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Balancer broken?

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Date Editor Before After
4/24/2016 11:40:12 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:36:35 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:34:50 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:31:34 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:31:15 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:29:40 PMGBrankTheEloIsALie before revert after revert
4/24/2016 11:29:08 PMGBrankTheEloIsALie before revert after revert
Before After
1 [quote]Both are O(n)[/quote] 1 [quote]Both are O(n)[/quote]
2 ...anything higher would simply blow up for the objective function of a (semi?-)brute-forced NP problem. 2 ...anything higher would simply blow up for the objective function of a (semi?-)brute-forced NP problem.
3 \n 3 \n
4 I haven't looked into the code (well, recently enough), but I'd wager a guess that the current cost function is below O(n) in the grand scheme of things because you don't actually evaluate the full function for every combination. 4 I haven't looked into the code (well, recently enough), but I'd wager a guess that the current cost function is below O(n) in the grand scheme of things because you don't actually evaluate the full function for every combination.
5 \n 5 \n
6 What I mean: If you only ever swap out two players when mutating your team composition, then you only need to subtract their elo difference from one team and add it to the other. I'm not currently in the state of mind to investigate whether you can cover all combinations with this efficiently, but if you can, then you can calculate the cost functions of N team compositions in O(N), not O(N * n) with n being the amount of players. 6 What I mean: If you only ever swap out two players when mutating your team composition, then you only need to subtract their elo difference from one team and add it to the other. I'm not currently in the state of mind to investigate whether you can cover all combinations with this efficiently, but if you can, then you can calculate the cost functions of N team compositions in O(N), not O(N * n) with n being the amount of players.
7 \n 7 \n
8 With variance/standard deviation, the change in the mean means (no pun intended) that you definitely need to recalculate the entire function. 8 With variance/standard deviation, the change in the mean means (no pun intended) that you definitely need to recalculate the entire function.
9 \n 9 \n
10 [quote]You can do addition and multiplication in the same cycle btw[/quote] 10 [quote]You can do addition and multiplication in the same cycle btw[/quote]
11 I know about FMA, but that's not the type of cost reduction I'm talking about. 11 I know about FMA, but that's not the type of cost reduction I'm talking about.
12 \n 12 \n
13 \n 13 \n
14 For the record: This is the function we're essentially talking about here. 14 For the record: This is the function we're essentially talking about here.
15 [quote]sum_(k=1)^(N)(sum_(l=1)^(N)((avg_(w, X_k)(T_k)-avg_(w, X_l)(T_l))² + c(sqrt(Var_(w, X_k)(T_k)) - sqrt(Var_(w, X_l)(T_l)))²)). 15 [quote]sum_(k=1)^(N)(sum_(l=1)^(N)((avg_(w, X_k)(T_k)-avg_(w, X_l)(T_l))² + c(sqrt(Var_(w, X_k)(T_k)) - sqrt(Var_(w, X_l)(T_l)))²)).
16 [/quote](quoted from the @Brackman post) 16 [/quote](quoted from the @Brackman post)
17 \n 17 \n
18 And all this is only for the "easy" case of two teams. 18 And all this is only for the "easy" case of two teams.
19 (And no, this type of discussion is actually [i]not[/i] why I chose this name...)