Unformatted text preview: Point of Interest
Coverage
Coverage
Chris Ellis
COT6410
10/6/08 Points of Interest (POI)
Points
Surveillance of locations or objects is
Surveillance
useful for study or intelligence gathering
useful
POI can be located in hostile/dangerous
POI
environments
environments
Autonomous tracking agents can be used Tracking Agents
Tracking
Remote or AI controlled
Mobile (land or air)
Equipped with cameras
Equipped
or sensor devices
or
Safer than deploying
Safer
humans to hostile
environments
environments (http://www.europeansecurity.com/imgbiblio/warrior_01s.jpg)
(www.iirobotics.com/catalog/images/roviomain.jpg) Concerns
Concerns
What if there are more POI than trackers?
How do we optimally place trackers to
How
cover the POI?
cover
P
P P
P
P P
P P
P P P P
P P
P Formalizing the Problem
Formalizing
Represent POI as vertices in a weighted,
Represent
fully connected graph
fully
Assign Euclidean distance between two
Assign
POI as weight of edge between
corresponding vertices
corresponding
Choose a fixed integer R as radius of
Choose
vision from Tracker
vision
Restrict Tracker placement to existing
Restrict
vertices
vertices Problem Declaration
Problem
Optimization Problem Given: a graph G = (V,E) and integer R
Problem: What is the minimum number of
Problem:
trackers with vision range R required to cover
all vertices v ∈ V?
A vertex is “covered” if a tracker is placed on
vertex
it, or there is an edge with weight < R
connected to another vertex with a tracker
located on it
located Problem Declaration
Problem
Decision Problem (POIC) Given: a graph G and integers R and K
Problem: Can all vertices v ∈ V be covered by K
Problem:
trackers with vision range R?
trackers POIC ∈ NP
POIC Oracle produces a set of vertices S ⊆ V ∋ S = K
Oracle
For each vertex v ∈ V if v ∈ S or v is incident to some
For
vertex in S with edge weight less than R, mark that
vertex as covered
vertex
If any vertex v ∈ V is not covered, return “No”,
If
otherwise return “Yes”
otherwise Problem Reduction
Problem
Reduce to Vertex Cover
Convert an arbitrary instance of VC to
Convert
POIC
POIC Given an instance of VC
Given
a graph G = (V,E) and integer K
graph Construct an instance of POIC
Construct
a graph G’ = (V’,E’) and integer K’, R
graph Problem Reduction
Problem
Let V’ = V
Let K’ = K
Let R = 2
For each edge e = (vi,vj) iin E construct an
n
edge e’ = (vi,vj) with weight 1
edge
For each edge in the compliment of G
For
construct an edge with weight eleventyconstruct
billion Caveat
Caveat
“True” instances of POIC are based on
True”
realworld locations
realworld
Vertices and edges conform to the
Vertices
constraints on points in 2D or 3D
Euclidean space
Euclidean
My definition of POIC does not Proof of Correctness
Proof
One to one mapping from v ∈ V to v’ ∈ V’
One
vi covers vk (VC sense) iff vi’ covers vk’
iff
(POIC sense)
(POIC
vi is covered by vk (VC sense) iff vi’ iis
s
iff
covered by vk’ (POIC sense)
covered
VC of G exists iff POIC of G’ exists ...
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Full Document
 Fall '10
 Dutton
 Pythagorean Theorem, euclidean distance, Euclidean space, vertices

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