Cyclic solutions to
the
round-robin Bridge tournament problem
Solutions for 1 through 23 Bridge tables (up to 92
players)
are presented
Combinatorial Optimization for Whist Tournaments (Whist
Tournament/Schedule)
Combinatorics and Computer Algorithm are shown at the end
(Includes the “Conveyor Belt Algorithm” for Simple
Partnerships/Player vs. Player)
Suppose you are organizing a Bridge round robin tournament
(Could also be tennis doubles, partnership golf, partnership
horseshoes, etc.), and to equally mix the
partnership/opponent pairs, you would like to have every player have
every other player as a partner once and as an opponent twice. For
example, if there are 8 players “A”, “B”,
“C”, “D”, “E”, “F”,
“G”, and “H”, you will have two tables of
Bridge players. Suppose you wish to arrange a sequential seating
arrangement such that after 7 rounds of Bridge:
Player “A” has had player “B” as a partner once
and as an opponent twice.
Player “A” has had player “C” as a partner once
and as an opponent twice.
Player “A” has had player “D” as a partner once
and as an opponent twice.
Player “A” has had player “E” as a partner once
and as an opponent twice.
Player “A” has had player “F” as a partner once
and as an opponent twice.
Player “A” has had player “G” as a partner once
and as an opponent twice.
Player “A” has had player “H” as a partner once
and as an opponent twice.
Player “B” has had player “A” as a partner once
and as an opponent twice.
Player “B” has had player “C” as a partner once
and as an opponent twice.
Player “B” has had player “D” as a partner once
and as an opponent twice.
Player “B” has had player “E” as a partner once
and as an opponent twice.
etc.
for all possible combinations.
The diagram above shows one of the 6 intrinsically
different position solutions for a 2-table round-robin tournament.
(There are multiple permutations within these 6 patterns.) To solve the
problem, start with player “A” at position “0”,
player “B” at position “1”, player
“C” at position “2”, etc. for the first round
of play. When the first round of play is over, player “A”
stays at position “0”, and all other players advance to the
next higher “position”. (With the exception of the person
who was at “7” who instead cycles back to position
“1”.) This same
advancement/cycle sequence is repeated for the third round, etc. The
full sequence of who sits where for each round then becomes:
Round
Player Player Player Player Player
Player Player Player
Nbr.
A
B
C
D
E
F
G H
1
0
1
2
3
4
5
6 7
2
0
2
3
4
5
6
7 1
3
0
3
4
5
6
7
1 2
4
0
4
5
6
7
1
2 3
5
0
5
6
7
1
2
3 4
6
0
6
7
1
2
3
4 5
7
0
7
1
2
3
4
5 6
Each entry in the above table shows the Bridge table
position/location that each person should go to on each of the 7 rounds
of the tournament. When all 7 rounds have finished, each player will
have had each other player as a partner once and as an opponent twice.
A shorthand summary for this particular 2-table solution plus the other
5 solutions is thus:
Solution
<---- Table 1 ----> <---- Table 2
---->
1 2 and 3 vs. 4 and
6 5 and 1 vs. 7 and 0
2 2 and 3 vs. 5 and
1 4 and 6 vs. 7 and 0
3 2 and 3 vs. 7 and
0 4 and 6 vs. 5 and 1
4 4 and 5 vs. 1 and
3 6 and 2 vs. 7 and 0
5 4 and 5 vs. 6 and
2 1 and 3 vs. 7 and 0
6 4 and 5 vs. 7 and
0 1 and 3 vs. 6 and 2
A Solution for 3 Bridge Tables (12 players)
The diagram below shows one of the table position
definitions that will solve the problem for 12 players. (There are 20
intrinsically different cyclical solutions - each of which has a large
number of permutations.)
Solution
<----- Table 1 -----> <----- Table 2
-----> <----- Table 3 ----->
1 2 and 3 vs. 1 and
6 8 and 10 vs. 4 and 7 5 and 9 vs.
11 and 0
The complete cycle sequence for these 12 players would be:
Round Plr
Plr Plr Plr Plr Plr Plr Plr Plr Plr Plr Plr
Nbr
A B C D E
F G H I J
K L
1 0 1 2
3 4 5 6 7
8 9 10 11
2 0 2 3
4 5 6 7 8
9 10 11 1
3 0 3 4
5 6 7 8 9
10 11 1 2
4 0 4 5
6 7 8 9 10
11 1 2 3
5 0 5 6
7 8 9 10 11
1 2 3 4
6 0 6 7
8 9 10 11 1
2 3 4 5
7 0 7 8
9 10 11 1 2
3 4 5 6
8 0 8 9 10
11 1 2 3 4
5 6 7
9 0 9 10 11
1 2 3 4 5
6 7 8
10
0 10 11 1 2
3 4 5 6 7
8 9
11
0 11 1 2 3
4 5 6 7 8
9 10
Again, when all 11 rounds have finished, each player will have had each
other player as a partner once and as an opponent twice.
There is considerable latitude in setting up a system that
will work. While the above examples list the players in alphabetical
order, this is not necessary. Players can start at any random position.
Also the table numbers can be permuted in any order. For 3 tables the
names “Table 1”, “Table 2”, and “Table
3” can be permuted in any of 3 x 2 x 1 = 6 ways. Also, the
north-south and east-west definitions can be swapped and the compass
directions of the tables can be rotated. Finally, there are solutions
using a complementary numbering system which has the effect of
reversing the cycle sequence. However, for any individual table, the
numeric pairs must be matched as above. (E.g. for Table 1 in the above
example, Positions 2 and 3 must be opposite each other, and the other 2
positions must be 1 & 6.)
The other 19 solutions are:
Solution
<----- Table 1 -----> <----- Table 2
-----> <----- Table 3 ----->
2 9 & 10 vs.
5 & 7 1 & 4 vs. 11
& 0 2 & 6 vs. 3
& 8
3 1 & 2
vs. 4 & 6 7 & 10
vs. 11 & 0 5 & 9
vs. 3 & 8
4 8 & 9
vs. 5 & 10 1 & 3
vs. 4 & 7 2 & 6
vs. 11 & 0
5 4 & 5
vs. 9 & 2 8 & 10
vs. 11 & 0 3 & 6
vs. 7 & 1
6 8 & 9
vs. 4 & 6 2 & 5
vs. 10 & 3 7 & 1
vs. 11 & 0
7 8 & 9
vs. 4 & 6 2 & 5
vs. 7 & 1 10 & 3
vs. 11 & 0
8 8 & 9
vs. 7 & 1 4 & 6
vs. 10 & 3 2 & 5
vs. 11 & 0
9 8 & 9
vs. 7 & 1 4 & 6
vs. 11 & 0 2 & 5
vs. 10 & 3
10 8 & 9 vs.
11 & 0 4 & 6 vs. 10
& 3 2 & 5 vs. 7
& 1
11 4 & 5 vs.
11 & 0 10 & 1 vs. 8
& 2 6 & 9 vs. 3
& 7
12 5 & 6
vs. 1 & 4 7 & 9
vs. 10 & 3 8 & 2
vs. 11 & 0
13 6 & 7 vs.
11 & 0 10 & 1 vs. 9
& 3 2 & 5 vs. 4
& 8
14 5 & 6
vs. 7 & 10 2 & 4
vs. 8 & 1 9 & 3
vs. 11 & 0
15 2 & 3
vs. 5 & 7 6 & 9
vs. 8 & 1 10 & 4
vs. 11 & 0
16 2 & 3
vs. 5 & 7 6 & 9
vs. 10 & 4 8 & 1
vs. 11 & 0
17 2 & 3 vs.
10 & 4 5 & 7
vs. 8 & 1 6 & 9
vs. 11 & 0
18 2 & 3 vs.
10 & 4 5 & 7 vs. 11
& 0 6 & 9 vs. 8
& 1
19 2 & 3 vs.
11 & 0 5 & 7
vs. 8 & 1 6 & 9
vs. 10 & 4
20 6 & 7
vs. 9 & 2 1 & 3
vs. 11 & 0 5 & 8
vs. 10 & 4
Other Table Configurations
There are cyclic solutions for larger numbers of
tables. (Number of players always equals 4 times the number of tables.)
If there is only 1 table (4 players), the answer is trivial. One player
remains at position “0” and the other 3 players sequence in
either a clockwise or counterclockwise direction.
A solution for 4 Bridge Tables (16 players) (Selected
from a total of 128 solutions)
Rapidly increasing numbers of solutions exist for all
larger numbers of Bridge tables. The solution below defines a
table-position numbering system that will allow each player to have
each other player as a partner once and as an opponent twice. Players
can be seated randomly, and thereafter sequence/cycle as in the earlier
examples
Table 1
12 and 13 vs. 3 and 9
Table 2 4
and 6 vs. 14 and 2
Table 3 7
and 11 vs. 5 and 10
Table 4 1
and 8 vs. 15 and 0
A solution for 5 Bridge Tables (20 players) (Selected
from a total of 1,710 solutions)
Table 1
14 and 15 vs. 19 and 0
Table 2 16 and
18 vs. 1 and 10
Table 3 4
and 7 vs. 6 and 13
Table 4 5
and 9 vs. 12 and 17
Table 5 2
and 8 vs. 3 and 11
A solution for 6 Bridge Tables (24 players) (Selected
from a total of 81,576 solutions)
Table 1
17 and 18 vs. 3 and 13
Table 2 20 and
22 vs. 6 and 9
Table 3 15
and 19 vs. 8 and 16
Table 4 5
and 10 vs. 2 and 11
Table 5 21
and 4 vs. 23 and 0
Table 6 7
and 14 vs. 1 and 12
A solution for 7 Bridge Tables (28 players) (Selected
from a total of 7,910,127 solutions)
Table 1
22 and 23 vs. 1 and 14
Table 2 5
and 7 vs. 8 and 17
Table 3 9
and 12 vs. 4 and 11
Table 4 21
and 25 vs. 10 and 18
Table 5 19
and 24 vs. 3 and 15
Table 6 20
and 26 vs. 6 and 16
Table 7
2 and 13 vs. 27 and 0
A solution for 8 Bridge Tables (32 players)
Table 1
24 and 25 vs. 31 and 0
Table 2 9 and
11 vs. 28 and 5
Table 3 10 and
13 vs. 22 and 26
Table 4 7
and 12 vs. 2 and 15
Table 5 21
and 27 vs. 20 and 29
Table 6 23
and 30 vs. 8 and 19
Table 7 4
and 14 vs. 3 and 17
Table 8
6 and 18 vs. 1 and 16
A solution for 9 Bridge Tables (36 players)
Table 1
26 and 27 vs. 11 and 22
Table 2 29 and
31 vs. 3 and 19
Table 3 9
and 12 vs. 6 and 20
Table 4 10
and 14 vs. 23 and 30
Table 5 28
and 33 vs. 4 and 16
Table 6 7
and 13 vs. 8 and 21
Table 7 24
and 32 vs. 25 and 34
Table 8 5
and 15 vs. 1 and 18
Table 9 2
and 17 vs. 35 and 0
A solution for 10 Bridge Tables (40 players)
Table 1
33 and 34 vs. 5 and 17
Table 2 29 and
31 vs. 11 and 24
Table 3 12 and
15 vs. 26 and 36
Table 4 9
and 13 vs. 6 and 22
Table 5 27
and 32 vs. 28 and 35
Table 6 10
and 16 vs. 4 and 18
Table 7 30
and 38 vs. 3 and 21
Table 8 37
and 7 vs. 8 and 23
Table 9 14
and 25 vs. 1 and 20
Table 10 2 and
19 vs. 39 and 0
A solution for 11 Bridge Tables (44 players)
Table 1
9 and 10 vs. 5 and 19
Table 2 33 and
35 vs. 4 and 20
Table 3 12 and
15 vs. 1 and 22
Table 4 13
and 17 vs. 30 and 38
Table 5 32
and 37 vs. 11 and 26
Table 6 36
and 42 vs. 34 and 41
Table 7 31
and 40 vs. 16 and 28
Table 8 29
and 39 vs. 2 and 21
Table 9 7
and 18 vs. 14 and 27
Table 10 8 and
25 vs. 6 and 24
Table 11 3
and 23 vs. 43 and 0
A solution for 12 Bridge Tables (48 players)
Table 1
35 and 36 vs. 19 and 31
Table 2 37 and
39 vs. 4 and 22
Table 3 12 and
15 vs. 5 and 21
Table 4 13
and 17 vs. 2 and 23
Table 5 40
and 45 vs. 32 and 43
Table 6 38
and 44 vs. 10 and 18
Table 7 34
and 41 vs. 16 and 30
Table 8 33
and 42 vs. 11 and 28
Table 9 46
and 9 vs. 8 and 27
Table 10 7 and
20 vs. 47 and 0
Table 11 14 and
29 vs. 6 and 26
Table 12 3
and 25 vs. 1 and 24
A solution for 13 Bridge Tables (52
players)
Table 1 41 and 42 vs. 39 and 48
Table 2 38 and 40 vs. 12 and 15
Table 3 13 and 17 vs. 2 and 25
Table 4 45 and 50 vs. 36 and 46
Table 5 43 and 49 vs. 14 and 31
Table 6 37 and 44 vs. 4 and 24
Table 7 10 and 18 vs. 1 and 26
Table 8 9 and 20 vs. 19 and 33
Table 9 35 and 47 vs. 6 and 28
Table 10 21 and 34 vs. 51 and 0
Table 11 7 and 22 vs. 3 and 27
Table 12 16 and 32 vs. 11 and 30
Table 13 5 and 23 vs. 8 and 29
A solution for 14 Bridge Tables (56
players)
Table 1 44 and 45 vs. 5 and 25
Table 2 41 and 43 vs. 15 and 18
Table 3 13 and 17 vs. 3 and 29
Table 4 48 and 53 vs. 40 and 47
Table 5 46 and 52 vs. 1 and 28
Table 6 12 and 20 vs. 38 and 50
Table 7 42 and 51 vs. 55 and 0
Table 8 39 and 49 vs. 6 and 30
Table 9 54 and 10 vs. 8 and 31
Table 10 9 and 22 vs. 14 and 33
Table 11 23 and 37 vs. 16 and 34
Table 12 21 and 36 vs. 19 and 35
Table 13 7 and 24 vs. 2 and 27
Table 14 11 and 32 vs. 4 and 26
A solution for 15 Bridge Tables (60 players)
Table 1 17 and 18 vs. 49 and 57
Table 2 56 and 58 vs. 2 and 29
Table 3 44 and 47 vs. 3 and 31
Table 4 46 and 50 vs. 9 and 25
Table 5 15 and 20 vs. 11 and 34
Table 6 42 and 48 vs. 41 and 51
Table 7 45 and 52 vs. 59 and 0
Table 8 13 and 22 vs. 14 and 35
Table 9 12 and 23 vs. 8 and 33
Table 10 43 and 55 vs. 7 and 26
Table 11 40 and 53 vs. 4 and 28
Table 12 10 and 24 vs. 16 and 36
Table 13 39 and 54 vs. 19 and 37
Table 14 21 and 38 vs. 5 and 27
Table 15 6 and 32 vs. 1 and 30
A solution for 16 Bridge Tables (64
players)
Table 1 60 and 61 vs. 18 and 20
Table 2 49 and 52 vs. 3 and 33
Table 3 46 and 50 vs. 43 and 54
Table 4 48 and 53 vs. 10 and 26
Table 5 17 and 23 vs. 13 and 25
Table 6 15 and 22 vs. 9 and 27
Table 7 51 and 59 vs. 6 and 34
Table 8 47 and 56 vs. 16 and 38
Table 9 45 and 55 vs. 21 and 40
Table 10 62 and 12 vs. 11 and 36
Table 11 44 and 58 vs. 14 and 37
Table 12 42 and 57 vs. 7 and 28
Table 13 24 and 41 vs. 8 and 35
Table 14 19 and 39 vs. 4 and 30
Table 15 5 and 29 vs. 2 and 31
Table 16 1 and 32 vs. 63 and 0
A solution for 17 Bridge Tables (68
players)
Table 1
65 and 66 vs. 49 and 60
Table 2 18 and
20 vs. 2 and 33
Table 3 50 and
53 vs. 6 and 36
Table 4 51
and 55 vs. 15 and 22
Table 5 52 and
57 vs. 54 and 63
Table 6 17 and
23 vs. 13 and 25
Table 7 56 and
64 vs. 7 and 30
Table 8 48
and 58 vs. 12 and 26
Table 9 46 and
59 vs. 1 and 34
Table 10 47 and 62
vs. 4 and 32
Table 11 45 and
61 vs. 19 and 41
Table 12 27 and
44 vs. 67 and 0
Table 13 10 and
28 vs. 21 and 42
Table 14 24 and
43 vs. 5 and 31
Table 15 9
and 29 vs. 8 and 37
Table 16 16 and
40 vs. 3 and 35
Table 17 14 and
39 vs. 11 and 38
A solution for 18 Bridge Tables (72
players)
Table 1 68 and 69 vs. 20 and 22
Table 2 55 and 58 vs. 52 and 57
Table 3 56 and 60 vs. 7 and 32
Table 4 17 and 23 vs. 5 and 33
Table 5 18 and 25 vs. 50 and 63
Table 6 59 and 67 vs. 1 and 36
Table 7 53 and 62 vs. 49 and 64
Table 8 51 and 61 vs. 10 and 30
Table 9 15 and 26 vs. 19 and 43
Table 10 54 and 66 vs. 27 and 46
Table 11 70 and 13 vs. 4 and 34
Table 12 12 and 28 vs. 29 and 47
Table 13 48 and 65 vs. 6 and 38
Table 14 24 and 45 vs. 8 and 39
Table 15 9 and 31 vs. 16 and 42
Table 16 21 and 44 vs. 2 and 35
Table 17 14 and 41 vs. 71 and 0
Table 18 11 and 40 vs. 3 and 37
A solution for 19 Bridge Tables (76
players)
Table 1 73 and 74 vs. 20 and 22
Table 2 57 and 60 vs. 59 and 63
Table 3 56 and 61 vs. 52 and 66
Table 4 17 and 23 vs. 24 and 47
Table 5 18 and 25 vs. 4 and 36
Table 6 64 and 72 vs. 29 and 49
Table 7 62 and 71 vs. 7 and 34
Table 8 55 and 65 vs. 15 and 26
Table 9 58 and 70 vs. 14 and 43
Table 10 54 and 67 vs. 5 and 35
Table 11 13 and 28 vs. 21 and 46
Table 12 53 and 69 vs. 6 and 40
Table 13 51 and 68 vs. 2 and 37
Table 14 12 and 30 vs. 75 and 0
Table 15 31 and 50 vs. 9 and 33
Table 16 27 and 48 vs. 10 and 32
Table 17 19 and 45 vs. 3 and 39
Table 18 16 and 44 vs. 8 and 41
Table 19 11 and 42 vs. 1 and 38
A solution for 20
Bridge Tables (80 players)
Table 1 76
and 77 vs. 23 and 25
Table 2 60 and 63 vs. 20 and 28
Table 3 22 and 26 vs. 52 and 72
Table 4 57 and 62 vs. 24 and 49
Table 5 61 and 67 vs. 55 and 68
Table 6 58 and 65 vs. 79 and 0
Table 7 66 and 75 vs. 11 and 44
Table 8 64 and 74 vs. 27 and 50
Table 9 59 and 70 vs. 3 and 41
Table 10 18 and 30 vs. 21 and 48
Table 11 17 and 31 vs. 4 and 38
Table 12 54 and 69 vs. 53 and 71
Table 13 78 and 15 vs. 10 and 34
Table 14 56 and 73 vs. 5 and 37
Table 15 13 and 32 vs. 9 and 35
Table 16 12 and 33 vs. 8 and 43
Table 17 29 and 51 vs. 14 and 45
Table 18 19 and 47 vs. 2 and 39
Table 19 7 and 36 vs. 16 and 46
Table 20 6 and 42 vs. 1 and 40
A solution for 21 Bridge Tables (84 players) (Added on July 16, 2008)
Table 1 80 and 81 vs. 23 and 25
Table 2 79 and 82 vs. 63 and 68
Table 3 22 and 26 vs. 65 and 77
Table 4 64 and 70 vs. 2 and 40
Table 5 59 and 66 vs. 16 and 57
Table 6 61 and 69 vs. 62 and 76
Table 7 19 and 28 vs. 24 and 50
Table 8 20 and 30 vs. 6 and 43
Table 9 67 and 78 vs. 55 and 72
Table 10 58 and 71 vs. 5 and 38
Table 11 60 and 75 vs. 12 and 34
Table 12 15 and 31 vs. 32 and 53
Table 13 56 and 74 vs. 10 and 35
Table 14 54 and 73 vs. 8 and 44
Table 15 13 and 33 vs. 21 and 49
Table 16 29 and 52 vs. 4 and 39
Table 17 27 and 51 vs. 83 and 0
Table 18 9 and 36 vs. 18 and 47
Table 19 7 and 37 vs. 1 and 41
Table 20 17 and 48 vs. 14 and 46
Table 21 11 and 45 vs. 3 and 42
A solution for 22 Bridge Tables (88 players) (Added on July 20, 2008)
Table 1 84 and 85 vs. 66 and 68
Table 2 25 and 28 vs. 22 and 26
Table 3 65 and 70 vs. 63 and 71
Table 4 67 and 73 vs. 8 and 46
Table 5 23 and 30 vs. 64 and 78
Table 6 74 and 83 vs. 7 and 50
Table 7 72 and 82 vs. 87 and 0
Table 8 20 and 31 vs. 27 and 54
Table 9 69 and 81 vs. 29 and 55
Table 10 62 and 75 vs. 13 and 36
Table 11 18 and 33 vs. 4 and 41
Table 12 61 and 77 vs. 15 and 34
Table 13 59 and 76 vs. 5 and 40
Table 14 86 and 17 vs. 14 and 48
Table 15 60 and 80 vs. 10 and 38
Table 16 58 and 79 vs. 16 and 49
Table 17 35 and 57 vs. 6 and 45
Table 18 32 and 56 vs. 1 and 43
Table 19 12 and 37 vs. 2 and 42
Table 20 24 and 53 vs. 3 and 44
Table 21 9 and 39 vs. 21 and 52
Table 22 19 and 51 vs. 11 and 47
A solution for 23 Bridge Tables (92 players) (Added on Aug. 16, 2008)
It
took nearly 4 weeks of search time on a Skulltrail computer,
simultaneously running 7 copies of the program, to find the following
solution. The complexity of the search increases rapidly with any
further increase in the number of bridge tables. Thus, the search will
not be continued to higher orders.
Table 1 89 and 90 vs. 68 and 70
Table 2 25 and 28 vs. 69 and 73
Table 3 72 and 77 vs. 67 and 75
Table 4 20 and 26 vs. 76 and 87
Table 5 23 and 30 vs. 29 and 57
Table 6 22 and 31 vs. 74 and 86
Table 7 78 and 88 vs. 62 and 82
Table 8 66 and 79 vs. 4 and 44
Table 9 71 and 85 vs. 13 and 38
Table 10 18 and 33 vs. 5 and 43
Table 11 64 and 80 vs. 35 and 59
Table 12 17 and 34 vs. 9 and 41
Table 13 63 and 81 vs. 2 and 45
Table 14 65 and 84 vs. 16 and 52
Table 15 15 and 36 vs. 19 and 53
Table 16 61 and 83 vs. 1 and 46
Table 17 37 and 60 vs. 6 and 48
Table 18 32 and 58 vs. 24 and 55
Table 19 12 and 39 vs. 11 and 50
Table 20 27 and 56 vs. 7 and 42
Table 21 10 and 40 vs. 8 and 49
Table 22 21 and 54 vs. 3 and 47
Table 23 14 and 51 vs. 91 and 0
Combinatorics and the
Computer Algorithm
The table below shows some of the combinatorics associated
with the round-robin problem. If you are going to have a tournament
without someone sitting out part of the time, the number of players
must be a multiple of four. In each game each player has a partner.
Thus the number of teams is exactly one half of the number of players.
A Bridge table has 4 positions for the players (Usually referred to as:
North, East, South, and West). Hence the number of tables is
one-quarter the number of players.
Nbr. Perm. In theory Qualifying For
each set Actual Nbr.
Nbr.
Nbr. Nbr. for
Nbr. Nbr. Comb Nbr.
Teams of teams Nbr Solutions
Players Teams
Tables Players for
Teams via Prgrm Table Combs. via Program
4
2
1
24
1
1
1
1
8
4
2
40,320
15
3
3
6
12
6 3
4.7900E+08
945
25
15 20
16
8 4
2.0923E+13
135,135
631
105 128
20 10
5 2.4329E+18
3.4459E+07
25,905
945 1,710
24 12
6 6.2045E+23
1.3749E+10
1,515,283
10,395 81,576
28 14
7 3.0489E+29 7.9059E+12
128,102,625
135,135 7,910,127
32 16
8 2.6313E+35
6.1903E+15
2,027,025
36 18
9 3.7199E+41
6.3327E+18
34,459,425
40 20
10 8.1592E+47
8.2008E+21
6.5473E+08
44 22
11 2.6583E+54
1.3113E+25
1.3749E+10
48 24
12 1.2414E+61
2.5374E+28
3.1623E+11
52 26
13 8.0858E+67
5.8436E+31
7.9059E+12
56 28
14 7.1100E+74
1.5795E+35
2.1346E+14
60 30
15 8.3210E+81
4.9518E+38
6.1903E+15 Probably >
64 32
16 1.2689E+89
1.7822E+42
1.9190E+17 Number of
68 34
17 2.4800E+96
7.2979E+45
6.3327E+18 Qualifying
72 36
18 6.1234E+103
3.3738E+49
2.2164E+20 Teams when
76 38
19 1.8855E+111
1.7487E+53
8.2008E+21 Nbr.Players
80 40
20 7.1569E+118
1.0098E+57
3.1983E+23 is > 60.
84 42
21 3.3142E+126
6.4620E+60
1.3113E+25
88
44 22
1.8548E+134
4.5590E+64
5.6386E+26
92 46
23 1.2438E+142
3.5300E+68
2.5374E+28
=
=
=
NbrPlayers! (NbrPlayers-2)!
/ (NbrTeams!) /
(NbrTeams-1)! /
(NbrTables!) /
(2^(NbrTeams-1)) (2^NbrTables)
If there were no constraints regarding the seating
arrangements for 92 players, you could fill the first table position
with any of 92 player-positions, the second table position with any of
the remaining 91 player-positions, etc. for all 92 positions. The
number of possible permutations is thus “NbrPlayers”
Factorial. By the time you get to 92 players, the result is billions of
times larger than the number of atoms in the known universe. How can I
claim this when no one has counted the number of atoms in the known
universe? Practical answer: “Pick a number” - any number
you want, for the number of atoms in the known universe.
The actual definition of a Bridge game requires that a
player-position must have a specified partner-position who sits on the
far side of a Bridge table. Thus the positions for the partners are
paired to form teams. For the solution of the round-robin problem, one
team combination is defined and held as a constant while the other team
positions are allowed to float. In the diagram below which shows
combinations for 8 players, position “0” is always paired
with position “7”. In general, if there are “N”
players involved, then the player-positions are identified by the
digits from “0” to “N-1”. Position
“0” is always teamed with position “N-1”

In the diagram above, note how player-positions 1 to 7 are
on the circumference of a circle while position “0” is
“fixed” outside the circle. The diagram shows 2 different
ways that an 8-player tournament can be paired into teams. If there
were no restrictions on pairing combinations, then you could
arbitrarily connect any two of the player/positions. Since we have
already isolated two players (In this case, positions “0”
and “7”), the total number of player-positions to be paired
is 2 less than “NbrPlayers”, which can be done in
FACT(NbrPlayers-2) ways. There would be considerable duplication of
these pairings as the teams could be permuted in FACT(NbrTeams-1) ways.
Thus we divide FACT(NbrPlayers-2) by FACT(NbrTeams-1) ways to adjust
for permutations of team positions. Finally, each team could swap the
position of the partners (north to south or east to west). We divide
again by 2^(NbrTeams-1) which gives us the theoretical number of ways
the player-positions could be matched into teams.
However, the round-robin solution requires that each
player must have each other player as a partner exactly once. In our
diagram, this will only happen if we can pair all the players in a way
that results in each line having a different span length (chord
distance). The player-position from “0” to “7”
has a defined length of “0”. (In any size diagram, the line
connecting position “0” with position “N-1”
will always have a length of zero.) In the left diagram above (red
lines), the line connecting position “2” to position
“3” has a length of “1”. Similarly, the line
connecting “4” and “6” has a length of
“2”. Finally, the line connecting “1” to
“5” has a length of “3”. (Note: This is a
“Shortest chord” length.) The blue lines show a
symmetrically similar solution. Both of these diagrams lead to
solutions to the round-robin tournament problem. There is still a third
combination to connect the player-positions such that all lengths are
different. (See if you can find it.)
One of the tasks of the computer program is to generate
these connections. The program also counts them as they are found. It
found 3 of these for the 8-player problem, 25 for the 12-player
problem, etc. By the time the computer program got to 28 players, it
found 128,102,625 different ways to match player positions such that
each player would have each other player as a partner exactly once.
The next step in the computer program was to permute each
of the above qualifying team candidates among the tables to see if each
player will have each other player as an opponent exactly twice. For
this, we use the above distance diagram again with the additional
definition that all distances between any position and position
“0” are defined as 0. We then match all opponent
combinations to see if each distance occurs exactly twice.
As an example, let’s use the left (red) solution of:
2 and 3 vs. 4 and 6 5 and 1 vs. 7 and 0
for the 8-player problem.
If we tabulate the opponents and distances for “2 and 3 vs. 4 and
6”, we have the following opponent matches and distances:
2 to 4, dist = 2
2 to 6, dist = 3
3 to 4, dist = 1
3 to 6, dist = 3
Similarly, “5 and 1 vs. 7 and 0” yields:
5 to 7, dist = 2
5 to 0, dist = 0
1 to 7, dist = 1
1 to 0, dist = 0
In the above listing, each of the distances (0, 1, 2, 3) occurred
exactly twice. Thus,
2 and 3 vs. 4 and 6 5 and 1 vs. 7 and 0
is a valid solution to the round-robin problem. It turns out that both
of the other two permutations of these teams also pass the
“opponent” test. Solutions 1, 2 and 3 (see the list for
8-player solutions above) are team permutations of the
“Red” diagram. The “Blue” diagram supplies 3
more solutions to the 8-player problem. As mentioned earlier, there is
still another “pairing” of the positions 1 through 6.
However, this candidate did not pass the above “opponent”
test.
By the time the problem was expanded to 28 players, the
computer found 128,102,625 different ways of pairing the players such
that each chord distance was different. Each of these 128,102,625
candidates was then run through a search tree that has 135,135
different permutations to see
if a total solution existed. 7,910,127 of these yielded valid position
definitions that solve a round robin tournament for 28 players. This
effort can be affectionately described as “Computationally
intensive”.
For larger numbers of players, the search was cut short
after a single solution was found. The search algorithm prunes off
branches whenever possible, but
there is a limit to what computers can do.
Finally, the computer program has a “verify”
function that sequences players through the positions as shown for the
8-player and 12-player solutions. This was done just to double check
that everything was valid. All solutions shown here passed this
“double check” test.
“Conveyor Belt
Algorithm” for Simple Partnerships
A related round-robin scheduling problem involves
sequencing tournament play for fixed teams. This would apply when the
individuals in each team form a permanent partnership, but you wish to
schedule team play so that each team plays against each other team
exactly once. The algorithm/scheduling sequence given below will also
work for any single-player game where each player plays against each
other player exactly once. Examples would include chess, singles
tennis, any player-vs.-player game, any team-vs.-team game, etc.
The algorithm will work for any number of teams. The
example shown below has an even number of teams, but if you have an odd
number, just add a “dummy” team named “Bye” to
make the total come out even. Anyone who plays against
“Bye” gets an automatic win for that round.
The example uses 8 teams. If you have more (or less) than
8 teams, just extend (or shorten) the length of the columns as needed.
Team positions are defined by the digits 0 to 7. Two columns are used
with the digits 1 to 4 in the right column. The left column has the
remaining digits (5-7) plus a “0” at the top. If you have a
larger number of teams, simply extend the columns. For example, if you
have 20 teams, the right column would have the numbers from 1 to 10
while the left column would work up from 11 to 19 with a
“0” at the top.
0 vs. 1
7 vs. 2
6 vs. 3
5 vs. 4
Initially, teams may be assigned to the 8 positions in any random
order. For our example:
Team A is at position 0
Team B is at position 1
Team C is at position 2
Team D is at position 3
Team E is at position 4
Team F is at position 5
Team G is at position 6
Team H is at position 7
For each round of play, the team that is at position 0
plays against the team at position 1, etc. for each row of the diagram.
Thus, first round opponents are:
Team A (at position 0) plays against Team B (at position 1)
Team H (at position 7) plays against Team C (at position 2)
Team G (at position 6) plays against Team D (at position 3)
Team F (at position 5) plays against Team E (at position 4)
For the next round, the teams move as follows: Whoever was
at position “0” stays there. (In fact the team at
“0” stays there for the entire tournament.) All other teams
advance one position except for the team at “7” which
cycles back to position “1”. Thus for round 2:
Team A stays at position 0
Team B moves from position 1 to position 2
Team C moves from position 2 to position 3
Team D moves from position 3 to position 4
Team E moves from position 4 to position 5
Team F moves from position 5 to position 6
Team G moves from position 6 to position 7
Team H moves from position 7 to position 1
The opponents for round 2 thus become:
Team A (at position 0) plays against Team H (at position 1)
Team G (at position 7) plays against Team B (at position 2)
Team F (at position 6) plays against Team C (at position 3)
Team E (at position 5) plays against Team D (at position 4)
For each subsequent round of play, Team A stays at
position “0”. All other teams advance one position. The
sequence is somewhat like a circular conveyor belt. A more accurate
name for the process is “cyclic solution”, but it is easier
to think of it as a circular conveyor belt moving along an assembly
line. After 7 rounds, each team will have played against each other
team exactly once. If you have 20 teams then 19 rounds will be required.
At each position on the moving assembly line
“something” happens. The “something” is that
whoever is at this position is matched against whoever is in the
“return” position of the conveyor belt going in the other
direction. The sequence will match each team against each other team
exactly once, and will work for any number of positions provided the
number of positions on the moving belt is an odd number. The
requirement for an “odd number” of positions along the
moving belt is why position “0” is defined as a stationary
position. If we start with an “even” number of positions
and “freeze” position “0”, we will be left with
an odd number for the teams that move.
Permission is granted to anyone who may wish to use
information on this web page for any not-for-profit use provided
acknowledgement is made to the author (Bill Butler) and a reference is
included to this web page. (
http://www.durangobill.com/BridgeCyclicSolutions.html)
Bill Butler
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