Random Number Between 2 Double Numbers

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Is it possible to generate a random number between 2 doubles?

Example:

public double GetRandomeNumber(double minimum, double maximum)
{
    return Random.NextDouble(minimum, maximum) 
}

Then I call it with the following:

double result = GetRandomNumber(1.23, 5.34);

Any thoughts would be appreciated.

13 Answers

Yes.

Random.NextDouble returns a double between 0 and 1. You then multiply that by the range you need to go into (difference between maximum and minimum) and then add that to the base (minimum).

public double GetRandomNumber(double minimum, double maximum)
{ 
    Random random = new Random();
    return random.NextDouble() * (maximum - minimum) + minimum;
}

Real code should have random be a static member. This will save the cost of creating the random number generator, and will enable you to call GetRandomNumber very frequently. Since we are initializing a new RNG with every call, if you call quick enough that the system time doesn't change between calls the RNG will get seeded with the exact same timestamp, and generate the same stream of random numbers.

I'm a bit late to the party but I needed to implement a general solution and it turned out that none of the solutions can satisfy my needs.

The accepted solution is good for small ranges; however, maximum - minimum can be infinity for big ranges. So a corrected version can be this version:

public static double NextDoubleLinear(this Random random, double minValue, double maxValue)
{
    // TODO: some validation here...
    double sample = random.NextDouble();
    return (maxValue * sample) + (minValue * (1d - sample));
}

This generates random numbers nicely even between double.MinValue and double.MaxValue. But this introduces another "problem", which is nicely presented in this post: if we use such big ranges the values might seem too "unnatural". For example, after generating 10,000 random doubles between 0 and double.MaxValue all of the values were between 2.9579E+304 and 1.7976E+308.

So I created also another version, which generates numbers on a logarithmic scale:

public static double NextDoubleLogarithmic(this Random random, double minValue, double maxValue)
{
    // TODO: some validation here...
    bool posAndNeg = minValue < 0d && maxValue > 0d;
    double minAbs = Math.Min(Math.Abs(minValue), Math.Abs(maxValue));
    double maxAbs = Math.Max(Math.Abs(minValue), Math.Abs(maxValue));

    int sign;
    if (!posAndNeg)
        sign = minValue < 0d ? -1 : 1;
    else
    {
        // if both negative and positive results are expected we select the sign based on the size of the ranges
        double sample = random.NextDouble();
        var rate = minAbs / maxAbs;
        var absMinValue = Math.Abs(minValue);
        bool isNeg = absMinValue <= maxValue ? rate / 2d > sample : rate / 2d < sample;
        sign = isNeg ? -1 : 1;

        // now adjusting the limits for 0..[selected range]
        minAbs = 0d;
        maxAbs = isNeg ? absMinValue : Math.Abs(maxValue);
    }

    // Possible double exponents are -1022..1023 but we don't generate too small exponents for big ranges because
    // that would cause too many almost zero results, which are much smaller than the original NextDouble values.
    double minExponent = minAbs == 0d ? -16d : Math.Log(minAbs, 2d);
    double maxExponent = Math.Log(maxAbs, 2d);
    if (minExponent == maxExponent)
        return minValue;

    // We decrease exponents only if the given range is already small. Even lower than -1022 is no problem, the result may be 0
    if (maxExponent < minExponent)
        minExponent = maxExponent - 4;

    double result = sign * Math.Pow(2d, NextDoubleLinear(random, minExponent, maxExponent));

    // protecting ourselves against inaccurate calculations; however, in practice result is always in range.
    return result < minValue ? minValue : (result > maxValue ? maxValue : result);
}

Some tests:

Here are the sorted results of generating 10,000 random double numbers between 0 and Double.MaxValue with both strategies. The results are displayed with using logarithmic scale:

0..Double.MaxValue

Though the linear random values seem to be wrong at first glance the statistics show that none of them are "better" than the other: even the linear strategy has an even distribution and the average difference between the values are pretty much the same with both strategies.

Playing with different ranges showed me that the linear strategy gets to be "sane" with range between 0 and ushort.MaxValue with a "reasonable" minimum value of 10.78294704 (for ulong range the minimum value was 3.03518E+15; int: 353341). These are the same results of both strategies displayed with different scales:

0..UInt16.MaxValue


Edit:

Recently I made my libraries open source, feel free to see the RandomExtensions.NextDouble method with the complete validation.

The simplest approach would simply generate a random number between 0 and the difference of the two numbers. Then add the smaller of the two numbers to the result.

You could use code like this:

public double getRandomNumber(double minimum, double maximum) {
    return minimum + randomizer.nextDouble() * (maximum - minimum);
}

Use a static Random or the numbers tend to repeat in tight/fast loops due to the system clock seeding them.

public static class RandomNumbers
{
    private static Random random = new Random();
    //=-------------------------------------------------------------------
    // double between min and the max number
    public static double RandomDouble(int min, int max) 
    {
        return (random.NextDouble() * (max - min)) + min;
    }
    //=----------------------------------
    // double between 0 and the max number
    public static double RandomDouble(int max) 
    {
        return (random.NextDouble() * max);
    }
    //=-------------------------------------------------------------------
    // int between the min and the max number
    public static int RandomInt(int min, int max) 
    {   
        return random.Next(min, max + 1);
    }
    //=----------------------------------
    // int between 0 and the max number
    public static int RandomInt(int max) 
    {
        return random.Next(max + 1);
    }
    //=-------------------------------------------------------------------
 }

See also : https://docs.microsoft.com/en-us/dotnet/api/system.random?view=netframework-4.8

Random rand = new Random();

rand.Next(1,5)/10; //Means between 0,1 and 0,5

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