Re: Generator of integer partitionts of n

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);
   }
}

[blokhead]

Changing $integer to 6 for example gives:

6 
5 1 
4 2 
4 1 1 
3 3 1   # oops 
3 2 1
3 1 1 1 
2 2 2 1 # oops
2 2 1 1 
2 1 1 1 1   
1 1 1 1 1 1

You can check how many partitions of N exist for the first many values of N at this site. For N=10, for instance, there should be 77 and your script returns less than 50 (many of them even adding up to more than 10). Update: Other than an extra 1 on a few partitions here and there, it seems to get all of them though.