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University of Denver Department of Mathematics Department of Computer Science. Applications Ad hoc Wireless networks Robot Route Planning in a terrain of varied types (ex: grassland, brush land, forest, water etc.) Geometric graphs Planar graph Unit disk graph. Geometric Routing.
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University of Denver Department of Mathematics Department of Computer Science
Applications Ad hoc Wireless networks Robot Route Planning in a terrain of varied types (ex: grassland, brush land, forest, water etc.) Geometric graphs Planar graph Unit disk graph Geometric Routing
General graph • A graph (network) consists of nodes and edges represented as G(V, E, W) e3(5) b a e4(2) e6(2) e1(1) e e5(2) d c e2(2)
b a e d c Planar Graphs • A Planar graph is a graph that can be drawn in the plane such that edges do not intersect Examples: Voronoi diagram and Delaunay triangulation
Topics: Minimum Disk Covering Problem (MDC) Minimum Forwarding Set Problem (MFS) Two-Hop Realizability (THP) Exact Solution to Weighted Region Problem (WRP) Raster and Vector based solutions to WRP Conclusion Questions? AGENDA
Topics: Minimum Disk covering Problem (MDC) Minimum Forwarding Set Problem (MFS) Two-Hop realizability (THP) Exact solution to Weighted Region Problem (WRP) Raster and vector based solutions to WRP Conclusion Questions?
1 1 . Minimum Disk Covering Problem (MDC) Cover Blue points with unit disks centered at Red points !! Use Minimum red disks!!
1 Other Variation Cover all Blues with unit disks centered at blue points !! Using Minimum Number of disks
Complexity • MDC is known to be NP-complete • Reference “Unit Disk Graphs”Discrete Mathematics 86 (1990) 165–177, B.N. Clark, C.J. Colbourn and D.S. Johnson.
Previous work (Cont…) A 108-approximation factor algorithm for MDC is known “Selecting Forwarding Neighbors in Wireless Ad-Hoc Networks” Jrnl: Mobile Networks and Applications(2004) Gruia Calinescu ,Ion I. Mandoiu ,Peng-Jun Wan Alexander Z. Zelikovsky
Previous method • Tile the plane with equilateral triangles of unit side • Cover Each triangle by solving a Linear program (LP) • Round the solution to LP to obtain a factor of 6 for each triangle
1 The method to cover triangle
Covering a triangle IF No blue points in a triangle- NOTHING TO DO!! IF∆ contains RED + BLUETHEN Unit disk centered at RED Covers the ∆ Assume BLUE + RED do not share a ∆
Covering a triangle cont… A T1 T3 B C T2
Covering a triangle cont… • Using Skyline of disks • cover each of the 3 sides with 2-approximation • combine the result to get: • 6-approximation for each ∆
Desired PropertyP • Skyline gives an approximation factor of 2 • No two discs intersect more than once inside a triangle • No Two discs are tangent inside the triangle
Unit disk intersects at most 18 triangles It can be easily verified that a Unit disk intersects at most 18 equilateral triangles in a tiling of a plane
Result 108-approximation • Covered each triangle with approximation factor of 6 • Optimal cover can intersect at most 18 triangles • Hence, 6 *18 = 108 - approximation
Improvements CAN WE • use a larger tile? • split the tile into two regions? • get better than 6-approximation by different tiling? • cover the plane instead of tiling?
Can we use a larger tile? • If tile is larger than a unit diameter !! • Unit disc inside Tile cannot cover the tile • Hence we cannot use previous method
Split the tile into two regions v0 n = 2m +1 n = 5; m = 2 v1 v4 v3 v2
Different shape Tile? • Each side with 2-approx. factor • Hence 8 for a square • Unit disk can intersect 14 such squares • 14 * 8 =112 • No Gain by such method
Different shape Tile? • Each side with 2-approx. factor • Hence 12 for a hexagon • Unit disk can intersect 12 such hexagons • 12 * 12 =144 • No Gain by such method
Our Approach • How about using a unit diameter hexagon as a tile • Split the tile into 3 regions around the hexagon • Does this give a better bound?
Hexagon- split it into 3 regions • Partition Hexagon into 3 regions (Similar to triangle) • Obtain 2-approximation for each side 6-approximation for hexagon • Unit disk intersects 12 hexagons • Hence, 6 * 12 = 72-approximation T1 T3 T2
Covering • Instead of tiling the plane, how about covering the plane?
Conclusion of MDC • Conjecture: A unit disk will intersect at least 12 tiles of any covering of R2 by unit diameter tiles • Each tile has an approximation of 6 by the known method • Cannot do better than 72 by the method used
Topics: Minimum Disk covering Problem (MDC) Minimum Forwarding Set Problem (MFS) Two-Hop realizability (THP) Exact solution to Weighted Region Problem (WRP) Raster and vector based solutions to WRP Conclusion Questions?
2. Minimum Forwarding Set Problem (MFS) s ONE-HOP REGION A Cover blue points with unit disks centered at red points, now all red points are inside a unit disk
Previous work (MFS) • Despite its simplicity, complexity is unknown • 3- and 6-approximation algorithms known • Algorithm is based on property P
Desired Property P Again • No two discs intersect more than once along their border inside a region Q • No Two discs are tangent inside a region Q • A disk intersect exactly twice along their border with Q P1 Q P3
PropertyP • Property P applies if the region is outside of disk radius Unit disk A s Q
Redundant points • Remove redundant points Redundant point x y s
Bell and Cover of node x • Remove points inside the Bell- Bell Elimination Algorithm (BEA)
Analysis • Assume points to be uniformly distributed • BEA eliminates all the points inside the disk of radius • Need about 75 points • Therefore exact solution
Distance of one-hop neighbors • Extra region
Topics: Minimum Disk covering Problem (MDC) Minimum Forwarding Set Problem (MFS) Two-Hop realizability (THP) Exact solution to Weighted Region Problem (WRP) Raster and vector based solutions to WRP Conclusion Questions?
b a c a b c s Degree of at most 2 • Two-hop to bipartite graph 2 1 3 1 2 3 4 4
a b d c 2 5 3 . Two-hop realizability • Result:A bipartite graph having a degree of at most 2 is two-hop realizable two-hop neighbors one-hop neighbors 1 3 4
Topics: Minimum Disk covering Problem (MDC) Minimum Forwarding Set Problem (MFS) Two-Hop realizability (THP) Exact solution to Weighted Region Problem (WRP) Raster and vector based solutions to WRP Conclusion Questions?
Objective - Find an optimal path from START to GOAL Complexity of WRP is unknown 4. Weighted region problem (WRP)
Planar Graphs • Planar sub-division considered as planar graph
Shortest path G(V, E, W) • Dijkstra algorithm finds a shortest path from a source vertex to all other vertices • Running time O(|V| log |V|+ |E|) • Linear time for planar graphs
WRP - General case Notations f = weight of face f e = weight of edge e, where e = f f’ ≤ min {f, f’} A weight of implies A path cannot cross that face or edge Note that all optimal paths must be piecewise linear!!
Snell’s Law Cost function Optimal point of incidence
Weight w R w v v R 0/1/ Special case WRP • Construct a critical graph G • Run Dijkstra on G Weight 0
Convex Polygon C • Exact path when s in C and t is arbitrary • Construct “Exact Weighted” Graph • Add edges that contribute to exact path • Run Dijkstra shortest path Algorithm
Critical edges C Critical points s t