Locally Delaunay Geometric
Graphs
Shakhar Smorodinsky
ETH Zurich
Joint work with
Rom Pinchasi, MIT
Geometric Graphs
A Geometric Graph is:
G=(V,E) +embedding in R2
K-locally Delaunay graphs
A Geometric graph is k-locally Delaunay if:
Can be embedded s.t. every edge is isolated
from its (k)-neighbors by some disc
1-locally Delaunay
Problem 1:
What is the maximum edge complexity of a
1-locally delaunay graph G=(V,E)?
Motivation?
Topology Control for
Sensor Networks
First observation: [S. Kapoor, X.Y. Li 03]:
G cannot contain a K3,3. Hence |E| = O(n5/3)
G contains no K3,3. Hence |E| = O(n5/3)
[Kovari, Sos, Turan]
Proof:
As a matter of fact: if G can be embeded
without a self-crossing C4 then |E| = O(n8/5) .
[Pinchasi, Radoicic 03].
Next improvement:
Can G contain a K2,2 ?
Yes!
Our contribution:
Thm [Pinchasi, S]: If G=(V,E) is 1-locally Delaunay
then |E|=O(n3/2)
Lets assume many things:
1) All edges makes a small angle with
the vector (0,1).
2) All edges cross the x-axis
3) For every edge e, the witnessing disc
is such that its center is left to e
If G=(V,E) is 1-locally Delaunay then
|E|=O(n3/2)
Proof:
If G=(V,E) is 1-locally Delaunay then
|E|=O(n3/2)
Proof: (cont)
Under these assumptions:
G contains no K2,2
If G=(V,E) is 1-locally Delaunay then
|E|=O(n3/2)
Proof: (cont)
Assume G contains K2,2:
A contradiction
Thm [Pinchasi, S]: If G=(V,E) is 2locally Delaunay then |E|=O(n)
Remark:
First observation: G contains no self
crossing copy of P3
Hence by
[Pach, Pinchasi, Tardos, Toth 03]
|E| = O(n log n)
If G=(V,E) is 2-locally Delaunay then
|E|=O(n)
Proof:
• Lets assume small angles between
edges
• Remove the (upper, lower) right most
and left most edge from every vertex
If G=(V,E) is 2-locally Delaunay then
|E|=O(n)
Proof: (cont)
Claim:
No edge survived!!!
Claim:
No edge survived!!!