branches in each A's list of B's, or remove its ends etcso that if I want to remove the last B in every A{0;1},{0;2},{0;3},{0;4},{0;5},{0;6}{1;1},{1;2},{1;3},{1;4}{2;1},{2;2},{2;3},{2;4},{2;5}would become{0;1},{0;2},{0;3},{0;4},{0;5}
{1;1},{1;2},{1;3}
{2;1},{2;2},{2;3},{2;4}I guess the question is do I need to figure out the cull pattern- each B may have different lengths...…
So u seem to have the same issue as me then? I have to explicitly change it to 0 1 2 3, the default is not this it is 0 1 3 2. Converse for a tru face it is as expected, default is 0 1 2.
Added by Steve Lewis at 7:00pm on December 24, 2013
ng (It's a bit similar to the Knapsack problem):
I have a Variable --> XandI Have fix numbers (can we call "pieces") 9,12,15,18
I'd like to reach the X, with the summing of these numbers and using the minimum pieces ,it can't be lower than X, but it can be higher, maximum with 3.After this it has to found the most optimal combination which mostly use the same pieces
E.G.
X=98
The wrong solution is like = 1pcs of 18 = 9pcs of 9
Sum of pieces are 10
OR
= 3pcs of 18 = 1pcs of 15 = 1pcs of 12 = 2pcs of 9
Sum of pieces are 7
The right solution in this case = 5pcs of 18 = 1pcs of 9
(5*18)+(1*9)=99 it's good beacuse it's over with maximum 3 and uses the minimum pieces
Then it sends to a list like18 : 5pcs15 : 0pcs12 : 0pcs9 : 1pcsCan somebody help me ? Or is it possible to make this ?
Thank you…
Added by Petrik Kollár at 1:09am on November 10, 2017