in reply to 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); } }

-Mark

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Re^2: Generator of integer partitionts of n
by blokhead (Monsignor) on Aug 28, 2004 at 07:02 UTC
    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.

    blokhead

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