Whilst it is true that no element is repeated, for the purposes of the OP, this solution will produce multiple results for the same three elements being summed in a different order. Some form of filtering, perhaps involving a numerical sort of element numbers, a pack to a string and a grep with a %seen hash, would be required to remove duplicates.

Update: Here's some code to show what I mean. All the faff with pack q{N*}, @{ $_ } is because we may be dealing with larger arrays where subscripts go into multiple digits. This code:-

use 5.026;
use warnings;

use Algorithm::Permute;

my @arr  = ( 1, 2, 3, 4, 5 );
my $perm = Algorithm::Permute->new( [ 0 .. $#arr ], 3 );

my @allPerms;
while ( my @res = $perm->next() )
{
    push @allPerms, [ sort { $a <=> $b } @res ];
}

my @uniqPerms = do {
   my %seen;
   map  { [ unpack q{N*}, $_ ] }
   grep { ! $seen{ $_ } ++ }
   sort
   map  { pack q{N*}, @{ $_ } }
   @allPerms
   };

say join q{+}, @{ $_ } for @uniqPerms;

[download]

Produces:-

0+1+2
0+1+3
0+1+4
0+2+3
0+2+4
0+3+4
1+2+3
1+2+4
1+3+4
2+3+4

[download]

Cheers,

JohnGG

In reply to Re^4: Sum of N elements in an M element array by johngg
in thread Sum of N elements in an M element array by abhay180

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