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NeilH
21-08-2007, 06:28 AM
The FIDE Swiss Pairing rule C.11 is as follows

"As long as x is less than p: increase x by 1. When pairing a remainder group undo all pairings of players moved down also. Restart at C3. "


For instance, with the score bracket with p = 1 and x = 0
Player A 3.5
Player B 3
Player C 3
Player D 3

The pairing of this heterogeneous score bracket would result in the pairing of
Player A vs Player B

With a remainder score bracket of
with p = 1 and x = 0
Player C 3
Player D 3

If rule C.11 is to be applied to the remainder group. What should happen?

1) Should the original heterogeneous score bracket be reinstated, but with x = 1 ?

OR

2) Should the original heterogeneous score bracket be reinstated, but once a pairing is found in this score bracket, the remainder score bracket should be repaired, but with x = 1 ?

After writing this post, I think the answer is 1. But I would be grateful if someone would confirm this or tell me the correct interpretation of this rule.

Thanks

Neil

drbean
23-08-2007, 01:03 AM
The FIDE Swiss Pairing rule C.11 is as follows

"As long as x is less than p: increase x by 1. When pairing a remainder group undo all pairings of players moved down also. Restart at C3. "

If rule C.11 is to be applied to the remainder group. What should happen?

1) Should the original heterogeneous score bracket be reinstated, but with x = 1 ?

OR

2) Should the original heterogeneous score bracket be reinstated, but once a pairing is found in this score bracket, the remainder score bracket should be repaired, but with x = 1 ?

Neil

Increasing x is not desirable, because players should get their preferences. So I think the preferred strategy would be to find another pairing of the moved down players that didn't increase x and then hope that with the different players in the remainder group, there wouldn't have to be an increase in x there either.

The alternative 1) may (or may not) follow naturally anyway as a consequence of trying to pair the original heterogenous group later when you back up after failing to pair the homogeneous group in C13 and C14. But I am not sure.

pax
23-08-2007, 01:05 PM
Increasing x is not desirable, because players should get their preferences. So I think the preferred strategy would be to find another pairing of the moved down players that didn't increase x and then hope that with the different players in the remainder group, there wouldn't have to be an increase in x there either.

It's not a matter of the preferred strategy, but rather the correct one.

NeilH
23-08-2007, 05:06 PM
It's not a matter of the preferred strategy, but rather the correct one.

I totally agree, but what is the correct interpretation. There must be someone out there who understands this rule.

Bill Gletsos
24-08-2007, 12:57 AM
I totally agree, but what is the correct interpretation. There must be someone out there who understands this rule.Your example is too general and non specific and makes too many assumptions.

The point is you dont suddenly just get to C11.
You have to proceed thru C1- C10 first.

In reality the following happens.
First you have to pair the 4 player group consisting of players A, B C and D.
However in this groups S1 consists solely of A and S2 of B, C and D.
You are assuming A and B are a colour match.
If so that means you can stop at C6 and restart at C2 with the homogeneous remainder group of C and D.
As such if C and D have played then no pairing is possible however the unpairing of the previous A nd B players would be carried out by C10 not C11.

Note that C10 only allows for the unpairing of the lowest moved down pairing but you only had one moved down pairing to start with.
C11 not only allows for change in x but also for the unpairing of all players moved down, the latter a fact that would be significant in a case where more than 1 player had moved down, but unimportant in your example.

Therefore if you get to C11 then x may increase (only if p > x), however when pairing a remainder group the unpairing of all the moved down will always occur.

NeilH
26-08-2007, 07:39 PM
Thanks Bill. I tried to make up an example, but obviously failed. I will try to find a specific example and post it here.

Bill Gletsos
26-08-2007, 08:02 PM
Thanks Bill. I tried to make up an example, but obviously failed. I will try to find a specific example and post it here.Dont waste your time, as I believe my answer covers all circumstances.

drbean
23-11-2007, 07:34 PM
The question NeilH asked is about how x is calculated. Whose x is the x in C11?

C11.
As long as x is less than p: increase x by 1. When pairing a remainder group undo all pairings of players moved down also. Restart at C3.

The choice NeilH offered us was:



If rule C.11 is to be applied to the remainder group. What should happen?
1) Should the original heterogeneous score bracket be reinstated, but with x = 1 ?
OR
2) Should the original heterogeneous score bracket be reinstated, but once a pairing is found in this score bracket, the remainder score bracket should be repaired, but with x = 1 ?


That is, is the x in C11 the heterogeneous bracket's x or the remainder group's x?

If the bracket is repaired, the remainder group that brought us to this point is dissolved. If the answer is 2), does that mean the x of all subsequent remainder groups in the bracket has to increase by 1 when these groups come into existence? Or does C11 repairing start again from scratch? In the latter case, how are x increases accumulated?

And does alternative 2) also mean that heterogeneous brackets never benefit from an x increase? Would this mean that a heterogeneous bracket with a non-zero x would be unpairable?

On the other hand, if the answer is 1), does that mean remainder groups never benefit from an x increase and so are unpairable if x is not 0, a situation we saw with a 7-member all-draw pairing table (http://chesschat.org/showthread.php?t=7139)? The same problem of x increases in post-C11 remainder groups occurs pairing the heterogeneous 2-bracket in the tournament discussed in the middle of http://chesschat.org/showthread.php?t=6925.

The definition of x in A8 is: The number of pairings which can be made in a score bracket, either homogeneous or heterogeneous, not fulfilling all color preferences.

Following this definition, I think this means the x in C11 is the whole heterogeneous bracket's, not the remainder group's.

There is a cryptic comment in B4 about deducting x:



B4.
As many players as possible receive their color preference.
(Whenever x of a score bracket is unequal to zero this rule
will have to be ignored. x is deducted by one each time a
color preference cannot be granted.)


x is the number of pairings not fulfilling preferences. Why would that number decrease when a preference can't be granted? You would expect it to increase.

I think this makes more sense if there is the idea that x has to be shared between one place and another, and in particular, by the heterogeneous bracket and the remainder group.



x [in the remainder group] is deducted by one each time a color preference cannot be granted [in the heterogeneous bracket from which the remainder group
comes.]


This keeps the total number of pairs where preferences cannot be granted constant across different alternative pairings of the heterogeneous bracket and remainder groups.

I think somewhere bartolin was saying the remainder group could use x not 'used up' by the heterogeneous bracket. I guess this is that idea. Remainder groups inherit the x not consumed by their parent heterogeneous brackets.

The flip side of this view is that we cannot limit p and x to the number of downfloaters in the heterogeneous bracket.

p and its decreases have also to be tracked across the heterogeneous bracket and its remainder groups. There may be no connection between the p of the heterogeneous bracket and the p of the remainder group. One could be larger or smaller.

When Games::Tournament::Swiss was reaching C11 with a remainder group, it was starting with the x left over by the heterogeneous bracket, but then if a pairing could not be found with that x value, it was increasing x freely through the appropriate number of C11 cycles, continuing to try to pair the remainder group until x reached the p of the group.

That strategy was of questionable legality, because C11 says to undo all pairings in the heterogeneous bracket when pairing a remainder group. That is, it says to dissolve the remainder group and start again with the heterogeneous bracket.

The depth-first strategy also precludes the finding of a later pairing which has a larger x value in the heterogenous bracket but a smaller total x when combined with the x of the remainder group.

In a move away from depth-first searching, Games::Tournament::Swiss, at the moment, is going on to the next pairing of the heterogenous bracket, in a breadth-first strategy, if the number of preference-violating matches in the remainder group is more than X minus the number of preference-violating matches accepted in the heterogeneous bracket. Where, X = B (or W) - Q. Here B,W and Q have the usual meanings of the corresponding lower-case letters, but are calculated BEFORE division into a heterogeneous bracket and separate remainder group.

It is this big X restriction that we are relaxing in C11, the X of the heterogeneous bracket and remainder group combined.

Later I will post the logs produced pairing the 7-member all-draw tourney at http://chesschat.org/showthread.php?t=7139 and in the http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm one at http://chesschat.org/showthread.php?t=6925.