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  1. #1
    CC Candidate Master
    Join Date
    Oct 2006
    Posts
    72

    calculation of x for heterogeneous bracket

    My question is: Should the calculation of x depend on whether the bracket is homogeneous or not and thus whether p is less than half the number of players?

    The handbook definition is not very clear, but x appears to represent the
    lower limit, ranging from zero to p, of the number of matches in the score
    bracket in which players will have their preferences unsatisfied. A8

    This is an undesirable state of affairs, but if there are more players with one preference than another, one or more players is going to have to play with their preference unsatisfied.

    At least this is the situation in a homogeneous bracket where p is half the number of players. And this seems to be the assumption of the calculation of x.

    If b >> w then x = b-q, else x = w-q.

    But what about the situation in a heterogeneous bracket where the number of players in S1 is less than the number in S2? In this situation, where not all of the players in S2 will be paired, should we be looking at subsets of S2 with p players?

    Thus if we have one player with a White preference in S1 and 3 players in S2, one with a Black preference and 2 with a White preference, is x not 1 but 0?

    The handbook says x can be calculated as above, but doesn't say anything about homogeneous and heterogeneous, even though the definition does.

  2. #2
    CC Rookie
    Join Date
    Dec 2008
    Location
    Roermond, Netherlands
    Posts
    3
    Quote Originally Posted by drbean
    My question is: Should the calculation of x depend on whether the bracket is homogeneous or not and thus whether p is less than half the number of players?

    The handbook definition is not very clear, but x appears to represent the
    lower limit, ranging from zero to p, of the number of matches in the score
    bracket in which players will have their preferences unsatisfied. A8

    This is an undesirable state of affairs, but if there are more players with one preference than another, one or more players is going to have to play with their preference unsatisfied.

    At least this is the situation in a homogeneous bracket where p is half the number of players. And this seems to be the assumption of the calculation of x.

    If b >> w then x = b-q, else x = w-q.

    But what about the situation in a heterogeneous bracket where the number of players in S1 is less than the number in S2? In this situation, where not all of the players in S2 will be paired, should we be looking at subsets of S2 with p players?

    Thus if we have one player with a White preference in S1 and 3 players in S2, one with a Black preference and 2 with a White preference, is x not 1 but 0?

    The handbook says x can be calculated as above, but doesn't say anything about homogeneous and heterogeneous, even though the definition does.
    In a heterogeneous pairinggroup x is calculated the same as in a homogeneous group.
    In the given example:
    P=1 W=3 B=1 Q=2 X=1
    The calculation makes that one player will not get the colour of his preference. It doensn't mather if it is the upfloater (the pairing to be made first) or the left over pairing to be made in a (homogeneous) restgroup.

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