For practice, I am using WinTD 4.20 to re-create a small (11-player) section of a scholastic tournament run last year.
In the second round, there were five players at zero, and two at one-half. So the top two in the ZERO score group were paired with the two players who drew in Round 1, leaving three players in the ZERO score group. Here are their ratings and first round colors:
601, White
554, White
504, Black
I expected the 504 player to get the BYE, as happened in the actual tournament, and the pairing to be 601 vs. 554. But WinTD transposes the bottom half according to 29E5a. The 80-point rule, gives the 554 player the BYE, and pairs 504 vs 601.
It just feels wrong to transpose a player to a BYE for the purpose of “maximizing the number of players who receive their due color.”
Nonetheless, it does comport with 29D1. Determination, which says, “Care must be taken… that the color consequences are acceptable (29E, 29E3).”
“29E3. Due Colors in succeeding rounds. As many players as possible are given their due colors in each succeeding round, so long as the pairings conform to the basic Swiss system rules.”
I believe the actual tournament was run with SwissSys.
SwissSys correctly implements rule 28L2: “In subsequent rounds, [the bye] is given to the lowest-rated player in the lowest score group, but not to an unrated player.” WinTD treats the full-point bye the same as any other odd man situation, as in rule 29D1 which you cited. Strictly speaking, this should be announced as a pairing variation at the start of the tournament.
Ken Ballou (wileycoyote) has said that he’s considering submitting an ADM to allow the WinTD pairings as a variation, but we haven’t discussed it in the Rules Committee.
Actually, to be more precise, the ADM I submitted this year would have modified 28L2 to allow the WinTD pairings as the rule (no variation involved). The ADM was referred to the rules committee, so it will come back at the next Delegates’ Meeting under “old business.”
I’m trying to figure out how there were five with 0 and 2 with 1/2 in an 11 player tournament. My advanced math tells me that leaves 4 people with 1. How did you get 4 winners and 5 losers? Usually if anything there would be more people with 1 than with 0, because someone might have had a bye.
None of this helps with the rules question, but I think the rules gurus have that part covered.