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Lucena
20-06-2006, 12:16 AM
Ok, I'm going to start a little experiment. I am going to create a mock (Fide Swiss Rules) tournament and people can comment on my pairings. I'll be using Swiss Perfect (trial version) and if anyone can be bothered scrutinising my pairings I would appreciate any feedback.

Garvinator
20-06-2006, 12:19 AM
Ok, I'm going to start a little experiment. I am going to create a mock (Fide Swiss Rules) tournament and people can comment on my pairings. I'll be using Swiss Perfect (trial version) and if anyone can be bothered scrutinising my pairings I would appreciate any feedback.
will do.

Lucena
20-06-2006, 03:45 PM
I'll come up with a list of "players" in the next couple of days, am a bit busy at the moment.

Desmond
20-06-2006, 03:48 PM
I'll come up with a list of "players" in the next couple of days, am a bit busy at the moment.

Can I be the under-rated bunny who wins the tourney with a picket fence please?:D

Lucena
20-06-2006, 04:45 PM
Can I be the under-rated bunny who wins the tourney with a picket fence please?:D

No, but one of the players will be named in your honour!

Garvinator
20-06-2006, 04:47 PM
I'll come up with a list of "players" in the next couple of days, am a bit busy at the moment.
If you want to really test the accuracy of sp pairings, then I would suggest about 20 players or so and 7 to 10 rounds.

That way you will get a good separation of score groups, therefore having to deal with float issues as well.

Basil
20-06-2006, 05:16 PM
I'd like to be one of the players. I get the jitters if something is happening without me. I'd be happy to be the middle of the road plonker who gets wiped out, on account of it being close to reality.

Garvinator
20-06-2006, 05:38 PM
I'd like to be one of the players. I get the jitters if something is happening without me. I'd be happy to be the middle of the road plonker who gets wiped out, on account of it being close to reality.
Gareth could almost design it as a mass chesschat tournament :whistle: :uhoh:

Ratings:

Ian Rogers - 7 2599!!
Igor Goldenberg - 7 2353!!
George Xie - 40 2316!!
Leonid Sandler - 7 2310!!
Michael Baron - 10 2282!
jeffrei - 0 2278?
Greg Canfell - 0 2276!!
Tristan Boyd - 11 2247!!
drug - 10 2239!!
Comrade Zukovsky - 0 2226!
Brett Tindall- 0 2221!
Brian Jones - 4 2124!!
Andrew Bird - 25 2098!!
bobby1972 - 28 2084!!
macavity - 31 2069!!
Gareth Charles - 27 2061!!
paulb - 7 2058!!
Goughfather - 7 2054!!
Laura Moylan - 23 2054!!
Ronald Yu - 7 2043!!
Jason Hu - 28 2020!!
Lee Jones - 3 1999!!
Kerry Stead - 7 1998!!
Jason Chan - 7 1991!!
firegoat7 - 0 1942!!
Kevin Bonham - 6 1941!!
pballard - 0 1877
rob - 11 1871!!
elevatorescapee - 7 1785!
Ian Rout - 17 1777!!
Barry Cox - 15 1741!!
pax - 0 1739?
Amiel Rosario - 16 1715!!
Shaun Press - 9 1711!
Candy Cane - 0 1672!!
The_Wise_Man - 1 1671!
Belthaser - 1 1647
JGB - 0 1640
David Richards - 0 1639!
Dozy - 7 1555!!
starter - 0 1565!
Arrogant-One - 8 1563!
Liberace - 0 1541!!
skip - 0 1534
Phil Bourke - 8 1500
Trent Parker - 29 1493!!
antichrist - 0 1492
Howard Duggan - 12 1437!!
Frosty - 5 1435!
bergil - 30 1434!!
Scott Colliver - 0 1415?
Paul Sike - 7 1406!!
themovingman - 23 1402!!
AES - 0 1375!
Careth - 14 1325!!
Matt Sweeney - 7 1307!!
Garvin Gray - 18 1253!!
EGOR - 17 1224!!
qpawn - 5 1177
bunta - 0 1021
alana - 30 994

Lucena
20-06-2006, 06:03 PM
I was actually thinking of running such a tournament for real but I figured there wouldn't be enough people who would play in it. If I am going to make up the results of the games myself (which is what I intend to do), I would rather not have the mock results be seen to reflect on the actual people in real life.

antichrist
20-06-2006, 07:39 PM
I was actually thinking of running such a tournament for real but I figured there wouldn't be enough people who would play in it. If I am going to make up the results of the games myself (which is what I intend to do), I would rather not have the mock results be seen to reflect on the actual people in real life.

Don't be chicken Gareth, bring it on. Let it reflect on their miserable lives and let all hell break loose.

CAn A/C play AC? I am going to challenge her to a CC game, thanks for the impetus.

Lucena
21-06-2006, 10:55 AM
Well, here's my list of players. I should make it clear that these experimental results and pairings are by no means intended to realistically replicate real world events. I could have just named the players "1", "2", etc but I thought that would be a bit boring.

1.|Fischer, Robert||2895|
2.|Kasparov, Garry||2886|
3.|Botvinnik, Mikhail||2885|
4.|Lasker, Emanuel||2878|
5.|Capablanca, Jose||2877|
6.|Alekhine, Alexander||2865|
7.|Karpov, Anatoly||2848|
8.|Steinitz, Wilhelm||2826|
9.|Tarrasch, Siegbert||2824|
10.|Korchnoi, Viktor||2814|
11.|Smyslov, Vassily||2800|
12.|Tal, Mikhail||2799|
13.|Chigorin, Mikhail||2797|
14.|Najdorf, Miguel||2797|
15.|Petrosian, Tigran V||2796|
16.|Bronstein, David||2792|
17.|Rubinstein, Akiba||2789|
18.|Keres, Paul||2786|
19.|Reshevsky, Sammy||2785|
20.|Nimzovitch, Aron||2780|
21.|Spassky, Boris||2773|
22.|Euwe, Max||2769|
23.|Bogolyubov, Yefim||2768|
24.|Timman, Jan||2768|
25.|Schlechter, Carl||2764|
26.|Fine, Reuben||2762|
27.|Stahlberg, Gideon||2762|
28.|Tartakover, Savielly||2761|
29.|Larsen, Bent||2755|
30.|Flohr, Salo||2754|
31.|Polgar, Judit||2746|
32.|Duras, Oldrich||2743|
33.|Kashdan, Isaac||2742|
34.|Eliskases, Erich||2733|
35.|Khalifman, Alexander||2724|
36.|Byrne, Robert||2720|
37.|Kasimdzhanov, Rustam||2708|
38.|Benko, Pal||2687|
39.|Browne, Walter||2678|
40.|Denker, Arnold||2677|
41.|Showalter, Jackson||2676|
42.|Canal, Esteban||2675|
43.|Shamkovich, Leonid||2675|
44.|Samisch, Friedrich||2665|
45.|Rossolimo, Nicolas||2663|
46.|Chiburdanidze, Maia||2661|
47.|Nehzmetdinov, Rashid||2660|
48.|Murey, Yakov||2658|
49.|Dake, Arthur||2655|
50.|Bukic, Enver||2647|
51.|Leonhardt, Paul||2639|
52.|Byrne, Donald||2633|
53.|Polgar, Zsuzsa||2633|
54.|Rauzer, Vsevolod||2627|
55.|Colle, Edgar||2619|
56.|Xie Jun, ||2619|
57.|Znosko-Borovsky, Eugene||2613|
58.|Alexander, Hugh||2610|
59.|Cramling, Pia||2605|
60.|Petrosian, Tigran L||2594|
61.|Lasker, Edward||2583|
62.|Bonch-Osmolovsky, Mikhail||2577|
63.|Skembris, Spyridon||2577|
64.|Nakamura, Hikaru||2563|
65.|Milner-Barry, P. S.||2552|
66.|Bednarsky, J||2544|
67.|Golombek, Harry||2543|
68.|Polgar, Sofia||2540|
69.|Polgar, Istvan||2529|
70.|Cheron, Andre||2521|
71.|Barden, Leonard||2497|
72.|Te Kolste, Jan||2497|
73.|Kosteniuk, Alexandra||2489|
74.|Skripchenko, Almira||2484|
75.|Krush, Irina||2478|
76.|Petrosian, Tigran A||2420|
77.|Duchamp, Marcel||2413|
78.|Bogart, Humphrey||Unrated|
79.|Charles, Ray||Unrated|
80.|Einstein, Albert||Unrated|
81.|Oppenheimer, Robert||Unrated|

The players have been rated according to their highest "historical ratings" given by Jeff Sonas as his chessmetrics site. I am not concerned with the accuracy of his methods of rating, I just wanted some plausible ratings.

Any feedback on the field? I acknowledge it is a bit top-heavy. It's not very well spread-out either in terms of gaps between each of the players ratings.

Is 9 rounds an acceptable length for such a tournament?

EDIT: Bill, how do you get the cross-table etc into that format with the scroll bar thing?

Desmond
21-06-2006, 11:09 AM
21.|Spassky, Boris||2773|
You are too, too kind

Garvinator
21-06-2006, 06:16 PM
Is 9 rounds an acceptable length for such a tournament?
for the numbers of players, i would recommend 11 or more rounds. Will help with proper sorting of the field.

Lucena
21-06-2006, 08:47 PM
for the numbers of players, i would recommend 11 or more rounds. Will help with proper sorting of the field.

I was considering that actually. In the end my reasoning was something like, "minimum number of rounds needed to spread the scores out with n players is approximately log (base 2) n. We have 80 or so players, log (base 2) 80 is between 6 and 7. So we have a couple of extra rounds just to be on the safe side. Does anyone else have an opinion on this?

PS If there are no further objections within 1 hour, I'll post the R1 draw at the end of that time.

Garvinator
22-06-2006, 02:38 AM
ok first tip Gareth for posting cross tables on here, disable smilies. That will get rid of the smiles you see.

Next, to post the 'fancy' looking tables ie cross tables, standings you see-

1) Bring up the cross table after round one in sp.
2) Then go to file - export view - text
3) Then save your new file in desktop (you can save the new file wherever you like, but you will find it easiest later if it is on the desktop ;) ).
4) Choose separator space.

Now you will have to cross table on your desktop as a file.
5) Double click on the file to open it.
6) Highlight the whole cross table.
7) Save a copy of that highlighted cross table using control+c and come back to chesschat
8) Create a new post in your assigned thread and paste the cross table to the new thread, remember to disable smilies.

Now to create the 'great' look of the cross table.

9) Highlight the cross table on chesschat, then in all the options you see above your post, click on #. It will put code before and after your highlights, just in the same way as 'quoting' does.

Now hit submit reply and you should have the good looking cross table.

Lucena
22-06-2006, 10:25 AM
Thanks - I'll give it a go soon!

Garvinator
22-06-2006, 04:55 PM
Do you want to fide rate this tournament?

Lucena
22-06-2006, 05:04 PM
Do you want to fide rate this tournament?

Well I was just treating it like a normal Fide rated tournament except I have supplied the ratings from Jeff Sonas's site. Why do you ask?

Lucena
22-06-2006, 07:06 PM
No Name Result Name

1 Showalter, J (41) 0:1 Fischer, R (1)
2 Kasparov, G (2) 1:0 Canal, E (42)
3 Shamkovich, L (43) 0:1 Botvinnik, M (3)
4 Lasker, E (4) 1:0 Samisch, F (44)
5 Rossolimo, N (45) 0:1 Capablanca, J (5)
6 Alekhine, A (6) 1:0 Chiburdanidze, M (46)
7 Nehzmetdinov, R (47) 0:1 Karpov, A (7)
8 Steinitz, W (8) .5:.5 Murey, Y (48)
9 Dake, A (49) 0:1 Tarrasch, S (9)
10 Korchnoi, V (10) 1:0 Bukic, E (50)
11 Leonhardt, P (51) 0:1 Smyslov, V (11)
12 Tal, M (12) 1:0 Byrne, D (52)
13 Polgar, Z (53) .5:.5 Chigorin, M (13)
14 Najdorf, M (14) 1:0 Rauzer, V (54)
15 Colle, E (55) 0:1 Petrosian, T (15)
16 Bronstein, D (16) 1:0 Xie Jun, (56)
17 Znosko-Borovsky, E (57) .5:.5 Rubinstein, A (17)
18 Keres, P (18) 1:0 Alexander, H (58)
19 Cramling, P (59) 0:1 Reshevsky, S (19)
20 Nimzovitch, A (20) 1:0 Petrosian, T (60)
21 Lasker, E (61) 0:1 Spassky, B (21)
22 Euwe, M (22) 1:0 Bonch-Osmolovsky, M (62)
23 Skembris, S (63) 0:1 Bogolyubov, Y (23)
24 Timman, J (24) 1:0 Nakamura, H (64)
25 Milner-Barry, P (65) 0:1 Schlechter, C (25)
26 Fine, R (26) 1:0 Bednarsky, J (66)
27 Golombek, H (67) .5:.5 Stahlberg, G (27)
28 Tartakover, S (28) .5:.5 Polgar, S (68)
29 Polgar, I (69) 0:1 Larsen, B (29)
30 Flohr, S (30) 1:0 Cheron, A (70)
31 Barden, L (71) 0:1 Polgar, J (31)
32 Duras, O (32) 1:0 Te Kolste, J (72)
33 Kosteniuk, A (73) 0:1 Kashdan, I (33)
34 Eliskases, E (34) 1:0 Skripchenko, A (74)
35 Krush, I (75) 0:1 Khalifman, A (35)
36 Byrne, R (36) 1:0 Petrosian, T (76)
37 Duchamp, M (77) 0:1 Kasimdzhanov, R (37)
38 Benko, P (38) 1:0 Bogart, H (78)
39 Charles, R (79) 0:1 Browne, W (39)
40 Denker, A (40) 1:0 Einstein, A (80)
41 Oppenheimer, R (81) 1:0 BYE

Lucena
22-06-2006, 07:15 PM
Place Name Feder Rtg Loc Score M-Buch. Buch. Progr.

1-36 Oppenheimer, Robert 1 0.5 0.5 1.0
Fischer, Robert 2895 1 0.0 0.0 1.0
Kasparov, Garry 2886 1 0.0 0.0 1.0
Botvinnik, Mikhail 2885 1 0.0 0.0 1.0
Lasker, Emanuel 2878 1 0.0 0.0 1.0
Capablanca, Jose 2877 1 0.0 0.0 1.0
Alekhine, Alexander 2865 1 0.0 0.0 1.0
Karpov, Anatoly 2848 1 0.0 0.0 1.0
Tarrasch, Siegbert 2824 1 0.0 0.0 1.0
Korchnoi, Viktor 2814 1 0.0 0.0 1.0
Smyslov, Vassily 2800 1 0.0 0.0 1.0
Tal, Mikhail 2799 1 0.0 0.0 1.0
Najdorf, Miguel 2797 1 0.0 0.0 1.0
Petrosian, Tigran V 2796 1 0.0 0.0 1.0
Bronstein, David 2792 1 0.0 0.0 1.0
Keres, Paul 2786 1 0.0 0.0 1.0
Reshevsky, Sammy 2785 1 0.0 0.0 1.0
Nimzovitch, Aron 2780 1 0.0 0.0 1.0
Spassky, Boris 2773 1 0.0 0.0 1.0
Euwe, Max 2769 1 0.0 0.0 1.0
Bogolyubov, Yefim 2768 1 0.0 0.0 1.0
Timman, Jan 2768 1 0.0 0.0 1.0
Schlechter, Carl 2764 1 0.0 0.0 1.0
Fine, Reuben 2762 1 0.0 0.0 1.0
Larsen, Bent 2755 1 0.0 0.0 1.0
Flohr, Salo 2754 1 0.0 0.0 1.0
Polgar, Judit 2746 1 0.0 0.0 1.0
Duras, Oldrich 2743 1 0.0 0.0 1.0
Kashdan, Isaac 2742 1 0.0 0.0 1.0
Eliskases, Erich 2733 1 0.0 0.0 1.0
Khalifman, Alexander 2724 1 0.0 0.0 1.0
Byrne, Robert 2720 1 0.0 0.0 1.0
Kasimdzhanov, Rustam 2708 1 0.0 0.0 1.0
Benko, Pal 2687 1 0.0 0.0 1.0
Browne, Walter 2678 1 0.0 0.0 1.0
Denker, Arnold 2677 1 0.0 0.0 1.0
37-46 Steinitz, Wilhelm 2826 0.5 0.5 0.5 0.5
Chigorin, Mikhail 2797 0.5 0.5 0.5 0.5
Rubinstein, Akiba 2789 0.5 0.5 0.5 0.5
Stahlberg, Gideon 2762 0.5 0.5 0.5 0.5
Tartakover, Savielly 2761 0.5 0.5 0.5 0.5
Murey, Yakov 2658 0.5 0.5 0.5 0.5
Polgar, Zsuzsa 2633 0.5 0.5 0.5 0.5
Znosko-Borovsky, Eugene 2613 0.5 0.5 0.5 0.5
Golombek, Harry 2543 0.5 0.5 0.5 0.5
Polgar, Sofia 2540 0.5 0.5 0.5 0.5
47-81 Showalter, Jackson 2676 0 1.0 1.0 0.0
Canal, Esteban 2675 0 1.0 1.0 0.0
Shamkovich, Leonid 2675 0 1.0 1.0 0.0
Samisch, Friedrich 2665 0 1.0 1.0 0.0
Rossolimo, Nicolas 2663 0 1.0 1.0 0.0
Chiburdanidze, Maia 2661 0 1.0 1.0 0.0
Nehzmetdinov, Rashid 2660 0 1.0 1.0 0.0
Dake, Arthur 2655 0 1.0 1.0 0.0
Bukic, Enver 2647 0 1.0 1.0 0.0
Leonhardt, Paul 2639 0 1.0 1.0 0.0
Byrne, Donald 2633 0 1.0 1.0 0.0
Rauzer, Vsevolod 2627 0 1.0 1.0 0.0
Colle, Edgar 2619 0 1.0 1.0 0.0
Xie Jun, 2619 0 1.0 1.0 0.0
Alexander, Hugh 2610 0 1.0 1.0 0.0
Cramling, Pia 2605 0 1.0 1.0 0.0
Petrosian, Tigran L 2594 0 1.0 1.0 0.0
Lasker, Edward 2583 0 1.0 1.0 0.0
Bonch-Osmolovsky, Mikhail 2577 0 1.0 1.0 0.0
Skembris, Spyridon 2577 0 1.0 1.0 0.0
Nakamura, Hikaru 2563 0 1.0 1.0 0.0
Milner-Barry, P. S. 2552 0 1.0 1.0 0.0
Bednarsky, J 2544 0 1.0 1.0 0.0
Polgar, Istvan 2529 0 1.0 1.0 0.0
Cheron, Andre 2521 0 1.0 1.0 0.0
Barden, Leonard 2497 0 1.0 1.0 0.0
Te Kolste, Jan 2497 0 1.0 1.0 0.0
Kosteniuk, Alexandra 2489 0 1.0 1.0 0.0
Skripchenko, Almira 2484 0 1.0 1.0 0.0
Krush, Irina 2478 0 1.0 1.0 0.0
Petrosian, Tigran A 2420 0 1.0 1.0 0.0
Duchamp, Marcel 2413 0 1.0 1.0 0.0
Bogart, Humphrey 0 1.0 1.0 0.0
Charles, Ray 0 1.0 1.0 0.0
Einstein, Albert 0 1.0 1.0 0.0

Bill Gletsos
22-06-2006, 07:22 PM
Gareth that doesnt look like you followed Garvin's instructions with regards the exporting of the views in post #16.

Lucena
22-06-2006, 11:04 PM
Gareth that doesnt look like you followed Garvin's instructions with regards the exporting of the views in post #16.

I wanted to post the results before the cross table, so I tried to adapt his instructions to the results. Likewise for the standings after that. I thought I followed his instructions accurately when I posted the cross table in the last post{EDIT: wonky cross table removed and reinstated below}. I'm happy with the fact that I can get the scroll bar now, but I'm guessing the ratings, scores etc are meant to be in columns so I'm not sure what I'm getting wrong that's making them out of line.

EDIT: Looks like the problem with the names etc not lining up occurred because the box "formatted output" and/or the box "column headers" wasn't checked when I exported. Everything seems to line up now.

Lucena
23-06-2006, 12:19 AM
No Name Feder Rtg 1

1 Oppenheimer, Robert 0:W
2 Fischer, Robert 2895 47:W
3 Kasparov, Garry 2886 48:W
4 Botvinnik, Mikhail 2885 49:W
5 Lasker, Emanuel 2878 50:W
6 Capablanca, Jose 2877 51:W
7 Alekhine, Alexander 2865 52:W
8 Karpov, Anatoly 2848 53:W
9 Tarrasch, Siegbert 2824 54:W
10 Korchnoi, Viktor 2814 55:W
11 Smyslov, Vassily 2800 56:W
12 Tal, Mikhail 2799 57:W
13 Najdorf, Miguel 2797 58:W
14 Petrosian, Tigran V 2796 59:W
15 Bronstein, David 2792 60:W
16 Keres, Paul 2786 61:W
17 Reshevsky, Sammy 2785 62:W
18 Nimzovitch, Aron 2780 63:W
19 Spassky, Boris 2773 64:W
20 Euwe, Max 2769 65:W
21 Bogolyubov, Yefim 2768 66:W
22 Timman, Jan 2768 67:W
23 Schlechter, Carl 2764 68:W
24 Fine, Reuben 2762 69:W
25 Larsen, Bent 2755 70:W
26 Flohr, Salo 2754 71:W
27 Polgar, Judit 2746 72:W
28 Duras, Oldrich 2743 73:W
29 Kashdan, Isaac 2742 74:W
30 Eliskases, Erich 2733 75:W
31 Khalifman, Alexander 2724 76:W
32 Byrne, Robert 2720 77:W
33 Kasimdzhanov, Rustam 2708 78:W
34 Benko, Pal 2687 79:W
35 Browne, Walter 2678 80:W
36 Denker, Arnold 2677 81:W
37 Steinitz, Wilhelm 2826 42:D
38 Chigorin, Mikhail 2797 43:D
39 Rubinstein, Akiba 2789 44:D
40 Stahlberg, Gideon 2762 45:D
41 Tartakover, Savielly 2761 46:D
42 Murey, Yakov 2658 37:D
43 Polgar, Zsuzsa 2633 38:D
44 Znosko-Borovsky, Eugene 2613 39:D
45 Golombek, Harry 2543 40:D
46 Polgar, Sofia 2540 41:D
47 Showalter, Jackson 2676 2:L
48 Canal, Esteban 2675 3:L
49 Shamkovich, Leonid 2675 4:L
50 Samisch, Friedrich 2665 5:L
51 Rossolimo, Nicolas 2663 6:L
52 Chiburdanidze, Maia 2661 7:L
53 Nehzmetdinov, Rashid 2660 8:L
54 Dake, Arthur 2655 9:L
55 Bukic, Enver 2647 10:L
56 Leonhardt, Paul 2639 11:L
57 Byrne, Donald 2633 12:L
58 Rauzer, Vsevolod 2627 13:L
59 Colle, Edgar 2619 14:L
60 Xie Jun, 2619 15:L
61 Alexander, Hugh 2610 16:L
62 Cramling, Pia 2605 17:L
63 Petrosian, Tigran L 2594 18:L
64 Lasker, Edward 2583 19:L
65 Bonch-Osmolovsky, Mikhail 2577 20:L
66 Skembris, Spyridon 2577 21:L
67 Nakamura, Hikaru 2563 22:L
68 Milner-Barry, P. S. 2552 23:L
69 Bednarsky, J 2544 24:L
70 Polgar, Istvan 2529 25:L
71 Cheron, Andre 2521 26:L
72 Barden, Leonard 2497 27:L
73 Te Kolste, Jan 2497 28:L
74 Kosteniuk, Alexandra 2489 29:L
75 Skripchenko, Almira 2484 30:L
76 Krush, Irina 2478 31:L
77 Petrosian, Tigran A 2420 32:L
78 Duchamp, Marcel 2413 33:L
79 Bogart, Humphrey 34:L
80 Charles, Ray 35:L
81 Einstein, Albert 36:L

Garvinator
23-06-2006, 01:15 AM
No Name Rtg

1. Fischer, Robert 2895
2. Kasparov, Garry 2886
3. Botvinnik, Mikhail 2885
4. Lasker, Emanuel 2878
5. Capablanca, Jose 2877
6. Alekhine, Alexander 2865
7. Karpov, Anatoly 2848
8. Steinitz, Wilhelm 2826
9. Tarrasch, Siegbert 2824
10. Korchnoi, Viktor 2814
11. Smyslov, Vassily 2800
12. Tal, Mikhail 2799
13. Chigorin, Mikhail 2797
14. Najdorf, Miguel 2797
15. Petrosian, Tigran V 2796
16. Bronstein, David 2792
17. Rubinstein, Akiba 2789
18. Keres, Paul 2786
19. Reshevsky, Sammy 2785
20. Nimzovitch, Aron 2780
21. Spassky, Boris 2773
22. Euwe, Max 2769
23. Bogolyubov, Yefim 2768
24. Timman, Jan 2768
25. Schlechter, Carl 2764
26. Fine, Reuben 2762
27. Stahlberg, Gideon 2762
28. Tartakover, Savielly 2761
29. Larsen, Bent 2755
30. Flohr, Salo 2754
31. Polgar, Judit 2746
32. Duras, Oldrich 2743
33. Kashdan, Isaac 2742
34. Eliskases, Erich 2733
35. Khalifman, Alexander 2724
36. Byrne, Robert 2720
37. Kasimdzhanov, Rustam 2708
38. Benko, Pal 2687
39. Browne, Walter 2678
40. Denker, Arnold 2677
41. Showalter, Jackson 2676
42. Canal, Esteban 2675
43. Shamkovich, Leonid 2675
44. Samisch, Friedrich 2665
45. Rossolimo, Nicolas 2663
46. Chiburdanidze, Maia 2661
47. Nehzmetdinov, Rashid 2660
48. Murey, Yakov 2658
49. Dake, Arthur 2655
50. Bukic, Enver 2647
51. Leonhardt, Paul 2639
52. Byrne, Donald 2633
53. Polgar, Zsuzsa 2633
54. Rauzer, Vsevolod 2627
55. Colle, Edgar 2619
56. Xie Jun, 2619
57. Znosko-Borovsky, Eugene 2613
58. Alexander, Hugh 2610
59. Cramling, Pia 2605
60. Petrosian, Tigran L 2594
61. Lasker, Edward 2583
62. Bonch-Osmolovsky, Mikhail 2577
63. Skembris, Spyridon 2577
64. Nakamura, Hikaru 2563
65. Milner-Barry, P. S. 2552
66. Bednarsky, J 2544
67. Golombek, Harry 2543
68. Polgar, Sofia 2540
69. Polgar, Istvan 2529
70. Cheron, Andre 2521
71. Barden, Leonard 2497
72. Te Kolste, Jan 2497
73. Kosteniuk, Alexandra 2489
74. Skripchenko, Almira 2484
75. Krush, Irina 2478
76. Petrosian, Tigran A 2420
77. Duchamp, Marcel 2413
78. Bogart, Humphrey
79. Charles, Ray
80. Einstein, Albert
81. Oppenheimer, Robert

Garvinator
23-06-2006, 01:17 AM
EDIT: Looks like the problem with the names etc not lining up occurred because the box "formatted output" and/or the box "column headers" wasn't checked when I exported. Everything seems to line up now.
yes, that is the part that I missed, forgot to tell you to tick the column headers box :eek:

Lucena
23-06-2006, 01:21 AM
So why has it put Oppenheimer first on the standings list and the cross table?

Garvinator
23-06-2006, 01:23 AM
So why has it put Oppenheimer first on the standings list and the cross table?
because of tie break options used. I have sent a reply to your email explaining a few things.

Garvinator
23-06-2006, 01:24 AM
No Name Rtg Total Result Name Rtg Total

1 Showalter, Jackson (41) 2676 [0] 0:1 Fischer, Robert (1) 2895 [0]
2 Kasparov, Garry (2) 2886 [0] 1:0 Canal, Esteban (42) 2675 [0]
3 Shamkovich, Leonid (43) 2675 [0] 0:1 Botvinnik, Mikhail (3) 2885 [0]
4 Lasker, Emanuel (4) 2878 [0] 1:0 Samisch, Friedrich (44) 2665 [0]
5 Rossolimo, Nicolas (45) 2663 [0] 0:1 Capablanca, Jose (5) 2877 [0]
6 Alekhine, Alexander (6) 2865 [0] 1:0 Chiburdanidze, Maia (46) 2661 [0]
7 Nehzmetdinov, Rashid (47) 2660 [0] 0:1 Karpov, Anatoly (7) 2848 [0]
8 Steinitz, Wilhelm (8) 2826 [0] .5:.5 Murey, Yakov (48) 2658 [0]
9 Dake, Arthur (49) 2655 [0] 0:1 Tarrasch, Siegbert (9) 2824 [0]
10 Korchnoi, Viktor (10) 2814 [0] 1:0 Bukic, Enver (50) 2647 [0]
11 Leonhardt, Paul (51) 2639 [0] 0:1 Smyslov, Vassily (11) 2800 [0]
12 Tal, Mikhail (12) 2799 [0] 1:0 Byrne, Donald (52) 2633 [0]
13 Polgar, Zsuzsa (53) 2633 [0] .5:.5 Chigorin, Mikhail (13) 2797 [0]
14 Najdorf, Miguel (14) 2797 [0] 1:0 Rauzer, Vsevolod (54) 2627 [0]
15 Colle, Edgar (55) 2619 [0] 0:1 Petrosian, Tigran V (15) 2796 [0]
16 Bronstein, David (16) 2792 [0] 1:0 Xie Jun, (56) 2619 [0]
17 Znosko-Borovsky, Eugene (57) 2613 [0] .5:.5 Rubinstein, Akiba (17) 2789 [0]
18 Keres, Paul (18) 2786 [0] 1:0 Alexander, Hugh (58) 2610 [0]
19 Cramling, Pia (59) 2605 [0] 0:1 Reshevsky, Sammy (19) 2785 [0]
20 Nimzovitch, Aron (20) 2780 [0] 1:0 Petrosian, Tigran L (60) 2594 [0]
21 Lasker, Edward (61) 2583 [0] 0:1 Spassky, Boris (21) 2773 [0]
22 Euwe, Max (22) 2769 [0] 1:0 Bonch-Osmolovsky, Mikhail (62) 2577 [0]
23 Skembris, Spyridon (63) 2577 [0] 0:1 Bogolyubov, Yefim (23) 2768 [0]
24 Timman, Jan (24) 2768 [0] 1:0 Nakamura, Hikaru (64) 2563 [0]
25 Milner-Barry, P. S. (65) 2552 [0] 0:1 Schlechter, Carl (25) 2764 [0]
26 Fine, Reuben (26) 2762 [0] 1:0 Bednarsky, J (66) 2544 [0]
27 Golombek, Harry (67) 2543 [0] .5:.5 Stahlberg, Gideon (27) 2762 [0]
28 Tartakover, Savielly (28) 2761 [0] .5:.5 Polgar, Sofia (68) 2540 [0]
29 Polgar, Istvan (69) 2529 [0] 0:1 Larsen, Bent (29) 2755 [0]
30 Flohr, Salo (30) 2754 [0] 1:0 Cheron, Andre (70) 2521 [0]
31 Barden, Leonard (71) 2497 [0] 0:1 Polgar, Judit (31) 2746 [0]
32 Duras, Oldrich (32) 2743 [0] 1:0 Te Kolste, Jan (72) 2497 [0]
33 Kosteniuk, Alexandra (73) 2489 [0] 0:1 Kashdan, Isaac (33) 2742 [0]
34 Eliskases, Erich (34) 2733 [0] 1:0 Skripchenko, Almira (74) 2484 [0]
35 Krush, Irina (75) 2478 [0] 0:1 Khalifman, Alexander (35) 2724 [0]
36 Byrne, Robert (36) 2720 [0] 1:0 Petrosian, Tigran A (76) 2420 [0]
37 Duchamp, Marcel (77) 2413 [0] 0:1 Kasimdzhanov, Rustam (37) 2708 [0]
38 Benko, Pal (38) 2687 [0] 1:0 Bogart, Humphrey (78) [0]
39 Charles, Ray (79) [0] 0:1 Browne, Walter (39) 2678 [0]
40 Denker, Arnold (40) 2677 [0] 1:0 Einstein, Albert (80) [0]
41 Oppenheimer, Robert (81) [0] 1:0 BYE

Garvinator
23-06-2006, 01:26 AM
No Name Rtg Total Result Name Rtg Total

1 Fischer, Robert (1) 2895 [1] : Euwe, Max (22) 2769 [1]
2 Bogolyubov, Yefim (23) 2768 [1] : Kasparov, Garry (2) 2886 [1]
3 Botvinnik, Mikhail (3) 2885 [1] : Timman, Jan (24) 2768 [1]
4 Schlechter, Carl (25) 2764 [1] : Lasker, Emanuel (4) 2878 [1]
5 Capablanca, Jose (5) 2877 [1] : Fine, Reuben (26) 2762 [1]
6 Larsen, Bent (29) 2755 [1] : Alekhine, Alexander (6) 2865 [1]
7 Karpov, Anatoly (7) 2848 [1] : Flohr, Salo (30) 2754 [1]
8 Tarrasch, Siegbert (9) 2824 [1] : Duras, Oldrich (32) 2743 [1]
9 Polgar, Judit (31) 2746 [1] : Korchnoi, Viktor (10) 2814 [1]
10 Smyslov, Vassily (11) 2800 [1] : Eliskases, Erich (34) 2733 [1]
11 Kashdan, Isaac (33) 2742 [1] : Tal, Mikhail (12) 2799 [1]
12 Khalifman, Alexander (35) 2724 [1] : Najdorf, Miguel (14) 2797 [1]
13 Petrosian, Tigran V (15) 2796 [1] : Byrne, Robert (36) 2720 [1]
14 Kasimdzhanov, Rustam (37) 2708 [1] : Bronstein, David (16) 2792 [1]
15 Browne, Walter (39) 2678 [1] : Keres, Paul (18) 2786 [1]
16 Reshevsky, Sammy (19) 2785 [1] : Benko, Pal (38) 2687 [1]
17 Oppenheimer, Robert (81) [1] : Nimzovitch, Aron (20) 2780 [1]
18 Spassky, Boris (21) 2773 [1] : Denker, Arnold (40) 2677 [1]
19 Polgar, Sofia (68) 2540 [.5] : Steinitz, Wilhelm (8) 2826 [.5]
20 Chigorin, Mikhail (13) 2797 [.5] : Znosko-Borovsky, Eugene (57) 2613 [.5]
21 Rubinstein, Akiba (17) 2789 [.5] : Golombek, Harry (67) 2543 [.5]
22 Stahlberg, Gideon (27) 2762 [.5] : Polgar, Zsuzsa (53) 2633 [.5]
23 Murey, Yakov (48) 2658 [.5] : Tartakover, Savielly (28) 2761 [.5]
24 Bonch-Osmolovsky, Mikhail (62) 2577 [0] : Showalter, Jackson (41) 2676 [0]
25 Canal, Esteban (42) 2675 [0] : Lasker, Edward (61) 2583 [0]
26 Nakamura, Hikaru (64) 2563 [0] : Shamkovich, Leonid (43) 2675 [0]
27 Samisch, Friedrich (44) 2665 [0] : Skembris, Spyridon (63) 2577 [0]
28 Bednarsky, J (66) 2544 [0] : Rossolimo, Nicolas (45) 2663 [0]
29 Chiburdanidze, Maia (46) 2661 [0] : Milner-Barry, P. S. (65) 2552 [0]
30 Cheron, Andre (70) 2521 [0] : Nehzmetdinov, Rashid (47) 2660 [0]
31 Te Kolste, Jan (72) 2497 [0] : Dake, Arthur (49) 2655 [0]
32 Bukic, Enver (50) 2647 [0] : Polgar, Istvan (69) 2529 [0]
33 Skripchenko, Almira (74) 2484 [0] : Leonhardt, Paul (51) 2639 [0]
34 Byrne, Donald (52) 2633 [0] : Barden, Leonard (71) 2497 [0]
35 Rauzer, Vsevolod (54) 2627 [0] : Kosteniuk, Alexandra (73) 2489 [0]
36 Petrosian, Tigran A (76) 2420 [0] : Colle, Edgar (55) 2619 [0]
37 Xie Jun, (56) 2619 [0] : Krush, Irina (75) 2478 [0]
38 Alexander, Hugh (58) 2610 [0] : Duchamp, Marcel (77) 2413 [0]
39 Bogart, Humphrey (78) [0] : Cramling, Pia (59) 2605 [0]
40 Petrosian, Tigran L (60) 2594 [0] : Charles, Ray (79) [0]
41 Einstein, Albert (80) [0] 1:0 BYE

Lucena
24-06-2006, 12:15 AM
Results should be up within a day or so.

Garvinator
24-06-2006, 12:21 AM
Well I was just treating it like a normal Fide rated tournament except I have supplied the ratings from Jeff Sonas's site. Why do you ask?
because i was thinking that with Robert Fischer in the draw, you will have some kind of tournament drama, so you could have said, no plans to fide rate the tournament and then report to us that Bobby has withdrawn over changed conditions :P :P :P tournament has to be realistic you know :whistle:

four four two
24-06-2006, 12:38 AM
Results should be up within a day or so.

To maintain historical accuracy Bobby should forfeit his first game...:D ;)

antichrist
24-06-2006, 01:36 AM
To maintain historical accuracy Bobby should forfeit his first game...:D ;)

I can remember that live - can you?

Bill Gletsos
24-06-2006, 01:56 AM
To maintain historical accuracy Bobby should forfeit his first game...:D ;)If you are referring to his match with Spassky in 1972 he forfeited the second game.

antichrist
24-06-2006, 03:10 PM
Reminds of a school friend who claimed her front gate was jammed so could not come in - did not wash well with the nuns. Had us all in fits of laughter.

ElevatorEscapee
24-06-2006, 05:35 PM
Reminds of a school friend who claimed her front gate was jammed so could not come in - did not wash well with the nuns. Had us all in fits of laughter.
You were taught by nuns? :lol:

pax
26-06-2006, 01:28 PM
You probably won't get many (any?) pairings with so many players (unless you play 15 rounds or so).

Lucena
26-06-2006, 02:50 PM
You probably won't get many (any?) pairings with so many players (unless you play 15 rounds or so).

Sorry, do you mean I won't get any pairings between the top players? I don't completely understand what you are saying.

Garvinator
26-06-2006, 04:03 PM
Sorry, do you mean I won't get any pairings between the top players? I don't completely understand what you are saying.
What I believe Pax was trying to say and what I raised earlier is that with 81 players and 9 rounds, there will usually be about 8 or so players in each score group after the 9 rounds. Therefore swissperfect will not be properly tested for finding correct pairings.

Hence why for 81 players, about 15 rounds is required so you get score groups with only a couple of players in each group. Then the conflicts begin between colours, floats, scoregroups and players having played each other etc. This is when you find out if the chosen pairing program is any good- by how accurately it follows the pairing rules it is designed for ie swiss perfect/dutch pairing rules.

Lucena
26-06-2006, 05:14 PM
Ok no problem, guess I'll just have to have more rounds then. Thanks Garvin and pax for the advice.

Garvinator
26-06-2006, 05:16 PM
Ok no problem, guess I'll just have to have more rounds then. Thanks Garvin and pax for the advice.
or you could just keep your top 32 players and get rid of the rest, surely that will be quicker to get you where you want to go :uhoh:

Lucena
26-06-2006, 09:01 PM
or you could just keep your top 32 players and get rid of the rest, surely that will be quicker to get you where you want to go :uhoh:

Nah, more interesting this way, although I guess it could take a while.

Garvinator
26-06-2006, 09:02 PM
Nah, more interesting this way, although I guess it could take a while.
and much more work to check out similar 'problems' in the same round :hand: over a total of 15 or so rounds.
If you think I am trying to cut down the amount of pairing checking that those of us who are going to check out the pairings are going to have to do, you are damn right ;)

Desmond
26-06-2006, 09:03 PM
Forgive my n00b question, but how are you deciding the results of the games?

Garvinator
26-06-2006, 09:04 PM
Forgive my n00b question, but how are you deciding the results of the games?
Gareth is making it up :P

Lucena
26-06-2006, 09:09 PM
Yeah, pretty much. Although I am making the result of each game reflect approximately the relative strength of the players, also I'm taking into account (in some cases) what happened historically between those 2 players.

pax
27-06-2006, 04:38 PM
You probably won't get many (any?) pairings with so many players (unless you play 15 rounds or so).

Gawd. I *meant* to say you won't get many challenging pairings with so few players. By that I mean pairings in which flaws in pairing algorithms are likely to be exposed, or similarly where human arbiters are more likely to make mistakes.

Lucena
28-06-2006, 02:25 AM
[
Gawd. I *meant* to say you won't get many challenging pairings with so few players. By that I mean pairings in which flaws in pairing algorithms are likely to be exposed, or similarly where human arbiters are more likely to make mistakes.

Yes, I get what you meant now.


pairing checking that those of us who are going to check out the pairings

Woohoo! :clap: I was hoping someone would help me out!


much more work to check out similar 'problems' in the same round :hand: over a total of 15 or so rounds. If you think I am trying to cut down the amount of pairing checking that those of us who are going to check out the pairings are going to have to do, you are damn right ;)

Fair enough then. I kind of like this tournament the way it is though, I will put this one on hiatus and start another one shortly, ie now.

[CODE]No Name Feder Rtg Loc

1. Laver, Rod 2810
2. Sampras, Pete 2800
3. McEnroe, John 2780
4. Agassi, Andre 2775
5. Borg, Bjorn 2770
6. Federer, Roger 2760
7. Becker, Boris 2750
8. Edberg, Stefan 2740
9. Courier, Jim 2735
10. Hewitt, Lleyton 2730
11. Nadal, Rafael 2725
12. Rafter, Pat 2720
13. Chang, Michael 2715
14. Nalbandian, David 2712
15. Roddick, Andy 2710
16. Cash, Pat 2680
17. El Aynaoui, Younes 2660
18. Philippousis, Mark 2650
19. Baghdatis, Marcos 2640

Don't be fooled, yes I know in tennis you don't allocate colours and can't agree to draws, but cut me some slack ok? Also, don't read too much into the ratings I have assigned the players - remember this is only an exercise!


No Name Result Name

1 Laver, R (1) 1:0 Hewitt, L (10)
2 Nadal, R (11) 0:1 Sampras, P (2)
3 McEnroe, J (3) 1:0 Rafter, P (12)
4 Chang, M (13) 0:1 Agassi, A (4)
5 Borg, B (5) 1:0 Nalbandian, D (14)
6 Roddick, A (15) 0:1 Federer, R (6)
7 Becker, B (7) 1:0 Cash, P (16)
8 El Aynaoui, Y (17) 0:1 Edberg, S (8)
9 Courier, J (9) 1:0 Philippousis, M (18)
10 Baghdatis, M (19) 1:0 BYE

Anyone got any ideas on how many rounds I should make it? Is 7 enough or would more be better?

Garvinator
28-06-2006, 02:54 AM
Anyone got any ideas on how many rounds I should make it? Is 7 enough or would more be better?
make it 9, give it a real work out :) and remember to send me the sp files for this new tournament ;)

Lucena
28-06-2006, 02:56 AM
make it 9, give it a real work out :) and remember to send me the sp files for this new tournament ;)
9 it is, then.

Garvinator
28-06-2006, 02:57 AM
9 it is, then.
and now I am done for the evening.

Garvinator
28-06-2006, 04:02 PM
Do you mind if we delete Baghdatis? That way there will be 18 players and no byes. Having a bye adds a whole new set of pairing difficulties.

pax
28-06-2006, 06:15 PM
Do you mind if we delete Baghdatis? That way there will be 18 players and no byes. Having a bye adds a whole new set of pairing difficulties.

Aren't "pairing difficulties" the whole point?

Since there are a whole raft of rules concerning bye allocation, these should be tested too.

Lucena
28-06-2006, 11:19 PM
Aren't "pairing difficulties" the whole point?
Indeed, having a bye was intentional.
Since there are a whole raft of rules concerning bye allocation, these should be tested too.
I figured that was a good idea.

Garvinator
28-06-2006, 11:24 PM
I figured that was a good idea.
No problem, bye stays in.

Are we going to round two? What pairings do you get Gareth?

Lucena
29-06-2006, 12:31 AM
No problem, bye stays in.

Are we going to round two? What pairings do you get Gareth?


No Name Result Name

1 Federer, R (6) : Laver, R (1)
2 Sampras, P (2) : Becker, B (7)
3 Edberg, S (8) : McEnroe, J (3)
4 Agassi, A (4) : Courier, J (9)
5 Baghdatis, M (19) : Borg, B (5)
6 Hewitt, L (10) : Roddick, A (15)
7 Nalbandian, D (14) : Nadal, R (11)
8 Rafter, P (12) : El Aynaoui, Y (17)
9 Cash, P (16) : Chang, M (13)
10 Philippousis, M (18) 1:0 BYE

Garvinator
29-06-2006, 01:12 AM
pairings are fine, of course it is only round 2 of 9. Results and your sp pairings for round three :)

Lucena
29-06-2006, 04:49 PM
No Name Result Name

1 Federer, R (6) .5:.5 Laver, R (1)
2 Sampras, P (2) 1:0 Becker, B (7)
3 Edberg, S (8) .5:.5 McEnroe, J (3)
4 Agassi, A (4) 1:0 Courier, J (9)
5 Baghdatis, M (19) 0:1 Borg, B (5)
6 Hewitt, L (10) 1:0 Roddick, A (15)
7 Nalbandian, D (14) 0:1 Nadal, R (11)
8 Rafter, P (12) 1:0 El Aynaoui, Y (17)
9 Cash, P (16) 0:1 Chang, M (13)
10 Philippousis, M (18) 1:0 BYE


No Name Result Name

1 Borg, B (5) : Sampras, P (2)
2 Laver, R (1) : Agassi, A (4)
3 McEnroe, J (3) : Federer, R (6)
4 Becker, B (7) : Edberg, S (8)
5 Courier, J (9) : Rafter, P (12)
6 Philippousis, M (18) : Hewitt, L (10)
7 Nadal, R (11) : Baghdatis, M (19)
8 Chang, M (13) : Nalbandian, D (14)
9 Roddick, A (15) : Cash, P (16)
10 El Aynaoui, Y (17) 1:0 BYE

Lucena
30-06-2006, 12:43 AM
Round 3 results


No Name Result Name

1 Borg, B (5) 0:1 Sampras, P (2)
2 Laver, R (1) .5:.5 Agassi, A (4)
3 McEnroe, J (3) 1:0 Federer, R (6)
4 Becker, B (7) .5:.5 Edberg, S (8)
5 Courier, J (9) 1:0 Rafter, P (12)
6 Philippousis, M (18) 0:1 Hewitt, L (10)
7 Nadal, R (11) 1:0 Baghdatis, M (19)
8 Chang, M (13) 1:0 Nalbandian, D (14)
9 Roddick, A (15) .5:.5 Cash, P (16)
10 El Aynaoui, Y (17) 1:0 BYE

Lucena
30-06-2006, 09:21 PM
Round 4 pairings


No Name Result Name

1 Sampras, P (2) : McEnroe, J (3)
2 Agassi, A (4) : Borg, B (5)
3 Nadal, R (11) : Laver, R (1)
4 Edberg, S (8) : Chang, M (13)
5 Hewitt, L (10) : Courier, J (9)
6 Federer, R (6) : Becker, B (7)
7 Rafter, P (12) : Baghdatis, M (19)
8 El Aynaoui, Y (17) : Roddick, A (15)
9 Cash, P (16) : Philippousis, M (18)
10 Nalbandian, D (14) 1:0 BYE

Garvinator
02-07-2006, 11:16 AM
No Name Rtg Total Result Name Rtg Total

1 SAMPRAS, Pete (2) 2800 [3] : MCENROE, John (3) 2780 [2.5]
2 AGASSI, Andre (4) 2775 [2.5] : BORG, Bjorn (5) 2770 [2]
3 NADAL, Rafael (11) 2725 [2] : LAVER, Rod (1) 2810 [2]
4 EDBERG, Stefan (8) 2740 [2] : CHANG, Michael (13) 2715 [2]
5 HEWITT, Lleyton (10) 2730 [2] : COURIER, Jim (9) 2735 [2]
6 FEDERER, Roger (6) 2760 [1.5] : BECKER, Boris (7) 2750 [1.5]
7 RAFTER, Pat (12) 2720 [1] : BAGHDATIS, Marcos (19) 2640 [1]
8 EL AYNAOUI, Younes (17) 2660 [1] : RODDICK, Andy (15) 2710 [.5]
9 CASH, Pat (16) 2680 [.5] : PHILIPPOUSIS, Mark (18) 2650 [1]
10 NALBANDIAN, David (14) 2712 [0] 1:0 BYE

Garvinator
02-07-2006, 11:19 AM
No Name Rtg Total 1 2 3 4

1 SAMPRAS, Pete 2800 3 9:W 12:W 5:W 2:
2 MCENROE, John 2780 2.5 13:W 6:D 11:W 1:
3 AGASSI, Andre 2775 2.5 10:W 7:W 4:D 5:
4 LAVER, Rod 2810 2 8:W 11:D 3:D 9:
5 BORG, Bjorn 2770 2 19:W 16:W 1:L 3:
6 EDBERG, Stefan 2740 2 14:W 2:D 12:D 10:
7 COURIER, Jim 2735 2 15:W 3:L 13:W 8:
8 HEWITT, Lleyton 2730 2 4:L 17:W 15:W 7:
9 NADAL, Rafael 2725 2 1:L 19:W 16:W 4:
10 CHANG, Michael 2715 2 3:L 18:W 19:W 6:
11 FEDERER, Roger 2760 1.5 17:W 4:D 2:L 12:
12 BECKER, Boris 2750 1.5 18:W 1:L 6:D 11:
13 RAFTER, Pat 2720 1 2:L 14:W 7:L 16:
14 EL AYNAOUI, Younes 2660 1 6:L 13:L 0:W 17:
15 PHILIPPOUSIS, Mark 2650 1 7:L 0:W 8:L 18:
16 BAGHDATIS, Marcos 2640 1 0:W 5:L 9:L 13:
17 RODDICK, Andy 2710 .5 11:L 8:L 18:D 14:
18 CASH, Pat 2680 .5 12:L 10:L 17:D 15:
19 NALBANDIAN, David 2712 0 5:L 9:L 10:L 0:

Lucena
02-07-2006, 02:37 PM
Sorry Garvin, I forgot to follow your advice in changing the settings so that the draw displays players' scores etc. I was meaning to do it before:(

Lucena
02-07-2006, 03:16 PM
I've changed the settings to include more information in the draw and results.


No Name Rtg Total Result Name Rtg Total

1 Sampras, Pete 2800 [3] 1:0 McEnroe, John 2780 [2.5]
2 Agassi, Andre 2775 [2.5] .5:.5 Borg, Bjorn 2770 [2]
3 Nadal, Rafael 2725 [2] 0:1 Laver, Rod 2810 [2]
4 Edberg, Stefan 2740 [2] 1:0 Chang, Michael 2715 [2]
5 Hewitt, Lleyton 2730 [2] .5:.5 Courier, Jim 2735 [2]
6 Federer, Roger 2760 [1.5] 1:0 Becker, Boris 2750 [1.5]
7 Rafter, Pat 2720 [1] 1:0 Baghdatis, Marcos 2640 [1]
8 El Aynaoui, Younes 2660 [1] .5:.5 Roddick, Andy 2710 [.5]
9 Cash, Pat 2680 [.5] 1:0 Philippousis, Mark 2650 [1]
10 Nalbandian, David 2712 [0] 1:0 BYE

Lucena
02-07-2006, 03:26 PM
No Name Rtg Total Result Name Rtg Total

1 Agassi, Andre 2775 [3] : Sampras, Pete 2800 [4]
2 Laver, Rod 2810 [3] : Edberg, Stefan 2740 [3]
3 McEnroe, John 2780 [2.5] : Hewitt, Lleyton 2730 [2.5]
4 Borg, Bjorn 2770 [2.5] : Federer, Roger 2760 [2.5]
5 Courier, Jim 2735 [2.5] : Nadal, Rafael 2725 [2]
6 Chang, Michael 2715 [2] : Rafter, Pat 2720 [2]
7 Becker, Boris 2750 [1.5] : El Aynaoui, Younes 2660 [1.5]
8 Baghdatis, Marcos 2640 [1] : Cash, Pat 2680 [1.5]
9 Nalbandian, David 2712 [1] : Philippousis, Mark 2650 [1]
10 Roddick, Andy 2710 [1] 1:0 BYE

Garvinator
02-07-2006, 04:33 PM
Ok Gareth,

Here is the first round where the dutch pairing rules really begin as now we are starting to get down to one and two players per score group.
Hence why both Pax and myself were suggesting having many rounds and only a few players :uhoh: .

The pairing info for each player below- this can be seen in sp by view- pairing info.

If you find the explanations confusing, dont worry, ask further questions if you need to. Almost everyone is confused when trying to work out dutch pairings the first time.



Place No Opponents Colours Float Score


1 2 : 11,7,5,3 BWBW D 4

2-4 1 : 10,6,4,11 WBWB u 3
4 : 13,9,1,5 BWBW D 3
8 : 17,3,7,13 BWBW d 3

5-9 3 : 12,8,6,2 WBWB U 2.5
5 : 14,19,2,4 WBWB U 2.5
6 : 15,1,3,7 BWBW 2.5
9 : 18,4,12,10 WBWB 2.5
10 : 1,15,18,9 BWBW 2.5

10-12 11 : 2,14,19,1 WBWW 2
12 : 3,17,9,19 BWBW 2
13 : 4,16,14,8 WBWB d 2

13-15 7 : 16,2,8,6 WBWB u 1.5
16 : 7,13,15,18 BWBW U 1.5
17 : 8,12,-,15 WB-W D 1.5

16-19 14 : 5,11,13,- BWB- uD 1
15 : 6,10,16,17 WBWB U 1
18 : 9,-,10,16 B-WB D 1
19 : -,5,11,12 -WBB 1

I think most of the columns are self-explanatory, except for the float column. What the float symbols mean:
Big D (D) - the player downfloated (went down to a lower score group) one round before
Little D (d)- the player downfloated (went down to the lower score group) two rounds before
Big U (U)- the player upfloated (went up to a higher score group) one round before
Little U (u)- the player upfloated (went up to a higher score group) two rounds before

These are important as there are very strict 'floating' rules in the dutch pairing rules.

Ok now time to explain each pairing:


Place No Opponents Colours Float Score


1 2 : 11,7,5,3 BWBW D 4

2-4 1 : 10,6,4,11 WBWB u 3
4 : 13,9,1,5 BWBW D 3
8 : 17,3,7,13 BWBW d 3

2 wants black

1 wants white, has played 4, has upfloated.
4 wants black, has played 1
8 wants black

Clearly in this score group it is not possible for each of the four players to receive their 'due' colour as three require black. So you try and give as many players as possible their colour.
2 v 1 looks best, but 1 has upfloated two rounds before, so cant go up this round.

4 v 2
1 v 8

2 is higher ranked in this round so keeps its colour alternation.
1 and 8 were due opposite colours.

Next group:


5-9 3 : 12,8,6,2 WBWB U 2.5
5 : 14,19,2,4 WBWB U 2.5
6 : 15,1,3,7 BWBW 2.5
9 : 18,4,12,10 WBWB 2.5
10 : 1,15,18,9 BWBW 2.5

3 wants white, has played 6
5 wants white, has played none of the others
6 wants black, has played 3
9 wants white, has played 10
10 wants black, has played 9

Now you divide the players into top half/ bottom half (referred to as S1/S2), when the score group has odd numbers, S1 is 'always' the smaller group.


S1 S2
3 6
5 9
10

So pairing S1 with S2, you look for the best colour matches across the S's.
3 and 6 look a great colour match, but have played either, so that doesnt work.
Next colour match for 3 is 10. So pairing is 3 v 10
Now that leaves player 5. Only 6 and 9 remain. 5 v 6 match for colours. Therefore, 9 downfloats.

3 v 10
5 v 6

This score group, the rules are a little different as you have 9 downfloating. When a player has downfloated, the whole new group is referred to as a heterogenous score group. A score group of all the same scores is a homogenous score group.


9 : 18,4,12,10 WBWB 2.5
10-12 11 : 2,14,19,1 WBWW 2
12 : 3,17,9,19 BWBW 2
13 : 4,16,14,8 WBWB d 2

9 wants white, has played 12
11 must be black
12 wants black, has played 9
13 wants white

Regarding S1 and S2 allocations, when downfloating a player, that player comprises S1 and the rest are S2.


S1: S2:
9 11
12
13

9 is paired first and cant play 12. So for colours, 9 v 11
That leaves only 13 v 12.

9 v 11
13 v 12

Next score group:

13-15 7 : 16,2,8,6 WBWB u 1.5
16 : 7,13,15,18 BWBW U 1.5
17 : 8,12,-,15 WB-W D 1.5

7 wants white, has played 16
16 wants black, has played 7
17 has had one more white than black, so is regarded as +1 and has a higher preference for black than a player who has had an even number of white/blacks. Also, 17 cant downfloat (well actually 17 can be downfloated, but it takes extraordinary conditions for this to happen)


S1: S2:
7 16
17

7 has played 16. So that leaves 7 v 17. 16 downfloats.

Final score group:


16 : 7,13,15,18 BWBW U 1.5

14 : 5,11,13,- BWB- uD 1
15 : 6,10,16,17 WBWB U 1
18 : 9,-,10,16 B-WB D 1
19 : -,5,11,12 -WBB 1

16 wants black, has played 15, 18
14 is -1 preference white, has had a bye.
15 wants white, has played 16
18 is -1 preference white and has had a bye
19 is -1 preference white and has had a bye


S1: S2:
16 14
15
18
19

Pair 16 first, 16 v 15 looks good, but they have played each other, next colour match is 19. So 19 v 16

After the downfloated player 16 has been paired, you are left with players who are all from the same score group. Now it is a homogenous score group and is divided as normal:


S1: S2:
14 15
18

14 is -1 preference white, has had a bye.
15 wants white, has played 16
18 is -1 preference white and has had a bye

As you can see, 14 and 18 have had byes, so that leaves 15 as the bye. 14 and 18 are paired together with the higher ranked player (14) getting their due colour, so 14 v 18. 15 bye

These match the sp pairings.

Bill Gletsos
02-07-2006, 06:27 PM
Ok now time to explain each pairing:


Place No Opponents Colours Float Score


1 2 : 11,7,5,3 BWBW D 4

2-4 1 : 10,6,4,11 WBWB u 3
4 : 13,9,1,5 BWBW D 3
8 : 17,3,7,13 BWBW d 3

2 wants black

1 wants white, has played 4, has upfloated.
4 wants black, has played 1
8 wants black

Clearly in this score group it is not possible for each of the four players to receive their 'due' colour as three require black. So you try and give as many players as possible their colour.
2 v 1 looks best, but 1 has upfloated two rounds before, so cant go up this round.

4 v 2
1 v 8

2 is higher ranked in this round so keeps its colour alternation.
1 and 8 were due opposite colours.You need to be more explicit in explaining these pairings because although you get it right this time you easily could not.

Since there is only one player on 4 points they float down to the next score group.
This results in a hetrogenous score group such that S1 consists of 1 player and S2 of 3 players. From your description above that is not at all clear.

Thus w=1, b=3, q=2, x=1, p=1.


S1 S2
2 1
4
8As you noted 2 playing 1 is no good because of rule B6 (upfloat 2 rounds previous).
Therefore apply C7 and do the transpositions in the order:
1-8-4 (still no good)
4-1-8
restarting at C6 in each case.

2 playing 4 is ok as x=1 (one pairing will not have colour match).
Thus we have the pairing 4 V 2.

That leaves the homogenous score group of 1 and 8 to be pairied. So retsrat at C1.
So w=1, b=1, q=1, x=0, p=1.

Pairing is thus 1 V 8.

Note that in the above if player 1 had not been an upfloat in in round 3 then the pairings would have been :
1 V 2 and 8 V 4.

These would also have been the pairings if round 5 was the last round as B6 (along with B2 and B5) does not apply in the last round.

Garvinator
02-07-2006, 06:31 PM
You need to be more explicit in explaining these pairings because although you get it right this time you easily could not.

Since there is only one player on 4 points they float down to the next score group.
This results in a hetrogenous score group such that S1 consists of 1 player and S2 of 3 players. From your description above that is not at all clear.

Thank you for adding it in this way. I did cover S1 and S2 in later score groups. I really was trying to cut down the amount of explanation required.

If you notice, I do refer to the heterogenous and homogenous score groups later on, but mainly I referred to it when a group changes from heterogenous to homogenous.

A small case of being slightly lazy:eek: on my part.

Also I was trying to avoid all the x, p, q and b talk as much as possible as it can be confusing to those who are coming into 'pairing talk' the first time.

Lucena
02-07-2006, 06:32 PM
Hmm, some good stuff for me to work through there to make sure I understand how the system works. Can I just ask though, when you mean the FIDE Dutch Rules, do you mean these (http://www.fide.com/official/handbook.asp?level=C0402)? And are those rules up to date?

Garvinator
02-07-2006, 06:40 PM
Hmm, some good stuff for me to work through there to make sure I understand how the system works. Can I just ask though, when you mean the FIDE Dutch Rules, do you mean these (http://www.fide.com/official/handbook.asp?level=C0402)? And are those rules up to date?
not quite, you want: http://www.fide.com/official/handbook.asp?level=C0401

Bill Gletsos
02-07-2006, 07:11 PM
Thank you for adding it in this way. I did cover S1 and S2 in later score groups. I really was trying to cut down the amount of explanation required.True but I thought it was important to explain it up front.

If you notice, I do refer to the heterogenous and homogenous score groups later on, but mainly I referred to it when a group changes from heterogenous to homogenous.

A small case of being slightly lazy:eek: on my part.;)

Also I was trying to avoid all the x, p, q and b talk as much as possible as it can be confusing to those who are coming into 'pairing talk' the first time.This I think is a fatal mistake since it is clear from previous discussions in other threads on this BB that even experienced arbiters are failing to do the x, p etc calculations and as such they get the pairings incorrect.
By failing to do/show the calculations you only highten the chance of error.

Lucena
02-07-2006, 07:41 PM
Hmm, some good stuff for me to work through there to make sure I understand how the system works. Can I just ask though, when you mean the FIDE Dutch Rules, do you mean these (http://www.fide.com/official/handbook.asp?level=C0402)? And are those rules up to date?


not quite, you want: http://www.fide.com/official/handbook.asp?level=C0401

Thanks. Are those completely up to date? (I seem to remember hearing that sometimes they don't update stuff properly on the FIDE website)

Bill Gletsos
02-07-2006, 07:43 PM
Thanks. Are those completely up to date? (I seem to remember hearing that sometimes they don't update stuff properly on the FIDE website)They are.

Lucena
02-07-2006, 10:40 PM
I seem to remember some discussions a while ago about the "ambiguity" of the "Dutch" pairing rules, ie they could be interpreted differently by different people. Is this currently the case with the Dutch pairings?

Lucena
03-07-2006, 04:50 PM
I'm going to go ahead with the next round - it would be good for me to go thru the calculations but (as I trust Bill and Garvin are correct) I think I'll go through all of them at the end, so I don't lose momentum in producing pairings.


No Name Rtg Total Result Name Rtg Total

1 Agassi, Andre 2775 [3] .5:.5 Sampras, Pete 2800 [4]
2 Laver, Rod 2810 [3] 1:0 Edberg, Stefan 2740 [3]
3 McEnroe, John 2780 [2.5] 1:0 Hewitt, Lleyton 2730 [2.5]
4 Borg, Bjorn 2770 [2.5] 0:1 Federer, Roger 2760 [2.5]
5 Courier, Jim 2735 [2.5] 1:0 Nadal, Rafael 2725 [2]
6 Chang, Michael 2715 [2] .5:.5 Rafter, Pat 2720 [2]
7 Becker, Boris 2750 [1.5] 1:0 El Aynaoui, Younes 2660 [1.5]
8 Baghdatis, Marcos 2640 [1] 0:1 Cash, Pat 2680 [1.5]
9 Nalbandian, David 2712 [1] 1:0 Philippousis, Mark 2650 [1]
10 Roddick, Andy 2710 [1] 1:0 BYE

Lucena
03-07-2006, 09:13 PM
No Name Rtg Total Result Name Rtg Total

1 Sampras, Pete 2800 [4.5] : Laver, Rod 2810 [4]
2 Courier, Jim 2735 [3.5] : McEnroe, John 2780 [3.5]
3 Federer, Roger 2760 [3.5] : Agassi, Andre 2775 [3.5]
4 Edberg, Stefan 2740 [3] : Borg, Bjorn 2770 [2.5]
5 Rafter, Pat 2720 [2.5] : Becker, Boris 2750 [2.5]
6 Hewitt, Lleyton 2730 [2.5] : Chang, Michael 2715 [2.5]
7 Cash, Pat 2680 [2.5] : Nalbandian, David 2712 [2]
8 Philippousis, Mark 2650 [1] : Roddick, Andy 2710 [2]
9 El Aynaoui, Younes 2660 [1.5] : Baghdatis, Marcos 2640 [1]
10 Nadal, Rafael 2725 [2] 1:0 BYE

Garvinator
03-07-2006, 11:17 PM
Ok Gareth, thanks for being a major pain in the butt this round :eek: ;) This was difficult.

I am not going to give a detailed explanation yet as to pairings for next round and how they came about.

Reason being is I really need Bill to see what he comes up with and the reasons for doing so. Kevin might even have a go ;)

Sp:

No Name Rtg Total Result Name Rtg Total

1 SAMPRAS, Pete (2) 2800 [4.5] : LAVER, Rod (1) 2810 [4]
2 COURIER, Jim (9) 2735 [3.5] : MCENROE, John (3) 2780 [3.5]
3 FEDERER, Roger (6) 2760 [3.5] : AGASSI, Andre (4) 2775 [3.5]
4 EDBERG, Stefan (8) 2740 [3] : BORG, Bjorn (5) 2770 [2.5]
5 RAFTER, Pat (12) 2720 [2.5] : BECKER, Boris (7) 2750 [2.5]
6 HEWITT, Lleyton (10) 2730 [2.5] : CHANG, Michael (13) 2715 [2.5]
7 CASH, Pat (16) 2680 [2.5] : NALBANDIAN, David (14) 2712 [2]
8 PHILIPPOUSIS, Mark (18) 2650 [1] : RODDICK, Andy (15) 2710 [2]
9 EL AYNAOUI, Younes (17) 2660 [1.5] : BAGHDATIS, Marcos (19) 2640 [1]
10 NADAL, Rafael (11) 2725 [2] 1:0 BYE

Swiss Master 5:


Table White - Black Results
----------------------------------------------------------------------------------------- round 6
1 Sampras ( 4.5) - Laver ( 4 ) 2- 1
2 Courier ( 3.5) - McEnroe ( 3.5) 9- 3
3 Federer ( 3.5) - Agassi ( 3.5) 6- 4
4 Edberg ( 3 ) - Borg ( 2.5) 8- 5
5 Rafter ( 2.5) - Becker ( 2.5) 12- 7
6 Hewitt ( 2.5) - Chang ( 2.5) 10- 13
7 Cash ( 2.5) - El Aynaoui ( 1.5) 16- 17
8 Roddick ( 2 ) - Nalbandian ( 2 ) 15- 14
9 Philippousis ( 1 ) - Baghdatis ( 1 ) 18- 19
Bye : 11 Nadal

From my manual pairings and further looking at the pairing rules, I think sp is more correct than sm5, which is why I want Bill to have a look at the pairings and give his opinion.

Sp pairing info:


No Opponents Colours Float Score


2 : 11,7,5,3,4 BWBWB D 4.5

1 : 10,6,4,11,8 WBWBW 4

3 : 12,8,6,2,10 WBWBW u 3.5
4 : 13,9,1,5,2 BWBWW Ud 3.5
6 : 15,1,3,7,5 BWBWB 3.5
9 : 18,4,12,10,11 WBWBW D 3.5

8 : 17,3,7,13,1 BWBWB 3

5 : 14,19,2,4,6 WBWBW u 2.5
7 : 16,2,8,6,17 WBWBW 2.5
10 : 1,15,18,9,3 BWBWB 2.5
12 : 3,17,9,19,13 BWBWB 2.5
13 : 4,16,14,8,12 WBWBW 2.5
16 : 7,13,15,18,19 BWBWB uD 2.5

11 : 2,14,19,1,9 WBWWB U 2
14 : 5,11,13,-,18 BWB-W d 2
15 : 6,10,16,17,- WBWB- uD 2

17 : 8,12,-,15,7 WB-WB d 1.5

18 : 9,-,10,16,14 B-WBB d 1
19 : -,5,11,12,16 -WBBW U 1

pax
04-07-2006, 06:07 PM
From my manual pairings and further looking at the pairing rules, I think sp is more correct than sm5, which is why I want Bill to have a look at the pairings and give his opinion.

I think the difference between the two pairings is in the way the bye is allocated.

There are essentially no problems until 16 is floated down into the '2' score group.

At that point, it is easy to notice that only 11 or 16 can receive the bye and allocate it to 11 and continue with the pairings, reaching:

16-14
18-15
17-19

As paired by SP. That is, in fact what I did when I first attempted a manual pairing.

But in fact, the pairing rules do not allow you to allocate the bye until the final score group. So instead you get:

16-14
11-15

And it is not until you get down to the final score group do you see that you cannot arrive at a legal pairing. You end up applying C13 twice (unpairing this previous pairing), and arriving at a merged score group of 16, 11, 14, 15, 17, 18 and 19.


It isn't entirely clear to me how this group should be divided. A6 suggests that S1 consists of *all* players moved from a higher group, but it also implies that S1 should always be smaller than or equal to S2 in size, which in this case is contradictory. If you treat this group as if it were homogeneous, you get the pairings:
16-17
15-14
18-19

As per SM5.

I would also point out that the SM5 pairing is strictly better than the SP pairing according to the pairing criterion B3 (sum of difference of scores).

Garvinator
04-07-2006, 07:07 PM
Ok time to start.

In typing this all out, right at the end was I aware that pax has replied about the 2.5 and below score groups. Further discussion on that to come.


I will include my foul ups in regards to how I paired the players the first time. I think it could be very instructive.


2 : 11,7,5,3,4 BWBWB D 4.5

1 : 10,6,4,11,8 WBWBW 4

Since player 2 has no other player in its score group, you downfloat it to be paired with the only player in next score group. So you get 2 v 1. They are due opposite colours, so no issues.


3 : 12,8,6,2,10 WBWBW u 3.5
4 : 13,9,1,5,2 BWBWW Ud 3.5
6 : 15,1,3,7,5 BWBWB 3.5
9 : 18,4,12,10,11 WBWBW D 3.5

3 is +1 for colours, has played 6.
4 is +1 for colours, also must be black, has played 9
6 is -1 for colours, has played 3
9 is +1 for colours.

There are three players due black, one due white. This means that not all four colour preferences can be allocated.

S1: S2:
3 6
4 9

3 and 6 have played, so have 4 and 9, so that means 3 v 9 and 4 v 6.

In the 3/9 pairing, they are both +1, so the higher ranked player(3) keeps their normal colour alternation, so 3 is black. 9 v 3
In the 4/6 pairing, 4 has had two whites in a row, so 4 must be black. 6 v 4



8 : 17,3,7,13,1 BWBWB 3

5 : 14,19,2,4,6 WBWBW u 2.5
7 : 16,2,8,6,17 WBWBW 2.5
10 : 1,15,18,9,3 BWBWB 2.5
12 : 3,17,9,19,13 BWBWB 2.5
13 : 4,16,14,8,12 WBWBW 2.5
16 : 7,13,15,18,19 BWBWB uD 2.5

What I had worked out was.
8 v 10
16 v 5
12 v 7
13 downfloats.

8: -1, played 7,13
5: +1, upfloated two rounds before
7: +1, played 8, 16
10: -1
12: -1, played 13
13: +1, played 8, 12, 16
16: -1, played 13, upfloated 2 rounds before, downfloated last round

4 players require black, 3 require white, this means only 3 pairings can have correct colour allocation. One player misses out.

My reasoning was as follows.

S1:
8

S2: 5,7,10,12,13,16

Pair 8 first, 5 cant upfloat, 8 and 7 have played, next colour match for 8 is 13 but they have played. So I paired 8 v 10 (next player in after 5 and 7).

Then 16 v 5, 12 v 7 follow for colours. 13 downfloats.

All looks fine except that if you look at the pairing section B, relative criteria:

(These are in descending priority. They should be fulfilled as much as possible. To comply with these criteria, transpositions or even exchanges may be applied, but no player should be moved down to a lower score bracket).

B.4 As many players as possible receive their colour preference. (Whenever x of a score bracket is unequal to zero this rule will have to be ignored. x is deducted by one each time a colour preference cannot be granted.)

B.5 No player shall receive an identical float in two consecutive rounds.

As you can see B4 ranks as a higher priority to B5.

As worked out in my pairings for this score group, 6 players are supposed to get correct colours. Under my worked out pairings, only 4 do.

Now if you pair:

8 v 5
12 v 7
10 v 13
16 downfloats.

All 6 players get their correct colour allocation. B5 is of a lesser priority to B4.

Now my pairings for the bottom score groups were wrong, because I had 13 instead of 16 as the dfloater.

Pax has given an answer and I am sure Bill will reply in the next few hours or so.

Garvinator
04-07-2006, 07:16 PM
Gareth,

What can be seen from this discussion of round 6 is how difficult it can be to pair players correctly. We are doing this in relative peace and quiet. Not in a tournament hall with plenty of players around and quite a few players asking, when will the pairings be out?

I dont think any arbiter would be faulted if they went with either set of pairings ie sp or sm5 in the rush of multiple round day tournament. But the question to be solved is: which set of pairings is most correct?

Bill Gletsos
04-07-2006, 09:03 PM
One thing I will point out is that if player 11 withdraws after round 5 then SP and SM5 still get respective round 6 pairings identical to those with player 11 included. Note in this scenario where player 11 is no longer in the event then C13 does not come into play.

Having player 11 out makes the pairing procedure much simpler.

However the question is still in this case which of SP or SM5 is pairing correctly.

Garvinator
04-07-2006, 10:04 PM
I think the difference between the two pairings is in the way the bye is allocated.

There are essentially no problems until 16 is floated down into the '2' score group.

At that point, it is easy to notice that only 11 or 16 can receive the bye and allocate it to 11 and continue with the pairings, reaching:

16-14
18-15
17-19

As paired by SP. That is, in fact what I did when I first attempted a manual pairing.

But in fact, the pairing rules do not allow you to allocate the bye until the final score group. So instead you get:

16-14
11-15

And it is not until you get down to the final score group do you see that you cannot arrive at a legal pairing. You end up applying C13 twice (unpairing this previous pairing), and arriving at a merged score group of 16, 11, 14, 15, 17, 18 and 19.


It isn't entirely clear to me how this group should be divided. A6 suggests that S1 consists of *all* players moved from a higher group, but it also implies that S1 should always be smaller than or equal to S2 in size, which in this case is contradictory. If you treat this group as if it were homogeneous, you get the pairings:
16-17
15-14
18-19

As per SM5.

I would also point out that the SM5 pairing is strictly better than the SP pairing according to the pairing criterion B3 (sum of difference of scores).

Hello Pax,

I think I might have some kind of an answer.

A3 states:

A.3 Score brackets

Players with equal scores constitute a homogeneous score bracket. Players who remain unpaired after the pairing of a score bracket will be moved down to the next score bracket, which will therefore be heterogeneous. When pairing a heterogeneous score bracket these players moved down are always paired first whenever possible, giving rise to a remainder score bracket which is always treated as a homogeneous one.
A heterogeneous score bracket of which at least half of the players have come from a higher score bracket is also treated as though it was homogeneous. (my bolding and italics).

So I imagine the procedure to be:

1) combine all the players 16, 11, 14, 15, 17, 18, 19
2) pair these players just like they are in the exact same score group
3) As the players are all the same score group, 11 is the lowest ranked player to not have had a bye, so 11 gets the bye.
4) Pair via normal pairing rules

which you have answered above.
-----------------------------------------------------------------------------------------------------------

But then, you pair as you first did, getting to the bottom score group and realise you dont have a legal pairing, so you apply c13 and combine the bottom two score groups. A legal pairing then forms (doesnt it?) of 17 v 19.

Then combine 18 with the next highest score group to give:
11, 14, 15, 18.

Now 11 appears and receives the bye. So that leaves 14, 15, 18.

S1:
14
S2:
15
18

14 has played 18, so that leaves 14 v 15 and 15 v 18.

14 v 15 appears correct, but (and now I am saving typing :uhoh: ) if you pair 14/15, 16/18 cant be paired, so you are left with 15/18, 16/14.

Therefore, this way you get:

16/14
15/18
17/19

which is the sp pairings.

Again, which is correct?

Bill Gletsos
04-07-2006, 10:50 PM
I will look at the simple case where player 11 has withdrawn as once this is understood it is easy to see how to apply it to the actual situation where he is still in the event.

Player 16 drops into the 2 point score group comprising 14 and 15.
So p=1, w=2, b=1, x=0 and q=2

Applying C6 we get:

S1 S2
16 14
15
Now 16 V 14 appears valid, however it leaves 15 as a downfloat in violation of B5.
So let us follow the pairing rules exactly as written.
As such in accordance with C7 try a transposition in S2 and restart at C6.

This gives:

S1 S2
16 15
14
This now violates the x=0 condition as well as violating B6 for 14.
Therefore carry out an exchange as per C8 and restart at C5.
We now get:

S1 S2
14 16
15
This is identical to the first attempt above so is no good.
Again carry out C7 and restart at C6.

This results in:

S1 S2
14 15
16
This gives X =0 but downfloating 16 violates B5.

We have run out of transpositions and exchanges.
Therefore we are left with two choices:
1) Pair 16 V 14 and downfloat 15 violating B5.
2) Pair 15 V 14 and downfloat 16 violating B5.

Now we are supposed to meet as many of criteria B as possible.

They are idenitical in all aspects execpt choice 2) meets B3 better than choice 1).
The pairing therefore is 15 v 14.

Now 16 downfloats to meet 17 where p=1, w=2, b=0, q=1 and x=1.

The pairing is 16 V 17 meets these conditions.

That leaves the 1 point score group where p=1, w=1, b=1, q=1 and x=0.
this is met by 18 V19.

In accordance with F1 of the rules the pairings are ordered as:
16 V 17
15 V 14
18 v 19

Simple isnt it. :uhoh: :hmm: :wall: :doh: :rolleyes: :whistle: :hand:

Garvinator
05-07-2006, 01:25 PM
We have run out of transpositions and exchanges.
Therefore we are left with two choices:
1) Pair 16 V 14 and downfloat 15 violating B5.
2) Pair 15 V 14 and downfloat 16 violating B5.

Now we are supposed to meet as many of criteria B as possible.

They are idenitical in all aspects execpt choice 2) meets B3 better than choice 1).
The pairing therefore is 15 v 14.

Now 16 downfloats to meet 17 where p=1, w=2, b=0, q=1 and x=1.

The pairing is 16 V 17 meets these conditions.

I find it a little strange that the pairing rules would encourage floating down a player (16) two score groups, when 16 does have a legal pairing in this current score group. Surely the needs of the player on the higher score (16) takes precedence over the player on the lower score when they have equal claims. This principle applies in other areas of dutch pairing rules ie colour allocation.

Also, does this section come into play from the floater selection rules: http://www.fide.com/official/handbook.asp?level=C0402B


10.3 If there is a choice as to which player floats to a lower group, the player chosen is the lowest numbered player in the score-group who has a compatible opponent in the lower score-group, after excluding the opponents of other floaters who have higher scores or higher pairing numbers than the proposed floater.

Regarding the allocation of the bye, I dont think this situation has still be answered:

1) Has sm5/sp gone through the score groups including (16,11,14,15) and then gotten to the bottom score group, then broken them down according to c13, going back up through the draw?
2) Has sm5/sp done the pairings the way you have and treated 16/14/15 differently according to the program?
3) Has sm5 treated the whole bottom seven players as one homogenous score group?

I am also surprised that the dutch pairing rules dont seem to have a big section of how to handle the bye. There are only brief mentions of it throughout the rules.

Garvinator
05-07-2006, 01:26 PM
Gareth,

Happy you included a bye now;)

Bill Gletsos
05-07-2006, 01:55 PM
I find it a little strange that the pairing rules would encourage floating down a player (16) two score groups, when 16 does have a legal pairing in this current score group. Surely the needs of the player on the higher score (16) takes precedence over the player on the lower score when they have equal claims.They dont have equal claims. One of them is a better fit with regards B3 and the rules state that the B rules should be fulfilled as much as possible.

This principle applies in other areas of dutch pairing rules ie colour allocation.No. You determine the pairings via Section C. Only after you have pairied the score group do you determine the colour allocation via Section E.

Also, does this section come into play from the floater selection rules: http://www.fide.com/official/handbook.asp?level=C0402BThey have no bearing on the Dutch rules.

Regarding the allocation of the bye, I dont think this situation has still be answered:

1) Has sm5/sp gone through the score groups including (16,11,14,15) and then gotten to the bottom score group, then broken them down according to c13, going back up through the draw?Yes.
Although to a human it is obvious that player 11 will most likely get the bye, according to the Dutch rules who gets it isnt determined at the time you are pairing player 16 with the score group consisting of players 11, 14 and 15.
It isnt till pairing later score groups and then having to unpair previous ones that you finally get to the situation of player 11 floating down to the lowest score group.

2) Has sm5/sp done the pairings the way you have and treated 16/14/15 differently according to the program?Not sure what you mean.
Following the pairing rules with the bye included you eventually get down to having to unpair the bottom group. Eventually the score group consisting of 16, 11, 14 and 15 is unpaired and it is clear that 11 needs to float down. That leaves you with having to pair 16, 14 and 15.
They are paired as I described in my post above.
You then proceed to pair the remaining score groups.
This will again lead to not being able to pair the final group so again they are unpaired.
Again you will eventually by following the rules have to float 11 down from the 16, 11, and 17 group.
This leaves 16 V 17 to pair.
having paired them as I showed above you get left with the bottom group.
All that is now required is for 11, 18 and 19 to be paired.
This group is finally able to be paired as 18 v 19 with 11 getting the bye.

3) Has sm5 treated the whole bottom seven players as one homogenous score group?I do not believe so.

I am also surprised that the dutch pairing rules dont seem to have a big section of how to handle the bye. There are only brief mentions of it throughout the rules.I dont think it needs to.

pax
05-07-2006, 01:56 PM
I will look at the simple case where player 11 has withdrawn as once this is understood it is easy to see how to apply it to the actual situation where he is still in the event.


I'm sure this is the wrong approach. Allocating the bye should, in principle, be the very last thing you do (or rather part of the last score group you pair). There are cases where pairing programs produce a very strange looking bye, which turns out to be correct when you strictly apply top-down pairings.

[aside: I do however think the Dutch pairings would be a whole lot simpler if allocating the bye was instead the *first* thing you did]

The way I think of it is to insert an extra player - the 'bye player', with a play history, but no colour preference.

Brian_Jones
05-07-2006, 02:03 PM
I have always maintained that:

1. The bye should always be done first
2. The pairings should be done from both top and bottom towards the middle.

Brian_Jones
05-07-2006, 02:06 PM
But maybe my rules were designed for big swisses.
The field in this tournament is tiny ie much much too small.

Bill Gletsos
05-07-2006, 02:06 PM
I'm sure this is the wrong approach.I was not suggesting that you allocate the bye first.

From my post #81 I made it clear that if player 11 was not in the draw at all SP and SM5 would still have gotten the pairings they did with player 11 included. In this scenario where player 11 is not in the round 6 draw at all it is clear that C13 does not come into effect.

Therefore my point was that we should first understand how the remaining group of players 16, 14, 15, 17, 18, 19 should be paired.
Then once we understand that, we can go on to understand how the inclusion of player 11 may impact those pairings.

Allocating the bye should, in principle, be the very last thing you do (or rather part of the last score group you pair).It is.

There are cases where pairing programs produce a very strange looking bye, which turns out to be correct when you strictly apply top-down pairings.Agreed.

Bill Gletsos
05-07-2006, 02:06 PM
I have always maintained that:

1. The bye should always be done first
2. The pairings should be done from both top and bottom towards the middle.Too bad that isnt the Dutch Rules.

Bill Gletsos
05-07-2006, 02:08 PM
But maybe my rules were designed for big swisses.
The field in this tournament is tiny ie much much too small.Makes no difference to the application of the rules.

pax
05-07-2006, 02:31 PM
I have always maintained that:

1. The bye should always be done first
2. The pairings should be done from both top and bottom towards the middle.

Yes, you may well maintain that (and on this, I believe Peter Parr agrees) but unfortunately FIDE stipulates differently for Dutch rules.

The Lim rules however do as you suggest.

Kevin Bonham
05-07-2006, 03:25 PM
I think a slightly more detailed discussion re B3 is desirable here.

Bill notes that the SM pairing satisfies B3 better than the SP pairing. However in my view the reason for that is not simply that it allows the 14-15 pairing (score difference 0) in preference to a pairing of one of these players with 16 (score difference 0.5). Rather, the issue (for me) is that while pairing 14-15 for a score difference of 0 creates a score difference of 1 in the game between 16 and 17, if you go the other way you are forced to pair a player on 2 with a player on 1 (score difference total 1.5, which is worse) because 15 has played both 17 and 16.

If there were more players on 1.5 and there was an option available to pair 16 with 14 and 15 with another player on 1.5, thus creating two pairs with score differences of 0.5 rather than one pair with score difference of 1, I would do so as it seems closer to the spirit of the system. But in this case that option is not available, and the cost of cutting the 1-point score difference for the player on 2.5 to half a point is the introduction of an unnecessary 1-point score difference for another player in the same score group (and, as it turns out, another 0.5 point issue further down.)

I do think there are some ambiguities in the interpretation of B3, in terms of to what extent you take account of the consequences of a particular player downfloating when deciding who to downfloat. For instance, does the fact that a player on 2.5 will ultimately be forced to play someone on 1.5 justify instead deciding in the first place to downfloat a different player from the 2.5 scoregroup? (I would invoke the "too much work for the DOP" argument and say "no".)

In my view the SM draw is at least clearly better than the SP draw.

Kevin Bonham
05-07-2006, 03:29 PM
I have always maintained that:

1. The bye should always be done first
2. The pairings should be done from both top and bottom towards the middle.

In terms of what would be ideal (rather than what is actually the rule) I agree with 1 but not necessarily 2. Because of the potential impact on ratings prizes I prefer mismatches to be shuffled towards the tail end of the field rather than the middle and pairing top-bottom usually acheives this better.

Garvinator
05-07-2006, 05:11 PM
No Name Rtg Total Result Name Rtg Total

1 SAMPRAS, Pete (2) 2800 [4.5] : LAVER, Rod (1) 2810 [4]
2 COURIER, Jim (9) 2735 [3.5] : MCENROE, John (3) 2780 [3.5]
3 FEDERER, Roger (6) 2760 [3.5] : AGASSI, Andre (4) 2775 [3.5]
4 EDBERG, Stefan (8) 2740 [3] : BORG, Bjorn (5) 2770 [2.5]
5 RAFTER, Pat (12) 2720 [2.5] : BECKER, Boris (7) 2750 [2.5]
6 HEWITT, Lleyton (10) 2730 [2.5] : CHANG, Michael (13) 2715 [2.5]
7 CASH, Pat (16) 2680 [2.5] : EL AYNAOUI, Younes (17) 2660 [1.5]
8 RODDICK, Andy (15) 2710 [2] : NALBANDIAN, David (14) 2712 [2]
9 PHILIPPOUSIS, Mark (18) 2650 [1] : BAGHDATIS, Marcos (19) 2640 [1]
10 NADAL, Rafael (11) 2725 [2] 1:0 BYE


Gareth, The sm5 pairings above seem to agree with the dutch pairing rules, so these should be the round 6 pairings.

Garvinator
07-07-2006, 02:23 PM
now we await Gareth's next installment.

Lucena
07-07-2006, 06:15 PM
now we await Gareth's next installment.

Ok, we seem to be agreed that the Swiss Master pairings are the right ones. I'll post the results shortly.

Lucena
08-07-2006, 06:42 PM
No Name Rtg Total Result Name Rtg Total

1 Sampras, Pete 2800 [4.5] 0:1 Laver, Rod 2810 [4]
2 Courier, Jim 2735 [3.5] 0:1 McEnroe, John 2780 [3.5]
3 Federer, Roger 2760 [3.5] .5:.5 Agassi, Andre 2775 [3.5]
4 Edberg, Stefan 2740 [3] .5:.5 Borg, Bjorn 2770 [2.5]
5 Rafter, Pat 2720 [2.5] .5:.5 Becker, Boris 2750 [2.5]
6 Hewitt, Lleyton 2730 [2.5] 0:1 Chang, Michael 2715 [2.5]
7 Cash, Pat 2680 [2.5] .5:.5 El Aynaoui, Younes 2660 [1.5]
8 Roddick, Andy 2710 [2] 0:1 Nalbandian, David 2712 [2]
9 Philippousis, Mark 2650 [1] 0:1 Baghdatis, Marcos 2640 [1]
10 Nadal, Rafael 2725 [2] 1:0 BYE

Lucena
09-07-2006, 11:01 PM
Here are the pairings according to SP:


No Name Rtg Total Result Name Rtg Total

1 Laver, Rod 2810 [5] : McEnroe, John 2780 [4.5]
2 Federer, Roger 2760 [4] : Sampras, Pete 2800 [4.5]
3 Agassi, Andre 2775 [4] : Edberg, Stefan 2740 [3.5]
4 Chang, Michael 2715 [3.5] : Courier, Jim 2735 [3.5]
5 Borg, Bjorn 2770 [3] : Cash, Pat 2680 [3]
6 Becker, Boris 2750 [3] : Nadal, Rafael 2725 [3]
7 Nalbandian, David 2712 [3] : Rafter, Pat 2720 [3]
8 Baghdatis, Marcos 2640 [2] : Roddick, Andy 2710 [2]
9 El Aynaoui, Younes 2660 [2] : Philippousis, Mark 2650 [1]
10 Hewitt, Lleyton 2730 [2.5] 1:0 BYE

Garvinator
09-07-2006, 11:22 PM
Gareth,

Would you like to have a go at some explanations for the pairings and how you came about them?

This round doesnt appear as anywhere near complicated as the previous one.

Lucena
10-07-2006, 03:28 AM
Gareth,

Would you like to have a go at some explanations for the pairings and how you came about them?

This round doesnt appear as anywhere near complicated as the previous one.

Yeah I'll have a go, to be honest I haven't verified the pairings in relation to the rules, but I'll give it a shot. It might be a while before I get around to it as I have some things to sort out.

Garvinator
10-07-2006, 11:34 AM
Yeah I'll have a go, to be honest I haven't verified the pairings in relation to the rules, but I'll give it a shot. It might be a while before I get around to it as I have some things to sort out.
the sp pairings may or may not be correct, that is what you have to work out ;)

Garvinator
16-07-2006, 12:39 AM
Gareth,

How is it going with the calcs?

drbean
16-10-2006, 02:58 PM
No Name Result Name

1 Federer, R (6) : Laver, R (1)
2 Sampras, P (2) : Becker, B (7)
3 Edberg, S (8) : McEnroe, J (3)
4 Agassi, A (4) : Courier, J (9)
5 Baghdatis, M (19) : Borg, B (5)
6 Hewitt, L (10) : Roddick, A (15)
7 Nalbandian, D (14) : Nadal, R (11)
8 Rafter, P (12) : El Aynaoui, Y (17)
9 Cash, P (16) : Chang, M (13)
10 Philippousis, M (18) 1:0 BYE

These are the pairings for round 2 after the 1-9 players beat the 10-18 ones in round 1 with 19 having a bye.

The second bracket with 0 points is made up of 10-18. This gives S1 10-13 and S2 14-18. Rules C1-6 produce the pairing 10&14, 11&15, 12&16 13&17. B1 & B2 are not violated. So the pairing is considered complete, even though 2 players do not receive their preference.

The real pairing (of SwissPerfect?) swaps 14 and 15 and 16 & 17.

My question is my pairing of 10&14, 11&15, 12&16 13&17 wrong? The pairing rules are not clear. They seem to imply it is not wrong.

Garvinator
20-10-2006, 02:00 PM
Presenting this pairing info:


6 : 19,-,14 W-B Ud 2.5

1 : -,9,13 -BW 2
4 : 16,12,7 WBW 2
5 : 17,11,2 BWW 2
8 : 21,14,19 WBW 2
11 : 24,5,21 WBW 2
14 : 27,8,6 BWW D 2
15 : 2,24,9 WB- 2
16 : 4,23,10 BWB u 2
17 : 5,26,12 WBW 2

The main question I have is, who is supposed to go up to play 6- sm5 gives 4, sp gives 1.

I can see cases for both, so which is it? I think it should be 4.

Garvinator
20-10-2006, 07:16 PM
any replies?

Oepty
20-10-2006, 10:03 PM
Garvin. I don't see any reason to make the pairing 6-4, instead of 6-1. Can you please explain why you would do so?
Scott

Garvinator
20-10-2006, 10:35 PM
Garvin. I don't see any reason to make the pairing 6-4, instead of 6-1. Can you please explain why you would do so?
Scott
At first glance 6-1 looks right and it isnt a bad? pairing. But I believe the reason is stronger colour preference.
Player 1 is 0 colour balance (BW) and so has a mild colour preference for black.
Player 4 is +1 colour balance (WBW) and so has a strong colour preference.

Therefore, 4 has the stronger colour preference.

Both pairings programs gave the non upfloater as white against 14.
If 4 is upfloated, 1 is then white against 14 and would be a +1 colour balance.
If 1 is upfloated, 4 is then white against 14 and would be a +2 colour balance.

As I said, swiss master 5 spat out 6-4 which alerted my attention to it.

Now I tried doing the pairings for this by doing the mathematics that Bill G has shown in the past and got confused :confused:

Kevin Bonham
21-10-2006, 09:25 PM
Now I tried doing the pairings for this by doing the mathematics that Bill G has shown in the past and got confused :confused:

I am also pretty hazy on correctly applying the exact mechanics of section C sometimes but I cannot yet see a firm reason in section C why the SP draw is wrong.

The interesting thing to me is that section B only says "As many players as possible receive their colour preference". It doesn't say anything about applying exchanges to grant as many players as possible a stronger colour preference. That concept of a stronger colour preference trumping a weaker one applies with specific reference to a pairing that is already formed (section E).

You'll still have two players not granted their colour preference no matter what you do. The difference is that for the pairing 6-1, then two strong colour preferences are not granted, while for 6-4, one strong and one weak preference are not granted. However, this is at the cost of not pairing the strongest player in the 2-group with the leader, and furthermore at the cost of the strongest player in the 2-group not receiving their weak colour preference and hence having an absolute colour preference next time. In "spirit of the system" terms it can be argued either way.

If I was doing it by hand without time to go through section C carefully I would definitely pair 6-1. Going through section C carefully might get a different answer, but I'm not as yet convinced. At a quick attempt I get the pairing 6-1 with all the rest moved down, then restart at C6, then pair the remaining eight players with transpositions per C7 until the pairing is complete - I can't see where in section C the pairing 6-1 gets undone.

Garvinator
21-10-2006, 10:49 PM
While I am attempting to frame a response which I dont have as yet, maybe Bill could be kind enough to do the maths to work this out.


You'll still have two players not granted their colour preference no matter what you do. The difference is that for the pairing 6-1, then two strong colour preferences are not granted, while for 6-4, one strong and one weak preference are not granted. However, this is at the cost of not pairing the strongest player in the 2-group with the leader (my bolding), and furthermore at the cost of the strongest player in the 2-group not receiving their weak colour preference and hence having an absolute colour preference next time. In "spirit of the system" terms it can be argued either way.
There is actually two players on 3 points, so 1 or 4 arent going to be playing the leader in this round (4). I didnt include all the other pairings, as it was this score group that was causing the concern.

I am at least happy that I havent missed something obvious and my original questions were fair enough. Still wondering though why the completely approved of- fide program- sm5 is the one saying 6-4?

Oepty
22-10-2006, 03:41 PM
At first glance 6-1 looks right and it isnt a bad? pairing. But I believe the reason is stronger colour preference.
Player 1 is 0 colour balance (BW) and so has a mild colour preference for black.
Player 4 is +1 colour balance (WBW) and so has a strong colour preference.

Therefore, 4 has the stronger colour preference.

Both pairings programs gave the non upfloater as white against 14.
If 4 is upfloated, 1 is then white against 14 and would be a +1 colour balance.
If 1 is upfloated, 4 is then white against 14 and would be a +2 colour balance.

As I said, swiss master 5 spat out 6-4 which alerted my attention to it.

Now I tried doing the pairings for this by doing the mathematics that Bill G has shown in the past and got confused :confused:

Garvin. I don't have the rules in front of me at the moment, but I think the idea making the pairing 6-4 is completely wrong. There is no reason that you shouldn't pair 6-1 even if they didn't have different colour preferences and I don't remember anything in the rules about upfloating a different player because they have a stronger color preference. The only reasons not to make the pairing 6-1 would be if they had played before, 1 had upfloated in either of the last 2 rounds or both 6 and 1 had the same absolute color preferences.
I too would like Bill to have a look at this example as he explains these things extremely well.
Scott

Bill Gletsos
22-10-2006, 10:23 PM
I have been and still am very busy, however at a quick glance I can see no reason why the 6-1 pairing is wrong and why SM5 pairs 6-4.

However Garvin it might be beneficial if you could email me the SM5 tournament file.

Kevin Bonham
22-10-2006, 10:39 PM
Garvin, if you feel like it then feel free to send me the SP file as well (I don't have SM5).

I wonder if it might be a float issue, but SP generally handles floats correctly in my experience.

Garvinator
23-10-2006, 10:04 AM
Geez, everyone wants me to send files :P I will send soon to both.

Kevin Bonham
23-10-2006, 01:35 PM
File received - thanks.

It's not a float issue. I still cannot see why SM has paired 6-4.

Garvinator
23-10-2006, 01:51 PM
File received - thanks.

It's not a float issue. I still cannot see why SM has paired 6-4.
Now awaiting Bill's reply. As I said previously, I wasnt even considering upfloating 4 until sm5 paired 6-4.

Bill Gletsos
23-10-2006, 02:15 PM
Now awaiting Bill's reply.I'll try and look at it tonight if I have a chance.

Brian_Jones
23-10-2006, 02:26 PM
Garvin, How are you ranking the players on 2 points?
You seem to be using original starting sequence.
Should you not be ranking according to the tie-break sequence?
Then the player with only two games drops down.

Bill Gletsos
23-10-2006, 02:52 PM
Garvin, How are you ranking the players on 2 points?
You seem to be using original starting sequence.They should be ranked according to A2 of the Dutch Rules. i.e. score, rating, FIDE Title, alphabetically.

Should you not be ranking according to the tie-break sequence?No.
Tie-breaks have no bearing on ranking order with regards the Dutch pairing rules.

Garvinator
23-10-2006, 02:53 PM
Garvin, How are you ranking the players on 2 points?
You seem to be using original starting sequence.
Should you not be ranking according to the tie-break sequence?
Then the player with only two games drops down.
Under the dutch pairing rules, players in each score group are ranked by rating:


A.2 Order

For pairing purposes only, the players are ranked in order of, respectively
a) score
b) rating
c) FIDE-title (IGM-WGM-IM-WIM-FM-WFM-no title)
d) alphabetically (unless it has been previously stated that this criterion has been replaced by another one)


A.6 Subgroups

To make the pairing, each score bracket will be divided into two subgroups, to be called S1 and S2.
In a heterogeneous score bracket S1 contains all players moved down from a higher score bracket.
In a homogeneous score bracket S1 contains the higher half (rounding downwards) of the number of players in the score bracket.
The number of players in S1 will be indicated by "p", indicating the number of pairings to be made.
In both cases S2 contains all other players of the score bracket.
In both S1 and S2 players are ordered according to A2. (my bolding)

So in this case:
S1 is 4 and the rest are S2.

Players would be paired in buchholz order instead of rating order if this was announced at the start of the tournament.

I see Bill and I have had a cross over of replies ;)

Bill Gletsos
23-10-2006, 02:58 PM
Players would be paired in buchholz order instead of rating order if this was announced at the start of the tournament.Except then you wouldnt be following the Dutch rules as they dont allow for tie-break replacing rating.

All that is allowed under A2 is that a tie-break method could be used instead of alphabetical.

Garvinator
23-10-2006, 03:12 PM
Time to throw another spanner in the works:

Same round, except that in the 1.5 score group, sp and sm5 disagree again and I think for the exact same reason as we are trying to work out for the 2.5/2 score group.

No Opponents Colours Float Score
2 : 15,10,5 BWB 3
7 : 20,13,4 BWB 3

6 : 19,-,14 W-B Ud 2.5

1 : -,9,13 -BW 2
4 : 16,12,7 WBW 2
5 : 17,11,2 BWW 2
8 : 21,14,19 WBW 2
11 : 24,5,21 WBW 2
14 : 27,8,6 BWW D 2
15 : 2,24,9 WB- 2
16 : 4,23,10 BWB u 2
17 : 5,26,12 WBW 2

3 : -,18,- -W- D 1.5
18 : -,3,22 -BW D 1.5
27 : 14,-,28 W-B d 1.5

9 : -,1,15 -W- 1
10 : 22,2,16 BBW 1
12 : 25,4,17 BWB 1
13 : 26,7,1 WBB 1
19 : 6,22,8 BWB 1
21 : 8,25,11 BWB 1
25 : 12,21,24 WBW 1
26 : 13,17,- BW- D 1

20 : 7,-,- W-- D 0.5
23 : -,16,- -B- D 0.5
28 : -,-,27 --W d 0.5

22 : 10,19,18 WBB U 0
24 : 11,15,25 BWB 0

Sp pairs:


27 v 3
10 v 18

sm5 pairs:


27 v 18
10 v 3

Oepty
23-10-2006, 05:56 PM
I am not sure whether it is for the same reason in the second example, but again I think that sm5 has just got it wrong.

For the group on 1.5, x = 0 and p = 1.
You get s1 and S2 as below
S1 S2
3 18
,, 27

The first pairing to try is 3-18, which is clearly illegal seeing they have already played.
This means you need to see whether pairing 3 - 27 works. It is legal as they have not played and pairing them does not force either player to have an illegal colour split. 3 -27 have opposite color preferences so we can float 18 down to the next score group.
Scott

Oepty
23-10-2006, 06:02 PM
Garvin. A question from me on the pairings. It appears that in round 3 14 was white against 6, with 14 on 2 points downfloating to play 6 on 1.5. This doesn't seem right to me, but that might because I don't have all the info I need. Can you please post the pairing info after round 2?
Scott

Garvinator
23-10-2006, 06:08 PM
Garvin. A question from me on the pairings. I am ruling all questions illegal except the ones I ask ;)


It appears that in round 3 14 was white against 6, with 14 on 2 points downfloating to play 6 on 1.5. This doesn't seem right to me, but that might because I don't have all the info I need. Can you please post the pairing info after round 2?


No Opponents Colours Float Score
2 : 15,10 BW 2
4 : 16,12 WB 2
5 : 17,11 BW 2
7 : 20,13 BW 2
14 : 27,8 BW 2

6 : 19,- W- D 1.5

Oepty
23-10-2006, 06:26 PM
Garvin, thanks. I am fairly sure that the pairing should have been reversed. 14 is the higher ranked player seeing they have more points than 6. 14 color preference should have been granted, so they should have been black. How was the pairing made? SP, SM5 or manually?
Scott

Bill Gletsos
23-10-2006, 06:26 PM
At a very quick glance SM5 looks correct to me and SP wrong.
The reason I say this is that player 3 was not actually a downfloat in the last round as he appears not to have been paired in round 3. As such SP should not have flagged him as a D.

Also note that 27 was not a downfloat in round 2 for the same reason.

Garvinator
23-10-2006, 06:35 PM
Garvin, thanks. I am fairly sure that the pairing should have been reversed. 14 is the higher ranked player seeing they have more points than 6. 14 color preference should have been granted, so they should have been black. How was the pairing made? SP, SM5 or manually?
Scott
Both sp and sm5 agreed for these pairings and so was played by the computer pairings.


E. COLOUR ALLOCATION RULES

For each pairing apply (with descending priority):

E.1 Grant both colour preferences.

E.2 Grant the stronger colour preference.


14 : 27,8 BW 2
6 : 19,- W- D 1.5

Both players are 'due' black, but 6 has the stronger colour preference, being +1, whereas 14 is 0 colour preference and so is only mild.

Oepty
23-10-2006, 06:40 PM
Both sp and sm5 agreed for these pairings and so was played by the computer pairings.




14 : 27,8 BW 2
6 : 19,- W- D 1.5

Both players are 'due' black, but 6 has the stronger colour preference, being +1, whereas 14 is 0 colour preference and so is only mild.

Garvin. Yes that is correct, I was wrong.

Garvinator
23-10-2006, 06:41 PM
At a very quick glance SM5 looks correct to me and SP wrong.
The reason I say this is that player 3 was not actually a downfloat in the last round as he appears not to have been paired in round 3. As such SP should not have flagged him as a D.

Also note that 27 was not a downfloat in round 2 for the same reason.
I remember you commenting on this bye situation previously. Do you know where or can you at least repeat it?

The relevant fide pairing rule seems to be.


A.5 Byes

Should the total number of players be (or become) odd, one player ends up unpaired.
This player receives a bye: no opponent, no colour, 1 point. A bye is considered to be a downfloat.

I think the only question is whether this:
A bye is considered to be a downfloat. is talking only of the 'left-over' player or of any players allowed byes during a tournament.

Oepty
23-10-2006, 06:42 PM
At a very quick glance SM5 looks correct to me and SP wrong.
The reason I say this is that player 3 was not actually a downfloat in the last round as he appears not to have been paired in round 3. As such SP should not have flagged him as a D.

Also note that 27 was not a downfloat in round 2 for the same reason.

Bill. I overlooked that, and I think this means SM5 is correct. Thanks, you have a far better grasp of these things than I do.
Scott

Oepty
23-10-2006, 06:47 PM
I remember you commenting on this bye situation previously. Do you know where or can you at least repeat it?

The relevant fide pairing rule seems to be.



I think the only question is whether this: is talking only of the 'left-over' player or of any players allowed byes during a tournament.

I think the context means means it is talking about only a forced bye as a result of the pairings, not a bye taken by choice.
Scott

Bill Gletsos
23-10-2006, 06:50 PM
1 point and 1/2 point byes should be treated as downfloats.
Zero point byes should not.

SP treats all byes as downfloats.
SM5 only treats 1 point and 1/2 point byes as downfloats.

Garvinator
23-10-2006, 06:54 PM
1 point and 1/2 point byes should be treated as downfloats.
Zero point byes should not.

SP treats all byes as downfloats.
SM5 only treats 1 point and 1/2 point byes as downfloats.


At a very quick glance SM5 looks correct to me and SP wrong.
The reason I say this is that player 3 was not actually a downfloat in the last round as he appears not to have been paired in round 3. As such SP should not have flagged him as a D.

Also note that 27 was not a downfloat in round 2 for the same reason.
In round 2, 27 was awarded a half point bye and so should be little d, which the pairing info provided lists as.

Bill Gletsos
23-10-2006, 06:58 PM
In round 2, 27 was awarded a half point bye and so should be little d, which the pairing info provided lists as.Ok, thanks for pointing that out.
I hadnt looked at the files just what was posted here, and I had incorrectly assumed it was a zero point bye.

That however makes no difference to the pairings. SM5 is correct and SP is wrong.

Garvinator
23-10-2006, 07:00 PM
Ok, thanks for pointing that out.
I hadnt looked at the files just what was posted here, and I had incorrectly assumed it was a zero point bye.

That however makes no difference to the pairings. S5 is correct and SP is wrong.
This is getting confusing ;) :eek: :uhoh:

Bill Gletsos
23-10-2006, 07:03 PM
This is getting confusing ;) :eek: :uhoh:Not at all.

3 has no downfloat.
neither does 27.
3 cannot play 18
3 V 27 means 18 downfloats but he downfloated last round. He therefore violates B5.
So you try 27 V 18 (no violations) and downfloat 3.
3 can downfloat because he hasnt been downfloated in the last two rounds. ;)
You then get 10 V 3.

Garvinator
23-10-2006, 07:08 PM
Not at all.
I was meaning confusing because still outstanding is:


6 : 19,-,14 W-B Ud 2.5

1 : -,9,13 -BW 2
4 : 16,12,7 WBW 2
5 : 17,11,2 BWW 2
8 : 21,14,19 WBW 2
11 : 24,5,21 WBW 2
14 : 27,8,6 BWW D 2
15 : 2,24,9 WB- 2
16 : 4,23,10 BWB u 2
17 : 5,26,12 WBW 2

which was the start of all this:

Sp paired:

6 v 1

and

sm5 paired:

6 v 4.

Desmond
23-10-2006, 09:07 PM
As a tournament player, I would just like to interject the following, hopefully when the arbiter is right in the middle of concerntrating, and probably about to solve the problem:

Is the next round out yet? Can you at least tell me who I'm playing? Did you know I've already had two blacks today? What time is the prize-giving going to be held?

Garvinator
23-10-2006, 10:02 PM
I usually respond ;) :P


Is the next round out yet?Yes, now go around the room and look for it.


Can you at least tell me who I'm playing? Always the number one seed


Did you know I've already had two blacks today?yes and you now get a third.



What time is the prize-giving going to be held?alot later if you dont shut up :P :lol:

Bill Gletsos
23-10-2006, 10:15 PM
Also it seems SP gets a different draw in round 3 from SM5 on boards 10-13.

SM5 wanted:
18 v 27
28 V 24
22 V 26
and 25 got the bye

SP gave:
28 v 27
18 v 22
25 v 24
and 26 got the bye


As such I assume you overrode SM5 to match SP in round 3.

Garvinator
23-10-2006, 10:26 PM
Also it seems SP gets a different draw in round 3 from SM5 on boards 10-13.

SM5 wanted:
18 v 27
28 V 24
22 V 26
and 25 got the bye

SP gave:
28 v 27
18 v 22
25 v 24
and 26 got the bye

As such I assume you overrode SM5 to match SP in round 3.
Pairings were done just before the round began without a sm5 check. 26 is a bye player who plays normally when there are odd numbers and sits out when there are even numbers. Sm5 still gives different pairings even with 26 declared as the bye player.

I am not the only person who is responsible for doing the pairings when I am playing in the same tournament ;).

Bill Gletsos
23-10-2006, 10:32 PM
Ok I've had a good look at it and I believe the correct pairings for round 4 are:

6 v 1
4 V 14
27 V 18
10 v 3

Bill Gletsos
23-10-2006, 10:34 PM
FWIW Swiss-Manager gets in Round 4:
6 v 1
4 V 14
27 V 18
10 v 3

Bill Gletsos
23-10-2006, 11:00 PM
BTW in round 3 Swiss-Manaager wanted on boards 10-13:
18 v 27
22 V 28
25 V 24
26 the bye

Kevin Bonham
24-10-2006, 02:17 AM
I think the context means means it is talking about only a forced bye as a result of the pairings, not a bye taken by choice.
Scott

Correct. This is the cause of the second problem. There is no provision for a player to get points for being absent in the system, because in the strict version of the system:

Players known in advance not to play in a particular round are not paired in that round and score zero (my emphasis)

(There was a debate about the misnomer "zero point bye" with AO sometime back.)

SM5 has correctly not recorded the absence as a downfloat although half-points were awarded for the absence. SP has incorrectly treated it as a standard bye, which it isn't.

Note also that the official rules state clearly that a bye (as distinct from an absence) is worth one point. That is why it is considered to be a downfloat - the player, a tailender, has received a free win and so does not deserve to be paired with players on scores below them for a while.

A bye awarded by virtue of lack of opponent is a downfloat whether you decide to award 1 or 1/2.

A "bye" on request due to absence should not be treated as a downfloat whether you decide to award 1/2 or 0.

drbean
19-01-2007, 03:10 PM
This is the pairings for Round 3:




No Name Result Name

1 Borg, B (5) : Sampras, P (2)
2 Laver, R (1) : Agassi, A (4)
3 McEnroe, J (3) : Federer, R (6)
4 Becker, B (7) : Edberg, S (8)
5 Courier, J (9) : Rafter, P (12)
6 Philippousis, M (18) : Hewitt, L (10)
7 Nadal, R (11) : Baghdatis, M (19)
8 Chang, M (13) : Nalbandian, D (14)
9 Roddick, A (15) : Cash, P (16)
10 El Aynaoui, Y (17) 1:0 BYE

I don't understand why 18 is paired above 13. Why is 13 downfloated?

The pairingtable at the end of Round 2 is:



Round 3 Pairing Groups
-----------------------------------------------
Place No Opponents Roles Score
1-3
2 11,7 WB 2
4 13,9 WB 2
5 14,19 BW 2
4-7
1 10,6 BW 1.5
3 12,8 BW 1.5
6 15,1 WB 1.5
8 17,3 WB 1.5
8-15
7 16,2 BW 1
9 18,4 BW 1
10 1,15 WB 1
11 2,14 BW 1
12 3,17 WB 1
13 4,16 BW 1
18 9,- W- 1
19 -,5 -B 1
16-19
14 5,11 WB 0
15 6,10 BW 0
16 7,13 WB 0
17 8,12 BW 0


7 is paired with 8, which was floated down from bracket 2.
9,10,11,12,13,18,19 are the Remainder Group.

S1 is 9,10,11. S2 is 12,13,18,19. But 11 and 18 both have a preference for Black. So C7 shuffles 18 and 19. And then they all have preferences satisfied.



next: Bracket [3] 8 7 9 10 11 12 13 18 19
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1
C4, S1 & S2: 8, 7 9 10 11 12 13 18 19
C5, Ordered: 8, 7 9 10 11 12 13 18 19
colors: 8:7
C6others: hetero
Remainder Group, Bracket 3: 9 10 11 12 13 18 19
C2, x=0
C3, p=3
C4, S1 & S2: 9 10 11, 12 13 18 19
C5, Ordered: 9 10 11, 12 13 18 19
C7, 12 13 19 18
colors: 12:9 10:13 19:11
C6others: homo: downfloating 18
Floating Down: 18 [3] 9 10 11 12 13 19 18 => [4] 18 14 15 16 17


Why does the published pairing downfloat 13?

Bill Gletsos
20-01-2007, 12:01 AM
This is the pairings for Round 3:



I don't understand why 18 is paired above 13. Why is 13 downfloated?

The pairingtable at the end of Round 2 is:



Round 3 Pairing Groups
-----------------------------------------------
Place No Opponents Roles Score
1-3
2 11,7 WB 2
4 13,9 WB 2
5 14,19 BW 2
4-7
1 10,6 BW 1.5
3 12,8 BW 1.5
6 15,1 WB 1.5
8 17,3 WB 1.5
8-15
7 16,2 BW 1
9 18,4 BW 1
10 1,15 WB 1
11 2,14 BW 1
12 3,17 WB 1
13 4,16 BW 1
18 9,- W- 1
19 -,5 -B 1
16-19
14 5,11 WB 0
15 6,10 BW 0
16 7,13 WB 0
17 8,12 BW 0


7 is paired with 8, which was floated down from bracket 2.
9,10,11,12,13,18,19 are the Remainder Group.

S1 is 9,10,11. S2 is 12,13,18,19. But 11 and 18 both have a preference for Black. So C7 shuffles 18 and 19. And then they all have preferences satisfied.



next: Bracket [3] 8 7 9 10 11 12 13 18 19
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1
C4, S1 & S2: 8, 7 9 10 11 12 13 18 19
C5, Ordered: 8, 7 9 10 11 12 13 18 19
colors: 8:7
C6others: hetero
Remainder Group, Bracket 3: 9 10 11 12 13 18 19
C2, x=0
C3, p=3
C4, S1 & S2: 9 10 11, 12 13 18 19
C5, Ordered: 9 10 11, 12 13 18 19
C7, 12 13 19 18
colors: 12:9 10:13 19:11
C6others: homo: downfloating 18
Floating Down: 18 [3] 9 10 11 12 13 19 18 => [4] 18 14 15 16 17


Why does the published pairing downfloat 13?Player 18 received the Bye in Round 2. According to A5 a bye is considered a downfloat.

As such your statement:

S1 is 9,10,11. S2 is 12,13,18,19. But 11 and 18 both have a preference for Black. So C7 shuffles 18 and 19. And then they all have preferences satisfied. is incorrect as your parings violate Dutch pairing rule B5 that no player shall receive an identical float in two consecutive rounds.

Hence the pairings of 9 V 12, 13 V 10 and 11 V 19 fail so reapply C7.

This gives 9 V 12, 18 V 10 and 11 V 13 which fails because of color with 11 V 13.
Therefore reapply C7 which results in 9 V 12, 18 v 10 and 11 V 19 with 13 the downfloat.

drbean
21-01-2007, 08:06 PM
I posted a question about the pairing of the 3rd bracket in Round 3:




Round 3 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score

...

8-15
7 16,2 BW 1
9 18,4 BW 1
10 1,15 WB 1
11 2,14 BW 1
12 3,17 WB 1
13 4,16 BW 1
18 9,- W- D 1
19 -,5 -B d 1



7 was upfloated. I thought the bracket should be paired:
12:9 10:13 19:11, downfloating 18.

But Bill Gletsos said:


Player 18 received the Bye in Round 2. According to A5 a bye is considered a downfloat.

As such your statement:
is incorrect as your parings violate Dutch pairing rule B5 that no player shall receive an identical float in two consecutive rounds.

Hence the pairings of 9 V 12, 13 V 10 and 11 V 19 fail so reapply C7.

This gives 9 V 12, 18 V 10 and 11 V 13 which fails because of color with 11 V 13.
Therefore reapply C7 which results in 9 V 12, 18 v 10 and 11 V 19 with 13 the downfloat.

Okay. Thanks for that. I wonder if this means you have to interpret 'is considered' extremely closely in C6.



If now p pairings are obtained in compliance with
B1 and B2 the pairing of this score bracket is considered
complete.


It's only complete if certain other conditions, eg the B5, and B6 mentioned in C10, are being waived.

This is the import of the preamble in B, also.



Relative Criteria
(These are in descending priority. They should be fulfilled
as much as possible. To comply with these criteria,
transpositions or even exchanges may be applied, but no
player should be moved down to a lower score bracket).


Actually that last part of the preamble leads into a question I have about the next round, Round 4. The pairing table for the first 2 brackets is:



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 WBW 3
2-3
3 12,8,6 BWB 2.5
4 13,9,1 WBW D 2.5


4 was paired with 1, who had 1.5, in Round 3 and drew with him.

The question is about B4 (as many players AS POSSIBLE receive their color preference) and B5 (no identical float in 2 consecutive rounds). B4 has higher priority than B5.

I would argue 2 should be paired with 4 and 3 downfloated. One of 2 or 4 doesn't get their color preference, but that is easily rectifiable as WBWWBB in following rounds. It looks like this approach would be giving B5 precedence to B4, but I think the AS POSSIBLE in B4 means this is arguable.

The downfloating of 4 twice can not be rectified, as there is no mechanism like color preferences to even things up, eg being upfloated.

The actual pairing is:



No Name Result Name

1 Sampras, P (2) : McEnroe, J (3)
2 Agassi, A (4) : Borg, B (5)


The last part of the preamble to the Relative Criteria says no player should be moved down to a lower score bracket to satisfy them. But someone is going to be downfloated anyway.

I don't know what my question is. I guess it is, is the pairing right?

Bill Gletsos
23-01-2007, 12:11 AM
Okay. Thanks for that. I wonder if this means you have to interpret 'is considered' extremely closely in C6.C6 is poorly worded.

As many of sections B1-B6 should be met as possible.


Actually that last part of the preamble leads into a question I have about the next round, Round 4. The pairing table for the first 2 brackets is:



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 WBW 3
2-3
3 12,8,6 BWB 2.5
4 13,9,1 WBW D 2.5


4 was paired with 1, who had 1.5, in Round 3 and drew with him.

The question is about B4 (as many players AS POSSIBLE receive their color preference) and B5 (no identical float in 2 consecutive rounds). B4 has higher priority than B5.

I would argue 2 should be paired with 4 and 3 downfloated.You can argue what you like but you would be wrong. ;)

One of 2 or 4 doesn't get their color preference, but that is easily rectifiable as WBWWBB in following rounds.What colours the player may get in future rounds is totally irrelevant.

It looks like this approach would be giving B5 precedence to B4, but I think the AS POSSIBLE in B4 means this is arguable.

The downfloating of 4 twice can not be rectified, as there is no mechanism like color preferences to even things up, eg being upfloated.

The actual pairing is:



No Name Result Name

1 Sampras, P (2) : McEnroe, J (3)
2 Agassi, A (4) : Borg, B (5)


The last part of the preamble to the Relative Criteria says no player should be moved down to a lower score bracket to satisfy them. But someone is going to be downfloated anyway.

I don't know what my question is. I guess it is, is the pairing right?The pairing is correct.

Pair the score group consusting of players 2, 3 and 4.
From C2 determine x via A8 and from C3 determine p via A6.

From A8 w = 2, b = 1 and q =2, therefore x = 0 and from A6 p = 1.


S1 only comprises 2, S2 consists of 3 and 4.

So 2 V 3 and downflloat 4.
2 V 3 satisfies the condition x = 0, however downfloating 4 violates B5.

However 2 V 4 and downfloating 3 although eliminating the problem of 4 with B6 creates a far greater totally unacceptable situation as it violates the x=0 condition.

The only valid pairing is 2 V 3 with 4 downfloating.

If in the above above scenario of players 2, 3 and 4, player 4 had a black colour preference then the situation would have been different.
In that case w = 1, b=2, q =1, x =0 and p = 1.

As before 2 v 3 and downfloat 4 meetes x = 0 but violates B5.
However now 2 V 4 and downfloat 3 means that the downfloating of 3 still does not violates B5 but now 2 V 4 now meets x = 0.

As such the correct paoring is 2 V 4 and 3 is downfloated.

drbean
26-01-2007, 01:11 AM
Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 WBW 3
2-3
3 12,8,6 BWB 2.5
4 13,9,1 WBW D 2.5

Paired as:

No Name Result Name

1 Sampras, P (2) : McEnroe, J (3)
2 Agassi, A (4) : Borg, B (5)



Ie, Criterion B4 (give preferences) has priority and B5
(no identical floats in 2 consecutive rounds) is dropped.




The pairing is correct.



Okay. No point in finessing the algorithm.



S1 only comprises 2, S2 consists of 3 and 4.

So 2 V 3 and downflloat 4.
2 V 3 satisfies the condition x = 0, however
downfloating 4 violates B5.

However 2 V 4 and downfloating 3 although eliminating
the problem of 4 with B6 creates a far greater totally
unacceptable situation as it violates the x=0
condition.

The only valid pairing is 2 V 3 with 4 downfloating.


Okay. Thanks. The simpler, the better.

My question now is how this is carried out in terms of
C1-14.

Does the situation of C1 apply? I thought that C1 only
applied at the start of pairing a bracket. After
checking for players who had played ALL of the others
or who had no compatible preferences with ANY other
player, you downfloated them.

Thereafter, you had to follow the sequence through C2,
C3, C4, C5, C6, C7, etc, ending up somehow at C6.

In this case, this would involve a one-player
homogenous remainder group, but that player would be
downfloated eventually in C6.

With a one-member remainder group, p=0. But I believe
the actions are the same, whether there is only one
member or more members.




next: Bracket [2] 2 3 4
colors: 2:3
C6others: Remainder Group, Bracket 2: 4
C2, x=0
C3, p=0 Homogeneous.
C4, S1 & S2: , 4
C5, Ordered: , 4 C6pairs: B5, NOK: player 4 floated Down 1 round ago
C7, last shuffle
C8, swap 0: , 4
C5, Ordered: , 4 C6pairs: B5, NOK: player 4 floated Down 1 round ago
C7, last shuffle
C8, swap 1: last swap



Bear with me as I look at following Player 4 through
this sequence.

We go through C5-8 transposing and exchanging the
members, but here there is only one member, so there is
very little to do.

It is only at this point however, that we know that 4
is going to have to be downfloated, and B5 is going to
be violated.

That is, only at this point, do we know we are going to
have to backtrack and look for a different partner for
2.

C10 looks like it might be the rule that we need. It
talks about undoing pairings of players that have
downfloated.

But above it is C9. Although there is nothing said
about how to get to C10, I thought you would fall
through following the sequence C4,5,6,7,8,9,10.

C9 says that criterion B6 and B5 are to be dropped for
downfloats, ie for players downfloated in the previous
round, or 4 in our case.

If C9 is applied first it means that the single member
in the remainder group gets downfloated to bracket 3.



C9, Dropping B5
C4, S1 & S2: , 4
C5, Ordered: , 4 C6pairs:
C6others: Floating Down: 4 [2] 4 => [3] 4 1 5 8 9 10 11 13
next: Bracket [3] 4 1 5 8 9 10 11 13


In this case, it's what we want. Without C9, we can't
drop Criteria B5 and B6 and downfloat 4.

But in Bill Gletsos' other scenario, where 4 has the
same preference as 3, it's not what we want.



If in the above above scenario of players 2, 3 and 4,
player 4 had a black colour preference then the
situation would have been different. In that case w =
1, b=2, q =1, x =0 and p = 1.

As before 2 v 3 and downfloat 4 meetes x = 0 but
violates B5. However now 2 V 4 and downfloat 3 means
that the downfloating of 3 still does not violates B5
but now 2 V 4 now meets x = 0.


But we are only going to be able to re-pair these 3
members of bracket 2 if C10 is applied before C9.

So alternatively to the above invocation of C9, we
could have:



C10,re-pairing Player 2 in Bracket 2
C7, 4 3 C6pairs: B4: x=0, table 1 NOK
C7, last shuffle
C10,again Dropping B6
C2, x=0
C3, p=1 Heterogeneous.
C4, S1 & S2: 2, 4 3
C5, ordered: 2, 3 4 C6pairs:
colors: 2:3
C6others: Remainder Group, Bracket 2: 4

et cetera


This looks long and involved, but done by a machine,
the right answer comes out at the end, if the
program is correct.

[As you have probably guessed, I'm writing a swiss
pairing program. See
http://www.xray.mpe.mpg.de/mailing-lists/modules/2007-01/msg00341.html

I'm going to release it as Free Software. I'm wondering
if it's okay if I use the tennis player tournament of this
thread in my tests.]

But my question here is, which of the procedures apply
in the downfloating of 4, C1 or C10?

Bill Gletsos
27-01-2007, 04:20 PM
Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 WBW 3
2-3
3 12,8,6 BWB 2.5
4 13,9,1 WBW D 2.5

Paired as:

No Name Result Name

1 Sampras, P (2) : McEnroe, J (3)
2 Agassi, A (4) : Borg, B (5)



Ie, Criterion B4 (give preferences) has priority and B5
(no identical floats in 2 consecutive rounds) is dropped.



Okay. No point in finessing the algorithm.That is your conclsuion, not mine.

The issue is not one of finessing the algorithm but ensuring that your algortithm do the pairings correctly.

Okay. Thanks. The simpler, the better.Again that is your conclusion. My point above remains.

My question now is how this is carried out in terms of
C1-14.

Does the situation of C1 apply? I thought that C1 only
applied at the start of pairing a bracket. After
checking for players who had played ALL of the others
or who had no compatible preferences with ANY other
player, you downfloated them.I believe your language here is too generic and not specific enough.

You dont downfloat just because as you put it there being 'no compatible preferences'.
e.g. It is entirely possible that within a score group no players can get their colour preferences. That does not automatically mean you downfloat them to the next score group. Provided they dont violate B1 or B2 then it more than likely you can pair them.

Thereafter, you had to follow the sequence through C2,
C3, C4, C5, C6, C7, etc, ending up somehow at C6.Following on from C6 the subsequent C rules will cause you to restart at C6 or earlier.

In this case, this would involve a one-player
homogenous remainder group, but that player would be
downfloated eventually in C6.Again this is a generalisation.
The pairing rules are simple.

You start with the highest score group and start at C1 and you proceed from there.

In the case in point where the highest score group contains only 1 player via C1 it is clear you downfloat them to the next score bracket and proceed from C2 onwards.

With a one-member remainder group, p=0. But I believe
the actions are the same, whether there is only one
member or more members.This isnt an accurate summation.

When pairing a score group you will have had the situation where the number of pairings equals p. The fact you may then have a remainder group after p pairings is complete does not mean that a one member remainder group represents p = 0. In fact having paired a score group, then p for the remainder group (be it made of single players or not) is determined when the remainder group is downfloated to the next score group in accordance with C3. Originally it wont be 0 but for a score group comprising of one palyer it will eventaully become 0 as explained below.




next: Bracket [2] 2 3 4
colors: 2:3
C6others: Remainder Group, Bracket 2: 4
C2, x=0
C3, p=0 Homogeneous.
C4, S1 & S2: , 4
C5, Ordered: , 4 C6pairs: B5, NOK: player 4 floated Down 1 round ago
C7, last shuffle
C8, swap 0: , 4
C5, Ordered: , 4 C6pairs: B5, NOK: player 4 floated Down 1 round ago
C7, last shuffle
C8, swap 1: last swap



Bear with me as I look at following Player 4 through
this sequence.

We go through C5-8 transposing and exchanging the
members, but here there is only one member, so there is
very little to do.

It is only at this point however, that we know that 4
is going to have to be downfloated, and B5 is going to
be violated.

That is, only at this point, do we know we are going to
have to backtrack and look for a different partner for
2.

C10 looks like it might be the rule that we need. It
talks about undoing pairings of players that have
downfloated.

But above it is C9. Although there is nothing said
about how to get to C10, I thought you would fall
through following the sequence C4,5,6,7,8,9,10.

C9 says that criterion B6 and B5 are to be dropped for
downfloats, ie for players downfloated in the previous
round, or 4 in our case.

If C9 is applied first it means that the single member
in the remainder group gets downfloated to bracket 3.



C9, Dropping B5
C4, S1 & S2: , 4
C5, Ordered: , 4 C6pairs:
C6others: Floating Down: 4 [2] 4 => [3] 4 1 5 8 9 10 11 13
next: Bracket [3] 4 1 5 8 9 10 11 13


In this case, it's what we want. Without C9, we can't
drop Criteria B5 and B6 and downfloat 4.I dont believe your logic is correct there.

In my previous post in pairing player 2, I started at C1 and never went past C9.
Having started at C1 when I hit C2 I am pairing a heterogenous score group.
The first time I hit C6 I meet the x = 0 condition but I violate B5. As a human I know that carrying out a transposition via C7 will violate the x = 0 condition, however a computer must check it. Hence it must carry out C7 and do a transposition and restart at C6. Now it will discover this fails the x = 0 test. As such the program now fall thru C7 and c8 and hit c9 and restart looking at the scorer group anew from C4. Now when I get to C6 I have the same situation as i originally had when I first hit C6 with x = 0 being true and no violation of B5 as I have dropped this criteria as per C9.
So I have the pairing 2 V 3 and from C6 I have the situation of in case of a heterogeneous score bracket: only players moved down were paired so far. Start at C2 with the homogeneous remainder group.
In all these calculations the situation has always been p = 1.
As such restaring at C2, I have a score group consisting of only player 4. Now again to a human it is obvious player 4 needs to downfloat, but to a computer it actually needs to work it all out.

So from C2 and applying A8, player 4 has a black colour preference and hence b = 1, w = 0, q = 1 and x =0.
Now from c3 and applying A6 p = 1.

So C4 gives S1 of player 4 and S2 is empty.
Apply C5 which essentially does nothing in this case.
Apply C6.
However with only one player it is impossible to create a pairing to satisfy p =1.

Now with only one player in the score group you will fall through C7, C8 etc finally getting to C14.

C14 says reduce p by 1 thus p now equals 0. In accordance with C14 if p = 0 then move the score bracket (the singular player 4) down to the next score group restart at C1.

In al of the above where we ended up with 4 as the lone member of the remainder score group it is obvious to a human that 4 needs to downfloat but to a computer it needs to work its way fully through the pairing rules.

But in Bill Gletsos' other scenario, where 4 has the
same preference as 3, it's not what we want.This isnt correct. In my scenario where player 4 had the same colour preference as 3 then you end up pairing 2 V 3 as follows.

Start at C1 dropping player 2 to the next score group amd progress thru C2 to C6. 2 V 3 as noted meets x = 0 but fails B5.
Therefore carry out a transposition as per C7 and restart at c6.
This gives 2 V 4 which meets x = 0 but does not violate B5.
As such I have have the pairing 2 V 4 and from C6 C6 I have the situation of in case of a heterogeneous score bracket: only players moved down were paired so far. Start at C2 with the homogeneous remainder group.
therefore as above start at C2 with the homogeneous remainder group consisting this time of player 3 and the logic follows as described above with player 4.


But we are only going to be able to re-pair these 3
members of bracket 2 if C10 is applied before C9.

So alternatively to the above invocation of C9, we
could have:



C10,re-pairing Player 2 in Bracket 2
C7, 4 3 C6pairs: B4: x=0, table 1 NOK
C7, last shuffle
C10,again Dropping B6
C2, x=0
C3, p=1 Heterogeneous.
C4, S1 & S2: 2, 4 3
C5, ordered: 2, 3 4 C6pairs:
colors: 2:3
C6others: Remainder Group, Bracket 2: 4

et cetera


This looks long and involved, but done by a machine,
the right answer comes out at the end, if the
program is correct.None of this is relevant as shown by my comments above.


But my question here is, which of the procedures apply
in the downfloating of 4, C1 or C10?See above.

drbean
31-01-2007, 03:12 PM
The issue is not one of finessing the algorithm but ensuring that your algortithm do the pairings correctly.



Sorry. Yes, I agree. Ensuring the pairings are done correctly is what's important. What I meant to say is that the program should follow the standard practice. By the algorithm, I meant the procedure defined by 04. FIDE Swiss Rules | 04.1. Swiss System Based on Rating. The program should just try to implement that algorithm, rather than embody a different algorithm.




Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 WBW 3
2-3
3 12,8,6 BWB 2.5
4 13,9,1 WBW D 2.5






In my previous post in pairing player 2, I started at C1 and never went past C9.
Having started at C1 when I hit C2 I am pairing a heterogenous score group.
The first time I hit C6 I meet the x = 0 condition but I violate B5.



In other words:



next, Bracket 2: 2 3 4
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1 Heterogeneous.
C4, S1 & S2: 2, 3 4
C5, ordered: 2, 3 4 C6pairs:
Roles: E1 2:3
C6others: Remainder Group, Bracket 2: 4


Pairing 2 & 3 meets the x=0 condition, but this results in player 4 being downfloated, which violates B5. So we can't go ahead and just pair 2 & 3.

It appears that the human arbiter is able to look ahead (or has to look ahead) from C6 to the pairing of the next (remainder) group. That is just as the arbiter is about to accept 2:3, they see that this results in a problem pairing the players in the remainder group.

This kind of lookahead and branching is something I want to minimize, so I wonder if this is the only point where I will have to do it.

Bill Gletsos talks about how the computer can work out it can continue with the pairing of Bracket 2, as a result of this knowledge, but doesn't say how it will acquire this knowledge that what is about to happen to player 4 will violate B5.




As a human I know that carrying out a transposition via C7 will violate the x = 0 condition, however a computer must check it. Hence it must carry out C7 and do a transposition and restart at C6. Now it will discover this fails the x = 0 test. As such the program now fall thru C7 and c8 and hit c9 and restart looking at the scorer group anew from C4. Now when I get to C6 I have the same situation as i originally had when I first hit C6 with x = 0 being true and no violation of B5 as I have dropped this criteria as per C9.



I need to remember B5 is being dropped for the whole of the heterogeneous group, the players paired first AND the remainder group.




So I have the pairing 2 V 3 and from C6 I have the situation of in case of a heterogeneous score bracket: only players moved down were paired so far. Start at C2 with the homogeneous remainder group.

In all these calculations the situation has always been p = 1.
As such restaring at C2, I have a score group consisting of only player 4. Now again to a human it is obvious player 4 needs to downfloat, but to a computer it actually needs to work it all out.



So the algorithm is able to cope with the problem, but the fact that there is a problem at that point is something that the human arbiter is only aware of by looking ahead.

To handle lookaheads like this is going to require searching and branching.

Incidentally in a one person remainder group, doesn't p=0? p is the top half of the players in the group. Half of one, rounded down, is zero.

Thanks for this help in showing me how the FIDE Rules work in practice.

Bill Gletsos
31-01-2007, 11:41 PM
Pairing 2 & 3 meets the x=0 condition, but this results in player 4 being downfloated, which violates B5. So we can't go ahead and just pair 2 & 3.

It appears that the human arbiter is able to look ahead (or has to look ahead) from C6 to the pairing of the next (remainder) group. That is just as the arbiter is about to accept 2:3, they see that this results in a problem pairing the players in the remainder group.Incorrect. The human arbiter is not attempting to pair the remainder group.

The human arbiter has tried the pairing 2 V 3 with 4 downfloating. This meets X = 0 but violates B5. The human arbiter knows that 2 V 4 violates x = 0. As such 2 V 3 is the correct pairing with 4 downfloating as it is the only one that does not violate x = 0.

This kind of lookahead and branching is something I want to minimize, so I wonder if this is the only point where I will have to do it.

Bill Gletsos talks about how the computer can work out it can continue with the pairing of Bracket 2, as a result of this knowledge, but doesn't say how it will acquire this knowledge that what is about to happen to player 4 will violate B5.You seem to have misunderstood the situation. Coming thru C6 the first time you try pairing 2 V 3. This results in meeting x = 0. It also results in having to downfloat 4. However the human and indeed the computer knows that it downfloated plater 4 in the previous round. Therefore the human and the computer knows immediately that downfloating player 4 violates B5. No look ahead at pairing player 4 is required to know that downfloating player 4 violates B5.

I need to remember B5 is being dropped for the whole of the heterogeneous group, the players paired first AND the remainder group.Actually C9 makes no distinction in dropping B5 based on whether the group is heterogeneous or homogeneous.

So the algorithm is able to cope with the problem, but the fact that there is a problem at that point is something that the human arbiter is only aware of by looking ahead.

To handle lookaheads like this is going to require searching and branching.I'm not sure what you are saying is right as I dont believe you understood the situation I described above.

In the case of the tennis player group as shown you try 2 v3 and get x = 0 being true but player 4 violating B5.
You need to try a transposition to see if there is a better pairing so you get 2 V 4 with 3 downfloating. This violates x = 0. The fact the downfloat of 3 doesnt violate B5 is immaterial. All this however has at no stage attempted to do anything with the downfloated player be it 3 or 4 and it is clear 2 V 3 is the pairing as it is the only one meeting x = 0.

However in the case of players 3 and 4 having the same colour preference then as noted in a previous post 2 V 3 and downfloating 4 meets x = 0 and violates B5 but after a transposition 2 V 4 and downfloating 3 meets x = 0 and does not violate B5. Again no attempt to do anything with the downfloated player be it 3 or 4 was done and in this scenario it is clear the pairing is 2 V 4 and downfloat 3 as this is a superior pairing to 2 V 3 and downfloat 4 as although both meet x = 0 downfloating 3 does not violate B5.

Note in an the original scenario but where player 2 had already played players 3 & 4 then you would end up with a paiting of 4 V 3 and downfloating 2 to the next score group.

Incidentally in a one person remainder group, doesn't p=0? p is the top half of the players in the group. Half of one, rounded down, is zero.That is a valid point however on a close inspection of the rules it could be argued that they always assume there is at least one player in S1.
However before looking at that lets assume that for a one player remainder group your p = 0 is correct.

As such after getting the pairing 2 V 3 in C6 then you have a remainder group of 1 player and in accordance with C6 restart at C2.

This time when you hit C6, S1 has no players and p = 0. As such you need to make zero pairings. This is easily satisfied and in accordance with C6 you follow the part tht says: "in case of a homogeneous score bracket: remaining players are moved down to the next score bracket. With this score bracket restart at C1."

However if S1 is always considered to have at least on player in it then in a one player remainder group p = 1. In that case as I stated in a previous post, in applying C6 with only one player it is impossible to create a pairing to satisfy p = 1. Therefore with only one player in the score group you will fall through C7, C8 etc finally getting to C14. Application of C14 will lead to a restart at C1.

The bottom line of all this is that it is totally immaterial in a one player group whether p is 0 or 1 as it makes no real difference as you will always end up downfloating the player and restarting at C1.

Thanks for this help in showing me how the FIDE Rules work in practice.No problem, glad to help.

drbean
02-02-2007, 12:38 AM
Incorrect. The human arbiter is not attempting to pair the remainder group.


If the arbiter is only concentrating on C6 ie the pairs in S1 and S2, how do they know what is going to happen to those players not paired? This is something which will only have to be decided at a later stage in the program. At least, that's what I thought I could do. The computer knows 4, in this case, is going to end up in the remainder group, but only when it starts to pair the remainder group, in C1 presumably, does it realize that 4 doesn't have a partner and so is going to be downfloated. At that stage, C6, ie the pairing of 2 and 3, is in the past, something that has already been decided. At least, that has been the way I've been trying to implement the algorithm, one step at a time.

It appears that, the same way as a chess player decides what to do, the arbiter needs to be able to foresee what is going to happen if a choice is made, at least at some particular points.

Here's a representation of the ways I have seen the parts of the algorithm relating to each other. On the left hand side is the stage the program is at at that particular point. On the right is a list of possible continuations from that point. At each stage it merely carries out the content described in each of the corresponding rules.



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

ERROR, [ undef, ERROR ],


You seem to have misunderstood the situation. Coming thru C6 the first time you try pairing 2 V 3. This results in meeting x = 0. It also results in having to downfloat 4. However the human and indeed the computer knows that it downfloated player 4 in the previous round. Therefore the human and the computer knows immediately that downfloating player 4 violates B5. No look ahead at pairing player 4 is required to know that downfloating player 4 violates B5.


The computer knows 4 was downfloated in the previous round, but is not able to look into the future.

It is lookahead in so far as nothing in C6 says anything about downfloating players from a homogeneous remainder group. Pairing the remainder group is
the responsibility of C2. Note in the representation above, I've made it C1 instead of C2, to catch this particular case.



In the case of the tennis player group as shown you try 2 v3 and get x = 0 being true but player 4 violating B5.
You need to try a transposition to see if there is a better pairing so you get 2 V 4 with 3 downfloating. This violates x = 0. The fact the downfloat of 3 doesnt violate B5 is immaterial.


OK.



All this however has at no stage attempted to do anything with the downfloated player be it 3 or 4 and it is clear 2 V 3 is the pairing as it is the only one meeting x = 0.


The arbiter has seen that the downfloating of 4 will violate B5, at a point where nothing has been done with 4.


On the question of the value of p in a one player bracket:


.. on a close inspection of the rules it could be argued that they always assume there is at least one player in S1.



It appears to assume players in S2 too. I think I will make C1 callable at some points extra to those in the above representation to get rid of 1-player brackets. This would force an attempt to downfloat, in this example, 4, which would be caught by B5.

I think C.04 could also be more explicit about awarding byes, the subject of my next message.

drbean
02-02-2007, 12:48 AM
In the same round as the problem with B5, which resulted in the downfloating of 4 two times in a row, there is another floating problem in the last bracket.



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score

...

13 4,16,14 WBW D 2
11-12
6 15,1,3 BWB 1.5
7 16,2,8 WBW U 1.5
13-16
12 3,17,9 BWB 1
17 8,12,- WB- D 1
18 9,-,10 B-W d 1
19 -,5,11 -WB 1
17-18
15 6,10,16 WBW 0.5
16 7,13,15 BWB 0.5
19
14 5,11,13 BWB U 0


14 played 13, who was downfloated in Round 3. 14 is now alone at the bottom. To pair with any player will mean that 14 has two upfloats in a row, which violates B5. But this is a Relative Criterion, so may be dropped if we have to. In any case, giving 14 the Bye would be nice.

The published pairings are:



7 Rafter, P (12) : Baghdatis, M (19)
8 El Aynaoui, Y (17) : Roddick, A (15)
9 Cash, P (16) : Philippousis, M (18)
10 Nalbandian, D (14) 1:0 BYE


So 14 gets the bye, which is convenient. But how did (Swiss Perfect and Swiss Manager?) arrive at these pairings? Starting with the fifth bracket, presumably first they tried:



next, Bracket 5: 12 17 18 19
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Homogeneous.
C4, S1 & S2: 12 17, 18 19
C5, ordered: 12 17, 18 19 C6pairs:
Roles: E1 12:18 E3 17:19
C6others: no non-paired players
next, Bracket 6: 15 16
C1, B1a NOK: 15 16 Floating Down: 15 16 [6] 15 16 => [7] 16 15 14
next, Bracket 7: 16 15 14
C1, B1,2 test: ok, no unpairables


C1 says:


If the score bracket contains a player for whom no opponent can be
found within this score bracket without violating B1 or B2 then:
* if this player was moved down from a higher score bracket apply
C12.
* if this score bracket is the lowest one apply C13.
* in all other cases: move this player down to the next score
bracket.


My interpretation of this is that it is something you check before you divide the bracket into S1 and S2 and pair corresponding members off. It doesn't mean if there are 3 players in the lowest score bracket you join the 2 lowest score brackets looking for a non-existent fourth player to avoid awarding a bye. It means that you check for a player who has played all the others, or who has an Absolute preference the same as the other players they haven't played.

So, I'm claiming 16, 15 in Bracket 6 fail this test, so they are floated down. Then 16 and 15 together with 14 in Bracket 7 pass the C1 test. The published pairing looks like it is saying they failed it again and this results in the application of C13 and the joining of Brackets 5,6,7, into one big bracket.

The published pairing might claim the provisional criteria B5 and B6 apply to C1 like they do to C6. (In this case, 14 'deserves' the Bye, because otherwise it will be upfloated twice in a row.) But after the waiving of B5 and B6, the algorithm returns only to B4, so I think B5 and B6 never apply to C1.

Or is perhaps C1 something that you can apply at any time? It doesn't fit into any particular place in the algorithm?

I would continue:



C2, x=0
C3, p=1 Homogeneous.
C4, S1 & S2: 15, 16 14
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs: B5, NOK: 14 floated Up, 1 rounds ago
C7, last shuffle
C8, swap 0: last swap
C9, Dropping B6
C4, S1 & S2: 15, 14 16
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs: B5, NOK: 14 floated Up, 1 rounds ago
C7, last shuffle
C8, swap 0: last swap
C9, Dropping B5
C4, S1 & S2: 15, 14 16
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs:
Roles: E1 14:15
C6others: Remainder Group, Bracket 7: 16
16
C1,
C13, only player 16 bye: 16
Pairing complete


So the algorithm falls through to C9 twice and finally B5 is waived. Like player 4, at the other end of the score table, who was downfloated twice in a row, 14 is upfloated twice in a row.

So again the question is, is the published pairing correct and why?

Bill Gletsos
02-02-2007, 02:39 AM
If the arbiter is only concentrating on C6 ie the pairs in S1 and S2, how do they know what is going to happen to those players not paired? This is something which will only have to be decided at a later stage in the program. At least, that's what I thought I could do. The computer knows 4, in this case, is going to end up in the remainder group, but only when it starts to pair the remainder group, in C1 presumably, does it realize that 4 doesn't have a partner and so is going to be downfloated. At that stage, C6, ie the pairing of 2 and 3, is in the past, something that has already been decided. At least, that has been the way I've been trying to implement the algorithm, one step at a time.No, in my opinion it is far simpler than this.

The computer knew it was pairing a score group comprising 3 players. It knew it was making one pairing (p = 1) therefore it can easily know it has one player left over. Therefore when it determines that the pairing 2 V 3 is valid, it can immediately downfloat player 4 and restart at C1.

It appears that, the same way as a chess player decides what to do, the arbiter needs to be able to foresee what is going to happen if a choice is made, at least at some particular points.True, but not in the example we were discussing. In your comments you seem to be asking a specific question about why a paring is such and such but when given the answer then go from making a specific statement to a generalisation.

In my opinion that isnt the easiest way for you to understand what is happening as a very minor change in conditions can lead to signifcant changes in the correct pairing,

A human knows that he has a score group of 3 players and in completing one pairing will have a player left over and must therefore downfloat them.Exactly as I mentioned above. As such the computer doesnt need to actually do anything other than downfloat the player and restart at C1.

Here's a representation of the ways I have seen the parts of the algorithm relating to each other. On the left hand side is the stage the program is at at that particular point. On the right is a list of possible continuations from that point. At each stage it merely carries out the content described in each of the corresponding rules.



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

ERROR, [ undef, ERROR ],


The computer knows 4 was downfloated in the previous round, but is not able to look into the future.

It is lookahead in so far as nothing in C6 says anything about downfloating players from a homogeneous remainder group. Pairing the remainder group is
the responsibility of C2. Note in the representation above, I've made it C1 instead of C2, to catch this particular case.As I said i think it is simpler than that.As soon as the computer pairs players 2 and 3 via C6 it knows it has a remainder group of 1 and must downfloat that player and can immediately restart at C1.


The arbiter has seen that the downfloating of 4 will violate B5, at a point where nothing has been done with 4.No. In checking the validity of the pairing 2 V 3 the computer lmpws that it must have to downfloat player 4 and needs to check if downfloating player 4 violates B5 or B6. In the original scenario there isnt. In the scenario where players 3 and 4 both have the same colour preference then 2 V 4 and downfloating 3 is the better paiiring as previously explained.

On the question of the value of p in a one player bracket:


It appears to assume players in S2 too. I think I will make C1 callable at some points extra to those in the above representation to get rid of 1-player brackets. This would force an attempt to downfloat, in this example, 4, which would be caught by B5.You shouldnt actually have to go through the process of downfloating player 4 to determine that he violates B5.
You should be able to do this as part of checking the validity of the 2 V 3 pairing.

Bill Gletsos
02-02-2007, 03:26 AM
In the same round as the problem with B5, which resulted in the downfloating of 4 two times in a row, there is another floating problem in the last bracket.



Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score

...

13 4,16,14 WBW D 2
11-12
6 15,1,3 BWB 1.5
7 16,2,8 WBW U 1.5
13-16
12 3,17,9 BWB 1
17 8,12,- WB- D 1
18 9,-,10 B-W d 1
19 -,5,11 -WB 1
17-18
15 6,10,16 WBW 0.5
16 7,13,15 BWB 0.5
19
14 5,11,13 BWB U 0


14 played 13, who was downfloated in Round 3. 14 is now alone at the bottom. To pair with any player will mean that 14 has two upfloats in a row, which violates B5. But this is a Relative Criterion, so may be dropped if we have to. In any case, giving 14 the Bye would be nice.

The published pairings are:



7 Rafter, P (12) : Baghdatis, M (19)
8 El Aynaoui, Y (17) : Roddick, A (15)
9 Cash, P (16) : Philippousis, M (18)
10 Nalbandian, D (14) 1:0 BYE


So 14 gets the bye, which is convenient. But how did (Swiss Perfect and Swiss Manager?) arrive at these pairings? Starting with the fifth bracket, presumably first they tried:



next, Bracket 5: 12 17 18 19
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Homogeneous.
C4, S1 & S2: 12 17, 18 19
C5, ordered: 12 17, 18 19 C6pairs:
Roles: E1 12:18 E3 17:19
C6others: no non-paired players
next, Bracket 6: 15 16
C1, B1a NOK: 15 16 Floating Down: 15 16 [6] 15 16 => [7] 16 15 14
next, Bracket 7: 16 15 14
C1, B1,2 test: ok, no unpairables


C1 says:


My interpretation of this is that it is something you check before you divide the bracket into S1 and S2 and pair corresponding members off. It doesn't mean if there are 3 players in the lowest score bracket you join the 2 lowest score brackets looking for a non-existent fourth player to avoid awarding a bye. It means that you check for a player who has played all the others, or who has an Absolute preference the same as the other players they haven't played.

So, I'm claiming 16, 15 in Bracket 6 fail this test, so they are floated down. Then 16 and 15 together with 14 in Bracket 7 pass the C1 test. The published pairing looks like it is saying they failed it again and this results in the application of C13 and the joining of Brackets 5,6,7, into one big bracket.

The published pairing might claim the provisional criteria B5 and B6 apply to C1 like they do to C6. (In this case, 14 'deserves' the Bye, because otherwise it will be upfloated twice in a row.) But after the waiving of B5 and B6, the algorithm returns only to B4, so I think B5 and B6 never apply to C1.

Or is perhaps C1 something that you can apply at any time? It doesn't fit into any particular place in the algorithm?

I would continue:



C2, x=0
C3, p=1 Homogeneous.
C4, S1 & S2: 15, 16 14
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs: B5, NOK: 14 floated Up, 1 rounds ago
C7, last shuffle
C8, swap 0: last swap
C9, Dropping B6
C4, S1 & S2: 15, 14 16
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs: B5, NOK: 14 floated Up, 1 rounds ago
C7, last shuffle
C8, swap 0: last swap
C9, Dropping B5
C4, S1 & S2: 15, 14 16
C5, ordered: 15, 16 14 C6pairs: B1a: table 1 NOK
C7, 14 16 C6pairs:
Roles: E1 14:15
C6others: Remainder Group, Bracket 7: 16
16
C1,
C13, only player 16 bye: 16
Pairing complete


So the algorithm falls through to C9 twice and finally B5 is waived. Like player 4, at the other end of the score table, who was downfloated twice in a row, 14 is upfloated twice in a row.

So again the question is, is the published pairing correct and why?I only looked at this very quickly but as far as I can see the published pairing is correct and the C1 test isnt the issue here.

In determing that 15 cannot play 16 because it violates B1 they both downfloat to the last score group and you start at C1.

Now in falling thru C2 to C6 the important thing is that the group is homogeneous.

Now it appears that you failed to notice that C9 only applies to downfloats, not upfloats.
As such any pairing of player 14 with either 15 or 16 violates B5 as an upfloat for player 14.

Also C10 does not apply as this isnt a homogeneous remainder group, it is simply a homogeneous group.

Therefore you will eventually fall thru C11 down to C13 (note C12 does not apply).

drbean
11-02-2007, 12:35 PM
Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
2 11,7,5 BWB 3
2-3
3 12,8,6 WBW 2.5
4 13,9,1 BWB D 2.5





The computer knew it was pairing a score group comprising 3 players. It knew it was making one pairing (p = 1) therefore it can easily know it has one player left over.







In checking the validity of the pairing 2 V 3 the computer lmpws that it must have to downfloat player 4 and needs to check if downfloating player 4 violates B5 or B6. In the original scenario there isnt. In the scenario where players 3 and 4 both have the same colour preference then 2 V 4 and downfloating 3 is the better paiiring as previously explained.
You shouldnt actually have to go through the process of downfloating player 4 to determine that he violates B5.
You should be able to do this as part of checking the validity of the 2 V 3 pairing.

Okay, I think I can check if a single unpaired member was downfloated or not. I see I was already checking if the unpaired members of a homogeneous group had been downfloated before. I can extend that to check if there is a single unpaired member in a heterogeneous group.



Next, Bracket 1: 2
C1, Floating Down: 2 [1] 2 => [2] 2 3 4
Next, Bracket 2: 2 3 4
C1, B1,2 test: ok, no unpairables
C2, x=0
C3, p=1 Heterogeneous.
C4, S1 & S2: 2, 3 4
C5, ordered: 2, 3 4
C6, B5: NOK. 4 floated Down 1 rounds ago
C7, 4 3
C6, B4: x=0, table 1 NOK
C7, last shuffle
C9, Dropping B6 for Downfloats
C4, S1 & S2: 2, 4 3
C5, ordered: 2, 3 4
C6, B5: NOK. 4 floated Down 1 rounds ago
C7, 4 3
C6, B4: x=0, table 1 NOK
C7, last shuffle
C9, Dropping B5 for Downfloats
C4, S1 & S2: 2, 4 3
C5, ordered: 2, 3 4
C6,
Roles: E1 2:3
C6others: Remainder Group, Bracket 2: 4
4
C1, Floating Down: 4 [2] 4 => [3] 4 1 5 8 9 10 11 13
Next, Bracket 3: 4 1 5 8 9 10 11 13



This is the same as the published pairing.

drbean
13-02-2007, 01:11 AM
Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
...

13-16
12 3,17,9 BWB 1
17 8,12,- WB- D 1
18 9,-,10 B-W d 1
19 -,5,11 -WB 1
17-18
15 6,10,16 WBW 0.5
16 7,13,15 BWB 0.5
19
14 5,11,13 BWB U 0

The published pairings:

7 Rafter, P (12) : Baghdatis, M (19)
8 El Aynaoui, Y (17) : Roddick, A (15)
9 Cash, P (16) : Philippousis, M (18)
10 Nalbandian, D (14) 1:0 BYE




I only looked at this very quickly but as far as I can see the published pairing is correct and the C1 test isnt the issue here.

In determing that 15 cannot play 16 because it violates B1 they both downfloat to the last score group and you start at C1.



That's the way I would do it too, but could C14 be used as an alternative here? According to C14, after attempting to pair with p=1, p is reduced to 0, and then the whole bracket is floated down to the next bracket.



Now it appears that you failed to notice that C9 only applies to downfloats, not upfloats.
As such any pairing of player 14 with either 15 or 16 violates B5 as an upfloat for player 14.


Oh, I made a mistake there. My analysis was wrong.



Therefore you will eventually fall thru C11 down to C13 (note C12 does not apply).

C12 applies to heterogeneous groups. C13 applies to the lowest score group. But the action in both cases starts the same way: undo the pairing in the previous score group and find a pairing which allows a pairing in the present bracket.

In our case, we have a final group made up of brackets 6 and 7. The penultimate group is bracket 5. We have to undo the pairing of 12&18 and 17&19, going through C7,8,9,11,14 until we are pairing only 2 players, not 4.



C14, Bracket 5, now p=1
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 18 19 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 17 18
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 18 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, last transposition
C8, exchange 1: last S1,S2 exchange
C9, Dropping B6 for Downfloats
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 18 19 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 17 18
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 18 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, last transposition
C8, exchange 1: last S1,S2 exchange
C9, Dropping B5 for Downfloats
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, 1 tables paired. E1 12&18
C6others: Floating remaining 17 19 Down. [5] 12 18 17 19 => [6] 19 17 15 1
4 16
Next, Bracket 6: 19 17 15 14 16
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Heterogeneous.
C4, S1 & S2: 19 17 & 15 14 16
C5, ordered: 17 19 & 15 16 14
C6, 2 tables paired. E1 17&15 E2 16&19
C6others: Remainder Group, Bracket 6: 14
C1,
C13, only player 14 bye: 14
Pairing complete


12 and 17 both have a preference for White, but 17 was downfloated in the previous round, so can't be downfloated until we drop B5 in C9.

This gives a pairing of 12 and 18, with 17 and 19 downfloating.

I don't see why the published pairing is of 12 and 19. 18 comes first in D1 transposition order. What is wrong with the pairing of 17&15 and 16&19?

Bill Gletsos
14-02-2007, 01:58 PM
C12 applies to heterogeneous groups. C13 applies to the lowest score group. But the action in both cases starts the same way: undo the pairing in the previous score group and find a pairing which allows a pairing in the present bracket.

In our case, we have a final group made up of brackets 6 and 7. The penultimate group is bracket 5. We have to undo the pairing of 12&18 and 17&19, going through C7,8,9,11,14 until we are pairing only 2 players, not 4.



C14, Bracket 5, now p=1
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 18 19 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 17 18
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 18 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, last transposition
C8, exchange 1: last S1,S2 exchange
C9, Dropping B6 for Downfloats
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 18 19 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 17 18
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, 19 18 17
C6, B5: NOK. 17 floated Down 1 rounds ago
C7, last transposition
C8, exchange 1: last S1,S2 exchange
C9, Dropping B5 for Downfloats
C4, S1 & S2: 12 & 19 18 17
C5, ordered: 12 & 17 18 19
C6, B1a: table 1 NOK
C7, 18 17 19
C6, 1 tables paired. E1 12&18
C6others: Floating remaining 17 19 Down. [5] 12 18 17 19 => [6] 19 17 15 1
4 16
Next, Bracket 6: 19 17 15 14 16
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Heterogeneous.
C4, S1 & S2: 19 17 & 15 14 16
C5, ordered: 17 19 & 15 16 14
C6, 2 tables paired. E1 17&15 E2 16&19
C6others: Remainder Group, Bracket 6: 14
C1,
C13, only player 14 bye: 14
Pairing complete


12 and 17 both have a preference for White, but 17 was downfloated in the previous round, so can't be downfloated until we drop B5 in C9.

This gives a pairing of 12 and 18, with 17 and 19 downfloating.

I don't see why the published pairing is of 12 and 19. 18 comes first in D1 transposition order. What is wrong with the pairing of 17&15 and 16&19?Again I have only looked at this quickly but it appears to me that the following is the correct sequence.

Originally you pair the score group with a score = 1.
Note p = 2 not p =1.
S1: 12, 17 and S2: 18, 19


S1 S2
12 18
17 19


This gives the pairings 12 v 18 & 17 V 19.

Now you cannot pair the players (15 7 16) in score group 0.5 because they violate B1.
In accordance with C1 you move both down to the 0 score group.

Now p = 1 , S1 = 15 and S2 = 16, 14.

No matter how you pair these the pairing of player 14 with either 15 or 16 violates B5 for upfloats.

Now only C10 allows you to drop B5 for upfloats, however although it is a homogeneous group it is not a homogeneous remainder group and therefore C10 does not apply as it only applies to a homogeneous remainder group.

C11 makes no difference and C12 will not apply as it only applies to hetergeneous groups.

Therefore C13 will result in the unpairing of all pairings in the score bracket with score = 1.

So in accordance with C13 the pairings 12 v 18 and 17 v 19 are undone and another set of pairings are tried with S1: 12, 17 and S2: 18 : 19

The next one to try is via transposition
So you get:


S1 S2
12 19
17 18


Because we already know that 14, 15 & 16 cannot be paired then clearly in line with C13 we get a group consisting of players 17, 18, 15, 16 and 14.

S1: 17, 18 S2: 15, 16 and 14.



S1 S2
17 15
18 16
14



On a completely seperate note, if by any reading of the C13 rule you end up merging the bottom three score groups, you also get the published pairings.

in that case you now have one large score group consisting of players 12, 17, 18, 19, 15, 16 and 14.


When you merge the groups on 1 point, 0.5 point and 0 points the order of the players is 12, 17, 18, 19, 15 and 16, not 12, 15, 16, 17, 18 and 19.

As such S1 is 12, 17 and 18 and S2 is 19, 15, 16 and 14.

This leads to the initial setup of :



S1 S2
12 19
17 15
18 16
14

drbean
22-02-2007, 03:46 PM
So in accordance with C13 the pairings 12 v 18 and 17 v 19 are undone and another set of pairings are tried with S1: 12, 17 and S2: 18 : 19

The next one to try is via transposition
So you get:


S1 S2
12 19
17 18



Here there seems to be a big step. We are about to downfloat 17 and 18, but 17 was downfloated in the round before, which is also a consideration.

Also does the above rationale accord with C13: 'Try to find another pairing in the penultimate score bracket which will allow a pairing in the lowest score bracket?' This is the next pairing by transposition, ie another pairing, but if the other pair is going to be downfloated, why not just downfloat 17 and 19?

Also it would be good to be able to incorporate C14 into this process, because it is about reducing p (and downfloating remaining players). C14: 'Decrease p by 1 (and if the original value of x was greater than zero decrease x by 1 as well).'

So the approach I took was to cycle through all the remaining transpositions and exchanges in C7 and C8 and then go to C14 via C9 and C11.



C7, [ C6PAIRS, C8, C9, REPEATC10 ],
C8, [ C5, C9, C10 ],
C9, [ \C4, C10, C11 ],
C11, [ C3, C13, C14 ],
C14, [ C1, C4 ],




Because we already know that 14, 15 & 16 cannot be paired then clearly in line with C13 we get a group consisting of players 17, 18, 15, 16 and 14.

S1: 17, 18 S2: 15, 16 and 14.



S1 S2
17 15
18 16
14



C13 is imprecise, I think. The wording 'Try to find another pairing' suggests that any way is okay. Clearly, it is going to be not 12, because it is the topranking member of S1, but if it were my choice, I would downfloat 12, because that would allow 17 not to be downfloated. It is only the relative importance of B4 and B5, that is, that C9 relaxes B5, before C11 relaxes B4, which prevents this.



On a completely seperate note, if by any reading of the C13 rule you end up merging the bottom three score groups, you also get the published pairings.

in that case you now have one large score group consisting of players 12, 17, 18, 19, 15, 16 and 14.


When you merge the groups on 1 point, 0.5 point and 0 points the order of the players is 12, 17, 18, 19, 15 and 16, not 12, 15, 16, 17, 18 and 19.

As such S1 is 12, 17 and 18 and S2 is 19, 15, 16 and 14.

This leads to the initial setup of :



S1 S2
12 19
17 15
18 16
14


This is interesting. I wonder if this would always produce the same pairings. That is, perhaps in practice, the two parts, 'Try to find another pairing in the penultimate score bracket which will allow a pairing in the lowest score bracket,' and, 'If in the penultimate score bracket p becomes zero (i.e. no pairing can be found which will allow a correct pairing for the lowest score bracket) then the two lowest score brackets are joined into a new lowest score bracket,' are equivalent.

Rather than fix this 'bug', I am going to go ahead and release the software, and if anyone wants to fix it, they can.