Mathematical_Recreations-Kraitchik-2e

Permutational Problems ~35
n = 17
1,17- 2,16 3,15- 4,14 5,13- 6,12 7,11- 8,10 9
2, 1- 3,17 4,16- 5,15 6,14- 7,13 8,12- 9,11 10
3, 2- 4, 1 5,17- 6,16 7,15- 8,14 9,13-10,12 11
4, 3- 5, 2 6, 1- 7,17 8,16- 9,15 10,14-11,13 12
5, 4- 6, 3 7, 2- 8, 1 9,17-10,16 11,15-12,14 13
6, 5- 7, 4 8, 3- 9, 2 10, 1-11,17 12,16-13,15 14
7, 6- 8, 5 9, 4-10, 3 11, 2-12, 1 13,17-14,16 15
8, 7- 9, 6 10, 5-11, 4 12, 3-13, 2 14, 1-15,17 16
9, 8-10, 7 11, 6-12, 5 13, 4-14, 3 15, 2-16, 1 17
10, 9-11, 8 12, 7-13, 6 14, 5-15, 4 16, 3-17, 2 1
11,10-12, 9 13, 8-14, 7 15, 6-16, 5 17, 4- 1, 3 2
12,11-13,10 14, 9-15, 8 16, 7-17, 6 1, 5- 2, 4 3
13,12-14,11 15,10-16, 9 17, 8- 1, 7 2, 6- 3, 5 4
14,13-15,12 16,11-17,10 1, 9- 2, 8 3, 7- 4, 6 5
15,14-16,13 17,12- 1,11 2,10- 3, 9 4, 8- 5, 7 6
16,15-17,14 1,13- 2,12 3,11- 4,10 5, 9- 6, 8 7
17,16- 1,15 2,14- 3,13 4,12- 5,11 6,10- 7, 9 8
These tables ltre not as satisfactory as they might be since
a particular player does not have the other players as opponents
the same number of times. For example, when
n = 8 the first player meets the second 4 times, the third twice
and the fourth not at all. A more equitable arrangement
would be obtained if we could replace conditions (2) and (3)
by the following:
(2') Each player has every other player as partner just once.
(3') Each player has every other player as opponent just
twice.
These conditions can be realized when n is of the form
4k or 4k + 1, but not otherwise.
For n = 4 and n = 5 the schedule is the same as in the preceding
table.