Yeah, in fact, I somehow calculated in assumption of
n
being the amount of elements in matrix, notn²
(assuming square matrix)But I am impressed to know that there are serial algorithms that approach
O(n²)
, thank you for sharing that infoYeah, they work by turning the problem into some crazy kind of group theory and attacking it that way. Every once in a while someone shaves the decimal down slightly, just by implementing the deep math in a more efficient way. A new approach will be needed if it is in fact possible to get down to O(n2), though. Strassen’s is a divide and conquer algorithm, and each step of the iteration looks like this:
S[1] = B[1, 2] - B[2, 2] S[2] = A[1, 1] + A[1, 2] S[3] = A[2, 1] + A[2, 2] S[4] = B[2, 1] - B[1, 1] S[5] = A[1, 1] + A[2, 2] S[6] = B[1, 1] + B[2, 2] S[7] = A[1, 2] - A[2, 2] S[8] = B[2, 1] + B[2, 2] S[9] = A[1, 1] - A[2, 1] S[10] = B[1, 1] + B[1, 2] P[1] = STRASSEN(A[1, 1], S[1]) P[2] = STRASSEN(S[2], B[2, 2]) P[3] = STRASSEN(S[3], B[1, 1]) P[4] = STRASSEN(A[2, 2], S[4]) P[5] = STRASSEN(S[5], S[6]) P[6] = STRASSEN(S[7], S[8]) P[7] = STRASSEN(S[9], S[10]) C[1..n / 2][1..n / 2] = P[5] + P[4] - P[2] + P[6] C[1..n / 2][n / 2 + 1..n] = P[1] + P[2] C[n / 2 + 1..n][1..n / 2] = P[3] + P[4] C[n / 2 + 1..n][n / 2 + 1..n] = P[5] + P[1] - P[3] - P[7] return C
In my copy of Introduction to Algorithms, it says something like “this is the most bullshit algorithm in the book and it’s not close” underneath. You can make it a bit neater by representing the multiplication operation as a 3-dimensional tensor, but at the end of the day it’s still just a stupid arithmetic trick. One that’s built into your GPU.