1 Introduction
An EPG graph is a graph that admits a representation in which its vertices are represented by paths of a grid , such that two vertices of are adjacent if and only if the corresponding paths have at least one common edge.
The study of EPG graphs has motivation related to the problem of VLSI design that combines the notion of edge intersection graphs of paths in a tree with a VLSI grid layout model, see Golumbic et al. (2009). The number of bends in an integrated circuit may increase the layout area, and consequently increase the cost of chip manufacturing. This is one of the main applications that instigate research on the EPG representations of some graph families when there are constraints on the number of bends in the paths used in the representation. Other applications and details on circuit layout problems can be found in Bandy and Sarrafzadeh (1990); Molitor (1991).
A graph is a EPG graph if it admits a representation in which each path has at most bends. As an example, Figure 1(a) shows a , Figure 1(b) shows an EPG representation where the paths have no bends and Figure 1(c) shows a representation with at most 1 bend per path. Consequently, is a EPG graph. More generally, EPG graphs coincide with interval graphs.
(a) The graph  (b) EPG representation of (edgeclique)  (b) (c) EPG representation of (clawclique) 
The bend number of a class of graphs is the smallest for which all graphs in the class have a EPG representation. Interval graphs have bend number Golumbic et al. (2009), trees have bend number Golumbic et al. (2009) and outerplanar graphs have bend number Heldt et al. (2014a). The bend number for the class of planar graphs is still open, but it is either or Heldt et al. (2014a).
The class of EPG graphs has been studied in several papers, such as Alcón et al. (2016); Asinowski and Suk (2009); Cohen et al. (2014); Golumbic et al. (2009); Heldt et al. (2014b); Pergel and Rza̧żewski (2017), among others. The investigations frequently approach characterizations with respect to the number of bends of the graph representation. Regarding the complexity of recognizing EPG graphs, only the complexity of recognizing few of these subclasses of EPG graphs have been determined: EPG graphs can be recognized in polynomial time, since it corresponds to the class of interval graphs, see Booth and Lueker (1976). In contrast, recognizing EPG and EPG graphs are complete problems, see Heldt et al. (2014b) and Pergel and Rza̧żewski (2017), respectively. Moreover, recognizing EPG graphs remains complete even for shaped paths in grid, see Cameron et al. (2016).
A collection of sets satisfies the Helly property when every subcollection of
that is pairwise intersecting has at least one common element. The study of the Helly property is useful in very diverse areas of science, and we can enumerate applications in semantics, code theory, computational biology, database, image processing, graph theory, optimization, and linear programming
Dourado et al. (2009).The Helly property can be applied to the EPG representation problem, where each path is considered to be a set of edges. A graph has a Helly EPG representation if there is a EPG representation of where each path has at most bends and this representation satisfies the Helly property. Figure 2(a) presents two EPG representations of a graph with five vertices. Figure 2(b) illustrates 3 pairwise intersecting paths (), containing a common edge, so it is a Helly EPG representation. In Figure 2(c), although the three paths are pairwise intersecting, there is no common edge in all 3 paths, and therefore they do not satisfy the Helly property. The Helly property related to EPG representations of graphs has been studied in Golumbic et al. (2009) and Golumbic et al. (2013). In particular, they have determined the strong Helly number of EPG graphs.
(a) A graph with 5 vertices  (b) EPG representation that satisfies the Helly property  (c) EPG representation that does not satisfy the Helly property 
Next, we describe some terminology and notation used.
The term grid is used to denote the Euclidean space of integer orthogonal coordinates. Each pair of integer coordinates corresponds to a point or vertex of the grid. The term edge of the grid, will be used to denote a pair of vertices that are at distance one in the grid. Two edges and are consecutive edges when they share exactly one point of the grid. A (simple) path in the grid is as a sequence of distinct edges , where consecutive edges are adjacent, i.e., contain a common vertex, whereas nonconsecutive edges are not adjacent. In this context, two paths only intersect if they have at least a common edge. The first and last edges of a path are called extremity edges.
The direction of an edge is vertical when the first coordinates of its vertices are equal, and is horizontal when the second coordinates are equal. A bend in a path is a pair of consecutive edges of that path, such that the directions of and are different. When two edges and form a bend, they are called bend edges. A segment is a set of consecutive edges with no bends. Two paths are said to be edgeintersecting, or simply intersecting, if they share at least one edge. Throughout the paper any time we say that two paths intersect, we mean that they edgeintersect.
EPG graphs are the class of intersection graphs of paths in a grid Golumbic et al. (2009). This class, consists of the class of graphs in which its vertices are represented by paths of a grid , such that two vertices in are adjacent if and only if their corresponding paths intersect. If every path in the representation has at most bends, we say that this graph has a EPG representation.When we say that this is a single bend representation.
A family of sets is pairwise intersecting if any two sets in the family intersect. A collection of nonempty sets satisfies the Helly property when every pairwise intersecting subcollection of has at least one element that is in every subset of .
The Helly property can be applied to the EPG representation problem, where each path is considered to be a set of edges. A graph has a Helly EPG representation if there is a EPG representation of where each path has at most bends and this representation satisfies the Helly property.
This paper studies graphs that have a HellyEPG representation. We prove that the problem of recognizing Helly EPG graphs is complete. In particular, we show that this results remains true even if the graphs are apex and degenerate.
2 Preliminaries
The study starts with the following lemma.
Lemma 1 (Golumbic et al. (2009)).
Every graph is an EPG graph.
We show that the result extends to HellyEPG graphs.
Lemma 2.
Every graph is a HellyEPG graph.
Proof.
Let be a graph with vertices and maximal cliques . We construct a HellyEPG representation of , using a grid of size . The rows correspond to the maximal cliques and are numbered . Each vertex contains segments from column to column . Each maximal clique is mapped into an edge . Each path contains all edges corresponding to the maximal cliques containing .
Moreover, two distinct paths intersect at the edges corresponding to the maximal cliques containing .
Let , consider the maximal cliques containing in ascending order of their indices. The path representing , starts at vertex of and descends column until , where is the first clique containing , then bends at vertex and proceeds to the right, traversing edge representing . Then proceeds further on row until reaching vertex , where it bends again, descending column until reaching vertex , where is the next maximal clique containing where it bends again. It proceeds in row , traversing edge , and so on, until all edges of corresponding to the maximal cliques containing have been traversed by .
It follows that two distinct paths, and , intersect exactly at the rows corresponding to the maximal cliques containing both and . Hence the representation is Helly. ∎
Figure 3 shows the grid and the path corresponding to the vertex , contained in maximal cliques and of .
Corollary 3.
Every graph containing maximal cliques admits a Helly EPG representation.
Next we examine the EPG representations of a few graphs that we employ in our constructions. In special, we consider EPG representations of ’s.
Definition 4.
Let be a grid and let be a 4star as depicted in Figure 4(a). Let be a collection of paths each containing exactly two edges of the star:

A true pie is a representation where each of forms a bend in .

A false pie is a representation where two of the paths do not contain bends, while the remaining two do not share an edge.
(a) 4star in grid (b) True pie (c) False pie Figure 4: EPG representation of the induced cycle of size 4 as pies with emphasis in center
Definition 5.
Consider a rectangle of any size with 4 corners at vertices , positioned as in Figure 5. A frame is a representation containing 4 paths , each having a bend in a different corner of a rectangle, and such that the subpaths , share at least one edge. While the subpaths and do not share edges.
(a) Points of the coordinates of bends of a frame  (c) Paths of a frame 
Lemma 6 (Golumbic et al. (2009)).
Every that is an induced subgraph of a graph corresponds, in any representation, to a true pie, a false pie, or a frame.
Definition 7.
A EPG representation is minimal when its set of edges does not properly contain another EPG representation.
The octahedral graph is the graph containing 6 vertices and 12 edges, depicted in Figure 6(a). Next, we consider representations of the octahedral.
The next lemma follows directly from the discussion presented in Heldt et al. (2014b).
Lemma 8.
The octahedral graph has a unique minimal EPG representation.
Proof.
The octahedral graph has in its construction induced cycles of size 4 (’s). Take an induced subgraph of the octahedral . The pairs of nonadjacent vertices of the induced are false twins whose neighborhoods are the remaining vertices of the induced . Each of the vertices outside the are adjacent to all vertices of the . Thus, if in a EPG representation of , the is represented as a frame, no single bend path can simultaneously intersect the 4 paths representing the vertices of the induced . Therefore, we conclude that the frame structure can not be part of an representation.
With the same reasoning, take a EPG representation of where the induced subgraph is represented as true pie or false pie. When adding the false twin vertices, which are neighbors of all vertices of taken from , both representations converge to the structure represented in Figure 6(b). ∎
(a) The octahedral graph  (b) EPG representation of the graph 
By Lemma 8, has a unique minimal EPG representation, up to isomorphisms, as depicted in Figure 6(b). The paths and do not satisfy the Helly property. Therefore Helly EPG.
Clearly, EPG and Helly EPG graphs coincide. Hence Helly EPG can be recognized in polynomial time Booth and Lueker (1976). However, Lemma 8 has proved that the class of Helly EPG is a proper subclass of EPG. It remains the question about the complexity of recognizing Helly EPG graphs. In this paper we answer such a question.
Figure 7 depicts graphs of the classes EPG, Helly EPG and EPG that distinguish these classes.
3 Membership in
Helly EPG recognition problem can be formally described as follows:
Helly EPG Recognition  

Input:  A graph and an integer , for some fixed .  
Goal: 

In this section, we show that Helly EPG Recognition, where is bounded by a polynomial function of , belongs to .
A (positive) certificate for the Helly EPG recognition consists of a grid , a set of bend paths of , which is in onetoone correspondence with the vertex set of , such that, for each pair of distinct paths if and only if the corresponding vertices are adjacent in . Furthermore, satisfies the Helly property.
The following are key concepts that make it easier to control the size of an EPG representation. A relevant edge of a path in a EPG representation is one which is either an extremity edge or a bend edge of the path. Therefore each path with at most bends can have up to relevant edges, and any EPG representation contains at most distinct relevant edges.
To show that there is a nondeterministic polynomialtime algorithm for Helly EPG recognition, it is enough to consider as certificate a EPG representation containing a collection of paths, , such that each path is given by its set of relevant edges along with the relevant edges, that intersects , of each path intersecting , where . The relevant edges for each path are given in the order that they appear in the path, so as to make straightforward checking that the edges correspond to a unique path with at most bends. This representation is also handy for checking that the paths form an intersection model for .
In order to verify in polynomial time that the input is in fact a positive certificate for the problem, we have to assert the following:

The sequence of relevant edges of a path determines in polynomial time;

Two paths intersect if and only if they intersect in some relevant edge;

The set of relevant edges satisfies the Helly property.
The following lemma states that condition (i) holds.
Lemma 9.
Each path can be determined uniquely in polynomial time by the sequence of its relevant edges.
Proof.
Consider the sequence of relevant edges of some path . Start from an extremity edge of . Let be the row (column) containing the last considered relevant edge. The next relevant edge in the sequence, must be also contained in row (column) . If is an extremity edge, the process is finished and the path has been determined. It contains all edges between the considered relevant edges in the sequence. Otherwise, if is a bend edge , the next relevant edge is the second bend edge of this same bend, which is contained in some column (row) . The process continues until the second extremity edge of is located.
With the above procedure, we can determine in time, whether path contains any given edge of the grid . Therefore, the sequence of relevant edges of uniquely determines . ∎
Next we assert property (ii).
Lemma 10.
Let be the set of paths in a EPG representation of , and let . Then , are intersecting paths if and only if their intersection contains at least one relevant edge.
Proof.
Assume that are intersecting and we show they contain a common relevant edge. Without loss of generality, suppose intersect at row i of the grid, in the EPG representation . The following are the possible cases that may occur:

Case 1: Neither nor contain bends in row i.
Then and are entirely contained in row i. Since they intersect, either overlap, or one of the paths contains the other. In any these situations, they intersect in a common extremity edge, that is a relevant edge.

Case 2: does not contain bends in i, but does.
If some bend edge of also belongs to , then intersect in a relevant edge. Otherwise, since intersect, the only possibility is that the intersection contains an extremity edge of or . Hence the paths intersect in relevant edge.

Case 3: Both , contain bends in i
Again, if the intersection occurs in some bend edge of or , the lemma follows. Otherwise, the same situation as above must occur, that is must intersect in same extremity edge.
In any of the cases, and intersect in some relevant edge. ∎
The two previous lemmas let us check that a certificate is an actual EPG representation of a given graph . The next lemma says we can also verify in polynomial time that the representation encoded in the certificate is a Helly representation. Fortunately we do not need to check every subset of intersecting paths of the representation to make sure they have a common intersection.
Lemma 11.
Let be a collection of paths encoded as a sequence of relevant edges that constitute a EPG representation of a graph . We can verify in polynomial time if has the Helly property.
Proof.
Let be the set of relevant edges of . Consider each triple of edges of . Let be the set of paths of containing at least two of the edges in the triple . By Gilmore’s Theorem Berge and Duchet (1975), has the Helly property if an only if the subset of paths corresponding to each triple has a nonempty intersection. By Lemma 10, it suffices to examine the intersections on relevant edges. Therefore a polynomial algorithm for checking if has the Helly property could examine each of the subsets , and for each relevant edge of a path in , to compute the number of paths in that contain . Then has the Helly property if and only if for every there exists some relevant edge that is present in all paths in , yielding a nonempty intersection. ∎
Corollary 12.
Let be a set a pairwise intersecting paths in a Helly EPG representation of a graph . Then the intersection of all paths of contains at least one relevant edge.
Note that the property described in Corollary 12 is a consequence of Gilmore’s Theorem, see Berge and Duchet (1975), and it applies only to representations that satisfy Helly’s property.
Lemma 13.
Let be a Helly EPG graph. Then admits a Helly EPG representation on a grid of size at most .
Proof.
Let be a EPG representation of a graph on a grid with the smallest possible size. Let be the set of paths of . Note that . A counting argument shows that there are at most relevant edges in . If has a pair of consecutive columns neither of which contains relevant edges of , and such that there is no relevant edge crossing from to , then we can contract each edge crossing from to into single vertices so as to obtain a new EPG representation of on a smaller grid, which is a contradiction. An analogous argument can be applied to pairs of consecutive rows of the grid. Therefore the grid is such that each pair of consecutive columns and of consecutive rows of has at least one relevant edge of or contains a relevant edge crossing it. Since is the smallest possible grid for representing then the first row and the first column of must both contain at least one point belonging to some relevant edge of . Thus, if is EPG then it admits a EPG representation on a grid of size at most . In addition, by Corollary 12, it holds that the contraction operation previously described preserves the Helly property, if any. Hence, letting be a Helly EPG representation of a graph on a grid with the smallest possible size it holds that has size at most . ∎
Theorem 14.
Helly EPG recognition is in , whenever is bounded by a polynomial function of .
Proof.
By Lemma 13 and the fact that is bounded by a polynomial function of , it follows that the collection can be encoded through its relevant edges with bits.
4 hardness
Now we will prove that Helly EPG graph recognition is complete. For this proof we follow the basic strategy described in the prior hardness proof of Heldt et al. (2014b). We set up a reduction from Positive (1 in 3)3SAT defined as follows:
Positive (1 in 3)3SAT  

Input: 


Goal: 

Positive (1 in 3)3SAT is a well known complete problem (see Garey and Johnson (1979), problem [L04], page 259). Positive (1 in 3)3SAT remains complete when the incidence graph of the input CNF (Conjunctive Normal Form) formula is a planar graph Mulzer and Rote (2008).
Given a formula that is an instance of Positive (1 in 3)3SAT we will present a polynomialtime construction of a graph such that is Helly EPG if and only if is satisfiable. This graph will contain an induced subgraph with 12 vertices (called clause gadget) for every clause , and an induced subgraph (variable gadget) for each variable , containing a special vertex , plus a base gadget with 55 additional vertices.
We will use a graph isomorphic to the graph presented in Figure 8, as a gadget to perform the proof. For each clause of of the target problem, we will have a clause gadget isomorphic to , denoted by .
The reduction of a formula from Positive (1 in 3)3SAT to a particular graph such that has a Helly EPG representation if only if is satisfiable, is given below.
Definition 15.
Let be a CNFformula with variable set and clause set with no negative literals, in which every clause has exactly three literals. The graph is constructed as follows:

For each clause create a clause gadget , isomorphic to graph ;

For each variable create a variable vertex that is adjacent to the vertex , or of , when is the first, second or third variable in , respectively;

For each variable vertex , construct a variable gadget formed by adding two copies of , and , and making adjacent to the vertices of the 2 triangles in and .

Create a vertex , that will be used as vertical reference of the construction, and add an edge from to each vertex of a clause gadget;

Create a bipartite graph with a particular vertex that is in the largest stable set. This vertex is nominated true vertex. Vertex is adjacent to all and also to ;

Create two graphs isomorphic to , and . The vertex is connected to each vertex of the triangle (a,b,c) in and ;

Create two graphs isomorphic to , and . The vertex is connected to each vertex of the triangle (a,b,c) in and ;

The subgraph induced by the set of vertices will be referred to as the base gadget.
Figure 9 illustrates how this construction works on a small formula.
Lemma 16.
Given a satisfiable instance of Positive (1 in 3)3SAT, the graph constructed from according to Definition 15 admits a Helly EPG representation.
Proof.
We will use the true pie and false pie structures to represent the clause gadgets , but the construction could also be done with the frame structure without loss of generality, see Figure 10.
(a) Based in false pie  (b) Based in true pie 
The variable gadgets will be represented by structures as of Figure 11.
The base gadget will be represented by the structure of Figure 12.
It is easy to see that the representations of the clause gadgets, variable gadgets, and base gadgets are all Helly EPG. Now we need to describe how these representations can be combined in order to construct a single bend representation .
Given an assignment that satisfies , we can construct a Helly EPG representation . First we will fix the representation structure of the base gadget in the grid to guide the single bend representation, see Figure 12. Next we will insert the variable gadgets with the following rule: if the variable related to the path had assignment True, then the adjacency between the path with is horizontal, and vertical otherwise. For example, for an assignment to variables of the formula that generated the gadget of Figure 9, it will give us a single bend representation (base gadget + variables gadget) according to the Figure 13(a).
When a formula of Positive (1in3)3sat has clauses whose format of assignment is or then we will use false pie to represent these clauses, but when the clause has format we will use true pie to represent this clause. To insert a clause gadget , we introduce a horizontal line in the grid between the horizontal rows used by the paths for the two false variables in . Then we connect the path of to vertically using the bend of . However, we introduce a vertical line in the grid, between the vertical line of the grid used by and the path to the true variable in , i.e. between and the path of the true variable . At the point where and cross, to insert the center of the clause gadget as can be seen in Figure 13(b). A complete construction of this single bend representation for the can be verified in Figure 14.
Note that when we join all these representations of gadgets that form we do not increase the number of bends. Then the representation is necessarily EPG. Let us show that it satisfies the Helly property.
A simple way to check that satisfies the Helly property is to note that the particular graph never forms triangles between variable, clause, and base gadgets. Thus, any triangle of is inside a variable, clause or base gadget. As we only use Helly EPG representations of such gadgets, is a Helly EPG representation of . ∎
Next, we consider the converse. Let be a Helly EPG representation of .
Definition 17.
Let be the graph shown in Figure 8, such that the 4cycle corresponds in to a false pie or true pie, then:

the center is the unique gridpoint of this representation which is contained in every path representing 4cycle ;

a central ray is an edgeintersection between two of the paths corresponding to vertices , respectively.
Note that every EPG representation of a satisfies the Helly property, see Lemma 6, and triangles have EPG representations that satisfy the Helly property, e.g. the one shown in Figure 1(b). The graph is composed by a 4cycle and eight cycles of size 3.
As has well known representations (see in Lemma 6), then we can start drawing the Helly EPG representation of from these structures. Figure 15 shows possible representations for .
If is represented by a pie, then the paths share the center of the representation. On the other hand, if is represented by a frame then the bends of the 4 paths correspond to the four distinct corners of a rectangle, i.e. all paths representing the vertices of have distinct bend points, see Golumbic et al. (2009).
(a) Representation with omitted clause gadgets  (b) Representation with associated with the clause in highlighted 
Next we examine the use of the frame structure.
Proposition 18.
In a EPG representation of a isomorphic to a frame, every path that represents a vertex of the intersects exactly two other paths and of the frame, so that one of the intersections is horizontal and the other is vertical.
Proposition 19.
Given a Helly EPG representation of a graph that has an induced whose representation is isomorphic to a frame. If there is a vertex of , outside the , that is adjacent to exactly two consecutive vertices of this , then the path representing shares at least one common edgeintersection with the paths representing both of these vertices.
Proof.
By assumption, has a triangle containing and two vertices of a . Therefore the path representing shares at least one common edge intersecting with the paths representing these neighbors, otherwise the representation does not satisfy the Helly property. ∎
By Proposition 18 and Proposition 19 we can conclude that for every vertex such that , when we use a frame to represent the , will have at least one common edgeintersection with the pair of paths representing its neighbors in . Figure 15(c) presents a possible Helly EPG representation of . Note that we can apply rotations and mirroring operations, while maintaining it as a Helly EPG representation of .
Definition 20.
In a single bend representation of a graph isomorphic to a frame, the paths that represent consecutive vertices in the are called consecutive paths and the segment that corresponds to the intersection between two consecutive paths is called side intersection.
Lemma 21.
In any single bend minimal representation of a graph isomorphic to , there are two paths in that have horizontal directions and the other two paths have vertical directions.
Proof.
If the is represented by a true pie or false pie, then each path of shares two central rays with two other paths of , where each central ray corresponds to one pair of consecutive vertices in .
As the vertices and are adjacent to pairs of consecutive vertices in so the paths and have to be positioned in each one of the different central rays, 2 are horizontal and 2 are vertical.
If the is represented by a frame then each path of the has a bend positioned in the corners of the frame. In the frame, the adjacency relationship of pairs of consecutive vertices in the is represented by the edgeintersection of the paths that constitute the frame. Thus, since a frame has two parts in the vertical direction and two parts in the horizontal direction, then there are two paths in that have horizontal direction and two that have vertical direction.
Note that, by minimality of the representation, no additional edge is needed on the different paths. ∎
Corollary 22.
In any single bend minimal representation of a graph isomorphic to , the following paths are on the same central ray or side intersection: and ; and ; and ; and .
(a) Based in false pie  (b) Based in true pie  (c) Based in frame 
Fact 23.
The following proposition helps us in understanding of the hardness proof.
Proposition 24.
In any single bend Helly representation of the graph presented in Figure 17(a), the path has obstructed extremities and bends.
Proof.
Consider consisting of a vertex together with two graphs, and , isomorphic to and a bipartite graph , such that: is a vertex of the largest stable set of the ; is adjacent to an induced cycle of size 3 of , and to an induced cycle of size 3 of , , see Figure 17(a).
We know that the paths belonging to the largest stable set of a always will bend into a false pie, see Fact 23. Since is part of the largest stable set of the , then has an obstructed bend, see Figure 17(b).
The vertex is adjacent to and , so that its path intersects the paths representing them. But in a single bend representations of a graph isomorphic to there are pairs of paths that always are on some segment of a central ray or a side intersection, see Corollary 22, and the representation of ( similarly has one these paths. Therefore, there is an edge in the set of paths that represent ( similarly in ) that has a intersection of 3 paths representing (and ), otherwise the representation would not be Helly. There is another different edge in the same central ray or side intersection that contains three other paths and one of them is not in the set of paths ( similarly . Thus in a single bend representation of , the paths that represent ( similarly must intersect in a bend edge or an extremity edge of , because intersects only one of the paths that are on some central ray or side intersection where ( similarly is. As the bend of is already obstructed by structure of , then ( similarly in ) must be positioned at an extremity edge of . This implies that has a condition of obstructed extremities, see Figure 17(b).
(a) The graph  (b)A EPG representation of 
∎
Definition 25.
We say that a segment is internally contained in a path if is contained in , and it does not intersect a relevant edge of .
Lemma 26.
If a graph , constructed according to Definition 15, admits a Helly EPG representation, then the associated CNFformula is a yesinstance of Positive (1 in 3)3sat.
Proof.
Suppose that has a Helly EPG representation, . From we will construct an assignment that satisfies .
First, note that in every single bend representation of a , the path of each vertex of the greater stable set, in particular (in ), has bends contained in a false pie (see Remark 23).
The vertex is adjacent to the vertices of a triangle of and . As the is positioned in the bend of , then in the representations of and are positioned at the extremities of , see Proposition 4.3.
Without loss of generality assume that is a horizontal segment in .
We can note in that: the number of paths with segment internally contained in is the number of clauses in ; the intersection between each in the gadget clause and each path indicates the variables composing the clause. Thus, we can assign to each variable the value True if the edge intersecting and is horizontal, and False otherwise.
In Lemma 21 it was shown that any minimal EPG representation of a clause gadget has two paths in with vertical direction and the other two paths have horizontal direction. Since intersects , it follows that in a single bend representation of we must connect two of these in order to represent a false assignment, and exactly one will represent a true assignment. Thus, from we construct an assignment to such that every clause has exactly one variable with a true value. ∎
Note that a EPG representation is Helly if and only if each clique is represented by an edgeclique (and not by a clawclique). More details on edgeclique and clawclique can be found in Golumbic et al. (2009). Thus, an alternative way to check that a representation is Helly is to note that all cliques are represented as edgecliques.
Some of the vertices of have highly constrained EPG representations. Vertex has its bend and both extremities obstructed by its neighbors in , and in the subgraphs. Vertex and each variable vertex must have one of its segments internally contained in , and also have its extremities and bends obstructed. Therefore, vertex and each variable vertex have only one segment each that can be used in an EPG representation to make them adjacent to the clause gadget. The direction of this segment, being either horizontal or vertical, can be used to represent the true or false value for the variable. The clause gadgets, on the other hand, are such that exactly two of its adjacencies to the variable vertices and to can be realized with a horizontal intersection whereas the other two must be realized with a vertical intersection. If we consider the direction used by as a truth assignment, we get that exactly one of the variables in each clause will be true in any possible representation of . Conversely, it is fairly straightforward to obtain a EPG representation for given a truth assignment for the formula .
Theorem 27.
Helly EPG graph recognition is complete.
We say that a apex graph is a graph that can be made planar by the removal of vertices. A degenerate graph is a graph in which every subgraph has a vertex of degree at most . Recall that Positive (1 in 3)3SAT remains complete when the incidence graph of the input formula is planar Mulzer and Rote (2008). Thus, the following corollary holds.
Corollary 28.
Helly EPG graph recognition is complete on apex and degenerate graphs.
Proof.
To prove that is 3degenerate we just apply the degenerate graphs recognition algorithm, consisting of repeatedly removing the vertices of minimum degree from the graph. Note that each vertex to be removed at each iteration of the algorithm always has degree at most three, and therefore the graphs constructed according to Definition 15 are degenerate.
Now, remind that Positive (1 in 3)3SAT remains complete when the incidence graph of is planar, see Mulzer and Rote (2008). Thus, from an instance of Planar Positive (1 in 3)3SAT, one can build the incidence graph , let us call it . The incidence graph is a bipartite graph where one of the partitions is related to the clauses of and the other partition is related to its variables. Each clause of is represented by a vertex and each variable of the clauses is represented by a vertex . An edge exists in if and only if the variable occurs in the clause .
(a) Incidence graph , in a bipartite representation  (b)Incidence graph in a planar representation 
Figure 18(a) depicts an incidence graph . Since the formula from which was constructed is an instance of Planar Positive (1in3)3SAT, we know that the incidence graph of the formula is planar Mulzer and Rote (2008). Figure 18(b) depicts a planar embedding of the incidence graph . Finally, by using the planar embedding of the incidence graph, we can appropriately replace the vertices representing variables and clauses in by variables gadgets and clauses gadgets. As each variable gadget, clause gadget and base gadget are planar, then something not planar may have arisen only from the intersection that was made between them. As assures that there is a planar arrangement between the intersections of the variable devices and clause devices, then from one can construct a graph such that the removal of and results into a planar graph, see Figura 19. Thus is 2apex. ∎
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