note
blokhead
[http://www.research.att.com/projects/OEIS?Anum=A000041|Sloane's Integer Sequences] gives the number for partitions of N for many values of N. <strike>The first solution given in this thread (and the one in its reply) doesn't work. For instance, with N=10, it only gives 42 when it should give 77 (among other obviously invalid partitions it lists).</strike>
<p>
Here's one using an iterator. Since the number of partitions of N grows exponentially with N, it might be best to not have the entire set of partitions in memory. This is based off the algorithm outlined [http://www2.toki.or.id/book/AlgDesignManual/BOOK/BOOK4/NODE153.HTM|here].
<code>
my $n = shift || 10;
my $iter = int_partitions($n);
my $total;
while ( my @part = $iter->() ) {
$total++;
print "@part\n";
}
print "There are $total partitions of $n\n";
sub int_partitions {
my $n = shift;
my @tally;
return sub {
if (!@tally) {
@tally = ( (0) x $n, 1 );
return ($n);
}
## last partition is $n 1's
return if $tally[1] == $n;
## take one away from smallest >1 part
my ($least) = grep { $tally[$_] } 2 .. $n;
$tally[$least]--;
$tally[$least-1]++;
$tally[1]++;
## collect multiple 1's into groups smaller than $least
while ( $least > 2 and $tally[1] > 1 ) {
my $move = ( sort { $a <=> $b } $least-1, $tally[1] )[0];
$tally[1] -= $move;
$tally[$move]++;
}
return map { ($_) x $tally[$_] } reverse 1 .. $n;
};
}
</code>
This returns partitions in decreasing lexicographic order. Cheers!
<p>
<b>Update:</b> Just for fun, here's code for iterating over all partitions of a set (also inspired from the article linked above):
<code>
my @set = @ARGV ? @ARGV : 1 .. 4;
my $iter = set_partitions(@set);
my $total;
while ( my @part = $iter->() ) {
$total++;
print join(" ", map { "[@$_]" } @part), $/;
}
print "There are $total partitions of @set\n";
sub set_partitions {
my @universe = @_;
my @growth;
return sub {
if (!@growth) {
@growth = (0) x @universe;
return [ @universe ];
}
return if $growth[-1] == $#growth;
my $i = $#growth;
$growth[$i--] = 0 while $growth[$i-1] == $growth[$i] - 1;
$growth[$i]++;
my @return;
push @{ $return[$growth[$_]] }, $universe[$_] for 0 .. $#growth;
return @return;
};
}
</code>
The internal order of parts is not significant. The parts are returned in ascending order of their smallest element. If the input set has duplicates, so will the output. I'd have to think about how to do this for multisets... Hey, this is fun! ;)
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blokhead
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