题
IM试图在C中实现一个正向离散的余弦变换(DCT)。代码是通过DCT()函数将单个像素的单个像素传输到转换矩阵,然后通过IDCT返回到原始像素值( ) 功能。请参阅随附的代码。我的输出表格IDCT是连续的244、116、244、116等的连续值,从IDCT值的外观看来,我的程序看起来不正常。每个功能都应该期望吗?显然,在IDCT之后,我应该在原始输入矩阵附近变得很漂亮。
谢谢
# include <stdio.h>
# define PI 3.14
void dct(float [][]); // Function prototypes
void idct(float [][]); // Function prototypes
void dct(float inMatrix[8][8]){
double dct,
Cu,
sum,
Cv;
int i,
j,
u,
h = 0,
v;
FILE * fp = fopen("mydata.csv", "w");
float dctMatrix[8][8],
greyLevel;
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 {
Cu = (1.0);
}
sum = 0.0;
for (i = 0; i < 8; i++) {
for (j = 0; j < 8; j++) {
// Level around 0
greyLevel = inMatrix[i][j];
dct = greyLevel * cos((2 * i + 1) * u * PI / 16.0) *
cos((2 * j + 1) * v * PI / 16.0);
sum += dct;
}
}
dctMatrix[u][v] = 0.25 * Cu * Cv * sum;
fprintf(fp, "\n %f", dctMatrix[u][v]);
}
fprintf(fp, "\n");
}
idct(dctMatrix);
}
void idct(float dctMatrix[8][8]){
double idct,
Cu,
sum,
Cv;
int i,
j,
u,
v;
float idctMatrix[8][8],
greyLevel;
FILE * fp = fopen("mydata.csv", "a");
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 {
Cu = (1.0);
}
// Level around 0
greyLevel = dctMatrix[u][v];
idct = (greyLevel * cos((2 * i + 1) * u * PI / 16.0) *
cos((2 * j + 1) * v * PI / 16.0));
sum += idct;
}
}
idctMatrix[i][j] = 0.25 * Cu * Cv * sum;
fprintf(fp, "\n %f", idctMatrix[i][j]);
}
fprintf(fp, "\n");
}
}
int main() {
float
testBlockA[8][8] = { {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} },
testBlockB[8][8] = {{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255} };
dct(testBlockB);
}
解决方案
在if语句中,简历常数分配中至少有两个错别字:
if (v == 0) {
Cv = 1.0 / sqrt(2.0);
} else {
Cu = (1.0); // << this should be Cv = 1.0
}
不过检查得太好了。使用 德国维基百科 关于余弦变换,遵循代码的工作...我不想花时间弄清楚您如何定义哪种转换常数。我想您需要确保使用正确的常数和逆函数:
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
void dct(float **DCTMatrix, float **Matrix, int N, int M);
void write_mat(FILE *fp, float **testRes, int N, int M);
void idct(float **Matrix, float **DCTMatrix, int N, int M);
float **calloc_mat(int dimX, int dimY);
void free_mat(float **p);
float **calloc_mat(int dimX, int dimY){
float **m = calloc(dimX, sizeof(float*));
float *p = calloc(dimX*dimY, sizeof(float));
int i;
for(i=0; i <dimX;i++){
m[i] = &p[i*dimY];
}
return m;
}
void free_mat(float **m){
free(m[0]);
free(m);
}
void write_mat(FILE *fp, float **m, int N, int M){
int i, j;
for(i =0; i< N; i++){
fprintf(fp, "%f", m[i][0]);
for(j = 1; j < M; j++){
fprintf(fp, "\t%f", m[i][j]);
}
fprintf(fp, "\n");
}
fprintf(fp, "\n");
}
void dct(float **DCTMatrix, float **Matrix, int N, int M){
int i, j, u, v;
for (u = 0; u < N; ++u) {
for (v = 0; v < M; ++v) {
DCTMatrix[u][v] = 0;
for (i = 0; i < N; i++) {
for (j = 0; j < M; j++) {
DCTMatrix[u][v] += Matrix[i][j] * cos(M_PI/((float)N)*(i+1./2.)*u)*cos(M_PI/((float)M)*(j+1./2.)*v);
}
}
}
}
}
void idct(float **Matrix, float **DCTMatrix, int N, int M){
int i, j, u, v;
for (u = 0; u < N; ++u) {
for (v = 0; v < M; ++v) {
Matrix[u][v] = 1/4.*DCTMatrix[0][0];
for(i = 1; i < N; i++){
Matrix[u][v] += 1/2.*DCTMatrix[i][0];
}
for(j = 1; j < M; j++){
Matrix[u][v] += 1/2.*DCTMatrix[0][j];
}
for (i = 1; i < N; i++) {
for (j = 1; j < M; j++) {
Matrix[u][v] += DCTMatrix[i][j] * cos(M_PI/((float)N)*(u+1./2.)*i)*cos(M_PI/((float)M)*(v+1./2.)*j);
}
}
Matrix[u][v] *= 2./((float)N)*2./((float)M);
}
}
}
int main() {
float
testBlockA[8][8] = { {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} },
testBlockB[8][8] = {{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255},
{255, 0, 255, 0, 255, 0, 255, 0},
{0, 255, 0, 255, 0, 255, 0, 255} };
FILE * fp = fopen("mydata.csv", "w");
int dimX = 8, dimY = 8;
int i, j;
float **testBlock = calloc_mat(dimX, dimY);
float **testDCT = calloc_mat(dimX, dimY);
float **testiDCT = calloc_mat(dimX, dimY);
for(i = 0; i<dimX; i++){
for(j = 0; j<dimY; j++){
testBlock[i][j] = testBlockB[i][j];
}
}
dct(testDCT, testBlock, dimX, dimY);
write_mat(fp, testDCT, dimX, dimY);
idct(testiDCT, testDCT, dimX, dimY);
write_mat(fp, testiDCT, dimX, dimY);
fclose(fp);
free_mat(testBlock);
free_mat(testDCT);
free_mat(testiDCT);
return 0;
}
编辑DCT基于Wiki中公式DCT-II的交叉生产。 IDCT基于公式DCT-III的交叉生产,其每个维度为2/N(因为这是文本中提到的DCT-II的倒数)。编辑我很确定逆DCT中的因素应为SQRT(2)而不是您的版本中的1/SQRT(2)。
其他提示
你没有
#include <math.h>
这可能意味着编译器正在假设有关数学功能的事情,例如它们都返回int
. 。请注意,您所调用的所有功能都需要在某个地方声明,C没有“内置” sin()
它不仅仅是内置的 printf()
(对于后者,您正确包括 stdin.h
, , 当然)。
另外,您可以使用 M_PI
包含在内 <math.h>
.
除了先前关于CV常数中错别字的答案(在DCT()和IDCT()函数中)外,您还使用了 逆DCT公式(第二) 错误。您必须在循环中每次都乘以简历和CU。因此,IDCT()的正确代码应为:
void idct(float dctMatrix[8][8]){
double idct,
Cu,
sum,
Cv;
int i,
j,
u,
v;
float idctMatrix[8][8],
greyLevel;
FILE * fp = fopen("mydata.csv", "a");
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); //mistake was here - the same is in dct()
}
greyLevel = dctMatrix[u][v];
// Multiply by Cv and Cu here!
idct = (greyLevel * Cu * Cv *
cos((2 * i + 1) * u * PI / 16.0) *
cos((2 * j + 1) * v * PI / 16.0));
sum += idct;
}
}
// not "* Cv * Cu" here!
idctMatrix[i][j] = 0.25 * sum;
fprintf(fp, "\n %f", idctMatrix[i][j]);
}
fprintf(fp, "\n");
}
}
在这种情况下,输出值接近255、0、255、0,等等。
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