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NeilH
22-09-2007, 06:05 PM
I have a query about the 'correct' pairing in the following circumstances.

Here are the top 2 score brackets:

Score Bracket 1 on 5 points
1 Knuth, Hannes #23, 3, 5, 4, 8 #WBWBB#

Score Bracket 2 on 3.5 points
6 Rohl, Rainer #13, 7, 17, 3, 12 #BWBWB#
7 Kesten, Sebastian #14, 6, 9, 8, 2 #WBBWB#
2 Friedrich, Paul-Ro # 9, 20, 10, 5, 7 #BWBBW#

Players are listed in rating order. I have ignored floats as they have no impact. All players have a White colour preference with 1 having an absolute White colour preference.

The two score brackets are merged into a hetrogeneous score bracket with p=1

Score Bracket 1
Sub Group 1
1 Knuth, Hannes #23, 3, 5, 4, 8 #WBWBB#
Sub Group 2
6 Rohl, Rainer #13, 7, 17, 3, 12 #BWBWB#
7 Kesten, Sebastian #14, 6, 9, 8, 2 #WBBWB#
2 Friedrich, Paul-Ro # 9, 20, 10, 5, 7 #BWBBW#

This gives a pairing of 1 v 6 with 7 and 2 going into a homogeneous remainder score bracket. As 7 and 2 are unpairiable they will both downfloat into the next score bracket.

However these score brackets were actually paired as 1 v 7 and 6 v 2.
I like this pairing as it limits the downfloats. However I don't understand why it is correct in this case.

Can anyone explain why this is the correct pairing?

Bill Gletsos
22-09-2007, 10:38 PM
I have a query about the 'correct' pairing in the following circumstances.

Here are the top 2 score brackets:

Score Bracket 1 on 5 points
1 Knuth, Hannes #23, 3, 5, 4, 8 #WBWBB#

Score Bracket 2 on 3.5 points
6 Rohl, Rainer #13, 7, 17, 3, 12 #BWBWB#
7 Kesten, Sebastian #14, 6, 9, 8, 2 #WBBWB#
2 Friedrich, Paul-Ro # 9, 20, 10, 5, 7 #BWBBW#

Players are listed in rating order. I have ignored floats as they have no impact. All players have a White colour preference with 1 having an absolute White colour preference.

The two score brackets are merged into a hetrogeneous score bracket with p=1

Score Bracket 1
Sub Group 1
1 Knuth, Hannes #23, 3, 5, 4, 8 #WBWBB#
Sub Group 2
6 Rohl, Rainer #13, 7, 17, 3, 12 #BWBWB#
7 Kesten, Sebastian #14, 6, 9, 8, 2 #WBBWB#
2 Friedrich, Paul-Ro # 9, 20, 10, 5, 7 #BWBBW#

This gives a pairing of 1 v 6 with 7 and 2 going into a homogeneous remainder score bracket. As 7 and 2 are unpairiable they will both downfloat into the next score bracket.

However these score brackets were actually paired as 1 v 7 and 6 v 2.
I like this pairing as it limits the downfloats. However I don't understand why it is correct in this case.

Can anyone explain why this is the correct pairing?After pairing 1 V 6 and then starting at C2 with the homogeneous remainder group you fall down thru C6-C9 all which make no difference as 7 V 2 violates B1. You then fall into C10. This causes the unpairing of 1 V 6 and leads to the pairing 1 V 7 and 6 V 2.

NeilH
22-09-2007, 10:47 PM
Thanks for that Bill. I was restarting at C1 with the Homogeneous, remainder score bracket rather than C2. I missed that bit of the rules.

Bill Gletsos
22-09-2007, 10:50 PM
Thanks for that Bill. I was restarting at C1 with the Homogeneous, remainder score bracket rather than C2. I missed that bit of the rules.I note however that it appears you have your 3.5 score group ranked in the wrong order.

S1 = 1 and S2 = 2, 6, 7 in that order.

Now after getting the pairing 1 V 2 it leaves you having to pair 6 & 7.
They however violate B1.
Hence as per the logic I described above you fall into C10.
1 v 6 is no better as noted above so you end up with 1 v7 and 6 v 2

NeilH
23-09-2007, 08:43 PM
This case is similar but slightly different as it generates a homogeneous remainder group with a single player.

The logical pairing is 2 v 1 and 4 v 3, but I can't see how the rules work to generate this pairing.

Score Bracket 1 (Score 5.5)
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

Score Bracket 2 (Score: 5)
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

Score Bracket 3 (Score: 4.5)
3 Ly, Moulthun 20, 10, 1, 8, 7, 2 #BWBWBW# # D #

Garvinator
23-09-2007, 08:53 PM
This case is similar but slightly different as it generates a homogeneous remainder group with a single player.

The logical pairing is 2 v 1 and 4 v 3, but I can't see how the rules work to generate this pairing.

Score Bracket 1 (Score 5.5)
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

Score Bracket 2 (Score: 5)
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

Score Bracket 3 (Score: 4.5)
3 Ly, Moulthun 20, 10, 1, 8, 7, 2 #BWBWBW# # D #

What pairings were generated?

NeilH
23-09-2007, 09:46 PM
What pairings were generated?

The actual pairing was 2 vs 1 and 3 vs 4

Bill Gletsos
23-09-2007, 10:05 PM
The actual pairing was 2 vs 1 and 3 vs 4That is indeed the correct pairing.

Bill Gletsos
23-09-2007, 10:23 PM
The reason why it is 3 V 4 and not 4 V 3 is because Lazarus and Ly have identical colour histories and hence E4 applies.
Lazarus having the higher score has the higher rank.

NeilH
24-09-2007, 12:46 AM
My questions is how we get these pairings rather than the colour allocation.

Starting with the scorebrackets:

Score Bracket 1 (Score 5.5)
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

Score Bracket 2 (Score: 5)
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

Score Bracket 3 (Score: 4.5)
3 Ly, Moulthun 20, 10, 1, 8, 7, 2 #BWBWBW# # D #

Applying C1 to Score bracket 1 means player 2 will downfloat to score bracket 2, giving a heterogeneous score bracket with

S1
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

S2
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

This gives the pairing 2 v 4, with player 1 downfloating into a homogeneous remainder score bracket containing only one player. Pairings restarts at C2, as Bill pointed out in his previous post.

This is where I am start to struggle!

If a score bracket has only 1 player then I think, is this case, C14 means the player downfloats to the next score bracket. Is this correct?

This gives a homogeneous score bracket with players 1 and 3 and pairing starts at C1.

C1 states that "if this player was moved down from a higher score bracket apply C12 ". But C12 doesn't apply as this is a homogeneous score bracket.
This isn't the lowest score bracket, so C13 doesn't apply. Therefore the players are moved down.

This means both players downfloat and can be paired in that score bracket. Therefore I don't get the correct pairing.

My understanding is obviously wrong, but where am I going wrong?

Bartolin
24-09-2007, 01:19 AM
Starting with the scorebrackets:

Score Bracket 1 (Score 5.5)
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

Score Bracket 2 (Score: 5)
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

Score Bracket 3 (Score: 4.5)
3 Ly, Moulthun 20, 10, 1, 8, 7, 2 #BWBWBW# # D #

Applying C1 to Score bracket 1 means player 2 will downfloat to score bracket 2, giving a heterogeneous score bracket with

S1
2 Antic, Dejan 18, 25, 14, 7, 5, 3 #WBWBWB# # D #

S2
1 Rogers, Ian 11, 16, 3, 4, 14, 5 #WBWBWB# # #
4 Lazarus, Benjamin 13, 27, 19, 1, 16, 15 #BWBWBW# # D D#

This gives the pairing 2 v 4, with player 1 downfloating into a homogeneous remainder score bracket containing only one player. Pairings restarts at C2, as Bill pointed out in his previous post.

This is where I am start to struggle!

If a score bracket has only 1 player then I think, is this case, C14 means the player downfloats to the next score bracket. Is this correct?

I would say, starting at C2 one has to stop at C6 and never gets to C14. In C3 we determine 'p' which is 0 in this case (homogeneous group -> S1 contains 0 players). In C6 we see that we have to pair 0 players, therefore we move player 1 down to the next score bracket (C6 dot 1).

Now it becomes easier again: 1 and 3 won't match, therefore we have to repair the penultimate score bracket according to C13. After applying it twice we have to find another opponent for player 2. Instead of 2-4 we get 2-1, moving player 4 to the next score bracket where it is paired with player 3.

Or isn't that the correct procedure?

Bartolin
24-09-2007, 01:31 AM
I would say, starting at C2 one has to stop at C6 and never gets to C14. In C3 we determine 'p' which is 0 in this case (homogeneous group -> S1 contains 0 players). In C6 we see that we have to pair 0 players, therefore we move player 1 down to the next score bracket (C6 dot 1).

Now it becomes easier again: 1 and 3 won't match, therefore we have to repair the penultimate score bracket according to C13. After applying it twice we have to find another opponent for player 2. Instead of 2-4 we get 2-1, moving player 4 to the next score bracket where it is paired with player 3.

Or isn't that the correct procedure?

Sorry, I completely missed that this was just the top of the pairing table. Of course, C13 doesn't apply, because it is not the last score bracket.

But nevertheless, one doesn't get to C14, since I think C12 must be applied -- even though the score bracket we have (1 and 3) isn't regarded as heterogeneous according to A3, last sentence. (I think, C12 refers to the fact, that we have players from a higher score bracket, not to the question how the score bracket is to be treated -- and it's only there that A3 says "it is treated as though it was homogeneous.) At least this interpretation seems to makes sense to me.

NeilH
25-09-2007, 05:17 PM
I find it odd that rule A3 is so specific about what is homogeneous or heterogeneous, yet for C12 you are suggesting that there should be a more literal interpretation. There must be a simplier answer to this question.

Bartolin
05-10-2007, 01:21 AM
I find it odd that rule A3 is so specific about what is homogeneous or heterogeneous, yet for C12 you are suggesting that there should be a more literal interpretation. There must be a simplier answer to this question.

Unfortunately, I don't see a simpler answer. Note that we are arriving at C12
via C1 dot 1, which explicitely asks whether the player in question was moved
down from a higher score bracket. Therefore it seems plausible that C12
applies to those groups.

But apart from the wording of C12, the sequence C1 -> C12 is designed for
cases like this, where we can find a "better" downfloater in the previous score
bracket. I don't think, the fact, that there is only one player at 4.5 points, is a
good reason, not to switch players 1 and 4.

Bartolin
10-10-2007, 03:42 AM
I have a question about the correct pairing in a tournament as well.

Pairing table for round 5:

Round 5 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
8 18,5,3,1 BWBW D 3.5
2
1 11,10,6,8 WBWB U 3
3-9
2 12,7,9,5 BWBW 2.5
4 14,9,7,17 BWBW 2.5
5 15,8,13,2 WBWB 2.5
7 17,2,4,9 WBWB 2.5
9 19,4,2,7 WBWW 2.5
16 6,19,11,10 WBWB 2.5
17 7,14,18,4 BWBB d 2.5
10-14
3 13,6,8,11 WBWB u 2
6 16,3,1,15 BWBW 2
11 1,12,16,3 BWBW 2
13 3,18,5,19 BWBW 2
15 5,20,10,6 BWWB 2
15-16
10 20,1,15,16 BWBW 1.5
20 10,15,14,12 WBBW D 1.5
17-19
12 2,11,19,20 WBWB dU 1
18 8,13,17,14 WBWW u 1
19 9,16,12,13 BWBB u 1
20
14 4,17,20,18 WBWB 0

I'm especially interested in the last three boards. IMHO the first
seven boards should be:

7-8
1-2
16-4
17-5
3-9
13-6
15-11

though I'm not sure about upfloating player 3. But I think it's possible
to pair the first seven boards without moving a player down to the
score bracket with players 10 and 20 (1.5 points).

So the tail of the pairing table is:

-------------------------------------------------------------------------
Place No Opponents Roles Float Score
15-16
10 20,1,15,16 BWBW 1.5
20 10,15,14,12 WBBW D 1.5
17-19
12 2,11,19,20 WBWB dU 1
18 8,13,17,14 WBWW u 1
19 9,16,12,13 BWBB u 1
20
14 4,17,20,18 WBWB 0

I think, B6 for upfloats should be waived for players
18 and 19 -- even if there is no explicite rule in C to do so.
(C10 only speaks about homogeneous remainder groups.)

But according to B (relative criteria), B6 should be waived.
After doing so, we have the problem that 10-18 and 19-20
violates B4 for player 10 (he wants black). I think, we should
therefore waive B5 for player 12 as well, ending with

12-10
20-18
19-14

(20-19 at the penultimate board would be even better
in view of auf B4, but then in the next score bracket 18-14
is not possible, therefore according to C12 one should
choose 20-18.)

Is that correct?

Garvinator
10-10-2007, 12:20 PM
What was paired by the computer and can you also provide a cross table for the previous four rounds?

Garvinator
10-10-2007, 12:22 PM
From a quick look this is what I came up with.

I agree with 7 v 8 and 1 v 2.

I dont think the 2.5 score group has been paired correctly, which has implications for the lower score groups.

4 14,9,7,17 BWBW 2.5
5 15,8,13,2 WBWB 2.5
9 19,4,2,7 WBWW 2.5
16 6,19,11,10 WBWB 2.5
17 7,14,18,4 BWBB d 2.5

S1 S2

4 wants black 9 wants black
5 wants white 16 wants white
17 wants white, has downfloated two rounds before.

So you have three players preferencing white, two preferencing black. This means you can get two matching colour pairings (four players should get their correct colour preferences).

4 v 16
5 v 9 seems to work for colours, but downfloating 17 violates B6. So that isnt ideal.

So lets try downfloating 16 as that leaves four players- two want white, two want black.

So this leaves:

S1 S2

4 wants black 9 wants black
5 wants white 17 wants white, has downfloated two rounds before.

4 v 9 and 4 v 17 have played, so that leaves 4 v 5. 9 v 17 is also a legal pairing.

So the pairings should be:

5 v 4
17 v 9
16 downfloats.

With this pairing combination, B4 and B6 are both met.

With 16 being the downfloater, this has a flow on effect for the 2 point score group.

Bill Gletsos
10-10-2007, 12:38 PM
From a quick look this is what I came up with.

I agree with 7 v 8 and 1 v 2.Thats unfortunate as it means you are both wrong.

Garvinator
10-10-2007, 12:55 PM
Thats unfortunate as it means you are both wrong.
OK I will have a closer look in a couple of hours ;)

Bill Gletsos
10-10-2007, 01:09 PM
OK I will have a closer look in a couple of hours ;)Actually ignore what I said as I was actually looking at two different parts of his pairings and yours. ;)

eclectic
10-10-2007, 01:18 PM
can't wait for a correct pairings shoutbox trivia quiz ... :rolleyes:

Garvinator
10-10-2007, 02:34 PM
Actually ignore what I said as I was actually looking at two different parts of his pairings and yours. ;)
Now I am even more confused that before :whistle:

Bill Gletsos
10-10-2007, 02:39 PM
Now I am even more confused that before :whistle:Sorry for the confusion. I was flicking between his post and yours and got confused. :doh:
Your post above regarding the pairings of the 2.5 scoregroup looks correct to me.

Bartolin
10-10-2007, 06:09 PM
From a quick look this is what I came up with.

Thanks for this analysis. Unfortunately, I don't have crosstables for the earlier rounds at the moment. The quoted pairing table arose when trying to reproduce the pairings of http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm with on older version of Games::Tournament::Swiss. It had some problems with earlier rounds as well, but I'm wondering, how the quoted pairing table should be handled.

I dont think the 2.5 score group has been paired correctly, which has implications for the lower score groups.

4 14,9,7,17 BWBW 2.5
5 15,8,13,2 WBWB 2.5
9 19,4,2,7 WBWW 2.5
16 6,19,11,10 WBWB 2.5
17 7,14,18,4 BWBB d 2.5

S1 S2

4 wants black 9 wants black
5 wants white 16 wants white
17 wants white, has downfloated two rounds before.

So you have three players preferencing white, two preferencing black. This means you can get two matching colour pairings (four players should get their correct colour preferences).

4 v 16
5 v 9 seems to work for colours, but downfloating 17 violates B6. So that isnt ideal.

So lets try downfloating 16 as that leaves four players- two want white, two want black.

So this leaves:

S1 S2

4 wants black 9 wants black
5 wants white 17 wants white, has downfloated two rounds before.

4 v 9 and 4 v 17 have played, so that leaves 4 v 5. 9 v 17 is also a legal pairing.

So the pairings should be:

5 v 4
17 v 9
16 downfloats.

With this pairing combination, B4 and B6 are both met.

That looks good. When I wrote my post, I wrongly assumed that we had a hetergenous group. But indeed we can exchange 5 and 9.

With 16 being the downfloater, this has a flow on effect for the 2 point score group.

The next score bracket would consist of:

16 6,19,11,10 WBWB 2.5
3 13,6,8,11 WBWB u 2
6 16,3,1,15 BWBW 2
11 1,12,16,3 BWBW 2
13 3,18,5,19 BWBW 2
15 5,20,10,6 BWWB 2

S1 S2

16 wants white 3 wants white, has upfloated two rounds before
6 wants black
11 wants black
13 wants black
15 wants white

16 already played against 6 and 11, so the pairing should be 16-13, leaving a homogeneous remainder group with

S1 S2

3 wants white 11 wants black
6 wants black 15 wants white

Since 3 already played against 6 and 11, the only possible pairings are 3-15 and 11-6.

Should one go back and try to pair 16-15 or even 16-3 in order to prevent this double violation of B4 in the remainder group?

I think, one should stay with 16-13, 3-15, 11-6, because otherwise one would get a violation of B4 for the pairing with the downfloated player (16) and that is considered worse than a double violation of B4 for lower players. Or am I wrong again?

The pairing of the last 3 boards seems to be unaffected by the pairing of the 2 point score bracket. Is 12-10, 20-18, 19-14 correct for those last three boards?

Garvinator
10-10-2007, 07:17 PM
From seeing the linked pairings, I wonder, is Round 5 the final round? If it is the final round, then B6 would be ignored for all players over 50%, so this would mean that player 17 can be downfloated and so you would get the pairings in the link.

Should one go back and try to pair 16-15 or even 16-3 in order to prevent this double violation of B4 in the remainder group?

I think, one should stay with 16-13, 3-15, 11-6, because otherwise one would get a violation of B4 for the pairing with the downfloated player (16) and that is considered worse than a double violation of B4 for lower players. Or am I wrong again?

The pairing of the last 3 boards seems to be unaffected by the pairing of the 2 point score bracket. Is 12-10, 20-18, 19-14 correct for those last three boards?
The pairings worked out for 16 and the 2 point score group are correct. The key variable to remember is to pair from the top down. This means you pair 16 first and then start pairing the 2 point score group together after 16 has been legally paired.

Bartolin
10-10-2007, 07:34 PM
From seeing the linked pairings, I wonder, is Round 5 the final round? If it is the final round, then B6 would be ignored for all players over 50%, so this would mean that player 17 can be downfloated and so you would get the pairings in the link.

Indeed, round 5 was the final round. So I guess, you are right about downfloating player 17.

But it was instructive to work out the "not last round" pairings as well.

Thanks

Christian

Bartolin
10-10-2007, 07:40 PM
Thanks for this analysis. Unfortunately, I don't have crosstables for the earlier rounds at the moment. The quoted pairing table arose when trying to reproduce the pairings of http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm with on older version of Games::Tournament::Swiss. It had some problems with earlier rounds as well [...]

To avoid confusion I should note that my pairing table is not the correct pairing table for round 5 of the linked tournament. Since Games::Tournament::Swiss had some problems with pairings of earlier rounds, I accepted the pairings given by Games::Tournament::Swiss and invented some results for the divergent pairings to get to round 5. And with that fictious pairing table my question arose.

Bill Gletsos
10-10-2007, 07:43 PM
Indeed, round 5 was the final round.This is essential fact that would have been useful to know when the question of pairings was first asked. :wall: :wall:

So I guess, you are right about downfloating player 17.No guessing about it. He is obviously correct.

But it was instructive to work out the "not last round" pairings as well.It would have been far better to provide all the information up front and then ask the question "what if it wasnt the last round".

Bill Gletsos
10-10-2007, 07:45 PM
Is 12-10, 20-18, 19-14 correct for those last three boards?In my opinion, no.

Bartolin
10-10-2007, 08:31 PM
This is essential fact that would have been useful to know when the question of pairings was first asked. :wall: :wall:

I don't think, that fact was that important. As I wrote in my first post, I was especially interested in the pairings of the last three boards. Those are not affected by the question whether it is the last round or not. (At least I don't see how they are affected.)

Furthermore, the answer given (assuming it was not the last round) suited my question -- so what's the problem? Though I must admit that my question wouldn't have been useful to get the pairings for the last round of a tournament. But that wasn't my aim.

Garvinator
10-10-2007, 08:44 PM
Bartolin,

The cross table you supplied earlier. I notice it is different to the cross table that would be generated from the pairings given in the link? Was this intentional?

Garvinator
10-10-2007, 08:46 PM
Thanks ChristianIs your last name Krause?

Bartolin
10-10-2007, 08:48 PM

No, no ;)

But I'm from Germany as well.

Bartolin
10-10-2007, 08:53 PM
Bartolin,

The cross table you supplied earlier. I notice it is different to the cross table that would be generated from the pairings given in the link? Was this intentional?

Yes, it was. As I tried to explain here:

http://www.chesschat.org/showpost.php?p=170416&postcount=27

I used some other pairings in earlier rounds to test Games::Tournament::Swiss. So my pairing table was "half invented".

I'm really sorry for the confusion I caused. I hope it is not considered bad practice to ask questions about notional situations.

Garvinator
10-10-2007, 08:58 PM
I'm really sorry for the confusion I caused. I hope it is not considered bad practice to ask questions about notional situations.
Bad practice, no. But it is certainly more polite to give ALL possible information. Had I known it was the final round, how I would have answered the question would have been different (not just the final answer).

Knowing whether it is the final round or not is even more important when dealing with some players who are above 50% and some who arent.

Bartolin
10-10-2007, 09:06 PM
Bad practice, no. But it is certainly more polite to give ALL possible information. Had I known it was the final round, how I would have answered the question would have been different (not just the final answer).

Knowing whether it is the final round or not is even more important when dealing with some players who are above 50% and some who arent.

Okay, I'll try to behave better in the future ;)

Bill Gletsos
10-10-2007, 09:41 PM
I don't think, that fact was that important.Of course it was important as you queried the the need to upfloat player 3 and whether you may need to or not could well be dependant on the pairings that came before.

As I wrote in my first post, I was especially interested in the pairings of the last three boards. Those are not affected by the question whether it is the last round or not. (At least I don't see how they are affected.)Yes but given you didnt get the first 7 boards correct irrespective of whether it was the final round or not then providing all the information is important.

Furthermore, the answer given (assuming it was not the last round) suited my question -- so what's the problem?None, except in future I'll think twice before wasting my time on your questions.

Though I must admit that my question wouldn't have been useful to get the pairings for the last round of a tournament. But that wasn't my aim.Yes but whether your aim is correct depends on what goes before and given you didnt get that correct then it can quite easily throw your other assumptions into complete disarray.

Bartolin
11-10-2007, 02:15 AM
Is 12-10, 20-18, 19-14 correct for those last three boards?

In my opinion, no.

So, I'll try again.

The tail of the pairing table:

-------------------------------------------------------------------------
Place No Opponents Roles Float Score
15-16
10 20,1,15,16 BWBW 1.5
20 10,15,14,12 WBBW D 1.5
17-19
12 2,11,19,20 WBWB dU 1
18 8,13,17,14 WBWW u 1
19 9,16,12,13 BWBB u 1
20
14 4,17,20,18 WBWB 0

Since 15 and 16 are not a match they are downfloated to the next score bracket. Therefore we have the following heterogenous score bracket:

S1 S2

10 wants black 12 wants white, has upfloated last round, has downfloated two rounds before
20 wants black 18 wants black, has upfloated two rounds before
19 wants white, has upfloated two rounds before

According to A8, x equals 0 for this score bracket.

Obviously we have to waive B5 for player 20 before being able to get 2 (=p) pairings. Furthermore we have to waive B6 for players 18 and 19 as well -- for the same reason.

The first try (10-18, 19-20) violates B4. Before accepting such a violation we should waive B5 first (for player 12). After doing that we would get 12-10 and 19-20, no longer violating B4.

The problem is, that 18 can't be downfloated to 14 because both played in round 4 already. Therefore we have to go back to the penultimate score bracket.

At first I thought, we had to continue at the situation where we left this score bracket (B6 and B5 being waived for upfloaters). If that were true, we could pair 12-10 and 20-18, downfloating 19 (as I asked in my original post).

Now, after getting the hint that those pairings aren't correct, I have another idea. Since we have to violate B4 anyway, we could as well accept the "earlier" violation of B4 (10-18, 19-20), thereby respecting B5 for player 12.

Speaking in terms of procedure C, after going back to the penultimate score bracket we have to increase x by 1 according to C11. After that, we go back to C3, thereby putting B6 and B5 back into validity. Now with x=1 we can either pair 10-18, 19-20, downfloating 12 or we could pair 19-10, 20-18, again downfloating 12.

Do we have to fulfil the color preferences in the match of the higher player (10) the same way we have to try to minize his score differences first? If not, I think we should take the pairing 10-18, 19-20, 12-14 since that's the first of those two pairing we come accross during appliance of C7.

So, are the pairings 10-18, 19-20, 12-14 correct?

drbean
15-10-2007, 05:48 PM
I agree with 7 v 8 and 1 v 2.

I dont think the 2.5 score group has been paired correctly, which has implications for the lower score groups.

4 14,9,7,17 BWBW 2.5
5 15,8,13,2 WBWB 2.5
9 19,4,2,7 WBWW 2.5
16 6,19,11,10 WBWB 2.5
17 7,14,18,4 BWBB d 2.5

S1 S2

4 wants black 9 wants black
5 wants white 16 wants white
17 wants white, has downfloated two rounds before.

So you have three players preferencing white, two preferencing black. This means you can get two matching colour pairings (four players should get their correct colour preferences).

4 v 16
5 v 9 seems to work for colours, but downfloating 17 violates B6. So that isnt ideal.

So lets try downfloating 16 as that leaves four players- two want white, two want black.

So this leaves:

S1 S2

4 wants black 9 wants black
5 wants white 17 wants white, has downfloated two rounds before.

4 v 9 and 4 v 17 have played, so that leaves 4 v 5. 9 v 17 is also a legal pairing.

So the pairings should be:

5 v 4
17 v 9
16 downfloats.

With this pairing combination, B4 and B6 are both met.

With 16 being the downfloater, this has a flow on effect for the 2 point score group.

Bring it on!

That pinpointed 2 separate bugs in Games::Tournament::Swiss! It now pairs Bracket 3's Remainder Group:

Next, Bracket 3's Remainder Group: 4 5 9 16 17
C1, B1,2 test: OK, no unpairables
C2, x=0
C3, p=2 Homogeneous.
C4, S1 & S2: 4 5 & 9 16 17
C5, ordered: 4 5 &
9 16 17
C6, B1a: table 1 NOK
C7, 16 9 17
C6, B6Down, table 3: 17 NOK. Floated Down 2 rounds ago
C7, 16 17 9
C6, B4: x=0, table 2 NOK
C7, 17 9 16
C6, B1a: table 1 NOK
C7, last transposition
C8, exchange a: 4 9, 5 16 17
C5, ordered: 4 9 &
5 16 17
C6, B6Down, table 3: 17 NOK. Floated Down 2 rounds ago
C7, 5 17 16
C6, 2 paired. OK E1 5&4 E1 17&9
C6others: Floating remaining 16 Down. [3's Remainder Group] 4 9 5 17 & [4] 3 6 11 13 15 16

That is the same as the above.

Incidentally the code sections above, are they output from Swiss Perfect?

drbean
15-10-2007, 06:44 PM
Pairing what is apparently a fictitious pairing table :-)

Thanks for this analysis. Unfortunately, I don't have crosstables for the earlier rounds at the moment. The quoted pairing table arose when trying to reproduce the pairings of http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm with on older version of Games::Tournament::Swiss. It had some problems with earlier rounds as well, ...

The next score bracket would consist of:

16 6,19,11,10 WBWB 2.5
3 13,6,8,11 WBWB u 2
6 16,3,1,15 BWBW 2
11 1,12,16,3 BWBW 2
13 3,18,5,19 BWBW 2
15 5,20,10,6 BWWB 2

16 already played against 6 and 11, so the pairing should be 16-13, leaving a homogeneous remainder group with

S1 S2

3 wants white 11 wants black
6 wants black 15 wants white

Since 3 already played against 6 and 11, the only possible pairings are 3-15 and 11-6.

This pairing hasn't reached C11 so far. It is only in C6,7. So there can't be any x decreases. In fact, I don't think the Procedures allows any x decreases in remainder groups. C10 kicks in first, and then in C11, the whole heterogeneous bracket from which the remainder group came is re-paired.

A remainder group never sees C11, I think.

At the moment, Games::Tournament::Swiss is first reaching C10, and undoing the pairing with 13. But it is not able to pair with 15, because they both have the same preference. So it then reaches C11, where x becomes 1.

It then repairs the whole heterogenous bracket. But seeing this is a heterogeneous bracket with only one member from the previous bracket, the same pairings are being considered again:

10, No more opponents for Player 16. Re-forming Bracket 4
C11, x=1,
C3, p=1 Heterogeneous.
C4, S1 & S2: 16 & 15 3 6 11 13
C5, ordered: 16 &
3 6 11 13 15
C6, B6Up, table 1: 3 NOK. Floated Up 2 rounds ago
C7, 6 3 11 13 15
C6, B1a: table 1 NOK
C7, 11 3 6 13 15
C6, B1a: table 1 NOK
C7, 13 3 6 11 15
C6, 1 paired. OK E1 16&13
C6others: Bracket 4's Remainder Group: 3 6 11 15
Remaindering 3 6 11 15. [4] 16 13 & [4's Remainder Group] 3 6 11 15

I wonder if B6 still should prevent pairing 16 and 3. Anyway, it's pairing up with 13 again!

And then it goes on to C12, ignoring 15, because x=0 still and tries to repair the 2.5 Bracket.

That looks like a bug. Or should it only increase x the next time it hits C11?

Look.

C10, Unpairing Player 16 and 13 in Bracket 4. Bracket 4's C10Repair Group: 16 13 3 6 15 11
C7, 15 3 6 11 13
C6, B4: x=0, table 1 NOK
C7, last transposition
C10, No more opponents for Player 16. Re-forming Bracket 4
C11, x=p=1 already, no more x increases in Bracket 4.
C12, Undoing Bracket 3 matches. Re-pairing Bracket 3,

Should one go back and try to pair 16-15 or even 16-3 in order to prevent this double violation of B4 in the remainder group?

I think, one should stay with 16-13, 3-15, 11-6, because otherwise one would get a violation of B4 for the pairing with the downfloated player (16) and that is considered worse than a double violation of B4 for lower players. Or am I wrong again?

Is that the double whammy? Being upfloated again AND not getting one's preference? :-)

Obviously, Games::Tournament::Swiss is doing it the right way. Or do I mean the wrong way?

Bartolin
23-10-2007, 02:07 AM
This pairing hasn't reached C11 so far. It is only in C6,7. So there can't be any x decreases. In fact, I don't think the Procedures allows any x decreases in remainder groups. C10 kicks in first, and then in C11, the whole heterogeneous bracket from which the remainder group came is re-paired.

A remainder group never sees C11, I think.

I think you are right. At least it seems that a remainder group never sees C11 without being re-transformed to a heterogeneous group.

At the moment, Games::Tournament::Swiss is first reaching C10, and undoing the pairing with 13. But it is not able to pair with 15, because they both have the same preference. So it then reaches C11, where x becomes 1.

It then repairs the whole heterogenous bracket. But seeing this is a heterogeneous bracket with only one member from the previous bracket, the same pairings are being considered again:

That sounds sensible to me.

The fact that C11 requires us to re-pair the downfloated players as well seems to imply that there is no special treatment of those players with respect to B4. (I asked that question above) :

Do we have to fulfil the color preferences in the match of the higher player (10) the same way we have to try to minize his score differences first? If not, I think we should take the pairing 10-18, 19-20, 12-14 since that's the first of those two pairing we come accross during appliance of C7.

But after pairing 16 and 13 again, there seems to be a problem:

And then it goes on to C12, ignoring 15, because x=0 still and tries to repair the 2.5 Bracket.

That looks like a bug. Or should it only increase x the next time it hits C11?

It seems to me, that x shouldn't equal zero at this point. Actually we just increased x to be able to get other pairings. And since we didn't "use this x increase" for re-pairing the downfloated player (16), it should be left over for the remainder group. At least everything else looks strange to me. (Wouldn't one end with a kind of circle otherwise?)

On the other hand, the procedure from C says that we have to recalculate x for the remainder group at C2. Actually we have to deal with two different 'x-es': One x for the whole group (including 16 and 13) and one x for the remainder group.

Intuitively, it seems sensible to me to "preserve" the increased x for our remainder group if it's not used for pairing the downfloated players.

drbean
26-10-2007, 09:51 AM
From a quick look this is what I came up with.

I agree with 7 v 8 and 1 v 2.

I dont think the 2.5 score group has been paired correctly, which has implications for the lower score groups.

4 14,9,7,17 BWBW 2.5
5 15,8,13,2 WBWB 2.5
9 19,4,2,7 WBWW 2.5
16 6,19,11,10 WBWB 2.5
17 7,14,18,4 BWBB d 2.5

S1 S2

4 wants black 9 wants black
5 wants white 16 wants white
17 wants white, has downfloated two rounds before.

So you have three players preferencing white, two preferencing black. This means you can get two matching colour pairings (four players should get their correct colour preferences).

4 v 16
5 v 9 seems to work for colours, but downfloating 17 violates B6. So that isnt ideal.

So lets try downfloating 16 as that leaves four players- two want white, two want black.

So this leaves:

S1 S2

4 wants black 9 wants black
5 wants white 17 wants white, has downfloated two rounds before.

4 v 9 and 4 v 17 have played, so that leaves 4 v 5. 9 v 17 is also a legal pairing.

So the pairings should be:

5 v 4
17 v 9
16 downfloats.

With this pairing combination, B4 and B6 are both met.

With 16 being the downfloater, this has a flow on effect for the 2 point score group.

This pairing is what Games::Tournament::Swiss is now producing, thanks to the bug correction. It is producing it both for the pairing table produced by its wrong pairings in previous rounds, and for the right pairing tables from http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm.

This is not the pairing produced in the actual tournament, where 17 is downfloated instead of 16.

I guess 17 is being downfloated despite this being its second downfloat, because of the last-round rule. Which says:

Note: B2, B5 and B6 do not apply when pairing players with a
score of over 50% in the last round.

What is the force of this rule. Is it MUST, SHOULD or CAN? http://www.faqs.org/rfcs/rfc2119.html

What is the reason for it? B2 is an absolute criterion, but it doesn't apply. The only reason I can see is to make it easier for the arbiters to produce pairings under difficult circumstances.

Why is it only for high-scorers? I would have thought their pairings are the ones you have to be more careful about.

Why isn't B3 mentioned? I can see why B4 isn't. B4 has the same effect as B2. Perhaps it's not mentioned because B3 has the same effect as B5 and B6.

I don't see any need for the last round rule.

Garvinator
26-10-2007, 11:28 AM
Note: B2, B5 and B6 do not apply when pairing players with a
score of over 50% in the last round.

What is the force of this rule. Is it MUST, SHOULD or CAN? http://www.faqs.org/rfcs/rfc2119.html It is a must criterion. B2 does not apply at all in the final round under dutch pairing rules.

What is the reason for it? B2 is an absolute criterion, but it doesn't apply. The only reason I can see is to make it easier for the arbiters to produce pairings under difficult circumstances.The reason for it is to try and produce as many pairings as possible with players on the same score between the leaders. Also it helps to reduce ridiculous last round pairings, which is not uncommon if B2 was applied.

Why is it only for high-scorers? I would have thought their pairings are the ones you have to be more careful about.Answered above I believe.

Why isn't B3 mentioned? I can see why B4 isn't. B4 has the same effect as B2. Perhaps it's not mentioned because B3 has the same effect as B5 and B6.Can you explain this further? Each of the B's are for different reasons and cover different situations. B3 doesnt have the same effect as B5 or B6.

B3 covers score difference between different pairing options, B5 and B6 covers float status.

I don't see any need for the last round rule.I can see why it is there and is very useful for tournaments in Australia, where fields of 50-100 players are common in 7 rounds and having rating spreads of 2000 points.

drbean
26-10-2007, 11:46 AM
The next score bracket would consist of:

16 6,19,11,10 WBWB 2.5
3 13,6,8,11 WBWB u 2
6 16,3,1,15 BWBW 2
11 1,12,16,3 BWBW 2
13 3,18,5,19 BWBW 2
15 5,20,10,6 BWWB 2

S1 S2

16 wants white 3 wants white, has upfloated two rounds before
6 wants black
11 wants black
13 wants black
15 wants white

16 already played against 6 and 11, so the pairing should be 16-13, leaving a homogeneous remainder group with

S1 S2

3 wants white 11 wants black
6 wants black 15 wants white

Since 3 already played against 6 and 11, the only possible pairings are 3-15 and 11-6.

That's what Games::Tournament::Swiss is now producing.

Should one go back and try to pair 16-15 or even 16-3 in order to prevent this double violation of B4 in the remainder group?

I think, one should stay with 16-13, 3-15, 11-6, because otherwise one would get a violation of B4 for the pairing with the downfloated player (16) and that is considered worse than a double violation of B4 for lower players. Or am I wrong again?

Perhaps after C11 x increases, 16 should be paired with 3. But I think that the pairing is the one you get just from following the procedures in order. The way Games::Tournament::Swiss is doing it now is:

16 is first paired and then the remainder group is then paired.

C6PAIRS, 1 2 paired. OK
C6PAIRS, E1 5&4 E1 17&9
C6OTHERS, Floating remaining 16 Down. [3's Remainder Group] 4 9 5 17 & [4] 3 6 11 13 15 16
C1, B1,2 test: OK, no unpairables
C5, 3 6 11 13 15
C6PAIRS, B4: x=0, table 1 NOK
C7, 6 3 11 13 15
C6PAIRS, B1a: table 1 NOK
C7, 11 3 6 13 15
C6PAIRS, B1a: table 1 NOK
C7, 13 3 6 11 15
C6PAIRS, B56: OK.
C6PAIRS, 1 paired. OK
C6PAIRS, E1 16&13
C6OTHERS, Remaindering 3 6 11 15.
[4] 16 13 & [4's Remainder Group] 3 6 11 15

But we will find the remainder group can't be paired because of B4 violations.

C5, 3 6
C5, & 11 15
C6PAIRS, B1a: table 1 NOK
C7, 15 11
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, exchange a in 4's Remainder Group
C8, 3 11, 6 15
C6PAIRS, B1a: table 1 NOK
C7, 15 6
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, last S1,S2 exchange in 4's Remainder Group

And then we run through those again after C9 relaxations.

Then a different partner is chosen for 16, through C10 and C11. But in this case, 15 is the only possible partner left. (We have already tried pairing 16 with the others in Bracket 4(2).) And pairing with 15 would violate B4.

C10, Unpairing 16 and 13 in Bracket 4(2)
C10, Bracket 4's C10Repair Group: 16 13 3 11 15 6
C7, 15 3 6 11 13
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C10, No more opponents for Player 16
C10, Giving up on Bracket 4's C10 Repair Group (2C10Repair)

Because we are now repairing all the downfloated players in Bracket 4(2) (although actually in this case with only one downfloated player, we are looking at the same pairings as in C10), I think we should NOT raise x at this point. We give all the other possible pairings of the downfloaters a chance at leading to pairing of the remainder group WITHOUT x raises and thus WITHOUT B4 violations.

C11, Deleting matches in Bracket 4(2)
C11, Bracket 4's C11 Repairing: 16 15 3 6 11 13
C7, 15 3 6 13 11
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C6PAIRS, B4: x=0, table 1 NOK
C7, 6 3 11 13 15
C6PAIRS, B1a: table 1 NOK
C7, 11 3 6 13 15
C6PAIRS, B1a: table 1 NOK
C7, 13 3 6 11 15
C6PAIRS, B56: OK.
C6PAIRS, 1 paired. OK
C6PAIRS, E1 16&13
C6OTHERS, Remaindering 3 6 11 15.
[4 (post-C11)] 16 13 & [4 (post-C11)'s Remainder Group] 3 6 11 15

Note that 3 was passed over again. I think this is correct. I don't know what the first C7 transposition is doing in there. Perhaps that's a bug. For some reason, I think we have to restart at C7.

But because the pairings being considered are the same, we see the same failure as before.

C5, 3 6
C5, & 11 15
C6PAIRS, B1a: table 1 NOK
C7, 15 11
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, exchange a in 4 (post-C11)'s Remainder Group
C8, 3 11, 6 15
C6PAIRS, B1a: table 1 NOK
C7, 15 6
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, last S1,S2 exchange in 4 (post-C11)'s Remainder Group

And then we run through them again for C9 relaxations. Whether post C10 and C11 we need to do that or not, is debatable, I guess.

Now x is increased to 1, and we start waiving B4 violations. I said before that remainder groups never get past C10 and that they are always reincorporated into their heterogeneous brackets, but I've now changed my mind. I think I was wrong.

I think that a remainder group of a C11 repairing of a heterogeneous bracket makes it past C10 into C11, where x increases are applied to allow the pairing of the remainder group, and thus the repairing of the whole heterogeneous bracket.

C11, Trying next pairing in 4 (post-C11)'s Remainder Group(2C11RepairRemainder)
C7, 15 11
C6PAIRS, B4: x=1, table 2 NOK
C7, last transposition
C8, exchange a in 4 (post-C11)'s Remainder Group
C8, 3 11, 6 15
C6PAIRS, B1a: table 1 NOK
C7, 15 6
C6PAIRS, B4: x=1, table 2 NOK
C7, last transposition
C8, last S1,S2 exchange in 4 (post-C11)'s Remainder Group
C11, Trying next pairing in 4 (post-C11)'s Remainder Group(2C11RepairRemainder)
C7, 15 11
C6PAIRS, B56: OK.
C6PAIRS, 1 2 paired. OK
C6PAIRS, E3 3&15 E4 11&6

Again that C7 in there means we aren't considering 3&11 and 6&15. I'm trying to avoid getting caught in an infinite loop by going back to the same place we started at, I guess.

Anyway, this confirms the theoretical derivation of the same result.

This is different than the actual pairing at http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm, because there 17 was downfloated, rather than 16.

This work on C10,11 is all recent, so there are still bugs, I think. I'm trying to get C10,11 to work with C12. If any earlier versions of Games::Tournament::Swiss got pairings correct that involved these procedures, then it was coincidence.

drbean
26-10-2007, 03:04 PM
The tail of the pairing table:

-------------------------------------------------------------------------
Place No Opponents Roles Float Score
15-16
10 20,1,15,16 BWBW 1.5
20 10,15,14,12 WBBW D 1.5
17-19
12 2,11,19,20 WBWB dU 1
18 8,13,17,14 WBWW u 1
19 9,16,12,13 BWBB u 1
20
14 4,17,20,18 WBWB 0

I'll consider the actual pairingtable for round 5. The floats are different and some of the opponents in the 3rd round are different.

15-16
10 20,1,15,16 BWBW 1.5
20 10,15,18,12 WBBW D 1.5
17-19
12 2,11,17,20 WBWB uU 1
18 8,13,20,14 WBWW 1
19 9,16,14,13 BWBB 1
20
14 4,17,19,18 WBWB 0

Floating 10 and 20 down.

S1 S2

10 wants black 12 wants white, has upfloated last round, has downfloated two rounds before
20 wants black 18 wants black, has upfloated two rounds before
19 wants white, has upfloated two rounds before

According to A8, x equals 0 for this score bracket.

Obviously we have to waive B5 for player 20 before being able to get 2 (=p) pairings. Furthermore we have to waive B6 for players 18 and 19 as well -- for the same reason.

The first try (10-18, 19-20) violates B4. Before accepting such a violation we should waive B5 first (for player 12). After doing that we would get 12-10 and 19-20, no longer violating B4.

The problem is, that 18 can't be downfloated to 14 because both played in round 4 already. Therefore we have to go back to the penultimate score bracket.

12&10 and 19&20 is the pairing Games::Tournament::Swiss is also getting first:
It then tries to pair 14 and 18 in the last bracket.

C1, NOK. 10 20: floating down from 5
C1, [5] & [6] 12 18 19 10 20
C1, Bracket 5 (1.5) dissolved.
C5, ordered: 10 20
& 12 18 19
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B4: x=0, table 1 NOK
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B4: x=0, table 1 NOK
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B6Up, table 1: 12 NOK. Floated Up 2 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B4: x=0, table 1 NOK
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C10, Dropping B6 for Upfloats in Bracket 6(1)
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B5Up, table 1: 12 NOK. Floated Up 1 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B4: x=0, table 1 NOK
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C10, Dropping B5 for Upfloats in Bracket 6(1)
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B56: OK.
C6PAIRS, 1 2 paired. OK
C6PAIRS, E1 12&10 E1 19&20
C6OTHERS, Floating remaining 18 Down. [6] 10 20 12 19 & [7] 14 18

But 14 and 18 can't be matched, so 18 is floated back via C13 and C7 continues at the next transposition after 12 19 18, which is 18 12 19.

At first I thought, we had to continue at the situation where we left this score bracket (B6 and B5 being waived for upfloaters). If that were true, we could pair 12-10 and 20-18, downfloating 19 (as I asked in my original post).

In the real pairing table, 20 and 18 have already met, so that is not a possibility. Note C10 is reached without a remainder group being found, and float checks are turned back on again after reaching C11.

C1, NOK. 14 18 in last bracket, 7 (0).
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B4: x=0, table 1 NOK
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C10, Float checks all dropped in Bracket 6(1)
C11, x=1
C11, All float checks on in Bracket 6 (1)
[C5, ordered: 10 20
& 12 18 19 It doesn't actually pass C5]
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition

Let's take a break from the pairing process here.

Now, after getting the hint that those pairings aren't correct, I have another idea. Since we have to violate B4 anyway, we could as well accept the "earlier" violation of B4 (10-18, 19-20), thereby respecting B5 for player 12.

I think it is better just to follow the order of the procedures, because backtracking will take place in C11-13.

Speaking in terms of procedure C, after going back to the penultimate score bracket we have to increase x by 1 according to C11. After that, we go back to C3, thereby putting B6 and B5 back into validity. Now with x=1 we can either pair 10-18, 19-20, downfloating 12 or we could pair 19-10, 20-18, again downfloating 12.

Games::Tournament::Swiss is finding the 10&18, 19&20 pairing first, but this may be due to the slightly different pairing tables.

C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B5Down, table 2: 20 NOK. Floated Down 1 rounds ago
C7, 19 12 18
C6PAIRS, B1a: table 2 NOK
C7, 19 18 12
C6PAIRS, B1a: table 2 NOK
C7, last transposition
C6PAIRS, B1a: table 2 NOK
C7, 12 19 18
C6PAIRS, B6Up, table 1: 12 NOK. Floated Up 2 rounds ago
C7, 18 12 19
C6PAIRS, B1a: table 2 NOK
C7, 18 19 12
C6PAIRS, B56: OK.
C6PAIRS, 1 2 paired. OK
C6PAIRS, E2 10&18 E1 19&20
C6OTHERS, Floating remaining 12 Down. [6] 10 20 18 19 & [7] 14 12

Do we have to fulfil the color preferences in the match of the higher player (10) the same way we have to try to minize his score differences first? If not, I think we should take the pairing 10-18, 19-20, 12-14 since that's the first of those two pairing we come accross during appliance of C7.

Here this is a heterogeneous group, but not a remainder group, so the differences in treatment of downfloaters and that of original members is not so evident. The color clashes are counted in the test of each transposition. It doesn't make a difference whether they're in S1 or S2 as long as the total number of clashes is less than x (xprime).

So, are the pairings 10-18, 19-20, 12-14 correct?

Well, that's what Games::Tournament::Swiss is getting at the moment, so I think it's OK.

But look at the transpositions which are rejected because the the downfloaters downfloated before. As you argued before, there is no point in rejecting a transposition because the downfloater is downfloating twice, particularly where there is no alternative pairing for this player in the bracket above, as is the case here.

The fact that a rejection is only being recorded here, when the same repeated downfloating is occuring at each of the transpositions is an artifact of the way the test is being conducted and reported. B56 violations are only tested if B1,2,4 tests pass.

This probably doesn't change the pairings that Games::Tournament::Swiss is producing. It only means that the cycles through C9 for heterogeneous groups is a waste of time.

Is that correct?

drbean
30-10-2007, 01:18 AM
Note: B2, B5 and B6 do not apply when pairing players with a
score of over 50% in the last round.

It is a must criterion. B2 does not apply at all in the final round under dutch pairing rules.

The reason for it is to try and produce as many pairings as possible with players on the same score between the leaders. Also it helps to reduce ridiculous last round pairings, which is not uncommon if B2 was applied.

I see that B3 is very important and can thus overrule even B2 in the last round.

Why isn't B3 mentioned? I can see why B4 isn't. B4 has the same effect as B2. Perhaps it's not mentioned because B3 has the same effect as B5 and B6.

Can you explain this further? Each of the B's are for different reasons and cover different situations. B3 doesnt have the same effect as B5 or B6.

B3 covers score difference between different pairing options, B5 and B6 covers float status.

When Swiss arbiters developed the swiss tournament, they traded shorter tournaments for determining partners on the fly. In a round robin tournament, you know your partner in the last round even before the tournament starts, and presumably color is also similarly determinable.

At the same time, they maintained the fairness of the roundrobin by having players play others with the same score.

The fly in the ointment of the swiss tournament is that it is not always possible to pair players with someone with the same score. One reason is B1 of course. The other reason is an odd number of players in the same bracket.

Another concern is a balance in playing the 2 colors. The consequence of these problems is floating, which is another word for score differences, kind of.

All I meant was that B3 was about floating and score differences and B5 and B6 are also about floating and score differnces. The aim of both is to control them.

Of course they are all related. B2 color clashes can lead to floating, which leads to B5 and B6.

But B4 and B2 are both about controlling color balance and so stand in some sort of opposition to B3, B5 and B6.

It seems the FIDE rules do a better job with color preferences than floating.

I can see why it is there and is very useful for tournaments in Australia, where fields of 50-100 players are common in 7 rounds and having rating spreads of 2000 points.

Okay. I'll try to implement it.

Bartolin
30-10-2007, 02:31 AM
I guess 17 is being downfloated despite this being its second downfloat, because of the last-round rule. Which says:

Note: B2, B5 and B6 do not apply when pairing players with a
score of over 50% in the last round.

[...]

What is the reason for it? B2 is an absolute criterion, but it doesn't apply. The only reason I can see is to make it easier for the arbiters to produce pairings under difficult circumstances.

Why is it only for high-scorers? I would have thought their pairings are the ones you have to be more careful about.

Why isn't B3 mentioned? I can see why B4 isn't. B4 has the same effect as B2. Perhaps it's not mentioned because B3 has the same effect as B5 and B6.

I don't see any need for the last round rule.

As I understand it, the raison d'être for the last round rule is to allow for as much direct matches for the top places as possible. It would be in the spirit of the Swiss System to determine the tournament winner with direct matches of the leading players. Without the last round rule it would be more likely that the winner didn't play against the second placed player.

The Surfers Paradise Open Division we discussed at
http://chesschat.org/showthread.php?t=7095 might be a good example of this -- at least if Swiss Manager 5 is correct with pairing 2 vs. 5. The pairings of Swiss Perfect (2 vs. 4, 5 vs. 1) don't give player 5 (who is third place before the last round) a chance to defeat the leader to share the first place.

drbean
27-11-2007, 02:48 PM
Here is a pairing of the 2.5-Bracket [4] of the final round of the http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm tournament, paired on the basis of big X pairings of heteroBrackets and their remainder groups, discussed at http://chesschat.org/showthread.php?p=174878#post174878.

This is not the pairing played in the tournament, where 17 was floated down instead of 16, although it had already been floated down 2 rounds before. The tournament followed the Rules, which remove B5,6 checks in the final round from the pairing procedure.

17 7,14,18,4 BWBB d 2.5

16 6,19,11,10 WBWB 2.5

3 13,6,8,11 WBWB u 2
6 16,3,1,15 BWBW 2
11 1,12,16,3 BWBW 2
13 3,18,5,19 BWBW 2
15 5,20,10,6 BWWB 2

3 has already played all but Player 15 of the original members of the 2-Bracket.
6 has played 2 of them and the downfloated 16. The other players have played 1 each. 11 has also played the downfloated 16.

At the same time x=0, which restricts our choices at first. It looks like this bracket should be difficult to pair.

Starting from where the bracket reaches C11:

C11, Bracket 4's C11 Repairing: 16 15 3 6 11 13, with X=1

C4, S1: 16 & S2: 15 3 6 11 13
C5, ordered: 16
& 3 6 11 13 15
C6PAIRS, B6Up, table 1: 3 NOK. Floated Up 2 rounds ago
C7, 6 3 11 13 15
C6PAIRS, B1a: table 1 NOK
C7, 11 3 6 13 15
C6PAIRS, B1a: table 1 NOK
C7, 13 3 6 11 15
C6PAIRS, B56: OK.
2-Bracket (4) tables 1 paired. OK
E1 16&13
0 of 1 X points used. 1 left for remainder group
C6OTHERS, Remaindering 3 6 11 15.
[4] 16 13 & [4's Remainder Group] 3 6 11 15
C2, x=1
[etc]

Going through the C7,8,9 permutations of the remainder group, no pairing is found, so we end up at C11 again.

C11, Repairing of 16 13 in 2-Bracket [4] failed pairing 3 11 15 6.
Trying next pairing with X=1
C7, 15 3 6 11 13
C6PAIRS, B56: OK.
2-Bracket (4) tables 1 paired. OK
E3 16&15
1 of 1 X points used. 0 left for remainder group
C6OTHERS, Remaindering 3 6 11 13.
[4] 16 15 & [4's Remainder Group] 3 6 11 13

But 3 has played all the other players in the remainder group, so it is unpairable.

C11, Repairing of 16 15 in 2-Bracket [4] failed pairing 3 6 11 13.
Trying next pairing with X=1
C9, No pairing with float checks on. Dropping B6 for Downfloats
in 2-Bracket [4]

Here is something I am less confident of. Up to this point in the C11 repairing, only pairings which pass downfloat checks have been accepted and allowed to generate remainder groups.

Now before increasing x, we are going to go through the same pairings in the bracket with the same x=1, except with B6 downfloat checks turned off.

And in fact, we go through them a 3rd time, with B5 downfloat checks turned off.

Up to the point the bracket reached C11, they had already been turned off. I am not too sure if float checks are supposed to be turned on again and gradually relaxed again after reaching C11.

Despite this relaxation the 2nd and 3rd time through, the heteroBracket pairings that are accepted and considered complete are the same as the 1st time through.

Because of that, I am not going to copy them here.

But then the C11 repairing gets to C10 upfloat check waiving and the different pairing of 3&16 is considered. This pairing was never considered even in the earlier C10 cycle, pre-C11, because 3 and 16 both want White.

C9, No pairing with all Downfloat checks dropped in 2-Bracket [4]
C10, No more pairings. Dropping B6 for Upfloats in 2-Bracket [4]
C4, S1: 16 & S2: 15 3 6 11 13
C5, ordered: 16
& 3 6 11 13 15
C6PAIRS, B56: OK.
2-Bracket (4) tables 1 paired. OK
E4 16&3
1 of 1 X points used. 0 left for remainder group
C6OTHERS, Remaindering 6 11 13 15.
[4] 16 3 & [4's Remainder Group] 6 11 13 15
C2, x=0

But this remainder group is also unpairable, and there are no more possibilities. (I don't know why B5 upfloat waiving does not take place. Perhaps that's a bug.)

C11, Repairing of 2-Bracket [4] failed. No more pairings with X=1
Retrying with X=2. All float checks on in 2-Bracket [4]
C4, S1: 16 & S2: 3 6 13 15 11
C5, ordered: 16
& 3 6 11 13 15
C6PAIRS, B6Up, table 1: 3 NOK. Floated Up 2 rounds ago
C7, 6 3 11 13 15
C6PAIRS, B1a: table 1 NOK
C7, 11 3 6 13 15
C6PAIRS, B1a: table 1 NOK
C7, 13 3 6 11 15
C6PAIRS, B56: OK.
2-Bracket (4) tables 1 paired. OK
E1 16&13
0 of 2 X points used. 2 left for remainder group
C6OTHERS, Remaindering 3 6 11 15.
[4] 16 13 & [4's Remainder Group] 3 6 11 15
C2, x=2
C3, p=2. Homogeneous.
C4, S1: 3 6 & S2: 11 15
C5, ordered: 3 6
& 11 15
C6PAIRS, B1a: table 1 NOK
C7, 15 11
C6PAIRS, B56: OK.
2Remainder-Bracket (4's Remainder Group) tables 1 2 paired. OK
E3 3&15 E4 11&6

Note that the remainder group uses both points of the 2 big X points available.

Note that we turned float checking back on, so the 16&3 pairing was not accepted. If we hadn't, its remainder group would have been pairable with 6&11 and 13&15, and we would have had a different pairing of the bracket instead of the one we have produced.

Note, in addition, that the pairing that we have produced is the same as earlier published versions of Games::Tournament::Swiss were producing.
.

Bartolin
27-11-2007, 05:47 PM
Here is a pairing of the 2.5-Bracket [4] of the final round of the http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm tournament, paired on the basis of big X pairings of heteroBrackets and their remainder groups, discussed at http://chesschat.org/showthread.php?p=174878#post174878.

Again an interesting example.

Here is something I am less confident of. Up to this point in the C11 repairing, only pairings which pass downfloat checks have been accepted and allowed to generate remainder groups.

Now before increasing x, we are going to go through the same pairings in the bracket with the same x=1, except with B6 downfloat checks turned off.

And in fact, we go through them a 3rd time, with B5 downfloat checks turned off.

Up to the point the bracket reached C11, they had already been turned off. I am not too sure if float checks are supposed to be turned on again and gradually relaxed again after reaching C11.

I'm not sure as well. But my gut feeling is that it is correct to turn them on again. That seems to meet the requirement "they should be fulfilled as much as possible" about the relative criteria from section B. (If we have to violate B4, then the other criteria should be fulfilled nevertheless.)

As you wrote this decision is an important one, since it influences the pairings:

Note that we turned float checking back on, so the 16&3 pairing was not accepted. If we hadn't, its remainder group would have been pairable with 6&11 and 13&15, and we would have had a different pairing of the bracket instead of the one we have produced.

Bill Gletsos
27-11-2007, 07:42 PM
Personally I dont agree with the arguments being made here.

As far as the hypothetical question goes of not treating it as the last round and therefore downfloating 16 instead of 17 then the correct pairings are:

16 v 13
3 v 15
11 V 6

BTW having decided on those pairings I checked them under Swiss Master 5 and it gives the same pairings.

Bill Gletsos
27-11-2007, 07:49 PM
Also the paiings shown on the web page match those generated by SM5 for only the first 4 rounds.

However with SM5 treating round 5 as the last then there is a difference compared to those on the web page.

Published were.

7 v 8
1 v 2
16 v 4
5 v 9
17 v 3
13 v 6
15 v 11
10 v 18
19 v 20
12 v 14

SM5 were:

7 v 8
1 v 2
16 v 4
5 v 9
17 v 6
3 v 15
13 v 11
10 v 18
19 v 20
12 v 14

Bartolin
27-11-2007, 10:27 PM
Personally I dont agree with the arguments being made here.

Does that mean that you don't agree with the "big X method" (sharing the value of x between the heterogenous group and its remainder group)? But how do we get a value of x greater 0 for the remainder group (3,6,11,15 after pairing 16 v 13)?

As far as the hypothetical question goes of not treating it as the last round and therefore downfloating 16 instead of 17 then the correct pairings are:

16 v 13
3 v 15
11 V 6

BTW having decided on those pairings I checked them under Swiss Master 5 and it gives the same pairings.

Very interesting, thanks!

Also the pairings shown on the web page match those generated by SM5 for only the first 4 rounds.

However with SM5 treating round 5 as the last then there is a difference compared to those on the web page.

Published were.

7 v 8
1 v 2
16 v 4
5 v 9
17 v 3
13 v 6
15 v 11
10 v 18
19 v 20
12 v 14

SM5 were:

7 v 8
1 v 2
16 v 4
5 v 9
17 v 6
3 v 15
13 v 11
10 v 18
19 v 20
12 v 14

This is interesting as well. As I may have said somewhere else, the pairings at the web page were generated using an older version of Swiss Chess http://www.swiss-chess.de/home_en.htm. Unfortunately I don't know exactly which version.

drbean
29-11-2007, 09:30 AM
Personally I dont agree with the arguments being made here.

As far as the hypothetical question goes of not treating it as the last round and therefore downfloating 16 instead of 17 then the correct pairings are:

16 v 13
3 v 15
11 V 6

BTW having decided on those pairings I checked them under Swiss Master 5 and it gives the same pairings.

Thanks for going to the trouble of doing these by hand
and with Swiss Master 5.

You, Swiss Master and Games::Tournament::Swiss all get the
same results. Which is good news.

drbean
04-12-2010, 02:02 PM
Place No Opponents Roles Float Score
15-16
10 20,1,15,16 BWBW 1.5
20 10,15,14,12 WBBW D 1.5
17-19
12 2,11,19,20 WBWB dU 1
18 8,13,17,14 WBWW u 1
19 9,16,12,13 BWBB u 1
20
14 4,17,20,18 WBWB 0

Bill Gletsos, Bartolin, Swiss Master 5 were
pairing this:

10&18
19&20
12&14

Games::Tournament::Swiss had been pairing it this way too.

But Games::Tournament:Swiss version 0.17 and some
earlier versions started pairing it:

12&10
20&18
19&14

Now it is pairing it as above. I guess that is the correct pairing.

antichrist
04-03-2011, 11:23 AM
Originally Posted by mowcop
Hi David, I would agree that the pairings you've put up seem natural, but that doesn't mean the ones posted are incorrect, or that yours are technically correct. I'm not the arbiter so there isn't much I can do but if I was I'd be fairly happy with the pairings as they are produced by SP.

8 Christopher Wallis 2232 [2.5] : Richard Voon 1758 [2.5]

9 Dusan Stojic 2173 [2.5] : Sarah Anton 1650 [2.5]

10 Jim Papadinis 1668 [2.5] : David J Garner 2170 [2.5]

11 Carl Gorka 2082 [2] : Michael Addamo 1641 [2.5]

12 Bosko Mijatovic 1783 [2.5] : Ian Stone 1818 [2]

13 Kerry Stead 1923 [2.5] : Laurence Matheson 2043 [2]

If we use the Dutch pairings to order the heterogeneous group it is important to start with a correct classification.

a) There are 11 players paired in the 2.5 group (heterogeneous)
b) Matheson is the floater from the 2 group.

We split them up into two groups S1 +S2 ,removing the floater who is paired down . S1 is always rounded down (not up)

So we get....
S1 (5 players)
Christopher Wallis 2232 [2.5]
Dusan Stojic 2173 [2.5]
David J Garner 2170 [2.5]
Carl Gorka 2082 [2]
and

S2 (5players)
Ian Stone 1818 [2]
Bosko Mijatovic 1783 [2.5]
Richard Voon 1758 [2.5]
Sarah Anton 1650 [2.5]

S3 (1 Player)

I made a mistake here I thought Anton was lower rated then Adammo.
I presume that on order Adammo should float down into a bracket with Matheson, who should get the color he is due.

Nevertheless, my point about the pairings is correct I think. The pairing of Mijatovic with Stone is incorrect. How is it possible?

Stead should NEVER be floating down in this situation because he is the 5th player in a group of 11.

If the floater is taken from the middle it will be Stone or Mijatovic, not Stead.

If the floater is taken from the bottom it will be Adammo (not anton)

The floating rules ought to be explained by someone like Grant.

It is of course important to know which pairing system is being used.

Originally Posted by mowcop
I'm not happy about floating Sarah down as usually a middle player in the pool drops.
I am wrong about Anton, I should have suggested Adammo because he is the lowest. You are correct on this Carl.

Originally Posted by mowcop
That leaves Chris, Dusan, David and myself to follow colours against the opponents cited, leaving Ian Stone and Kerry Stead as next highest rated.
Stone should be classified below Stead in the group split which should place him at the top of S2, not the bottom of S1. He then gets the correct color. I may be wrong, but that is how I see it, so I would like to understand this issue..

Originally Posted by mowcop
As Ian is due black it seems natural for him to stay up to even the numbers on 2.5 and Kerry drops, though this is harsh on Laurence who gets another black.
This does not seem right.

Originally Posted by mowcop
So that just leaves matching the bottom half players and I think the only difference I'd have made would have been to swap Bosko and Jim Papadinis.
I think we ought to be clear in understanding where Stone is placed in the pairing order. If he is in S2 he can never be paired against Mijatovic, nor could he be paired against Papadinis. It looks to me like Stone was placed in S1 and that is why he is paired against Mijatovic. Is it wrong for the arbiter/swiss perfect to pair Stone from the S1 group? This seems an important question to answer.

Originally Posted by mowcop
However, this is not based on a detailed look at the pairings as it's not really my job.
Yes fair point. But I still think the pairing is wrong. I accept I may be wrong, but I am trying to understand the why?

Bill Gletsos
04-03-2011, 03:37 PM
Originally Posted by mowcop
Hi David, I would agree that the pairings you've put up seem natural, but that doesn't mean the ones posted are incorrect, or that yours are technically correct. I'm not the arbiter so there isn't much I can do but if I was I'd be fairly happy with the pairings as they are produced by SP.

8 Christopher Wallis 2232 [2.5] : Richard Voon 1758 [2.5]

9 Dusan Stojic 2173 [2.5] : Sarah Anton 1650 [2.5]

10 Jim Papadinis 1668 [2.5] : David J Garner 2170 [2.5]

11 Carl Gorka 2082 [2] : Michael Addamo 1641 [2.5]

12 Bosko Mijatovic 1783 [2.5] : Ian Stone 1818 [2]

13 Kerry Stead 1923 [2.5] : Laurence Matheson 2043 [2]

If we use the Dutch pairings to order the heterogeneous group it is important to start with a correct classification.

a) There are 11 players paired in the 2.5 group (heterogeneous)
b) Matheson is the floater from the 2 group.

We split them up into two groups S1 +S2 ,removing the floater who is paired down . S1 is always rounded down (not up)

So we get....
S1 (5 players)
Christopher Wallis 2232 [2.5]
Dusan Stojic 2173 [2.5]
David J Garner 2170 [2.5]
Carl Gorka 2082 [2]
and

S2 (5players)
Ian Stone 1818 [2]
Bosko Mijatovic 1783 [2.5]
Richard Voon 1758 [2.5]
Sarah Anton 1650 [2.5]

S3 (1 Player)

I made a mistake here I thought Anton was lower rated then Adammo.
I presume that on order Adammo should float down into a bracket with Matheson, who should get the color he is due.

Nevertheless, my point about the pairings is correct I think. The pairing of Mijatovic with Stone is incorrect. How is it possible?

Stead should NEVER be floating down in this situation because he is the 5th player in a group of 11.

If the floater is taken from the middle it will be Stone or Mijatovic, not Stead.

If the floater is taken from the bottom it will be Adammo (not anton)

The floating rules ought to be explained by someone like Grant.

It is of course important to know which pairing system is being used.

Originally Posted by mowcop
I'm not happy about floating Sarah down as usually a middle player in the pool drops.
I am wrong about Anton, I should have suggested Adammo because he is the lowest. You are correct on this Carl.

Originally Posted by mowcop
That leaves Chris, Dusan, David and myself to follow colours against the opponents cited, leaving Ian Stone and Kerry Stead as next highest rated.
Stone should be classified below Stead in the group split which should place him at the top of S2, not the bottom of S1. He then gets the correct color. I may be wrong, but that is how I see it, so I would like to understand this issue..

Originally Posted by mowcop
As Ian is due black it seems natural for him to stay up to even the numbers on 2.5 and Kerry drops, though this is harsh on Laurence who gets another black.
This does not seem right.

Originally Posted by mowcop
So that just leaves matching the bottom half players and I think the only difference I'd have made would have been to swap Bosko and Jim Papadinis.
I think we ought to be clear in understanding where Stone is placed in the pairing order. If he is in S2 he can never be paired against Mijatovic, nor could he be paired against Papadinis. It looks to me like Stone was placed in S1 and that is why he is paired against Mijatovic. Is it wrong for the arbiter/swiss perfect to pair Stone from the S1 group? This seems an important question to answer.

Originally Posted by mowcop
However, this is not based on a detailed look at the pairings as it's not really my job.
Yes fair point. But I still think the pairing is wrong. I accept I may be wrong, but I am trying to understand the why?Firegoat7 is simply wrong.

The pairings for the players on 2.5 (including Gorka and Stone) are in accordance with the Dutch rule and are correct.

Bill Gletsos
04-03-2011, 03:59 PM
There is no hetrogenous group.
The group is homogeneous and consists solely of the the players on 2.5 (incuding Gorka and Stone).

Wallis, Stojic, Garnert, Gorka, Stead Mijakovic ana Papadinis all have a white colour preference. The other five have black colour preferences.

So what you get is:

q=6
w=6
b=5
p=5
x=0 (this means all 5 pairings must be colour matches)

S1
Wallis
Stojic
Garner
Gorka

S2
Stone
Mijatovic
Voon
Anton

The down float must be a player who has a white colour pref (as there are more white colour prefs than blacks).

Papadinis was a down float in round 3 as was Mijatovic who had the bye in round 3. therefore if they are down floated they violate rule B6. Hence Stead is the first possible down float.

Therefore S1 becomes

Wallis
Stojic
Garner
Gorka
Stone

and S2
Mijatovic
Voon
Papdinis
Anton

A straight S1 V S2 causes multiple colour preference violations.

So you need to carry out the necessary transpositions until you get all 5 pairings being colour matches.

Mijatovic cannot play Wallis, Stojic or Gorka as wrong colour pref and he has already played Garner. He there gets paired against Stone.

Voon is a colour match for Wallis.
Papadinis has a white colour pref so is matched against Garner.
Anton is a colour match for Stojioc and Addamo for Gorka.

The pairings are therefore

Wallis - Voon
Stojic - Anton
Mijatovic - Stone

Thus endeth the lesson.

antichrist
05-03-2011, 05:35 PM
FYI, my preference for white is just as strong as Matheson's for black (I entered the tournament in round 2 & was given a 0.5 bye for round 1) ... and my colour history is -BWB.

Of course this may further complicate matters as far as the technicalities are concerned ...
-------------------------------------------------

It is not clear to me what colour Bill allocated to him, only we know he floats down

Kevin Bonham
05-03-2011, 07:31 PM
It is not clear to me what colour Bill allocated to him, only we know he floats down

Stead being white in the pairing Stead - Matheson is correct. Both Stead and Matheson have colour history -BWB (neither played the first round) and therefore both would ideally get white. But Stead is ranked above Matheson because although Matheson has the higher rating, Stead has the higher score, which has priority (see A2 at http://www.fide.com/component/handbook/?id=83&view=article). Therefore Stead alternates to the previous round and is white, while Matheson is black.

Also it's clear to me that Matheson is the correct opponent for Stead. Matheson is the second highest rated player in the 2/4 group. The highest is Rujevic who was upfloated in round 3 and therefore cannot upfloat in round 5. The 2 group plus Stead has a massive imbalance of players seeking white over black (I think it's 9 to 4) so the pairing involving Stead is not required to satisfy both player's colour preferences. While firegoat says Matheson shouldn't have black again, ideally that is true for most of the 2 group and in practice some of them will unavoidably get a second black in a row.

[EDIT:Second paragraph was premature as I was not aware of the badly documented A7d change. See discussion below.]

05-03-2011, 07:37 PM
Bill, any explanation for my pairing against Matheson??
I get the whole colour balancing with the group on 2.5, meaning I float down to balance colours (although on principle it seems odd given that it is an odd numbered round & a large number of players in the grouing have a mild colour preference ... but they're the rules it seems).

What I don't get is why I am paired with Matheson?
As the floater, I should get the highest seeded player in the next score group, providing no other pairing criteria are violated ...
so Rujevic (up float in round 3) is not possible.

Why does A7d not also apply?
d.While pairing an odd-number round players having a strong colour preference (players who have had an odd number of games before by any reason) shall be treated like players having an absolute colour preference as long as this does not result in additional downfloaters (GA 2001)

Both Matheson & myself had a bye in round 1 & our colour history is -BWB. Why does the rule about absolute colour preferences apparently not apply in this case?

The alternative pairing would see me playing white against S Stojic ... any reason why this is not the case??

Of course its all academic, as the pairings have been published, but it would be good to know why ...

**edit** of course Kevin explained it all while I was typing up the question ... :hmm:

ER
05-03-2011, 07:54 PM
**edit** of course Kevin explained it all while I was typing up the question ... :hmm:
So Kerry are you satisfied with Kevin's explanation?

Kevin Bonham
05-03-2011, 08:20 PM
Actually I'm curious about that A7d business Kerry mentions because I've actually never seen that rule (?) before.

Sometimes it has turned out in the past that stuff like that appearing on the FIDE website isn't actually officially adopted, strangely enough.

If A7d is officially part of the system then it is more recent than the pairing module in SP which would therefore be out of date on that issue.

05-03-2011, 08:20 PM
So Kerry are you satisfied with Kevin's explanation?
Nothing wrong with the explanation, just seems odd in the rules that if it is at all possible to keep colours balanced that this becomes a priority (over top half v bottom half), as with the 2.5 score group ... and then this is just a quickly ignored when a score group has to have a number of colour imbalances, as with the 2 score group.
Just seems somewhat illogical to prioritise it so much in one circumstance & then drop in in another, using Bill's formula (A8 from pairing rules) for x using w, b & q ...

* edit * Kevin, we really need to stop writing our posts at the same time ...

Kevin Bonham
05-03-2011, 08:49 PM
Nothing wrong with the explanation, just seems odd in the rules that if it is at all possible to keep colours balanced that this becomes a priority (over top half v bottom half), as with the 2.5 score group ... and then this is just a quickly ignored when a score group has to have a number of colour imbalances, as with the 2 score group.

It's not ignoring it in the latter case but realising that it cannot allocate all colours fairly within the group it has created. After Stead is downfloated to the 2 group it temporarily treats Stead plus all the 2/4ers as a single heterogeneous group. So within that group, if exchanges are applied to fix Stead's colour, that just moves the colour problem elsewhere in the group leaving the same number of colour problems.

I'm not quite sure why the system isn't written to consider colour equalisation in the pairings involving downfloaters to a heterogeneous group as more important than colour equalisation in the rest of the group, and therefore worth applying exchanges to achieve. Perhaps the problem with that is an increased incidence of cushy downfloats/harsh upfloats.

* edit * Kevin, we really need to stop writing our posts at the same time ...

:lol: That one was the same minute!

Oepty
05-03-2011, 09:24 PM
Actually I'm curious about that A7d business Kerry mentions because I've actually never seen that rule (?) before.

Sometimes it has turned out in the past that stuff like that appearing on the FIDE website isn't actually officially adopted, strangely enough.

If A7d is officially part of the system then it is more recent than the pairing module in SP which would therefore be out of date on that issue.

What is on the website is definitely different to the 2005 rules in the 3rd Edition of Reuben's book. I can't remember seeing this change before but I suspect Bill would know when it happened and if it has been officially changed. I wonder why the changes have GA 2001 by them.
The last sentence of A7c was previously A7d and A7d and A7e seem to be completely new. If the changes are correct it probably will have quite an impact on Australian tournaments as there is a lot of half points given. Really makes SP obsolete
Scott

Craig_Hall
05-03-2011, 10:23 PM
Purely a guess, but it seems more likely that it's meant to be GA 2010, not GA 2001, but one can never tell with FIDE.

Kevin Bonham
05-03-2011, 10:31 PM
Purely a guess, but it seems more likely that it's meant to be GA 2010, not GA 2001, but one can never tell with FIDE.

I'm pretty sure it is not GA 2010 since it was not on the agenda for the GA or for the Pairings Committee, and it is not something I have seen as an outcome of that recent FIDE congress. Can't find any 2001 documentation online, not even sure if there was a GA in that year!

Oepty
05-03-2011, 10:35 PM
Purely a guess, but it seems more likely that it's meant to be GA 2010, not GA 2001, but one can never tell with FIDE.

You might be right but I have looked in the minutes for the 2010 general assembly and can not find anything on it. Still as you say it is FIDE so almost anything is possible
Scott

Oepty
05-03-2011, 10:40 PM
I'm pretty sure it is not GA 2010 since it was not on the agenda for the GA or for the Pairings Committee, and it is not something I have seen as an outcome of that recent FIDE congress. Can't find any 2001 documentation online, not even sure if there was a GA in that year!

I found a report to the USCF from the 2001 general assembly but nothing about pairing rules in it. I think they hold a general assembly every year.
Scott

Kevin Bonham
05-03-2011, 10:42 PM
As an aside it's worth noting that not everything in the rules in the 2nd and 3rd editions of Reuben's book is official as of the time those volumes were published either (unlike the 1st edition which was completely FIDE-kosher but is now very dated). See Gijssen's review of 2nd ed: http://www.chesscafe.com/text/review298.pdf

Kevin Bonham
05-03-2011, 10:50 PM
I think they hold a general assembly every year.

Yes, they do it seems. Certainly there was one in 2001 in Greece.

Bill Gletsos
06-03-2011, 12:36 AM
As far as I can determine although A7d & A7e were decided upon at the 2001 General Assembly they did not appear on the FIDE website until late in 2010.
Prior to appearing on the FIDE website in 2010 they do not appear to have been generally published anywhere else.

Interestingly what is shown as A7d was mentioned in the minutes of the Swiss Pairings Committee of 13th October 2009 at the FIDE Congress in Greece.

However in those minutes it was referred to as A7c.
Also what currently appears on the FIDE website as A7d was changed at that meeting to be:

While pairing an odd-number round players having a strong colour preference (players who have had a bye before) shall be treated like players having an absolute colour preference (according to B2a and B2b) as long as this does not result in additional downfloaters.

As such it appears that even now the FIDE website is not listing the last version as the changes approved at the recent FIDE Congress in 2010 in Khanty-Mansiysk are missing.

ER
06-03-2011, 06:47 AM
Nothing wrong with the explanation, just seems odd in the rules that if it is at all possible to keep colours balanced that this becomes a priority (over top half v bottom half), as with the 2.5 score group ... and then this is just a quickly ignored when a score group has to have a number of colour imbalances, as with the 2 score group.
Just seems somewhat illogical to prioritise it so much in one circumstance & then drop in in another, using Bill's formula (A8 from pairing rules) for x using w, b & q ...

* edit * Kevin, we really need to stop writing our posts at the same time ...

Thanks, :) good luck Monday night!

Oepty
06-03-2011, 08:42 AM
As far as I can determine although A7d & A7e were decided upon at the 2001 General Assembly they did not appear on the FIDE website until late in 2010.
Prior to appearing on the FIDE website in 2010 they do not appear to have been generally published anywhere else.

Interestingly what is shown as A7d was mentioned in the minutes of the Swiss Pairings Committee of 13th October 2009 at the FIDE Congress in Greece.

However in those minutes it was referred to as A7c.
Also what currently appears on the FIDE website as A7d was changed at that meeting to be:

As such it appears that even now the FIDE website is not listing the last version as the changes approved at the recent FIDE Congress in 2010 in Khanty-Mansiysk are missing.

Bill, so you are saying it took FIDE 9 years to tell everyone they had changed the rules. That is awful even by FIDE's poor standards.
Scott

Kevin Bonham
06-03-2011, 11:49 AM
Bill, so you are saying it took FIDE 9 years to tell everyone they had changed the rules. That is awful even by FIDE's poor standards.
Scott

It's a bit like "if a tree falls in the forest and no one hears it does it make a sound?" - If FIDE changes a rule and doesn't document it (beyond perhaps the minutes of the Pairing Committee) has the rule even been changed?

Note also the two different wordings of the rule in question:

"While pairing an odd-number round players having a strong colour preference (players who have had a bye before) shall be treated like players having an absolute colour preference (according to B2a and B2b) as long as this does not result in additional downfloaters."

vs

"d.While pairing an odd-number round players having a strong colour preference (players who have had an odd number of games before by any reason) shall be treated like players having an absolute colour preference as long as this does not result in additional downfloaters".

(The second wording is clearly better as there is no reason to limit the reasons why a player would have had an odd number of games for the rule to apply.)

So it looks like it may be the case that the pairing Stead-Matheson is actually wrong, but wrong under a change that postdates the SP pairing module and that was actually made 10 years ago but with FIDE only recently publishing it. But there may also be more to come about this.

Grant Szuveges in the other place poses the following test:

I also would like to put a challenge to every single arbiter in Australia: see if you can work out the "correct" pairings for round 6 of the MCC Championship manually WITHOUT using a computer. I highly doubt that any of you could do it. Post your pairings here and then we will let the computer do the draw and see how many people actually got them all correct....

The fact is that even if there are differences between what an arbiter gets and what SP (or any other program) gets, that does not mean the arbiter is wrong. It may mean that the pairing program was wrong. Many pairing programs have been shown to have produced wrong pairings here, including SP many times. But there have also been many cases where human arbiters have overridden SP, usually under pressure from player complaints, and then it has turned out that SP was right and they were wrong.

It's not impossible to get it right manually. It just takes a long time.

Oepty
06-03-2011, 12:36 PM
It's a bit like "if a tree falls in the forest and no one hears it does it make a sound?" - If FIDE changes a rule and doesn't document it (beyond perhaps the minutes of the Pairing Committee) has the rule even been changed?

Note also the two different wordings of the rule in question:

"While pairing an odd-number round players having a strong colour preference (players who have had a bye before) shall be treated like players having an absolute colour preference (according to B2a and B2b) as long as this does not result in additional downfloaters."

vs

"d.While pairing an odd-number round players having a strong colour preference (players who have had an odd number of games before by any reason) shall be treated like players having an absolute colour preference as long as this does not result in additional downfloaters".

(The second wording is clearly better as there is no reason to limit the reasons why a player would have had an odd number of games for the rule to apply.)

So it looks like it may be the case that the pairing Stead-Matheson is actually wrong, but wrong under a change that postdates the SP pairing module and that was actually made 10 years ago but with FIDE only recently publishing it. But there may also be more to come about this.

Grant Szuveges in the other place poses the following test:

The fact is that even if there are differences between what an arbiter gets and what SP (or any other program) gets, that does not mean the arbiter is wrong. It may mean that the pairing program was wrong. Many pairing programs have been shown to have produced wrong pairings here, including SP many times. But there have also been many cases where human arbiters have overridden SP, usually under pressure from player complaints, and then it has turned out that SP was right and they were wrong.

It's not impossible to get it right manually. It just takes a long time.

I agree with Kevin if A7d is actaully part of the rules then Stead and Matheson should not be playing, but as we are not sure about what the rules really are the pairings should stand.

As far as Grant's challenge goes I think that there are alot of arbiters in the country who could do manual pairings for the tournament correctly. Some aribters have had to manually pair far larger tournaments in the past. It in some cases took hours to do, but it was done and done well. I would hope I could get the pairings correct myself and I am sure Bill would. I would be willing to attempt to do the pairings for round 6 if enough information is given publically. It would be an interesting test.

As Kevin said getting it right is not the same thing as getting the same as what SP gives, SP sometimes gets things wrong.
Scott

Bill Gletsos
06-03-2011, 04:52 PM
Over on the toolbox Kerry says:

The part I don't get is that if colour balancing is so important, then I should logically be playing the highest seeded player on 2/4 who is due black (that does not violate other criteria). This rules out Rujevic (floated up in round 3) & Matheson (due white with -BWB history), so should logically leave Svetosar Stojic as the next available player ...As I pointed out over there Kerry's first sentence isnt actually true.

It works this way.

Kerry floats down to the 2 point score group.
Kerry is the only player in S1, the other 12 players are in S2 as per A6.

S1

S2
Rujevic
Matheson
Stojic
Dowling
Fitzmaurice
Beaumont
Lycett
Krunic
Wyss
Kovacevic
Beattie
Dale

Kerry has a white colour preference as do 7 of the other 12 players.

So from A6 and A8 we have:

p=1
w=8
b=5
q=7
x=1

Since x=1 it means that for the one pairing we are doing it does not have to be a colour match.

Rujevic cannot be a match since he violates B6.
Stead v Matheson is correct if the previously unknown A7d is ignored.
Note if Matheson had been white in round 1 then Stead v Matheson would still be the correct pairing even if A7d were applied.

ER
06-03-2011, 05:22 PM
hehe, you guys are all going bonkers, except those of you who are already there!
I mean after all this concentrated crap who would want to go and do FIDE Arbiter courses?
Better go learn and play Chess!
Or even better do what I do and start roaming around Aus (best country in the world by far) under the pretext of playing Chess! :P

06-03-2011, 05:27 PM
Over on the toolbox Kerry says:
As I pointed out over there Kerry's first sentence isnt actually true.

It works this way.

Kerry floats down to the 2 point score group.
Kerry is the only player in S1, the other 12 players are in S2 as per A6.

S1

S2
Rujevic
Matheson
Stojic
Dowling
Fitzmaurice
Beaumont
Lycett
Krunic
Wyss
Kovacevic
Beattie
Dale

Kerry has a white colour preference as do 7 of the other 12 players.

So from A6 and A8 we have:

p=1
w=8
b=5
q=7
x=1

Since x=1 it means that for the one pairing we are doing it does not have to be a colour match.

Rujevic cannot be a match since he violates B6.
Stead v Matheson is correct if the previously unknown A7d is ignored.
Note if Matheson had been white in round 1 then Stead v Matheson would still be the correct pairing even if A7d were applied.

While I agree with the numbers you have used, I'm still not clear about the result (ie: the Stead-Matheson pairing).

Using the numbers in A6 & A8 gives x=1, however it seems odd to use the colour preferences for the entire score group on 2 if the number of pairings to be made (p) is one (as S1 is made up entirely of the downfloater{s} from the score above).
Lets say that the colour balance was different & for whatever reason, EVERYONE is due white ... then our numbers would be:
p=1
w=13
b=0
q=7
x=6
If we are only worried about one pairing at the moment (given that there is only one player in S1), why does it matter that 6 pairings will not follow colour matching? They appear to be independent numbers ...
The formula seems to consider things if you are pairing the entire group of players (both S1 & S2) rather than simply finding an opponent for the 'floater' in S1 ...

Seeing as we know about A7d, why is it being ignored?? Does it matter if it was previously unknown?

06-03-2011, 05:36 PM
hehe, you guys are all going bonkers, except those of you who are already there!
I mean after all this concentrated crap who would want to go and do FIDE Arbiter courses?
Better go learn and play Chess!
Or even better do what I do and start roaming around Aus (best country in the world by far) under the pretext of playing Chess! :P
Its about minimising areas that fall into the 20% that you can get wrong in the test at the arbiters course :P

ER
06-03-2011, 05:53 PM
Its about minimising areas that fall into the 20% that you can get wrong in the test at the arbiters course :P

Fine it's OK for passing the test then. Actually, in Maths I used to pass with 55 and 60 %s by just remembering formulae which I had not the slightest clue how to operate by just sticking them in the answer space, showing that I know which applied to which prob and claiming headaches, toothaches etc for not being able to complete the paper.
Of course I 'd get some 30 - 40 %s by doing the piss easy trigo and algebra crappy ones in the beginning.
Right so far!
However, when it comes to the real life imagine you being the arbiter or whoever does the draw for next round and you have to face ALL THIS (scroll up and have a look :eek: :doh: )!
http://i266.photobucket.com/albums/ii269/theogrit/Large%20Smilies/1lg003prayer1.gif
Also imagine that's not only a couple of players who complain or query but 7-8 or even more!
I bet you, showing them the FA certificate won't convince them! :) :P

Bill Gletsos
06-03-2011, 05:54 PM
While I agree with the numbers you have used, I'm still not clear about the result (ie: the Stead-Matheson pairing).

Using the numbers in A6 & A8 gives x=1, however it seems odd to use the colour preferences for the entire score group on 2 if the number of pairings to be made (p) is one (as S1 is made up entirely of the downfloater{s} from the score above).
Lets say that the colour balance was different & for whatever reason, EVERYONE is due white ... then our numbers would be:
p=1
w=13
b=0
q=7
x=6
If we are only worried about one pairing at the moment (given that there is only one player in S1), why does it matter that 6 pairings will not follow colour matching? They appear to be independent numbers ...
The formula seems to consider things if you are pairing the entire group of players (both S1 & S2) rather than simply finding an opponent for the 'floater' in S1 ...You might think that but that isnt how it works.

Seeing as we know about A7d, why is it being ignored?? Does it matter if it was previously unknown?Did the arbiter give any indication they would override the computer pairings?

06-03-2011, 07:16 PM
You might think that but that isnt how it works.

Did the arbiter give any indication they would override the computer pairings?
As far as your question is concerned, I'm not sure. I missed round 1 & most pre-round announcements since then have simply been in regards to turning off mobile phones, where to find arbiter's contact details if needed between rounds, etc rather than anything to do with following the computer pairings or over-riding them manually.

As for your first point, why use the formulas for x & p again if they aren't needed? Surely the other rules (A7, B5, B6, etc) also need to be followed? Or are you saying (based on your question) that the computer got it wrong by not considering A7?

Kevin Bonham
06-03-2011, 08:29 PM
If we are only worried about one pairing at the moment (given that there is only one player in S1), why does it matter that 6 pairings will not follow colour matching? They appear to be independent numbers ...

It matters because the point of the formula is to establish to what extent colour issues in a group can be fixed. For that single pairing it doesn't matter if x is anything from 1 to 6, because in all those cases it will do the same thing (keep the first allowable single pairing even if it is a colour mismatch). But if x is zero that makes a difference to how it handles the pairing.

Suppose that in the 13 player group (Stead + 12 on 2) the split of players due white vs players due black is actually 7-6 not 9-4. In this case x=0, which means that in the one pairing to be made, a pairing not satisfying the colour preferences of both players is not allowed. In this case it would find a different opponent for Stead.

The reason for this is that there is a prospect in that case of avoiding any colour mismatches in that group, since with Stead paired against a black-seeking opponent, the split of the remainder is 6-5 and it should be possible to pair the rest of the group without colour issues by downfloating one of the white-seekers to the 1.5 group.

But if x>0 then there are so many white-seekers that if you change Stead's opponent so that he plays a black-seeker, the rest of the heterogeneous group will still have a colour mismatch somewhere, and you won't be able to pair the remainder of the 2 group without one.

The point of the value of x is basically "Can exchanges help reduce the number of colour clashes in this group? To the extent they can do it, to the extent they can't don't bother trying".

The formula seems to consider things if you are pairing the entire group of players (both S1 & S2) rather than simply finding an opponent for the 'floater' in S1 ...

Yes, which is odd. I mentioned a possible reason why in #64.

Surely the other rules (A7, B5, B6, etc) also need to be followed? Or are you saying (based on your question) that the computer got it wrong by not considering A7?

SP was programmed before this new A7d text was (apparently) added so it's apparently a case of SP's pairing module being out of date.

06-03-2011, 09:47 PM
It matters because the point of the formula is to establish to what extent colour issues in a group can be fixed. For that single pairing it doesn't matter if x is anything from 1 to 6, because in all those cases it will do the same thing (keep the first allowable single pairing even if it is a colour mismatch). But if x is zero that makes a difference to how it handles the pairing.

Suppose that in the 13 player group (Stead + 12 on 2) the split of players due white vs players due black is actually 7-6 not 9-4. In this case x=0, which means that in the one pairing to be made, a pairing not satisfying the colour preferences of both players is not allowed. In this case it would find a different opponent for Stead.

The reason for this is that there is a prospect in that case of avoiding any colour mismatches in that group, since with Stead paired against a black-seeking opponent, the split of the remainder is 6-5 and it should be possible to pair the rest of the group without colour issues by downfloating one of the white-seekers to the 1.5 group.

But if x>0 then there are so many white-seekers that if you change Stead's opponent so that he plays a black-seeker, the rest of the heterogeneous group will still have a colour mismatch somewhere, and you won't be able to pair the remainder of the 2 group without one.

The point of the value of x is basically "Can exchanges help reduce the number of colour clashes in this group? To the extent they can do it, to the extent they can't don't bother trying".

Yes, which is odd. I mentioned a possible reason why in #64.

SP was programmed before this new A7d text was (apparently) added so it's apparently a case of SP's pairing module being out of date.

Ah, that kind of makes sense ... so its important if x=0, so that a colour balance can be maintained, but if its so lop-sided one way or the other, then it doesn't make much difference because by definition you're going to have colour imbalances in the pairings regardless of what you do.

Is it possible to modify the programming of SP's pairing module to reflect the 'new' A7d?

Garvinator
06-03-2011, 09:56 PM
Is it possible to modify the programming of SP's pairing module to reflect the 'new' A7d?No, my understanding is that the sp program is under copyright and the ACF paid a fee to use it and distribute.

There are quite a few pairing laws that sp does not recognise as it is a 1998 program, but tournaments still use it because they get it for free.

Garvinator
06-03-2011, 10:03 PM
After inputting all the results into SM5, this is what I get for Round 5, 2 point score group:

13 Stead, Kerry (2.5) - Stojic, Svetozar ( 2 ) 17- 16
14 Lycett, Garry (2 ) - Rujevic, Mirko ( 2 ) 26- 5
15 Matheson, Laurence (2 ) - Krunic, Michael ( 2 ) 14- 30
16 Dowling, John (2 ) - Kovacevic, Paul ( 2 ) 18- 33
17 Fitzmaurice, Garth (2 ) - Beattie, Roger ( 2 ) 21- 38
18 Wyss, Felix (2 ) - Beaumont, David ( 2 ) 32- 24
19 Ly, Thai (1.5) - Dale, Ari ( 2 ) 20- 41
20 Chmiel, Rad (1.5) - Palma, Mario ( 1.5) 39- 43
21 Ogden, Marcus W (1.5) - Roberts, Jono ( 1.5) 40- 59
22 Renzies, Elliott (1.5) - Brown, Kevin M ( 1.5) 53- 42

Kevin Bonham
06-03-2011, 10:11 PM
After inputting all the results into SM5, this is what I get for Round 5, 2 point score group:

Which looks consistent with the mystery rule - unless it is right for the wrong reasons somehow.

Carl Gorka
06-03-2011, 10:51 PM
Just to mix things up, I am no longer in the 2.5 player pool, nor is Ian Stone. I won our postponed game tonight:D

Garvinator
06-03-2011, 11:33 PM
Just to mix things up, I am no longer in the 2.5 player pool, nor is Ian Stone. I won our postponed game tonight:DAhh this is what I like to see. All this work done on the pairings and it goes up in smoke just like that :)

Bill Gletsos
06-03-2011, 11:36 PM
One thing is clear.
Firegoat7 is clueless when it comes to understanding the FIDE Swiss Dutch system.
Trying to explain it to him is a total waste of time.

Garvinator
06-03-2011, 11:36 PM
Which looks consistent with the mystery rule - unless it is right for the wrong reasons somehow.I would find the 'right for the wrong reasons' concept hard to believe since A7 is being debated as the possible source of some of the pairing issues, then SM5 produces a set of pairings consistent with the speculation.

Shows once again that SM5 is proving itself to be best by test for the current rules. It is amazing though, each time these pairing scenarios come up and we manual pair the scenarios to an inch of their life, SM5 produces the correct pairings each and every time, and sometimes for some bizarre reasons.

Bill Gletsos
06-03-2011, 11:37 PM
Ahh this is what I like to see. All this work done on the pairings and it goes up in smoke just like that :)Not really as I doubt they will change the published draw.

Carl Gorka
07-03-2011, 08:21 AM
Not really as I doubt they will change the published draw.

Correct

Kevin Bonham
07-03-2011, 09:48 AM
Shows once again that SM5 is proving itself to be best by test for the current rules. It is amazing though, each time these pairing scenarios come up and we manual pair the scenarios to an inch of their life, SM5 produces the correct pairings each and every time, and sometimes for some bizarre reasons.

Might be worth testing this further by subjecting this one (http://chesschat.org/showthread.php?p=292728#post292728) to further examination.

Garvinator
07-03-2011, 10:02 AM
Might be worth testing this further by subjecting this one (http://chesschat.org/showthread.php?p=292728#post292728) to further examination.It has already been given the SM5 treatment in post one, or are you wanting to look now at something else?

Kevin Bonham
07-03-2011, 10:05 AM
It has already been given the SM5 treatment in post one, or are you wanting to look now at something else?

Well no one posting on the thread agreed with SM5 but that said it was not looked at very closely. Maybe if we look more closely we will see SM5 is right, maybe not. I'll have a go at some stage.

antichrist
10-03-2011, 09:17 AM
Grant from dark side of moon

I'll have a crack at the top few boards (given that they won be impacted by the deferred Ly-Dale game):
1: Morris-Cheng
2: Zelesco-Johansen
3: Sirota-Jager
4: Dragicevic-Gorka
7: Voon-Tan

Well, Kerry and Carl are not chicken! They have both stuck their necks out and had a go at my challenge - but what about the rest of you arbiters??? Bill, what about you? Id like to see if anyone who defends the logic of the Dutch Swiss pairing rules can actually do the draw without a computer... Any more takers???

AC
Step right up gentlemen! I would like to see Brian Jones have a go as well.

Bill Gletsos
10-03-2011, 09:25 AM
Grant from dark side of moon

I'll have a crack at the top few boards (given that they won be impacted by the deferred Ly-Dale game):
1: Morris-Cheng
2: Zelesco-Johansen
3: Sirota-Jager
4: Dragicevic-Gorka
7: Voon-Tan

Well, Kerry and Carl are not chicken! They have both stuck their necks out and had a go at my challenge - but what about the rest of you arbiters??? Bill, what about you? Id like to see if anyone who defends the logic of the Dutch Swiss pairing rules can actually do the draw without a computer... Any more takers??? I have demonstrated many times that I know how to do a manual pairing.
I dont need to prove anything.

antichrist
10-03-2011, 09:32 AM
I have demonstrated many times that I know how to do a manual pairing.
I dont need to prove anything.

WE know that, but we want to see if you get the same pairing as the computer doing the Dutch Swissing pairing rules? that is the challenge.

Even I can do manual pairing so you are not saying much.

Kevin Bonham
10-03-2011, 09:57 AM
WE know that, but we want to see if you get the same pairing as the computer doing the Dutch Swissing pairing rules? that is the challenge.

No, the challenge is whether the computer doing the pairing can get the same pairing as Bill.

Really if they want CC regulars to get interested in doing a full manual pairing then more of the discussion needs to occur over here.

antichrist
10-03-2011, 11:30 AM
I have demonstrated many times that I know how to do a manual pairing.
I dont need to prove anything.

From Dark Side of MoonOriginally Posted by antichrist

Bill that is a cop out worthy of mop and bucket brigade of Grant and Firegoat. I can do a simple manual pairing, you do it and get same result as Dutch Swiss pairing on computer first go!

I would like to see that.

Firegoat
People who talk big after the facts are like chess players who say they should have won if only X. Do or no do, that is reality. Excuses are for the weak.

cheers,
-----------------------------------
AC
Well Bill you have convinced no one so far

Bill Gletsos
10-03-2011, 11:44 AM
From Dark Side of MoonOriginally Posted by antichrist

Bill that is a cop out worthy of mop and bucket brigade of Grant and Firegoat. I can do a simple manual pairing, you do it and get same result as Dutch Swiss pairing on computer first go!

I would like to see that.

Firegoat
People who talk big after the facts are like chess players who say they should have won if only X. Do or no do, that is reality. Excuses are for the weak.

cheers,
-----------------------------------
AC
Well Bill you have convinced no one so farThe toolbox is mainly the domain of idiots.
Trying to enlighten them is nothing more than a waste of my time.

antichrist
10-03-2011, 09:30 PM
The toolbox is mainly the domain of idiots.
Trying to enlighten them is nothing more than a waste of my time.

Now Bill, you called the wrong horse there. The one issued the challenge there was Grant, a responsible member and club organiser. Now if you accept this mission Bill...........................

Grant Szuveges
11-03-2011, 08:09 AM
No, the challenge is whether the computer doing the pairing can get the same pairing as Bill.

Really if they want CC regulars to get interested in doing a full manual pairing then more of the discussion needs to occur over here.

Hi Kevin (and regular chesschat users)

As people should know Im not aligned to either forum - I use them both. So, I will put the same challenge out over here:

Can anyone actually do the draw manually (without using a computer) and get it correct in accordance with the Dutch pairing rules????

The reason I have put the challenge out there is that I believe that the Dutch pairng rules are illogical and not upholding the spirit in which the Swiss system is supposed to be about. If they were more simple and more logical, then more people would be able to follow them and thus do the draw manually.

So, Id like to see some more people have a go at doing the pairings and we can see who does and doesnt get them right (assuming that Swissperfect will get them "right" - which it should)

Grant Szuveges
11-03-2011, 08:12 AM
I have demonstrated many times that I know how to do a manual pairing.
I dont need to prove anything.

Bill, you dont "need" to prove anything - but if you can demonstate that you can do this, you will silence a lot of critics..... You will prove wrong a lot of people on the other forum who dont believe you... You will gain a lot of street cred....

Grant Szuveges
11-03-2011, 08:17 AM
but we want to see if you get the same pairing as the computer doing the Dutch Swissing pairing rules? that is the challenge.

Yes, THAT is the challenge

Grant Szuveges
11-03-2011, 08:20 AM
[QUOTE=antichrist]

Bill that is a cop out worthy of mop and bucket brigade of Grant and Firegoat.
[QUOTE]

Not sure how I qualify to be part of the mop and bucket brigade..... What does this mean anyway?

Kevin Bonham
11-03-2011, 10:08 AM
Can anyone actually do the draw manually (without using a computer) and get it correct in accordance with the Dutch pairing rules????

The reason I have put the challenge out there is that I believe that the Dutch pairng rules are illogical and not upholding the spirit in which the Swiss system is supposed to be about.

Well I can tell you that when I do the pairings manually using the system as written I will also get pairings which you will consider to be illogical and not upholding the spirit in which the Swiss system is supposed to be about. That applies whether the pairings I get are exactly the same as SP's or not - even if they are not quite the same because of a mistake or differing interpretation by me or by it, they will differ from your intuition in the same way. The main thing people don't "get" about the system is that it cares more about colour than about seeding. This is because it aims first and foremost to use data from the event being paired, rather than from outside it.

As such I'm not sure what you're aiming to prove here.

You can see an example of me manually pairing part of a draw (and getting the same result as Swiss Perfect) here (http://chesschat.org/showpost.php?p=303322&postcount=14).

I'm going away for the state champs today so won't have much time over the weekend but next week sometime I will have a crack at doing an MCC champs draw by hand - and post how long it took me.

Not sure how I qualify to be part of the mop and bucket brigade..... What does this mean anyway?

A reference to a CC injoke from about five years ago in which the poster then known as starter (now known variously as MOZ or ursogr8) was posting obfuscatory silly stuff and Bill started responding with "Alert! Mop and bucket required" (to clean up all the dribble presumably); several others joined in.

Oepty
11-03-2011, 07:02 PM
Grant, I said I would do pairings for the next round earlier in the thread expecting to have plenty of time being unemployed, but I ended up getting a weeks work so did not have time. I will try and do a manual pairing of the next round next week if I am not working.
Scott

ER
11-03-2011, 07:42 PM
A reference to a CC injoke from about five years ago in which the poster then known as starter (now known variously as MOZ or ursogr8) was posting obfuscatory silly stuff and Bill started responding with "Alert! Mop and bucket required" (to clean up all the dribble presumably); several others joined in.

Did it work??? :doh: :hmm: :lol:

Grant Szuveges
12-03-2011, 03:27 PM
Well I can tell you that when I do the pairings manually using the system as written I will also get pairings which you will consider to be illogical and not upholding the spirit in which the Swiss system is supposed to be about. That applies whether the pairings I get are exactly the same as SP's or not - even if they are not quite the same because of a mistake or differing interpretation by me or by it, they will differ from your intuition in the same way. The main thing people don't "get" about the system is that it cares more about colour than about seeding. This is because it aims first and foremost to use data from the event being paired, rather than from outside it.

I'm going away for the state champs today so won't have much time over the weekend but next week sometime I will have a crack at doing an MCC champs draw by hand - and post how long it took me.

Hi Kevin

You have described the Dutch system perfectly here (by saying that it cares more about colour than seeding). This is exactly what I dont like about it. In my opinion seeding should always come before colour - because this was the original idea of the Swiss system.

My issue is not with any individual person or with Swissperfect. My issue is with the Dutch system as I believe that seeding should come before colour - as I believe that this is more logical.

The point about using data from the event rather than from outside of it is interesting and I see the point, but I believe that upholding the seeding order (despite the fact that it isnt from within the tournament itself) is more important.

A practical example: If I was on 3/4 in a tournament, I would much rather be playing against a 1600 than a 2000. Even if I had 2 blacks in a row to play the 1600, I would feel much more comfortable than playing white (the "correct" colour) agains the 2000. It doesnt seem right to me that 2 2000s would play eachother and 2 1600s would play eachother if they were all on 3/4. I know that that is not the rules of the Dutch system, but its just my opinion.

I "get" the Dutch system, but I simply dont agree with it or think it is logical.

Thanks for offering to have a crack at the MCC Champs draw - curious to see how many people get it right. Cheers

Oepty
12-03-2011, 05:27 PM
Grant, I have now got myself unconfused about the progression of the tournament, I did not realise that there was no game on Monday so I thought the round 6 pairing would have already been published. It appears it has not so I have had a go at the round 6 pairings which I don't really think were that difficult. Have dealt with far worse before. As I have not seen the result of the postponed Thai Ly - Ari Dale game I have paired with that game being a draw. If that results becomes known I can redo the pairings with the result, if it is not a draw, taken into account. I will PM you the draw I have done as soon as I typed it up, only have it on paper for the moment.
Scott

Oepty
14-03-2011, 09:42 PM
Blast it, the MCC round 6 pairings are out and I forgot to send mine to Grant so I guess I will have to do round 7 pairings to show that they can be done. It was a bit impossible to get the same pairings anyway with me not knowing Bobby Cheng was taking a bye.
Scott

Kevin Bonham
07-04-2011, 12:39 AM
The reason I have put the challenge out there is that I believe that the Dutch pairng rules are illogical and not upholding the spirit in which the Swiss system is supposed to be about. If they were more simple and more logical, then more people would be able to follow them and thus do the draw manually.

So, Id like to see some more people have a go at doing the pairings and we can see who does and doesnt get them right (assuming that Swissperfect will get them "right" - which it should)

OK I have now had a go at this for round 9 and posted it as a new thread.

After I did so Garvin posted the Swiss Master 5 pairings, SM5 being a generally superior program to SP.

The upshot was:

* There were only two areas of difference between my draw and SP's, both well down the crosstable, one affecting two pairings and one affecting several. In one of these SP was right while I made a clerical error. In another one, we were both wrong - SP made an inferior exchange in a scoregroup, while I made the correct exchange but then overlooked a half-point bye being a downfloat in a lower group.

* On the very bottom board both SP and I got the colours wrong. In my case, just another clerical error (I decided for some reason that 6 is an odd number.)

* There is a big issue with the first 3 pairings with SP and I both getting a pairing different to the SM5 pairing. The SM5 pairing seems more in "the spirit" of the system but it is difficult extracting it literally from the rules as written.

I believe that I've shown clearly that it is possible for a human to manually pair results following the Dutch system and get results very similar to those that are obtained by computer programs, with differences attributable to human error rather than inability to understand the system.

Of course this won't happen if instead of following the Dutch system you just follow your own ideas of how Swiss pairings "should" work.

Oepty
07-04-2011, 02:43 AM
OK I have now had a go at this for round 9 and posted it as a new thread.

After I did so Garvin posted the Swiss Master 5 pairings, SM5 being a generally superior program to SP.

The upshot was:

* There were only two areas of difference between my draw and SP's, both well down the crosstable, one affecting two pairings and one affecting several. In one of these SP was right while I made a clerical error. In another one, we were both wrong - SP made an inferior exchange in a scoregroup, while I made the correct exchange but then overlooked a half-point bye being a downfloat in a lower group.

* On the very bottom board both SP and I got the colours wrong. In my case, just another clerical error (I decided for some reason that 6 is an odd number.)

* There is a big issue with the first 3 pairings with SP and I both getting a pairing different to the SM5 pairing. The SM5 pairing seems more in "the spirit" of the system but it is difficult extracting it literally from the rules as written.

I believe that I've shown clearly that it is possible for a human to manually pair results following the Dutch system and get results very similar to those that are obtained by computer programs, with differences attributable to human error rather than inability to understand the system.

Of course this won't happen if instead of following the Dutch system you just follow your own ideas of how Swiss pairings "should" work.

Also I did a 100% correct pairing of round 7 of this tournament which I put on the other BB. It ended up not being the pairing used because of players taking byes and my revised pairings did contain 2 errors. I think it is quite clear humans are able to follow the Dutch Pairings rules and get pairings as good as any computer program. There might be a few issues with the rules clarity but a computer programmer is going to have the same issues understanding the rules that anyone doing manual pairings is going to have and is just as likely to get them wrong.
Scott

Oepty
09-04-2011, 12:35 PM
I have been looking at some SP files I have of previous tournaments to try and help me understand pairings more and to see if SP has made any mistakes. I am getting totally lost in trying to pair the bottom part of this round.

Place No Opponents Colours Float Score

1 5 : 16,11,12,4,3,1,6 BWBWBBW D 6

2 3 : 13,10,7,1,5,12,9 BWBBWWB D 5.5

3 9 : 19,6,14,7,8,2,3 BWBWWBW Ud 5

4-6 1 : 15,8,6,3,12,5,4 BWBWBWB 4.5
2 : 18,7,11,12,10,9,8 WBBWBWW uD 4.5
4 : 12,18,13,5,17,8,1 WBWBWBW 4.5

7-11 6 : 14,9,1,8,13,7,5 WBWBWWB U 4
12 : 4,15,5,2,1,3,18 BWWBWBW D 4
13 : 3,19,4,10,6,21,7 WBBWBWB 4
17 : 10,20,8,19,4,14,11 BWBWBWB 4
18 : 2,4,15,11,7,10,12 BWBWBWB U 4

12-18 7 : 20,2,3,9,18,6,13 BWWBWBW 3
8 : 21,1,17,6,9,4,2 WBWWBWB U 3
10 : 17,3,21,13,2,18,15 WBWBWBW U 3
11 : -,5,2,18,14,19,17 -BWBBWW 3
14 : 6,-,9,20,11,17,16 B-WBWBW U 3
19 : 9,13,16,17,20,11,- WWBBWB- D 3
21 : 8,16,10,15,-,13,20 BWBW-BB D 3

19-20 15 : 1,12,18,21,16,-,10 WBWBW-B D 2.5
16 : 5,21,19,-,15,20,14 WBW-BWB D 2.5

21 20 : 7,17,-,14,19,16,21 WB-WBBW U 1

SP gives what I think are correct pairings for boards 1-5 of
5-9
3-4
1-2
17-6
13-12

This leaves 18 to float down.
I think 18 should play 8, and the rest of the 3 group gets paired as
18-8
14-7
19-10
21-11
This however means there cannot be a legal pairing in the bottom 3 players as they have all already had the bye.
It seems that now we need to C13 on the 3 scoregroup to float down some players including 1 who can have the bye but I am not sure how to do it exactly.

The SP Pairings for the bottom part are
18-20
19-7
21-14
15-8
16-11
10 bye
It seems to come from SP putting all these players into 1 big group and pairing that. I doubt that is what SP should be doing.
Scott

Bill Gletsos
09-04-2011, 12:46 PM
I have been looking at some SP files I have of previous tournaments to try and help me understand pairings more and to see if SP has made any mistakes. I am getting totally lost in trying to pair the bottom part of this round.

Place No Opponents Colours Float Score

1 5 : 16,11,12,4,3,1,6 BWBWBBW D 6

2 3 : 13,10,7,1,5,12,9 BWBBWWB D 5.5

3 9 : 19,6,14,7,8,2,3 BWBWWBW Ud 5

4-6 1 : 15,8,6,3,12,5,4 BWBWBWB 4.5
2 : 18,7,11,12,10,9,8 WBBWBWW uD 4.5
4 : 12,18,13,5,17,8,1 WBWBWBW 4.5

7-11 6 : 14,9,1,8,13,7,5 WBWBWWB U 4
12 : 4,15,5,2,1,3,18 BWWBWBW D 4
13 : 3,19,4,10,6,21,7 WBBWBWB 4
17 : 10,20,8,19,4,14,11 BWBWBWB 4
18 : 2,4,15,11,7,10,12 BWBWBWB U 4

12-18 7 : 20,2,3,9,18,6,13 BWWBWBW 3
8 : 21,1,17,6,9,4,2 WBWWBWB U 3
10 : 17,3,21,13,2,18,15 WBWBWBW U 3
11 : -,5,2,18,14,19,17 -BWBBWW 3
14 : 6,-,9,20,11,17,16 B-WBWBW U 3
19 : 9,13,16,17,20,11,- WWBBWB- D 3
21 : 8,16,10,15,-,13,20 BWBW-BB D 3

19-20 15 : 1,12,18,21,16,-,10 WBWBW-B D 2.5
16 : 5,21,19,-,15,20,14 WBW-BWB D 2.5

21 20 : 7,17,-,14,19,16,21 WB-WBBW U 1

SP gives what I think are correct pairings for boards 1-5 of
5-9
3-4
1-2
17-6
13-12

This leaves 18 to float down.
I think 18 should play 8,Your pairing of 18 V 8 violates B5.

and the rest of the 3 group gets paired as
18-8
14-7
19-10
21-11
This however means there cannot be a legal pairing in the bottom 3 players as they have all already had the bye.
It seems that now we need to C13 on the 3 scoregroup to float down some players including 1 who can have the bye but I am not sure how to do it exactly.

The SP Pairings for the bottom part are
18-20
19-7
21-14
15-8
16-11
10 bye
It seems to come from SP putting all these players into 1 big group and pairing that. I doubt that is what SP should be doing.
ScottI doubt it is correct as I have seen this happen before and SP was not correct in those situations either.

Kevin Bonham
09-04-2011, 01:13 PM
18 playing 8 is a problem because 8 just upfloated.

Under the outdated rules as SP is applying them, S1 contains 18 (W), and S2 contains 7 (B), 8 (B), 10 (B), 11 (B abs), 14 (B), 19 (W), 21 (W abs). p=1 x=1 so the pairing need not be a colour match. 18 should therefore play 19 (18-19).

Under A7e which SP does not include, 14 becomes due white to reduce the value of x to zero. But now no opponent can be found for 18 in the group in compliance with B1 through B6 (note that where the Handbook says compliance with B1 and B2 it is out of date since this has been changed). Therefore increase x to 1, at which point changing 14 to a due W no longer reduces the value of x so I think 14 returns to due B. Again, 18 ends up playing 19.

In the 3 group you cannot make 3 pairings (no matter who 18 plays, since one must get the bye and another must downfloat) but you can make 2. Also 21 can't downfloat so must be paired within the 3 group.

I'm not sure this is correct but after a lot of mucking around I got:

18-19
21-7
14-8
16-11
15-20
10 bye

Whatever SP is doing to get the hideous pairing 18 vs 20 I am confident it is wrong.

Kevin Bonham
09-04-2011, 01:16 PM
I am getting the impression that SP does not apply C14 correctly or maybe even at all. It seems that whenever it can't make p pairings it panics.

Oepty
09-04-2011, 01:25 PM
Your pairing of 18 V 8 violates B5.

I know, it was deliberate, but if not 8 who do you pair 18 with? You need to pair 18 with someone with a black preference to keep x=1. They have either played 18 (7, 10 and 11) or floated up in the previous round (8 and 14).
Scott

Kevin Bonham
09-04-2011, 01:32 PM
I know, it was deliberate, but if not 8 who do you pair 18 with? You need to pair 18 with someone with a black preference to keep x=1.

You only have to satisfy x=1 for the p (1) pairing you make before moving on to the remainder group. That this may cause x to become 2 in the remainder group (although this actually only happens under the old rules, with A7e added 14 becomes due white and x is 1) doesn't matter.

Oepty
09-04-2011, 01:40 PM
18 playing 8 is a problem because 8 just upfloated.

Under the outdated rules as SP is applying them, S1 contains 18 (W), and S2 contains 7 (B), 8 (B), 10 (B), 11 (B abs), 14 (B), 19 (W), 21 (W abs). p=1 x=1 so the pairing need not be a colour match. 18 should therefore play 19 (18-19).

This is what I thought and would have done until I got a similar situation in the MCC championships. Floats were not an issue in that case though. It was like this, PLayer 1 was floating down to a group with 6 players. Players 1 and 2 had white preferences, not absolute and players 3-7 had black preferences so x =1. 1 could play 2, but SP paired 1 against 3. I could not see any reason it did this except that SP was maintaining the colour balance across the whole group.

Under A7e which SP does not include, 14 becomes due white to reduce the value of x to zero. But now no opponent can be found for 18 in the group in compliance with B1 through B6 (note that where the Handbook says compliance with B1 and B2 it is out of date since this has been changed). Therefore increase x to 1, at which point changing 14 to a due W no longer reduces the value of x so I think 14 returns to due B. Again, 18 ends up playing 19.

In the 3 group you cannot make 3 pairings (no matter who 18 plays, since one must get the bye and another must downfloat) but you can make 2. Also 21 can't downfloat so must be paired within the 3 group.

I'm not sure this is correct but after a lot of mucking around I got:

18-19
21-7
14-8
16-11
15-20
10 bye

Whatever SP is doing to get the hideous pairing 18 vs 20 I am confident it is wrong.

The arbiter in this case actually changed the pairings from the SP pairings but got different pairings to these but I am not sure how. Your go looks to be reasonable but I have no confidence to say it is correct.
Scott

Oepty
09-04-2011, 01:41 PM
You only have to satisfy x=1 for the p (1) pairing you make before moving on to the remainder group. That this may cause x to become 2 in the remainder group (although this actually only happens under the old rules, with A7e added 14 becomes due white and x is 1) doesn't matter.

I was not trying to use the new rules in A7.
Scott

Kevin Bonham
09-04-2011, 02:03 PM
This is what I thought and would have done until I got a similar situation in the MCC championships. Floats were not an issue in that case though. It was like this, PLayer 1 was floating down to a group with 6 players. Players 1 and 2 had white preferences, not absolute and players 3-7 had black preferences so x =1. 1 could play 2, but SP paired 1 against 3. I could not see any reason it did this except that SP was maintaining the colour balance across the whole group.

I can't see any literal warrant in the rules for doing so given that strange note in brackets after B4 ("Whenever x of a score bracket is unequal to zero this rule will have to be ignored. x is deducted by one each time a colour preference cannot be granted.")

I have great trouble understanding that note in B4. After all when x isn't zero you can still have as many players as possible, as opposed to fewer, receiving their colour preference. And in section C, x isn't deducted by one when a colour preference can't be granted; rather it starts at the number of initially ungrantable colour preferences and then increases by one every time it becomes impossible to pair with the number of colour clashes you were aiming for.

Oepty
09-04-2011, 02:16 PM
I can't see any literal warrant in the rules for doing so given that strange note in brackets after B4 ("Whenever x of a score bracket is unequal to zero this rule will have to be ignored. x is deducted by one each time a colour preference cannot be granted.")

I have great trouble understanding that note in B4. After all when x isn't zero you can still have as many players as possible, as opposed to fewer, receiving their colour preference. And in section C, x isn't deducted by one when a colour preference can't be granted; rather it starts at the number of initially ungrantable colour preferences and then increases by one every time it becomes impossible to pair with the number of colour clashes you were aiming for.

I actually did what SP did and paired 1 against 3 as well. I did so after reading the rules quite a few times and changing my mind on the situation. I cannot remember exactly what I read that changed my mind.

As far as the note for B4 goes, I have taken the decreasing of x to mean that if x = n then you do a pairing that does not meet colour requirements then x = n-1 for the rest of the pairings for the group still to be done, if there is still some to be done. In practice I guess you already do this automatically. The first part of the note makes no sense to me either.
Scott