Now imagine that each time one of the dials changes, all the dials to the right of it can be remade. For example, each dial could exclude any numbers that exist on any dial to its left.That is, instead of looping over the static AoA, you can replace the arrayrefs with codrefs that filter out any elements that have already been used by outer loops.
New update: In Re: Efficient Unique Nested Combinations, FFRANK points out that my "I think you want" is not the correct algorithm. A corrected solution is posted in reply thereto.
Update: I think you just want to ensure that each element is greater-than-or-equal-to the previous one, so:
#!perl use warnings; use strict; use Algorithm::Loops 'NestedLoops'; my @symbol = ( [ 'a', 'b', 'c' ], [ 'a', 'b', 'c' ], [ 'a', 'b', 'c' ], [ 'c', 'd', 'e' ], ); my $combos = NestedLoops([ $symbol[0], map { my $_hold = $_; sub { [grep {$_ ge $_[$#_]} @{$symbol[$_hold]}] } } 1..$#symbol ]); my @result; print "@result\n" while @result = $combos->();
In reply to Re: Efficient Unique Nested Combinations
by Roy Johnson
in thread Efficient Unique Nested Combinations
by FFRANK
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