Given:
1. The nearest neighbor of ππ then ππ-ππ is a Delaunay edge.
2. In a 3D set of points, if we know that consecutive points ie... ππ-πi+1 are nearest neighbors.
3. The 3D points do not form a straight line
Assumption:
Each Delaunay tesselation (3D) has at least 2 nearest neighbor edges.
Is my assumption true? If not can you please explain to me the possible exceptions?
Thanks,
Pranav
Dear Pranav Khade
Are you trying to play chess in 3D?
Your idea is a good one, but your assumptions are not.
You need to give a clear definition of paths, so I suggest for you to start in one 3D box, it includes 8 points. I prefer to give each point the following notation
P(i, j, k), so the locations of the 8 points are at
(0,0,0) (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1),(0,1,1) and (1,1,1).
Study this cube carefully, define each Delanoy edge (axioms of the path), and then add another box, which means 12 points, etc.
If you find the closed formula that allows you to calculate all possible paths from the starting point at the origin to the farthest point at the upper corner of the rectangular box, then you are on the right track.
I wish you good luck.
Issam Kaddoura Thank you for your answer. I am writing an algorithm that analyses the protein shape which lead me to an observation that can only be true if my assumption is right so I wanted to verify is there is such theorm/rule in existence.
I will do the exercise you have suggested and will try to figure out why I am getting such an observation if my assumption is not true.
Thank you!
Pranav
- Badges
- Science topic
- Similar topics
- Computer Science and Engineering
- Computer Graphics and Animation
- 3D