#CROSSOUT PS4 CODE#

Nonetheless, here we'll try to fix history in code.Ĭhallenge: write code to detect the red curb, tuning the 1.1 factor to look the best. Philosophically, they say that you are better off taking in events as they have actually happened. Suppose you are visiting Stanford and you park your car here, and get a parking ticket. By adjusting the * factor, we can zero in on the look we want. The larger the value, the higher the bar is set to detect red pixels. The specific value, 1.1, can be tuned by trying different values, until we get the look we want. The fix is to multiply the avg by some factor to make it more restrictive, like this: If the test is (pixel.getRed() > avg) we get all the areas that have every a tiny red cast, which is not restrictive enough. 160), this checks if the red is relatively large compared to the other two colors. Rather than checking the literal value of the red (e.g. Then compare the red value to the average to decide if the pixel is reddish. For each pixel, first compute the average value for that pixel.

This is a better way to select the red parts of the sign. Then try the flip experiment back and forth - awesome!.* 1.1 adjustment factor, more/less restrictive to get look we want.This works very well, picking out pixels with red cast.Or in terms of code, the test will look like (pixel.getRed() > avg) - this will add the 'relative to the other colors' quality that the previous test was missing. So here the first pixel is reddish and the other two are not. The avg gives us a handy way to tell if a color is high relative to the others: A pixel is reddish if the red value is over the avg. Here are the three pixels again, but now the avg value is show as a line drawn across the bars.

Recall this code which, inside a loop, computes the average of the red, blue, and green values and stores that number in a variable 'avg'.Īvg = (pixel.getRed() + pixel.getGreen() + pixel.getBlue())/3 The problem with our earlier strategy was that it just looked at absolute numbers, (red > 160), failing to take into account how red relates to the other two colors. Find the tallest bar relative to the others. Look at the graphs to figure which shade dominates for each pixel. Q: What color is the strongest for each pixel? Suppose these three values are graphed as bar graphs like this: Each pixel has the familiar red/green/blue values. Next we'll improve the test so it can distinguish the red and white parts of the sign. Recall that pure white is (255, 255, 255), so in whitish areas, all three values are high, so our (red > 160) test gets those too - the white letters inside the sign are a clear example of this problem. The problem with that approach is that the red value is high in two kinds of cases - the red part of the sign we want, but also parts of the scene that are near white. Trying to change the stop sign to be blue, the test (red > 160) gets both white and red areas. Flip try flipping - get opposite parts of the image Restrictive: When adjusting a test, is the goal to make it more or less restrictive?.With a >, the 160 is like a hurdle we raise or lower.If we change it to 220 or 250, will the if-test select more or fewer pixels than at 160? What if we set it to 100? 50? Nothing magic about 160 - adjust it to get the result we want.The test will be true or false for each pixel.

Crossout ps4