A local characterization for perfect plane near-triangulations
AA local characterization for perfect plane near-triangulations
Sameera M Salam a , Jasine Babu b , K Murali Krishnan a a Department of Computer Science and Engineering, National Institute of Technology Calicut, Kerala, India 673601 b Department of Computer Science and Engineering, Indian Institute of Technology, Palakkad, Kerala, India 678557
Abstract
We derive a local criterion for a plane near-triangulated graph to be perfect. It is shown that a planenear-triangulated graph is perfect if and only if it does not contain either a vertex, an edge or a triangle, theneighbourhood of which has an odd hole as its boundary. The characterization leads to an O ( n ) algorithmfor checking perfectness of plane near-triangulations. Keywords:
Plane near-triangulated graphs, Plane triangulated graphs, Perfect graphs.
1. Introduction
A plane embedding of a planar graph G is said to be a plane near-triangulation if all its faces, exceptpossibly the exterior face, are triangles. It was known even before the strong perfect graph theorem (Chud-novsky 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 amethod to determine whether a given planar graph of n vertices is perfect in O ( n ) time. Later, Cornu´ejolset al. [3] discovered an algorithm to recognize perfect graphs in O ( n ) time, using the strong perfect graphtheorem.Though structural characterizations for perfect plane triangulations were attempted in the literature (seefor example, Benchetrit and Bruhn [1]), a local characterization for perfect plane triangulations (or planenear-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-) triangulatedgraph to be perfect. The characterization leads to an O ( n ) algorithm for checking perfectness of a planenear-triangulation. No quadratic time algorithm seems to be known in the literature for testing perfectnessof plane triangulations.If a plane near-triangulation G contains a cut vertex or an edge separator, we can split G into two inducedsubgraphs such that G is perfect if and only if each of the induced subgraphs is perfect. Consequently, itsuffices to consider plane triangulations that are both 2-connected and have no edge separators.A triangle ∆ in G consisting of vertices x, y, z , is a separating triangle in G if the interior of ∆ as wellas the exterior of ∆ contain at least one vertex. Let Int (∆) and
Ext (∆) denote the set of vertices in theinterior and exterior of ∆. It is not hard to see that G is perfect if and only if the subgraphs induced by Int (∆) ∪ { x, y, z } and Ext (∆) ∪ { x, y, z } are perfect. Thus, a separating triangle in G splits G into twoinduced 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 separatoror a separating triangle [6]. It is easy to see that the closed neighbourhood N [ x ] of any internal vertex x ina W-triangulation G will induce a wheel; for otherwise G will contain a separating triangle. It was shownby Salam et al. [6] that a plane W-triangulation that does not contain any induced wheel on five vertices is Email addresses: [email protected] (Sameera M Salam), [email protected] (Jasine Babu), [email protected] (K Murali Krishnan)
Preprint submitted to arXiv July 30, 2019 a r X i v : . [ c s . D M ] J u l ot perfect if and only if it contains either a vertex or a face, the boundary of the exterior face of the closedneighbourhood of which, induces an odd hole. However, their proof strategy was crucially dependent on thegraph 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 isnot perfect, then there exists a small connected induced subgraph X in G whose local neighbourhood hasan odd hole as the boundary of its exterior face. We will show that X will either be a vertex, an edge or afacial 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-triangulationin 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 thefacial triangle consisting of vertices x, y and z has an odd hole as its exterior boundary, and hence we canchoose 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. Figure 1: The local neighbourhood of the facial triangle ∆ = { x, y, z } has an odd hole in its 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 thereexists a vertex, an edge or a triangle, the exterior boundary of the local neighbourhood of which, is an oddhole. In Section 3, we describe an O ( n ) 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 definedas an odd hole such that there is no other odd hole in C ∪ Int ( C ). Let C be a minimal odd hole in G andlet S = { a , a , . . . , 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 ∈ S for i > s may be inferred as reference to the vertex a i mod ( s +1) .With this notation, we have a i a i +1 ∈ E ( G ), for 0 ≤ i ≤ s . Since C is an odd hole, s must be even and s ≥
4. Throughout the paper, the length of a path (respectively, cycle) will be the number of edges in thepath (respectively, cycle). 2et 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 vertices. Moreover, for any vertex b ∈ 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 thatvertex would be | S | , which is odd. The number of vertices in H cannot be two, as in that case the twovertices 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 itis 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 assumethat a is adjacent to b . For contradiction, assume that the neighbours of b on C are not consecutive. Thatis, for some 0 < i < j < k < l ≤ s + 1, we have { a , a , . . . a i − } ⊆ N ( b ), { a j , a j +1 , . . . , a k − } ⊆ N ( b ) and a l mod ( s +1) ∈ N ( b ); but { a i , a i +1 , . . . , a j − } ∩ N ( b ) = ∅ and { a k , a k +1 , . . . , a l − } ∩ N ( b ) = ∅ (see Figure 2).Note that, it is possible to have a l mod ( s +1) = a .Let P , P , P , P and P be subpaths of C defined as follows: P = a a . . . a i − , P = a i − a i a i +1 . . . a j , P = a j a j +1 . . . a k − , P = a k − a k a k +1 . . . a l and P = a l a l +1 . . . a s a . Note that, the union of thesefive paths is C . We know that the closed neighbourhood of b induces a wheel. Let R denote the cycleforming the exterior boundary of this wheel. Since G is an even W-triangulation, R is of even length.Let Q = x x . . . x p , Q = y y . . . y q be subpaths of R that are vertex disjoint from C , such that theconcatenation of the paths P , a i − x , Q , x p a j , P , a k − y , Q , y q a l (in that order) forms a subpath of R .The path P has at least two edges, because i < j . It is easy to see that ba i − P a j b is an induced cycle in G and since C is assumed to be a minimal odd hole, P must be of even length. Similarly, the path P alsohas at least two edges and is of even length. Note that, the edge set ( E ( C ) \ E ( P )) ∪ E ( Q ) ∪ { a i − x , x p a j } is another induced cycle in G . Hence, the path Q must be of odd length. Similarly, the path Q is alsoof odd length. But, this implies that the induced cycle formed by the edge set ( E ( C ) \ ( E ( P ) ∪ E ( P ))) ∪ E ( Q ) ∪ E ( Q ) ∪ { a i − x , x p a j , a k − y , y q a l } is an odd hole. This contradicts the minimality of C and hencewe can conclude that the neighbours of b on C must be consecutive. a a i − a i a j a k − a k a l − a l P P P P P x x p y y q a s Q Q b Figure 2: b is a vertex with non consecutive neighbours in C Lemma 2.2. H is a -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 . It is easy to see that b must have at least oneneighbour on C . By Claim 2.1, neighbours of b on C are consecutive. Without loss of generality, let3 ( b ) ∩ V ( C ) = { a , a , . . . , a i } , for some 0 ≤ 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 , a , . . . , a i should be consecutive vertices in R as well. Let H and H be two connected components of H \ b . Let v be a neighbour of b in H and v be a neighbour of b in H . Both v and v are vertices that belong to the cycle R and they cannot be consecutive on R . Let P and P denote the two edge disjoint paths between v and v in R such that their union is R . As v and v are not connected in H \ b , both P and P must intersect C . These intersections happen on vertices in V ( C ) ∩ V ( R ) ⊆ N ( b ). Hence there exist 0 ≤ j, k ≤ i such that a j ∈ V ( P ) ∩ V ( C ) and a k ∈ V ( P ) ∩ V ( C ).Note that, if we delete v and v from R , a j and a k get disconnected from each other. However, this isimpossible, since vertices N ( b ) ∩ V ( C ) = { a , a , . . . , a i } are known to be consecutive on R . Hence, H is2-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 (cid:48) be the cycle forming the boundary of the exterior face of H . If C (cid:48) is a triangle, then it must be afacial triangle in G , as G is assumed to contain no separating triangles. In this case, the exterior face of theclosed neighbourhood of C (cid:48) is the odd hole C , and Theorem 1.1 is immediate. Hence, we assume hereafterthat C (cid:48) is not a triangle.Let T = { b , b , . . . , b t } , t ≥ C (cid:48) , listed in clockwise order. To simplify the notation,reference to a vertex b j ∈ T for j > t may be inferred as reference to the vertex b j mod ( t +1) . It is easy to seethat every vertex in T is a neighbour of at least one vertex in S and vice versa. Let G (cid:48) = ( V (cid:48) , E (cid:48) ) be thesubgraph of G with V (cid:48) = S ∪ T and E (cid:48) = { uv : u ∈ S, v ∈ T } ∪ E ( C ) ∪ E ( C (cid:48) ). Note that, b i b i +1 ∈ E (cid:48) , forall 0 ≤ i ≤ t ; but E (cid:48) excludes chords in C (cid:48) . Since G is a plane near-triangulation, the following observationis immediate. Observation 2.4.
For every ≤ i ≤ s and ≤ j ≤ t : The neighbours of a i in C (cid:48) must be b l , b l +1 . . . b m for some consecutive integers l, l + 1 . . . m . The neighbours of b j in C must be a k , a k +1 . . . a r for some consecutive integers k, k + 1 . . . r . a i and a i +1 must have a common neighbour in C (cid:48) . b j and b j +1 must have a common neighbour in C . The minimum degree of any vertex of G (cid:48) is at least . Consider a vertex a i ∈ S . Suppose b l , b l +1 , . . . b m are the neighbours of a i in C (cid:48) . If there is an edgebetween some non-consecutive vertices b p and b q for some l ≤ p, q ≤ m , then a i , b p and b q will form aseparating triangle in G , a contradiction. Hence we have: Observation 2.5.
For any a i ∈ S , there exists an edge between two neighbours of a i in C (cid:48) if and only ifthey are consecutive in C (cid:48) . Lemma 2.6.
For any vertex b j ∈ T , deg G (cid:48) ( b j ) is either or an even number greater than .Proof. Let deg G (cid:48) ( b j ) (cid:54) = 3 and be odd. Since deg C (cid:48) ( b j ) = 2, the neighbours of b j in C form a path (say Z ) ofeven length ( ≥ b j in C be a i , a i +1 , . . . , a k in the clockwise order (see Figure 3(a)).Note that a k a i / ∈ 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 ≥
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 ∈ T , deg G (cid:48) ( a i ) is either or an even number greater than . i a k b b l Z a k b j (a) (b) a j a i P Q RY X Y Figure 3: (a) The degree of vertex b i is odd. (b) The degree of vertex a i is odd Proof.
Suppose deg G (cid:48) ( a i ) >
3. With no loss of generality, let the neighbours of a i in C (cid:48) be b , b , . . . , b l , l ≥ l is even. Then, the length ofthe path b , b . . . , b l is even. Let the neighbours of b in the clockwise direction in C be a j , a j +1 , . . . , a i and the neighbours of b l in the clockwise direction in C be a i , a i +1 , . . . , a k . We may assume without loss ofgenerality that j < i .Let P be the path in C from a j to a k through a i . As deg G (cid:48) ( b ) and deg G (cid:48) ( b l ) are even (by Lemma 2.6),the length of path P is even, ≥
2. Let X be the path a j b b . . . b l a k . Since l is assumed to be even, deg G (cid:48) ( a i )is odd, and hence X must be of even length. Note that, a k (cid:54) = a j , as in that case the path P will be thewhole (odd) cycle C which is impossible as P has even length. Thus, the edges a j b l and a k b cannot bepresent in G (cid:48) as otherwise we would have a k = a j . Hence, unless a k a j is an edge in G (cid:48) , then X is an inducedpath in G . However, a k a j cannot be an edge in G (cid:48) as otherwise X along with the edge a k a j induces an oddhole, distinct from C , with vertices chosen only from S and Int ( C ), which is impossible. Thus, we concludethat X is an induced path in G (cid:48) . Consequently, the cycle formed by replacing even length path P in C withthe 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 ∈ S with deg G (cid:48) ( a i ) = 4 (respectively deg G (cid:48) ( a i ) >
4) willbe called an α vertex (respectively α (cid:48) vertex). Similarly a vertex b j ∈ T with deg G (cid:48) ( b j ) = 4 (respectively deg G (cid:48) ( b j ) >
4) will be called an β vertex (respectively β (cid:48) vertex). Every degree three vertex in S (respec-tively T ) will be called a γ vertex (respectively δ vertex). Let V α , V α (cid:48) , V β , V β (cid:48) , V γ and V δ denote the set of α, α (cid:48) , β, β (cid:48) , γ and δ vertices respectively and N α , N α (cid:48) , N β , N β (cid:48) , N γ and N δ denote the number of α, α (cid:48) , β, β (cid:48) , γ and δ vertices respectively (See Figure 4). a a a a a a a a a a a b b b b b b b C C (cid:48) V α = { a , a , a , a } V α (cid:48) = { a } V β = { b , b } V β (cid:48) = { b , b , b } V δ = { b , b } V γ = { a , a , a , a , a , a } S = { a , a , . . . , a } T = { b , b , . . . , b } Figure 4: An illustration of the notation.
With the above notation, the following observation is immediate.5 bservation 2.8. Every α and α (cid:48) vertex (respectively β and β (cid:48) vertex) has exactly two neighbours in T (respectively S )that are not δ vertices (respectively γ vertices). Every γ vertex (respectively δ vertex) must have exactly one neighbour in T (respectively S ). Moreoverthe neighbour must be a β (cid:48) vertex (respectively α (cid:48) vertex). Lemma 2.9. N γ and N δ are even.Proof. Let G (cid:48)(cid:48) be the bipartite subgraph of G (cid:48) with vertex set V (cid:48)(cid:48) = V δ ∪ V α (cid:48) and edge set E (cid:48)(cid:48) = { a i b j : a i ∈ V α (cid:48) , b j ∈ V δ } . Consider any vertex a i ∈ V α (cid:48) . We know that in G (cid:48) , a i has exactly two neighbours in S . Sinceby Lemma 2.7, deg G (cid:48) ( a i ) is even, it follows that in G (cid:48) , a i has an even number of neighbours from T . FromObservation 2.8, a i has exactly two neighbours in T that are not in V δ . Hence, the number of edges from a i to V δ in G (cid:48) must be even. Therefore, in the bipartite graph G (cid:48)(cid:48) , every vertex in V α (cid:48) has an even degree. ByObservation 2.8, deg G (cid:48)(cid:48) ( b j ) = 1, for each b j ∈ V δ . Consequently, N δ = | V δ | = (cid:80) a i ∈ V α (cid:48) deg G (cid:48)(cid:48) ( a i ) is an evennumber.The proof for the claim that N γ is even is similar.Note that | S | = N γ + N α + N α (cid:48) . By Lemma 2.9, N γ is even. As C is an odd hole, we have: Corollary 2.10. N α + N α (cid:48) is odd. Lemma 2.11. Every vertex in S is a neighbour of a β or a β (cid:48) vertex. Every vertex in T is a neighbour of a α or a α (cid:48) vertex. N α + N α (cid:48) = N β + N β (cid:48) .Proof. The first two parts are easy to see. To prove the third part, consider the bipartite subgraph G (cid:48)(cid:48) =( V (cid:48)(cid:48) , E (cid:48)(cid:48) ) of G (cid:48) with vertex set V (cid:48)(cid:48) = V α ∪ V α (cid:48) ∪ V β ∪ V β (cid:48) and edge set E (cid:48)(cid:48) = { a i b j : a i ∈ V α ∪ V α (cid:48) , b j ∈ V β ∪ V β (cid:48) } .By Observation 2.8, every α and α (cid:48) vertex has exactly two neighbours in T that are either β vertices or β (cid:48) vertices. Similarly, each β and β (cid:48) vertex have exactly two neighbours in S which are either α or α (cid:48) vertices.Thus, G (cid:48)(cid:48) is a 2-regular bipartite graph. Hence, N α + N α (cid:48) = N β + N β (cid:48) .Combining Lemma 2.9, Corollary 2.10 and Lemma 2.11, we see that | T | = N δ + N β + N β (cid:48) is odd. Thismeans that C (cid:48) is an odd cycle. As C (cid:48) cannot be an odd hole and we have assumed that it is not a triangle,we have: Observation 2.12. C (cid:48) is an odd cycle with at least one chord in G connecting vertices in C (cid:48) . b i b j a l a k a x a y P Q Figure 5: b i ∈ V β (cid:48) and b j ∈ V β . The following lemma shows that any chord in C (cid:48) must have a δ vertex as one of its end points.6 emma 2.13. If b i b j is a chord in C (cid:48) then { b i , b j } ∩ V δ (cid:54) = ∅ .Proof. Suppose { b i , b j } ∩ V δ = ∅ and i < j . Let a k , a k +1 , . . . , a l be the neighbours of b i in C and let a x , a x +1 , . . . , a y be the neighbours of b j in C , both taken in clockwise order (see Figure 5). As b i b j is achord, there exists at least one vertex between b i and b j in C (cid:48) (in both directions). Hence a l (cid:54) = a x (otherwisethe vertices a x , b i , b j will form a separating triangle). Similarly, a k (cid:54) = a y . As b i and b j have even number ofneighbours 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, assumethat 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 } formsan odd hole consisting of vertices chosen only from S and Int ( C ), a contradiction.By Lemma 2.13, C (cid:48) contains at least one chord connecting a vertex b j ∈ V δ to some other vertex in C (cid:48) .By Observation 2.8, the neighbour of b j in C must be an α (cid:48) vertex. Hence, Corollary 2.14.
There exists at least one α (cid:48) vertex in C . The next lemma shows that if a chord in C (cid:48) connects two δ vertices, then their neighbours in C will beadjacent. Lemma 2.15.
Let { b i , b j } ⊆ V δ . Let a x and a y be the neighbours of b i and b j respectively in C . If b i b j isa chord in C (cid:48) (in G ), then a x a y is an edge in C .Proof. Let { b i , b j } ⊆ V δ and suppose b i b j is a chord in G connecting vertices of C (cid:48) . Let a x and a y be theunique neighbours of b i and b j respectively in C (Observation 2.8). It follows that a x (cid:54) = a y , as otherwisethe vertices b i , a x , b j will form a separating triangle in G (See Figure 6(a)). Since C is an odd hole, exactlyone 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 isnot 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 holeconsisting of vertices chosen only from S and Int ( C ) (see Figure 6(b), (c)), a contradiction. b j a yb i a x = a y b i b j b j a yb i a x a x (b)(a) (c) PP Figure 6: (a) a x = a y (b) a x a y is not an edge in G (c) a x a y is an edge Next we bound the number of α and α (cid:48) vertices in C . A bound on the number of β and β (cid:48) verticesfollows from this. Lemma 2.16. N α + N α (cid:48) ≤ Proof.
Suppose N α + N α (cid:48) >
3. As N α + N α (cid:48) is odd (by Corollary 2.10), N α + N α (cid:48) ≥
5. By Corollary 2.14,an α (cid:48) vertex a i must exist in C . By the definition of an α (cid:48) vertex and by Lemma 2.7, we know that a i hasan an even number ( ≥
6) of neighbours in G (cid:48) , of which exactly two are on C . Without loss of generality, let b q , b q +1 , . . . , b r ( r ≥ 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 . . . , a i and a i , a i +1 , . . . , a y respectively. Clearly, a x and a y are in V α ∪ V α (cid:48) .If x = y , then V α ∪ V α (cid:48) = { a x , a i } , contradicting our assumption that N α + N α (cid:48) ≥
5. Further, since a x , a i a y are the only vertices in V α ∪ V α (cid:48) in the clockwise subpath of C from a x to a y , at least two internalvertices of the clockwise subpath of C from a y to a x must belong to V α ∪ V α (cid:48) , since N α + N α (cid:48) ≥ b p , b p +1 , . . . , b q and b r , b r +1 , . . . , b s be the neighbours of a x and a y in C (cid:48) respectively in clockwise order.Note that b p and b s are distinct, non-adjacent vertices in C (cid:48) . This is because N β + N β (cid:48) = N α + N α (cid:48) ≥ b p and b s in C be a g , a g +1 , . . . , a x and a y , a y +1 , . . . , a h respectively. As N α + N α (cid:48) ≥ a g (cid:54) = a h . Since b p , b q , b r and b s are elements of V β ∪ V β (cid:48) , their degrees mustbe 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 evenlength. Consequently, as C is an odd hole, the (chordless) path L from a h to a g in clockwise order in C (asshown in Figure 7(a)) must be of odd length ( ≥ 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 . . . b q a i b r b r +1 . . . 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 oflength at least seven. If we prove that this cycle is chordless, it will be an odd hole, consisting of verticeschosen 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 . ByObservation 2.5, there cannot be any chord in P connecting any two vertices in the set { b p , b p +1 , . . . b q } orany two vertices in the set { b r , b r +1 , . . . , 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 ( β ) ∪ V ( β (cid:48) ), by Lemma 2.13, there will not be a chord betweenthem. Further, as there is no edge connecting a x and a y , no chord exists between a vertex in the set { b p +1 , . . . b q − } and a vertex in the set { b r +1 . . . b s − } (by Lemma 2.15). Hence, chords in P can exist onlybetween a vertex in the set { b p , b q } and a vertex in the set { b r +1 , . . . , b s − } or between a vertex in the set { b r , b s } and a vertex in the set { b p +1 , . . . , b q − } . We systematically rule out these possibilities below. • Case 1
There exists a b p b k chord in C (cid:48) for some b k ∈ { b r +1 . . . , b s − } (Figure 7(b)): Let L (cid:48) bethe (chordless) path from a y to a g in the clockwise direction in C . Since b s is of even degree (byLemma 2.6), and L (cid:48) is obtained by combining the path from a y to a h along C with the odd lengthpath L , path L (cid:48) should be of even length. Consequently, the path L (cid:48) 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. • Case 2
There exists a b s b k chord in C (cid:48) for some b k ∈ { b p +1 . . . , b q − } : This case is symmetric to Case1. • Case 3
There exists a b q b k chord in C (cid:48) for some b k ∈ { b r +1 . . . , b s − } (Figure 7(c)): Let the neighboursof a h in C (cid:48) be b s , b s +1 , . . . , b m and the neighbours of b m on C be a h , a h +1 , . . . , a j . Note that since N α + N α (cid:48) ≥
5, by part (3) of Lemma 2.11, C (cid:48) should contain at least five vertices. Hence we see that b m (cid:54) = b p . Consider the path Q = a x b q b q +1 . . . b r a y b s . . . b m a j . As a i and a h are of even degree (byLemma 2.7), the path Q is of even length. As b q , b r , b s and b m are elements in V ( β ) ∪ V ( β (cid:48) ), theyhave even number of neighbours in C (by Lemma 2.6). Hence the path from a x to a j in clockwisedirection in C is of even length. Consequently, as C is an odd hole, the path M from a j to a x in C inclockwise 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 ), acontradiction. Note that this is true even if a j = a g (i.e., N α + N α (cid:48) = 5).Thus it suffices to prove that the path Q is chordless. We rule out each of the following possible casesof chords appearing in Q .(a) There exists a chord connecting a vertex in the set { b q +1 , . . . , b r } and a vertex in the set { b s , . . . , b m } in C (cid:48) : As b q b k is a chord in C (cid:48) , 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 ∈ { b s +1 , . . . , b m − } (Figure 7(d)): Let b q b l be a chord in C (cid:48) . Let M (cid:48) 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 ( β ) ∪ V ( β (cid:48) ), 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 path8 (cid:48) must have even length ( ≥ M (cid:48) 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. • Case 4
There exists a b r b k chord in C (cid:48) for some b k ∈ { b p +1 . . . , b q − } : This case is symmetric to theCase 3.Hence we conclude that N α + N α (cid:48) ≤ b p b q b r a g a i a y a h a g a y a h b p b q b r b m a j (a) b k LM a x a x b p b q b r b s b m a g a x a i a y a h a j b l M (cid:48) b s a g a x a i a y a h b p b q b r b k b s L (cid:48) (b)(c) (d) b k a i b s Figure 7: N α + N α (cid:48) ≥ By part (3) of Lemma 2.11, We have:
Corollary 2.17. N β + N β (cid:48) ≤ . Since N α + N α (cid:48) ≤ C is an odd hole, C contains at least one γ vertex which must be adjacent to a β (cid:48) vertex in C (cid:48) (by part 2 of Observation 2.8). Consequently we have: Corollary 2.18.
There exists at least one β (cid:48) vertex in C (cid:48) . That is, N β (cid:48) ≥ . Lemma 2.19.
Let a j be an α (cid:48) vertex in C . Let b x , b x +1 , . . . , b y be the neighbours of a j in C (cid:48) in clockwiseorder. Then at least one among b x and b y must be a β vertex. roof. Suppose that b x and b y are β (cid:48) vertices. Let a i , a i +1 . . . , a j and a j , a j +1 . . . , a k ( j ≥ i + 3, k ≥ 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 ( ≥
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 ( ≥ C is an odd hole, we see that a i (cid:54) = a k . By Lemma 2.13, thereis 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 atleast one among b x and b y must be a β vertex. a j a k a i b x b y b z P a j b y b z b x a k a i (b) (a) Figure 8: (a) N β (cid:48) = 3 and N δ (cid:54) = 0 (b) b x is a β (cid:48) vertex and both a i and a j are α (cid:48) vertices. Since there exists at least one α (cid:48) vertex in C (by Corollary 2.14), we have: Corollary 2.20. N β ≥ . As N β ≥ N β (cid:48) ≥ N β + N β (cid:48) ≥
2. Moreover, by part(3) of Lemma 2.11 and Corollary 2.10, N β + N β (cid:48) (= N α + N α (cid:48) ) is odd and by Corollary 2.17, N β + N β (cid:48) ≤ Observation 2.21. N α + N α (cid:48) = N β + N β (cid:48) = 3 . Lemma 2.22.
Let b x be a β (cid:48) vertex in C (cid:48) and a i , a i +1 , . . . , a j be the neighbours of b x in C in clockwiseorder. If N α (cid:54) = 0 then at least one among a i and a j must be an α vertex.Proof. For the sake of contradiction assume that a i and a j are α (cid:48) vertices (see Figure 8(b)). Let b z , b z +1 . . . b x and b x , b x +1 , . . . , b y be the neighbours of a i and a j in C respectively in clockwise order. Let a k be an α vertex in C . By Observation 2.21, a i , a j and a k must be the only vertices in C that are not γ vertices.Hence b y and b z must be neighbours of a k and b y b z must be an edge in C (cid:48) .As a i has two neighbours in C (cid:48) which are not δ vertices ( b x and b z ), we conclude using part (1) ofObservation 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 α vertex. Lemma 2.23. If N α = 0 , then there exist a β (cid:48) vertex b i and a δ vertex b x satisfying the following:(a) b i is the unique β (cid:48) vertex in C (cid:48) .(b) b i b x is a chord in C (cid:48) .(c) Every vertex in C is adjacent to either b i or b x . roof. (a) By Corollary 2.18, there exists at least one β (cid:48) vertex in C (cid:48) . Also by Corollary 2.20, there existsat least one β vertex in C (cid:48) . As N β + N β (cid:48) = 3 (by Observation 2.21), it is enough to prove that N β (cid:48) (cid:54) = 2.Assume that N β (cid:48) = 2. Let b i and b j be two consecutive β (cid:48) vertices in clockwise order in C . Then b i and b j must have a common neighbour (say a k ) in C . As N α = 0 and N α + N α (cid:48) > a k must be an α (cid:48) vertex. But this is not possible by Lemma 2.19. Therefore, N β (cid:48) = 1.(b) By Observation 2.21 and part (a), N β = 2. Let b j and b k be the β vertices and b i be the β (cid:48) vertexin C (cid:48) . Let a p , a p +1 , . . . , a q ( q ≥ p + 3) be the neighbours of b i in C arranged in clockwise order (seefigure 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 inclockwise order. Note that a p , a q and a r are α (cid:48) vertices by assumption. Hence no two vertices fromthe set { b i , b j , b k } are consecutive on C (cid:48) . Note that b j , b j +1 , . . . , b k are the neighbours of a r in C (cid:48) Then the vertices a p , b i , a q , b j , b j +1 , . . . , b k , a p forms an odd cycle (say C (cid:48)(cid:48) ). Since C (cid:48)(cid:48) contains verticesonly in C and Int ( C ), it cannot be an odd hole. Hence, there must be at least one chord inside C (cid:48)(cid:48) . However, there is no chord between any two vertices in { b i , b j , b k } and between any two verticesin { b j , b j +1 , . . . , b k } (by Observation 2.5). Hence the only possibility for the chord is b i b x such that x ∈ { j + 1 , . . . , k − } (see Figure 9(b)).(c) By Part (a), since b i is the unique β (cid:48) vertex in C (cid:48) , every γ vertex in C must be adjacent to b i (part2 of Observation 2.8). Moreover as N α = 0, b i must have two α (cid:48) vertices (say a p and a q ) in C asneighbours. Consequently, as N α (cid:48) = 3 (Observation 2.21), the only one vertex that is not a neighbourof b i in C is an α (cid:48) vertex, say a r (see Figure 9(b)). By part (b) there exists a chord b i b x between the β (cid:48) vertex b i and a δ vertex b x in C (cid:48) . As b x is a δ vertex, it has exactly one neighbour on C . Thisneighbour 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 ) willform a separating triangle in G . Hence a r is the neighbour of b x in C . Corollary 2.24. If N α = 0 , then there exist a chord b i b x in C (cid:48) such that the boundary of the exterior faceof 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 α = 0. That is, thereexists an edge in G , the exterior boundary of its local neighbourhood is the hole C . Our goal is to obtain asimilar local characterization when N α > b j b k a p a q a r b i b j b k a p a q a r b x (a) (b) b i Figure 9: N α (cid:48) = 3 For the rest of the paper, we use the following notation. Let A be the set of all vertices in C of type α or α (cid:48) and let B be the set of all vertices in C (cid:48) of type β or β (cid:48) . 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 (cid:48) . Thereexists at least one β vertex in B (by Corollary 2.20) and at least one α (cid:48) vertex in A (by Corollary 2.14). We11 z b y a i a j a k (a) a i a k b x a j b z b x b y (b) Figure 10: (a) N β = 1, N α (cid:48) = 1 (b) N β = 2, N α (cid:48) = 2 and b x b z is not an edge in C (cid:48) fix b y to be a β vertex in B and a i to be an α (cid:48) 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 ∈ B be a β vertex. Then the boundary of the exterior face of the subgraph induced bythe closed neighbourhood of B \ { b y } in G is the odd hole C .Proof. Since b y is a β 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 haveanother common neighbour b t for any t ∈ { 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 bea 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 . Thelemma follows since C is assumed to be the boundary of the exterior face of the subgraph induced by theclosed neighbourhood of B .Since Corollary 2.18 guarantees that C (cid:48) contains a β vertex, we now have a local characterization for theodd hole C in the sense that C will be the exterior boundary of the local neighbourhood of the two elementset { b x , b z } . We now strengthen Lemma 2.25 by showing that there exists an edge in G (connecting twovertices in C (cid:48) ) whose local neighbourhood has C as its exterior boundary. Lemma 2.26.
There exists two vertices b p and b q in C (cid:48) such that: • b p b q is an edge in G . • The boundary of the exterior face of the closed neighbourhood of b p b q is the odd hole C .Proof. If N α = 0, then the result follows by Corollary 2.24. Hence, assume that N α > b y to be a β vertex. Therefore, if b x and b z are adjacent in C (cid:48) , setting b p b q = b x b z suffices (by Lemma 2.25).If b x b z are not adjacent in C (cid:48) , the common neighbour of b x and b z on C must be an α (cid:48) vertex. ByLemma 2.19, either b x or b z is a β vertex and by Corollary 2.18, at least one of them must be a β (cid:48) vertex.Hence, without loss of generality, we may assume that b x is a β vertex and b z is a β (cid:48) vertex. By Lemma 2.22,at least one of the neighbours of b z must be an α vertex. Therefore, the common neighbour of b z and b y in C must be an α vertex. This implies that b z b y is an edge of C (cid:48) (see Figure 10(b)). Since b x is a β 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 de-composed into induced 2-connected subgraphs containing no separating triangles or edge separators without12ffecting the perfectness of the graph, it was sufficient to limit the attention to W-triangulations. We notedthat if any internal vertex x in a W-triangulation has on odd degree, the result was immediate becausethe exterior face of the closed neighbourhood of x would have been an odd hole. Consequently, the core ofthe problem was to characterize odd holes in even W-near-triangulated (induced) subgraphs of the originalplane near-triangulation. If an even W-near-triangulation G is non-perfect, we considered a minimal oddhole C in G such that there is no other odd hole in C ∪ Int ( C ). Then we considered the cycle C (cid:48) formingthe boundary of the subgraph H induced by the vertices in Int ( C ). We showed that either (i) C (cid:48) 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 (cid:48) such that the boundary of the exterior face of the closedneighbourhood 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 thefollowing: • a vertex x (of odd degree), the exterior boundary of the local neighbourhood of which, is an odd hole. • an edge xy , the exterior boundary of the local neighbourhood of which, is an odd hole. • 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 1.1into 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-triangulatedsubgraph G (cid:48) 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 couldbe a separating triangle in G . Hence, identification of the separating triangles in G becomes a significantalgorithmic consideration. It turns out this task is easy due to the algorithm by Kant [5] that identifies theseparating 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-perfectnear-triangulation G , we can find an induced subgraph X which is either a vertex, an edge or a triangle, suchthat 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 exampleof a graph illustrating this fact. In this example, the exterior boundary of the local neighbourhood of theedge ip induces the odd hole abcde . However, the subgraph induced by the neighbours of the edge ip (or anyother edge or face) does not induce an odd hole. The requirement to check the exterior boundary of the localneighbourhood of each edge and each face results in quadratic time complexity for the algorithm describedin the next section. If a W-near-triangulation G does not contain any induced wheel on five vertices, then itis known that G is not perfect if and only if it contains a vertex or a face whose neighbours induce an oddhole [6]. Consequently, for this restricted class of graphs, checking perfectness requires only sub-quadratictime.
3. Recognition of perfect plane near-triangulations
In this section, we describe an O ( n ) 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 knowntechniques for handling planar graphs. For the sake of clarity, the steps of the algorithm and its analysisare 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 separatingtriangles in a plane triangulation, G can be split into its 4-connected blocks. It is well known that a planetriangulation is 4-connected if and only if it is free of separating triangles. Consequently, the 4-connected13 cde fghi j k l mno p b Figure 11: The neighbours of the edge ip does not induce an odd hole though the boundary of its local neighbourhood inducesan odd hole. blocks identified by Kant’s algorithm are precisely the maximal induced W-triangulations in G . We will calleach such 4-connected block a W-component of G [6]. We have already seen that G is perfect if and onlyif all its W-components are perfect. Hence, our task reduces to the problem of finding a quadratic timealgorithm for checking whether a given plane triangulation without separating triangles is perfect. We splitour task into four stages, as described below. Let H be a plane triangulation on n vertices, without separating triangles. Suppose we are given theadjacency list of H as input. We will first compute a planar straight line embedding of H on a n × n grid,using the linear time algorithm of Schnyder [7]. This essentially gives us the ( x, y ) coordinates of each vertexof H . Using these coordinates, the slopes of all the edges of H can be computed in O ( m ) time, where m isthe number of edges of H . Note that, for a planar graph, m = O ( n ). It would be useful to preprocess theadjacency list of H at this point, so that for each vertex v , its neighbours are arranged in the descendingorder of the slopes of edges incident at v . This preprocessing can be done in O ( m log n ) time. The nextthree stages involve verifying the conditions of perfectness mentioned in Theorem 1.1 one by one. If any vertex u of H is of odd degree, the open neighbourhood N H ( u ) induces an odd hole and H is notperfect. Checking the degrees of all vertices can be easily done in time linear in n . If every vertex of H isof even degree, we need to check the remaining conditions to verify the perfectness of H . 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 ) ∪ N ( v ) ∪ { u, v } and construct an indicator array A uv of length n , where A uv [ i ] = 1 if vertex v i ∈ S uv and zero otherwise. The construction of these arrays can be done in O ( mn ) = O ( n ) time in total.Now, we construct m plane subgraphs of H , one corresponding to each edge of H . The graph H uv willbe the induced subgraph of H on the vertex set S uv . We store the coordinate position information of eachvertex of H uv , by copying the same from H . To get the sorted adjacency list of H uv , we start with a copyof 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 theinformation mentioned above, takes only O ( m ( m + n )) = O ( n ) time in total.14ow, we describe a procedure to check if the exterior boundary of an induced subgraph H uv constructedabove is an odd hole. It is easy to see that the exterior boundary of H uv is an induced cycle. So, it suffices toidentify the vertices on the exterior boundary and check the parity of their count. Since we have coordinatesof vertices from a straight line drawing, this is easy. First, identify the left most vertex of H uv in the straightline embedding. This only involves identifying a vertex of H uv with the smallest x coordinate. Clearly, thisvertex is on the exterior face of H uv . Let this vertex be l . Recall that the adjacency list of each vertex issorted in the decreasing order of slopes. The edge with the largest slope incident at vertex l must be onthe exterior boundary of H uv . The other endpoint of this edge can be identified as the next vertex on theboundary 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 theadjacency list of x , then y is the next vertex on the boundary of H uv . When this procedure encounters theinitial vertex l again, we would have identified all the vertices on the boundary of H . Thus, identifying theexterior boundary of H uv and checking whether it is an odd hole, can be done in O ( m + n ) time. For checkingthe boundaries of all subgraphs H uv we constructed, total time required is only O ( m ( m + n )) = O ( n ).If this check fails to find an odd hole, we have to check the boundary of the local neighbourhood of eachtriangle, as described below. 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 theexterior boundary of the closed neighbourhood of a triangle uvw forms an odd hole can be done in a wayvery similar to the method we discussed in the previous subsection. For each triangle uvw , we will define aset S uvw = N ( u ) ∪ N ( v ) ∪ N ( w ) ∪ { u, v, w } and an indicator array A uvw for S uvw . Then, we can constructplane subgraphs H uvw , the induced subgraph of H on the vertex set S uvw . The method of checking whetherthe boundary of H uvw is an odd hole or not, is the same as the method described earlier for H uv . Thenumber of subgraphs to be processed is the number of faces of H , which is linear in n . Hence, the timerequired for checking the boundaries of each such subgraph is again O ( n ) only. The method described above for recognizing planar perfect graphs can be extended to recognize perfectplane near-triangulations by the simple modifications described below, without affecting the complexity ofthe algorithm. The strategy is to triangulate the plane near-triangulation; use the algorithm of Kant [5] toidentify the W-components of the triangulated graph; and retrieve the W-components of the original planenear-triangulation.Let G ( V, E ) be a plane near-triangulation. If G is not 2-connected, in linear time we can find the2-connected components of G using depth first search, and work on each component. Hence, we assumewithout loss of generality that G is 2-connected.We can embed G into an n × n grid using the algorithm of Schnyder [7] in O ( n ) time, preprocess theadjacency list of G as described in Subsection 3.1 in O ( m log n ) time and identify the vertices on the boundaryof exterior face of G by using the method discussed in Subsection 3.3 in O ( n ) time. Let v , v , . . . , v k , v be the cycle forming the the boundary of exterior face of G . Construct a plane triangulation G (cid:48) from G byadding a new vertex v on the exterior face of G and adding edges from v to v i for every i ∈ { , , . . . , k } .The adjacency list of G can be modified to get the adjacency list of G (cid:48) in linear time. Note that, as G (cid:48) is atriangulation, it contains no edge separators. We can use the linear time algorithm of Kant [5] for identifyingseparating triangles in the plane triangulation G (cid:48) and decompose G (cid:48) 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 externalface of G , forming a separating triangle v i v j v in G (cid:48) . Conversely, a separating triangle in G (cid:48) containing v must be of the form v v i v j for some i, j ∈ { , , . . . , k } , with v i v j forming a chord connecting two verticeson the exterior face of G . Moreover, every separating triangle in G will be a separating triangle in G (cid:48) andevery separating triangle in G (cid:48) that does not contain v is a separating triangle in G . Hence, we have thefollowing observation. 15 bservation 3.1. Let G (cid:48) , G (cid:48) , . . . , G (cid:48) r be the W-components of G (cid:48) . Let G i = G (cid:48) i if v / ∈ V ( G (cid:48) i ) and G i = G (cid:48) i \ { v } otherwise. Then, G , G , . . . , G r are precisely the W-components of G . As a consequence of Observation 3.1, it suffices to decompose G (cid:48) into its W-components and removethe vertex v from each W-component to recover the W-components of G . Removing v from all the W-components of G (cid:48) 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 ) 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 )time.
4. Acknowledgment
We thank Sunil Chandran, IISc. Bangalore and Ajit A. Diwan, IIT Bombay for discussions and sugges-tions. We thank the latter also for the example in Figure 11.
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