Discrete Cosine Transform (Algorithm and Program)

15 Jun, 2022
Image Compression : Image is stored or transmitted with having pixel value. It can be compressed by reducing the value its every pixel contains. Image compression is basically of two types : 1. Lossless compression : In this type of compression, after recovering image is exactly become same as that was before applying compression techniques and so, its quality didn’t gets reduced. 2. Lossy compression : In this type of compression, after recovering we can’t get exactly as older data and that’s why the quality of image gets significantly reduced. But this type of compression results in very high compression of image data and is very useful in transmitting image over network.Discrete Cosine Transform is used in lossy image compression because it has very strong energy compaction, i.e., its large amount of information is stored in very low frequency component of a signal and rest other frequency having very small data which can be stored by using very less number of bits (usually, at most 2 or 3 bit). To perform DCT Transformation on an image, first we have to fetch image file information (pixel value in term of integer having range 0 – 255) which we divides in block of 8 X 8 matrix and then we apply discrete cosine transform on that block of data.After applying discrete cosine transform, we will see that its more than 90% data will be in lower frequency component. For simplicity, we took a matrix of size 8 X 8 having all value as 255 (considering image to be completely white) and we are going to perform 2-D discrete cosine transform on that to observe the output. 
Algorithm : Let we are having a 2-D variable named matrix of dimension 8 X 8 which contains image information and a 2-D variable named dct of same dimension which contain the information after applying discrete cosine transform. So, we have the formula dct[i][j] = ci * cj (sum(k=0 to m-1) sum(l=0 to n-1) matrix[k][l] * cos((2*k+1) *i*pi/2*m) * cos((2*l+1) *j*pi/2*n) where ci= 1/sqrt(m) if i=0 else ci= sqrt(2)/sqrt(m) and similarly, cj= 1/sqrt(n) if j=0 else cj= sqrt(2)/sqrt(n) and we have to apply this formula to all the value, i.e., from i=0 to m-1 and j=0 to n-1Here, sum(k=0 to m-1) denotes summation of values from k=0 to k=m-1. In this code, both m and n is equal to 8 and pi is defined as 3.142857. 
C++
Java
C#
// CPP program to perform discrete cosine transform#include <bits/stdc++.h>using namespace std;#define pi 3.142857const int m = 8, n = 8; // Function to find discrete cosine transform and print itint dctTransform(int matrix[][n]){    int i, j, k, l;     // dct will store the discrete cosine transform    float dct[m][n];     float ci, cj, dct1, sum;     for (i = 0; i < m; i++) {        for (j = 0; j < n; j++) {             // ci and cj depends on frequency as well as            // number of row and columns of specified matrix            if (i == 0)                ci = 1 / sqrt(m);            else                ci = sqrt(2) / sqrt(m);            if (j == 0)                cj = 1 / sqrt(n);            else                cj = sqrt(2) / sqrt(n);             // sum will temporarily store the sum of            // cosine signals            sum = 0;            for (k = 0; k < m; k++) {                for (l = 0; l < n; l++) {                    dct1 = matrix[k][l] *                           cos((2 * k + 1) * i * pi / (2 * m)) *                           cos((2 * l + 1) * j * pi / (2 * n));                    sum = sum + dct1;                }            }            dct[i][j] = ci * cj * sum;        }    }     for (i = 0; i < m; i++) {        for (j = 0; j < n; j++) {            printf("%f\t", dct[i][j]);        }        printf("\n");    }} // Driver codeint main(){    int matrix[m][n] = { { 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 } };    dctTransform(matrix);    return 0;} //This code is contributed by SoumikMondal
// Java program to perform discrete cosine transform import java.util.*; class GFG{    public static int n = 8,m = 8;    public static double pi = 3.142857;         // Function to find discrete cosine transform and print it    static strictfp void dctTransform(int matrix[][])    {        int i, j, k, l;          // dct will store the discrete cosine transform        double[][] dct = new double[m][n];          double ci, cj, dct1, sum;          for (i = 0; i < m; i++)        {            for (j = 0; j < n; j++)            {                // ci and cj depends on frequency as well as                // number of row and columns of specified matrix                if (i == 0)                    ci = 1 / Math.sqrt(m);                else                    ci = Math.sqrt(2) / Math.sqrt(m);                                     if (j == 0)                    cj = 1 / Math.sqrt(n);                else                    cj = Math.sqrt(2) / Math.sqrt(n);                  // sum will temporarily store the sum of                // cosine signals                sum = 0;                for (k = 0; k < m; k++)                {                    for (l = 0; l < n; l++)                    {                        dct1 = matrix[k][l] *                               Math.cos((2 * k + 1) * i * pi / (2 * m)) *                               Math.cos((2 * l + 1) * j * pi / (2 * n));                        sum = sum + dct1;                    }                }                dct[i][j] = ci * cj * sum;            }        }          for (i = 0; i < m; i++)        {            for (j = 0; j < n; j++)                System.out.printf("%f\t", dct[i][j]);            System.out.println();        }    }         // driver program    public static void main (String[] args)    {        int matrix[][] = { { 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 } };        dctTransform(matrix);    }} // Contributed by Pramod Kumar
// C# program to perform discrete cosine transformusing System; class GFG{     // Function to find discrete cosine transform and print it  public static void dctTransform(int[,] matrix, int m, int n)  {    double pi = 3.142857;          int i, j, k, l;     // dct will store the discrete cosine transform    double[,] dct = new double[m, n];    double ci, cj, dct1, sum;    for (i = 0; i < m; i++)    {      for (j = 0; j < n; j++)      {         // ci and cj depends on frequency as well as        // number of row and columns of specified matrix        if (i == 0)          ci = 1 / Math.Sqrt(m);        else          ci = Math.Sqrt(2) / Math.Sqrt(m);        if (j == 0)          cj = 1 / Math.Sqrt(n);        else          cj = Math.Sqrt(2) /Math.Sqrt(n);         // sum will temporarily store the sum of        // cosine signals        sum = 0;        for (k = 0; k < m; k++)        {          for (l = 0; l < n; l++)          {            dct1 = matrix[k, l] *              Math.Cos((2 * k + 1) * i * pi / (2 * m)) *              Math.Cos((2 * l + 1) * j * pi / (2 * n));            sum = sum + dct1;          }        }        dct[i,j] = ci * cj * sum;      }    }     for (i = 0; i < m; i++)    {      for (j = 0; j < n; j++)      {        Console.Write(dct[i,j]);      }      Console.WriteLine();    }  }     // Driver code  static void Main()  {    const int m = 8, n = 8;    int[,] matrix = new int[m, n] {      { 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 }    };    dctTransform(matrix, m, n);  }} // This code is contributed by SoumikMondal
Output:  
2039.999878    -1.168211    1.190998    -1.230618    1.289227    -1.370580    1.480267    -1.626942    
-1.167731       0.000664    -0.000694    0.000698    -0.000748    0.000774    -0.000837    0.000920    
1.191004       -0.000694    0.000710    -0.000710    0.000751    -0.000801    0.000864    -0.000950    
-1.230645       0.000687    -0.000721    0.000744    -0.000771    0.000837    -0.000891    0.000975    
1.289146       -0.000751    0.000740    -0.000767    0.000824    -0.000864    0.000946    -0.001026    
-1.370624       0.000744    -0.000820    0.000834    -0.000858    0.000898    -0.000998    0.001093    
1.480278       -0.000856    0.000870    -0.000895    0.000944    -0.001000    0.001080    -0.001177    
-1.626932       0.000933    -0.000940    0.000975    -0.001024    0.001089    -0.001175    0.001298
Complexity Analysis:
Time Complexity: O((nm)2),  We used 4 nested loops in the dctTransform() function.
Auxiliary Space: O(nm),  We used n*m space for storing the discrete cosine transform.
This article is contributed by Aditya Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
SoumikMondal
sainitushar438
Mathematical
Matrix
Mathematical
Matrix
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