PDA

View Full Version : SP SP error



Garvinator
28-10-2007, 08:49 PM
No that is not stuttering in the title ;) it comes from the 2007 Surfers Paradise Open Division.

I think I have now seen the worst performance that Swiss imperfect has to offer. If someone can beat this, I will be astounded.

On to the gory.

Final round Open Division. Round running very late, I only had Swiss imperfect available as my own laptop had crashed (so swiss master 5 wasnt available).

Round 7 came out by SIP as:


No Name Loc Total Result Name Loc Total

1 ANTIC, Dejan (2) 2486 [5.5] .5:.5 LANE, Gary W (4) 2405 [4]
2 FROEHLICH, Peter (5) 2365 [4.5] .5:.5 SCHMALTZ, Roland (1) 2568 [5]
3 LY, Moulthun (7) 2280 [4.5] 1:0 DRAGICEVIC, Domagoj (11) 2139 [4]
4 KLEIN, Felix (8) 2212 [4] 1:0 LAZARUS, Benjamin (13) 2058 [4]
5 THOMAS, Brian (14) 1874 [4] 1:0 NAKAUCHI, Gene (15) 1870 [4]
6 SALES, Jesse Noel (6) 2312 [4] 1:0 LIU, Yi (29) 1510 [3.5]
7 LOVEJOY, David (16) 1868 [3.5] 0:1 SOLOMON, Stephen J (3) 2447 [3.5]
8 STOJIC, Dusan (10) 2142 [3] 1:0 MULLER, Jonas (21) 1677 [3]
9 VAN PELT, Michael (24) 1654 [3] 0:1 JONES, Lee R (12) 2086 [3]
10 KORENEVSKI, Oleg (20) 1695 [2.5] .5:.5 GRIGG, Sam (22) 1676 [3]
11 STOKES, Mark C (23) 1668 [2.5] 1:0 STAHNKE, Alexander (27) 1615 [2.5]
12 ALKIN, John (26) 1616 [2.5] 1:0 CHELEBICHANIN, Nenad (25) 1642 [2.5]
13 KINDER, Jessica (28) 1588 [2.5] .5:.5 CHUNG, Francisco (18) 1761 [2]
14 WELLER, Tony (17) 1788 [2] .5:.5 SZUVEGES, Narelle S (19) 1699 [2]
15 LYONS, Kieran C (30) 1439 [2] 1:0 DELMASTRO, Joe (34) [1.5]
16 DILLA, Edsil (9) 2176 [1.5] 0:1 CRISTUTA, Roy (32) 1333 [1.5]
17 CIGELJ, David (33) 1210 [1] 0:1 SRETENOVIC, Laz (31) 1430 [1] without the scores obviously ;).

I had a brief skim and saw all the 'jumbled up' scores and decided that I wasnt going to do a manual check and delay the final round. So I put up the SIP pairings. I was loath to attempt a manual pairing because it looked like from the pairings above that it would take some investigation to sort through it all. Yes, the final round apply pairings box did appear :P

Play started a few minutes later. Stephen Solomon came up and asked me had Antic played Froehlich. I checked the computer and said it doesnt look like it. Stephen replied then that the pairings seemed wrong.

I then had a brief look at and it was clear that SIP had fouled up.

This was the pairing info after round 6 for the boards in question:


2 : 19,9,7,1,3,15 WBWBWB D 5.5

1 : 18,10,6,2,8,7 BWBWBW D 5

5 : 22,14,11,8,28,3 BWBWBB 4.5
7 : 24,16,2,14,22,1 BWBWBB U 4.5

4 : 21,11,19,13,15,12 WBWBWB 4
6 : 23,13,1,15,19,22 WBWBWB 4
8 : 25,15,3,5,1,11 WBWBWB 4
11 : 28,4,5,3,10,8 BWWBBW u 4
13 : 30,6,26,4,18,10 BWBWBW 4
14 : 31,5,27,7,24,25 WBWBWB D 4
15 : 32,8,34,6,4,2 BWBWBW U 4

Now it is abundently clear that 2 should play 5 and 1 and 7 join the 4 point score group.

I checked this with SM5 when I got home and it agrees with 2 and 5.

What sip seems to have done is just moved all four players into the 4 point score group and paired as normal (top half v bottom half).

In my opinion, this is pitiful for a recognised pairing program to pair like this. I can accept that part of the problem does lie with me, but with previous experience on here of seeing manual pairings go wrong and with the last round starting very late, I didnt want to go with manual pairings and be wrong because it was rushed.

My opinion is that the ACF Council needs to consider getting a general licence for a different pairing program. A pairing program that has fouled things up like this needs a serious review and scrutiny. I understand that SP is widely used in Australia, but when it makes simple errors like this, it needs to be changed.

Bill Gletsos
28-10-2007, 09:34 PM
What sip seems to have done is just moved all four players into the 4 point score group and paired as normal (top half v bottom half).No that clearly isnt the case as 5 wouldnt play 1 in that case.
At a quick glance what appears to have happened with SP is the following.

S1 = 2 and 1 and S2 = 5 and 7
p = 2, w = 3, b = 1, q = 2 and x = 1

Note that it is impossible to get p = 2, no matter what you do so you end up falling down to C14 where p is reduced by 1 to 1 and x is also reduced by 1 to 0

2 v 5 violates the x = 0 condition
In fact the only way to get the x = 0 condition from S1 and S2 is 5 v 1
This means 2 and 7 float down to the 4 point group.

It is not immediately obvious how SM5 gets the 2 v 5 pairing according to the C pairing rules.

Garvinator
28-10-2007, 10:14 PM
No that clearly isnt the case as 5 wouldnt play 1 in that case.
At a quick glance what appears to have happened with SP is the following.

S1 = 2 and 1 and S2 = 5 and 7
p = 2, w = 3, b = 1, q = 2 and x = 1

Note that it is impossible to get p = 2, no matter what you do so you end up falling down to C14 where p is reduced by 1 to 1 and x is also reduced by 1 to 0

2 v 5 violates the x = 0 condition
In fact the only way to get the x = 0 condition from S1 and S2 is 5 v 1
This means 2 and 7 float down to the 4 point group.

It is not immediately obvious how SM5 gets the 2 v 5 pairing according to the C pairing rules.
Ok I will have a better look now ;)

Garvinator
28-10-2007, 10:32 PM
It is not immediately obvious how SM5 gets the 2 v 5 pairing according to the C pairing rules.
I think it is.


S1: S2:

2 5
1 7

Now 2 has played 1 and 7, so that only leaves 5 v 2. 1 v 7 have already played so they downfloat.

SM5 gives a full pairing of:


1 Peter Froehlich ( 4.5) - Dejan Antic ( 5.5) 5- 2
2 Gary Lane ( 4 ) - Roland Schmaltz ( 5 ) 4- 1
3 Moulthun Ly ( 4.5) - Domagoj Dragicevic ( 4 ) 7- 11
4 Felix Klein ( 4 ) - Ben Lazarus ( 4 ) 8- 13
5 Brian Thomas ( 4 ) - Gene Nakauchi ( 4 ) 14- 15
6 Jesse Noel Sales ( 4 ) - Yi Liu ( 3.5) 6- 29
7 David Lovejoy ( 3.5) - Stephen Solomon ( 3.5) 16- 3
8 Dusan Stojic ( 3 ) - Jonas Muller ( 3 ) 10- 21
9 Michael Van Pelt ( 3 ) - Lee Jones ( 3 ) 24- 12
10 Oleg Korenevski ( 2.5) - Sam Grigg ( 3 ) 20- 22
11 Mark Stokes ( 2.5) - Alex Stahnke ( 2.5) 23- 27
12 John Alkin ( 2.5) - Nenad Chelebichanin ( 2.5) 26- 25
13 Jessica Kinder ( 2.5) - Francisco Chung ( 2 ) 28- 18
14 Tony Weller ( 2 ) - Narelle Szuveges ( 2 ) 17- 19
15 Kieran Lyons ( 2 ) - Joe Delamstro ( 1.5) 30- 34
16 Edsil Dilla ( 1.5) - Roy Cristuta ( 1.5) 9- 32
17 David Cigelj ( 1 ) - Laz Sretenovic ( 1 ) 33- 31

Bill Gletsos
29-10-2007, 12:03 AM
I think it is.


S1: S2:

2 5
1 7

Now 2 has played 1 and 7, so that only leaves 5 v 2. 1 v 7 have already played so they downfloat.Just stating it does not make it so.
You havent made the slightest attempt to show how that fits in with the section C pairing rules.

Get back to me when you can. :hand:

Bartolin
29-10-2007, 01:19 AM
No that clearly isnt the case as 5 wouldnt play 1 in that case.
At a quick glance what appears to have happened with SP is the following.

S1 = 2 and 1 and S2 = 5 and 7
p = 2, w = 3, b = 1, q = 2 and x = 1

Note that it is impossible to get p = 2, no matter what you do so you end up falling down to C14 where p is reduced by 1 to 1 and x is also reduced by 1 to 0

2 v 5 violates the x = 0 condition
In fact the only way to get the x = 0 condition from S1 and S2 is 5 v 1
This means 2 and 7 float down to the 4 point group.

It is not immediately obvious how SM5 gets the 2 v 5 pairing according to the C pairing rules.

I guess it has something to do with B3. The pairing 2-4 instead of 5-2 could be seen as a violation of B3 for the leading player (player 2). And since B3 is more important than B4, SM5 could have decided to accept 5-2 even though it violates B4.

[There is a lot of discussion about the interpretation of B3 in the thread "Another sp error" http://chesschat.org/showthread.php?t=6619]

By the way, intuitively I think that a violation of a mild color preference (as will happen when accepting 5-2) shouldn't be a big problem. The rules introduced the concept of "strong" vs. "mild" color preferences in A7 (b and c). But somehow this concept doesn't seem to be used in the other part of C. I think that a violation of a mild color preference (as in the case in question) is a worse reason to move the leading player down than a violation of a strong color preference would be. But this consideration doesn't seem to play a role. Or do I overlook something?

Bill Gletsos
29-10-2007, 01:28 AM
I guess it has something to do with B3.No it doesnt.

The pairing 2-4 instead of 5-2 could be seen as a violation of B3 for the leading player (player 2). And since B3 is more important than B4, SM5 could have decided to accept 5-2 even though it violates B4.You are pairing the players S1 = 2, 1 and S2 = 5, 7 not worrying about the next score group.
It is clear that the value of p and of x is of prime importance.

Bartolin
29-10-2007, 01:59 AM
You are pairing the players S1 = 2, 1 and S2 = 5, 7 not worrying about the next score group.

I'm afraid, I don't fully understand this.

Wouldn't that mean, that the implications of B3 change dramatically once p is reduced by C14? If I don't have to care about players who are about to downfloat, it would suddenly be best to pair 5 and 7, once p becomes 1. In this case the score difference of the two players paired would be minimal (zero). So, 5-7 would be the best pair with respect to B3 and it should be taken since B3 is more important than B4. [Note that C8 allows exchanges since the group is considered homogenous according to A3. Furthermore B2 isn't in effect since this is the last round.]

But to pair 5-7 and move 1 and 2 down would be just ridiculous.

Once again, I seem to be unable to grasp how B3 is incorporated in the pairing procedure from C.

drbean
29-10-2007, 10:35 AM
2 : 19,9,7,1,3,15 WBWBWB D 5.5

1 : 18,10,6,2,8,7 BWBWBW D 5

5 : 22,14,11,8,28,3 BWBWBB 4.5
7 : 24,16,2,14,22,1 BWBWBB U 4.5

4 : 21,11,19,13,15,12 WBWBWB 4
6 : 23,13,1,15,19,22 WBWBWB 4
8 : 25,15,3,5,1,11 WBWBWB 4
11 : 28,4,5,3,10,8 BWWBBW u 4
13 : 30,6,26,4,18,10 BWBWBW 4
14 : 31,5,27,7,24,25 WBWBWB D 4
15 : 32,8,34,6,4,2 BWBWBW U 4

Now it is abundently clear that 2 should play 5 and 1 and 7 join the 4 point score group.



Can we see the whole pairing table? Recursive C12 pairing of heterogeneous brackets may force backtracking that requires unintuitive pairings going back several brackets, the same way that C13 and C14 did in http://chesschat.org/showthread.php?t=6619

Garvinator
29-10-2007, 11:03 AM
Can we see the whole pairing table? Recursive C12 pairing of heterogeneous brackets may force backtracking that requires unintuitive pairings going back several brackets, the same way that C13 and C14 did in http://chesschat.org/showthread.php?t=6619

2 : 19,9,7,1,3,15 WBWBWB D 5.5

1 : 18,10,6,2,8,7 BWBWBW D 5

5 : 22,14,11,8,28,3 BWBWBB 4.5
7 : 24,16,2,14,22,1 BWBWBB U 4.5

4 : 21,11,19,13,15,12 WBWBWB 4
6 : 23,13,1,15,19,22 WBWBWB 4
8 : 25,15,3,5,1,11 WBWBWB 4
11 : 28,4,5,3,10,8 BWWBBW u 4
13 : 30,6,26,4,18,10 BWBWBW 4
14 : 31,5,27,7,24,25 WBWBWB D 4
15 : 32,8,34,6,4,2 BWBWBW U 4

3 : 20,12,8,11,2,5 BWBWBW 3.5
16 : 33,7,29,21,25,19 WBWBWB D 3.5
29 : 12,20,16,22,26,28 BWBWBW 3.5

10 : 27,1,24,25,11,13 WBWBWB d 3
12 : 29,3,25,27,21,4 WBWBWW 3
21 : 4,28,30,16,12,18 BWBWBW 3
22 : 5,31,9,29,7,6 WBWBWW 3
24 : 7,33,10,34,14,30 WBBWBW 3

20 : 3,29,33,17,31,27 WBWBWB 2.5
23 : 6,30,32,26,-,33 BWBW-B D 2.5
25 : 8,32,12,10,16,14 BWBWBW U 2.5
26 : 9,17,13,23,29,32 WBWBWB 2.5
27 : 10,18,14,12,34,20 BWBWBW 2.5
28 : 11,21,17,9,5,29 WBWBWB 2.5

17 : 34,26,28,20,-,31 BWBW-B d 2
18 : 1,27,31,33,13,21 WBWBWB 2
19 : 2,34,4,30,6,16 BWBWBW U 2
30 : 13,23,21,19,9,24 WBWBWB 2

9 : 26,2,22,28,30,34 BWBWBW 1.5
32 : 15,25,23,31,33,26 WBWBWW u 1.5
34 : 17,19,15,24,27,9 WBWBWB 1.5

31 : 14,22,18,32,20,17 BWBWBW 1
33 : 16,24,20,18,32,23 BWBWBW Ud 1

drbean
30-10-2007, 12:37 AM
2 : 19,9,7,1,3,15 WBWBWB D 5.5

1 : 18,10,6,2,8,7 BWBWBW D 5

5 : 22,14,11,8,28,3 BWBWBB 4.5
7 : 24,16,2,14,22,1 BWBWBB U 4.5

4 : 21,11,19,13,15,12 WBWBWB 4
6 : 23,13,1,15,19,22 WBWBWB 4
8 : 25,15,3,5,1,11 WBWBWB 4
11 : 28,4,5,3,10,8 BWWBBW u 4
13 : 30,6,26,4,18,10 BWBWBW 4
14 : 31,5,27,7,24,25 WBWBWB D 4
15 : 32,8,34,6,4,2 BWBWBW U 4

3 : 20,12,8,11,2,5 BWBWBW 3.5
16 : 33,7,29,21,25,19 WBWBWB D 3.5
29 : 12,20,16,22,26,28 BWBWBW 3.5

10 : 27,1,24,25,11,13 WBWBWB d 3
12 : 29,3,25,27,21,4 WBWBWW 3
21 : 4,28,30,16,12,18 BWBWBW 3
22 : 5,31,9,29,7,6 WBWBWW 3
24 : 7,33,10,34,14,30 WBBWBW 3

20 : 3,29,33,17,31,27 WBWBWB 2.5
23 : 6,30,32,26,-,33 BWBW-B D 2.5
25 : 8,32,12,10,16,14 BWBWBW U 2.5
26 : 9,17,13,23,29,32 WBWBWB 2.5
27 : 10,18,14,12,34,20 BWBWBW 2.5
28 : 11,21,17,9,5,29 WBWBWB 2.5

17 : 34,26,28,20,-,31 BWBW-B d 2
18 : 1,27,31,33,13,21 WBWBWB 2
19 : 2,34,4,30,6,16 BWBWBW U 2
30 : 13,23,21,19,9,24 WBWBWB 2

9 : 26,2,22,28,30,34 BWBWBW 1.5
32 : 15,25,23,31,33,26 WBWBWW u 1.5
34 : 17,19,15,24,27,9 WBWBWB 1.5

31 : 14,22,18,32,20,17 BWBWBW 1
33 : 16,24,20,18,32,23 BWBWBW Ud 1

Thanks for this pairing table. It smoked out more bugs in Games::Tournament::Swiss. Without the complete pairing table it is not possible for it to pair even the top tables.

It is now pairing the first tables:



Round 7: 2 (5.5), 1 (5), 5 7 (4.5), 4 6 8 11 13 14 15 (4), 3 16 29 (3.5), 10 12 21 22 24 (3), 20 23 25 26 27 28 (2.5), 17 18 19 30 (2), 9 32 34 (1.5), 31 33 (1),
C1, NOK. 2 only member in 1 (5.5).
C1, Floating 2 down to 2 (5)
C1, [1] & [2] 1 2
C1, Bracket 1 (5.5) dissolved. Pairing Bracket 2
C1, NOK. 1 2 B1a/B2a incompatible in 2 (5)
C1, Floating 1 2 down to 3 (4.5)
C1, [2] & [3] 5 7 1 2
C1, Bracket 2 (5) dissolved. Pairing Bracket 3
C1, NOK. 7 B1a/B2a incompatible in 3 (4.5)
C1, Floating 7 down to 4 (4)
C1, [3] 5 1 2 & [4] 4 6 8 11 13 14 15 7
C2, x=0
C3, p=1. Homogeneous.
C4, S1: 2 & S2: 1 5
C5, ordered: 2
& 1 5
C6PAIRS, B1a: table 1 NOK
C7, 5 1
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, exchange a in 3
C8, 1, 2 5
C5, ordered: 1
& 2 5
C6PAIRS, B1a: table 1 NOK
C7, 5 2
C6PAIRS, B5Down, table 1: 2 NOK. Floated Down 1 rounds ago
C7, last transposition
C8, last S1,S2 exchange in 3
C9, Dropping B6 for Downfloats
C4, S1: 2 & S2: 1 5
C5, ordered: 2
& 1 5
C6PAIRS, B1a: table 1 NOK
C7, 5 1
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, exchange a in 3
C8, 1, 2 5
C5, ordered: 1
& 2 5
C6PAIRS, B1a: table 1 NOK
C7, 5 2
C6PAIRS, B5Down, table 1: 2 NOK. Floated Down 1 rounds ago
C7, last transposition
C8, last S1,S2 exchange in 3
C9, Dropping B5 for Downfloats
C4, S1: 2 & S2: 1 5
C5, ordered: 2
& 1 5
C6PAIRS, B1a: table 1 NOK
C7, 5 1
C6PAIRS, B4: x=0, table 1 NOK
C7, last transposition
C8, exchange a in 3
C8, 1, 2 5
C5, ordered: 1
& 2 5
C6PAIRS, B1a: table 1 NOK
C7, 5 2
C6PAIRS, B56: OK.
C6PAIRS, 1 paired. OK
C6PAIRS, E1 5&1
C6OTHERS, Floating remaining 2 Down. [3] 1 5 & [4] 4 6 8 11 13 14 15 7 2


In Bracket 3, 7 is floated down because it has played 1 and 2 and it can't play 5 because they both have an Absolute preference for White.

5 and 2 both have a preference for White, but 5 can only be paired with the remaining 1, with a preference for Black, after B5 for downfloats is waived for 2. 2 already downfloated in the previous round, so it can't downfloat again until this is done.

Am I correct here? Is this the proper way to carry out a B5 check? It seems to be the same sort of logic as here: http://chesschat.org/showpost.php?p=142260&postcount=158

Garvinator
30-10-2007, 01:43 AM
In Bracket 3, 7 is floated down because it has played 1 and 2 and it can't play 5 because they both have an Absolute preference for White.

5 and 2 both have a preference for White, but 5 can only be paired with the remaining 1, with a preference for Black, after B5 for downfloats is waived for 2. 2 already downfloated in the previous round, so it can't downfloat again until this is done.

Am I correct here? Is this the proper way to carry out a B5 check? It seems to be the same sort of logic as here:
Ok I am going to have to admit that I dont understand this and a few previous issues. I have asked another arbiter regarding all this for his opinion and awaiting reply.

Why are you worried about float status ie B5, B6?
Round 7 is the final round, so B2, B5 and B6 don't apply.

Now I commented previously that I thought it was obvious that 5 v 2 is the correct pairing and 1 and 7 are downfloated. I did not refer to all the x's, p's etc for the following reason.

In the 2,1,5,7 group. 2 has played 1 and 7. This means 2 only has 5 as a legal pairing. 2 is the highest scorer and so is paired first. I would suspect that this is how SM5 has paired this group. I can now see how 5 v 1 is possible as it is a better colour match.

Bill, I did not quite understand your post. The main point I didn't understand was, what do you believe the pairings should be?

I just put the 5 v 1 pairing into SM5 and then auto paired the rest of the pairings and it has produced the same pairings as SP. Very interesting.

So the question comes down to, what is the correct pairing: 5 v 2 or 5 v 1? SM5 says 5 v 2, SP says 5 v 1.

I am relieved that I did not manual pair this round in a rush, as the matter is not so clear cut.

Bartolin
30-10-2007, 02:16 AM
In Bracket 3, 7 is floated down because it has played 1 and 2 and it can't play 5 because they both have an Absolute preference for White.

We have a problem here, because the round in question is the last round. According to part B of the pairing rules, B2 doesn't apply for the last round for players with a score of over 50 %. But that's the case with player 7. Therefore you cannot move player 7 down that easily.

Otherwise your (or Games::Tournament::Swiss's) reasoning would be correct, I think.

Bartolin
30-10-2007, 02:45 AM
Thanks for this pairing table. It smoked out more bugs in Games::Tournament::Swiss. Without the complete pairing table it is not possible for it to pair even the top tables.

This is a bit off topic, but ... Is there an easy way to feed Games::Tournament::Swiss with a pairing table, without providing a list of players (aka "league.yaml")? I guess, I could just create such a list with fake entries ("name: Player_1, rating: 5000, name: Player_2, rating: 4900, etc."). But maybe this step can be avoided somehow?

Garvinator
30-10-2007, 08:51 AM
I now understand the situation with the top board pairings, I partly understood it previously, but it just seemed wrong to not paired the person on the highest score with his first legal opponent.

So, if we do have this correct, then why has SM5 paired 5 v 2 when it has been right so often before?

drbean
30-10-2007, 01:35 PM
This is a bit off topic, but ... Is there an easy way to feed Games::Tournament::Swiss with a pairing table, without providing a list of players (aka "league.yaml")? I guess, I could just create such a list with fake entries ("name: Player_1, rating: 5000, name: Player_2, rating: 4900, etc."). But maybe this step can be avoided somehow?

Hey, I guess the arbiters aren't interested in the mechanics of the programming, so probably don't particularly want such messages here, but, quick answer, look at the start of http://cpansearch.perl.org/src/DRBEAN/Games-Tournament-Swiss-0.14/t/29682_2.t, my test of whether Games::Tournament::Swiss is pairing the http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm tournament correctly. There a list of Contestant objects is created. You can write the code to do this in a couple of lines. The contestants get arbitrary names, ratings and titles of course.

It would be useful to have a little script that took just a pairing table and produced pairings and a record of how the pairings were produced.

Bartolin
30-10-2007, 06:40 PM
I now understand the situation with the top board pairings, I partly understood it previously, but it just seemed wrong to not paired the person on the highest score with his first legal opponent.

So, if we do have this correct, then why has SM5 paired 5 v 2 when it has been right so often before?

I'm still not convinced that 5 v 1 is the correct pairing.

If players 2 and 1 had the same score (e.g. if player 2 had only 5 points as well), then 5 v 1 looks okay. Following the analysis of Bill Gletsos (http://chesschat.org/showpost.php?p=172327&postcount=2) 5 v 1 would be preferable to 5 v 2, because the latter violates B4. Is there an easy way run SM5 with a manipulated pairing table like that? I guess, it would agree with 5 v 1 as well.

But in fact player 2 has 5.5 points. As pointed out in other discussions we have to check for all relative criteria B3-B6 before accepting pairings of a score bracket (via C6). In this case (since it's the last round) we can ignore B6 and B5. That leaves us with B3 and B4. As pointed out, B4 favours 5 v 1. So the only reason to choose 5 v 2 instead of 5 v 1 could be B3. And indeed 5 v 2 minimizes the score difference for the leading player 2.

Actually I'm not sure what B3 means in a situation like this:

S1 = 2 and 1 and S2 = 5 and 7
p = 1, and x = 0

Following procedure C one gets:


5 v 2; 1 and 7 down -> violates B4, score difference between paired players 1.0
(after transposition): 5 v 1; 2 and 7 down -> B4 fulfilled, score difference between paired players 0.5, leading player gets opponent with 4 points
(after exchange): 7 v 5; 1 and 2 down -> violates B4, score difference between paired players 0.0, leading player gets opponent with 4 points
There aren't any other possibilities

If B3 really only applies to the actual pairings, the 3rd try seems to be best.

If B3 takes into account the score differences of downfloated players and if it gives special weight to the leading player, then the first try seems to be best.

Only if B3 is ignored (or is interpreted in a way that the first and the second try are equivalent -- but I don't see how to interprete it like that) the second is the best.

However, I don't believe that B3 is just irrelevant or has no meaning at all in this situation.

Any thoughts?

Bill Gletsos
30-10-2007, 07:51 PM
I'm afraid, I don't fully understand this.

Wouldn't that mean, that the implications of B3 change dramatically once p is reduced by C14? If I don't have to care about players who are about to downfloat, it would suddenly be best to pair 5 and 7, once p becomes 1. In this case the score difference of the two players paired would be minimal (zero). So, 5-7 would be the best pair with respect to B3 and it should be taken since B3 is more important than B4. [Note that C8 allows exchanges since the group is considered homogenous according to A3. Furthermore B2 isn't in effect since this is the last round.]

But to pair 5-7 and move 1 and 2 down would be just ridiculous.

Once again, I seem to be unable to grasp how B3 is incorporated in the pairing procedure from C.This is a very good questioning post and highlights the ambiguity of the rules.

Bill Gletsos
30-10-2007, 07:55 PM
I'm still not convinced that 5 v 1 is the correct pairing.

If players 2 and 1 had the same score (e.g. if player 2 had only 5 points as well), then 5 v 1 looks okay. Following the analysis of Bill Gletsos (http://chesschat.org/showpost.php?p=172327&postcount=2) 5 v 1 would be preferable to 5 v 2, because the latter violates B4. Is there an easy way run SM5 with a manipulated pairing table like that? I guess, it would agree with 5 v 1 as well.

But in fact player 2 has 5.5 points. As pointed out in other discussions we have to check for all relative criteria B3-B6 before accepting pairings of a score bracket (via C6). In this case (since it's the last round) we can ignore B6 and B5. That leaves us with B3 and B4. As pointed out, B4 favours 5 v 1. So the only reason to choose 5 v 2 instead of 5 v 1 could be B3. And indeed 5 v 2 minimizes the score difference for the leading player 2.

Actually I'm not sure what B3 means in a situation like this:

S1 = 2 and 1 and S2 = 5 and 7
p = 1, and x = 0

Following procedure C one gets:


5 v 2; 1 and 7 down -> violates B4, score difference between paired players 1.0
(after transposition): 5 v 1; 2 and 7 down -> B4 fulfilled, score difference between paired players 0.5, leading player gets opponent with 4 points
(after exchange): 7 v 5; 1 and 2 down -> violates B4, score difference between paired players 0.0, leading player gets opponent with 4 points
There aren't any other possibilities

If B3 really only applies to the actual pairings, the 3rd try seems to be best.

If B3 takes into account the score differences of downfloated players and if it gives special weight to the leading player, then the first try seems to be best.

Only if B3 is ignored (or is interpreted in a way that the first and the second try are equivalent -- but I don't see how to interprete it like that) the second is the best.

However, I don't believe that B3 is just irrelevant or has no meaning at all in this situation.

Any thoughts?Yes.
This is an excellent post. :clap: :clap: :clap:

Garvinator
31-10-2007, 02:43 AM
Is there an easy way run SM5 with a manipulated pairing table like that? I guess, it would agree with 5 v 1 as well.
SM5 pairs 5 v 1 once player 2 is reduced to 5 points.

Denis_Jessop
31-10-2007, 07:40 PM
I have had a look at this matter from the point of view so far only of the pairing of the score bracket containing players 2, 1, 5 and 7.

It seems to me that some of the arguments and observations fail to give effect to all of the relevant provisions in their entirety.

First, I observe again that the FIDE Swiss Rules aim to set out a legislative scheme to enable the pairing of players in a Swiss Tournament. In interpreting this scheme the object must be to find ways to make it work and not to find ways in which it may not work. Moreover, all of the provisions must be read together and made to work together as far as possible. Thus in, say, interpreting B3, one cannot just look at it in isolation. For present purposes Part A deals with definitions, Part B with Pairing Criteria and Part C with Pairing Procedures: that is, each deals with a different aspect of the pairing process and all must be read together to achieve the pairings for, say, a round.

Looking first at B3 which seems to be giving Bartolin such problems, I point out that it uses the words "... two players paired against each other...". It is the only para of Part B to use those words. Its operation must be either to a particular pairing or to the overall pairings. The latter approach is virtually impracticable. Nor is it consistent with the restrictive reference to "two players". Where it applies to a particular pairing, the question must be asked "how did that pairing come about?" The answer can only be "upon the application of the Pairing Procedures in Part C".

Thus Bill's initial analysis is right in that the pairing of 5 and 1 is the only pairing that satisfies part C (x=0; p=1) and so must be made irrespective of B3. Incidentally, the pairing of 5 and 1 is better than that of 5 and 2 when B3 is applied (even though it is not here applicable) as the score difference of 5 and 1 is 0.5 and that of 5 and 2 is 1. As Bartolin points out, the pairing of 5 and 7 is even better (a difference of 0) but, like that of 5 and 2, that pairing is not a valid pairing under C and so the players cannot be said to have been "paired against each other" for the purposes of B3.

Another aspect of B that should not be overlooked is that B3+ are not mandatory. Thus B3 itself says "...should be as small as possible..." which contains two discretions - "should" (not must) and "as possible". Also the introductory words to the Relative Criteria in B say "They should be fulfilled as much as possible." again leaving room for departure. By contrast, the provisions of Part C are quite specific and so, if there is a situation in which there is a clear inconsistency between B and C, I believe that C prevails. But I do not believe that, on the correct interpretation of B3, such an inconsistency exists in the present case.

One possibly unintended (but perhaps not)* result of this is that, while B gives preference to equality of scores over colour preference by putting B3 before B4, the application of C can turn this around as it works so as to give colour preference the priority, at least in the present case. In this regard, it should not be overlooked that C only does this upon application of "x", "p" and so on that are defined in Part A, further emphasising how the provisions of one Part or provision in it cannot be read in isolation.


I am rather puzzled by Bartolin's raising the issue of players 1 and 2 being treated as being on the same score and even more puzzled by the results of Garvin's experiment with SM5 as there is no justification at all under the FIDE Rules for doing that, nor, I think, should it affect the situation as it has no bearing on the values of x or p or, as far as I can see, on the application of B3 were it relevant .

In short, though the rules are not as clear as they might be ( and I would not like to be the person trying to make them clearer), my view is that, if carefully analysed, they work - at least in the present case.

DJ

* there is often great argument over which element should have preference so the FIDE drafters, if aware of what they were doing, may have been having a bob each way:doh:

Bartolin
31-10-2007, 08:55 PM
Thanks for this analysis! There are some points I'll have to think about.


I have had a look at this matter from the point of view so far only of the pairing of the score bracket containing players 2, 1, 5 and 7.

As a sidenote: I also did only look at the first four players.



Looking first at B3 which seems to be giving Bartolin such problems, I point out that it uses the words "... two players paired against each other...". It is the only para of Part B to use those words. Its operation must be either to a particular pairing or to the overall pairings. The latter approach is virtually impracticable. Nor is it consistent with the restrictive reference to "two players". Where it applies to a particular pairing, the question must be asked "how did that pairing come about?" The answer can only be "upon the application of the Pairing Procedures in Part C".

Okay, I understand your reasoning, though I'm still not fully convinced.

But to another point: I think we had worked out earlier, that in C6 (and thus "upon the application of the Pairing Procedures in Part C") we have to check pairings with their compliance not only with B1 and B2, but also with B3-B6 (cmp. http://chesschat.org/showpost.php?p=167088&postcount=22 ).

It seems to me, that B3 clearly refers to a comparision of different possible pairings. As you said, all of those pairings have to be gained by following procedure C.


Thus Bill's initial analysis is right in that the pairing of 5 and 1 is the only pairing that satisfies part C (x=0; p=1) and so must be made irrespective of B3. Incidentally, the pairing of 5 and 1 is better than that of 5 and 2 when B3 is applied (even though it is not here applicable) as the score difference of 5 and 1 is 0.5 and that of 5 and 2 is 1. As Bartolin points out, the pairing of 5 and 7 is even better (a difference of 0) but, like that of 5 and 2, that pairing is not a valid pairing under C

But why is 5 vs 7 not a valid pairing under C?

As I wrote above (following the analysis of Bill Gletsos) we arrive at the following situation after reaching C14 the first time:

S1 = 2 and 1 and S2 = 5 and 7
p = 1, and x = 0

Now after one transposition according to C7 we gain 5 v 1, downfloating 2 and 7. We return to C6 with this pairing.

My question is now: Is it correct to stop at this point? Clearly the pairing is in compliance with B1 and B4. B2, B5 and B6 are irrelevant because of the last round rule.

But what about B3? I think B3 requires us to compare the actual pairing with some other pairings. But why should we only compare it with the pairings we came along until this point? Why don't we compare it with the pairings that would come next when following procedure C?

If we don't stop at this point, we have to make an exchange next (C8). After that, we suddenly get the pairing 7 v 5 (1 and 2 down). This is better according to B3 (at least to your interpretation of B3, regarding only the two players paired against each other).

It's true, that this pairing violates B4, but since B3 is more important than B4, maybe we just have to go on. The next relevant step in procedure C would be to increase x by 1 according to C11.

In the next cycle through C 6-8 suddenly 2 v 5 and 7 v 5 would become valid pairings. And since there are no more problems with x=0, 7 v 5 would look like the best pairing.

Again, I think the critical question is whether B3 requires us to compare the actual pairing only with the pairings we came along before or also with those which would arise after continuing with procedure C.


I am rather puzzled by Bartolin's raising the issue of players 1 and 2 being treated as being on the same score and even more puzzled by the results of Garvin's experiment with SM5 as there is no justification at all under the FIDE Rules for doing that, nor, I think, should it affect the situation as it has no bearing on the values of x or p or, as far as I can see, on the application of B3 were it relevant .

I didn't want to suggest that one should do such a thing like modifying the pairing table. I only asked myself why SM5 paired 2 v 5 instead of 1 v 5 (as SP did). I hypothesized that SM5 did that because it applied B3 in a certain way (namely trying to minimize the score difference of the leading player) whereas SP didn't apply B3 in that way. To test my hypothesis I suggested to run the same pairing table with exactly one difference: player 2 on 5 points instead of 5.5. That way, the two pairings 2 v 5 and 1 v 5 would make no difference regarding to the score difference of the leading player(s).

I think it's interesting that SM5 converted to the 1 v 5 pairing in this case. But once again, that was just a try to test my hypothesis.

Bartolin
31-10-2007, 09:21 PM
Again, I think the critical question is whether B3 requires us to compare the actual pairing only with the pairings we came along before or also with those which would arise after continuing with procedure C.

Sorry for replying to myself. But I just wanted to add, that the sentence from part B (regarding the relative criteria B3-B6


To comply with these criteria, transpositions or even exchanges may be applied, but no player should be moved down to a lower score bracket.

seems to imply, that we are allowed to apply further transpositions and exchanges to find better pairings. That makes me think that we have a good reason to compare the actual pairing with later pairings as well.

Denis_Jessop
31-10-2007, 09:50 PM
Isn't the answer, briefly, that once you apply the transposition that gives the valid paring of 5 and 1, C6 deems the pairing of that score bracket to be complete and 2 and 7 then form S1 of the score bracket containing all the players on 4 as S2? There is no room to apply the exchange that would allow the pairing of 5 and7. That is, there is nothing in C that allows you to unpair a complete pairing on the ground that it may violate B3 nor, in this case, does the 5 and 1 pairing necessarily violate B3 because of its non-mandatory (=discretionary) wording.

The danger with the "looking ahead approach" is that what is then being done in effect is to make a provisional draw for the whole field every time you pair a score bracket; but that is not how Part C is structured. And if one were to do this with B3 then presumably the same would have to be done with all of B. In my view that clearly is not how the scheme is meant to work. The way C prescribes it, one pairs downwards according to the provisions of C and then one starts unravelling from the bottom only if a problem reveals itself there.

Incidentally, to get a little off-topic, the rules are still interesting in relation to the question about whether any Part B criteria apart from B1 and B2 apply when pairing under C6. I don't recall anyone having mentioned the introductory words to Part C that say "...apply the following procedures..until an acceptable pairing is obtained. Afterwards the colour allocation rules (E) are used..." . Where does B4 stand in all of this? One could try to draw a distinction between "colour preference" and "colour allocation" but for the fact that the colour allocation rules in E rely almost entirely on colour preferences.

DJ

Garvinator
31-10-2007, 09:50 PM
I really think that this one should be thrown to the fide pairings committee for a definitive answer. A fide endorsed program (and a program that has been tested previously on here as being the most accurate) is saying that 5 v 2 is the correct pairing, but depending on who is answering and when, we are about 50-50 on which is the 'correct' pairing.

I would say that 5 v 2 is most 'in the spirit' (dangerous words I know) of the swiss system.

Bartolin
31-10-2007, 11:15 PM
Isn't the answer, briefly, that once you apply the transposition that gives the valid paring of 5 and 1, C6 deems the pairing of that score bracket to be complete and 2 and 7 then form S1 of the score bracket containing all the players on 4 as S2? There is no room to apply the exchange that would allow the pairing of 5 and7. That is, there is nothing in C that allows you to unpair a complete pairing on the ground that it may violate B3

Well, indeed there is nothing like that in C. But wouldn't that argumentation lead to the conclusion that B3 has no use at all? At the moment I don't see, how B3 can ever influence the pairings if we argue that we are not allowed to unpair a complete pairing in order to check for better B3 compliant pairings.

And furthermore: Doesn't the sentence from B (regarding the relative criteria) "To comply with these criteria, transpositions or even exchanges may be applied, but no player should be moved down to a lower score bracket." explicitly allows us to apply exchanges? If we are allowed to do so to comply with B6, B5 and B4 why not to comply with B3 (which is "more important")?

I guess, your answer will be: "Because Part C doesn't say so explicitly, whereas it has rules for B4-B6." I'm inclined to think that we have to read C6 as if it refers to B3-B6 as well -- and thereby allows us to test further transpositions, exchanges and increases of x to get the "right" pairing.


... nor, in this case, does the 5 and 1 pairing necessarily violate B3 because of its non-mandatory (=discretionary) wording.

I still think that it violates B3, if it has to be applied to the players actually paired and if we are allowed to make further transpositions and exchanges.


The danger with the "looking ahead approach" is that what is then being done in effect is to make a provisional draw for the whole field every time you pair a score bracket; but that is not how Part C is structured. And if one were to do this with B3 then presumably the same would have to be done with all of B. In my view that clearly is not how the scheme is meant to work. The way C prescribes it, one pairs downwards according to the provisions of C and then one starts unravelling from the bottom only if a problem reveals itself there.

I agree on the point, that Part C isn't meant to "look ahead" for the whole field. But maybe we are allowed to look ahead for the players in the actual score bracket? That is, we can stop when we see that moving a player down would result in a double downfloat for this player. And maybe we can also stop when we see, that the score difference for the player to be downfloated will become very high (as in this case, when we move the leading player down).

Finally, I still wonder with ggrayggray, why two FIDE approved programs end with two different pairings in such a simple case (only four players).

Bartolin
31-10-2007, 11:34 PM
Incidentally, to get a little off-topic, the rules are still interesting in relation to the question about whether any Part B criteria apart from B1 and B2 apply when pairing under C6. I don't recall anyone having mentioned the introductory words to Part C that say "...apply the following procedures..until an acceptable pairing is obtained. Afterwards the colour allocation rules (E) are used..." . Where does B4 stand in all of this? One could try to draw a distinction between "colour preference" and "colour allocation" but for the fact that the colour allocation rules in E rely almost entirely on colour preferences.

Maybe I missunderstood the rules or your question (or both), but ...

I understand those introductory words to Part C in the following way: With Part C we determine the pairings (in the sense "who plays against whom"). We don't determine who gets white and who gets black in Part C. Only after determining all single pairings we go to Part E and actually allocate the colors. (Applying the procedure in Part C makes sure that we don't cause serious problems by doing so.)

By the way, I finally found out, where the concept of "strong" vs. "mild" color preferences from A7 is used again. It's in E2! ;-)

Garvinator
01-11-2007, 12:20 AM
Finally, I still wonder with ggrayggray, why two FIDE approved programs end with two different pairings in such a simple case (only four players).Swiss Perfect 98 (that is the full title of the program) is not fide approved. It is an acf approved pairing program. As you can guess from the full title, it was developed in 1998. No more updates are provided by the program author, so I would not be surprised if it is out of date regarding the current pairing rules.

Btw, just because something is fide approved does not mean it is any good :whistle:

Bill Gletsos
01-11-2007, 02:15 AM
I really think that this one should be thrown to the fide pairings committee for a definitive answer. A fide endorsed program (and a program that has been tested previously on here as being the most accurate) is saying that 5 v 2 is the correct pairing, but depending on who is answering and when, we are about 50-50 on which is the 'correct' pairing.I have to admit to being a little mischevious in this thread.

My post #2 was simply there to make people think about the situation and not as a statement that SM5 was incorrect.

In fact to that end I was very specific with my wording when I said:

It is not immediately obvious how SM5 gets the 2 v 5 pairing according to the C pairing rules.
as opposed to saying:

It is not immediately obvious to me how SM5 gets the 2 v 5 pairing according to the C pairing rules.

You in fact did pickup on my lack of a definitive statement in my post #2 when you said in post #12

Bill, I did not quite understand your post. The main point I didn't understand was, what do you believe the pairings should be?

I also deliberately threw a spanner in the works with my post #7 to try and make people really question the pairings and in particluar what I had said in post #2. I wanted people to think and challenge what I said in post #2. ;)

As such I was very pleased when Bartolin in post #8 immediately questioned my post #7 and I acknowledged this along with the fact I thought his post #17 was an excellent post.

His logic is still a little unclear (flawed?) as he seemingly fails to appreciate the significance of the value of x as well as not applying C11 in that post.

He nearly gets is right in post #22 but gets sidetracked by the 5 V 7 pairing which is clearly incorrect.

Essentially you do not stop where my post #2 stops.

After all it is clear that keeping player 2 in the score group rather than downfloating him is better in relation to B3.

You carry on thru C7 etc but do not stop when you first get the 2 V 5 pairing as this violates the x = 0 condition.
Eventually you fall down to C11 and increase x by 1 to 1 and restart.

Now you will again get to the 2 v 5 pairing but this time x = 1 so it is allowable.

Carrying on to look for any other valid pairing such as 5 V 7 is rubbish as it will result in B3 for player 2 being worse.

The correct pairing is indeed 5 V 2.

Garvinator
01-11-2007, 04:27 AM
The correct pairing is indeed 5 V 2.
So, in your opinion, Stephen's questioning of the pairings was correct, but not as simple as it first appeared :uhoh:

Bill Gletsos
01-11-2007, 09:24 AM
So, in your opinion, Stephen's questioning of the pairings was correct, but not as simple as it first appeared :uhoh:Correct.
It was entirely reasonable for him to question why the highest scoring player was not playing his highest scoring unplayed opponent.
However it is not that easy for an arbiter to very quickly determine that the published pairings were incorrect.

Denis_Jessop
01-11-2007, 11:02 AM
Correct.
It was entirely reasonable for him to question why the highest scoring player was not playing his highest scoring unplayed opponent.
However it is not that easy for an arbiter to very quickly determine that the published pairings were incorrect.

Bill

Given that you say that your analysis in post #2 was incorrect in that you now say that pairing 5 and 2 is correct and pairing 5 and 1 is not, could you now please explain to a simple country boy like me why it is so?

I don't see anything in the intervening posts that establishes the reason.

DJ

Brian_Jones
01-11-2007, 12:09 PM
Bill. Given that you say that your analysis in post #2 was incorrect in that you now say that pairing 5 and 2 is correct and pairing 5 and 1 is not, could you now please explain to a simple country boy like me why it is so? I don't see anything in the intervening posts that establishes the reason. DJ

To any experienced tournament player, 5 v 2 is as plain as the nose on your face (just a metaphor nothing personal) :)

Garvinator
01-11-2007, 12:28 PM
To any experienced tournament player, 5 v 2 is as plain as the nose on your face (just a metaphor nothing personal) :)
But sometimes obvious pairings are not always correct. As I said earlier, 5 v 2 is most in the spirit of the swiss system.

Just for those who have not read my first post in this thread. Round 7 running late, only swiss perfect available and I clicked pair.

Out comes the top boards with 5.5 playing 4 and other jumbled up score pairings. I could have attempted to confirm the pairings, but I did not want to attempt a manual pairing with players waiting around. So I just went with the SP pairings. I thought there was a much greater chance of me getting the pairings wrong than SP. Had I had SM5 available, I would have gone with those pairings.

Denis_Jessop
01-11-2007, 02:15 PM
To any experienced tournament player, 5 v 2 is as plain as the nose on your face (just a metaphor nothing personal) :)

I see that you belong to the school of intuitive pairing. I was more concerned about the rules but perhaps they don't matter ;)

DJ

Brian_Jones
01-11-2007, 02:22 PM
Just for those who have not read my first post in this thread. Round 7 running late, only swiss perfect available and I clicked pair. As you must. No fault of yours Garvin. The players cannot be kept waiting.


Had I had SM5 available, I would have gone with those pairings. This is now a job for the ACF. Nationally, we have to stop using SP98. :evil:

I have been recommending that the ACF adopt SM5 for some time now but nobody else seems prepared to support me?

Basil
01-11-2007, 03:15 PM
I see that you belong to the school of intuitive pairing. I was more concerned about the rules but perhaps they don't matter ;)

DJ
Denis, did you ever do you own advocacy? I sense you could have been rather deadly on the soap box regardless of the merit of your client's position :lol:

Garvinator
01-11-2007, 03:25 PM
This is now a job for the ACF. Nationally, we have to stop using SP98. :evil:

I have been recommending that the ACF adopt SM5 for some time now but nobody else seems prepared to support me?
I thought you were supporting Swiss Manager :P

Have you read my post one of this thread. Now that further study has been done, my thread starter was a bit strong, but the point still remains.

I think one of the issues though is cost. To get a general licence for SM5 would cost thousands I would imagine.

Compared to the current free situation with sp98, which is good enough for most general tournaments.

Of course we could go to Swis Sys :uhoh:

Denis_Jessop
01-11-2007, 05:02 PM
As you must. No fault of yours Garvin. The players cannot be kept waiting.

This is now a job for the ACF. Nationally, we have to stop using SP98. :evil:

I have been recommending that the ACF adopt SM5 for some time now but nobody else seems prepared to support me?

This is a newie on me too as I always thought you were an SM man. My attempts to get the ACF interested in a change last year were not very productive. Now, as a loyal ACF man, I am barracking for SP in this matter :wall: :evil: :D :hmm: et al.

DJ

Bartolin
02-11-2007, 04:37 AM
Bill

Given that you say that your analysis in post #2 was incorrect in that you now say that pairing 5 and 2 is correct and pairing 5 and 1 is not, could you now please explain to a simple country boy like me why it is so?

I don't see anything in the intervening posts that establishes the reason.

DJ

I know you didn't ask me, but since Bill didn't answer until now and since we discussed the pairing earlier I will try to describe the pairing process as I understand it now. I hope I get it right this time.

First the crucial point: In contrast to post #7, but in accordance with post #29, B3 does not mean that we have to take into account only the players actually paired. We have to look at the players which are about to downfloat as well. Therefore 5 v 1 and downfloating 2 is worse than 5 v 2 and downfloating 1 according to B3. The wording of B3 seems to be "suboptimal".

But now to the pairing process: Though we start with S1=(2,1), S2=(5,7), p=2 and x=1 we are not able to get two pairings (since 2 and 1 both want to play 5). Therefore we fall down to C14, reducing p to 1 and x to 0 (cmp. post #2).

Now we have S1=(2,1), S2=(5,7), p=1 and x=0.

The first pairings we have to try are 2 v 5 and 1 v 7. But 2 v 5 violates B4. Since x=0 we are not allowed to accept this pairing (yet).

After one transposition we get 2 v 7 and 1 v 5. But 1 v 5 violates B3 (as described above) -- its better to keep the leading player 2 in this score bracket. So again, we have to go on.

After one exchange we get 2 v 1 and 5 v 7. But 5 v 7 violates B3 and B4.

We can't apply more transpositions or exchanges, so we have to go on. C9 and C10 don't apply, so we and with C11, increasing x to 1 and restarting at C3.

Now we have S1=(2,1), S2=(5,7), p=1 and x=1.

The first pairing we have to try are again 2 v 5 and 1 v 7. This time x=1 and B4 is ignored. Therefore we can accept this pairing (2 v 5, moving 1 and 7 down) finally.

[In contrast to what I said earlier I now think, that we don't have to compare this pairing with later pairings for B3 compliance.]

Brian_Jones
02-11-2007, 06:58 AM
My attempts to get the ACF interested in a change last year were not very productive.

Does anybody on the ACF actually play chess?

Basil
02-11-2007, 10:51 AM
Does anybody on the ACF actually play chess?
:lol:

Denis_Jessop
02-11-2007, 02:34 PM
Does anybody on the ACF actually play chess?


The answer to this depends on the definition of "play".

For example, I think that they all know the moves and the Laws of Chess unlike some posters in another place.

In that sense they all play. But whether they actually makes moves in anger or in competition on a board against a human opponent is a question that is not beyond doubt.

As for knowing how the FIDE Swiss Rules work, it is questionable whether anyone really knows though some know more than others. That is what makes a thread like this such fun. :cool:

I must agree that the temptation to say that "that looks like a good draw" in the sense that it has a certain artistic elegance and balance is very strong. It also saves a lot of work if you are trying to verify a computer-made draw manually. ;)

I could ask what is the point of the question when I recall that CEOs in other sports, not to mention business, just move around regardless on the supposition that if you can manage an ice cream factory you can manage a law firm. :D


DJ

Denis_Jessop
02-11-2007, 03:01 PM
I know you didn't ask me, but since Bill didn't answer until now and since we discussed the pairing earlier I will try to describe the pairing process as I understand it now. I hope I get it right this time.

First the crucial point: In contrast to post #7, but in accordance with post #29, B3 does not mean that we have to take into account only the players actually paired. We have to look at the players which are about to downfloat as well. Therefore 5 v 1 and downfloating 2 is worse than 5 v 2 and downfloating 1 according to B3. The wording of B3 seems to be "suboptimal".

But now to the pairing process: Though we start with S1=(2,1), S2=(5,7), p=2 and x=1 we are not able to get two pairings (since 2 and 1 both want to play 5). Therefore we fall down to C14, reducing p to 1 and x to 0 (cmp. post #2).

Now we have S1=(2,1), S2=(5,7), p=1 and x=0.

The first pairings we have to try are 2 v 5 and 1 v 7. But 2 v 5 violates B4. Since x=0 we are not allowed to accept this pairing (yet).

After one transposition we get 2 v 7 and 1 v 5. But 1 v 5 violates B3 (as described above) -- its better to keep the leading player 2 in this score bracket. So again, we have to go on.

After one exchange we get 2 v 1 and 5 v 7. But 5 v 7 violates B3 and B4.

We can't apply more transpositions or exchanges, so we have to go on. C9 and C10 don't apply, so we and with C11, increasing x to 1 and restarting at C3.

Now we have S1=(2,1), S2=(5,7), p=1 and x=1.

The first pairing we have to try are again 2 v 5 and 1 v 7. This time x=1 and B4 is ignored. Therefore we can accept this pairing (2 v 5, moving 1 and 7 down) finally.

[In contrast to what I said earlier I now think, that we don't have to compare this pairing with later pairings for B3 compliance.]

Thanks very much for that. Your analysis of the pairing process seems right. I had thought from the beginning, applying the Jones Test before it was enunciated, that pairing 2 and 5 looked right.

On the other hand I don't agree with your second paragraph as stated. I don't think you can simply say that, apparently in furtherance of B3, downfloating 2 is worse than downfloating 1 unless you make a complete draw and find out which result gives overall the higher number of pairings complying with B3. That is, if you take that approach, you must have regard to all players, not just 1 and 2. I don't see B3 etc as working that way. But also I don't see that your second para is relevant as what it avers does not arise in the application of the provisions of C in any case.

DJ

Basil
02-11-2007, 03:01 PM
For example, I think that they all know the moves and the Laws of Chess unlike some posters in another place.
OK Denis - that's three. Paying $50 HCDs the lot. Your only difficulty is repeating the form next week. That's the trouble when you set the bar high.

Spiny Norman
02-11-2007, 03:50 PM
As a club president, I'd be interested to know "where do we go from here"? We use SP98 and I also find it very convenient to publish results out of SP98 by using Jon Paxman's SP2HTML tool. We'd be interested in changing if the new solution gave more correct pairings than SP98, however there are other factors for the average club, such as ease of publishing results. I know nothing about SM5 ... so:
-- what does it cost?
-- where does one buy it?
-- if we are using a non-ACF-recommended application, are there any implications (are we out on a limb if someone complains about one of its pairings?)
Further guidance and observations welcomed.

drbean
02-11-2007, 05:14 PM
After all it is clear that keeping player 2 in the score group rather than downfloating him is better in relation to B3.



I thought that there was a rule of thumb you could apply here pairing in the final round. You can only float down players which have played every other player in the bracket (assuming an even number of players), I thought. Because B2 is no longer a violation of C1 in the final round.

There is no player which has played all other players in this new first bracket. So there is no player which can be floated down.

But 3 of them have played each of the other 2, so we have 3 players all looking to play the same one player, in this case, 5.




You carry on thru C7 etc but do not stop when you first get the 2 V 5 pairing as this violates the x = 0 condition.
Eventually you fall down to C11 and increase x by 1 to 1 and restart.

Now you will again get to the 2 v 5 pairing but this time x = 1 so it is allowable.

Carrying on to look for any other valid pairing such as 5 V 7 is rubbish as it will result in B3 for player 2 being worse.

The correct pairing is indeed 5 V 2.

But 1 and 7 have already played each other and are so unpairable. And we can't float them, because we haven't reached C14 yet.

Thus, we can't accept 5&2. At least not yet.

However looking ahead, we see that 2 has no possible partners besides 5. Which of the three of 2,1 and 7 get 5 will depend on what part of the C7 cycle we are in when we reach C14, I think.

Perhaps because after C14 we restart at C1, the first player in S1, ie 2 will get 5.

And we have to float down 1 and 7.

Bill Gletsos
02-11-2007, 05:21 PM
I thought that there was a rule of thumb you could apply here pairing in the final round. You can only float down players which have played every other player in the bracket (assuming an even number of players), I thought. Because B2 is no longer a violation of C1 in the final round.

There is no player which has played all other players in this new first bracket. So there is no player which can be floated down.

But 3 of them have played each of the other 2, so we have 3 players all looking to play the same one player, in this case, 5.To me this seems totally irrlevant to the point of mine you quoted.


But 1 and 7 have already played each other and are so unpairable. And we can't float them, because we haven't reached C14 yet.

Thus, we can't accept 5&2. At least not yet.

However looking ahead, we see that 2 has no possible partners besides 5. Which of the three of 2,1 and 7 get 5 will depend on what part of the C7 cycle we are in when we reach C14, I think.

Perhaps because after C14 we restart at C1, the first player in S1, ie 2 will get 5.

And we have to float down 1 and 7.I believe most of the above is wrong.

This stems from your comment "because we haven't reached C14 yet."

It is clear from my post #2 that we did indeed hit C14.

At a quick glance Bartolin appears to have it pretty much right in his post #40.

Denis_Jessop
02-11-2007, 07:36 PM
As a club president, I'd be interested to know "where do we go from here"? We use SP98 and I also find it very convenient to publish results out of SP98 by using Jon Paxman's SP2HTML tool. We'd be interested in changing if the new solution gave more correct pairings than SP98, however there are other factors for the average club, such as ease of publishing results. I know nothing about SM5 ... so:
-- what does it cost?
-- where does one buy it?
-- if we are using a non-ACF-recommended application, are there any implications (are we out on a limb if someone complains about one of its pairings?)
Further guidance and observations welcomed.

My feeling - personal not necessarily a prediction of what the ACF will do - is that the ACF will stick with SP for the moment as far as clubs go because it is cheap, it allows importation of players from the ACF master list, most people now know how to use it and generally it gets things right. For major events like Australian Championships, SM5 or SManager may be used either with or without SP as they are FIDE approved. This has been happening in some events already. The ACF has not investigated the cost of using either SM5 or SM though the latter is very expensive at about $200 for a single licence. Whether bulk licences are available for either program I don't know. I suppose the initiative for change in this matter should come from the ACF President.

DJ

Brian_Jones
03-11-2007, 09:00 AM
The ACF has not investigated the cost of using either SM5 or SM though the latter is very expensive at about $200 for a single licence. Whether bulk licences are available for either program I don't know. I suppose the initiative for change in this matter should come from the ACF President. DJ

The initiative has already come from the chess professionals (organisers and arbiters).

ACE purchased a Swiss-Manager license for use at the 2007 Sydney International Open (SIO).

But I shudder at the thought of another Australian Championship using Swiss Perfect (as was done at Brisbane). We'll need a day betweeen rounds to do the pairings manually and recheck a million times! :D

Denis_Jessop
03-11-2007, 10:51 AM
The initiative has already come from the chess professionals (organisers and arbiters).

ACE purchased a Swiss-Manager license for use at the 2007 Sydney International Open (SIO).

But I shudder at the thought of another Australian Championship using Swiss Perfect (as was done at Brisbane). We'll need a day betweeen rounds to do the pairings manually and recheck a million times! :D

Ditto. I acknowledged that. I was referring to an ACF initiative for general club use. ;) :hmm:

DJ

Bill Gletsos
03-11-2007, 11:01 AM
But I shudder at the thought of another Australian Championship using Swiss Perfect (as was done at Brisbane). We'll need a day betweeen rounds to do the pairings manually and recheck a million times! :DThe arbiters overrode the SP pairings in the last round and it was the manual pairing that was wrong.