Question

I’m looking for an algorithm which can take two sets of integers (both positive and negative) and find subsets within each that have the same sum.

The problem is similar to the subset sum problem except that I’m looking for subsets on both sides.

Here’s an example:

List A {4, 5, 9, 10, 1}

List B {21, 7, -4, 180}

So the only match here is: {10, 1, 4, 9} <=> {21, 7, -4}

Does anyone know if there are existing algorithms for this kinda problems?

So far, the only solution I have is a brute force approach which tries every combination but it performs in Exponential time and I’ve had to put a hard limit on the number of elements to consider to avoid it from taking too long.

The only other solution I can think of is to run a factorial on both lists and look for equalities there but that is still not very efficient and takes exponentially longer as the lists get bigger.

Answer 1

What others have said is true:

1. This problem is NP-complete. An easy reduction is to usual subset-sum. You can show this by noting that a subset of A sums to a subset of B (not both empty) only if a non-empty subset of A union (-B) sums to zero.

2. This problem is only weakly NP-complete, in that it’s polynomial in the size of the numbers involved, but is conjectured to be exponential in their logarithms. This means that the problem is easier than the moniker ‘NP-complete’ might suggest.

3. You should use dynamic programming.

So what am I contributing to this discussion? Well, code (in Perl):

@a = qw(4 5 9 10 1); @b = qw(21 7 -4 180); %a = sums( @a ); %b = sums( @b ); for $m ( keys %a ) {     next unless exists $b{$m};     next if $m == 0 and (@{$a{0}} == 0 or @{$b{0}} == 0);     print 'sum(@{$a{$m}}) = sum(@{$b{$m}})\n'; }  sub sums {     my( @a ) = @_;     my( $a, %a, %b );     %a = ( 0 => [] );     while( @a ) {         %b = %a;         $a = shift @a;         for my $m ( keys %a ) {             $b{$m+$a} = [@{$a{$m}},$a];         }     %a = %b;     }     return %a; } 

It prints

sum(4 5 9 10) = sum(21 7) sum(4 9 10 1) = sum(21 7 -4) 

so, notably, there is more than one solution that works in your original problem!

Edit: User itzy correctly pointed out that this solution was wrong, and worse, in multiple ways!! I’m very sorry about that and I’ve hopefully addressed these concerns in the new code above. Nonetheless, there is still one problem, namely that for any particular subset-sum, it only prints one of the possible solutions. Unlike the previous problems, which were straight-up errors, I would classify this as an intentional limitation. Best of luck and beware of bugs!