There are several ways to solve your problem, each has its own advantages and disadvantages. First I’d like to note that you already got quite a few of responses that do the following: they generate a random number, then check somehow whether it was already used in the array, and if it was already used, … Read more
You need to share the Random() instance across the whole class: public class Numbers { Random randnum; public Numbers() { randnum = new Random(); randnum.setSeed(123456789); } public int random(int i){ return randnum.nextInt(i); } }
If you have a shuffled desk, into which you wish to shuffle a batch of new cards (and you know that none of the cards are duplicates), then I think the following is valid. ForEach card in batch: gap = random(deck.size() + 1) # choose a gap between cards, before first, or after last. deck.insertAt(gap,card) … Read more
Something like a Mersenne Twister (from Boost.Random) is deterministic.
I think this is completely unneccessary, because as the Javadoc you quote clearly states: Default-constructed SecureRandom instances seed themselves. The person who wrote this probably didn’t know that. They might also actually decrease security by forcing a fixed seed length that could be less-than-ideal for the RNG implementation. Finally, assuming the snippet is posted unaltered, … Read more
Well one blink at the implementation would show you that he CAN’T have any reason for that claim at all. Why? Because that’s how the set seed function looks like: synchronized public void setSeed(long seed) { seed = (seed ^ multiplier) & mask; this.seed.set(seed); haveNextNextGaussian = false; } And that’s exactly what’s called from the … Read more
Use this to create random digits with any specified length public string RandomDigits(int length) { var random = new Random(); string s = string.Empty; for (int i = 0; i < length; i++) s = String.Concat(s, random.Next(10).ToString()); return s; }
This basically boils down to a limit and it makes sense. Consider the combinations of numbers between 0 and 10 and count the various differences you can make. For example, there is one combination with a difference of 9 — (0, 9). There are 5 with a difference of 5: [0, 5], [1, 6], [2, … Read more
This page at Wolfram MathWorld discusses how to get a power-law distribution from a uniform distribution (which is what most random number generators provide). The short answer (derivation at the above link): x = [(x1^(n+1) – x0^(n+1))*y + x0^(n+1)]^(1/(n+1)) where y is a uniform variate, n is the distribution power, x0 and x1 define the … Read more
Depending on your use case, another option is the math/rand package. Don’t do this if you’re generating numbers that need to be completely unpredictable. It can be helpful if you need to get results that are reproducible, though — just pass in the same seed you passed in the first time. Here’s the classic “seed … Read more