Non-repetitive random number C#

Question

To generate Random numbers from 1- 20 I need to pick selective and it should not be repetitive.

How to do this in C#

Note I need to loop through as like this

Random rnd = new Random()
rnd.Next(1,20)
for(int i =0; i<=20;i++)
{

}

For all the loops number should be 1 to 20

Answer 1

What exactly do you mean by "should not be repetitive"? If you mean that you don't want to get any duplicates, then you should basically take a list of the numbers 1-20, shuffle them, and then grab one at a time from the head of the list. For an efficient way to shuffle a list, see this Stack Overflow answer.

If you just mean that your current attempt gives 5, 5, 5, 5, 5, 5, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2 etc then chances are you're creating a new instance of Random each time you pick a number: don't do that. Each time you create an instance, it will use the current time as the "seed" for the random number generator (unless you specify one explicitly). That means if you create several instances in quick succession, each will get the same seed and therefore give the same sequence of numbers.

Instead, use a single instance of Random and reuse it. (Note that it's not thread-safe though, which is a pain.) For instance:

private static readonly Random Rng = new Random();

public int NextNumber()
{
    return Rng.Next(20) + 1;
}

That won't be thread-safe, but let us know if that's a problem. An alternative is sometimes to pass the Random into the method (which would normally be more complicated, of course):

public int NextNumber(Random rng)
{
    return rng.Next(20) + 1;
}

then the caller can reuse the instance appropriately.

If you want a thread-safe way of generating random numbers, you might want to look at my StaticRandom class in MiscUtil.

(Note that using rng.Next(1, 21) would also work fine - I happen to prefer the version above as I think it reduces the guesswork about inclusive/exclusive boundaries, but it's a matter of personal taste.)

Answer 2

From MSDN

"One way to improve randomness is to make the seed value time-dependent."

Another fact

you should "create one Random to generate many random numbers over time." This will enhance the random generation

Answer 3

This method will generate all the numbers, and no numbers will be repeated:

/// <summary>
/// Returns all numbers, between min and max inclusive, once in a random sequence.
/// </summary>
IEnumerable<int> UniqueRandom(int minInclusive, int maxInclusive)
{
    List<int> candidates = new List<int>();
    for (int i = minInclusive; i <= maxInclusive; i++)
    {
        candidates.Add(i);
    }
    Random rnd = new Random();
    while (candidates.Count > 0)
    {
        int index = rnd.Next(candidates.Count);
        yield return candidates[index];
        candidates.RemoveAt(index);
    }
}

You can use it like this:

Console.WriteLine("All numbers between 0 and 20 in random order:");
foreach (int i in UniqueRandom(0, 20)) {
    Console.WriteLine(i);
}

Answer 4

I did one this way awhile back. I don't know how it compares to the other methods presented as far as efficiency, randomness, etc. But it seems to work:

List<int> integers = new List<int>() { 1, 2, 3, 4, 5, 6,7, 8, 9, 10, 11, 12 };

Random rnd = new Random();

var ints = from i in integers
           orderby rnd.Next(integers.Count)
           select i;

Answer 5

An IEnumerable implementation, based on Hallgrim's answer:

public class UniqueRandoms : IEnumerable<int>
{
    Random _rand = new Random();
    List<int> _candidates;

    public UniqueRandoms(int maxInclusive)
        : this(1, maxInclusive)
    { }

    public UniqueRandoms(int minInclusive, int maxInclusive)
    {
        _candidates = 
            Enumerable.Range(minInclusive, maxInclusive - minInclusive + 1).ToList();
    }

    public IEnumerator<int> GetEnumerator()
    {
        while (_candidates.Count > 0)
        {
            int index = _rand.Next(_candidates.Count);
            yield return _candidates[index];
            _candidates.RemoveAt(index);
        }
    }

    System.Collections.IEnumerator System.Collections.IEnumerable.GetEnumerator()
    {
        return GetEnumerator();
    }
}