Generating N numbers that sum to 1
The task you are trying to accomplish is tantamount to drawing a random point from the N-dimensional unit simplex.
http://en.wikipedia.org/wiki/Simplex#Random_sampling might help you.
A naive solution might go as following:
public static double[] getArray(int n)
{
double a[] = new double[n];
double s = 0.0d;
Random random = new Random();
for (int i = 0; i < n; i++)
{
a [i] = 1.0d - random.nextDouble();
a [i] = -1 * Math.log(a[i]);
s += a[i];
}
for (int i = 0; i < n; i++)
{
a [i] /= s;
}
return a;
}
To draw a point uniformly from the N-dimensional unit simplex, we must take a vector of exponentially distributed random variables, then normalize it by the sum of those variables. To get an exponentially distributed value, we take a negative log
of uniformly distributed value.
Get n random numbers, calculate their sum and normalize the sum to 1 by dividing each number with the sum.
This is relatively late, but to show the ammendment to @Kobi's simple and straightforward answer given in this paper pointed to by @dreeves which makes the sampling uniform. The method (if I understand it clearly) is to
- Generate n-1 distinct values from the range [1, 2, ... , M-1].
- Sort the resulting vector
- Add 0 and M as the first and last elements of the resulting vector.
- Generate a new vector by computing xi - xi-1 where i = 1,2, ... n. That is, the new vector is made up of the differences between consecutive elements of the old vector.
- Divide each element of the new vector by M. You have your uniform distribution!
I am curious to know if generating distinct random values and normalizing them to 1 by dividing by their sum will also produce a uniform distribution.
Get n random numbers, calculate their sum and normalize the sum to 1 by dividing each number with the sum.
Expanding on Kobi's answer, here's a Java function that does exactly that.
public static double[] getRandDistArray(int n) {
double randArray[] = new double[n];
double sum = 0;
// Generate n random numbers
for (int i = 0; i < randArray.length; i++) {
randArray[i] = Math.random();
sum += randArray[i];
}
// Normalize sum to 1
for (int i = 0; i < randArray.length; i++) {
randArray[i] /= sum;
}
return randArray;
}
In a test run, getRandDistArray(5)
returned the following
[0.1796505603694718, 0.31518724882558813, 0.15226147256596428, 0.30954417535503603, 0.043356542883939767]