71 lines
1.8 KiB
C
71 lines
1.8 KiB
C
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#include "hsv.h"
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#include <stdio.h>
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#include <stdlib.h>
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#include <math.h>
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#define FAIL -1
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#define OK 0
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int test_conv(int r, int g, int b, double tolerance) {
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vec3 rgb = {r/255.0, g/255.0, b/255.0};
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double hsv[3];
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vec3 out;
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int test[3];
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rgb_to_hsv(rgb, hsv);
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hsv_to_rgb(hsv, out);
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test[0] = out[0]*255.0;
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test[1] = out[1]*255.0;
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test[2] = out[2]*255.0;
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if(fabs(rgb[0] - out[0]) > tolerance || fabs(rgb[1] - out[1]) > tolerance || fabs(rgb[2] - out[2]) > tolerance) {
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fprintf(stderr, "FAIL:\t#%02X%02X%02X -> #%02X%02X%02X\nIN:\t(%.12f %.12f %.12f)\nOUT:\t(%.12f %.12f %.12f)\nDIF:\t(%.12f %.12f %.12f)\n",
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r, g, b,
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test[0], test[1], test[2],
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rgb[0], rgb[1], rgb[2],
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out[0], out[1], out[2],
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rgb[0]-out[0], rgb[1]-out[1], rgb[2]-out[2]);
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return FAIL;
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} else {
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return OK;
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}
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}
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int main() {
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for(int r = 0; r < 256; r++) {
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for(int g = 0; g < 256; g++) {
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for(int b = 0; b < 256; b++) {
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if(test_conv(r, g, b, 0.001) != OK) {
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return -1;
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}
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}
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}
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}
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}
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// To RGB steps:
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// 1. C = V * S <- maps to 3 in inv
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// 2. H' = H * 6 <- maps to 7 in inv
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// 3. X = C * (1 - abs(H' - 2*floor(H'/2) - 1))
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// 4. M = V - C <- maps to 4 in inv
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// 5. R = V, G = X + M, B = M
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// To HSV steps:
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// 1. V = max(R, G, B) <- maps to 5,R in inv
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// 2. M = min(R, G, B) <- maps to 5,B in inv
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// 3. S = C / V <- maps to 1 in inv
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// 4. C = V - M <- maps to 4 in inv
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// 5. X = G - M
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// 6. H' = X/C
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// 7. H = H'/6 <- maps to 2 in inv
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//
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// So out of the two algos, the only step that doest match exactly so far is
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// RGB: X = C * (1 - abs(H' - 2*floor(H'/2) - 1))
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// HSV: X = G - M, H' = X/C
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//
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// After floor():
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// RGB: X = C * (1 - abs(H' - 1))
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//
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// After abs() assuming H' = [0, 1]:
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// RGB: X = C * (1 - (1 - H'))
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// RGB: X = C * H'
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