1 University of Denver Department of Mathematics Department of Computer Science

Geometric Routing June 8 2006

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:

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)2. Minimum Forwarding Set Problem (MFS)3. Two-Hop realizability (THP)4. Exact solution to Weighted Region Problem


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!!

1


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Other Variation

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

1


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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.


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Previous work Problem has a constant factor

approximation algorithmShown By

Hervé Brönnimann, Michael T. Goodrich(1994) “Almost optimal set covers in finite VC-dimension”


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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


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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


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The Method to cover triangle

1


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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 ∆


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Covering a triangle Method

A

C

B

T’1

T’2

T’3


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Covering a triangle

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 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


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Result 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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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?


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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 regions?

NO!

We obtain two intersection inside the tile (Violates

Property P)

T’1

T’2


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Split the tile into two regions Using any cut

v0

v1

v2

v3

v4

n = 2m +1

n = 5; m = 2


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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? 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 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?


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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’2

T’3


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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


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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

x


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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


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Desired Property P Again1. No two discs intersect more than once along their

border inside a region Q2. No Two discs are tangent inside the a region Q 3. A disk intersect exactly twice along their border

with Q 1

3


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Property P Property P applies if the region is outside

of disk radius 2

Unit disk 2

Q


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Bell and Cover of x Remove points inside the Bell- Bell

Elimination Algorithm (BEA)


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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

2


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Result of Running BEA Graph

80

82

84

86

88

90

92

94

96

98

100

50 55 65 75 80 90 97 120 160 200

Number of Points

% success

50 82.5652 84.255 85.1260 89.9865 91.2370 94.3175 95.4277 97.7680 97.7585 97.7690 98.1295 99.0297 99.1

100 99.11120 99.65140 99.93160 99.94180 99.99200 99.99


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Distance of one-hop neighbors

Extra region


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Approximation factor


38





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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


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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



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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

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.


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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 edgesQuickTime™ and a

TIFF (LZW) decompressorare needed to see this picture.

2

9

5

43

∞

∞


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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


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Preliminaries 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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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


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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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Dijkstra’s Example(1)

s

u v

yx

10

5

1

2 39

4 67

2

s

u v

yx

10

5

1

2 39

4 67

2

Dijkstra(G,s)01 for each vertex v V[G]02 d [v] = 03 v. Parent = UNKNOWN04 d[s] = 005 S 06 Q V[G]07 while Q ≠ Ø08 u extractMin(Q)09 S S {u} 10 for each v adjacent(u) do11 Relax(u, v, G)


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Dijkstra’s Example (2)u v

s

yx

10

5

1

2 39

4 67

2

s

u v

yx

10

5

1

2 39

4 67

2



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Dijkstra’s Example (3)

u v

yx

10

5

1

2 39

4 67

2

u v

yx

10

5

1

2 39

4 67

2



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Dijkstra’s 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 (V) (1) (V 2)

binary search tree

(log V) (log V) (E log V)

Fibonacci heap

(log V) (1) (amort.)

(V logV + E)


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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

Yes!!

a b

c d

e

Is this Planar?

Examples: Voronoi diagram and Delaunay triangulation


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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 Dijkstra’s algorithm complexity for planar

graph is O(n log n)


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WRP - Algorithms VECTOR BASED

The WRP: Finding shortest path… O(n8L) [2]

Path planning 0/1/ WRP…[1]

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]


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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


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WRP - General case Notationsf = 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


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WRP - Local Optimality … Snell’s law Light seeks the path of minimum time Light obeys Snell’s law Index of refraction for a region

speed of traveling through that region Shortest paths in WRP must also obey

Snell’s law Lemma 3.2 [2]:

- Let e = min {f, f’}, f, f’< ∞- Optimal path passing through edge e implies path obeys Snell’s law at edge e


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t ( x1,y1)

e

f (f)

f′(f′)

s(,-y)

c(x,)

θ

θ′

f≥f′>

WRP - Local optimality

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

F(x) = f x2 + y2 + f ' (x −x)2 + y

2

F '(x) =

f

x

x2 + y2= f '

(x −x)

(x −x)2 + y2⇒

f sinθ = f 'sinθ '


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t ( x1,y1)

e

f (f)

f′(f′)

s(,-y)

θc(f,e)

θc(f′,e)

f≥f′>

x x′

WRP - Local optimality If path uses the edge

eF(x, x ') = f x2 + y

2 +e(x'−x) + f ' (x −x')2 + y2

subjectto x≤x'

F '(x,x') =

f

x

x2 + y2=e = f '

(x −x')

(x −x')2 + y2⇒

f sinθ =e = f 'sinθ '

θc( f ,e) =sin−e f

and θc( f ',e) =sin−e f '


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Local Optimality Criterion of path p

a) If p passes through e p obeys Snell’s law

b) If p shares a segment on e p enter and exit e at a critical angle.

DefinitionRoot: Marching back from some point x

along p, root r is the first vertex or critical point of entry on an edgey2 y3 y2 is the root of x

x


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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


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Locally f-free path pIf 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 = min{pf(x), pf’(x)}

s

xf´

f

Locally f ´- free path

Locally f- free path


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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

y

r

edge sequence

e1

ek

e

.

.

.

Frontier point zz y’ f

f’

Lemma 4.3 [2]

a) p to (y, y’) from root r does not intersect

b) z (y, y’) |rz| < |rx| where x (y,y’) ^ x z


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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

y

r

e1

ek

e

.

.

.

y’ ff’

r’

x x’

Root r is either a vertex or a critical point of entry


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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 Snell’s law whenever we cross the boundary

Keep track of location of wave-front where it hits the edges -> 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)


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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


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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 Dijkstra’s algorithm on visibility graph

Previous algorithm solves this special case, but restriction on weight allows a better algorithm


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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


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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

v wR

vw

H

R


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Optimal paths

Various local optimal segment characteristics

Grey - Obstacles; light - 0-regions

0-entrance points


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Critical graph From the observation we made,

construct a critical graph G’(V’,E’) from input G(V, E) as follows1. V’ = V {s, t}2. 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 q

p

r

s

x


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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/∞ WRP reduces to- Construct the critical graph G´ - Run Dijkstra’s 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


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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


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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 Snell’s law

Stop when the two refraction rays first encounter a “topological fork”

u

v

topological fork of two rays

pq

e1

e2

e3

e4

e5


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Trace a critical refraction path from u to v through the cone.

A unique locally optimal path from v to u will stay inside the refraction cone that is split by u

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 > critical angle3. Both rays encounter outer boundary

u

v

critical entry

u

v

θc

θa

θb


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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 Dijkstra’s algorithm on path-net

u

v

u

v

critical entry

θc


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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)


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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


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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 Dijkstra’s algorithm

Raster-based algorithms

8 connected 16 connected 32 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-connectivity:√2 8-connectivity:

(√2+1)/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 errorConnectivi

tyMaximum Elongatio

n

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


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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

Documents

1 University of Denver Department of Mathematics Department of Computer Science