I am fine with round robin gone. I understand the issue. But can there be configurable options of tiebreaking methods in swiss? The FIDE rules state that a fast game is played. That can be done as an option for tiebreaking. For example, a 5|0 or 3|0 game is played, or the Sonnebon Berger score is used, depending on creator choice.
@odoaker2015
i agree with you that swiss tournaments won't work with number of rounds close to the number od paticipants. because of the rule that no player can play another player twice. (the point difference is not important. that rule can be broken as often and as hard as necessary)
@TacticalBert
regarding your dispute: let's try it:
6 player tournament; 5 rounds. (spoiler: not even 4 are possible)
round 1.:
A-D 1:0
E-B 0:1
C-F 1:0
round 2:
B-A 1/2
D-C 1-0
F-E 1/2
round 3:
C-B 1/2
A-F 0-1
E-D 1-0
round 4:
???
A has to play C or E
B has to play D or F
C has to play A or E
D has to play B or F
E has to play A or C
F has to play B or D
A, C and E have to play each other.
B, D and F have to play each other.
it's already over. no more pairings possible without breaking the rule that no player can play another one twice. (so round robin is also no longer possible).
Is that enough proof? Or still not convinced?
i agree with you that swiss tournaments won't work with number of rounds close to the number od paticipants. because of the rule that no player can play another player twice. (the point difference is not important. that rule can be broken as often and as hard as necessary)
@TacticalBert
regarding your dispute: let's try it:
6 player tournament; 5 rounds. (spoiler: not even 4 are possible)
round 1.:
A-D 1:0
E-B 0:1
C-F 1:0
round 2:
B-A 1/2
D-C 1-0
F-E 1/2
round 3:
C-B 1/2
A-F 0-1
E-D 1-0
round 4:
???
A has to play C or E
B has to play D or F
C has to play A or E
D has to play B or F
E has to play A or C
F has to play B or D
A, C and E have to play each other.
B, D and F have to play each other.
it's already over. no more pairings possible without breaking the rule that no player can play another one twice. (so round robin is also no longer possible).
Is that enough proof? Or still not convinced?
@simon_h81 You don't have to convince me... *sigh* I never said your example was NOT possible, I am just advocating it is perfectly possible to have a 5-round Swiss tournament with 6 players under FIDE rules. Like you say, one of the problems - in contrary to round-robin - is that it is a bit result-dependant. Maybe you missed my post #211 where I pointed out what you are trying to point out to me?
In round-robin, the pairing of e.g. round 3 would of course be different. Even better: before the start of the tournament the pairings of all rounds are already known.
In round-robin, the pairing of e.g. round 3 would of course be different. Even better: before the start of the tournament the pairings of all rounds are already known.
actually i really missed that post (was just reading the whole thread). i apologize for that. but first it seemed like you were saying that it's alwas possible to get all the rounds played.
but after rereading couple of posts i have to admit it was rather @Molurus who didn't accept odoaker's experience as proof and wanted further proof.
but after rereading couple of posts i have to admit it was rather @Molurus who didn't accept odoaker's experience as proof and wanted further proof.
@simon_h81
In my perception this was a discussion about what the FIDE regulations are concerning Swiss tournaments. The only proof for that we need are the actual FIDE regulations.
Odoaker insisted he wanted to prove something by trying this in a software implementation, which, in my view, is a ridiculous way of trying to prove what the FIDE regulations are, because there simply is no guarantee that the software actually does abide by the FIDE regulations. So if the software refuses to pair after round X, that could mean one of two things: the FIDE rules don't allow it, or, the software doesn't handle the specifications for moving players to a lower score group. So the simple observation that odoaker's software refuses to pair after some number of rounds really doesn't prove anything.
So I really wasn't asking for proof. I was asking for an argument, and objected to the idea that what his software does or doesn't do constitutes proof of anything. As far as I'm concerned, his software was and still is completely irrelevant to the point discussed.
In my perception this was a discussion about what the FIDE regulations are concerning Swiss tournaments. The only proof for that we need are the actual FIDE regulations.
Odoaker insisted he wanted to prove something by trying this in a software implementation, which, in my view, is a ridiculous way of trying to prove what the FIDE regulations are, because there simply is no guarantee that the software actually does abide by the FIDE regulations. So if the software refuses to pair after round X, that could mean one of two things: the FIDE rules don't allow it, or, the software doesn't handle the specifications for moving players to a lower score group. So the simple observation that odoaker's software refuses to pair after some number of rounds really doesn't prove anything.
So I really wasn't asking for proof. I was asking for an argument, and objected to the idea that what his software does or doesn't do constitutes proof of anything. As far as I'm concerned, his software was and still is completely irrelevant to the point discussed.
i think i don't really understand what you want proof for?
@simon_h81
The point of discussion at that point was whether or not the FIDE rules would allow to have a swiss tournament with N players and N-1 rounds. (Which effectively would be a round robin.)
Disregarding constraints regarding colors played, I really don't see why not, if I look at what the FIDE rule book says about this. Odoaker seems convinced that FIDE doesn't allow this, and was trying to prove this by saying that his software doesn't do it. Which to me, is just a flawed way of proving anything regarding FIDE rules.
The point of discussion at that point was whether or not the FIDE rules would allow to have a swiss tournament with N players and N-1 rounds. (Which effectively would be a round robin.)
Disregarding constraints regarding colors played, I really don't see why not, if I look at what the FIDE rule book says about this. Odoaker seems convinced that FIDE doesn't allow this, and was trying to prove this by saying that his software doesn't do it. Which to me, is just a flawed way of proving anything regarding FIDE rules.
So basically... I don't want proof for anything. :) I was looking for an argument why the FIDE rules wouldn't allow this.
I will not comment on this any more. Everyone can form their own opinion on this. I have said more than just my opinion on this, that is enough! And the software I use is recommended by Deutscher Schachbund and is absolutely error-free in terms of the application of the FIDE rules! This software is used by all chess clubs in Germany for hosting tournaments! I will not say more about this!
it's not fide who doesn't allow it. it's the system itself that prohibits a swiss tournament with n participants and n-1 rounds (round robin) to work properly.
look at my example tournament in #302
look at my example tournament in #302
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