CODEWITHSUNDEEP
c++c++11randomc++14
user2930006
user2930006

Reputation: 233

std::discrete_distribution of a specified range of random numbers

I know I can use std::discrete_distribution like this:

std::default_random_engine generator;
std::discrete_distribution<int> distribution {2,2,1,1,2,2,1,1,2,2};

int p[10]={};

for (int i=0; i<100; ++i) {
    int number = distribution(generator);
    ++p[number];
}

However, this is going to generate the numbers in the range of 0-9 as per the weight specified.

What can I do to generate the numbers within a user specified range, say for example, 24-33 or 95-104 etc but still using the distribution specified in discrete_distribution ?

Upvotes: 0

Views: 345

Answers (2)

niceman
niceman

Reputation: 2673

when you have a function that returns a random number in range [r1,r2), and you want [min,max) mathematics to help.

First we calculate d=(max-min)/(r2-r1) and then we multiply the range [r1,r2) by d so we get [d*r1,d*r2) which we'll make [r1',r2'), now you calculate diff=abs(min-r1)' and sum that to the range [r1',r2'), we used abs because we want the difference without sign, and that my friend will get you the range you want, let's assume the function that returns in range [r1,r2) is somerand() then : 

const int r1=0,r2=9
int myrand(const int min,const int max){
    const int d=(max-min)/(r2-r1);
    const int localrand=d*somerand();
    return localrand + abs(min - d*r1);
}

you can check that the function returns min when somerand() returns r1 and your function returns max when somerand() returns r2.

The solution above won't work with e.g. : r1=0,r2=2,min=0,max=3, in short max-min should be divisible by r2-r1.

Upvotes: -3

Izaya
Izaya

Reputation: 409

You can just add 24 or 95 to the number that is generated. At the beginning you have numbers from 0 to 9, when you add 24 to them you have numbers from 24 to 33.

Upvotes: 5

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