Easy.
2 gets white against 4, 1 gets black against 3, and 5 and 6 get byes.
It is “a bit” tougher if you give everybody actual games.
Obviously 6 players can play 5 rounds without having to have been paired more than once, as a RR table will show, but unfortunately, 0-pt BYES would need to be given while a couple of players are “waiting” to get a game.
The point is that if you use the 6-player RR table (p. 295 in you handy rulebook), then you don’t need to give any byes.
a) pair the first round normally.
b) assign pairing numbers so that this pairing is 3-6, 5-4, 1-2 (line 1 in the table on p. 295)
c) in every subsequent round, pair the top score group to find the best game for the top-ranked player.
d) use the pairing for the top-ranked player to select a line from the table
e) use those pairings (adjusting colors, as needed).
6 players, 5 rounds, no byes.
In every round, the top-ranked player gets the “best” pairing according to Swiss principles. Lower ranked players get strange looking pairings early on - and, of course, the top-ranked player eventually gets some pretty strange looking pairings (after all, he does have to play everyone, sometime).
In the event above, you could take the out of assigning 2 byes - but try to take this event to 5 rounds and see if you like the result.
[quote=“sloan”]
The point is that if you use the 6-player RR table (p. 295 in you handy rulebook), then you don’t need to give any byes.
a) pair the first round normally.
b) assign pairing numbers so that this pairing is 3-6, 5-4, 1-2 (line 1 in the table on p. 295)
c) in every subsequent round, pair the top score group to find the best game for the top-ranked player.
d) use the pairing for the top-ranked player to select a line from the table
e) use those pairings (adjusting colors, as needed). 6 players, 5 rounds, no byes.
I would be very careful about adjusting colors. Colors are going to be 3-2 (or 2-3) in any event–flip the wrong one and you get 4-1. When you try to fix it in a later round, you can weave a tangled (i.e, unsolvable) web very easily. Even a 3-2 to 2-3 flip could cause a problem later.
In every round, the top-ranked player gets the “best” pairing according to Swiss principles. Lower ranked players get strange looking pairings early on - and, of course, the top-ranked player eventually gets some pretty strange looking pairings (after all, he does have to play everyone, sometime).
[quote]
I agree that you should be careful with colors. In this particular case (6 players, 5 rounds) - I would simply run a straight round-robin, and worry about colors only if some players withdraw. However, the usual application of this procedure is something like 8 players and 5 rounds. In that case, not all the lines in the RR table will eventually be played (and, you can’t really predict which ones will be left out - at least, not easily). So, if you follow the colors in the table too religiously you can create a color imbalance where none would exist if you played the entire table. So…you do have to adjust some colors along the way.
Small turnouts for a 4 or 5 round event or sections are not so rare in my area. In a previous event with 5 rounds and 7 players, the software appeared to “know” that it should use Rule 29L logic. I assume that the # of players and rounds “gave” it this information. The second round pairings were strange and unSwisslike, so several strong players protested and they were manually changed. I believe the software was trying to use 29 L logic to get workable pairings in this scenario. I was surprised that neither the TD nor the experienced players were aware that “pure” Swiss pairings were virtually impossible in this situation. So, I believe the software is aware of this problem and tries to address it in certain situations.
Some scholastic tournaments I’ve run have had 6 players in a 4 round Swiss. Because of trying to avoid teammates or siblings the last round becomes impossible to pair correctly even with teammates being allowed to play each other or having somebody get 3 whites or blacks in a row. Somebody ends out playing another player a second time.
I think as long as there aren’t match ups that you’re trying to avoid, one should be able to get legal pairings. I played in a 4 round tournament that had 6 players. The make up of the tournament was very strange with a GM, 2 IMs, and three untitled players rated 1700, 1474 & 1256. If a tournament like this can get through 4 rounds with no pairing hiccups anything can.
After two rounds it looked like this:
1 2597 B4 (1.0) W2 (1.5)
2 2472 W5 (1.0) B1 (1.5)
3 2394 B6 (1.0) W4 (2.0)
4 1700 W1 (0.0) B3 (0.0)
5 1474 B2 (0.0) W6 (1.0)
6 1256 W3 (0.0) B5 (0.0)
Somebody asked me what I thought the 3rd round pairings would be. I was expecting pairings that would stay closer to the score groups and look like this:
#2 (1.5) vs #3 (2)
#5 (1) vs #1 (1.5)
#4 (0) vs #6 (0)
However having those pairings and colors in round three would have led to one player getting 3 blacks and another player getting 3 whites in round 4. Instead Swiss-Sys paired:
#1 (1.5) vs #3 (2.0) (#1 had a 2nd white in a row)
#2 (1.5) vs #6 (0.0) (#6 had a 2nd Black in a row)
#4 (0.0) vs #5 (1.0)
The last round everyone got their 2nd of whatever color they needed and the score group match ups were reasonable:
#3 (2.5) vs #2 (2.5)
#5 (1.0) vs #1 (2.0)
#6 (0.0) vs #4 (0.5)
If you look carefully you’ll notice a distinct pattern. The titled players drew each other, and pounded the untitled players (shark bait!) The fish fed upon each other. However with 6 prizes guaranteed everyone went home with something.
This more colorful write up on this tournament can be found at:
WinTD does not know how many rounds there are until you click the “Last Round?” box when pairing.
What could the TD instuct/set WinTD to do if he knew of this situation?
How should he explain this pairing problem to the players involved?
Hmm, what software is this? I’m inclined to doubt that either WinTD or SwisSys would bother including this logic. Who would want to computer-pair 7 players to begin with? Pairing programs are intended for large events.
Besides, round 2 is not the problem:
Theorem 1. In a 6-player Swiss, any legal pairings can be used in rounds 1 and 2, and there will still be legal pairings for rounds 3, 4, and 5.
Theorem 2. In a 6-player Swiss, after rounds 1 and 2 are complete, there are exactly four possible sets of legal pairings in round 3, and in exactly three of these four, there will still be legal pairings in round 4.
Theorem 3. In a 6-player Swiss, if there are legal pairings in round 4, there will also be legal pairings in round 5.
Note: “Legal pairings” just means don’t pair the same players twice (and no byes, forfeits, or withdrawals). Color problems and wild score differences don’t count.
The proofs are left as exercises for the reader.
Polly: In your example, the computer (or TD) simply lucked out. By chance, he/she/it missed the “bad” pairing mentioned in Theorem 2.
Bill Smythe
I believe Swiss-Sys will look ahead if the number of rounds are known. In my example even my marginal hand paired guesses for round 3 allow for legal pairings even though two players will end out with 3 of one color in round 4. The tournament was a year and half ago so I would guess Swiss 7.4 was being used. I recreated the tournament using exact ratings from 12/07 and paired using Swiss 8.5 and still got the same 3rd and 4th round pairings.
I don’t see a senario in round 3 that would come up with the crappy pairings that doesn’t allow legal pairings of some sort being done. Maybe no upsets, or lots of upsets and no draws would cause that one bad pairing to come up. Perhaps Steve could tell us if it was dumb luck or the nature of all teh grand master draws that caused somewhat formal pairings.
And…even though WinTD doesn’t know the number of rounds, I believe it will look ahead at least 1 round (iff you do not check the “last round?” box).
Try my original example:
Run the experiment this way: Assign ratings in, say, 50-point intervals to players 1 through 6, let SwisSys pair round 1 (I assume it will make the same pairings as above, except that colors may be reversed), enter the above results for round 1, let SwisSys pair round 2 (does it again make the same pairings?), and enter the above results for round 2. Then I’m interested in seeing what SwisSys does in round 3, where the only “natural” pairings are the disastrous pairings above.
Hmm. Does it do this regardless of tournament size? Seems to me such a check would be pointless for, say, 100 players.
Bill Smythe
SwissSys 8.5 makes the same pairings, and falls into the same trap. Specifying number of rounds = 4 makes no difference. The problem is that the pairing rules generate a contradiction. Do you violate the rules in round 3, or follow them and possibly be forced to violate them in round 4?
It’s very interesting that in my example Swiss-Sys was willing to break up the score groups in round 3 by having one 0 play a 1 and the other 0 playing 1.5 and giving players consecutive colors, but then having all the colors and score groups working out for the last round. There were no true upsets in the tournament until the last round. The #1-3 players taking draws amongst each other aren’t true upsets. #4 & #5 drawing each other in round 3 isn’t really an upset.
It seems like an upset of #1 in the 2nd round causes problems in the 4th round, that only are avoided in round 3 by splitting up the 1 and 0 score groups. In Bill’s example I switched #4 and #5. I still ended out with two players getting 3 of one color, but at least no one played each other twice.
[code]
7 or 8 players
Rd Pairings
1: 1-8 7-2 6-3 5-4
2: 8-5 4-6 3-7 2-1
3: 2-8 1-3 7-4 6-5
4: 8-6 5-7 4-1 3-2
5: 3-8 2-4 1-5 7-6
6: 8-7 6-1 5-2 4-3
7: 4-8 3-5 2-6 1-7
W/D plr Reversals
1 6-2 2-5 5-3
2 7-1 5-3 3-4
3 6-2
4 5-3 7-1
5 8-4 6-2
6 8-3 7-1
7 8-4 5-1 1-6 6-2
8 none[/code]
The RR and dropout reversal tables for 8 players are shown above, but they assume all rounds will be played in order. If you were to play for example (in order) rounds 1, 4, 6, 3, & 2; would the above table be useless?
The order the rounds are played in doesn’t matter (except possibly in FIDE norm reporting). The actual rounds played do matter. If you played only rounds 1, 2, 3, 5 and 7, then player 8 ends up with four blacks in five rounds. That is why Ken said you can use the table but may have to adjust colors.
There is one advantage of the row order in the RR tables. In general, the colors for each player alternate, with the occasional two colors in a row. If you play fewer rounds, but always use consecutive rows in the table, then you shouldn’t get into too much color trouble. That would mean that all of the omitted rows need to be at the bottom and/or the top of the table. As it happens, those rows contain the matchups that are least likely to occur in a normal swiss, barring upsets: 1 vs. N, 1 vs. N-1 etc., for N players.
Using this scheme for 8 players and 5 rounds, you would first drop the 7th row of the table (1 vs. 8 ), and then the 1st row (1 vs. 7). This still equalizes colors (3:2 or 2:3) for all 8 players.
7 or 8 players
Rd Pairings
1: 4-8 5-3 6-2 7-1
2: 8-7 1-6 2-5 3-4
3: 3-8 4-2 5-1 6-7
4: 8-6 7-5 1-4 2-3
5: 2-8 3-1 4-7 5-6
6: 8-5 6-4 7-3 1-2
7: 1-8 2-7 3-6 4-5 This method has at least a couple of potential problems. Depending on the number of rounds and players, there would probably be cases where the colors would not equalize for every player. More importantly, with upsets, the top two players might never be paired. But maybe it would work to use this approach until the last round. At that point, you could take a look at things and use one of the “omitted” rows instead, and/or play with the colors.
The only benefit of this method, other than simplicity, is to minimize the color problems. The procedure that Dr. Sloan outlined (find the best match for the top player in each round) may well be preferable in most, or all, cases. The only hesitation I have is (a) being able to deal with the color problems, and (b) knowing which is preferable: better matchups on the top board, or better colors for everyone. Or can we have both?
Please note that it’s not my procedure. It’s the one printed in the rulebook.
“It’s not just a good idea, it’s the Law”
I have to confess that I find this entire thread bewildering, since my understanding was that you shouldn’t attempt to run an n-round Swiss tournament with fewer than 2^(n – 1) players. Of course having only six players is going to break a four-round Swiss – for four rounds, you need at least eight players. So play a three-round Swiss. That’s certainly enough to establish a winner, as I think Smythe Dakota’s example shows. If you really feel the need to play four rounds, and you have only six players, then don’t use the Swiss system.