尝试实现dct 8 * 8矩阵的逆 [英] Trying to implement The inverse of a dct 8*8 matrix
问题描述
我设法计算出8 * 8矩阵的dct，但在进行逆运算时遇到了麻烦。谁能看一下这段代码，然后告诉我我现在在做什么。我应该获得与以前完全相同的值，但即时通讯获得不同的值。我正在读取一个csv文件的输入，然后将其放到另一个csv文件中。它在c
I have managed to calculate the dct of an 8*8 matrix and I am having trouble doing the inverse. Can anyone have a look at this code and tell me what I am doing now. I should be getting the exact same values as before but im getting different values. I am reading in the input from a csv file and out putting it to anther csv file. Its programmed in c
void idct_func(float inMatrix[8][8]){
double idct,
Cu,
sum,
Cv;
int i,
j,
u,
v;
float idctMatrix[8][8],
greyLevel;
FILE * fp = fopen("mydata.csv", "r");
FILE * wp = fopen("idct.csv", "w");
fprintf(fp, "\n Inverse DCT");
for (i = 0; i < 8; ++i) {
for (j = 0; j < 8; ++j) {
sum = 0.0;
for (u = 0; u < 8; u++) {
for (v = 0; v < 8; v++) {
if (u == 0)
Cu = 1.0 / sqrt(2.0);
else
Cu = 1.0;
if (v == 0)
Cv = 1.0 / sqrt(2.0);
else
Cv = (1.0);
// Level around 0
greyLevel = idctMatrix[u][v];
idct = (greyLevel * cos((2 * i + 1) * u * M_PI / 16.0) *
cos((2 * j + 1) * v * M_PI / 16.0));
sum += idct;
}
}
idctMatrix[i][j] = 0.25 * Cu * Cv * sum;
fprintf(wp, "\n %f", idctMatrix[i][j]);
}
fprintf(wp, "\n");
}
原始矩阵为：
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255},
{255, 255, 255, 255, 255, 255, 255, 255}};
dct是：
2040 0 -0 0 0 0 -0 -0
0 0 0 0 -0 0 -0 0
-0 0 -0 0 0 0 0 0
0 -0 -0 -0 0 -0 -0 0
0 0 -0 0 -0 -0 -0 0
0 -0 -0 -0 -0 0 -0 -0
-0 -0 -0 0 0 0 0 -0
-0 0 0 0 -0 0 -0 0
计算出的IDct应该与原始ID
the idct calculated should be the same as the original
推荐答案
您正在尝试计算IDCT就地使用 idctMatrix [] [] 作为输入和输出，因此在此过程中将修改输入。错了您需要为输入（或输出）提供一个单独的二维数组。
You're trying to calculate the IDCT in-place, using idctMatrix[][] as both input and output and hence the input is being modified in the process. That's wrong. You need to have a separate 2-dimensional array for input (or output).
看来，与u和v有关的比例因子 Cu 和 Cv 应用于u和v循环之外。
It also appears that the u- and v-dependent scaling factors Cu and Cv are applied outside of the u- and v- loops. That would be wrong too.
编辑：如果您不懂单词，这是代码：
EDIT: Here's the code if you can't understand the words:
#include <stdio.h>
#include <math.h>
#ifndef M_PI
#define M_PI 3.14159265358979324
#endif
void Compute8x8Dct(const double in[8][8], double out[8][8])
{
int i, j, u, v;
double s;
for (i = 0; i < 8; i++)
for (j = 0; j < 8; j++)
{
s = 0;
for (u = 0; u < 8; u++)
for (v = 0; v < 8; v++)
s += in[u][v] * cos((2 * u + 1) * i * M_PI / 16) *
cos((2 * v + 1) * j * M_PI / 16) *
((i == 0) ? 1 / sqrt(2) : 1) *
((j == 0) ? 1 / sqrt(2) : 1);
out[i][j] = s / 4;
}
}
void Compute8x8Idct(const double in[8][8], double out[8][8])
{
int i, j, u, v;
double s;
for (i = 0; i < 8; i++)
for (j = 0; j < 8; j++)
{
s = 0;
for (u = 0; u < 8; u++)
for (v = 0; v < 8; v++)
s += in[u][v] * cos((2 * i + 1) * u * M_PI / 16) *
cos((2 * j + 1) * v * M_PI / 16) *
((u == 0) ? 1 / sqrt(2) : 1.) *
((v == 0) ? 1 / sqrt(2) : 1.);
out[i][j] = s / 4;
}
}
void Print8x8(const char* title, const double in[8][8])
{
int i, j;
printf("%s\n", title);
for (i = 0; i < 8; i++)
{
for (j = 0; j < 8; j++)
printf("%8.3f ", in[i][j]);
printf("\n");
}
}
int main(void)
{
double pic1[8][8], dct[8][8], pic2[8][8];
int i, j;
for (i = 0; i < 8; i++)
for (j = 0; j < 8; j++)
#if 01
pic1[i][j] = 255;
#else
pic1[i][j] = (i ^ j) & 1;
#endif
Print8x8("pic1:", pic1);
Compute8x8Dct(pic1, dct);
Print8x8("dct:", dct);
Compute8x8Idct(dct, pic2);
Print8x8("pic2:", pic2);
return 0;
}
其输出为：
pic1:
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
dct:
2040.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
0.000 0.000 0.000 0.000 0.000 0.000 0.000 0.000
pic2:
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
255.000 255.000 255.000 255.000 255.000 255.000 255.000 255.000
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