{"id":696,"date":"2014-07-12T18:07:47","date_gmt":"2014-07-12T17:07:47","guid":{"rendered":"http:\/\/blog.blockos.org\/?p=696"},"modified":"2014-09-21T17:04:15","modified_gmt":"2014-09-21T16:04:15","slug":"10-tons-histogram","status":"publish","type":"post","link":"https:\/\/blog.blockos.org\/?p=696","title":{"rendered":"10 tons histogram."},"content":{"rendered":"

What is an image histogram?
\nWell, it is fairly simple. Consider we have a grayscale image. Its histogram is an array that stores the number of occurence of each grey level. It gives us a representation of the tone distribution.<\/p>\n

\"Lena<\/a>

Source image and its histogram.<\/p><\/div>\n

If an image is dark, there will be a peak near the lowest tone values. Or the other hand brighter image will have peaks near the maximum tone values.<\/p>\n

\"Dark<\/a>

Dark image<\/p><\/div>\n

\"Bright<\/a>

Bright image with associated histogram.<\/p><\/div>\n

The image histogram can also give us informations about its contrast. An image with a low contrast will have a peak around its median value and most of the histogram values will be concentrated around that peak. Alternatively a highly contrasted image will have peaks around its minimal and maximal tone value as long as sparse low values in between.<\/p>\n

\"Low<\/a>

Low contrast image<\/p><\/div>\n

\"High<\/a>

High contrast image<\/p><\/div>
\nIt seems logical that if we stretch the histogram over the whole tone range, we will increase the image constrast. This operation is called histogram equalization<\/strong>. Before going any further, let’s introduce some basic definitions. The first one is the probability density function<\/strong>:<\/p>\n

\\(P^{[l_s,l_f]}_l = \\frac{h_l}{H^{[l_s,l_f]}}\\)<\/p>\n

With \\(0 \\le l_s < l_f \\le l_{max}[\/latex]. Usually [latex]l_{max} = 255[\/latex].<\/p>\n

[latex]H^{[m,n]}\\) is introduced for brevity. It’s the sum of the histogram value over the range \\([m,n]\\).<\/p>\n

\\(H^{[m,n]} = \\sum_{i=m}^{n}h_i\\)<\/p>\n

Next is the probability distribution function<\/strong>.<\/p>\n

\\(C^{[l_s,l_f]}_l = \\frac{\\sum_{i=l_s}^{l}h_i}{H^{[l_s,l_f]}}\\)<\/p>\n

Which can be rewritten as:<\/p>\n

\\(C^{[l_s,l_f]}_l = \\sum_{i=l_s}^{l}P^{[l_s,l_f]}_i\\)<\/p>\n

The mean of the histogram values is also called the image brightness.<\/p>\n

\\(l_m = \\frac{\\sum_{i=l_s}^{l_f}i \\times h_i}{H^{[l_s,l_f]}}\\)<\/p>\n

To put it simply,<\/p>\n

\\(l_m = \\sum_{i=l_s}^{l_f}i P^{[l_s,l_f]}_i\\)<\/p>\n

The aim of histogram equalization is to produce a histogram where each tone has the same number of occurence (uniform distribution). This is equivalent to saying that any tone \\(l\\) has the same density.<\/p>\n

\\(P’^{[l_s,l_f]}_{l} = \\frac{1}{l_f – ls}\\)<\/p>\n

Which gives us the ideal probability distribution function.<\/p>\n

\\(C’^{[l_s,l_f]}_{l} = \\sum_{i=l_s}^{l}P’^{[l_s,l_f]}_i = \\frac{l-ls}{lf – ls}\\)<\/p>\n

A way to acheive this is to find \\(l’\\) that minimizes the difference between \\(C’^{[l_s,l_f]}_{l’}\\) and \\(C^{[l_s,l_f]}_{l}\\). In other words, we want to find \\(l’\\) so that:<\/p>\n

\\(C^{[l_s,l_f]}_{l} = C’^{[l_s,l_f]}_{l’}\\)<\/p>\n

Which gives us:<\/p>\n

\\(l’ = l_s + \\big\\langle(l_f – l_s) C^{[l_s,l_f]}_l \\big\\rangle\\)<\/p>\n

Where \\(\\langle x \\rangle\\) is the nearest integer to \\(x\\).<\/p>\n

This technique is sometimes referred as the classical histogram equalization<\/strong> or CHE<\/strong>. The drawback of this method is that it changes the image brightness.<\/p>\n

The following source code is performing CHE<\/strong> with \\(ls = 0\\) and \\(l_f = l_{max}\\).
\nThis code is using the CImg<\/a> library.<\/p>\n

\r\n    \/\/ We will load a dark grayscale version of the \r\n    \/\/ classical lena image.\r\n    CImg<unsigned char> source;\r\n    source.load("dark_lena.png");\r\n\t\r\n    unsigned long int lut[256];\r\n    unsigned long int histogram[256];\r\n\r\n    \/\/ Build image histogram.\r\n    memset(histogram, 0, sizeof(unsigned long int)*256);\r\n    cimg_forXY(source, x, y)\r\n    {\r\n        histogram[source(x,y,0,0)]++;\r\n    }\r\n\r\n    \/\/ Compute probality distribution function.\r\n    unsigned long int sum = 0;\r\n    for(int i=0; i<256; i++)\r\n    {\r\n        sum += histogram[i];\r\n        lut[i] = floor(255.0 * sum \/ (source.width() * source.height()) + 0.5);\r\n    }\r\n    \r\n    \/\/ Perform CHE\r\n    cimg_forXY(source, x, y)\r\n    {\r\n        source(x,y,0,0) = lut[(int)source(x, y, 0, 0)];\r\n    }\r\n  \r\n<\/pre>\n
\"CHE\"<\/a>
Result of CHE source code.<\/p><\/div>\n
References:<\/h2>\n