1 Introduction
A plane embedding of a planar graph is said to be a plane neartriangulation if all its faces, except possibly the exterior face, are triangles. It was known even before the strong perfect graph theorem (Chudnovsky et al. (2006)) that a planar graph is not perfect if and only if it contains an induced odd hole (Tucker (1973)). Algorithmic recognition of planar perfect graphs was subsequently studied by Hsu (1987) and Grötschel et al. (1984). Though structural characterizations for perfect plane triangulations were attempted in the literature (see for example, Benchetrit and Bruhn (2015)), a local characterization for perfect plane triangulations (or plane neartriangulations) does not seem to be known. An attempt in this direction was initiated by Salam et al. (2019). In this work, we extend their results to obtain a local characterization for a plane (near) triangulated graph to be perfect.
If a plane neartriangulation contains a cut vertex or an edge separator, we can split into two induced subgraphs such that is perfect if and only if each of the induced subgraphs is perfect. Consequently, it suffices to consider plane triangulations that are both connected and have no edge separators.
A triangle in consisting of vertices , is a separating triangle in if the interior of as well as the exterior of contain at least one vertex. Let and denote the set of vertices in the interior and exterior of . It is not hard to see that is perfect if and only if the subgraphs induced by and are perfect. Thus, a separating triangle in splits into two induced subgraphs such that is perfect if and only if both the induced subgraphs are perfect. Consequently, we assume hereafter that does not contain any separating triangles as well.
A triangulation is a connected plane neartriangulation that does not contain any edge separator or a separating triangle Salam et al. (2019). It is easy to see that the closed neighbourhood of any internal vertex in a triangulation will induce a wheel; for otherwise will contain a separating triangle. It was shown by Salam et al. (2019) that a plane triangulation that does not contain any induced wheel on five vertices is not perfect if and only if it contains either a vertex or a face, the boundary of the exterior face of the local neighbourhood of which, is an odd hole. However, their proof strategy was crucially dependent on the graph being free of induced wheels on five vertices.
Given a triangulation , our objective is to show that if is not perfect, then there exists a small connected induced subgraph in whose local neighbourhood (the subgraph induced by and its neighbours) has an odd hole as the boundary of its exterior face. We will show that will either be a vertex, an edge or a facial triangle of .
First, observe that if an internal vertex of has odd degree, then the local neighbourhood of is a wheel, whose exterior boundary is an odd hole. Thus, the nontrivial graphs to consider are triangulations in which all internal vertices are of even degree. An even triangulation is a triangulation in which every internal vertex has even degree Salam et al. (2019).
Figure 1 shows an even triangulation that is not perfect Salam et al. (2019). Here, the closed neighbourhood of the facial triangle consisting of vertices and has an odd hole as its exterior boundary, and hence we can choose . Note that, for this particular graph, no smaller substructure (a vertex or an edge) exists, whose local neighbourhood has an odd hole as its exterior boundary.
In Section 2, we prove that every nonperfect even triangulation contains a subset of vertices , which consists of either the endpoints of an edge or a facial triangle, such that the induced subgraph has an odd hole as the boundary of its exterior face. This yields:
Theorem 1.1.
Let be a plane neartriangulated graph. is not perfect if and only if there exists a vertex, an edge or a triangle, the exterior boundary of the local neighbourhood of which, is an odd hole.
2 Perfect plane neartriangulations
Let be an even nonperfect neartriangulated graph. A minimal odd hole in is defined as an odd hole such that there is no other odd hole in . Let be a minimal odd hole in and let be the set of vertices in , listed in clockwise order. To avoid cumbersome notation, hereafter a reference to a vertex for may be inferred as reference to the vertex . With this notation, we have , for . Since is an odd hole, must be even and . Let be the subgraph of induced by the vertices in . Since is a plane neartriangulation, is nonempty. We show next that is a connected plane neartriangulation.
Claim 1.
has at least vertices. Moreover, for any vertex , the neighbours of on (if any) are consecutive vertices of .
Proof.
Since is an even triangulation, cannot be a single vertex as otherwise the degree of that vertex would be , which is odd. The number of vertices in cannot be two, as in that case the two vertices must be adjacent, with even degree and having exactly two common neighbours on . However, this will contradict the parity of the number of vertices in . Thus, has at least vertices. (Note that it is possible for to have exactly three vertices as in Figure 1).
Let be a vertex of with at least one neighbour on . Without loss of generality, we may assume that is adjacent to . For contradiction, assume that the neighbours of on are not consecutive. That is, for some , we have , and ; but and (see Figure 2). Note that, it is possible to have .
Let and be subpaths of defined as follows: , , , and . Note that, the union of these five paths is . We know that the closed neighbourhood of induces a wheel. Let denote the cycle forming the exterior boundary of this wheel. Since is an even triangulation, is of even length. Let , be subpaths of that are vertex disjoint from , such that the concatenation of the paths (in that order) forms a subpath of .
The path has at least two edges, because . It is easy to see that is an induced cycle in and since is assumed to be a minimal odd hole, must be of even length. Similarly, the path also has at least two edges and is of even length. Note that, the edge set is another induced cycle in . Hence, the path must be of odd length. Similarly, the path is also of odd length. But, this implies that the induced cycle formed by the edge set is an odd hole. This contradicts the minimality of and hence we can conclude that the neighbours of on must be consecutive. ∎
Lemma 2.1.
is a connected neartriangulation with at least three vertices.
Proof.
Since is a plane neartriangulation, it is easy to see that is a plane near triangulation. Hence, by Claim 1, it only remains to prove that has no cut vertices.
For contradiction, suppose is a cut vertex in . It is easy to see that must have at least one neighbour on . By Claim 1, neighbours of on are consecutive. Without loss of generality, let , for some . Since is a triangulation, the neighbourhood of induces a wheel with as the center. Let be the induced cycle formed by the neighbours of in . Since has no separating triangles, should be consecutive vertices in as well. Let and be two connected components of . Let be a neighbour of in and be a neighbour of in . Both and are vertices that belong to the cycle and they cannot be consecutive on . Let and denote the two edge disjoint paths between and in such that their union is . As and are not connected in , both and must intersect . These intersections happen on vertices in . Hence there exist such that and . Note that, if we delete and from , and get disconnected from each other. However, this is impossible, since vertices are known to be consecutive on . Hence, is connected. ∎
Consequently from Lemma 2.1 we have:
Corollary 2.1.
The boundary of the exterior face of has at least three vertices.
Let be the cycle forming the boundary of the exterior face of . If is a triangle, then it must be a facial triangle in , as is assumed to contain no separating triangles. In this case, the exterior face of the closed neighbourhood of is the odd hole , and Theorem 1.1 is immediate. Hence, we assume hereafter that is not a triangle.
Let , be the vertices of , listed in clockwise order. To simplify the notation, reference to a vertex for may be inferred as reference to the vertex . It is easy to see that every vertex in is a neighbour of at least one vertex in and vice versa. Let be the subgraph of with and . Note that, , for all ; but excludes chords in . Since is a plane neartriangulation, the following observation is immediate.
Observation 1.
For every and :

The neighbours of in must be for some consecutive integers .

The neighbours of in must be for some consecutive integers .

and must have a common neighbour in .

and must have a common neighbour in .

The minimum degree of any vertex of is at least .
Consider a vertex . Suppose are the neighbours of in . If there is an edge between some nonconsecutive vertices and for some , then and will form a separating triangle in , a contradiction. Hence we have:
Observation 2.
For any , there exists an edge between two neighbours of in if and only if they are consecutive in .
Lemma 2.2.
For any vertex , is either or an even number greater than .
Proof.
Let and be odd. Since , the neighbours of in form a path (say ) of even length (). Let the neighbours of in be in the clockwise order. Note that . Further, the length of the subpath of from to in clockwise direction, must be an odd number , since is an odd hole (see Figure 3(a)). Hence the cycle formed by replacing the path in with the edges will be an odd hole, distinct from , with vertices chosen only from and . This contradicts the minimality of .
∎
Lemma 2.3.
For any vertex , is either or an even number greater than .
Proof.
Suppose . With no loss of generality, let the neighbours of in be , in the clockwise order. For contradiction, suppose is even. Let the neighbours of in the clockwise direction in be and the neighbours of in the clockwise direction in be (see Figure 3(b)). We may assume without loss of generality that .
Let be the path in from to through . As and are even (by Lemma 2.2), the length of path is even, . Let be the path . Since is assumed to be even, is odd, and hence must be of even length. Note that, , as in that case the path will be the hole (odd) cycle which is impossible as has even length. Thus, the edges and cannot be present in as otherwise we would have . Hence we conclude that is an induced path in .
Consequently, the cycle formed by replacing even length path in with the even length induced path will be an odd hole, distinct from , with vertices chosen only from and . This contradicts the choice of . ∎
We introduce some notation. A vertex with (respectively ) will be called an vertex (respectively vertex). Similarly a vertex with (respectively ) will be called an vertex (respectively vertex). Every degree three vertex in (respectively ) will be called a vertex (respectively vertex). Let and denote the set of and vertices respectively and and denote the number of and vertices respectively (See Figure 4).
With the above notation, the following observation is immediate.
Observation 3.

Every and vertex (respectively and vertex) has exactly two neighbours in (respectively ) that are not vertices (respectively vertices).

Every vertex (respectively vertex) must have exactly one neighbour in (respectively ). Moreover the neighbour must be a vertex (respectively vertex).
Lemma 2.4.
and are even.
Proof.
Let be the bipartite subgraph of with vertex set and edge set . Consider any vertex . We know that in , has exactly two neighbours in . Since by Lemma 2.3, is even, it follows that in , has an even number of neighbours from . From Observation 3, has exactly two neighbours in that are not in . Hence, the number of edges from to in must be even. Therefore, in the bipartite graph , every vertex in has an even degree. By Observation 3, , for each . Consequently, is an even number.
The proof for the claim that is even is similar. ∎
Note that . By Lemma 2.4, is even. As is an odd hole, we have:
Corollary 2.2.
is odd.
Observation 4.

Every vertex in is a neighbour of a or a vertex.

Every vertex in is a neighbour of a or a vertex.

.
Proof.
The first two parts of the observation are easy to see. To prove the third part, consider the bipartite subgraph of with vertex set and edge set . By Observation 3, every and vertex has exactly two neighbours in that are either vertices or vertices. Similarly, each and vertex have exactly two neighbours in which are either or vertices. Thus, is a regular bipartite graph. Hence, . ∎
Combining Lemma 2.4, Corollary 2.2 and Observation 4, we see that is odd. This means that is an odd cycle. As cannot be an odd hole and we have assumed that it is not a triangle, we have:
Observation 5.
is an odd cycle with least one chord in connecting vertices in .
The following lemma shows that any chord in must have a vertex as one of its end points.
Lemma 2.5.
If is a chord in then .
Proof.
Suppose and . Let be the neighbours of in and let be the neighbours of in , both taken in clockwise order (see Figure 5). As is a chord, there exists at least one vertex between and in (in both directions). Hence (otherwise the vertices will form a separating triangle). Similarly, . As and have even number of neighbours in , it is easy to see that either the clockwise path from to or the clockwise path from to in must be even, for otherwise cannot be an odd hole. With no loss of generality, assume that the length of is even. Then the vertices in the path along with the edges forms an odd hole consisting of vertices chosen only from and , a contradiction. ∎
Since by Lemma 2.5 contains at least one chord connecting a vertex to some other vertex in , by Observation 3, there must be a vertex which is a neighbour of in . Hence,
Corollary 2.3.
There exists at least one vertex in .
The next lemma shows that if a chord in connects two vertices, then their neighbours in will be adjacent.
Lemma 2.6.
Let . Let and be the neighbours of and respectively in . If is a chord in (in ), then is an edge in .
Proof.
Let and suppose is a chord in connecting vertices of . Let and be the unique neighbours of and respectively in (Observation 3). It follows that , as otherwise the vertices will form a separating triangle in (See Figure 6(a)). Since is an odd hole, exactly one of the two paths from to through vertices in must be of even length. With no loss of generality, assume that the path from to in the clockwise direction is even (see Figure 6(b)). Then if is not an edge in , the edges in the path together with the edges and forms an odd hole consisting of vertices chosen only from and (see Figure 6(b), (c)), a contradiction. ∎
Next we bound the number of and vertices in . A bound on the number of and vertices follows from this.
Lemma 2.7.
Proof.
Suppose . As is odd (by Corollary 2.2), . By Corollary 2.3, an vertex must exist in . By the definition of an vertex and by Observation 2.3, we know that has an an even number () of neighbours in , of which exactly two are on . Without loss of generality, let () be the neighbours of in (see Figure 7(a)). Let the neighbours of and in in clockwise order be and respectively. Clearly, and are in . If , then , contradicting our assumption that . Further, since and are the only vertices in in the clockwise subpath of from to , at least two internal vertices of the clockwise subpath of from to must belong to , since .
Let and be the neighbours of and in respectively in clockwise order. Note that and are distinct, nonadjacent vertices in . This is because (part (3) of Observation 4). Let the neighbours of and in be and respectively. As , . Since and are elements of , their degrees must be even (by Lemma 2.2). Hence the path from to through and along vertices in is of even length. Consequently, as is an odd hole, the (chordless) path from to in clockwise order in (as shown in Figure 7(a)) must be of odd length (). We will show that there exists another odd hole in , consisting of vertices chosen only from and , contradicting the choice of .
Consider the path . Since and are of even degree (by Lemma 2.3), is of even length (see Figure 7(a)). This path, together with the with the path , form an odd cycle of length at least seven. If we prove that this cycle is chordless, it will be an odd hole, consisting of vertices chosen only from and , a contradiction to the choice of as desired.
Hence we analyze the possible chords in for the cycle formed by edges of and the edges of . By Observation 2, there cannot be any chord in connecting any two vertices in the set or any two vertices in the set . For the same reason, a chord also cannot exist. Moreover, since as and are elements in , there will not be a chord between them by Lemma 2.5. Further, by Lemma 2.6, no chord exists between a vertex in the set and a vertex in the set . Hence possible chords in can exist only between a vertex in the set and a vertex in the set or between a vertex in the set and a vertex in the set . We systematically rule out these possibilities below.

Case 1 There exists a chord in for some (Figure 7(b)): Let be the (chordless) path from to in the clockwise direction in . Since is of even degree (by Lemma 2.2), and is obtained by combining the path from to along with the odd length path , path should be of even length. Consequently, the path along with the edges and will induce an odd hole consisting of vertices chosen only from and , a contradiction.

Case 2 There exists a chord in for some : This case is symmetric to the Case 1.

Case 3 There exists a chord in for some (Figure 7(c)): Let the neighbours of in be and the neighbours of on be . Note that since , by part (3) of Observation 4, should contain at least five vertices. Hence we see that . Consider the path . As and are of even degree (by Lemma 2.3), the path is of even length. As and are elements in , they have even number of neighbours in (by Lemma 2.2). Hence the path from to in clockwise direction in is of even length. Consequently, as is an odd hole, the path from to in in clockwise order is of odd length. Suppose that the path is chordless. Then combining the path with yields an odd hole (see Figure 7(c)) consisting of vertices chosen only from and , a contradiction. Note that this is true even if (i.e., ).
Thus it suffices to prove that the path is chordless. We rule out each of the following possible cases of chords appearing in .

There exists a chord connecting a vertex in the set and a vertex in the set in : As is a chord in , this is impossible, as otherwise cannot be planar.

There exists a chord or a chord: This possibility is ruled out by Lemma 2.5.

There exists a chord such that (Figure 7(d)): Let be a chord in . Let be the path from to in clockwise direction in . Since and are elements in , they have even number of neighbours in (by Lemma 2.2). Hence the path from to in clockwise direction in is of odd length. Consequently, as is an odd hole, the path must have even length (). Hence the path along with the edges and will induce an odd hole consisting of vertices chosen only from and , a contradiction.
Thus we conclude that the path is chordless, as required.


Case 4 There exists a chord in for some : This case is symmetric to the Case 3.
Hence we conclude that . ∎
By part (3) of Observation 4, We have:
Corollary 2.4.
.
Since and is an odd hole, contains at least one vertex which must be adjacent to a vertex in (by part 2 of Observation 3). Consequently we have:
Corollary 2.5.
There exists at least one vertex in . That is, .
Lemma 2.8.
Let be an vertex in . Let be the neighbours of in in clockwise order. Then at least one among and must be a vertex.
Proof.
Suppose that and are vertices. Let and (, ) be the neighbours of and respectively in , considered in clockwise order (see Figure 8(a)). As and have even number () of neighbours in (by Lemma 2.2), the path from to through in is chordless and has even length (). Since is an odd hole, we see that . By Lemma 2.5, there is no chord connecting and in . Replacing the path in by the edges and induces an odd hole in consisting of vertices chosen only from and , a contradiction. Hence at least one among and must be a vertex. ∎
Since there exists at least one vertex in (by Corollary 2.3), we have:
Corollary 2.6.
.
As and (by Corollary 2.6 and Corollary 2.4), we have . Moreover, by part (3) of Observation 4 and Corollary 2.2, is odd and by Corollary 2.4, . Thus we have:
Observation 6.
.
Lemma 2.9.
Let be a vertex in and be the neighbours of in in clockwise order. If then at least one among and must be an vertex.
Proof.
For the sake of contradiction assume that and are vertices (see Figure 8(b)). Let and be the neighbours of and in respectively in clockwise order. Let be an vertex in . By Observation 6, , and must be the only vertices in that are not vertices. Hence and must be neighbours of and must be an edge in .
As has two neighbours in which are not vertices ( and ), we conclude using part (1) of Observation 3 that cannot be a neighbour of . Similarly, cannot be a neighbour of . Moreover, there cannot be a chord between and or between and (by Lemma 2.5). Consequently, the vertices should induce an odd hole which contains vertices only in and , a contradiction. Hence at least one among and must be an vertex. ∎
Lemma 2.10.
If , then there exist a vertex and a vertex satisfying the following:

is the unique vertex in .

is a chord in .

Every vertex in is adjacent to either or .
Proof.

By Corollary 2.5, there exists at least one vertex in . Also by Corollary 2.6, there exists at least one vertex in . As (by Observation 6), it is enough to prove that . Assume that . Let and be two consecutive vertices in clockwise order in . Then and must have a common neighbour (say ) in . As and , must be an vertex. But this is not possible by Lemma 2.8. Therefore, .

By Observation 6 and part (a), . Let and be the vertices in . Let () be the neighbours of in arranged in clockwise order (see figure 9(a)). Let and be the neighbours of and let and be the neighbours of in in clockwise order. Then the vertices forms an odd cycle (say ). Since contains vertices only in and , it cannot be an odd hole. Hence, there must be at least one chord inside . However, there is no chord between any two vertices in
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