Number of rounds in a Swiss tournament?

[Ajedrecista]

Hello:

I have doubts about in which subforum this thread must be placed. Moderators: feel free to move it to a more suitable subforum, if any.

I searched over the Internet and I found that Swiss tournaments with the same number of players have different number of rounds:

Aeroflot rapid Swiss (2013)

European Individual Chess Championship (2014)

Aeroflot had 257 players and 9 rounds; EICC had 257 players (plus two that did not play) and 11 rounds. With such a high number of players, is it really important the number of rounds (between not absurd bounds, of course)? There were more than 170 teams in the last Olympiad and there were 11 rounds, so I see a wide variety.

I found the following tables:

http://www.wizards.com/dci/downloads/Swiss_Pairings.pdf

http://wizards.custhelp.com/app/answers ... events-faq

First link:

Code: Select all

CHART FOR APPROPRIATE NUMBER OF ROUNDS OF SWISS TO SELECT THE TOP 8 PLAYERS FOR SINGLE ELIMINATION:

Number of Players    Number of Rounds
-----------------    ----------------
  17– 32 players     5 rounds of Swiss
  33– 64             6 rounds of Swiss
  65–128             7 rounds of Swiss
 129–226             8 rounds of Swiss
 227–409             9 rounds of Swiss
 410+               10 rounds of Swiss

------------------------

Second link:

Code: Select all

 Players      Rounds
====================
   2             1	
   3-   4        2	
   5-   8        3	
   9-  16        4	
  17-  32        5	
  33-  64        6	
  65- 128        7
 129- 212        8
 213- 384        9
 385- 672       10
 673-1248       11
1249-2272       12
2273+           13

Please note that these two sources use a 3 - 1 - 0 score system instead of the classical 1 - 0.5 - 0 in chess. It is clear that with 128 players or less, the formula is: rounds = ceiling[ln(players)/ln(2)]. But I have not found a clear relation for players > 128. Excel adjusts very well these series with exponential fits, but their expressions are somewhat weird. I have even tried with binary, octal and hexadecimal but I do not see patterns.

After this introduction: are there formulæ like rounds = rounds(players)? Or is it the same if there are 9, 10 or 11 rounds in a Swiss tournament with 250 players? Just an example.

This thread is not probably off-topic if you replace players by engines. Any help is appreciated. Thanks in advance.

Regards from Spain.

Ajedrecista.

[cdani]

Hi!
There is not a standard rule. Basically it's the organizer's decision. I know a lot of cases and, as an arbiter, I read all the rules, and it's always like this.

[bob]

Hello:

I have doubts about in which subforum this thread must be placed. Moderators: feel free to move it to a more suitable subforum, if any.

I searched over the Internet and I found that Swiss tournaments with the same number of players have different number of rounds:

Aeroflot rapid Swiss (2013)

European Individual Chess Championship (2014)

Aeroflot had 257 players and 9 rounds; EICC had 257 players (plus two that did not play) and 11 rounds. With such a high number of players, is it really important the number of rounds (between not absurd bounds, of course)? There were more than 170 teams in the last Olympiad and there were 11 rounds, so I see a wide variety.

I found the following tables:

http://www.wizards.com/dci/downloads/Swiss_Pairings.pdf

http://wizards.custhelp.com/app/answers ... events-faq

First link:

Code: Select all

CHART FOR APPROPRIATE NUMBER OF ROUNDS OF SWISS TO SELECT THE TOP 8 PLAYERS FOR SINGLE ELIMINATION:

Number of Players    Number of Rounds
-----------------    ----------------
  17– 32 players     5 rounds of Swiss
  33– 64             6 rounds of Swiss
  65–128             7 rounds of Swiss
 129–226             8 rounds of Swiss
 227–409             9 rounds of Swiss
 410+               10 rounds of Swiss

------------------------

Second link:

Code: Select all

 Players      Rounds
====================
   2             1	
   3-   4        2	
   5-   8        3	
   9-  16        4	
  17-  32        5	
  33-  64        6	
  65- 128        7
 129- 212        8
 213- 384        9
 385- 672       10
 673-1248       11
1249-2272       12
2273+           13

Please note that these two sources use a 3 - 1 - 0 score system instead of the classical 1 - 0.5 - 0 in chess. It is clear that with 128 players or less, the formula is: rounds = ceiling[ln(players)/ln(2)]. But I have not found a clear relation for players > 128. Excel adjusts very well these series with exponential fits, but their expressions are somewhat weird. I have even tried with binary, octal and hexadecimal but I do not see patterns.

After this introduction: are there formulæ like rounds = rounds(players)? Or is it the same if there are 9, 10 or 11 rounds in a Swiss tournament with 250 players? Just an example.

This thread is not probably off-topic if you replace players by engines. Any help is appreciated. Thanks in advance.

Regards from Spain.

Ajedrecista.

Here's the gist:

if you have N players, where N is a power of two, or if not rounded up to a power of two, then you need log2(N) rounds to have a chance of finding a winner. And there are no guarantees. If you want to try to identify the 2nd best and 3rd best, you need another round. But if you go too far, the last rounds become meaningless since the swiss system will pair all the top scores together until they have all played, meanwhile the lower players are playing each other and they have to rise as well. And as those lower players reach the top, they get swatted back down, maybe, depending on the number of rounds.

Swiss works pretty well, but we have seen some REALLY ugly problems with say a 9 round swiss and 32 players. By the time you get through 5 rounds the best have played and can't be paired against each other a second time. So the last 4 rounds are almost meaningless, except when you get the occasional upset where a very low-ranked/seeded player beats one of the current top group and shuffles things a bit. Upsets do happen, and have happened in past events.

My personal choice would be N+2 or N+2 rounds (N = log2(players)). If you go too far beyond that it is messy. You can finagle the pairings so that in the first 4 rounds, the top half only play the bottom half of the seeds, so that the last 5 rounds you get the "real tournament" going and the games are interesting, but that is not always easy to do.

With 16 players, rather than 9 round I would personally prefer a single-round-robin event. Then pairing and seeding doesn't have a lot of influence in final standings.

Just random thoughts from someone that directed human tournaments for many years...

[Ozymandias]

I don't remember who, but I was told that you need as many rounds as you would if the tournament format were a knock-out.

Code: Select all

  2=1
  4=2
  8=2
 16=3
 32=4
 64=5
128=6
256=7
512=8

Following this rule, only the biggest swiss tournaments (more than 1024 players) really need 10 rounds. For the rest of them, is overkill.

[michiguel]

I don't remember who, but I was told that you need as many rounds as you would if the tournament format were a knock-out.

Code: Select all

  2=1
  4=2
  8=2
 16=3
 32=4
 64=5
128=6
256=7
512=8

Following this rule, only the biggest swiss tournaments (more than 1024 players) really need 10 rounds. For the rest of them, is overkill.

That is a number that indicates the bare minimum. It is certainly better to play more. Particularly in tournaments where there is a big parity (for instance, close strong tournaments like National Championships where people had to qualify).

Miguel

[Ozymandias]

The good thing about the bare minimum is that it can be determined, anything else above that, is open to interpretation.

[bob]

I don't remember who, but I was told that you need as many rounds as you would if the tournament format were a knock-out.

Code: Select all

  2=1
  4=2
  8=2
 16=3
 32=4
 64=5
128=6
256=7
512=8

Following this rule, only the biggest swiss tournaments (more than 1024 players) really need 10 rounds. For the rest of them, is overkill.

You missed it a bit. 8 requires 3 rounds. Otherwise after one round you have 4 winners and 4 losers, after the second round you have 2 with 2 wins out of the original 4 winners, but you have no overall winner. A 3rd round (log2(Num_Players)) solves that partially. log2(N)+1 is even safer since chess has draws.

[Ozymandias]

Absolutely right, after the 2nd line, everything is +1.

[kaissa]

Hello:

Second link:

Code: Select all

 Players      Rounds
====================
   2             1	
   3-   4        2	
   5-   8        3	
   9-  16        4	
  17-  32        5	
  33-  64        6	
  65- 128        7
 129- 212        8
 213- 384        9
 385- 672       10
 673-1248       11
1249-2272       12
2273+           13

After this introduction: are there formulæ like rounds = rounds(players)? Or is it the same if there are 9, 10 or 11 rounds in a Swiss tournament with 250 players? Just an example.

Here's the gist:

if you have N players, where N is a power of two, or if not rounded up to a power of two, then you need log2(N) rounds to have a chance of finding a winner. And there are no guarantees. If you want to try to identify the 2nd best and 3rd best, you need another round.

Please bear in mind that Swiss system is not a type of tournament that tries to rank the players in the tournament; it just gives a large amount of players the chance to play the same number of rounds in a limited time compared to round-robin.

Your second link is the minimum number of rounds you need to have. For 16 players, you have 4 rounds which will only show the strongest and weakest player in that tournament; the classification between 2nd and 15th is meaningless.

If you want to have some players qualify for another tournament, than you need to add more rounds at this stage; let's call it the second phase. The more rounds you have the better chance of finding the strongest players. Kazic gave a rule of extra 2 rounds for extra 1 place in 70s. So according to him if you need top 3 places in a 16 player tournament, you need 4 + ( 2 extra places * 2 rounds ) = 8 rounds to find the top 3 and the bottom 3 players. You can see the limitation if you want to have more players qualify as this Swiss system is getting close to 15 rounds of round robin tournament. The more rounds you have in the second phase, the more confidence you will have in finding the strongest and weakest players.

Another formula used is R = ( P + 7*Q ) / 5 where R is rounds needed, P is number of players and Q is number of qualifications. http://www.chesscafe.com/text/geurt125.pdf In the above example this formula gives 7,4 rounds to find the top 3.

I am not going to talk about accelareted pairings but as you can see Swiss system is a system that tries to give large number of players to play in a tournament in a short time period instead of ranking these players in strength. So as long as you do not use it as a qualification tournament you are fine.

The last USSR championship in 1991 was a 64 player 11 round Swiss tournament. According to the first formula, only the top 3 and bottom 3 were ranked. The second formula states that you need 14,2 rounds to find the strongest and weakest players. Take a pick.

Best,