矩形矩阵复杂度

Question

I have to compute complexity of algorithm which searches an element in a rectangular matrix. It uses divide and conquer. For square matrix my time functions becomes a*T(n/b)+O(n^2). But for rectangular matrix I don't know how to denote the division if it has to be divided into 4 sub matrices. Will it be a*T(m*n/4) + O(n)?

Answer 1

You didn't describe you algorithm so it's not possible to answer the question precisely. But I'll try assuming the algorithm is like:

1. If m = 1 or n = 1 then process matrix in O(m * n) time.
2. Else ( m > 1 and n > 1 ) divide matrix into four matricies of size not more then [m / 2]x[n / 2] , where [y] is a minimal integer not less then y .
3. Call the algorithm with each of these matricies. Then "merge" results in O(m * n) time.

In this case recurrent equation for time complexity will be

1. T(1, n) = O(n)
2. T(m, 1) = O(m)
3. T(m, n) = 4 * T([m / 2], [n / 2]) + O(m * n)

Lets solve it

T(m, n) = O(m * n) +
          4 * O([m / 2] * [n / 2]) +
          4 ^ 2 * O([m / 4] * [n / 4]) +
          ... +
          4 ^ L * O([m / 2 ^ L] * [n / 2 ^ L]),  where L = [log(min(m, n))]
T(m, n) = O(m * n + ... + 4 ^ L * [m / 2 ^ L] * [n / 2 ^ L]) =
        = O(m * n + ... + 2 ^ L * [m / 2 ^ L] * 2 ^ L * [n / 2 ^ L] =
        = O(m * n + ... + m * n  (L times)) =
        = O(L * m * n) = O(m * n * [log(min(m, n))])

So the answer to your question is

T(m, n) = O(m * n * [log(min(m, n))]), where log stands for a binary logarithm and [y] stands for ceil function.