I missed that "BA" in the list. The last page of Algorithm::Loops talks specifically about this problem and shows a few ways to do it. A fairly minor update to my code is one such way:
#!/usr/bin/perl -w use strict; use Algorithm::Loops qw( NestedLoops NextPermute ); my @items= ( 1..5 ); my $choose= 2; my $iter= NestedLoops( [ [ 0..$#items ], ( sub { [ $_+1 .. $#items ] } ) x ($choose-1), ], ); my @choice; while( @choice= @items[$iter->()] ) { do { print "@choice\n"; # replace above line with your code } while( NextPermute(@choice) ); }
produces
1 2 2 1 1 3 3 1 1 4 4 1 1 5 5 1 2 3 3 2 2 4 4 2 2 5 5 2 3 4 4 3 3 5 5 3 4 5 5 4
- tye
In reply to Re^3: Searching for a Permutation Algorithm for nPr where n != r (A::Loops)
by tye
in thread Searching for a Permutation Algorithm for nPr where n != r
by ozboomer
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