Perform Histogram Equalization on the Gray level distribution shown in the table. Draw the histograms of the original and equalized images. Comment on the contrast of the image before because grain equalization and after histogram equalization.

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Gray Levels } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { No. of Pixels } & 0 & 100 & 400 & 50 & 200 & 50 & 200 & 0 \\ \hline \end{array} $$

Solution:

$ n=\sum n_x=1000\\ $

$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Gray lenel } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline n_k & 0 & 100 & 400 & 50 & 200 & 50 & 200 & 0 \\ \hline \hline P_f=\frac{n_K}{n} & 0 & 0.1 & 0.4 & 0.05 & 0.2 & 0.05 & 0.2 & 0 \\ \hline \text { CDF }\left(S_k\right) & 0 & 0.1 & 0.5 & 0.55 & 0.75 & 0.8 & 1 & 1 \\ \hline(L-1) S_K & 0 & 0.7 & 3.5 & 3.85 & 5.25 & 5.6 & 7 & 7 \\ \hline \text { Rounding } & 0 & 1 & 4 & 4 & 5 & 6 & 7 & 7 \\ \hline & & & & & & & & \\ \hline \end{array} $

$ \begin{array}{|c|c|c|} \hline \text { old gray level } & \text { No. of pixel } & \text { New gray level. } \\ \hline 0 & 0 & 0 \\ \hline 1 & 100 & 1 \\ \hline 2 & 400 & 4 \\ \hline 3 & 50 & 4 \\ \hline 4 & 200 & 5 \\ \hline 5 & 50 & 6 \\ \hline 6 & 200 & 7 \\ \hline 7 & 0 & 7 \\ \hline \end{array} $

$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { New graylerel } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { eq. No. of pixels } & 0 & 100 & 0 & 0 & 450 & 200 & 50 & 200 \\ \hline \end{array} $

In original histogram the pixels are evenly distributed so contrast is less.

Whereas in equalized image histogram the maximum pixels are on the higher side.