Here is a different algorithm that seems to be simpler:
my $integer = 5;
my @p;
part( 2*$integer, $integer, 0);
sub part
{
my ($n, $k, $t) = @_;
$p[$t] = $k;
print( join " ", @p[1..$#p], "\n") if $n == $k;
for (my $j = $k<$n$k ? $k : $n$k; $j >= 1; $j) {
part( $n$k, $j, $t+1);
}
}
which results in
1004% perl part.pl
5
4 1
3 2
3 1 1
2 2 1
2 1 1 1
1 1 1 1 1
Update: Thanks to blokhead for catching my error! I had the correct algorithm, but blew it on the print statement. The last valid element of @p is at index $t. Here is the corrected code:
my $integer = 5;
my @p;
part( 2*$integer, $integer, 0);
sub part
{
my ($n, $k, $t) = @_;
$p[$t] = $k;
print( join " ", @p[1..$t], "\n") if $n == $k;
for (my $j = $k<$n$k ? $k : $n$k; $j >= 1; $j) {
part( $n$k, $j, $t+1);
}
}
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