How can I generate a random number from 1 - 39 without making a number overlap?

Question

#include <iostream>
#include <stdlib.h>
#include <time.h>
using namespace std;

void randnum()
{
    int random;
    srand((unsigned int)time(0));
    for(int i=1;i<=5;i++)
    {
        random=(rand()%39)+1;
        cout<<random<<endl;
    }
}

int main()
{
    cout<<"Five random number is here"<<endl;
    randnum();
    system ("PAUSE");
    return 0;
}

I am randomly doing this to practice C++. I always get confused in setting the range of random generator (is my method correct? from 1-39). Also how can I prevent numbers from overlapping one another? That is, if I am outputting 5 different numbers from 1-39, it can always 5 different numbers like 4,5,2,7,12 instead of 4,5,2,4,12 (4 is used twice here)

Answer 1

Yes, the method of getting a random number between 1 and 39 is correct.

To ensure non-overlapping numbers, two algorithms come to mind:

1. keep a set of already-served numbers and skip when they are picked a second time, or
2. create a list of all candidate numbers and randomly reorder them, then serve them in order

Answer 2

The method is correct. To prevent numbers from overlapping one another, for me the best solution would be to create a vector of already generated numbers, and every time you generate a new random number, see if it is already in the vector. If it is, regenerate. If it isn't, then add it to the vector and move on.

Answer 3

random=(rand()%39)+1;

This could lead to a duplicate.

---

I always get confused in setting the range of random generator (is my method correct from 1-39?)

The number of elements in a range [begin,end] (where the bracket means "inclusive") is:

 count = end - begin + 1

If you need one of 0-based count elements, then you perform:

rand() % count

Because the starting element may not be 0, you actually perform the following to get a value in the range:

rand() % count + begin

---

Also how can I prevent numbers from overlapping one another?

In this case, one of the easier solutions would be to use a vector. Its not as efficient as other answers (like the one @Retired Ninja suggested), but its easier to understand. Something like shown below.

The code below just dumps the result of the shuffle (which is not random because it repeats across runs based on the seed used). It should not be hard for you to adapt it to the first 5 elements (we can't give you all the answers).

ShuffledRange range (1, 39);
...

$ ./tt.exe 
29 33 8 37 9 32 38 24 16 14 36 7 10 31 34 39 27 11 6 4 35 1 19 20 18 15 5 12 22
21 3 30 17 25 2 28 23 26 13 

If you specify a seed (the default is 0), then you will get a different sequence:

ShuffledRange range (1, 39, 2);
...

$ ./tt.exe 
12 20 28 6 7 15 32 17 35 11 18 31 27 4 23 36 25 24 22 1 33 2 37 39 21 9 38 13 5 3
14 10 8 34 16 19 29 26 30

The code below needs C++ 11 because of random_shuffle. Visual Studio 2012 should be fine with C++ 11. I'm not sure about Visual Studio 2010.

GCC will need:

$ g++ -Wall -Wextra -std=c++11 tt.cpp -o tt.exe

And Mac OS X:

$ g++ -Wall -Wextra -std=c++11 -stdlib=libc++ tt.cpp -o tt.exe

---

class ShuffledRange
{
public:

    explicit ShuffledRange(unsigned int low, unsigned int high, int seed=0)
      : m_numbers(move(create_numbers(low,high,seed))), m_it(m_numbers.begin()) { }

    unsigned int GetCount() const {
        return static_cast<unsigned int>(m_numbers.size());
    }

    bool HasNext() const {
        return m_it != m_numbers.end();
    }

    unsigned int GetNext()
    {
        if(!HasNext())
            throw std::runtime_error("No numbers left");

        unsigned int temp = *m_it++;
        return temp;
    }

protected:                

    vector<unsigned int> create_numbers(unsigned int low, unsigned int high, int seed)
    {
        if(high < low)
            throw std::runtime_error("Bad range of elements");

        vector<unsigned int> temp;
        temp.reserve(high - low + 1);

        for(unsigned int i = low; i <= high; i++)
            temp.push_back(i);

        srand(seed);
        random_shuffle(temp.begin(), temp.end());

        return temp;
    }

private:

    vector<unsigned int> m_numbers;
    vector<unsigned int>::iterator m_it;
};

int main(int argc, char* argv[])
{
    ShuffledRange range(1, 39);

    while(range.HasNext())
        cout << range.GetNext() << " ";

    cout << endl;

    return 0;
}

---

A hint....

int main()
{
    cout<<"Five random number is here"<<endl;
    randnum();
    system ("PAUSE");
    return 0;
}

If you place a breakpoint (F9) on main's closing brace (i.e., the }), then you won't need the system ("PAUSE");. Visual Studio will break and wait for you. Once you've inspected the values, then press F5 to finish the program.

Answer 4

Try create a vector containing the numbers 1 -39, shuffle them, and pick the first 5. Then you have 5 non-repeating random numbers.

random_shuffle() function is implemented in the C++ library. Check here: http://www.cplusplus.com/reference/algorithm/random_shuffle/