University of Denver

1

• University of Denver
• Department of Mathematics
• Department of Computer Science

2
Routing in Geometric Settings

• 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

3
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

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?

5
1 - Minimum Disk Covering Problem (MDC)
Cover Blue points with unit disks centered at
Red points !! Use Minimum red disks!!
6
Other Variation
Cover all Blues with unit disks centered at blue
points !! Using Minimum Number of disks
7
Complexity

• MDC is known to be NP-complete
• Reference Unit Disk GraphsDiscrete
  Mathematics 86 (1990) 165177, B.N. Clark, C.J.
  Colbourn and D.S. Johnson.

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

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

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

11
The Method to cover triangle
12
Covering 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 ?
13
Covering a triangle Method
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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 ?

15
Desired Property P

• No two discs intersect more than once inside a
  triangle
• No Two discs are tangent inside the triangle

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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
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

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?

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

20
Split the tile into two regions?

• NO!
• We obtain two intersection inside the tile
  (Violates Property P)

T1
T2
21
Split the tile into two regions Using any cut
v0
n 2m 1 n 5 m 2
v1
v4
v3
v2
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

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

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?

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

T1
T3
T2
26
Covering

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

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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

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?

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
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

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
32
Property P

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

Unit disk
Q
33
Bell and Cover of x

• Remove points inside the Bell- Bell Elimination
  Algorithm (BEA)

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

35
Result of Running BEA

• Graph

36
Distance of one-hop neighbors

• Extra region

37
Approximation factor
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?

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

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Degree of at most 2

• If G is realizable then sub-graph G is
  realizable

one-hop region
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?

42
4 - Weighted region problem (WRP)

• Find a optimal path from START to GOAL in a given
  subdivision
• Goal is to find a path with minimum total cost

43
Weighted region problem- Planar Graphs

• WRP finds a shortest path on a planar
  sub-division from source s to destination t
• Planar sub-division considered as planar graph
• Edges are the boundary of faces of sub-division
• Vertices are the intersection of edges

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

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)
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General Shortest path G(V, E, W)

• W E ? non-negative weight
• Source s 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

47
Dijkstra's Algorithm

• Greedy algorithm
• Basic idea
• Algorithm maintains two sets of nodes, Solved S
  and Unsolved U
• Each iteration takes a node from U and moves it
  into S
• Algorithm terminates when U becomes empty
• Choosing a node v from U is by a greedy choice

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Dijkstras Example(1)
Dijkstra(G,s) 01 for each vertex v Î VG 02 d
v 03 v. Parent UNKNOWN 04 ds 0 05
S Æ 06 Q VG 07 while Q ? Ø 08 u
extractMin(Q) 09 S S È u 10 for each v
Î adjacent(u) do 11 Relax(u, v, G)
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Dijkstras Example (2)
Dijkstra(G,s) 01 for each vertex v Î VG 02 d
v 03 v. Parent UNKNOWN 04 ds 0 05
S Æ 06 Q VG 07 while Q ? Ø 08 u
extractMin(Q) 09 S S È u 10 for each v
Î adjacent(u) do 11 Relax(u, v, G)
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Dijkstras Example (3)
u
v
1
8
9
Dijkstra(G,s) 01 for each vertex v Î VG 02 d
v 03 v. Parent UNKNOWN 04 ds 0 05
S Æ 06 Q VG 07 while Q ? Ø 08 u
extractMin(Q) 09 S S È u 10 for each v
Î adjacent(u) do 11 Relax(u, v, G)
10
9
2
3
0
4
6
7
5
5
7
2
y
x
51
Dijkstras Correctness

• We will prove that whenever u is added to S,
  u.d() d(s,u), i.e., that d is minimum, and that
  equality is maintained thereafter
• Proof (by contradiction)
• Note that "v, v.d() ³ d(s,v)
• Let u be the first vertex picked such that there
  is a shorter path than u.d(), i.e., that Þ u.d()
  gt d(s,u)
• We will show that this assumption leads to a
  contradiction

52
Dijkstras Running Time

• Extract-Min executed V time
• Decrease-Key executed E time
• Time V TExtract-Min E TDecrease-Key
• T depends on different Q implementations

Q T(Extract-Min) T(Decrease-Key) Total
array O(V) O(1) O(V 2)
binary search tree O(log V) O(log V) O(E log V)
Fibonacci heap O(log V) O(1) (amort.) O(V logV E)
53
Planar Graphs

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

b
a
e
d
c
Yes!!
Is this Planar?
Examples Voronoi diagram and Delaunay
triangulation
54
Planar Graphs

• n - e f 2
• n, e, f of vertices, edges, faces
• Implies, of edges and faces is O(n)
• Planar graph has at most 3n - 6 edges, n 3
• Every planar subdivision can be considered as a
  planar graph
• Dijkstras algorithm complexity for planar graph
  is O(n log n)

55
WRP - Algorithms

• VECTOR BASED
• The WRP Finding shortest path O(n8L) 2
• Path planning 0/1/? WRP1
• A new algorithm for computing shortest path in
  WRP O(n3) 3

• RASTER BASED
• Cross country Movement Planning 4
• Planning Shortest Paths among 2D and 3D weighted
  Regions Using Framed-Subspaces 5

56
WRP - General case

• Input
• Planar straight line subdivision is specified by
  faces, vertices, and edges
• Two points s and t, source and destination,
  respectively
• Assumption- all faces are triangles- s and t
  are vertices
• Output
• ?-optimal path from s to t is specified by users
• path within a factor of (1 ?) from the optimal

57
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!!

Short cut
58
WRP - Local Optimality

• Snells law
• Light seeks the path of minimum time
• Light obeys Snells law
• Index of refraction for a region ? speed of
  traveling through that region
• Shortest paths in WRP must also obey Snells law
• Lemma 3.2 2 - Let ?e min ?f, ?f, ?f,
  ?flt 8- Optimal path passing through edge e
  implies path obeys Snells law at edge e

59
WRP - Local optimality

• If path passes through edge e, Cost function from
  s to t

60
WRP - Local optimality

• If path uses the edge e

61
Local Optimality Criterion of path p

• If p passes through e ? p obeys Snells law
• If p shares a segment on e ? p enter and exit e
  at a critical angle.
• Definition
• Root Marching back from some point x along p,
  root r is the first vertex or critical point of
  entry on an edge

y2
y3
y2 is the root of x
x
62
Local Optimality Criterion of path p

• (1) Between two consecutive vertices v, v on
  path p there is at most one critical point of
  entry to an edge and at most one critical point
  of exit OR
• (2) Path p can be modified such that (1) holds
  without altering the length of the path
• Therefore following cannot happen

63
Locally f-free path p

• If p connecting s to x ? interior(f ? f) does
  not go through face f then we say p is locally
  f-free path - pf(x)
• A locally optimal path from s to x minpf(x),
  pf(x)

f
Locally f - free path
x
f
Locally f- free path
s
64
Intervals of a locally f-free path p on e

• p with root r to some point y on an edge e may
  cross set of edges ? (edge sequence) as shown

f
y
e
f
edge sequence ?
ek
. . .

• Lemma 4.3 2
• p to (y, y) from root r does not intersect
• ?z ?(y, y) ? rz lt rx where x ?(y,y) x ? z

e1
r
65
Intervals of optimality

• y,y forms an interval of optimality for a root
  r
• Such intervals form a covering of e and have
  mutually disjoint interiors

f
y
e
f
ek
. . .
e1
r
Root r is either a vertex or a critical point of
entry
66
Algorithm

• Triangulate planar subdivision
• Add s and t (source and destination) as vertices
• Simulate propagation of wave-front of light
  starting from s by applying Snells law whenever
  we cross the boundary
• Keep track of location of wave-front where it
  hits the edges -gt Intervals of optimality
• Events are intervals of optimality
• Events are stored in a priority queue sorted by
  distance from s
• Number of events is bounded within O(n4)

67
Algorithm complexity

• Number of intervals is bounded within O(n4)
• When two intervals overlap, find tie point
• Complexity of find tie point is O(k2 log
  nNW/?)k of events O(n4)n of
  verticesN largest integer coordinate of any
  vertexW Largest weight of any face or edge?
  Error bound specified in the input
• Complexity of the algorithm is O(n8)
• A O(n3) algorithm using path-net exists

68
0/1/? -WRP

• Weights restricted to 0/1/?
• Without 0-regions, visibility graph can be
  constructed in O(n log n K)K of edges in
  visibility graph
• Run Dijkstras algorithm on visibility graph
• Previous algorithm solves this special case, but
  restriction on weight allows a better algorithm

69
0/1/? - Least risk paths

• Find a least risk path in a polygon with k
  threats- risk ? distance in the view of threat-
  threats have unlimited range of line of sight
• Treat exterior boundary as ?-regionTreat line of
  sight as 1-regionThose not visible to any threat
  as 0-region
• ?

70
0/1/? -WRP Observations

• A segment is locally optimal path if its interior
  is free from obstacles and 0-regions
• Shortest path from v to a region R not containing
  v is a segment vw from v to a point w
• if w is a vertex of R edges incident on w lies
  inside half-plane H, a line through w
  perpendicular to vw
• if w is on an edge such that vw is perpendicular
  to that edge

71
Optimal paths

Various local optimal segment characteristics Gre
y - Obstacles light - 0-regions
0-entrance points
72
Critical graph

• From the observation we made, construct a
  critical graph G(V,E) from input G(V, E) as
  follows1. V V ? s, t2. E E ?
  locally optimal segments with vertices as
  endpoints ?locally optimal segments with
  0-entrance points, store one of end points of the
  edge

73
Critical graph Example

• New edges added are (pq) and (rs) with weights
  pq and rx.
• It is easy to add x as node, but this increases
  the node complexity and hence increase the
  running time of the algorithm

74
Query Critical graph

• Number of edges in critical graph is O(n2) since
  we add edges but not vertices
• Complexity of finding a shortest path in 0/1/8
  WRP reduces to- Construct the critical graph G
  - Run Dijkstras algorithm on G with O(n log n
  K), where K is number of edges
• Therefore complexity is O(n2) since K is bounded
  by O(n2)

75
General WRP-improved algorithm

• WRP algorithm of complexity O(n8) is not
  practical
• A new algorithm is based on constructing a
  relatively sparse graph path-net
• Path-net guarantees an ?-optimal path between two
  query points
• User defined parameter k controls the density of
  the graph
• Varying k, we can get close to optimal

76
Path-net Algorithm O(n3)

• Construct path-net G(V, E) from a given
  input G(V, E), k is user defined integer
• Construct k evenly-spaced cones around each
  vertex of V Maximum number of cones is kn
• For each vertex v, find a possible sub-path from
  v either to another vertex u or to a critical
  point of entry on some edge of subdivision.
• There is at most one sub-path per cone
• Example k 5

v
360/5 72o
77
Topological Fork

• The edge sequence of one ray first starts to
  differ from the edge sequence of the other
• Locally optimal sub-path from v goes through
  boundaries of the cone and traverse egdes,
  obeying Snells law
• Stop when the two refraction rays first encounter
  a topological fork

p
q
topological fork of two rays
e4
e5
e3
e1
e2
78
Topological Fork

• Topological fork must be one of the following
  types1. Two rays split at a vertex2. Both rays
  incident on the same edge and any one
  incident angle gt critical angle3. Both rays
  encounter outer boundary

A unique locally optimal path from v to u will
stay inside the refraction cone that is split by u
?a
?c
?b
Trace a critical refraction path from u to v
through the cone.
79
Path-net Edges

• Find optimal sub-paths, path-net edges
• Calculate the weight of path-net edges
• Add path-net edges and vertices into path-net
  graph,
• G(V, E)
• Run Dijkstras algorithm on path-net

80
Path-net complexity

• Path-net has O(kn) nodes, since each cone can
  create at most one link
• Propagation of two rays of a cone takes
• O(n2) for all vertices - Locally optimal
  path can cross each edge
• at most n times (Lemma 7.1 in 2)- An
  edge sequence has O(n2) crossed edges
• Overall complexity is O(kn3) to build a data
  structure with size O(kn)

81
Path-net complexity

• What if s or t changes?
• Connect new s, t to existing path-net using
  similar techniques used for construction of
  path-net and run path-net algorithm

82
Path-net Implementation

• Part of WRP-Solve Compare various approaches to
  route planning
• ModSAF Military simulation system
• 3 compares the result of Path-net algorithm to
  1. Grid based algorithm 2. Edge
  subdivision algorithm
• in both real and simulated data set

83
Raster-based algorithms

• Transform weighted planar graph to uniform
  rectangular grid
• Assign a suitable weight to each grid cell
• Weight in a cell cost/unit distance traveled in
  that cell
• Make a graph with nodes and edges
• - nodes raster cells
• - edges the possible paths between the
  nodes
• Find the optimal path by running Dijkstras
  algorithm

8 connected
32 connected
16 connected
84
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-connectivityv2
• 8-connectivity(v21)/5

85
n-connected Raster

• Standard 4-connected raster
• A number of geometric distortions exists
• - To reduce the distortions, increase
• connectivity
• More-connected raster 8-connected,
  16-connected, 32-, 64, and 128-connected
• - Computation of the cost of an edge is
• complicated

86
n-connected Raster

• Table shows connectivity to distortion error

Connectivity Maximum Elongation ? Maximum Deviation ?
4 1.41421 0.50000
8 1.08239 0.20710
16 1.02749 0.11803
32 1.01308 0.08114
64 1.00755 0.06155
128 1.00489 0.04951
87
Extended Raster-based algorithms

• Define nodes at the sides of the raster cells
• Allowed directions increases as of nodes
• increases

88
Extended Raster-based algorithms

• Solves the problem of intersecting paths and
  possibly expensive angles
• Drawback Search graph is dense when
  intermediate nodes increases

89
Quadtree Based Raster algorithms

• The uniform areas are grouped together
• Smaller of nodes and edges in the graph
• Efficient memory and algorithms
• Drawback Non-optimal paths possible

90
Vector vs. Raster comparisons

• .

91
Vector vs. Raster comparisons