n:\u00a0 Some number 0..m<\/li>\n<\/ul>\nHere is the pseudocode that you can port to your favorite language:<\/p>\n
f(n,m):\r\n a=5\u03c0n\/3m + \u03c0\/2\r\n r=sin(a) * 192 + 128\r\n r=max(0,min(255,y))\r\n g=sin(a - 2\u03c0\/3) * 192 + 128\r\n g=max(0,min(255,y))\r\n b=sin(a - 4\u03c0\/3) * 192 + 128\r\n b=max(0,min(255,y))\r\n return (((((0xFF << 8) || r) << 8) || g) << 8) || b<\/pre>\nBe sure to pre-calculate constants, such as 5\u03c0\/3, \u03c0\/2, 2\u03c0\/3, and 4\u03c0\/3<\/p>\n
The last line uses bit shifting and bitwise-or, so that you get a 32-bit color value in “ARGB” format:<\/p>\n
0xaarrggbb<\/pre>\n <\/p>\n
<\/p>\n
<\/span>What Are We Trying to Accomplish?<\/span><\/h2>\n![\"\"](\"https:\/\/justinparrtech.com\/JustinParr-Tech\/wp-content\/uploads\/ColorFunction-Figures-01.png\")
<\/p>\n
Given an integer value “n”, we want to build a function “f” that returns a 32-bit color value in “ARGB” format.<\/p>\n
\n- For lower values of “n”, we want to return a color closer to Red<\/li>\n
- As “n” increases, we want to return Yellow, Green, Cyan, Blue, and eventually Magenta<\/li>\n
- We want r, g, and b color values from 0 to 255 (0x00 to 0xFF in hex)<\/li>\n
- We want “f” to return a single 32-bit integer in the form:
\n0xFFrrggbb<\/p>\n