c++ - 遵循 Glynn 计算永久值的公式的矩阵乘法问题

Question

标签： c++mathmatrix

Answer 1

我认为您确实需要检查两次您对方程式的理解。请记住，每个求和或乘法都是一个 for 循环。因此，通过使用原始论文中的符号，您会得到：

#include <vector>
#include <cassert>

using mat = std::vector<std::vector<double>>;

double permanent(const mat& input_matrix, const mat& sign_matrix)
{
    int m = input_matrix.size();
    assert(m > 2);
    assert(input_matrix[0].size() == m);
    assert(sign_matrix.size() == m);
    int cols = sign_matrix[0].size();
    assert(cols == 1 << (m - 1));

    double result = 0;
    for (int t = 0; t < cols; ++t) {
        double delta = 1;
        for (int k = 0; k < m; ++k) {
            delta *= sign_matrix[k][t];
        }
        double p = 1;
        for (int j = 0; j < m; j++) {
            double s = 0;
            for (int i = 0; i < m; i++) {
                s += sign_matrix[i][t] * input_matrix[i][j];
            }
            p *= s;
        }
        result += delta * p;
    }
    return result / cols;
}

int main()
{
    mat A = { 
        { 1, 4, 7, }, 
        { 2, 5, 8, }, 
        { 3, 6, 9, },
    };
    mat sign_mat = {
        { 1,  1,  1,  1, },
        { 1, -1,  1, -1, },
        { 1,  1, -1, -1, },
    };
    auto perA = permanent(A, sign_mat);
}