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Bartolin
16-09-2007, 07:03 PM
Hi,

I would like to pick up a question discussed in the thread "Another sp error?" (http://chesschat.org/showthread.php?t=6619), namely: How should B3-B6 be incorporated into the pairing procedure given by C, especially C6?

There were a lot of useful remarks about this question before the thread moved in another direction. There is still an unanswered question waiting in that thread and I don't want to stop that discussion due to asking a completely different question. Therefore I opened this new thread. If this is regarded as bad practice, I beg your pardon.

Back to my question about B3-B6. Basically, there were two different positions:

1. C6 could be interpreted literal. That would mean that only B1 and B2 must be
fulfilled -- violations of B3 to B6 are irrelevant, since they aren't mentioned
in C6. (compare http://chesschat.org/showpost.php?p=167008&postcount=15,
http://chesschat.org/showpost.php?p=167079&postcount=21)

2. C6 could be seen as badly worded; it really should refer to B1 to B6 instead of just B1 to B2. This view is conform with section B, which states that "[B3 to B6] 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." Bill Gletsos said that all FIDE approved programs act this way. (cp. http://chesschat.org/showpost.php?p=167088&postcount=22 and http://chesschat.org/showpost.php?p=167092&postcount=25).

If one takes the second position (and I think, it's the more convincing one), there remains the question, how B3 is to be incorporated in a pairing algorithm (http://chesschat.org/showpost.php?p=167008&postcount=15).

It has been pointed out that B3 is relevant only for heterogeneous groups or for groups that are actually heterogeneous, but are treated as homogeneous according to A3 (http://chesschat.org/showpost.php?p=167071&postcount=17).
Does one have to ignore the procedure from C for those groups and instead:
1. compute all possible pairings for those groups (by allowing transpositions and exchanges) first
2. check which fits best with B3 (let's assume, B3 refers to the sum of score differences)
3. if there are several pairings which are "best" according to B3, check which fits best with B4
4. if there are several pairings which fit with B4, check which fits best with B5
5. if there are several pairings which fit with B5, check which fits best with B6
6. if there are several pairings which fit with B6, return to C, apply the procedure and check, which one is obtained first?

This doesn't sound like a smart algorithm to me.

But enough of theoretical questions, here is an example. It's from a real tournament and trying to re-pair it, the above questions arose.


Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
8 18,5,3 BWB D 3
2
1 11,10,6 WBW u 2.5
3-8
2 12,7,9 BWB 2
4 14,9,7 BWB 2
5 15,8,13 WBW 2
7 17,2,4 WBW 2
9 19,4,2 WBW 2
17 7,14,12 BWB D 2
9-14
3 13,6,8 WBW U 1.5
6 16,3,1 BWB 1.5
10 20,1,15 BWB 1.5
11 1,12,16 BWB d 1.5
15 5,20,10 BWW 1.5
16 6,19,11 WBW 1.5
15-17
13 3,18,5 BWB 1
19 9,16,14 BWB 1
20 10,15,18 WBB 1
18
12 2,11,17 WBW U 0.5
19-20
14 4,17,19 WBW 0
18 8,13,20 WBW 0

In the tournament the pairings were: 8-1, 2-5, 4-17, 9-7, 11-3, 6-15, 10-16, 13-19, 20-12 and 18-14.

The first 8 pairings are reasonably clear, but how to get the last two pairings.

20 is downfloated to 12, forming a score bracket (a homogeneous one, according to A3, last sentence). Unfortunately, 12 was upfloated two times. Therefore, 20-12 doesn't seem to fit according to C1-C9. Next, C10 doesn't seem to be relevant, since we don't have a homogeneous remainder group. Next, C11 applies, but increasing x doesn't help. Next C12 doesn't apply. Neither does C13. Therefore we arrive at C14, decrease p by 1 (to 0) and move both players down to a joined score bracket 20,12,14,18.

First question: Doesn't that contradicts with B, saying about Relative Criteria B3-B6: "no player should be moved down to a lower score bracket"? According to B, 20-12 seems to be a correct pairing -- even if one doesn't arrive there according to C. At what point of the pairing procedure should a computer program consider this?

But let's assume, 20 and 12 are to be moved down. Now S1 becomes 20,12 and S2 14,18. The first pairings one arrives with are 20-14 and 18-12. Those are fine as long one ignores B3. (For 20-14 there is a score difference of 1, for 12-18 the score difference is 0.5.) Better pairings are of course 20-12 (sic!) and 18-14. But again: At what point of the pairing procedure should a computer program consider this?

By the way: I tried to re-pair the given tournament with Games::Tournament::Swiss (http://search.cpan.org/~drbean/Games-Tournament-Swiss-0.08/) and it gave the pairings 20-14 and 18-12 for the last two boards. That is because it takes C6 literal and ignores B3.

I hope my questions aren't stupid. I don't have experience with the pairing rules, but I really want to understand them. (And I want to improve the pairing algorithm of Games::Tournament::Swiss.)

Thanks

Christian

Bill Gletsos
16-09-2007, 07:26 PM
But enough of theoretical questions, here is an example. It's from a real tournament and trying to re-pair it, the above questions arose.


Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
8 18,5,3 BWB D 3
2
1 11,10,6 WBW u 2.5
3-8
2 12,7,9 BWB 2
4 14,9,7 BWB 2
5 15,8,13 WBW 2
7 17,2,4 WBW 2
9 19,4,2 WBW 2
17 7,14,12 BWB D 2
9-14
3 13,6,8 WBW U 1.5
6 16,3,1 BWB 1.5
10 20,1,15 BWB 1.5
11 1,12,16 BWB d 1.5
15 5,20,10 BWW 1.5
16 6,19,11 WBW 1.5
15-17
13 3,18,5 BWB 1
19 9,16,14 BWB 1
20 10,15,18 WBB 1
18
12 2,11,17 WBW U 0.5
19-20
14 4,17,19 WBW 0
18 8,13,20 WBW 0

In the tournament the pairings were: 8-1, 2-5, 4-17, 9-7, 11-3, 6-15, 10-16, 13-19, 20-12 and 18-14.

The first 8 pairings are reasonably clear, but how to get the last two pairings.

20 is downfloated to 12, forming a score bracket (a homogeneous one, according to A3, last sentence). Unfortunately, 12 was upfloated two times. Therefore, 20-12 doesn't seem to fit according to C1-C9. Next, C10 doesn't seem to be relevant, since we don't have a homogeneous remainder group. Next, C11 applies, but increasing x doesn't help. Next C12 doesn't apply. Neither does C13. Therefore we arrive at C14, decrease p by 1 (to 0) and move both players down to a joined score bracket 20,12,14,18.

First question: Doesn't that contradicts with B, saying about Relative Criteria B3-B6: "no player should be moved down to a lower score bracket"? According to B, 20-12 seems to be a correct pairing -- even if one doesn't arrive there according to C. At what point of the pairing procedure should a computer program consider this?

But let's assume, 20 and 12 are to be moved down. Now S1 becomes 20,12 and S2 14,18. The first pairings one arrives with are 20-14 and 18-12. Those are fine as long one ignores B3. (For 20-14 there is a score difference of 1, for 12-18 the score difference is 0.5.) Better pairings are of course 20-12 (sic!) and 18-14. But again: At what point of the pairing procedure should a computer program consider this?

By the way: I tried to re-pair the given tournament with Games::Tournament::Swiss (http://search.cpan.org/~drbean/Games-Tournament-Swiss-0.08/) and it gave the pairings 20-14 and 18-12 for the last two boards. That is because it takes C6 literal and ignores B3.

I hope my questions aren't stupid. I don't have experience with the pairing rules, but I really want to understand them. (And I want to improve the pairing algorithm of Games::Tournament::Swiss.)

Thanks

ChristianThe last two pairing are correct.
20 down floats to the score group that only contains player 12.
They are a match.
By definirion B3 will be violated as it is a hetrogeneous group.
A volation of b5-B6 in this case is irrelevant as there is no other player in the score group and you certainly do not drop both 20 and 12 to the next lower group.
Remeber B3-B6 are relative criteria not absolute.
Only in the case where 20 V 12 violates B1 or B2 would they both drop to the next score group.

drbean
17-09-2007, 01:45 AM
How should B3-B6 be incorporated into the pairing procedure given by C, especially C6?

Back to my question about B3-B6. Basically, there were two
different positions:

1. C6 could be interpreted literal. That would mean that
only B1 and B2 must be fulfilled.

2. C6 could be seen as badly worded; it really should refer
to B1 to B6 instead of just B1 to B2.



I think differentiating between B and C is like trying
to distinguish between the spirit and the letter of the law.
They're supposed to be united.

Whether they are united or not is something you can argue
over.




It has been pointed out that B3 is relevant only for
heterogeneous groups or for groups that are actually
heterogeneous, but are treated as homogeneous according to
A3.


There are 2 sorts of heterogenous groups too. Ones that are
made by floating down unpaired players from a higher bracket,
as in C6 and C14, and others in the last bracket resulting
from pairing with the penultimate bracket as in C13.

The first sort, the procedure is quite explicit about how to
pair, I think. Although the vague wording, 'try to find a
different opponent' is used in C10 about undoing the pairing
of a heterogeneous bracket, it says you are supposed to
restart at C7.

The effect is to minimize the B3 inequality, the result also
of S1 being just the players from the higher bracket.

The second sort, the same vague wording is used, but there
is no indication where to start again. There is no explicit
help on minimizing B3 inequality in C13.




Does one have to ignore the procedure from C for those
groups and instead:
1. compute all possible pairings for those groups (by
allowing transpositions and exchanges) first
2. check which fits best with B3 (let's assume, B3 refers to
the sum of score differences)
3. if there are several pairings which are "best" according
to B3, check which fits best with B4
4. if there are several pairings which fit with B4, check
which fits best with B5
5. if there are several pairings which fit with B5, check
which fits best with B6
6. if there are several pairings which fit with B6, return
to C, apply the procedure and check, which one is obtained
first?

This doesn't sound like a smart algorithm to me.



I think a smart idea might be to apply the general
principles of the procedure in reverse and try to pair the
lowest players first. In the example from
http://chesschat.org/showthread.php?t=6619



1 : 6,4,2,5 WBWB D 3.5
2 : 7,3,1,4 BWBW D 3.5

3 : 8,2,6,7 WBWB D 2.5
6 : 1,5,3,9 BWBW D 2.5

4 : 9,1,7,2 BWBB U 2
5 : 10,6,8,1 WBWW U 2
8 : 3,9,5,10 BWBW D 2

7 : 2,10,4,3 WBWW U 1
9 : 4,8,10,6 WBWB U 1

10 : 5,7,9,8 BWBB U 0


pair 10 with the rest ranked in reverse order. This would
give 10&6. Then pair 9, 9&7, then 8, et cetera,

The same way that the procedure has B3 inequality reduction
built in going down, it will have going up. If everything is
done in reverse.

I remember reading there was an alternative to the Dutch
procedure which paired in both directions at the same time.
This procedure would seem to have some connection here.




.. here is
an example. It's from a real tournament



This appears to be from
http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm

and answers Kevin Bonham's request for an example about
minimizing B3 inequality in the last 2 brackets by avoiding
C13 amalgamation.







Round 4 Pairing Groups
-------------------------------------------------------------------------
Place No Opponents Roles Float Score
1
8 18,5,3 BWB D 3
2
1 11,10,6 WBW u 2.5
3-8
2 12,7,9 BWB 2
4 14,9,7 BWB 2
5 15,8,13 WBW 2
7 17,2,4 WBW 2
9 19,4,2 WBW 2
17 7,14,12 BWB D 2
9-14
3 13,6,8 WBW U 1.5
6 16,3,1 BWB 1.5
10 20,1,15 BWB 1.5
11 1,12,16 BWB d 1.5
15 5,20,10 BWW 1.5
16 6,19,11 WBW 1.5
15-17
13 3,18,5 BWB 1
19 9,16,14 BWB 1
20 10,15,18 WBB 1
18
12 2,11,17 WBW U 0.5
19-20
14 4,17,19 WBW 0
18 8,13,20 WBW 0



In the tournament the pairings were: 8-1, 2-5, 4-17, 9-7, 11-3, 6-15, 10-16, 13-19, 20-12 and 18-14.

The first 8 pairings are reasonably clear, but how to get the last two pairings.

20 is downfloated to 12, forming a score bracket (a homogeneous one, according to A3, last sentence). Unfortunately, 12 was upfloated two times. Therefore, 20-12 doesn't seem to fit according to C1-C9. Next, C10 doesn't seem to be relevant, since we don't have a homogeneous remainder group. Next, C11 applies, but increasing x doesn't help. Next C12 doesn't apply. Neither does C13. Therefore we arrive at C14, decrease p by 1 (to 0) and move both players down to a joined score bracket 20,12,14,18.


First question: Doesn't that contradicts with B, saying about Relative Criteria B3-B6: "no player should be moved down to a lower score bracket"? According to B, 20-12 seems to be a correct pairing -- even if one doesn't arrive there according to C. At what point of the pairing procedure should a computer program consider this?




The last two pairing are correct.
20 down floats to the score group that only contains player 12.
They are a match.
By definirion B3 will be violated as it is a hetrogeneous group.
A volation of b5-B6 in this case is irrelevant as there is no other player in the score group and you certainly do not drop both 20 and 12 to the next lower group.
Remeber B3-B6 are relative criteria not absolute.
Only in the case where 20 V 12 violates B1 or B2 would they both drop to the next score group.

What is the importance of there being no other player in the
bracket? That there is no better choice than the one that
we are considering?

There must be an implicit waiving of B5 for upfloats. There
is no indication of that in the procedure. There is no
written procedure for waiving B5 for upfloats except when
a homogeneous remainder group cannot be paired.



But let's assume, 20 and 12 are to be moved down. Now S1 becomes 20,12 and S2 14,18. The first pairings one arrives with are 20-14 and 18-12. Those are fine as long one ignores B3. (For 20-14 there is a score difference of 1, for 12-18 the score difference is 0.5.) Better pairings are of course 20-12 (sic!) and 18-14. But again: At what point of the pairing procedure should a computer program consider this?

By the way: I tried to re-pair the given tournament with Games::Tournament::Swiss (http://search.cpan.org/~drbean/Games-Tournament-Swiss-0.08/) and it gave the pairings 20-14 and 18-12 for the last two boards. That is because it takes C6 literal and ignores B3.



Players with 3 different scores are in the same score
bracket. B3 differences would be minimalized if there were
only 2 different score groups. It seems like a failure of
C1-6 to minimize score inequality. I'm just following the
rules, but you say I've got to bend them, be flexible :-)

Bill Gletsos
17-09-2007, 02:08 AM
What is the importance of there being no other player in the
bracket? That there is no better choice than the one that
we are considering?

There must be an implicit waiving of B5 for upfloats. There
is no indication of that in the procedure. There is no
written procedure for waiving B5 for upfloats except when
a homogeneous remainder group cannot be paired.As I have said numerous times it all comes down to poor (some would say abysmal) wording of the rules, especially in section C.

Just consder C6 as saying that the pairing is considered complete provided B1 & B2 are absolutely not violated and that as many of B3-B6 as possible are not violated (remembering that B3-B6 are all relative and ranked in importance).

Bill Gletsos
17-09-2007, 02:20 AM
Also a critical part of section B is the last sentence of:

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).

Bartolin
17-09-2007, 03:41 AM
I remember reading there was an alternative to the Dutch
procedure which paired in both directions at the same time.
This procedure would seem to have some connection here.

I guess you mean the system described at http://www.fide.com/official/handbook.asp?level=C0402. Is it called "Lim System"? There are four different systems described at http://www.fide.com/official/handbook.asp?level=C04. The Dutch System is the first one (under C04.1).



.. here is an example. It's from a real tournament

This appears to be from
http://www.lsvmv.de/turniere/erg/eon_2007a_paar.htm


Indeed it is. Round 1 to 3 are paired identically by Games::Tournament::Swiss 0.08.




There must be an implicit waiving of B5 for upfloats. There
is no indication of that in the procedure. There is no
written procedure for waiving B5 for upfloats except when
a homogeneous remainder group cannot be paired.


As I have said numerous times it all comes down to poor (some would say abysmal) wording of the rules, especially in section C.

Just consder C6 as saying that the pairing is considered complete provided B1 & B2 are absolutely not violated and that as many of B3-B6 as possible are not violated (remembering that B3-B6 are all relative and ranked in importance).

IMHO that sounds convincing. So the waiving of B5 for upfloats is integrated in B itself, saying that "no player should be moved down to a lower score bracket" -- and thereby integrated in C6.

Bill, would you agree that C6 should be read as follows (though, I guess, my wording could be improved further):


If now p pairings are obtained in compliance with B1 and B2, thereby violating B3-B6 as little as possible, the pairing of this score bracket is considered complete.
If one or more of B3-B6 are violated try to find "better" pairings which minimize those violations while still being in compliance with B1 and B2. Do so by applying C7 and C8 as needed. If there are no better pairings regarding B3-B6, use the first set of pairings.


So in my example, one gets to the point that 20 and 12 form one score bracket. The one and only pairing possible (20-12) is okay, since B5 (generally speaking: all of B3-B6) must be dropped before moving down a player and since there is no alternative to 20-12.

Bill Gletsos
17-09-2007, 08:50 PM
Bill, would you agree that C6 should be read as follows (though, I guess, my wording could be improved further):

If now p pairings are obtained in compliance with B1 and B2, thereby violating B3-B6 as little as possible, the pairing of this score bracket is considered complete.
If one or more of B3-B6 are violated try to find "better" pairings which minimize those violations while still being in compliance with B1 and B2. Do so by applying C7 and C8 as needed. If there are no better pairings regarding B3-B6, use the first set of pairings.
More than just C7 and C8 should be applied.
If C6 leads to a violation of B3-B6 then the following parts of section C should be applied, however in doing this no player should be moved down to a lower score bracket to cause B3-B6 to be satisfied (in line with my post #5 above).

drbean
19-09-2007, 10:51 PM
20 is downfloated to 12, forming a score bracket (a
homogeneous one, according to A3, last sentence).
Unfortunately, 12 was upfloated two times. Therefore, 20-12
doesn't seem to fit according to C1-C9. Next, C10 doesn't
seem to be relevant, since we don't have a homogeneous
remainder group. Next, C11 applies, but increasing x doesn't
help. Next C12 doesn't apply. Neither does C13. Therefore we
arrive at C14, decrease p by 1 (to 0) and move both players
down to a joined score bracket 20,12,14,18.



More than just C7 and C8 should be
applied. If C6 leads to a violation of B3-B6 then the
following parts of section C should be applied, however in
doing this no player should be moved down to a lower score
bracket to cause B3-B6 to be satisfied .

Perhaps the way to make sure no pair is downfloated if they
only satisfy B1 and B2 is to run the B1, B2 checks again for
pairs which have failed the further B4,5,6 compliance checks
after we have passed C14 the first time. This will give them
a second chance at staying in the bracket. With p (pprime)
now being less than the number of players in S1, these are
the pairs which are about to be downfloated.

This is separate from the problem of a single unpaired,
previously-downfloated player in a heterogeneous group,
discussed by Bill Gletsos at
http://chesschat.org/showpost.php?p=142260&postcount=158.

Bartolin
20-09-2007, 01:02 AM
Perhaps the way to make sure no pair is downfloated if they
only satisfy B1 and B2 is to run the B1, B2 checks again for
pairs which have failed the further B4,5,6 compliance checks
after we have passed C14 the first time. This will give them
a second chance at staying in the bracket. With p (pprime)
now being less than the number of players in S1, these are
the pairs which are about to be downfloated.


Am I right, assuming that you are talking about how to design
the pairing algorithm?

Actually I'm not sure, if I understood your suggestion. What
will be the consequence of a second check for B1 and B2
compliance being "positiv"? If there are only two players
(as in my example at the beginning of this thread) then it's
clear they should be paired. But if there are (let's say) 4 players,
how should one proceed after arriving at C14. Which pairings
should be checked for B1/B2 compliance?

Just brainstorming: What about a procedure like:
After arriving at C6 for the first time, whenever a set of pairings
is compliant with B1/B2, we do the following:
Checking whether the pairings are fully compliant with B3-B6.
If so, the pairings are considered "perfect" and one proceeds as
given in C6 dot 1 and dot2.
If the pairings violate on or more of B3-B6, we first check
whether it is the first set of pairings we examine. If so, we put
it aside as the "best set of pairing until now". If there is already
a "best set of pairings until now", we compare the current
pairings with that "best set". If the current pairings are "better"
(obviously one needs a kind of measurement for "B3-B6 compliance"),
they are promoted to "best set of pairings until now" and the
old ones are dropped.
After comparing the current pairings as described we move on
with C7-C14. Whenever we arrive at a new set of pairings which
is compliant with B1/B2, we compare it with the "best set of pairings
until now".
If C requires to move down one or more players, we stop
instantly and apply the "best set of pairings until now" instead.

You got my idea? Regarding the measurement for "B3-B6 compliance",
maybe one can specify penalty points for violation of Bx?

drbean
21-09-2007, 09:55 AM
What will be the consequence of a second check for B1 and B2
compliance being "positiv"? If there are only two players
(as in my example at the beginning of this thread) then it's
clear they should be paired.



OK. If there are only 2 players in the bracket, check for
bare B1, B2 compliance BEFORE going to C14, because they are
both about to be downfloated.

(This will not be necessary in a homogeneous remainder
group, because C10 will take care of this for us, so that
only B1,B2 compliance is being tested. (The problem we are
talking about here is the fact that B5,6 upfloating waiving is
never explicity sanctioned except in the case of a
homogeneous remainder group.))

Come to think of it, however, NONE of C7-C14 are relevant in
our progressive relaxation of B3-B6, IN THE CASE OF a
bracket of only 2 players.

We have to accept a B1,2-compliant pairing ANYWAY. There is
NO progressive relaxation. B3-6 are IRRELEVANT in the case
of a bracket of 2 players.



But if there are (let's say) 4 players,
how should one proceed after arriving at C14. Which pairings
should be checked for B1/B2 compliance?

Just brainstorming: What about a procedure like:
After arriving at C6 for the first time, whenever a set of pairings
is compliant with B1/B2, we do the following:
Checking whether the pairings are fully compliant with B3-B6.
If so, the pairings are considered "perfect" and one proceeds as
given in C6 dot 1 and dot2.
If the pairings violate on or more of B3-B6, we first check
whether it is the first set of pairings we examine. If so, we put
it aside as the "best set of pairing until now". If there is already
a "best set of pairings until now", we compare the current
pairings with that "best set". If the current pairings are "better"
(obviously one needs a kind of measurement for "B3-B6 compliance"),
they are promoted to "best set of pairings until now" and the
old ones are dropped.
After comparing the current pairings as described we move on
with C7-C14. Whenever we arrive at a new set of pairings which
is compliant with B1/B2, we compare it with the "best set of pairings
until now".
If C requires to move down one or more players, we stop
instantly and apply the "best set of pairings until now" instead.


This looks like 'memoization.' Paraphrasing Jamie Zawinski,
"Some people, when confronted with a problem, think I know,
I'll use 'memoization'. Now they have two problems."

I suppose a pairing that is a straight pairing of S1 and S2 is
a better pairing than one that has gone through
transpositions and exchanges. (I think that the aim of
dividing a bracket into an S1 and S2 and pairing one half
against the other is either to maximize the difference in
the ratings of the pair at each table, or to minimize the
differences between the rating gaps at each table. The
players in the bracket may have equal scores, but this
doesn't mean they have equal chances of winning their
games.)

There is a performance reason to support memoization. Rather
than do the same computation twice, it is better to store
the result of doing it the first time.

At this point, I am more interested in getting a 'correct'
rendition of the procedure than a fast one, however.

If you aren't concerned about transpositions and exchanges,
then the last pairing before reaching C9 is as good as
the first one.

It would also be good to avoid special-casing. Rather than allow
a 2-player bracket a special privilege that shortcuts the
C9-C14 process, it would be good to find a general procedure
that is clean and which can be applied to all brackets.

Bartolin
21-09-2007, 06:29 PM
OK. If there are only 2 players in the bracket, check for
bare B1, B2 compliance BEFORE going to C14, because they are
both about to be downfloated.

(This will not be necessary in a homogeneous remainder
group, because C10 will take care of this for us, so that
only B1,B2 compliance is being tested. (The problem we are
talking about here is the fact that B5,6 upfloating waiving is
never explicity sanctioned except in the case of a
homogeneous remainder group.))

Come to think of it, however, NONE of C7-C14 are relevant in
our progressive relaxation of B3-B6, IN THE CASE OF a
bracket of only 2 players.

We have to accept a B1,2-compliant pairing ANYWAY. There is
NO progressive relaxation. B3-6 are IRRELEVANT in the case
of a bracket of 2 players.


Yes, I think so, too. But I also agree that it would be good to have a generic procedure, which doesn't have to look at special cases.



This looks like 'memoization.' Paraphrasing Jamie Zawinski,
"Some people, when confronted with a problem, think I know,
I'll use 'memoization'. Now they have two problems."

I suppose a pairing that is a straight pairing of S1 and S2 is
a better pairing than one that has gone through
transpositions and exchanges. (I think that the aim of
dividing a bracket into an S1 and S2 and pairing one half
against the other is either to maximize the difference in
the ratings of the pair at each table, or to minimize the
differences between the rating gaps at each table. The
players in the bracket may have equal scores, but this
doesn't mean they have equal chances of winning their
games.)

IMHO, this point is important. I'm not sure about your suggestion,
that a straight pairing of S1 and S2 is a better pairing than one
that has gone through transpositions and exchanges. Part B of
the rules state, that B3-B6 should be fulfilled as much as possible.
So if a straight pairing of S1 and S2 is less compliant with B3-B6,
it is worse than another pairing, which is gained after some
transpositions or exchanges, but which is more B3-B6 compliant.

My idea was not only to use 'memoization' (didn't know that word,
thanks), but mainly to check for best B3-B6 compliance.

Let's take the examle I cited in this thread
(http://chesschat.org/showpost.php?p=167067&postcount=16)
from a Q-A sequence of Kevin Bonham and Geurt Gijssen:


[...] We have 4 players with 2 point and 4 players with 1 point.
The players with 2 points are A, B, C and D. The players with 1 point
are E, F, G and H.

It is possible to make the pairings A-C and B-D. Going to the next group the
arbiter discovers that E played already against F, G and H and the pairing F-G is
possible.

Well, what to do? Point C13 of the Pairing Procedure says that the 8 players
should form one group and the most likely pairings will be in that case:

A * E, B * F, C * G and D * H. But these pairings violate one of the Criteria of
Chapter B, which says that if possible players with the same score should play
against each other and good pairings could be: A * C, B * E, D * F and G * H.
Without any doubt, this is the best pairing. But there is a conflict in the
regulations. The criteria of Chapter B are correct, but, and this is the point, the
pairing procedures of Chapter C say something else. The problem is that the
pairing procedures cannot cover all situations, and I was also told that it is very
difficult to program for the computer. I am afraid that we have to accept a
situation that the computer produces pairings we shall simply accept. One thing is
sure: the computer is objective. There is no discussion that we have to change the
pairings only in cases in which the computer violates the criteria of Chapter B.
An example: Two players are leading, they did not play in any previous round,
the colours fit and the computer does not pair them against each other. And there
are of course more cases like this.

I think, in this case, my idea would be helpful. As Geurt said, the most likely
pairings would be A-E, B-F, C-G, D-H, that is S1 vs. S2. But it would be better to
apply some exchanges and transpositions until one gets S1: A, B, D, G and
S2: C, E, F, H. The resulting pairings A-C, B-E, D-F, G-H would be recognized as
more B3-B6 compliant, therefore it would replace A-E, B-F, C-G, D-H as the "best
pairing".


There is a performance reason to support memoization. Rather
than do the same computation twice, it is better to store
the result of doing it the first time.

At this point, I am more interested in getting a 'correct'
rendition of the procedure than a fast one, however.

As noted above, I don't think my suggestion is only performance related.


If you aren't concerned about transpositions and exchanges,
then the last pairing before reaching C9 is as good as
the first one.

The fact a set of pairings needs to be more compliant
with B3-B6 to replace the "best set of pairings until now" has a
built-in preference for earlier generated pairings. That is, it doesn't
really consider later pairings "as good as" earlier ones.

Does that make sense?

drbean
23-09-2007, 08:12 PM
We are talking about pairs which don't satisfy the Relative
Criteria, but do satisfy the Absolute Criteria. They
shouldn't be downfloated (normally? ever?), but what do we
do with them, when? Is it possible to rank pairings as
better than others? Where are those pairings found? At the
end of the pairing process, or earlier on?




IMHO, this point is important. I'm not sure about your
suggestion, that a straight pairing of S1 and S2 is a better
pairing than one that has gone through transpositions and
exchanges. Part B of the rules state, that B3-B6 should be
fulfilled as much as possible. So if a straight pairing of
S1 and S2 is less compliant with B3-B6, it is worse than
another pairing, which is gained after some transpositions
or exchanges, but which is more B3-B6 compliant.



I agree.

This indicates I think that saving pairings is not what we
should be saving. A later pairing having gone through C7,8
will be better.

Then with C9-14 we start to drop the relative criteria.
These pairings are not 'good', but some of them will be the
same that were already considered. So perhaps they can be
saved?




My idea was not only to use 'memoization' (didn't know that word,
thanks), but mainly to check for best B3-B6 compliance.



I guess this is an important issue which I don't understand.




Let's take the examle I cited in this thread
(http://chesschat.org/showpost.php?p=167067&postcount=16)
from a Q-A sequence of Kevin Bonham and Geurt Gijssen:

I think, in this case, my idea would be helpful. As Geurt
said, the most likely pairings would be A-E, B-F, C-G, D-H,
that is S1 vs. S2. But it would be better to apply some
exchanges and transpositions until one gets S1: A, B, D, G
and S2: C, E, F, H. The resulting pairings A-C, B-E, D-F,
G-H would be recognized as more B3-B6 compliant, therefore
it would replace A-E, B-F, C-G, D-H as the "best pairing".



I think this issue has already been settled by Dennis
Jessop, but the problem of representing this situation as a
pairing table which could be computer-analyzed interested
me.

I tried constructing a pairing table that could be paired
with my 'pair' script. The idea of 2 groups, the first of
which is pairable without downfloats and the second (last)
which is unpairable, forcing a re-pairing of the first, or
even an amalgamation, is pretty much the same as in
the table of http://chesschat.org/showthread.php?t=6619,
which was for a real tournament.

I don't know if it is possible to generate a well-formed
table that has the form of the example of Kevin Bonham and
Geurt Gijssen.

With only 4 players, there needs to be at least 3 rounds
before an unpairable group is generated, I think. Everyone
has to play everyone else first.

Mine isn't well-formed, but the script only checks for the
right roles and opponents.

What I did was have the unpairable bracket, 15-18 play each other
over 3 rounds, except for one pair in the 3rd round, who I
found partners for elsewhere.

To do this, I had two shadow groups, 1-6 and 11-14, which provided
partners for the penultimate and final brackets, 7-10 and
15-18.



1 2,4,6 WBW 2
2 1,5,7 BWB 2
3 4,6,8 WBW 2
4 3,1,9 BWB 2
5 6,2,10 WBW 2
6 5,3,1 BWB 2

7 8,10,2 WBW 1.5
8 7,9,3 BWB 1.5
9 10,8,4 WBW 1.5
10 9,7,5 BWB 1.5

11 13,14,12 BWB 2
12 14,13,11 BBW 2
13 11,12,15 WWB 2
14 12,11,16 WBW 2

15 18,17,13 WBW 1
16 17,18,14 BWB 1
17 16,15,18 WWB 1
18 15,16,17 BBW 1


The pairable penultimate bracket, 7-10 needed such partners
too. I don't know if its shadow group is the minimum size
necessary, but I needed a big group, I think, to allow the 2
shadow groups to form one pairable score bracket.

C11, x=2,
C3, p=2 Homogeneous.
C4, S1 & S2: 7 8 & 9 10
C5, ordered: 7 8 &
9 10
C6, 2 paired. E4 9&7 E4 8&10
C6others: no non-paired players
Next, Bracket 3: 15 16 17 18
C1, NOK. 17 18
C13, Undoing Bracket 2 matches. Re-pairing Bracket 2. p=2. Bracket 2: 7 9 10 8 & Bracket 3: 15 16 17 18

The penultimate bracket is paired (x=2), but the
final bracket is unpairable, so in C13, the penultimate
bracket is re-paired.

C7, 10 9
C6, B1a: table 1 2 NOK
C7, last transposition
C8, exchange a: 7 9, 8 10
C5, ordered: 7 9 &
8 10
C6, B1a: table 1 2 NOK
C7, 10 8
C6, B1a: table 1 2 NOK
C7, last transposition
C8, last S1,S2 exchange
C9, B5,6 already dropped for Downfloats in Bracket 2.
C11, x=p=2 already, no more x increases in Bracket 2.
C14, Bracket 2, now p=1
C4, S1 & S2: 7 8 & 9 10
C5, ordered: 7 8 &
9 10
C6, 2 paired. E4 9&7
C6others: Floating remaining 8 10 Down. [2] 7 9 10 8 => [3] 10 8 15 16 17 18
Next, Bracket 3: 10 8 15 16 17 18
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Heterogeneous.
C4, S1 & S2: 10 8 & 15 16 17 18
C5, ordered: 8 10 &
15 16 17 18
C6, 2 paired. E1 8&15 E4 10&16
C6others: Remainder Group, Bracket 3: 17 18
C1, NOK. 17 18

But, what's this? The downfloating 8 and 10 should pair with
the unpairable 17 and 18, but they're pairing up with 15,
and 16, which are the pairable 2 in the final bracket!

And 17 and 18 form a remainder group, because the bracket is
heterogeneous!

And there the script starts to emit smoke, and dies.




The fact a set of pairings needs to be more compliant
with B3-B6 to replace the "best set of pairings until now" has a
built-in preference for earlier generated pairings. That is, it doesn't
really consider later pairings "as good as" earlier ones.

Does that make sense?

I agree if you're talking about a pairing reached through
C9-14. In those pairings, Relative Criteria requirements
have been waived.

But the players in the pairing won't have changed from the
first time round the pairing was considered in the
C6,C7,C8-cycle stage. The actual pairings being generated
don't change in each cycle through the later C9-C14 cycles.
Only the criteria change.

If you're talking about the C6-8 cycle, I think the most
recent pairing is better.

Bartolin
23-09-2007, 09:40 PM
I think this issue has already been settled by Dennis
Jessop, but the problem of representing this situation as a
pairing table which could be computer-analyzed interested
me.

I tried constructing a pairing table that could be paired
with my 'pair' script. The idea of 2 groups, the first of
which is pairable without downfloats and the second (last)
which is unpairable, forcing a re-pairing of the first, or
even an amalgamation, is pretty much the same as in
the table of http://chesschat.org/showthread.php?t=6619,
which was for a real tournament.

I don't know if it is possible to generate a well-formed
table that has the form of the example of Kevin Bonham and
Geurt Gijssen.

With only 4 players, there needs to be at least 3 rounds
before an unpairable group is generated, I think. Everyone
has to play everyone else first.

Mine isn't well-formed, but the script only checks for the
right roles and opponents.

What I did was have the unpairable bracket, 15-18 play each other
over 3 rounds, except for one pair in the 3rd round, who I
found partners for elsewhere.

To do this, I had two shadow groups, 1-6 and 11-14, which provided
partners for the penultimate and final brackets, 7-10 and
15-18.



1 2,4,6 WBW 2
2 1,5,7 BWB 2
3 4,6,8 WBW 2
4 3,1,9 BWB 2
5 6,2,10 WBW 2
6 5,3,1 BWB 2

7 8,10,2 WBW 1.5
8 7,9,3 BWB 1.5
9 10,8,4 WBW 1.5
10 9,7,5 BWB 1.5

11 13,14,12 BWB 2
12 14,13,11 BBW 2
13 11,12,15 WWB 2
14 12,11,16 WBW 2

15 18,17,13 WBW 1
16 17,18,14 BWB 1
17 16,15,18 WWB 1
18 15,16,17 BBW 1


The pairable penultimate bracket, 7-10 needed such partners
too. I don't know if its shadow group is the minimum size
necessary, but I needed a big group, I think, to allow the 2
shadow groups to form one pairable score bracket.

C11, x=2,
C3, p=2 Homogeneous.
C4, S1 & S2: 7 8 & 9 10
C5, ordered: 7 8 &
9 10
C6, 2 paired. E4 9&7 E4 8&10
C6others: no non-paired players
Next, Bracket 3: 15 16 17 18
C1, NOK. 17 18
C13, Undoing Bracket 2 matches. Re-pairing Bracket 2. p=2. Bracket 2: 7 9 10 8 & Bracket 3: 15 16 17 18

The penultimate bracket is paired (x=2), but the
final bracket is unpairable, so in C13, the penultimate
bracket is re-paired.

C7, 10 9
C6, B1a: table 1 2 NOK
C7, last transposition
C8, exchange a: 7 9, 8 10
C5, ordered: 7 9 &
8 10
C6, B1a: table 1 2 NOK
C7, 10 8
C6, B1a: table 1 2 NOK
C7, last transposition
C8, last S1,S2 exchange
C9, B5,6 already dropped for Downfloats in Bracket 2.
C11, x=p=2 already, no more x increases in Bracket 2.
C14, Bracket 2, now p=1
C4, S1 & S2: 7 8 & 9 10
C5, ordered: 7 8 &
9 10
C6, 2 paired. E4 9&7
C6others: Floating remaining 8 10 Down. [2] 7 9 10 8 => [3] 10 8 15 16 17 18
Next, Bracket 3: 10 8 15 16 17 18
C1, B1,2 test: ok, no unpairables
C2, x=1
C3, p=2 Heterogeneous.
C4, S1 & S2: 10 8 & 15 16 17 18
C5, ordered: 8 10 &
15 16 17 18
C6, 2 paired. E1 8&15 E4 10&16
C6others: Remainder Group, Bracket 3: 17 18
C1, NOK. 17 18

But, what's this? The downfloating 8 and 10 should pair with
the unpairable 17 and 18, but they're pairing up with 15,
and 16, which are the pairable 2 in the final bracket!

And 17 and 18 form a remainder group, because the bracket is
heterogeneous!

And there the script starts to emit smoke, and dies.

Oops. Shouldn't it have went as follows:

[...]
C1, NOK. 17 18
-> apply C13: Undo penultimate Bracket, try to find new pairings in
that Bracket.
-> Apply some transpositions until we have S1 8,10 S2 17,18,15,16
-> pair 8-17,20-18
-> remainder group 15,16
-> pair 15-16

Could it possibly be a bug with the numbering of the score brackets?
Your script wrote:



C6others: Floating remaining 8 10 Down. [2] 7 9 10 8 => [3] 10 8 15 16 17 18
Next, Bracket 3: 10 8 15 16 17 18
[...]
C6others: Remainder Group, Bracket 3: 17 18
C1, NOK. 17 18


Here we have "Bracket 3" twice. Shouldn't the last score bracket be number 4?

But maybe it's another problem.

Bartolin
23-09-2007, 10:58 PM
We are talking about pairs which don't satisfy the Relative
Criteria, but do satisfy the Absolute Criteria. They
shouldn't be downfloated (normally? ever?)

I think: They should be downfloated only if C explicitely says so.
As suggested by Bill Gletsos (http://chesschat.org/showpost.php?p=168059&postcount=7
C6 should be read as if it contained the following phrase:
"If C6 leads to a violation of B3-B6 then the following parts of
section C should be applied, however in doing this no player
should be moved down to a lower score bracket to cause B3-B6
to be satisfied."


[...] but what do we
do with them, when? Is it possible to rank pairings as
better than others? Where are those pairings found? At the
end of the pairing process, or earlier on?




IMHO, this point is important. I'm not sure about your
suggestion, that a straight pairing of S1 and S2 is a better
pairing than one that has gone through transpositions and
exchanges. Part B of the rules state, that B3-B6 should be
fulfilled as much as possible. So if a straight pairing of
S1 and S2 is less compliant with B3-B6, it is worse than
another pairing, which is gained after some transpositions
or exchanges, but which is more B3-B6 compliant.

I agree.

This indicates I think that saving pairings is not what we
should be saving. A later pairing having gone through C7,8
will be better.

Again, I'm not sure. Couldn't it be, that transpositions lead
to better B5,B6 compliance but to more inequalities regarding
score differences? I guess this can only happen in a large
jointed last score bracket?!

Generally speaking, I would neither earlier nor later pairings
expect to be "better" automatically. (Though I'm not absolutely
sure about this. I remember, you argued elsewhere that the
procedure in C tends to take B3-B6 into account by itself.)
Therefore I suggested to compute a penalty value for
"B3-B6 noncompliance" for each pairing one comes across.


Then with C9-14 we start to drop the relative criteria.
These pairings are not 'good', but some of them will be the
same that were already considered. So perhaps they can be
saved?

I guess, C9 becomes superfluous if we take the way I described
, because we want to check all pairings anyway -- until we get one
which is fully B3-B6 compliant.

But regarding C10-14: Don't we do something else here, except
dropping relative criteria? We're changing pairings already made
in another score bracket as well. Therefore I think the penalty
values computed up to here can't be used anymore.



My idea was not only to use 'memoization'
(didn't know that word, thanks), but mainly to check for best B3-B6 compliance.


I guess this is an important issue which I don't understand.

Hmm, I don't know how to express it better. I'll try again: Let's assume
we arrive at C6 and the current set of pairings is in compliance with
B1,B2 but not fully compliant with B3-B6. Our goal (according to the
'instruction' in B) is to search for a set of pairings which is most
compliant with B3-B6. Since we can't be sure that applying C7 and C8
automatically results in "better" pairings (for instance in case of a large
jointed last score bracket inequalities in score differences could arise),
it seems sensible to store a penalty value for each pairing and compare
it with the value of the best pairing we got so far. Only if the later pairing
has a smaller penalty value, it is considered "better".

Therefore the stored penalty value is not only used for 'memoization', but
also to be able to check whether a newly generated pairing is really "better"
than the best we had so far.

But maybe, I'm off the track.

drbean
26-09-2007, 08:40 PM
As suggested by Bill Gletsos
(http://chesschat.org/showpost.php?p=168059&postcount=7
C6 should be read as if it contained the following phrase:
"If C6 leads to a violation of B3-B6 then the following
parts of section C should be applied, however in doing this
no player should be moved down to a lower score bracket to
cause B3-B6 to be satisfied."



I guess this is pretty close to the right way of rewording
C6. There is no explicit statement in C6 that B3-B6 Relative
Criteria should be considered in the pairing of the bracket.
It's only in the later C9-14 procedures which talk about
relaxation of the Relative Criteria that it becomes clear
you were supposed to be considering them.

I can remember being confused. I think this confusion might
be reflected in a lack of clearness in the way that
Games::Tournament::Swiss implements C6.




Again, I'm not sure. Couldn't it be, that transpositions lead
to better B5,B6 compliance but to more inequalities regarding
score differences? I guess this can only happen in a large
jointed last score bracket?!



I wonder if the concern with ordering by pairing number
isn't an attempt to reduce a similar sort of inequality.
Within a bracket, instead of pairing the top player with the
bottom player and 2 players in the middle together, the top
player is paired with a player in the middle of the bracket.
The bottom player is also paired with a player in the
middle.

Transpositions and exchanges are going to reduce this
'equality' of ranking difference.

Note that in C8, the exchange procedures of D2 are not
applied to heterogeneous groups. This prevents consideration
of pairings that partner players with opponents originating
from within the same higher bracket and pairing of players
together from within the lower. Is the intention of C to
keep score differences uniform?

Alternatively, it may just be due to the fact that there
will be no pairings of players together with partners who
also came from the higher bracket. If they could be
partnered, they would have been partnered in the higher
bracket. Unless pprime has been reduced trying to pair the
final bracket. And then we want them to find partners from
the final bracket.




Generally speaking, I would neither earlier nor later
pairings expect to be "better" automatically. (Though I'm
not absolutely sure about this. I remember, you argued
elsewhere that the procedure in C tends to take B3-B6 into
account by itself.) Therefore I suggested to compute a
penalty value for "B3-B6 noncompliance" for each pairing one
comes across.



This idea of calculating values for the pairings that we
consider in the first C6-8 cycle suggests there is an
alternative, but equivalent, procedure to C04, which doesn't
require cycling through the same C6-8 cycle, looking at the
same pairings a number of times deciding if they meet our
requirements.

After the first cycle doesn't find a pairing that satisfies
the Absolute Criteria and the Relative Criteria, the C04
procedure involves repetition of the same B6-8 cycle after,
first, relaxation of the B6 Criterion in C9, then relaxation
of B5 in C9, then, p relaxations of the B4 preference in
C11. Finally we go through the same B6-8 cycle p times,
prepared to accept one less pair each time in C14. Thus, we
may go through the B6-8 cycle a maximum 2p+2 times.

C10,12,13 are some special rules for special types of
brackets.




I guess, C9 becomes superfluous if we take the way I
described , because we want to check all pairings anyway --
until we get one which is fully B3-B6 compliant.



So we store some representation of which Criteria are
satisfied by the pairing, or which aren't, the first time
through, and then if we come to the end of the C6-C8 cycle
without having found a perfect Relative-Criteria-compliant
pairing, we have a way of deciding which of the pairings is
best?




But regarding C10-14: Don't we do something else here,
except dropping relative criteria? We're changing pairings
already made in another score bracket as well. Therefore I
think the penalty values computed up to here can't be used
anymore.



Perhaps we would just use different cycles for different
types of score bracket.




Hmm, I don't know how to express it better. I'll try again:
Let's assume we arrive at C6 and the current set of pairings
is in compliance with B1,B2 but not fully compliant with
B3-B6. Our goal (according to the 'instruction' in B) is to
search for a set of pairings which is most compliant with
B3-B6. Since we can't be sure that applying C7 and C8
automatically results in "better" pairings (for instance in
case of a large jointed last score bracket inequalities in
score differences could arise), it seems sensible to store a
penalty value for each pairing and compare it with the value
of the best pairing we got so far. Only if the later pairing
has a smaller penalty value, it is considered "better".



You have to have some way of converting Relative Criteria
compliance into a score. The more compliances a pairing has
and the more important compliances it has, the better the
pairing. Is it possible to rank all the possible
compliances?

We can count xprime and pprime. Can we count B5,6
compliance?

Bartolin
28-09-2007, 12:13 AM
I wonder if the concern with ordering by pairing number
isn't an attempt to reduce a similar sort of inequality.
Within a bracket, instead of pairing the top player with the
bottom player and 2 players in the middle together, the top
player is paired with a player in the middle of the bracket.
The bottom player is also paired with a player in the
middle.

Transpositions and exchanges are going to reduce this
'equality' of ranking difference.

Note that in C8, the exchange procedures of D2 are not
applied to heterogeneous groups. This prevents consideration
of pairings that partner players with opponents originating
from within the same higher bracket and pairing of players
together from within the lower. Is the intention of C to
keep score differences uniform?

Alternatively, it may just be due to the fact that there
will be no pairings of players together with partners who
also came from the higher bracket. If they could be
partnered, they would have been partnered in the higher
bracket. Unless pprime has been reduced trying to pair the
final bracket. And then we want them to find partners from
the final bracket.

I think the second point is the more important one. In a heteregenous group we already know that players from S1 can't or shouldn't be paired against each other. Therefore exchanges aren't useful.



Generally speaking, I would neither earlier nor later
pairings expect to be "better" automatically. (Though I'm
not absolutely sure about this. I remember, you argued
elsewhere that the procedure in C tends to take B3-B6 into
account by itself.) Therefore I suggested to compute a
penalty value for "B3-B6 noncompliance" for each pairing one
comes across.


This idea of calculating values for the pairings that we
consider in the first C6-8 cycle suggests there is an
alternative, but equivalent, procedure to C04, which doesn't
require cycling through the same C6-8 cycle, looking at the
same pairings a number of times deciding if they meet our
requirements.

After the first cycle doesn't find a pairing that satisfies
the Absolute Criteria and the Relative Criteria, the C04
procedure involves repetition of the same B6-8 cycle after,
first, relaxation of the B6 Criterion in C9, then relaxation
of B5 in C9, then, p relaxations of the B4 preference in
C11. Finally we go through the same B6-8 cycle p times,
prepared to accept one less pair each time in C14. Thus, we
may go through the B6-8 cycle a maximum 2p+2 times.

C10,12,13 are some special rules for special types of
brackets.


I guess, C9 becomes superfluous if we take the way I
described , because we want to check all pairings anyway --
until we get one which is fully B3-B6 compliant.


So we store some representation of which Criteria are
satisfied by the pairing, or which aren't, the first time
through, and then if we come to the end of the C6-C8 cycle
without having found a perfect Relative-Criteria-compliant
pairing, we have a way of deciding which of the pairings is
best?

Yes, that was my idea. But it seems to be a (maybe fruitless) attempt to invent an (cleaner) alternative to C04, which arrives at those results C04 is meant to arrive at.




But regarding C10-14: Don't we do something else here,
except dropping relative criteria? We're changing pairings
already made in another score bracket as well. Therefore I
think the penalty values computed up to here can't be used
anymore.

Perhaps we would just use different cycles for different
types of score bracket.

Ahh, another attempt to reconstruct C04? ;-)



You have to have some way of converting Relative Criteria
compliance into a score. The more compliances a pairing has
and the more important compliances it has, the better the
pairing. Is it possible to rank all the possible
compliances?

We can count xprime and pprime. Can we count B5,6
compliance?

I think we could invent some ranking procedure. But to do that, the vague wording of B regarding the Relative Criteria ("These are in descending priority. They should be fulfilled as much as possible.") must be translated to a specific ranking algorithm. It must be clear what is "better":

* one score difference of 2 and one of 0 OR two score differences of 1
* one violation of B5 OR instead two violations of B6
* ...

One way to go might be to assign four numerical values to each set of pairings. The first value must be a representation of B3 violation (for instance the sum of score differences over all pairings), the second value stands for B4 violation, the third for B5 violation (for instance number of single pairings not fulfilling B5), the forth for B6 violation.

When comparing different sets of pairings, we start with the value of B3 violation and choose the set of pairings with the lowest value. If there is more than one set of pairings with that value, we compare the second value, and so on.

But again, this is just brainstorming.

drbean
12-10-2007, 01:29 PM
I think the second point is the more important one. In a heteregenous group we already know that players from S1 can't or shouldn't be paired against each other. Therefore exchanges aren't useful.

Yes, that was my idea. But it seems to be a (maybe fruitless) attempt to invent an (cleaner) alternative to C04, which arrives at those results C04 is meant to arrive at.

Ahh, another attempt to reconstruct C04? ;-)

I think we could invent some ranking procedure. But to do that, the vague wording of B regarding the Relative Criteria ("These are in descending priority. They should be fulfilled as much as possible.") must be translated to a specific ranking algorithm. It must be clear what is "better":

* one score difference of 2 and one of 0 OR two score differences of 1
* one violation of B5 OR instead two violations of B6
* ...

One way to go might be to assign four numerical values to each set of pairings. The first value must be a representation of B3 violation (for instance the sum of score differences over all pairings), the second value stands for B4 violation, the third for B5 violation (for instance number of single pairings not fulfilling B5), the forth for B6 violation.

When comparing different sets of pairings, we start with the value of B3 violation and choose the set of pairings with the lowest value. If there is more than one set of pairings with that value, we compare the second value, and so on.

But again, this is just brainstorming.

C04 is like the wheels in an odometer. One wheel spins around one complete revolution for each single step that the wheel on the left of it (the procedure below it) takes.

C04 is like doing an arithmetical operation on some numbers, then doing the same operation in order on all the numbers lower than these numbers and then stopping when the answer is the same as the answer we first got.