Java Math.random period

Question

I'm working on a big school project about random numbers, but I can't find the period for Math.random() . I have version 7.0.800.15 installed and I'm working an a windows 10 computer. I've tried determining the period with a simple program which saves the first values of:

double num = Math.random(); 

in an array and then loops until it finds the same values in a row again, thus a period would have passed, but with no result, the period is too long.

So my question is: What is the period of Math.random() on my version? Or: Is there a way to determine the period using a simple program?

Edit: took away a source pointing to a page about JavaScript, it was not relevant

Answer 1

Java's Math.Random uses a linear congruential generator with a modulus of 2^48. The period of such pseudorandom generator with well-chosen parameters is equal to the modulus. Apparently the parameters in Java are sanely chosen, so in practise the period is 2^48.

Sources: https://en.wikipedia.org/wiki/Linear_congruential_generator http://www.javamex.com/tutorials/random_numbers/java_util_random_algorithm.shtml#.WKX-gRJ97dQ

Answer 2

The wiki on linear congruential generator cites Java (java.util.Random) as having a modulus of 2 ⁴⁸ . That is likely the period but you may need to read more about these types of random generators.

This question ( How good is java.util.Random? ) also cites the same period.

Answer 3

Just to add to the other answers and to comment a little more generally on random number generators and writing a program to determine what the period is, beware of the Birthday Paradox and the Gambler's Fallacy . If you generate some value x , the next number is still just as likely to be x as any other number, and the number of numbers you need to generate before you're likely to have a duplicate is actually surprisingly small (meaning that you could, in principle, start seeing some duplicates before the end of the period, which complicates writing a program to test this).

The probability of a duplicate for probabilities up to 50% or so can be approximated by sqrt(2m * p(n)) where p(n) is the probability you're trying to calculating and m is the number of choices. For a 32-bit integer, sqrt(2m * p(n)) = sqrt(2 * 2^32 * 0.5) = sqrt(2^32) = 65,536 . There you have it - once you generate 65,536 numbers there's approximately a 50-50 chance you've generated a duplicate.

Once you've generated 2^32 + 1 values, the Pigeonhole Principle specifies that you must have generated at least one duplicate (assuming, of course, that you're generating a 32-bit number).

You may also be interested in this question on whether you can count on random numbers to be unique.