Hopefully I'm not giving anything away with this post (and Stewart Reuben is too busy at Hastings to notice), but here is the question on Swiss Pairings that was given as part of the arbiters exam in Dresden. NB I got the answer partly wrong (as did a number of other arbiters), and when I give the answer in a day or 2 (in the comments section) I'll explain the error in my thinking.
Part A: You have a 15 player tournament, the players being identified by their seeding numbers (1 through to 15). The top seed is White in round 1. What are the first round pairings?
The results of round 1 (in board order) are 1-0, 0.5-0.5, 1-0, 0-1, 1-0, 0.5-0.5, 1-0, 1-0
Part B: Player 13 withdraws from the event after round 1. What are the pairings for round 2?
Subscribe to:
Post Comments (Atom)
12 comments:
Were you allowed paper? :P
Ok I'll try.
Part A: Round 1 pairings would be:
1v8 1:0
9v2 0.5-0.5
3v10 1-0
11v4 0-1
5v12 1-0
13v6 0.5-0.5
7v14 1-0
with player 15 having a full-point bye for round 1.
Part B: Pairings for Round 2 (by my guess)
4v1
7v3
15v5
2v9
6v11
8v12
10v14
Yes, we were allowed paper, so it would be a good idea to use it!
Your part A is correct, but not part B (and not in the same way I got part B wrong either).
For Part B how about
5-1
15-3
4-7
2-9
6-8
10-12
14-11
Getting closer ( and closer than I did), but still not entirely correct.
5 v 1
7 v 3
4 v 15
9 v 6
2 v 8
10 v 12
14 v 11
2 v 9 is incorrect for Part B as they have already played in round one, so can not be paired again.
For my pairing answer, it should read 6 v 9, not 9 v 6.
6 had black in round one, 9 had white in round one, so when paired together, 6 is due white and 9 is due black.
Er yes, 2 and 9 have played, I missed that.
We have a winner! Garvin has given the correct pairings (once he swapped 6v9).
Ian Rout gets some sympathy for pairing 2v9 in round 2, as Stewart Reuben told me that a very experienced International Arbiter did exactly the same thing when attempting to answer the question.
Where I (and everyone else) went wrong was in the pairing of the upfloat for player 2. Possibly as a result of learning the Lim system previously I went looking for an opponent for 2 that had a compatible colour preference. As a result I paired 2v11 followed by 8v12 and 10v14.
However the first step in working out pairings is to calculate the magic w,b,p and q numbers. In the case of the group (S1) containing player 2 (on 0.5) and the rest (S2) (all on 0), q=1 which means that you can have 1 pairing where both players do NOT get their colour preferences. Therefore 2 v 8 is both an acceptable, and indeed required pairing, leaving 10v12 and 14v11 as the pairings from the bottom group.
In my view Ian's pairing of the first group (5-1, 15-3, 4-7) is preferable to Garvin's (5-1, 7-3, 4-15) as in the first case 3,4 and 7 are all granted colour preference while in the second case 7 and 4 are but 3 are not.
Relative Criteria heading + rule B4 makes it clear that transpositions and even exchanges may be applied to ensure as many players as possible receive their colour preference.
Actually you are right. So fixated on the bottom set of pairings I didn't check Garvin's top set.
The correct answer is actually
5v1 15v3 4v7 6v9 2v8 10v12 14v11
as given by Stewart Reuben (and agreed with by Swiss Perfect).
KB said:
In my view Ian's pairing of the first group (5-1, 15-3, 4-7) is preferable to Garvin's (5-1, 7-3, 4-15) as in the first case 3,4 and 7 are all granted colour preference while in the second case 7 and 4 are but 3 are not.
Garvin's reply:
Since your post, I put this situation into Swiss Master 5.5 and it gives the pairings KB lists for 1 point score group.
P.S. Have just seen Shaun's reply. Shaun you can add Swiss Master as to agreeing with the 1 point score group that KB lists.
Post a Comment