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Chang
29-09-2006, 03:03 AM
Hi,

I don’t understand the FIDE Swiss Rule in….
....
A6. Subgroups
To make the pairing, each score bracket will be divided into two subgroups, to be called S1 and S2. In a heterogeneous score bracket S1 contains all players moved down from a higher score bracket. In a homogeneous score bracket S1 contains the higher half (rounding downwards) of the number of players in the score bracket.

The number of players in S1 will be indicated by "p", indicating the number of pairings to be made. In both cases S2 contains all other players of the score bracket. In both S1 and S2 players are ordered according to A2……
......

Suppose in the second round two one-point players are moved down to half point bracket which originally has 6 players. How many players would be in S1? Two or Four? I think in this case is two and these two would be provisionally paired with the first two of half point players. The rest 4 half point players would form another homogeneous bracket. Am I correct? Thank you for your reply.

Regards,

Chang
29-09-2006, 03:09 AM
I just realize that it's impossible to have 2 one-point players moved down in the second round. Anyway, I think you understand my point. Sorry...

Denis_Jessop
29-09-2006, 05:51 PM
Hi,

I don’t understand the FIDE Swiss Rule in….
....
A6. Subgroups
To make the pairing, each score bracket will be divided into two subgroups, to be called S1 and S2. In a heterogeneous score bracket S1 contains all players moved down from a higher score bracket. In a homogeneous score bracket S1 contains the higher half (rounding downwards) of the number of players in the score bracket.

The number of players in S1 will be indicated by "p", indicating the number of pairings to be made. In both cases S2 contains all other players of the score bracket. In both S1 and S2 players are ordered according to A2……
......

Suppose in the second round two one-point players are moved down to half point bracket which originally has 6 players. How many players would be in S1? Two or Four? I think in this case is two and these two would be provisionally paired with the first two of half point players. The rest 4 half point players would form another homogeneous bracket. Am I correct? Thank you for your reply.

Regards,
Hi Chang

If you don't understand the FIDE Swiss Rules you are one of a substantial majority of chess players :)

To answer your specific question, when pairing a heterogeneous score bracket, S1 contains the player or players moved down and S2 contains the players in the score group down to which they were moved. (Digression: I expect many \$HCDs for that grammatical tour de force.:rolleyes: ) That is, S1 and S2 do not have to contain the same number of players.

DJ