Comment reconstruire le graphe de visibilité d’un polygone? (Reconstructing visibility graphs of...

Comment reconstruire le graphe de visibilité d’un polygone?

(Reconstructing visibility graphs of polygons)

Jérémie Chalopin, Shantanu DasLIF, Aix-Marseille University

Yann Disser, Matúš Mihalák, Peter WidmayerETH Zurich, Suisse

12èmes Journées Graphes et Algorithmes, Marseille, 2010

10 Nov 2010 JGA 2010 @ Marseille 2

Robots in Polygons

• Robots move from vertex to vertex inside a polygon

• The polygon is – simple (no holes)

– finite

– of size n = # vertices

• Visibility restricted to straight lines.

Objectives:• Build a map of the polygon

• Rendezvous (Gathering of robots)


Visibility Graph of Polygons

Gvis(P):

• Vertices of Gvis <=> vertices of P

• Edge (u,v) in Gvis iff

u and v can see each-other.

(if the line u to v is inside P)

Objective:

• Construct Gvis

(up to isomorphism)


The Robot Model

• The robots sees a subset of the vertices.

• It can decide to move to one of the visible vertices

• It repeats its observation from its new location.

• The robot can not measure distances or angles.

• The robot can not distinguishable vertices.


Local Orientation

• Robot sees the visible vertices in CCW order.

There is a local orientation on edges of Gvis.

• After each move, the robot can look-back to its previous location.

Each edge of Gvis is labeled from

both sides.

1

2

3


Polygons vs. Graphs

Exploration of Polygons ≈ Exploration of edge-labeled graphs


Polygons vs. Graphs II


Property [YK 1996]: A robot exploring an edge-labeled graph of

known size n, can not always reconstruct the graph.


Polygons vs. Graphs III

Property [YK 1996]: A robot exploring an edge-labeled graph of

known size n, can not always reconstruct the graph.

Theorem: A robot exploring a polygon can always reconstruct the

visibility graph of the polygon if it knows an upper bound on n.



Graph Exploration

• The view of an exploring robot:

Minimum-base


Graph Exploration II

• The minimum-base of a graph G:

The smallest graph B such that G covers B

Property [YK 1996]: A robot exploring an edge-labeled graph can

construct the minimum-base if it knows an upper bound on n.


Exploring visibility graphs

• If C1,C2,…,Cp are the classes of vertices in G

vis

• | Ci | = q = n/p

• Classes repeat periodically on the boundary.

• How to find the internal edges (chords)?


Properties of Polygons

• Every polygon has an ear!

( If a,b,c appear in this order on the boundary: b is an ear iff a sees c.)

• Removing an ear of a simple polygon leaves a smaller polygon.

(visibility relationships are maintained)

• Every sub-polygon of four or more vertices has a chord.


Properties of Polygons II

It is easy to recognize an ear! Check for the paths:

(1, -1) (-2, 2) (1, -1)

(-1, 1) (2, -2) (-1, 1) 12

1

-2

Lemma:

If v is an ear,

every vertex in the class of v is an ear.

-1

-1


Deconstructing Polygons

Choose a class Ci of ears.

Remove all Ci vertices from P

(i.e. remove a vertex from B)


Deconstructing Polygons-II

Choose a class Ci of ears.

Remove all Ci vertices from P


Repeat until a single class remains!


Deconstructing Polygons III

Choose a class Ci of ears. Remove all Ci vertices from P


Repeat until a single class remains!

Remaining vertices form a clique!

Lemma:

There is a unique class C* which forms a clique.


Using class C*

C* corresponds to a vertex with q-1 self loops in the minimum-base.

=>

Robot can compute n = p*q

Lemma:

There is a unique class C* which forms a clique.


Solving Rendezvous

Solving rendezvous is easy!

1. Compute minimum-base.

2. Identify C*

3. Go to any vertex of C*

Rendezvous:

Position the robots s.t. they are mutually visible to each other.


Constructing Gvis

Edges incident to C* – Can be identified easily.

– Clique edges partitions P

– Each class appears once in each part

Same holds for any other class that forms a clique

How to identify the remaining edges?


Identifying Adjacencies

Identify edges (vi,v

i+k) of increasing

distances k = 2, 3, ..., n/2.

• Is the next unidentified vertex

vj = v

i+k or not?

• Easy, if in different classes.

• Let y be the number of dist. (k-1) backward-edges of v

i+k

• Go to vj and look back (LB)

We can show that

vj = v

i+k <=> LB = -(y+1)


Complexity of the Algorithm

• The complexity is dominated by cost of constructing minimum-base

• Walk along the boundary and identify the neighbors of each vertex.

• Use distinguishing paths to identify classes (n-1 paths of length n)

• Iteratively obtain distinguishing paths for k =1 to n

• Cost = O(n3m) moves.

(Additional cost = O(n2) moves)


Summary

• A robot moving in a polygon P that – knows an UB on n (#vertices of P)

– and can look back

• Is able to – compute the value of n

– construct the visibility graph

– solve rendezvous

• Why visibility graphs?– It is not possible to determine the exact shape of the polygon.

– Visibility graphs provide sufficient topological information.


Related Results

• With angle measurements at each vertex

A robot moving only on the boundary and knowing n, can reconstruct the polygon. [Disser et al. 2010]

• Convexity Detection + Look-back

A robot knowing n can construct the visibility graph [Bilo et al. 2009]

• Only Convexity Detection

A robot can knowing n construct the visibility graph (in exponential time) [(unpublished)]

• Impossibility

A robot moving on the boundary can not construct visibility graph even if it knows n [Bilo et al. 2009]

• Distance Measurements to visible vertices

Can the robot construct the visibility graph?


Merci de votre attention!

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