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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 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); } }


In reply to Re: Generator of integer partitionts of n by kvale
in thread Generator of integer partitionts of n by chiburashka

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