtPA2XtCD ~&lrPrhl2E?/~&l ^|/successor@l ~&hl, ^|/~& nleq-<&l+ * ^\~&r ~&l|| product@rnmhPX+-
- cast %nL
main = smooth<2,3,5>* nrange(1,20)--<1691,1000000></lang>
- Output:
The great majority of time is spent calculating the millionth Hamming number.
< 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 2125764000, 519312780448388736089589843750000000000000000000000000000000000000000000000000000000>
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