Round Robin Tournament Scheduling

Round robin schedules

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sezuh(Guest)

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on: July 18, 2011, 03:59:12 PM
Hi,
I am looking for the table that shows the threesome pairings for a total of 51 players, Each player plays with all 50 possible partners exactly once.  
thanks
 :)


Ian Wakeling

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Reply #1 on: July 19, 2011, 03:54:12 AM
That would be a tournament with 25 rounds.  I would be interested to know what sport you would like to apply this to.

Below each row is a single round with the 17 threesomes.

1  2 27 |  3  7 26 | 28 51 32 |  6 11 50 | 31 25 36 |  8  9 38 | 33 13 34 | 22 21 48 | 47 23 46 |  4 12 20 | 29 45 37 |  5 15 43 | 30 18 40 | 14 16 49 | 39 24 41 | 17 10 44 | 42 19 35
1  3 28 |  2  7 51 | 27 26 32 |  4  8 22 | 29 47 33 |  9 10 39 | 34 14 35 | 23 17 49 | 48 24 42 |  5 13 21 | 30 46 38 |  6 16 44 | 31 19 41 | 15 12 50 | 40 25 37 | 18 11 45 | 43 20 36
1  4 29 |  3  8 47 | 28 22 33 |  5  9 23 | 30 48 34 | 10 11 40 | 35 15 36 | 24 18 50 | 49 25 43 |  2 12 45 | 27 20 37 |  6 14 17 | 31 42 39 | 16 13 51 | 41 26 38 | 19  7 46 | 44 21 32
1  5 30 |  4  9 48 | 29 23 34 |  6 10 24 | 31 49 35 | 11  7 41 | 36 16 32 | 25 19 51 | 50 26 44 |  3 13 46 | 28 21 38 |  2 15 18 | 27 43 40 | 12 14 47 | 37 22 39 | 20  8 42 | 45 17 33
1  6 31 |  5 10 49 | 30 24 35 |  2 11 25 | 27 50 36 |  7  8 37 | 32 12 33 | 26 20 47 | 51 22 45 |  4 14 42 | 29 17 39 |  3 16 19 | 28 44 41 | 13 15 48 | 38 23 40 | 21  9 43 | 46 18 34
1  7 32 |  2 26 28 | 27  3 51 |  8 12  6 | 33 31 37 | 11 16 30 | 36  5 41 | 13 14 43 | 38 18 39 |  9 17 25 | 34 50 42 | 10 20 48 | 35 23 45 | 19 21 29 | 44  4 46 | 22 15 49 | 47 24 40
1  8 33 |  3 22 29 | 28  4 47 |  7 12 31 | 32  6 37 |  9 13  2 | 34 27 38 | 14 15 44 | 39 19 40 | 10 18 26 | 35 51 43 | 11 21 49 | 36 24 46 | 20 17 30 | 45  5 42 | 23 16 50 | 48 25 41
1  9 34 |  4 23 30 | 29  5 48 |  8 13 27 | 33  2 38 | 10 14  3 | 35 28 39 | 15 16 45 | 40 20 41 |  7 17 50 | 32 25 42 | 11 19 22 | 36 47 44 | 21 18 31 | 46  6 43 | 24 12 51 | 49 26 37
1 10 35 |  5 24 31 | 30  6 49 |  9 14 28 | 34  3 39 | 11 15  4 | 36 29 40 | 16 12 46 | 41 21 37 |  8 18 51 | 33 26 43 |  7 20 23 | 32 48 45 | 17 19 27 | 42  2 44 | 25 13 47 | 50 22 38
1 11 36 |  6 25 27 | 31  2 50 | 10 15 29 | 35  4 40 |  7 16  5 | 32 30 41 | 12 13 42 | 37 17 38 |  9 19 47 | 34 22 44 |  8 21 24 | 33 49 46 | 18 20 28 | 43  3 45 | 26 14 48 | 51 23 39
1 12 37 |  7  6 33 | 32  8 31 | 13 17 11 | 38 36 42 | 16 21 35 | 41 10 46 | 18 19 48 | 43 23 44 |  2 20 29 | 27  4 45 | 14 22  5 | 39 30 47 | 15 25 28 | 40  3 50 | 24 26 34 | 49  9 51
1 13 38 |  8  2 34 | 33  9 27 | 12 17 36 | 37 11 42 | 14 18  7 | 39 32 43 | 19 20 49 | 44 24 45 |  3 21 30 | 28  5 46 | 15 23  6 | 40 31 48 | 16 26 29 | 41  4 51 | 25 22 35 | 50 10 47
1 14 39 |  9  3 35 | 34 10 28 | 13 18 32 | 38  7 43 | 15 19  8 | 40 33 44 | 20 21 50 | 45 25 46 |  4 17 31 | 29  6 42 | 12 22 30 | 37  5 47 | 16 24  2 | 41 27 49 | 26 23 36 | 51 11 48
1 15 40 | 10  4 36 | 35 11 29 | 14 19 33 | 39  8 44 | 16 20  9 | 41 34 45 | 21 17 51 | 46 26 42 |  5 18 27 | 30  2 43 | 13 23 31 | 38  6 48 | 12 25  3 | 37 28 50 | 22 24 32 | 47  7 49
1 16 41 | 11  5 32 | 36  7 30 | 15 20 34 | 40  9 45 | 12 21 10 | 37 35 46 | 17 18 47 | 42 22 43 |  6 19 28 | 31  3 44 | 14 24 27 | 39  2 49 | 13 26  4 | 38 29 51 | 23 25 33 | 48  8 50
1 17 42 | 12 11 38 | 37 13 36 | 18 22 16 | 43 41 47 | 21 26 40 | 46 15 51 | 23 24 28 | 48  3 49 |  4  6 39 | 29 14 31 |  7 25 34 | 32  9 50 | 19  2 10 | 44 35 27 | 20  5 33 | 45  8 30
1 18 43 | 13  7 39 | 38 14 32 | 17 22 41 | 42 16 47 | 19 23 12 | 44 37 48 | 24 25 29 | 49  4 50 |  5  2 40 | 30 15 27 |  8 26 35 | 33 10 51 | 20  3 11 | 45 36 28 | 21  6 34 | 46  9 31
1 19 44 | 14  8 40 | 39 15 33 | 18 23 37 | 43 12 48 | 20 24 13 | 45 38 49 | 25 26 30 | 50  5 51 |  6  3 41 | 31 16 28 |  9 22 36 | 34 11 47 | 17  2 35 | 42 10 27 | 21  4  7 | 46 32 29
1 20 45 | 15  9 41 | 40 16 34 | 19 24 38 | 44 13 49 | 21 25 14 | 46 39 50 | 26 22 31 | 51  6 47 |  2  4 37 | 27 12 29 | 10 23 32 | 35  7 48 | 18  3 36 | 43 11 28 | 17  5  8 | 42 33 30
1 21 46 | 16 10 37 | 41 12 35 | 20 25 39 | 45 14 50 | 17 26 15 | 42 40 51 | 22 23 27 | 47  2 48 |  3  5 38 | 28 13 30 | 11 24 33 | 36  8 49 | 19  4 32 | 44  7 29 | 18  6  9 | 43 34 31
1 22 47 |  3  4 33 | 28  8 29 | 17 16 43 | 42 18 41 | 23  2 21 | 48 46 27 | 26  6 45 | 51 20 31 |  9 11 44 | 34 19 36 | 12  5 39 | 37 14 30 | 24  7 15 | 49 40 32 | 25 10 38 | 50 13 35
1 23 48 |  4  5 34 | 29  9 30 | 18 12 44 | 43 19 37 | 22  2 46 | 47 21 27 | 24  3 17 | 49 42 28 | 10  7 45 | 35 20 32 | 13  6 40 | 38 15 31 | 25  8 16 | 50 41 33 | 26 11 39 | 51 14 36
1 24 49 |  5  6 35 | 30 10 31 | 19 13 45 | 44 20 38 | 23  3 42 | 48 17 28 | 25  4 18 | 50 43 29 | 11  8 46 | 36 21 33 | 14  2 41 | 39 16 27 | 22  7 40 | 47 15 32 | 26  9 12 | 51 37 34
1 25 50 |  6  2 36 | 31 11 27 | 20 14 46 | 45 21 39 | 24  4 43 | 49 18 29 | 26  5 19 | 51 44 30 |  7  9 42 | 32 17 34 | 15  3 37 | 40 12 28 | 23  8 41 | 48 16 33 | 22 10 13 | 47 38 35
1 26 51 |  2  3 32 | 27  7 28 | 21 15 42 | 46 17 40 | 25  5 44 | 50 19 30 | 22  6 20 | 47 45 31 |  8 10 43 | 33 18 35 | 16  4 38 | 41 13 29 | 24  9 37 | 49 12 34 | 23 11 14 | 48 39 36


sezuh(Guest)

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Reply #2 on: July 19, 2011, 01:50:47 PM
Hi Ian,thanks ever so much for your help,appreciated very very very..........much.i've been trying to solve this for nearly 4 years
but not with computer with pen and papers!!! :)
Code: [Select]
I would be interested to know what sport you would like to apply this toits not for sport its like a puzzle "Social golfer problem or Kirkman's School girl problem"there are 5 of them thanks to you 1 down and 4 to go.... :)
1) foursome pairings for a total of 52 players, Each player plays with all 51 possible partners exactly once,with 17 rounds
2) foursome pairings for a total of 40 players, Each player plays with all 39 possible partners exactly once,with 13 rounds
3) fivesome pairings for a total of 45 players, Each player plays with all 44 possible partners exactly once,with 11 rounds
I wish known this site earlier my curiosity killing me not knowing if those problems solveable or not..?
Forever greatful for your wonderful solution.
sezuh


Ian Wakeling

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Reply #3 on: July 20, 2011, 03:47:51 AM
Hi Sezuh,

Many thanks for your reply.  While I have used a computer to generate the schedule above, the program I used was just implementing a known mathematical construction.  All of the generalisations of Kirkman's problem where there are 6i+3 school girls have solutions that have been known for some time.

Regarding the other three examples you give.  I have a book that says design (1) exists, but sadly it doesn't provide details of a construction.  There is an example of the 40 player schedule (2) that can be found here on this message board.  The 45 player problem (3), is as far as I am aware, an open problem, so nobody knows if it exits or not.

If you register on this site you will be able to reply to the message here and also contact me off-line if you want to.

Hope that helps.

Ian.


sezuh

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Reply #4 on: July 20, 2011, 12:24:56 PM
Ian,
thanks for your time and wonderful help,much appreciated..i have another request if you dont  mind its similar to this problem
but with extra constraint;
 WhatI am looking for the table that shows the fivesome pairings for a total of 50 players, Each player plays with all 49 possible partners exactly 4 times,with 49 rounds, do you think this is possible to solve? :)
thanks again for welcoming me to the forum.
have a nice evening.... :)
« Last Edit: July 20, 2011, 12:25:28 PM by sezuh »
Kind regards
sezuh


Ian Wakeling

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Reply #5 on: July 21, 2011, 03:24:02 AM
sezuh,

Yes this is possible and has a simple cyclic construction.  If you took the following as the 1st round:

(∞ 0 4 25 29)
(1 5 8 34 48)
(2 15 27 37 46)
(3 13 14 16 44)
(6 21 35 36 47)
(7 12 18 24 40)
(9 10 22 32 41)
(11 19 39 42 43)
(17 20 26 31 33)
(23 28 30 38 45)

Then all the remaining rounds could be generated by adding 1 mod(49) to the previous round.  So the first two fivesomes of round 2 would be

(∞ 1 5 26 30)
(2 6 9 35 0)

Are you doing anything like that when you look for a solution with pen and paper?

Regards,

Ian.


sezuh

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Reply #6 on: July 21, 2011, 08:09:18 AM
Ian, thanks ever so much for the advice and help, I'm going to try and complete it if I can
in my day off.
Code: [Select]
Are you doing anything like that when you look for a solution with pen and paper? no ,normally I get the combination
from La jolla covering or if its small combination I try to do it myself then try to divide it to groups which is very much time consuming
for my part .i can not thank you enough for this help.
have a nice day :)
Kind regards
sezuh


Ian Wakeling

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Reply #7 on: July 22, 2011, 03:39:10 AM
Sezuh,

You don't need to have the complete schedule of 49 rounds to verify that it will work.  Consider the 50th player represented by infinity who will always occur in the first fivesome, the other four players in that fivesome cycle through all the numbers 0..48, so it is easy to see that the 50th player occurs exactly 4 times each with the other 49 players.

The 49 players who are represented by the numbers mod 49, have 49*48/2=1176 possible pairs among themselves.  These pairs can be divided into 24 subsets of 49 pairs, where each subset contains all pairs of players who have the same difference modulo 49.  Consider the pair (5 8) which occurs in the second fivesome and the pairs that will be developed from it over the next 48 rounds by adding one each time.  These will be (5,8),(6 9),(7 10),..,(45 48),(46 0),(47 1),(48 2),(0 3),(1 4),..,(4 7).  These are all of the pairs that differ by 3 mod 49.  The same subset of 49 pairs, but in a different order, will also be generated by the pair (13 16) in the 4th fivesome, the pair (39 42) in the 8th fivesome and the pair (17 20) in the 9th fivesome.  So I have verified that the complete schedule will contain these 49 pairs at least 4 times each.  To show the complete schedule is fully balanced, you just need to examine the differences of all the pairs taken from the 10 starter blocks and verify that all 24 differences mod 49 are covered exactly 4 times each.

Hope that helps.

Ian.


sezuh

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Reply #8 on: July 22, 2011, 12:06:13 PM
Hi Ian,
thanks for the wonderful help, I completed the schedule.
but you mind tell how do you come up with first round numbers and what the logic behind it ?
can you give me the first round for  the table that shows the fivesome pairings for a total of 40 players, Each player plays with all 39 possible partners exactly 4 times, with 39 rounds ?  is it also cycle through all the numbers 0..38 like the problem above?
I realy appreciate your help with out your I don't think I coud have solved it.
Here the result if any one interested;
Code: [Select]
50 49 4 25 29
1 5 8 34 48
2 15 27 37 46
3 13 14 16 44
6 21 35 36 47
7 12 18 24 40
9 10 22 32 41
11 19 39 42 43
17 20 26 31 33
23 28 30 38 45
50 1 5 26 30
 2 6 9 35 49
3 16 28 38 47
4 14 15 17 45
7 22 36 37 48
8 13 19 25 41
10 11 23 33 42
12 20 40 43 44
18 21 27 32 34
24 29 31 39 46
50 2 6 27 31
3 7 10 36 1
4 17 29 39 48
5 15 16 18 46
8 23 37 38 49
9 14 20 26 42
11 12 24 34 43
13 21 41 44 45
19 22 28 33 35
25 30 32 40 47
50 3 7 28 32
4 8 11 37 2
5 18 30 40 49
6 16 17 19 47
9 24 38 39 1
10 15 21 27 43
12 13 25 35 44
14 22 42 45 46
20 23 29 34 36
26 31 33 41 48
50 4 8 29 33
5 9 12 38 3
6 19 31 41 1
7 17 18 20 48
10 25 39 40 2
11 16 22 28 44
13 14 26 36 45
15 23 43 46 47
21 24 30 35 37
27 32 34 42 49
50 5 9 30 34
6 10 13 39 4
7 20 32 42 2
8 18 19 21 49
11 26 40 41 3
12 17 23 29 45
14 15 27 37 46
16 24 44 47 48
22 25 31 36 38
28 33 35 43 1
50 6 10 31 35
7 11 14 40 5
8 21 33 43 3
9 19 20 22 1
12 27 41 42 4
13 18 24 30 46
15 16 28 38 47
17 25 45 48 49
23 26 32 37 39
29 34 36 44 2
50 7 11 32 36
8 12 15 41 6
9 22 34 44 4
10 20 21 23 2
13 28 42 43 5
14 19 25 31 47
16 17 29 39 48
18 26 46 49 1
24 27 33 38 40
30 35 37 45 3
50 8 12 33 37
9 13 16 42 7
10 23 35 45 5
11 21 22 24 3
14 29 43 44 6
15 20 26 32 48
17 18 30 40 49
19 27 47 1 2
25 28 34 39 41
31 36 38 46 4
50 9  13 34 38
10 14 17 43 8
11 24 36 46 6
12 22 23 25 4
15 30 44 45 7
16 21 27 33 49
18 19 31 41 1
20 28 48 2 3
26 29 35 40 42
32 37 39 47 5
50 10 14 35 39
11 15 18 44 9
12 25 37 47 7
13 23 24 26 5
16 31 45 46 8
17 22 28 34 1
19 20 32 42 2
21 29 49 3 4
27 30 36 41 43
33 38 40 48 6
50 11 15 36 40
12 16 19 45 10
13 26 38 48 8
14 24 25 27 6
17 32 46 47 9
18 23 29 35 2
20 21 33 43 3
22 30 1 4 5
28 31 37 42 44
34 39 41 49 7
50 12 16 37 41
13 17 20 46 11
14 27 39 49 9
15 25 26 28 7
18 33 47 48 10
19 24 30 36 3
21 22 34 44 4
23 31 2 5 6
29 32 38 43 45
35 40 42 1 8
50 13 17 38 42
14 18 21 47 12
15 28 40 1 10
16 26 27 29 8
19 34 48 49 11
20 25 31 37 4
22 23 35 45 5
24 32 3 6 7
30 33 39 44 46
36 41 43 2 9
50 14 18 39 43
15 19 22 48 13
16 29 41 2 11
17 27 28 30 9
20 35 49 1 12
21 26 32 38 5
23 24 36 46 6
25 33 4 7 8
31 34 40 45 47
37 42 44 3 10
50 15 19 40 44
16 20 23 49 14
17 30 42 3 12
18 28 29 31 10
21 36 1 2 13
22 27 33 39 6
24 25 37 47 7
26 34 5 8 9
32 35 41 46 48
38 43 45 4 11
50 16 20 41 45
17 21 24 1 15
18 31 43 4 13
19 29 30 32 11
22 37 2 3 14
23 28 34 40 7
25 26 38 48 8
27 35 6 9 10
33 36 42 47 49
39 44 46 5 12
50 17 21 42 46
18 22 25 2 16
19 32 44 5 14
20 30 31 33 12
23 38 3 4 15
24 29 35 41 8
26 27 39 49 9
28 36 7 10 11
34 37 43 48 1
40 45 47 6 13
50 18 22 43 47
19 23 26 3 17
20 33 45 6 15
21 31 32 34 13
24 39 4 5 16
25 30 36 42 9
27 28 40 1 10
29 37 8 11 12
35 38 44 49 2
41 46 48 7 14
50 19 23 44 48
20 24 27 4 18
21 34 46 7 16
22 32 33 35 14
25 40 5 6 17
26 31 37 43 10
28 29 41 2 11
30 38 9 12 13
36 39 45 1 3
42 47 49 8 15
50 20 24 45 49
21 25 28 5 19
22 35 47 8 17
23 33 34 36 15
26 41 6 7 18
27 32 38 44 11
29 30 42 3 12
31 39 10 13 14
37 40 46 2 4
43 48 1 9 16
50 21 25 46 1
22 26 29 6 20
23 36 48 9 18
24 34 35 37 16
27 42 7 8 19
28 33 39 45 12
30 31 43 4 13
32 40 11 14 15
38 41 47 3 5
44 49 2 10 17
50 22 26 47 2
23 27 30 7 21
24 37 49 10 19
25 35 36 38 17
28 43 8 9 20
29 34 40 46 13
31 32 44 5 14
33 41 12 15 16
39 42 48 4 6
45 1 3 11 18
50 23 27 48 3
24 28 31 8 22
25 38 1 11 20
26 36 37 39 18
29 44 9 10 21
30 35 41 47 14
32 33 45 6 15
34 42 13 16 17
40 43 49 5 7
46 2 4 12 19
50 24 28 49 4
25 29 32 9 23
26 39 2 12 21
27 37 38 40 19
30 45 10 11 22
31 36 42 48 15
33 34 46 7 16
35 43 14 17 18
41 44 1 6 8
47 3 5 13 20
50 25 29 1 5
26 30 33 10 24
27 40 3 13 22
28 38 39 41 20
31 46 11 12 23
32 37 43 49 16
34 35 47 8 17
36 44 15 18 19
42 45 2 7 9
48 4 6 14 21
50 26 30 2 6
27 31 34 11 25
28 41 4 14 23
29 39 40 42 21
32 47 12 13 24
33 38 44 1 17
35 36 48 9 18
37 45 16 19 20
43 46 3 8 10
49 5 7 15 22
50 27 31 3 7
28 32 35 12 26
29 42 5 15 24
30 40 41 43 22
33 48 13 14 25
34 39 45 2 18
36 37 49 10 19
38 46 17 20 21
44 47 4 9 11
1 6 8 16 23
50 28 32 4 8
29 33 36 13 27
30 43 6 16 25
31 41 42 44 23
34 49 14 15 26
35 40 46 3 19
37 38 1 11 20
39 47 18 21 22
45 48 5 10 12
2 7 9 17 24
50 29 33 5 9
30 34 37 14 28
31 44 7 17 26
32 42 43 45 24
35 1 15 16 27
36 41 47 4 20
38 39 2 12 21
40 48 19 22 23
46 49 6 11 13
3 8 10 18 25
50 30 34 6 10
31 35 38 15 29
32 45 8 18 27
33 43 44 46 25
36 2 16 17 28
37 42 48 5 21
39 40 3 13 22
41 49 20 23 24
47 1 7 12 14
4 9 11 19 26
50 31 35 7 11
32 36 39 16 30
33 46 9 19 28
34 44 45 47 26
37 3 17 18 29
38 43 49 6 22
40 41 4 14 23
42 1 21 24 25
48 2 8 13 15
5 10 12 20 27
50 32 36 8 12
33 37 40 17 31
34 47 10 20 29
35 45 46 48 27
38 4 18 19 30
39 44 1 7 23
41 42 5 15 24
43 2 22 25 26
49 3 9 14 16
6 11 13 21 28
50 33 37 9 13
34 38 41 18 32
35 48 11 21 30
36 46 47 49 28
39 5 19 20 31
40 45 2 8 24
42 43 6 16 25
44 3 23 26 27
1 4 10 15 17
7 12 14 22 29
50 34 38 10 14
35 39 42 19 33
36 49 12 22 31
37 47 48 1 29
40 6 20 21 32
41 46 3 9 25
43 44 7 17 26
45 4 24 27 28
2 5 11 16 18
8 13 15 23 30
50 35 39 11 15
36 40 43 20 34
37 1 13 23 32
38 48 49 2 30
41 7 21 22 33
42 47 4 10 26
44 45 8 18 27
46 5 25 28 29
3 6 12 17 19
9 14 16 24 31
50 36 40 12 16
37 41 44 21 35
38 2 14 24 33
39 49 1 3 31
42 8 22 23 34
43 48 5 11 27
45 46 9 19 28
47 6 26 29 30
4 7 13 18 20
10 15 17 25 32
50 37 41 13 17
38 42 45 22 36
39 3 15 25 34
40 1 2 4 32
43 9 23 24 35
44 49 6 12 28
46 47 10 20 29
48 7 27 30 31
5 8 14 19 21
11 16 18 26 33
50 38 42 14 18
39 43 46 23 37
40 4 16 26 35
41 2 3 5 33
44 10 24 25 36
45 1 7 13 29
47 48 11 21 30
49 8 28 31 32
6 9 15 20 22
12 17 19 27 34
50 39 43 15 19
40 44 47 24 38
41 5 17 27 36
42 3 4 6 34
45 11 25 26 37
46 2 8 14 30
48 49 12 22 31
1 9 29 32 33
7 10 16 21 23
13 18 20 28 35
50 40 44 16 20
41 45 48 25 39
42 6 18 28 37
43 4 5 7 35
46 12 26 27 38
47 3 9 15 31
49 1 13 23 32
2 10 30 33 34
8 11 17 22 24
14 19 21 29 36
50 41 45 17 21
42 46 49 26 40
43 7 19 29 38
44 5 6 8 36
47 13 27 28 39
48 4 10 16 32
1 2 14 24 33
3 11 31 34 35
9 12 18 23 25
15 20 22 30 37
50 42 46 18 22
43 47 1 27 41
44 8 20 30 39
45 6 7 9 37
48 14 28 29 40
49 5 11 17 33
2 3 15 25 34
4 12 32 35 36
10 13 19 24 26
16 21 23 31 38
50 43 47 19 23
44 48 2 28 42
45 9 21 31 40
46 7 8 10 38
49 15 29 30 41
1 6 12 18 34
3 4 16 26 35
5 13 33 36 37
11 14 20 25 27
17 22 24 32 39
50 44 48 20 24
45 49 3 29 43
46 10 22 32 41
47 8 9 11 39
1 16 30 31 42
2 7 13 19 35
4 5 17 27 36
6 14 34 37 38
12 15 21 26 28
18 23 25 33 40
50 45 49 21 25
46 1 4 30 44
47 11 23 33 42
48 9 10 12 40
2 17 31 32 43
3 8 14 20 36
5 6 18 28 37
7 15 35 38 39
13 16 22 27 29
19 24 26 34 41
50 46 1 22 26
47 2 5 31 45
48 12 24 34 43
49 10 11 13 41
3 18 32 33 44
4 9 15 21 37
6 7 19 29 38
8 16 36 39 40
14 17 23 28 30
20 25 27 35 42
50 47 2 23 27
48 3 6 32 46
49 13 25 35 44
1 11 12 14 42
4 19 33 34 45
5 10 16 22 38
7 8 20 30 39
9 17 37 40 41
15 18 24 29 31
21 26 28 36 43
50 48 3 24 28
49 4 7 33 47
1 14 26 36 45
2 12 13 15 43
5 20 34 35 46
6 11 17 23 39
8 9 21 31 40
10 18 38 41 42
16 19 25 30 32
22 27 29 37 44
Kind regards
sezuh


Ian Wakeling

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Reply #9 on: July 23, 2011, 02:31:31 AM
Hi Sezuh,

Coming up with first round numbers is definitely the hard part.  There is a book that I can recommend called The CRC Handbook of Combinatorial Designs. Specifically there is a section on difference families and 1-rotational designs that deals with these problems.  It gives the following for 20 players in fivesomes:

(∞ 0 2 3 14) (1 5 6 10 12) (4 7 8 13 16) (9 11 15 17 18)

and the following for 40 players:

(∞ 0 3 9 27) (2 5 13 26 32) (1 4 20 29 36) (8 16 25 30 35) (10 11 12 14 22) (7 15 17 31 33) (6 21 28 34 38) (18 19 23 24 37)

that's it for fivesomes but the book has a few more resolvable designs for block sizes of 6, 7 and 8.  I suspect if you followed some of the literature mentioned in the Handbook you would find the logic you are looking for, but it is likely to involve some complex combinatorial mathematics.

On the other hand, my example above for 50 players was found with a computer search that broadly speaking is similar to the one described in
Morales, Luis B.. (2001). Two New 1-rotational (36,9,8) and (40,10,9) RBIBDs. Journal of Combinatorial Mathematics and Combinatorial Computing 36 (2001), 119-126. 38.  The problem is tractable because instead of searching for the complete schedule that you have posted above, it reduces to that of finding the set of 10 base blocks.

Kind regards,

Ian.
« Last Edit: July 23, 2011, 02:40:03 AM by Ian »


sezuh

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Reply #10 on: July 24, 2011, 12:12:20 PM
Thanks Ian,
that is very good of you for those new scheduling list very much appreciated.
If anyone interested on those two list ,i dont mind posting them on here .
have a good day.
Kind regards
sezuh