| |
BGonline.org Forums
Predrawing 5 rounds of Swiss with 32+ players?
Posted By: Rich Munitz In Response To: Pittsburgh Swiss Format (Bob Koca)
Date: Friday, 18 September 2009, at 6:52 a.m.
In Madison I had briefly discussed with the staff my belief that it should be possible to pregenerate drawsheets to determine several initial rounds of Swiss play with the goal that people could start their next match the moment the opponent was ready rather than waiting for a redraw at a preset time once all results were known. But the devil is in the details as they pointed out and dealing with a draw size that is unknown ahead of time and the requirement that two players who have played once must not play again makes it a mess. And as was mentioned, many people simply like the idea of being able to walk out if they finish early and know exactly what time to show up for the next round.
But with events like Pittsburgh possibly looking for a speedier way to manage Swiss, I gave it some further thought.
I now believe that it should be possible to do a single random draw and generate 5 initial rounds of Swiss play in advance for all even tournament sizes >= 32 (34 may be an exception). Meaning opponent match-up is deterministic so that the winner and loser of a match move to a well defined new slot on the drawsheet independently of all other outcomes in the tournament.
The 8-player Swiss brackets method suggested by UBK works fine as a starting point. The problem is what to do with the extra players.
Odd numbers of players are death to any Swiss tournament under time pressure since someone is always sitting out without a match. So lets assume odd numbers can be avoided, either by shifting players between divisions, or by picking a player at random to be locked in the janitor's closet for the weekend, or whatever the airlines do to get rid of people they sold seats to that don't exist. If we are desparate, we can simply add an additional player named Bye to the event and follow the even rules, noting that some players will either have to be awarded free wins, or make up matches later if they survive.
Say you do create groups of 8 and have people left over. The remainder will be 2, 4 or 6. It is necessary to have at least 4 players in a group to ensure that each player can play 3 matches without having to play the same player twice. And it is possible to construct 3 rounds of play with either 4 or 6 players such that the match-ups either have the same record, or at most a difference in one win/loss (though the 4-group is in the end nothing more than a round robin, so if it works out faster for the 2-0 to play the 0-2 last, so be it). If there are 2 extra players, combine them with a group of 8 (total 10) and then break that into a group of 4 and a group of 6. So in this manner, any even tournament size can have 3 rounds of Swiss determined by one random draw and managed with multiple 8 and at most one 6 and one 4-player drawsheets.
Next, if you have at least 32 players (but not exactly 34), you'd have at least 4 groups of 8. So it is possible to construct groups of 4 to generate an additional 2 rounds of match-ups and ensure that each group of 4 is made of people who were each in a different group of 8. You'd rank the end positions 1-8 (or 6 or 4) in each initial drawsheet in order of decreasing wins, and then feed all players into a list such that the position of rank R in drawsheet X of N feeds into slot (R-1)*N + X. There will be up to 6 gaps in this list in known positions. Starting with slot 1, you'd direct each group of 4 consecutive non-empty slots into a new drawsheet of 4 players. If N is not divisible by 4, there are two extra players and one group will need to have 6. In that case, the first group of four whose first slot has a rank of 2 (from the previous drawsheet) will be extended to 6. If that first slot of this special group of 6 happens to be the feed in from a 4 or 6 player drawsheet, the sheets should be renumbered to ensure that slot is filled from an 8-player sheet. These steps will ensure that all players in this group of 6 have an identical record (2-1). (Except possibly in the case of a group of 4 round robin ending with records of 3-0,1-2,1-2,1-2)
If there are exactly 38 players (34 is already broken above), there is another special situation because there needs to be a special group of 6 created for rounds 4 and 5, but there will only be 5 total groups from the original 3 rounds. Thus the special group of 6 must have two players from the same original group. However, as we know these two players must feed into slots 1 and 6, I have verified that a drawsheet of 6 playing 2 matches each can be created in advance to ensure that these two players will never play each other and all matches will be amongst identical records, or at worst one different. We know this by having ensured that the 6 players in this sheet start with identical records.
What this solution fails to account for is attrition after round 4 as is done in Madison. That being anyone with an 0-4 record does not get to play round 5. I haven't worked that out, but believe that this is solvable, at least partially. One in 8 players should have a 0-4 record after 3 rounds and with 32 players, there should be 4. It seems that in the case where we have that ugly 6 player field for rounds 4 and 5 be the bottom 6 which should contain at least 4 0-3 players that will play off first match, eliminating 2 and leaving 4. Those 4 can then play round 5.
| |
BGonline.org Forums is maintained by Stick with WebBBS 5.12.