View Full Version : Burstein Pairing System

Garvinator

05-09-2009, 12:29 AM

Does anyone know much about the burstein pairing system? From my understanding it is used in the Olympiad, but apart from that I do not see it being used much.

The fide handbook does mention it here: http://www.fide.com/fide/handbook?id=86&view=article

Whilst at first glance it looks confusing, it seems to have some good and bad points and could be worth a go in a tournament. Swiss Master 5 does give it as a default pairing option to use when pairing players, instead of just the standard dutch pairing system.

Kevin Bonham

05-09-2009, 02:33 PM

FIDE really need to clear up their nomenclature. Not only is "Sonnenborn-Berger" actually correctly spelled "Sonneborn-Berger" but the system they incorrectly call "Sonnenborn-Berger" is correctly called "Neustadtl".

I think Burstein based on Buchholz is a promising system. The problem with Burstein based on Neustadtl is that the Neustadtl tiebreak is verging on pseudoscience - it actually measures how erratically a player has performed rather than how well they have performed.

Garvinator

05-09-2009, 02:49 PM

I think Burstein based on Buchholz is a promising system. The problem with Burstein based on Neustadtl is that the Neustadtl tiebreak is verging on pseudoscience - it actually measures how erratically a player has performed rather than how well they have performed.

Swiss Master 5's default for the Burstein System is Buchholz.

I think this system is certainly worth a decent trial, but one of the main issues with trying something new is that chess players are not keen on something new and unfamiliar, especially when it comes to how the players are paired in the tournament.

Also, how the players are paired under Burstein certainly is not as obvious at first glance as under dutch pairing.

Kevin Bonham

05-09-2009, 03:06 PM

I wonder how Burstein based on Buchholz deals with a case in which a player withdraws and hence all their opponents get worse Buchholz scores than would otherwise be the case. If an affected opponent had an otherwise high Buchholz score, that could help that player fly under the radar, but if they had an otherwise low-ish Buccholz score that could make it even lower and cause them to receive much harsher pairings.

Denis_Jessop

05-09-2009, 04:04 PM

FIDE really need to clear up their nomenclature. Not only is "Sonnenborn-Berger" actually correctly spelled "Sonneborn-Berger" but the system they incorrectly call "Sonnenborn-Berger" is correctly called "Neustadtl".

I think Burstein based on Buchholz is a promising system. The problem with Burstein based on Neustadtl is that the Neustadtl tiebreak is verging on pseudoscience - it actually measures how erratically a player has performed rather than how well they have performed.

Actually, the Neustadtl or Sonne(n)born-Berger system was invented by Oscar Gelbfuhs in about 1873. Neustadtl didn't invent it (or at least suggest it) until 1882. That's the upshot of what is in Wikipedia. Cecil Purdy wrote a bit about the nomenclature if the system in "Chess World" but I can't find it just now.

DJ

Kevin Bonham

05-09-2009, 04:24 PM

Actually, the Neustadtl or Sonne(n)born-Berger system was invented by Oscar Gelbfuhs in about 1873. Neustadtl didn't invent it (or at least suggest it) until 1882. That's the upshot of what is in Wikipedia. Cecil Purdy wrote a bit about the nomenclature if the system in "Chess World" but I can't find it just now.

According to the Oxford Companion, Neustadtl's system is a simplified version of Gelbfuhs' system, which was designed for tournaments where not everyone had played the same number of games:

Neustadtl: sum of defeated opponents' scores, plus half the sum of drawn opponents' scores.

Gelbfuhs: sum of defeated opponents' scores (each expressed as a fraction), plus half the sum of drawn opponents' scores (each expressed as a fraction).

For a tournament where everyone has played in all games, each player's Neustadtl score is n times their Gelbfuhs score, where n is the number of rounds.

For a tournament where not everyone has played all games Neustadtl obviously doesn't work very well.

Also according to Olimpbase (http://www.olimpbase.org/index.html?http%3A%2F%2Fwww.olimpbase.org%2Farchiv e%2Fnews_archive.html%3Fstart_from%3D25%26ucat%3D% 26archive%3D1184441969%26subaction%3Dlist-archive%26id%3D%26), Gelbfuhs proposed that his score be used not as a tiebreaker for outright score, but as a substitute for it.

The Olimpbase article suggests the system should be called "Gelbfuhs-Neustadtl".

Denis_Jessop

05-09-2009, 09:06 PM

According to the Oxford Companion, Neustadtl's system is a simplified version of Gelbfuhs' system, which was designed for tournaments where not everyone had played the same number of games:

Neustadtl: sum of defeated opponents' scores, plus half the sum of drawn opponents' scores.

Gelbfuhs: sum of defeated opponents' scores (each expressed as a fraction), plus half the sum of drawn opponents' scores (each expressed as a fraction).

For a tournament where everyone has played in all games, each player's Neustadtl score is n times their Gelbfuhs score, where n is the number of rounds.

For a tournament where not everyone has played all games Neustadtl obviously doesn't work very well.

Also according to Olimpbase (http://www.olimpbase.org/index.html?http%3A%2F%2Fwww.olimpbase.org%2Farchiv e%2Fnews_archive.html%3Fstart_from%3D25%26ucat%3D% 26archive%3D1184441969%26subaction%3Dlist-archive%26id%3D%26), Gelbfuhs proposed that his score be used not as a tiebreaker for outright score, but as a substitute for it.

The Olimpbase article suggests the system should be called "Gelbfuhs-Neustadtl".

Yes - I think that is right. Perhaps the thing could be settled by calling it the "Geneusonberg System" :D :cool:

DJ

Garvinator

05-09-2009, 09:16 PM

Is there any danger that someone might actually post something relevant about how the pairings are done, instead of the origins of some of the 'tiebreak' systems came from ;)

aransandraseg

05-09-2009, 09:26 PM

This is getting ridiculous. How do you guys know all of this. Focus on playing chess. Don't worry about these crazy distractions.

Garvinator

05-09-2009, 09:28 PM

This is getting ridiculous. How do you guys know all of this. Focus on playing chess. Don't worry about these crazy distractions.

Well my question to start off with is as an arbiter if anyone else can add some more detail about the pairing system used in the Olympiad ie good and bad points etc.

Garvinator

06-09-2009, 01:43 PM

Interesting, looking at the 2008 Olympiad pairings, which uses the burstein pairings, the teams are paired dutch style for the first round ie 1 v 32, 33 v 2, 3 v 34 but the swiss master 5 program when pairing by buchholz pairs the players for the first round as 1 v 64, 2 v 63, 3 v 62 etc.

Is anyone able to comment on which is the correct method and would obtain the 'best' results if players inside a score group are being paired by SB and buchholz tiebreaks?

This is getting ridiculous. How do you guys know all of this. Focus on playing chess. Don't worry about these crazy distractions.

lol :clap: I think he is correct! then again :hmm:

Denis_Jessop

06-09-2009, 02:53 PM

Is there any danger that someone might actually post something relevant about how the pairings are done, instead of the origins of some of the 'tiebreak' systems came from ;)

The problem is that almost nobody knows anything about the Burstein System or how it works in practice. But some of us know a bit about trivia vaguely related to it and can't resist the temptation to add to the CCF fount of knowledge. :) If you start an esoteric thread we don't hold you responsible for the consequences even though on CCF these are reasonably foreseeable ;) :( :D = :confused:

DJ

Kevin Bonham

06-09-2009, 07:52 PM

Is anyone able to comment on which is the correct method and would obtain the 'best' results if players inside a score group are being paired by SB and buchholz tiebreaks?

For the first round it really should not make a lot of difference as the system has heaps of time to sort them out. I'm thinking that 1-64, 2-63, 3-62 etc produces lots of upsets in the middle whereas 1-33, 2-34 etc should not produce as many upsets. But while in 1-33 (etc) the upsetters could be anywhere in the field, in 1-64 (etc) they will be closer to the midfield. If the top seeds are beating utter patzers in round 2 who just got lucky in round 1, then most likely the top seeds' Buchholz will suffer after round about 4, and then need correction with much harder opponents. For this reason I'd rather go with 1-64 (etc) from the start. Effectively 1-64 says that the even score of participants at the start of the tournament is incorrect, and it tries to correct it in a manner that quickly stretches the field. That's just theory though, I haven't tried it in practice.

I don't even want to think about "SB" since it is silly and the Burstein system shouldn't be used (and probably won't work reliably) in conjunction with it.

Garvinator

07-09-2009, 01:36 AM

For this reason I'd rather go with 1-64 (etc) from the start.

Since your post, I have thought about this quite a bit. I can see a couple of issues with using 1-64, but they are not to do with the burstein system per se, but more to do with the type of players we have in most tournaments in Australia.

One of the concerns with dutch swiss tournaments is the beatings new players get from meeting players in the first couple of rounds. If the tournament is paired burstein, then this effect would be even worse to the point where the number 1 seed (most likely rated 2200 or so) would be meeting a complete beginner or similar.

While pairing burstein might help for those in the middle of the draw and are the ones are the ones who can be 'let down' by a dutch monster swiss, not sure if this off-sets the potential for increased mis-matches for the top and bottom seeds.

Acceleration was used in the last olympiad, most likely to try and off-set this effect (I know there were other reasons for the acceleration), but my point here is to do with the issues outlined above. Unfortunately as a pairing system, this would create two new issues for a general tournament:

1) A new pairing system, Burstein

2) Acceleration. Tournament organisers and players in Australia are generally acceleration shy, so to introduce two very new concepts in a prize money tournament where a financial loss is a strong possibility would most likely be just too risky.

If pairing the first round without acceleration, I think I would prefer to go with 1-33 etc just to play it a bit safe. This is how the first round pairings were done at the last olympiad.

I think the burstein system would be good to trial in a few blitz tournaments, where many rounds can be scheduled and the players are only paying $10 entry. It would give useful data for comparison/analysis without having to risk the bank to do it :)

... are only paying $10 entry. It would give useful data for comparison/analysis without having to risk the bank to do it :)

Consider paying them to make sure someone turns up :P

Kevin Bonham

07-09-2009, 12:18 PM

One of the concerns with dutch swiss tournaments is the beatings new players get from meeting players in the first couple of rounds. If the tournament is paired burstein, then this effect would be even worse to the point where the number 1 seed (most likely rated 2200 or so) would be meeting a complete beginner or similar.

Good point. I was thinking of it in terms of advantages for sorting the field and so on but this is a very good reason not to use 1-64 (etc) in a typical weekend tournament. Indeed the player at the bottom of the table, if they keep losing and there are no upsets, will get the same run of beatings they would get in a normal swiss plus an extra beating from player 1 in the first round. If there are upsets then the bottom player can have an even harder time.

Garvinator

12-09-2009, 05:56 PM

In Stewart Reuben's third edition handbook, he comments that the burstein pairing system only starts after rounds 3 or 4 of a seeded swiss with players pairied using their buchholz scores. For the early rounds, it is normal dutch rules.

Now knowing this, I really do want to give this system a go as I think it could have a lot of merit.

Kevin Bonham

12-09-2009, 06:01 PM

In Stewart Reuben's third edition handbook, he comments that the burstein pairing system only starts after rounds 3 or 4 of a seeded swiss with players pairied using their buchholz scores. For the early rounds, it is normal dutch rules.

I am guessing that the reason for this is that it takes a few rounds for the Buchholz score to actually mean anything, so if you use Burstein pairings for the first few rounds then you are most likely really just doing a standard Swiss but with top v bottom instead of top half v bottom half.

Garvinator

12-09-2009, 08:41 PM

I am guessing that the reason for this is that it takes a few rounds for the Buchholz score to actually mean anything, so if you use Burstein pairings for the first few rounds then you are most likely really just doing a standard Swiss but with top v bottom instead of top half v bottom half.

Correct and if it was top v bottom then it would have the issues we have already highlighted.

Pepechuy

07-04-2011, 09:13 AM

I have been manually simulating it in a few small tournaments (the results are taken from real round-robin tournaments).

I do it exactly the way it is described in the handbook, so the first round pairing is first against last.

Since the Burstein system does not float more than one player from one score group to another, it quickly gets several score groups merged once the leaders have played each other. In the last round of my simulations, it was not uncommon to merge all the players in a single group.

My impression is: the priority of the Burstein system is not to give the tournament leaders the strongest available opposition (which the other systems do). The priority is to provide equal opposition (taking into account the whole tournament) to players with the same score.

Also, please note that the Olympiad is not played on a "pure" Burstein system. Besides pairing the first round the "usual" way, the Olympiad makes it a hybrid with the Lim system. First pair from top to median, then bottom to median and at the end the median score group. The definition of "median score group" is different from the Lim system (for Lim it is the number of played rounds divided by two, for the Olympiad it is the score of the "median player"), but my guess is that it should be almost always the same.

If I were a tournament director, I would use it in the following cases:

1. A very even field. I remember the Nordic zonals used to be 18 players, 9 round swiss sytem. And none of those players was weak.

2. If no ratings are available. I am thinking of a childrens tournament with many newcomers. In this case it does not matter pairing first against last in the first round.

Pepechuy

25-08-2011, 04:26 AM

FIDE really need to clear up their nomenclature. Not only is "Sonnenborn-Berger" actually correctly spelled "Sonneborn-Berger" but the system they incorrectly call "Sonnenborn-Berger" is correctly called "Neustadtl".

I think Burstein based on Buchholz is a promising system. The problem with Burstein based on Neustadtl is that the Neustadtl tiebreak is verging on pseudoscience - it actually measures how erratically a player has performed rather than how well they have performed.

There is a problem with using primarily Buchholz for the Burstein system.

I will not argue that Buchholz is a better tiebreak system than Sonneborn-Berger. However, the Burstein pairing system is using Sonneborn-Berger not as a tiebreaker, but as a ranking. It is not unusual to have heterogeneous pairing groups, because there are floaters or because score groups are merged (and the highest scoring player is not automatically placed higher in the ranking list).

In an heterogeneous group, using Buchholz as ranking does not make much sense. Ranking by score and then by Buchholz might be better, but I think that destroys the original intent of the Burstein system.

forlano

31-08-2011, 12:59 AM

I have been manually simulating it in a few small tournaments (the results are taken from real round-robin tournaments).

I liked your example.

Recently I had a look at this system. Below you can find the pairing in the correct order for the case of 8 players.

These are the differences with the Dutch Systems:

1) Section D of Dutch system (transpositions and exchange) are replaced by the example pairing (listed below). The logic is that (1,N) has higher priority over (2,N-1) and so on being N the number of players in the score group. Thus, before to unpair, for example (1,N), all other pairs should be tried.

The examples and this logic is easier to undenstand and to perform by hands: just destroy the last done pairs and restart without to ruin the top pairs.

The section D is instead a mess and does not describe really what to do in the general case (how many arbiters know how to choose in the right order 4 players out of 8 when performing exchange in a group of 16 players?). It seems that even some endorsed FIDE program has not yet realised how to do this "elementary" task. The arbiter are lucky because such problem is very rare (almost impossible) to face in a real tournament.

2) The paranoia floater of Dutch system is absent in the Burstein system. Just avoid to downfloat two times in a row the same player.

3) Exists only omogenous group in Burstein (no S1 with floaters and S2 with a much greater number of players).

4) Try to accomodate the maximun numbers of colors while generate the pairing in the correct order (the same of Ducth).

5) In case of odd number of players the Burstein assign immediately the BYE.

In the first rounds (first and second) the buchholz/sonneborn are likely to be the same. In this case the rating is used and the system looks like a simplified Dutch.

At moment there are not pairing program that run it. If I have understood SwissMaster run a Dutch using as criteria the buchholz to sort the players in S1 and S2. But maybe I am wrong on this point.

The system looks sound and merits to be tested in individual tournament. This prevent to have sistematic effect at the boundary of two score groups when the upfloater (the higher rated) risk to be the same of the previous round.This is typical of systems based on rating.

Moreover the higher rated player in the Dutch (and similar system) is constantly treated as the strongest of the tournament. This is not likely the case if there are many top players spreaded in the range of 30-40 elo points. In these case it is better to ask directly to the tournament who is the strongest and sort them according to it.

Regards,

Luigi

1 8 2 7 3 6 4 5 #1

1 8 2 7 3 5 4 6 #2

1 8 2 7 3 4 5 6 #3

1 8 2 6 3 7 4 5 #4

1 8 2 6 3 5 4 7 #5

1 8 2 6 3 4 5 7 #6

1 8 2 5 3 7 4 6 #7

1 8 2 5 3 6 4 7 #8

1 8 2 5 3 4 6 7 #9

1 8 2 4 3 7 5 6 #10

1 8 2 4 3 6 5 7 #11

1 8 2 4 3 5 6 7 #12

1 8 2 3 4 7 5 6 #13

1 8 2 3 4 6 5 7 #14

1 8 2 3 4 5 6 7 #15

1 7 2 8 3 6 4 5 #16

1 7 2 8 3 5 4 6 #17

1 7 2 8 3 4 5 6 #18

1 7 2 6 3 8 4 5 #19

1 7 2 6 3 5 4 8 #20

1 7 2 6 3 4 5 8 #21

1 7 2 5 3 8 4 6 #22

1 7 2 5 3 6 4 8 #23

1 7 2 5 3 4 6 8 #24

1 7 2 4 3 8 5 6 #25

1 7 2 4 3 6 5 8 #26

1 7 2 4 3 5 6 8 #27

1 7 2 3 4 8 5 6 #28

1 7 2 3 4 6 5 8 #29

1 7 2 3 4 5 6 8 #30

1 6 2 8 3 7 4 5 #31

1 6 2 8 3 5 4 7 #32

1 6 2 8 3 4 5 7 #33

1 6 2 7 3 8 4 5 #34

1 6 2 7 3 5 4 8 #35

1 6 2 7 3 4 5 8 #36

1 6 2 5 3 8 4 7 #37

1 6 2 5 3 7 4 8 #38

1 6 2 5 3 4 7 8 #39

1 6 2 4 3 8 5 7 #40

1 6 2 4 3 7 5 8 #41

1 6 2 4 3 5 7 8 #42

1 6 2 3 4 8 5 7 #43

1 6 2 3 4 7 5 8 #44

1 6 2 3 4 5 7 8 #45

1 5 2 8 3 7 4 6 #46

1 5 2 8 3 6 4 7 #47

1 5 2 8 3 4 6 7 #48

1 5 2 7 3 8 4 6 #49

1 5 2 7 3 6 4 8 #50

1 5 2 7 3 4 6 8 #51

1 5 2 6 3 8 4 7 #52

1 5 2 6 3 7 4 8 #53

1 5 2 6 3 4 7 8 #54

1 5 2 4 3 8 6 7 #55

1 5 2 4 3 7 6 8 #56

1 5 2 4 3 6 7 8 #57

1 5 2 3 4 8 6 7 #58

1 5 2 3 4 7 6 8 #59

1 5 2 3 4 6 7 8 #60

1 4 2 8 3 7 5 6 #61

1 4 2 8 3 6 5 7 #62

1 4 2 8 3 5 6 7 #63

1 4 2 7 3 8 5 6 #64

1 4 2 7 3 6 5 8 #65

1 4 2 7 3 5 6 8 #66

1 4 2 6 3 8 5 7 #67

1 4 2 6 3 7 5 8 #68

1 4 2 6 3 5 7 8 #69

1 4 2 5 3 8 6 7 #70

1 4 2 5 3 7 6 8 #71

1 4 2 5 3 6 7 8 #72

1 4 2 3 5 8 6 7 #73

1 4 2 3 5 7 6 8 #74

1 4 2 3 5 6 7 8 #75

1 3 2 8 4 7 5 6 #76

1 3 2 8 4 6 5 7 #77

1 3 2 8 4 5 6 7 #78

1 3 2 7 4 8 5 6 #79

1 3 2 7 4 6 5 8 #80

1 3 2 7 4 5 6 8 #81

1 3 2 6 4 8 5 7 #82

1 3 2 6 4 7 5 8 #83

1 3 2 6 4 5 7 8 #84

1 3 2 5 4 8 6 7 #85

1 3 2 5 4 7 6 8 #86

1 3 2 5 4 6 7 8 #87

1 3 2 4 5 8 6 7 #88

1 3 2 4 5 7 6 8 #89

1 3 2 4 5 6 7 8 #90

1 2 3 8 4 7 5 6 #91

1 2 3 8 4 6 5 7 #92

1 2 3 8 4 5 6 7 #93

1 2 3 7 4 8 5 6 #94

1 2 3 7 4 6 5 8 #95

1 2 3 7 4 5 6 8 #96

1 2 3 6 4 8 5 7 #97

1 2 3 6 4 7 5 8 #98

1 2 3 6 4 5 7 8 #99

1 2 3 5 4 8 6 7 #100

1 2 3 5 4 7 6 8 #101

1 2 3 5 4 6 7 8 #102

1 2 3 4 5 8 6 7 #103

1 2 3 4 5 7 6 8 #104

1 2 3 4 5 6 7 8 #105

Pepechuy

31-08-2011, 08:45 AM

I liked your example.

Recently I had a look at this system. Below you can find the pairing in the correct order for the case of 8 players.

These are the differences with the Dutch Systems:

1) Section D of Dutch system (transpositions and exchange) are replaced by the example pairing (listed below). The logic is that (1,N) has higher priority over (2,N-1) and so on being N the number of players in the score group. Thus, before to unpair, for example (1,N), all other pairs should be tried.

The examples and this logic is easier to undenstand and to perform by hands: just destroy the last done pairs and restart without to ruin the top pairs.

The section D is instead a mess and does not describe really what to do in the general case (how many arbiters know how to choose in the right order 4 players out of 8 when performing exchange in a group of 16 players?). It seems that even some endorsed FIDE program has not yet realised how to do this "elementary" task. The arbiter are lucky because such problem is very rare (almost impossible) to face in a real tournament.

That is actually easy. There is a total order among all the exchanges.

I think the big problem with the Dutch system is that the rules are not written with a description of WHAT it is trying to achieve; but with a set of steps on HOW to achieve it. They tell you which path to follow, but not where are you going.

It is possible to write the Dutch system purely on descriptive terms of the correct pairing, and I think that would be easier to understand than the current wording. Actually I can do it, but barely have time. Maybe on my next vacation.

I also think that the Burstein system can be improved by modifying rule

6.5

"In each SG maximum number of players should get their due colors."

with an analogous of rules A7d and A7e of the Dutch system, because not every pairing which breaks a colour preference is the same: it is worse to break an strong colour preference than a mild colour preference (and, in the last round, the worst is to break an absolute colour preference). Instead of simply maximising the number of pairings which give both players their due colours, try also to minimise the number of absolute (last round only) and strong colour preferences broken.

Also, all the systems that allow a last round colour exception can be improved as follows. They all put "having three times in a row the same colour" and "having a colour difference less than -2 or more than +2" in the same bucket... and the last one is clearly worse. The exception should first allow having three times in a row the same colour, and only if that is not enough allow a colour difference of -3 or +3

forlano

31-08-2011, 12:39 PM

It is possible to write the Dutch system purely on descriptive terms of the correct pairing, and I think that would be easier to understand than the current wording. Actually I can do it, but barely have time. Maybe on my next vacation.

This is a very challenging task.

I would be satisfied if only section D could be described in a more clear way.

Said in other terms, let's suppose I am able to generate all the pairing for 10 players in a scoregroup (as the ones listed above for the Burstein system). Is there a criteria that permits me to sort them in the right order for any N according to the Dutch system? So, given two pairings how the arbiter can recognise the one coming before? For Burstein it's immediate.

Moreover, If I follow the instruction of section D I will generate more and more time the same pairing that has been already processed. The section D is a very silly way to suggest to the arbiter to obtain something.

I also think that the Burstein system can be improved by modifying rule

6.5

"In each SG maximum number of players should get their due colors."

I would not do so. As you said there is a clear aim of the system. The natural pairing (1,N), (2,N-1), etc..., is the ideal one. When you try to accomodate more colors you start to generate more pairings. Each new pairing bring you in the opposite direction of the original target.

If you insist to give much more attention to the colors (absolute, mild preference...) this means to continue to generate more pairings. It is not evident that at the end the original purpose could be reached.

Regards,

Luigi

Pepechuy

01-09-2011, 05:33 AM

This is a very challenging task.

I would be satisfied if only section D could be described in a more clear way.

Said in other terms, let's suppose I am able to generate all the pairing for 10 players in a scoregroup (as the ones listed above for the Burstein system). Is there a criteria that permits me to sort them in the right order for any N according to the Dutch system? So, given two pairings how the arbiter can recognise the one coming before? For Burstein it's immediate.

Moreover, If I follow the instruction of section D I will generate more and more time the same pairing that has been already processed. The section D is a very silly way to suggest to the arbiter to obtain something.

Ok, here is the description of section D, DUTCH SYSTEM. In a score bracket to be paired, the players are ranked 1,2,...,n

Since you are well versed in pairings, go to the section called THE ORDER, the PRELIMINARIES are for the general audience which may require detailed explanations.

PRELIMINARIES

Lets construct a simple graph (no loops and no parallel edges).

Each vertex corresponds to a player. Add a vertex for the bye, in case it is the last score group and the number of players is odd.

Put an edge between two vertices if it the corresponding players can be paired. That means that they have not faced each other, and do not have the same absolute colour preference (I am assuming it is not the last round). In case of the bye, put and edge with all the players who have not received a point without playing.

Of all the matchings in the this graph, lets find the biggest one. Assume this matching has p edges.

There are p possible pairings (2p is less of equal to n).

Note: there might be several maximum matchings which have p edges.

Note: in the last round, if allowing a colour allocation exception generates a bigger matching, then it should be taken (but not if it gives the same p).

Note: if it is the last score bracket and 2p is strictly smaller than n, then it is necessary to change the downfloaters from the last score group, and if that does not work then it is necessary to merge it with the next to last score group.

THE ORDER

Lets write the pairings the way you did in your post... for example

1 3 2 5 4 7...

means 1 against 3, 2 against five, 4 against 7, etc.

In each pair the higher ranked player is listed first (so 1 3 5 2 is incorrect).

The pairs are ordered putting by the higher ranked player (so 1 3 4 7 2 5 is incorrect).

The remaining unpaired players (which will be floated down to the next score group) are simply listed in order at the end.

Now, assume we have two possible pairings listed the way described above. You need to know which one comes first (here we are either ignoring colours and floating, or assuming both pairings fulfill the same colour and floating requirements).

Separate each pairing list in two parts: the first part has only the first odd p places in the list, the second part has all the remaining places. Exchanges deal with the first part, transpositions deal with the second part.

EXCHANGES

For each of the two pairings, count how many times numbers greater than p appear in the first p odd places of the list. The one which has less appearances comes first.

If they have the same number of appearances, then sum the first p odd places in the list... in the example

1 3 2 5 4 7...

it means 1+2+4+...

If one of the two pairings gives a smaller sum than the other, then it comes first.

If they both give the same sum, the look at the first difference between the two pairings in the sublist of the first p odd places. The one that has the smallest number (highest ranked player) comes first.

If they have exactly the same sublist, then they are tied in exchanges, and we need to look for the smallest transposition.

TRANSPOSITIONS.

Just compare the two sublists which remove the first p odd places. Compare them place by place, find the first difference. The one that has the lowest number (highest ranked player) comes first.

I would not do so. As you said there is a clear aim of the system. The natural pairing (1,N), (2,N-1), etc..., is the ideal one. When you try to accomodate more colors you start to generate more pairings. Each new pairing bring you in the opposite direction of the original target.

If you insist to give much more attention to the colors (absolute, mild preference...) this means to continue to generate more pairings. It is not evident that at the end the original purpose could be reached.

Regards,

Luigi

You have a good point here. (In this last paragraph Luigi is talking about the BURSTEIN system and a suggestion of mine to "improve" it).

losboba

03-09-2011, 04:14 AM

If they both give the same sum, the look at the first difference between the two pairings in the sublist of the first p odd places. The one that has the smallest number (highest ranked player) comes first.

That's not completely correct. Let's suppose we have ten players.

The first pairing is: 1 4 2 3 5 8 6 7 9 10 (odds: 1 2 5 6 9)

The second one is: 1 2 3 6 4 5 7 10 8 9 (odds: 1 3 4 7 8)

Which pairing comes first?

Same number of exchanges (two), same sum for the odds (23). If the quoted suggestion is applied, we should choose the first pairing. Right?

Wrong. In the first pairing we get 3 and 4 in the "even" group (i.e. they are the exchanged players). In the second pairing, the exchanged players are 2 and 5. The rule says: "When applying an exchange between S1 and S2 the difference between the numbers exchanged should be as small as possible. When differences of various options are equal take the one concerning the lowest player of S1". The lowest player of S1 is 5, therefore we should take firstly the second pairing.

The quoted suggestion might help, for instance, when having to choose between (odds: 1 2 5 6 9) and (odds: 1 2 5 7 8). Unfortunately, right now, there is nothing in the rules stating that it's possible to apply it. Of course, it is a sensible choice, but only after adding something like "Then take the one concerning the highest player of S2" to the rule, there would no room for protest.

Cheers,

Roberto

Pepechuy

03-09-2011, 05:30 AM

That's not completely correct. Let's suppose we have ten players.

The first pairing is: 1 4 2 3 5 8 6 7 9 10 (odds: 1 2 5 6 9)

The second one is: 1 2 3 6 4 5 7 10 8 9 (odds: 1 3 4 7 8)

Which pairing comes first?

Same number of exchanges (two), same sum for the odds (23). If the quoted suggestion is applied, we should choose the first pairing. Right?

Wrong. In the first pairing we get 3 and 4 in the "even" group (i.e. they are the exchanged players). In the second pairing, the exchanged players are 2 and 5. The rule says: "When applying an exchange between S1 and S2 the difference between the numbers exchanged should be as small as possible. When differences of various options are equal take the one concerning the lowest player of S1". The lowest player of S1 is 5, therefore we should take firstly the second pairing.

The quoted suggestion might help, for instance, when having to choose between (odds: 1 2 5 6 9) and (odds: 1 2 5 7 8). Unfortunately, right now, there is nothing in the rules stating that it's possible to apply it. Of course, it is a sensible choice, but only after adding something like "Then take the one concerning the highest player of S2" to the rule, there would no room for protest.

Cheers,

Roberto

Then I am understanding something wrong. Could you please tell me which is the correct order of the following four exchanges? (I give only the odds)

1 3 4 6 9

1 3 4 7 8

1 2 5 6 9

1 2 5 7 8

I do not understand what do you mean when you, in your example, you say

"The lowest player of S1 is 5, therefore we should take firstly the second pairing."

Why is 5 the lowest player of S1?

Greetings,

José.

losboba

03-09-2011, 08:55 AM

Then I am understanding something wrong. Could you please tell me which is the correct order of the following four transpositions? (I give only the odds)

Sure. But I'd rather start with the second part of your question:

I do not understand what do you mean when you, in your example, you say

"The lowest player of S1 is 5, therefore we should take firstly the second pairing."

Why is 5 the lowest player of S1?

When we have to pair ten players (*), numbered from 1 to 10 (11), we apply the part of rule A.6 that says: "In a homogeneous score bracket (**) S1 contains the higher half (rounding downwards) of the number of players in the score bracket" .

Therefore, we have S1 = [1 2 3 4 5], which is the basis for all the exchanging procedures. We also have S2 = [6 7 8 9 10 (11)].

As you can see, 5 is the lowest player of S1 and 6 is the highest player of S2.

(*) or eleven - by the way, examples with an odd number of players are more meaningful, but I digress

(**) it is a homogeneous score bracket, otherwise we wouldn't talk of exchanges

Now let's go back to your first question, i.e.

the correct order of the following four transpositions:

1 3 4 6 9

1 3 4 7 8

1 2 5 6 9

1 2 5 7 8

(I don't know if "transpositions" is a good term - probably I would call them "pairings" or, more simply, S1)

(a) 1 3 4 6 9: 2 exchanges, sum=23, 2-5 out, 6-9 in

(b) 1 3 4 7 8: 2 exchanges, sum=23, 2-5 out, 7-8 in

(c) 1 2 5 6 9: 2 exchanges, sum=23, 3-4 out, 6-9 in

(d) 1 2 5 7 8: 2 exchanges, sum=23, 3-4 out, 7-8 in

Remembering what I said in my previous post (we should add "Then take the one concerning the highest player of S2" in the D.2 rule), the correct order is: (a) (b) (c) (d). In fact, (a) precedes (b) because the exchange concerns the highest player of S2 (6). And (a) precedes (c) because the exchange concerns the lowest player of S1 (5)

Hope it helps.

Cheers,

Roberto

Pepechuy

03-09-2011, 11:30 AM

Sure. But I'd rather start with the second part of your question:

When we have to pair ten players (*), numbered from 1 to 10 (11), we apply the part of rule A.6 that says: "In a homogeneous score bracket (**) S1 contains the higher half (rounding downwards) of the number of players in the score bracket" .

Therefore, we have S1 = [1 2 3 4 5], which is the basis for all the exchanging procedures. We also have S2 = [6 7 8 9 10 (11)].

As you can see, 5 is the lowest player of S1 and 6 is the highest player of S2.

(*) or eleven - by the way, examples with an odd number of players are more meaningful, but I digress

(**) it is a homogeneous score bracket, otherwise we wouldn't talk of exchanges

Now let's go back to your first question, i.e.

(I don't know if "transpositions" is a good term - probably I would call them "pairings" or, more simply, S1)

(a) 1 3 4 6 9: 2 exchanges, sum=23, 2-5 out, 6-9 in

(b) 1 3 4 7 8: 2 exchanges, sum=23, 2-5 out, 7-8 in

(c) 1 2 5 6 9: 2 exchanges, sum=23, 3-4 out, 6-9 in

(d) 1 2 5 7 8: 2 exchanges, sum=23, 3-4 out, 7-8 in

Remembering what I said in my previous post (we should add "Then take the one concerning the highest player of S2" in the D.2 rule), the correct order is: (a) (b) (c) (d). In fact, (a) precedes (b) because the exchange concerns the highest player of S2 (6). And (a) precedes (c) because the exchange concerns the lowest player of S1 (5)

Hope it helps.

Cheers,

Roberto

Ok, thanks, I begin to understand. "Transpositions" is an incorrect term, sorry for causing confusion... it should read "exchanges". I edited my post afterwards, but I see you answered the original wording.

I can see that examples with an odd number of players are more meaningful, because player 1 could then be exchanged (to be floated down). In general, if n=2p then player 1 can not be exchanged, but if n>2p then player 1 can become a downfloater by being exchanged.

Now let me reword my ordering of exchanges. I think there is a clear agreement about counting the number of exchanged players, and adding the ranking of the players than end up ranked higher than their respective opponent (it is not what rule D2 says but it is equivalent).

What I had wrong was when two pairings had the same number of players exchanged, and their sum was equal. Lets call q the number of players exchanged.

For each of the two pairings, compare them backwards, beginning at place p-q and ending at place 1. Find the first difference. The one with the smallest number (highest ranked player) comes first.

In case they are equal in all those p-q places, if I understand correctly what you say, then:

A. This case is not specified in rule D2

B. You suggest comparing the remaining q places forwards the way it was in my original description.

Greetings,

José.

losboba

03-09-2011, 10:12 PM

I can see that examples with an odd number of players are more meaningful, because player 1 could then be exchanged (to be floated down). In general, if n=2p then player 1 can not be exchanged, but if n>2p then player 1 can become a downfloater by being exchanged.

Exactly.

Now let me reword my ordering of exchanges. I think there is a clear agreement about counting the number of exchanged players, and adding the ranking of the players than end up ranked higher than their respective opponent (it is not what rule D2 says but it is equivalent).

Exactly.

What I had wrong was when two pairings had the same number of players exchanged, and their sum was equal. Lets call q the number of players exchanged.

For each of the two pairings, compare them backwards, beginning at place p-q and ending at place 1. Find the first difference. The one with the smallest number (highest ranked player) comes first.

Exactly. It means that a lower ranked player (from the original S1) has been exchanged.

In case they are equal in all those p-q places, if I understand correctly what you say, then:

A. This case is not specified in rule D2

Exactly. Not yet.

B. You suggest comparing the remaining q places forwards the way it was in my original description.

Exactly.

Actually, it is more inside trading than a suggestion. ;)

Cheers,

Roberto

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