python - python 实现中的 Strassen 算法错误

Question

我通过 Strassen 算法和 Python 3 中的朴素嵌套 for 循环实现得到了不同的矩阵乘法输出。

代码:

def new_matrix(r, c):
    """Create a new matrix filled with zeros."""
    matrix = [[0 for row in range(r)] for col in range(c)]
    return matrix


def direct_multiply(x, y):
    if len(x[0]) != len(y):
        return "Multiplication is not possible!"
    else:
        p_matrix = new_matrix(len(x), len(y[0]))
        for i in range(len(x)):
            for j in range(len(y[0])):
                for k in range(len(y)):
                    p_matrix[i][j] += x[i][k] * y[k][i]
    return p_matrix


def split(matrix):
    """Split matrix into quarters."""
    a = b = c = 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 add_matrix(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 subtract_matrix(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 strassen(x, y, n):
    # base case: 1x1 matrix
    if n == 1:
        z = [[0]]
        z[0][0] = x[0][0] * y[0][0]
        return z
    else:
        # split matrices into quarters
        a, b, c, d = split(x)
        e, f, g, h = split(y)

        # p1 = a*(f-h)
        p1 = strassen(a, subtract_matrix(f, h), n/2)

        # p2 = (a+b)*h
        p2 = strassen(add_matrix(a, b), h, n/2)

        # p3 = (c+d)*e
        p3 = strassen(add_matrix(c, d), e, n/2)

        # p4 = d*(g-e)
        p4 = strassen(d, subtract_matrix(g, e), n/2)

        # p5 = (a+d)*(e+h)
        p5 = strassen(add_matrix(a, d), add_matrix(e, h), n/2)

        # p6 = (b-d)*(g+h)
        p6 = strassen(subtract_matrix(b, d), add_matrix(g, h), n/2)

        # p7 = (a-c)*(e+f)
        p7 = strassen(subtract_matrix(a, c), add_matrix(e, f), n/2)

        z11 = add_matrix(subtract_matrix(add_matrix(p5, p4), p2), p6)

        z12 = add_matrix(p1, p2)

        z21 = add_matrix(p3, p4)

        z22 = add_matrix(subtract_matrix(subtract_matrix(p5, p3), p7), p1)

        z = new_matrix(len(z11)*2, len(z11)*2)
        for i in range(len(z11)):
            for j in range(len(z11)):
                z[i][j] = z11[i][j]
                z[i][j+len(z11)] = z12[i][j]
                z[i+len(z11)][j] = z21[i][j]
                z[i+len(z11)][j+len(z11)] = z22[i][j]

        return z


a = [[11,11,11,11],[22,22,22,22],[33,33,33,33],[44,44,44,44]]
b = [[101,181,119,113],[22,22,22,22],[33,33,33,33],[44,44,44,44]]

print(f"a = {a}")
print(f"b = {b}")

print(f"Using Strassen's algorithm:\na*b = {strassen(a, b, 4)}")

print(f"Using naive algorithm:\na*b = {direct_multiply(a, b)}")

输出:

$ python3 strassen.py

a = [[11, 11, 11, 11], [22, 22, 22, 22], [33, 33, 33, 33], [44, 44, 44, 44]]

b = [[101, 181, 119, 113], [22, 22, 22, 22], [33, 33, 33, 33], [44, 44, 44, 44]]

Using Strassen's algorithm:

a*b = [[2200, 3080, 2398, 2332], [4400, 6160, 4796, 4664], [6600, 9240, 7194, 6996], [8800, 12320, 9592, 9328]]

Using naive algorithm:

a*b = [[2200, 2200, 2200, 2200], [6160, 6160, 6160, 6160], [7194, 7194, 7194, 7194], [9328, 9328, 9328, 9328]]

有人能帮忙吗？

Answer 1

应该是

p_matrix[i][j] += x[i][k] * y[k][j]

在您的direct_multiply 函数中。您通过这种方式将一行中的第 k 元素与一列中的第 k 元素相乘，然后累加。就像您进行矩阵乘法一样。

输出

a = [[11, 11, 11, 11], [22, 22, 22, 22], [33, 33, 33, 33], [44, 44, 44, 44]]
b = [[101, 181, 119, 113], [22, 22, 22, 22], [33, 33, 33, 33], [44, 44, 44, 44]]
Using Strassen's algorithm:
a*b = [[2200, 3080, 2398, 2332], [4400, 6160, 4796, 4664], [6600, 9240, 7194, 6996], [8800, 12320, 9592, 9328]]
Using naive algorithm:
a*b = [[2200, 3080, 2398, 2332], [4400, 6160, 4796, 4664], [6600, 9240, 7194, 6996], [8800, 12320, 9592, 9328]]