python - 使用英特尔 MKL 计算 `trans(a)*inv(b)*a` 的正确方法（后续问题）

问题描述

这是另一个具有相同标题的后续问题（我对其进行了重大编辑，但有人告诉我这应该是另一个问题 - 我想不出另一个标题）。

yn = trans(a)*inv(zt)*a + trans(b)*inv(zl)*b

其中a和b是 m×n 实矩阵，zt是zlm×m 复数矩阵。生成的复数矩阵yn是 n×n。

这是我的做法：

zt <- inv(zt)
zl <- inv(zl)
c <- zt*a
yn <- trans(a)*c
c <- zl*b
yn <- trans(b)*c + yn

C代码：

#include <math.h>
#include <complex.h>
#include <stdlib.h>
#include <stdio.h>
#include <mkl_types.h>
#define MKL_Complex16 _Complex double //overwrite type
#include <mkl.h>
#include <mkl_lapacke.h>

int print_zmatrix_file(int m, int n, _Complex double* a, int lda, FILE* fp)
{
    int i, j;
    for( i = 0; i < m; i++ )
    {
        for( j = 0; j < n; j++ )
        {
            fprintf(fp, "(%.6f%+.6fj)", creal(a[i*lda+j]), cimag(a[i*lda+j]) );
            if (j < n - 1) fprintf(fp, ",");
        }
        fprintf(fp, "\n");
    }
    return 0;
}

int calc_yn(
    _Complex double* yn, double* a, double *b, _Complex double* zl,
    _Complex double* zt, int m, int n)
{
    lapack_int* ipiv = (MKL_INT*) malloc(sizeof(lapack_int)*m);
    LAPACKE_zgetrf(LAPACK_ROW_MAJOR, m, m, zt, m, ipiv);
    LAPACKE_zgetri(LAPACK_ROW_MAJOR, m, zt, m, ipiv);
    LAPACKE_zgetrf(LAPACK_ROW_MAJOR, m, m, zl, m, ipiv);
    LAPACKE_zgetri(LAPACK_ROW_MAJOR, m, zl, m, ipiv);
    free(ipiv);
    const double alpha = 1.0;
    const double beta = 0.0;
    lapack_complex_double* c = (lapack_complex_double*) malloc(
        sizeof(lapack_complex_double)*(m*n));
    // c <- zt*a
    cblas_zgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans,
                m, n, m,
                &alpha, zt, m, a, n,
                &beta, c, n);
    FILE* fp = fopen("c1.csv", "w");
    print_zmatrix_file(m, n, c, n, fp);
    fclose(fp);
    // yn <- aT*c
    cblas_zgemm(CblasRowMajor, CblasTrans, CblasNoTrans,
                n, n, m,
                &alpha, a, n, c, n,
                &beta, yn, n);
    // c <- zl*b
    cblas_zgemm(CblasRowMajor, CblasNoTrans, CblasNoTrans,
                m, n, m,
                &alpha, zl, m, b, n,
                &beta, c, n);
    FILE* fp2 = fopen("c2.csv", "w");
    print_zmatrix_file(m, n, c, n, fp2);
    fclose(fp2);
    // yn <- bT*c + yn
    cblas_zgemm(CblasRowMajor, CblasTrans, CblasNoTrans,
                n, n, m,
                &alpha, b, n, c, n,
                &alpha, yn, n);
    free(c);
    return 0;
}

int main()
{
    int m = 2;
    int n = 3;
    _Complex double* yn = (_Complex double*) malloc(sizeof(_Complex double)*(n*n));
    double a[] = {
        0.5, 0.0, 0.5,
        0.5, 0.5, 0.0
    };
    double b[] = {
        1.0, 0.0, -1.0,
        1.0, -1.0, 0.0
    };
    _Complex double zt[] = {
        (0.004 + 0.09*I), (-0.004 - 0.12*I),
        (-0.004 - 0.12*I), (0.005 + 0.11*I)
    };
    _Complex double zl[] = {
        (0.1 + 2.13*I), (-124.004 - 800.12*I),
        (-124.004 - 800.12*I), (0.4 + 4.08*I)
    };
    calc_yn(yn, a, b, zl, zt, m, n);
    FILE* fp = fopen("yn.csv", "w");
    print_zmatrix_file(n, n, yn, n, fp);
    fclose(fp);
    free(yn);
    return 0;
}
// compile command (MKLROOT is defined by a bash script that is shipped together with intel's MKL):
//gcc -std=c11 -DMKL_ILP64 -m64 -g -o test.a test.c -L${MKLROOT}/lib/intel64 -Wl,--no-as-needed -lmkl_intel_ilp64 -lmkl_intel_thread -lmkl_core -liomp5 -lpthread -lm -ldl

上一个问题中的代码在mallocto中有错误yn（它使用sizeof(_Complex double*)而不是sizeof(_Complex double)）。纠正该错误后，代码编译并成功运行。运行后，我将结果与我使用 SciPy 得到的结果进行了比较。他们不同意。

import numpy
from scipy import linalg

a = numpy.array([[0.5, 0.0, 0.5],
                 [0.5, 0.5, 0.0]])
b = numpy.array([[1.0, 0.0, -1.0],
                 [1.0, -1.0, 0.0]])
zt = numpy.array([[0.004 + 0.09j, -0.004 - 0.12j],
                  [-0.004 - 0.12j, 0.005 + 0.11j]])
zl = numpy.array([[0.1 + 2.13j, 124.004 - 800.12j],
                  [124.004 - 800.12j, 0.4 + 4.08j]])

c1 = numpy.matmul(linalg.inv(zt), a)
m1 = numpy.matmul(a.T, c1)
c2 = numpy.matmul(linalg.inv(zl), b)
m2 = numpy.matmul(b.T, c2)
yn = m1 + m2

yn_file = numpy.genfromtxt('yn.csv', delimiter=',', dtype=numpy.complex128)
c1_file = numpy.genfromtxt('c1.csv', delimiter=',', dtype=numpy.complex128)
c2_file = numpy.genfromtxt('c2.csv', delimiter=',', dtype=numpy.complex128)

numpy.max(numpy.abs(yn)) #0.004958820819049211
numpy.max(numpy.abs(yn_file)) #60.4590237745794

numpy.max(numpy.abs(c1)) #25.549314567403204
numpy.max(numpy.abs(c1_file)) #41.278805716697306

numpy.max(numpy.abs(c2)) #0.0012411403762584482
numpy.max(numpy.abs(c2_file)) #0.03292682468747935

我的 C 代码或 Python 代码有问题。为什么我得到不同的结果？

编辑：根据@Bwebb 建议进一步测试。他注意到Python 代码中-124.004 - 800.12i出现了复制粘贴错误。+124.004 - 800.12i更正不会改变结果

为了更容易测试，我使用了以下矩阵：

a = numpy.array([[1.0, 0.0],
                 [0.0, 1.0]])
b = numpy.array([[0.0, -1.0],
                 [-1.0, 0.0]])
zt = a
zl = b

这导致

yn = [[1.0, -1.0]
      [-1.0, 1.0]]

Python 代码给出了该结果，但 C 代码给出了

yn = [[0.0 + 2.0j, 1.0 + 2.0j]
      [-1.0 + 2.0j, 0.0 + 0.0j]]

这让我得出结论，C 代码是错误的，但我不知道在哪里。

标签： pythoncscipyintel-mkl

python - 使用英特尔 MKL 计算 `trans(a)inv(b)a` 的正确方法（后续问题）

问题描述

解决方案

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