Generate a random even number inside a range?

Question

Here's the format I'm following:

 int randomNum = rand.nextInt((max - min) + 1) + min;

So here's my code. I'm trying to get a random even number between 1 and 100:

 Random rand = new Random(); 
 int randomNum = rand.nextInt((100 - 2) + 1) + 2;

I saw one solution is to multiply the number by 2, but that breaches the intended range.

I thought of doing a bunch of if statements to solve this, but it seems overly complex. Is there a relatively simple solution?

Answer 1

Just generate a number from 0 to 49 and then multiply it by 2.

Random rand = new Random(); 
int randomNum = rand.nextInt(100/2) *2;

To do it in a range just add the range in:

int randomNum = startOfRange+rand.nextInt((endOfRange-startOfRange)/2) *2;

Note that startOfRange should be an even number or converted into an even number.

Answer 2

Here´s a tiny example on how to do it

static Random rand = new Random();

public static void main(String[] args) {
    for(int i = 0;i<100;++i)
        System.out.println(generateEvenNumber(0, 100));

}

private static int generateEvenNumber(int min, int max) {
    min = min % 2 == 1 ? min + 1 : min; // If min is odd, add one to make sure the integer division can´t create a number smaller min;
    max = max % 2 == 1 ? max - 1 : max; // If max is odd, subtract one to make sure the integer division can´t create a number greater max;
    int randomNum = ((rand.nextInt((max-min))+min)+1)/2; // Divide both by 2 to ensure the range
    return randomNum *2; // multiply by 2 to make the number even
}

Answer 3

In Java 1.7 or later, I would use ThreadLocalRandom:

import java.util.concurrent.ThreadLocalRandom;

// Get even random number within range [min, max]
// Start with an even minimum and add random even number from the remaining range
public static int randEvenInt(int min, int max) {
    if (min % 2 != 0) ++min;
    return min + 2*ThreadLocalRandom.current().nextInt((max-min)/2+1);
}

// Get odd random number within range [min, max]
// Start with an odd minimum and add random even number from the remaining range
public static int randOddInt(int min, int max) {
    if (min % 2 == 0) ++min;
    return min + 2*ThreadLocalRandom.current().nextInt((max-min)/2+1);
}

The reason to use ThreadLocalRandom is explained here. Also note, that the reason we +1 to the input to ThreadLocalRandom.nextInt() is to make sure the max is included in the range.

Answer 4

After you get the random number between 1 and 100 and multiply it by 2, you can get the number % 100. Something like:

int random = rand.nextInt(100);
random = (random * 2) % 100;

This way number will always be less than 100 and even.

Answer 5

Another approach (maybe not the optimal one):

1. Create an array (or a List) with all the even numbers between 1 and 100. (This is easy with a loop)
2. When you need a random even number just take a random number from that array or list (using random with length or size).

Answer 6

Best way to generate even random number between 2 and 1000

Random random = new Random();
        int Low = 2;
        int High = 1000;
        int randomNumber = random.nextInt(High-Low) + Low;
        if(randomNumber % 2 != 0){
            randomNumber = randomNumber + 1;
        }

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Answer 7

private int getRandomNoOdd(int maxValue) {//min value is 0, including zero, max value can potentially be reached by randomness
    int random = 0;
    while (true) {
        random = (int) (Math.random() * (maxValue + 1));
        if (random % 2 != 0)
            continue;
        break;
    }
    return random;
}

Answer 8

You can also get a random number from a range then, if % 2 != 0, add 1.