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The following snippet shows how to generate uniform random partitions of a number fast.

Take the following definitions.

use strict; 
my %npart; sub cntpart1 { my($n, $m) = @_; $n = 0+$n; $m = 0+$m; my $c
+ = \$npart{$n." ".$m}; defined($$c) and return $$c; $n <= 0 and retur
+n $$c = 1; my $s = 0; for my $k (1 .. ($m < $n ? $m : $n)) { $s += cn
+tpart1($n - $k, $k); } $$c = $s; } 
sub randpart1 { my($n, $m) = @_; $n <= 0 and return; my($s, $k) = 0; f
+or my $j (1 .. ($m < $n ? $m : $n)) { my $p = cntpart1($n - $j, $j); 
+rand($s += $p) < $p and $k = $j; } $k, randpart1($n - $k, $k); } 
sub randpart { my($n) = @_; randpart1($n, $n); }
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Then randpart($n) generates a random partition with uniform probablity among all partitions of the positive integer $n.

As an example, run this.

perl -we 'use strict; my %npart; sub cntpart1 { my($n, $m) = @_; $n = 
+0+$n; $m = 0+$m; my $c = \$npart{$n." ".$m}; defined($$c) and return 
+$$c; $n <= 0 and return $$c = 1; my $s = 0; for my $k (1 .. ($m < $n 
+? $m : $n)) { $s += cntpart1($n - $k, $k); } $$c = $s; } sub randpart
+1 { my($n, $m) = @_; $n <= 0 and return; my($s, $k) = 0; for my $j (1
+ .. ($m < $n ? $m : $n)) { my $p = cntpart1($n - $j, $j); rand($s += 
+$p) < $p and $k = $j; } $k, randpart1($n - $k, $k); } sub randpart { 
+my($n) = @_; randpart1($n, $n); } for (1 .. 10000) { print join(" ", 
+randpart($ARGV[0])), "\n"; }' 6 | sort | uniq -c
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This quickly generates ten thousand random partitions of 6, and then sort | uniq builds a frequency table so you can see each of the eleven possible partitions are generated approximately the same number of times.

The function is fast even for larger $n values too. Internally it works by cntpart1($n, $m) calculating the number of partitions of $n with no partition greater than $m, and this count is memoized.

Update: You may want to add a no warnings "recursion";

Update 2008 sep 28: Limbic~Region referred me to his code RFC: Integer::Partition::Unrestricted which computes the number of partitions of any integer really fast. I'll have to read its implementation on whether it can help here.

In reply to Generate uniform random partitions of a number by ambrus

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