Discrete Cosine Transform (Algorithm and Program)
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-1
Here, 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.
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
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