{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…
TREE B
{0} n=1 {0;1} n=4
{1} n=1 {0;4} n=4
{2} n=1 {1;1} n=4
{1;2} n=4
{1;3} n=4
{1;4} n=4
{2;1} n=2
{2;2} n=4
{2;3} n=4
{2;4} n=4
Both trees are generated from sliders, so could have any number of branches, although they are tied together. Tree A is a set of division points on a line, Tree B is a set of intersections from lines generated radially from the first (in this case three) points. I am trying to perform a "closest point" operation between the first tree and the second tree-- only, I do not want them to cross list, or long or short list. I want the {0} point to operate with those entries in the 2nd tree that start with {0,x}. So it would look like
{0} --> closest point with {0;1},{0;4}
{1} --> closest point with {1;1},{1,2},{1,3},{1,4} etc
I cannot figure out how this works. What I am visually trying to do is cast rays from a string of points so that they stop when they encounter another curve. I am having trouble picking through the intersection events to get what I want. Check the attached files for some clarity. THANK YOU…
Added by Joshua Jordan at 12:06am on February 5, 2012