原理：分塊矩陣乘法，進行8次矩陣乘法，時間複雜度為 $\theta(n^3) = \theta(n^}) $ , 改進後僅需要7次乘法, 時間複雜度為 $\theta(n^})$
具體推到見演算法導論中利用主定理推導時間複雜度
def matrix_divide(a):
rows = len(a)
mid = rows // 2
a11 = [[0]*mid for _ in range(mid)]
a12 = [[0]*mid for _ in range(mid)]
a21 = [[0]*mid for _ in range(mid)]
a22 = [[0]*mid for _ in range(mid)]
for i in range(mid):
for j in range(mid):
a11[i][j] = a[i][j]
a12[i][j] = a[i][mid+j]
a21[i][j] = a[mid+i][j]
a22[i][j] = a[mid+i][mid+j]
return a11, a12, a21, a22
def matrix_add(a, b):
rows = len(a)
c = [[0]*rows for _ in range(rows)]
for i in range(rows):
for j in range(rows):
c[i][j] = a[i][j] + b[i][j]
return c
def matrix_sub(a, b):
rows = len(a)
c = [[0]*rows for _ in range(rows)]
for i in range(rows):
for j in range(rows):
c[i][j] = a[i][j] - b[i][j]
return c
def matrix_merge(c11, c12, c21, c22):
rows = len(c11)
n = rows * 2
c = [[0]*n for _ in range(n)]
for i in range(rows):
for j in range(rows):
c[i][j] = c11[i][j]
c[i][rows+j] = c12[i][j]
c[rows+i][j] = c21[i][j]
c[rows+i][rows+j] = c22[i][j]
return c
def strassen(a, b):
n = len(a)
c = [[0] for _ in range(n)]
if n == 1:
c[0][0] = a[0][0]*b[0][0]
return c
a11, a12, a21, a22 = matrix_divide(a)
b11, b12, b21, b22 = matrix_divide(b)
s1 = matrix_sub(b12, b22)
s2 = matrix_add(a11, a12)
s3 = matrix_add(a21, a22)
s4 = matrix_sub(b21, b11)
s5 = matrix_add(a11, a22)
s6 = matrix_add(b11, b22)
s7 = matrix_sub(a12, a22)
s8 = matrix_add(b21, b22)
s9 = matrix_sub(a11, a21)
s10 = matrix_add(b11, b12)
p1 = strassen(a11, s1)
p2 = strassen(s2, b22)
p3 = strassen(s3, b11)
p4 = strassen(a22, s4)
p5 = strassen(s5, s6)
p6 = strassen(s7, s8)
p7 = strassen(s9, s10)
c11 = matrix_add(p5, matrix_sub(p4, matrix_sub(p2, p6)))
c12 = matrix_add(p1, p2)
c21 = matrix_add(p3, p4)
c22 = matrix_add(p5, matrix_sub(p1, matrix_add(p3, p7)))
return matrix_merge(c11, c12, c21, c22)
def main():
a = [[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]]
b = [[5,5,5,5],[6,6,6,6],[7,7,7,7],[8,8,8,8]]
c = strassen(a, b)
print(c)
if __name__ == '__main__':
main()
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