Set value in an array[3], that the sum of all three values is always 1.0

Question

So I'm working on a Civ like game right now and want to have a setting how taxes are split in gold income, science and luxury goods.

private double[] rate = new double[] { 0.4, 0.6, 0.0 }; 

now I wanted to add a methode to change one of the values to a higher or lower value but maintain the overall sum of 1.0 (100%) but I'm stuck here with how to take or add values evenly from the other two rates

    public void setRate (int index, double r) {
    if (r < 0.0) {
        throw new IllegalArgumentException("Invalid Rate");
    }

    double diff = r - rate[index];
    if (diff > 0) {

        // ?

    } else if (diff < 0) {

        // ?

    }       
    rate[index] = r;
}

EDIT for amitp:

public void setRate (int index, double r) {
    if (rate[index] = 1.0) {
         for (int i = 0; i < 3; i++) {
              if (i != index) {
                  rate[i] = (rate[index] - r) / 2;
             }
        }
    } else { 
        for (int i = 0; i < 3; i++) {
            if (i != index) {
                rate[i] = rate[i] * (1.0 - r) / (1.0 - rate[index]);
            }
        rate[index] = r;
        }
    }

Answer 1

I think the simplest approach is to rescale the values.

The values other than the one you're setting currently sum to 1.0 - rate[index]. You want them to sum to 1.0 - r. You can therefore scale each of them by (1.0 - new_value) / (1.0 - old_value) to preserve the sum.

public void setRate (int index, double r) {
    for (int i = 0; i < 3; i++) {
        if (i != index) {
            rate[i] = rate[i] * (1.0 - r) / (1.0 - rate[index]);
        }
    }
    rate[index] = r;
}

This solution avoids:

1. Different code for positive and negative changes
2. Having to handle underflow (something + diff going below 0.0) and overflow (something + diff going above 1.0)
3. A value going from zero to non-zero, or vice versa. If the player set something at 0, then they probably want to keep it there.

Side note: a geometric interpretation is that your three values are coordinates in a cube. You constrain x + y + z = 1.0 so that puts you on an equilateral triangle embedded in the cube. Thinking about it this way helped me come up with the above solution.

Answer 2

You could divide diff by two and then subtract it or add it accordingly to the other two.

double diff = r - rate[index];
double change = diff/2 * -1; //we need to reverse the operation

for(int i = 0 ; i < 3; i++)
{
    if(index != i) rate[i] += change;
}

Still you should be careful for number precision errors. For that you could use an int[] where all values sum up to 100.

Answer 3

One way to do this is to add to the indexes' value by its self multiplied by diff. rate[i] = rate[i] + rate[i]*diff where i is the other to values other than index

    double d = 1 - rate[index]       
    double diff = rate[index] - r;
    rate[index] = r;
    index = (index + 1) % 3;           // increment and wrap around index
    rate[index] += diff*rate[index]/d;
    index = (index + 1) % 3;           // increment and wrap around index
    rate[index] += diff*rate[index]/d;

Hope that helps.