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Well, for any given target node you build a string of the correct number of left and right moves (or zeroes and ones) and then any permutation is one of the admissible solutions (and all of them).

UPDATE: Just as illustration, using a module, and having to filter out duplicates in the permutations, it could work like this:

use strict; use warnings; use Algorithm::Permute; my $node = '5-2'; my ($all, $right) = split /-/, $node; my @path = ((0) x ($all-$right),(1) x $right); my %pathes; Algorithm::Permute::permute { my $key = join '', @path; if( not exists $pathes{$key} ) { $pathes{$key} = 1; my ($l, $r) = (0,0); my $path = "($l-$r) ".join( " ", map{ "(".(++$l)."-".($r+=$_). +")" } @path); print "$path\n"; } } @path;

In reply to Re^3: compute paths in Pascal's triangle (aka Tartaglia's one) by hdb
in thread compute paths in Pascal's triangle (aka Tartaglia's one) by Discipulus

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