CMPEN/EE455: Digital Image Processing I. Computer Project # 4:

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1 CMPEN/EE455: Digital Image Processing I Computer Project # 4: Image Enhancement: Histogram Modification Taylor Blackford, Taiwei Yue, Huiyuan Zhou Date: November/02/2015 Objectives The purpose of this project is to realize image enhancement using histogram modification. We investigated various techniques to change the histogram of the original image to meet different requirements. We also developed a function named as cdfer.m ourselves to help us accomplish this project. This function is programed to help us calculate the pdf and cdf of the input image. The original image truck.gif to be processed are shown in Fig. 1. The first problem of this project focuses on obtaining histogram p r(r) and cdf c r (r) of the original image. We solve this problem by means of the function cdfer.m which not only return the values of histogram p r(r) and cdf c r (r) but also plot corresponding pictures of them. Problem 2 asks us to realize the contrast stretching using the method described. To deal with this problem, we first built up the transformation function and then apply it to the original image. In order to solve problem 3 which asks us to apply histogram equalization (HE) processing to original image get a contrast enhanced output image, we follow the procedure taught in class and explained in class note L-19. Problem 4 requires us to apply power-law (gamma) transformation to original image to see the effect of different values of the constant γ on output images. To solve this, we simply change the grey level intensity of every pixel in the image based on the Eq. (3.2-3) in G&W textbook and plot the results. We compared all of the images and histograms and cdfs produced to see how different techniques enhance the images. Details of the method we adopted to solve each problem are presented in the following section.

2 Figure. 1 Original image truck.gif Methods For problem 1, it asks us to compute and plot the image s histogram p r(r) and cdf c r (r). In our project, we programed a function named cdfer.m to fulfill this function. The flow chart for the programed function is shown below as Figure 2. Start read image f False i<=m; j<=n; level<=255 True Finish Hist(level)=Hist(level)+1 i=i+1;j=j+1;level=level+1 Figure. 2. The flow chart of the programed function cdfer.m.

3 We first read the image and mark it as f. Then we build up function hist(x) to count the number of the points have the value of x. Then the function hist(*) gives the histogram of the image f. After we have the function hist, we continue on counting the total number of points that have the intensity value from 0 to x. After that we divide it by (M*N) which, finally, gives us the cdf function. The processing is based on the definition of the histogram and cdf. The flow chart for the cdf is shown below as figure 3. Start read image f Fals i<=m; j<=n; level<=255 Tru cfd(x) = sum(hist(1: x) /(M N) Hist(level)=Hist(level)+ 1 Finish Figure. 3. The flow chart of the cdf programming. For problem 2, we apply the method mentioned in the Lecture 18. The transformation is called piecewise-linear transformation s = T(r). And the figure for the transformation is shown in Fig. 4 below. In Fig. 5, we show the flow chart of the programming for Problem 2. s L-1 L1 L2 L-1 r Figure. 4. The plot of piecewise linear transformation

4 Start read image f Calculate cdf Cdf(i)<=0.15 Cdf(i+1)>.15 Cdf(i)<0.85 Cdf(i+1)>=.85 True True L1=i L2=i+1 Calculate parameter m=(l-1)/(l2-l1) B=-m*L1 f(m,n)<l1 f(m,n)>l2 L1<=f(m,n)<=L2 f_p2=0 f_p2=255 f_p2=m*f(m,n)+b f_p2 Finish Figure. 5 The flow chart for problem 2

5 Based on the lecture, we have the math equation for the piecewise linear transformation which is: 0, 0 r < L 1 T(r) = { mr + b, L 1 r L 2 L 1, L 2 < r L 1 where m = L 1 L 2 L 1 and b = ml 1. According to the requirements of this problem, all pixels with grey levels below 15% in the cdf to black which means set their value as 0. Setting all pixels with grey level above 85% in the cdf to white which means giving the value to 255. After appropriate calculating, we ve obtained the value for L 1 and L 2 as L 1 = 71 and L 2 = 127. Meanwhile, the values for m and b are m = , b = e + 02, respectively. To solve problem 3, we first compute the pdf p r (r) and the cdf c r (r) of the original image. Then, as illustrated on class notes L19-5, we construct the transformation based on the following equation: s = [(L 1) c r (r)] (1) where the bracket means to find the nearest integer value around (L 1) c r (r). To better illustrate the algorithm involved in our code, the flow chart of problem 3 is shown in Fig. 6 (a). The transformation function s = T(r) is presented in Fig. 6 : (*) (a) Figure. 6 (a) flow chart for problem 3; the histogram equalization transformation function.

6 (a) (c) Figure. 7 (a) flow charts of problem 4; transformation function with γ = 5 and (c) γ = 0.2. To solve problem 4, following the Eq. (3.2-3) in the textbook (shown as the following Eq. (2) in our report), we first calculate the appropriate constant c to make the maximum grey level in the transformed new image according to Eq. (3) where L 1 is the largest grey level and max(r) is to find the maximum grey level value among pixels in the original image. This is to make sure that the transformed new image has the maximum grey level value of L 1 and to make the new image easy to plot. s = cr γ (2) c = (L 1)/max (r) γ (3) Then, we set γ to be equal to 5 and 0.2 respectively to construct the transformation equation. To better illustrate our algorithm, a flow chart is shown in Fig. 7 (a) and the corresponding transformation functions are shown in Fig. 7 and (c) with γ = 5 and γ = 0.2, respectively.

7 Results and Analysis Problem 1 In Fig. 8 (a) and we show the initial image of the truck and the histogram and cdf of the truck image, respectively. The initial truck image is a 256 by 256 image 8-bit digital image. Based on Fig. 8, most of the gray levels of pixels in the image locate around 127, and very little number of pixels intensity values are located higher than 150. As the result, the cdf of the image already reach 1 around 150. It should be noted that the very narrow white lines in the histogram and cdf pictures are due to image distortion caused by out computer when we plot the picture in Matlab. (a) Figure. 8 (a) the initial image of truck; the histogram and cdf of initial image of truck. Problem 2 For problem 2, the results are shown below in Fig. 9. Fig. 9 (a) shows the initial image of truck and Fig. 9 gives the image of f_p2 which is the stretched output image with value of the gray level of each pixel directly calculated by the math function s = T(r). However, the discrete image should have an integer value for gray level. In our code, we use floor function to find the corresponding integer value. And the result with interger gray level is shown in Fig. 9 (c). It is clear that and (c) do not

8 have big difference. At the same time, it is quite obvious that the result enhance the contrast of the initial image. However, we also notice that the enhanced image lose some details. The histogram and the cdf function for the enhanced image is shown below as Fig. 10. Due to the transformation function we construct for the image enhancement, the cdf of the result will not start from 0, which is due to the fact that in the new image all the pixels with gray level below 15% in the cdf of the initial image are all set to be equal to 0. And the new cdf does not reach 1 smoothly, which is because we set the all pixels in the initial image to 1 when their gray level above 85% in the cdf. As we can see the cdf in Fig. 10 at 255 has a sharp increasing. Another obvious difference is that the slope of cdf in Fig. 10 is much smoother compared with that in Fig. 10 (a) of the original image. (a) (c) Figure. 9. (a) the initial image of truck; the image of f_p2; (c) the image of ff. (a) Figure. 10. (a) the histogram and cdf for initial image; the histogram and cdf for enhanced image.

9 Problem 3 The results of problem 3 are shown in Fig. 11. Fig. 11 (a) shows the output image after applying HE on the original one and Fig. 11 shows the corresponding histogram p r(r) and cdf c r (r) of the output image. Comparing Fig. 8 and Fig. 11 we can see that instead of concentrating with the grey level interval from around 50 to 150, in the new image the histogram spread within a much wider range which is from 0 to 255. This behavior demonstrate the effectiveness of the HE transformation. However, just as illustrated in the textbook, the discrete HE cannot yield a perfect uniform histogram p r(r) like its continuous counterpart. Even though, by comparing Fig. 8 (a) and Fig. 11 (a) we can see an obvious image contrast enhancement. (a) Figure. 11 (a) the image after histogram equlization; histogram and cdf of the new image. Problem 4

10 (a) Figure. 12 (a) The image after histogram equlization; histogram and cdf of the new image. (a) Figure. 13 (a) The image after histogram equlization; histogram and cdf of the new image. The results of problem 4 are presented in Fig. 12 and Fig. 13 with the constant γ = 5 and γ = 0.2, respectively. We can easily tell from comparing Fig. 12 (a), and Fig. 13 (a) that when γ = 5 the image looks much darker than that when γ = 0.2. This phenomenon can be explained by considering the corresponding transformation functions and histograms. From Fig. 7 and Fig. 7 (c) we can see that when γ = 5, the transformation tries to map intensity value located at higher region to lower region. In contrast, when γ = 0.2, the transformation tends to map intensity value located at lower region to higher region. This can be also illustrated by comparing histograms shown in Fig. 12 and Fig. 13. we can see that when γ = 5, the power-law transformation tends to push histogram towards lower region

11 which is approximately from 0 to 100. On the contrary, when γ = 0.2, the transformation tends to push histogram to higher region from around 200 to 255. Problem 5 In problem 1, we practice programming our own function cdfer.m to plot the histogram p r(r) and cdf c r (r) of the original image. The purpose of problem 2 is to apply piecewise-linear transformation to the original image to realize the contrast stretching task. We started by constructing the mathematical function of the transformation. Based on this function, then, we calculate the intensity value of each pixels and finally we plot the results. By comparing the output image to the original one, we can see an obvious improvement in terms of the image contrast. In problem 3 we use HE to accomplish the contrast enhancement of the original image. The results shows that the histogram of the output image spread over a much larger range compared to that of the original one with almost uniform shape. Although, the perfect uniform histogram cannot obtained by discrete HE, we can still see a good contrast enhancement by comparing the output image with the original one. In problem 4, we investigate features of the power-law (gamma) transformation. Specifically, we investigate the effect of different values of constant γ on the output image. From the results we can see that when the value of γ is below 1, the transformation tend to push histogram of the image towards high intensity value region resulting in a brighter looking output image. In contrast, if the value of γ is above 1, a darker looking output image is obtained due the fact that the transformation now push the histogram to the region with low intensity values. Conclusions From the results of problem 1, we see that a majority of the grey levels are centered around 127 bits, and most do no extend past 150. Almost 1/3 of the grey scale is unused! The cdf shows a moderate level of gain over the range of used grey levels. Problem 2 stretched the used grey scale. We first set the upper 15% and lower 15% to their respective extreme. Then we stretched the inner 70% to cover the full 0 to 255 range. This new image has much higher contrast that the original, but there is a loss of detail on the truck. The new histogram looks more

12 discrete than continuous, which makes sense since it was stretched from a smaller range. The cdf shows a good rate at which it increases. Overall this method increases the contrast greatly. Problem 3 equalized the image. At first glance, the image also shows increased contrast compared to the original, but also shows the detail on the truck. Looking at the histogram it also looks somewhat discrete like the previous part, but more like the original curve. The cdf is almost perfectly linear with a constant gain, which shows it was correctly equalized. This method had the best increase of contrast while still keeping necessary details. Problem 4 applied two different gamma correction coefficients to the image. Appling a gamma coefficient simply raises the pixels grey level to the gamma power. The first coefficient is 5, which causes the image to be much darker, which the second coefficient,.2, brightens the image greatly. Neither of these corrections do much for the contrast, but mostly help with the brightness of the image. Overall the equalization of the image did the best job and increasing the contrast. It made larger components easier to detect while keeping the details inside them visible also. However, there are also other applications of histogram modifications other than contrast enhancement depending on the specific purpose of image processing.

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