QR 分解cpu程序

1. 代码

Makefile

EXE := hello_qr

SRC_QR = qr_main.c qr_func.c
$(EXE): $(SRC_QR)
        gcc $^ -o $@ -lm

.PHONY: clean
clean:
        -rm -rf $(EXE)

qr_main.c

#include "stdio.h"

int maqr(double* a,int m,int n,double* q);

int main()
{
    int i,j;
    static double q[4][4],a[4][3]={ {1.0,1.0,-1.0},
       {2.0,1.0,0.0},{1.0,-1.0,0.0},{-1.0,2.0,1.0}};

    i=maqr(a,4,3,q);
    if (i!=0)
      { printf("MAT Q IS:\n");
        for (i=0; i<=3; i++)
          { for (j=0; j<=3; j++)
              printf("%7.4f ",q[i][j]);
            printf("\n");
          }
        printf("\n");
        printf("MAT R IS:\n");
        for (i=0; i<=3; i++)
          { for (j=0; j<=2; j++)
              printf("%7.4f ",a[i][j]);
            printf("\n");
          }
        printf("\n");
      }

        return 0;
}

qr_func.c

#include "math.h"
#include "stdio.h"

int maqr(double* a, int m, int n, double* q)
{
        int i,j,k,l,nn,p,jj;
        double u,alpha,w,t;
        if (mu) u=w;
                }
                alpha=0.0;
                for (i=k; i<=m-1; i++)
                {
                        t=a[i*n+k]/u; alpha=alpha+t*t;}
                        if (a[l]>0.0) u=-u;
                        alpha=u*sqrt(alpha);
                        if (fabs(alpha)+1.0==1.0)
                        {
                                printf("fail\n"); return(0);
                        }
                                u=sqrt(2.0*alpha*(alpha-a[l]));
                                if ((u+1.0)!=1.0)
                                {
                                        a[l]=(a[l]-alpha)/u;
                                        for (i=k+1; i<=m-1; i++)
                                        {
                                                p=i*n+k;
                                                a[p]=a[p]/u;
                                        }
                                        for (j=0; j<=m-1; j++)
                                        {
                                                t=0.0;
                                                for (jj=k; jj<=m-1; jj++)
                                                        t=t+a[jj*n+k]*q[jj*m+j];
                                                for (i=k; i<=m-1; i++)
                                                {
                                                        p=i*m+j;
                                                        q[p]=q[p]-2.0*t*a[i*n+k];
                                                }
                                        }
                                        for (j=k+1; j<=n-1; j++)
                                        {
                                                t=0.0;
                                                for (jj=k; jj<=m-1; jj++)
                                                        t=t+a[jj*n+k]*a[jj*n+j];
                                                for (i=k; i<=m-1; i++)
                                                {
                                                        p=i*n+j;
                                                        a[p]=a[p]-2.0*t*a[i*n+k];
                                                }
                                        }
                                        a[l]=alpha;

                                        for (i=k+1; i<=m-1; i++)
                                                a[i*n+k]=0.0;
                                }
                }
                for (i=0; i<=m-2; i++)
                        for (j=i+1; j<=m-1;j++)
                        {
                                p=i*m+j;
                                l=j*m+i;
                                t=q[p];
                                q[p]=q[l];
                                q[l]=t;
                        }
                return(1);
}

2. 运行效果