VPG and EPG bend-numbers of Halin Graphs
aa r X i v : . [ m a t h . C O ] J a n VPG and EPG bend-numbers of Halin Graphs
Mathew C. Francis Computer Science Unit, Indian Statistical Institute, Chennai
Abhiruk Lahiri
Department of Computer Science and Automation, Indian Institute of Science, Bangalore
Abstract
A piecewise linear simple curve in the plane made up of k + 1 line segments, each of which is either horizontalor vertical, with consecutive segments being of different orientation is called a k -bend path . Given a graph G ,a collection of k -bend paths in which each path corresponds to a vertex in G and two paths have a commonpoint if and only if the vertices corresponding to them are adjacent in G is called a B k -VPG representation of G . Similarly, a collection of k -bend paths each of which corresponds to a vertex in G is called an B k -EPGrepresentation of G if any two paths have a line segment of non-zero length in common if and only if theircorresponding vertices are adjacent in G . The VPG bend-number b v ( G ) of a graph G is the minimum k suchthat G has a B k -VPG representation. Similarly, the EPG bend-number b e ( G ) of a graph G is the minimum k such that G has a B k -EPG representation. Halin graphs are the graphs formed by taking a tree with nodegree 2 vertex and then connecting its leaves to form a cycle in such a way that the graph has a planarembedding. We prove that if G is a Halin graph then b v ( G ) ≤ b e ( G ) ≤
2. These bounds are tight.In fact, we prove the stronger result that if G is a planar graph formed by connecting the leaves of anytree to form a simple cycle, then it has a VPG-representation using only one type of 1-bend paths and anEPG-representation using only one type of 2-bend paths.
1. Introduction
The notion of edge intersection graphs of paths on a grid (EPG graphs) was first introduced by Golumbic,Lipshteyn and Stern [GLS09]. A graph G is said to be an EPG graph if its vertices can be mapped to pathsin a grid graph (a graph with vertex set { ( x, y ) : x, y ∈ N } and edge set { ( x, y )( x ′ , y ′ ) : | x − x ′ | + | y − y ′ | = 1 } )in such a way that two vertices in G are adjacent if and only if the paths corresponding to them share anedge of the grid graph. If in addition, none of the paths used have more than k bends (90-degree turns), then G is said to be a B k -EPG graph. We could also think of the paths on the grid graph to be just piecewiselinear curves on the plane and this gives rise to the following equivalent definition for EPG graphs. A k -bendpath is piecewise linear curve made up of k + 1 horizontal and vertical line segments in the plane such thatany two consecutive line segments of the curve are of different orientation. A graph G = ( V, E ) is a B k -EPGgraph if there exists a k -bend path P u corresponding to each vertex u ∈ V ( G ) such that any two paths P u and P v have a horizontal or vertical line segment of non-zero length in common if and only if uv ∈ E ( G ). Thecollection of k -bend paths { P u } u ∈ V ( G ) is said to be a B k -EPG representation of G . The EPG bend-number b e ( G ) of a graph G is the minimum k such that G has a B k -EPG representation. Clearly, B -EPG graphsare the well known class of interval graphs. Several papers have explored the EPG bend-number of differentclasses of graphs. The EPG bend-number of trees and cycles have been determined in [GLS09]. Bounds onthe the EPG bend-number of planar graphs, outerplanar graphs and complete bipartite graphs have beendetermined in [HKU12] and [HKU14].Later in 2012, Asinowski et al. [ACG +
11, ACG +
12] introduced the study of vertex intersection graphsof paths on a grid (VPG graphs). For defining VPG graphs, we will as before talk about k -bend paths in Email addresses: [email protected] (Mathew C. Francis), [email protected] (Abhiruk Lahiri) Partially supported by the DST INSPIRE Faculty Award IFA12-ENG-21.
Preprint submitted to arXiv.org September 26, 2018 he plane, instead of dealing with paths on a grid graph. A collection of k -bend paths { P u } u ∈ V ( G ) is saidto be a B k -VPG representation of a graph G , if for u, v ∈ V ( G ), we have uv ∈ E ( G ) if and only if P u and P v have at least one point in common. A graph is said to be simply a VPG graph if it is a B k -VPG graphfor some k . Several relationships between VPG graphs and other known graph classes have been studiedin [ACG + + B -VPG graphs are a special case of segmentgraphs (intersection graphs of line segments in the plane), which are known as 2-DIR graphs. L-VPG graphsare intersection graphs of L-shaped paths, which is clearly a subclass of B -VPG graphs. The subclass ofL-VPG graphs which have L-VPG representations in which no two paths cross (have a common internalpoint) are called L-contact graphs [CKU]. More results and a literature survey of graph classes that admitcontact VPG representations can be found in [AF].A tree-union-cycle graph is formed by taking a tree and then connecting its leaves to form a cycle in sucha way that the graph has a planar embedding. A Halin graph is a tree-union-cycle graph with no degree 2vertices. These graphs were first introduced by Rudolf Halin in his study of minimally 3-connected graphs[Hal71].In this paper we show that Halin graphs are B -VPG graphs as well as B -EPG graphs. In fact,we show the stronger results that any tree-union-cycle graph has a L-VPG representation and a B -EPGrepresentation with one type of 2-bend paths. We also demonstrate Halin graphs that are not B -VPGgraphs and Halin graphs that are not B -EPG graphs. Note that L-contact graphs are subclass of 2-degenerate graphs [CKU]. As Halin graphs have minimum degree three, it is not possible to obtain aL-contact representation for Halin graphs. B -VPG representation of Halin graphs and IO-graphs (2-connected planar graphs in which the interiorvertices form a (possibly empty) independent set) was studied in [BD]. It was shown in that paper thatall IO-graphs are in L-VPG and that all Halin graphs have a B -VPG representation in which every vertexother than one is represented using an L-shaped path and one vertex is represented using a L-shaped path.As any IO-graph can be easily seen to be an induced subgraph of some tree union cycle graph, our resultsshow that both these graph classes are subclasses of L-VPG.
2. Preliminaries
A simple graph G = T ∪ C , where T is a tree and C is a simple cycle induced by all the leaves of the tree T such that G is planar, is called a tree-union-cycle graph . A Halin graph is a tree-union-cycle graph withno degree 2 vertex.Let us consider a tree-union-cycle graph G = T ∪ C , where the cycle C is of length k . Let C be c c . . . c k − c . Let S = V ( G ) \ V ( C ) be the set of internal vertices of the tree T . Let us also assume the set S is not a singleton set, i.e. the graph G is not a wheel graph. Designate an internal vertex r of T , to bethe root of the tree T . Once a root is fixed, we define the following notations.1. For any two vertices u, v ∈ T , u is said to be an ancestor of v if u lies on the path rT v . On the otherhand if v lies on the path rT u then u is called a descendant of v . If uv ∈ T and v is a descendant of u then we often call v is a child of u and u is the parent of v .2. For any vertex u ∈ T , let L r ( u ) be the set of all leaves that are descendants of u . If u is a leaf thendefine L r ( u ) = { u } . Clearly if u is a descendant of v then L r ( u ) ⊆ L r ( v ).3. For each vertex u ∈ T , define the height of the vertex h r ( u ) as the length of the path rT u . Let hca ( u, v ) = min { h r ( x ) : x ∈ uT v } . Lemma 1.
Let u, v ∈ S , then u is an ancestor of v or v is an ancestor of u if and only if L r ( u ) ∩ L r ( v ) = ∅ .Proof. Let u is an ancestor of v . Clearly L r ( v ) ⊆ L r ( u ). Hence L r ( u ) ∩ L r ( v ) = ∅ . Similarly it can beshown if v is an ancestor of u then L r ( u ) ∩ L r ( v ) = ∅ . Conversely, let L r ( u ) ∩ L r ( v ) = ∅ . Hence there exists c i ∈ L r ( u ) ∩ L r ( v ). This implies that u and v both are on the path c i T r . Hence one of them is an ancestorof the other. This completes the proof.
Observation 1.
For any u ( = r ) ∈ S , L r ( u ) does not contain all the leaves of T . roof. For the sake of contradiction, assume if L r ( u ) = L r ( r ) = V ( C ). Then it implies that the degree of r is one, which contradicts the fact that r is an internal vertex.The following lemma can be easily seen. Lemma 2.
Fix a vertex r ∈ S , as the root of the tree T . Then the vertices in L r ( u ) appear consecutively inthe cyclic ordering c , c , . . . , c k − , c . Lemma 3.
Let c i , c i +1 be two leaves and r be an internal vertex on the path c i T c i +1 . If we fix root as r , then for any u ∈ S , the vertices in L r ( u ) appear consecutively in the linear ordering c i +1 , c i +2 , . . . ,c k − , c , c , . . . , c i .Proof. Suppose there exists a vertex u ∈ S such that the vertices in L r ( u ) do not appear consecutively inthe linear ordering c i +1 , c i +2 , . . . , c k − , c , c , . . . , c i . Then it follows from Lemma 2 that c i , c i +1 ∈ L r ( u ). If u = r , then all leaves are descendant of u . Hence u = r . If v is a vertex such that c i +1 ∈ L r ( v ), then v is onthe path c i +1 T r . Similarly, if w is a vertex such that c i ∈ L r ( w ), then w is on the path c i T r . Therefore, u is an internal vertex of both c i +1 T r and c i T r . Since r ∈ c i T c i +1 , c i T r and c i +1 T r have no internal vertexin common. This implies that such a u cannot exist.Now we prove the following lemma which is used to define a linear ordering of the vertices on the cycle. Lemma 4.
Fix a vertex r ∈ S as the root of T . Then there exists an index i , such that all the internalvertices except one on the path c i T c i +1 are of degree two.Proof. Let us consider two consecutive leaves c i and c i +1 of T such that h r ( hca ( c i , c i +1 )) is maximum. Weclaim that this pair of leaves c i and c i +1 satisfy the statement of the lemma. Let u = hca ( c i , c i +1 ). If u isthe common parent of c i and c i +1 , then our claim holds trivially.For the sake of contradiction, assume u is not the common parent of c i and c i +1 and at least one of thepaths from uT c i and uT c i +1 has an internal vertex of degree at least three. Let that vertex be v , lying on thepath uT c i . Consider L r ( v ). Clearly c i is in L r ( v ). We know that the degree of v is at least three. So v hasat least one leaf other than c i as its descendant. From Lemma 2, we know that the vertices in L r ( v ) appearconsecutively in the cyclic ordering of leaves. So either of c i − or c i +1 is in L r ( v ). We have c i +1 / ∈ L r ( v ).Otherwise, v is a common ancestor of both c i and c i +1 with h r ( v ) > h r ( u ), which contradicts the fact that u = hca ( c i , c i +1 ). So c i +1 / ∈ L r ( v ). Therefore, we can conclude that c i − ∈ L r ( v ). But then we have found apair of leaves c i − and c i such that h r ( hca ( c i − , c i )) ≥ h r ( v ) > h r ( u ). This contradicts our initial choice ofthe pair of leaves c i , c i +1 . So there is no internal vertex with degree more than two on the path uT c i . Witha similar argument we can conclude that there is no internal vertex with degree more than two on the path uT c i +1 . This concludes the proof.
3. Halin graphs are L-VPG graphs
Let c i , c i +1 be two leaves satisfying Lemma 4. Define, b j = c ( i + j ) mod k , for 0 ≤ j ≤ k −
1. Fix r ′ = hca ( c i , c i +1 ) to be the new root of T .The following observations are easy consequences of our choice of root and Lemma 3. Observation 2.
Every internal vertex of both the paths b T r ′ and b T r ′ has exactly one descendant leaf. Observation 3.
For any u ∈ S , the vertices in L r ′ ( u ) appear consecutively in the linear ordering b , b , . . . ,b k − . Let us define an L-shaped curve on the plane as the set of points given by L ([ x , x ] , [ y , y ]) = { ( x, y ) | x = x and y ∈ [ y , y ] or y = y and x ∈ [ x , x ] } . By definition, a graph is an L-VPG graph if one can associatean L-shaped curve L u to each vertex u of the graph such that uv is an edge of the graph if and only if L u ∩ L v = ∅ . Now we show that G is an L-VPG graph by defining the L-shaped curve L u to be associatedwith each vertex u ∈ V ( G ). For any vertex u ∈ V ( G ) with L u = L ([ x , x ] , [ y , y ]), we define l x ( u ) = x , r x ( u ) = x , l y ( u ) = y and r y ( u ) = y . Define h = max { h r ′ ( u ) | u ∈ V ( G ) } .Define L b = L ([1 . , k − , [0 , h − h r ′ ( b ) + 2]). 3efine L b = L ([1 , , [1 , h − h r ′ ( b ) + 2]).Define L b k − = L ([ k − , k ] , [0 , h − h r ′ ( b k − ) + 2]).For every leaf b i other than b , b or b k − , define L b i = L ([ i, i + 1] , [1 , h − h r ′ ( b i ) + 2]).For every internal vertex v , define L v = ([min { l x ( w ) | w ∈ L r ′ ( v ) } , max { l x ( w ) | w ∈ L r ′ ( v ) } ] , [ h − h r ′ ( v )+1 , h − h r ′ ( v ) + 2]).Now we prove that the collection of L-shapes {L u } u ∈ V ( G ) forms a valid one bend V P G representation of G . We first state some observations that will be needed. Observation 4.
For any u ∈ S and leaves b i , b j , b k , if l x ( b i ) < l x ( b j ) < l x ( b k ) and b i , b k ∈ L r ′ ( u ) then b j ∈ L r ′ ( u ) .Proof. Suppose for the sake of contradiction that there exist u ∈ S and leaves b i , b j , b k with l x ( b i ) < l x ( b j ) Let u, v ∈ V ( G ) be distinct vertices such that u is an ancestor of v . Then, (a) l y ( v ) / ∈ [ l y ( u ) , r y ( u )] , (b) l x ( v ) ∈ [ l x ( u ) , r x ( u )] , and (c) l y ( u ) ∈ [ l y ( v ) , r y ( v )] if and only if u is the parent of v .Proof. According to our construction, for any internal vertex w , l y ( w ) = h − h r ′ ( w ) + 1. When u is anancestor of v clearly u is an internal vertex. Suppose first that v is a leaf. Clearly, l y ( v ) = 0 or 1. But h r ′ ( u )can be at most h − l y ( u ) ≥ 2. Hence l y ( v ) / ∈ [ l y ( u ) , r y ( u )]. If v is an internal vertex then h r ′ ( u ) < h r ′ ( v ). Hence l y ( v ) < l y ( u ), implies l y ( v ) / ∈ [ l y ( u ) , r y ( u )]. This completes the proof of (a). Now, weprove the Observation 5(b). Since u is an ancestor of v , we have L r ′ ( v ) ⊆ L r ′ ( u ). From our construction, l x ( u ) = min { l x ( w ) | w ∈ L r ′ ( u ) } ≤ l x ( v ) and r x ( u ) = max { l x ( w ) | w ∈ L r ′ ( u ) } ≥ l x ( v ). So we have l x ( v ) ∈ [ l x ( u ) , r x ( u )]. This proves (b). In order to prove (c), first assume u to be the parent of v . Then u isalways an internal vertex. From our construction we have l y ( u ) = r y ( v ). This implies l y ( u ) ∈ [ l y ( v ) , r y ( v )].Conversely, let l y ( u ) ∈ [ l y ( v ) , r y ( v )]. From our construction we have r y ( v ) = h − h r ′ ( v ) + 2. Since u is anancestor of v , we have h r ′ ( u ) ≤ h r ′ ( v ) − 1. If h r ′ ( u ) < h r ′ ( v ) − 1, then we have r y ( v ) < l y ( u ). But thiscontradicts the fact that l y ( u ) ∈ [ l y ( v ) , r y ( v )]. We can thus conclude that h r ′ ( u ) = h r ′ ( v ) − 1, implying that u is the parent of v . This completes the proof of (c). Observation 6. Let u, v ∈ V ( G ) such that u is an internal vertex and v is a leaf. Then, (a) l y ( v ) / ∈ [ l y ( u ) , r y ( u )] , and (b) l x ( v ) ∈ [ l x ( u ) , r x ( u )] if and only if u is an ancestor of v .Proof. According to our construction, l y ( u ) = h − h r ′ ( u ) + 1. For any leaf v , l y ( v ) = 0 or 1. Also we know, h r ′ ( u ) ≤ h − 1. Hence l y ( v ) < l y ( u ). This proves (a). In order to prove (b), first assume that u is an ancestorof v . From Observation 5(b), we have l x ( v ) ∈ [ l x ( u ) , r x ( u )]. Conversely, assume l x ( v ) ∈ [ l x ( u ) , r x ( u )].Suppose that u is not an ancestor of v . Then v / ∈ L r ′ ( u ). (Note that this means that u = r ′ .) From ourconstruction, l x ( u ) = min { l x ( w ) | w ∈ L r ′ ( u ) } and r x ( u ) = max { l x ( w ) | w ∈ L r ′ ( u ) } . Let b i ∈ L r ′ ( u ) besuch that l x ( u ) = l x ( b i ) and let b j ∈ L r ′ ( u ) be such that r x ( u ) = l x ( b j ). So we have found three leaves b i , v and b j such that l x ( b i ) ≤ l x ( v ) ≤ l x ( b j ). Note that from our construction, for any two leaves w, w ′ , we have l x ( w ) = l x ( w ′ ) if and only if w = w ′ . Since b i , b j ∈ L r ′ ( u ) and v / ∈ L r ′ ( u ), we can conclude that the vertices b i , b j are distinct from v . Therefore, we have l x ( b i ) < l x ( v ) < l x ( b j ). This contradicts Observation 4. Hence u is an ancestor of v . This completes the proof. 4 bservation 7. Let u and v be both internal vertices. If u is not an ancestor of v , or v is not an ancestorof u , then [ l x ( v ) , r x ( v )] ∩ [ l x ( u ) , r x ( u )] = ∅ .Proof. Let u and v be internal vertices such that neither of them is an ancestor of the other. Hence from theLemma 1, we have L r ′ ( u ) ∩ L r ′ ( v ) = ∅ . For the sake of contradiction, assume [ l x ( v ) , r x ( v )] ∩ [ l x ( u ) , r x ( u )] = ∅ .Then either l x ( u ) ∈ [ l x ( v ) , r x ( v )] or l x ( v ) ∈ [ l x ( u ) , r x ( u )]. Let us consider the case when l x ( u ) ∈ [ l x ( v ) , r x ( v )].From our construction, l x ( u ) = min { l x ( w ) | w ∈ L r ′ ( u ) } , l x ( v ) = min { l x ( w ) | w ∈ L r ′ ( v ) } and r x ( v ) =max { l x ( w ) | w ∈ L r ′ ( v ) } . Let b i , b j and b k be leaves such that l x ( b i ) = l x ( v ), l x ( b j ) = l x ( u ) and l x ( b k ) = r x ( v ). Clearly, b i , b k ∈ L r ′ ( v ) and b j ∈ L r ′ ( u ). Since L r ′ ( u ) ∩ L r ′ ( v ) = ∅ , we know that b j / ∈ L r ′ ( v ). Thisimplies that b j is distinct from both b i and b k . Since l x ( v ) ≤ l x ( u ) ≤ r x ( v ), we have l x ( b i ) ≤ l x ( b j ) ≤ l x ( b k ).From our construction, it is clear that for any two leaves w, w ′ , we have l x ( w ) = l x ( w ′ ) if and only if w = w ′ .Therefore, since b j is distinct from b i , b k , we have l x ( b i ) < l x ( b j ) < l x ( b k ). Since we also have b i , b k ∈ L r ′ ( v )and b j / ∈ L r ′ ( v ), we have a contradiction to Observation 4. The proof for the case when l x ( v ) ∈ [ l x ( u ) , r x ( u )]is similar. Observation 8. Let u and v be both leaves. Then u and v are consecutive leaves if and only if either: (a) l x ( u ) ∈ [ l x ( v ) , r x ( v )] and l y ( v ) ∈ [ l y ( u ) , r y ( u )] , or (b) l x ( v ) ∈ [ l x ( u ) , r x ( u )] and l y ( u ) ∈ [ l y ( v ) , r y ( v )] .Proof. Suppose that u = b i and v = b j are consecutive leaves. Without loss of generality we assume that i < j . Clearly, i < k − 1. First, consider the case when i > 0. Then, we have j > 1. From the construction,it is clear that l x ( v ) = r x ( u ) and l y ( u ) ∈ [ l y ( v ) , r y ( v )]. Hence it satisfies (b). If i = 0 (that is u = b ), either v = b or v = b k − . If u = b and v = b , then l x ( u ) ∈ [ l x ( v ) , r x ( v )] and l y ( v ) ∈ [ l y ( u ) , r y ( u )]. Hence itsatisfies (a). If u = b and v = b k − , then l x ( v ) = r x ( u ) and l y ( u ) = l y ( v ). In this case, (b) is satisfied. Soat least one of (a) or (b) is true when u and v are consecutive leaves. Conversely, let us assume u = b i and v = b j are not consecutive leaves. Again, we assume without loss of generality that i < j . First consider thecase when i > 0. From the construction, it is clear that if j > i + 1, then [ l x ( u ) , r x ( u )] ∩ [ l x ( v ) , r x ( v )] = ∅ .This implies that neither of (a) or (b) is satisfied. Now consider the case when u = b and v = b or k − l y ( u ) / ∈ [ l y ( v ) , r y ( v )] and l x ( u ) / ∈ [ l x ( v ) , r x ( v )]. Hence neither of (a) or (b) is satisfied. Thisconcludes the proof.Now we are ready to show the main result of this section. Claim 1. The collection of L-shapes {L u } u ∈ V ( G ) forms a valid one bend VPG representation of G .Proof. We prove that for distinct u, v ∈ V ( G ), uv ∈ E ( G ) if and only if L u ∩ L v = ∅ . It is easy to see that L u ∩ L v = ∅ if and only if L u , L v satisfy at least one of the following two conditions:(1) l x ( u ) ∈ [ l x ( v ) , r x ( v )] and l y ( v ) ∈ [ l y ( u ) , r y ( u )], or(2) l x ( v ) ∈ [ l x ( u ) , r x ( u )] and l y ( u ) ∈ [ l y ( v ) , r y ( v )].Therefore, we only need to show that for distinct u, v ∈ V ( G ), uv ∈ E ( G ) if and only if L u , L v satisfyeither condition (1) or condition (2). Note that we will be done if we show that L u , L v satisfy eithercondition (1) or condition (2) if and only if u is a parent of v , v is a parent of u , or u, v are consecutiveleaves.Let u be the parent of v . From Observation 5(c), we have l y ( u ) ∈ [ l y ( v ) , r y ( v )]. Also, from Observa-tion 5(b), we have l x ( v ) ∈ [ l x ( u ) , r x ( u )]. Therefore, L u , L v satisfy condition (2). Similarly, it can be shownthat when v is the parent of u , L u , L v satisfy the condition (1). When u, v are consecutive leaves, fromObservation 8, we have that L u , L v satisfy either condition (1) or condition (2).Conversely, we show that if u is not a parent of v , v is not a parent of u and u, v are not two consecutiveleaves, then L u , L v do not satisfy either condition (1) or condition (2). Suppose first that neither of u, v is an ancestor of the other. If u is an internal vertex and v is a leaf, then by Observation 6(a), we have l y ( v ) / ∈ [ l y ( u ) , l y ( u )], implying that L u , L v do not satisfy condition (1). By Observation 6(b), we have l x ( v ) / ∈ [ l x ( u ) , r x ( u )], implying that L u , L v do not satisfy condition (2) either. Similarly, it can be shown that when v is an internal vertex and u is a leaf, L u , L v do not satisfy either condition (1) or condition (2). Let u and v both be internal vertices. Then, by Observation 7, we have l x ( u ) / ∈ [ l x ( v ) , r x ( v )] and l x ( v ) / ∈ [ l x ( u ) , r x ( u )].5herefore, L u , L v do not satisfy either condition (1) or condition (2). Let us now look at the case when u and v are both leaves. Then, Observation 8 implies that L u , L v do not satisfy either condition (1) orcondition (2).Suppose that u is an ancestor of v . Then, by Observation 5(a), we know that L u , L v do not satisfycondition (1). Also, since u is not the parent of v , we know by Observation 5(c) that L u , L v do not satisfycondition (2) either. If v is an ancestor of u , then the same line of reasoning shows that L u , L v do not satisfyeither condition (1) or condition (2).This completes the proof of Claim 1.We now show that there are Halin graphs that have VPG bend-number more than 0. v v v v v v Figure 1: A Halin graph which has no B -VPG representation. Claim 2. There exists a Halin graph that is not B -VPG.Proof. Consider the Halin graph with six vertices and nine edges, shown in Figure 1. For the sake ofcontradiction, assume that this graph is in B -VPG. Then it has a VPG representation in which each vertex v is represented by a horizontal or vertical line segment, which we denote by S v . If ( x , y ) and ( x , y ) are thetwo endpoints of a segment S v in this representation, we define l x ( v ) = min { x , x } , r x ( v ) = max { x , x } , l y ( v ) = min { y , y } and r y ( v ) = { y , y } . Clearly, if S v is a horizontal segment, we have l y ( v ) = r y ( v ) andif it is a vertical segment, we have l x ( v ) = r x ( v ).Notice that v , v , v induce a K in the graph. Clearly, some two among S v , S v and S v have thesame orientation, i.e., either horizontal or vertical. Because the vertices are all symmetric to each other,we can assume without loss of generality that S v and S v are both horizontal. (We can argue in a similarfashion if they are both vertical segments.) As they intersect, we have l y ( v ) = l y ( v ) and [ l x ( v ) , r x ( v )] ∩ [ l x ( v ) , r x ( v )] = ∅ . Let [ l x ( v ) , r x ( v )] ∩ [ l x ( v ) , r x ( v )] = [ a, b ]. Again by symmetry between v and v ,we can assume without loss of generality that l x ( v ) ≤ l x ( v ). Note that v has one neighbour v , whichhas no adjacency with v and v has one neighbour v , which has no adjacency with v . This implies that l x ( v ) = a and r x ( v ) = b and also that there exists a point ( e, l y ( v )) ∈ S v with l x ( v ) ≤ e < a . If S v isa vertical line segment, then this means that r x ( v ) = l x ( v ) = e < a . Suppose that S v is a horizontal linesegment. Then l y ( v ) = r y ( v ) = l y ( v ). Since S v contains the point ( e, l y ( v )), it cannot be the case that r x ( v ) ≥ a as that would imply that the point ( a, l y ( v )) ∈ S v which would be a contradiction to the factthat S v ∩ S v = ∅ . Therefore, we can conclude that r x ( v ) < a . Therefore, irrespective of whether S v ishorizontal or vertical, we have r x ( v ) < a . Arguing very similarly, we can also conclude that l x ( v ) > b . So,we have r x ( v ) < a ≤ b < l x ( v ) and therefore, S v ∩ S v = ∅ . But this contradicts the fact that v and v are adjacent in the graph. Theorem 1. Tree-union-cycle graphs are in B -VPG but there is even a Halin graph that is not in B -VPG.Proof. The proof follows directly from Claims 1 and 2. Corollary 1. If H is any Halin graph, then b v ( H ) ≤ . This bound is tight. 4. Halin graphs are B -EPG graphs We relabel the vertices in C as a , a , . . . , a k − and reassign the root of the tree T at a new internalvertex. Recall that every internal vertex in the path b T b except one has degree two. Let b ′ be the internalvertex of T adjacent to b and let b ′ be the internal vertex of T that is adjacent to b (note that it is possible6o have b ′ = b ′ ). If b ′ has degree two, then define a j = b ( j +1) mod k , for 0 ≤ j ≤ k − 1, and define r = b ′ .If b ′ has degree more than two, then we define a j = b ( k − j ) mod k , for 0 ≤ j ≤ k − 1, and define r = b ′ . Forthe rest of the section, wherever we use r it will mean our new choice of the root. The ancestor-descendantrelationship and height is also defined with respect to this new root.The following is an easy consequence of our choice of root and Lemma 3. Observation 9. For any vertex u ∈ S , the vertices in L r ( u ) appear consecutively in a , a , . . . , a k − . Observation 10. r is the parent of a . Denote by a ′ the parent of a k − . Observation 11. Either a ′ has degree two or a ′ = r .Proof. If r = b ′ , from our labelling, we have a k − = b . Clearly b ′ has been chosen as a root because degreeof b ′ is two. Hence in this case a ′ has degree two. If r = b ′ , from our labelling, we have a k − = b . In thiscase we know that b ′ has degree more than two. Since exactly one vertex of the path b ′ T b ′ has degree otherthan two in T , we must either have b ′ = b ′ , in which case b ′ = a ′ has degree 2, or b ′ = b ′ , in which case a ′ = r . This completes the proof. (a) (b) Figure 2: (a) Four different types of C-shaped curves. (b) Four different types of S-shaped curves In a B -EPG representation of a graph, each vertex is represented by a piecewise linear curve in theplane made up of three line segments each of which is horizontal or vertical with consecutive segments beingof different orientation. It is not hard to see that such a curve can have any of eight different shapes—i.e.,it can be a C-shaped curve and its rotations, or an S-shaped curve and its rotations and reflections (seeFigure 2). We show that G has an EPG representation using C-shaped curves of one type. Later, we showthat G also has an EPG representation using S-shaped curves.A C-shaped curve is defined as a set of points given by C ( x , x , x , y , y ) = { ( x, y ) | x = x and y ∈ [ y , y ] } ∪ { ( x, y ) | y = y and x ∈ [ x , x ] } ∪ { ( x, y ) | y = y and x ∈ [ x , x ] } . Our aim is to associatea C-shaped curve C u to each vertex u ∈ V ( G ) such that uv ∈ E ( G ) if and only if C u ∩ C v contains atleast a horizontal or a vertical line segment of non-zero length. For any vertex u ∈ V ( G ) with C u = C ( x , x , x , y , y ), define l x ( u ) = x , p x ( u ) = x , q x ( u ) = x , l y ( u ) = y and r y ( u ) = y . Define h =max { h r ( u ) | u ∈ V ( G ) } . Let ǫ = 1 / a i other than a and a k − , define C a i = C ( i, i + 1 + ǫ, i + 3 ǫ, , h − h r ( a i ) + 1).Define C a = C (0 , ǫ, k − ǫ, , h + 1).Define C a k − = C ( k − , k + ǫ, k − ǫ, , h + 1).For every internal vertex v other than r and a ′ , define C v = C (min { i + 2 ǫ | a i ∈ L r ( v ) } , max { i + 3 ǫ | a i ∈ L r ( v ) } , min { i + 3 ǫ | a i ∈ L r ( v ) } , h − h r ( v ) , h − h r ( v ) + 1).If a ′ = r , define C a ′ = C ( k − ǫ, k − ǫ, k − ǫ, , h − h r ( a ′ ) + 1) and define C r = C (2 ǫ, k − ǫ, , h, h + 1).If a ′ = r , define C r = C (2 ǫ, k − ǫ, k − ǫ, h, h + 1). (Note that we can get a valid representationof G in which no two curves cross more than once by setting p x ( a ′ ) = k − ǫ . But here, we are using p x ( a ′ ) = k − ǫ so that the proof becomes a little bit simpler.)Our aim is now to prove that {C u } u ∈ V ( G ) forms a valid B -EPG representation of G . We first make someobservations. Observation 12. Let u be the parent of v . (a) If v / ∈ { a , a k − } , then l y ( u ) = r y ( v ) and | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , q x ( v )] | > , If v ∈ { a , a k − } and u = r , then r y ( u ) = r y ( v ) and | [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] | > , and (c) If v = a k − and u = r , then l y ( u ) = l y ( v ) and | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] | > .Proof. If v / ∈ { a , a k − } , we cannot have u = a ′ and a ′ = r . Therefore, from the construction, we have l y ( u ) = h − h r ( u ) and r y ( v ) = h − h r ( v ) + 1. Clearly, since u is the parent of v we have h r ( v ) = h r ( u ) + 1.This implies l y ( u ) = r y ( v ). First, let us consider the case when v is an internal vertex. Since u is the parentof v , L r ( v ) ⊆ L r ( u ). From the construction, we have l x ( u ) = min { i + 2 ǫ | a i ∈ L r ( u ) } ≤ min { i + 2 ǫ | a i ∈ L r ( v ) } = l x ( v ) and q x ( v ) = min { i + 3 ǫ | a i ∈ L r ( v ) } ≤ max { i + 3 ǫ | a i ∈ L r ( u ) } = p x ( u ). (Notethat this holds also when v = a ′ , since in that case a ′ = r , and therefore L r ( a ′ ) = { a k − } . Also notethat it is possible that u = r .) Clearly from our construction we have l x ( u ) < p x ( u ) and l x ( v ) < q x ( v ).Since we have l x ( u ) ≤ l x ( v ) < q x ( v ) ≤ p x ( u ), we conclude that | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , q x ( v )] | > 1. Let usconsider the case when v = a j is a leaf. Clearly, we have v ∈ L r ( u ). Then from the construction, we have l x ( u ) = min { i + 2 ǫ | a i ∈ L r ( u ) } , p x ( u ) = max { i + 3 ǫ | a i ∈ L r ( u ) } , l x ( v ) = j and q x ( v ) = j + 3 ǫ . Since v ∈ L r ( u ), it implies that l x ( u ) ≤ j + 2 ǫ and p x ( u ) ≥ j + 3 ǫ . Hence [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , q x ( v )] contains theinterval [ j + 2 ǫ, j + 3 ǫ ]. This implies that | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , q x ( v )] | > 1. This completes the proof of (a).If u = r and v = a , from the construction we have r y ( u ) = h + 1 = r y ( v ). Again from the constructionwe have [ l x ( u ) , q x ( u )] = [2 ǫ, k − ǫ ] if u = r = a ′ and [ l x ( u ) , q x ( u )] = [2 ǫ, 1] if u = r = a ′ . Also wehave [ l x ( v ) , q x ( v )] = [0 , k − ǫ ]. Hence [2 ǫ, ⊂ [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )], which implies | [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] | > 1. Now consider the case u = r and v = a k − . As u is the parent of v , we have u = r = a ′ .From the construction we have r y ( u ) = h + 1 = r y ( v ), [ l x ( u ) , q x ( u )] = [2 ǫ, k − ǫ ] and [ l x ( v ) , q x ( v )] =[ k − , k − ǫ ]. As [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] = [ k − , k − ǫ ], we have | [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] | > v = a k − and u = r , then from the construction we have l y ( v ) = 0 and l y ( u ) = 0. From theconstruction [ l x ( u ) , p x ( u )] = [ k − ǫ, k − ǫ ] and [ l x ( v ) , p x ( v )] = [ k − , k + ǫ ]. This implies that[ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = [ k − ǫ, k − ǫ ]. Hence | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] | > 1. This completesproof of (c). Observation 13. Let u and v be two vertices such that neither of them is an ancestor of the other. Fur-thermore, let u be an internal vertex. Then, (a) [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = ∅ , and (b) If v = a , then [ l x ( u ) , max { p x ( u ) , q x ( u ) } ] ∩ [ l x ( v ) , max { p x ( v ) , q x ( v ) } ] = ∅ .Proof. Let us first prove (a). From Lemma 1, we have L r ( u ) ∩ L r ( v ) = ∅ . First we consider the case when both u and v are internal vertices. Let us define l min ( u ) = min { i | a i ∈ L r ( u ) } and l max ( u ) = max { i | a i ∈ L r ( u ) } .From Observation 9 and L r ( u ) ∩ L r ( v ) = ∅ , we conclude [ l min ( u ) , l max ( u )] ∩ [ l min ( v ) , l max ( v )] = ∅ . Againfrom our construction, we have [ l x ( w ) , p x ( w )] = [ l min ( w ) + 2 ǫ, l max ( w ) + 3 ǫ ], when w ∈ S . Hence weconclude [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = ∅ (note that ǫ = 1 / ǫ < v = a j is a leaf. As v is not a descendant of u , we have v / ∈ L r ( u ). Following Observation 9 this impliesthat j / ∈ [ l min ( u ) , l max ( u )]. From the construction we have [ l x ( v ) , p x ( v )] = [ j, j + 1 + ǫ ]. Again, from theconstruction we have [ l x ( u ) , p x ( u )] = [ l min ( u ) + 2 ǫ, l max ( u ) + 3 ǫ ]. Hence [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = ∅ as ǫ = 1 / 4. This completes the proof of (a).Let us now prove (b). Clearly, u = r and v = r , because neither of u or v is a descendant of theother. This tells us that u, v / ∈ { a , r } . From the construction, we know that for every vertex w / ∈ { a , r } ,max { p x ( w ) , q x ( w ) } = p x ( w ). So it is sufficient to show that, [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = ∅ . This followsfrom (a). Observation 14. Let u and v be both leaves. If u and v are consecutive leaves then either: (a) l y ( u ) = l y ( v ) and | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] | > , or (b) r y ( u ) = r y ( v ) and | [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] | > .Proof. Suppose u = a i and v = a i +1 are consecutive leaves where 0 ≤ i ≤ k − 2. From the construction itis clear that l y ( u ) = l y ( v ) = 0 for all i . Again from the construction we have [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] =[ i + 1 , i + 1 + ǫ ]. Hence they satisfy (a). When u = a and v = a k − , from the construction it is clear that r y ( u ) = r y ( v ) = h + 1. Then [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] = [ k − , k − ǫ ]. Hence they satisfy (b).8e are now ready to prove that {C u } u ∈ V ( G ) forms a valid B -EPG representation of G . Claim 3. For distinct u, v ∈ V ( G ) , uv ∈ E ( G ) if and only if C u ∩ C v contains a horizontal or a vertical linesegment of non-zero length.Proof. It is easy to see that C u ∩ C v contains a horizontal or a vertical line segment of non-zero length if andonly if C u , C v satisfies at least one of the following.(1) l y ( u ) = l y ( v ) and | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] | > 1, or(2) r y ( u ) = r y ( v ) and | [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] | > 1, or(3) l y ( u ) = r y ( v ) and | [ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , q x ( v )] | > 1, or(4) l y ( v ) = r y ( u ) and | [ l x ( v ) , p x ( v )] ∩ [ l x ( u ) , q x ( u )] | > 1, or(5) l x ( u ) = l x ( v ) and | [ l y ( u ) , r y ( u )] ∩ [ l y ( v ) , r y ( v )] | > u, v ∈ V ( G ), uv ∈ E ( G ) if and only if C u , C v satisfy any one of the conditions (1)to (5), thereby completing the proof. Note that we only need to prove that C u , C v satisfy any of theconditions (1) to (5) if and only if u is the parent of v , or v is the parent of u , or u, v are consecutiveleaves.If u is the parent of v , then from Observation 12, C u and C v satisfy at least one of the conditions (1),(2) or (3). When u, v are consecutive leaves, from Observation 14 we have that C u , C v satisfy condition (1)or (2).Conversely, we show that if u is not a parent of v , v is not a parent of u and u, v are not two consecutiveleaves, then C u , C v do not satisfy any one of the conditions (1) to (5).Suppose first that one of u, v is an internal vertex. Without loss of generality assume that u is an internalvertex.Suppose further that neither of u, v is an ancestor of the other. Clearly, in this case neither u nor v is r .By Observation 13(a), we know that C u , C v do not satisfy condition (1). Observation 13(a) also tells us that l x ( u ) = l x ( v ) and therefore C u , C v do not satisfy condition (5). If v = a , then by Observation 13(b), C u , C v do not satisfy any of the conditions (2), (3) and (4). Let us suppose that v = a . Then r y ( v ) = h + 1. Itis clear from our construction that since u is an internal vertex other than r , r y ( u ) = h + 1 and therefore, r y ( u ) = r y ( v ). This implies C u , C v do not satisfy condition (2). Also, from our construction, there is novertex w such that l y ( w ) = h + 1, and therefore C u , C v do not satisfy condition (3). Again, we have l y ( v ) = 0and no vertex w such that r y ( w ) = 0. Therefore C u , C v do not satisfy condition (4).Suppose that one of u, v is an ancestor of the other. Without loss of generality assume that u is anancestor of v , but not the parent. Note that in this case, u = a ′ and v = a . Clearly, h r ( v ) ≥ h r ( u ) + 2. Let v = a k − . From the construction it is clear that r y ( v ) = h − h r ( v ) + 1 < h − h r ( u ) = l y ( u ) and therefore, wehave [ l y ( u ) , r y ( u )] ∩ [ l y ( v ) , r y ( v )] = ∅ . It follows that C u , C v do not satisfy conditions (1) to (5). Let us considerthe case when v = a k − . Suppose u = r . From the construction it is clear that l y ( v ) = l y ( u ), r y ( v ) = r y ( u ), l y ( v ) = r y ( u ) and r y ( v ) = l y ( u ). Also from the construction it is clear that l x ( v ) = l x ( u ). Hence C u , C v do notsatisfy conditions (1) to (5). Suppose that u = r and v = a k − . Then we have l x ( u ) = l x ( v ), l y ( u ) = l y ( v ), l y ( u ) = r y ( v ) and l y ( v ) = r y ( u ). Therefore, C u and C v do not satisfy conditions (1), (3), (4) or (5). Also,since q x ( u ) = 1 < k − l x ( v ) (recall that that r = u = a ′ ), we have [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] = ∅ ,implying that C u and C v do not satisfy condition (2).Suppose u and v are two nonadjacent leaves. Then l x ( u ) = l x ( v ). This implies C u , C v do not satisfycondition (5). Again l y ( u ) = 0 and l y ( v ) = 0. Hence, l y ( u ) = r y ( v ) and l y ( v ) = r y ( u ). This implies C u , C v do not satisfy conditions (3) and (4). As u, v are nonadjacent leaves, from the construction we have[ l x ( u ) , p x ( u )] ∩ [ l x ( v ) , p x ( v )] = ∅ . This implies C u , C v do not satisfy condition (1). If u, v / ∈ { a , a k − } , fromthe construction we have [ l x ( u ) , q x ( u )] ∩ [ l x ( v ) , q x ( v )] = ∅ . If either of u or v belongs to { a , a k − } then r y ( u ) = r y ( v ). This implies C u , C v do not satisfy condition (2).This completes the proof of Claim 3.Now, we prove that there are Halin graphs that do not have B -EPG representations.In a B -EPG representation, a one bend curve looks either like an L-shaped curve, or its rotations. Letus denote a one bend curve as L ′ ( x , x , y , y , ( x i , y j )) = { ( x, y ) | y = y j and x ∈ [ x , x ] } ∪ { ( x, y ) | x = x i and y ∈ [ y , y ] } , where i, j ∈ { , } . Clearly an L-shaped curve is L ′ ( x , x , y , y , ( x , y )). Suppose thatthere is a B -EPG representation for a graph H in which each vertex u ∈ V ( H ) is represented by a one-bend9 v v v v v v v v v v v Figure 3: A Halin graph which has no B -EPG representation. curve L ′ u . With respect to this representation, for any vertex u ∈ V ( H ) define L ′ u = L ′ ( x , x , y , y , ( x i , y j )),where i, j ∈ { , } . Also define l x ( u ) = x , r x ( u ) = x , l y ( u ) = y and r y ( u ) = y . Let us denote the bendpoint ( x i , y j ) by b u . Let us also denote the x -coordinate of b u by x ( b u ) and y -coordinate of b u by y ( b u ). Itis easy to see that L ′ u ∩ L ′ v contains a horizontal or a vertical line segment of non-zero length if and only if L ′ u , L ′ v satisfy at least one of the following conditions.(1) x ( b u ) = x ( b v ) and | [ l y ( u ) , r y ( u )] ∩ [ l y ( v ) , r y ( v )] | > 1, or(2) y ( b u ) = y ( b v ) and | [ l x ( u ) , r x ( u )] ∩ [ l x ( v ) , r x ( v )] | > {L ′ u } u ∈ V ( H ) form a valid B -EPG representation of the graph H , for every u, v ∈ V ( H ), uv ∈ E ( H ) if and only if L ′ u , L ′ v satisfy at least one of the conditions (1) or (2).First we prove the following observation for later use. Observation 15. Suppose that the curves {L ′ u i } i ∈{ , , , } form a B -EPG representation of the chordlesscycle on four vertices denoted by u u u u u . If L ′ u , L ′ u and L ′ u , L ′ u both satisfy condition (1) or bothsatisfy condition (2), then b u = b u .Proof. Let us first consider the case when L ′ u , L ′ u and L ′ u , L ′ u both satisfy condition (1). In this case, wehave x ( b u ) = x ( b u ) = x ( b u ). As u and u are nonadjacent, L ′ u and L ′ u do not satisfy condition (1),which implies that | [ l y ( u ) , r y ( u )] ∩ [ l y ( u ) , r y ( u ] | ≤ 1. Therefore, we have either r y ( u ) ≤ l y ( u ) or r y ( u ) ≤ l y ( u ). We can assume without loss of generality that r y ( u ) ≤ l y ( u ) (as the cycle can be relabeledto make this true if it is not already the case). Clearly, this implies l y ( u ) < r y ( u ) ≤ l y ( u ) < r y ( u ). Nowsuppose that x ( b u ) = x ( b u ) ( = x ( b u ) = x ( b u )). As u and u are nonadjacent, L ′ u and L ′ u shouldnot satisfy condition (1), implying that either r y ( u ) ≤ l y ( u ) or r y ( u ) ≤ l y ( u ). If r y ( u ) ≤ l y ( u ), wehave by our previous observation that r y ( u ) < l y ( u ). But this would imply that L ′ u , L ′ u do not satisfyeither condition (1) or condition (2), which contradicts the fact that u and u are adjacent. Similarly, if r y ( u ) ≤ l y ( u ), we have r y ( u ) < l y ( u ), which means that L ′ u , L ′ u will not satisfy either condition (1)or condition (2), which is a contradiction as u and u are adjacent. Therefore, we can conclude that x ( b u ) = x ( b u ). Since this means that x ( b u ) = x ( b u ) and x ( b u ) = x ( b u ), it must be the case that both L ′ u , L ′ u and L ′ u , L ′ u satisfy condition (2). Then y ( b u ) = y ( b u ) and y ( b u ) = y ( b u ) together implies y ( b u ) = y ( b u ). As x ( b u ) = x ( b u ) we conclude that b u = b u .We can argue similarly for the case when L ′ u , L ′ u and L ′ u , L ′ u both satisfy condition (2).We consider H to be the Halin graph with thirty one vertices and fifty edges, shown in the Figure 3. Weshow that this graph has no B -EPG representation, thereby completing the proof the Claim 4. Claim 4. There exists a Halin graph that cannot be represented as an edge intersection graph of paths on agrid such that each path has at most one bend.Proof. For the sake of contradiction assume that H has a B -EPG representation. This implies, with everyvertex v ∈ V ( G ), we can associate a one bend curve L ′ v , such that uv ∈ E ( G ) if and only if L ′ u , L ′ v satisfy atleast one of the conditions (1) or (2). Note that the vertex v has five neighbours v , v , . . . , v . Clearly, eachpair of curves L ′ v , L ′ v i satisfies at least one of the conditions (1) or (2) for each value of i ∈ { , , . . . , } . Notethat we can assume that a majority of these pairs of curves satisfy condition (2) (as if that is not the case we10an reflect the whole collection of curves along the line x = y to obtain another valid representation of thegraph in which this is true). So we assume without loss of generality that each pair L ′ v , L ′ v i , for 1 ≤ i ≤ y ( b v ) = y ( b v i ) and | [ l x ( v ) , r x ( v )] ∩ [ l x ( v i ) , r x ( v i )] | > 1, for each i ∈ { , , } .As { v , v , v } form an independent set in the graph, there exists a vertex, say v , such that [ l x ( v ) , r x ( v )] ( [ l x ( v ) , r x ( v )]. This means that if any pair of curves L ′ v , L ′ v i , where i ∈ { , , } , satisfy condition (2), then L ′ v , L ′ v i also satisfy condition (2), which is a contradiction as none of the vertices v , v or v are adjacentto v . Therefore, for each i ∈ { , , } , the curves L ′ v , L ′ v i must satisfy the condition (1). Consider thecycle v v v v v . Then from the Observation 15 we have b v = b v . Also v v v v v form a cycle. Againapplying Observation 15 we have b v = b v . This implies that | [ l y ( v i ) , r y ( v i )] ∩ [ l y ( v j ) , r y ( v j )] | > i, j ∈ { , , } . As we have x ( b v ) = x ( b v ) = x ( b v ) = x ( b v ), this means that L ′ v i , L ′ v j satisfycondition (1), which is a contradiction to the fact that { v , v , v } form an independent set in the graph.Hence the proof. Theorem 2. Tree-union-cycle graphs are in B -EPG. There are Halin graphs that are not in B -EPG.Proof. The proof follows directly from Claims 3 and 4. Corollary 2. If H is any Halin graph, then b e ( H ) ≤ . This bound is tight. 5. Concluding remarks We showed that the tree-union-cycle graph G has a B -EPG representation using C-shaped curves. Itcan be seen that G has a B -EPG representation using S-shaped curves too.An S-shaped curve is defined as a set of points given by S ( x , x , x , y , y ) = { ( x, y ) | x = x and y ∈ [ y , y ] } ∪ { ( x, y ) | y = y and x ∈ [ x , x ] } ∪ { ( x, y ) | y = y and x ∈ [ x , x ] } . Our aim is to associatean S-shaped curve S u to each vertex u ∈ V ( G ) such that uv ∈ E ( G ) if and only if S u ∩ S v containsat least a horizontal or a vertical line segment of non-zero length. For any vertex u ∈ V ( G ) with S u = S ( x , x , x , y , y ), define l x ( u ) = x , m x ( u ) = x , r x ( u ) = x , l y ( u ) = y and r y ( u ) = y .We choose the root r and label the leaves a , a , . . . , a k − as described in the previous section. Define h = max { h r ( u ) | u ∈ V ( G ) } . Let ǫ = 1 / h .For every leaf a i other than a or a k − , define S a i = S ( i − − ǫ, i, i + 1 − ǫ, , h − h r ( a i ) + 1).Define S a = S ( − ǫ, , k − ǫ, , h + 1).Define S a k − = S ( k − − ǫ, k − , k − ǫ, , h + 1).For every internal vertex v other than r and a ′ , define S v = S (min { i | a i ∈ L r ( v ) } , max { i + ( h − h r ( v )) ǫ | a i ∈ L r ( v ) } , max { i + 1 − ǫ | a i ∈ L r ( v ) } , h − h r ( v ) , h − h r ( v ) + 1).If a ′ = r , define S a ′ = S ( k − − ǫ, k − ǫ, k − ǫ, , h − h r ( a ′ ) + 1) and define S r = S ( ǫ, k − ǫ, k − ǫ, h, h + 1).If a ′ = r , define S r = S ( ǫ, k − − ǫ, k − ǫ, h, h + 1).It is easy to see that S u ∩ S v contains a horizontal or a vertical line segment of non-zero length if andonly if S u , S v satisfies at least one of the following.(1) l y ( u ) = l y ( v ) and | [ l x ( u ) , m x ( u )] ∩ [ l x ( v ) , m x ( v )] | > 1, or(2) r y ( u ) = r y ( v ) and | [ m x ( u ) , r x ( u )] ∩ [ m x ( v ) , r x ( v )] | > 1, or(3) l y ( u ) = r y ( v ) and | [ l x ( u ) , m x ( u )] ∩ [ m x ( v ) , r x ( v )] | > 1, or(4) l y ( v ) = r y ( u ) and | [ l x ( v ) , m x ( v )] ∩ [ m x ( u ) , r x ( u )] | > 1, or(5) m x ( u ) = m x ( v ) and | [ l y ( u ) , r y ( u )] ∩ [ l y ( v ) , r y ( v )] | > u, v ∈ V ( G ), uv ∈ E ( G ) if and only if S u , S v satisfy at least one of the conditions (1) to (5). Thus the collection of S-shaped curves {S v } v ∈ V ( G ) forms a B -EPG representation of the graph G . 11 eferences [ACG + 11] Andrei Asinowski, Elad Cohen, Martin Charles Golumbic, Vincent Limouzy, Marina Lipshteyn,and Michal Stern. String graphs of k -bend paths on a grid. Electronic Notes in Discrete Mathe-matics , 37:141–146, 2011.[ACG + 12] Andrei Asinowski, Elad Cohen, Martin Charles Golumbic, Vincent Limouzy, Marina Lipshteyn,and Michal Stern. Vertex intersection graphs of paths on a grid. J. Graph Algorithms Appl. ,16(2):129–150, 2012.[AF] Nieke Aerts and Stefan Felsner. Vertex contact graphs of paths on a grid. In Graph-TheoreticConcepts in Computer Science - 40th International Workshop, WG 2014, Nouan-le-Fuzelier,France, June 25-27, 2014. Revised Selected Papers , pages 56–68.[BD] Therese Biedl and Martin Derka. 1-string B -VPG representations of planar partial 3-trees andsome subclasses. In Proceedings of the 27th Canadian Conference on Computational Geometry,CCCG 2015, Kingston, Ontario, Canada, 2015 .[CKU] Steven Chaplick, Stephen G. Kobourov, and Torsten Ueckerdt. Equilateral L-contact graphs.In Graph-Theoretic Concepts in Computer Science - 39th International Workshop, WG 2013,L¨ubeck, Germany, June 19-21, 2013, Revised Papers , pages 139–151.[GLS09] Martin Charles Golumbic, Marina Lipshteyn, and Michal Stern. Edge intersection graphs ofsingle bend paths on a grid. Networks , 54(3):130–138, 2009.[Hal71] Rudolf Halin. Studies on minimally n -connected graphs. Combinatorial Mathematics and itsApplications , pages 129–136, 1971.[HKU12] Daniel Heldt, Kolja B. Knauer, and Torsten Ueckerdt. On the bend-number of planar andouterplanar graphs. In LATIN , pages 458–469, 2012.[HKU14] Daniel Heldt, Kolja B. Knauer, and Torsten Ueckerdt. Edge-intersection graphs of grid paths:The bend-number.