When using the Swiss System For a given number of players too few will not produce a winner and at the other extreem too many rounds will waste time and make pairings impossible.
Does a formula (or better yet) a chart exist for a proper range for round counts given numbers of players?
2^r where r is the number of rounds gives you the maximum number of players in a section that would yield a single perfect score winner. If you consider draws, then 2^r+1 would normally suffice. The scholastic nationals use the last formula to determine if accelerated pairings are used in a championship section. Of course, there is no formula for non-perfect score winners in a section. An organizer canât control that.
The other post gave you the formula, and as noted this cannot help draws for non-perfect scores, but if you want a chart for perfect scores:
0-2: 1
3-4: 2
5-8: 3
9-16: 4
17-32: 5
33-64: 6
65-128: 7
129-256: 8
So many easy prize divisions have been blown by the leader losing in the last round leaving a 6-way tie.
Thanks to everyone, very helpfull.
The US Open is a 9 round tournament. Last year there were 488 players, and heading into the final round there were 7 players at 7, and only 2 of them won, so the title was shared by 2 players at 8-1. The co-champions did not face each other, so you could argue that it could have been a 10 or possibly even an 11 round event without ugly pairings on the top boards at the end. (To me, âugly pairingsâ means means that in the final round some of the top players are paired two score-groups down because theyâve already faced others in the top 2 score groups.)
But draws are more common among higher rated players, so what works in the US Open might not work in a less strong event, especially a scholastic event.
I guess I donât even understand the question. The only way to be sure to have a clear winner is to use tiebreaks, whether mathematical or some sort of playoff. Mr. Nietmanâs formula of 2^(n+1) will usually work to have no more than one perfect score, but Iâm not sure that is either necessary or desirable.
There is always a way to pair another round, even in the worst case scenario of a four round, three player tournament. A tournament shouldnât end just because one player is in clear first, or continue just because there are players who are tied for the lead. And it saddens me to describe playing chess as âwast[ing] timeâ. You just need to plan how long you want the tournament to last and what to do with tied players at the end, bearing in mind that a lot of players will also have budgeted time and wonât want to wait around for a playoff, or even to play at an accelerated time control. Just announce n advance so there are no surprises.
I am not trying to find perfection. We have had arguments at my club about setting up tournaments. We are a small group. We set up âcontingenciesâ, X number of players = Y number of rounds.
We were setting up our blitz tournament contingencies, 4 players = triple RR, ect. When we come to 15 players we, ran out of time for a RR. So I went with two swiss tournaments, 4 rounds each, to which someone said why not one tourn at 8 rounds. I said the later pairings would be squirrelly if not illegal.
So I asked. Hence tHe wording in my original post, surely there are upper and lower practical limits.
For a Blitz Swiss tournament, you could always hold 2 games per round, one as White, one as Black. That makes color-balancing a non-issue when pairing.
With a 15 players to 8 rounds, one thing you can do is to use âdeceleratedâ pairings, which gives extra points (for pairing purposes) to the bottom half players (starting in R2; R1 is paired normally). This delays the inevitable match-ups between the top players so they occur later rather than earlier. With 15 to 8 with standard Swiss methods there almost certainly would be legal pairings, but round 8 would have a player with an even score or better likely getting the bye, and the top score probably having to drop at least 2.0 points to find an opponent.