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Garvinator
08-07-2007, 09:32 PM
This occurred over the weekend and looked very strange indeed.

Pairing info from sp:


No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0

Sp paired round 5 as:

1 v 7
9 v 2
3 v 5
10 v 6
4 v 8

Sm5 paired round 5 as:

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

drbean
25-08-2007, 10:36 PM
This occurred over the weekend and looked very strange indeed.

Pairing info from sp:


No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0

Sp paired round 5 as:

1 v 7
9 v 2
3 v 5
10 v 6
4 v 8



While I don't know if sp is wrong or not, it's what you get when all players become amalgamated into one score group through repeated application of C13.

S1 is 1,2,3,6,4 and S2 is the others. S2 goes through some transpositions.




C5, ordered: 1 2 3 6 4 & 5 8 7 9 10
C6, B1a: table 1 NOK
C7, 8 5 7 9 10
C6, B1a: table 3 NOK
C7, 8 5 9 7 10
C6, B2a: table 5 NOK
C7, 8 5 9 10 7
C6, B1a: table 5 NOK
C7, 8 5 10 7 9
C6, B1a: table 5 NOK
C7, 8 5 10 9 7
C6, B1a: table 4 NOK
C7, 8 7 5 9 10
C6, B1a: table 2 NOK
C7, 8 9 5 7 10
C6, B2a: table 5 NOK
C7, 8 9 5 10 7
C6, B1a: table 5 NOK
C7, 8 9 7 5 10
C6, B1a: table 3 NOK
C7, 8 9 10 5 7
C6, B1a: table 4 NOK
C7, 8 9 10 7 5
C6, B4: x=0, table 3 NOK
C7, 8 10 5 7 9
C6, B1a: table 5 NOK
C7, 8 10 5 9 7
C6, B1a: table 4 NOK
C7, 8 10 7 5 9
C6, B1a: table 3 NOK
C7, 8 10 9 5 7
C6, B1a: table 4 NOK
C7, 8 10 9 7 5
C6, B4: x=0, table 3 NOK
C7, 7 5 8 9 10
C6, B1a: table 3 NOK
C7, 7 5 9 8 10
C6, B2a: table 5 NOK
C7, 7 5 9 10 8
C6, B4: x=0, table 2 NOK
C7, 7 8 5 9 10
C6, B1a: table 4 NOK
C7, 7 8 5 10 9
C6, B1a: table 5 NOK
C7, 7 8 9 5 10
C6, B1a: table 4 NOK
C7, 7 8 9 10 5
C6, B4: x=0, table 2 NOK
C7, 7 9 5 8 10
C6, B2a: table 5 NOK
C7, 7 9 5 10 8
C6, 5 tables paired. E1 1&7 E1 9&2 E1 3&5 E1 10&6 E1 4&8
C6others: no non-paired players
Pairing complete



This is output from Games::Tournament::Swiss, which itself however reveals some bugs getting this result.




Sm5 paired round 5 as:

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7




We might be able to make some inferences about what score groups were joined by C13 to get this Sm5 result. The fact 10 is paired with 6 suggests that they were in the same or adjacent score groups. Though 6 could have been downfloated twice to a position 2 scoregroups below its original position.

When 1 and 2 were downfloated from Group 1 because they had already met, Group 2 became a homogeneous group. In this situation, it would have been very possible for 2 to have been downfloated again. In fact, Sm5 pairs it with 8 which was in Group 3.

This means Group 2 was joined with the lower groups through application of C13. Or does it?

A rule of thumb seems to be to look at the pairings and see what board it is on, and look at the pairingtable and decide how high its possible board would be without C13.




Player Board Pairingtable
1 1 1
2 2 1
3 1 1or2
4 4 2
5 4 2
6 3 1or2
7 5 4
8 2 2
9 5 4
10 3 5



10 is the only one which is higher in the pairings than possible from the pairing table without C13 joining. Perhaps Groups 3-5 were joined.

Kevin Bonham
26-08-2007, 04:11 AM
While I don't know if sp is wrong or not, it's what you get when all players become amalgamated into one score group through repeated application of C13.

Yes, which SP seems to love doing no matter how blatantly it violates B3, which is supposed to be "fulfilled as much as possible", in the process.

Denis_Jessop
26-08-2007, 12:45 PM
Without getting too deeply into all of the possibilities, it is clear that the pairing 10 v 6 is correct (and is made by both programs) since 10 has already played 4 of those immediately above him (5, 8, 7 and) and can't be paired against the other (4) because of the 3 colours rule. But to make that pairing it is necessary, applying C13, to form a score bracket of only players below the top 2. Then, having successfully paired the bottom player, we revert to the usual pairing procedures.

Thus S1 = players 1 and 2.and S2 = players 3 and 6. This seems to me to give the top board pairing as 1v3 and the second as 8v2. 2 cannot be paired against 6 who has already been paired against 10 nor can 1 be paired against 6 for the same reason. Thus an opponent for 2 must be found from the next score bracket containing 4, 5 and 8. Both 4 and 5 have upfloated the previous round and so 2 plays 8. This leaves the pairings of 4v5 and 9v7 which seem to me to comply with the colour rules and, especially with the principle that players with equal scores play each other.

As has been noted, it seems that the SP draw could only have been achieved if all players had been paired as one score bracket . That was not necessary to make a pairing for 10 which was the primary purpose of applying C13. There is another possible explanation for the SP draw. That is, 1v3 is not a proper colour match but 1v8 is. Yet, if that pairing is made the draw comes down to 4v7 on bottom board which is an invalid pairing. However 1v7 is no better a colour match that 1v3 and is not the correct pairing as it also involves an upfloat by 7 who upfloated the previous round. Thus it seems fairly clear that the "one bracket" approach was used and this is just wrong.

DJ

Denis_Jessop
26-08-2007, 07:49 PM
Since posting #4, I have had another look at this thanks to some e-mails from Shaun Press. He correctly pointed out to me that it is not wrong for SP to treat the whole field as one score bracket as that is what happens when successive applications of C13 are made.


Where I think SP goes wrong is in its treatment of that score bracket. On my understanding of the rules, what should happen is that the highest players - nos 1 and 2 form S1 and the rest S2. When S1 is paired v S2 and beyond, the SM5 draw will emerge. It would seem that SP has tried just to pair the top half against the bottom half and that is its mistake.

DJ

drbean
26-08-2007, 09:50 PM
Without getting too deeply into all of the possibilities, it is clear that the pairing 10 v 6 is correct (and is made by both programs) since 10 has already played 4 of those immediately above him (5, 8, 7 and) and can't be paired against the other (4) because of the 3 colours rule.

Yes, it is clear 10 can't play anyone in the bottom 3 brackets for these reasons. And 6 is the best choice of the other 4 players, because of the 2 players which are color compatible (2, and 6), it has the lower score.



But to make that pairing it is necessary, applying C13, to form a score bracket of only players below the top 2. Then, having successfully paired the bottom player, we revert to the usual pairing procedures.


The FIDE rules don't seem to allow for leaving unpaired players in a higher score bracket and then going back after pairing players in a lower bracket.

However, undoing pairings in a higher bracket perhaps amounts to the same thing.



Thus S1 = players 1 and 2.and S2 = players 3 and 6. This seems to me to give the top board pairing as 1v3 and the second as 8v2. 2 cannot be paired against 6 who has already been paired against 10 nor can 1 be paired against 6 for the same reason. Thus an opponent for 2 must be found from the next score bracket containing 4, 5 and 8. Both 4 and 5 have upfloated the previous round and so 2 plays 8.

This makes sense.

The first time through following the FIDE rules, 1 and 2 are floated down from Bracket 1 because they have already played each other.

After waiving the color preference rule B4 and the floating rules B5, B6, the pairing in Bracket 2 is

E4 1&3 E4 6&2

But although this is great for 1,2,3, and 6, it leaves 10 without anyone to play. And we have to get to the last bracket and C13 before we find that out.



This leaves the pairings of 4v5 and 9v7 which seem to me to comply with the colour rules and, especially with the principle that players with equal scores play each other.


Yes, they're a match made in heaven :-)



As has been noted, it seems that the SP draw could only have been achieved if all players had been paired as one score bracket . That was not necessary to make a pairing for 10 which was the primary purpose of applying C13.



I think that because the FIDE rules only allow backtracking to the penultimate bracket, the fact that 2 (from Bracket 1) is paired with 8 (from Bracket 3) after having already been paired with 6 in Bracket 2 seems to me to require that there was one big score bracket.

(Though 2 (and 6) could have been floated down across Bracket 2 to Bracket 3, I guess, meaning that there were 2 brackets, 1-2 and 3-5.)

This is not to say that I agree with the sp pairing. sp seems to have treated the one big score bracket as a homogeneous group, but instead of S1: 1,2,3,6,4 and S2: 5,8,7,9,10, perhaps it should be S1: 1,2 and S2: 3,6,4,5,8,7,9,10



There is another possible explanation for the SP draw. That is, 1v3 is not a proper colour match but 1v8 is. Yet, if that pairing is made the draw comes down to 4v7 on bottom board which is an invalid pairing.



I got lost here. I can't see the inference here.



However 1v7 is no better a colour match that 1v3 and is not the correct pairing as it also involves an upfloat by 7 who upfloated the previous round.


That's right. C9 describes how to drop the downfloat criterion, but there is no description of how the upfloat one is dropped, even though it is also a relative, rather than absolute, criterion. It's certainly not preferred, but it might be legal as a last resort.

I think 1v7 is the better color match than 1v3.



Thus it seems fairly clear that the "one bracket" approach was used and this is just wrong.
DJ

Games::Tournament::Swiss at the moment is joining all the brackets into one bracket and with an S1 of 1,2, first pairing:

1&8
9&2

and then for the remainder group, pairing:

10&3
6&7
4&5

Pairing 9&2 violates B5 for upfloats, as do 10&3 and 6&7. Hmm.

Denis_Jessop
26-08-2007, 11:22 PM
In view of the complexities of this thing and the uncertainty about how SP approached it, I have has another look at it and have come up with yet another view based on discussions with a colleague.

Basically I now revert to my original view that it is not correct to finish up with a single score bracket.

To begin with one looks at 1v2 - an invalid pairing. So 1 & 2 drop to join 3 & 6 to form a score bracket to be a paired 1v3 and 2v6 as 1 cannot play either 2 or 6. After making further pairings, a situation arises where the last s/b is 7 & 10 who can't be paired. Then C13 is applied. It operates so that eventually there is a situation in which the penultimate s/b is 1,2,3,6 and the lowest s/b is 4,5,8,7,9,10. C13 then requires the ps/b to be unpaired and an attempt made to re-pair it so as to allow a valid pairing of the ls/b. That cannot be done while p=2. The merging of the 2 brackets into one can only occur if p becomes zero. As it stands p = 2 and the only way p can be reduced is on application of C14. This is the step that I overlooked as perhaps have everyone else. C14 applies so as to decrease p by 1. When this is done, the attempt to pair the ps/b with p=1 is satisfied by pairing 1v3 and leaving 2 and 6 to be paired with the others. Although the computer may have to do this by a laborious set of calculations taking it only a split second, humans can equally see immediately that the only valid pairing for 10 is 2 or 6. Eventually the pairings made by SM5 are reached.

It is not absolutely clear what SP has done. It may have ignored C14 and paired the whole group as one with top half v bottom half. On the other hand it should be noted that the SP draw gives colour matches for all pairings whereas SM5 has them for 4 pairings only (1v3 is not a colour match). So SP may have done some fiddling to come up with perfect colour matches though from what starting point I don't know.

DJ

drbean
02-09-2007, 12:50 AM
Since posting #4, I have had another look at this thanks to some e-mails from Shaun Press. He correctly pointed out to me that it is not wrong for SP to treat the whole field as one score bracket as that is what happens when successive applications of C13 are made.

Where I think SP goes wrong is in its treatment of that score bracket. On my understanding of the rules, what should happen is that the highest players - nos 1 and 2 form S1 and the rest S2. When S1 is paired v S2 and beyond, the SM5 draw will emerge. It would seem that SP has tried just to pair the top half against the bottom half and that is its mistake.

DJ

If all the score brackets get joined into one bracket, is it a homogeneous group or a heterogeneous one? If it is heterogeneous, then I agree 1 and 2 form S1 and the rest S2. If it is homogeneous, then the top half should be paired with the bottom half.

A3 says: "A heterogeneous score bracket of which at least half of the players have come from a higher score bracket is also treated as though it was homogeneous."

Here if 3 score brackets are joined, I don't think it is clear whether it should be treated as homogeneous or not.

A3 also says: "When pairing a heterogeneous score bracket these players moved down are always paired first whenever possible, giving rise to a remainder score bracket which is always treated as a homogeneous one."

If just 1 and 2 are paired, then I guess the rest of the big bracket, being a remainder group, should be treated as homogeneous.

Denis_Jessop
02-09-2007, 05:15 PM
If all the score brackets get joined into one bracket, is it a homogeneous group or a heterogeneous one? If it is heterogeneous, then I agree 1 and 2 form S1 and the rest S2. If it is homogeneous, then the top half should be paired with the bottom half.

A3 says: "A heterogeneous score bracket of which at least half of the players have come from a higher score bracket is also treated as though it was homogeneous."

Here if 3 score brackets are joined, I don't think it is clear whether it should be treated as homogeneous or not.

A3 also says: "When pairing a heterogeneous score bracket these players moved down are always paired first whenever possible, giving rise to a remainder score bracket which is always treated as a homogeneous one."

If just 1 and 2 are paired, then I guess the rest of the big bracket, being a remainder group, should be treated as homogeneous.

If there is just one score bracket as a result of repeated applications of C13, that bracket must be homogeneous as the definition in A3 of a heterogeneous score bracket either can't apply because nobody has been moved down, or, in this case, all players but no. 10 must have moved down and so the bracket must be treated as homogeneous under the >50% rule in A3.

But see my post #7 where I say that in my view (further revised) you don't get to the single big score bracket situation anyway.

DJ

drbean
03-09-2007, 03:37 PM
Basically I now revert to my original view that it is not correct to finish up with a single score bracket.


I changed my mind too about a minor matter. I said there was no rule to drop the upfloat criterion. I made a mistake. It's part of C10.

In the following, you make a persuasive case for there being no need to form a single score group.



To begin with one looks at 1v2 - an invalid pairing. So 1 & 2 drop to join 3 & 6 to form a score bracket to be a paired 1v3 and 2v6 as 1 cannot play either 2 or 6. After making further pairings, a situation arises where the last s/b is 7 & 10 who can't be paired. Then C13 is applied. It operates so that eventually there is a situation in which the penultimate s/b is 1,2,3,6 and the lowest s/b is 4,5,8,7,9,10.


10 has played all except 4 in the lowest score bracket, and can't play 4 because they both have an absolute preference for White.



C13 then requires the ps/b to be unpaired and an attempt made to re-pair it so as to allow a valid pairing of the ls/b. That cannot be done while p=2. The merging of the 2 brackets into one can only occur if p becomes zero. As it stands p = 2 and the only way p can be reduced is on application of C14. This is the step that I overlooked as perhaps have everyone else. C14 applies so as to decrease p by 1. When this is done, the attempt to pair the ps/b with p=1 is satisfied by pairing 1v3 and leaving 2 and 6 to be paired with the others. Although the computer may have to do this by a laborious set of calculations taking it only a split second, humans can equally see immediately that the only valid pairing for 10 is 2 or 6. Eventually the pairings made by SM5 are reached.


I think this is correct. Because 2 and 6 are also a compatible pair, perhaps the SwissPerfect algorithm was greedy and paired 2 and 6 also. Perhaps there is no other p=1 pairing than this one, where you get one for free.

I am interested in the mechanics of keeping 2 and 6 from pairing up with each other and getting them into the last bracket with the minimum of special casing. If they go into a Bracket 2 remainder group by themselves, they become paired again.



It is not absolutely clear what SP has done. It may have ignored C14 and paired the whole group as one with top half v bottom half. On the other hand it should be noted that the SP draw gives colour matches for all pairings whereas SM5 has them for 4 pairings only (1v3 is not a colour match). So SP may have done some fiddling to come up with perfect colour matches though from what starting point I don't know.

DJ

I think it is probably FIDE-approved fiddling!

But I agree it probably took a wrong turning with 2 and 6. Perhaps even if it did pair them in a Bracket 2 remainder group, joining that group and the last group would have still given the Sm5 answer.

Garvinator
03-09-2007, 04:24 PM
If there was ever to be an arbiters exam, this situation would be a good question to have for pairings.

drbean
04-09-2007, 05:04 PM
If there was ever to be an arbiters exam, this
situation would be a good question to have for pairings.

I think it's a good question because the fact player 10 has so
few choices for partners allows you to reason about the later
FIDE rules like C13 and C14, something which you don't often get
a chance to do.

At the moment, Games::Tournament::Swiss's pairing of the round is
consistent with Denis Jessop's observation that C14's decreasing
of p to 1 in the second bracket allows you to avoid joining all
10 players into one big score bracket.



No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


However, it is not producing the same pairings as Sm5.

Instead it is producing 1&3, 10&2, 6&7, 4&8, 9&5

It pairs the last 4 tables like this:



Next, Bracket 3: 6 2 4 8 9 10 7 5
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=4 Homogeneous.
C4, S1 & S2: 2 6 4 5 & 8 9 10 7
C5, ordered: 2 6 4 5 &
8 7 9 10
C6, B1a: table 3 4 NOK
C7, 8 7 10 9
C6, B2a: table 3 NOK
C7, 8 9 7 10
C6, B1a: table 2 3 4 NOK
C7, 8 10 7 9
C6, B1a: table 3 NOK
C7, 8 10 9 7
C6, B1a: table 3 NOK
C7, 7 8 9 10
C6, B1a: table 1 3 4 NOK
C7, 9 8 7 10
C6, B1a: table 3 4 NOK
C7, 9 8 10 7
C6, B2a: table 3 4 NOK
C7, 9 7 8 10
C6, B1a: table 4 NOK
C7, 9 7 10 8
C6, B1a: table 4 NOK
C7, 9 10 8 7
C6, B2a: table 4 NOK
C7, 9 10 7 8
C6, B1a: table 3 4 NOK
C7, 10 8 7 9
C6, B1a: table 3 NOK
C7, 10 8 9 7
C6, B1a: table 3 NOK
C7, 10 7 8 9
C6, Tables 1 2 3 4 paired. E1 10&2 E2 6&7 E1 4&8 E1 9&5


Sm5 paired round 5 as:

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

What about floats here?

Bill Gletsos
04-09-2007, 07:53 PM
I think it's a good question because the fact player 10 has so
few choices for partners allows you to reason about the later
FIDE rules like C13 and C14, something which you don't often get
a chance to do.

At the moment, Games::Tournament::Swiss's pairing of the round is
consistent with Denis Jessop's observation that C14's decreasing
of p to 1 in the second bracket allows you to avoid joining all
10 players into one big score bracket.



No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


However, it is not producing the same pairings as Sm5.

Instead it is producing 1&3, 10&2, 6&7, 4&8, 9&5

It pairs the last 4 tables like this:



Next, Bracket 3: 6 2 4 8 9 10 7 5
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=4 Homogeneous.
C4, S1 & S2: 2 6 4 5 & 8 9 10 7
C5, ordered: 2 6 4 5 &
8 7 9 10
C6, B1a: table 3 4 NOK
C7, 8 7 10 9
C6, B2a: table 3 NOK
C7, 8 9 7 10
C6, B1a: table 2 3 4 NOK
C7, 8 10 7 9
C6, B1a: table 3 NOK
C7, 8 10 9 7
C6, B1a: table 3 NOK
C7, 7 8 9 10
C6, B1a: table 1 3 4 NOK
C7, 9 8 7 10
C6, B1a: table 3 4 NOK
C7, 9 8 10 7
C6, B2a: table 3 4 NOK
C7, 9 7 8 10
C6, B1a: table 4 NOK
C7, 9 7 10 8
C6, B1a: table 4 NOK
C7, 9 10 8 7
C6, B2a: table 4 NOK
C7, 9 10 7 8
C6, B1a: table 3 4 NOK
C7, 10 8 7 9
C6, B1a: table 3 NOK
C7, 10 8 9 7
C6, B1a: table 3 NOK
C7, 10 7 8 9
C6, Tables 1 2 3 4 paired. E1 10&2 E2 6&7 E1 4&8 E1 9&5


Sm5 paired round 5 as:

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

What about floats here?You have a fatal flaw in your logic.


I think it's a good question because the fact player 10 has so
few choices for partners allows you to reason about the later
FIDE rules like C13 and C14, something which you don't often get
a chance to do.

At the moment, Games::Tournament::Swiss's pairing of the round is
consistent with Denis Jessop's observation that C14's decreasing
of p to 1 in the second bracket allows you to avoid joining all
10 players into one big score bracket.



No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


However, it is not producing the same pairings as Sm5.

Instead it is producing 1&3, 10&2, 6&7, 4&8, 9&5

It pairs the last 4 tables like this:



Next, Bracket 3: 6 2 4 8 9 10 7 5
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=4 Homogeneous.
C4, S1 & S2: 2 6 4 5 & 8 9 10 7
C5, ordered: 2 6 4 5 &
8 7 9 10
C6, B1a: table 3 4 NOK
C7, 8 7 10 9
C6, B2a: table 3 NOK
C7, 8 9 7 10
C6, B1a: table 2 3 4 NOK
C7, 8 10 7 9
C6, B1a: table 3 NOK
C7, 8 10 9 7
C6, B1a: table 3 NOK
C7, 7 8 9 10
C6, B1a: table 1 3 4 NOK
C7, 9 8 7 10
C6, B1a: table 3 4 NOK
C7, 9 8 10 7
C6, B2a: table 3 4 NOK
C7, 9 7 8 10
C6, B1a: table 4 NOK
C7, 9 7 10 8
C6, B1a: table 4 NOK
C7, 9 10 8 7
C6, B2a: table 4 NOK
C7, 9 10 7 8
C6, B1a: table 3 4 NOK
C7, 10 8 7 9
C6, B1a: table 3 NOK
C7, 10 8 9 7
C6, B1a: table 3 NOK
C7, 10 7 8 9
C6, Tables 1 2 3 4 paired. E1 10&2 E2 6&7 E1 4&8 E1 9&5


Sm5 paired round 5 as:

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

What about floats here?Your pairings of 10&2, 6&7, 4&8, 9&5significantly violate B3.
The SM5 pairings dont.

Bartolin
04-09-2007, 08:33 PM
You have a fatal flaw in your logic.

Your pairings of 10&2, 6&7, 4&8, 9&5significantly violate B3.
The SM5 pairings dont.

May I ask how B3 is to interpreted if there are more than one pairing to be made within one score bracket (as in this case)? Does B3 imply that the sum of differences should be as small as possible or does it aim at a kind of "uniform distribution of differences"?

I guess in this case it doesn't make a difference, but there could be other cases.

Best regards,

Christian

drbean
06-09-2007, 12:20 PM
Your pairings of [1&3,] 10&2, 6&7, 4&8, 9&5significantly violate B3.
The SM5 pairings dont.

The score differences of the above pairing are
1 + 3.5 + 1.5 + 0 + 1 = 7

The SM5 pairings of
1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

have score differences of 1 + 1.5 + 2.5 + 0 + 0 = 5, which
is two points better.

I guess the pairing of 10&2 is a glaring difference. One has
the top score and the other has the bottom score.

2.5 is the minimum difference of any match in which 10
participates. There is no other player who can play 10 whose
score is closer than 2.5 points away.

So clearly the SM5 pairing is better than my pairing.

I don't know the correct way to check pairings for B3
complicance. In C. Pairing Procedures, there is no mention
of how to incorporate B3 into the pairing process.

C6 says if p pairings are obtained in compliance with
B1 and B2 the pairing of this score bracket is considered
complete. Do I have to ignore this and check all the
possible pairings? This could be computationally expensive.

I thought that if I took care of B5 and B6, B3 would look
after itself. Perhaps B5 and B6 need to be relaxed in
stages.

First, only allow consecutive floats one bracket above or
below, then 2.

To avoid the need to check all the possible pairings after
waiving B5 and B6, I would have to have some way
of knowing how the differences are going to vary without
knowing the actual pairings themselves.

Then I would have some way of knowing when to stop.

The existence of a heterogeneous group is an undesirable
pairing of players with different scores. It's the same
undesirability that B3 is talking about.

The one big score group of all players is undesirable
because of the B3 criterion.

B3 comes before B5 and B6, so it is more important than
them.

But B5,6 and B3 are factors that vary along different
dimensions. B3 is just about the pairings for the next
round. B5,6 are about B3 problems that are becoming chronic
over different rounds.

By definition, I don't need to be concerned about B3 in a
homogeneous group. In the normal kind of heterogeneous
group, B3 will take care of itself too.

It's after C13 and C14 are applied multiple times that the
problems start. This is not really a B5,6 (ie C9 and C10)
problem.

Even then, all the transposition and exchange possibilities
are being considered before C14's reduction of p to zero
(the necessary condition for C13 and C14's joining of score
groups).

So it seems the rules already have consideration of B3
built in.

And the fact that I am not getting the SM5 pairing is a bug
in transposition and exchange procedures.

But wait. After C13 or C14 is applied twice, we have players
in the same bracket whose scores differ by 2 or more points.
It's then that B3 considerations are not built in. We have
to try and keep the players with different scores separate
(kind of like brackets within brackets).

For that reason, it might be good to try and pair the
players with high scores first. That suggests treating
the consolidated brackets as heterogeneous rather than as
homogeneous, a question raised at
http://chesschat.org/showpost.php?p=166708&postcount=8
and http://chesschat.org/showpost.php?p=166761&postcount=9
in this same thread.

Bartolin
07-09-2007, 08:08 PM
It might be interesting to read a passage from "An Arbiters's Notebook" (Nr. 47, February 2002) by Geurt Gijssen. [1] On page 6 starts the following question/answer passage:


Question: Dear Geurt, I have a problem with interpreting the pairing
rules of the Dutch System for doing Swiss draws as approved by FIDE. Sections B and C
appear to be in conflict with each other. I was playing in an event recently where
there were four players in the bottom score group. It was not possible to form two
legal pairs out of these four players because too many of them had played each
other, but there were two players within that group who could have played each
other. The arbiters applied section C13 which states that where no pairings can be
found to allow a correct pairing entirely within the lowest score bracket, then the
two lowest score brackets are joined. This meant that the bottom score group was
joined with the next score group up (also containing four players), and that every
player in the next score group up therefore played a player on a lower score.

However, section B3 says that the difference of scores of two players paired
against each other shall be as small as possible and ideally zero. Section B also
says that this criterion "should be fulfilled as much as possible" and that
transpositions and exchanges can be applied to achieve this. The pairing obtained
using section C13 was not the best possible in this way. It would have been
possible (without changing the colours of any player) for two players in the
lowest score group to play each other, two players in the second-lowest score
group to play each other, and two players from the second-lowest score group to
each play a player from the lowest score group. This would have meant that the
difference in scores between opponents existed on two boards rather than on four.

How should an arbiter make sense of this apparent conflict when there is nothing
in the rules to say which of sections B and C should take priority in this case?
Kevin Bonham (Australia)

Answer Answer Dear Kevin, congratulations, you scored a full point. For people who do
not know too much about the Swiss system and its regulations, I will give a very
brief explanation.

In Chapter B of the Swiss Pairing regulations the Pairing Criteria are described.
Here are some of these criteria: two players shall play only once against each
other; no player shall play three times with the same colors in a row; the
difference of the scores of two players paired against each other should be as
small as possible and ideally zero.

In Chapter C there is a description of the Pairing Procedures. In this chapter you
can find what to do if the number of players with the same score is odd; what to
do if there are 10 players in a group and it is not possible to make 5 pairings, but
only 4 pairings; what to do after the arbiter has finished the pairings for all
players and then he discovers that the players in the last group (this is the group
of players with the lowest score) cannot be paired. Let me explain.

We have 4 players with 2 point and 4 players with 1 point. The players with 2
points are A, B, C and D. The players with 1 point are E, F, G and H.

It is possible to make the pairings A-C and B-D. Going to the next group the
arbiter discovers that E played already against F, G and H and the pairing F-G is
possible.

Well, what to do? Point C13 of the Pairing Procedure says that the 8 players
should form one group and the most likely pairings will be in that case:

A * E, B * F, C * G and D * H. But these pairings violate one of the Criteria of
Chapter B, which says that if possible players with the same score should play
against each other and good pairings could be: A * C, B * E, D * F and G * H.
Without any doubt, this is the best pairing. But there is a conflict in the
regulations. The criteria of Chapter B are correct, but, and this is the point, the
pairing procedures of Chapter C say something else. The problem is that the
pairing procedures cannot cover all situations, and I was also told that it is very
difficult to program for the computer. I am afraid that we have to accept a
situation that the computer produces pairings we shall simply accept. One thing is
sure: the computer is objective. There is no discussion that we have to change the
pairings only in cases in which the computer violates the criteria of Chapter B.
An example: Two players are leading, they did not play in any previous round,
the colours fit and the computer does not pair them against each other. And there
are of course more cases like this.

And you probably will not like this: the most important thing is that the pairings
at the top are completely correct. The pairings in the bottom (groups), especially
at the end of a tournament, are not of the highest priority.


As I understand, the answer says that there are cases when the pairing criteria from section B and the pairing procedure from section C don't fit completely. In those cases the violation of relative criteria B3 to B6 are acceptable.

But there is another interesting aspect of the example cited (also in respect to the question about penultimate score brackets at http://chesschat.org/showthread.php?t=6856): Both question and answer seem to imply that the lowest and the penultimate score bracket should be joined at once, since it's impossible to pair the lowest score bracket. But what about C14? Wouldn't one get satisfactory results by decreasing p to 1 for the penultimate score bracket, thereby pairing A with C and downfloating B and D, and afterwards pairing S1 (B, D) with S2 (E, F, G, H) and so on? I guess one would end with A-C, B-E, D-F, G-H -- just the pairings which comply with B3. Maybe C14 was neglected in the answer to simplify matters and to highlight possible conflicts between pairing procedure (C) and pairing criteria (B)?

Best regards,

Christian

[1] http://www.chesscafe.com/text/geurt47.pdf

Bill Gletsos
07-09-2007, 08:33 PM
The score differences of the above pairing are
1 + 3.5 + 1.5 + 0 + 1 = 7

The SM5 pairings of
1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

have score differences of 1 + 1.5 + 2.5 + 0 + 0 = 5, which
is two points better.

I guess the pairing of 10&2 is a glaring difference. One has
the top score and the other has the bottom score.

2.5 is the minimum difference of any match in which 10
participates. There is no other player who can play 10 whose
score is closer than 2.5 points away.

So clearly the SM5 pairing is better than my pairing.More importantly two of the Sm5 pairings have zero score difference with regards B3 where as yours only has one.


I don't know the correct way to check pairings for B3
complicance. In C. Pairing Procedures, there is no mention
of how to incorporate B3 into the pairing process.

C6 says if p pairings are obtained in compliance with
B1 and B2 the pairing of this score bracket is considered
complete. Do I have to ignore this and check all the
possible pairings? This could be computationally expensive.

I thought that if I took care of B5 and B6, B3 would look
after itself. Perhaps B5 and B6 need to be relaxed in
stages.C6 is badly worded (as are numerous other sections of the Swiss rules).
All aspects of section B should be considered with regards C6.

It is clear that the following is true.

There are 4 types of score brackets.

1) A heterogeneous group.

2) A pure homogeneous group.

3) A homogeneous remainder group.

4) A homogeneous group that is actually hetrogeneous but treated a homogeneous as described in A3.

B3 would only apply in 1) & 4) above.


And the fact that I am not getting the SM5 pairing is a bug
in transposition and exchange procedures.I'm do not believe that is true, as you pairing is due to valid transpositions carried out prior to any exchanges. Your problem appears to be totally due to ignoring the implications of B3.

Kevin Bonham
07-09-2007, 08:43 PM
As I understand, the answer says that there are cases when the pairing criteria from section B and the pairing procedure from section C don't fit completely. In those cases the violation of relative criteria B3 to B6 are acceptable.

I don't see where he says that at all. He says that you have to change the pairings when the computer violates criteria of section B. He doesn't say whether that means the absolute criteria only or the relative criteria as well so I am assuming he means where the computer needlessly violates any of them (of course, sometimes you can't help it.)

My interpretation is to prefer section B including the relative criteria over section C in the case of any conflict between them, precisely because this avoids the absurd creation of joined-together mega-groups creating mismatches on many boards at once.

By the way where Geurt wrote:


And you probably will not like this: the most important thing is that the pairings at the top are completely correct.

...he was wrong - I do like it - one should comply with B3 as much as possible in each scoregroup as it is moved down whatever the impacts on lower score groups.


May I ask how B3 is to interpreted if there are more than one pairing to be made within one score bracket (as in this case)? Does B3 imply that the sum of differences should be as small as possible or does it aim at a kind of "uniform distribution of differences"?

This is a very good question. I may have some further comments on this if I can generate a genuine case where there is conflict between different possible aims! (eg a case where one must choose between a double downfloat and two single downfloats for players in the same score bracket).

A common situation arises near the tail end of the field where one has to choose between, say, pairing a player on 2 with a player on 0 and a player with 1 and a player with 1, or pairing 2-1 and 1-0. In this case I pair 2-1 and 1-0 because it minimises the difference affecting the higher scoring player.

If anyone can find an example with more players where one might have to choose between, say, one 2-2 and one 2-0 or two 2-1s, I'd be very interested to see it.

Bill Gletsos
07-09-2007, 08:44 PM
It might be interesting to read a passage from "An Arbiters's Notebook" (Nr. 47, February 2002) by Geurt Gijssen. [1] On page 6 starts the following question/answer passage:I was aware of this and i probably should have referred to it.

As I understand, the answer says that there are cases when the pairing criteria from section B and the pairing procedure from section C don't fit completely. In those cases the violation of relative criteria B3 to B6 are acceptable.Actually it is more to do with the Swiss rules being badly worded especially in section C.

But there is another interesting aspect of the example cited (also in respect to the question about penultimate score brackets at http://chesschat.org/showthread.php?t=6856): Both question and answer seem to imply that the lowest and the penultimate score bracket should be joined at once, since it's impossible to pair the lowest score bracket. But what about C14? Wouldn't one get satisfactory results by decreasing p to 1 for the penultimate score bracket, thereby pairing A with C and downfloating B and D, and afterwards pairing S1 (B, D) with S2 (E, F, G, H) and so on? I guess one would end with A-C, B-E, D-F, G-H -- just the pairings which comply with B3. Maybe C14 was neglected in the answer to simplify matters and to highlight possible conflicts between pairing procedure (C) and pairing criteria (B)?I believe Geurt just ignored C14 so as to simplify his example and explanation.

Bartolin
07-09-2007, 09:24 PM
As I understand, the answer says that there are cases when the pairing criteria from section B and the pairing procedure from section C don't fit completely. In those cases the violation of relative criteria B3 to B6 are acceptable.

I don't see where he says that at all. He says that you have to change the pairings when the computer violates criteria of section B. He doesn't say whether that means the absolute criteria only or the relative criteria as well so I am assuming he means where the computer needlessly violates any of them (of course, sometimes you can't help it.)

My interpretation is to prefer section B including the relative criteria over section C in the case of any conflict between them, precisely because this avoids the absurd creation of joined-together mega-groups creating mismatches on many boards at once.

Thanks for pointing that out. Maybe I misunderstood his reasoning about accepting computer generated pairings:


I am afraid that we have to accept a situation that the computer produces pairings we shall simply accept.

But you are right, he didn't narrow his mention of section B:


There is no discussion that we have to change the
pairings only in cases in which the computer violates the criteria of Chapter B.


If anyone can find an example with more players where one might have to choose between, say, one 2-2 and one 2-0 or two 2-1s, I'd be very interested to see it.

I'll try to find an example.

Christian

Denis_Jessop
07-09-2007, 09:28 PM
I'm not sure how productive a discussion on the application of Sections B and C of the Swiss Rules will be.

But I point out that the pairing criteria in B3 - B6 are expressed to be "relative", to be "in descending priority" and to "be fulfilled as much as possible".

That is a clear implication that it is recognised that situations will arise in which it is not possible to fulfill all of B3 - B6. In such a case the requirements are dropped in reverse order (beginning with B6). The only situations in which the question whether to drop B6 - B3 as contemplated by the words quoted above will arise in the application of the pairing procedures in C.

It follows that B6 - B3 are prima facie to be applied but their application is subject to the need to achieve valid pairings. If this cannot be done without dropping one or more of B6 - B3 then that is the course to be adopted. In that respect B3 - B6 are subject to C.

Were this not so, all the words quoted above would be meaningless and that is not a proper approach to interpretation of the provisions unless the quoted words can be demonstrated to have no possible application. In fact, in making pairings, B and C have to be read together so that pairings are made in accordance with C applying the criteria in B, dropping the latter (B3 - B6) only if it is necessary to make a valid pairing under C.

The situation is not helped by some ambiguous drafting as in C6. There I believe that all the major pairing programs apply B3 - B6 as well as B1 and B2. But there is some sort of an argument for the view that C6 means what it says and that valid pairings for S1 are made without reference to B3 - B6 on the basis that the best opponents according to the spirit of the Swiss rules for the players floating down are the highest players in the group below, subject only to B1 and B2. That is, the literal interpretation accords with the main purpose and object of the Swiss Rules.

DJ

Bill Gletsos
07-09-2007, 11:07 PM
The situation is not helped by some ambiguous drafting as in C6. There I believe that all the major pairing programs apply B3 - B6 as well as B1 and B2.Correct.

But there is some sort of an argument for the view that C6 means what it says and that valid pairings for S1 are made without reference to B3 - B6 on the basis that the best opponents according to the spirit of the Swiss rules for the players floating down are the highest players in the group below, subject only to B1 and B2. That is, the literal interpretation accords with the main purpose and object of the Swiss Rules.Close inspection of any FIDE approved program that does pairings according to the "Dutch" rules, shows that the literal interpretation of C6 is not valid.

Also although C6 refers to B1 and B2 it really should be referring to B1 to B6. In fact Stewart Reuben tries to rectify this in the section of his book The Chess Organiser's Handbook Second Edition on the Dutch rules by renumbering B1 & B2 as just B1 with sections (a)-(e) and renumbering B3 to B6 as B2 with sections (a)-(d).

Bartolin
07-09-2007, 11:30 PM
May I ask how B3 is to interpreted if there are more than one pairing to be made within one score bracket (as in this case)? Does B3 imply that the sum of differences should be as small as possible or does it aim at a kind of "uniform distribution of differences"?

This is a very good question. I may have some further comments on this if I can generate a genuine case where there is conflict between different possible aims! (eg a case where one must choose between a double downfloat and two single downfloats for players in the same score bracket).

A common situation arises near the tail end of the field where one has to choose between, say, pairing a player on 2 with a player on 0 and a player with 1 and a player with 1, or pairing 2-1 and 1-0. In this case I pair 2-1 and 1-0 because it minimises the difference affecting the higher scoring player.

If anyone can find an example with more players where one might have to choose between, say, one 2-2 and one 2-0 or two 2-1s, I'd be very interested to see it.

What about the following situation at the bottom of the table (I guess, it could arise in reality -- but I don't have a proof for that assumption):

A 2.5 points -- played vs. B and C already
B 2 points -- played vs. A already
C 2 points -- played vs. A already
D 1.5 points
E 1 points -- played vs. F already
F 1 points -- played vs. E already

If I'm not mistaken, in this case players A,B,C form one score bracket. Player A must be moved down because he already played against B and C. It's possible to pair A and D, but in that case B and C have to be paired against E and F, since the latter have already played against each other. Another possibility is to move A even further down, playing E (or F). In that case the other pairings would be B-C and D-E(F).

So basically we have two cases:

1. A-D, B-E, C-F which results in a score difference of 1 for each pairing (3 total) and

2. A-E, B-C, D-F which results in a score difference of 1.5 for pairing A-E, 0 for B-C and 0.5 for D-F (2 total)

Now I think, different interpretations of B3 could lead to different results

a) If the main criterion is to minimize the sum of score differences, the second case would be "better".

b) If the main criterion is to minimize the impact on the highest placed player, the first case would be "better".

c) If the main criterion is to minimize the maximum of score differences, again the first case would be "better".

And so on -- I guess one could think of even more interpretations of B3.

If you were looking for something else, I'm sorry for the noise.

Christian

Denis_Jessop
07-09-2007, 11:33 PM
Close inspection of any FIDE approved program that does pairings according to the "Dutch" rules, shows that the literal interpretation of C6 is not valid.

I take it that what you are saying is that, in approving computer programs that apply all of B to C6, FIDE is indirectly ruling that C6 should be interpreted that way. It's not exactly the best means of doing it but I think the better view is that one anyway even though the other view is at least arguable, in my opinion.

DJ

Bill Gletsos
07-09-2007, 11:52 PM
I take it that what you are saying is that, in approving computer programs that apply all of B to C6, FIDE is indirectly ruling that C6 should be interpreted that way. It's not exactly the best means of doing it but I think the better view is that one anyway even though the other view is at least arguable, in my opinion.One would assume that the pairing program endorsed by the Dutch Federation that carries out the FIDE "Dutch" pairing rules would do the pairings correcty especially with reagrds something as fundamental as C6.

However the bottom line is that I am prepared to accept the view of a member of the FIDE Swiss Pairing Committee over that of people who are just theorising (myself included).

NeilH
08-09-2007, 07:04 AM
Has any one got a cross table for this tournaments, eg colours, opponents, result for each round. As I would like to pair this tournament with the software I am developing to see how it coupes with this complex situation

Denis_Jessop
08-09-2007, 01:35 PM
What about the following situation at the bottom of the table (I guess, it could arise in reality -- but I don't have a proof for that assumption):

A 2.5 points -- played vs. B and C already
B 2 points -- played vs. A already
C 2 points -- played vs. A already
D 1.5 points
E 1 points -- played vs. F already
F 1 points -- played vs. E already

If I'm not mistaken, in this case players A,B,C form one score bracket. Player A must be moved down because he already played against B and C. It's possible to pair A and D, but in that case B and C have to be paired against E and F, since the latter have already played against each other. Another possibility is to move A even further down, playing E (or F). In that case the other pairings would be B-C and D-E(F).

So basically we have two cases:

1. A-D, B-E, C-F which results in a score difference of 1 for each pairing (3 total) and

2. A-E, B-C, D-F which results in a score difference of 1.5 for pairing A-E, 0 for B-C and 0.5 for D-F (2 total)

Now I think, different interpretations of B3 could lead to different results

a) If the main criterion is to minimize the sum of score differences, the second case would be "better".

b) If the main criterion is to minimize the impact on the highest placed player, the first case would be "better".

c) If the main criterion is to minimize the maximum of score differences, again the first case would be "better".

And so on -- I guess one could think of even more interpretations of B3.

If you were looking for something else, I'm sorry for the noise.

Christian

In my view this is an example of what I said about the relative operation of B and C.

That is, the correct procedure is as follows.

A B and C are the one score bracket. A is moved to the next score bracket as A cannot be found an opponent in the first score bracket (see C1) and B is paired against C. The next score bracket contains A and D who are then paired. The lowest score bracket contains E and F who canot be validly paired. Applying C13 the penultimate score bracket (A and D) is undone and A is paired against E or F and then D against E or F (depending on colour preference which is not stated in this example - nor is the "float"situation). The pairing B v C is not undone. B3 is essentially irrelevent to this situation or, alternatively, it would not affect the pairings as it applies so as to render the difference of the scores "as small as possible". That means as small as possible having regard to the operation of the pairing procedures in C. Those procedures do not allow the pairing B v C to be undone.

DJ

Bartolin
08-09-2007, 06:27 PM
In my view this is an example of what I said about the relative operation of B and C.

That is, the correct procedure is as follows.

A B and C are the one score bracket. A is moved to the next score bracket as A cannot be found an opponent in the first score bracket (see C1) and B is paired against C. The next score bracket contains A and D who are then paired. The lowest score bracket contains E and F who canot be validly paired. Applying C13 the penultimate score bracket (A and D) is undone and A is paired against E or F and then D against E or F (depending on colour preference which is not stated in this example - nor is the "float"situation). The pairing B v C is not undone. B3 is essentially irrelevent to this situation or, alternatively, it would not affect the pairings as it applies so as to render the difference of the scores "as small as possible". That means as small as possible having regard to the operation of the pairing procedures in C. Those procedures do not allow the pairing B v C to be undone.

DJ

I think I understand your point about the relative operation of B and C.

But just to grasp the way C1 to 14 work exactly, I have one last question about your answer. (It's kind of a new question, but I think its relevant to this example):



A B and C are the one score bracket. A is moved to the next score bracket as A cannot be found an opponent in the first score bracket (see C1) and B is paired against C.

Why is A moved down to the next score bracket? C1 says, C12 shall be applied in a case like this (A was already moved down once). Well, in the previous score bracket (which containes just A) no other pairing is possible.

But my question is now: How to procede after C12? Does one return directly to C1 and moves down A (as you indicated)? Or does one go to C13 (which doesn't apply since it's not the lowest score bracket) and afterwards to C14?

In the latter case, p is decreased to zero, therefor the entire score bracket is moved down, giving rise to a new score bracket of A, B, C, D. And now there are different possibilities to proceed (namely A-D, B-E, C-F or A-E, B-C, D-F -- compare my first example).

Christian

Denis_Jessop
08-09-2007, 08:38 PM
I think I understand your point about the relative operation of B and C.

But just to grasp the way C1 to 14 work exactly, I have one last question about your answer. (It's kind of a new question, but I think its relevant to this example):



Why is A moved down to the next score bracket? C1 says, C12 shall be applied in a case like this (A was already moved down once). Well, in the previous score bracket (which containes just A) no other pairing is possible.

But my question is now: How to procede after C12? Does one return directly to C1 and moves down A (as you indicated)? Or does one go to C13 (which doesn't apply since it's not the lowest score bracket) and afterwards to C14?

In the latter case, p is decreased to zero, therefor the entire score bracket is moved down, giving rise to a new score bracket of A, B, C, D. And now there are different possibilities to proceed (namely A-D, B-E, C-F or A-E, B-C, D-F -- compare my first example).

Christian


I was assuming from your example that A had not been moved down form a higher score bracket.

If A had been moved down then, as you say, in pairing the score bracket A B and C, there will be no opponent for A and so the previous score bracket from which A was moved down will be undone so as to see if pairings can be made in the higher score bracket that will enable another player to be moved down (C12) to enable a pairing with either B or C. This last statement is not in the rules but is the only reason to go through this procedure.

If C12 cannot be used to allow another player to drop down then A will remain in that score bracket with B and C and will drop down as described in my other post. Although that seems not to be consistent with C1, that provision must apply so that, if dot point 1 does not enable a pairing to be made with B or C by dropping a player other than A, A must then be regarded as within dot point 3 - "in all other cases". There is no other provision that can be used as C13 specifically applies only to the lowest score bracket and so cannot be used for any other purpose.


DJ

Bartolin
08-09-2007, 10:12 PM
I was assuming from your example that A had not been moved down form a higher score bracket.

If A had been moved down then, as you say, in pairing the score bracket A B and C, there will be no opponent for A and so the previous score bracket from which A was moved down will be undone so as to see if pairings can be made in the higher score bracket that will enable another player to be moved down (C12) to enable a pairing with either B or C. This last statement is not in the rules but is the only reason to go through this procedure.

If C12 cannot be used to allow another player to drop down then A will remain in that score bracket with B and C and will drop down as described in my other post. Although that seems not to be consistent with C1, that provision must apply so that, if dot point 1 does not enable a pairing to be made with B or C by dropping a player other than A, A must then be regarded as within dot point 3 - "in all other cases". There is no other provision that can be used as C13 specifically applies only to the lowest score bracket and so cannot be used for any other purpose.

Firstly, I thought, there was an own score bracket for A in the beginning. So let's say the situation we started from was:

Bracket n: A 2.5
Bracket n+1: B,C 2.0
Bracket n+2: D 1.5
Bracket n+3: E,F 1.0

So A was moved down once (to bracket n+1) because there obviously wasn't any pairing in bracket n. Or am I wrong, assuming that A get's its own score bracket? If so, let's assume for the sake of my argument, that in score bracket n-1 where only players X and Y with 3 points which where paired against each other.

Secondly, I thought, that C12 in my question above applied to bracket n und therefore there was no other possibility than moving down A to B and C. So, step by step, I think the following happens:

1. A is moved down to bracket n+1
2. C1 dot 1 applies so C12 is used next.
3. It's not possible to get another pairing from bracket n (according to C12) since there is only player A in bracket n.

Now the question is, where to continue. You argue, one should directly go back from C12 to C1 and apply C1 dot 3.

4. Since C12 doesn't apply, one returns to C1 and applies C1 dot 3 (as there is no other alternative).
5. A is moved down to bracket n+2, while B is paired against C in bracket n+1.
6. and so on as you described

But isn't there another possibility:

4. Since C12 doesn't apply, one continues down to C13.
5. C13 doesn't apply since it's not the lowest score bracket. Therefore one continues C14.
6. According to C14 one decreases p (for bracket n+1) to zero. Afterwards, the whole bracket n+1 is moved down to n+2 (players A,B,C,D) and one restarts at C1.

Christian

Bill Gletsos
08-09-2007, 11:05 PM
But isn't there another possibility:

4. Since C12 doesn't apply, one continues down to C13.
5. C13 doesn't apply since it's not the lowest score bracket. Therefore one continues C14.
6. According to C14 one decreases p (for bracket n+1) to zero. Afterwards, the whole bracket n+1 is moved down to n+2 (players A,B,C,D) and one restarts at C1.Unfortunately if you go from C1 to C12 and then to C14 then you have never actually calculated p as that is determined in C3.

Bartolin
08-09-2007, 11:25 PM
Unfortunately if you go from C1 to C12 and then to C14 then you have never actually calculated p as that is determined in C3.

Interesting point. Though I don't have calculated p if it is possible to find another pairing in the previous group, either. I mean, C12 refers in sentence 2 to p already -- though one can arrive there without visiting C3 (that is without calculating p).

So you state, the correct sequence is the one described by Denis. I'm not yet absolutely convinced that this strictly follows from the rules, but I accept that the rules are to be interpreted this way.

Thanks

Christian

Bill Gletsos
09-09-2007, 02:09 AM
Actually I think what you are highlighting is just the abysmal wording of the Swiss Rules, especially Section C.

drbean
12-09-2007, 09:49 AM
Bill Gletsos's message about returning to C1 from C12,
rather than going on to C14, being the right path, because p
had not been calculated, something that only happens in C3,
was shocking news.

The first 5 procedures form a group, and are often, but not
always, in invariant order. So perhaps there are some which
can be moved into C1 without change in the pairings that
would be produced.

An inadequacy of the Rules is the lack of a box-and-arrows
diagram, something that would reduce ambiguity, although the
instructions about moving around the procedures provide a
verbal equivalent perhaps.

Here is my interpretation of the ways in and out of the
procedures. The first procedure on each line is the state
you are in. The procedures following it in the brackets are
the possible states you can move to from the present state.



START, [ C1, LAST ],
C1, [ C2, NEXT, C13, C12 ],
C2, [ C3 ],
C3, [ C4 ],
C4, [ C5 ],
C5, [ C6PAIRS ],
C6PAIRS, [ COLORS, C6OTHERS, C7, NEXT ],
COLORS, [ C6OTHERS ],
C6OTHERS, [ NEXT, C1, C7 ],
C7, [ C6PAIRS, C8, C9, REPEATC10 ],
C8, [ C5, C9, C10 ],
C9, [ C4, C10, C11 ],
C10, [ C7, C2, C11 ],
REPEATC10,[ C2, C11 ],
C11, [ C3, C13, C14 ],
C12, [ PREV, C14 ],
C13, [ C7, C1 ],
C14, [ C1, C4, C13 ],
NEXT, [ C1, LAST ],
PREV, [ C1 ],
LAST, [ LAST ],
ERROR, [ ERROR ],


If there is a problem in moving C3 into C1, it would be
because there is an exit into C2 or C3 from a state other
than C1. I have those in C10, REPEATC10 and C11.

I have a question about this 'p' C3 is talking about. Which
p is it? The p of A6 or the p (which I am calling pprime) of
C14. The p of C14 (ie pprime) is reduced in C14, but the p
of C3 and A6 is half the total number in the bracket or the
number moved down.

This has implications for pairing in C6, I think. I think
you have to attempt to pair all the players in S1, but
choose only pprime of them.

Kevin Bonham
17-09-2007, 12:45 AM
And this is a last-round draw I didn't like the look of and overrode:


No Name Feder Rtg Loc

1. Bonham, Kevin TAS 1960
2. Rout, Ian C ACT 1867
3. Frame, Nigel TAS 1794
4. Gibbs, Glen B TAS 1748
5. Dyer, Alastair TAS 1738
6. Ivkovic, Milutin TAS 1488
7. Horton, Vincent TAS 1419
8. Martin, Janice TAS 1397
9. Horton, Russell TAS 1350
10. Richards, Graham TAS 1331
11. Kruup, Andrew TAS 1327
12. Fifield, Andrew TAS 1313
13. Sturges, Tony (Thelston) TAS 1248
14. Krasnicki, Alina TAS


Place No Opponents Colours Float Score


1 1 : 8,6,5,2,4 BWBWB D 4

2-3 2 : 9,7,3,1,6 WBWBW 3.5
4 : 11,3,9,7,1 WBWBW U 3.5

4-6 3 : 10,4,2,6,5 BWBWB D 3
5 : 12,8,1,11,3 BWWBW Ud 3
7 : 14,2,10,4,8 BWBWB 3

7-9 6 : 13,1,11,3,2 WBWBB u 2.5
9 : 2,12,4,8,13 BWBWB uD 2.5
10 : 3,13,7,-,14 WBW-B D 2.5

10-12 8 : 1,5,12,9,7 WBWBW d 2
11 : 4,14,6,5,12 BWBWB u 2
12 : 5,9,8,13,11 WBBWW u 2

13-14 13 : 6,10,14,12,9 BWBBW Ud 1
14 : 7,11,13,-,10 WBW-W Ud 1

SP's draw:


No Name Total Result Name Total

1 Bonham, Kevin [4] : Frame, Nigel [3]
2 Gibbs, Glen B [3.5] : Rout, Ian C [3.5]
3 Horton, Vincent [3] : Dyer, Alastair [3]
4 Ivkovic, Milutin [2.5] : Fifield, Andrew [2]
5 Kruup, Andrew [2] : Richards, Graham [2.5]
6 Horton, Russell [2.5] : Krasnicki, Alina [1]
7 Sturges, Tony (Thelston) [1] : Martin, Janice [2]


I changed the bottom three boards to:

Horton, Russell [2.5] - Richards, Graham [2.5]
Martin, Janice [2] - Krasnicki, Alina [1]
Sturges, Tony [1] - Kruup, Andrew [2]

as this is clearly more B3-compliant. This looks very much like the same issue as in the Australian Championships 2001-2 draw that was the subject of my letter to Gijssen.

Garvinator
17-09-2007, 01:28 AM
Cross Table please Kevin. Seeing what SM5 gives.

Bill Gletsos
17-09-2007, 02:14 AM
Cross Table please Kevin. Seeing what SM5 gives.SP is clearly wrong.

At a quick glance Kevin's manual changes look correct.

Kevin Bonham
17-09-2007, 11:28 PM
Cross Table please Kevin. Seeing what SM5 gives.


No Name Feder Rtg Loc 1 2 3 4 5 6

1. Bonham, Kevin TAS 1960 8:D 6:W 5:W 2:W 4:D 3:D
2. Rout, Ian C ACT 1867 9:W 7:W 3:D 1:L 6:W 4:W
3. Frame, Nigel TAS 1794 10:W 4:W 2:D 6:D 5:L 1:D
4. Gibbs, Glen B TAS 1748 11:W 3:L 9:W 7:W 1:D 2:L
5. Dyer, Alastair TAS 1738 12:D 8:W 1:L 11:D 3:W 7:W
6. Ivkovic, Milutin TAS 1488 13:W 1:L 11:W 3:D 2:L 12:D
7. Horton, Vincent TAS 1419 14:W 2:L 10:W 4:L 8:W 5:L
8. Martin, Janice TAS 1397 1:D 5:L 12:W 9:D 7:L 14:W
9. Horton, Russell TAS 1350 2:L 12:W 4:L 8:D 13:W 10:W
10. Richards, Graham TAS 1331 3:L 13:W 7:L :D 14:W 9:L
11. Kruup, Andrew TAS 1327 4:L 14:W 6:L 5:D 12:D 13:W
12. Fifield, Andrew TAS 1313 5:D 9:L 8:L 13:W 11:D 6:D
13. Sturges, Tony (Thelston) TAS 1248 6:L 10:L 14:W 12:L 9:L 11:L
14. Krasnicki, Alina TAS 7:L 11:L 13:L :W 10:L 8:L

I did not have time to do a full check to see if that was the best possible pairing but it was obvious to me that (at least on my interpretation) the SP pairing given in my previous post was wrong and the one I found was a significant improvement, so I was happy with that.

Garvinator
17-09-2007, 11:58 PM
I did not have time to do a full check to see if that was the best possible pairing but it was obvious to me that (at least on my interpretation) the SP pairing given in my previous post was wrong and the one I found was a significant improvement, so I was happy with that.
Was aware of that :) I also noticed when putting pairings, results into SM5,that it gave different pairings in rounds 4 and 5.

Anyways, SM5 gives:


1 Bonham ( 4 ) - Frame ( 3 ) 1- 3
2 Gibbs ( 3.5) - Rout ( 3.5) 4- 2
3 Horton, Vincent ( 3 ) - Dyer ( 3 ) 7- 5
4 Horton, Russell ( 2.5) - Richards ( 2.5) 9- 10
5 Ivkovic ( 2.5) - Fifield ( 2 ) 6- 12
6 Martin ( 2 ) - Krasnicki ( 1 ) 8- 14
7 Kruup ( 2 ) - Sturges ( 1 ) 11- 13

Kevin Bonham
18-09-2007, 12:16 AM
Looks like I got the colours the wrong way round in Kruup vs Sturges - which would have been avoided had I read the colours for the previous rounds off the "pairing info" field not the crosstable (where it looks like the bold confused me).

Interested to see what SM5 gives for round 5 as the Frame-Dyer pairing (with a player on 3 being floated down then playing a 2 because he has already played both 2.5s) looked rather odd but I wasn't certain it was wrong so let it stand.

Garvinator
18-09-2007, 12:30 AM
Interested to see what SM5 gives for round 5 as the Frame-Dyer pairing (with a player on 3 being floated down then playing a 2 because he has already played both 2.5s) looked rather odd but I wasn't certain it was wrong so let it stand.

1 Frame ( 3 ) - Bonham ( 3.5) 3- 1
2 Gibbs ( 3 ) - Rout ( 2.5) 4- 2
3 Ivkovic ( 2.5) - Dyer ( 2 ) 6- 5
4 Martin ( 2 ) - Horton, Vincent ( 2 ) 8- 7
5 Horton, Russell ( 1.5) - Richards ( 1.5) 9- 10
6 Sturges ( 1 ) - Kruup ( 1.5) 13- 11
7 Fifield ( 1.5) - Krasnicki ( 1 ) 12- 14

drbean
18-09-2007, 04:30 PM
Place No Opponents Colours Float Score


1 1 : 8,6,5,2,4 BWBWB D 4

2-3 2 : 9,7,3,1,6 WBWBW 3.5
4 : 11,3,9,7,1 WBWBW U 3.5

4-6 3 : 10,4,2,6,5 BWBWB D 3
5 : 12,8,1,11,3 BWWBW Ud 3
7 : 14,2,10,4,8 BWBWB 3

7-9 6 : 13,1,11,3,2 WBWBB u 2.5
9 : 2,12,4,8,13 BWBWB uD 2.5
10 : 3,13,7,-,14 WBW-B D 2.5

10-12 8 : 1,5,12,9,7 WBWBW d 2
11 : 4,14,6,5,12 BWBWB u 2
12 : 5,9,8,13,11 WBBWW u 2

13-14 13 : 6,10,14,12,9 BWBBW Ud 1
14 : 7,11,13,-,10 WBW-W Ud 1

SP's draw:


No Name Total Result Name Total

1 Bonham, Kevin [4] : Frame, Nigel [3]
2 Gibbs, Glen B [3.5] : Rout, Ian C [3.5]
3 Horton, Vincent [3] : Dyer, Alastair [3]
4 Ivkovic, Milutin [2.5] : Fifield, Andrew [2]
5 Kruup, Andrew [2] : Richards, Graham [2.5]
6 Horton, Russell [2.5] : Krasnicki, Alina [1]
7 Sturges, Tony (Thelston) [1] : Martin, Janice [2]




Although this format is good for display at the tournament,
it's difficult to make sense of the pairings because there
are no pairing numbers. You have to look at 2 other tables
to see what is happening.

Games::Tournament::Swiss, with some bugs, is at the moment
pairing the top 3 tables the same.

From the 4th it pairs,
4: Ivkovic and Horton, 6 and 9, both on 2.5,
5: Richards and Fifield, 10 and 12, on 2,5 and 2.
6: Martin and Krasnicki, 8 and 14, on 2 and 1.
7: Kruup and Sturges, 11 and 13, on 2 and 1.

Which looks like the same B3 inequality as:




I changed the bottom three boards to:

Horton, Russell [2.5] - Richards, Graham [2.5]
Martin, Janice [2] - Krasnicki, Alina [1]
Sturges, Tony [1] - Kruup, Andrew [2]



The top 3 players are downfloated to the 3rd bracket, where
all players are paired. In the 4th bracket, 6 and 9 are
paired and 10 is downfloated to the penultimate bracket.

The pairing of 10 with 8 means the downfloating of 11 and 12
to the last bracket where there is no possible pairing.
Applying C13 (and C10) to the penultimate bracket gives the
pairing above.

SM's pairing:







1 Bonham ( 4 ) - Frame ( 3 ) 1- 3
2 Gibbs ( 3.5) - Rout ( 3.5) 4- 2
3 Horton, Vincent ( 3 ) - Dyer ( 3 ) 7- 5
4 Horton, Russell ( 2.5) - Richards ( 2.5) 9- 10
5 Ivkovic ( 2.5) - Fifield ( 2 ) 6- 12
6 Martin ( 2 ) - Krasnicki ( 1 ) 8- 14
7 Kruup ( 2 ) - Sturges ( 1 ) 11- 13





Apparently SM is not satisfied with the above analysis, even
though the B3 inequality is the same. It has undone the
pairing of the 4th bracket and floated down 6 instead.

Bill Gletsos
18-09-2007, 05:20 PM
The top 3 players are downfloated to the 3rd bracket, where
all players are paired. In the 4th bracket, 6 and 9 are
paired and 10 is downfloated to the penultimate bracket..This is incorrect.
w = 3, b = 0, p =1, q= 2 and x =1
Hence since x = 1 you need a pairing that is not a colour match.
If you pair 6 & 9 then the downfloat of 10 violates B5.
Pairing 9 and 10 and downfloating 10 causes no violation of B5 and is the preferred pairing.

The only situation where 10 would downfloat would be if 6 had a black colour preference. Lets assume 9 wants B instead of W because his history is BWBWW.
Therefore in this situation w = 2, b = 1, p = 1, q = 2 and x = 0

Now although 6 V 9 still causes the downfloat of 10, it is valid as it is the first match where x = 0.
Thus the procedure is C6, note x = 0 but B% violated, try c7 for all transpositions (still no better pairing), try C8 for all exchanges and repeat all transpositions for each exchange (still no better pairing), apply C9 and drop B6 and B5 respectively.
Now you get 6 V 9 and can ignore the B5 violation in line with c9.

Note if in the original 2.5 score bracket either 9 or 10 had a black colour preference then x would be 0 in which case you would have the pairing 9 v 10 with 6 downfloating.

drbean
18-09-2007, 08:48 PM
This is incorrect...
If you pair 6 & 9 then the downfloat of 10 violates B5.
Pairing 9 and 10 and downfloating 10 causes no violation of
B5 and is the preferred pairing.



There was a discussion about the similar situation of a
heterogeneous group producing a remainder group with only
one player who had already downfloated in the previous round
at http://chesschat.org/showpost.php?p=142260&postcount=158

The failure of Games::Tournament::Swiss to downfloat 6
instead of 10 here touched 2 different bugs.

Those fixed, it is now producing the same pairings as SM.

drbean
21-09-2007, 01:30 PM
Let's go back to Dennis Jessop's earlier analysis of the
original problem with SP's pairing of the original pairing
table, and why C14 allows us to stop at 2 big score groups:
1,2,3,6 and 4,5,8,7,9,10 by reduction of p (pprime) in the
first bracket to 1, avoiding having to put all into one big
bracket as has apparently been done by SP:

I don't have a problem with that, but it means rethinking
a solution I had for avoiding downfloating of pairs that
satisfy B1,2, just because they don't satisfy B3-6.




No Opponents Colours Float Score
1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0







To begin with one looks at 1v2 - an invalid pairing. So 1 &
2 drop to join 3 & 6 to form a score bracket to be a paired
1v3 and 2v6 as 1 cannot play either 2 or 6. After making
further pairings, a situation arises where the last s/b is 7
& 10 who can't be paired. Then C13 is applied. It operates
so that eventually there is a situation in which the
penultimate s/b is 1,2,3,6 and the lowest s/b is
4,5,8,7,9,10. C13 then requires the ps/b to be unpaired and
an attempt made to re-pair it so as to allow a valid pairing
of the ls/b. That cannot be done while p=2. The merging of
the 2 brackets into one can only occur if p becomes zero. As
it stands p = 2 and the only way p can be reduced is on
application of C14. This is the step that I overlooked as
perhaps have everyone else. C14 applies so as to decrease p
by 1. When this is done, the attempt to pair the ps/b with
p=1 is satisfied by pairing 1v3 and leaving 2 and 6 to be
paired with the others. Although the computer may have to do
this by a laborious set of calculations taking it only a
split second, humans can equally see immediately that the
only valid pairing for 10 is 2 or 6. Eventually the pairings
made by SM5 are reached.



This is resulting in the downfloating of 2 and 6 into the
last score bracket, EVEN THOUGH on their run through C6
they had been paired.

I still think this is the correct interpretation,
but it poses a problem for the handling of this pairing
table: http://chesschat.org/showthread.php?t=6900

In particular, as Bill Gletsos noted there, the 12&20
pairing should be accepted, even though B5 is not satisfied.
12 had been upfloated in the previous round and 20 is
downfloating from the 5th bracket. But no player should be
moved down to a lower bracket to satisfy B3-6.

The problem there was that there is no explicit sanctioning
of B5,6 waiving for upfloats, except for homogeneous
remainder groups in C10. I thought we could get around that
by letting through all pairings whether they satisfy B3-6 or
not, just before reducing p (pprime) in C14.

The problem for that solution is that it will prevent us
downfloating 2 and 6 here.

Despite the similarity (B1,2-compliant pairings facing
downfloating), in the first case, we want to downfloat the
pair and in the second we don't.

Bartolin
21-09-2007, 06:47 PM
This is resulting in the downfloating of 2 and 6 into the
last score bracket, EVEN THOUGH on their run through C6
they had been paired.

I still think this is the correct interpretation,
but it poses a problem for the handling of this pairing
table: http://chesschat.org/showthread.php?t=6900

In particular, as Bill Gletsos noted there, the 12&20
pairing should be accepted, even though B5 is not satisfied.
12 had been upfloated in the previous round and 20 is
downfloating from the 5th bracket. But no player should be
moved down to a lower bracket to satisfy B3-6.

The problem there was that there is no explicit sanctioning
of B5,6 waiving for upfloats, except for homogeneous
remainder groups in C10. I thought we could get around that
by letting through all pairings whether they satisfy B3-6 or
not, just before reducing p (pprime) in C14.

The problem for that solution is that it will prevent us
downfloating 2 and 6 here.

Despite the similarity (B1,2-compliant pairings facing
downfloating), in the first case, we want to downfloat the
pair and in the second we don't.

But isn't this difference due to the fact, that in the first case we
decide to downfloat 2 and 6 after applying C13 and C14. That
means, we are unpairing the penultimate score bracket (the first
score bracket) and trying to find better pairings there. At one point
we reduce p to 1 (due to C14) and after that, we are obliged to
downfloat 2 and 6 (since only 1 and 3 are paired in the penultimate
score bracket. I think your idea about checking B1-B2 compliance
before downfloating a pair shouldn't be applied when downfloating
players due to already getting p pairings.

drbean
21-09-2007, 08:04 PM
I want to ask about how to pair a big last bracket that
amalgamates 2 or more of the original bottom brackets, and
think about how to reduce B3 inequality in this bracket.




No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0



In pairing this original pairing table of ggrayggray's that
started this thread, the SM5 pairings of

1 v 3
8 v 2
10 v 6
4 v 5
9 v 7

had score differences of 1 + 1.5 + 2.5 + 0 + 0 = 5, which
was two points better than what Games::Tournament::Swiss was
able to produce.

B3 is the most important of the relative criteria, and even
though it is built in, I believe, to C7-14, going down, it is
not after reaching C13, and we have a big last score bracket.

After reaching this point, the C rules seem to lose
direction. I don't think SP is necessarily violating rules,
but people seem to be unhappy with B3 differences.

Why does SM cope with B3 differences better than SP? If
there is some algorithm or recognized procedure, then the
program could just carry it out until it got the answer.

I wrote a message here somewhere about how the total minimum
difference must be greater than the sum of the score
differences from one score group with an odd number of
players to the next with an odd number, though I didn't see
that here, this time.

If there is no procedure, the program could continue until
this limit or another, better one is reached, or all the possible
valid pairings had been produced. This could take a long
time, if the bracket is large.

Alternatively, the arbiter could specify the difference they
were prepared to accept and the program could continue until
a pairing giving this total difference was reached.

If there was no such pairing, the arbiter would run the
program again with a higher difference.

Another alternative is for the program to keep a running
record of differences produced by the valid pairings it has
found and decide, if the values don't differ much to stop
with a pairing with the best value.

If some other pattern is seen, it might be able to make a
more intelligent choice.

Meanwhile, some ways of pairing last brackets may minimize
differences more than others. The problem with the pairing
table was that the last player, 10 could play none of the
last 5 players above it.

Just as when moving down, in a heterogeneous group, the
downfloating players are paired first because this minimizes
score differences, it makes sense, having reached bottom and
going back up, to pair the bottom players first with those
closest to them in score.

One way to do this is to turn S1 and S2 on their head. Make S1
the bottom players and S2 the rest. And reverse rank S1 and
S2. Then proceed as usual with C6 and C7, etc.

Trying this procedure out, we get:

Next, Bracket 3: 6 2 5 8 4 9 7 10
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=4 Homogeneous.
C4, S1 & S2: 4 9 7 10 & 6 2 5 8

If the bracket is not considered homogeneous we could pair
10 first, and then deal with 7 and 9.

C5, ordered: 10 7 9 4 &
5 8 6 2

The first row is S1, and now 10, with the lowest score, has
first place, in it. The last place in S2, is 2, with the
highest score.

Of course, the process will have to continue through C6 and
C7 until 10 can be paired with 6.

C6, Bracket tables 1 2 3 4 paired. E1 10&6 E2 8&7 E1 9&2 E1 4&5 B3=6

The B3 difference value of 6 is not as good as SM's.

But if we don't stop with this first pairing that satisfies
all the B criteria except B3, but continue to find others
with possibly different B3 values, we get these other
pairings.

C6, E1 10&2 E2 8&7 E1 9&5 E1 4&6 B3=6
C6, E1 10&2 E2 6&7 E1 9&5 E1 4&8 B3=6
C6, E1 10&6 E2 8&7 E1 9&2 E1 4&5 B3=6
C6, E1 10&2 E2 8&7 E1 9&5 E1 4&6 B3=6
C6, E1 10&2 E2 6&7 E1 9&5 E1 4&8 B3=6
C6, E1 9&2 E1 10&6 E1 4&5 E2 8&7 B3=6
C6, E1 10&2 E1 4&6 E1 9&5 E2 8&7 B3=6
C6, E1 10&2 E2 6&7 E1 9&5 E1 4&8 B3=6
C6, E1 9&2 E1 10&6 E1 4&5 E2 8&7 B3=6
C6, E2 2&5 E1 10&6 E1 4&8 E1 9&7 B3=4
C6, E4 8&2 E1 10&6 E1 4&5 E1 9&7 B3=4

The above pairing is the SM pairing.

C6, E1 9&2 E1 10&6 E1 4&5 E2 8&7 B3=6
C6, E1 10&2 E4 8&6 E1 4&5 E1 9&7 B3=4
C6, E1 9&2 E1 10&6 E1 4&5 E2 8&7 B3=6
C6, E1 10&2 E1 4&6 E1 9&5 E2 8&7 B3=6
C6, E1 10&2 E2 6&7 E1 9&5 E1 4&8 B3=6
C6, E2 2&5 E1 10&6 E1 4&8 E1 9&7 B3=4
C6, E4 8&2 E1 10&6 E1 4&5 E1 9&7 B3=4
C6, E1 10&2 E4 8&6 E1 4&5 E1 9&7 B3=4
C6, E1 10&2 E2 6&7 E1 4&8 E1 9&5 B3=6
C6, E4 8&2 E1 10&6 E1 4&5 E1 9&7 B3=4
C6, E1 9&2 E1 10&6 E1 4&5 E2 8&7 B3=6
C6, E1 10&2 E1 4&6 E1 9&5 E2 8&7 B3=6
C6, E1 10&2 E4 8&6 E1 4&5 E1 9&7 B3=4
C6, E1 10&2 E1 4&6 E1 9&5 E2 8&7 B3=6
C6, E4 8&2 E1 10&6 E1 4&5 E1 9&7 B3=4
C6, E1 10&2 E4 8&6 E1 4&5 E1 9&7 B3=4
C6, E1 10&2 E1 4&6 E1 9&5 E2 8&7 B3=6
C6, E1 10&2 E1 4&8 E1 9&5 E2 6&7 B3=6
C6, E2 2&5 E1 10&6 E1 4&8 E1 9&7 B3=4
C6, E2 2&5 E1 10&6 E1 4&8 E1 9&7 B3=4
C6, E1 10&2 E1 4&6 E2 8&7 E1 9&5 B3=6
C6, E1 10&2 E2 6&7 E1 4&8 E1 9&5 B3=6
C6, E1 10&2 E1 4&5 E4 6&8 E1 9&7 B3=4
C6, E1 10&2 E1 4&6 E2 8&7 E1 9&5 B3=6
C6, E1 10&2 E1 4&6 E2 8&7 E1 9&5 B3=6
C6, E1 10&2 E1 4&5 E4 6&8 E1 9&7 B3=4
C6, E1 10&2 E1 4&5 E4 6&8 E1 9&7 B3=4
C6, E1 10&2 E1 9&5 E1 4&8 E2 6&7 B3=6
C6, E1 10&2 E1 9&5 E1 4&8 E2 6&7 B3=6
C6, E1 10&2 E1 4&6 E2 8&7 E1 9&5 B3=6
C6, E1 10&2 E1 4&8 E2 6&7 E1 9&5 B3=6

I stopped here and didn't check all the possible valid
pairings, but it looks like SM's pairing is not going to be
improved on. 4 and 6 are the only B3 values being found.

I wonder how SM decides it has the best pairing.

drbean
23-09-2007, 06:51 PM
The problem we are discussing is the downfloating of pairs
that satisfy B1,2. The Rules say pairs should not be
downfloated to satisfy the other, relative criteria. This
means pairs should not be downfloated in general (always?)?
But as indicated by Dennis Jessop
(http://chesschat.org/showpost.php?p=166345&postcount=7), we
sometimes want to downfloat pairs to prevent amalgamation of
the penultimate and final score brackets.

Another complication is that the application of B5,6 for
upfloats is never explicitly waived (except in the case of
remainder groups).

I brought up the contradiction of the desirable downfloating
above, and the UNdesirable downfloating of
http://chesschat.org/showthread.php?t=6900, as noted by Bill
Gletsos.

Bartolin says:


[I]n the first case we decide to downfloat 2
and 6 after applying C13 and C14. That means, we are
unpairing the penultimate score bracket (the first score
bracket) and trying to find better pairings there. At one
point we reduce p to 1 (due to C14) and after that, we are
obliged to downfloat 2 and 6 (since only 1 and 3 are paired
in the penultimate score bracket. I think your idea about
checking B1-B2 compliance before downfloating a pair
shouldn't be applied when downfloating players due to
already getting p pairings.

The problem with this is that I don't think it will solve
the case of 12 and 20 in
http://chesschat.org/showthread.php?t=6900

This is because of C's lack of an explicit waiving of B5,6 in
the case of upfloats, except for remainder groups.

The idea of checking B1,2 compliance again within C6 is to
let through those pairs that fail B5,6, AFTER C14 has
already reduced pprime. Before C14, C7-12 are efforts to
find a better pairing that will allow us not to float down
such pairs. C14 is a recognition of the failure of these
efforts.

I see the instruction (?) not to float down pairs which
satisfy just B1,2 as an if-all-else-fails measure directed
at the C procedures. 'If C fails to work, whatever you do,
don't downfloat pairs satisfying B1,2.'

I don't think pprime being less than p absolves you of the
responsibility to accept the pair.

At the same time, we want to downfloat 2 and 6 in the
penultimate score bracket in
http://chesschat.org/showthread.php?t=6619.

Perhaps the way to do it, is to start reviewing pairs which
we are about to float down after we reach C14 in the
bracket, EXCEPT in the case of re-pairing the penultimate
bracket.

This way we can have our cake and eat it, satisfying both
Dennis Jessop and Bill Gletsos.

An alternative might be to slip in a waiving of B5,6 for
upfloats somewhere before C14.

This might be more in the spirit of the Rules. It still
won't solve the problem posed by the instruction (?) not to
downfloat pairs which can be paired.

But perhaps I am being overly sensitive. Perhaps,'To comply
with these criteria, transpositions or even exchanges may be
applied, but no player should be moved down to a lower score
bracket).' is not an instruction never to downfloat pairs.
Perhaps we are free to downfloat pairs to pair lower
brackets.

As in, 'Try to find another pairing in the penultimate score
bracket which will allow a pairing in the lowest score
bracket. If in the penultimate score bracket p becomes zero
(i.e. no pairing can be found which will allow a correct
pairing for the lowest score bracket) then the two lowest
score brackets are joined into a new lowest score bracket.'

Bartolin
23-09-2007, 09:18 PM
I still have to digest what you said.

But maybe we have different understandings of 'p'.


The idea of checking B1,2 compliance again within C6 is to
let through those pairs that fail B5,6, AFTER C14 has
already reduced pprime. Before C14, C7-12 are efforts to
find a better pairing that will allow us not to float down
such pairs. C14 is a recognition of the failure of these
efforts.

I see the instruction (?) not to float down pairs which
satisfy just B1,2 as an if-all-else-fails measure directed
at the C procedures. 'If C fails to work, whatever you do,
don't downfloat pairs satisfying B1,2.'

I don't think pprime being less than p absolves you of the
responsibility to accept the pair.

I agree that the "instruction" from B, not to float down pairs which
satisfy just B1,B2, should be read as a kind of "fallback" instruction.

But I'm not sure, I would like to sign your last sentence. In an earlier
post (http://chesschat.org/showpost.php?p=167525&postcount=34)
you asked:


I have a question about this 'p' C3 is talking about. Which
p is it? The p of A6 or the p (which I am calling pprime) of
C14. The p of C14 (ie pprime) is reduced in C14, but the p
of C3 and A6 is half the total number in the bracket or the
number moved down.

This has implications for pairing in C6, I think. I think
you have to attempt to pair all the players in S1, but
choose only pprime of them.

Maybe also http://chesschat.org/showpost.php?p=168386&postcount=45
is relevant to this discussion.

I think, the 'p' from C14 is the same 'p' as in C3 and A6. I think, 'p' is
computed once (in C3) according to A6, afterwards it is (if necessary)
reduced by C14. Whenever C6 or C12 or Cx speak about 'p', they are
referring to the current value of 'p'.

I don't see, why we need a 'pprime' at all. But it's quite possible that I
miss something here.

But back to your argument:


I don't think pprime being less than p absolves you of the
responsibility to accept the pair.

I think, the idea of C14 is just that: If it's not possible to get 'p' pairings
than ignore one player (initially the last one in S1) and try to get at least
'p-1', 'p-2' ... pairings. The last player are to be downfloated afterwards.


But perhaps I am being overly sensitive. Perhaps,'To comply
with these criteria, transpositions or even exchanges may be
applied, but no player should be moved down to a lower score
bracket).' is not an instruction never to downfloat pairs.
Perhaps we are free to downfloat pairs to pair lower
brackets.

I would understand it as follows:

Do not downfloat players only because you can't generate pairings which
are fully compliant with B3-B6. Use the instructions from C7 and below to try to find
pairings which fulfill B3-B6 as good as possible. If C fails to generate pairings
which comply with all of B3-B6, accept the set of pairing which is the
best, relatively spoken. But if C explicitely says to downfloat players, then do so!

What would that mean for the examples in question:



No Opponents Colours Float Score
1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0




To begin with one looks at 1v2 - an invalid pairing. So 1 &
2 drop to join 3 & 6 to form a score bracket to be a paired
1v3 and 2v6 as 1 cannot play either 2 or 6. After making
further pairings, a situation arises where the last s/b is 7
& 10 who can't be paired. Then C13 is applied. It operates
so that eventually there is a situation in which the
penultimate s/b is 1,2,3,6 and the lowest s/b is
4,5,8,7,9,10. C13 then requires the ps/b to be unpaired and
an attempt made to re-pair it so as to allow a valid pairing
of the ls/b. That cannot be done while p=2. The merging of
the 2 brackets into one can only occur if p becomes zero. As
it stands p = 2 and the only way p can be reduced is on
application of C14. This is the step that I overlooked as
perhaps have everyone else. C14 applies so as to decrease p
by 1. When this is done, the attempt to pair the ps/b with
p=1 is satisfied by pairing 1v3 and leaving 2 and 6 to be
paired with the others. Although the computer may have to do
this by a laborious set of calculations taking it only a
split second, humans can equally see immediately that the
only valid pairing for 10 is 2 or 6. Eventually the pairings
made by SM5 are reached.

If p becomes 1 for the penultimate score bracket we get the
pairing 1-3. We are now allowed (obliged?) to downfloat 2 and 6
to the last score bracket. This is because C6 says:


Pair the highest player of S1 against the highest one of S2, the second highest one of S1 against the second highest on e of S2, etc. If now p pairings are obtained in compliance with B1 and B2 the pairing of this score bracket is considered complete.

* in case of a homogeneous score bracket: remaining players are moved down to the next score bracket. With this score bracket restart at C1.

At this point we only care about one pairing, that is 1-3. All other players (2,6)
are moved down.

Second example:



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
8 18,5,3 BWB D 3
2
1 11,10,6 WBW u 2.5
3-8
2 12,7,9 BWB 2
4 14,9,7 BWB 2
5 15,8,13 WBW 2
7 17,2,4 WBW 2
9 19,4,2 WBW 2
17 7,14,12 BWB D 2
9-14
3 13,6,8 WBW U 1.5
6 16,3,1 BWB 1.5
10 20,1,15 BWB 1.5
11 1,12,16 BWB d 1.5
15 5,20,10 BWW 1.5
16 6,19,11 WBW 1.5
15-17
13 3,18,5 BWB 1
19 9,16,14 BWB 1
20 10,15,18 WBB 1
18
12 2,11,17 WBW U 0.5
19-20
14 4,17,19 WBW 0
18 8,13,20 WBW 0


Here we have 20 and 12 in one score bracket. The pairing 20-12
violates B5 for upfloats, but that's no problem because of the 'fallback'
clause in B. Since C doesn't require us to downfloat 20 and 12 explicitely,
the fallback clause in B takes over and we accept 20-12.

drbean
25-09-2007, 07:53 PM
We have been discussing internal inconsistencies in the
Rules that have become apparent in the work to implement
routines that downfloat pairs that satisfy the Absolute
Criteria and Relative Criteria to pair the final bracket,
and that do NOT downfloat pairs at other times, even when
they violate some of the Relative Criteria.

Bartolin says:




I agree that the "instruction" from B, not to float down
pairs which satisfy just B1,B2, should be read as a kind of
"fallback" instruction.

But I'm not sure, I would like to sign your last sentence.
(which was that the instruction (?) not to downfloat pairs
maintains its force even when pprime is less than p,
post-C14.)



OK. I guess you are right there. If we can't downfloat pairs
under any circumstances, then we can't avoid the undesirable
amalgamation of the penultimate and final brackets of the
pairing table, the situation that started this thread off.

C14 seems the appropriate mechanism to downfloat pairs. It
is a last resort. C13 describes this situation, with its
reference to p becoming zero.




I think, the idea of C14 is just that: If it's not possible
to get 'p' pairings than ignore one player (initially the
last one in S1) and try to get at least 'p-1', 'p-2' ...
pairings. The last player are to be downfloated afterwards.



You mean the last player in S1 is the one to be downfloated,
or paired in a (heterogeneous?) remainder group?

The fact that we have had to decrease pprime, other than in
the situation above of reaching C14 through C13, means that
it has not been possible to pair one or more of the members
of S1. These are the players to ignore. We also face the
undesirable downfloating of these players 2 score places in
the case of a heterogeneous group. We have not been able to
pair them with an S2 member.

In the case that we are reducing pprime in the penultimate
score group having reached C14 through C13, in a situation
where every pair satisfies all the criteria, the last player
of S1 seems to be the appropriate one, I agree.

Note:

[Meanwhile, in the general case, reaching C14 in ways other
than through C13, perhaps there is never a member of S1 who
it has been impossible to pair. That is, perhaps the only
way to reach C14 is through C13.

If there had been a player which it was impossible to pair,
they would have been found in C1.

No, it might be possible to pair A individually and pair B
individually, but not pair both at the same time. Such
situations will not be found in C1, I think.]

But, you raise the question of which players to ignore. We
have p players in S1, but only pprime are going to be
paired. Which ones?

The ones forming the first pprime pairs. But if we seek
partners for only the first pprime players we face the
danger of not getting pprime pairs, or of overlooking the
best pairing. After C14, the procedure resumes at C4 where
S1 and S2 are ordered.

Perhaps the first player in S1 is unpairable. If we take
only the first pprime members of S1 and try to pair them,
then pprime is going to reach zero, before we find pprime
matches.

Unless I make a mistake in my reasoning, we should maintain
S1 at p players and seek to pair with S2 but take only the
first pprime pairs.

Do we also need to consider the different ways we can choose
pprime pairs from p pairs?




Do not downfloat players only because you can't
generate pairings which are fully compliant with B3-B6. Use
the instructions from C7 and below to try to find pairings
which fulfill B3-B6 as good as possible. If C fails to
generate pairings which comply with all of B3-B6, accept the
set of pairing which is the best, relatively spoken. But if
C explicitely says to downfloat players, then do so!



I agree, but we need a procedure to do this that is as
consistent with C (in spirit and letter) as possible and in
a way that is clean and easily understood.

The problem is that C never explicitly waives application of
B5,6 for upfloats except in the case of heterogeneous groups.
This was the reason that Games::Tournament::Swiss
downfloated 20 and 12, the problem you noted at
http://chesschat.org/showthread.php?t=6900

Therefore I propose that we introduce a waiving of ALL
Relative Criteria (actually, it's only a waiving of B5,6 for
upfloats in homogeneous groups) as a pre-C14 procedure.

I don't consider this a big rewriting of C. Perhaps it is a
mere oversight that this is not part of C. Perhaps the
formulators considered it common sense and didn't require
explicit mention.

The fact the more experienced here have not commented
perhaps means they don't see it as an important issue.

We waive all Relative Criteria application and run through
all the C6-13 stages again. And then we go to C14.

What do you think?

Bartolin
27-09-2007, 11:42 PM
I think, the idea of C14 is just that: If it's not possible
to get 'p' pairings than ignore one player (initially the
last one in S1) and try to get at least 'p-1', 'p-2' ...
pairings. The last player are to be downfloated afterwards.

You mean the last player in S1 is the one to be downfloated,
or paired in a (heterogeneous?) remainder group?

I wrongly assumed, the sequence C14 -> C4 -> C5 -> C6 worked another way (I thought the number of players in S1 had to be decreased, not only the number of pairings to be made. Then, I thought, we had to do some exchanges (D2) until the right players are in S1.) But after reading your reply, I changed my view. Now I think your description is correct.


The fact that we have had to decrease pprime, other than in
the situation above of reaching C14 through C13, means that
it has not been possible to pair one or more of the members
of S1. These are the players to ignore. We also face the
undesirable downfloating of these players 2 score places in
the case of a heterogeneous group. We have not been able to
pair them with an S2 member.

In the case that we are reducing pprime in the penultimate
score group having reached C14 through C13, in a situation
where every pair satisfies all the criteria, the last player
of S1 seems to be the appropriate one, I agree.

You are right, this is a very special case.


Note:

[Meanwhile, in the general case, reaching C14 in ways other
than through C13, perhaps there is never a member of S1 who
it has been impossible to pair. That is, perhaps the only
way to reach C14 is through C13.

If there had been a player which it was impossible to pair,
they would have been found in C1.

No, it might be possible to pair A individually and pair B
individually, but not pair both at the same time. Such
situations will not be found in C1, I think.]

I agree


But, you raise the question of which players to ignore. We
have p players in S1, but only pprime are going to be
paired. Which ones?

The ones forming the first pprime pairs. But if we seek
partners for only the first pprime players we face the
danger of not getting pprime pairs, or of overlooking the
best pairing. After C14, the procedure resumes at C4 where
S1 and S2 are ordered.

Perhaps the first player in S1 is unpairable. If we take
only the first pprime members of S1 and try to pair them,
then pprime is going to reach zero, before we find pprime
matches.

Unless I make a mistake in my reasoning, we should maintain
S1 at p players and seek to pair with S2 but take only the
first pprime pairs.

As I said above, I think you got it right.


Do we also need to consider the different ways we can choose pprime pairs from p pairs?

At first, I thought it would be okay to always choose the first pprime pairs. At least I don't see anything in the rules which implies that it is more important to pair the higest rated players.

But at second thought, I'm wondering whether there could arise problems with score differences (B3) later.

Let's take the following example:

Player A 5.0 (played against B, C, D, E, G)
Player B 4.5 (played (against A, C)
Player C 4.5 (played (against A, B, D, E)
Player D 4.0 (played against A, C)
Player E 4.0 (played against A, C)
Player F 4.0
Player G 3.5 (played against A)
Player H 3.5

Now, A is moved down to B and C. All of them are moved down to D, E, F to form one large score bracket. It's not possible to get three pairings with S1 (A,B,C) and S2 (D,E,F) since both, A and C must play against F. Therefore (after some transpositions) we reach C14 and pprime is decreased to 2.

Now the first shot is:

A-D -- bad pairing
B-E -- good pairing
C-F -- good pairing

This satisfies C6, therefore A and D are moved down, and we get

B-E, C-F, A-G, D-H

But I guess, it would be better to have

A-F, B-E, C-G, D-H

because this way the score difference for the highest player A is minimized (cp. http://www.chesschat.org/showpost.php?p=167073&postcount=18 and http://www.chesschat.org/showpost.php?p=66586&postcount=12).

Hmm, sounds difficult.



Do not downfloat players only because you can't
generate pairings which are fully compliant with B3-B6. Use
the instructions from C7 and below to try to find pairings
which fulfill B3-B6 as good as possible. If C fails to
generate pairings which comply with all of B3-B6, accept the
set of pairing which is the best, relatively spoken. But if
C explicitely says to downfloat players, then do so!


I agree, but we need a procedure to do this that is as
consistent with C (in spirit and letter) as possible and in
a way that is clean and easily understood.

The problem is that C never explicitly waives application of
B5,6 for upfloats except in the case of heterogeneous groups.
This was the reason that Games::Tournament::Swiss
downfloated 20 and 12, the problem you noted at
http://chesschat.org/showthread.php?t=6900

Therefore I propose that we introduce a waiving of ALL
Relative Criteria (actually, it's only a waiving of B5,6 for
upfloats in homogeneous groups) as a pre-C14 procedure.

I guess, you mean, it's a waiving of B5,6 for all groups, ecxept for homogeneous remainder groups? Homogeneous remainder groups are handled by C10.


I don't consider this a big rewriting of C. Perhaps it is a
mere oversight that this is not part of C. Perhaps the
formulators considered it common sense and didn't require
explicit mention.

The fact the more experienced here have not commented
perhaps means they don't see it as an important issue.

We waive all Relative Criteria application and run through
all the C6-13 stages again. And then we go to C14.

What do you think?

In general, I like your proposal. Only one problem bothers me -- but maybe you have a good solution for this one as well? Let's assume we have the following situation:

Player A -- 3 points
Player B -- 2.5 points, upfloated once (last round)
Player C -- 2.5 points, upfloated once (2 rounds ago)

This should fall through C6-C13, because there is no waiving for B5,B6 in this case. But after waiving all Relative Criteria, one get's A-B which is inferior to A-C (A-B violates B5 while A-C only violates B6).

Maybe we shouldn't just waive all Relative Criteria at once for this second run through C6-C13, but instead waive them one at a time? Of course that would increase the amount of computations necessary to get the correct pairings. My former idea to keep a kind of measure of B3-B6 violation was meant to have an criterion to decide which pairing to choose in such a situation, instead of going through C6-C13 a second time.

Garvinator
30-09-2007, 05:01 PM
Since this thread seems to have gone quiet.

Came from tournament today. 5 round tournament. Pairing info after round 4


No Opponents Colours Float Score

1 : 6,4,2,3 BWBW D 3.5
2 : 7,3,1,5 WBWB D 3.5

4 : 9,1,7,10 WBWB D 3

3 : 8,2,6,1 BWBB U 2
5 : 10,6,8,2 BWBW U 2
7 : 2,10,4,8 BWBW 2

6 : 1,5,3,9 WBWB D 1
8 : 3,9,5,7 WBWB 1
9 : 4,8,10,6 BWBW U 1
10 : 5,7,9,4 WBWW U 1


Swiss Perfect, Swiss Master 5 and Swiss Manager were all giving different round 5 pairings.

Bill Gletsos
30-09-2007, 06:05 PM
Swiss Perfect, Swiss Master 5 and Swiss Manager were all giving different round 5 pairings.And they were?

Garvinator
30-09-2007, 06:40 PM
And they were?
Apologies. Thought I would leave that until had a few replies to see what each respondent gave as answers without seeing the computer pairings.

Anyways:

Swiss Perfect:


No Name Loc Total Result Name Loc Total

1 KINDER, Jessica (2) 1624 [3.5] : EDWARDS, Regina (4) 1409 [3]
2 BONELL, Sarah (8) 646 [1] : JULE, Alexandra (1) 1756 [3.5]
3 LYONS, Kieran C (3) 1466 [2] : LEE, Isabelle (10) 588 [1]
4 YUM, Winnie (9) 634 [1] : SIMMONDS, Leteisha (5) 985 [2]
5 KARIBASIC, Melanie (6) 820 [1] : YUM, Jenny (7) 781 [2]

Swiss Master 5.5


1 Jessica Kinder ( 3.5) - Regina Edwards ( 3 ) 2- 4
2 Jenny Yum ( 2 ) - Alexandra Jule ( 3.5) 7- 1
3 Kieran Lyons ( 2 ) - Isabelle Lee ( 1 ) 3- 10
4 Winnie Yum ( 1 ) - Leteisha Simmonds ( 2 ) 9- 5
5 Melanie Karibasic ( 1 ) - Sarah Bonell ( 1 ) 6- 8

Swiss Manager


Bo.SNo. Name Pts Res. Pts Name SNo.

1 2 Jessica Kinder 3½ 3 Regina Edwards 4
2 5 Leteisha Simmonds 2 3½ Alex Jule 1
3 3 Kieran Lyons 2 1 Winnie Yum 9
4 6 Melanie Karibasic 1 2 Jenny Yum 7
5 8 Sarah Bonell 1 1 Isabelle Lee 10

I went with the Swiss Manager pairings as they 'seemed' most correct. I chose them as B3 has higher priority than B5. Also there was a small amount of 'swiss logic' in my decision ie Number 1 seed meets seeds 2,3,4,5 and 6 with the SwissManager draw.

Now that I am at home and have all the time to decide, I think the SM5 pairings are correct. If the draw should be 7 v 1, I am not certain how SwissManager came up with its pairings.

Btw, all three programs come up with the same pairings as above (for their respective programs) if round 5 is regarded as not the final round.

I think SP has just combined the bottom two score groups and paired top half v bottom half.

Bill Gletsos
30-09-2007, 08:36 PM
Having had a quick look at it I'm virtually certain that the pairings generated by SM5 are correct.

drbean
02-10-2007, 03:21 PM
I wrote about the consequences of using C14 to reduce p in
the penultimate score bracket and prevent amalgamation with
the final square bracket. This use needs to be reconciled
with the relative criteria rubric that says not to downfloat
players to satisfy the relative criteria. A complication is
the fact the procedures in C don't explicitly allow repeat
upfloat check dropping in all circumstances, even though the
spirit of the rubric requires it.


I wrongly assumed, the sequence C14 -> C4 ->
C5 -> C6 worked another way (I thought the number of players
in S1 had to be decreased, not only the number of pairings
to be made. Then, I thought, we had to do some exchanges
(D2) until the right players are in S1.) But after reading
your reply, I changed my view. Now I think your description
is correct.


What I was arguing there was that the reduction of p
(pprime) in 14 doesn't mean that we should change S1 and S2,
That is, it doesn't mean each time through C14, that S1 is
one player shorter.

No, I think the pairings being considered after each time
through C14 would be the same, just our standards would be
lower. That is, we would be prepared to choose a pairing
that before we rejected.

But this is just my interpretation and perhaps the intent of
the formulators was to force pairing of shorter S1s and
longer S2s.


At first, I thought it would be okay to
always choose the first pprime pairs. At least I don't see
anything in the rules which implies that it is more
important to pair the higest rated players.

But at second thought, I'm wondering whether there could
arise problems with score differences (B3) later.

Let's take the following example:

Player A 5.0 (played against B, C, D, E, G)
Player B 4.5 (played (against A, C)
Player C 4.5 (played (against A, B, D, E)
Player D 4.0 (played against A, C)
Player E 4.0 (played against A, C)
Player F 4.0
Player G 3.5 (played against A)
Player H 3.5

Now, A is moved down to B and C. All of them are moved down
to D, E, F to form one large score bracket. It's not
possible to get three pairings with S1 (A,B,C) and S2
(D,E,F) since both, A and C must play against F.


S1 is A,B,C and S2 is D,E,F

DEF
DFE
EDF
EFD
FDE
FED

Whichever way we transpose DEF, one of D and E will have to be
first or last. But this will produce an impossible pairing, because A
and C are first and last.


Therefore (after some transpositions) we
reach C14 and pprime is decreased to 2.


Now the first shot is:

A-D -- bad pairing
B-E -- good pairing
C-F -- good pairing

This satisfies C6, therefore A and D are moved down, and we
get

B-E, C-F, A-G, D-H


G had already played A in your pairing table, but it
doesn't change the force of the argument, I think.

With B3 score difference of 0.5 + 0.5 + 1.5 + 0.5 = 3




But I guess, it would be better to have

A-F, B-E, C-G, D-H


With B3 score difference of 1 + 0.5 + 1 + 0.5 = 3, the same
but with less variation in the differences. I guess we need
to calculate the variance too.



because this way the score difference for the highest player A is minimized (cp. http://www.chesschat.org/showpost.php?p=167073&postcount=18 and http://www.chesschat.org/showpost.php?p=66586&postcount=12).


That is good material. I think the case of one double
downfloat over two brackets being preferred over two single
downfloats from one score bracket to the next is an
inadequacy in the C procedures. A pairing with higher p
(pprime) will always be preferred over a pairing with a
lower.

Perhaps instead of taking the first pairing we find after
going back to C6,7,9 from C14, we choose the best B3 one?
Another idea might be to not consider the bracket as
homogeneous. That way, we pair just A first, and avoid it
being floated down again.

This seems to be argument against my interpretation of the p
in C3 as being the same p that it was all the way through!

Perhaps we could do both, if this was going to be a double
downfloat.

Anway, it would be good to have your problem, which is
another response to Kevin Bonham in the first page above, as
a pairing table that could be fed to the pairing programs.
And the 2005 Kevin Bonham pairing table of the second page
above would be good to have as a test that a program would
have to pass.

I have some tests for Games::Tournament::Swiss, but
obviously not enough.



In general, I like your proposal [for an explicit dropping
of B5,6 for upfloats for all brackets, not just homogeneous
remainder groups]. Only one problem bothers me -- but maybe
you have a good solution for this one as well? Let's assume
we have the following situation:

Player A -- 3 points
Player B -- 2.5 points, upfloated once (last round)
Player C -- 2.5 points, upfloated once (2 rounds ago)

This should fall through C6-C13, because there is no waiving
for B5,B6 in this case. But after waiving all Relative
Criteria, one get's A-B which is inferior to A-C (A-B
violates B5 while A-C only violates B6).

Maybe we shouldn't just waive all Relative Criteria at once
for this second run through C6-C13, but instead waive them
one at a time? Of course that would increase the amount of
computations necessary to get the correct pairings.


I guess this means we should do this the same way as in C9,
first B6 for upfloats is dropped (while B5 remains in force)
and then the whole C6,7,8 cycle is performed again, and then
B5 for upfloats is dropped, and no upfloat or downfloat
criteria are considered at all.



My former idea to keep a kind of measure of B3-B6 violation was
meant to have an criterion to decide which pairing to choose
in such a situation, instead of going through C6-C13 a
second time.



I still think that is a good idea that needs to be kept in
mind. At the moment the way I calculate the D1
transpositions is to skip all those which clearly will not
result in p pairs. That is if ABC cannot be paired with DEF,
because A and D have the same preferences, then, there is no
point in checking whether ABC can be paired with DFE, the
next transposition.

The algorithm instead skips ahead to EDF. That would have to
change to keep violation measures for all transpositions.



Another thing I am doing when checking a pairing is stopping
when I find the first violation of the first criterion in
force.

So a violation of B1 stops the test and the next
transposition is found in C7. If none of the pairs have met
before (B1) but both have the same absolute preference (B2),
the test is also stopped and the next transposition is
found. The reason the pairing failed and the first pair
which failed are reported.

The next thing that can go wrong with a pairing is not
having enough preference-satisfying matches (B4). The number
of pairs where preferences are not satisfied is reported.

All this is good for the computer and good for the user. They
can see what is most importantly wrong with that pairing.
But what about if the first thing wrong is a violation of
the B5,6 criteria?

The order of waiving of the criteria is None B6Down B5Down
B6Up B5Up All, but in what order should the tests for the
criteria in force be carried out?

I have been carrying them out in the above order, but
perhaps I should be carrying them out in reverse order. That
is the most important test is that which will be waived
last. So the user wants to hear about a B5Up violation,
rather than a B6Down violation. Or do they?

But I also have another, though related, problem.

The different problem is where to drop upfloat criteria
testing. I thought that before C14 was the right place. This
was the last possible place if the problem in
http://chesschat.org/showpost.php?p=167972&postcount=2
was to be solved.

But perhaps it is better to follow the way the C procedures
sequence the relaxing of B4 and B5,6.

There is no overlapping of this relaxation. It is all or
nothing. No increase in x is allowed until ALL of the
downfloat criteria checks (and upfloat criteria too in the
case of heterogeneous groups) have been dropped.

All C6,7,8,9,10 cycles are completed before C11 is reached
the first time.

I could imagine there being some more complex, phased
dropping of the checks, especially seeing that the
preference concept includes a recovery mechanism, allowing
a buffering of preference awarding in one round with relaxation
of B4 and a preferential granting of these preferences through
B2 in the next.

There is no such buffering in repeat downfloat check
relaxation.

This clear sequence of criteria check dropping suggests ALL
B5,6 checks should be dropped before C11, which implements
B4 relaxation, rather than before C14.

However, I suddenly wonder if it is possible for there to be
score differences and thus the possibility of upfloats in a
bracket, unless it is a heterogeneous bracket.

Perhaps this hand-wringing over the failure to explicitly
waive upfloat checks except in the case of heterogeneous
groups is unnecessary.

Bill Gletsos
02-10-2007, 04:12 PM
Actually in the example I believe the correct pairing is A-F and B-D with C and E moving down which gives a complete pairing of:

A-F, B-D, C-G and E-H

Bartolin
02-10-2007, 09:26 PM
Actually in the example I believe the correct pairing is A-F and B-D with C and E moving down which gives a complete pairing of:

A-F, B-D, C-G and E-H

Yes, I think that is even better than A-F, B-E, C-G, D-H. But I don't fully understand how to identify this pairing as the correct one. I think, the procedure from C (Pairing procedure) is clear until one has one score bracket with S1=A,B,C S2=D,E,F. Then, after some tries with C7-C12 p is decreased by 1 according to C14. But how to procede after p becomes 2?

The first set of pairing which is tried after C14 should be A-D, B-E, C-F, which gives 2 (=p) correct pairings and should therefore be okay according to the rules in C. But it would mean to downfloat player A, which seems wrong.

Bill Gletsos
02-10-2007, 09:32 PM
The first set of pairing which is tried after C14 should be A-D, B-E, C-F, which gives 2 (=p) correct pairings and should therefore be okay according to the rules in C.Incorrect. The B section still applies.

Bartolin
02-10-2007, 10:13 PM
Incorrect. The B section still applies.

Well, I expected to get an answer like this. But I don't see how an arbiter (or a computer program) is expected to incorporate section B in the pairing procedure. Is there another way than just taking all possible sets of pairings of S1=A,B,C S2=D,E,F with p=2 in account and comparing their "B-compliance"?

If that's the only way, I wonder what the so called "procedure" in section C is invented for. (I mean, if I can't use it to identify the correct set of pairings, what is it good for?)

And a second question: Why is A-F, B-E, C-G, D-H worse than A-F, B-D, C-G, E-H, exactly? The "B-compliance" of those sets of pairings is identically, I think (though I would say intuitively that the higher player (D) shouldn't be downfloated, I don't see a rule like this in section B). Is there something in section C, why the first one is worse?

Bill Gletsos
02-10-2007, 10:38 PM
Well, I expected to get an answer like this. But I don't see how an arbiter (or a computer program) is expected to incorporate section B in the pairing procedure.Where as it may well be difficult for an arbiter to determine it for a large event, it shouldnt be for a correctly written program.

Is there another way than just taking all possible sets of pairings of S1=A,B,C S2=D,E,F with p=2 in account and comparing their "B-compliance"?No, because compliance with the B section is important. Of course when the score group includes players on different scores then B3 is going to have an impact obn determining the correct pairings.

If that's the only way, I wonder what the so called "procedure" in section C is invented for. (I mean, if I can't use it to identify the correct set of pairings, what is it good for?)Of course you can use section C to determine the correct set of pairings. You just have to do it correctly.

And a second question: Why is A-F, B-E, C-G, D-H worse than A-F, B-D, C-G, E-H, exactly? The "B-compliance" of those sets of pairings is identically, I think (though I would say intuitively that the higher player (D) shouldn't be downfloated, I don't see a rule like this in section B). Is there something in section C, why the first one is worse?Thats because you simply got it wrong. The transposition of DEF is done in an order as determined by section D and FDE occurs before FED. With respect to the information provided in the example, although they are identical with regards section B, FED occurs first and is therefore the correct one.

Bartolin
02-10-2007, 11:26 PM
Is there another way than just taking all possible sets of pairings of S1=A,B,C S2=D,E,F with p=2 in account and comparing their "B-compliance"?

No, because compliance with the B section is important. Of course when the score group includes players on different scores then B3 is going to have an impact obn determining the correct pairings.

Thanks for this reply! Actually I supposed it to work this way, but I wanted to be sure. I hope my understanding of the interaction of sections B and C is a bit better now.

On the other hand, my earlier question about the interpretation of B3 (see http://chesschat.org/showpost.php?p=166888&postcount=14) is not fully answered, I think. At least it's not obvious, why B-E, C-F, A-G, D-H is worse than A-F, B-D, C-G, E-H according to section B. The sum of point differences is 3 in both cases and the first solution is reached first by following the procedure from C.



Of course you can use section C to determine the correct set of pairings. You just have to do it correctly.

Point taken.



And a second question: Why is A-F, B-E, C-G, D-H worse than A-F, B-D, C-G, E-H, exactly? The "B-compliance" of those sets of pairings is identically, I think (though I would say intuitively that the higher player (D) shouldn't be downfloated, I don't see a rule like this in section B). Is there something in section C, why the first one is worse?
Thats because you simply got it wrong. The transposition of DEF is done in an order as determined by section D and FDE occurs before FED. With respect to the information provided in the example, although they are identical with regards section B, FED occurs first and is therefore the correct one.

I see.

That implies that drbean is right with his interpretation of C14, that a reduction of p doesn't result in a reduction of the number of players in S1 (this number stays at the original value of p) but in a reduction of pairings to be made (see http://chesschat.org/showpost.php?p=169636&postcount=56).

Bill Gletsos
03-10-2007, 12:09 AM
On the other hand, my earlier question about the interpretation of B3 (see http://chesschat.org/showpost.php?p=166888&postcount=14) is not fully answered, I think. At least it's not obvious, why B-E, C-F, A-G, D-H is worse than A-F, B-D, C-G, E-H according to section B. The sum of point differences is 3 in both cases and the first solution is reached first by following the procedure from C.Incorrect.

As was noted S1 = A, B, C and S2 = D, E, F

That can be paired with p = 2. G isnt in the picture at this stage.

Bartolin
03-10-2007, 12:26 AM
Incorrect.

As was noted S1 = A, B, C and S2 = D, E, F

That can be paired with p = 2. G isnt in the picture at this stage.

Well, it can. But there are different possible pairings.

Following procedure C (namely C6) the first set of pairings we get is: A-D, B-E, C-F. With p=2, this would be okay according to C6. The pairings found (B-E, C-F) don't violate B1, B2 and let's assume they don't violate B4, B5 and B6 neither.

So, the only reason, not to take this two pairings (B-E, C-F) and to downfloat A and D must be, that there are better pairings with respect to B3.

You said, the correct pairing for this score bracket is A-F, B-D, downfloating C and E.

Why is A-F, B-D (downfloating C and E) better than B-E, C-F (downfloating A and D)?

Bill Gletsos
03-10-2007, 01:18 AM
Well, it can. But there are different possible pairings.

Following procedure C (namely C6) the first set of pairings we get is: A-D, B-E, C-F. With p=2, this would be okay according to C6. The pairings found (B-E, C-F) don't violate B1, B2 and let's assume they don't violate B4, B5 and B6 neither.

So, the only reason, not to take this two pairings (B-E, C-F) and to downfloat A and D must be, that there are better pairings with respect to B3.

You said, the correct pairing for this score bracket is A-F, B-D, downfloating C and E.

Why is A-F, B-D (downfloating C and E) better than B-E, C-F (downfloating A and D)?Because the pairing of player A is the most important and A-F minimizes the difference for player A, whereas downfloating him clearly increases the difference.

Bartolin
03-10-2007, 01:35 AM
Because the pairing of player A is the most important and A-F minimizes the difference for player A, whereas downfloating him clearly increases the difference.

That was my primary thought as well. But is there something in the rules giving special relevance to the higher player explicitely? What if the original situation would have been (A has 4.5 points only, like B and C):

Player A 4.5 (played against B, C, D, E, G)
Player B 4.5 (played (against A, C)
Player C 4.5 (played (against A, B, D, E)
Player D 4.0 (played against A, C)
Player E 4.0 (played against A, C)
Player F 4.0
Player G 3.5 (played against A)
Player H 3.5

Would A-F, B-D (downfloating C and E) still be better than B-E, C-F (downfloating A and D)? (Remember, following procedure C, we get B-E, C-F first!)

Generally speaking: Should the relative criteria (B3-B6) be read as: A violation of one of the relative criteria for a higher player is always treated as being worse than a violation of the same criterion for one (or more?) lower player(s)?

Sorry for insisting, but I really want to understand the general principles.

Bill Gletsos
03-10-2007, 02:03 AM
That was my primary thought as well. But is there something in the rules giving special relevance to the higher player explicitely? What if the original situation would have been (A has 4.5 points only, like B and C):

Player A 4.5 (played against B, C, D, E, G)
Player B 4.5 (played (against A, C)
Player C 4.5 (played (against A, B, D, E)
Player D 4.0 (played against A, C)
Player E 4.0 (played against A, C)
Player F 4.0
Player G 3.5 (played against A)
Player H 3.5I see that as irrelevant as your aim is to still pair player A first as he is the highest ranked.

Would A-F, B-D (downfloating C and E) still be better than B-E, C-F (downfloating A and D)? (Remember, following procedure C, we get B-E, C-F first!)Irrelevant as explained above.

Generally speaking: Should the relative criteria (B3-B6) be read as: A violation of one of the relative criteria for a higher player is always treated as being worse than a violation of the same criterion for one (or more?) lower player(s)?Generally that is probably true however you have to be careful with wording like that.
e.g. Player 1 and 2 are on 4 points and both have a W colour pref. They have also played each other. Players 3 and 4 are on 3.5 and have colour prefs of W, B respectively. Now p = 2, w = 3, b = 1, q = 2 and x = 1. As such 1 V 3 and 2 V 4 are the correct pairings even though 1 v 3 violates B4 whilst 1 V 4 and 2 V 3 violates B4 bit on a lower board. This is because x <> 0 allows for B4 to be ignored and it is ignored for the first occurence.

Bartolin
03-10-2007, 03:30 AM
What I was arguing there was that the reduction of p
(pprime) in 14 doesn't mean that we should change S1 and S2,
That is, it doesn't mean each time through C14, that S1 is
one player shorter.

No, I think the pairings being considered after each time
through C14 would be the same, just our standards would be
lower. That is, we would be prepared to choose a pairing
that before we rejected.

But this is just my interpretation and perhaps the intent of
the formulators was to force pairing of shorter S1s and
longer S2s.

The idea of decreasing S1 and increasing S2 seems wrong to me. For instance,
for heterogeneous groups there is no mechanism to get a specific player back
to S1, once he went to S2. Furthermore D2 speaks only about S1 and S2 having
the same number of members or S2 having one member more than S1. The
possibility of a large difference (e.g. S1 with 2 players, S2 with 5 players) isn't
recognized there.

Also Bill Gletsos' post (http://chesschat.org/showpost.php?p=169737&postcount=61)
fits with unaltered S1 and S2 groups (he speaks about S2 = DEF after decreasing p
to 2).


With B3 score difference of 1 + 0.5 + 1 + 0.5 = 3, the same
but with less variation in the differences. I guess we need
to calculate the variance too.

Or we have to give more weight to score differences for the higher players (see http://chesschat.org/showpost.php?p=169754&postcount=65).


The order of waiving of the criteria is None B6Down B5Down
B6Up B5Up All, but in what order should the tests for the
criteria in force be carried out?

Reading this, I'm asking myself: Why gives section C this strange order
(B6Down, B5Down, B6Up, B5Up) instead of B6 (Down and Up), B5 (Down and Up).
Reading section B, I would expect that.

At the moment it seems to me, that B6 and B5 for downfloats are pretty irrelevant in this
context. They are only relevant if we have players from different score groups in
one bracket. Usually (except for an almagated last score bracket) we arrive at such
a situation only when players are moved down from a higher bracket. But in this
case they are expected to get a downfloat, and if it's their second downfloat, that
can't be hindered (otherwise they would/should have been paired in the higher
score bracket)! Obviously a violation of B5 or B6 for downfloats isn't reason
enough to fiddle with the pairings made in the previous score bracket.
Therefore it would be wrong to dismiss a pairing only because a player from a
higher score bracket gets a second downfloat. I guess, this holds true even for
remainder groups, which arise after pairing only some of the downfloated players.

This leaves us with B6 and B5 only relevant in case of an amalgated last score
bracket. But does it really make a difference there? Or are those B6,B5 checks
irrelevant at all? Can you imagine a situation where they make a difference?

In all other cases, the "relevant" order is of waiving is "B6Up, B5Up, Others", as
one would expect after reading section B.

But why does C10 only speaks about homogeneous remainder groups? Maybe,
the intention of C10 isn't primary to waive B6 and B5, but to give a procedure for
those homogeneous remainder groups. Maybe it's main purpose is to give the
"unusual" instruction to redo one of the pairings of the previous score bracket,
before waiving B6 and B5? Maybe waiving B6 and B5 is mentioned explicitely,
because it should be applied after fiddling with the last score bracket?

If one reads C9, C10 and C11 this way, a new interpretation arises: It follows from
section B, that the criteria B3-B6 should be waived step by step (starting
with B6). C10 and C11 only give some instructions for special cases (namely for
homogeneous remainder groups). The purpose of C9 isn't clear to me at the
moment. The "natural" points for waiving then seem to be
* for B6 and B5 before C11,
* for B4 before C12,
* for B3 before C14

What do you think about that?



This clear sequence of criteria check dropping suggests ALL
B5,6 checks should be dropped before C11, which implements
B4 relaxation, rather than before C14.


That's my conclusion as well.


However, I suddenly wonder if it is possible for there to be
score differences and thus the possibility of upfloats in a
bracket, unless it is a heterogeneous bracket.

Perhaps this hand-wringing over the failure to explicitly
waive upfloat checks except in the case of heterogeneous
groups is unnecessary.

I think so, too. But "heterogeneous groups" must include those, which have
players from different score brackets, but are treated as homogeneous for
some reasons (e.g. A3).

Bartolin
03-10-2007, 06:17 PM
At the moment it seems to me, that B6 and B5 for downfloats are pretty irrelevant in this
context. They are only relevant if we have players from different score groups in
one bracket. Usually (except for an almagated last score bracket) we arrive at such
a situation only when players are moved down from a higher bracket. But in this
case they are expected to get a downfloat, and if it's their second downfloat, that
can't be hindered (otherwise they would/should have been paired in the higher
score bracket)! Obviously a violation of B5 or B6 for downfloats isn't reason
enough to fiddle with the pairings made in the previous score bracket.
Therefore it would be wrong to dismiss a pairing only because a player from a
higher score bracket gets a second downfloat. I guess, this holds true even for
remainder groups, which arise after pairing only some of the downfloated players.

This leaves us with B6 and B5 only relevant in case of an amalgated last score
bracket. But does it really make a difference there? Or are those B6,B5 checks
irrelevant at all? Can you imagine a situation where they make a difference?


Thinking about this a bit longer, I still can't imagine a situation where a pairing
is rejected only because a player receives a second downfloat (B6 or B5).
Once a player is moved down to a lower score bracket, he needs to receive a
downfloat. So, violations of B6 or B5 for downfloats could only be avoided by
moving down another player in the first place. But section C doesn't say we should
do such a thing (but note, that it tells us to do such a thing in the special case of
a homogeneous remainder group regarding upfloats (C10)).

Is there anything else C9 is used for, except for giving us instructions, to drop
B6 and B5 as soon as possible (since violations of those rules must be accepted).

[Maybe this is old hat for you, but until yesterday I didn't see it that way.]

Bill Gletsos
03-10-2007, 06:43 PM
Thinking about this a bit longer, I still can't imagine a situation where a pairing
is rejected only because a player receives a second downfloat (B6 or B5).
Once a player is moved down to a lower score bracket, he needs to receive a
downfloat. So, violations of B6 or B5 for downfloats could only be avoided by
moving down another player in the first place. But section C doesn't say we should
do such a thing (but note, that it tells us to do such a thing in the special case of
a homogeneous remainder group regarding upfloats (C10)).

Is there anything else C9 is used for, except for giving us instructions, to drop
B6 and B5 as soon as possible (since violations of those rules must be accepted).

[Maybe this is old hat for you, but until yesterday I didn't see it that way.]Whilst pairing a score group if it is correctly determined that a player will float down to the next group even if this causes a violation of B5 or B6, then obviously when pairing the score group to which he has been downfloated it is clear he must still be in violation of B5 or B6. That is simply to be expected and as such when you reach C6 for this score group it does not mean the pairing in this score bracket is invalidated and someone else should have been downfloated.
After all if someone else who would not have violated B5 or B6 could have been moved down (without violating B1, B2, B3 or B4) in the original score group then this should have happened in the first place because if this is indeed possible then the pairing of the original score group was incorrect.

Bartolin
03-10-2007, 07:10 PM
Whilst pairing a score group if it is correctly determined that a player will float down to the next group even if this causes a violation of B5 or B6, then obviously when pairing the score group to which he has been downfloated it is clear he must still be in violation of B5 or B6. That is simply to be expected and as such when you reach C6 for this score group it does not mean the pairing in this score bracket is invalidated and someone else should have been downfloated.
After all if someone else who would not have violated B5 or B6 could have been moved down (without violating B1, B2, B3 or B4) in the original score group then this should have happened in the first place because if this is indeed possible then the pairing of the original score group was incorrect.

So one has to check violations of B5 and B6 for downfloaters while computing pairings in the original score bracket? That's interesting, because I originally assumed, one should check the pairings only. But now, thinking about it, it sounds quite obvious.

But one doesn't have to check for violations of B5 and B6 across more than two score brackets, does one? Let's assume the following example

Player A 2.0
Player B 2.0 (played against E, F)
Player C 2.0 (played against D)
Player D 1.5 (played against C, F)
Player E 1.5 (Played against B)
Player F 1.0 (played against D, downfloated last round)
Player G 1.0
Player H 0.5

* First try would be A-B, moving down C
* -> C-E, moving down D
* -> D-G, moving down F (though there will be a violation of B5 for player F, but there is no other possibilty in the this or the previous score bracket)
* -> F-H results in a violation of B5 for player F

Does one need to go backwards and try to find other pairings in order to avoid the violation of B5 for player F?

If we change the pairing in the first score bracket to A-C, we will get pairings without any violation of B5:
* A-C, moving down B
* -> B-D, moving down E
* -> E-F, moving down G, no violation of B5
* -> G-H

Is the second pairing (A-C, B-D, E-F, G-H) correct or the first one (A-B, C-E, D-G, F-H)?

Bartolin
03-10-2007, 07:27 PM
Does one need to go backwards and try to find other pairings in order to avoid the violation of B5 for player F?

Speaking with myself, it seems to me that C 9 says, one should just accept this violation of B5. Or am I wrong?

Denis_Jessop
03-10-2007, 08:18 PM
I think I may have said this before but the situation re B and C is that B prescribes Pairing Criteria and C prescribes Pairing Procedures.

As both relate to pairing, they interact and have to be applied together. It is not right to apply one and ignore the other as the two deal with different aspects of the same thing - achieving a set of valid pairings in accordance with both B and C (and A, for that matter).

DJ

Bartolin
03-10-2007, 08:25 PM
I think I may have said this before but the situation re B and C is that B prescribes Pairing Criteria and C prescribes Pairing Procedures.

As both relate to pairing, they interact and have to be applied together. It is not right to apply one and ignore the other as the two deal with different aspects of the same thing - achieving a set of valid pairings in accordance with both B and C (and A, for that matter).

DJ

Yes, you definitely said that before. I'm trying to keep it in mind, but I still don't fully understand, what it means for this situation. If I accept the pairing F-H above, that seems to be a violation of B, if I reject that pairing and start changing pairings in the first score bracket, that seems to be a violation of C (but maybe it is not?). I don't see how to comply with both of them.

Bill Gletsos
04-10-2007, 01:02 AM
So one has to check violations of B5 and B6 for downfloaters while computing pairings in the original score bracket? That's interesting, because I originally assumed, one should check the pairings only. But now, thinking about it, it sounds quite obvious.

But one doesn't have to check for violations of B5 and B6 across more than two score brackets, does one? Let's assume the following example

Player A 2.0
Player B 2.0 (played against E, F)
Player C 2.0 (played against D)
Player D 1.5 (played against C, F)
Player E 1.5 (Played against B)
Player F 1.0 (played against D, downfloated last round)
Player G 1.0
Player H 0.5

* First try would be A-B, moving down C
* -> C-E, moving down D
* -> D-G, moving down F (though there will be a violation of B5 for player F, but there is no other possibilty in the this or the previous score bracket)
* -> F-H results in a violation of B5 for player F

Does one need to go backwards and try to find other pairings in order to avoid the violation of B5 for player F?

If we change the pairing in the first score bracket to A-C, we will get pairings without any violation of B5:
* A-C, moving down B
* -> B-D, moving down E
* -> E-F, moving down G, no violation of B5
* -> G-H

Is the second pairing (A-C, B-D, E-F, G-H) correct or the first one (A-B, C-E, D-G, F-H)?Given your examples totally ignore the issue of colour prefs and the calculation of x and its implact on the possible pairings, then there may well be issues that are being overlooked and assumprions made on your part that are incorrect.
However for simplicitys sake lets assume that each and every pairing generated in your first try is a colour match. In that case the first try is correct.

However if A wants white and B wants white and C wants black then A V C is correct and B downfloats then the next pairing is B v D and E downfloats.
Now if E wants white and F wants black then irrespective of G's colour pref x = 0 so E V F is correct (its a colour match) and G downfloats is ok, but if E wants white and F wants white and G wants black then x still equals 0, so E V G is correct (as E V F violates x = 0) and F downfloats. This B5 violation is irrelevant as the B4 violation caused by an E V F pairing is more significant (and is not allowed due to x being = 0).

This however could be further complicated if E wants white and F wants white and G wants white. In this case x = 1 so a non coolur match is permiited. Therefore E V F is the correct pairing even though E V F violates B4 (as does E V G), beacuse E V F does not violate B5 where as E V G does.

As such the issue of colour prefs is highly significant all the way through the pairings and really should not be omitted from theoretical examples as doing so can render the examples completely worthless.

Unless of course some kind soul goes to the trouble of highlighting their importance. :whistle:

Bartolin
04-10-2007, 01:17 AM
Given your examples totally ignore the issue of colour prefs and the calculation of x and its implact on the possible pairings, then there may well be issues that are being overlooked and assumprions made on your part that are incorrect.
However for simplicitys sake lets assume that each and every pairing generated in your first try is a colour match. In that case the first try is correct.

Yes, I should have noted, that I assumed colour prefs to have no impact (to keep the example as simple as possible). That means, my example should be designed as follows:

Player A 2.0 -- wants White
Player B 2.0 -- wants Black -- (played against E, F)
Player C 2.0 -- wants Black -- (played against D)
Player D 1.5 -- wants White -- (played against C, F)
Player E 1.5 -- wants White -- (Played against B)
Player F 1.0 -- wants Black -- (played against D, downfloated last round)
Player G 1.0 -- wants Black
Player H 0.5 -- wants White

Therefore as well A-B, C-E, D-G, F-H (my first try) as A-C, B-D, E-F, G-H (my second variation) fulfills all colour preferences.

So, I understood you correctly, that the violation of B5 doesn't matter in this case and A-B, C-E, D-G, F-H is the correct pairing?


However if A wants white and B wants white and C wants black then A V C is correct and B downfloats then the next pairing is B v D and E downfloats.
Now if E wants white and F wants black then irrespective of G's colour pref x = 0 so E V F is correct (its a colour match) and G downfloats is ok, but if E wants white and F wants white and G wants black then x still equals 0, so E V G is correct (as E V F violates x = 0) and F downfloats. This B5 violation is irrelevant as the B4 violation caused by an E V F pairing is more significant (and is not allowed due to x being = 0).

Yes, I guess I understand your reasoning.


As such the issue of colour prefs is highly significant all the way through the pairings and really should not be omitted from theoretical examples as doing so can render the examples completely worthless (unless of course someone goes to the trouble of highlighting the importance of the colour prefs).

Well, I'm sorry for causing this trouble. But I tried to keep my example as simple as possible by designing it in a way that colour preferences were unimportant. I really should have noted that.

Bill Gletsos
04-10-2007, 03:09 AM
Yes, I should have noted, that I assumed colour prefs to have no impact (to keep the example as simple as possible). That means, my example should be designed as follows:

Player A 2.0 -- wants White
Player B 2.0 -- wants Black -- (played against E, F)
Player C 2.0 -- wants Black -- (played against D)
Player D 1.5 -- wants White -- (played against C, F)
Player E 1.5 -- wants White -- (Played against B)
Player F 1.0 -- wants Black -- (played against D, downfloated last round)
Player G 1.0 -- wants Black
Player H 0.5 -- wants White

Therefore as well A-B, C-E, D-G, F-H (my first try) as A-C, B-D, E-F, G-H (my second variation) fulfills all colour preferences.

So, I understood you correctly, that the violation of B5 doesn't matter in this case and A-B, C-E, D-G, F-H is the correct pairing?Correct.

Yes, I guess I understand your reasoning.Good.

Well, I'm sorry for causing this trouble. But I tried to keep my example as simple as possible by designing it in a way that colour preferences were unimportant. I really should have noted that.The problem is that trying to simplify it is more likely to cause confusion rather than promote understanding as the colour prefs can cause signficant differences to the pairings often for different reasons as I explained in my previous post.

Bartolin
04-10-2007, 03:17 AM
So, I understood you correctly, that the violation of B5 doesn't matter in this case and A-B, C-E, D-G, F-H is the correct pairing?
Correct.
Thanks again for this information!


The problem is that trying to simplify it is more likely to cause confusion rather than promote understanding as the colour prefs can cause signficant differences to the pairings often for different reasons as I explained in my previous post.

I'll try to give better examples in the future ;)

drbean
04-10-2007, 10:49 AM
I think I may have said this before but the situation re B and C is that B prescribes Pairing Criteria and C prescribes Pairing Procedures.

As both relate to pairing, they interact and have to be applied together. It is not right to apply one and ignore the other as the two deal with different aspects of the same thing - achieving a set of valid pairings in accordance with both B and C (and A, for that matter).

DJ

If this is the case, then the C Procedures become a heuristic rather than an algorithm. A cynical definition of a heuristic is something that doesn't work ;-)

If making pairing completely mechanical requires more than tweaking C, then does the idea that C represents the flow of control through the pairing process have to be abandoned too?

If the flow of control is more complicated than can be represented in a C-like diagram, the relation of the parts to the whole becomes complex.

The style of http://www.fide.com/official/handbook.asp?level=C0402B may be more programmable. I wonder why the style of http://www.fide.com/official/handbook.asp?level=C0401 became the norm.

Is it because it is more mechanical, or more precise?

Games::Tournament::Swiss may be boxing above its class.

drbean
04-10-2007, 08:53 PM
Since this thread seems to have gone quiet.

Came from tournament today. 5 round tournament. Pairing info after round 4


No Opponents Colours Float Score

1 : 6,4,2,3 BWBW D 3.5
2 : 7,3,1,5 WBWB D 3.5

4 : 9,1,7,10 WBWB D 3

3 : 8,2,6,1 BWBB U 2
5 : 10,6,8,2 BWBW U 2
7 : 2,10,4,8 BWBW 2

6 : 1,5,3,9 WBWB D 1
8 : 3,9,5,7 WBWB 1
9 : 4,8,10,6 BWBW U 1
10 : 5,7,9,4 WBWW U 1


Swiss Perfect, Swiss Master 5 and Swiss Manager were all giving different round 5 pairings.

I just want to note the differences with the original
pairing table that started the thread. That table was:



No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


Was round 4 replayed with a different pairing? Was the
original pairing wrong?

Pairing this new table posed a number of problems for
Games::Tournament::Swiss that are still not completely
solved.

Round 5: 1 2 (3.5), 4 (3), 3 5 7 (2), 6 8 9 10 (1),
Next, Bracket 1: 1 2
C1, NOK. 1 2 Bracket 1. Floating 1 Down, 2 Down, [1] 1 2 => [2] 2 1 4
Next, Bracket 2: 2 1 4

The 2 players in the first bracket are unpairable, so are
floated down to Bracket 2. But there, 1 is still unpairable.
According to C1, we are supposed to continue at C12,
repairing the first bracket. But there are no repairings of
the first bracket.

There is no indication in C12 where to go if the previous
bracket cannot be repaired. Games::Tournament::Swiss was
continuing to C14, which resulted in the whole second
bracket being floated down, even though 4 was pairable with
2.

I think it should proceed to C12 only if some players
were NOT moved down from the higher score bracket.

It should just follow the 'or else' clause in this case:
"[I]n all other cases: move this player down to the next
score bracket." So, 1 should just fall straight through
Bracket 2, on its way from Bracket 1 to Bracket 3.

In Bracket 2:



C1, NOK. 1 Bracket 2. Floating 1 Down, [2] 2 4 => [3] 1 3 5 7
C2, x=1
C3, p=1 Homogeneous.
C4, S1 & S2: 2 & 4
C5, ordered: 2 &
4
C6, B5Down, table 1: 2 NOK. Floated Down 1 rounds ago
C7, last transposition
C8, last S1,S2 exchange:
C9, Dropping B6 for Downfloats
C4, S1 & S2: 2 & 4
C5, ordered: 2 &
4
C6, B5Down, table 1: 2 NOK. Floated Down 1 rounds ago
C7, last transposition
C8, last S1,S2 exchange:
C9, Dropping B5 for Downfloats
C4, S1 & S2: 2 & 4
C5, ordered: 2 &
4
C6, 1 paired. E4 2&4 B3 M=0.50 V=0.25
C6others: no non-paired players


Seeing that there are only 2 players in the bracket, perhaps
the B6,5 tests can be waived. Only B1,2 apply because none
of the Relative Criteria are supposed to cause a downfloating.

We will see in Bracket 3 however, that we may sometimes want to
float down the 2 players of a 2-player group.

Next, Bracket 3: 1 3 5 7



C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=1 Heterogeneous.
C4, S1 & S2: 1 & 3 5 7
C5, ordered: 1 &
3 5 7
C6, B1a: table 1 NOK
C7, 5 3 7
C6, B5Down, table 1: 1 NOK. Floated Down 1 rounds ago
C7, 7 3 5
C6, B5Down, table 1: 1 NOK. Floated Down 1 rounds ago
C7, last transposition
C9, Dropping B6 for Downfloats
C4, S1 & S2: 1 & 3 5 7
C5, ordered: 1 &
3 5 7
C6, B1a: table 1 NOK
C7, 5 3 7
C6, B5Down, table 1: 1 NOK. Floated Down 1 rounds ago
C7, 7 3 5
C6, B5Down, table 1: 1 NOK. Floated Down 1 rounds ago
C7, last transposition
C9, Dropping B5 for Downfloats
C4, S1 & S2: 1 & 3 5 7
C5, ordered: 1 &
3 5 7
C6, B1a: table 1 NOK
C7, 5 3 7
C6, 1 paired. E4 5&1 B3 M=1.00 V=1.25
C6others: Bracket 3's Remainder Group: 3 7
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1 Homogeneous.
C4, S1 & S2: 3 & 7
C5, ordered: 3 &
7
C6, 1 paired. E1 3&7 B3 M=0.67 V=0.83
C6others: no non-paired players


The fact there was a remainder group is important, because
we are going to find the last bracket was unpairable.
Following C13, we need to find a pairing in Bracket 3 which
will allow the pairing of the last bracket.

But do we need to re-pair 3 and 7 or 1,3,5 and 7?

Next, Bracket 4: 6 8 9 10

Here in the last bracket, as before with 10, one player,
here 9, has already played all the others. So following
C13, we try to repair bracket 3.



C1, NOK. 9
C13, Undoing Bracket 3 matches. Re-pairing Bracket 3. p=1. Bracket 3: 3 7 & Bracket 4: 6 8 9 10
C7, last transposition
C8, last S1,S2 exchange:
C10,re-pairing Player 1 in Bracket 3
C7, 5 7 3
C6, 1 paired. E4 5&1 B3 M=1.00 V=1.25
C6others: Bracket 3's Remainder Group: 7 3
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1 Homogeneous.
C4, S1 & S2: 3 & 7
C5, ordered: 3 &
7
C6, 1 paired. E1 3&7 B3 M=0.67 V=0.83


Note that, to repair the remainder group, it is going
through C10 trying to repair 1, the downfloated player. But
5&1 and 3&7 are the same pairing as before. This looks like
a mistake.

At the moment Games::Tournament::Swiss is joining all of the
players in Bracket 3 and 4, which appears to be what Swiss
Manager and Swiss Perfect are doing too.

Its first 15 alternative pairings are:

C6, 4 paired. E1 6&7 E1 8&1 E4 5&9 E1 3&10 SwissPerfect
C6, 4 paired. E1 6&10 E1 8&1 E4 5&9 E1 3&7
C6, 4 paired. E1 6&10 E1 8&1 E4 7&9 E1 3&5
C6, 4 paired. E1 6&7 E1 8&10 E4 1&9 E1 3&5
C6, 4 paired. E1 6&7 E1 8&1 E4 5&9 E1 3&10
C6, 4 paired. E1 6&10 E1 8&1 E1 3&9 E4 7&5
C6, 4 paired. E1 6&10 E1 8&1 E4 7&9 E1 3&5
C6, 4 paired. E1 6&7 E1 8&10 E1 3&9 E4 1&5 SwissManager
C6, 4 paired. E1 6&7 E1 8&10 E4 1&9 E1 3&5
C6, 4 paired. E1 6&10 E1 8&1 E1 3&9 E4 5&7
C6, 4 paired. E1 6&10 E1 8&1 E4 5&9 E1 3&7
C6, 4 paired. E1 6&7 E1 8&1 E1 3&10 E4 9&5
C6, 4 paired. E1 6&7 E4 1&9 E1 8&10 E1 3&5
C6, 4 paired. E1 6&7 E1 3&9 E1 8&10 E4 1&5
C6, 4 paired. E1 6&7 E4 1&9 E1 8&10 E1 3&5

The Swiss Master 5.5 is not in the first 15.


1 Jessica Kinder ( 3.5) - Regina Edwards ( 3 ) 2- 4
2 Jenny Yum ( 2 ) - Alexandra Jule ( 3.5) 7- 1
3 Kieran Lyons ( 2 ) - Isabelle Lee ( 1 ) 3- 10
4 Winnie Yum ( 1 ) - Leteisha Simmonds ( 2 ) 9- 5
5 Melanie Karibasic ( 1 ) - Sarah Bonell ( 1 ) 6- 8

Its second table is 1 and 7. It looks like it has only
joined Bracket 3's remainder group with Bracket 4.

[Last minute note: SwissManager could also have just
joined Bracket 3's remainder group. It paired 1&5.]

But perhaps it used C10 to pair 1 and 7 and leave 3 and 5 to
be joined with the last bracket.

Note this question of whether to repair the whole of the
second-last bracket when it is homogeneous is one bartolin
and I debated at:
Linkname: 'penultimate bracket' ambiguous - Chess Chat Forum
URL: http://chesschat.org/showthread.php?t=6856



I went with the Swiss Manager pairings as they 'seemed' most correct. I chose them as B3 has higher priority than B5. Also there was a small amount of 'swiss logic' in my decision ie Number 1 seed meets seeds 2,3,4,5 and 6 with the SwissManager draw.

Now that I am at home and have all the time to decide, I think the SM5 pairings are correct. If the draw should be 7 v 1, I am not certain how SwissManager came up with its pairings.


The question of who to pair with 1 is important, I agree.
But it looks like 1 could still be paired with 5, even if
you just joined the last 2 score brackets. So the existence
of such a pairing isn't proof that the last 2 brackets
weren't joined.



Btw, all three programs come up with the same pairings as above (for their respective programs) if round 5 is regarded as not the final round.

I think SP has just combined the bottom two score groups and paired top half v bottom half.

I think that is the case, seeing it is the first pairing
obtained by Games::Tournament::Swiss, which did join the
bottom 2 brackets.

Bill Gletsos
04-10-2007, 09:18 PM
Since this thread seems to have gone quiet.

Came from tournament today. 5 round tournament. Pairing info after round 4


No Opponents Colours Float Score

1 : 6,4,2,3 BWBW D 3.5
2 : 7,3,1,5 WBWB D 3.5

4 : 9,1,7,10 WBWB D 3

3 : 8,2,6,1 BWBB U 2
5 : 10,6,8,2 BWBW U 2
7 : 2,10,4,8 BWBW 2

6 : 1,5,3,9 WBWB D 1
8 : 3,9,5,7 WBWB 1
9 : 4,8,10,6 BWBW U 1
10 : 5,7,9,4 WBWW U 1


Swiss Perfect, Swiss Master 5 and Swiss Manager were all giving different round 5 pairings.
I just want to note the differences with the original
pairing table that started the thread. That table was:



No Opponents Colours Float Score


1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


Was round 4 replayed with a different pairing? Was the
original pairing wrong?I believe they are totally different events.

Bill Gletsos
04-10-2007, 09:36 PM
Pairing this new table posed a number of problems for
Games::Tournament::Swiss that are still not completely
solved.

Round 5: 1 2 (3.5), 4 (3), 3 5 7 (2), 6 8 9 10 (1),
Next, Bracket 1: 1 2
C1, NOK. 1 2 Bracket 1. Floating 1 Down, 2 Down, [1] 1 2 => [2] 2 1 4
Next, Bracket 2: 2 1 4

The 2 players in the first bracket are unpairable, so are
floated down to Bracket 2. But there, 1 is still unpairable.
According to C1, we are supposed to continue at C12,
repairing the first bracket. But there are no repairings of
the first bracket.

There is no indication in C12 where to go if the previous
bracket cannot be repaired. Games::Tournament::Swiss was
continuing to C14, which resulted in the whole second
bracket being floated down, even though 4 was pairable with
2.This is just wrong.

After 1 & 2 downfloat to the 3 point score group you end up with the homogeneous score group of 1, 2 and 4.

It is clear that player 1 cannot play either players 2 or 4.
Hence according to C1 player 1 drops to the next score group and you proceed to pair players 2 and 4
S1 = 2, S2 = 4
w =0, b =2, p =1, q =1, x = 1

Hence 2 V 4 is a valid pairing.

The remaining pairings logically follow on after that.

As I noted previosuly the SM5 pairings are correct.

Garvinator
04-10-2007, 10:26 PM
I believe they are totally different events.
DrBean, Bill is correct. Two completely different events.

Bartolin
05-10-2007, 12:41 AM
The 2 players in the first bracket are unpairable, so are
floated down to Bracket 2. But there, 1 is still unpairable.
According to C1, we are supposed to continue at C12,
repairing the first bracket. But there are no repairings of
the first bracket.

There is no indication in C12 where to go if the previous
bracket cannot be repaired. Games::Tournament::Swiss was
continuing to C14, which resulted in the whole second
bracket being floated down, even though 4 was pairable with
2.

I think it should proceed to C12 only if some players
were NOT moved down from the higher score bracket.

It should just follow the 'or else' clause in this case:
"[I]n all other cases: move this player down to the next
score bracket." So, 1 should just fall straight through
Bracket 2, on its way from Bracket 1 to Bracket 3.

See also Denis Jessop's post
http://www.chesschat.org/showpost.php?p=167123&postcount=29
which highlights that one should return to C1 dot 3 after a failure of C12.

drbean
13-10-2007, 11:19 PM
The order of downfloat check waiving is: (B6Down, B5Down,
B6Up, B5Up)



Reading this, I'm asking myself: Why gives section C this
strange order (B6Down, B5Down, B6Up, B5Up) instead of B6
(Down and Up), B5 (Down and Up).


I guess the problems of making the procedures manageable
forces this arrangement. Otherwise, you are forced to mix
C9 and C10 together. C10 is about heterogeneous groups
with downfloating players who are being paired with players
who face upfloating again. That's why upfloat check waiving
is handled there.



At the moment it seems to me, that B6 and B5 for downfloats
are pretty irrelevant in this context. They are only
relevant if we have players from different score groups in
one bracket.


I haven't thought much about floats in homogeneous brackets.

I thought the reason for the float criterion was that the
person who had a downfloat had had an easier game than the
others at the same level in its bracket.

And the opposite for upfloats.

To have to give someone an easier game is unavoidable, but
for someone to have 2 easier games than others would not be
good if it can be avoided.

It's the same kind of thinking as avoiding awarding byes
twice to the same person.

Because there can be people with different float histories
with the same scores, I think the 2 issues are separate, but
not independent.

In a homogeneous bracket, someone who had an upfloat in the
round before should be paired with someone who had a
downfloat, you might argue.

I think I was mistaken about what a downfloat was. I thought
that pairing someone who had had a downfloat with someone
who had had an upfloat (or a downfloat?), was to float them
again, EVEN IF THEY HAD THE SAME SCORES.

If this is not the case, and if in a homogeneous group there
is no floating, there is no need to carry out B5,6 checks at
all in C6 in homogeneous groups. And so we don't have to
waive them in C9, or C10. We can go straight to x increases
in C11.



Usually (except for an almagated last score bracket) we
arrive at such a situation only when players are moved down
from a higher bracket. But in this case they are expected to
get a downfloat, and if it's their second downfloat, that
can't be hindered (otherwise they would/should have been
paired in the higher score bracket)!


They've only been downfloated because C14 has been reached
and p has been reduced. That is, there was no other
alternative to downfloating them, as you said.

But if these downfloat repetitions are unavoidable, then
there is no need to carry out B5,6Down checks in a
heterogeneous group. And there is no need to waive them in
C9.

(But would this reasoning apply for downfloats 2 rounds before?)

In this case, C9 becomes irrelevant to the pairings in any
group (but that's a strange conclusion) and C10 becomes
specific to homogeneous groups, and my idea that there
needed to be some extra waiving of upfloating criteria
checks that the Procedures didn't explicitly specify for
homogeneous groups was a MISTAKE.



Obviously a violation of B5 or B6 for downfloats isn't
reason enough to fiddle with the pairings made in the
previous score bracket. Therefore it would be wrong to
dismiss a pairing only because a player from a higher score
bracket gets a second downfloat.

The Procedures do engage in backtracking. C10 and C12 are
examples of that.

But wait! Although downfloating players is unavoidable, in
the case that we have a choice, which players would it be
good to downfloat? Obviously not the ones who have already
been downfloated before. We should choose the ones who have
not been downfloated before.

Bill Gletsos said such proactiveness is required:

http://chesschat.org/showpost.php?p=142260&postcount=158

But perhaps C10 and C11's backtracking and repairing of the
bracket before will handle this case?

But if this backtracking and repairing does solve the
problem, we WILL need some kind of downfloater retention in
homogeneous groups to avoid ending up with the same pairings
as before in the cycle through C7,8 and creating an infinite
loop. Because there is nothing in the Procedures that you could
call proactive, in the sense of going ahead of where you are
at the moment before actually getting there.

I just realized that the checking I have been doing is
completely wrong. The floating in a homogeneous group
happens after reaching C14, and p (pprime) starts being cut.

It is those players who are NOT paired and who MUST be
downfloated who should not be the players in the bracket who
downfloated in the 2 previous rounds.

That explains why there is nothing about the Relative
Criteria in C6.


I guess, this holds true
even for remainder groups, which arise after pairing only
some of the downfloated players.

This leaves us with B6 and B5 only relevant in case of an
amalgated last score bracket. But does it really make a
difference there? Or are those B6,B5 checks irrelevant at
all? Can you imagine a situation where they make a
difference?

In all other cases, the "relevant" order is of waiving is
"B6Up, B5Up, Others", as one would expect after reading
section B.

But why does C10 only speaks about homogeneous remainder
groups? Maybe, the intention of C10 isn't primary to waive
B6 and B5, but to give a procedure for those homogeneous
remainder groups. Maybe it's main purpose is to give the
"unusual" instruction to redo one of the pairings of the
previous score bracket, before waiving B6 and B5? Maybe
waiving B6 and B5 is mentioned explicitely, because it
should be applied after fiddling with the last score
bracket?


In the instruction to redo the pairing of the previous score
bracket, it's more specifically, the heterogeneous bracket
from which the remainder group came.

I agree the whole sequence from C10 to C12 is focused on
heterogenous groups, more than homogeneous ones. Clearly,
these are the problematic ones. The idea is to stop floating
that has started to occur, by extraordinary measures.

C10 is a halfway house before C12. C10 is about
repairing the last player in S1 in a heterogenous bracket.
C12 is about going back to the bracket before that, if the
repairing of the heterogeneous bracket doesn't work. But why
isn't there a procedure to re-pair two or more players from
S1 who were floated down?

Whether the heterogeneous bracket from which the remainder
group came is the previous score bracket or not is
important I think, because in C13, an unpairable final
bracket has to be joined with the previous score bracket.

Although it would be good to limit this to the remainder
group.

In this pairing table, for the 4th round of a tournament
where ALL the matches in previous rounds have ended in draws
or byes is:



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1-3
5 2,4,- WW- D 2
6 3,-,7 B-B d 2
7 -,1,6 -BW d 2
4-7
1 4,7,2 WWB u 1.5
2 5,3,1 BBW 1.5
3 6,2,4 WWB 1.5
4 1,5,3 BBW 1.5


At the moment, Games::Tournament::Swiss cannot pair this.

Round 4: 5 6 7 (2), 1 2 3 4 (1.5),

5 and 6 are paired and 7 is downfloated again.

C6, 1 paired. OK E1 6&5
C6others: Floating remaining 7 Down. [1] 5 6 & [2] 1 2 3 4 7
Next, Bracket 2: 1 2 3 4 7
C1, B1,2 test: OK, no unpairables
C2, x=0
C3, p=1 Heterogeneous.
C4, S1 & S2: 7 & 1 2 3 4
C5, ordered: 7 &
1 2 3 4
C6, B1a: table 1 NOK
C7, 2 1 3 4
C6, B4: x=0, table 1 NOK
C7, 3 1 2 4
C6, 1 paired. OK E1 3&7
C6others: Bracket 2's Remainder Group: 1 2 4
Remaindering 1 2 4. [2] 7 3 & [2's Remainder Group] 1 2 4
Next, Bracket 2's Remainder Group: 1 2 4

If it was simple, it could just give 4 the bye, and pair 1
and 2. But 1 and 2 have already met, and there is C13.

C1, NOK. 1
C13, p=1. Undoing Bracket 2 matches. Re-pairing Bracket 2. Bracket 2: 7 3 & Bracket 2's Remainder Group: 1 2 4
C7, last transposition
C8, last S1,S2 exchange:
C9, B5,6 already dropped for Downfloats in Bracket 2.
C10, Bracket 2 not remainder or C10repair group. Passing to C11
C11, x=p=1 already, no more x increases in Bracket 2.
C12, Heterogenous group 2 now homogeneous
No repairings of previous score bracket
C13, Not last group. Passing to C14
C14, Bracket 2, now p=0
Moving all Group 2's Remainder Group members back to 2. 1 2 4 => Bracket 2: 7 3 1 2 4

And Games::Tournament::Swiss appears to gets stuck in a
infinite loop. If the interpretation of the previous bracket
in C13 is to mean Bracket 1, then this might not happen.

It is deciding this is not the last bracket, because it is
trying to pair 7 and 3. Perhaps it should be trying to join
7 and 3 with 5 and 6, rather than 1,2 and 4.

Perhaps this is not such a good example. It may just be
a bug.



If one reads C9, C10 and C11 this way, a new interpretation
arises: It follows from section B, that the criteria B3-B6
should be waived step by step (starting with B6). C10 and
C11 only give some instructions for special cases (namely
for homogeneous remainder groups). The purpose of C9 isn't
clear to me at the moment. The "natural" points for waiving
then seem to be
* for B6 and B5 before C11,
* for B4 before C12,
* for B3 before C14

What do you think about that?


I'm now waiving B6,5 in C10 for all kinds of brackets. But
it appears I have an incorrect interpretation of what it
means to do a B6,5 check.

B4 for preference-filling is relaxed in C11, so this is the
appropriate point.

B3 is about score differences. C14 is the point at which a
player faces a downfloat, so this again is appropriate.

I think the natural associations are B6,5 with C9, C10, B4
with C11, and B3 with C14. These are not accidental associations.

C10 and C12 are as you say, special cases.



I think so, too. But "heterogeneous groups" must include those, which have
players from different score brackets, but are treated as homogeneous for
some reasons (e.g. A3).

That's something the Rules say, but people seem to be unhappy
about it meaning B3 score differences can be disregarded.

Bartolin
23-10-2007, 02:50 AM
I think I was mistaken about what a downfloat was. I thought
that pairing someone who had had a downfloat with someone
who had had an upfloat (or a downfloat?), was to float them
again, EVEN IF THEY HAD THE SAME SCORES.

I'm quite sure that this interpretation is wrong. According to A4 floats are only possible in heterogeneous groups (when opponents have different scores).



If this is not the case, and if in a homogeneous group there
is no floating, there is no need to carry out B5,6 checks at
all in C6 in homogeneous groups. And so we don't have to
waive them in C9, or C10. We can go straight to x increases
in C11.

I think that's true.


But if these downfloat repetitions are unavoidable, then
there is no need to carry out B5,6Down checks in a
heterogeneous group. And there is no need to waive them in C9.

(But would this reasoning apply for downfloats 2 rounds before?)

In this case, C9 becomes irrelevant to the pairings in any
group (but that's a strange conclusion) ...

But at the moment it seems correct to me. As you suggested further down your post, we have to check for B5,B6Down violations while pairing the previous score bracket -- that is, before downfloating the players in the first place.

Again, maybe the purpose of C9 is just to waive those unavoidable (intentional) violations of B5,B6Down as early as possible.


... and C10 becomes
specific to homogeneous groups, and my idea that there
needed to be some extra waiving of upfloating criteria
checks that the Procedures didn't explicitly specify for
homogeneous groups was a MISTAKE.

I'm afraid, I don't understand this part. Do you really mean that C10 becomes specific to homogenous groups?


But wait! Although downfloating players is unavoidable, in
the case that we have a choice, which players would it be
good to downfloat? Obviously not the ones who have already
been downfloated before. We should choose the ones who have
not been downfloated before.

Bill Gletsos said such proactiveness is required:

http://chesschat.org/showpost.php?p=142260&postcount=158


Yes, it seems to me that we have to check before actually moving down a player.



But perhaps C10 and C11's backtracking and repairing of the
bracket before will handle this case?

But if this backtracking and repairing does solve the
problem, we WILL need some kind of downfloater retention in
homogeneous groups to avoid ending up with the same pairings
as before in the cycle through C7,8 and creating an infinite
loop. Because there is nothing in the Procedures that you could
call proactive, in the sense of going ahead of where you are
at the moment before actually getting there.

I just realized that the checking I have been doing is
completely wrong. The floating in a homogeneous group
happens after reaching C14, and p (pprime) starts being cut.

It is those players who are NOT paired and who MUST be
downfloated who should not be the players in the bracket who
downfloated in the 2 previous rounds.


That seems sensible to me.


That explains why there is nothing about the Relative
Criteria in C6.

I think the Relative Criteria should be considered in C6 nevertheless.



Whether the heterogeneous bracket from which the remainder
group came is the previous score bracket or not is
important I think, because in C13, an unpairable final
bracket has to be joined with the previous score bracket.

Although it would be good to limit this to the remainder
group.

Is there anybody else who can enlighten us about this point?

drbean
25-10-2007, 07:42 PM
According to A4 floats are only possible in heterogeneous groups (when opponents have different scores).

But at the moment it seems correct to me. As you suggested further down your post, we have to check for B5,B6Down violations while pairing the previous score bracket -- that is, before downfloating the players in the first place.


When Bill Gletsos was discussing B5,6 violations, at http://chesschat.org/showpost.php?p=142260&postcount=158, I didn't understand that that was the normal case. I thought it was a special case.

As you said before, in a heterogeneous bracket, there is no way to avoid the downfloaters being paired with a player with a different score.

So in the bracket above, efforts have to be made to ensure that the players who will downfloat haven't already downfloated.



Again, maybe the purpose of C9 is just to waive those unavoidable (intentional) violations of B5,B6Down as early as possible.


I agree, that's the first Relative Criteria waiving. But I think B65 checks have to be carried out in C6, at least the first time round, even before p has been decreased, ie when p is half the number of players in the bracket. Because, if the number of players is odd, one player is going to be downfloated, and we don't want it to be one which downfloated previously.




drbean wrote:

In this case, C9 becomes irrelevant to the pairings in any group (but that's a strange conclusion) and C10 becomes specific to homogeneous groups, and my idea that there needed to be some extra waiving of upfloating criteria checks that the Procedures didn't explicitly specify for homogeneous groups was a MISTAKE.

I'm afraid, I don't understand this part. Do you really mean that C10 becomes specific to homogenous groups?


Why is there only a reference to waiving upfloat criteria checks when repairing a heterogeneous bracket in C10 and not to waiving them when repairing a homogeneous bracket, even though it is (apparently) necessary to do this to avoid (under some circumstances) violating the proviso that none of the Relative Criteria are cause for downfloating players?

Perhaps it is because downfloating such players never results in violating the Criteria (in homogeneous groups). I don't remember why I thought C10 becoming specific to homogeneous groups would result in that situation.

But, in any case, to downfloat a player who has upfloated in a previous round can only result in it having an upfloat and a downfloat, not 2 downfloats. Players only face a downfloat in a homogeneous group, not an upfloat. And an upfloated player can only have another upfloat if it is in a heterogeneous group. So there is never any need to carry out upfloat checks, except in heterogeneous groups.

Is that a mistake?

Finally I made a mistake. I said:

C10 is a halfway house before C12. C10 is about repairing the last player in S1 in a heterogenous bracket. C12 is about going back to the bracket before that, if the repairing of the heterogeneous bracket doesn't work. But why isn't there a procedure to re-pair two or more players from S1 who were floated down?

In fact, that is C11.

Bartolin
25-10-2007, 10:48 PM
Trying to understand C10 better, I'd like to ask the following: How are the players we have to take to C2 (when C10 dot 1 applies)? Does C10 dot 1 speak

about the players of the remainder group only,
about the players of the remainder group plus the lowest moved down player plus his opponent or
about all players of the heterogenous group?

I think, it's number 2.

If that is correct, how can we prevent that we have exactly the same situation when we visit C10 the next time? Let's say we restart at C2 and pair the lowest downfloated player against the same opponent again (or does C10 dot 1 prohibits that?). Then all attempts to pair the remainder group would be exactly the same as last time. And in C10 we are suddenly speaking about the same "lowest moved down player" again.

drbean
28-10-2007, 01:04 AM
C10.
In case of a homogeneous remainder group: undo the pairing of
the lowest moved down player paired and try to find a
different opponent for this player by restarting at C7.

If no alternative pairing for this player exists then drop
criterion B6 first and then B5 for upfloats and restart at
C2.



Trying to understand C10 better, I'd like to ask the following: How are the players we have to take to C2 (when C10 dot 1 applies)? Does C10 dot 1 speak

about the players of the remainder group only,
about the players of the remainder group plus the lowest moved down player plus his opponent or
about all players of the heterogenous group?

I think, it's number 2.


It could mean just the first alternative, ie the remainder group. (Or could it? Seeing the remainder group's homogeneous, none of the players face an upfloat and if they can't be paired they will be downfloated, neither of which would be violations of B5,6 for upfloats. Thus dropping upfloat checks doesn't make sense.)

But I read it to mean alternative pairings which were not compliant with B5,6 because they were of the lowest downfloated player with partners who had already upfloated. This is answer number two.

I don't think it could be answer number three, because that is what C11 is talking about with the repairing of the whole heterogeneous bracket.

However if we are talking about the groups that reach C10, I think both post-C10 repair brackets and remainder groups of such brackets reach C10.

If we have a C10 repair group going through the C6,7 cycle, and there are no partners for the lowest downfloated member, the C10 repair group will end up at C10.

On the other hand, if we succeed in pairing the last downfloated player with another member of the remainder group, instead of with the partner that was first given to it, we then have to see if the rest of the remainder group plus the separated partner can be paired. This remainder group of a C10 repair group might also end up at C10.

In the case of a remainder group ending up at C10, it only means that particular pairing of the lowest downfloated member failed, so we go to C7 and find the next member to try. In the case of a C10 repair group, it means there is no pairing of the last downfloated player which will allow pairing of the whole heterogeneous group, and so we will have to go on to C11. But wait! We will go through that whole cycle again to try pairings with previously upfloated players in the remainder group. That is, the first time we only try pairings with remainder group members who haven't upfloated, and then if on the first and second run through a partner can't be found that allows the remainder group to be paired, a partner who has upfloated more recently can be tried.

Is the reason we go back to C2 because we also want possible pairings that were rejected before because of upfloat violations to get a second chance too? That is, we don't just want to consider the ones that come after the first compliant pairing of the downfloaters, but which failed to allow pairing of the whole heterogeneous bracket? I am not too sure.

If on trying to pair this C10 repair group, you end up at C10 again with all float checks waived and the last transposition tried, you know that the C10 repair group has failed to find an alternative pairing. You have to then go on to C11. That is the C10 repair group can run its natural course through C6,7.9(?),10 and the third time it reaches C10, without any more alternative players, you know it has failed.

If a remainder group hits C10 instead of the whole C10 repair group, then the next alternative player from the remainder group would be tried, by returning to C7.



If that is correct, how can we prevent that we have exactly the same situation when we visit C10 the next time? Let's say we restart at C2 and pair the lowest downfloated player against the same opponent again (or does C10 dot 1 prohibits that?). Then all attempts to pair the remainder group would be exactly the same as last time. And in C10 we are suddenly speaking about the same "lowest moved down player" again.

After the pairing in a group, I don't throw away the position reached in the transposition sequence. I continue at the next position after that, pairing the C10 repair group. And then I have an "escape hatch" for C10 repair groups after the last transposition is reached in C7. They are transported to C10 if there are no more transpositions, where a check is made of whether all float checks have been waived. If they have not, they are returned to C4 (not C2, because that resets p). If they have, they go on to C11.

I think that escape hatch is not a good thing. I don't know if it is necessary. It would be better for the last transposition to go through C9 and for C10 to know whether this was the last transposition or not. but I guess you have to test to see if there is a last transposition or not. I think the reason I have it is that you don't know if you have reached the last transposition or not until you have tried to get it.

Actually I don't know if that's true. The last transposition is the first transposition backwards. 1 2 3 4 and 4 3 2 1.

drbean
12-11-2007, 07:04 PM
In response to this pairing table,




No Opponents Colours Float Score

1 : 6,4,2,3 BWBW D 3.5
2 : 7,3,1,5 WBWB D 3.5

4 : 9,1,7,10 WBWB D 3

3 : 8,2,6,1 BWBB U 2
5 : 10,6,8,2 BWBW U 2
7 : 2,10,4,8 BWBW 2

6 : 1,5,3,9 WBWB D 1
8 : 3,9,5,7 WBWB 1
9 : 4,8,10,6 BWBW U 1
10 : 5,7,9,4 WBWW U 1



the different pairing programs were giving these pairings:




Swiss Perfect:


No Name Loc Total Result Name Loc Total

1 KINDER, Jessica (2) 1624 [3.5] : EDWARDS, Regina (4) 1409 [3]
2 BONELL, Sarah (8) 646 [1] : JULE, Alexandra (1) 1756 [3.5]
3 LYONS, Kieran C (3) 1466 [2] : LEE, Isabelle (10) 588 [1]
4 YUM, Winnie (9) 634 [1] : SIMMONDS, Leteisha (5) 985 [2]
5 KARIBASIC, Melanie (6) 820 [1] : YUM, Jenny (7) 781 [2]

Swiss Master 5.5


1 Jessica Kinder ( 3.5) - Regina Edwards ( 3 ) 2- 4
2 Jenny Yum ( 2 ) - Alexandra Jule ( 3.5) 7- 1
3 Kieran Lyons ( 2 ) - Isabelle Lee ( 1 ) 3- 10
4 Winnie Yum ( 1 ) - Leteisha Simmonds ( 2 ) 9- 5
5 Melanie Karibasic ( 1 ) - Sarah Bonell ( 1 ) 6- 8

Swiss Manager


Bo.SNo. Name Pts Res. Pts Name SNo.

1 2 Jessica Kinder 3 3 Regina Edwards 4
2 5 Leteisha Simmonds 2 3 Alex Jule 1
3 3 Kieran Lyons 2 1 Winnie Yum 9
4 6 Melanie Karibasic 1 2 Jenny Yum 7
5 8 Sarah Bonell 1 1 Isabelle Lee 10

...

I think SP has just combined the bottom two score groups and paired top half v bottom half.

That's what Games::Tournament::Swiss was doing too, but having just suddenly decided the way it was handling repairing of penultimate brackets was wrong, it has now started to produce the Swiss Manager pairings.

I thought that C13 repairing of penultimate brackets came before C14 reduction of p and stood in the way of it. Of course if you are going to reduce p, you are going to float down more players than before. And this would be a quick solution.

But I thought that this quick solution would prevent you finding a different solution that didn't involve the reduction of p. I thought that the language in C13 about reduction of p meant another pairing that would float down different players had to be sought before you were allowed to reduce p.

You would continue hitting C13 until there were no more failing pairings that didn't involve reducing p and then you would be allowed to go to C14 and start reducing p.

But now I realized, Where are these other failing pairings going to come from? C10,11,12 are productive sources of other pairings, but we have come to the end of the line. There ARE no other pairings.

C14 is also not a source of untried pairings. All it is doing is talking about reducing p. Neither it or C13 are sources of untried pairings.

As a result, anything about reduction of p that is allowed in C14 flows back up to C13 without the need to actually go there.

We don't have to exhaust any possibilities inherent in C13 before we are allowed to go on to C14 and reduce p. All the functionality of C14 is provided in C13.

The idea that they are in series suggested by their numbering is false. They really are in parallel.

As you can see, this argument is rhetorical, rather than logical, so I would be interested in views that counter it.

The pairing:




Round 5: 1 2 (3.5), 4 (3), 3 5 7 (2), 6 8 9 10 (1),
START, Go
NEXT, 3.5-Bracket [1]: 1 2
C1, NOK. 1 2 B1a/B2a incompatible in 1 (3.5)
C1, Floating 1 2 down to 3-Bracket [2]
C1, [1] & [2] 4 1 2
C1, 3.5-Bracket [1] dissolved
NEXT, 3-Bracket [2]: 4 1 2
C1, NOK. 1 B1a/B2a incompatible in 2 (3)
C1, Floating 1 down to 2-Bracket [3]
C1, [2] 4 2 & [3] 3 5 7 1
C2, x=1, xprime=1
C3, p=1. Homogeneous.
C4, S1: 2 & S2: 4
C5, ordered: 2
& 4



and after some relaxation of the B5,6 restrictions,




C6PAIRS, B56: OK.
C6PAIRS, 3-Bracket (2) tables 1 paired. OK
C6PAIRS, E4 2&4
NEXT, 2-Bracket [3]: 3 5 7 1
C1, B1,2 test: OK, no unpairables
C2, x=1, xprime=1
C3, p=1. Heterogeneous.
C4, S1: 1 & S2: 3 5 7
C5, ordered: 1
& 3 5 7



and after some more relaxation of the B5,6 restrictions,




C6PAIRS, B1a: table 1 NOK
C7, 5 3 7
C6PAIRS, B56: OK.
C6PAIRS, 2-Bracket (3) tables 1 paired. OK
C6PAIRS, E4 5&1
C6OTHERS, Remaindering 3 7.
[3] 1 5 & [3's Remainder Group] 3 7
C2, x=0, xprime=0
C3, p=1. Homogeneous.
C4, S1: 3 & S2: 7
C5, ordered: 3
& 7
C6PAIRS, B56: OK.
C6PAIRS, 2Remainder-Bracket (3's Remainder Group) tables 1 paired. OK
C6PAIRS, E1 3&7
NEXT, 1-Bracket [4]: 6 8 9 10
C1, NOK. 9 B1a/B2a incompatible in 4 (1)
C1, NOK. in last bracket, 4 (1).
C13, Undoing 2Remainder-Bracket [3's Remainder Group] matches



So here we are at C13. What Games::Tournament::Swiss was doing was trying to repair the 2Remainder-Bracket and thus force C10,11,12 repairing of the 2-Bracket.

What it now does is just reduce p in the 2Remainder-Bracket, which instantly results in its joining with the last 1-Bracket.




C13, penultimate p=0.
C13, Joining Bracket 3's Remainder Group, 4.
C13, [3's Remainder Group] 3 7 => [4] 6 8 9 10
C1, B1,2 test: OK, no unpairables
C2, x=0, xprime=0
C3, p=2. Heterogeneous.
C4, S1: 3 7 & S2: 6 8 9 10
C5, ordered: 3 7
& 6 8 9 10
C6PAIRS, B1a: table 1 NOK
C7, 8 6 9 10
C6PAIRS, B1a: table 1 NOK
C7, 9 6 8 10
C6PAIRS, B56: OK.
C6PAIRS, 1-Bracket (4) tables 1 2 paired. OK
C6PAIRS, E1 3&9 E1 6&7
C6OTHERS, Remaindering 8 10.
[4] 3 7 9 6 & [4's Remainder Group] 8 10
C2, x=0, xprime=0
C3, p=1. Homogeneous.
C4, S1: 8 & S2: 10
C5, ordered: 8
& 10
C6PAIRS, B56: OK.
C6PAIRS, 1Remainder-Bracket (4's Remainder Group) tables 1 paired. OK
C6PAIRS, E1 8&10
Pairing complete



But there still remains niggles about whether this is consistent with the rules. It's only the end of the line because it's the last bracket.

No players were floated down to the last bracket before we found it unpairable. But what if there had been players floated down? In that case, isn't there the possibility that we could have floated down a different set of players which resulted in the last bracket becoming pairable?

So, this is a nicer pairing than joining 8 players into one bracket, but is that just luck?

Kevin Bonham
26-04-2011, 06:33 PM
I think it would be good to post examples of SP draws that appear to be clearly wrong to this thread.

MCC Anzac Day Weekender round 7:

Pairing info after round 6:


Place No Opponents Colours Float Score


1 1 : 9,5,2,4,3,10 BWBWBW D 6

2-3 3 : 11,7,4,15,1,2 BWBWWB u 4
5 : 13,1,8,2,14,4 BBWWBB U 4

4-6 4 : 12,14,3,1,2,5 WBWBBW uD 3.5
7 : 14,3,12,13,15,6 WBWBBW u 3.5
8 : 15,2,5,11,6,9 BWBWBB D 3.5

7-9 2 : 10,8,1,5,4,3 WBWBWW d 3
10 : 2,13,6,14,-,1 BWBW-B Ud 3
14 : 7,4,11,10,5,15 BWBBWW 3

10-11 6 : -,-,10,12,8,7 --WBWB 2.5
11 : 3,9,14,8,-,13 WBWB-W Ud 2.5

12-14 12 : 4,15,7,6,9,- BWBWW- uD 2
13 : 5,10,-,7,-,11 WB-W-B D 2
15 : 8,12,9,3,7,14 WBWBWB d 2

List of players in seeding order

1.Baron
2. Hamilton
3. Dowling
4. Stead
5. Davis
6. Hain
7. Cavezza
8. Bekker
9. Ashlock
10. Gibson
11. Chew Lee
12. Gluzman
13. Hughes
14. Ying
15. Cotton

Player 9 withdrew before the final round. The three byes in round 5 were half-point byes while the sole bye in round 6 was full point. Previous byes were all either half point or full point.

SP produced the following round 7 draw:


No Name Total Result Name Total

1 cavezza, (7) [3.5] : baron, (1) [6]
2 davis, (5) [4] : dowling, (3) [4]
3 bekker, (8) [3.5] : stead, (4) [3.5]
4 cotton, (15) [2] : hamilton, (2) [3]
5 gibson, (10) [3] : gluzman, (12) [2]
6 hughes, (13) [2] : ying, (14) [3]
7 hain, (6) [2.5] : chew lee, (11) [2.5]

The bottom three pairings were overridden (by Grant, and I agreed) to:

Gibson (3) vs Chew Lee (2.5)
Hain (2.5) vs Ying (3)
Hughes (2) vs Gluzman (2)

...as there is no reason to downfloat the entire 3 group to the 2 group. Note that no pairing can be made within the 3 group since while two of them have not played each other, both those two have just gone WW and are not above 50%. However pairing some of the 3 group with the 2.5 group is possible, and there is no reason why 2 pairings cannot be made within the heterogeneous group formed by downfloating all the 3s to the 2.5 group.

I am not sure that override is the best possible but with limited time available at least it is an improvement.

Paul Cavezza
26-04-2011, 08:22 PM
Hi there-

I can't made sense of the above but make sure you've got the correct result for Gary Bekker in round 5 I think it was V Hain. There was a re-draw which might not have been posted here because it was assumed GB won when it was a draw.

Also the list of seeds there is incorrect I think- I'm (cavezza) above Hain if memory serves

Kevin Bonham
26-04-2011, 09:25 PM
Hi there-

I can't made sense of the above but make sure you've got the correct result for Gary Bekker in round 5 I think it was V Hain.

I have Hain - Bekker 1/2-1/2 in round 5. I didn't see that anywhere; I reconstructed rounds 3 and 5 from the other information.

As for the order of seeds I was going on what Grant posted in the MCC thread in posts 77 and 81. The pairing info I came out with is exactly the same as what he posted.

Paul Cavezza
26-04-2011, 11:39 PM
Yeah draw is the correct result.

Haven't seen that seeding/order before. Crosstable certainly didn't look like that at the club.

good luck!

MichaelBaron
27-04-2011, 02:20 AM
What I want to know is - will all the errors be corrected for the tourney to be included into the May Fide rating list? :)