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« Last post by Ian Wakeling on October 18, 2022, 01:39:42 PM »
Consider the 18 player tournament, and count the pairs in the schedule in different ways. With 8 rounds and 4 tables, there are 32 games. Each game has 2 partner pairs and 4 opposition pairs, so over the tournament 64 partner pairs and 128 opposition pairs, or a total of 192 pairs. With 18 players there are a possible 153 distinct player pairs, so the ideal arrangement, with no player pairs who never meet, or player pairs who meet 3 or more times, has:
114 pairs who meet once(x)
39 pairs who meet twice(y)
this follows because these numbers are the only solutions to:
x + y = 153
x*1 + y*2 = 192
As we have agreed, the best schedule will have all 39 pairs who meet twice of the type where 2 players partner once and oppose once. Therefore, by subtraction we can discover how the players pairs who meet once are distributed:
64 - 39 = 25 pairs who partner once
128 - 39 = 89 pairs who oppose once.
So putting this together the ideal parameters are as follows:
Never meet = 0
Partner once = 25
Oppose once = 89
Oppose twice = 0
Partner once + Oppose once = 39
Meet 3+ times = 0
This is not that far away from your Excel sheet. So I went looking, and I can find such a schedule by computer search:
(13 6 5 1) (14 3 7 16) ( 2 10 15 11) (17 12 18 9) ( 4 8)
( 6 11 12 15) ( 2 4 13 16) (17 8 10 3) ( 7 18 1 9) (14 5)
(14 8 1 12) (11 17 7 13) (15 6 16 9) ( 5 2 4 18) (10 3)
(13 1 10 16) ( 5 11 9 14) ( 3 4 18 12) ( 7 6 2 8) (15 17)
(15 17 1 4) (11 13 8 3) (18 2 14 10) (16 5 12 7) ( 9 6)
(18 14 15 13) ( 6 4 10 7) ( 5 17 16 8) ( 1 2 9 3) (12 11)
( 3 5 10 15) (11 18 16 6) ( 8 9 14 4) ( 2 12 1 17) ( 7 13)
( 3 17 6 14) ( 9 12 10 13) (15 18 7 8) ( 4 5 11 1) ( 2 16)
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