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I will assume the original matrix has $n-1$ rows, $n$ columns, and the rows are linearly independent (this is easy to check; and if it is not the case, then the problem is trivial). Adding a new row $ …

When you are dealing with binary matrices, this is the problem of graph canonization. … When you have non-binary matrices, you can think of this as a graph with labels on the edges. …

You could always try the A* algorithm. It might not be optimal but it is something you could try, and depending on how big the matrix is it might be good enough. You'll need to design an appropriate …

The problem can be solved with dynamic programming in $O(RC)$ time. I'll start by showing how to solve it in $O(RC^2)$ time. Let $M[\cdot,\cdot]$ denote the input matrix. Let $A[i,j]$ denote the to …

I expect that solving the problem exactly might be NP-hard. However, here is an approach that I expect will probably be good enough for practical purposes. It uses a subroutine to solve the following …

There is no algorithm with worst-case running time better than $O(n^2)$. There is a simple adversary argument to prove this. Consider any algorithm that is purported to be correct, and that always i …

It is easy to achieve $O(n^3 \lg n)$ running time by accumulating partial sums, if all entries of $A$ are positive. Define the $n\times n$ matrix $B$ so that $$B[i,j] = \sum_{r=0}^i \sum_{s=0}^j A[r …

Yes, absolutely. There is a straightforward way to do what you want, if you work through the details of the algorithm given at the link in your question. The trick is to use persistent data structur …

For instance, matrices form a star semiring. …

No, it's not true. You can use Gaussian elimination to invert a matrix in $O(n^3)$ time, but there are other algorithms that are even faster. The complexity of a problem is the running time of the f …

Here is one approach. It won't survive JPEG compression, though, so probably you can do better. Let $m$ be an integer large enough that $2^m$ is significantly larger than both the width and height o …

One approach is to use integer linear programming, and hope that an ILP solver will be able to find a solution. This will probably still be exponential in $n$ in the worst case, but it might be faste …

This gives you a sequence of candidate matrices, one per step. Choose the best one, i.e., the one where the sum of desired elements is the largest. …

Yes, it is solvable in $O(\max(m,n)^3)$ time. In fact, it should be solvable in $O(\min(m,n)^2 \max(m,n))$ time -- e.g., if $m<n$, it should be solvable in $O(m^2 n)$ time. Let $N_\text{down}(i,j)$ …

Yes, you can solve it faster than trying all possible permutations. The biggest such rectangle must span consecutive rows. So, for all $i,j$ with $i \le j$, check what is the biggest rectangle you c …