Random Integer Generator

Occupy the pit first. Revise later

I met a question yesterday, Given int Rand (1) = 0 or 1-uniformly distributed, write a function to implement Rand-uniformly distribut Ed

Because the undergraduate time probability did not learn, hung. Come back online A look at the information found there are similar topics-Given rand (5) = {1, 2, 3, 4, 5}, for Rand (7).

The code to ask for Rand7 is

 Public Static int random7 () {       int val = (RANDOM5 ()-1) * 5 + (RANDOM5 ()-1);          return Val < 21? Val% 7 + 1 : Random7 ();}

The idea is actually to map 1-5 of the numbers to 1-7 of the range, as long as it is uniformly distributed on the line, we can not care whether the 1-7 probability is 4%, 2, or 1%, as long as they are equal, the surplus part of us to abandon. For example, the calculation range is 1-10.

• We can take 1, 2, 3, 4, 5, 6, 7, Discard 8-10 (discard the method is to call once again the random function, the program may be dead loop, but from the probability of look dead cycle chance is very small)
• You can also take 4, 5, 6, 7, 8, 9, 10, discard 1-3, but return val-3 when calculating the answer.

The above code is a solution, we can also have other solutions. such as the following,

 Public Static int random7 () {       int val = (RANDOM5 ()-1) + (RANDOM5 ()-1) * 5 + (RANDOM5 ()-1) *;          return val < 119? Val% 7 + 1 : Random7 ();}

The principle is that as long as the number of RANDOM5 () is evenly mapped to a series of more than 7 consecutive numbers.

So back to this topic, Given rand (1) implements Rand (29), where rand1 () = 0 or 1, only two numbers at a time, so we can use the following code to map to 0-31, and then discard 30 and 31:

 Public Static int rand29 () {       int val = rand1 () + RAND1 () * 2 + rand1 () * 4 + RAND1 () * 8 + rand1 () *;            return Val < 29? val:rand29 ();}

Reference:

Http://www.growingwiththeweb.com/2014/03/given-random5-implement-random7.html

http://www.careercup.com/question?id=12426697

Random Integer Generator