Sum-subset with a fixed subset size

Question

The sum-subset problem states:

Given a set of integers, is there a non-empty subset whose sum is zero?

This problem is NP-complete in general. I'm curious if the complexity of this slight variant is known:

Given a set of integers, is there a subset of size k whose sum is zero?

For example, if k = 1, you can do a binary search to find the answer in O(log n). If k = 2, then you can get it down to O(n log n) (e.g. see Find a pair of elements from an array whose sum equals a given number). If k = 3, then you can do O(n^2) (e.g. see Finding three elements in an array whose sum is closest to a given number).

Is there a known bound that can be placed on this problem as a function of k?

As motivation, I was thinking about this question How do you partition an array into 2 parts such that the two parts have equal average? and trying to determine if it is actually NP-complete. The answer lies in whether or not there is a formula as described above.

Barring a general solution, I'd be very interested in knowing an optimal bound for k=4.

Answer 1

For k=4, space complexity O(n), time complexity O(n² * log(n))

Sort the array. Starting from 2 smallest and 2 largest elements, calculate all lesser sums of 2 elements (a[i] + a[j]) in the non-decreasing order and all greater sums of 2 elements (a[k] + a[l]) in the non-increasing order. Increase lesser sum if total sum is less than zero, decrease greater one if total sum is greater than zero, stop when total sum is zero (success) or a[i] + a[j] > a[k] + a[l] (failure).

The trick is to iterate through all the indexes i and j in such a way, that (a[i] + a[j]) will never decrease. And for k and l, (a[k] + a[l]) should never increase. A priority queue helps to do this:

1. Put key=(a[i] + a[j]), value=(i = 0, j = 1) to priority queue.
2. Pop (sum, i, j) from priority queue.
3. Use sum in the above algorithm.
4. Put (a[i+1] + a[j]), i+1, j and (a[i] + a[j+1]), i, j+1 to priority queue only if these elements were not already used. To keep track of used elements, maintain an array of maximal used 'j' for each 'i'. It is enough to use only values for 'j', that are greater, than 'i'.
5. Continue from step 2.

For k>4

If space complexity is limited to O(n), I cannot find anything better, than use brute force for k-4 values and the above algorithm for the remaining 4 values. Time complexity O(n^(k-2) * log(n)).

For very large k integer linear programming may give some improvement.

Update

If n is very large (on the same order as maximum integer value), it is possible to implement O(1) priority queue, improving complexities to O(n²) and O(n^(k-2)).

If n >= k * INT_MAX, different algorithm with O(n) space complexity is possible. Precalculate a bitset for all possible sums of k/2 values. And use it to check sums of other k/2 values. Time complexity is O(n^(ceil(k/2))).

Answer 2

The problem of determining whether 0 in W + X + Y + Z = {w + x + y + z | w in W, x in X, y in Y, z in Z} is basically the same except for not having annoying degenerate cases (i.e., the problems are inter-reducible with minimal resources).

This problem (and thus the original for k = 4) has an O(n^2 log n)-time, O(n)-space algorithm. The O(n log n)-time algorithm for k = 2 (to determine whether 0 in A + B) accesses A in sorted order and B in reverse sorted order. Thus all we need is an O(n)-space iterator for A = W + X, which can be reused symmetrically for B = Y + Z. Let W = {w1, ..., wn} in sorted order. For all x in X, insert a key-value item (w1 + x, (1, x)) into a priority queue. Repeatedly remove the min element (wi + x, (i, x)) and insert (wi+1 + x, (i+1, x)).

Answer 3

The solution for k=4 in O(n^2log(n))

Step 1: Calculate the pairwise sum and sort the list. There are n(n-1)/2 sums. So the complexity is O(n^2log(n)). Keep the identities of the individuals which make the sum.

Step 2: For each element in the above list search for the complement and make sure they don't share "the individuals). There are n^2 searches, each with complexity O(log(n))

EDIT: The space complexity of the original algorithm is O(n^2). The space complexity can be reduced to O(1) by simulating a virtual 2D matrix (O(n), if you consider space to store sorted version of the array).

First about 2D matrix: sort the numbers and create a matrix X using pairwise sums. Now the matrix is ins such a way that all the rows and columns are sorted. To search for a value in this matrix, search the numbers on the diagonal. If the number is in between X[i,i] and X[i+1,i+1], you can basically halve the search space by to matrices X[i:N, 0:i] and X[0:i, i:N]. The resulting search algorithm is O(log^2n) (I AM NOT VERY SURE. CAN SOMEBODY CHECK IT?).

Now, instead of using a real matrix, use a virtual matrix where X[i,j] are calculated as needed instead of pre-computing them.

Resulting time complexity: O( (nlogn)^2 ).

PS: In the following link, it says the complexity of 2D sorted matrix search is O(n) complexity. If that is true (i.e. O(log^2n) is incorrect), then the finally complexity is O(n^3).

Answer 4

Question that is very similar:

Is this variant of the subset sum problem easier to solve?

It's still NP-complete.

If it were not, the subset-sum would also be in P, as it could be represented as F(1) | F(2) | ... F(n) where F is your function. This would have O(O(F(1)) + O(F(2)) + O(F(n))) which would still be polynomial, which is incorrect as we know it's NP-complete.

Note that if you have certain bounds on the inputs you can achieve polynomial time.

Also note that the brute-force runtime can be calculated with binomial coefficients.

Answer 5

To build on awesomo's answer... if we can assume that numbers are sorted, we can do better than O(n^k) for given k; simply take all O(n^(k-1)) subsets of size (k-1), then do a binary search in what remains for a number that, when added to the first (k-1), gives the target. This is O(n^(k-1) log n). This means the complexity is certainly less than that.

In fact, if we know that the complexity is O(n^2) for k=3, we can do even better for k > 3: choose all (k-3)-subsets, of which there are O(n^(k-3)), and then solve the problem in O(n^2) on the remaining elements. This is O(n^(k-1)) for k >= 3.

However, maybe you can do even better? I'll think about this one.

EDIT: I was initially going to add a lot proposing a different take on this problem, but I've decided to post an abridged version. I encourage other posters to see whether they believe this idea has any merit. The analysis is tough, but it might just be crazy enough to work.

We can use the fact that we have a fixed k, and that sums of odd and even numbers behave in certain ways, to define a recursive algorithm to solve this problem.

First, modify the problem so that you have both even and odd numbers in the list (this can be accomplished by dividing by two if all are even, or by subtracting 1 from numbers and k from the target sum if all are odd, and repeating as necessary).

Next, use the fact that even target sums can be reached only by using an even number of odd numbers, and odd target sums can be reached using only an odd number of odd numbers. Generate appropriate subsets of the odd numbers, and call the algorithm recursively using the even numbers, the sum minus the sum of the subset of odd numbers being examined, and k minus the size of the subset of odd numbers. When k = 1, do binary search. If ever k > n (not sure this can happen), return false.

If you have very few odd numbers, this could allow you to very quickly pick up terms that must be part of a winning subset, or discard ones that cannot. You can transform problems with lots of even numbers to equivalent problems with lots of odd numbers by using the subtraction trick. The worst case must therefore be when the numbers of even and odd numbers are very similar... and that's where I am right now. A uselessly loose upper bound on this is many orders of magnitudes worse than brute-force, but I feel like this is probably at least as good as brute-force. Thoughts are welcome!

EDIT2: An example of the above, for illustration.

{1, 2, 2, 6, 7, 7, 20}, k = 3, sum = 20.
Subset {}:
 {2, 2, 6, 20}, k = 3, sum = 20
 = {1, 1, 3, 10}, k = 3, sum = 10
 Subset {}:
  {10}, k = 3, sum = 10
  Failure
 Subset {1, 1}:
  {10}, k = 1, sum = 8
  Failure
 Subset {1, 3}:
  {10}, k = 1, sum = 6
  Failure
Subset {1, 7}:
 {2, 2, 6, 20}, k = 1, sum = 12
 Failure
Subset {7, 7}:
 {2, 2, 6, 20}, k = 1, sum = 6
 Success

Answer 6

The time complexity is trivially O(n^k) (number of k-sized subsets from n elements).

Since k is a given constant, a (possibly quite high-order) polynomial upper bounds the complexity as a function of n.