I need to make a C program that creates/generates a schedule table for tournament "each against each other".
There are 16 teams (1 to 16 numbers) and 15 rounds. The table should contain the round where i-th and j-th team are playing agains one another and must be a 2D array, shown in rows and columns (see below).
Also if i == j then a[i][j] = 0, because no team plays against itself in any round.
The conditions of the task aren't very clear to me.
I've explained above the way I'm understanding it.
However, after hours of Googling seems like its basically round robin tournament.
I guess it should look like this:
teams 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 1 0 3 4 5 6 7 8 9 10 11 12 13 14 15 2
3 2 3 0 5 6 7 8 9 10 11 12 13 14 15 1 4
4 3 4 5 0 7 8 9 10 11 12 13 14 15 1 2 6
5 4 5 6 7 0 9 10 11 12 13 14 15 1 2 3 8
6 5 6 7 8 9 0 11 12 13 14 15 1 2 3 4 10
7 6 7 8 9 10 11 0 13 14 15 1 2 3 4 5 12
8 7 8 9 10 11 12 13 0 15 1 2 3 4 5 6 14
9 8 9 10 11 12 13 14 15 0 2 3 4 5 6 7 1
10 9 10 11 12 13 14 15 1 2 0 4 5 6 7 8 3
11 10 11 12 13 14 15 1 2 3 4 0 6 7 8 9 5
12 11 12 13 14 15 1 2 3 4 5 6 0 8 9 10 7
13 12 13 14 15 1 2 3 4 5 6 7 8 0 10 11 9
14 13 14 15 1 2 3 4 5 6 7 8 9 10 0 12 11
15 14 15 1 2 3 4 5 6 7 8 9 10 11 12 0 13
16 15 2 4 6 8 10 12 14 1 3 5 7 9 11 13 0
I don't know where to start, literally the only thing I can do is fill the main diagonal with zeros.
Here is my code, but it doesn't output quite well the table:
#include<stdio.h>
void print(int a[16][16]);
void schedule(int a[16][16]);
void main(){
int a[16][16], i, j;
schedule(a);
print(a);
}
void schedule(int a[16][16]){
for (int i = 0; i < 16; i++){
for (int j = 0; j < 16; j++)
if (i == j)
a[i][j] = 0;
}
}
void print(int a[16][16]){
int k = 0;
printf("teams: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16\n");
for (int i = 0; i < 16; i++){
k++;
if (i < 10)
printf(" %d ", k);
if (i >= 10)
printf(" %d ", k);
for (int j = 0; j < 16; j++){
if (a[i][j] == 0 && j < 10){
printf("%d ", a[i][j]);
}
else if (a[i][j] == 0 && j >= 10) {
printf(" %d ", a[i][j]);
}
if (j < 10)
printf(" ");
else
printf(" ");
}
printf("\n");
}
}
From looking at the table (by rows or columns) I noticed the numbers, that normally would be before the number in a[i][0], are at the end of the row + the number "covered" with 0. In other words, it seems to me that first row is shifting to the left, but that doesn't help me to come up with an algorithm.
Use this for filling the array:
const int NofTeams=16;
int a[NofTeams][NofTeams];
int i,j;
for(i = 0; i< NofTeams-1; i++)
{
for(j =0; j < NofTeams-1; j++)
{
if(i==j)
{
/* edge cases */
a[i][NofTeams-1]=((i+j)+(i+j)/NofTeams)%NofTeams;
a[NofTeams-1][j]=((i+j)+(i+j)/NofTeams)%NofTeams;
a[i][j]=0;
} else
{
a[i][j]=((i+j)+(i+j)/NofTeams)%NofTeams;
}
}
}
/* corner cases; [0][0] is diagonal */
a[0][NofTeams-1]=NofTeams-1;
a[NofTeams-1][NofTeams-1]=0;
a[NofTeams-1][0]=NofTeams-1;
The trick is the sequence ((i+j)+(i+j)/NofTeams)%NofTeams.
This is for never printing too high a round number (maximum is number of teams minus 1):
%NofTeams.
This is for jumping the number of teams (which after % ends up as an unwanted 0):
+(i+j)/NofTeams.
This, in combination with the others, is for the basic round robin:
(i+j) (first of the two).
Otherwise the trick is to only fill the non-edge cases with the loops and do the edge cases specifically, i.e. the basic table-making concept.
Output (simple 2D table printing, with first row and first column according to your desired output):
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
2 1 0 3 4 5 6 7 8 9 10 11 12 13 14 15 2
3 2 3 0 5 6 7 8 9 10 11 12 13 14 15 1 4
4 3 4 5 0 7 8 9 10 11 12 13 14 15 1 2 6
5 4 5 6 7 0 9 10 11 12 13 14 15 1 2 3 8
6 5 6 7 8 9 0 11 12 13 14 15 1 2 3 4 10
7 6 7 8 9 10 11 0 13 14 15 1 2 3 4 5 12
8 7 8 9 10 11 12 13 0 15 1 2 3 4 5 6 14
9 8 9 10 11 12 13 14 15 0 2 3 4 5 6 7 1
10 9 10 11 12 13 14 15 1 2 0 4 5 6 7 8 3
11 10 11 12 13 14 15 1 2 3 4 0 6 7 8 9 5
12 11 12 13 14 15 1 2 3 4 5 6 0 8 9 10 7
13 12 13 14 15 1 2 3 4 5 6 7 8 0 10 11 9
14 13 14 15 1 2 3 4 5 6 7 8 9 10 0 12 11
15 14 15 1 2 3 4 5 6 7 8 9 10 11 12 0 13
16 15 2 4 6 8 10 12 14 1 3 5 7 9 11 13 0