The following actually happened in one section of a recent OTB event. There were 6 players in the section, and 3 rounds. Names and ratings (but not rating differences) have been changed to protect the guilty. The crosstable is shown in rating order, not score order, to make the pairings easier to justify or to argue about.
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name score colors
Alice 1577 1.0 bwb W4 L2 L6
Bob 1507 2.0 wbw W5 W1 L3
Cathy 1275 3.0 bwb W6 W4 W2
David 1237 1.0 wbw L1 L3 W5
Ellen 1196 0.0 bwb L2 L6 L4
Frank unr. 2.0 wbw L3 W5 W1
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Beautiful, no? All the colors have alternated perfectly, and the pairings seem natural. There have been a few upsets, but nothing out of the ordinary.
This tournament had 3 rounds. Quiz question: If there had been a fourth round, what is the best way to pair the fourth round?
Assuming that 2’s round three opponent was 3 (not himself):
3-1
6-2
house player - 4
5 gets a bye
Better than a 5-4 rematch on the bottom (least significant) board.
Granted, but I was dealing with the actual situation, not the situation I would have wanted to deal with. Prevention is for the future. Damage control is for the present.
I have a fair amount of experience dealing with pairing problems due to errors made by multiple (very highly experienced) pairing TDs (which sometimes was me, nobody is absolutely perfect), and doing so under extreme time pressure that doesn’t allow spending time detailing how to avoid it in the future (once by doing something that the rulebook explicitly says should not be done - and having the appropriate US Chess committees say that it was the right thing to do and thus showing why the rulebook systemically uses “should not” instead of “must not”). Saying that something should have been avoided doesn’t do anything to deal with what actually happened.
The time for working on preventing future problems is after the damage control is done for the current problem (one caveat - other current but un-posted pairings with the same problem should probably be fixed before the current problem’s damage control is completed).
So, even if Bill didn’t intend it that way, this is a good pairing question about what to do when you find yourself up a certain creek while lacking a paddle.
If you have access to Doc Brown and Marty McFly then you may be able to prevent the problem from occurring. If you don’t have such access then you need to address what actually happened.
I don’t know what the “best” solution is, but I would follow the advice of my mentor (the guy who taught me most of what I know about directing) and pair it as a round robin with one round omitted. That neatly sidesteps any late-round pairing difficulties. The only thing that would require any thought would be which round to omit, and the obvious choice would be the round that pairs highest-rated vs. lowest-rated (or highest-rated vs. unrated, if we assume that the unrated player is inexperienced – although that’s not a safe assumption these days).
That’s what I figured most people would say, but I disagree with “neatly”.
If you look at the combinatorics, you’ll see that there are EXACTLY four ways to pair round 3, i.e. there are four different possible sets of pairings for round 3 involving all 6 players without a rematch. Of these four sets of possible round 3 pairings, ONE falls into the 6-player trap in round 4, the other three do not.
I prefer not to choose a pairing set based on what it does to the lowest-rated player. Instead, I prefer to use whichever one looks more Swiss-like overall. After all, the event was originally advertised as a Swiss, no?
Trying to fit the pairings into a round-robin table is overkill, if you ask me. There are only 6 players. How difficult can it be to shuffle 6 makeshift pairing cards, or to draw a few hexagons on a piece of scratch paper?
In the specific case that generated this thread, I noticed right away that all the colors alternated perfectly in round 2 and in round 3:
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name score colors
Alice 1577 1.0 bwb W4 L2 L6
Bob 1507 2.0 wbw W5 W1 L3
Cathy 1275 3.0 bwb W6 W4 W2
David 1237 1.0 wbw L1 L3 W5
Ellen 1196 0.0 bwb L2 L6 L4
Frank unr. 2.0 wbw L3 W5 W1
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That’s a bright red flag for me. Having studied the combinatorics of a 6-player Swiss, I knew instantly that a pairing set that alternates all colors in rounds 2 and 3 automatically falls into the 6-player trap in round 4.
So what to do? If you are using a pairing program, let it (tentatively) make round 3 pairings. If the pairings it makes alternate all the colors in both rounds, cancel the pairings and revert to the situation immediately after round 2. Go into “settings” and change the color-alternation transposition limit from 80 to 0. That might do the trick! Then press the “make pairings” button. If the pairings are different now, then voila! you’re done. If they’re the same, you’ll just have to pair by hand.
There are other ways to fall into the trap besides allowing color alternation to work “too well”. Therefore, once you have tentative round 3 pairings, a quick scratch-paper check where you attempt to make round 4 pairings – any round 4 pairings, regardless of score or color, as long as they don’t pair the same players twice – should tell you all you need to know.
Incidentally, if the 6-player event also has a 5th round, that round will never create problems as long as round 4 worked. In round 5, just pair each player against the one player he has not already faced.
A manual check is so much simpler, and keeps the TD in the loop so much better, that I will never understand why so many TDs recommend the artificial use of round-robin tables. It’s like using a sledge hammer to crack a peanut.
Having not studied any sort of combinatorics, I find the round-robin solution “so much simpler” – and if there are 5 rounds, it’s a no-brainer: it’s a round-robin by definition, so there’s no reason to spend any time at all making pairings. Just use the ones that are already made for you in the round-robin tables.
If there are 5 rounds, and you know for sure that the number of players will be 6 throughout (i.e. no late entrants and no drop-outs), then, sure, use the round-robin tables.
But if there is a reasonable possibility of additional (late) entrants, I’d prefer to remain flexible.
Or, if there are 6 players but only 4 rounds, I would prefer not to be forced to choose, before many results are in, which “round” from the table to omit. Again, I would prefer to remain flexible.
Incidentally, my “study” of 6-player Swiss combinatorics did not come from any book, or any classroom. It came from just working things out on paper.
My first encounter with the 6-player trap came in about 1970, as a player, not as a TD. In fact, it came as a player in a different section from the one I was in. There was an “elite” section for players rated 2000 or higher, advertised as a 5-round Swiss with 3 rounds Saturday and 2 rounds Sunday, but only 6 players showed up for that section.
The TD blithely made Swiss-style pairings on Saturday, and noticed (or somebody pointed out to him) late Saturday evening that no pairings existed for rounds 4 and 5.
His solution? Run 3 rounds on Sunday (instead of 2). Two games were played in the early morning, two in the afternoon, and two in the late evening. The two players that had to play in the early morning and late evening, with a long wait in the afternoon, were not pleased.
I had a TD who had 12 players, 10 rounds (I think one round per night). While a RR-Hybrid Swiss would ordinarily be the way to go with that, unfortunately, one player had two 1/2 byes (thus forcing two other full byes) which pretty much would wreck that idea. (You certainly can’t do the normal thing which is to naturally pair the top ranked player and pick the RR round to match—I guess you would probably have to find the lowest ranked player who hadn’t played the 1/2 point bye and pick the round where they would play). Of course, by the time you get to rounds 9 and 10 (or probably 7) the top ranked player has likely played everyone from 2 to 7 so even if pairings exist, they aren’t likely competitive.
Never run an R-round tournament with N players if R is close to N (unless R = N-1).
Never allow byes in a round-robin.
Never expect competitive pairings throughout a round robin.
I have not worked out the combinatorics for N players, R rounds except for N=6, even without complications like byes, drop-outs, etc.
It would be interesting to know the answer to the following:
What is the formula for calculating R, given N, that makes the following statement true: “If there are N players in an R-round tournament, a trap like the 6-player trap is possible if, and only if, the number of rounds is at least R.”
For simplicity you may assume that N is even and that there are no byes or drop-outs.
We already know that, if N=6, then R=4.
I’m sure somebody in the world knows the answer.
If you replace “the” with “a” then R = 2*(int(N/2+.5))-(any even number less of than N/2).
For 16 players you can run into the trap at:
14 (three players cycled through the other 13 in the first 13 rounds)
12 (five players cycled through the other 11 in the first 11 rounds)
10 (seven players cycled through the other 9 in the first 9 rounds)
For 10 players you can run into the trap at:
8 (three players cycled through the other 7 in the first 7 rounds)
6 (five players cycled through the other 5 in the first 5 rounds - think five 14xx and five 11xx and all draws)