elementary; you can see how I picked out 2 of the pentagon's sides individually. This is by far not the best way to do this.
The proper way is to use a an index pattern to pick out sides 1 & 2, 2 & 3, 3 & 4, 4 & 5, and 5 & 1, and then pass these pairs to the fillet routine. There is a way to do this but I couldn't remember the method. …
ve a Vertex [V] connected to four other Vertexs [N1-N4].
Each of the has a Value:
V ... 1
N1 ... 5
N2 ... 3
N3 ... 8
N4 ... 11
The Average Filter would set the Value of [V] to
(1+5+3+8+11)/5 = 5,6
The Median Filter would Sort Values and pick the middle one
1,3, [5], 8, 11
Hope that helped...…
{0;1;0}N=6
{0;1;1}N=6
{0;1;2}N=5
{0;2;0}N=7
{0;2;1}N=8
{0;2;2}N=9
Can you shift and wrap any of the paths A B or C?
Say if I wanted to shift and wrap B by 1 to get the following...
{0;0;0}N=7
{0;0;1}N=8
{0;0;2}N=9
{0;1;0}N=3
{0;1;1}N=2
{0;1;2}N=5
{0;2;0}N=6
{0;2;1}N=6
{0;2;2}N=5…
, branches, and trees.
This is currently how I think of it:
A 'point' component with 5 points will generally be compared to a 1-dimensional array with 5 indices.
If we flip this "path" then it becomes a 2-dimensional array looking like this [5][1].
Although it seems like Grasshopper does not iterate through these 'arrays' as one would assume.
Also, the question arises when we have a "tree" whose "paths" looks like:
{0;0;0} (5)
{0;0;1} (5)
{0;1;0} (5)
{0;1;1} (5)
{1;0;0} (5)
{1;0;1} (5)
{1;1;0} (5)
{1;1;1} (5)
For example.
Anyone have any insight here? Or where insight may be found.
Thanks!…