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23. Mayo 2015, 04:36:12
Walter Montego 
And instead it's 8-0-0?
There's a ghost in the machine.
I was thinking you should all have 9 of those S-B points 2 × 4 + 1, but I am not sure how it works. Maybe I'll read the explanation again. After reading it, I am confused and think you should all have 5 points, but that's not right since you all have 5 from wins alone. So I'm going to try to apply what I read and see if I can come up with 11 each. I do know that I am right about you three being equal no matter what it is if each of you defeated the other in a rotation as you have done.

I can't figure it out for 11 points. I'll stick with 9 each, but when I look at the Section 2 score my 2 × 3 + 1 = 7 works, but that could just be a coincidence that matches the correct score. I cannot see how player 4 Faith got only 1 S-B point though she won two games.

23. Mayo 2015, 05:09:59
rod03801 
Asunto: Re:
Modificado por rod03801 (23. Mayo 2015, 05:10:30)
Walter Montego: It's every win from the section.
just for ME. I won against players #1 #2 #4 #6 and #7
so thats 5 + 2 + 3 + 0 + 1 = 11

For Hqrls . HE won against players #2, #4, #5, #6 and #7
so that's 2 + 3 + 5 + 0 + 1 = 11

For Pedro. HE won against players #2, #3, #4, #6 and #7
so that's 2 + 5 + 3 + 0 + 1 = 11

Obviously the results are WRONG

23. Mayo 2015, 15:38:58
Walter Montego 
Asunto: Re:
rod03801: So you get a point for each win that an opponent of yours scored if you defeated that opponent? I see the totals now from your explanation. Thank you. And this explains the half point for draws, which for this particular tournament cannot happen. Yes, this tie breaking method is a great one for the computer to solve. Tedious and time consuming to do by hand.

As ThunderGr posts, Fencer should be told so he can restart the tournament with the correct people in it.

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