et tree structure.
Now let's say, wen have this original tree structure:
{0;0}
0 A
1 B
{0;1}
0 C
1 D
{0;2}
0 E
1 F
{0;3}
0 G
1 H
{0; just joking :D} ... so we have 4 paths and in each of them there is a list with whatever any two objects ... So the source mask should be {0;x}(y), where x = 0,1,2,3?! or is x = 3?! and y = 0,1 or is y=1 ... what's "static", what's "dynamic"? ... How to understand what?! ...
So let's say I want to get a tree, that looks that way:
{0;1}
0 A
1 B
2 C
3 D
{0;2}
0 E
1 F
2 G
3 H
What do I have to understand here? ... Thank you for the help and support in advance ...…
the same icon. The difference is that the original bend component made use of points to calculate the lines and the new "angle" component uses lines. Knowing this I created two lines from David Lister's original 3 points. Line A is point 1 to 2, and Line B is point 2 to 3, which I thought would produce the same angles to work from as the original 3 point bend component. Hope this explains it a bit. I look forward to seeing what you do with it :D Cheers!…
Added by David heaton at 3:18am on November 30, 2015
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
me list... as far as i can see in my geometry model there are multiple relations within the list of indexes... so i thought i´d like to flip the value - which is the relating index number - with the corresponding index so as to create branches with lists of indexes that relate to the value re-occuring within the flow of the original list... has anyone an idea on how to accomplish this?
example:
index 0 value 3
index 1 value 3
index 2 value 3
flipped and branched:
branch 3 index 0 value 0
branch 3 index 1 value 1
branch 3 index 2 value 2
greetings
ante…
Added by Ante Ljubas at 12:45pm on October 22, 2010
goes into the N input of the [Duplicate] component. It asks for an Integer and I want 3 paths for each branch so I put 3 in a string and fed it in. If you could offer any more clarification or guidance I would appriciate it. I set the O boolean to False. I'm sure it some some careless error on my part. By the way, your site, fancywires.com is amazing.
Thanks for your help,
Jeff…
s:
3
2
1
0
For Simplify I copied this from Grasshopper component help:Simplify a tree by removing the overlap shared amongst all branches. Imagine a tree with six branches: A = {0;1;0}B = {0;1;1}C = {0;1;2}D = {0;2;0}E = {0;2;1}F = {0;3;0}As you can see all these branches share the same first index (0). Thus if you were to Simplify this tree, you'd end up with: A = {1;0}B = {1;1}C = {1;2}D = {2;0}E = {2;1}F = {3;0}…