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| Math Central | Quandaries & Queries |
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Question from EARL: HI, I HAVE 28 GOLFERS AND I NEED A SCHEDULE FOR 5 DIFFERENT DAYS THERE WILL BE 7 GROUPS OF 4 PLAYERS EACH DAY. ALL PLAYERS WOULD LIKE TO PLAY WITH ONE ANOTHER ONCE AND NOT PLAY WITH EACH MORE THAN ONCE IF POSSIBLE OR LIMIT THE TIMES A GOLFER PLAYS WITH ANOTHER TO A MINIMUM IF POSSIBLE. (PLEASE SEND AN EXAMPLE OF THE SCHEDULING OF THE GROUPS FOR THE TEE OFF ORDER) |
Each player plays with only 3 others each day.
Thus over 5 days [s]he cannot play with more than 15
others.
With an "incomplete design" like this it is very easy to avoid any two players playing together twice.On the first day they can play
| a | b | c | d |
| e | f | g | h |
| i | j | k | l |
| m | n | o | p |
| q | r | s | t |
| u | v | w | x |
| y | z | A | B |
On the second day slide the second row up 1, the third row up 2, and the last row up 3:
| a | f | k | p |
| e | j | o | t |
| i | n | s | x |
| m | r | w | B |
| q | v | A | d |
| u | z | c | h |
| y | b | g | l |
Do this again on each succeeding day. You can do this for up to seven days before anybody plays with the same person. This works because 28/4 = 7 is prime and greater than or equal to 5.
If you wanted everybody to play with everybody else you would need at least 9 days so that a player could play the other (28-1) = 27 three at a time. Finding designs that let each pair play together exactly once is mach harder and cannot always be done.
Good Hunting!
RD
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