drbean

06-11-2007, 07:14 PM

This is a conceptual tournament. All games are drawn. The only way to get ahead, is to get a bye.

I am happy with all the pairings for round 1-3.

The problem in the 4th round is C10,11,12 conspire to never let a remainder group increase x.

The pairing table for the 4th round is:

Round 4 Pairing Groups

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

Place No Opponents Roles Float Score

1-3

5 2,4,- WW- D 2

6 3,-,7 B-B d 2

7 -,1,6 -BW d 2

4-7

1 4,7,2 WWB u 1.5

2 5,3,1 BBW 1.5

3 6,2,4 WWB 1.5

4 1,5,3 BBW 1.5

In the 1st bracket, 6 and 7 have played each other, so one of them must play 5. Because 6 is color compatible, 6 plays 5 and 7 is downfloated, despite having downfloated 2 rounds ago.

In the 1.5-Bracket, 7 has already played 1 and is color compatible with 3, so 7 and 3 are paired, leaving a remainder group of 1,2 and 4.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E1 7&3

C5, ordered: 1

& 2 4

In this 1.5-Bracket Remainder Group, 1 has played both 2 and 4. So 2 and 4 should play each other. Even though they both have a strong preference for Black.

At least, that is the way I thought this round should be paired.

But Games::Tournament::Swiss is producing a different pairing.

To get to C11 and x=1, first C10 has to be tried and 7 and 3 have to be unpaired!

C10, Unpairing 7 and 3 in Bracket 2(1.5)

C10, Bracket 2's C10Repair Group: 7 3 2 4 1

C7, 4 1 2 3

C6PAIRS, B4: x=0, table 1 NOK

C7, last transposition

We already tested 7 and 2, but they were color incompatible, and 4 is a B4 violation too.

C10, No more opponents for Player 7

C10, Giving up on lowest 7 in 1.5-Bracket [2]

C11, Deleting all matches in 1.5-Bracket [2]

C11, Bracket 2's C11 Repairing: 7 4 1 2 3, with x=1

C4, S1: 7 & S2: 4 1 2 3

C5, ordered: 7

& 1 2 3 4

Because there is only one downfloater in the 1.5-Bracket, the pairings being considered are the same. x is different however.

After dropping B6 for downfloats:

C9, Dropping B6 for Downfloats

C4, S1: 7 & S2: 4 1 2 3

C5, ordered: 7

& 1 2 3 4

C6PAIRS, B1a: table 1 NOK

C7, 2 1 3 4

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E2 2&7

C6OTHERS, Remaindering 1 3 4.

[2] 7 2 & [2's Remainder Group] 1 3 4

Now 2 can be paired with 7. Which leaves 1, 3 and 4 to be paired. But, in a similar situation to that before, 4 has played both 1 and 3. So 1 and 3 should be paired. But, they both have the same Strong preference for White.

And to get to C11 and x=1, first C10 has to be tried again.

Or does it? Is the increased x property inherited by all remainder groups of the repaired 1.5-Bracket?

C11 says:

As long as x is less than p: increase x by 1. When pairing a remainder group undo all pairings of players moved down also. Restart at C3.

This gives the impression that remainder groups may inherit x increases. It says 'also.' (Or does that 'also' just mean repair all the members of the heterogeneous group?)

But, C10 stands in the way of C11. At least for a bracket which has not received an x increase.

I thought that no remainder group ever got to C11. They were all fielded by C10. And so no remainder group ever got x increases.

Perhaps at least for brackets which are receiving x increases, C10 and C11 (and C12?) don't stand in sequence, but are alternative parallel states in the pairing procedure.

What happens if we DON'T increase x in the remainder group. It looks like we

may end up joining both brackets into one big group, because if we pair 7 with

4, the last player 7 can be paired with in the final bracket, we have to pair

1,2 and 3. But 2 and 3 have already played each other. And 1 has played 2. So

we have to pair 1 and 3, which are color incompatible. But we have not reached

C11 yet, so x=0 still.

Games::Tournament::Swiss is producing such a pairing after trying to repair 2- Bracket [1].

C10, Passing 2's Remainder Group (1.5Remainder) to C11.

C11, Repairing of 7 3 in 1.5-Bracket [2] failed pairing 2 4 1.

C11, Deleting all matches in 1.5-Bracket [2]. Repairing with x=1

C7, 4 1 2 3

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E2 4&7

And (cutting out lots of repeated cycles through C10 and C11 here),

C10, Passing 2's Remainder Group (1.5Remainder) to C11.

C11, Repairing of 7 4 in 1.5-Bracket [2] failed pairing 2 3 1.

C11, Deleting all matches in 1.5-Bracket [2]. Repairing with x=1

C7, last transposition

C10, No more opponents for Player 7

C10, Float checks all dropped in Bracket 2(1.5)

C10, Giving up on lowest 7 in 1.5-Bracket [2]

C11, x=p=1, no more x increases in 1.5-Bracket [2].

Giving up on C11 Repair. Trying C12

C12, Deleting matches in 2-Bracket [1], home of 7

C12, 2-Bracket [1] C12 Repairing: 7 6 5

To cut it short here, the C12 repairing fails, perhaps incorrectly, and Games:: Tournament::Swiss is joining the 2 brackets

C12, Repairing of 2-Bracket [1] failed to pair 1.5 [2]. Go to C13

C13, Undoing 2-Bracket [1] matches

C13, Unfloating 7 back from 2.

C13, penultimate p=0.

C13, Joining Bracket 1, 2.

C13, [1] 6 5 7 => [2] 7 4 2 3 1

C5, ordered: 5 6 7

& 1 2 3 4

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 2 3 paired. OK

C6PAIRS, E1 5&1 E1 2&6 E1 7&3

Pairing complete

So what do you think? The present solution isn't ideal. But automatic increases of x in remainder groups seems problematic too. Should pairing without them be tried first?

I am happy with all the pairings for round 1-3.

The problem in the 4th round is C10,11,12 conspire to never let a remainder group increase x.

The pairing table for the 4th round is:

Round 4 Pairing Groups

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

Place No Opponents Roles Float Score

1-3

5 2,4,- WW- D 2

6 3,-,7 B-B d 2

7 -,1,6 -BW d 2

4-7

1 4,7,2 WWB u 1.5

2 5,3,1 BBW 1.5

3 6,2,4 WWB 1.5

4 1,5,3 BBW 1.5

In the 1st bracket, 6 and 7 have played each other, so one of them must play 5. Because 6 is color compatible, 6 plays 5 and 7 is downfloated, despite having downfloated 2 rounds ago.

In the 1.5-Bracket, 7 has already played 1 and is color compatible with 3, so 7 and 3 are paired, leaving a remainder group of 1,2 and 4.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E1 7&3

C5, ordered: 1

& 2 4

In this 1.5-Bracket Remainder Group, 1 has played both 2 and 4. So 2 and 4 should play each other. Even though they both have a strong preference for Black.

At least, that is the way I thought this round should be paired.

But Games::Tournament::Swiss is producing a different pairing.

To get to C11 and x=1, first C10 has to be tried and 7 and 3 have to be unpaired!

C10, Unpairing 7 and 3 in Bracket 2(1.5)

C10, Bracket 2's C10Repair Group: 7 3 2 4 1

C7, 4 1 2 3

C6PAIRS, B4: x=0, table 1 NOK

C7, last transposition

We already tested 7 and 2, but they were color incompatible, and 4 is a B4 violation too.

C10, No more opponents for Player 7

C10, Giving up on lowest 7 in 1.5-Bracket [2]

C11, Deleting all matches in 1.5-Bracket [2]

C11, Bracket 2's C11 Repairing: 7 4 1 2 3, with x=1

C4, S1: 7 & S2: 4 1 2 3

C5, ordered: 7

& 1 2 3 4

Because there is only one downfloater in the 1.5-Bracket, the pairings being considered are the same. x is different however.

After dropping B6 for downfloats:

C9, Dropping B6 for Downfloats

C4, S1: 7 & S2: 4 1 2 3

C5, ordered: 7

& 1 2 3 4

C6PAIRS, B1a: table 1 NOK

C7, 2 1 3 4

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E2 2&7

C6OTHERS, Remaindering 1 3 4.

[2] 7 2 & [2's Remainder Group] 1 3 4

Now 2 can be paired with 7. Which leaves 1, 3 and 4 to be paired. But, in a similar situation to that before, 4 has played both 1 and 3. So 1 and 3 should be paired. But, they both have the same Strong preference for White.

And to get to C11 and x=1, first C10 has to be tried again.

Or does it? Is the increased x property inherited by all remainder groups of the repaired 1.5-Bracket?

C11 says:

As long as x is less than p: increase x by 1. When pairing a remainder group undo all pairings of players moved down also. Restart at C3.

This gives the impression that remainder groups may inherit x increases. It says 'also.' (Or does that 'also' just mean repair all the members of the heterogeneous group?)

But, C10 stands in the way of C11. At least for a bracket which has not received an x increase.

I thought that no remainder group ever got to C11. They were all fielded by C10. And so no remainder group ever got x increases.

Perhaps at least for brackets which are receiving x increases, C10 and C11 (and C12?) don't stand in sequence, but are alternative parallel states in the pairing procedure.

What happens if we DON'T increase x in the remainder group. It looks like we

may end up joining both brackets into one big group, because if we pair 7 with

4, the last player 7 can be paired with in the final bracket, we have to pair

1,2 and 3. But 2 and 3 have already played each other. And 1 has played 2. So

we have to pair 1 and 3, which are color incompatible. But we have not reached

C11 yet, so x=0 still.

Games::Tournament::Swiss is producing such a pairing after trying to repair 2- Bracket [1].

C10, Passing 2's Remainder Group (1.5Remainder) to C11.

C11, Repairing of 7 3 in 1.5-Bracket [2] failed pairing 2 4 1.

C11, Deleting all matches in 1.5-Bracket [2]. Repairing with x=1

C7, 4 1 2 3

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 paired. OK

C6PAIRS, E2 4&7

And (cutting out lots of repeated cycles through C10 and C11 here),

C10, Passing 2's Remainder Group (1.5Remainder) to C11.

C11, Repairing of 7 4 in 1.5-Bracket [2] failed pairing 2 3 1.

C11, Deleting all matches in 1.5-Bracket [2]. Repairing with x=1

C7, last transposition

C10, No more opponents for Player 7

C10, Float checks all dropped in Bracket 2(1.5)

C10, Giving up on lowest 7 in 1.5-Bracket [2]

C11, x=p=1, no more x increases in 1.5-Bracket [2].

Giving up on C11 Repair. Trying C12

C12, Deleting matches in 2-Bracket [1], home of 7

C12, 2-Bracket [1] C12 Repairing: 7 6 5

To cut it short here, the C12 repairing fails, perhaps incorrectly, and Games:: Tournament::Swiss is joining the 2 brackets

C12, Repairing of 2-Bracket [1] failed to pair 1.5 [2]. Go to C13

C13, Undoing 2-Bracket [1] matches

C13, Unfloating 7 back from 2.

C13, penultimate p=0.

C13, Joining Bracket 1, 2.

C13, [1] 6 5 7 => [2] 7 4 2 3 1

C5, ordered: 5 6 7

& 1 2 3 4

C6PAIRS, B56: OK.

C6PAIRS, 1.5-Bracket (2) tables 1 2 3 paired. OK

C6PAIRS, E1 5&1 E1 2&6 E1 7&3

Pairing complete

So what do you think? The present solution isn't ideal. But automatic increases of x in remainder groups seems problematic too. Should pairing without them be tried first?