5 8, and then the following values are obtain as the last one (8) plus 3, then this last one (11) plus 5, and then this last one (16) plus 8, and then it starts again: 24+3, 27+5, 32+8...
Thanks
…
Added by Jesus Galvez at 5:17am on November 27, 2012
{4}-0;3
{5}-6;7
{6}-5;7
{7}-5;6
Here it can be shown that there are two subgraphs containing 0,1,2,3,4 and 5,6,7. How can I use spiderweb (either using scripting or the components) to give me this result when I have many more vertices??
Thanks,
Sam…
Branch Address in the form {0;1;2;3;4;5;6;7;8;9;.......;N}
Try Grafting the input (I) Create Path by right clicking on it and selecting graft from the context menu…
in the desired order.
0 = 0
1 = 1
2 = 6
3 = 7
4 = 8
5 = 9
6 = 12
7 = 13
8 = 2
9 = 3
10 = 4
11 = 5
12 = 10
13 = 11
Where the first number is the index and the second number is the actual sorting key. Then you sort these keys while sorting your curves in parallel using the A input of the Sort component.
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
points 0, X-1, (2*x)-1, (3*X)-1, (4*X)-1, (5*X)-1 and then
1, X, (2*x), (3*X), (4*X), (5*X)
2, X+1, (2*x)+1, (3*X)+1, (4*X)+1, (5*X)+1
and so on till
5, X+4, (2*x)+4, (3*X)+4, (4*X)+4, (5*X)+4
How can I do this best?
Thanks,
Niels…