Killer TD

[Triumph]

Greetings,

…

The officers will be meeting this week to discuss reopening the Tuesday night events.

And now, the newsletter…

Game Of The Week

This week’s game comes from the Colorado Springs Chess Club’s event: CSCC Killer Bishop Rapid Online (3SS, G/10+10). It was nice to see that the computer can run these events without having to be present. I was unable to play in the event due to the surprise Memorial Day blizzard.

Of course, the computer can be unpredictable at times. I had asked it to run a four-round event like we had been doing the past several weeks. However, the computer called things quits after three rounds. It could have been because of the lighter holiday turnout that didn’t need more than 3 rounds to determine a winner, or it could have been because the computer is a ruthless TD.

When it comes to running a Swiss tournament, the computer will show no mercy:

No half point byes
No announcements
No warnings
Limited late entrants
One second late = no pairing
Starts your clock immediately
Bans Jeff Fox from round 1
Doesn’t follow advertisement
Leaves early
Doesn’t let you take prizes home

I would fire the bum, but it works cheap and shows ups on time. At least, we get a couple of rounds in to find some wild positions like this game from round 2.

Black to move

See the diagram and answer here:

cschess.webs.com/apps/photos/ph … =206209086

Killer TD

chessvideos.tv/chess-game-r … ?id=115488

[Event "CSCC Killer Bishop Rapid Online "]

[Site “https://cschess.webs.com/”]

[Date “2020.05.24”]

[Round “2.1”]

[White “mrpicklez”]

[Black “waynehatcher”]

[Result “0-1”]

[ECO “B09”]

[WhiteElo “1795”]

[BlackElo “1741”]

[TimeControl “600+10”]

[EndTime “19:03:39 PDT”]

[Termination “waynehatcher won by resignation”]

1. e4 d6 2. d4 Nf6 3. Nc3 g6 4. f4 Bg7 5. Nf3 c5 6. Bb5+ Bd7 7. e5 Ng4 8. e6

fxe6 9. Ng5 Bxb5 10. Nxe6 Bxd4 11. Nxd4 Bd7 12. Nf3 Nf6 13. O-O O-O 14. Nd5 Nc6

15. Nxf6+ Rxf6 16. Ng5 h6 17. Ne4 Rf8 18. Qd5+ Kg7 19. Bd2 Qc8 20. Bc3+ Kh7 21.

Ng5+ hxg5 22. Qxg5 Rf5 23. Qg3 Rh5 24. Rae1 Qf8 25. Qf3 Qf5 26. h3 Qd5 27. Qe3

Rf8 28. g4 Bxg4 29. Qg3 Bxh3 30. Rf3 Bf5 31. Kf2 Rh3 32. Qg2 Qxf3+ 33. Qxf3

Rxf3+ 34. Kxf3 Kg8 35. Rh1 Kf7 0-1

This Week In Chess

On May 24th, the Colorado Springs Chess Club held the CSCC Killer Bishop Rapid Online event (4SS, G/10+10).

chess.com/tournament/live/c … ne-1237640

Place, Player, Score

1 “#1waynehatcher 1754)” 2.5

2 “#2KingVed (1489)” 2.0

3 “#3jfoxhoot (1542)” 2.0

4 “#4mrpicklez (1742)” 1.0

5 “#5linuxguy1 (1526)” 1.0

6 “#6HermitCrab0 (1486)” 1.0

7 “#6CosmicNovaGalaxy (1298)” 1.0

8 “#8Navajo36us80917 (1119)” 0.5

9 “-MoxRuby (Unrated)” 1.0

[fpawn]

You need minimum 8 players at the start of round 1 for a 4 round tournament on Chess.com. Otherwise the system automatically truncates it to 3 rounds, even if there are adequate latejoins. That’s a bit silly in my opinion, but it is what it is.

[wintdoan]

@fpawn

You need minimum 8 players at the start of round 1 for a 4 round tournament on Chess.com. Otherwise the system automatically truncates it to 3 rounds, even if there are adequate latejoins. That’s a bit silly in my opinion, but it is what it is.

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

[jwiewel]

@wintdoan

@fpawn

You need minimum 8 players at the start of round 1 for a 4 round tournament on Chess.com. Otherwise the system automatically truncates it to 3 rounds, even if there are adequate latejoins. That’s a bit silly in my opinion, but it is what it is.

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

Tom knows the “pairing trap” but for those that don’t:

Call any player A, with the player’s round 1 opponent called B and round 2 opponent called C. Call C’s round one opponent D. Call B’s round 2 opponent E (can’t be D as otherwise E and F would play each other in both round 1 and round 2).
That gives the following pairings (ignoring color information)
Rd 1) A-B, C-D, E-F
Rd 2) A-C, B-E, D-F
At this point A, D and E have each played two of B, C and F. A has three possible round 3 opponents (D, E and F) and only F (1 of 3 possible opponents) gives the chance of blowing up round 4. If A plays F then D can play either B or E and only B (1 of 2 possible opponents) finishes blowing up round 4.
The problem pairing is below (one of six possible round 3 pairings)
A-F, B-D, C-E
After this, and looking at only players not yet played, A has only D or E to play in round 4, D has only A or E and E has only A or D, so if any two of those three play each other then the third player has to play a rematch.

Mathematically that seems to give a 16 2/3% chance of a problem (really a 25% chance since there are only four possible pairings without a rematch), but differing likely results based on ratings and colors would probably reduce that.

Safe round three pairings are: A-D, B-F, C-E / A-E, B-D, C-F / A-F, B-C, D-E

[dennisk]

@wintdoan

@fpawn

You need minimum 8 players at the start of round 1 for a 4 round tournament on Chess.com. Otherwise the system automatically truncates it to 3 rounds, even if there are adequate latejoins. That’s a bit silly in my opinion, but it is what it is.

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

I actually had this happen to me once (having to do a duplicate pairing in the final round of a small section), and was later advised by a more experienced TD to pair such a section as a round robin with (in the case of 6 players for 4 rounds) one round missing. Of course, no matter which round you choose to omit, you run the risk of unhappiness among those who would have had favorable pairings in that round.

[fpawn]

@wintdoan

@fpawn

You need minimum 8 players at the start of round 1 for a 4 round tournament on Chess.com. Otherwise the system automatically truncates it to 3 rounds, even if there are adequate latejoins. That’s a bit silly in my opinion, but it is what it is.

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

I have seen this pairing trap in practice. The Colorado Springs online tournament saw three players latejoin before round 2, which would have eliminated any issues. Just last Saturday, I had a 7 player tournament, originally scheduled for 4 rounds, also get truncated to 3 rounds. We even got one latejoin to make an even 8.

IMHO the most egregious scenario is that Chess.com automatically aborts any swiss tournament with three players, instead of running a quad with a bye. I joined a youth event at the last second to avoid having three unhappy kids.

Michael Aigner

[Smythe_Dakota]

@wintdoan

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

Actually, it’s 25%, not 10%.

In the specific case of 6 players and 4 or 5 rounds, care must begin when you pair round 3. There are four possible sets of pairings in round 3, one of which results in a trap in round 4. The other three will work out OK.

For 6 players with 4 or 5 rounds, you can use my hexagon trick. In Jeff’s example:

@jwiewel

Call any player A, with the player’s round 1 opponent called B and round 2 opponent called C. Call C’s round one opponent D. Call B’s round 2 opponent E [ and the remaining player F] … …
That gives the following pairings (ignoring color information)
Rd 1) A-B, C-D, E-F
Rd 2) A-C, B-E, D-F

– which yields the following hexagon:

— A — B
– / ------ \

• C ------- E
  – \ ------ /
  — D — F

(I hope this renders OK on your browser or phone screen.) Each player is placed between the two players he has already played in rounds 1 and 2. To put it another way, the six vertices of the hexagon represent players, and the six sides represent pairings for rounds 1 and 2.

There are now four possible ways to pair round 3. In one of these, every player is paired against the opponent diametrically opposite himself / herself on the hexagon. If you draw the lines on the hexagon (A-F, B-D, C-E), it will look like a giant asterisk inside the hexagon. This pairing is the troublesome one, that will leave you without pairings in round 4.

Any of the other three possible pairing sets will work. In each of these, only one player is paired against a diametrically opposite opponent. If you draw it out, the other two pairings will be parallel to each other, and perpendicular (if your hexagon is equilateral and equiangular) to the diametric pairing. For example, you could pair A-F (diametric), B-C (non-diameter chord), and D-E (non-diameter chord). Alternatively, you could start with B-D (diametric) and proceed from there, or with C-E (diametric) and proceed from there. Just make sure that, when you draw out the pairings, you have only one diametric pairing, and two parallel non-diametric chords.

If you follow the above rule, you will always have round 4 pairings, and if there are 5 rounds, you will have round 5 pairings too.

Bill Smythe

[wintdoan]

@Smythe Dakota

@wintdoan

You can’t really guarantee that a 4 round 6 player tournament can be run by a Swiss without duplicated pairings. I could see someone deciding that, if a computer is running this, avoiding the 10% or so chance of the Swiss blowing up in round 4 is simpler than dealing with the unhappy customers.

Actually, it’s 25%, not 10%.

In the specific case of 6 players and 4 or 5 rounds, care must begin when you pair round 3. There are four possible sets of pairings in round 3, one of which results in a trap in round 4. The other three will work out OK.

Four?? I believe it’s three. (Each player has played two others, leaving three possible opponents, but once you fix one pairing, the other two are forced).

But they’re way short of being equally likely. The problematical round three pairing for the highest ranked player will almost certainly be the lowest rated potential opponent. If you have the (most likely) round two pairings of 1-2, 3-4, 5-6, then the problem pairing in round three has 1 v 6.

[Smythe_Dakota]

@wintdoan

Four?? I believe it’s three. (Each player has played two others, leaving three possible opponents, but once you fix one pairing, the other two are forced).

There are three possible round 3 pairings that don’t fall into the round 4 trap, but there is that nasty fourth one that does.

It is not true in round 3 that “once you fix one pairing, the other two are forced”. Once you fix one pairing, there may be two ways to pair the remaining four players, one good way and one bad way. In other words, “the other two [ pairings ] are forced” only in the sense that “there is only one way to make the remaining two pairings that avoids the round 4 trap”.

Let’s renumber the hexagon (from my previous post) to follow your “most likely” scenario of 1-4, 2-5, 3-6 in round 1, followed by 1-2, 3-4, 5-6 in round two:

— 1 — 4
– / ------ \

• 2 ------- 3
  – \ ------ /
  — 5 — 6

If you start with a “diametrically opposite” pairing (1-6 or 2-3 or 4-5), there will still be two ways to make the remaining two pairings. For example, if you start with 1-6 as you mentioned, you can then pair 2-3 and 4-5, or you can pair 2-4 and 3-5. The first way is the bad way (note that it involves 3 “diametrically opposite” pairings), while the second way is the good way (there is only one diametrically opposite pairing and two non-diametric parallel chords).

… then the problem pairing in round three has 1 v 6.

You could equally well say that “the problem pairing in round three has 2 v 3” or that it “has 4 v 5”. These, together with 1 v 6, are the three diametrically opposite pairings. As long as you have only one of these three, you’re OK. If you have two (and hence all three), you’re cooked.

… But they’re way short of being equally likely. …

Based on typical Swiss pairings, I’ll take your word for it. You will always have one diametrically opposite pairing. Just make sure you have ONLY one.

Bill Smythe

[jwiewel]

@wintdoan

@Smythe Dakota

…

In the specific case of 6 players and 4 or 5 rounds, care must begin when you pair round 3. There are four possible sets of pairings in round 3, one of which results in a trap in round 4. The other three will work out OK.

Four?? I believe it’s three. (Each player has played two others, leaving three possible opponents, but once you fix one pairing, the other two are forced).

But they’re way short of being equally likely. The problematical round three pairing for the highest ranked player will almost certainly be the lowest rated potential opponent. If you have the (most likely) round two pairings of 1-2, 3-4, 5-6, then the problem pairing in round three has 1 v 6.

After
Rd 1 1-4, 5-2, 3-6
Rd 2 2-1, 4-3, 6-5
The four possible round three pairings are
A>1-6, 3-2, 5-4 (if the spread between 3 and 6 is less than 80 in the class section, they drew in round one, 6 beat 5, and he remaining games were won by the higher rated player - alternatively the underrated/unrated 6 is really as strong as 3 and 4 while 5 is much weaker and the game between 3 and 6 ended in a draw while 3 did not beat 4)
If any one of the three games is paired this way then the other two games have two ways they can be paired, one as in B, C or D and the other as in A

B>1-6, 4-2, 3-5 (3 drew 6 in round 1, lost to 4 in round 2 and 6 did not lose to 5)
C>1-3, 6-2, 5-4 (1 and 3 are 2-0)
D>1-5, 3-2, 6-4 (3 drew 6, 5 beat 6, 4 beat 3)

Having only six players usually happens to me in a class-based section and there is a higher than normal chance of the necessary one or two upsets happening. I am guessing I see it closer to 20% of the times that it could happen while somebody used to much larger spreads among six players might see it less than 10% of the time.
In a three round tournament all four pairings are equally valid and can be done based on the circumstances. In a four-round (or five-round) tournament then pairing A must be avoided and only one of the three remaining ones should be used in round 3.

[brennanprice]

The procedure of Rule 29L is fully justified and critical in a four-round, six-player event.

[Smythe_Dakota]

That procedure is fully justified, but other approaches are possible, as set forth in the posts above. What is critical is that the TD doesn’t just blindly go ahead and pair it as a Swiss (with or without a pairing program), thus possibly painting himself into a corner.

The worst thing to do is to make all the colors alternate perfectly in both round 2 and round 3. Then you are automatically skrewed in round 4.

Bill Smythe