Strassen矩阵乘法索引超出范围

Question

我需要你的帮助。我的计划是使用python实现Strassen-Algorithm，但我的问题是它仅适用于相同大小的矩阵，并且如果一个矩阵大于另一个矩阵，则会出现索引超出范围的错误。

我认为解决问题的方法是用零填充新矩阵，但说实话我真的不知道这是否可行或如何实现。

def showMatrix(m):
    for line in m:
        print('|', end='')
        i = 0
        for value in line:
            if (i > 0):
                print(' ', end='')
            print(value, end='')
            i = i + 1
        print("|")

def add_m(a, b):
    if type(a) == int:
        d = a + b
    else:
        d = []
        for i in range(len(a)):
            c = []
            for j in range(len(a[0])):
                c.append(a[i][j] + b[i][j])
            d.append(c)
    return d

def sub_m(a, b):
    if type(a) == int:
        d = a - b
    else:
        d = []
        for i in range(len(a)):
            c = []
            for j in range(len(a[0])):
                c.append(a[i][j] - b[i][j])
            d.append(c)
    return d


def MatMult(A, B):
    n = len(A)
    C = [[0 for i in range(n)] for j in range(n)]
    for i in range(n):
        for k in range(n):
            for j in range(n):
                C[i][j] += A[i][k] * B[k][j]
    return C

def split(matrix): #splits matrix in quarters
    a = matrix 
    b = matrix 
    c = matrix 
    d = matrix 
    while(len(a) > len(matrix)/2):
        a = a[:len(a)//2]
        b = b[:len(b)//2]
        c = c[len(c)//2:]
        d = d[len(d)//2:]
    while(len(a[0]) > len(matrix[0])/2):
        for i in range(len(a[0])//2):
            a[i] = a[i][:len(a[i])//2]
            b[i] = b[i][len(b[i])//2:]
            c[i] = c[i][:len(c[i])//2]
            d[i] = d[i][len(d[i])//2:]
    return a,b,c,d


def strassen(a, b,q):
    print('Parameter a:')
    showMatrix(a)
    print('Parameter b:')
    showMatrix(b)
    q = len(a) 
    if q == 1:
        return MatMult(a, b)
    else:
        a11, a12, a21, a22 = split(a)
        b11, b12, b21, b22 = split(b)

        # p1 = (a11+a22) * (b11+b22)
        p1 = strassen(add_m(a11,a22), add_m(b11,b22), q/2)

        # p2 = (a21+a22) * b11
        p2 = strassen(add_m(a21,a22), b11, q/2)

        # p3 = a11 * (b12-b22)
        p3 = strassen(a11, sub_m(b12,b22), q/2)

        # p4 = a22 * (b12-b11)
        p4 = strassen(a22, sub_m(b21,b11), q/2)

        # p5 = (a11+a12) * b22
        p5 = strassen(add_m(a11,a12), b22, q/2)

        # p6 = (a21-a11) * (b11+b12)
        p6 = strassen(sub_m(a21,a11), add_m(b11,b12), q/2)

        # p7 = (a12-a22) * (b21+b22)
        p7 = strassen(sub_m(a12,a22), add_m(b21,b22), q/2)


        # c11 = p1 + p4 - p5 + p7
        c11 = add_m(sub_m(add_m(p1, p4), p5), p7)

        # c12 = p3 + p5
        c12 = add_m(p3, p5)

        # c21 = p2 + p4
        c21 = add_m(p2, p4)

        # c22 = p1 + p3 - p2 + p6
        c22 = add_m(sub_m(add_m(p1, p3), p2), p6)

        c = [[0 for i in range(0, q)] for i in range(0, q)]
        for i in range(len(c11)):
            for j in range(len(c11)):
                c[i][j]                   = c11[i][j]
                c[i][j+len(c11)]          = c12[i][j]
                c[i+len(c11)][j]          = c21[i][j]
                c[i+len(c11)][j+len(c11)] = c22[i][j]

        return c

a, b = [ [3, 2, 1],[1, 0, 2]], [[1, 2],[0, 1],[4, 0]]
result = strassen(a,b,2)
showMatrix(result)

结果应该是[[7，8]，[9，2]]，但就像我说的那样，我得到的索引超出范围错误。