How to generate three unequal numbers?

Question

Here the number generation :

for( int i=0; i<3; i++) {
    int randomNumbers = random.nextInt(10) + 1; 
}

And here I can receive the same digits.

For example: 5,7,5;

But I need to receive only different numbers: 5,1,9.

Is it possible maybe with some little piece of code or I should write some method for it?

Do a while loop instead and check if you have already generated that number. If not, increase your counter — DJSchaffner, Feb 15 '20 at 10:35

Andronicus · Accepted Answer · 2020-02-15T18:14:14.337

5

You can use a Set which holds only unique values:

Set<Integer> randomInts = new HashSet();
while(randomInts.size() < 3) {
    randomInts.add(random.nextInt(10) + 1);
}

Of course, you need to face oversampling, but this approach scales very easily if performance isn't the most important.

edited Feb 15 '20 at 18:14

answered Feb 15 '20 at 10:41

Andronicus

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score 5 · Answer 2 · answered Feb 15 '20 at 10:47

5

Here's a way that avoids rejection sampling: generate one number up to 10, one up to 9, and up to 8. Avoid repeats by incrementing. This guarantees a uniform distribution.

int a = random.nextInt(10) + 1;
int b = random.nextInt(9) + 1;
if(b >= a) { b++; }
int c = random.nextInt(8) + 1;
if(c >= (a > b ? b : a)) { c++; }
if(c >= (a > b ? a : b)) { c++; }

answered Feb 15 '20 at 10:47

kaya3

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Bit shorter, not sure if better or worse: `c >= a || c >= b` and `c >= a && c >= b`. – Kelly Bundy Feb 15 '20 at 17:39

score 2 · Answer 3 · answered Feb 15 '20 at 10:50

2

In my point of view, using a while loop is possible to generate random numbers with different digits. Use while loop to check whether the number was generated earlier and if it was then generate new number.

answered Feb 15 '20 at 10:50

Asela Pathirage

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How to generate three unequal numbers?

3 Answers3