about random values [duplicate]
Possible Duplicate:
Getting N random numbers that the sum is M
Hi I have a question that:
how can i get random values where the sum of all of them is 1
.
like {0.5,0.5}
or {0.25,0.25,0.5}
and more.
also the number of these values are different every time ,once can be 2
and once can be 3
like the example above!
thanks.
I'll outline the basic algorithm for you:
decide how many random numbers you will generate that will be summed.
generate that many random numbers.
decide what they should all add up to.
divide number from the previous step by the sum of the random numbers.
divide each random number by the number from the previous step.
Basically what you'll be doing is generating a bunch of unbounded random numbers and then adjusting them all so they all add up to some specific number.
BTW: In the very unlikely event that all the random numbers you generate are zero, you'll have a divide by zero error with this algorithm. So you should trap for that in your implementation and retry the random number generation in a loop until you get a non-zero sum of random numbers.
In general you can generate an array of random size:
java.util.Random rand = new java.util.Random();
final int MAX_SIZE = 100;
double[] a = new double[1 + rand.nextInt(MAX_SIZE)];
(Here I'm assuming you will not want arrays larger than 100 elements.)
Then you fill the array with random positive numbers:
for (int i = 0; i < a.length; ++ i)
{
a[i] = rand.nextDouble();
}
Then you normalize the array (divide each element by the total sum). First we compute the total sum:
double sum = 0.0;
for (int i = 0; i < a.length; ++ i)
{
sum += a[i];
}
Then divide each array element by the sum:
for (int i = 0; i < a.length; ++ i)
{
a[i] /= sum;
}
If you want shorter code, you can combine the loop that accumulates the sum with the loop that fills the array with random positive integers. Here is the resulting code:
java.util.Random rand = new java.util.Random();
final int MAX_SIZE = 100;
double[] a = new double[1 + rand.nextInt(MAX_SIZE)];
double sum = 0.0;
for (int i = 0; i < a.length; ++ i)
{
a[i] = rand.nextDouble();
sum += a[i];
}
for (int i = 0; i < a.length; ++ i)
{
a[i] /= sum;
}
Well, I think that the easiest way would be something like this:
To generate two numbers:
r1 = new RandomNumber
r2 = 1 - r1
TO generate three numbers:
r1 = new RandomNumber
r2 = r1 / 2
r3 = 1 - r1
r1 = r1 / 2
I do not know if there is any simpler way of doing this though and the code above is just pseudo code.
Like @dsg but shorter. It allows you to specify what the sum should be.
public static double[] randomDoubles(int size, double sum) {
double total=0, doubles[]=new double[size];
for(int i=0;i<size;i++) total += doubles[i] = Math.random();
for(int i=0;i<size;i++) doubles[i] *= sum/total;
return doubles;
}
How about for n random numbers:
r1 = random(0,1)
r2 = random(0, 1-r1)
r3 = random(0, 1-r1-r2)
...
r(n) = 1 - r1-r2-f3...-r(n-1)
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