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






   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.




   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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