Hi,
Three questions about accelerated pairings:
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Is there a preference on whether to use 28R1 or 28R2? The book seems to give a preference to 28R2, but it is not explicit about this. 28R3 is clearly labeled Variation, so no question about that.
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Is the first round pairing with 28R2 really equal to that of 28R1, as the book states? It seems to me that there is ambiguity here where the odd player is concerned. In 28R1, the odd player in the top half would play the top rated player in the bottom half, as per normal pairing procedures for the odd man. However, 28R2 prescribes dividing the whole player list into four groups of roughly the same size. In doing this, you get groups that vary by at most by one player. Note that it is indeterminate which groups get the most players (it would make sense, perhaps that the higher rated groups would, but this again would not give the same results as 28R1). An example:
10 players rated: 1510, 1509, 1508, 1507, 1506, 1505, 1504, 1503, 1502, 1501
According to 28R1 pair as: 1510-1508, 1507-1509, 1506-1505, 1502-1504, 1503-1501
According to 28R2, you could have groups of size 3-3-2-2, 3-2-3-2, 2-3-2-3, 2-2-3-3, 3-2-2-3, 2-3-3-2. Some don’t even make sense, as the odd player would be in the wrong place or create an imbalance. Only 2-3-3-2 gives you a pairing as per 28R1. I think the problem is that the rule prematurely asks to split the score groups. It should say instead “For the first round, calculate as per 28R1”, or something like that, if that’s the real intent.
- 29E2 only mentions color allocation when the top half plays the lower half (it is explicit about this). I guess you could use the TD tip for top board in other sections to deal with the “effective” lower score group, but perhaps what to should be mentioned explicitly for accelerated pairings, specially where it concerns the odd player. With the odd player as in my example above, taking 29E2 to apply literally to accelerated pairings as well would leave players 1501 through 1504 without a color determination from the rules.
Thanks,
Luis