Generate Unique Random Numbers Between 1 and 100

Generate unique random numbers between 1 and 100

For example: To generate 8 unique random numbers and store them to an array, you can simply do this:

var arr = [];

while(arr.length < 8){

    var r = Math.floor(Math.random() * 100) + 1;

    if(arr.indexOf(r) === -1) arr.push(r);

}

console.log(arr);

Generate 'n' unique random numbers within a range

If you just need sampling without replacement:

>>> import random
>>> random.sample(range(1, 100), 3)
[77, 52, 45]

random.sample takes a population and a sample size k and returns k random members of the population.

If you have to control for the case where k is larger than len(population), you need to be prepared to catch a ValueError:

>>> try:
...   random.sample(range(1, 2), 3)
... except ValueError:
...   print('Sample size exceeded population size.')
... 
Sample size exceeded population size

Generating unique random numbers (integers) between 0 and 'x'

Use the basic Math methods:

• Math.random() returns a random number between 0 and 1 (including 0, excluding 1).
• Multiply this number by the highest desired number (e.g. 10)

• Round this number downward to its nearest integer

  Math.floor(Math.random()*10) + 1
  

Example:

//Example, including customisable intervals [lower_bound, upper_bound)
var limit = 10,
    amount = 3,
    lower_bound = 1,
    upper_bound = 10,
    unique_random_numbers = [];

if (amount > limit) limit = amount; //Infinite loop if you want more unique
                                    //Natural numbers than exist in a
                                    // given range
while (unique_random_numbers.length < limit) {
    var random_number = Math.floor(Math.random()*(upper_bound - lower_bound) + lower_bound);
    if (unique_random_numbers.indexOf(random_number) == -1) { 
        // Yay! new random number
        unique_random_numbers.push( random_number );
    }
}
// unique_random_numbers is an array containing 3 unique numbers in the given range

Generate a unique random number between 1 and 10?

1. Setup a list with all the valid values at Awake
2. Get a random value from the list
3. Remove the value from the list to avoid duplicates
4. BONUS - You could also extend the class to set any min and max numbers

public class RandomGenerator : MonoBehaviour
{
    public int minNumber = 1;
    public int maxNumber = 10;

    private List<int> _validNumbers;

    public int number;

    private void Awake()
    {
        _validNumbers = new List<int>();
        for (int i = minNumber; i <= maxNumber; i++)
            _validNumbers.Add(i);
    }

    private void Update()
    {
        if (Input.GetKeyDown(KeyCode.A))
        {
            if (_validNumbers.Count == 0)
                Debug.Log("No valid Numbers");
            else
                number = GetRandomNumber();
        }
    }

    private int GetRandomNumber()
    {
        var nextIndex = Random.Range(0, _validNumbers.Count - 1);
        var result    = _validNumbers[nextIndex];
        _validNumbers.RemoveAt(nextIndex);
        return result;
    }
}

EDIT AFTER COMMENTS:

This question is very similar to this other question. But Unity.Random is different than System.Random.

The answers offered in the other question work here too. But we have more choices here.

Generating Unique Random Numbers in Java

• Add each number in the range sequentially in a list structure.
• Shuffle it.
• Take the first 'n'.

Here is a simple implementation. This will print 3 unique random numbers from the range 1-10.

import java.util.ArrayList;
import java.util.Collections;

public class UniqueRandomNumbers {
    
    public static void main(String[] args) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        for (int i=1; i<11; i++) list.add(i);
        Collections.shuffle(list);
        for (int i=0; i<3; i++) System.out.println(list.get(i));
    }
}

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The first part of the fix with the original approach, as Mark Byers pointed out in an answer now deleted, is to use only a single Random instance.

That is what is causing the numbers to be identical. A Random instance is seeded by the current time in milliseconds. For a particular seed value, the 'random' instance will return the exact same sequence of pseudo random numbers.

Unique (non-repeating) random numbers in O(1)?

Initialize an array of 1001 integers with the values 0-1000 and set a variable, max, to the current max index of the array (starting with 1000). Pick a random number, r, between 0 and max, swap the number at the position r with the number at position max and return the number now at position max. Decrement max by 1 and continue. When max is 0, set max back to the size of the array - 1 and start again without the need to reinitialize the array.

Update:
Although I came up with this method on my own when I answered the question, after some research I realize this is a modified version of Fisher-Yates known as Durstenfeld-Fisher-Yates or Knuth-Fisher-Yates. Since the description may be a little difficult to follow, I have provided an example below (using 11 elements instead of 1001):

Array starts off with 11 elements initialized to array[n] = n, max starts off at 10:

+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|10|
+--+--+--+--+--+--+--+--+--+--+--+
                                ^
                               max    

At each iteration, a random number r is selected between 0 and max, array[r] and array[max] are swapped, the new array[max] is returned, and max is decremented:

max = 10, r = 3
           +--------------------+
           v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 7| 8| 9| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 9, r = 7
                       +-----+
                       v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 1| 2|10| 4| 5| 6| 9| 8| 7: 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 8, r = 1
     +--------------------+
     v                    v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 5| 6| 9| 1: 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

max = 7, r = 5
                 +-----+
                 v     v
+--+--+--+--+--+--+--+--+--+--+--+
| 0| 8| 2|10| 4| 9| 6| 5: 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

...

After 11 iterations, all numbers in the array have been selected, max == 0, and the array elements are shuffled:

+--+--+--+--+--+--+--+--+--+--+--+
| 4|10| 8| 6| 2| 0| 9| 5| 1| 7| 3|
+--+--+--+--+--+--+--+--+--+--+--+

At this point, max can be reset to 10 and the process can continue.

How to generate unique random numbers (that don't repeat)?

One straightforward way to do non-repeating 'random' (psudeorandom) whole numbers in a modest range is to create a list using range(1, n), then random.shuffle() the list, and then take as many numbers as you want from the list using pop() or a slice.

import random

max = 11
l = list(range(1, max))  # the cast to list is optional in Python 2
random.shuffle(l)

Now every time you want a random number, just l.pop().

Another is to use random.sample() -- see https://docs.python.org/3/library/random.html

how to generate unique random numbers with a specific range

Solution 1:

I read Jon Skeet's comment, of course, this is the easiest solution:

List<Integer> list = new ArrayList<>();
for (int i = 0; i < 255; i++) {
     list.add(i);
}
//and here is the point. Java already have this implemented for you
Collections.shuffle(list);

Or in Java 8 declarative style:

List<Integer> list= IntStream.range(0, 255)
    .boxed()
    .collect(Collectors.toList());
Collections.shuffle(list);

or

List<Integer> list = new ArrayList<>();
IntStream.range(0, 255).forEach(list::add);
Collections.shuffle(list);

Solution 2 (picking up on your solution):

You need to generate number for each cell, and check if this number already exists:

 short [] array =new short[255];
 Random rand = new Random();

 for (int i=0; i<array.length; i++) {
     int random_integer = -1;

     //generate integer while it exists in the array
     while(exists(random_integer, array)) {
         random_integer = rand.nextInt(255);
     }

     array[i] = random_integer;
}

And now, let's check whether it exists:

public boolean exists(int number, int[] array) {
    if (number == -1)
        return true; 

    for (int i=0; i<array.length; i++) {
        if (number == array[i])
            return true;
    }
    return false;
}

Of course, you can use a hashmap for example, to speed up the exists() method, i.e. to lower the complexity from O(n) to O(1);

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