OOn guarding polygons with holes
Sharareh Alipour
Institute for Research in Fundamental Sciences, [email protected]
Abstract
There is an old conjecture by Shermer [7] that in a polygon with n vertices and h holes, b n + h c vertex guards are sufficient to guard the entire polygon. The conjecture is proved for h = 1 byShermer [7] and Aggarwal [5] seperately. In this paper, we prove a theorem similar to the Shermer’sconjecture for a special case where the goal is to guard the vertices of the polygon (not the entirepolygon) which is equivalent to finding a dominating set for the visibility graph of the polygon. Ourproof also guarantees that the selected vertex guards also cover the entire outer boundary (outerperimeter of the polygon) as well. Theory of computation → Design and analysis of algorithms;Theory of computation → Computational geometry
Keywords and phrases
Art Gallery Problem, Polygon with Holes, Triangulation
Digital Object Identifier
A set S of points is said to guard a polygon if, for every point p in the polygon, there is some q ∈ S such that the line segment between p and q is inside the polygon.The art gallery problem asks for the minimum number of guards that are sufficient toguard any polygon with n vertices. There are numerous variations of the original problemthat are also referred to as the art gallery problem. In some versions guards are restricted tothe perimeter, or even to the vertices of the polygon which are called vertex guards. Someversions require only the perimeter or a subset of the perimeter to be guarded. The versionin which guards must be placed on vertices and only vertices need to be guarded is equivalentto the minimum dominating set problem for the visibility graph of the polygon.In graph theory, for a given graph G with vertex set V ( G ), U ⊆ V ( G ) is a dominating setfor G if every vertex v ∈ V ( G ) \ U has a neighbor in U . Minimum dominating set problem isto find a dominating set V ∗ ⊆ V ( G ) such that the size of V ∗ (denoted by | V ∗ | ) is minimumamong all dominating sets. Related results
Chvátal’s art gallery theorem [2] states that b n c vertex guards are always sufficient andsometimes necessary to guard a simple polygon with n vertices. Later, Fisk [3] gave a shortproof for Chvátal’s art gallery theorem.O’Rourke [6] proved that any polygon P with n vertices and h holes can be guarded withat most b n +2 h c vertex guards. Note that n is the total number of vertices of the polygonincluding the boundary and holes. But Shermer conjectured that any polygon P with n vertices and h holes can always be guarded with b n + h c vertex guards. This conjecture hasbeen proved by Shermer [7] and Aggarwal [5] independently for h = 1. For h >
1, theconjecture is still open for more than 35 years. However Hoffmann, Kaufmann and Kriegelin [4] and Bjorling-Sachs and Souvaine in [1] proved Shermer’s conjecture for point guards(i.e. the guards can be chosen from any points inside or on the boundary of the polygon). © Sharareh Alipour;licensed under Creative Commons License CC-BY42nd Conference on Very Important Topics (CVIT 2016).Editors: John Q. Open and Joan R. Access; Article No. 23; pp. 23:1–23:5Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . C G ] F e b Our result
In this paper, we prove that every polygon with holes has a special kind of triangulationto be specified shortly. Next by using this theorem, we prove that b n + h c vertex guards aresufficient to guard the boundary of a polygon with n vertices and h holes. By boundary of P we mean the outer perimeter of P . As far as we know this version has not been studied. In this section, we present some basic definitions and a theorem in order to prove our mainresult. It has been proved that every polygon with (or without) holes can be triangulatedand this triangulation is not always unique. (cid:73)
Definition 1.
In a given polygon P with h holes, a triangle ∆ in a triangulation of P iscalled a special triangle if one of its edges is an edge of a hole and the apex vertex is a vertexof the polygon not on that hole (see Figure 1). P h h e f ∆ ab cd ∆ Figure 1
A polygon with 2 holes. In this example, ∆ and ∆ are special triangles. (cid:73) Theorem 2.
Every polygon with holes has a triangulation with a special triangle.
Note that according to the definition of special triangle, one edge of a special triangle isalways on a hole and its apex vertex is on the boundary or on a different hole. In thefollowing we explain the proof of Theorem 2.
Consider a triangulation of our polygon. If the number of holes is bigger than one, then theremust exists an edge of this triangulation from a vertex of one of these holes to a vertex of theboundary of the polygon. If we make this edge into two parallel edges of very small distancefrom each other, we can make this hole into a part of the boundary, hence reduce the numberof holes. A special triangle for this reduced polygon is also a special triangle for the originalpolygon. By induction on the number of holes, therefore it is enough to consider only thecase when we have a polygon with one hole. Let us assume that this polygon with one holehas n vertices, and the existence of a special triangle for a polygon with one hole is provedfor the case when the number of vertices is less than n . Note that the smallest possible n is6 that happens when we have a triangle with a triangle as hole inside. Any triangulation . Alipour 23:3 for this particular polygon has a special triangle and in fact 3 special triangles. With thisassumption, we can assume that non-adjacent vertices of the boundary can not see eachother. Since if they do, the chord connecting them divide the boundary into two smaller partswhere the hole is inside one of them. A special triangle for this smaller instance of a polygonwith one hole, is also a special triangle for the original polygon. Hence a triangulation forthe polygon does not have a triangle whose vertices are all on the boundary. This impliesthat any vertex of the boundary must see at least one vertex of the hole. Assume that we donot have a special triangle, we want to reach to a contradiction.Let B be a vertex of the boundary. According to the previous argument, it will see avertex on the hole, say H . Without loss of generality, we may assume that the ray B H ina counter clockwise rotational sweep, sees part of the edge H H of the hole.Since a special triangle does not exist, this rotating ray will hit an obstacle that preventsit from seeing the entire edge H H . Let H be the point on the edge H H obtained byrotating this ray until it hits an obstacle. This obstacle is either a vertex from the boundaryor a vertex from the hole. Assume that it is a vertex B from the boundary (we will discussthe second possibility shortly). Since the vertices on the boundary do not see non-adjacentvertices on the boundary, B must be adjacent to B . Then B also sees the portion H H of H H and even more. Repeating the sweeping procedure with the ray B H , we will hitanother obstacle, unless B H H is a special triangle. The sequence of obstacles obtainedthis way can not be all the vertices of the boundary, because they keep seeing larger andlarger portions of the edge H H and hence they are different and we only have a finitenumber of vertices of the boundary. So we will reach a vertex H of the hole after say k ≥ B , . . . , B k are consecutive vertices on the boundary and the angles B i − B i B i +1 are all less than π , since they are obtained by counter clockwise sweeps. Thevertex B k will see a portion of the edge of the hole with the end-point H , say H H . Sinceboth of the edges of the hole with end-point H are to the right of the ray B k H , thereforewe still need to rotate this ray counter clockwise along the edge H H and hence if we hit aboundary vertex obstacle, say B k +1 , the angle B k − B k B k +1 is less than π .Notice that the obstacles encountered for a vertex of the boundary B by rotating itscorresponding ray counter-clockwise, can not all be among the vertices of the hole. In thiscase the rotating ray will make a full rotation of 2 π , and this is impossible since the maximumrotational angle that this ray can have is less than the angle of the vertex B , which isdefinitely less than 2 π degrees.Now since we are assuming that no special triangle exists, the process of sweeping raysand hitting obstacles will be continued forever. Producing a sequence of vertices of theboundary and holes. Since after reaching a boundary vertex, we can not get only verticesfrom the hole, the sequence of boundary vertices that are consecutive vertices of the polygonmust come back to the starting point. Hence we go along all the vertices of the boundary,however the outer angles B i − B i B i +1 are all less than π . This is a contradiction. Now as a result of Theorem 2, we present our main theorem. (cid:73)
Theorem 3.
For a given polygon P with n vertices and h holes, b n + h c vertex guards arealways sufficient to guard the vertices of P and also the entire boundary. Proof.
The proof is by induction on the number of holes. Chvátal’s theorem implies thatwhen h = 0, b n c vertex guards are sufficient to guard the entire polygon. Suppose that C V I T 2 0 1 6 P h a a Figure 2
We split the vertex a into 2 vertices a and a . Now the polygon has 1( h −
1) hole and16( n + 1) vertices. the theorem is proved for h − P with n vertices and h holes. According to Theorem 2, there is a triangulation with a special triangle∆. Suppose that ∆ has a vertex a , on the boundary or a hole and another edge, bc on someother hole. We split a into two vertices a and a . So, P is changed into a polygon P with h − n + 1 vertices (See Figure 2). According to the induction assumption, we canchoose b n +1+ h − c vertex guards that guard the vertices of polygon P and the boundary of P . Since a and a , are guarded in P , then all vertices of P are guarded by at most b n + h c vertex guards of P . Also by induction the boundary of P is guarded. On the other handthe boundary of P is a subset of the boundary of P , so the boundary of P is guarded too.Note that if we have two vertex guards on a and a , in P they are combined into one vertexguard. (cid:74)(cid:73) Remark 4.
Note that in fact this proof, gives something slightly more. The guards willcover not only the entire outer perimeter of the polygon, but also the perimeter of holes withan exception of at most h segments on them. The reason is that the segment that is thebase of the special triangle that was used in the proof above is not necessarily guarded, so asimilar induction on the number of holes will prove our claim. In this paper, we have proved a theorem similar to the Shermer’s conjecture for a special caseof the Art Gallery problem where we only need to guard the vertices of the polygon. Theproof is based on the existence of a special triangle in the polygon. The proposed algorithmis simple and easy to implement. In future work one can possibly extend this idea for thegeneral case.Also for a given connected graph G , it has been proved that the size of minimumdominating set of G is ≤ n G , our proof gives an upper bound of b n + h c for the size of minimum dominating set of G . References Iliana Bjorling-Sachs and Diane L. Souvaine. An efficient algorithm for guard placement inpolygons with holes.
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