How many rounds in a tournament?

So I guess some more fundamental questions are:

  1. Do Swiss tournaments have a fixed or variable number of rounds?
  2. How exactly does a Swiss tournament work on OGS?

Note that on Wikipedia, Swiss tournaments are described as having a fixed number of rounds. However, Sensei’s library describes a different format that has a variable number of rounds.

It seems that the wiki documentation for OGS implies that the Swiss tournaments have a variable number of rounds, so maybe this format is similar to how it is described on Sensei’s library?

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For N players, a Swiss tournament requires at least log2(N) rounds to determine a winner. In practice, this can still result in ties, so it is better to do K*log2(N) but K need not be larger than 2.

IIRC, the actual formula used by OGS is O(N) instead of O(log2 N) which results in a ridiculous number of rounds for large Swiss tournaments.


My understanding is that there are at least two types of Swiss tournament:

  1. Fixed length: the tournament lasts a fixed number of rounds, which may be set as a function of the initial number of players or just chosen some other way. If this number is set too low relative to the number of players, it is possible for the tournament to end with more than one player having won all of their games.
  2. Variable length: the tournament ends when there is only one player left that has won all of their games. With a pairing method that always assigns players with all wins with each other (except for maybe one odd one left out), this would most likely take no more than ceil(log2(N)) rounds.

Knowing the constant here is crucial. Hopefully, it is only a fraction of N, rather than a multiple.

Does this mean that Swiss tournaments are actually fixed length, but with the number of rounds being a function of the initial number of players?

I suppose that yet another possibility is that after each round, the number of played rounds is compared to a function of the number of remaining participants (those that have not chosen to drop out), and ending the tournament if the former is greater than the latter.

Because tournaments have a max number of participants, it should be possible to give in advance the max number of rounds at least. That can be a useful information for a player schedule or for avoiding tournaments requiring too much.

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In an effort to find an answer for the question of “How many rounds does a Swiss tournament last?” I found an old thread (actually started by me as well), where I got a precise answer from @matburt

However, it seems that this answer might be a bit dated, since it is from 2014, and the concept of “qualified places” seems to now be gone.


Or is still one since we are not yet

as far as OGS tournament operation is concerned


I don’t think “qualified places” was ever selectable. The formula gives the right number of rounds with Q=1.


I wanted to confirm this, since back a long time ago (in 2014), I believe the interface said something about qualified places on the page of Swiss tournament. I even asked about what this phrase meant back then:

However, now the page for a Swiss tournament does not mention anything about “qualified places”, so I wasn’t sure if the concept was done away with or just no longer displayed on the page.

So, once a Swiss tournament starts, is the number of rounds fixed (computed from the number of players that started the tournament)? Or does the number of remaining rounds get adjusted if players leave?

For the tournament that I linked to earlier, there were 128 players to start. If we use P=64 and Q=1 in the formula (P + 7 x Q)/5, we get 27.

So, will that tournament last 27 rounds? Or will it be shorter than that since some people have left?

@flovo, is that tournament fixed at 27 rounds?


@teapoweredrobot suggested elsewhere that Swiss tournaments on OGS might simply end as soon as a winner had been determined.

However, I’ve found an example of a Swiss tournament where that did not happen:

  • There were 6 players in this tournament.
  • After the first round, 3 had won their games, and the other 3 lost their games.
  • After the second round, only one player (who would go on to eventually become the champion) still had a perfect record, since the other two players that won in the first round lost their second round games.
  • However, the tournament continued for two more rounds, ending after four rounds.
  • The champion won all four of their games, while no one else won more than two.

@flovo, note that this tournament also contradicts the (P + 7) / 5 formula (assuming Q = 1). The tournament lasted 4 rounds, while the formula specifies only 2.6.


But, stop me if I am wrong, to win you can’t be sure until other players can’t catch up by winning the next ones and you losing them even if you have till now a perfect score.


Right, if you arrange a Swiss tournament as having a large enough number of rounds, it could be possible for a player leading as the only one with a perfect score to be later overtaken by others, if they start losing games allowing others to catch back up.

However, the purpose of my example is to contradict the speculation that an OGS Swiss tournament would end as soon as only one player had a perfect score.

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In my defence I made no claims about the inscrutable nature of OGS tournaments, the details of which are known only to Him. It’s just that my understanding of the Swiss system is that it should finish when the winner is determined.

I should say though that I’ve never known an IRL Go tournament use the Swiss system.

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Ok. Well as you, I dunno really how this “swiss” system run on OGS but usually, it’s just everyone against everyone, and wait the last game to determine the winner. And that will justify at least one of the 2 rounds in the example of tournament you gave, the last one not being necessary.

Sorry, I don’t mean to put you on the defensive. I mainly just tagged you since I thought you might be interested to see this counter-example.

I just want to understand how long a Swiss tournament might last. I think @Groin summed it up well above:

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I don’t think this is quite right as that would be a round robin which would only be for the smallest tournament IRL.

In think this is it. IRL you have some fixed number of rounds and the TD will arrange the pairings so that a winner is not determined too soon but ensuring that the final round is decisive. However on OGS the pairings are done automatically and there are more disqualifications (losses on time for example) than IRL so I guess it can’t work like that.
(MacMahon tournaments and pairings are also odd on OGS)

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Ok I reviewed my classics

Quite detailed.
Using this rules gives the following table

Rounds players number of places validated
6 26 1
6 19 2
6 12 3
7 17 3
8 24 3
9 31 3
An other approximation is:

The number of rounds for a Knock out plus 2 rounds per validated place

From looking through a few pages of history, it seems like 6-person Swiss tournaments always last four rounds, even if people drop out.

Here’s an example that had only 3 participants left after round two, and just 2 for the final round:

For more tournament pedant trivia, it’s interesting that Hacker_Denay played no moves, was “disqualified” in round two, had the fourth-highest score, and still won the third-place trophy.

Here’s a 4-person tournament that had three rounds:


I looked around some more and the tournaments I can find seem to be consistent with the formula floor((P+2*7)/5) limited to a maximum of 8 rounds.

This formula predicts that P=21 would be the first tournament with 7 rounds, and P=26 would be where they reach the max of 8 rounds.

EDIT: found one with P=21, and it worked! (From a three-year-old question here.)


Looks like the max is NOT 8 after all since this one fits floor((P+2*7)/5):

(In my previous example with P=36 it must have stopped early because all but one of the players were disqualified.)

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