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copied!<p>There are many ways to do it. The methods suggested by jeff7 are OK, but some drawbacks are:</p> <ul> <li>method 1 have parameters N and M, that you must choose, and you must also convert it to another colorspace.</li> <li>method 2 answered can be very slow, since you should compute a 16.7 Milion bins histogram and sort it by frequency (to obtain the 64 higher frequency values)</li> </ul> <p>I like to use an algorithm based on the <strong>Most Significant Bits</strong> to use in a RGB color and convert it to a 64 color image. If you're using C/OpenCV, you can use something like the function below.</p> <p>If you're working with gray-level images I recommed to use the LUT() function of the OpenCV 2.3, since it is faster. There is a tutorial on how to use LUT to reduce the number of colors. See: <a href="http://opencv.itseez.com/doc/tutorials/core/how_to_scan_images/how_to_scan_images.html">Tutorial: How to scan images, lookup tables...</a> However I find it more complicated if you're working with RGB images. </p> <pre><code>void reduceTo64Colors(IplImage *img, IplImage *img_quant) { int i,j; int height = img->height; int width = img->width; int step = img->widthStep; uchar *data = (uchar *)img->imageData; int step2 = img_quant->widthStep; uchar *data2 = (uchar *)img_quant->imageData; for (i = 0; i < height ; i++) { for (j = 0; j < width; j++) { // operator XXXXXXXX & 11000000 equivalent to XXXXXXXX AND 11000000 (=192) // operator 01000000 >> 2 is a 2-bit shift to the right = 00010000 uchar C1 = (data[i*step+j*3+0] & 192)>>2; uchar C2 = (data[i*step+j*3+1] & 192)>>4; uchar C3 = (data[i*step+j*3+2] & 192)>>6; data2[i*step2+j] = C1 | C2 | C3; // merges the 2 MSB of each channel } } } </code></pre>

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