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.

not smart enough to prove it mathematically but doesn't seem possible... if you minimize overlapping/duplicate pairings through the first 3 rounds there's no way to make it work in round 4.

3*4 - aaaabbbbccc*4 rounds

Three groups; A1(234)B1(234)C1(234)


Four days


First day; you don't specify time limits. You imply 12 holes a day but don't mention holes/day holes/course or time implications. Nor rounds/day


Is it 18 holes in a masters?? My local ground has 11 holes and a putting green.


I'd rather go to the driving green, or better yet the bar. Yawn.

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:
Suppose Player 2 only plays once with Player 3 and once with Player 4. Then the match-ups would like this:
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:
And we see that Players 2 and 3 play each other four times, which is a contradiction.

It is therefore not possible.

fmp