LDPC生成器矩阵使用减少的行梯形式 [英] LDPC generator matrix using reduced row-echelon form

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问题描述





我的代码仅适用于小矩阵。但实际上我需要代码才能使用更大的矩阵



Hi,

My code is working for small matrix only. But actually i need codes to work for Bigger matrix

From H matrix i need to get Systematic matrix

Systematic matrix, Hsys = [P/Identity]
Generator Matrix =[Identity/PTranspose]




Actual Source Code







#include <stdio.h>
#include <stdlib.h>
 
int main()
{
 
int i,j,sum=0,k,r2,j2,i2;
 
int r,c,n;
int temp=0;
 
k = 3;
r = 3;
c = 6;
n = 6;
int msg[3] = {0,1,1};
int H[3][6] = {1,1,0,0,1,0,1,0,0,1,0,1,1,1,1,0,0,1};
 
 
printf("H Matrix is: \n\n");
for(i=0;i<r;i++)
{
    for(j=0;j<c;j++)
        printf("%d\t",H[i][j]);
    printf("\n");
}
 
for(i=0;i<n-k;i++)  
{
    j=n-k+i;  
        if(H[i][j] != 1) 
        {
            for(i2=i+1;i2<r;i2++)
            {
                if(H[i2][j] == 1) 
                {
                    for(j2=0;j2<c;j2++)  
                    {
                        temp = H[i2][j2];
                        H[i2][j2] = H[i][j2];
                        H[i][j2] = temp;
                    }
                    break;
                }
                if(i2 == r-1)
                    printf("\nERROR..!! The whole of column %d has NO 1(at row %d)",j+1,i+1);
            }
        }
        for(i2 = 0;i2<r;i2++)  
        {
            if(i2 != i && H[i2][j] == 1) 
                {
                    for(j2=0;j2<c;j2++)  
                    {
                        H[i2][j2] = abs(H[i][j2] - H[i2][j2]); 
                    }
                }
        }
}
printf("\nH Matrix in different form is: \n\n");
for(i=0;i<r;i++)
{
    for(j=0;j<c;j++)
        printf("%d\t",H[i][j]);
    printf("\n");
}
 
printf("\n\nGenerator Matrix\n\n");
 
int G[10][10] = {0};
for (i=0;i<k;i++)
    for(j=0;j<k;j++)
        if(i == j)
        G[i][j] = 1;
 
for(i=0;i<r;i++)
    for(j=0;j<k;j++)
        G[j][k+i] = H[i][j];
 
for(i=0;i<r;i++)
{
    for(j=0;j<c;j++)
        printf("%d\t",G[i][j]);
    printf("\n");
}
 
int C[10];
int s = 0;
for(j=0;j<n;j++)
{
    for(i=0;i<k;i++)
    {
        s = s+msg[i]*G[i][j];
    }
    C[j] = s%2;
    s = 0;
}
printf("\n\nCode Word is: \n\n");
for(i=0;i<n;i++)
    printf("%d\t",C[i]);
}





我的尝试:





What I have tried:

I have tried Matrix[3][6], ya it`s working fine.

Now my requirement to move into Bigger matrix [48][96]. My code is not working for Bigger matrix. 







{0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
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推荐答案

不工作是什么意思?



但是,如果你将尺寸从3,6改为48,96你必须调整所有受影响的值。所以我建议第一步用定义替换文字数字以确保所有设置都正确:

What does "is not working" mean?

However, if you change the dimension from 3,6 to 48,96 you have to adjust all affected values. So I suggest as first step to replace the literal numbers by defines to ensure that all are set properly:
#define DIM 3 /* 48 */
#define DIM2 (2 * DIM)

k = DIM;
r = DIM;
c = DIM2;
n = DIM2;
int msg[DIM] = {0,1,1};
int H[DIM][DIM2] = {1,1,0,0,1,0,1,0,0,1,0,1,1,1,1,0,0,1};

/* ... */

int G[DIM][DIM2] = {0};

/* ... */

int C[DIM2];

请注意,我也使用了<$ c $的定义c> G 和 C 代码中大小为10的数组。如果您没有更改这些代码,则代码将无法按预期工作。

Note that I have used the defines also for the G and C arrays which have sizes of 10 in your code. If you have not changed those the code won't work as expected.


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