1 / 90

University of Denver Department of Mathematics Department of Computer Science

University of Denver Department of Mathematics Department of Computer Science. Geometric graphs Planar graph Unit disk graph Applications Ad hoc wireless networks Robot route planning moving in a terrain of varied types (e.g. grassland, brush land, forest, water etc.

damon
Download Presentation

University of Denver Department of Mathematics Department of Computer Science

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. University of Denver Department of Mathematics Department of Computer Science

  2. Geometric graphs Planar graph Unit disk graph Applications Ad hoc wireless networks Robot route planning moving in a terrain of varied types (e.g. grassland, brush land, forest, water etc Routing in Geometric Settings Geometric Routing June 8 2006

  3. 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 Geometric Routing June 8 2006

  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? Geometric Routing June 8 2006

  5. 1 1 - Minimum Disk Covering Problem (MDC) Cover Blue points with unit disks centered at Red points !! Use Minimum red disks!! Geometric Routing June 8 2006

  6. 1 Other Variation Cover all Blues with unit disks centered at blue points !! Using Minimum Number of disks Geometric Routing June 8 2006

  7. 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. Geometric Routing June 8 2006

  8. Previous work • Problem has a constant factor approximation algorithm Shown ByHervé Brönnimann, Michael T. Goodrich(1994) “Almost optimal set covers in finite VC-dimension” Geometric Routing June 8 2006

  9. Previous work (Cont…) “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 Presented 108-approximation factor Geometric Routing June 8 2006

  10. The 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 Geometric Routing June 8 2006

  11. 1 The Method to cover triangle Geometric Routing June 8 2006

  12. 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 ∆ Geometric Routing June 8 2006

  13. T’1 T’3 A T’2 B C Covering a triangle Method Geometric Routing June 8 2006

  14. Covering a triangle • Using Skyline of disks • cover each of the 3 sides with 2-approximation • combine the result to get: • 6-approximation for each ∆ Geometric Routing June 8 2006

  15. Desired Property P • No two discs intersect more than once inside a triangle • No Two discs are tangent inside the triangle Geometric Routing June 8 2006

  16. Unit circle intersects at most 18 triangles in a tiling It can be easily verified that a Unit disk intersects at most 18 equilateral triangles in a tiling of a plane Geometric Routing June 8 2006

  17. Result 108-approximation • Covered each triangle with approximation factor of 6 • Optimal cover can intersect at most 18 triangles • Hence, 6 *18 = 108 - approximation Geometric Routing June 8 2006

  18. 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? Geometric Routing June 8 2006

  19. 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 Geometric Routing June 8 2006

  20. Split the tile into two regions? • NO! • We obtain two intersection inside the tile (Violates Property P) T’1 T’2 Geometric Routing June 8 2006

  21. Split the tile into two regions Using any cut v0 n = 2m +1 n = 5; m = 2 v1 v4 v3 v2 Geometric Routing June 8 2006

  22. 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 Geometric Routing June 8 2006

  23. 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 Geometric Routing June 8 2006

  24. Our Approach • How about using a unit diameter hexagon as a tile • Combine result above with a splitting into 3 regions around the hexagon • Does it give a better bound? Geometric Routing June 8 2006

  25. 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 T’1 T’3 T’2 Geometric Routing June 8 2006

  26. Covering • Instead of tiling the plane, how about covering the plane Geometric Routing June 8 2006

  27. 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 Geometric Routing June 8 2006

  28. 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? Geometric Routing June 8 2006

  29. 2 - Minimum Forwarding Set Problem (MFS) x s Cover blue points with unit disks centered at red points, now all red points are inside a unit disk Geometric Routing June 8 2006

  30. Previous work (MFS) • Despite its simplicity, complexity is unknown • In “Selecting Forwarding Neighbors in Wireless Ad-Hoc Networks” , Calinescu et al.Present • 3- and 6-approximation with complexity O(n log2n) and O(n log n), respectively • Algorithm is based on property P Geometric Routing June 8 2006

  31. Desired Property P Again • No two discs intersect more than once along their border inside a region Q • No Two discs are tangent inside the a region Q • A disk intersect exactly twice along their border with Q 1 3 Geometric Routing June 8 2006

  32. Property P • Property P applies if the region is outside of disk radius Unit disk Q Geometric Routing June 8 2006

  33. Bell and Cover of x • Remove points inside the Bell- Bell Elimination Algorithm (BEA) Geometric Routing June 8 2006

  34. Result • Assume points to be uniformly distributed • Bell elimination eliminates all the points inside the disk of radius • Need about 75 points • Therefore exact solution Geometric Routing June 8 2006

  35. Result of Running BEA • Graph Geometric Routing June 8 2006

  36. Distance of one-hop neighbors • Extra region Geometric Routing June 8 2006

  37. Approximation factor Geometric Routing June 8 2006

  38. 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? Geometric Routing June 8 2006

  39. 3-Two-hop realizability • Embed a bipartite graph G as a two-hop problem • Show a solution to MFS is a solution to vertex cover problem of G • Hence MFS is as hard as vertex cover problem Geometric Routing June 8 2006

  40. Degree of at most 2 • If G is realizable then sub-graph G’ is realizable one-hop region Geometric Routing June 8 2006

  41. 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? Geometric Routing June 8 2006

  42. Find a optimal path from START to GOAL in a given subdivision Goal is to find a path with minimum total cost 4 - Weighted region problem (WRP) Geometric Routing June 8 2006

  43. 2 ∞ 3 9 4 5 ∞ Weighted region problem- Planar Graphs • WRP finds a shortest path on a planar sub-division from source s to destinationt • Planar sub-division considered as planar graph • Edges are the boundary of faces of sub-division • Vertices are the intersection of edges Geometric Routing June 8 2006

  44. WRP - weighted region problem • Travel allowed through faces and edges • To use Dijkstra algorithm, need to augment the given planar sub-division • Integer weight = { 0,1 …W,  } is assigned to regions/edges • A special case: Weights restricted to 0/1/ • Vector and Raster based algorithms exist Geometric Routing June 8 2006

  45. Preliminaries • 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) Geometric Routing June 8 2006

  46. General Shortest path G(V, E, W) • W: E  non-negative weight • Sources and destination t are vertices of G • Find a shortest path from s to t • Path goes through only edges and vertices • Dijkstra algorithm finds a shortest path from a source vertex to all other vertices Geometric Routing June 8 2006

  47. Dijkstra's Algorithm • Greedy algorithm • Basic idea • Algorithm maintains two sets of nodes, Solved Sand Unsolved U • Each iteration takes a node from Uand moves it into S • Algorithm terminates when U becomes empty • Choosing a node v from U is by a greedy choice Geometric Routing June 8 2006

  48. u v u v 1 1 ¥ ¥ 10 ¥ 10 10 9 9 2 3 2 3 s 0 s 0 4 6 4 6 7 7 5 5 ¥ ¥ 5 ¥ 2 2 x y x y Dijkstra’s Example(1) Dijkstra(G,s) 01foreach vertex vÎV[G] 02 d [v] = ¥ 03 v. Parent = UNKNOWN 04 d[s] = 0 05 S ¬ Æ 06 Q ¬ V[G] 07 while Q ≠ Ø 08 u¬extractMin(Q) 09 S¬ SÈ {u} 10 foreach v Îadjacent(u)do 11 Relax(u, v, G) Geometric Routing June 8 2006

  49. u v u v 1 1 8 13 8 14 10 10 9 9 2 3 2 3 s 0 s 0 4 6 4 6 7 7 5 5 5 7 5 7 2 2 x y x y Dijkstra’s Example (2) Dijkstra(G,s) 01foreach vertex vÎV[G] 02 d [v] = ¥ 03 v. Parent = UNKNOWN 04 d[s] = 0 05 S ¬ Æ 06 Q ¬ V[G] 07 while Q ≠ Ø 08 u¬extractMin(Q) 09 S¬ SÈ {u} 10 foreach v Îadjacent(u)do 11 Relax(u, v, G) Geometric Routing June 8 2006

  50. u v 1 8 9 10 9 2 3 0 4 6 7 5 5 7 2 x y Dijkstra’s Example (3) u v 1 8 9 Dijkstra(G,s) 01foreach vertex vÎV[G] 02 d [v] = ¥ 03 v. Parent = UNKNOWN 04 d[s] = 0 05 S ¬ Æ 06 Q ¬ V[G] 07 while Q ≠ Ø 08 u¬extractMin(Q) 09 S¬ SÈ {u} 10 foreach v Îadjacent(u)do 11 Relax(u, v, G) 10 9 2 3 0 4 6 7 5 5 7 2 x y Geometric Routing June 8 2006

More Related