Sodos and McMahon scores
wilemien
(see the discussion on sensei's library for all problems involved,
they are a concequence of the fact that the sodoms score can be equal
between players with different amount of wins)
and came to the remarkable conclusion that sodos in principle only
works after a "number of wins" "swiss score" or similar tie breaker.
(the problem being that in swiss and round robin this method is used
to decide the rank before the tie breaking fase. starts)
in McMahon tournaments the followup of tie breakers then could be:
0 McMahon Score
1 SOMS: Sum of (all) Opponents' McMahon scores
2 SOSOMS ( Sums of (all) Opponents' Soms scores
3 Swiss score
4 SODOMS Sum of Defeated Opponents McMahon scores
5 Face to face result
6 random tie breaker
the including of number of wins tiebreaker means that lower started
players have an advantage in comparison with the same followup without
it.
The nice thing is that the sodoms score after the number of wins
tiebreaker loses all problems connected with it.
Please comment.
Ben Finney
> I was puzzeling how sodos and McMahon scores can be united.
> (see the discussion on sensei's library for all problems involved,
URL to that discussion please?
--
\ “Not to be absolutely certain is, I think, one of the essential |
`\ things in rationality.” —Bertrand Russell |
_o__) |
Ben Finney
Robert Jasiek
<wile...@googlemail.com> wrote:
>I was puzzeling how sodos and McMahon scores can be united. [...]
>and came to the remarkable conclusion that sodos in principle only
>the including of number of wins tiebreaker means that lower started
>players have an advantage in comparison with the same followup without
>it.
>The nice thing is that the sodoms score after the number of wins
>tiebreaker loses all problems connected with it.
Can you please explain these statements by providing reasons or maybe
an example?
wilemien
> On Tue, 9 Mar 2010 02:40:15 -0800 (PST), wilemien
>
Discussions:
http://senseis.xmp.net/?SODOS
http://senseis.xmp.net/?SODOS%2FDiscussion
and
http://www.britgo.org/organisers/mcmahonpairing.html#sodos
Example:
(copied from http://senseis.xmp.net/?SODOS%2FDiscussion )
Here is an example of a 3 round tournament illustrating my concerns
about using SoDoS as a tie breaker.
We have a shodan Alan winning 2 games playing:
Bob(1k)-, Cath(1d)+ Dave(1k)+
We have a 1kyu Juliet winning 3 games playing:
Karen(1d)+ Lionel(1k)+ Martin(1k)+
It is not necessary to show the entire tournament results table. Here
is a summary of the key information for Alan's and Juliet's opponents'
scores in UK and EU styles.
Alan's opponents:
UK MMS Euro MMS
Name Wins initial final initial final
Bob(1k) 2 -1 1 19 21
Cath(1d) 1 0 1 20 21
Dave(1k) 2 -1 1 19 21
Juliet's opponents:
UK MMS Euro MMS
Name Wins initial final initial final
Karen(1d) 0 0 0 20 20
Lionel(1k) 2 -1 1 19 21
Martin(1k) 2 -1 1 19 21
Suppose we use SoDoS as the one and only tie breaker. Then we can
construct the portion of the final ranklist showing both Alan's and
Juliet's position. The column MMSi is the initial McMahon Pairing
score, and MMSf is the final McMahon Pairing score.
Then the final position of Alan and Juliet in the UK scale is:
Wins MMSi MMSf SoDoS WHO CONTRIBUTES TO SODOS
Alan(1d) 2 0 2 1+1=2 Cath+Dave
Juliet(1k) 3 -1 2 0+1+1=2 Karen+Lionel+Martin
Alan and Juliet are ranked equal, and they split a box of chocolate.
However in the EU scale:
Wins MMSi MMSf SoDoS WHO CONTRIBUTES TO SODOS
Juliet(1k) 3 19 22 20+21+21=62 Karen+Lionel+Martin
Alan(1d) 2 20 22 21+21=42 Cath+Dave
Now Juliet is ahead of Alan and gets all the chocolate!
I am not worrying here about which result is better!. All I care about
is that they are different.
-- end of example--
If SODOMS is used after number of wins the SODOMS tie break is not
used in this case Juliet gets the chocolate because she has more wins.
Because after the number of wins only players with an equal number of
wins have to be compared there is no problem where the zeropoint of
the MM score anymore.
In the example the zeropoint for the UK is at 1 dan, for the EU it is
20kyu.
The difference between the SODOS of
Juliet-EU and Juliet-UK is 3 times this difference for Alan the
difference is is 2 times.
this only depends on the number of wins.
If the players have the same number of wins then both their scores
have the same difference.
-- a resulting problem
A resulting problem is that using swiss score before sodoms score is
that it in theory can lead to players trying to start on a lower rank
(to have more chance of more wins) but because of the tiebreakers
before this, final mcMahon score, SOMS and SOSOMS i guess this is only
theoretical
To Robert - > I will explain it in sensei's after the discussion here.
Robert Jasiek
<wile...@googlemail.com> wrote:
>Here is an example
Thank you for the detailed and convincing example!
>Because after the number of wins only players with an equal number of
>wins have to be compared there is no problem where the zeropoint of
>the MM score anymore. [...]
>If the players have the same number of wins then both their scores
>have the same difference. [...]
>A resulting problem is that using swiss score before sodoms score is
>that it in theory can lead to players trying to start on a lower rank
>(to have more chance of more wins)
Are these claims already available as proofs in general? :)
Nick Wedd
<76d48d76-32f5-4b28...@d27g2000yqf.googlegroups.com>,
wilemien <wile...@googlemail.com> writes
>On 9 Mar, 14:55, Robert Jasiek <jas...@snafu.de> wrote:
>> On Tue, 9 Mar 2010 02:40:15 -0800 (PST), wilemien
>>
>> <wilem...@googlemail.com> wrote:
>> >I was puzzeling how sodos and McMahon scores can be united. [...]
>> >and came to the remarkable conclusion that sodos in principle only
>> >works after a "number of wins" "swiss score" or similar tie breaker. [...]
>> >the including of number of wins tiebreaker means that lower started
>> >players have an advantage in comparison with the same followup without
>> >it.
>> >The nice thing is that the sodoms score after the number of wins
>> >tiebreaker loses all problems connected with it.
>>
>> Can you please explain these statements by providing reasons or maybe
>> an example?
>
>Discussions:
>http://senseis.xmp.net/?SODOS
>http://senseis.xmp.net/?SODOS%2FDiscussion
>
>and
>http://www.britgo.org/organisers/mcmahonpairing.html#sodos
I was recently asked to update a BGA document which mentions this point.
My updating is incomplete, and has not yet been put on the BGA web site.
But my draft version is at
http://www.weddslist.com/oh/handbookch9.html#x13-870009.6
Nick
--
Nick Wedd ni...@maproom.co.uk
wilemien
> On Tue, 9 Mar 2010 10:45:59 -0800 (PST), wilemien
>
> >Here is an example
>
> Thank you for the detailed and convincing example!
I only copied it from sensei's library ( idid not even change the
layout.
>
> >Because after the number of wins only players with an equal number of
> >wins have to be compared there is no problem where the zeropoint of
> >the MM score anymore. [...]
> >If the players have the same number of wins then both their scores
> >have the same difference. [...]
> >A resulting problem is that using swiss score before sodoms score is
> >that it in theory can lead to players trying to start on a lower rank
> >(to have more chance of more wins)
>
> Are these claims already available as proofs in general? :)
What do you mean?
(which claims and what do you want to have proved?)
If you mean the theoretical option for sandbagging:
Suppose Juliet is in fact a 1 dan. ( she only pretended to be a 1 kyu)
In in the example Julliet's opponents were on average lower ranked
than Alans ( the difference is small)
Suppose for the sake of the argument that if she had started at her
proper rank and would have the same opponents as Alan and the same
result.
then Juliet and Alan would still tie. in the swiss score (number of
wins)
So by sandbagging she has gained an advantage in the number of wins.
(i agree it is all a bit contived , but it is about a theoretical
option)
wilemien
> In message
> <76d48d76-32f5-4b28-82fa-b7ec9b547...@d27g2000yqf.googlegroups.com>,
> wilemien <wilem...@googlemail.com> writes
Thanks i will have a good read through it and comment more later. (do
you have deadlines and so?)
Comments after a quick read.
make an prefered list of tie breakers. (and put all tiebreakers in
that order)
(may seem an increadable long list of tiebreaking (no tie will ever
reach the last rung) but is handy as list for organisers to refer to)
1) add an M where it is about Mc Mahon scores
so SOS -> SOMS
SODOS -> SODOMS
(Has the advantage that you can add the swiss variants (where it is
about winpoints as seperate tie breakers)
2) per tiebreaker /score what is the influence of a bye,(different
versions) no show of opponent and so on)
2) Have a study of the AGA tournaments standards at
http://www.usgo.org/tournaments/TournamentStandards/
(although i think i found there were some small errors in it)
.
For general tournament rules
US chess federation's official rules of chess (5th edition)
If you live in or bnearby london we could meet up and discuss it. (i
caould even lend you the chess tournament book)
Robert Jasiek
<wile...@googlemail.com> wrote:
>> Are these claims already available as proofs in general?
See below.
>> >Because after the number of wins only players with an equal number of
>> >wins have to be compared there is no problem where the zeropoint of
>> >the MM score anymore. [...]
Definitions:
Let Z be the MMS zero-point and a positive, negative or zero natural
number.
Let T = (T1, T2,...) be the ordered sequence of final result criteria.
Let Pk, Pl be arbitrary, fixed players.
Conjecture:
Consider the final results of an arbitrary MM tournament with these
characteristics:
- Z
- i < j
- Ti := Number of Wins Score
- Tj := SODOS
Then the player order is invariant under Z.
Proof:
Please provide some!
>> >If the players have the same number of wins then both their scores
>> >have the same difference.
Conjecture:
Consider the final results of an arbitrary MM tournament with these
characteristics:
- Z
- i < j
- Ti := Number of Wins Score
- Tj := SODOS
Then | SODOS(Pk) - SODOS(Pl) | is constant under Z.
Proof:
Please provide some!
>> >A resulting problem is that using swiss score before sodoms score is
>> >that it in theory can lead to players trying to start on a lower rank
>> >(to have more chance of more wins)
I suppose you suggest that your example, if worked out carefully
formally, would prove this?
marc
interesting and I have learned several things, that I'd like to
summarize as follows:
* SODOS makes more sense as the number of rounds / number of players
ratio grows
* SODOS can be used as a tiebreaker if "number of wins" is used as a
more significative tiebreaker
On 11 Març, 13:05, wilemien <wilem...@googlemail.com> wrote:
wilemien
> On Thu, 11 Mar 2010 03:59:10 -0800 (PST), wilemien
>
i use different definitions and so but here is a proof ( no garantee
for correctness given)
Definitions:
Pi <==> Player i
NWi <==> number of wins by player i
LOij <==> the players j losing to Pi
MMi is an MM rating with an arbitrary O point.
MMi(Pj) is the score of player j under MMrating i
SODOMS(MMi, Pi) = Sumj(MMi(LOij))
assumptions
MM0 is an MM rating with an arbitrary O point.
MM1 is an MMrating differing from an fixed amount C from the rating
from MM0
For all players
MM0(Pi) = MM1(Pi) + C
Proof:
Step 1
SODOMS(MM0, Pi) = Sumj(MM0(LOij)
Sumj(MM0(LOij) = Sumj(MM1(LOij)+C)
Sumj(MM1(LOij)+C) = Sumj(MM1(LOij) + Nw(Pi) x C
Sumj(MM1(LOij) + Nw(Pi) x C = SODOMS(MM1,Pi)+ Nw(Pi) x C
Step 2
SODOMS(MM0, Pi) - SODOMS(MM0, Pj) = SODOMS(MM1,Pi)+ Nw(Pi) x C -
[SODOMS(MM1,Pj)+ Nw(Pj) x C]
SODOMS(MM0, Pi) - SODOMS(MM0, Pj) = SODOMS(MM1,Pi)- SODOMS(MM1,Pi)+
Nw(Pi) x C - Nw(Pj) x C
SODOMS(MM0, Pi) - SODOMS(MM0, Pj)= SODOMS(MM1,Pi)- SODOMS(MM1,Pi)+ C x
(Nw(Pi) - Nw(Pj))
Given that Nw(Pi) = Nw(pj) (Both players have the same number of wins)
SODOMS(MM0, Pi) - SODOMS(MM0, Pj) = SODOMS(MM1,Pi)- SODOMS(MM1,Pi)
Robert Jasiek
Except for a few index typos etc., you have proven the second
conjecture, thank you very much! Let me write down your proof slightly
corrected:
Proposition 1:
Given the final results of an arbitrary McMahon tournament with an
arbitrary, fixed, natural-number zero-point and with an order of
placement criteria in that Number of Wins Score precedes SODOS. Then,
for every two players having the same number of wins, their SODOS
difference is constant under all possible zero-point values.
Proof by wilemien:
Definitions:
Pi :<==> Player i
NWi :<==> number of wins by player i
LOij :<==> the players j losing to Pi
MMi is an MMscore with an arbitrary 0 point.
MMi(Pj) is MMi of player j.
Let MM0 be an MMscore with an arbitrary 0 point.
MM1 := MM0 - C; C is arbitrary, fixed, natural-number.
Basic equations:
SODOS(MMi, Pi) = Sum_j(MMi(LOij))
For each Pi: MM0(Pi) = MM1(Pi) + C
NW(Pi) = NW(Pj) (*)
Step 1:
SODOS(MM0, Pi)
= Sum_j(MM0(LOij)
= Sum_j(MM1(LOij) + C)
= Sum_j(MM1(LOij) + NW(Pi) * C
= SODOS(MM1,Pi) + NW(Pi) * C
Step 2:
SODOS(MM0, Pi) - SODOS(MM0, Pj)
= SODOS(MM1,Pi) + NW(Pi) * C - ( SODOS(MM1,Pj) + NW(Pj) * C )
= SODOS(MM1,Pi) - SODOS(MM1,Pj) + NW(Pi) * C - NW(Pj) * C
= SODOS(MM1,Pi) - SODOS(MM1,Pj) + C * (NW(Pi) - NW(Pj))
(*) = SODOS(MM1,Pi) - SODOS(MM1,Pj).
QED.
>> Then the player order is invariant under the zero-point value.
And what can we say about this?
wilemien
Correct.
Butbe careful about the destinction between SODOMS and SODOS
(the first is about McMahon scores the second about winscores)
I think there is a relation between any tiebreaker and the number of
rounds.
and how higher the number of rounds the more prominent SODOS/ SODOMS
becomes. but it is more that tiebreakers give more value to early wins
becomes less reliable. and other tiebreakers break less ties as the
number of rounds grow.
in Roundrobin (what can seen as the ultimate number of rounds /
numbers of players ratio)
tiebreakers like SOS and SOSOS don't work at all because they are
equal for people with the same winscore.
And the argument that early wins lead to a more difficult tournament
loses much of its value. (because every player plays every other
player)
wilemien
QED
Robert Jasiek
<wile...@googlemail.com> wrote:
>> >> Then the player order is invariant under the zero-point value.
>>
>> And what can we say about this?
>
>QED
Ah, no, not that easy. I tried to prove it but that has turned out to
be more nasty than I hoped. For players within the same MMS - NW
group, we can say something. But what about players with different MMS
in general? Before I see the maths, I remain sceptical.
wilemien
> On Fri, 12 Mar 2010 07:36:12 -0800 (PST), wilemien
>
> >> >> Then the player order is invariant under the zero-point value.
>
> >> And what can we say about this?
>
> >QED
>
> Ah, no, not that easy. I tried to prove it but that has turned out to
> be more nasty than I hoped. For players within the same MMS - NW
> group, we can say something. But what about players with different MMS
> in general? Before I see the maths, I remain sceptical.
???
Or are you discussing the Whole McMahon system?
Players with different MMS (MMi(Pj)) are not in the tiegroup in the
first place.
The winner is choosen by
1 Highest MMS score (Whatever 0 point is used) is not even a tie
breaker it is THE way to decide who wins...)
... maybe other tie breakers
2 Number of wins
... maybe other tiebreakers
3 SODOMS
... maybe other tiebreakers
4 Random
Therefore SODOMS is not used on players with different MMS scores
Robert Jasiek
<wile...@googlemail.com> wrote:
>Or are you discussing the Whole McMahon system?
The entire players field.
>Players with different MMS (MMi(Pj)) are not in the tiegroup in the
>first place.
The problem is whether the order of all players remains the same for
1. MM0 - 2. Number of Wins Score - 3. SODOS
versus
1. MM1 - 2. Number of Wins Score - 3. SODOS
Note the different zero-points of MM0 versus MM1 !
>The winner is choosen by
Sure.
>Therefore SODOMS is not used on players with different MMS scores
SODOS is applied to all players of the players field since it is one
of the tiebreakers!
wilemien
> On Fri, 12 Mar 2010 11:38:25 -0800 (PST), wilemien
>
Still don't see the problem
From my proof:
assumptions
MM0 is an MM rating with an arbitrary O point.
MM1 is an MMrating differing from an fixed amount C from the rating
from MM0
For all players
MM0(Pi) = MM1(Pi) + C
and now you want a proof that
MM0(Pi) > MM0(pj) -> MM1(Pi) > MM1(pj) ?
it is easier to prove the stronger:
MM0(Pi) - MM0(pj) = MM1(Pi) - MM1(pj).
MM1(Pi) + C - MM1(Pj) - C = MM1(Pi) - MM1(pj)
QED
Robert Jasiek
<wile...@googlemail.com> wrote:
>MM0 is an MM rating with an arbitrary O point.
>MM1 is an MMrating differing from an fixed amount C from the rating
>from MM0
>
>For all players
>MM0(Pi) = MM1(Pi) + C
>
>and now you want a proof that
>
>MM0(Pi) > MM0(pj) -> MM1(Pi) > MM1(pj) ?
No.
I want to prove that the Final player order is the same. Your MM0 and
MM1 values are at the Tournament Start, right?
wilemien
> On Fri, 12 Mar 2010 14:01:59 -0800 (PST), wilemien
>
> >MM0 is an MM rating with an arbitrary O point.
> >MM1 is an MMrating differing from an fixed amount C from the rating
> >from MM0
>
> >For all players
> >MM0(Pi) = MM1(Pi) + C
>
> >and now you want a proof that
>
> >MM0(Pi) > MM0(pj) -> MM1(Pi) > MM1(pj) ?
>
> No.
>
> I want to prove that the Final player order is the same. Your MM0 and
> MM1 values are at the Tournament Start, right?
The MMi(Pj) value is set at the beginning of the tournament (initial
McMahon score) but MM0(Pi) = MM1(Pi) + C is true all the time.
Or am i missing something?
Robert Jasiek
<wile...@googlemail.com> wrote:
>The MMi(Pj) value is set at the beginning of the tournament (initial
>McMahon score) but MM0(Pi) = MM1(Pi) + C is true all the time.
>Or am i missing something?
Sure. We now have to assume MM0(Pi) = MM0(Pj) + E for some E,
calculate the two players' _final_ MM0_scores due to some suitable
assumptions and then compare that to their _final_ MM1_scores! We also
will have to consider C and the difference D of their number of wins.
wilemien
> On Fri, 12 Mar 2010 14:54:37 -0800 (PST), wilemien
>
> >The MMi(Pj) value is set at the beginning of the tournament (initial
> >McMahon score) but MM0(Pi) = MM1(Pi) + C is true all the time.
> >Or am i missing something?
>
> Sure. We now have to assume MM0(Pi) = MM0(Pj) + E for some E,
> calculate the two players' _final_ MM0_scores due to some suitable
> assumptions and then compare that to their _final_ MM1_scores! We also
> will have to consider C and the difference D of their number of wins.
Do not see the problem
Definitions:
Pi :<==> Player i
NWi :<==> number of wins by player i
MMi is an MMscore with an arbitrary 0 point.
MIi(Pj) is the initial MMi of player j.
MEi(Pj) is the Final MMi of player j.
Let MM0 be an MMscore with an arbitrary 0 point.
MM1 := MM0 - C; C is arbitrary, fixed, natural-number.
also
MI1
Basic equations:
1) For each Pi: MM0(Pi) = MM1(Pi) + C (impies also
MI0(Pi) = MI1(Pi) + C
ME0(Pi) = ME1(Pi) + C )
also
MEi(Pj) = MIi(Pj) +NWj (players j MMi endscore his beginscotre plus
number of wins)
i really don't see the problem
From 1 allready follows that
ME0(Pi) -ME0(Pj) = ME1(Pi)- ME1(Pj)
Robert Jasiek
<wile...@googlemail.com> wrote:
>Do not see the problem
If something is trivial, it must be written down!:)
> For each Pi: MM0(Pi) = MM1(Pi) + C (impies also [...]
> ME0(Pi) = ME1(Pi) + C )
Ah... this I was missing. Let me work out your sketch in detail:
Proposition 2:
Given the final results of an arbitrary McMahon tournament with an
arbitrary, fixed, natural-number zero-point and with the order MMS -
Number of Wins Score - SODOS as the placement criteria. Then the final
player order is invariant under a changing zero-point.
Definitions:
Pi :<==> player i
NW(Pi) :<==> number of wins of Pi
MI0(Pi) is an initial MMscore of Pi with an arbitrary, fixed
zero-point.
MF0(Pi) is the final MMscore of Pi given MI0(Pi)
MI1 := MM0 - C; C is arbitrary, fixed, natural-number.
Proof (idea of step 1 by wilemien):
Step 1:
For a Pi, let us have MI0(Pi) and his NW(Pi). Then we know:
1) MI0(Pi) = MI1(Pi) + C
// by definition
2) MF0(Pi) = MI0(Pi) + NW(Pi)
// his final MMS is NW greater than his initial MMS
3) MF1(Pi) = MI1(Pi) + NW(Pi)
// his final MMS is NW greater than his initial MMS
For a different Pj, let us have MI0(Pj) and his NW(Pj). Then we know:
4) MI0(Pj) = MI1(Pj) + C
5) MF0(Pj) = MI0(Pj) + NW(Pj)
6) MF1(Pj) = MI1(Pj) + NW(Pj)
It follows:
MF0(Pi) - MF0(Pj)
= MI0(Pi) + NW(Pi) - ( MI0(Pj) + NW(Pj) ) // (2) & (5)
= MI1(Pi) + C + NW(Pi) - ( MI1(Pj) + C + NW(Pj) ) // (1) & (4)
= MI1(Pi) + NW(Pi) - ( MI1(Pj) + NW(Pj) ) // arithmetics
= MF1(Pi) - MF1(Pj) // (3) & (6)
Step 2:
The MM0 - NW groups are in the same order as the MM1 - NW groups: In
both cases, the order of the two placement criteria and the NW
criterion are the same. So it follows from Step 1.
Step 3:
The third placement criterion SODOS does not change the player order:
This follows from the following, proven proposition 1:
"
Given the final results of an arbitrary McMahon tournament with an
arbitrary, fixed, natural-number zero-point and with an order of
placement criteria in that Number of Wins Score precedes SODOS. Then,
for every two players having the same number of wins, their SODOS
difference is constant under all possible zero-point values.
"
QED.
Comments:
So now we know also from a theoretical POV that the McMahon system
does make some sense in principle at all:) For every order of
placement criteria in a McMahon tournament, similarly one should prove
that it keeps the basic sense of McMahon! In particular, what about
proofs for MMS - SOS, for MMS - SOS - SOSOS, for MMS - DCn, for MMS -
DCi etc.?:)
wilemien
> On Sun, 14 Mar 2010 16:05:15 -0700 (PDT), wilemien
>
> If something is trivial, it must be written down!:)
That is true it will become a long article.
No That is not true there is only reasonable sense that MMS- NW
(number of wins)-SODOMS (sum of defeated opponents MM score) is more
reasonabvle than just MMS- NW (number of wins)-SODOMS (sum of
defeated opponents MM score)
PS don't equate SOS (Sum of opponent wins, a swiss tiebreaking
mechanism)) and SOMS (som of opponents MM scores ) same fore SODOS and
SODOMS
Robert Jasiek
<wile...@googlemail.com> wrote:
>PS don't equate SOS (Sum of opponent wins, a swiss tiebreaking
>mechanism)) and SOMS (som of opponents MM scores ) same fore SODOS and
>SODOMS
I don't, but also I am used to using SOS, SOSOS, SODOS for both Swiss
and McMahon, where it is understood that the definition differs to fit
the applied tournament system; this convention has been used for
decades now. Your different naming convention also makes sense but
SODOMS is a bit nasty to pronounce. So don't expect everybody to use
it.
wilemien
> On Tue, 16 Mar 2010 04:32:51 -0700 (PDT), wilemien
>
the problem is that you can use both SOS and SOMS in a McMahon
tournament.
maybe an idea to call the others Swiss Sos and Swiss SOSOS ed. (then
nobody has to think about GO-morra)
a possible long list of tie breakers could be
1 McMahon score
2 SOMS
3 SOSOMS
4 Swiss Score
5 Swiss SOS
6 Swiss SOSOS
7 SODOS
8 Swiss SODOS
9 Direct comparison
10 Progress (McMahon)
11 Progress (swiss)
12 Another tiebreaker
14 Nigiri
I was puzzeling about pairing rules (in the topgroup)
what about elimination followed by king of the hill pairings as a
general guide line?
Robert Jasiek
<wile...@googlemail.com> wrote:
>the problem is that you can use both SOS and SOMS in a McMahon
>tournament.
Since it is never done, it is no problem. Why might one want to do it?
>10 Progress (McMahon)
>11 Progress (swiss)
What are these?
>I was puzzeling about pairing rules (in the topgroup)
Everybody is... This is one of the things really in need of urgent
discussion.
>what about elimination followed by king of the hill pairings as a
>general guide line?
What do you mean here, please?
wilemien
> On Tue, 16 Mar 2010 12:10:41 -0700 (PDT), wilemien
>
progress = CUSS see http://senseis.xmp.net/?CUSS
Pairing idea
at start make the topgroup a kind of double elimination system.
(That is normally allready the case)
and after the single winner is pro visionally decided. (after around
log(n)/ log(2) rounds)
change to an kind of king on the hill system.
Strict "King Of the Hill" pairing is #1 plays #2, #3 plays #4, etc.
with no regard to previous pairings, thus unlimited repeat pairings
are allowed. (shortly mentioned at http://en.wikipedia.org/wiki/Swiss-system_tournament
and http://www.math.toronto.edu/jjchew/software/tsh/doc/pairing.html
<-- interesting website)
for Go tournaments I would change it to
Forget the winscores and let #1 plays the highest ranked player he
has't played yet
and do the same for the rest of for players in order they are
provisionally ranked.
(the ranking can change during this rounds)
Robert Jasiek
<wile...@googlemail.com> wrote:
>Strict "King Of the Hill" pairing is #1 plays #2, #3 plays #4, etc.
This is also called adjacency pairing is considered by the
international chess federations and many go players to be the worst
possible pairing method.