Inspired by a recent question, here a fantasy:

The Superduperquantumcomputer "Deep Ponder" has cracked Chess. Assume a 9 round Swiss tournament with 2^9 players, White always wins. Will the pairing computer (definitely not super quantum) go berserk? I don't think so, one can always float and the problems will only occur for the topmost players.

Anyone like to do the experiment? (Would do it myself - I have enough time...but no Swiss program.)

  • If you want to do the experiment yourself, you can download Vega from vegachess.com - you'll need to use the Linux version (unless you want to pay the fee to register the software) since the free version that runs on Windows is limited to 30 players.
    – patbarron
    Jul 6, 2021 at 19:33
  • 1
    I cannot tell what your question is. Can you rephrase please? Jul 6, 2021 at 20:41
  • 1
    @HaukeReddmann I too find the question unclear. What exactly do you mean by "go berzerk", what behaviour are you looking for? In practice no Swiss program would have a problem with this. Sure it's a little weird that one player will get 9 Whites in a row, but I don't see a reason for a pairing algorithm to be challenged by these or any other results. And by "most close to this extreme case" - by what measure? Format, actual results that occurred ... Jul 6, 2021 at 21:38
  • 2
    I'd also add that -- if this question isn't trivial -- we should still see whether it's really about chess in any way, as opposed to e.g. the algorithmic structure of Swiss pairing (which might be better asked on Math.SE). Jul 6, 2021 at 21:41
  • 1
    @MobeusZoom: As Brian already said, this problem is extremely improbable unless the number of players is low compared to the number of rounds. (Thus I don't expect this to ever happen in an event where arbiters are needed.) Still, I already encountered this in a little youth tournament of our club, the program said "Unpairable". I overruled and paired by hand to best judgement. The kiddies didn't object :-) Jul 10, 2021 at 7:30

1 Answer 1


Assume a 9 round Swiss tournament with 2^9 players, White always wins. Will the pairing computer (definitely not super quantum) go berserk?

Thank you for clarifying the difference between NA (National Arbiter) exam you took in Germany and the full FA (FIDE Arbiter) exam. This kind of stuff is covered in the FA exam.

The program won't go berserk. Here is the explanation.

The pairing system used by Swiss Manager is the FIDE proprietary JaVaFo pairing engine. It implements the FIDE (Dutch) pairing system. This is one of four approved pairing systems:


I guess you might think that the pairing program would go berserk because with only two results possible for each game there would only be 2, 3 or 4 score groups and as more players play each other and so are ineligible to play each other the program would "struggle" to find acceptable pairings which meet the rules.

Here is what article A.2 of C.04.3 FIDE (Dutch) System says:

For pairings purposes only, the players are ranked in order of, respectively

a: score

b: pairing numbers assigned to the players accordingly to the initial ranking list and subsequent modifications depending on possible late entries or rating adjustments

In other words, in every score group there is ALWAYS an ordering.

There then follow more paragraphs of complicated rules which, in truth, were generated from the original algorithms used by the first FIDE pairing program. Here is what the official JaVaFo website has to say:

A bit of history
In its original presentation, the wording of the Dutch System was more similar to a software algorithm than a classical set of rules. There was a good reason behind that. In fact, the legend says that first came the Swiss Master program that was implementing the Dutch System, then the rules to be published by FIDE were modelled on the behaviour of such program.

So, there is a well defined algorithm which works in exactly the same way regardless of the scores and composition of players. There IS a problem which can arise with the Swiss system but it arises in the exact OPPOSITE way to what you describe. If the number of players is only slightly bigger than the number of rounds (say 2 or 3 bigger) then it can happen in the last round, penultimate round in rare circumstances, that it is impossible to generate a legal pairing (no two players playing each other if they have played each other previously).

Moving on to the subject of expirmental evidence.

2^9 = 512 and I so I don't think anybody who isn't being paid to run a tournament with roughly 500 players is going to enter 500 players into a tournament file.

Just to satisfy the curiosity of those (perhaps you included) who want to see what a tournament where white always won would look like I've constructed such a tournament with 2^6 (=64) players playing a 9 round Swiss. To make it directly comparable to a 512 player 9 round Swiss I should probably have stopped after 6 rounds, but continuing to 9 let's you see what would happen if more rounds were played.

I've uploaded it to Chess-results here.

  • Many THX for making the experiment! It agrees with what I expected (indeed few players will lead to a problem, many players maximally will give a lot of floating) but expecting and knowing is different. (Concerning your other remark, as a German NA I must NOT know Swiss, and especially not so much to answer this "unpractical" question. I learnt the basics in my NA seminar just for the sake of it, but it's only in the curriculum of the next level FA, so I wasn't examined on that.) Jul 7, 2021 at 8:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.