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1  Schedules / Requests / Re: Help Schedule 14 bridge teams for 8 months on: 09/20/17 at 21:48:44 
Started by Nbridge  Post by Nbridge  
Thx it took me a bit of figuring out that "match" will become "table" but in the end I think I got it with the grid you have Thx for your help ! 

2  Schedules / Requests / Re: Help Schedule 14 bridge teams for 8 months on: 09/17/17 at 17:04:15 
Started by Nbridge  Post by Ian Wakeling  
Use the standard roundrobin that you can find by following the 'schedules' link near the top of this page and selecting 14 items. This will give you the first 13 rounds, another copy will take you to 26 rounds, then take the final 6 rounds from the top of a third copy. If you don't want to have the same opponents in the same order for each copy, then make a different assignment of the 14 teams to the numbers 1 to 14 for each copy.  
3  Schedules / Requests / Help Schedule 14 bridge teams for 8 months on: 09/17/17 at 09:42:35 
Started by Nbridge  Post by Nbridge  
Please help  newbie bridge league coordinator I need help figuring out how to create a grid for 14 teams for party bridge ( 28 players). Each team is fixed and we plan on playing 4 rounds each month for 8 months. Want to make sure everyone plays everyone as much as possible . Is there a formula I can use to come up with such a grid / table movement I have a grid for 16 teams and 20 teams that out league previously used ( if anyone needs those lol) but I cant for the life of me figure out how to translate that into a 14 team version. Attaching a pic of what my chart of 16 teams looks like 

4  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/13/17 at 04:29:56 
Started by Ben_J  Post by Ian Wakeling  
This is exactly the point of the paper by Warut Suksompong that you have linked to above, we need the construction from Theorem 4.1 to handle an odd number (n) of competitors, then you can guarantee a rest time of (n3)/2 games. Figures 4 and 5 in the paper give the schedules for n=5 and n=7 respectively, if I have understood correctly then this is the schedule for n=9 (perhaps you can check it with your spreadsheet): (1 2) (3 4) (5 6) (7 8) (1 9) (3 2) (5 4) (7 6) (8 9) (3 1) (5 2) (7 4) (6 8) (3 9) (5 1) (7 2) (4 8) (6 9) (5 3) (7 1) (2 8) (4 6) (5 9) (7 3) (1 8) (2 6) (4 9) (7 5) (3 8) (1 6) (2 4) (7 9) (5 8) (3 6) (1 4) (2 9) It is intended that these are unravelled from left to right and from top to bottom, but I think if you took these in reverse order it would give you a workable schedule with the 'close' fights at the end. 

5  Schedules / Requests / Re: Euchre Tables, 28 and 32 players, 12 Rounds on: 09/12/17 at 21:46:15 
Started by k48038  Post by k48038  
The 28 player schedule you sent is a HUGE improvement over what I was able to come up with using my trial and error manual method. For example, my result was that on average, each individual player had 6.2 other players that he did not share a table with over the course of the 12 rounds. So, after 12 rounds there were 6 other players that I never sat at the same table as. AND, everyone had at least 4 players that they did not play with and some had as many as 8 players that they never played with (this range of 4 to 8 resulted in the 6.2 average). The pairings that you sent greatly improved both of those numbers. The average went down from 6.2 to 2.1. Meaning that there are only, on average, 2 other players that I do not play at the same table with all night. And, in your scenario some people actually played at the same table as every other player at least once, and the worstcase was that there were 4 people they did not play with. My BEST case was 4 and your WORST case was 4. Your range was 1 to 4 giving me the 2.1 average. Thanks so much. I have not had the chance to run the same analysis on the 32 player pairings that you've provided, but I hope and expect improvement there, too. Thanks again! 

6  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/12/17 at 15:23:27 
Started by Ben_J  Post by Ben_J  
.....aaaand of course now I'm going back through it  I'm finding mistakes in my work... most of this byhand stuff I was working on more than a year ago... All the more reason to have been looking for algorithms and proper solvers for the problem  fewer mistakes that way. The solution for 10players worked perfectly by the way. I am not sure that simply removing matches as suggested to get the oddplayer solutions works as well  still testing that now... 

7  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/12/17 at 09:54:16 
Started by Ben_J  Post by Ben_J  
You're right  I didn't get your gist the first time. My mistake. Proofofconcept with 8players yields the following: 20 matches where one player has a 2match break, 13 matches where one player has a 3match break, 18 matches where one player has a 4match break, 1 matches where one player has a 5match break as compared to my byhand solution in which: 17 matches where one player has a 2match break, 21 matches where one player has a 3match break, 13 matches where one player has a 4match break, 1 matches where one player has a 5match break Beyond that, the solutions are surprisingly similar, structurally, in terms of the 'flow' of who fights who when; suggesting that I was mentally on the right track to begin with. I'll move forward to work this solution for 10person pool, and see what comes up. Thank you again. 

8  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/12/17 at 09:10:39 
Started by Ben_J  Post by Ian Wakeling  
Thanks for posting the link  I will take a look at the paper. Ian 

9  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/12/17 at 09:04:49 
Started by Ben_J  Post by Ian Wakeling  
I don't think I explained clearly enough, as there is no need to move matches around. If you take a full roundrobin schedule, apply any permutation to the competitor numbers, then you will still have a full roundrobin schedule. In the image I show the 8 player cyclic schedule in the top half, then below that I have applied the following permutation to the numbers : Leave competitors 1, 2 & 3 unchanged. Transform 4 to 5 5 to 7 6 to 8 7 to 6 8 to 4 the overall effect of these changes is to make the last round 4 fights between closely seeded competitors. I would be interested to know how this measures up if you unravel it as I have suggested above and then paste in to your spreadsheet. My gut feel is that your manual approach has given a close to optimal solution for 8 competitors, so I am not hopeful on improving on it. But perhaps the same approach applied to 10 competitors will offer an improvement. 

10  Schedules / Requests / Re: Single RR, 510 Competitors, ONE Venue Efficie on: 09/12/17 at 09:04:27 
Started by Ben_J  Post by Ben_J  
So, for example  I don't know why I didn't find this before in my searching, but this paper is exactly what I'm trying to figure out: http://cs.stanford.edu/~warut/roundrobin.pdf That said, having just read it quickly, it does not have a 'solver', per se, it simply proves that a few things are possible in terms of the things I'm trying to figure out. 
