now I want to combine some branches together ,the rule is : For path{2} contain number 2 and 5, then conbine the two paths together ,and for path{5} includes only 2&5,no other number ,so it's end .For path{3}, includes number 3&6 ,so we go to path{6}, path{6} includes 3&6&18, then wo go to path{18} , path{18} contains a new number 27, so we check path{27} ,path{27} includes only 27&18, no new numbers ,so it is end.
With this logic, path{2}&{5} become one tree finally , the contains is 2&5 ,and so path{3}&{6} &{18} &{27}(the contents is 3,6,18,27), and so others .
so what I want is:
{2}(2,5)+{5}(2,5)={2/5/anything}(2,5) ## the new path index doesnot matter{3}(3,6)+{6}(3,6,18)+{18}(18,27)+{27}(27,18)={3/6/18/27/?}(3,6,18,27) ``````etc
I tried path mapper, but I donot think it can do the trick this time. may be I just miss something very visible?? Awaiting for your kind help~Thanks in advance.…
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Added by Paul Cowell at 9:06am on October 31, 2016
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
es du module Galapagos inclus dans Grasshopper. inscrivez vous et venez avec votre pique-nique. C'est gratuit mais l'inscription est obligatoire. Contact : cadlantique@aol.com horaire de 18:30h a 20:30h ou plus ... A bientôt !…