algorithm - From an interview: Removing rows and columns in an n×n matrix to maximize the sum of remaining values

Question

Given an n×n matrix of real numbers. You are allowed to erase any number (from 0 to n) of rows and any number (from 0 to n) of columns, and after that the sum of the remaining entries is computed. Come up with an algorithm which finds out which rows and columns to erase in order to maximize that sum.

Answer 1

The problem is NP-hard. (So you should not expect a polynomial-time algorithm for solving this problem. There could still be (non-polynomial time) algorithms that are slightly better than brute-force, though.) The idea behind the proof of NP-hardness is that if we could solve this problem, then we could solve the the clique problem in a general graph. (The maximum-clique problem is to find the largest set of pairwise connected vertices in a graph.)

Specifically, given any graph with n vertices, let's form the matrix A with entries a[i][j] as follows:

• a[i][j] = 1 for i == j (the diagonal entries)
• a[i][j] = 0 if the edge (i,j) is present in the graph (and i≠j)
• a[i][j] = -n-1 if the edge (i,j) is not present in the graph.

Now suppose we solve the problem of removing some rows and columns (or equivalently, keeping some rows and columns) so that the sum of the entries in the matrix is maximized. Then the answer gives the maximum clique in the graph:

1. Claim: In any optimal solution, there is no row i and column j kept for which the edge (i,j) is not present in the graph. Proof: Since a[i][j] = -n-1 and the sum of all the positive entries is at most n, picking (i,j) would lead to a negative sum. (Note that deleting all rows and columns would give a better sum, of 0.)

2. Claim: In (some) optimal solution, the set of rows and columns kept is the same. This is because starting with any optimal solution, we can simply remove all rows i for which column i has not been kept, and vice-versa. Note that since the only positive entries are the diagonal ones, we do not decrease the sum (and by the previous claim, we do not increase it either).

All of which means that if the graph has a maximum clique of size k, then our matrix problem has a solution with sum k, and vice-versa. Therefore, if we could solve our initial problem in polynomial time, then the clique problem would also be solved in polynomial time. This proves that the initial problem is NP-hard. (Actually, it is easy to see that the decision version of the initial problem — is there a way of removing some rows and columns so that the sum is at least k —?is in NP, so the (decision version of the) initial problem is actually NP-complete.)