SESH Posted April 20, 2015 Share Posted April 20, 2015 (edited) This function returns a random number with minimum min and maximum max with normal distribution. More info here public static int rand(int min, int max) { int n; int mean = (min + max) / 2; int std = (max - mean) / 3; Random r = new Random(); do { double val = r.nextGaussian() * std + mean; n = (int) Math.round(val); } while (n < min || n > max); return n; } Edited April 20, 2015 by SESH Quote Link to comment Share on other sites More sharing options...
marselo Posted April 23, 2015 Share Posted April 23, 2015 I love this, I've been planning on writing this. Glad I found your thread man, well done 1 Quote Link to comment Share on other sites More sharing options...
lare96 Posted April 29, 2015 Share Posted April 29, 2015 Not all that great, since that while loop can potentially block forever... private final Random random = new Random(); public int exclusive(int min, int max) { if (max <= min) { max = min + 1; } return random.nextInt((max - min)) + min; } public int exclusive(int range) { return exclusive(0, range); } public int inclusive(int min, int max) { if (max < min) { max = min + 1; } return exclusive((max - min) + 1) + min; } public int inclusive(int range) { return inclusive(0, range); } Quote Link to comment Share on other sites More sharing options...
marselo Posted April 29, 2015 Share Posted April 29, 2015 Not all that great, since that while loop can potentially block forever... private final Random random = new Random(); public int exclusive(int min, int max) { if (max <= min) { max = min + 1; } return random.nextInt((max - min)) + min; } public int exclusive(int range) { return exclusive(0, range); } public int inclusive(int min, int max) { if (max < min) { max = min + 1; } return exclusive((max - min) + 1) + min; } public int inclusive(int range) { return inclusive(0, range); } Well what you did there is too complicated, what you can do is this: static Random r = new Random(); public static int rand (int mean, int stdDev){ return (int) (r.nextGaussian()*stdDev+mean); } Quote Link to comment Share on other sites More sharing options...
lare96 Posted April 29, 2015 Share Posted April 29, 2015 Well what you did there is too complicated, what you can do is this: static Random r = new Random(); public static int rand (int mean, int stdDev){ return (int) (r.nextGaussian()*stdDev+mean); } How is what I did complicated? I gave methods that support both inclusive and exclusive intervals. Quote Link to comment Share on other sites More sharing options...
SESH Posted May 3, 2015 Author Share Posted May 3, 2015 (edited) Not all that great, since that while loop can potentially block forever... private final Random random = new Random(); public int exclusive(int min, int max) { if (max <= min) { max = min + 1; } return random.nextInt((max - min)) + min; } public int exclusive(int range) { return exclusive(0, range); } public int inclusive(int min, int max) { if (max < min) { max = min + 1; } return exclusive((max - min) + 1) + min; } public int inclusive(int range) { return inclusive(0, range); } It would be incredibly, incredibly unlikely for it to even run through the while loop twice. Edited May 3, 2015 by SESH Quote Link to comment Share on other sites More sharing options...