On odd sizes, if you allow for the person whose "#", so to speak, has turned up to have a 'bye', then I suspect it works pretty well.

In other words, if player 2 in a 7 player tournament has a bye in round 5, player 3 a bye in round 3 and player 4 a bye in round 1, and writing code to allow byes in odd sized tourneys, the same kind of logic as in the even case should fit.

This chart might explain it better:

1: 7-6-5-4-3-2
2: 6-5-4-3-B-1
3: 5-4-B-2-1-7
4: B-3-2-1-7-6
5: 3-2-1-7-6-B
6: 2-1-7-B-5-4
7: 1-B-6-5-4-3


David.

In reply to Re^2: Algorithm problem: Order matches by difference between players by dwm042
in thread Algorithm problem: Order matches by difference between players by Dirk80

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