This sounds like a good dynamic programming exercise : 
use List::Util qw/sum/;
use Memoize; memoize('sum_to');
use strict;

sub sum_to { 
    # given : $n, @ary
    # return : two array refs, the first one summing to $n
    #          the second one has the remaining elements
    #          return nothing if it's impossible
    my $n = shift;
    my @ary = @_;
    return [[],\@ary] if $n==0;
    return if ($n<0 || @ary==0);
    for my $elem (@ary) {
        my %seen;
        my @left = grep { $_ != $elem || $seen{$_}++ } @ary;
        if (my $found = sum_to($n-$elem,@left)) {
            return [ [ $elem, @{ $found->[0] } ], 
                     [ @{ $found->[1] } ] 
                   ];
        }
    }
    return;
}

my @nums   = map int rand 1000, 1..100;
my $sum    = sum @nums;
my $target = int ( $sum / 2);

while ($target > 0) {
    my $found  = sum_to($target,@nums) or next;
    my $first  = $target       ." == ". join '+', @{ $found->[0] };
    my $second = ($sum-$target)." == ". join '+', @{ $found->[1] };
    print join "\n",$first, $second;
    die "there was a problem" unless eval $first && eval $second;
    last;
} continue {
    $target--;
}

[download]
I think this might fail in the case where there are two ways to sum to a given number, and only one of those ways is right -- to account for this, sum_to shd probably return an array of solutions.

In reply to Re: Divide array of integers into most similar value halves by bduggan
in thread Divide array of integers into most similar value halves by Pepe

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