ep is to understan the logics of what you want to do, in your case, build 4 point surfaces (u also need to know the right direction to build the surfaces). Then you can write an hipotetic list (by hand in a paper) of what you want. In your case the list was (0, 1, 3, 2) (2, 3, 5, 4) (4, 5, 7, 6), etc... if you can imagine building 2 lists, each one with the sequences (0, 2, 4, 6, etcc) and (1, 3, 5, 7, etc..) then you can manage with shift and graft to finally have four lists. A( 0 1 2 3 ...) B (1 3 5 etc..) C(3 5 7 etc..) D (2 4 6 etc..). And to achieve the 2 first lists, you need to get the odd and the pair numbers. The cull pattern does that amazingy well. With a pattern True-False you get de pair numbers, and with the False-True pattern you get de odd numbers.
Hope it was clear enough…
Added by Pep Tornabell at 5:32am on November 19, 2009
I've attached a rh and ghx file. http://we.tl/ghsd0UTxji - wetransfer link. I'm guessing the trick is to correctly define the Range (R) / Domain in the Random component, so that the circles' radi can vary between the fixed and static domain of 10 or 12 or 14 or 18 or 20. Meaning that the radi can only be 10, 12, 14, 18 or 20 - and not, lets say.. 11, 13, 15, 16, 17 or 19 for that matter. Thanks in advance, i hope someone can figure it out, because i can't seem to find anything online about how to set up such as specific domain.
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another example could be:
index 3 value 6
index 4 value 6
index 5 value 6
flipped and branched:
branch 6 index 0 value 3
branch 6 index 1 value 4
branch 6 index 2 value 5
Added by Ante Ljubas at 12:50pm on October 22, 2010
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.…