Another sp error?

[Garvinator]

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]

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]

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]

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]

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]

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]

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]

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]

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]

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]

If there was ever to be an arbiters exam, this situation would be a good question to have for pairings.

[drbean]

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]

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]

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]

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.