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  1. #1
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    Incorporation of B3-B6 into pairing procedure C (Swiss Dutch)

    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=...&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=...&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=...&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.

    Code:
                    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...nt-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

  2. #2
    Illuminati Bill Gletsos's Avatar
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    Quote Originally Posted by Bartolin
    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.

    Code:
                    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...nt-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
    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.
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  3. #3
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    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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

    Code:
     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.

    Quote Originally Posted by Bartolin

    .. 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.

    Quote Originally Posted by Bartolin



    Code:
                    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?
    Quote Originally Posted by Bill Gletsos
    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.

    Quote Originally Posted by Bartolin
    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...nt-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 :-)

  4. #4
    Illuminati Bill Gletsos's Avatar
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    Quote Originally Posted by drbean
    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).
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  5. #5
    Illuminati Bill Gletsos's Avatar
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    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).
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  6. #6
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    Quote Originally Posted by drbean
    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).

    Quote Originally Posted by drbean
    .. 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.

    Quote Originally Posted by Bill Gletsos
    Quote Originally Posted by drbean
    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.

  7. #7
    Illuminati Bill Gletsos's Avatar
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    Quote Originally Posted by Bartolin
    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).
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  8. #8
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    Quote Originally Posted by bartolin
    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.
    Quote Originally Posted by Bill Gletsos
    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=...&postcount=158.

  9. #9
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    Quote Originally Posted by drbean
    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?

  10. #10
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    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin
    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.' [JOKE] Paraphrasing Jamie Zawinski,
    "Some people, when confronted with a problem, think I know,
    I'll use 'memoization'. Now they have two problems." [/JOKE]

    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.

  11. #11
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    Quote Originally Posted by drbean
    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.

    Quote Originally Posted by drbean
    This looks like 'memoization.' [JOKE] Paraphrasing Jamie Zawinski,
    "Some people, when confronted with a problem, think I know,
    I'll use 'memoization'. Now they have two problems." [/JOKE]

    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".

    Quote Originally Posted by drbean
    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.

    Quote Originally Posted by drbean
    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?

  12. #12
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    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?

    Quote Originally Posted by Bartolin

    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?

    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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.

    Code:
     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.

    Quote Originally Posted by Bartolin

    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.

  13. #13
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    Quote Originally Posted by drbean
    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.

    Code:
     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.

  14. #14
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    Quote Originally Posted by drbean
    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."

    Quote Originally Posted by drbean
    [...] 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?
    Quote Originally Posted by drbean
    Quote Originally Posted by Bartolin
    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.

    Quote Originally Posted by drbean
    Quote Originally Posted by Bartolin
    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.

  15. #15
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    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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?

    Quote Originally Posted by Bartolin

    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.

    Quote Originally Posted by Bartolin

    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?

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