Question

How do I divide a large positive integer n into m parts uniformly randomly.
Post-condition: Adding up all the m parts should give n.

Below is my attempt(in java like pseudocode), but I don’t think it will give me uniformly random distribution.
I am first finding the average part avg by dividing n/m. Then I am generating m-1 random numbers which are around avg in magnitude(by alternately generating random numbers between 0 & avg, and *avg & 2*avg*. Then I am subtracting the sum of these m-1 numbers from original number n and setting that as the m‘th part.

Assume that the function rand(x, y) returns a random number uniformly between x and y.

int[] divideUniformlyRandomly(int n, int m)
{
    int[] res = new int[m];
    int avg = n / m;
    int sum = 0;
    bool alternator = false;
    for(int i = 0; i < m - 1; i++)
    {
        if(alternator == false)
        {
            res[i] = rand(0, avg);
            alternator = true;
        }
        else
        {
            res[i] = rand(avg, 2*avg);
            alternator = false;
        }
        sum += res[i];
    }
    res[m-1] = n - sum;
    return res;
}

Answer 1

public double[] divideUniformlyRandomly(double number, int part) {
    double uniformRandoms[] = new double[part];
    Random random = new Random();

    double mean = number / part;
    double sum = 0.0;

    for (int i=0; i<part / 2; i++) {
        uniformRandoms[i] = random.nextDouble() * mean;

        uniformRandoms[part - i - 1] = mean + random.nextDouble() * mean;

        sum += uniformRandoms[i] + uniformRandoms[part - i -1];
    }
    uniformRandoms[(int)Math.ceil(part/2)] = uniformRandoms[(int)Math.ceil(part/2)] + number - sum;

    return uniformRandoms;
}