Swiss Pairing Minimum Numbers?

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John Upham
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Swiss Pairing Minimum Numbers?

Post by John Upham » Sat May 14, 2022 5:28 pm

I should know this but I'll ask anyway...

For an n round Swiss paired individual tournament what is the minimum number of players that will allow a viable pairing solution for all n rounds?

Explicitly for a five round event what is the minimum number of players?

Many thanks for your help.
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Roger de Coverly
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Re: Swiss Pairing Minimum Numbers?

Post by Roger de Coverly » Sat May 14, 2022 5:42 pm

John Upham wrote:
Sat May 14, 2022 5:28 pm
Explicitly for a five round event what is the minimum number of players?
It's probably 8 for a 5 round Swiss, or 7 with a bye.

The Berks & Bucks tournament historically had very small sections from time to time. Certainly 7 even if had the side effect that a last round bye could push the recipient up to second or third. Perhaps the arbiter checked at round 4 that the pairings allowed a round 5.

Think of it as a seeded all play all.

Sean Hewitt would sometimes run his nine round all play alls that way, if he had twelve or fourteen entrants for nine rounds.

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IM Jack Rudd
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Re: Swiss Pairing Minimum Numbers?

Post by IM Jack Rudd » Sat May 14, 2022 6:24 pm

When I last ran a 7-player five-round Swiss, I wasn't sure whether it could jam, but I didn't take the chance: I split it into two groups, sizes 4 and 3, ran them both as all-play-alls, and then ran the last two rounds of the event as a Swiss.

Francis Fields
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Re: Swiss Pairing Minimum Numbers?

Post by Francis Fields » Thu May 19, 2022 4:29 pm

The minimum number of chess players for a Swiss is 18. This is a calculation I considered some time ago.

The range of grades (or strong players) can change this - though only by a small number of players.

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Matt Mackenzie
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Re: Swiss Pairing Minimum Numbers?

Post by Matt Mackenzie » Thu May 19, 2022 6:08 pm

That may be a practical minimum rather than a theoretical one.
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John Upham
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Re: Swiss Pairing Minimum Numbers?

Post by John Upham » Thu May 19, 2022 7:04 pm

Francis Fields wrote:
Thu May 19, 2022 4:29 pm
The minimum number of chess players for a Swiss is 18. This is a calculation I considered some time ago.

The range of grades (or strong players) can change this - though only by a small number of players.
Is that regardless of the number of rounds?
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David Sedgwick
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Re: Swiss Pairing Minimum Numbers?

Post by David Sedgwick » Thu May 19, 2022 10:09 pm

John Upham wrote:
Thu May 19, 2022 7:04 pm
Francis Fields wrote:
Thu May 19, 2022 4:29 pm
The minimum number of chess players for a Swiss is 18. This is a calculation I considered some time ago.

The range of grades (or strong players) can change this - though only by a small number of players.
Is that regardless of the number of rounds?
John, I would suggest that you look at some earlier examples of Mr Fields's contributions to this Forum before seeking to enter into a serious discussion with him.

See for instance viewtopic.php?f=2&t=10160&p=229096#p229096 and the comment by Leonard Barden.

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John Upham
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Re: Swiss Pairing Minimum Numbers?

Post by John Upham » Thu May 19, 2022 10:11 pm

David Sedgwick wrote:
Thu May 19, 2022 10:09 pm


John, I would suggest that you look at some earlier examples of Mr Fields's contributions to this Forum before seeking to enter into a serious discussion with him.

See for instance viewtopic.php?f=2&t=10160&p=229096#p229096 and the comment by Leonard Barden.

The name did ring a bell in that respect: thanks for the reminder!

Talking of nonsense contributions, whatever happened to the ellipsis gentleman....?

I miss his "contributions"....
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Mike Gunn
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Re: Swiss Pairing Minimum Numbers?

Post by Mike Gunn » Fri May 20, 2022 11:56 am

Top tip: write down the Berger table pairings for the number of players you have and select a new set for each round. This avoids blockage in the later rounds.

Roger de Coverly
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Re: Swiss Pairing Minimum Numbers?

Post by Roger de Coverly » Fri May 20, 2022 12:37 pm

There's a working example at the moment. The Eng;ish Women's Championship has seven rounds with ten players.

It might be easier with pairing cards than with using a computer program, but a method of ensuring that a earlier round pairing doesn't lock out future rounds would be to test the existence of pairings for later rounds against the premise that all games in the current and future rounds round are draws or byes. With a small number of players many of the usual rules for a Swiss are likely to be broken so as to give priority to the rule that you don't play the same opponent more than once.

Another way of expressing this is to check that with n rounds to go there are at least n pairings that don't break the duplicate pairing rule, all other rules being suspended.

SeanCoffey
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Re: Swiss Pairing Minimum Numbers?

Post by SeanCoffey » Sat May 21, 2022 5:26 pm

I'm not quite sure what the OP is asking; as has been pointed out above, there is always a viable pairing of all rounds by, for example, taking rounds from a Berger table, so it comes down to which Swiss pairing algorithm you're using.

If the question is what the numbers have to be to ensure that problems can't possibly arise, this seems relevant:

Alexander Rosa, Walter D. Wallis,
Premature sets of 1-factors or how not to schedule round robin tournaments,
Discrete Applied Mathematics,
Volume 4, Issue 4,
1982,
Pages 291-297,
ISSN 0166-218X,
https://doi.org/10.1016/0166-218X(82)90051-8.
(https://www.sciencedirect.com/science/a ... 8X82900518)

1-factor -> round, K_2n -> set of 2n players, premature set of k 1-factors = set of k rounds that don't allow any other rounds to be paired. This ignores colour restrictions and other complications: players can be paired if and only if they haven't previously been paired.

For a 5-round Swiss, 6 players may have problems: if the first 3 rounds are, in effect, a Scheveningen between A, B, C and D, E, F, then round 4 can't be paired. Similarly, a 7-round Swiss with 10 players can have problems after round 5.

Though this paper doesn't say so explicitly, it seems that for a 5-round Swiss, 8 is the minimum number without byes to guarantee that no problems can arise. For 8 players, a 6-round Swiss can generate a problem (corollary 2.2 here).

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Re: Swiss Pairing Minimum Numbers?

Post by SeanCoffey » Sat May 21, 2022 6:35 pm

Reading a little further, a more relevant paper is:

Rolf Rees, Walter D. Wallis,
The spectrum of maximal sets of 1-factors,
Discrete Mathematics,
Volume 97,
1991,
Pages 357-369
https://core.ac.uk/download/pdf/82613754.pdf

A premature set of k 1-factors = a set of k rounds that don't allow _all_ other rounds of a round robin to be paired, contrary to what I carelessly wrote above. A maximal set of k 1-factors = a set of k rounds that don't allow _any_ other rounds to be be paired.

The second paper's Theorem 1.0. gives a complete characterization of the k's and n's for which maximal sets exist. From this, there can be no problem pairing a 5-round Swiss with 8 players (ignoring the colour assignment restrictions).

Roger de Coverly
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Re: Swiss Pairing Minimum Numbers?

Post by Roger de Coverly » Sat May 21, 2022 7:00 pm

SeanCoffey wrote:
Sat May 21, 2022 6:35 pm
From this, there can be no problem pairing a 5-round Swiss with 8 players (ignoring the colour assignment restrictions).

Would there not be round 4 pairings which prevented round 5 ones?

Alex McFarlane
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Re: Swiss Pairing Minimum Numbers?

Post by Alex McFarlane » Sat May 21, 2022 7:37 pm

Roger de Coverly wrote:
Sat May 21, 2022 7:00 pm
Would there not be round 4 pairings which prevented round 5 ones?
No.

If A,B,C,D had played each of E,F,G,H then each group can be paired within itself. e.g. A-B, C-D, E-F, G-H

Your problems are when you have an odd number in each subgroup as they cannot then be paired together.

Francis Fields
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Re: Swiss Pairing Minimum Numbers?

Post by Francis Fields » Thu May 26, 2022 5:12 pm

The point of the Swiss pairing system ( as believed to be devised by Arpad Elo ) is to aim to pair players of similar strength. This requires more players than a theoretical minimum.