Generate random values with fixed sum in C++ [closed]

Answer 1

My C++ is very (very very) rusty. So let's assume you already know how to get a random number between x and y with the function random(x,y). Then here is some psuedocode in some other c derived language:

int n = ....; // your input
int sum = ....; // your input
int[] answer = new int[n];
int tmpsum = 0;
for (int i=0; i < n; i++) {
    int exactpart = sum/n;
    int random = (exactpart/2) + random(0,exactpart);
    int[i] = tmpsum + random > sum ? sum - tmpsum : random;
    tmpsum += int[i];
}
int[n-1] += sum - tmpsum;

Answer 2

First generate n random variables. Then sum them up: randomSum. Calculate coefficient sum/randomSum. Then multiply all random variables with that coefficient.

Integers would pose a problem... Rounding too (probably)

Answer 3

You can generate n numbers with a normal distribution then normalize them to your sum

Answer 4

you have a loop, where the number of iterations is equal to the number of random numbers you want minus 1. for the first iteration, you find a random number between 0 and the sum. you then subtract that random number from the sum, and on the next iteration you get another random number and subtract that from the sub sum minus the last iteration

its probably more easy in psuedocode

int sum = 10;
int n = 5; // 5 random numbers summed to equal sum
int subSum = sum;
int[] randomNumbers = new int[n];

for(int i = 0; i < n - 2; i++)
{
    randomNumbers[i] = rand(0, subSum); // get random number between 0 and subSum 
    subSum -= randomNumbers[i];
}

randomNumbers[n - 1] = subSum; // leftovers go to the last random number

Answer 5

You can generate n values defined by this : ((Sum - sumOfGeneratedValues) / n - (numberOfGeneatedValue)) -+X (With X maximal deviance)

Example :

SUM = 100 N = 5 +- 10

---

Rand between 100 - 0 / 5 - 0 --> 20 +-10 (So bewteen 10 and 30)

Value1 = 17;

Rand between 100 - 17 / 5 - 1 ---> 21 +-10 (So between 11 and 31)

... etc

Deviance would make your random uniform :)