without How to generate random integers within a specific range in Java?




random number generator java without repetition (24)

How do I generate a random int value in a specific range?

I have tried the following, but those do not work:

Attempt 1:

randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.

Attempt 2:

Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.

Let us take an example.

Suppose I wish to generate a number between 5-10:

int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);

Let us understand this...

Initialize max with highest value and min with the lowest value.

Now, we need to determine how many possible values can be obtained. For this example, it would be:

5, 6, 7, 8, 9, 10

So, count of this would be max - min + 1.

i.e. 10 - 5 + 1 = 6

The random number will generate a number between 0-5.

i.e. 0, 1, 2, 3, 4, 5

Adding the min value to the random number would produce:

5, 6, 7, 8, 9, 10

Hence we obtain the desired range.


It's better to use SecureRandom rather than just Random.

public static int generateRandomInteger(int min, int max) {
    SecureRandom rand = new SecureRandom();
    rand.setSeed(new Date().getTime());
    int randomNum = rand.nextInt((max - min) + 1) + min;
    return randomNum;
}

Just use the Random class:

Random ran = new Random();
// Assumes max and min are non-negative.
int randomInt = min + ran.nextInt(max - min + 1);


If you want to try the answer with the most votes above, you can simply use this code:

public class Randomizer
{
    public static int generate(int min,int max)
    {
        return min + (int)(Math.random() * ((max - min) + 1));
    }

    public static void main(String[] args)
    {
        System.out.println(Randomizer.generate(0,10));
    }
}

It is just clean and simple.


Use:

minimum + rn.nextInt(maxValue - minvalue + 1)

Here is a simple sample that shows how to generate random number from closed [min, max] range, while min <= max is true

You can reuse it as field in hole class, also having all Random.class methods in one place

Results example:

RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert

Sources:

import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}

This methods might be convenient to use:

This method will return a random number between the provided min and max value:

public static int getRandomNumberBetween(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt(max - min) + min;
    if (randomNumber == min) {
        // Since the random number is between the min and max values, simply add 1
        return min + 1;
    } else {
        return randomNumber;
    }
}

and this method will return a random number from the provided min and max value (so the generated number could also be the min or max number):

public static int getRandomNumberFrom(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt((max + 1) - min) + min;

    return randomNumber;
}

Here's a helpful class to generate random ints in a range with any combination of inclusive/exclusive bounds:

import java.util.Random;

public class RandomRange extends Random {
    public int nextIncInc(int min, int max) {
        return nextInt(max - min + 1) + min;
    }

    public int nextExcInc(int min, int max) {
        return nextInt(max - min) + 1 + min;
    }

    public int nextExcExc(int min, int max) {
        return nextInt(max - min - 1) + 1 + min;
    }

    public int nextIncExc(int min, int max) {
        return nextInt(max - min) + min;
    }
}

int random = minimum + Double.valueOf(Math.random()*(maximum-minimun)).intValue();

Or take a look to RandomUtils from Apache Commons.


The Math.Random class in Java is 0-based. So, if you write something like this:

Random rand = new Random();
int x = rand.nextInt(10);

x will be between 0-9 inclusive.

So, given the following array of 25 items, the code to generate a random number between 0 (the base of the array) and array.length would be:

String[] i = new String[25];
Random rand = new Random();
int index = 0;

index = rand.nextInt( i.length );

Since i.length will return 25, the nextInt( i.length ) will return a number between the range of 0-24. The other option is going with Math.Random which works in the same way.

index = (int) Math.floor(Math.random() * i.length);

For a better understanding, check out forum post Random Intervals (archive.org).


Use:

Random ran = new Random();
int x = ran.nextInt(6) + 5;

The integer x is now the random number that has a possible outcome of 5-10.


Another option is just using Apache Commons:

import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;

public void method() {
    RandomData randomData = new RandomDataImpl();
    int number = randomData.nextInt(5, 10);
    // ...
 }

When you need a lot of random numbers, I do not recommend the Random class in the API. It has just a too small period. Try the Mersenne twister instead. There is a Java implementation.


private static Random random = new Random();    

public static int getRandomInt(int min, int max){
  return random.nextInt(max - min + 1) + min;
}

OR

public static int getRandomInt(Random random, int min, int max)
{
  return random.nextInt(max - min + 1) + min;
}

You can edit your second code example to:

Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum =  rn.nextInt(range) + minimum;

With java-8 they introduced the method ints(int randomNumberOrigin, int randomNumberBound) in the Random class.

For example if you want to generate five random integers (or a single one) in the range [0, 10], just do:

Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();

The first parameter indicates just the size of the IntStream generated (which is the overloaded method of the one that produces an unlimited IntStream).

If you need to do multiple separate calls, you can create an infinite primitive iterator from the stream:

public final class IntRandomNumberGenerator {

    private PrimitiveIterator.OfInt randomIterator;

    /**
     * Initialize a new random number generator that generates
     * random numbers in the range [min, max]
     * @param min - the min value (inclusive)
     * @param max - the max value (inclusive)
     */
    public IntRandomNumberGenerator(int min, int max) {
        randomIterator = new Random().ints(min, max + 1).iterator();
    }

    /**
     * Returns a random number in the range (min, max)
     * @return a random number in the range (min, max)
     */
    public int nextInt() {
        return randomIterator.nextInt();
    }
}

You can also do it for double and long values.

Hope it helps! :)


In Java 1.7 or later, the standard way to do this 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(min, max + 1);

See the relevant JavaDoc. This approach has the advantage of not needing to explicitly initialize a java.util.Random instance, which can be a source of confusion and error if used inappropriately.

However, conversely there is no way to explicitly set the seed so it can be difficult to reproduce results in situations where that is useful such as testing or saving game states or similar. In those situations, the pre-Java 1.7 technique shown below can be used.

Before Java 1.7, the standard way to do this is as follows:

import java.util.Random;

/**
 * Returns a pseudo-random number between min and max, inclusive.
 * The difference between min and max can be at most
 * <code>Integer.MAX_VALUE - 1</code>.
 *
 * @param min Minimum value
 * @param max Maximum value.  Must be greater than min.
 * @return Integer between min and max, inclusive.
 * @see java.util.Random#nextInt(int)
 */
public static int randInt(int min, int max) {

    // NOTE: This will (intentionally) not run as written so that folks
    // copy-pasting have to think about how to initialize their
    // Random instance.  Initialization of the Random instance is outside
    // the main scope of the question, but some decent options are to have
    // a field that is initialized once and then re-used as needed or to
    // use ThreadLocalRandom (if using at least Java 1.7).
    // 
    // In particular, do NOT do 'Random rand = new Random()' here or you
    // will get not very good / not very random results.
    Random rand;

    // nextInt is normally exclusive of the top value,
    // so add 1 to make it inclusive
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return randomNum;
}

See the relevant JavaDoc. In practice, the java.util.Random class is often preferable to java.lang.Math.random().

In particular, there is no need to reinvent the random integer generation wheel when there is a straightforward API within the standard library to accomplish the task.


Generate a random number for the difference of min and max by using the nextint(n) method and then add min number to the result:

Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);

ThreadLocalRandom equivalent of class java.util.Random for multithreaded environment. Generating a random number is carried out locally in each of the threads. So we have a better performance by reducing the conflicts.

int rand = ThreadLocalRandom.current().nextInt(x,y);

x,y - intervals e.g. (1,10)


 rand.nextInt((max+1) - min) + min;

As of Java 7, you should no longer use Random. For most uses, the random number generator of choice is now ThreadLocalRandom.

For fork join pools and parallel streams, use SplittableRandom.

Joshua Bloch. Effective Java. Third Edition.

Starting from Java 8

For fork join pools and parallel streams, use SplittableRandom that is usually faster, has a better statistical independence and uniformity properties in comparison with Random.

To generate a random int in the range [0, 1_000]:

int n = new SplittableRandom().nextInt(0, 1_001);

To generate a random int[100] array of values in the range [0, 1_000]:

int[] a = new SplittableRandom().ints(100, 0, 1_001).parallel().toArray();

To return a Stream of random values:

IntStream stream = new SplittableRandom().ints(100, 0, 1_001);

public static Random RANDOM = new Random(System.nanoTime());

public static final float random(final float pMin, final float pMax) {
    return pMin + RANDOM.nextFloat() * (pMax - pMin);
}

Forgive me for being fastidious, but the solution suggested by the majority, i.e., min + rng.nextInt(max - min + 1)), seems perilous due to the fact that:

  • rng.nextInt(n) cannot reach Integer.MAX_VALUE.
  • (max - min) may cause overflow when min is negative.

A foolproof solution would return correct results for any min <= max within [Integer.MIN_VALUE, Integer.MAX_VALUE]. Consider the following naive implementation:

int nextIntInRange(int min, int max, Random rng) {
   if (min > max) {
      throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
   }
   int diff = max - min;
   if (diff >= 0 && diff != Integer.MAX_VALUE) {
      return (min + rng.nextInt(diff + 1));
   }
   int i;
   do {
      i = rng.nextInt();
   } while (i < min || i > max);
   return i;
}

Although inefficient, note that the probability of success in the while loop will always be 50% or higher.





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