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  1. #1
    CC Candidate Master
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    'penultimate bracket' ambiguous

    C13 talks about unpairing the penultimate score bracket, if
    the lowest score bracket cannot be paired.

    In discussion with bartolin, I was wondering whether, if the
    penultimate score bracket was a heterogeneous group with a
    remainder group, the bracket which should be re-paired would
    be the remainder group, or the whole heterogeneous group from
    which the remainder group came.

    The pairing table which he came up with:

    Code:
                    Round 3 Pairing Groups
    -------------------------------------------------------------------------
    Place  No  Opponents     Roles     Float Score
    1-2
           1   7,2            BW            1.5
           2   8,1            WB            1.5
    3-10
           3   9,11           BW            1
           4   10,12          WB            1
           5   11,9           BW            1
           6   12,10          WB            1
           9   3,5            WB            1
           10  4,6            BW            1
           11  5,3            WB            1
           12  6,4            BW            1
    11-12
           7   1,8            WB            0.5
           8   2,7            BW            0.5
    This is somewhat similar to the table of ggrayggray's
    http://www.chesschat.org/showthread.php?t=6619, in that the
    last score bracket is unpairable as it is, if not as
    elegantly.

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

    C6, Bracket tables 1 2 paired. E1 4&1 E1 2&3
    C6others: Remainder Group, Bracket 2: 5 6 9 10 11 12

    The remainder group is paired as:

    C6, Bracket tables 1 2 3 paired. E1 6&5 E1 9&10 E1 11&12
    C6others: no non-paired players
    Next, Bracket 3: 7 8

    But because 7 and 8 have already played each other, they can't
    be paired, and C13 applies.

    The question is: Is the remainder group, 5 6 9 10 11 12
    unpaired, or is the whole of group 2, 1 2 3 5 6 9 10 11 12
    unpaired?

    He took the position that only the remainder group is
    unpaired. I took the position that the whole group is
    unpaired.

    See

    Linkname: #29073: Problems with C14 in FIDE.pm
    URL: http://rt.cpan.org/Ticket/Display.html?id=29073

    down towards the end.

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

    C13.
    In case of the lowest score bracket: the pairing of the
    penultimate score bracket is undone. Try to find another
    pairing in the penultimate score bracket which will allow a
    pairing in the lowest score bracket. If in the penultimate
    score bracket p becomes zero (i.e. no pairing can be found
    which will allow a correct pairing for the lowest score
    bracket) then the two lowest score brackets are joined into a
    new lowest score bracket. Because now another score bracket
    is the penultimate one C13 can be repeated until an
    acceptable pairing is obtained.

    The whole question of the status of remainder groups is
    unclear to me. Are they new brackets that are interposed into
    the original ones, or is their existence only virtual?

  2. #2
    CC Candidate Master
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    Quote Originally Posted by drbean
    C13 talks about unpairing the penultimate score bracket, if
    the lowest score bracket cannot be paired.

    In discussion with bartolin, I was wondering whether, if the
    penultimate score bracket was a heterogeneous group with a
    remainder group, the bracket which should be re-paired would
    be the remainder group, or the whole heterogeneous group from
    which the remainder group came.

    [...]

    He took the position that only the remainder group is
    unpaired. I took the position that the whole group is
    unpaired.

    [...]

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

    C13.
    In case of the lowest score bracket: the pairing of the
    penultimate score bracket is undone. Try to find another
    pairing in the penultimate score bracket which will allow a
    pairing in the lowest score bracket. If in the penultimate
    score bracket p becomes zero (i.e. no pairing can be found
    which will allow a correct pairing for the lowest score
    bracket) then the two lowest score brackets are joined into a
    new lowest score bracket. Because now another score bracket
    is the penultimate one C13 can be repeated until an
    acceptable pairing is obtained.

    The whole question of the status of remainder groups is
    unclear to me. Are they new brackets that are interposed into
    the original ones, or is their existence only virtual?
    Since no one answered until now, I will try to reformulate my argument that only the remainder group is to be unpaired. Maybe someone familiar with the pairing procedure can confirm my reasoning or can point out, what I'm missing?

    I think the relevant part of the rules is A3, 3rd sentence, part 2:

    When pairing a heterogeneous score bracket these players moved down are always paired first whenever possible, giving rise to a remainder score bracket which is always treated as a homogeneous one.
    According to this, the remainder group seems to be a "regular" score bracket. It comes into existence only after C6 dot 2 applies. If one deals with the following score bracket, any reference to "the previous" or "the penultimate" score bracket must refer to this new score bracket. The fact that C6 dot 2 says "Start at C2 with the homogeneous remainder group." instead of "Start at C2 with the homogeneous score bracket, formed by the remainder group" (or something like this) is IMHO an example of imprecise wording.

    Also the introduction to C states:

    Starting with the highest score bracket apply the following procedures to all score brackets until an acceptable pairing is obtained.
    Therefore I would assume that all rules apply to "score brackets" and to "score brackets" only. Therefore any reference to "remainder groups" are to be interpreted as references to "score brackets formed by a remainder group" (or likewise).

    What do you think?

    Christian

  3. #3
    CC Grandmaster Denis_Jessop's Avatar
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    I agree that this looks like yet another case of bad drafting in C. The references in it to "group" should be to "bracket" as it is clear from A3 that a remainder score group (sic) is to be treated as a score bracket for pairing purposes. I think this is reasonably clear, though not well expressed, when A3 is read with the second dot point in C6.

    DJ
    ...I don't want to go among mad people Alice remarked, "Oh, you can't help that," said the Cat: we're all mad here. I am mad. You're mad." "How do you know I'm mad?" said Alice. "You must be," said the Cat ,"or you wouldn't have come here."

  4. #4
    CC Candidate Master
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    remainder group tie precedent in C10,11

    Although this is not a conclusive argument, C10,11 set a
    precedent for including the remainder group with the group
    from which it came when re-pairing a remainder group.

    They say:

    Code:
    C10.
    In case of a homogeneous remainder group: undo the pairing of
    the lowest moved down player paired and try to find a
    different opponent for this player by restarting at C7.
    
    If no alternative pairing for this player exists then drop
    criterion B6 first and then B5 for upfloats and restart at
    C2.
    The homogeneous remainder group is being repaired, but the
    group from which it came is being repaired (partially) too.

    Code:
    C11.
    As long as x is less than p: increase x by 1. When pairing a
    remainder group undo all pairings of players moved down also.
    Restart at C3.
    The remainder group is not only being repaired, the whole
    bracket is being repaired.

    Incidentally, it looks like there is a significance to the
    different use of the words bracket and group. I thought they
    were interchangeable, but the term 'score group' never
    appears.

    S1 and S2 are referred to as 'subgroups'. The other only use
    of 'group' is to refer to 'remainder groups'.

    There is no reference to a 'group' in C13, only the
    penultimate 'score bracket.'

    Although this is a precedent for regarding the 'group' to be
    repaired in C13 as the whole bracket, rather than the
    'remainder group', there may be good reason to ignore it in
    this case, because the pairing has reached the bottom
    bracket, and is now going back up.

    Perhaps pairing just the remainder group means less
    likelihood of disturbing the good pairings at the top.

    If the remainder group can be paired without pairing the
    whole bracket, fewer higher-placed players' matches will be
    affected.

  5. #5
    CC Candidate Master
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    Quote Originally Posted by drbean
    Although this is not a conclusive argument, C10,11 set a
    precedent for including the remainder group with the group
    from which it came when re-pairing a remainder group.
    Well, but one could turn your argument against you and say: C10 and C11 explicitely state that one or more pairings outside the current score bracket (C10,11 say 'group') shall be undone. Therefore the 'normal' modus seems to be to deal with the current score bracket only -- and not with the score bracket the remainder group arose from.

    But my impression is, that we won't get much farther by interpreting the (obviously) inferior wording of the rules word-for-word. Nevertheless, I will add one more point in favour of the view that any remainder group should be treated as a score bracket of its own.

    C6 says:

    in case of a heterogeneous score bracket: only players moved down were paired so far. Start at C2 with the homogeneous remainder group.
    So it talks (imprecisely, I think) about a "homogeneous remainder group". But directely after that, C2 refers to A8 which says:

    Definition of "x"

    The number of pairings which can be made in a score bracket, either homogeneous or heterogeneous, not fulfilling all colour preferences, is represented by the symbol x.
    Here we have a reference to a "score bracket" again.

    But I think the point you made about not disturbing the good pairings at the top ...

    Quote Originally Posted by drbean
    Perhaps pairing just the remainder group means less
    likelihood of disturbing the good pairings at the top.

    If the remainder group can be paired without pairing the
    whole bracket, fewer higher-placed players' matches will be
    affected.
    ... is a very strong one. Even without interpreting the rules word-for-word I would argue, it's better to re-pair only the remainder group and leave the pairings of the higher-placed players as they are.

    But on the other hand, I'm no arbiter at all and don't have any experience with interpreting the pairing rules. Since we need an answer to this question to make progress with the bug report at http://rt.cpan.org/Public/Bug/Display.html?id=29073 , I would be very happy to get another statement of one of the experienced arbiters around.

    Thanks

    Christian

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