How to do matrix conversions by row and columns toggles?

Question

I have got a square matrix consisting of elements either 1 or 0. An ith row toggle toggles all the ith row elements (1 becomes 0 and vice versa) and jth column toggle toggle

Answer 1

Algorithm

Simplify the problem from "Try to transform A into B" into "Try to transform M into 0", where M = A xor B. Now all the positions which must be toggled have a 1 in them.

Consider an arbitrary position in M. It is affected by exactly one column toggle and exactly one row toggle. If its initial value is V, the presence of the column toggle is C, and the presence of the row toggle is R, then the final value F is V xor C xor R. That's a very simple relationship, and it makes the problem trivial to solve.

Notice that, for each position, R = F xor V xor C = 0 xor V xor C = V xor C. If we set C then we force the value of R, and vice versa. That's awesome, because it means if I set the value of any row toggle then I will force all of the column toggles. Any one of those column toggles will force all of the row toggles. If the result is the 0 matrix, then we have a solution. We only need to try two cases!

Pseudo-code

function solve(Matrix M) as bool possible, bool[] rowToggles, bool[] colToggles:
    For var b in {true, false}
        colToggles = array from c in M.colRange select b xor Matrix(0, c)
        rowToggles = array from r in M.rowRange select colToggles[0] xor M(r, 0)
        if none from c in M.colRange, r in M.rowRange
                where colToggle[c] xor rowToggle[r] xor M(r, c) != 0 then
            return true, rowToggles, colToggles
        end if
    next var
    return false, null, null
end function

Analysis

The analysis is trivial. We try two cases, within which we run along a row, then a column, then all cells. Therefore if there are r rows and c columns, meaning the matrix has size n = c * r, then the time complexity is O(2 * (c + r + c * r)) = O(c * r) = O(n). The only space we use is what is required for storing the outputs = O(c + r).

Therefore the algorithm takes time linear in the size of the matrix, and uses space linear in the size of the output. It is asymptotically optimal for obvious reasons.