Elegant way to draw n random values in C++11? - c++

Question

For my program, I needed so far to draw one random value in [0..k[ from time to time, and using C++11 <random> features works really well. My current code is something like
class Random
{
public:
Random() : rng( rd() ) { }
inline int getRandNum( int limit ) { return ( numbers(rng) % limit ); }
private:
std::random_device rd;
std::mt19937 rng;
std::uniform_int_distribution<int> numbers;
};
Now, I need to draw in a row n different values in [0..k[. I was looking for something in <random> allowing that, but either I am not able to find it, or such a thing does not exist yet. Is there a clever, more elegant way to proceed than calling my getRandNum function and repeat until I get n different values?
EDIT: to give an idea, in my program k is some thousands and n some tens.

Answer 1

This solution is not C++ specific but can be easily implemented in any language.
What you want is essentially shuffle numbers 0 to k and pick the first n numbers, where n <= k.
This can be done using a reservoir sampling algorithm. See this wikipedia link for the pseudocode.
Note that it is possible to get the n numbers without storing all k numbers and shuffling them. That is, it is possible to just use O(n) space, where n is the number of random numbers you wish to obtain, instead of O(k). The time complexity for this algorithm is O(k), if we assume generating the random number takes O(1) time.

Answer 2

If k is several thousands and n is tens, then a permutation generation is really not the best choise. But calling getRandNum is not what you want too, because it can return the same value several times.
One option is to generate random sequence all at once, checking that the numbers don't repeat. The easiest (and may be even the most efficient) way to achieve this is to use a set.
Like so:
#include <vector>
#include <set>
#include <iostream>
#include <random>
class Random
{
public:
Random() : rng( rd() ) { }
inline int getRandNum( int limit ) { return ( numbers(rng) % limit ); }
std::set<int> getRandSequence(int limit, int n);
private:
std::random_device rd;
std::mt19937 rng;
std::uniform_int_distribution<int> numbers;
};
std::set<int> Random::getRandSequence(int limit, int n)
{
std::set<int> generatedSequence;
while (generatedSequence.size() < n) //size() for set is O(1) if I'm not mistaken
generatedSequence.insert(getRandNum(limit));
return generatedSequence;
}
int main()
{
Random r;
auto sequence = r.getRandSequence(1000, 10);
std::cout << "Seq;uence: " << std::endl;
for (int number : sequence)
std::cout << number << std::endl;
std::cout << "End" << std::endl;
return 0;
}
Ideone demo.
By the way, random_device creation is expensive, but uniform_int_distribution creation, as far as I remember, is not. So this might be even more efficient:
std::set<int> Random::getRandSequence(int limit, int n)
{
std::uniform_int_distribution<int> uiniformDistribution(0, limit);
std::set<int> generatedSequence;
while (generatedSequence.size() < n)
generatedSequence.insert(uiniformDistribution(rng));
return generatedSequence;
}
Besides, when you get a uniform distribution and then apply % limit to it, you don't get a uniform distribution anymore.

Answer 3

std::random_device rd; // obtain a random number from hardware
std::mt19937 eng(rd()); // seed the generator
std::uniform_int_distribution<> distr(0, 1500); // define the range
for(int a=0; a<limit; a++){
cout << distr(eng); //draw random nubmer