In essence this problem requires that you generate the subsets of the original set. Say you have the set (1,2,3,4). You want to find all these subsets 

1
1 2
1 3
1 4 
1 2 3
1 2 4
1 3 4
1 2 3 4
2
2 3
2 4
2 3 4
3
3 4
4

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One neat way to get all the combinations is to use a binary approach where the presence of absence of a binary 1 is used to represent "grab this element" from the set. The number of possible combinations is (2**n)-1. It is easier to see in code: 
my @list = (1,2,3,4);
my $n = 2**@list -1;
for $b(1..$n) {
    printf "%2d: %04b ", $b, $b;
    my $i = $b;
    for (@list) {
        print "$_ " if $i%2;
        $i >>= 1;
        
    }
    print "\n";
}
__DATA__
 1: 0001 1    (get element 0)
 2: 0010 2    (get element 1)
 3: 0011 1 2  (get elements 0 and 1)
 4: 0100 3    (get element 2)
 5: 0101 1 3  (etc)
 6: 0110 2 3 
 7: 0111 1 2 3 
 8: 1000 4 
 9: 1001 1 4 
10: 1010 2 4 
11: 1011 1 2 4 
12: 1100 3 4 
13: 1101 1 3 4 
14: 1110 2 3 4 
15: 1111 1 2 3 4

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In reply to Re: Recursion problem by tachyon-II
in thread Recursion problem by someone202

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