Another pairings question

Running Chess Tournaments
1

The following standings were for the sixth and final round of the Open section at a significant prize tournament in the first half of 2010. First place prize money was $1000. All of the players with 3.0 were in contention for at least a share of second place.

4.0 Player A (2400)
3.5 Player B (2200)
3.0 Player C (2250) - reentry
3.0 Player D (2200)
3.0 Player E (2150)
3.0 Player F (2125)
2.5 Player G (2050)
2.0 Player H (2100)

A can only play D or F
B can play C, D or E
C had black vs F in 3-day, reentered and then played A and G in 2-day
D can play anyone except E
E can only play B or C
F faced white C (who subsequently reentered), then played B and E
G played A, B and C
H played B and D

Ignoring colors, here are four possible pairings.

Swiss-sys pairings:
A-D
B-E
C-F (players meet for second time)
G-H

Alternative #1:
A-D
B-C
E-G
F-H (less desirable because 3.0 faces a 2.0)

Alternative #2:
A-F (no doubt F would be very unhappy)
B-D
C-E
G-H

Alternative #3:
A-F (transpose D with F)
B-E
C-D
G-H

The computer’s pairings were eventually posted. However, USCF rule 28S says players “should not” be paired again if one of them had reentered. Now I know that there were at least two legal alternate pairings to preserve the 3.0 score group. What do you think?

Several players complained about the pairings and the two senior TDs took a closer look at them. They found Alternative #1 but the TDs judged it to be inferior because it breaks the 3.0 score group. Working quietly on my own, I saw Alternative #2, but didn’t want to get mixed up in the dispute. Neither the TDs nor the complainants came up with Alternatives #2 or #3.

Still, Player F was entitled to get his proper pairing in the money round. The difference between facing an experienced master (C) for the second time in the tournament or a mid expert (H) means a lot to a 2125 rated player.

What should have happened in this case? Given that this was the money round, I believe my opponent should have formally appealed the pairings. Facing an opponent twice is quirky enough to get a second opinion. I believe the TDs would have tried contacting a special arbiter. Ironically though, the apparently correct pairing A-F is even “worse” than C-F.

Michael Aigner

2

Personally, I’d have gone with alternative 2 and the 75-point transposition from D to F.

28S1 references 27A1, so the importence for the player who did not re-enter to avoid playing the re-entering player is put at the highest priority. It cannot say “must not” rather than “should not” because there are some cases where 27A1 must be violated (the example given was more rounds than potential opponents).

3

Clearly the pairing rules are not complex enough.

4

I agree with Jeff’s pairings. I have to wonder if the “previous pairing number” had been correctly set for player C’s reentry. If so, I think that SwissSys pairing C and F again should be reported to Thad Suits as a bug.

5

When making that report, include whether or not the sections were merged from different computers. It seems like re-entries are handled more cleanly if the player is listed as a re-entry in a section being paired on the same computer and in the same event as the computer pairing the section the player left.

My first guess without knowing anything about the event is that the re-entries were done in a different area on a different computer.

6

Why is Alternative #2, which affects three boards (two transpositions), superior to Alternative #3, which changes only two boards (one transposition)?

That is a possibility. The TDs used two computers at different times during the weekend.

7

This thread hints at two important things:

  1. When a section or a tournament is this small, when the number of rounds is also small, it works out well to start out with a RR table pairing for a few (if not all) rounds.

  2. Those pairing programs are nice but the TD is still responsible for checking the final output. And sometimes you just have to bite the bullet and decide which pairing rule to break.

8

I prefer alternative #2 over alternative #3 because #2 assigns B a higher ranked opponent than C. Ideally, after pairing A and F, you’d like to pair B and C, but that does not allow you to pair the rest of the 3.0 score group (since D and E have met). In that case, pairing B-D and C-E is closer to the natural pairings than is B-E and C-D.

9

I removed some players who had byes or withdrew. There were 15 players total, with 8 in both 3-day and 2-day schedules (including the one who played both). That’s not too small.

A major contributing factor was the inability of the two TDs or the complaining players to find Alternatives #2 or #3. Should they have found these under time pressure before the last round?

You’re right that TDs today often are not as proficient (experienced) at doing hand pairings. How can they be when all tournaments are run by computer? Is it fair these days to expect a local or even senior TD to solve a complex pairings problem when they’ve rarely done any hand pairings? I have no doubt that if the TDs in this case had seen Alternatives #2 or #3, they would have changed the pairings.

By the way, one of the TDs has 20+ years of experience directing, including before the popularity of computer programs.

Michael Aigner

10

#2 follows example 3 under 29E7. Both alt #2 and alt #3 are valid and it comes down to a person’s preference.

11

I think it’s a poor idea to pair by computer when the tournament (or section) is small (say, fewer than 10 or 15 players). There are all kinds of already-played-each-other situations that a human can anticipate better ahead of time.

An extreme example is the famous Dennis Keen trap, which can occur in a 6-player event of 4 or 5 rounds. It is possible to pair rounds 1-3 in such a way that there are NO pairings in round 4 which avoid repeat pairings for all players. As far as I know neither of the two major pairing programs will avoid this trap.

Go ahead, try it. Let’s say the results are as follows after 2 rounds:

1 1900 wb 1.0 W4 L2
2 1800 bw 2.0 W5 W1
3 1700 wb 2.0 W6 W4
4 1600 bw 0.0 L1 L3
5 1500 wb 0.0 L2 L6
6 1400 bw 1.0 L3 W5

What are your pairings in round 3?

Bill Smythe

12

And that is why some of the pairing questions on some of the TD exams can’t be solved with a pairings program.

13

I ran into this problem recently with SwissSys, although to be fair I had two of the players “teamed”, so that might be the reason it made that pairing.

Alex Relyea

14

It may have been the “team” that caused your particular situation, but I have seen SwissSys walk blindly into that trap before with no restrictions on pairing players.

By the way, it is quite worthwhile to work through Bill Smythe’s example by hand to understand why the third round pairings are absolutely crucial. The first two rounds don’t matter; the trap occurs when you pair the third round. The director must be careful not to “partition” the players into two groups of three, where each player in one group has met all the players in the other group. If this happens, in the fourth round, two players must meet again.

Suppose you have partitioned your players into two sets, {A, B, C} and {D, E, F} where each of players A, B, and C have faced all of D, E, and F. (That also means each of players D, E, and F have faced all of A, B, and C.) In order to avoid pairing players a second time, you must pair {A, B, C} among themselves, and {D, E, F} among themselves. That, of course, is impossible; you will be forced to pair two players again in the fourth round.

15

Ironically, it seems to actually have been a mistake for Player F to appeal the computer pairings that were posted, because he would face a 2400 rated IM instead of playing a 2250 for the second time.

Michael Aigner

16

It is ironic that one of the most plausible ways to fall into this trap is to make all the colors work perfectly in the first three rounds. That gives you your partition – {A, B, C} are the players who started with white, {D, E, F} with black.

For those interested in the math, in round 3 of a 6-player tournament there are always exactly four possible sets of pairings, exactly one of which falls into the trap.

Bill Smythe

17

That could be avoided with a swiss guided round-robin. Select pairing numbers which will give traditional swiss pairings in the first round then select the round from the RR table that pairs the most number of perfect scores.

18

The above is true in any Swiss when pairing the second to last round when it’s one round short of a round robin. The reason it happens mostly in a 6 player 4 round Swiss is that the plurality of Swiss tournaments with an even number of rounds seem to be 4 rounds. It doesn’t happen when the number of rounds is odd. The following are examples of trap rounds:

round 5 of an 8 player 6 round Swiss
round 7 of a 10 player 8 round Swiss
round 9 of a 12 player 10 round Swiss etc.

If there are 3 players who have yet to play each other going into that round and two of them aren’t paired, you have left yourself with an impossibility for the next round because when 2 of them play each other it leaves the third with only the 2 that just got paired as his only potential new opponents. I also used to fall into that trap. Now I know to look for it. On the rare cases where someone notices a 3rd round pairing irregularity, I point out that if I didn’t do it that 2 players would have to play again in the last round. That always seems to satisfy the one asking the question.

As for teamed players, it makes it slightly more difficult to pair but still possible. SwissSys would have made the same error without the teamed players. I know because I’ve had to make manual corrections.

19

I’m not exactly sure what you mean, but it doesn’t seem to be exactly correct.

Actually for any odd N, if you have 2N players and divide them into two groups of N each of which has played all of the other group in rounds 1 to N, then you won’t have a legal pairing in round N+1. The problem occurs because you can’t pair everyone within their own group of N since it is odd.

N-# of entries- Round with pairing problem
1-------2----------------2
3-------6----------------4
5------10----------------6
7------14----------------8

The time you need to fix the problem is round N.

20

Not true. With 6 players, the trap occurs with equal likelihood in either a 4- or a 5-round event. In fact, if you can pair round 4, you can automatically pair round 5 – just pair each player against the one player he hasn’t already played.

Perhaps you meant that the trap can occur only in an even-numbered round. That’s true for 6 players, I’m not sure about more.

Bill Smythe