使用矩阵,加速框架,iOS

Question

我有两个矩阵:A和B.

1. 我该如何存放？
2. 如何使用Accelerate框架计算矩阵A的逆矩阵？
3. 如何找到A*B的产品？
4. 如何使用Accelerate框架转置矩阵A？

谢谢你回答我的问题!

帮助文件

#import <Foundation/Foundation.h>
#include <Accelerate/Accelerate.h>

@interface Working_with_matrices : NSObject
-(int)invert_matrix:(int) N andWithMatrix:(double*) matrix;
@end

实施文件

#import "Working_with_matrices.h"
#include <Accelerate/Accelerate.h>

@implementation Working_with_matrices
-(int) matrix_invert:(int) N andWithMatrix:(double*)matrix
{    
int error=0;
int *pivot = malloc(N*N*sizeof(int));
double *workspace = malloc(N*sizeof(double));

dgetrf_(&N, &N, matrix, &N, pivot, &error);

if (error != 0) {
    NSLog(@"Error 1");
    return error;
}

dgetri_(&N, matrix, &N, pivot, workspace, &N, &error);

if (error != 0) {
    NSLog(@"Error 2");
    return error;
}

free(pivot);
free(workspace);
return error;
}

从main函数调用我的代码

#import <Foundation/Foundation.h>
#import "Working_with_matrices.h"

int main(int argc, const char * argv[])
{
int N = 3;
double A[9];
Working_with_matrices* wm=[[Working_with_matrices alloc]init];

A[0] = 1; A[1] = 1; A[2] = 7;
A[3] = 1; A[4] = 2; A[5] = 1;
A[6] = 1; A[7] = 1; A[8] = 3;
[wm invert_matrix:N andWithMatrix:A];
//        [ -1.25  -1.0   3.25 ]
// A^-1 = [  0.5    1.0  -1.5  ]
//        [  0.25   0.0  -0.25 ] 
for (int i=0; i<9; i++) 
{
    NSLog(@"%f", A[i]);
}
return 0;
}

Answer 1

我仍然有点使用加速框架,但我会尽我所能.

1. 加速框架期望矩阵作为一维数组传递.因此,如果您有一个4x4矩阵,第一行将放在数组的索引0-3中,第二行将放在索引4-7中,依此类推.
2. 我从来没有这样做,但这个答案看起来是一个很好的起点./sf/answers/792504961/

3. 您要使用的方法是vDSP_mmul单精度或vDSP_mmulD双精度.您可能希望查看它的文档以更好地了解如何使用它,但这是一个让您入门的示例.

   float *matrixA;  //set by you
   float *matrixB;  //set by you
   float *matrixAB; //the matrix that the answer will be stored in
   
   vDSP_mmul( matrixA, 1, matrixB, 1, matrixAB, 1, 4, 4, 4 );
   // the 1s should be left alone in most situations
   // The 4s in order are:
   //     the number of rows in matrix A
   //     the number of columns in matrix B
   //     the number of columns in matrix A and the number of rows in matrix B.