I don't see the problem. Once you have enough bits to resolve every pairwise difference you're just reduced to the problem of comparing the sums of two lists o

I don't see the problem. Once you have enough bits to resolve every pairwise difference you're just reduced to the problem of comparing the sums of two lists of n bit integers and I'm pretty sure that's in P. If it's not then that should be the main point here, sqrt has nothing to do with it. EDIT: You also know what the largest possible error is on each pairwise delta so if you add log(N + 1) bits you handle even the worst case where one sum is +N x maxerror and the other -N x maxerror.