university of denver department of mathematics department of computer science

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University of Denver

Department of Mathematics

Department of Computer Science

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Geometric RoutingGeometric Routing

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

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General graphGeneral graph

A graph (network) consists of nodes and edges represented as G(V, E, W)

a b

c d

ee1(1)

e3(5)

e4(2)

e2(2)

e6(2)

e5(2)

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Planar Graphs Planar Graphs

A Planar graph is a graph that can be drawn in the plane such that edges do not intersect

a b

c d

e

Examples: Voronoi diagram and Delaunay triangulation

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AGENDAAGENDA

Topics:1. Minimum Disk Covering Problem (MDC)

2. Minimum Forwarding Set Problem (MFS)

3. Two-Hop Realizability (THP)

4. Exact Solution to Weighted Region Problem (WRP)

5. Raster and Vector based solutions to WRP Conclusion Questions?

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Topics:1. Minimum Disk covering Problem (MDC)


3. Two-Hop realizability (THP)

4. Exact solution to Weighted Region Problem (WRP)

5. Raster and vector based solutions to WRP Conclusion Questions?

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1 . Minimum Disk Covering Problem (MDC)

Cover Blue points with unit disks centered at Red points !! Use Minimum red disks!!

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Other VariationOther Variation

Cover all Blues with unit disks centered at blue points !! Using Minimum Number of disks

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ComplexityComplexity

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.

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Previous work (Cont…)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

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Previous methodPrevious 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

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The method to cover triangleThe method to cover triangle

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Covering a triangleCovering a triangle

IF No blue points in a triangle- NOTHING TO DO!!

IF ∆ contains RED + BLUE THEN Unit disk centered

at RED Covers the ∆

Assume BLUE + RED do not share a ∆

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Covering a triangle cont…Covering a triangle cont…

A

CB

T1

T2

T3

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Covering a triangle cont…Covering a triangle cont…

1. Using Skyline of disks

2. cover each of the 3 sides with 2-approximation

3. combine the result to get:

6-approximation for each ∆

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Desired PropertyDesired Property P

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

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Unit disk intersects at most 18 trianglesUnit 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

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Result 108-approximationResult 108-approximation

Covered each triangle with approximation factor of 6

Optimal cover can intersect at most 18 triangles

Hence, 6 *18 = 108 - approximation

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ImprovementsImprovements

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?

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Can we use a larger tile?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

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Split the tile into two regionsSplit the tile into two regions

v0

v1

v2

v3

v4

n = 2m +1

n = 5; m = 2

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Different shape Tile?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

2

1

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Different shape Tile?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

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Our ApproachOur 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?

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Hexagon- split it into 3 regionsHexagon- 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

T2

T3

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CoveringCovering

Instead of tiling the plane, how about covering the plane?

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Conclusion of MDCConclusion 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

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2. Minimum Forwarding Set Problem (MFS)2. Minimum Forwarding Set Problem (MFS)

Cover blue points with unit disks centered at red points, now all red points are inside a unit disk

s

A

ONE-HOP REGION

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Previous work (MFS)Previous work (MFS) Despite its simplicity, complexity is

unknown

3- and 6-approximation algorithms known Algorithm is based on property P

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Desired Property Desired Property P P AgainAgain

1. No two discs intersect more than once along their border inside a region Q

2. No Two discs are tangent inside a region Q

3. A disk intersect exactly twice along their border with Q

P1

P3

Q

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PropertyProperty P

Property P applies if the region is outside of disk radius 2

Unit disk 2

Q

s

A

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Redundant pointsRedundant points Remove redundant points

s

Redundant point

xy

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Bell and Cover of node Bell and Cover of node xx Remove points inside the Bell- Bell

Elimination Algorithm (BEA)

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AnalysisAnalysis

Assume points to be uniformly distributed BEA eliminates all the points inside the

disk of radius Need about 75 points Therefore exact solution

2

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80828486889092949698

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50 60 75 85 97 140 200Number of Points

% success

Empirical result

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Distance of one-hop neighbors

Extra region

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Approximation factorApproximation factor

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Degree of at most 2Degree of at most 2

Two-hop to bipartite graph

s

ab

c

1 23

4

1

2

3

a

b

c

4

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3 . Two-hop realizability3 . Two-hop realizability

Result:A bipartite graph having a degree of at most 2 is two-hop realizable

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2

3

a

b

c

4d

5one-hop neighbors two-hop neighbors

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4. Weighted region problem (WRP)4. Weighted region problem (WRP)

Objective - Find an optimal path from START to GOAL Complexity of WRP is unknown

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

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Planar GraphsPlanar Graphs

Planar sub-division considered as planar graph

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

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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

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WRP - General case 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!!

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t ( x1,y1)

e

f (f)

f′ (f′)

s ( 0,-y0)

c (x, 0)

θ

θ′

f ≥ f′ > 0

Snell’s LawSnell’s Law

Cost function

f sinθ = α f ' sinθ '

Optimal point of incidence

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0/1/0/1/ Special case WRP Special case WRP

v wR

Construct a critical graph G Run Dijkstra on G

Weight 0

v w R

Weight

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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

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Critical points

θC

Critical edges

s

t

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Snell points

θ1

x

s

t

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Border points

t

s

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5. WRP - General case

-optimal path

• -optimal path from s to t is specified by users• path within a factor of (1+ ) from the optimal

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Transform weighted planar graph to uniform rectangular grid Make a graph with nodes and edges

- nodes : raster cells

- edges : the possible paths between the nodes Find the optimal path by running Dijkstra’s algorithm

Raster-based algorithms

8 connected 16 connected 32 connected

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Raster-based algorithms…

Advantages- Simple to

implement- Well suited for

grid input data- Easy to add other

cost criteria

Drawbacks- Errors in distance

estimate, since we measure grid distance instead of Euclidean distance

- Error factor :

4-connectivity:√2

8-connectivity:(√2+1)/5

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Distortions - bends in raster paths

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Approximate by a straight lineApproximate by a straight line

Reduce deviation errors

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Compare vector vs. Raster

Raster 1178.68 50 secs

Straight 1130.56 65 secs

Vector(=.1)

1128.27 < 1 sec

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ConclusionConclusion

Improved approximation to MDC Bell elimination algorithm Two-hop realizability Exact solutions to special cases of WRP Straight optimal raster paths

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university of denver department of mathematics department of computer science

Documents

unit disks

triangleunit disk

hexagonunit disk

squareunit disk

larger tile

unit diameter

unit disc

red blue