Divide a number into random parts

Question

I need to divide a number into a collection of random parts. So that the sum of all parts is equal to the number.
E.g.:
public void divideRandomly ( int number, int parts)
int[] returnArray = new int[parts];
//do random division calculations
return returnArray;
Lets say the number = 20
And parts = 4
I would then like to return an array of ints containing e.g. 3, 11, 5, 1
OR e.g. 4,8,2,6
Does anyone have a clue as to how I could do this?
Cheers

Answer 1

Be aware that given the wrong parameters this solution
can take a long time to find a list. This seems misleading, as the termination condition is based
on randomness. Although, it is possible to predict an average
termination time.
Basically your algorithm starts by generating initial random values
in the range [0, maxValue]. You limit the size of the maximum
generated value, although I'm not sure why. Obviously maxValue
has to be greater than sum / length, otherwise it would be
impossible to generate any combination values that add up to sum.
Then your algorithm asks if the sum of those initial random values
is equal to the target sum. If not, you cycle through the list replacing
the the current position with a new random value and updating the
current sum.
As an example, let's suppose that maxValue is equal to sum, and
that sum = 1000, and length (numParts) is 4.
Assuming the minValue is 1, not 0, then the total number of
permutations of 4 numbers that add up to 1000 is:
165,668,499
However, the total number of permutations is:
1000^4 = 1,000,000,000,000
In terms of probability, the chance of finding a solution on each try is:
165,668,499 / 1,000,000,000,000 = 0.000165668499
Thus, you'll need approximately 6036 loops on average to find a solution for 1,000 split 4 ways.
For 2,000 split 4 ways, number of perms is:
2000^4 = 16,000,000,000,000
The perms of 4 numbers that add up to 2000 is:
1,329,336,999
Probability of finding a solution on each try is:
0.0000830835624375
Average number of tries is: 12036
I'm impressed that the random approach is that fast.