A local characterization for perfect plane near-triangulations

TL;DR

This paper presents a local criterion to determine the perfectness of plane near-triangulated graphs, leading to an efficient $O(n^2)$ algorithm for checking their perfectness.

Contribution

It introduces a novel local characterization for perfect plane near-triangulations and provides an efficient algorithm based on this criterion.

Findings

A plane near-triangulated graph is perfect if and only if it lacks certain local configurations.

The paper provides an $O(n^2)$ algorithm for testing perfectness in these graphs.

The characterization simplifies the verification process for perfectness in plane near-triangulations.

Abstract

We derive a local criterion for a plane near-triangulated graph to be perfect. It is shown that a plane near-triangulated graph is perfect if and only if it does not contain either a vertex, an edge or a triangle, the neighbourhood of which has an odd hole as its boundary. The characterization leads to an $O(n^2)$ algorithm for checking perfectness of plane near-triangulations.

Full text

††journal: arXiv

1 Introduction

A plane embedding of a planar graph $G$ is said to be a plane near-triangulation if all its faces, except possibly the exterior face, are triangles. It was known even before the strong perfect graph theorem (Chudnovsky et al. [2]) that a planar graph is not perfect if and only if it contains an induced odd hole (Tucker [8]). Algorithmic recognition of planar perfect graphs was subsequently studied by Hsu [4] who discovered a method to determine whether a given planar graph of $n$ vertices is perfect in $O(n^{3})$ time. Later,  Cornuéjols et al. [3] discovered an algorithm to recognize perfect graphs in $O(n^{9})$ time, using the strong perfect graph theorem.

Though structural characterizations for perfect plane triangulations were attempted in the literature (see for example, Benchetrit and Bruhn [1]), a local characterization for perfect plane triangulations (or plane near-triangulations) does not seem to be known. An attempt in this direction was initiated by Salam et al. [6]. In this work, we extend their results to obtain a local characterization for a plane (near-) triangulated graph to be perfect. The characterization leads to an $O(n^{2})$ algorithm for checking perfectness of a plane near-triangulation. No quadratic time algorithm seems to be known in the literature for testing perfectness of plane triangulations.

If a plane near-triangulation $G$ contains a cut vertex or an edge separator, we can split $G$ into two induced subgraphs such that $G$ is perfect if and only if each of the induced subgraphs is perfect. Consequently, it suffices to consider plane triangulations that are both $2$-connected and have no edge separators.

A triangle $\Delta$ in $G$ consisting of vertices $x,y,z$, is a separating triangle in $G$ if the interior of $\Delta$ as well as the exterior of $\Delta$ contain at least one vertex. Let $Int(\Delta)$ and $Ext(\Delta)$ denote the set of vertices in the interior and exterior of $\Delta$. It is not hard to see that $G$ is perfect if and only if the subgraphs induced by $Int(\Delta)\cup\{x,y,z\}$ and $Ext(\Delta)\cup\{x,y,z\}$ are perfect. Thus, a separating triangle in $G$ splits $G$ into two induced subgraphs such that $G$ is perfect if and only if both the induced subgraphs are perfect. Consequently, we assume hereafter that $G$ does not contain any separating triangles as well.

A W-triangulation is a $2$-connected plane near-triangulation that does not contain any edge separator or a separating triangle [6]. It is easy to see that the closed neighbourhood $N[x]$ of any internal vertex $x$ in a W-triangulation $G$ will induce a wheel; for otherwise $G$ will contain a separating triangle. It was shown by Salam et al. [6] that a plane W-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 closed neighbourhood of which, induces an odd hole. However, their proof strategy was crucially dependent on the graph being free of induced wheels on five vertices.

Let $X$ be a set of vertices in a graph $G$. The local neighbourhood of $X$ is defined as the subgraph $G[N[X]]$ induced by $X$ and its neighbours in $G$. Given a W-triangulation $G$, our objective is to show that if $G$ is not perfect, then there exists a small connected induced subgraph $X$ in $G$ whose local neighbourhood has an odd hole as the boundary of its exterior face. We will show that $X$ will either be a vertex, an edge or a facial triangle of $G$.

First, observe that if an internal vertex $x$ of $G$ has odd degree, then the local neighbourhood of $X=\{x\}$ is a wheel, whose exterior boundary is an odd hole. Thus, the non-trivial graphs to consider are W-triangulations in which all internal vertices are of even degree. An even W-triangulation is a W-triangulation in which every internal vertex has even degree [6].

Figure 1 shows an even W-triangulation $G$ that is not perfect [6]. Here, the local neighbourhood of the facial triangle consisting of vertices $x,y$ and $z$ has an odd hole as its exterior boundary, and hence we can choose $X=\{x,y,z\}$. 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 non-perfect even W-triangulation $G$ contains a subset of vertices $X$, which consists of either the endpoints of an edge or a facial triangle, such that the induced subgraph $G[N[X]]$ has an odd hole as the boundary of its exterior face. This yields:

Theorem 1.1**.**

Let $G=(V,E)$ be a plane near-triangulated graph. $G$ 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.

In Section 3, we describe an $O(n^{2})$ algorithm that uses Theorem 1.1 to check whether a plane near-triangulation $G$ of $n$ vertices is perfect.

2 Perfect plane near-triangulations

Let $G=(V,E)$ be an even non-perfect W-near-triangulated graph. A minimal odd hole $C$ in $G$ is defined as an odd hole such that there is no other odd hole in $C\cup Int(C)$. Let $C$ be a minimal odd hole in $G$ and let $S=\{a_{0},a_{1},\ldots,a_{s}\}$ be the set of vertices in $C$, listed in clockwise order. To avoid cumbersome notation, hereafter a reference to a vertex $a_{i}\in S$ for $i>s$ may be inferred as reference to the vertex $a_{i\bmod(s+1)}$. With this notation, we have $a_{i}a_{i+1}\in E(G)$, for $0\leq i\leq s$. Since $C$ is an odd hole, $s$ must be even and $s\geq 4$. Throughout the paper, the length of a path (respectively, cycle) will be the number of edges in the path (respectively, cycle).

Let $H$ be the subgraph of $G$ induced by the vertices in $Int(C)$. We study the structure of $H$ in detail, in order to derive our perfectness characterization. Since $G$ is a plane near-triangulation, $H$ is non-empty. We show next that $H$ is a $2$-connected plane near-triangulation.

Claim 2.1**.**

$H$* has at least $3$ vertices. Moreover, for any vertex $b\in V(H)$, the neighbours of $b$ on $C$ (if any) are consecutive vertices of $C$.*

Proof.

Since $G$ is an even W-triangulation, $H$ cannot be a single vertex as otherwise the degree of that vertex would be $|S|$, which is odd. The number of vertices in $H$ cannot be two, as in that case the two vertices must be adjacent, with even degree and having exactly two common neighbours on $C$. However, this will contradict the parity of the number of vertices in $C$. Thus, $H$ has at least $3$ vertices. (Note that it is possible for $H$ to have exactly three vertices as in Figure 1).

Let $b$ be a vertex of $H$ with at least one neighbour on $C$. Without loss of generality, we may assume that $a_{0}$ is adjacent to $b$. For contradiction, assume that the neighbours of $b$ on $C$ are not consecutive. That is, for some $0<i<j<k<l\leq s+1$, we have $\{a_{0},a_{1},\ldots a_{i-1}\}\subseteq N(b)$, $\{a_{j},a_{j+1},\ldots,a_{k-1}\}\subseteq N(b)$ and $a_{l\bmod(s+1)}\in N(b)$; but $\{a_{i},a_{i+1},\ldots,a_{j-1}\}\cap N(b)=\emptyset$ and $\{a_{k},a_{k+1},\ldots,a_{l-1}\}\cap N(b)=\emptyset$ (see Figure 2). Note that, it is possible to have $a_{l\bmod(s+1)}=a_{0}$.

Let $P_{1},P_{2},P_{3},P_{4}$ and $P_{5}$ be subpaths of $C$ defined as follows: $P_{1}=a_{0}a_{1}\ldots a_{i-1}$, $P_{2}=a_{i-1}a_{i}a_{i+1}\ldots a_{j}$, $P_{3}=a_{j}a_{j+1}\ldots a_{k-1}$, $P_{4}=a_{k-1}a_{k}a_{k+1}\ldots a_{l}$ and $P_{5}=a_{l}a_{l+1}\ldots a_{s}a_{0}$. Note that, the union of these five paths is $C$. We know that the closed neighbourhood of $b$ induces a wheel. Let $R$ denote the cycle forming the exterior boundary of this wheel. Since $G$ is an even W-triangulation, $R$ is of even length. Let $Q_{1}=x_{1}x_{2}\ldots x_{p}$, $Q_{2}=y_{1}y_{2}\ldots y_{q}$ be subpaths of $R$ that are vertex disjoint from $C$, such that the concatenation of the paths $P_{1},a_{i-1}x_{1},Q_{1},x_{p}a_{j},P_{3},a_{k-1}y_{1},Q_{2},y_{q}a_{l}$ (in that order) forms a subpath of $R$.

The path $P_{2}$ has at least two edges, because $i<j$. It is easy to see that $ba_{i-1}P_{2}a_{j}b$ is an induced cycle in $G$ and since $C$ is assumed to be a minimal odd hole, $P_{2}$ must be of even length. Similarly, the path $P_{4}$ also has at least two edges and is of even length. Note that, the edge set $(E(C)\setminus E(P_{2}))\cup E(Q_{1})\cup\{a_{i-1}x_{1},x_{p}a_{j}\}$ is another induced cycle in $G$. Hence, the path $Q_{1}$ must be of odd length. Similarly, the path $Q_{2}$ is also of odd length. But, this implies that the induced cycle formed by the edge set $(E(C)\setminus(E(P_{2})\cup E(P_{4})))\cup E(Q_{1})\cup E(Q_{2})\cup\{a_{i-1}x_{1},x_{p}a_{j},a_{k-1}y_{1},y_{q}a_{l}\}$ is an odd hole. This contradicts the minimality of $C$ and hence we can conclude that the neighbours of $b$ on $C$ must be consecutive. ∎

Lemma 2.2**.**

$H$* is a $2$-connected near-triangulation with at least three vertices.*

Proof.

Since $G$ is a plane near-triangulation, it is easy to see that $H$ is a plane near-triangulation. Hence, by Claim 2.1, it only remains to prove that $H$ has no cut vertices.

For contradiction, suppose $b$ is a cut vertex in $H$. Since $H$ is a near-triangulation, $b$ must be a vertex on the exterior face of $H$. (Any internal vertex $v$ of $H$ has the property that each pair of its neighbours lie on a cycle. Thus $v$ cannot be a cut vertex of $H$.) Moreover, as $H$ is the subgraph induced by the vertices in $Int(C)$ and as $G$ is a near-triangulation, every vertex on the exterior boundary of $H$ must be adjacent to some vertex in $C$. Hence $b$ must have at least one neighbour on $C$ By Claim 2.1, neighbours of $b$ on $C$ are consecutive. Without loss of generality, let $N(b)\cap V(C)=\{a_{0},a_{1},\ldots,a_{i}\}$, for some $0\leq i<s$. Since $G$ is a W-triangulation, the neighbourhood of $b$ induces a wheel with $b$ as the center. Let $R$ be the induced cycle formed by the neighbours of $b$ in $G$. Since $G$ has no separating triangles, $a_{0},a_{1},\ldots,a_{i}$ should be consecutive vertices in $R$ as well. Let $H_{1}$ and $H_{2}$ be two connected components of $H\setminus b$. Let $v_{1}$ be a neighbour of $b$ in $H_{1}$ and $v_{2}$ be a neighbour of $b$ in $H_{2}$. Both $v_{1}$ and $v_{2}$ are vertices that belong to the cycle $R$ and they cannot be consecutive on $R$. Let $P_{1}$ and $P_{2}$ denote the two edge disjoint paths between $v_{1}$ and $v_{2}$ in $R$ such that their union is $R$. As $v_{1}$ and $v_{2}$ are not connected in $H\setminus b$, both $P_{1}$ and $P_{2}$ must intersect $C$. These intersections happen on vertices in $V(C)\cap V(R)\subseteq N(b)$. Hence there exist $0\leq j,k\leq i$ such that $a_{j}\in V(P_{1})\cap V(C)$ and $a_{k}\in V(P_{2})\cap V(C)$. Note that, if we delete $v_{1}$ and $v_{2}$ from $R$, $a_{j}$ and $a_{k}$ get disconnected from each other. However, this is impossible, since vertices $N(b)\cap V(C)=\{a_{0},a_{1},\ldots,a_{i}\}$ are known to be consecutive on $R$. Hence, $H$ is $2$-connected. ∎

Consequently from Lemma 2.2 we have:

Corollary 2.3**.**

The boundary of the exterior face of $H$ has at least three vertices.

Let $C^{\prime}$ be the cycle forming the boundary of the exterior face of $H$. If $C^{\prime}$ is a triangle, then it must be a facial triangle in $G$, as $G$ is assumed to contain no separating triangles. In this case, the exterior face of the closed neighbourhood of $C^{\prime}$ is the odd hole $C$, and Theorem 1.1 is immediate. Hence, we assume hereafter that $C^{\prime}$ is not a triangle.

Let $T=\{b_{0},b_{1},\ldots,b_{t}\}$, $t\geq 3$ be the vertices of $C^{\prime}$, listed in clockwise order. To simplify the notation, reference to a vertex $b_{j}\in T$ for $j>t$ may be inferred as reference to the vertex $b_{j\bmod(t+1)}$. It is easy to see that every vertex in $T$ is a neighbour of at least one vertex in $S$ and vice versa. Let $G^{\prime}=(V^{\prime},E^{\prime})$ be the subgraph of $G$ with $V^{\prime}=S\cup T$ and $E^{\prime}=\{uv:u\in S,v\in T\}\cup E(C)\cup E(C^{\prime})$. Note that, $b_{i}b_{i+1}\in E^{\prime}$, for all $0\leq i\leq t$; but $E^{\prime}$ excludes chords in $C^{\prime}$. Since $G$ is a plane near-triangulation, the following observation is immediate.

Observation 2.4**.**

For every $0\leq i\leq s$ and $0\leq j\leq t$:

The neighbours of $a_{i}$ in $C^{\prime}$ must be $b_{l},b_{l+1}\ldots b_{m}$ for some consecutive integers $l,l+1\ldots m$. 2. 2.

The neighbours of $b_{j}$ in $C$ must be $a_{k},a_{k+1}\ldots a_{r}$ for some consecutive integers $k,k+1\ldots r$. 3. 3.

$a_{i}$* and $a_{i+1}$ must have a common neighbour in $C^{\prime}$.* 4. 4.

$b_{j}$* and $b_{j+1}$ must have a common neighbour in $C$.* 5. 5.

The minimum degree of any vertex of $G^{\prime}$ is at least $3$.

Consider a vertex $a_{i}\in S$. Suppose $b_{l},b_{l+1},\ldots b_{m}$ are the neighbours of $a_{i}$ in $C^{\prime}$. If there is an edge between some non-consecutive vertices $b_{p}$ and $b_{q}$ for some $l\leq p,q\leq m$, then $a_{i},b_{p}$ and $b_{q}$ will form a separating triangle in $G$, a contradiction. Hence we have:

Observation 2.5**.**

For any $a_{i}\in S$, there exists an edge between two neighbours of $a_{i}$ in $C^{\prime}$ if and only if they are consecutive in $C^{\prime}$.

Lemma 2.6**.**

For any vertex $b_{j}\in T$, $deg_{G^{\prime}}(b_{j})$ is either $3$ or an even number greater than $3$.

Proof.

Let $deg_{G^{\prime}}(b_{j})\neq 3$ and be odd. Since $deg_{C^{\prime}}(b_{j})=2$, the neighbours of $b_{j}$ in $C$ form a path (say $Z$) of even length ($\geq 2$). Let the neighbours of $b_{j}$ in $C$ be $a_{i},a_{i+1},\ldots,a_{k}$ in the clockwise order (see Figure 3(a)). Note that $a_{k}a_{i}\notin E(G)$. Further, since $C$ is an odd hole, the length of the subpath $Y$ of $C$ from $a_{k}$ to $a_{i}$ in clockwise direction, must be an odd number $\geq 3$. Hence the cycle formed by replacing the path $Z$ in $C$ with the edges $a_{i}b_{j},b_{j}a_{k}$ will be an odd hole, distinct from $C$, with vertices chosen only from $S$ and $Int(C)$. This contradicts the minimality of $C$.

∎

Lemma 2.7**.**

For any vertex $a_{i}\in S$, $deg_{G^{\prime}}(a_{i})$ is either $3$ or an even number greater than $3$.

Proof.

Suppose $deg_{G^{\prime}}(a_{i})>3$. With no loss of generality, let the neighbours of $a_{i}$ in $C^{\prime}$ be $b_{0},b_{1},\ldots,b_{l}$, $l\geq 1$ in the clockwise order (see Figure 3(b)). For contradiction, suppose $l$ is even. Then, the length of the path $b_{0},b_{1}\ldots,b_{l}$ is even. Let the neighbours of $b_{0}$ in the clockwise direction in $C$ be $a_{j},a_{j+1},\ldots,a_{i}$ and the neighbours of $b_{l}$ in the clockwise direction in $C$ be $a_{i},a_{i+1},\ldots,a_{k}$. Vertices $b_{0}$ and $b_{l}$ cannot be adjacent in $G^{\prime}$, as otherwise $a_{i}b_{0}b_{l}$ would be a separating triangle in $G$ which is a contradiction. Hence, $b_{l}\neq b_{t}$ and $b_{l+1}\neq b_{0}$. Now, by Observation 2.3, $b_{l}$ and $b_{l+1}$ have a common neighbor in $C$. This common neighbor is different from $a_{i}$, since $b_{l+1}$ is not a neighbour of $a_{i}$. Thus, $deg_{G^{\prime}}(b_{l})>3$. Similarly, $deg_{G^{\prime}}(b_{0})>3$. Hence, we may assume without loss of generality that $j<i$.

Let $P$ be the path in $C$ from $a_{j}$ to $a_{k}$ through $a_{i}$. As $deg_{G^{\prime}}(b_{0})$ and $deg_{G^{\prime}}(b_{l})$ are even (by Lemma 2.6), the length of path $P$ is even, $\geq 2$. Let $X$ be the path $a_{j}b_{0}b_{1}\ldots b_{l}a_{k}$. Since $l$ is assumed to be even, $deg_{G^{\prime}}(a_{i})$ is odd, and hence $X$ must be of even length. Note that, $a_{k}\neq a_{j}$, as in that case the path $P$ will be the whole (odd) cycle $C$ which is impossible as $P$ has even length. Thus, the edges $a_{j}b_{l}$ and $a_{k}b_{0}$ cannot be present in $G^{\prime}$ as otherwise we would have $a_{k}=a_{j}$. Hence, unless $a_{k}a_{j}$ is an edge in $G^{\prime}$, then $X$ is an induced path in $G$. However, $a_{k}a_{j}$ cannot be an edge in $G^{\prime}$ as otherwise $X$ along with the edge $a_{k}a_{j}$ induces an odd hole, distinct from $C$, with vertices chosen only from $S$ and $Int(C)$, which is impossible. Thus, we conclude that $X$ is an induced path in $G^{\prime}$. Consequently, the cycle formed by replacing even length path $P$ in $C$ with the even length induced path $X$ will be an odd hole, distinct from $C$, with vertices chosen only from $S$ and $Int(C)$. This contradicts the choice of $C$. ∎

We introduce some notation. A vertex $a_{i}\in S$ with $deg_{G^{\prime}}(a_{i})=4$ (respectively $deg_{G^{\prime}}(a_{i})>4$) will be called an $\alpha$ vertex (respectively $\alpha^{\prime}$ vertex). Similarly a vertex $b_{j}\in T$ with $deg_{G^{\prime}}(b_{j})=4$ (respectively $deg_{G^{\prime}}(b_{j})>4$) will be called an $\beta$ vertex (respectively $\beta^{\prime}$ vertex). Every degree three vertex in $S$ (respectively $T$) will be called a $\gamma$ vertex (respectively $\delta$ vertex). Let $V_{\alpha},V_{\alpha^{\prime}},V_{\beta},V_{\beta^{\prime}},V_{\gamma}$ and $V_{\delta}$ denote the set of $\alpha,\alpha^{\prime},\beta,\beta^{\prime},\gamma$ and $\delta$ vertices respectively and $N_{\alpha},N_{\alpha^{\prime}},N_{\beta},N_{\beta^{\prime}},N_{\gamma}$ and $N_{\delta}$ denote the number of $\alpha,\alpha^{\prime},\beta,\beta^{\prime},\gamma$ and $\delta$ vertices respectively (See Figure 4).

With the above notation, the following observation is immediate.

Observation 2.8**.**

Every $\alpha$ and $\alpha^{\prime}$ vertex (respectively $\beta$ and $\beta^{\prime}$ vertex) has exactly two neighbours in $T$ (respectively $S$) that are not $\delta$ vertices (respectively $\gamma$ vertices). 2. 2.

Every $\gamma$ vertex (respectively $\delta$ vertex) must have exactly one neighbour in $T$ (respectively $S$). Moreover the neighbour must be a $\beta^{\prime}$ vertex (respectively $\alpha^{\prime}$ vertex).

Lemma 2.9**.**

$N_{\gamma}$* and $N_{\delta}$ are even.*

Proof.

Let $G^{\prime\prime}$ be the bipartite subgraph of $G^{\prime}$ with vertex set $V^{\prime\prime}=V_{\delta}\cup V_{\alpha^{\prime}}$ and edge set $E^{\prime\prime}=\{a_{i}b_{j}\colon a_{i}\in V_{\alpha^{\prime}},b_{j}\in V_{\delta}\}$. Consider any vertex $a_{i}\in V_{\alpha^{\prime}}$. We know that in $G^{\prime}$, $a_{i}$ has exactly two neighbours in $S$. Since by Lemma 2.7, $deg_{G^{\prime}}(a_{i})$ is even, it follows that in $G^{\prime}$, $a_{i}$ has an even number of neighbours from $T$. From Observation 2.8, $a_{i}$ has exactly two neighbours in $T$ that are not in $V_{\delta}$. Hence, the number of edges from $a_{i}$ to $V_{\delta}$ in $G^{\prime}$ must be even. Therefore, in the bipartite graph $G^{\prime\prime}$, every vertex in $V_{\alpha^{\prime}}$ has an even degree. By Observation 2.8, $deg_{G^{\prime\prime}}(b_{j})=1$, for each $b_{j}\in V_{\delta}$. Consequently, $N_{\delta}=|V_{\delta}|=\sum_{a_{i}\in V_{\alpha^{\prime}}}{deg_{G^{\prime\prime}}(a_{i})}$ is an even number.

The proof for the claim that $N_{\gamma}$ is even is similar. ∎

Note that $|S|=N_{\gamma}+N_{\alpha}+N_{\alpha^{\prime}}$. By Lemma 2.9, $N_{\gamma}$ is even. As $C$ is an odd hole, we have:

Corollary 2.10**.**

$N_{\alpha}+N_{\alpha^{\prime}}$* is odd.*

Lemma 2.11**.**

Every vertex in $S$ is a neighbour of a $\beta$ or a $\beta^{\prime}$ vertex. 2. 2.

Every vertex in $T$ is a neighbour of a $\alpha$ or a $\alpha^{\prime}$ vertex. 3. 3.

$N_{\alpha}+N_{\alpha^{\prime}}=N_{\beta}+N_{\beta^{\prime}}$.

Proof.

The first two parts are easy to see. To prove the third part, consider the bipartite subgraph $G^{\prime\prime}=(V^{\prime\prime},E^{\prime\prime})$ of $G^{\prime}$ with vertex set $V^{\prime\prime}=V_{\alpha}\cup V_{\alpha^{\prime}}\cup V_{\beta}\cup V_{\beta^{\prime}}$ and edge set $E^{\prime\prime}=\{a_{i}b_{j}\colon a_{i}\in V_{\alpha}\cup V_{\alpha^{\prime}},b_{j}\in V_{\beta}\cup V_{\beta^{\prime}}\}$. By Observation 2.8, every $\alpha$ and $\alpha^{\prime}$ vertex has exactly two neighbours in $T$ that are either $\beta$ vertices or $\beta^{\prime}$ vertices. Similarly, each $\beta$ and $\beta^{\prime}$ vertex have exactly two neighbours in $S$ which are either $\alpha$ or $\alpha^{\prime}$ vertices. Thus, $G^{\prime\prime}$ is a $2$-regular bipartite graph. Hence, $N_{\alpha}+N_{\alpha^{\prime}}=N_{\beta}+N_{\beta^{\prime}}$. ∎

Combining Lemma 2.9, Corollary 2.10 and Lemma 2.11, we see that $|T|=N_{\delta}+N_{\beta}+N_{\beta^{\prime}}$ is odd. This means that $C^{\prime}$ is an odd cycle. As $C^{\prime}$ cannot be an odd hole and we have assumed that it is not a triangle, we have:

Observation 2.12**.**

$C^{\prime}$* is an odd cycle with at least one chord in $G$ connecting vertices in $C^{\prime}$.*

The following lemma shows that any chord in $C^{\prime}$ must have a $\delta$ vertex as one of its end points.

Lemma 2.13**.**

If $b_{i}b_{j}$ is a chord in $C^{\prime}$ then $\{b_{i},b_{j}\}\cap V_{\delta}\neq\emptyset$.

Proof.

Suppose $\{b_{i},b_{j}\}\cap V_{\delta}=\emptyset$ and $i<j$. Let $a_{k},a_{k+1},\ldots,a_{l}$ be the neighbours of $b_{i}$ in $C$ and let $a_{x},a_{x+1},\ldots,a_{y}$ be the neighbours of $b_{j}$ in $C$, both taken in clockwise order (see Figure 5). As $b_{i}b_{j}$ is a chord, there exists at least one vertex between $b_{i}$ and $b_{j}$ in $C^{\prime}$ (in both directions). Hence $a_{l}\neq a_{x}$ (otherwise the vertices $a_{x},b_{i},b_{j}$ will form a separating triangle). Similarly, $a_{k}\neq a_{y}$. As $b_{i}$ and $b_{j}$ have even number of neighbours in $C$, it is easy to see that either the clockwise path $P$ from $a_{l}$ to $a_{x}$ or the clockwise path $Q$ from $a_{y}$ to $a_{k}$ in $C$ must be even, for otherwise $C$ cannot be an odd hole. With no loss of generality, assume that the length of $P$ is even. Then the vertices in the path $P$ along with the edges $\{a_{x}b_{j},b_{j}b_{i},b_{i}a_{l}\}$ forms an odd hole consisting of vertices chosen only from $S$ and $Int(C)$, a contradiction. ∎

By Lemma 2.13, $C^{\prime}$ contains at least one chord connecting a vertex $b_{j}\in V_{\delta}$ to some other vertex in $C^{\prime}$. By Observation 2.8, the neighbour of $b_{j}$ in $C$ must be an $\alpha^{\prime}$ vertex. Hence,

Corollary 2.14**.**

There exists at least one $\alpha^{\prime}$ vertex in $C$.

The next lemma shows that if a chord in $C^{\prime}$ connects two $\delta$ vertices, then their neighbours in $C$ will be adjacent.

Lemma 2.15**.**

Let $\{b_{i},b_{j}\}\subseteq V_{\delta}$. Let $a_{x}$ and $a_{y}$ be the neighbours of $b_{i}$ and $b_{j}$ respectively in $C$. If $b_{i}b_{j}$ is a chord in $C^{\prime}$ (in $G$), then $a_{x}a_{y}$ is an edge in $C$.

Proof.

Let $\{b_{i},b_{j}\}\subseteq V_{\delta}$ and suppose $b_{i}b_{j}$ is a chord in $G$ connecting vertices of $C^{\prime}$. Let $a_{x}$ and $a_{y}$ be the unique neighbours of $b_{i}$ and $b_{j}$ respectively in $C$ (Observation 2.8). It follows that $a_{x}\neq a_{y}$, as otherwise the vertices $b_{i},a_{x},b_{j}$ will form a separating triangle in $G$ (See Figure 6(a)). Since $C$ is an odd hole, exactly one of the two paths from $a_{x}$ to $a_{y}$ through vertices in $C$ must be of even length. With no loss of generality, assume that the path $P$ from $a_{y}$ to $a_{x}$ in the clockwise direction is even (see Figure 6(b)). Then if $a_{x}a_{y}$ is not an edge in $C$, the edges in the path $P$ together with the edges $a_{x}b_{i},b_{i}b_{j}$ and $b_{j}a_{y}$ forms an odd hole consisting of vertices chosen only from $S$ and $Int(C)$ (see Figure 6(b), (c)), a contradiction. ∎

Next we bound the number of $\alpha$ and $\alpha^{\prime}$ vertices in $C$. A bound on the number of $\beta$ and $\beta^{\prime}$ vertices follows from this.

Lemma 2.16**.**

$N_{\alpha}+N_{\alpha^{\prime}}\leq 3$**

Proof.

Suppose $N_{\alpha}+N_{\alpha^{\prime}}>3$. As $N_{\alpha}+N_{\alpha^{\prime}}$ is odd (by Corollary 2.10), $N_{\alpha}+N_{\alpha^{\prime}}\geq 5$. By Corollary 2.14, an $\alpha^{\prime}$ vertex $a_{i}$ must exist in $C$. By the definition of an $\alpha^{\prime}$ vertex and by Lemma 2.7, we know that $a_{i}$ has an an even number ($\geq 6$) of neighbours in $G^{\prime}$, of which exactly two are on $C$. Without loss of generality, let $b_{q},b_{q+1},\ldots,b_{r}$ ($r\geq q+3$) be the neighbours of $a_{i}$ in $T$ (see Figure 7(a)). Let the neighbours of $b_{q}$ and $b_{r}$ in $C$ in clockwise order be $a_{x},a_{x+1}\ldots,a_{i}$ and $a_{i},a_{i+1},\ldots,a_{y}$ respectively. Clearly, $a_{x}$ and $a_{y}$ are in $V_{\alpha}\cup V_{\alpha^{\prime}}$. If $x=y$, then $V_{\alpha}\cup V_{\alpha^{\prime}}=\{a_{x},a_{i}\}$, contradicting our assumption that $N_{\alpha}+N_{\alpha^{\prime}}\geq 5$. Further, since $a_{x},a_{i}$ and $a_{y}$ are the only vertices in $V_{\alpha}\cup V_{\alpha^{\prime}}$ in the clockwise subpath of $C$ from $a_{x}$ to $a_{y}$, at least two internal vertices of the clockwise subpath of $C$ from $a_{y}$ to $a_{x}$ must belong to $V_{\alpha}\cup V_{\alpha^{\prime}}$, since $N_{\alpha}+N_{\alpha^{\prime}}\geq 5$.

Let $b_{p},b_{p+1},\ldots,b_{q}$ and $b_{r},b_{r+1},\ldots,b_{s}$ be the neighbours of $a_{x}$ and $a_{y}$ in $C^{\prime}$ respectively in clockwise order. Note that $b_{p}$ and $b_{s}$ are distinct, non-adjacent vertices in $C^{\prime}$. This is because $N_{\beta}+N_{\beta^{\prime}}=N_{\alpha}+N_{\alpha^{\prime}}\geq 5$ (part (3) of Lemma 2.11). Let the neighbours of $b_{p}$ and $b_{s}$ in $C$ be $a_{g},a_{g+1},\ldots,a_{x}$ and $a_{y},a_{y+1},\ldots,a_{h}$ respectively. As $N_{\alpha}+N_{\alpha^{\prime}}\geq 5$, $a_{g}\neq a_{h}$. Since $b_{p},b_{q},b_{r}$ and $b_{s}$ are elements of $V_{\beta}\cup V_{\beta^{\prime}}$, their degrees must be even (by Lemma 2.6). Hence the path from $a_{g}$ to $a_{h}$ through $a_{x},a_{i}$ and $a_{y}$ along vertices in $C$ is of even length. Consequently, as $C$ is an odd hole, the (chordless) path $L$ from $a_{h}$ to $a_{g}$ in clockwise order in $C$ (as shown in Figure 7(a)) must be of odd length ($\geq 1$). We will show that there exists another odd hole in $G$, consisting of vertices chosen only from $S$ and $Int(C)$, contradicting the choice of $C$.

Consider the path $P=a_{g}b_{p}b_{p+1}\ldots b_{q}a_{i}b_{r}b_{r+1}\ldots b_{s}a_{h}$. Since $a_{x}$ and $a_{y}$ are of even degree (by Lemma 2.7), $P$ is of even length (see Figure 7(a)). This path, together with the with the path $L$, 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 $S$ and $Int(C)$, a contradiction to the choice of $C$ as desired.

Hence we analyze the possible chords in $G$ for the cycle formed by edges of $P$ and the edges of $L$. By Observation 2.5, there cannot be any chord in $P$ connecting any two vertices in the set $\{b_{p},b_{p+1},\ldots b_{q}\}$ or any two vertices in the set $\{b_{r},b_{r+1},\ldots,b_{s}\}$. For the same reason, a $b_{q}b_{r}$ chord also cannot exist. Moreover, since as $b_{q},b_{p},b_{r}$ and $b_{s}$ are elements in $V(\beta)\cup V(\beta^{\prime})$, by Lemma 2.13, there will not be a chord between them. Further, as there is no edge connecting $a_{x}$ and $a_{y}$, no chord exists between a vertex in the set $\{b_{p+1},\ldots b_{q-1}\}$ and a vertex in the set $\{b_{r+1}\ldots b_{s-1}\}$ (by Lemma 2.15). Hence, chords in $P$ can exist only between a vertex in the set $\{b_{p},b_{q}\}$ and a vertex in the set $\{b_{r+1},\ldots,b_{s-1}\}$ or between a vertex in the set $\{b_{r},b_{s}\}$ and a vertex in the set $\{b_{p+1},\ldots,b_{q-1}\}$. We systematically rule out these possibilities below.

Case 1 There exists a $b_{p}b_{k}$ chord in $C^{\prime}$ for some $b_{k}\in\{b_{r+1}\ldots,b_{s-1}\}$ (Figure 7(b)): Let $L^{\prime}$ be the (chordless) path from $a_{y}$ to $a_{g}$ in the clockwise direction in $C$. Since $b_{s}$ is of even degree (by Lemma 2.6), and $L^{\prime}$ is obtained by combining the path from $a_{y}$ to $a_{h}$ along $C$ with the odd length path $L$, path $L^{\prime}$ should be of even length. Consequently, the path $L^{\prime}$ along with the edges $a_{g}b_{p},b_{p}b_{k}$ and $b_{k}a_{y}$ will induce an odd hole consisting of vertices chosen only from $S$ and $Int(C)$, a contradiction.

• 2.

Case 2 There exists a $b_{s}b_{k}$ chord in $C^{\prime}$ for some $b_{k}\in\{b_{p+1}\ldots,b_{q-1}\}$: This case is symmetric to Case 1.

• 3.

Case 3 There exists a $b_{q}b_{k}$ chord in $C^{\prime}$ for some $b_{k}\in\{b_{r+1}\ldots,b_{s-1}\}$ (Figure 7(c)): Let the neighbours of $a_{h}$ in $C^{\prime}$ be $b_{s},b_{s+1},\ldots,b_{m}$ and the neighbours of $b_{m}$ on $C$ be $a_{h},a_{h+1},\ldots,a_{j}$ . Note that since $N_{\alpha}+N_{\alpha^{\prime}}\geq 5$, by part (3) of Lemma 2.11, $V_{\beta}\cup V_{\beta^{\prime}}$ should contain at least five vertices. Hence we see that $b_{m}\neq b_{p}$. Consider the path $Q=a_{x}b_{q}b_{q+1}\ldots b_{r}a_{y}b_{s}\ldots b_{m}a_{j}$. As $a_{i}$ and $a_{h}$ are of even degree (by Lemma 2.7), the path $Q$ is of even length. As $b_{q},b_{r},b_{s}$ and $b_{m}$ are elements in $V(\beta)\cup V(\beta^{\prime})$, they have even number of neighbours in $C$ (by Lemma 2.6). Hence the path from $a_{x}$ to $a_{j}$ in clockwise direction in $C$ is of even length. Consequently, as $C$ is an odd hole, the path $M$ from $a_{j}$ to $a_{x}$ in $C$ in clockwise order is of odd length. Suppose that the path $Q$ is chordless. Then combining the path $Q$ with $M$ yields an odd hole (see Figure 7(c)) consisting of vertices chosen only from $S$ and $Int(C)$, a contradiction. Note that this is true even if $a_{j}=a_{g}$ (i.e., $N_{\alpha}+N_{\alpha^{\prime}}=5$).

Thus it suffices to prove that the path $Q$ is chordless. We rule out each of the following possible cases of chords appearing in $Q$.

• (a)

There exists a chord connecting a vertex in the set $\{b_{q+1},\ldots,b_{r}\}$ and a vertex in the set $\{b_{s},\ldots,b_{m}\}$ in $C^{\prime}$: As $b_{q}b_{k}$ is a chord in $C^{\prime}$, this is impossible, as otherwise $G$ cannot be planar.

• (b)

There exists a $b_{q}b_{m}$ chord or a $b_{q}b_{s}$ chord: This possibility is ruled out by Lemma 2.13.

• (c)

There exists a chord $b_{q}b_{l}$ such that $b_{l}\in\{b_{s+1},\ldots,b_{m-1}\}$ (Figure 7(d)): Let $b_{q}b_{l}$ be a chord in $C^{\prime}$ . Let $M^{\prime}$ be the path from $a_{h}$ to $a_{x}$ in clockwise direction in $C$. Since $b_{q},b_{r}$ and $b_{s}$ are elements in $V(\beta)\cup V(\beta^{\prime})$, they have even number of neighbours in $C$ (by Lemma 2.6). Hence the path from $a_{x}$ to $a_{h}$ in clockwise direction in $C$ is of odd length. Consequently, as $C$ is an odd hole, the path $M^{\prime}$ must have even length ($\geq 2$). Hence the path $M^{\prime}$ along with the edges $a_{x}b_{q},b_{q}b_{l}$ and $b_{l}a_{h}$ will induce an odd hole consisting of vertices chosen only from $S$ and $Int(C)$, a contradiction.

Thus we conclude that the path $Q$ is chordless, as required.

• 4.

Case 4 There exists a $b_{r}b_{k}$ chord in $C^{\prime}$ for some $b_{k}\in\{b_{p+1}\ldots,b_{q-1}\}$: This case is symmetric to the Case 3.

Hence we conclude that $N_{\alpha}+N_{\alpha^{\prime}}\leq 3$. ∎

By part (3) of Lemma 2.11, We have:

Corollary 2.17**.**

$N_{\beta}+N_{\beta^{\prime}}\leq 3$.

Since $N_{\alpha}+N_{\alpha^{\prime}}\leq 3$ and $C$ is an odd hole, $C$ contains at least one $\gamma$ vertex which must be adjacent to a $\beta^{\prime}$ vertex in $C^{\prime}$ (by part 2 of Observation 2.8). Consequently we have:

Corollary 2.18**.**

There exists at least one $\beta^{\prime}$ vertex in $C^{\prime}$. That is, $N_{\beta^{\prime}}\geq 1$.

Lemma 2.19**.**

Let $a_{j}$ be an $\alpha^{\prime}$ vertex in $C$. Let $b_{x},b_{x+1},\ldots,b_{y}$ be the neighbours of $a_{j}$ in $C^{\prime}$ in clockwise order. Then at least one among $b_{x}$ and $b_{y}$ must be a $\beta$ vertex.

Proof.

Suppose that $b_{x}$ and $b_{y}$ are $\beta^{\prime}$ vertices. Let $a_{i},a_{i+1}\ldots,a_{j}$ and $a_{j},a_{j+1}\ldots,a_{k}$ ($j\geq i+3$, $k\geq j+3$) be the neighbours of $b_{x}$ and $b_{y}$ respectively in $C$, considered in clockwise order (see Figure 8(a)). As $b_{x}$ and $b_{y}$ have even number ($\geq 4$) of neighbours in $C$ (by Lemma 2.6), the path $P$ from $a_{i}$ to $a_{k}$ through $a_{j}$ in $C$ is chordless and has even length ($\geq 6$). Since $C$ is an odd hole, we see that $a_{i}\neq a_{k}$. By Lemma 2.13, there is no chord connecting $b_{x}$ and $b_{y}$ in $G$. Replacing the path $P$ in $C$ by the edges $a_{i}b_{x},b_{x}a_{j},a_{j}b_{y}$ and $b_{y}a_{k}$ induces an odd hole in $G$ consisting of vertices chosen only from $S$ and $Int(C)$, a contradiction. Hence at least one among $b_{x}$ and $b_{y}$ must be a $\beta$ vertex. ∎

Since there exists at least one $\alpha^{\prime}$ vertex in $C$ (by Corollary 2.14), we have:

Corollary 2.20**.**

$N_{\beta}\geq 1$.

As $N_{\beta}\geq 1$ and $N_{\beta^{\prime}}\geq 1$ (by Corollary 2.20 and Corollary 2.17), we have $N_{\beta}+N_{\beta^{\prime}}\geq 2$. Moreover, by part (3) of Lemma 2.11 and Corollary 2.10, $N_{\beta}+N_{\beta^{\prime}}(=N_{\alpha}+N_{\alpha^{\prime}})$ is odd and by Corollary 2.17, $N_{\beta}+N_{\beta^{\prime}}\leq 3$. Thus we have:

Observation 2.21**.**

$N_{\alpha}+N_{\alpha^{\prime}}=N_{\beta}+N_{\beta^{\prime}}=3$.

Lemma 2.22**.**

Let $b_{x}$ be a $\beta^{\prime}$ vertex in $C^{\prime}$ and $a_{i},a_{i+1},\ldots,a_{j}$ be the neighbours of $b_{x}$ in $C$ in clockwise order. If $N_{\alpha}\neq 0$ then at least one among $a_{i}$ and $a_{j}$ must be an $\alpha$ vertex.

Proof.

For the sake of contradiction assume that $a_{i}$ and $a_{j}$ are $\alpha^{\prime}$ vertices (see Figure 8(b)). Let $b_{z},b_{z+1}\ldots b_{x}$ and $b_{x},b_{x+1},\ldots,b_{y}$ be the neighbours of $a_{i}$ and $a_{j}$ in $C$ respectively in clockwise order. Let $a_{k}$ be an $\alpha$ vertex in $C$. By Observation 2.21, $a_{i}$, $a_{j}$ and $a_{k}$ must be the only vertices in $C$ that are not $\gamma$ vertices. Hence $b_{y}$ and $b_{z}$ must be neighbours of $a_{k}$ and $b_{y}b_{z}$ must be an edge in $C^{\prime}$.

As $a_{i}$ has two neighbours in $C^{\prime}$ which are not $\delta$ vertices ($b_{x}$ and $b_{z}$), we conclude using part (1) of Observation 2.8 that $b_{y}$ cannot be a neighbour of $a_{i}$. Similarly, $b_{z}$ cannot be a neighbour of $a_{j}$. Moreover, there cannot be a chord between $b_{x}$ and $b_{y}$ or between $b_{z}$ and $b_{x}$ (by Lemma 2.13). Consequently, the vertices $a_{i},b_{x},a_{j},b_{y},b_{z}$ should induce an odd hole which contains vertices only in $C$ and $Int(C)$, a contradiction. Hence at least one among $a_{i}$ and $a_{j}$ must be an $\alpha$ vertex. ∎

Lemma 2.23**.**

If $N_{\alpha}=0$, then there exist a $\beta^{\prime}$ vertex $b_{i}$ and a $\delta$ vertex $b_{x}$ satisfying the following:

• (a)

$b_{i}$* is the unique $\beta^{\prime}$ vertex in $C^{\prime}$.*

• (b)

$b_{i}b_{x}$* is a chord in $C^{\prime}$.*

• (c)

Every vertex in $C$ is adjacent to either $b_{i}$ or $b_{x}$.

Proof.

• (a)

By Corollary 2.18, there exists at least one $\beta^{\prime}$ vertex in $C^{\prime}$. Also by Corollary 2.20, there exists at least one $\beta$ vertex in $C^{\prime}$. As $N_{\beta}+N_{\beta^{\prime}}=3$ (by Observation 2.21), it is enough to prove that $N_{\beta^{\prime}}\neq 2$. Assume that $N_{\beta^{\prime}}=2$. Let $b_{i}$ and $b_{j}$ be two consecutive $\beta^{\prime}$ vertices in clockwise order in $C$. Then $b_{i}$ and $b_{j}$ must have a common neighbour (say $a_{k}$) in $C$. As $N_{\alpha}=0$ and $N_{\alpha}+N_{\alpha^{\prime}}>0$, $a_{k}$ must be an $\alpha^{\prime}$ vertex. But this is not possible by Lemma 2.19. Therefore, $N_{\beta^{\prime}}=1$.

• (b)

By Observation 2.21 and part (a), $N_{\beta}=2$. Let $b_{j}$ and $b_{k}$ be the $\beta$ vertices and $b_{i}$ be the $\beta^{\prime}$ vertex in $C^{\prime}$. Let $a_{p},a_{p+1},\ldots,a_{q}$ ($q\geq p+3$) be the neighbours of $b_{i}$ in $C$ arranged in clockwise order (see figure 9(a)). Let $a_{q}$ and $a_{r}$ be the neighbours of $b_{j}$ and let $a_{r}$ and $a_{p}$ be the neighbours of $b_{k}$ in $C$ in clockwise order. Note that $a_{p},a_{q}$ and $a_{r}$ are $\alpha^{\prime}$ vertices by assumption. Hence no two vertices from the set $\{b_{i},b_{j},b_{k}\}$ are consecutive on $C^{\prime}$. Note that $b_{j},b_{j+1},\ldots,b_{k}$ are the neighbours of $a_{r}$ in $C^{\prime}$ Then the vertices $a_{p},b_{i},a_{q},b_{j},b_{j+1},\ldots,b_{k},a_{p}$ forms an odd cycle (say $C^{\prime\prime}$). Since $C^{\prime\prime}$ contains vertices only in $C$ and $Int(C)$, it cannot be an odd hole. Hence, there must be at least one chord inside $C^{\prime\prime}$. However, there is no chord between any two vertices in $\{b_{i},b_{j},b_{k}\}$ and between any two vertices in $\{b_{j},b_{j+1},\ldots,b_{k}\}$ (by Observation 2.5). Hence the only possibility for the chord is $b_{i}b_{x}$ such that $x\in\{j+1,\ldots,k-1\}$ (see Figure 9(b)).

• (c)

By Part (a), since $b_{i}$ is the unique $\beta^{\prime}$ vertex in $C^{\prime}$, every $\gamma$ vertex in $C$ must be adjacent to $b_{i}$ (part 2 of Observation 2.8). Moreover as $N_{\alpha}=0$, $b_{i}$ must have two $\alpha^{\prime}$ vertices (say $a_{p}$ and $a_{q}$) in $C$ as neighbours. Consequently, as $N_{\alpha^{\prime}}=3$ (Observation 2.21), the only one vertex that is not a neighbour of $b_{i}$ in $C$ is an $\alpha^{\prime}$ vertex, say $a_{r}$ (see Figure 9(b)). By part (b) there exists a chord $b_{i}b_{x}$ between the $\beta^{\prime}$ vertex $b_{i}$ and a $\delta$ vertex $b_{x}$ in $C^{\prime}$. As $b_{x}$ is a $\delta$ vertex, it has exactly one neighbour on $C$. This neighbour cannot be $a_{p}$ (or $a_{q}$) as otherwise the vertices $b_{i},b_{x}$ and $a_{p}$ (respectively $b_{i},b_{x}$ and $a_{q}$) will form a separating triangle in $G$. Hence $a_{r}$ is the neighbour of $b_{x}$ in $C$.

∎

Corollary 2.24**.**

If $N_{\alpha}=0$, then there exist a chord $b_{i}b_{x}$ in $C^{\prime}$ such that the boundary of the exterior face of the local neighbourhood of $b_{i}b_{x}$ is the odd hole $C$.

Note that Corollary 2.24 gives a local characterization for the odd hole $C$ when $N_{\alpha}=0$. That is, there exists an edge in $G$, the exterior boundary of its local neighbourhood is the hole $C$. Our goal is to obtain a similar local characterization when $N_{\alpha}>0$.

For the rest of the paper, we use the following notation. Let $A$ be the set of all vertices in $C$ of type $\alpha$ or $\alpha^{\prime}$ and let $B$ be the set of all vertices in $C^{\prime}$ of type $\beta$ or $\beta^{\prime}$. By Observation 2.21, $|A|=|B|=3$. Let $a_{i},a_{j}$ and $a_{k}$ (respectively $b_{x},b_{y}$ and $b_{z}$) be the vertices of $A$ (respectively $B$) listed in clockwise order in $G^{\prime}$. There exists at least one $\beta$ vertex in $B$ (by Corollary 2.20) and at least one $\alpha^{\prime}$ vertex in $A$ (by Corollary 2.14). We fix $b_{y}$ to be a $\beta$ vertex in $B$ and $a_{i}$ to be an $\alpha^{\prime}$ vertex in $A$. The next lemma shows that every vertex in $C$ must be a neighbour of either $b_{x}$ or $b_{z}$ (or both).

Lemma 2.25**.**

Let $b_{y}\in B$ be a $\beta$ vertex. Then the boundary of the exterior face of the subgraph induced by the closed neighbourhood of $B\setminus\{b_{y}\}$ in $G$ is the odd hole $C$.

Proof.

Since $b_{y}$ is a $\beta$ vertex, $b_{y}$ must have exactly two neighbours in $C$ - say $a_{i}$ and $a_{j}$ in clockwise order. Further, we have $j=i+1$ and $a_{i}a_{j}$ will be an edge in $C$. (See Figure 10(a)). Moreover, $a_{i}$ and $a_{j}$ cannot have another common neighbour $b_{t}$ for any $t\in\{x,z\}$ as otherwise $b_{t},a_{i}$ and $a_{j}$ will form a separating triangle with $b_{y}$ in the interior. As each of $a_{i}$ and $a_{j}$ must have a neighbour in $B$ distinct from $b_{y}$ (by Observation 2.8), we may assume without loss of generality that $b_{x}$ is a neighbour of $a_{i}$ and $b_{z}$ is a neighbour of $a_{j}$. Thus, each neighbour of $b_{y}$ in $C$ is either a neighbour of $b_{x}$ or a neighbour of $b_{z}$. Since every vertex in $C$ must be a neighbour $b_{x}$ or $b_{y}$ or $b_{z}$ (Lemma 2.11), it follows that every vertex in $C$ is a neighbour of $b_{x}$ or $b_{z}$. The lemma follows since $C$ is assumed to be the boundary of the exterior face of the subgraph induced by the closed neighbourhood of $B$. ∎

Since Corollary 2.18 guarantees that $C^{\prime}$ contains a $\beta$ vertex, we now have a local characterization for the odd hole $C$ in the sense that $C$ will be the exterior boundary of the local neighbourhood of the two element set $\{b_{x},b_{z}\}$. We now strengthen Lemma 2.25 by showing that there exists an edge in $G$ (connecting two vertices in $C^{\prime}$) whose local neighbourhood has $C$ as its exterior boundary.

Lemma 2.26**.**

There exists two vertices $b_{p}$ and $b_{q}$ in $C^{\prime}$ such that:

$b_{p}b_{q}$* is an edge in $G$.*

• 2.

The boundary of the exterior face of the closed neighbourhood of $b_{p}b_{q}$ is the odd hole $C$.

Proof.

If $N_{\alpha}=0$, then the result follows by Corollary 2.24. Hence, assume that $N_{\alpha}>0$.

Recall that we have assumed $b_{y}$ to be a $\beta$ vertex. Therefore, if $b_{x}$ and $b_{z}$ are adjacent in $C^{\prime}$, setting $b_{p}b_{q}=b_{x}b_{z}$ suffices (by Lemma 2.25).

If $b_{x}b_{z}$ are not adjacent in $C^{\prime}$, the common neighbour of $b_{x}$ and $b_{z}$ on $C$ must be an $\alpha^{\prime}$ vertex. By Lemma 2.19, either $b_{x}$ or $b_{z}$ is a $\beta$ vertex and by Corollary 2.18, at least one of them must be a $\beta^{\prime}$ vertex. Hence, without loss of generality, we may assume that $b_{x}$ is a $\beta$ vertex and $b_{z}$ is a $\beta^{\prime}$ vertex. By Lemma 2.22, at least one of the neighbours of $b_{z}$ must be an $\alpha$ vertex. Therefore, the common neighbour of $b_{z}$ and $b_{y}$ in $C$ must be an $\alpha$ vertex. This implies that $b_{z}b_{y}$ is an edge of $C^{\prime}$ (see Figure 10(b)). Since $b_{x}$ is a $\beta$ vertex, every neighbour of $b_{x}$ in $C$ must be a neighbour of $b_{y}$ or $b_{z}$. Hence, by Lemma 2.25, setting $b_{p}b_{q}=b_{z}b_{y}$ suffices to complete the proof. ∎

Let us now consolidate our observations so far. Recall that our objective was to prove Theorem 1.1, to obtain a local characterization for a plane near-triangulation to be perfect. Since the graph can be decomposed into induced $2$-connected subgraphs containing no separating triangles or edge separators without affecting the perfectness of the graph, it was sufficient to limit the attention to W-triangulations. We noted that if any internal vertex $x$ in a W-triangulation has on odd degree, the result was immediate because the exterior face of the closed neighbourhood of $x$ would have been an odd hole. Consequently, the core of the problem was to characterize odd holes in even W-near-triangulated (induced) subgraphs of the original plane near-triangulation. If an even W-near-triangulation $G$ is non-perfect, we considered a minimal odd hole $C$ in $G$ such that there is no other odd hole in $C\cup Int(C)$. Then we considered the cycle $C^{\prime}$ forming the boundary of the subgraph $H$ induced by the vertices in $Int(C)$. We showed that either (i) $C^{\prime}$ is a (non-separating) triangle, the exterior face of the closed neighbourhood of which is the odd hole $C$ or (ii) there exists an edge $b_{p}b_{q}$ connecting vertices in $C^{\prime}$ such that the boundary of the exterior face of the closed neighbourhood of the edge $b_{p}b_{q}$ is the odd hole $C$ (Lemma 2.26). Thus we have:

Corollary 2.27**.**

A W-triangulation $G$ is not-perfect if and only if $G$ contains at least one among the following:

a vertex $x$ (of odd degree), the exterior boundary of the local neighbourhood of which, is an odd hole.

• 2.

an edge $xy$, the exterior boundary of the local neighbourhood of which, is an odd hole.

• 3.

a facial triangle $xyz$, the exterior boundary of the local neighbourhood of which, is an odd hole.

From this, Theorem 1.1 is immediate.

We make note of a few details about Theorem 1.1 which are of significance while translating Theorem 1.1 into an algorithm for checking whether a plane near-triangulation is perfect. Let $G$ be a plane near-triangulation that is not perfect. Let $X=xyz$ be a facial triangle in an induced even W-near-triangulated subgraph $G^{\prime}$ of $G$, such that exterior boundary of the local neighbourhood of $X$ induces an odd hole in $G$. It must be noted that $X$ need not necessarily be a facial triangle in the original graph $G$, but could be a separating triangle in $G$. Hence, identification of the separating triangles in $G$ becomes a significant algorithmic consideration. It turns out this task is easy due to the algorithm by Kant [5] that identifies the separating triangles and the W-triangulated subgraphs of $G$ in linear time.

The second point to note is the following. We cannot conclude from Theorem 1.1 that in a non-perfect near-triangulation $G$, we can find an induced subgraph $X$ which is either a vertex, an edge or a triangle, such that the neighbours of $X$ induces an odd hole in $G$. That is, it is necessary to inspect the exterior boundary of the local neighbourhood of each vertex, edge and face to detect an odd hole. Figure 11 gives an example of a graph illustrating this fact. In this example, the exterior boundary of the local neighbourhood of the edge $ip$ induces the odd hole $abcde$. However, the subgraph induced by the neighbours of the edge $ip$ (or any other edge or face) does not induce an odd hole. The requirement to check the exterior boundary of the local neighbourhood of each edge and each face results in quadratic time complexity for the algorithm described in the next section. If a W-near-triangulation $G$ does not contain any induced wheel on five vertices, then it is known that $G$ is not perfect if and only if it contains a vertex or a face whose neighbours induce an odd hole [6]. Consequently, for this restricted class of graphs, checking perfectness requires only sub-quadratic time.

3 Recognition of perfect plane near-triangulations

In this section, we describe an $O(n^{2})$ algorithm to determine whether a given plane near-triangulation $G$ of $n$ vertices is perfect. The algorithm essentially checks the conditions of Theorem 1.1, using well known techniques for handling planar graphs. For the sake of clarity, the steps of the algorithm and its analysis are presented in detail.

Initially, we assume that $G$ is a plane triangulation. Later, we will describe how to handle near-triangulations as well.

Given a plane triangulation $G$, using the linear time algorithm of Kant [5] for identifying separating triangles in a plane triangulation, $G$ can be split into its $4$-connected blocks. It is well known that a plane triangulation is $4$-connected if and only if it is free of separating triangles. Consequently, the $4$-connected blocks identified by Kant’s algorithm are precisely the maximal induced W-triangulations in $G$. We will call each such $4$-connected block a W-component of $G$ [6]. We have already seen that $G$ is perfect if and only if all its W-components are perfect. Hence, our task reduces to the problem of finding a quadratic time algorithm for checking whether a given plane triangulation without separating triangles is perfect. We split our task into four stages, as described below.

3.1 Preprocessing

Let $H$ be a plane triangulation on $n$ vertices, without separating triangles. Suppose we are given the adjacency list of $H$ as input. We will first compute a planar straight line embedding of $H$ on a $n\times n$ grid, using the linear time algorithm of Schnyder [7]. This essentially gives us the $(x,y)$ coordinates of each vertex of $H$. Using these coordinates, the slopes of all the edges of $H$ can be computed in $O(m)$ time, where $m$ is the number of edges of $H$. Note that, for a planar graph, $m=O(n)$. It would be useful to preprocess the adjacency list of $H$ at this point, so that for each vertex $v$, its neighbours are arranged in the descending order of the slopes of edges incident at $v$. This preprocessing can be done in $O(m\log{n})$ time. The next three stages involve verifying the conditions of perfectness mentioned in Theorem 1.1 one by one.

3.2 Checking vertex degrees

If any vertex $u$ of $H$ is of odd degree, the open neighbourhood $N_{H}(u)$ induces an odd hole and $H$ is not perfect. Checking the degrees of all vertices can be easily done in time linear in $n$. If every vertex of $H$ is of even degree, we need to check the remaining conditions to verify the perfectness of $H$.

3.3 Checking the boundary of the local neighbourhood of each edge

The next step is to check if the boundary of the local neighbourhood of any edge of $H$ is an odd hole. For each edge $uv$ of $H$, define $S_{uv}=N(u)\cup N(v)\cup\{u,v\}$ and construct an indicator array $A_{uv}$ of length $n$, where $A_{uv}[i]=1$ if vertex $v_{i}\in S_{uv}$ and zero otherwise. The construction of these arrays can be done in $O(mn)=O(n^{2})$ time in total.

Now, we construct $m$ plane subgraphs of $H$, one corresponding to each edge of $H$. The graph $H_{uv}$ will be the induced subgraph of $H$ on the vertex set $S_{uv}$. We store the coordinate position information of each vertex of $H_{uv}$, by copying the same from $H$. To get the sorted adjacency list of $H_{uv}$, we start with a copy of the sorted adjacency list of $H$. Then, mark the lists of vertices not in $S_{uv}$ as deleted. For each vertex $x$ in $S_{uv}$, go through the adjacency list $x$ in order, and when an edge $xv_{i}$ such that $A_{uv}[i]=0$ is encountered, then mark the edge as deleted. It takes only $O(m+n)$ time for obtaining the sorted adjacency list of $H_{uv}$, in this manner. The construction of subgraphs $H_{uv}$ corresponding to each edge $uv$ of $H$ along with the information mentioned above, takes only $O(m(m+n))=O(n^{2})$ time in total.

Now, we describe a procedure to check if the exterior boundary of an induced subgraph $H_{uv}$ constructed above is an odd hole. It is easy to see that the exterior boundary of $H_{uv}$ is an induced cycle. So, it suffices to identify the vertices on the exterior boundary and check the parity of their count. Since we have coordinates of vertices from a straight line drawing, this is easy. First, identify the left most vertex of $H_{uv}$ in the straight line embedding. This only involves identifying a vertex of $H_{uv}$ with the smallest $x$ coordinate. Clearly, this vertex is on the exterior face of $H_{uv}$. Let this vertex be $l$. Recall that the adjacency list of each vertex is sorted in the decreasing order of slopes. The edge with the largest slope incident at vertex $l$ must be on the exterior boundary of $H_{uv}$. The other endpoint of this edge can be identified as the next vertex on the boundary of $H_{uv}$. After identifying a new vertex $x$ on the boundary, it is easy to identify the next one. Suppose W is the vertex identified before $x$. If $y$ is the vertex that succeeds W (in the cyclic order) in the adjacency list of $x$, then $y$ is the next vertex on the boundary of $H_{uv}$. When this procedure encounters the initial vertex $l$ again, we would have identified all the vertices on the boundary of $H$. Thus, identifying the exterior boundary of $H_{uv}$ and checking whether it is an odd hole, can be done in $O(m+n)$ time. For checking the boundaries of all subgraphs $H_{uv}$ we constructed, total time required is only $O(m(m+n))=O(n^{2})$.

If this check fails to find an odd hole, we have to check the boundary of the local neighbourhood of each triangle, as described below.

3.4 Checking the boundary of the local neighbourhood of each triangle

Since $H$ is free of separating triangles, triangles of $H$ are precisely its faces. A listing of all the faces of $H$ can be done in linear time, by traversing the adjacency list of every vertex once. Checking whether the exterior boundary of the closed neighbourhood of a triangle $uvw$ forms an odd hole can be done in a way very similar to the method we discussed in the previous subsection. For each triangle $uvw$, we will define a set $S_{uvw}=N(u)\cup N(v)\cup N(w)\cup\{u,v,w\}$ and an indicator array $A_{uvw}$ for $S_{uvw}$. Then, we can construct plane subgraphs $H_{uvw}$, the induced subgraph of $H$ on the vertex set $S_{uvw}$. The method of checking whether the boundary of $H_{uvw}$ is an odd hole or not, is the same as the method described earlier for $H_{uv}$. The number of subgraphs to be processed is the number of faces of $H$, which is linear in $n$. Hence, the time required for checking the boundaries of each such subgraph is again $O(n^{2})$ only.

3.5 Handling plane near-triangulations

The method described above for recognizing planar perfect graphs can be extended to recognize perfect plane near-triangulations by the simple modifications described below, without affecting the complexity of the algorithm. The strategy is to triangulate the plane near-triangulation; use the algorithm of Kant [5] to identify the W-components of the triangulated graph; and retrieve the W-components of the original plane near-triangulation.

Let $G(V,E)$ be a plane near-triangulation. If $G$ is not $2$-connected, in linear time we can find the $2$-connected components of $G$ using depth first search, and work on each component. Hence, we assume without loss of generality that $G$ is $2$-connected.

We can embed $G$ into an $n\times n$ grid using the algorithm of Schnyder [7] in $O(n)$ time, preprocess the adjacency list of $G$ as described in Subsection 3.1 in $O(m\log n)$ time and identify the vertices on the boundary of exterior face of $G$ by using the method discussed in Subsection 3.3 in $O(n)$ time. Let $v_{1},v_{2},\ldots,v_{k},v_{1}$ be the cycle forming the the boundary of exterior face of $G$. Construct a plane triangulation $G^{\prime}$ from $G$ by adding a new vertex $v_{0}$ on the exterior face of $G$ and adding edges from $v_{0}$ to $v_{i}$ for every $i\in\{1,2,\ldots,k\}$. The adjacency list of $G$ can be modified to get the adjacency list of $G^{\prime}$ in linear time. Note that, as $G^{\prime}$ is a triangulation, it contains no edge separators. We can use the linear time algorithm of Kant [5] for identifying separating triangles in the plane triangulation $G^{\prime}$ and decompose $G^{\prime}$ into W-components.

Note that $v_{i}v_{j}$ is an edge separator of $G$ if and only if it is a chord connecting two vertices on the external face of $G$, forming a separating triangle $v_{i}v_{j}v_{0}$ in $G^{\prime}$. Conversely, a separating triangle in $G^{\prime}$ containing $v_{0}$ must be of the form $v_{0}v_{i}v_{j}$ for some $i,j\in\{1,2,\ldots,k\}$, with $v_{i}v_{j}$ forming a chord connecting two vertices on the exterior face of $G$. Moreover, every separating triangle in $G$ will be a separating triangle in $G^{\prime}$ and every separating triangle in $G^{\prime}$ that does not contain $v_{0}$ is a separating triangle in $G$. Hence, we have the following observation.

Observation 3.1**.**

Let $G_{1}^{\prime},G_{2}^{\prime},\ldots,G_{r}^{\prime}$ be the W-components of $G^{\prime}$. Let $G_{i}=G_{i}^{\prime}$ if $v_{0}\notin V(G_{i}^{\prime})$ and $G_{i}=G_{i}^{\prime}\setminus\{v_{0}\}$ otherwise. Then, $G_{1},G_{2},\ldots,G_{r}$ are precisely the W-components of $G$.

As a consequence of Observation 3.1, it suffices to decompose $G^{\prime}$ into its W-components and remove the vertex $v_{0}$ from each W-component to recover the W-components of $G$. Removing $v_{0}$ from all the W-components of $G^{\prime}$ requires only traversing the adjacency lists of all the graphs once and hence possible in $O(n)$ time. It is easy to verify that the $O(n^{2})$ method described earlier for checking perfectness of the W-components of a plane triangulation suffices for handling the W-components of the plane near-triangulation $G$ as well. Thus, it follows that checking the perfectness of plane near-triangulations requires only $O(n^{2})$ time.

4 Acknowledgment

We thank Sunil Chandran, IISc. Bangalore and Ajit A. Diwan, IIT Bombay for discussions and suggestions. We thank the latter also for the example in Figure 11.