Syntactic Confectionery Delight PerlMonks

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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 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

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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