Imagine an HSV cylinderFor all you Java developers, here are the JavaFX colors:Why not just evenly distribute them through the spectrum, like so:Now, scatter N points randomly within this space.Again, the most obvious solutions of traversing HSL space or RGB space will probably work just fine.I was unable to find this kind of algorithm here. The options object accepts the following properties: hue – Controls the hue of the generated color. The following code is in my onCreate () method: Random rand = new Random(); // Java 'Color' class takes 3 floats, from 0 to 1. float r = rand.nextFloat(); float g = rand.nextFloat(); float b = rand.nextFloat(); Color randomColor = new Color(r, g, b); Deterministic.I mention it because for certain classes of problems, this type of solution can work better than brute force.the following is the unsorted kelly colors according to the order above.Now you should have N points representing N colours that are about as different as possible within the colour space you’re interested in.And here are the RGB values in hex and 8-bit-per-channel representations:There are several color lists to consider:Here’s an idea.
For Kelly’s and Boynton’s list, I’ve already made the conversion to RGB (with the exception of white and black, which should be obvious). Your transparency is probably fixed, so for each color you need to generate three random numbers. So what you can do is start at color 0 (HEX 000000, black), then add a number each time (e.g. Generating random colors using randomColor JavaScript Library. It works by defining a piecewise linear function on the RGB cube. During refinement you can choose whether or not to randomize the search through the color space and (more importantly) change the perceptual thershold to generate more- or less-distinct colors. import java.awt.Color; import java.util.Random; /** * Contains a method to generate N visually distinct colors and helper methods. This defines a square cross section ring within the space.This is trivial in MATLAB (there is an hsv command):I would try to fix saturation and lumination to maximum and focus on hue only.
But I'm probably doing something wrong since I don't know what newTable.Rows.Count is.The range is 0 to 765, but there are 16,777,216 possible combinations between the minimum and the maximum.If you generate numbers in the range of 0 to 31, then multiply the result by 8 you will get numbers in the range 0 to 248 where each number will be at least 8 different than any other number. By avoiding R G and B values that are similar, you ensure that the resulting colors won't be similar. That doesn't avoid creating identical colors - if that's an issue then you need a process to check whether or not a color has already been generated before accepting the new color.I thought there was a way to ensure alpha = 255:The lower the sum of R+B+G, the darker it is:My point here is that's done only ONE time in the function; everything afterward will then use the List(Of Structure) which is in memory.Try the little test program I showed there.I would like to try this but I don't know how you declared newtable?You should be able to select a range of visibly distinct colors and eliminate any which are too dark.ok frank I tried your random maxvalue of 256^3The brightness of the 3 constiuents RGB vary when you look at them individually.
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