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Garvinator
27-10-2009, 05:22 PM
6 : 22,11,23,16,4 BWBWB d 2
11 : 25,6,18,8,24 WBWBW 2
16 : -,5,24,6,- -BWB- uD 2
17 : 1,20,8,24,19 WBWBW U 2
20 : 13,17,19,15,26 BWWBW u 2
25 : 11,-,9,-,18 B-W-B D 2
27 : 15,3,12,23,14 WBWB- d 2

15 : 27,24,22,20,28 BWBWB Ud 1.5
19 : 2,14,20,21,17 BWBWB D 1.5
21 : 4,18,-,19,23 BW-BW 1.5

Using dutch pairing rules, what should the pairings be for the score groups above.

Swiss perfect gives:

6 v 20
16 v 11
25 v 17
19 v 27
15 v 21

Swiss Master 5

27 v 11
16 v 20
25 v 17
6 v 19
15 v 21

forlano
11-11-2009, 06:49 PM
Swiss perfect gives:

6 v 20
16 v 11
25 v 17
19 v 27
15 v 21

Swiss Master 5

27 v 11
16 v 20
25 v 17
6 v 19
15 v 21

Hello,

can you post the entire .swf file generated by swissmaster?
Luigi

Garvinator
11-11-2009, 11:32 PM
Hello,

can you post the entire .swf file generated by swissmaster?
Luigi
Umm why? And will you take another two weeks to reply to this post?

Kevin Bonham
12-11-2009, 01:27 AM
Opening and not necessarily correct bid: at a quick look I cannot see why the SP draw is wrong. I can see that either 6 or 27 should be downfloated (since any other downfloat will violate B4 or B5 while downfloating either of these will only violate B6) but I can't see why you'd downfloat 6 instead of 27 since as far as I can tell you get p pairings in accordance with B1 and B2 right away and 27 is kicked downstairs.

I wonder if SM5 has been tripped up by what I assume is a pairing originally made and then replaced with a forfeit involving 27 in a previous round.

Eric Barentsen
12-11-2009, 04:24 AM
Opening and not necessarily correct bid: at a quick look I cannot see why the SP draw is wrong. I can see that either 6 or 27 should be downfloated (since any other downfloat will violate B4 or B5 while downfloating either of these will only violate B6) but I can't see why you'd downfloat 6 instead of 27 since as far as I can tell you get p pairings in accordance with B1 and B2 right away and 27 is kicked downstairs.

I wonder if SM5 has been tripped up by what I assume is a pairing originally made and then replaced with a forfeit involving 27 in a previous round.

Except for #27 all players in scoregroup 2 have a strong preference for a color.
S1 S2
#6 strong white #17 strong black
#11 strong black #20 strong black
# 16 strong white #25 strong white
#27 light white

In scoregroup 1,5 the color preference is:
#15 strong white
#19 strong white
#21 light black

In scoregroup 2:
P=3 W=4 B=3 Q=4 X=0
3 pairings according to colorpreference should be made. This means a white seeker will have to float down.
Seen the colorpreference in scoregroup 1.5
P=2 W=3 B=1 Q=2 X=1
One whiteseeker should be given black. This is preferable a light preference so #27 will be the best to float down.
The pairing will be:
6-20
16-11
25-17
19-27
15-21

forlano
12-11-2009, 06:56 AM
Umm why? And will you take another two weeks to reply to this post?

I saw your post when I answered. I and my friends are working on a program that performs the pairing with the dutch rules (we apperead on this forum in the recent past). This program save even the explanation of what it has done to arrive at a given pairing. Moreover it is compatible with the swissmaster file format.
So in order to see its answer and its reasons I need the whole swissmaster file.
If you want to have another opinion please post that file. I will try to answer in less than two weeks :) .

Kind regards,
Luigi

drbean
05-11-2010, 01:06 PM
[QUOTE=Garvinator]
6 : 22,11,23,16,4 BWBWB d 2
11 : 25,6,18,8,24 WBWBW 2
16 : -,5,24,6,- -BWB- uD 2
17 : 1,20,8,24,19 WBWBW U 2
20 : 13,17,19,15,26 BWWBW u 2
25 : 11,-,9,-,18 B-W-B D 2
27 : 15,3,12,23,14 WBWB- d 2

15 : 27,24,22,20,28 BWBWB Ud 1.5
19 : 2,14,20,21,17 BWBWB D 1.5
21 : 4,18,-,19,23 BW-BW 1.5

I have a question about the pairing of 27 in round 5 against 14.

There is no color for 27 in that round.

Does that mean that one of 27 or 14 forfeited in that round?

If that is the case, doesn't it mean 27 v 14 is a possible pairing in later rounds, and the 14 shouldn't appear?

It is confusing for the human arbiter to have the 14 there. But seeing Player 14 isn't in these 2 brackets, perhaps it doesn't matter.

Denis_Jessop
05-11-2010, 07:21 PM
I would think that you are right. There are several other cases of no colour and no player indicated by " - " in the quoted part of the table so why 14 is there is not at all clear.

DJ

Garvinator
05-11-2010, 07:37 PM
I have a question about the pairing of 27 in round 5 against 14.

There is no color for 27 in that round.

Does that mean that one of 27 or 14 forfeited in that round?

If that is the case, doesn't it mean 27 v 14 is a possible pairing in later rounds, and the 14 shouldn't appear?

It is confusing for the human arbiter to have the 14 there. But seeing Player 14 isn't in these 2 brackets, perhaps it doesn't matter.
14 did not appear for round five (the round you question) and did not appear for any future rounds.

But in more normal circumstances, under the fide pairing rules, as 14 forfeited to 27, they could be paired in later rounds.

Bill Gletsos
05-11-2010, 09:54 PM
The reason why player 27 has player 14 listed in round 5 is that it shows a historical record of the fact that 27 was scheduled to play 14 in that round but the result was a forfeit.

As such 27 & 14 could end up being paired in a future round as per the FIDE pairing rules.

The '-" in the pairing numbers list for players 16, 25 & 21 indicate that they were not paired for that particular round or had a bye.

Kevin Bonham
07-03-2011, 10:07 AM
since as far as I can tell you get p pairings in accordance with B1 and B2 right away and 27 is kicked downstairs.

Note that this logic is not necessarily correct since the p pairings have to be in accordance with B1 through B6 (that was not codified at the time of this thread but generally understood to be an error.)

I am suggesting this one be looked at more closely as a test of Garvin's hypothesis that SM5 is always right.

Denis_Jessop
07-03-2011, 10:34 AM
Note that this logic is not necessarily correct since the p pairings have to be in accordance with B1 through B6 (that was not codified at the time of this thread but generally understood to be an error.)

I am suggesting this one be looked at more closely as a test of Garvin's hypothesis that SM5 is always right.

Also Geurt Gijssen says that he has never known SM5 to be wrong.

DJ

Kevin Bonham
07-03-2011, 10:43 AM
Also Geurt Gijssen says that he has never known SM5 to be wrong.

I have known Geurt Gijssen to be wrong, rather often actually. Though I don't recall knowing him to be wrong on the matter of pairings.

Kevin Bonham
07-03-2011, 10:54 PM
6 : 22,11,23,16,4 BWBWB d 2
11 : 25,6,18,8,24 WBWBW 2
16 : -,5,24,6,- -BWB- uD 2
17 : 1,20,8,24,19 WBWBW U 2
20 : 13,17,19,15,26 BWWBW u 2
25 : 11,-,9,-,18 B-W-B D 2
27 : 15,3,12,23,14 WBWB- d 2

15 : 27,24,22,20,28 BWBWB Ud 1.5
19 : 2,14,20,21,17 BWBWB D 1.5
21 : 4,18,-,19,23 BW-BW 1.5

Using dutch pairing rules, what should the pairings be for the score groups above.

I'll have a crack at this one manually in full detail. It turns out to be relevant to the thread I brought it up on.

2 point group: due white: 6, 16, 25, 27
due black: 11, 17, 20

Initial S1 = 6(W), 11(B), 16(W)
Initial S2 = 17(B), 20(B), 25(W), 27(B)

x=0 so three pairings to be made and all colours to match. Note 11 has played 25, 17 has played 20. 27 and 6 are preferred floaters by B6 (all others are covered by B5 or B4).

First attempt (6,17), (11, 20), (16,25) fails as 11, 20 both due black.

Attempt to resolve by applying transposition within S2.

All transpositions within S2 fail because 11 has played the only white-seeker in S2.

Exchange 17 for 16.

New S1 = 6(W), 11(B), 17(B)
New S2 = 16 (W), 20(B), 25(W), 27(W)

Note we can't have 16 first or 20 second so we will have to start with 20 first and 16 second.

6(W) - 20 (B)
16(W) - 11 (B)
25(W) - 17 (B)
27(W) downfloats

Pair heterogeneous scoregroup 27 (2), 15, 19, 21 (1.5 each)

Three whiteseekers and one blackseeker so x=1

S1 is 27(W)
S2 is 15 (W), 19(W), 21(B)

27 vs 15 upfloats 15, disallowed, apply transposition

Pairing 19 and 27 is valid as x=1. 19 has stronger colour preference so 19-27 and 27 does not get colour preference (but see below).

leaving 15(W)-21(B)

So I get the same pairing as SP and Eric. Certainly by the rules as written.

Note that may not be the end of it, because the mysterious A7e applies here. A7e supposedly says:

While pairing an even-numbered round players having a mild colour preference (players who have had an even number of games before by any reason) shall be treated and counted as if they would have a mild colour preference of that kind (white resp. Black) which reduces the value of x (see A.8) as long as this does not result in additional downfloaters, (GA 2001)

...and 27 is one of these.

Now the questions here are:

(i) What does this actually mean? It seems almost nonsensical because it seems to be saying if they're a mild, treat them as a mild, so what.
(ii) Does it actually say what it is supposed to say?

I was just talking to Bill about this on the phone and we were discussing whether it's possible that A7e is misworded and is actually somehow supposed to make the system consider 27 to have a strong colour preference and prefer 6 over 27 as the downfloater, though it's very hard to see what would literally warrant undoing the previous scoregroup.

An interpretation which just occurred to me as I write this post is that A7e means that if you have an imbalanced (2+ more seeking one colour than the other, so x>0) group which includes a player with an even number of games (like 27 in this case) and that player is in the majority colour preference for that group, treat them as having the opposite mild colour preference and reduce the value of x in that manner. In other words, give them two of the same colour in a row in preference to colour issues for the others. That is the only way I can see that A7e as currently worded in the Handbook makes any sense.

If my interpretation is correct then in this case A7e has no effect on 27's pairing in the 2 group as the value of x cannot be reduced for this group. When 27 downfloats to the 1.5 group it makes 27 a weak black-seeker, which is irrelevant since 27 gets black anyway.

If that is correct then the SM5 pairing is probably a glitch involving bye colours or categorisation of byes re float status.

Garvinator
08-03-2011, 01:31 AM
If that is correct then the SM5 pairing is probably a glitch involving bye colours or categorisation of byes re float status.
I had been wondering about this. It is extremely unusual to have four players in a seven player score group marked as having already downfloated and then also those same four having the higher colour preference.

So I am thinking it could be an issue with how half point byes are treated between the two programs.

The answer could lie in trying to set the circumstances in how the sm5 pairings could come about.