Quote:
Originally Posted by merlot128
With 12 golfers playing 4 rounds in 3 groups of 4, is it possible for everyone to play with everyone else?
Day 1 - 1 2 3 4 - 5 6 7 8 - 9 10 11 12
Day 2 - ????
Day 3 - ????
Day 4 - ????
Any advice would be appreciated.
Suppose it is possible.
Now, each player must play with exactly one other person twice, and with the other 10 players once. We can renumber the players and rearrange the days so that the match-ups look like this:
Code:
1 2 3 4 / __ __ __ __ / __ __ __ __
1 5 6 7 / __ __ __ __ / __ __ __ __
1 8 9 10 / __ __ __ __ / __ __ __ __
1 11 12 10 / __ __ __ __ / __ __ __ __
Suppose Player 2 only plays once with Player 3 and once with Player 4. Then the match-ups would like this:
Code:
1 2 3 4 / __ __ __ __ / __ __ __ __
1 5 6 7 / 2 __ __ __ / 3 __ __ __
1 8 9 10 / 2 __ __ __ / 3 __ __ __
1 11 12 10 / 2 __ __ __ / 3 __ __ __
And we see there is no room for Player 4. Hence, Player 2 will play twice with either Player 3 or Player 4. Without loss of generality, assume Player 2 plays twice with Player 3. Then both Player 2 and Player 3 can only play once with Player 4. Therefore, the match-ups look like this:
Code:
1 2 3 4 / __ __ __ __ / __ __ __ __
1 5 6 7 / 4 __ __ __ / 2 3 __ __
1 8 9 10 / 4 __ __ __ / 2 3 __ __
1 11 12 10 / 4 __ __ __ / 2 3 __ __
And we see that Players 2 and 3 play each other four times, which is a contradiction.
It is therefore not possible.