Hamming numbers

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

Hamming numbers are numbers of the form

H = 2^i \cdot 3^j \cdot 5^k, \; \mathrm{where} \; i, j, k \geq 0.

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:

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

References

  1. Hamming numbers
  2. Smooth number
  3. Hamming problem from Dr. Dobb's CodeTalk (dead link as of Sep 2011; parts of the thread here and here).

Contents

[edit] Ada

Works with: GNAT

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.

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;
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):

   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."/");

and then use it like this:

   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))));

[edit] 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
}

[edit] 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.

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

[edit] AWK

 
# syntax: GAWK -f HAMMING_NUMBERS.AWK
BEGIN {
for (i=1; i<=20; i++) {
printf("%d ",hamming(i))
}
printf("\n1691: %d\n",hamming(1691))
exit(0)
}
function hamming(limit, h,i,j,k,n,x2,x3,x5) {
h[0] = 1
x2 = 2
x3 = 3
x5 = 5
for (n=1; n<=limit; n++) {
h[n] = min(x2,min(x3,x5))
if (h[n] == x2) { x2 = 2 * h[++i] }
if (h[n] == x3) { x3 = 3 * h[++j] }
if (h[n] == x5) { x5 = 5 * h[++k] }
}
return(h[limit-1])
}
function min(x,y) {
return((x < y) ? x : y)
}
 

output:

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

[edit] BBC BASIC

      @% = &1010
FOR h% = 1 TO 20
PRINT "H("; h% ") = "; FNhamming(h%)
NEXT
PRINT "H(1691) = "; FNhamming(1691)
END
 
DEF FNhamming(l%)
LOCAL i%, j%, k%, n%, m, x2, x3, x5, h%()
DIM h%(l%) : h%(0) = 1
x2 = 2 : x3 = 3 : x5 = 5
FOR n% = 1 TO l%-1
m = x2
IF m > x3 m = x3
IF m > x5 m = x5
h%(n%) = m
IF m = x2 i% += 1 : x2 = 2 * h%(i%)
IF m = x3 j% += 1 : x3 = 3 * h%(j%)
IF m = x5 k% += 1 : x5 = 5 * h%(k%)
NEXT
= h%(l%-1)

Output:

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

[edit] Bracmat

Translation of: D
( ( hamming
= x2 x3 x5 n i j k min
. tbl$(h,!arg) { This creates an array. Arrays are always global in Bracmat. }
& 1:?(0$h)
& 2:?x2
& 3:?x3
& 5:?x5
& 0:?n:?i:?j:?k
& whl
' ( !n+1:<!arg:?n
& !x2:?min
& (!x3:<!min:?min|)
& (!x5:<!min:?min|)
& !min:?(!n$h) { !n is index into array h }
& (  !x2:!min
& 2*!((1+!i:?i)$h):?x2
|
)
& (  !x3:!min
& 3*!((1+!j:?j)$h):?x3
|
)
& (  !x5:!min
& 5*!((1+!k:?k)$h):?x5
|
)
)
& !((!arg+-1)$h) (tbl$(h,0)&) { We delete the array by setting its size to 0 }
)
& 0:?I
& whl'(!I+1:~>20:?I&put$(hamming$!I " "))
& out$
& out$(hamming$1691)
& out$(hamming$1000000)
);

Output:

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

[edit] C

Using a min-heap to keep track of numbers. Does not handle big integers.

#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;
}

[edit] 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 <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 (1 << 24)
#define MAX_POW 1024
int n_hams = 0, idx[NK] = {0}, fac[] = { 2, 3, 5, 7, 11};
 
/* k-smooth numbers are stored as their exponents of each factor;
v is the log of the number, for convenience. */

typedef struct {
int e[NK];
double v;
} ham_t, *ham;
 
ham_t *hams, values[NK] = {{{0}, 0}};
double inc[NK][MAX_POW];
 
/* most of the time v can be just incremented, but eventually
* floating point precision will bite us, so better recalculate */

inline
void _setv(ham x) {
int i;
for (x->v = 0, i = 0; i < NK; i++)
x->v += inc[i][x->e[i]];
}
 
inline
int _eq(ham a, ham b) {
int i;
for (i = 0; i < NK && a->e[i] == b->e[i]; i++);
 
return i == NK;
}
 
ham get_ham(int n)
{
int i, ni;
ham h;
 
n--;
while (n_hams < n) {
for (ni = 0, i = 1; i < NK; i++)
if (values[i].v < values[ni].v)
ni = i;
 
*(h = hams + ++n_hams) = values[ni];
 
for (ni = 0; ni < NK; ni++) {
if (! _eq(values + ni, h)) continue;
values[ni] = hams[++idx[ni]];
values[ni].e[ni]++;
_setv(values + ni);
}
}
 
return hams + n;
}
 
void show_ham(ham h)
{
static mpz_t das_ham, tmp;
int i;
 
mpz_init_set_ui(das_ham, 1);
mpz_init_set_ui(tmp, 1);
for (i = 0; i < NK; i++) {
mpz_ui_pow_ui(tmp, fac[i], h->e[i]);
mpz_mul(das_ham, das_ham, tmp);
}
gmp_printf("%Zu\n", das_ham);
}
 
int main()
{
int i, j;
hams = malloc(sizeof(ham_t) * MAX_HAM);
 
for (i = 0; i < NK; i++) {
values[i].e[i] = 1;
inc[i][1] = log(fac[i]);
_setv(values + i);
 
for (j = 2; j < MAX_POW; j++)
inc[i][j] = j * inc[i][1];
}
 
printf(" 1,691: "); show_ham(get_ham(1691));
printf(" 1,000,000: "); show_ham(get_ham(1e6));
printf("10,000,000: "); show_ham(get_ham(1e7));
return 0;
}
Output:
     1,691: 2125764000
 1,000,000: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
10,000,000: 16244105063830431823239 ..<a gadzillion digits>.. 000000000000000000000

[edit] C++

[edit] C++11 For Each Generator

 
#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;
}
};
 

[edit] 5-Smooth

 
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;
}
 

Produces:

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

[edit] 7-Smooth

 
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;
}
 

Produces:

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

[edit] C#

Translation of: D
using System;
using System.Numerics;
using System.Linq;
 
namespace Hamming {
 
class MainClass {
 
public static BigInteger Hamming(int n) {
BigInteger two = 2, three = 3, five = 5;
var h = new BigInteger[n];
h[0] = 1;
BigInteger x2 = 2, x3 = 3, x5 = 5;
int i = 0, j = 0, k = 0;
 
for (int index = 1; index < n; index++) {
h[index] = BigInteger.Min(x2, BigInteger.Min(x3, x5));
if (h[index] == x2) x2 = two * h[++i];
if (h[index] == x3) x3 = three * h[++j];
if (h[index] == x5) x5 = five * h[++k];
}
return h[n - 1];
}
 
public static void Main(string[] args) {
Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).ToList().Select(x => Hamming(x))));
Console.WriteLine(Hamming(1691));
Console.WriteLine(Hamming(1000000));
}
}
}
Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

[edit] Generic version for any set of numbers

The algorithm is similar to the one above.

using System;
using System.Numerics;
using System.Linq;
 
namespace Hamming {
 
class MainClass {
 
public static BigInteger[] Hamming(int n, int[] a) {
var primes = a.Select(x => (BigInteger)x).ToArray();
var values = a.Select(x => (BigInteger)x).ToArray();
var indexes = new int[a.Length];
var results = new BigInteger[n];
results[0] = 1;
for (int iter = 1; iter < n; iter++) {
results[iter] = values[0];
for (int p = 1; p < primes.Length; p++)
if (results[iter] > values[p])
results[iter] = values[p];
for (int p = 0; p < primes.Length; p++)
if (results[iter] == values[p])
values[p] = primes[p] * results[++indexes[p]];
}
return results;
}
 
public static void Main(string[] args) {
foreach (int[] primes in new int[][] { new int[] {2,3,5}, new int[] {2,3,5,7} }) {
Console.WriteLine("{0}-Smooth:", primes.Last());
Console.WriteLine(string.Join(" ", Hamming(20, primes)));
Console.WriteLine(Hamming(1691, primes).Last());
Console.WriteLine(Hamming(1000000, primes).Last());
Console.WriteLine();
}
}
}
}
Output:
5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

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

[edit] Fast version

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

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

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

--Mike Lorenz

using System;
using System.Linq;
using System.Numerics;
 
namespace HammingFast {
 
class MainClass {
 
private static int[] _primes = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 };
 
public static BigInteger Big(int[] exponents) {
BigInteger val = 1;
for (int i = 0; i < exponents.Length; i++)
for (int e = 0; e < exponents[i]; e++)
val = val * _primes[i];
return val;
}
 
public static int[] Hamming(int n, int nprimes) {
var hammings = new int[n, nprimes]; // array of hamming #s we generate
var hammlogs = new double[n]; // log values for above
var primelogs = new double[nprimes]; // pre-calculated prime log values
var indexes = new int[nprimes]; // intermediate hamming values as indexes into hammings
var listheads = new int[nprimes, nprimes]; // intermediate hamming list heads
var listlogs = new double[nprimes]; // log values of list heads
for (int p = 0; p < nprimes; p++) {
listheads[p, p] = 1; // init list heads to prime values
primelogs[p] = Math.Log(_primes[p]); // pre-calc prime log values
listlogs[p] = Math.Log(_primes[p]); // init list head log values
}
for (int iter = 1; iter < n; iter++) {
int min = 0; // find index of min item in list heads
for (int p = 1; p < nprimes; p++)
if (listlogs[p] < listlogs[min])
min = p;
hammlogs[iter] = listlogs[min]; // that's the next hamming number
for (int i = 0; i < nprimes; i++)
hammings[iter, i] = listheads[min, i];
for (int p = 0; p < nprimes; p++) { // update each list head if it matches new value
bool equal = true; // test each exponent to see if number matches
for (int i = 0; i < nprimes; i++) {
if (hammings[iter, i] != listheads[p, i]) {
equal = false;
break;
}
}
if (equal) { // if it matches...
int x = ++indexes[p]; // set index to next hamming number
for (int i = 0; i < nprimes; i++) // copy each hamming exponent
listheads[p, i] = hammings[x, i];
listheads[p, p] += 1; // increment exponent = mult by prime
listlogs[p] = hammlogs[x] + primelogs[p]; // add log(prime) to log(value) = mult by prime
}
}
}
 
var result = new int[nprimes];
for (int i = 0; i < nprimes; i++)
result[i] = hammings[n - 1, i];
return result;
}
 
public static void Main(string[] args) {
foreach (int np in new int[] { 3, 4, 5 }) {
Console.WriteLine("{0}-Smooth:", _primes[np - 1]);
Console.WriteLine(string.Join(" ", Enumerable.Range(1, 20).Select(x => Big(Hamming(x, np)))));
Console.WriteLine(Big(Hamming(1691, np)));
Console.WriteLine(Big(Hamming(1000000, np)));
Console.WriteLine();
}
}
}
}
Output:
5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

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

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

[edit] C# Enumerator Version

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

using System;
using System.Collections.Generic;
using System.Linq;
using System.Numerics;
 
namespace HammingTest
{
class HammingNode
{
public double log;
public int[] exponents;
public HammingNode next;
public int series;
}
 
class HammingListEnumerator : IEnumerable<BigInteger>
{
private int[] primes;
private double[] primelogs;
private HammingNode next;
private HammingNode[] values;
private HammingNode[] indexes;
 
public HammingListEnumerator(IEnumerable<int> seeds)
{
// Ensure our seeds are properly ordered, and generate their log values
primes = seeds.OrderBy(x => x).ToArray();
primelogs = primes.Select(x => Math.Log10(x)).ToArray();
// Start at 1 (log(1)=0, exponents are all 0, series = none)
next = new HammingNode { log = 0, exponents = new int[primes.Length], series = primes.Length };
// Set all exponent sequences to the start, and calculate the first value for each exponent
indexes = new HammingNode[primes.Length];
values = new HammingNode[primes.Length];
for(int i = 0; i < primes.Length; ++i)
{
indexes[i] = next;
values[i] = AddExponent(next, i);
}
}
 
// Make a copy of a node, and increment the specified exponent value
private HammingNode AddExponent(HammingNode node, int i)
{
HammingNode ret = new HammingNode { log = node.log + primelogs[i], exponents = (int[])node.exponents.Clone(), series = i };
++ret.exponents[i];
return ret;
}
 
private void GetNext()
{
// Find which exponent value is the lowest
int min = 0;
for(int i = 1; i < values.Length; ++i)
if(values[i].log < values[min].log)
min = i;
 
// Add it to the end of the 'list', and move to it
next.next = values[min];
next = values[min];
 
// Find the next node in an allowed sequence (skip those that would be duplicates)
HammingNode val = indexes[min].next;
while(val.series < min)
val = val.next;
 
// Keep the current index, and calculate the next value in the series for that exponent
indexes[min] = val;
values[min] = AddExponent(val, min);
}
 
// Skip values without having to calculate the BigInteger value from the exponents
public HammingListEnumerator Skip(int count)
{
for(int i = count; i > 0; --i)
GetNext();
 
return this;
}
 
// Calculate the BigInteger value from the exponents
internal BigInteger ValueOf(HammingNode n)
{
BigInteger val = 1;
for(int i = 0; i < n.exponents.Length; ++i)
for(int e = 0; e < n.exponents[i]; e++)
val = val * primes[i];
return val;
}
 
public IEnumerator<BigInteger> GetEnumerator()
{
while(true)
{
yield return ValueOf(next);
GetNext();
}
}
 
System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
{
return this.GetEnumerator();
}
}
 
class Program
{
static void Main(string[] args)
{
foreach(int[] primes in new int[][] {
new int[] { 2, 3, 5 },
new int[] { 2, 3, 5, 7 },
new int[] { 2, 3, 5, 7, 9}})
{
HammingListEnumerator hammings = new HammingListEnumerator(primes);
System.Diagnostics.Debug.WriteLine("{0}-Smooth:", primes.Last());
System.Diagnostics.Debug.WriteLine(String.Join(" ", hammings.Take(20).ToArray()));
System.Diagnostics.Debug.WriteLine(hammings.Skip(1691 - 20).First());
System.Diagnostics.Debug.WriteLine(hammings.Skip(1000000 - 1691).First());
System.Diagnostics.Debug.WriteLine("");
}
}
}
}
 
Output:
5-Smooth:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

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

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

[edit] Clojure

This version implements Dijkstra's merge solution, so is closely related to the Haskell version.

(defn smerge [xs ys]
(lazy-seq
(let [x (first xs),
y (first ys),
[z xs* ys*]
(cond
(< x y) [x (rest xs) ys]
(> x y) [y xs (rest ys)]
 :else [x (rest xs) (rest ys)])]
(cons z (smerge xs* ys*)))))
 
(def hamming
(lazy-seq
(->> (map #(*' 5 %) hamming)
(smerge (map #(*' 3 %) hamming))
(smerge (map #(*' 2 %) hamming))
(cons 1))))

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.

[edit] 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]

[edit] Common Lisp

Maintaining three queues, popping the smallest value every time.

(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))))

A much faster method:

(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)))
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

[edit] D

[edit] 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.

import std.stdio, std.bigint, std.algorithm, std.range, core.memory;
 
auto hamming(in uint n) pure nothrow /*@safe*/ {
immutable BigInt two = 2, three = 3, five = 5;
auto h = new BigInt[n];
h[0] = 1;
BigInt x2 = 2, x3 = 3, x5 = 5;
size_t i, j, k;
 
foreach (ref el; h.dropOne) {
el = min(x2, x3, x5);
if (el == x2) x2 = two * h[++i];
if (el == x3) x3 = three * h[++j];
if (el == x5) x5 = five * h[++k];
}
return h.back;
}
 
void main() {
GC.disable;
iota(1, 21).map!hamming.writeln;
1_691.hamming.writeln;
1_000_000.hamming.writeln;
}
Output:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

Runtime is about 1.6 seconds with LDC2.

[edit] Alternative Version 1

This keeps O(n2 / 3) numbers in memory, but over-computes a sequence by a factor of about Θ(n2 / 3), calculating extra multiples past that as well. Incurs an extra O(logn) factor of operations per each number produced (reinserting its multiples into a tree). Doesn't stop when the target number is reached, instead continuing until it is no longer needed:

Translation of: Java
import std.stdio, std.bigint, std.container, std.algorithm, std.range,
core.memory;
 
BigInt hamming(in int n)
in {
assert(n > 0);
} body {
auto frontier = redBlackTree(2.BigInt, 3.BigInt, 5.BigInt);
auto lowest = 1.BigInt;
foreach (immutable _; 1 .. n) {
lowest = frontier.front;
frontier.removeFront;
frontier.insert(lowest * 2);
frontier.insert(lowest * 3);
frontier.insert(lowest * 5);
}
return lowest;
}
 
void main() {
GC.disable;
writeln("First 20 Hamming numbers: ", iota(1, 21).map!hamming);
writeln("hamming(1691) = ", 1691.hamming);
writeln("hamming(1_000_000) = ", 1_000_000.hamming);
}
Output:
First 20 Hamming numbers: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
hamming(1691) = 2125764000
hamming(1_000_000) = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

About 3.2 seconds run time with LDC2.

[edit] Alternative Version 2

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

Translation of: C
import std.stdio: writefln;
import std.bigint: BigInt;
import std.conv: text;
import std.numeric: gcd;
import std.algorithm: copy, map;
import core.stdc.stdlib: calloc;
import std.math: log; // ^^
 
// Number of factors.
enum NK = 3;
 
enum MAX_HAM = 10_000_000;
static assert(gcd(NK, MAX_HAM) == 1);
 
enum int[NK] factors = [2, 3, 5];
 
 
/// K-smooth numbers (stored as their exponents of each factor).
struct Hamming {
double v; // Log of the number, for convenience.
ushort[NK] e; // Exponents of each factor.
 
// log can't be used in CTFE yet.
//public static __gshared immutable double[factors.length] inc =
// factors[].map!log.array;
public static __gshared immutable double[factors.length] inc;
 
nothrow pure @nogc static this() {
// factors[].map!log.copy(inc[]);
foreach (immutable i, immutable f; factors)
inc[i] = f.log;
}
 
bool opEquals(in ref Hamming y) const pure nothrow @nogc {
//return this.e == y.e; // Too much slow.
foreach (immutable i; 0 .. this.e.length)
if (this.e[i] != y.e[i])
return false;
return true;
}
 
void update() pure nothrow @nogc {
//this.v = dotProduct(inc, this.e); // Too much slow.
this.v = 0.0;
foreach (immutable i; 0 .. this.e.length)
this.v += inc[i] * this.e[i];
}
 
string toString() const {
BigInt result = 1;
foreach (immutable i, immutable f; factors)
result *= f.BigInt ^^ this.e[i];
return result.text;
}
}
 
// Global variables.
__gshared Hamming[] hams;
__gshared Hamming[NK] values;
 
nothrow @nogc static this() {
// Slower than calloc if you don't use all the MAX_HAM items.
//hams = new Hamming[MAX_HAM];
 
auto ptr = cast(Hamming*)calloc(MAX_HAM, Hamming.sizeof);
static const err = new Error("Not enough memory.");
if (!ptr)
throw err;
hams = ptr[0 .. MAX_HAM];
 
foreach (immutable i, ref v; values) {
v.e[i] = 1;
v.v = Hamming.inc[i];
}
}
 
 
ref Hamming getHam(in size_t n) nothrow @nogc
in {
assert(n <= MAX_HAM);
} body {
// Most of the time v can be just incremented, but eventually
// floating point precision will bite us, so better recalculate.
__gshared static size_t[NK] idx;
__gshared static int n_hams;
 
for (; n_hams < n; n_hams++) {
{
// Find the index of the minimum v.
size_t ni = 0;
foreach (immutable i; 1 .. NK)
if (values[i].v < values[ni].v)
ni = i;
 
hams[n_hams] = values[ni];
hams[n_hams].update;
}
 
foreach (immutable i; 0 .. NK)
if (values[i] == hams[n_hams]) {
values[i] = hams[idx[i]];
idx[i]++;
values[i].e[i]++;
values[i].update;
}
}
 
return hams[n - 2];
}
 
 
void main() {
foreach (immutable n; [1691, 10 ^^ 6, MAX_HAM])
writefln("%8d: %s", n, n.getHam);
}

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.

[edit] 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.

import std.stdio: writefln;
import std.bigint: BigInt;
import std.conv: text;
import core.stdc.stdlib: malloc, calloc, free;
import std.math: log; // ^^
 
// Number of factors.
enum NK = 3;
 
__gshared immutable int[NK] primes = [2, 3, 5];
__gshared immutable double[NK] lnPrimes;
 
// log can't be used in CTFE yet.
static this() pure nothrow @safe @nogc {
foreach (immutable i, immutable f; primes)
lnPrimes[i] = f.log;
}
 
/// K-smooth numbers (stored as their exponents of each factor).
struct Hamming {
double ln; // Log of the number.
ushort[NK] e; // Exponents of each factor.
Hamming* next;
size_t n;
 
// Recompute the logarithm from the exponents.
void recalculate() pure nothrow @safe @nogc {
this.ln = 0.0;
foreach (immutable i, immutable ei; this.e)
this.ln += lnPrimes[i] * ei;
}
 
string toString() const {
BigInt result = 1;
foreach (immutable i, immutable f; primes)
result *= f.BigInt ^^ this.e[i];
return result.text;
}
}
 
Hamming getHam(in size_t n) nothrow @nogc
in {
assert(n && n != size_t.max);
} body {
static struct Candidate {
typeof(Hamming.ln) ln;
typeof(Hamming.e) e;
 
void increment(in size_t n) pure nothrow @safe @nogc {
e[n] += 1;
ln += lnPrimes[n];
}
 
bool opEquals(in ref Hamming y) const pure nothrow @safe @nogc {
// return this.e == y.e; // Slow.
return !((this.e[0] ^ y.e[0]) |
(this.e[1] ^ y.e[1]) |
(this.e[2] ^ y.e[2]));
}
 
int opCmp(in ref Candidate y) const pure nothrow @safe @nogc {
return (ln > y.ln) ? 1 : (ln < y.ln ? -1 : 0);
}
}
 
static struct HammingIterator { // Not a Range.
Candidate cand;
Hamming* base;
size_t primeIdx;
 
this(in size_t i, Hamming* b) pure nothrow @safe @nogc {
primeIdx = i;
base = b;
cand.e = base.e;
cand.ln = base.ln;
cand.increment(primeIdx);
}
 
void next() pure nothrow @safe @nogc {
base = base.next;
cand.e = base.e;
cand.ln = base.ln;
cand.increment(primeIdx);
}
}
 
HammingIterator[NK] its;
Hamming* head = cast(Hamming*)calloc(Hamming.sizeof, 1);
Hamming* freeList, cur = head;
Candidate next;
 
foreach (immutable i, ref it; its)
it = HammingIterator(i, cur);
 
for (size_t i = cur.n = 1; i < n; ) {
auto leastReferenced = size_t.max;
next.ln = double.max;
foreach (ref it; its) {
if (it.cand == *cur)
it.next;
if (it.base.n < leastReferenced)
leastReferenced = it.base.n;
if (it.cand < next)
next = it.cand;
}
 
// Collect unferenced numbers.
while (head.n < leastReferenced) {
auto tmp = head;
head = head.next;
tmp.next = freeList;
freeList = tmp;
}
 
if (!freeList) {
cur.next = cast(Hamming*)malloc(Hamming.sizeof);
} else {
cur.next = freeList;
freeList = freeList.next;
}
 
cur = cur.next;
version (fastmath) {
cur.ln = next.ln;
cur.e = next.e;
} else {
cur.e = next.e;
cur.recalculate; // Prevent FP error accumulation.
}
 
cur.n = i++;
cur.next = null;
}
 
auto result = *cur;
version (leak) {}
else {
while (head) {
auto tmp = head;
head = head.next;
tmp.free;
}
 
while (freeList) {
auto tmp = freeList;
freeList = freeList.next;
tmp.free;
}
}
 
return result;
}
 
void main() {
foreach (immutable n; [1691, 10 ^^ 6, 10_000_000])
writefln("%8d: %s", n, n.getHam);
}

The output is the same as the second alternative version.

[edit] Eiffel

 
note
description : "Initial part, in order, of the sequence of Hamming numbers"
math : "[
Hamming numbers, also known as regular numbers and 5-smooth numbers, are natural integers
that have 2, 3 and 5 as their only prime factors.
]"

computer_arithmetic :
"[
This version avoids integer overflow and stops at the last representable number in the sequence.
]"

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

source : "This problem was posed in Edsger W. Dijkstra, A Discipline of Programming, Prentice Hall, 1978"
date : "8 August 2012"
authors : "Bertrand Meyer", "Emmanuel Stapf"
revision : "1.0"
libraries : "Relies on SORTED_TWO_WAY_LIST from EiffelBase"
implementation : "[
Using SORTED_TWO_WAY_LIST provides an elegant illustration of how to implement
a lazy scheme in Eiffel through the use of object-oriented data structures.
]"

warning : "[
The formatting (<lang>) specifications for Eiffel in RosettaCode are slightly obsolete:
`note' and other newer keywords not supported, red color for manifest strings.
This should be fixed soon.
]"

 
class
APPLICATION
 
create
make
 
feature {NONE} -- Initialization
 
make
-- Print first 20 Hamming numbers, in order, and the 1691-st one.
local
Hammings: like hamming
-- List of Hamming numbers, up to 1691-st one.
do
Hammings := hamming (1691)
across 1 |..| 20 as i loop
io.put_natural (Hammings.i_th (i.item)); io.put_string (" ")
end
io.put_new_line; io.put_natural (Hammings.i_th (1691)); io.put_new_line
end
 
feature -- Basic operations
 
hamming (n: INTEGER): ARRAYED_LIST [NATURAL]
-- First `n' elements (in order) of the Hamming sequence,
-- or as many of them as will not produce overflow.
local
sl: SORTED_TWO_WAY_LIST [NATURAL]
overflow: BOOLEAN
first, next: NATURAL
do
create Result.make (n); create sl.make
sl.extend (1); sl.start
across 1 |..| n as i invariant
-- "The numbers output so far are the first `i' - 1 Hamming numbers, in order".
-- "Result.first is the `i'-th Hamming number."
until sl.is_empty loop
first := sl.first; sl.start
Result.extend (first); sl.remove
across << 2, 3, 5 >> as multiplier loop
next := multiplier.item * first
overflow := overflow or next <= first
if not overflow and then not sl.has (next) then sl.extend (next) end
end
end
end
end
 
Output:
1
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36
2125764000

[edit] Factor

Translation of: Scala
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 ;
 <hamming-iterator> 20 next-n .
 <hamming-iterator> 1691 nth-from-now .
 <hamming-iterator> 1000000 nth-from-now .
Translation of: Haskell

Lazy lists aren quite slow in Factor, but still.

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

[edit] Forth

Works with: Gforth version 0.7.0

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

\ 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.
Output:
2^0*3^0*5^0 2^1*3^0*5^0 2^0*3^1*5^0 2^2*3^0*5^0 2^0*3^0*5^1 2^1*3^1*5^0 2^3*3^0*5^0 2^0*3^2*5^0 2^1*3^0*5^1 2^2*3^1*5^0 2^0*3^1*5^1 2^4*3^0*5^0 2^1*3^2*5^0 2^2*3^0*5^1 2^3*3^1*5^0 2^0*3^0*5^2 2^0*3^3*5^0 2^1*3^1*5^1 2^5*3^0*5^0 2^2*3^2*5^0 
2^5*3^12*5^3 
2^55*3^47*5^64

A smaller, less capable solution is presented here. It solves two out of three requirements and is ANS-Forth compliant.

2000 cells constant /hamming
create hamming /hamming allot
( n1 n2 n3 n4 n5 n6 n7 -- n3 n4 n5 n6 n1 n2 n8)
: min? >r dup r> min >r 2rot r> ;
 
: hit? ( n1 n2 n3 n4 n5 n6 n7 n8 -- n3 n4 n9 n10 n1 n2 n7)
>r 2dup = \ compare number with found minimum
if -rot drop 1+ hamming over cells + @ r@ * rot then
r> drop >r 2rot r>
; \ if so, increment and rotate
 
: hamming# ( n1 -- n2)
1 hamming ! >r \ set first cell and initialize parms
0 5 over 3 over 2
r@ 1 ?do \ determine minimum and set cell
dup min? min? min? dup hamming i cells + !
2 hit? 5 hit? 3 hit? drop
loop \ find if minimum equals value
2drop 2drop 2drop hamming r> 1- cells + @
; \ clean up stack and fetch hamming number
 
: test
cr 21 1 ?do i . i hamming# . cr loop
1691 hamming# . cr
;

[edit] Fortran

Works with: Fortran version 90 and later

Using big_integer_module from here [1]

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

[edit] F#

This example does not show the output mentioned in the task description on this page (or a page linked to from here). Please ensure that it meets all task requirements and remove this message.
Note that phrases in task descriptions such as "print and display" and "print and show" for example, indicate that (reasonable length) output be a part of a language's solution.


 
type LazyList<'a> = Cons of 'a * Lazy<LazyList<'a>>
 
let rec inf_map f = function
| Cons(x, g) -> Cons(f x, lazy(inf_map f (g.Force())))
 
let rec (-|-) (Cons(x, f) as xs) (Cons(y, g) as ys) =
if x < y then Cons(x, lazy(f.Force() -|- ys))
elif x > y then Cons(y, lazy(xs -|- g.Force()))
else Cons(x, lazy(f.Force() -|- g.Force()))
 
let rec hamming =
Cons(1, lazy(let x = inf_map ((*) 2) hamming
let y = inf_map ((*) 3) hamming
let z = inf_map ((*) 5) hamming
x -|- y -|- z))
 
[<EntryPoint>]
let main args =
let a = ref hamming
let i = ref 0
while !i < 100 do
match !a with
| Cons (x, f) ->
printf "%d, " x
a := f.Force()
i := !i + 1
0

[edit] FunL

Translation of: Scala
native scala.collection.mutable.Queue
 
val hamming =
q2 = Queue()
q3 = Queue()
q5 = Queue()
 
def enqueue( n ) =
q2.enqueue( n*2 )
q3.enqueue( n*3 )
q5.enqueue( n*5 )
 
def stream =
val n = min( min(q2.head(), q3.head()), q5.head() )
 
if q2.head() == n then q2.dequeue()
if q3.head() == n then q3.dequeue()
if q5.head() == n then q5.dequeue()
 
enqueue( n )
n # stream()
 
for q <- [q2, q3, q5] do q.enqueue( 1 )
 
stream()
Translation of: Haskell
val hamming = 1 # merge( map((2*), hamming), merge(map((3*), hamming), map((5*), hamming)) )
 
def
merge( inx@x:_, iny@y:_ )
| x < y = x # merge( inx.tail(), iny )
| x > y = y # merge( inx, iny.tail() )
| otherwise = merge( inx, iny.tail() )
 
println( hamming.take(20) )
println( hamming(1690) )
println( hamming(2000) )
Output:
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
2125764000
8100000000

[edit] Go

[edit] Concise version using dynamic-programming

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

[edit] Longer version using dynamic-programming and logarithms

More than 10 times faster.

package main
 
import (
"flag"
"fmt"
"log"
"math"
"math/big"
"os"
)
 
var (
// print the whole sequence or just one element?
 
seqMode = flag.Bool("s", false, "sequence mode")
// precomputed base-2 logarithms for 3 and 5
lg3, lg5 float64 = math.Log2(3), math.Log2(5)
 
// state of the three multiplied sequences
front = [3]cursor{
{0, 0, 1}, // 2
{1, 0, lg3}, // 3
{2, 0, lg5}, // 5
}
 
// table for dynamic-programming stored results
table [][3]int16
)
 
type cursor struct {
f int // index (0, 1, 2) corresponding to factor (2, 3, 5)
i int // index into table for the entry being multiplied
lg float64 // base-2 logarithm of the multiple (for ordering)
}
 
func (c *cursor) val() [3]int16 {
x := table[c.i]
x[c.f]++ // multiply by incrementing the exponent
return x
}
 
func (c *cursor) advance() {
c.i++
// skip entries that would produce duplicates
for (c.f < 2 && table[c.i][2] > 0) || (c.f < 1 && table[c.i][1] > 0) {
c.i++
}
x := c.val()
c.lg = float64(x[0]) + lg3*float64(x[1]) + lg5*float64(x[2])
}
 
func step() {
table = append(table, front[0].val())
front[0].advance()
// re-establish sorted order
if front[0].lg > front[1].lg {
front[0], front[1] = front[1], front[0]
if front[1].lg > front[2].lg {
front[1], front[2] = front[2], front[1]
}
}
}
func show(elem [3]int16) {
z := big.NewInt(1)
for i, base := range []int64{2, 3, 5} {
b := big.NewInt(base)
x := big.NewInt(int64(elem[i]))
z.Mul(z, b.Exp(b, x, nil))
}
fmt.Println(z)
}
 
func main() {
log.SetPrefix(os.Args[0] + ": ")
log.SetOutput(os.Stderr)
flag.Parse()
if flag.NArg() != 1 {
log.Fatalln("need one positive integer argument")
}
var ordinal int // ordinal of last sequence element to compute
_, err := fmt.Sscan(flag.Arg(0), &ordinal)
if err != nil || ordinal <= 0 {
log.Fatalln("argument must be a positive integer")
}
table = make([][3]int16, 1, ordinal)
for i, n := 1, ordinal; i < n; i++ {
if *seqMode {
show(table[i-1])
}
step()
}
show(table[ordinal-1])
}
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

[edit] Haskell

[edit] The classic version

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

Runs in about a second on Ideone.com. The nested unions' effect is to produce the minimal value at each step, with duplicates removed. As Haskell evaluation model is on-demand, the three map expressions are in effect iterators, maintaining hidden pointers back into the shared named storage with which they were each created (a name is a pointer/handle in Haskell; to name is to point at, to refer to, to take a handle on).

The amount of operations is constant for each number produced, so the time complexity should be O(n). Empirically, it is slightly above that and worsening, suggestive of extra cost of bignum arithmetics. Using triples representation with logarithm values for comparisons amends 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 O(n2 / 3) in length – behind the scenes, as long as there's no re-use evident in the code.

[edit] Avoiding generation of duplicates

The classic version can be sped up quite a bit (above twice, and with better orders of growth, even closer to 1.0) by avoiding generation of duplicate values:

hamming = 1:foldl u [] [5,3,2] where
u s n = ar where
ar = merge s (n:map (n*) ar)
merge [] b = b
merge a@(x:xs) b@(y:ys)
| x < y = x:merge xs b
| otherwise = y:merge a ys
 
main = do
print $ take 20 hamming
print $ hamming !! 1690
print $ hamming !! (1000000-1)

[edit] Explicit multiples reinserting

This is a common approach which explicitly maintains an internal buffer of O(n2 / 3) elements, removing the numbers from its front and reinserting their 2- 3- and 5-multiples in order. It overproduces the sequence, stopping when the n-th number is no longer needed instead of when it's first found. Also overworks by maintaining this buffer in total order where just heap would be sufficient. Worse, this particular version uses a sequential list for its buffer. That means O(n2 / 3) operations for each number, instead of O(1) 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:

hamm n = drop n $ iterate (\(_,(a:t))-> (a,union t [2*a,3*a,5*a])) (0,[1])
Output:
*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]

Runs too slowly to reach 1,000,000, with orders of growth above ~‍ ‍(n‍ ‍1.7‍ ‍) and worsening. Last two lines show the internal buffer's length for several sample n‍ ‍s.

[edit] 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 O(n0.7) empirical time and space complexity:

-- 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 = let (r,t) = nthHam 1000000 in print t >> print (trival t)
 
lg3 = logBase 2 3; lg5 = logBase 2 5
logval (i,j,k) = fromIntegral i + fromIntegral j*lg3 + fromIntegral k*lg5
trival (i,j,k) = 2^i * 3^j * 5^k
estval n = (6*lg3*lg5* 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 :: Int -> (Double, (Int, Int, Int))
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) = ( fromIntegral $ 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 -- 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 /lg5) ], let p = fromIntegral k*lg5,
j <- [ 0 .. floor ((hi-p)/lg3) ], let q = fromIntegral j*lg3 + p,
let (i,frac) = pr (hi-q) ; r = hi-frac ] -- r = i + q
-- f 0 z == (sum $ map fst z, concat $ map snd z)
where pr = properFraction
f !c [] = (c,[]) -- code as a loop
f !c ((c1,b1):r) = let (cr,br) = f (c+c1) r -- to prevent space leak
in case b1 of { [v] -> (cr,v:br)
 ; _ -> (cr, br) }
Output:
-- time: 0.01s    memory: 3688 kB
(55,47,64)
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

[edit] 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.

# 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

[edit] J

Solution:
A concise tacit expression using a (right) fold:

hamming=: {. (/:~@~.@] , 2 3 5 * {)/@(1x ,~ i.@-)

Example usage:

   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

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

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

  • sort and remove duplicates
 /:~@~.@]
  • produce (the next) 3 elements by selection and multiplication:
 2 3 5 * {

or

 2 3 5 * LHA { RHA
  • the RH part forms an array of descending indices and the initial Hamming number 1
 (1x ,~ i.@-)

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)

   ({. (/:~@~.@] , 2 3 5 * {)/@(1x ,~ i.@-)) 7
1 2 3 4 5 6 8

This starts using a descending sequence with 1 appended:

    (1x ,~ i.@-) 7
6 5 4 3 2 1 0 1

and then the fold expression is inserted between these list elements and the result computed:

   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

(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.)

[edit] Java

Works with: Java version 1.5+

Has a common shortcoming of overproducing the sequence by about O(n2 / 3) entries, until the n-th number is no longer needed, instead of stopping as soon as it is reached. See Haskell for an illustration.

Inserting the top number's three multiples deep into the priority queue as it does, incurs extra cost for each number produced. To not worsen the expected algorithm complexity, the priority queue should have (amortized) O(1) implementation for both insertion and deletion operations but it looks like it's O(logn) in Java.

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

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

 
import java.math.BigInteger;
import java.util.*;
 
// [email protected]
 
public class HammingTriple implements Comparable<HammingTriple> {
 
// Precompute a couple of constants that we need all the time
private static final BigInteger two = BigInteger.valueOf(2);
private static final BigInteger three = BigInteger.valueOf(3);
private static final BigInteger five = BigInteger.valueOf(5);
private static final double logOf2 = Math.log(2);
private static final double logOf3 = Math.log(3);
private static final double logOf5 = Math.log(5);
 
// The powers of this triple
private int a, b, c;
 
public HammingTriple(int a, int b, int c) {
this.a = a; this.b = b; this.c = c;
}
 
public String toString() {
return "[" + a + ", " + b + ", " + c + "]";
}
 
public BigInteger getValue() {
return two.pow(a).multiply(three.pow(b)).multiply(five.pow(c));
}
 
public boolean equals(Object other) {
if(other instanceof HammingTriple) {
HammingTriple h = (HammingTriple) other;
return this.a == h.a && this.b == h.b && this.c == h.c;
}
else { return false; }
}
 
// Return 0 if this == other, +1 if this > other, and -1 if this < other
public int compareTo(HammingTriple other) {
// equality
if(this.a == other.a && this.b == other.b && this.c == other.c) {
return 0;
}
// this dominates
if(this.a >= other.a && this.b >= other.b && this.c >= other.c) {
return +1;
}
// other dominates
if(this.a <= other.a && this.b <= other.b && this.c <= other.c) {
return -1;
}
 
// take the logarithms for comparison
double log1 = this.a * logOf2 + this.b * logOf3 + this.c * logOf5;
double log2 = other.a * logOf2 + other.b * logOf3 + other.c * logOf5;
 
// are these different enough to be reliable?
if(Math.abs(log1 - log2) > 0.0000001) {
return (log1 < log2) ? -1: +1;
}
 
// oh well, looks like we have to do this the hard way
return this.getValue().compareTo(other.getValue());
// (getting this far should be pretty rare, though)
}
 
public static BigInteger computeHamming(int n, boolean verbose) {
if(verbose) {
System.out.println("Hamming number #" + n);
}
long startTime = System.currentTimeMillis();
 
// The elements of the search frontier
PriorityQueue<HammingTriple> frontierQ = new PriorityQueue<HammingTriple>();
int maxFrontierSize = 1;
 
// Initialize the frontier
frontierQ.offer(new HammingTriple(0, 0, 0)); // 1
 
while(true) {
if(frontierQ.size() > maxFrontierSize) {
maxFrontierSize = frontierQ.size();
}
// Pop out the next Hamming number from the frontier
HammingTriple curr = frontierQ.poll();
 
if(--n == 0) {
if(verbose) {
System.out.println("Time: " + (System.currentTimeMillis() - startTime) + " ms");
System.out.println("Frontier max size: " + maxFrontierSize);
System.out.println("As powers: " + curr.toString());
System.out.println("As value: " + curr.getValue());
}
return curr.getValue();
}
 
// Current times five, if at origin in (a,b) plane
if(curr.a == 0 && curr.b == 0) {
frontierQ.offer(new HammingTriple(curr.a, curr.b, curr.c + 1));
}
// Current times three, if at line a == 0
if(curr.a == 0) {
frontierQ.offer(new HammingTriple(curr.a, curr.b + 1, curr.c));
}
// Current times two, unconditionally
curr.a++;
frontierQ.offer(curr); // reuse the current HammingTriple object
}
}
}
 
Hamming number #1000000
Time: 650 ms
Frontier max size: 10777
As powers: [55, 47, 64]
As value: 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

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

[edit] Alternative

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

 
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;
}
}
 
$ 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
$ 

[edit] JavaScript

Works with: JavaScript version 1.7
Works with: Firefox version 2
Translation of: Ruby

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

Note the use of for (x in obj) to iterate over the properties of an object, versus for each (y in obj) to iterate over the values of the properties of an object.

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

[edit] Fast & complete version

Translation of: C#

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

--Mike Lorenz

<html>
<head></head>
<body>
<div id="main"></div>
</body>
<script src="http://code.jquery.com/jquery-latest.min.js"></script>
<script src="http://peterolson.github.com/BigInteger.js/BigInteger.min.js"></script>
<script type="text/javascript">
var _primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37];
 
function log(text) {
$('#main').append(text + "\n");
}
 
function big(exponents) {
var i, e, val = bigInt.one;
for (i = 0; i < exponents.length; i++)
for (e = 0; e < exponents[i]; e++)
val = val.times(_primes[i]);
return val.toString();
}
 
function hamming(n, nprimes) {
var i, iter, p, q, min, equal, x;
 
var hammings = new Array(n); // array of hamming #s we generate
hammings[0] = new Array(nprimes);
for (p = 0; p < nprimes; p++) {
hammings[0][p] = 0;
}
 
var hammlogs = new Array(n); // log values for above
hammlogs[0] = 0;
 
var primelogs = new Array(nprimes); // pre-calculated prime log values
var listlogs = new Array(nprimes); // log values of list heads
for (p = 0; p < nprimes; p++) {
primelogs[p] = listlogs[p] = Math.log(_primes[p]);
}
 
var indexes = new Array(nprimes); // intermediate hamming values as indexes into hammings
for (p = 0; p < nprimes; p++) {
indexes[p] = 0;
}
 
var listheads = new Array(nprimes); // intermediate hamming list heads
for (p = 0; p < nprimes; p++) {
listheads[p] = new Array(nprimes);
for (q = 0; q < nprimes; q++) {
listheads[p][q] = 0;
}
listheads[p][p] = 1;
}
 
for (iter = 1; iter < n; iter++) {
min = 0;
for (p = 1; p < nprimes; p++)
if (listlogs[p] < listlogs[min])
min = p;
hammlogs[iter] = listlogs[min]; // that's the next hamming number
hammings[iter] = listheads[min].slice();
for (p = 0; p < nprimes; p++) { // update each list head if it matches new value
equal = true; // test each exponent to see if number matches
for (i = 0; i < nprimes; i++) {
if (hammings[iter][i] != listheads[p][i]) {
equal = false;
break;
}
}
if (equal) { // if it matches...
x = ++indexes[p]; // set index to next hamming number
listheads[p] = hammings[x].slice(); // copy hamming number
listheads[p][p] += 1; // increment exponent = mult by prime
listlogs[p] = hammlogs[x] + primelogs[p]; // add log(prime) to log(value) = mult by prime
}
}
}
 
return hammings[n - 1];
}
 
$(document).ready(function() {
var i, nprimes;
var t = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,1691,1000000];
 
for (nprimes = 3; nprimes <= 4; nprimes++) {
var start = new Date();
log('<h1>' + _primes[nprimes - 1] + '-Smooth:' + '</h1>');
log('<table>');
for (i = 0; i < t.length; i++)
log('<tr>' + '<td>' + t[i] + ':' + '</td><td>' + big(hamming(t[i], nprimes)) + '</td>');
var end = new Date();
log('<tr>' + '<td>' + 'Elapsed time:' + '</td><td>' + (end-start)/1000 + ' seconds' + '</td>');
log('</table>');
}
});
</script>
</html>
Output:
5-Smooth:

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

7-Smooth:

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

[edit] Kotlin

Translation of: Java
import java.math.BigInteger;
import java.util.PriorityQueue;
 
val Three = BigInteger.valueOf(3)
val Five = BigInteger.valueOf(5)
 
fun updateFrontier(x : BigInteger, pq : PriorityQueue<BigInteger>) {
pq add(x shiftLeft(1))
pq add(x multiply(Three))
pq add(x multiply(Five))
}
 
fun hamming(n : Int) : BigInteger {
val frontier = PriorityQueue<BigInteger>()
updateFrontier(BigInteger.ONE, frontier)
var lowest = BigInteger.ONE
for (i in 1 .. n-1) {
lowest = frontier.poll() ?: lowest
while (frontier.peek() equals(lowest))
frontier.poll()
updateFrontier(lowest, frontier)
}
return lowest
}
 
fun main(args : Array<String>) {
System.out print("Hamming(1 .. 20) =")
for (i in 1 .. 20)
System.out print(" ${hamming(i)}")
System.out println("\nHamming(1691) = ${hamming(1691)}")
System.out println("Hamming(1000000) = ${hamming(1000000)}")
}
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

[edit] Liberty BASIC

LB has unlimited precision integers.

 
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
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

[edit]

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

[edit] 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)

[edit] 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]];
HammingList[20]
-> {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36}
HammingList[1691] // Last
-> 2125764000
HammingList[1000000] // Last
->519312780448388736089589843750000000000000000000000000000000000000000000000000000000

[edit] 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

[edit] Nimrod

Library: bigints
import bigints, math
 
proc hamming(limit): BigInt =
var
h = newSeq[BigInt](limit)
x2 = initBigInt(2)
x3 = initBigInt(3)
x5 = initBigInt(5)
i, j, k = 0
for i in 0..h.high: h[i] = initBigInt(1)
 
for n in 1 .. < limit:
h[n] = min(x2, x3, x5)
if x2 == h[n]:
inc i
x2 = h[i] * 2
if x3 == h[n]:
inc j
x3 = h[j] * 3
if x5 == h[n]:
inc k
x5 = h[k] * 5
 
result = h[h.high]
 
for i in 1 .. 20:
echo hamming(i)
 
echo hamming(1691)
echo hamming(1_000_000)

Output:

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

[edit] Oz

[edit] Lazy Version

Translation of: Haskell
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}}


[edit] Strict Version

The strict version uses iterators and a priority queue. Note that it can calculate other variations of the hamming numbers too. By changing K, it will calculate the p(K)-smooth numbers. (E.g. K = 3, it will use the first three primes 2,3 and 5, thus resulting in the 5-smooth numbers, see [2])

 
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
 

Strict version made by pietervdvn; do what you want with the code.

[edit] PARI/GP

This is a basic implementation; finding the millionth term requires 3 seconds and 54 MB. Much better algorithms exist.

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)
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

[edit] 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";
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>

[edit] Perl 6

Works with: Rakudo version 2012.01

The limit scaling is not required, but it cuts down on a bunch of unnecessary calculation.

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

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

[edit] PL/I

(subscriptrange):
Hamming: procedure options (main); /* 14 November 2013 */
declare (H(3000), t) fixed (15);
declare (i, j, k, m, n) fixed binary;
declare swaps bit (1);
 
on underflow ;
 
m = 0; n = 12;
do k = 0 to n;
do j = 0 to n;
do i = 0 to n;
m = m + 1;
H(m) = 2**i * 3**j * 5**k;
end;
end;
end;
/* sort */
swaps = '1'b;
do while (swaps); /* Cocktail-shaker sort is adequate, because values are largely sorted */
swaps = '0'b;
do i = 1 to m-1, i-1 to 1 by -1;
if H(i) > H(i+1) then /* swap */
do; t = H(i); H(i) = H(i+1); H(i+1) = t; swaps = '1'b; end;
end;
end;
do i = 1 to m;
put skip data (H(i));
end;
put skip data (H(1653));
end Hamming;

Results:

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;

[edit] Prolog

[edit] Generator idiom

%% 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.

SWI Prolog 6.2.6 produces (in about 7 ideone seconds):

 ?- time( main(1000000) ).
 [1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36]
 [2125764000,2147483648]
 519312780448388736089589843750000000000000000000000000000000000000000000000000000000
 % 10,017,142 inferences

[edit] 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).

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))).
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 .

[edit] Python

[edit] Version based on example from Dr. Dobb's CodeTalk

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
Output:
>>> list(islice(hamming2(), 20))
[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36]
>>> list(islice(hamming2(), 1690, 1691))
[2125764000]
>>> list(islice(hamming2(), 999999, 1000000))
[519312780448388736089589843750000000000000000000000000000000000000000000000000000000]

[edit] Another implementation of same approach

This version uses a lot of memory, it doesn't try to limit memory usage.

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)

[edit] "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).

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()

Results are the same as before.

[edit] Non-sharing recursive generator

Another formulation along the same lines, but greatly simplified, found here. Lacks data sharing, i.e. calls self recursively thus creating a separate copy of the data stream fed to the tee() call, again and again, instead of using its own output. This gravely impacts the efficiency. Not to be used.

from heapq import merge
from itertools import tee
 
def hamming_numbers():
last = 1
yield last
 
a,b,c = tee(hamming_numbers(), 3)
 
for n in merge((2*i for i in a), (3*i for i in b), (5*i for i in c)):
if n != last:
yield n
last = n

[edit] Cyclic generator method #2.

Cyclic generator method #2. Considerably faster due to early elimination (before merge) of duplicates. Currently the faster Python version. Direct copy of Haskell code.

from itertools import islice, chain, tee
 
def merge(r, s):
# This is faster than heapq.merge.
rr = r.next()
ss = s.next()
while True:
if rr < ss:
yield rr
rr = r.next()
else:
yield ss
ss = s.next()
 
def p(n):
def gen():
x = n
while True:
yield x
x *= n
return gen()
 
def pp(n, s):
def gen():
for x in (merge(s, chain([n], (n * y for y in fb)))):
yield x
r, fb = tee(gen())
return r
 
def hamming(a, b = None):
if not b:
b = a + 1
seq = (chain([1], pp(5, pp(3, p(2)))))
return list(islice(seq, a - 1, b - 1))
 
print hamming(1, 21)
print hamming(1691)[0]
print hamming(1000000)[0]

[edit] Qi

This example is incomplete. Parts 2 & 3 of task missing. Please ensure that it meets all task requirements and remove this message.
Translation of: Clojure
(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)
Output:
[1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36]

[edit] R

Recursively find the Hamming numbers below 231. Works for larger limits, however to find the 1000000th value, one needs guess the correct limit

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])

[edit] Racket

 
#lang racket
(require racket/stream)
(define first stream-first)
(define rest stream-rest)
 
(define (merge s1 s2)
(define x1 (first s1))
(define x2 (first s2))
(cond [(= x1 x2) (merge s1 (rest s2))]
[(< x1 x2) (stream-cons x1 (merge (rest s1) s2))]
[else (stream-cons x2 (merge s1 (rest s2)))]))
 
(define (mult k) (λ(x) (* x k)))
 
(define hamming
(stream-cons
1 (merge (stream-map (mult 2) hamming)
(merge (stream-map (mult 3) hamming)
(stream-map (mult 5) hamming)))))
 
(for/list ([i 20] [x hamming]) x)
(stream-ref hamming 1690)
(stream-ref hamming 999999)
 

Output:

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

[edit] Raven

Translation of: Liberty Basic
define hamming use $limit
[ ] as $h
1 $h 0 set
2 as $x2 3 as $x3 5 as $x5
0 as $i 0 as $j 0 as $k
1 $limit 1 + 1 range each as $n
$x3 $x5 min $x2 min $h $n set
$h $n get $x2 = if
$i 1 + as $i
$h $i get 2 * as $x2
$h $n get $x3 = if
$j 1 + as $j
$h $j get 3 * as $x3
$h $n get $x5 = if
$k 1 + as $k
$h $k get 5 * as $x5
$h $limit 1 - get
 
1 21 1 range each as $lim
$lim hamming print " " print
"\n" print
 
"Hamming(1691) is: " print 1691 hamming print "\n" print
 
# Raven can't handle > 2^31 using integers
#
#"Hamming(1000000) is: " print 1000000 hamming print "\n" print
Output:
1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 27 30 32 36 
Hamming(1691) is: 2125764000

[edit] REXX

[edit] concise version

This REXX program was a direct translatation from my old REXX subroutine to compute UGLY numbers.
The program computes enough Hamming numbers (just one past the current number).

/*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 ──► twentieth #s*/
call hamming 1691 /*show the 1,691st Hamming number*/
call hamming 1000000 /*show the one millionth number.*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────HAMMING subroutine──────────────────*/
hamming: procedure; parse arg x,y; if y=='' then y=x; w=length(y)
#2=1; #3=1; #5=1; @.=0; @.1=1
 
do n=2 for y-1
@.n = min(2*@.#2, 3*@.#3, 5*@.#5) /*pick the minimum of three pigs.*/
if 2*@.#2 == @.n then #2 = #2+1 /*# already defined? Use next #.*/
if 3*@.#3 == @.n then #3 = #3+1 /*" " " " " " */
if 5*@.#5 == @.n then #5 = #5+1 /*" " " " " " */
end /*n*/
do j=x to y /*W is used to align the index. */
say 'Hamming('right(j,w)") =" @.j /*list 'em, Dano.*/
end /*j*/
 
say right( 'length of last Hamming number =' length(@.y), 70); say
return

output when using the default input(s):

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

[edit] unrolled version

This REXX version is roughly twice as fast as the first version.

/*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 ──► twentieth #s*/
call hamming 1691 /*show the 1,691st Hamming number*/
call hamming 1000000 /*show the one millionth number.*/
exit /*stick a fork in it, we're done.*/
/*──────────────────────────────────HAMMING subroutine──────────────────*/
hamming: procedure; parse arg x,y; if y=='' then y=x; w=length(y)
#2=1; #3=1; #5=1; @.=0; @.1=1
 
do n=2 for y-1
_2 = @.#2 + @.#2 /*this is faster than 2 * @.#2 */
_3 = 3 * @.#3
_5 = 5 * @.#5
m =_2 /*assume a minimum (of the three)*/
if _3 < m then m =_3 /*is this less than the minimum? */
if _5 < m then m =_5 /* " " " " " " */
@.n = m /*now, assign the next Hamming #.*/
if _2 == m then #2 = #2 + 1 /*# already defined? Use next #.*/
if _3 == m then #3 = #3 + 1 /*" " " " " " */
if _5 == m then #5 = #5 + 1 /*" " " " " " */
end /*n*/
do j=x to y /*W is used to align the index. */
say 'Hamming('right(j,w)") =" @.j /*list 'em, Dano.*/
end /*j*/
 
say right( 'length of last Hamming number =' length(@.y), 70); say
return

output is identical to the first version.

[edit] Ruby

Translation of: Scala
Works with: Ruby version 1.8.6
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
Works with: Ruby version 1.9.1

This method does not require a library module.

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

And the "main" part of the task

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"
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

[edit] Run BASIC

 
dim h(1000000)
for i =1 to 20
print hamming(i);" ";
next i
 
print
print "Hamming List First(1691) =";chr$(9);hamming(1691)
print "Hamming List Last(1000000) =";chr$(9);hamming(1000000)
 
end
 
function hamming(limit)
h(0) =1
x2 = 2: x3 = 3: x5 =5
i = 0: j = 0: k =0
for n =1 to limit
h(n) = min(x2, min(x3, x5))
if x2 = h(n) then i = i +1: x2 =2 *h(i)
if x3 = h(n) then j = j +1: x3 =3 *h(j)
if x5 = h(n) then k = k +1: x5 =5 *h(k)
next n
hamming = h(limit -1)
end function
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

[edit] 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)
}

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:

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)
}

Usage:

scala> new Hamming take 20 toList
res87: List[BigInt] = List(1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36)

scala> new Hamming drop 1690 next
res88: BigInt = 2125764000

scala> new Hamming drop 999999 next
res89: BigInt = 519312780448388736089589843750000000000000000000000000000000000000000000000000000000

There's also a fairly mechanical translation from Haskell using purely functional lazy streams

Translation of: Haskell
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)))
}

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

[edit] 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)
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

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

[edit] Smalltalk

Works with: GNU Smalltalk

This is a straightforward implementation of the pseudocode snippet found in the Python section. Smalltalk supports arbitrary-precision integers, but the implementation is too slow to try it with 1 million.

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.

[edit] Tcl

This uses coroutines to simplify the description of what's going on.

Works with: Tcl version 8.6
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"
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):

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]"

[edit] Ursala

Smooth is defined as a second order function taking a list of primes and returning a function that takes a natural number n to the n-th smooth number with respect to them. An elegant but inefficient formulation based on the J solution is the following.

#import std
#import nat
 
smooth"p" "n" = ~&z take/"n" nleq-< (rep(length "n") ^Ts/~& product*K0/"p") <1>

This test program

main = smooth<2,3,5>* nrange(1,20)

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.

#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>
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>

[edit] zkl

var BN=Import("zklBigNum");  // only needed for large N
fcn hamming(N){
h:=List.createLong(N+1); (0).pump(N+1,h.write,Void); // fill list with stuff
h[0]=1;
#if 1 // regular (64 bit) ints
x2:=2; x3:=3; x5:=5; i:=j:=k:=0;
#else // big ints
x2:=BN(2); x3:=BN(3); x5:=BN(5); i:=j:=k:=0;
#endif
foreach n in ([1..N]){
z:=(x2<x3) and x2 or x3; z=(z<x5) and z or x5; h[n]=z;
if (h[n] == x2) { x2 = h[i+=1]*2 }
if (h[n] == x3) { x3 = h[j+=1]*3 }
if (h[n] == x5) { x5 = h[k+=1]*5 }
}
return(h[N-1])
}
[1..20].apply(hamming).println();
hamming(1691).println();
Output:
L(1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,27,30,32,36)
2125764000

While the other algorithms save [lots of] space, run time still sucks when n > 100,000 so memory usage might as well too. Change the #if 0 to 1 and

Output:
hamming(0d1_000_000).println();
519312780448388736089589843750000000000000000000000000000000000000000000000000000000

[edit] Direct calculation through triples enumeration

OK, I was wrong, calculating the nth Hamming number can be fast and efficient.

Translation of: Haskell
as direct a translation as I can, except using a nested for loop instead of list comprehension (which makes it easier to keep the count).
 
#-- 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
 
var BN=Import("zklBigNum");
var lg3 = (3.0).log()/(2.0).log(), lg5 = (5.0).log()/(2.0).log();
fcn logval(i,j,k){ lg5*k + lg3*j + i }
fcn trival(i,j,k){ BN(2).pow(i) * BN(3).pow(j) * BN(5).pow(k) }
fcn estval(n){ (6.0*lg3*lg5*n).pow(1.0/3) } #-- estimated logval, base 2
fcn rngval(n){
if(n > 500000) return(2.4496 , 0.0076); #-- empirical estimation
if(n > 50000) return(2.4424 , 0.0146); #-- correction, base 2
if(n > 500) return(2.3948 , 0.0723); #-- (dist,width)
if(n > 1) return(2.2506 , 0.2887); #-- around (log $ sqrt 30),
return(2.2506 , 0.5771); #-- says WP
}
 
fcn nthHam(n){ // -> (Double, (Int, Int, Int)) #-- n: 1-based: 1,2,3...
d,w := rngval(n); #-- correction dist, width
hi  := estval(n.toFloat()) - d; #-- hi > logval > hi-w
c,b := band(hi,w); #-- total count, the band
s  := b.sort(fcn(a,b){ a[0]>b[0] }); #-- sorted decreasing, result
m  := c - n; #-- m 0-based from top
nb  := b.len(); #-- |band|
res := s[m]; #-- result
 
if(w >= 1) throw(Exception.Generic("Breach of contract: (w < 1): " + w));
if(m < 0) throw(Exception.Generic("Not enough triples generated: " +c+n));
if(m >= nb)throw(Exception.Generic("Generated band is too narrow: " +m+nb));
return(res);
}
 
fcn band(hi,w){ //--> #-- total count, the band
b := Sink(List); cnt := 0;
foreach k in ([0 .. (hi/lg5).floor()]){ p := lg5*k;
foreach j in ([0 .. ((hi-p)/lg3).floor()]){ q := lg3*j + p;
i,frac := (hi-q).modf(); r := hi-frac; #-- r = i + q
cnt+=(i+1); #-- total count
if(frac<w) b.write(T(r,T(i,j,k))); #-- store it, if inside band
}
}
return(cnt,b.close());
}
fcn printHam(n){
r,t:=nthHam(n); i,j,k:=t; h:=trival(i,j,k);
println("Hamming(%,d)-->2^%d * 3^%d * 5^%d-->\n%s".fmt(n,i,j,k,h));
}
 
printHam(1691); //(5,12,3), 10 digits
printHam(0d1_000_000); //(55,47,64), 84 digits
printHam(0d10_000_000); //(80,92,162), 182 digits, 80 zeros at end
printHam(0d1_000_000_000); //(1334,335,404), 845 digits
Output:
Hamming(1,691)-->2^5 * 3^12 * 5^3-->
2125764000
Hamming(1,000,000)-->2^55 * 3^47 * 5^64-->
519312780448388736089589843750000000000000000000000000000000000000000000000000000000
Hamming(10,000,000)-->2^80 * 3^92 * 5^162-->
162441050638304318232392153117595750351085388205966408633356724833252116013682098127901554107666015625 <80 zeros>
Hamming(1,000,000,000)-->2^1334 * 3^335 * 5^404-->
621607575556524486163081633287207200394705651908965270659163240.......
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