Create different seeds for different instances of "Random"

Question

People usually ask why they get always the same numbers when they use Random. In their case, they unintenionally create a new instance of Random each time (instead of using only one instance), which of course leads to the same numbers the whole time. But in my case, I do need several instances of Random which return different number streams.

Using hard-coded seeds is a bad idea in my opinion since you get the same values again and again after restarting the program. What about this:

int seed1 = (int)DateTime.Now.Ticks - 13489565;
int seed2 = (int)DateTime.Now.Ticks - 5564;

I know this looks silly and naive, but it would avoid the same values after every restart and both of the seeds should differ. Or maybe

Random helper = new Random();
int seed1 = helper.Next(1, int.MaxValue);
int seed2 = helper.Next(1, int.MaxValue);

As you can see, I am a bit uncreative here and need your help. Thanks.

Answer 1

Jon Skeet, suggests using a secondary Random object and a lock to create a Random object factory. For example:

    public static Random NewRandom() 
    { 
        lock (globalLock) 
        { 
            return new Random(secondaryRandom.Next()); 
        } 
    } 

Take a look at the complete source code in Jon Skeet's blog.

Answer 2

I would use your second approach:

int randomCount = 10;
Random seeder = new Random();
var randoms = Enumerable.Range(0, randomCount)
    .Select(i => new Random(seeder.Next()))
    .ToList();

This uses linq to create a list of 10 differently-seeded Random instances. You'll therefore need using System.Linq; to have access to the Select extension method.

You could also reuse the seeder as one of your instances:

int randomCount = 10;
Random seeder = new Random();
var randoms = Enumerable.Range(0, randomCount - 1)
    .Select(i => new Random(seeder.Next()))
    .Concat(new [] { seeder })
    .ToList();

Answer 3

You could use a cryptographic generator to create your seeds e.g.

public static Random CreateRandom()
{
    using (var rng = new System.Security.Cryptography.RNGCryptoServiceProvider())
    {
        byte[] bytes = new byte[4];
        rng.GetNonZeroBytes(bytes);
        int seed = BitConverter.ToInt32(bytes, 0);
        return new Random(seed);
    }
}

Of course, if you just need ints then you could use RNGCryptoServiceProvider to generate them directly, although System.Random is probably faster.