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...…
nts me this:
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3, 4, 5], [0, 1, 2, 3, 4, 5, 6], [0, 1, 2, 3, 4, 5, 6, 7]]
this is what I wanted but how to convert this to tree in grasshopper?
In grasshopper I just get:
8x IronPython.Runtime.List…
ctivity of vertices ordered sequentially, the order defining the direction of the normal, using 0 1 3 2 Causes an error this way. If it a quad face it seems odd to me that you would label the vertices in such an order, as an engineer, i have never seen it done as 0 1 3 2, it could be 3 2 1 0, 2 1 0 3 etc but going 0 1 3 2 is not acceptable, i will do a bit more reading on this.…
Added by Steve Lewis at 5:18pm on December 24, 2013
e), {1;2}(line), {1;3}(line)... and on the other side to have {0;0}(all lines except {0}(0)), {0;1} (all lines except {0}(1)), {0;2}(all lines except {0}(2)), {0;3}(all lines except {0}(3)), {1;0} (all lines except {1}(0)), {1;1} (all lines except {1}(1)), {1;2} (all lines except {1}(2)) ,{1;3} (all lines except {1}(3))...The first tree is easy to achieve, simply grafting a branch for each element, and the other, what I've done is to copy all lines of each tree ({0},{1},{2},{3}), to have them in all branches of each tree ({0;0}(elements of {0}), {0;1}(elements of {0}),,{1;0}(elements of {1}), {1;1}(elements of {1})..., and then remove in the first branch({0;1} the first element(0), in the second branch the second element, the third branch the third element...And so correctly you compare each line with all the other within each branched tree.Aaaaapufff XD…