drbean

05-09-2007, 10:13 AM

C13 talks about unpairing the penultimate score bracket, if

the lowest score bracket cannot be paired.

In discussion with bartolin, I was wondering whether, if the

penultimate score bracket was a heterogeneous group with a

remainder group, the bracket which should be re-paired would

be the remainder group, or the whole heterogeneous group from

which the remainder group came.

The pairing table which he came up with:

Round 3 Pairing Groups

-------------------------------------------------------------------------

Place No Opponents Roles Float Score

1-2

1 7,2 BW 1.5

2 8,1 WB 1.5

3-10

3 9,11 BW 1

4 10,12 WB 1

5 11,9 BW 1

6 12,10 WB 1

9 3,5 WB 1

10 4,6 BW 1

11 5,3 WB 1

12 6,4 BW 1

11-12

7 1,8 WB 0.5

8 2,7 BW 0.5

This is somewhat similar to the table of ggrayggray's

http://www.chesschat.org/showthread.php?t=6619, in that the

last score bracket is unpairable as it is, if not as

elegantly.

The pairing of the first 2 tables is clear, I think.

C6, Bracket tables 1 2 paired. E1 4&1 E1 2&3

C6others: Remainder Group, Bracket 2: 5 6 9 10 11 12

The remainder group is paired as:

C6, Bracket tables 1 2 3 paired. E1 6&5 E1 9&10 E1 11&12

C6others: no non-paired players

Next, Bracket 3: 7 8

But because 7 and 8 have already played each other, they can't

be paired, and C13 applies.

The question is: Is the remainder group, 5 6 9 10 11 12

unpaired, or is the whole of group 2, 1 2 3 5 6 9 10 11 12

unpaired?

He took the position that only the remainder group is

unpaired. I took the position that the whole group is

unpaired.

See

Linkname: #29073: Problems with C14 in FIDE.pm

URL: http://rt.cpan.org/Ticket/Display.html?id=29073

down towards the end.

I don't think the wording of C13 makes it clear:

C13.

In case of the lowest score bracket: the pairing of the

penultimate score bracket is undone. Try to find another

pairing in the penultimate score bracket which will allow a

pairing in the lowest score bracket. 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. Because now another score bracket

is the penultimate one C13 can be repeated until an

acceptable pairing is obtained.

The whole question of the status of remainder groups is

unclear to me. Are they new brackets that are interposed into

the original ones, or is their existence only virtual?

the lowest score bracket cannot be paired.

In discussion with bartolin, I was wondering whether, if the

penultimate score bracket was a heterogeneous group with a

remainder group, the bracket which should be re-paired would

be the remainder group, or the whole heterogeneous group from

which the remainder group came.

The pairing table which he came up with:

Round 3 Pairing Groups

-------------------------------------------------------------------------

Place No Opponents Roles Float Score

1-2

1 7,2 BW 1.5

2 8,1 WB 1.5

3-10

3 9,11 BW 1

4 10,12 WB 1

5 11,9 BW 1

6 12,10 WB 1

9 3,5 WB 1

10 4,6 BW 1

11 5,3 WB 1

12 6,4 BW 1

11-12

7 1,8 WB 0.5

8 2,7 BW 0.5

This is somewhat similar to the table of ggrayggray's

http://www.chesschat.org/showthread.php?t=6619, in that the

last score bracket is unpairable as it is, if not as

elegantly.

The pairing of the first 2 tables is clear, I think.

C6, Bracket tables 1 2 paired. E1 4&1 E1 2&3

C6others: Remainder Group, Bracket 2: 5 6 9 10 11 12

The remainder group is paired as:

C6, Bracket tables 1 2 3 paired. E1 6&5 E1 9&10 E1 11&12

C6others: no non-paired players

Next, Bracket 3: 7 8

But because 7 and 8 have already played each other, they can't

be paired, and C13 applies.

The question is: Is the remainder group, 5 6 9 10 11 12

unpaired, or is the whole of group 2, 1 2 3 5 6 9 10 11 12

unpaired?

He took the position that only the remainder group is

unpaired. I took the position that the whole group is

unpaired.

See

Linkname: #29073: Problems with C14 in FIDE.pm

URL: http://rt.cpan.org/Ticket/Display.html?id=29073

down towards the end.

I don't think the wording of C13 makes it clear:

C13.

In case of the lowest score bracket: the pairing of the

penultimate score bracket is undone. Try to find another

pairing in the penultimate score bracket which will allow a

pairing in the lowest score bracket. 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. Because now another score bracket

is the penultimate one C13 can be repeated until an

acceptable pairing is obtained.

The whole question of the status of remainder groups is

unclear to me. Are they new brackets that are interposed into

the original ones, or is their existence only virtual?