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The first solution is to use the java.util.Random class: import java.util.Random; Random rand = new Random(); // Obtain a number between [0 - 49]. int n = rand.nextInt(50); // Add 1 to the result to get a number from the required range. // (i.e., [1 - 50]). n += 1; Another solution is using Math.random():
Java 7+. In Java 1.7 or later, the standard way to do this (generate a basic non-cryptographically secure random integer in the range [min, max]) is as follows: import java.util.concurrent.ThreadLocalRandom; // nextInt is normally exclusive of the top value, // so add 1 to make it inclusive. int randomNum = ThreadLocalRandom.current().nextInt ...
The NSA and Intel’s Hardware Random Number Generator. To make things easier for developers and help generate secure random numbers, Intel chips include a hardware-based random number generator known as RdRand. This chip uses an entropy source on the processor and provides random numbers to software when the software requests them.
To generate a 6-digit number: Use Random and nextInt as follows: Random rnd = new Random(); int n = 100000 + rnd.nextInt(900000); Note that n will never be 7 digits (1000000) since nextInt(900000) can at most return 899999.
With Java 8+ you can use the ints method of Random to get an IntStream of random values then distinct and limit to reduce the stream to a number of unique random values. ThreadLocalRandom.current().ints(0, 100).distinct().limit(5).forEach(System.out::println);
Do note that java.util.Random is actually a pseudo-random number generator based on the rather weak linear congruence formula. You mentioned the need for cryptography; you may want to investigate the use of a much stronger cryptographically secure pseudorandom number generator in that case (e.g. java.security.SecureRandom).
To get a random character containing all digits 0-9 and characters a-z, we would need a random number between 0 and 35 to get one of each character. BigInteger provides a constructor to generate a random number, uniformly distributed over the range 0 to (2^numBits - 1). Unfortunately 35 is not a number which can be received by 2^numBits - 1.
The first (100) is the number I am ADDING to the random one. This means that the new output number will be the random number + 100. numGen.nextInt() is the value of the random number itself, and because I put (100) in its parentheses, it is any number between 1 and 100. So when I add 100, it becomes a number between 101 and 200.
import java.util.Random; If you want to test it out try something like this. Random rn = new Random(); for(int i =0; i < 100; i++) { int answer = rn.nextInt(10) + 1; System.out.println(answer); } Also if you change the number in parenthesis it will create a random number from 0 to that number -1 (unless you add one of course like you have then ...
Random number generators use mathematical formulas that transfer set of numbers to another one. If, for example, you take a constant number N and another number n_0, and then take the value of n mod N (the modulo operator), you will get a new number n_1, which looks as it if is unrelated to n_0. Now, repeat the same process with n_1 and you'll ...