How to get random double value from random byte array values?

Question

I would like to use RNGCryptoServiceProvider as a source of random numbers. Since it can output them as an array of byte values, how can I convert them to 0 to 1 double values, while maintaining uniform results?

Answer 1

byte[] result = new byte[8];
rng.GetBytes(result);
return (double)BitConverter.ToUInt64(result,0) / ulong.MaxValue;

Answer 2

This is how I do it.

private static readonly System.Security.Cryptography.RNGCryptoServiceProvider _secureRng;
public static double NextSecureDouble()
{
  var bytes = new byte[8];
  _secureRng.GetBytes(bytes);
  var v = BitConverter.ToUInt64(bytes, 0);
  // We only use the 53-bits of integer precision available in a IEEE 754 64-bit double.
  // The result is a fraction, 
  // r = (0, 9007199254740991) / 9007199254740992 where 0 <= r && r < 1.
  v &= ((1UL << 53) - 1);
  var r = (double)v / (double)(1UL << 53);
  return r;
}

Coincidentally 9007199254740991 / 9007199254740992 is ~= 0.99999999999999988897769753748436, that will return the method Random.NextDoubleas the maximum value (see https://msdn.microsoft.com/en-us/library/system.random.nextdouble(v=vs.110).aspx ).

(max - min)/sqrt (12).

1000 2%.

10000 1%.

.

[Test]
public void Randomness_SecureDoubleTest()
{
  RunTrials(1000, 0.02);
  RunTrials(10000, 0.01);
}

private static void RunTrials(int sampleSize, double errorMargin)
{
  var q = new Queue<double>();

  while (q.Count < sampleSize)
  {
    q.Enqueue(Randomness.NextSecureDouble());
  }

  for (int k = 0; k < 1000; k++)
  {
    // rotate
    q.Dequeue();
    q.Enqueue(Randomness.NextSecureDouble());

    var avg = q.Average();

    // Dividing by n−1 gives a better estimate of the population standard
    // deviation for the larger parent population than dividing by n, 
    // which gives a result which is correct for the sample only.

    var actual = Math.Sqrt(q.Sum(x => (x - avg) * (x - avg)) / (q.Count - 1));

    // see http://stats.stackexchange.com/a/1014/4576

    var expected = (q.Max() - q.Min()) / Math.Sqrt(12);

    Assert.AreEqual(expected, actual, errorMargin);
  }
}

Answer 3

BitConverter.ToDouble(...). Double. Thre - , .

Answer 4

BitConverter Double:

byte[] random_bytes = new byte[8];  // BitConverter will expect an 8-byte array
new RNGCryptoServiceProvider().GetBytes(random_bytes);

double my_random_double = BitConverter.ToDouble(random_bytes, 0);