Making groups of 30 people

Question

I would like to write a program that can provide series of groups of 3 persons, so that no person is in the group with the same person twice. So we cannot have 123 and then 124 in two different series!

For example: 9 people. It can be formed four times (absolute max):
Series 1:
Group 1: 1 2 3
Group 2: 4 5 6
Group 3: 7 8 9

Series 2:
Group 1: 1 5 9
Group 2: 2 6 7
Group 3: 3 4 8

Series 3:
Group 1 3 5 7
Group 2: 1 6 8
Group 3: 2 4 9

Series 4:
Group 1: 3 6 9
Group 2: 2 5 8
Group 3: 1 4 7

But just with 12 people I find it hard to do it by hand. It is possible to form 4-5 series of 12 people (absolute max).

I've just no idea how to write this program. I cannot find a systematic way to do it besides just "trying" with pen and paper. I would like to do it with 30 people. With 30 people may form 13-14 series. (absolute max)

Answer 1

Ok, here is a backtracking solution using c++:

#include <stdio.h>

using namespace std;

int v[100], n;
int ma[100][100];

void init(int k)
{
    v[k] = 0;
}

bool solutionReached( int k ) 
{
    if (k == n + 1)
        return true;
    return false;
}

void printSolution( int k ) 
{
    for (int i = 1; i < k; i++)
    {
        printf("%i ", v[i]);
        if (i % 3 == 0)
        {
            printf("\n");
        }
    }

    for (i = 1; i < n; i++)
    {
        if (i % 3 == 1)
        {
            for (int j = i + 1; j < i + 3; j++)
            {
                ma[v[i]][v[j]] = 1;
                ma[v[j]][v[i]] = 1;
            }
        }

        for (int j = i + 1; j % 3 == 0; j++)
        {
            ma[v[i]][v[j]] = 1;
            ma[v[j]][v[i]] = 1;
        }
    }

    for (i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            //printf("%d ", ma[i][j]);
        }
        //printf("\n");
    }

    printf("\n");
}

bool hasSuccesor( int k ) 
{
    if(v[k] < n)
    {
        v[k]++;
        return true;
    }
    return false;
}

bool isValid( int k ) 
{
    for (int i = 1; i < k; i++)
    {
        if (v[i] == v[k])
        {
            return false;
        }

        /*if (ma[v[i]][v[k]] == 1)
        {
            return false;
        }*/
    }

    for (i = 1; i < k; i++)
    {
        if (i % 3 == 1)
        {
            for (int j = i + 1; j < i + 3; j++)
            {
                if (ma[v[i]][v[j]] == 1)
                {
                    return false;
                }
            }
        }

        for (int j = i + 1; j % 3 == 0; j++)
        {
            if (ma[v[i]][v[j]] == 1)
            {
                return false;
            }
        }
    }

    return true;
}

void bkt(int k)
{
    if(solutionReached(k))
        printSolution(k);
    else
    {
        init(k);
        while(hasSuccesor(k))
            if(isValid(k))
                bkt(k + 1);
    }
}

int main(int argc, char* argv[])
{
    n = 9;
    bkt(1);

    return 0;
}

If you want to experiment, change n = 9 to any number that divides by 3, like 12, 15, 21, but it will take a lot of time for even small numbers (>15) (it depends on computer).

Edit: I redo it so that "no person is in the group with the same person twice", but I only could find 3 groups instead of 4, for 9 people.

Ex: - for 9 program gives:

1 2 3 
4 5 6 
7 8 9 

1 4 7 
2 5 8 
3 6 9 

1 5 9 
2 6 7 
4 3 8 

For 12 program gives:

1 2 3 
4 5 6 
7 8 9 
10 11 12 

1 4 7 
2 5 10 
3 8 11 
6 9 12 

1 5 8 
2 4 12 
3 9 10 
7 6 11 

For 15:

1 2 3 
4 5 6 
7 8 9 
10 11 12 
13 14 15 

1 4 7 
2 5 8 
3 10 13 
6 11 14 
9 12 15 

1 5 9 
2 4 10 
3 6 15 
7 11 13 
8 12 14 

1 6 8 
2 7 14 
4 12 13 
5 10 15 
11 3 9 

For 18: (after a minut and a half - so there are more -)

1 2 3 
4 5 6 
7 8 9 
10 11 12 
13 14 15 

1 4 7 
2 5 8 
3 10 13 
6 11 14 
9 12 15 

1 5 9 
2 4 10 
3 6 15 
7 11 13 
8 12 14 

1 6 8 
2 7 14 
4 12 13 
5 10 15 
11 3 9 

1 2 3 
4 5 6 
7 8 9 
10 11 12 
13 14 15 
16 17 18 

1 4 7 
2 5 8 
3 6 9 
10 13 16 
11 14 17 
12 15 18 

1 5 9 
2 4 10 
3 7 11 
6 13 18 
8 15 17 
14 12 16 

1 6 8 
2 7 12 
3 4 13 
5 10 17 
9 14 18 
11 15 16 

1 10 14 
2 6 15 
3 5 12 
4 9 16 
7 13 17 
8 11 18 

If you want to save them to a file, include fstream and modify printSolution to:

        void printSolution( int k ) 
    {
        ofstream cout;
        cout.open("date.txt", ios::app);

        for (int i = 1; i < k; i++)
        {
            cout << v[i] << " ";
            if (i % 3 == 0)
            {
                cout << "\n";
            }
        }

        for (i = 1; i < n; i++)
        {
            if (i % 3 == 1)
            {
                for (int j = i + 1; j < i + 3; j++)
                {
                    ma[v[i]][v[j]] = 1;
                    ma[v[j]][v[i]] = 1;
                }
            }

            for (int j = i + 1; j % 3 == 0; j++)
            {
                ma[v[i]][v[j]] = 1;
                ma[v[j]][v[i]] = 1;
            }
        }
     }