3-degenerate induced subgraph of a planar graph
aa r X i v : . [ m a t h . C O ] J u l Yangyan Gu , H. A. Kierstead , Sang-il Oum ∗ ,Hao Qi † , and Xuding Zhu ‡ Department of Mathematics, Zhejiang Normal University, Jinhua, China School of Mathematical and Statistical Sciences, Arizona State University, Tempe, AZ, USA Discrete Mathematics Group, Institute for Basic Science (IBS), Daejeon, South Korea Department of Mathematical Sciences, KAIST, Daejeon, South Korea Institute of Mathematics, Academica Sinica, TaipeiEmail: , [email protected] , [email protected] , [email protected] , [email protected] July 28, 2020
Abstract
A graph G is d -degenerate if every non-null subgraph of G has a vertex ofdegree at most d . We prove that every n -vertex planar graph has a 3-degenerateinduced subgraph of order at least 3 n/ Keywords: planar graph; graph degeneracy.
Graphs in this paper are simple, having no loops and no parallel edges. For a graph G = ( V, E ), the neighbourhood of x ∈ V is denoted by N ( x ) = N G ( x ), the degree of x is denoted by d ( x ) = d G ( x ), and the minimum degree of G is denoted by δ ( G ). LetΠ = Π( G ) be the set of total orderings of V . For L ∈ Π, we orient each edge vw ∈ E as ( v, w ) if w < L v to form a directed graph G L . We denote the out-neighbourhood ,also called the back-neighbourhood , of x by N LG ( x ), the out-degree , or back-degree , of x by d LG ( x ). We write δ + ( G L ) and ∆ + ( G L ) to denote the minimum out-degree and the ∗ Supported by the Institute for Basic Science (IBS-R029-C1) † Partially supported by research grants NSFC 11771403, NSFC 11871439 and MOST 107-2811-M-001-1534. ‡ Partially supported by research grants NSFC 11971438, NSFC 11771403, ZJNSF LD19A01001 and111 project of the Ministry of Education. G L . We define | G | := | V | , called an order , and k G k := | E | .An ordering L ∈ Π( G ) is d -degenerate if ∆ + ( G L ) ≤ d . A graph G is d -degenerate ifsome L ∈ Π( G ) is d -degenerate. The degeneracy of G is min L ∈ Π( G ) ∆ + ( G L ). It is wellknown that the degeneracy of G is equal to max H ⊆ G δ ( H ).Alon, Kahn, and Seymour [4] initiated the study of maximum d -degenerate inducedsubgraphs in a general graph and proposed the problem on planar graphs. We studymaximum d -degenerate induced subgraphs of planar graphs. For a non-negative integer d and a graph G , let α d ( G ) = max {| S | : S ⊆ V ( G ) , G [ S ] is d -degenerate } and¯ α d = inf { α d ( G ) / | V ( G ) | : G is a non-null planar graph } . Let us review known bounds for ¯ α d . Suppose that G = ( V, E ) is a planar graph.For d ≥
5, trivially we have ¯ α d = 1 because planar graphs are 5-degenerate.For d = 0, a 0-degenerate graph has no edges and therefore α ( G ) is the size of amaximum independent set of G . By the Four Colour Theorem, G has an independentset I with | I | ≥ | V ( G ) | /
4. This bound is tight since α ( K ) = 1. Therefore ¯ α = 1 / α ≥ / α ≥ /
13 independently of the Four Colour Theorem by Cranston and Rabern in2016 [8].For d = 1, a 1-degenerate graph is a forest. Since K has no induced forest of ordergreater than 2, we have ¯ α ≤ /
2. Albertson and Berman [3] and Akiyama and Watan-abe [1] independently conjectured that ¯ α = 1 /
2. In other words, every planar graphhas an induced forest containing at least half of its vertices. This conjecture receivedmuch attention in the past 40 years; however, it remains largely open. Borodin [7]proved that the vertex set of a planar graph can be partitioned into five classes suchthat the subgraph induced by the union of any two classes is a forest. Taking thetwo largest classes yields an induced forest of order at least 2 | V ( G ) | /
5. So ¯ α ≥ / α . On the other hand, the conjectureof Albertson and Berman, Akiyama and Watanabe was verified for some subfamiliesof planar graphs. For example, C -free, C -free, or C -free planar graphs were shownin [17, 10] to be 3-degenerate, and a greedy algorithm shows that the vertex set of a3-degenerate graph can be partitioned into two parts, each inducing a forest. Hence C -free, C -free, or C -free planar graphs satisfy the conjecture. Moreover, Raspaudand Wang [13] showed that C -free planar graphs can be partitioned into two inducedforests, thus satisfying the conjecture.Now let us move on to the case that d = 2. The octahedron has 6 vertices and is4-regular, so a 2-degenerate induced subgraph has at most 4 vertices. Thus ¯ α ≤ / α ≥ /
2, which follows from the fact that G is 5-degenerate, and hence we can2reedily 2-colour G in an ordering that witnesses its degeneracy so that no vertex hasthree out-neighbours of the same colour, i.e., each colour class induces a 2-degeneratesubgraph.For d = 4, the icosahedron has 12 vertices and is 5-regular, so a 4-degenerate inducedsubgraph has at most 11 vertices. Thus ¯ α ≤ /
12. Again we conjecture that equalityholds. The best known lower bound is ¯ α ≥ /
9, which was obtained by Luko˘tka,Maz´ak and Zhu [12].In this paper, we study 3-degenerate induced subgraphs of planar graphs. Both theoctahedron and the icosahedron witness that ¯ α ≤ /
6. Here is our main theorem.
Theorem 1.1.
Every n -vertex planar graph has a -degenerate induced subgraph oforder at least n/ . We conjecture that the upper bounds for ¯ α d mentioned above are tight. Conjecture 1.1. ¯ α = 2 / , ¯ α = 5 / , and ¯ α = 11 / . The problem of colouring the vertices of a planar graph G so that colour classesinduce certain degenerate subgraphs has been studied in many papers. Borodin [7]proved that every planar graph G is acyclically 5-colourable, meaning that V ( G ) canbe coloured in 5 colours so that a subgraph of G induced by each colour class is 0-degenerate and a subgraph of G induced by the union of any two colour classes is 1-degenerate. As a strengthening of this result, Borodin [6] conjectured that every planargraph has degenerate chromatic number at most 5, which means that the vertices ofany planar graph G can be coloured in 5 colours so that for each i ∈ { , , , } , asubgraph of G induced by the union of any i colour classes is ( i − For sets X and Y , define Z = X ∪· Y to mean Z = X ∪ Y and X ∩ Y = ∅ . Let G = ( V, E ) be a graph with v, x, y ∈ V and X, Y ⊆ V . Then k v, X k is the number3f edges incident with v and a vertex in X and k X, Y k = P v ∈ X k v, Y k . When X and Y are disjoint, k X, Y k is the number of edges xy with x ∈ X and y ∈ Y . In general,edges in X ∩ Y are counted twice by k X, Y k . Let N ( X ) = S x ∈ X N ( x ) − X .We write H ⊆ G to indicate that H is a subgraph of G . The subgraph of G inducedby a vertex set A is denoted by G [ A ]. The path P with V ( P ) = { v , . . . , v n } and E ( P ) = { v v , . . . , v n − v n } is denoted by v · · · v n . Similarly the cycle C = P + v n v isdenoted by v · · · v n v .Now let G be a simple connected plane graph. The boundary of the infinite face isdenoted by B = B ( G ) and V ( B ( G )) is denoted by B = B ( G ). Then B is a subgraphof the outerplanar graph G [ B ]. For a cycle C in G , let int G [ C ] denote the subgraph of G obtained by removing all exterior vertices and edges and let ext G [ C ] be the subgraphof G obtained by removing all interior vertices and edges. Usually the graph G isclear from the text, and we write int[ C ] and ext[ C ] for int G [ C ] and ext G [ C ]. Letint( C ) = int[ C ] − V ( C ) and ext( C ) = ext[ C ] − V ( C ). Let N ◦ ( x ) = N ( x ) − B and N ◦ ( X ) = N ( X ) − B .For L ∈ Π, the up-set of x in L is defined as U L ( x ) = { y ∈ V : y > L x } and the down-set of x in L is defined as D L ( x ) = { y ∈ V : y < L x } . Note that for each L ∈ Π, y < L x means that y ≤ L x and y = x . For two sets X and Y , we say X ≤ L Y if x ≤ L y for all x ∈ X , y ∈ Y . In this section we phrase a stronger, more technical version of Theorem 1.1 that is moreamenable to induction. This is roughly analogous to the proof of the 5-ChoosabilityTheorem by Thomassen [14]. The strengthening takes three forms.If G = G ∪ G and G ∩ G = G [ A ], then we would like to join two 3-degeneratesubgraphs obtained from G and G by induction to form a 3-degenerate subgraph of G . The problem is that vertices from A and their neighbours may be in both subgraphs.Dealing with this motivates the following definitions.Let A ⊆ V = V ( G ). A subgraph H of G is ( k,A ) -degenerate if there exists anordering L ∈ Π( G ) such that A ≤ L V − A and d LH ( v ) ≤ k for every vertex v ∈ V ( H ) − A .Equivalently, every subgraph H ′ of H with V ( H ′ ) − A = ∅ has a vertex v ∈ V ( H ′ ) − A such that d H ′ ( v ) ≤ k . A subset Y of V is A -good if G [ Y ] is (3 ,A )-degenerate. We say asubgraph H is A -good if V ( H ) is A -good. Thus if A = ∅ then G is A -good if and onlyif G is 3-degenerate. Let f ( G ; A ) = max {| Y | : Y ⊆ V ( G ) is A -good } . Since ∅ is A -good, f ( G ; A ) is well defined.For an induced subgraph H of G and a set Y of vertices of H , we say Y is col-lectable in H if the vertices of Y can be ordered as y , y , . . . , y k such that for each i ∈{ , , . . . , k } , either y i / ∈ A and d H −{ y ,y ,...,y i − } ( y i ) ≤ V ( H ) −{ y , y , . . . , y i − } ⊆ A .4n order to build an A -good subset, we typically apply a sequence of operations ofdeleting and collecting. Deleting X ⊂ V means replacing G with G − X . An orderingwitnessing that Y is collectable is called a collection order. Collecting a sequence V , . . . , V s of disjoint subsets of V means first putting V at the end of L in a collectionorder for V , then putting V at the end of L − V in a collection order for V in G − V ,etc. Now if Y is collectable and V − Y is A -good then V is A -good. Definition 3.1.
A path v v . . . v ℓ of a plane graph G is admissible if ℓ > B ( G ) such that for each 1 < i < ℓ , G − v i has no path from v i − to v i +1 .A path of length 0 has only 1 vertex in its vertex set. Definition 3.2.
A set A of vertices of a plane graph G is usable in G if for eachcomponent G ′ of G , A ∩ V ( G ′ ) is the empty set or the vertex set of an admissible pathof G ′ . Lemma 3.1.
Let G be a plane graph and let A be a usable set in G . Then for eachvertex v of G , | N G ( v ) ∩ A | ≤ .Proof. This is clear from the definition of an admissible path.
Observation 3.1. If G is outerplanar and A is a usable set in G , then G is (2 ,A ) -degenerate. Observation 3.1 motivates the expectation that plane graphs with large boundarieshave large 3-degenerate induced subgraphs. Intuitively, we intend to prove that eachboundary vertex contributes 1 to f ( G ; A ) and each interior vertex contributes at least3 / f ( G ; A ). This is essentially true, with some easily detected exceptions. First wedefine a family of graphs for which special care is required to calculate f ( G ; A ).A set Z of vertices is said to be exposed if Z ⊆ B . We say that a vertex z is exposed if { z } is exposed. We say that deleting Y and collecting X exposes Z if Z ⊆ B ( G − Y − X ) − B . Definition 3.3.
Let Q = { Q , Q , Q +2 , Q , Q , Q +4 , Q ++4 } be the set of plane graphsshown in Figure 1. For a plane graph G , a cycle C of G is special if G C := int G [ C ] isisomorphic to a plane graph in Q , where C corresponds to the boundary. In this case, G C is also special .For a special cycle C of a plane graph G , we define T C := int G ( C ) , which is isomorphic to K , X C := { v ∈ V ( C ) : some facial cycle contains v and two vertices of T C } ,V C := V ( G C ) , Y C := X C ∪ V ( T C ) , and Y C := V C − Y C = V ( C ) − X C . Then V ( C ) = X C ∪ Y C . 5 v v x y z (a) Q v v v v x y z (b) Q v v v v x y z (c) Q +2 v v v v x y z (d) Q v v v v v x y z (e) Q v v v v v x y z (f) Q +4 v v v v v x y z (g) Q ++4 Figure 1: Plane graphs in Q defining special subgraphs G C where C corresponds to theboundary cycle. Solid black vertices denote vertices in X C . Observation 3.2.
Let C = v . . . v k v be a special cycle of G . If G C is (not onlyisomorphic but also equal to a plane graph) in Q , then the following hold.(a) T C = xyzx with N G ( x ) = { y, z, v , v } and N G ( y ) = { x, z, v , v } .(b) X C = { v , v , v } if G C = Q and X C = { v , v , v , v } if G C = Q .(c) Deleting any vertex in X C ∩ B exposes two vertices of T C .(d) After deleting any vertex v ∈ X C , V ( T C ) is collectable, except that if G C = Q ++4 and v = v then only { x, y } is collectable.(e) If Y C = ∅ then there is a facial cycle C ∗ containing Y C ∪{ v } for some v ∈ V ( T C ) .Moreover, v = z is unique, and if | Y C | = 2 , then C ∗ is unique.(f ) T C has at least two vertices v such that d G ( v ) = 4 . Note that vertices on C may have neighbours in ext( C ) or maybe contained in A .Thus we may not be able to collect vertices of C .A special cycle C is called exposed if X C ⊆ B ( G ). A special cycle packing of G isa set of exposed special cycles { C , . . . , C m } such that Y C i ∩ Y C j = ∅ for all i = j . Let τ ( G ) be the maximum cardinality of a special cycle packing and ∂ ( G ) = 34 | V ( G ) | + 14 ( | B | − τ ( G )) . We say that a special cycle packing of G is optimal if its cardinality is equal to τ ( G ). Theorem 3.2.
For all plane graphs G and usable sets A , f ( G ; A ) ≥ ∂ ( G ) . (3.1)Clearly | B | − τ ( G ) ≥ G with at least 2 vertices. The followingconsequence of Theorem 3.2 is the main result of this paper.6 orollary 3.3. Every n -vertex planar graph G (with n ≥ ) has an induced -degeneratesubgraph H with | V ( H ) | ≥ (3 n + 2) / . Suppose Theorem 3.2 is not true. Among all counterexamples, choose ( G ; A ) so that(i) | V ( G ) | is minimum,(ii) subject to (i), | A | is maximum, and(iii) subject to (i) and (ii), | E ( G ) | is maximum.We say that such a counterexample is extreme .If A ′ * V ( G ′ ), then we may abbreviate ( G ′ ; A ′ ∩ V ( G ′ )) by ( G ′ ; A ′ ), but still (ii)refers to | A ′ ∩ V ( G ′ ) | . We shall derive a sequence of properties of ( G ; A ) that leads toa contradiction. Trivially | V ( G ) | > G is connected (if G is the disjoint union of G and G , then f ( G ; A ) = f ( G ; A ) + f ( G ; A ) and ∂ ( G ) = ∂ ( G ) + ∂ ( G )). Lemma 4.1.
Let G be a plane graph and X be a subset of V ( G ) . If A is usable in G ,then A − X is usable in G − X .Proof. We may assume that G is connected and X = { v } . If v / ∈ A , then it is trivial.Let P = v v · · · v k be the admissible path in G such that A = V ( P ). If v = v or v = v k , then again it is trivial. If v = v i for some 0 < i < k , then by the definition ofadmissible paths, G − v i is disconnected, and v i − and v i +1 are in distinct components.Thus again A − { v } is usable in G − v .Suppose Y is a nonempty subset of V ( G ) and G [ Y ] is connected. Let C be anexposed special cycle of G ′ = G − Y . Then C satisfies one of the following conditions.(a) C is an exposed special cycle of G .(b) C is a non-exposed special cycle of G ; in this case X C ∩ ( B ( G ′ ) − B ) = ∅ .(c) C is not a special cycle of G ; in this case Y ⊆ int G ( C ), and so Y ∩ B = ∅ .A cycle C is type-a , -b , -c , respectively, if it satisfies condition (a), (b), (c), respectively.Let δ ( Y ) = ( , if G ′ has a type-c exposed special cycle , , otherwise . Lemma 4.2.
Let Y be a nonempty subset of V ( G ) such that G [ Y ] is connected. Let G ′ = G − Y . If C, C ′ are distinct exposed type-c special cycles of G ′ , then Y C ∩ Y C ′ = ∅ .Proof. Let C , C ′ be distinct exposed type-c special cycles of G ′ . Since B ∩ Y = ∅ and G [ Y ] is connected, there exists a facial cycle D of G ′ such that int G ( D ) = G [ Y ]. Then D is a facial cycle of both G ′ C and G ′ C ′ . Arguing by contradiction, suppose Y C ∩ Y C ′ = ∅ .Since V ( D ) ⊆ V ( G ′ C ) = Y C ∪ Y C and V ( D ) ⊆ V ( G ′ C ′ ) = Y C ′ ∪ Y C ′ , we have V ( D ) ⊆ V ( G ′ C ) ∩ V ( G ′ C ′ ) ⊆ Y C ∪ Y C ′ .
7y symmetry we may assume that | Y C ∩ V ( D ) | ≥ | Y C ′ ∩ V ( D ) | . Using | Y C | , | Y C ′ | ≤ ≤ | V ( D ) | ≤ | V ( G ′ C ) ∩ V ( G ′ C ′ ) | ≤ , Y C ⊆ V ( D ) , | Y C | = 2 , and Y C ′ ∩ V ( D ) = ∅ . We will show that H is isomorphic to H or H in Figure 2. Since | Y C | = 2,by Observation 3.2(e), D is the unique facial cycle in G ′ C such that there is a vertex˙ z ∈ V ( T C ) with Y C ∪ { ˙ z } ⊆ V ( D ). As ˙ z ∈ V ( D ) and ˙ z ∈ Y C , we have ˙ z ∈ Y C ′ . Since Y C ′ = ∅ , again by Observation 3.2(e), there is a unique vertex ¨ z ∈ V ( T C ′ ) such that Y C ′ ∪ { ¨ z } is contained in a facial cycle of G ′ C ′ . Then ¨ z ∈ Y C ′ ∩ Y C ⊆ V ( D ).First we show that | V ( G ′ C ) ∩ V ( G ′ C ′ ) | = 4. Assume to the contrary that | V ( G ′ C ) ∩ V ( G ′ C ′ ) | = 3. Since V ( D ) ⊆ V ( G ′ C ) ∩ V ( G ′ C ′ ), we conclude that | V ( D ) | = 3. Then ˙ z ¨ z isan edge, and the two inner faces of G ′ incident with ˙ z ¨ z are contained in V ( G ′ C ) ∩ V ( G ′ C ′ ).Since the intersection of any two inner faces of G ′ C has at most 2 vertices, we have | V ( G ′ C ) ∩ V ( G ′ C ′ ) | ≥
4, a contradiction.As V ( G ′ C ) ∩ V ( G ′ C ′ ) = Y C ∪ Y C ′ , we conclude that | Y C ′ | = 2 and | V ( C ) | = 5 = | V ( C ′ ) | .Let Q ∈ Q be the plane graph isomorphic to G ′ C . By inspection of Figure 1, G ′ C ′ is isomorphic to Q . We may assume that G ′ C = Q by relabelling vertices. Let u u ′ be an isomorphism from G ′ C to G ′ C ′ . Using uniqueness from Observation 3.2(e), z = ˙ z , z ′ = ¨ z , Y C = { v , v } and Y C ′ = { v ′ , v ′ } . To prove our claim let us divide our analysisinto two cases, resulting either in H or H . • If | V ( D ) | = 4, then Q = Q +4 and V ( G ′ C ) ∩ V ( G ′ C ) = V ( D ) = { v , v , v , z } . Since v , v ∈ Y C , we have v , z ∈ Y C ′ . Then v ′ = v , v ′ = z , and v = z ′ . As X C and X C ′ are exposed in G ′ , the cycle v v v v ′ v ′ v ′ v is in G ′ [ B ( G ′ )] and so H = H inFigure 2(a). • If | V ( D ) | = 3, then Q = Q ++4 , V ( D ) = { z, v , v } . By symmetry, we may assumethat z ′ = v . Then V ( G ′ C ) ∩ V ( G ′ C ′ ) = { z, v , v , v } , as C ′ contains all commonneighbors of z and z ′ in G ′ , which is a property of Q ++4 . Since v , v ∈ Y C and V ( G ′ C ) ∩ V ( G ′ C ′ ) ⊆ Y C ∪ Y C ′ , we deduce that v , z ∈ Y C ′ . By symmetry in G ′ C ′ ,we may assume that z = v ′ and v = v ′ . As X C and X C ′ are exposed in G ′ , thecycle v v v v ′ v ′ v ′ v is in G ′ [ B ( G ′ )]. So H = H + zz ′ = H in Figure 2(b).Notice that in both cases, v = v ′ ∈ B ( G ′ ) and v ∈ V ( D ). Set Y ′ = { v , x ′ , y ′ } and G ′′ = G − Y ′ . As V (int G ( D )) = Y , in G − v , we can collect both x ′ and y ′ and at leastone vertex of Y is exposed. Thus B ( G ′′ ) − B contains z, z ′ and ( B ( G ′′ ) − B ) ∩ Y = ∅ .So | B ( G ′′ ) − B | ≥ P be an optimal special cycle packing of G ′′ , and put P = { C ∗ ∈ P : C ∗ is a non-exposed special cycle of G } . Consider C ∗ ∈ P . As v = v ′ ∈ B ∩ Y ′ , there is no exposed type-c special cycle in G ′′ . Thus C ∗ is type-b, and so X C ∗ ∩ ( B ( G ′′ ) − B ) = ∅ . Let w ∈ X C ∗ ∩ ( B ( G ′′ ) − B ).8 v = v ′ v ′ v v = v ′ v ′ z = v ′ z ′ = v y ′ x ′ yx (a) H v v = v ′ v ′ v v = v ′ v ′ z = v ′ z ′ = v y ′ x ′ yx (b) H Figure 2: The isomorphism types H and H of H = G ′ [ V ( G ′ C ) ∪ V ( G ′ C ′ )] when | V ( D ) | =4 or | V ( D ) | = 3 in the proof of Lemma 4.2. Solid black vertices denote boundary verticesof G and thick edges represent edges in D .Since T C ∗ is connected, has a neighbour of w , and has no vertex from B ( G ′′ ), we have V ( T C ∗ ) ⊆ Y and X C ∗ ⊆ ( B ( G ′′ ) − B ) ∪ { v } .As P is a packing, 3 |P | ≤ | B ( G ′′ ) − B | +1. This implies that |P | ≤ | B ( G ′′ ) − B |− | B ( G ′′ ) − B | ≥
3. We now deduce that |P | ≤ | B ( G ′′ ) − B | − | B ( G ′′ ) | − ( | B | − − | B ( G ′′ ) | − | B | − . Therefore τ ( G ) ≥ τ ( G ′′ ) − |P | ≥ τ ( G ′′ ) − | B ( G ′′ ) | + | B | + 1 . Hence, using V ( G ) = V ( G ′′ ) ∪ Y ′ , ∂ ( G ) = 34 | V ( G ) | + 14( | B |− τ ( G )) ≤
34 ( | V ( G ′′ ) | +3)+ 14( | B ( G ′′ ) |− τ ( G ′′ ) −
1) = ∂ ( G ′′ )+2 . Now, as we have already collected x ′ , y ′ , we have f ( G ; A ) ≥ f ( G ′′ ; A ) + 2 ≥ ∂ ( G ′′ ) + 2 ≥ ∂ ( G ) . This contradicts the assumption that G is a counterexample. Lemma 4.3.
Let Y be a nonempty subset of V ( G ) such that G [ Y ] is connected and let G ′ = G − Y . Then ∂ ( G ) ≤ ∂ ( G ′ ) + 3 | Y | | B − B ( G ′ ) | + δ ( Y )4 . Moreover, if G has an exposed special cycle C such that Y C ∩ Y = ∅ and Y C ∩ Y C ′ = ∅ for any other exposed special cycle C ′ of G , then ∂ ( G ) ≤ ∂ ( G ′ ) + 3 | Y | | B − B ( G ′ ) | + δ ( Y ) − . roof. By Lemma 4.2, an optimal special cycle packing of G ′ has at most | B ( G ′ ) − B | type-b cycles, and δ ( Y ) type-c cycles. We can remove such cycles from the special cyclepacking of G ′ to obtain a special cycle packing of G . So τ ( G ) ≥ τ ( G ′ ) − | B ( G ′ ) − B | − δ ( Y ) = τ ( G ′ ) − | B ( G ′ ) | + | B | − | B − B ( G ′ ) | − δ ( Y ) . Plugging this into the definition of ∂ ( G ), we obtain ∂ ( G ) ≤ ∂ ( G ′ ) + 3 | Y | | B − B ( G ′ ) | + δ ( Y )4 . If G has an exposed special cycle C such that C is not a special cycle of G ′ and Y C is disjoint from Y C ′ for any other exposed special cycle C ′ of G , then we can add cycle C to the special cycle packing of G obtained above. So τ ( G ) ≥ τ ( G ′ ) − | B ( G ′ ) − B | − δ ( Y ) + 1 = τ ( G ′ ) − | B ( G ′ ) | + | B | − | B − B ( G ′ ) | − δ ( Y ) + 1 . Plug this into the definition of ∂ ( G ), we obtain ∂ ( G ) ≤ ∂ ( G ′ ) + 3 | Y | | B − B ( G ′ ) | + δ ( Y ) − . Lemma 4.4.
Every vertex v ∈ V − A satisfies d ( v ) ≥ .Proof. Suppose that d ( v ) ≤
3. Apply Lemma 4.3 with Y = { v } . Let G ′ = G − Y . Notethat if v is a boundary vertex, then δ ( Y ) = 0. So | B − B ( G ′ ) | + δ ( Y ) ≤
1. Therefore ∂ ( G ) ≤ ∂ ( G ′ ) + 34 + 14 . By the minimality of ( G ; A ), f ( G ′ ; A ) ≥ ∂ ( G ′ ). Therefore f ( G ; A ) = f ( G ′ ; A ) + 1 ≥ ∂ ( G ), a contradiction. Lemma 4.5.
There are no disjoint nonempty subsets X , Y of V ( G ) such that Y is aset of | X | interior vertices of G , G [ X ∪ Y ] is connected, and Y is collectable in G − X .Proof. Suppose not. Let G ′ = G − ( X ∪ Y ). We apply Lemma 4.3. Since | B − B ( G ′ ) | + δ ( X ∪ Y ) ≤ | X | , we have ∂ ( G ) ≤ ∂ ( G ′ ) + ( | X | + | Y | ) + | X | = ∂ ( G ′ ) + 4 | X | . As G is extreme, f ( G ′ ; A ) ≥ ∂ ( G ′ ). Hence f ( G ; A ) ≥ f ( G ′ ; A ) + | Y | = f ( G ′ ; A ) + 4 | X | ≥ ∂ ( G ′ ) + 4 | X | ≥ ∂ ( G ), a contradiction. Lemma 4.6.
For any two distinct special cycles C , C of G , Y C ∩ Y C = ∅ .Proof. Assume to the contrary that C , C are two special cycles of G with Y C ∩ Y C = ∅ .Observe that for each i = 1 , V ( T C i ) has two vertices of degree 4 and one vertex ofdegree 4, 5, or 6 in G .If T C and T C share an edge, say T C = xyz and T C = xyz ′ , then one of x, y , say x ,has degree 4. Since G is simple, z = z ′ . Let v be the other neighbor of x . By inspecting10ll graphs in Q , we deduce that each of z , z ′ is either adjacent to v or has degree atmost 5 in G . So in G − v , the set { x, y, z, z ′ } is collectable, contrary to Lemma 4.5.Assume T C and T C have a common vertex, say T C = xyz and T C = xy ′ z ′ . If noneof y , z , y ′ , z ′ have degree 6, then we can delete x and collect y , z , y ′ , z ′ , contrary toLemma 4.5. So we may assume that d G ( y ) = 6 and hence d G ( x ) = d G ( z ) = 4 and allthe faces incident to x are triangles because G C is isomorphic to Q ++4 . Thus we mayassume yy ′ , zz ′ ∈ E ( G ). By deleting y , we can collect x , z , z ′ , and y ′ , again contrary toLemma 4.5. (We collect y ′ ahead of z ′ if d G ( z ′ ) = 6 and collect z ′ ahead of y ′ otherwise.)Thus V ( T C ) ∩ V ( T C ) = ∅ .If X C ∩ V ( T C ) = ∅ , then for a vertex v of maximum degree in V ( T C ), afterdeleting v , we can collect the other two vertices of T C and two vertices of T C , contraryto Lemma 4.5. So X C ∩ V ( T C ) = ∅ and by symmetry, X C ∩ V ( T C ) = ∅ .If X C ∩ X C contains a vertex v , then by deleting v , we can collect two vertices fromeach of T C and T C , again contrary to Lemma 4.5 because V ( T C ) ∩ V ( T C ) = ∅ . Lemma 4.7. If C is a special cycle of G , then there is a vertex u ∈ X C such that V ( T C ) is collectable in G − u and G ′ = G − ( V ( T C ) ∪ { u } ) has no type-c special cycle.Proof. Suppose the lemma fails for some special cycle C of G with | E ( C ) | = k . Then G C is isomorphic to a graph Q ∈ Q . We may assume G C = Q . Then V ( T C ) iscollectable in G − v . Put Y = V ( T C ) ∪ { v } and G ′ = G − Y . Since G ′ has a type-cspecial cycle C ′ , G ′ C ′ has a facial cycle C ′′ with Y = V (int G ( C ′′ )).Then C ′′ consists of the subpath C − v from v to v k of length k − P from v k to v in G ′ . As G ′ C ′ is special, 3 ≤ | E ( C ′′ ) | ≤
5. So | E ( P ) | ≤ − ( k − ≤ N G ( v ) ⊆ V ( P ) ∪ { y, z } , so d G ( v ) ≤ | E ( P ) | + 3 ≤ − k ≤
7. If d G ( v ) ≤
6, thenafter deleting v we can collect Y : use the order x, y, v , z if d G ( v ) ≤
5; else d G ( v ) = 6and k ≤
4, so use the order x, y, z, v . This contradicts Lemma 4.5. Thus d G ( v ) = 7.So k = 3, | E ( P ) | = 4, | E ( C ′′ ) | = 5, G C = Q , and v is adjacent to all vertices of P .Setting u = v , and using symmetry between v and v , we see that v is also aninterior vertex with d G ( v ) = 7.Let P = v u u u v . Now G ′ C ′ is isomorphic to Q since C ′′ is a facial 5-cycle.Assume u u ′ is an isomorphism from Q to G ′ C ′ . Then C ′′ = z ′ v ′ v ′ v ′ v ′ z ′ .If there is w ∈ { v , v }∩{ v ′ , v ′ } then after deleting w we can collect { x, y, z, x ′ , y ′ , z ′ } ,contrary to Lemma 4.5. Else { v ′ , v ′ } = { u , u } and therefore z ′ = u , see Figure 3.After deleting { v , u } , we can collect { x, y, z, v , v , z ′ , x ′ , y ′ } , contrary to Lemma 4.5,as both v and v are interior vertices. Lemma 4.8. G has no special cycle.Proof. Assume to the contrary that C is a special cycle of G . By Lemma 4.7, there isa vertex u ∈ X C such that V ( T C ) is collectable in G − u and G ′ = G − ( V ( T C ) ∪ { u } )has no type-c special cycle. Observe that f ( G ; A ) ≥ f ( G ′ ; A ) + 3. So it suffices to showthat ∂ ( G ) ≤ ∂ ( G ′ ) + 3. Since G ′ has no type-c special cycles, every exposed specialcycle of G ′ is a special cycle of G . 11 = v ′ v ′ u = v ′ v = v ′ v = v ′ x ′ y ′ u = z ′ v y xz Figure 3: The graph int G ( C ′′ ) in the last part of the proof of Lemma 4.7 when z ′ = u .Note that d G ( v ) = 7 and v is an interior vertex.If u / ∈ B , then B ( G ′ ) = B and so B − B ( G ′ ) = ∅ . As δ ( V ( T C ) ∪ { u } ) = 0, we deducefrom Lemma 4.3 that ∂ ( G ) ≤ ∂ ( G ′ ) + · u ∈ B and so | B − B ( G ′ ) | = 1. If C is exposed in G ,then by Lemmas 4.3 and 4.6, ∂ ( G ) ≤ ∂ ( G ′ ) + · − .If C is not exposed in G , then X C has some interior vertex v . Since v is adjacent toa vertex of T C , v is exposed in G ′ . By Lemma 4.6, v / ∈ X C ′ for every exposed specialcycle C ′ of G ′ , because C ′ is a special cycle of G . Therefore, in an optimal special cyclepacking of G ′ , at most | B ( G ′ ) − B | − G . So, τ ( G ) ≥ τ ( G ′ ) − ( | B ( G ′ ) − B | −
1) = τ ( G ′ ) − ( | B ( G ′ ) | − | B | + 1) + 1 . Thus ∂ ( G ) = 34 | V ( G ) | + 14 ( | B | − τ ( G )) ≤
34 ( | V ( G ′ ) | + 4) + 14 ( | B | + ( − τ ( G ′ ) + | B ( G ′ ) | − | B | )) = ∂ ( G ′ ) + 3 . Lemma 4.9.
Let s be an integer. Let X and Y be disjoint subsets of V ( G ) such that Y is collectable in G − X . If | B ( G − ( X ∪ Y )) | ≥ | B ( G ) | + s , G [ X ∪ Y ] is connected,and ( X ∪ Y ) ∩ B ( G ) = ∅ , then s + | Y | < | X | .Proof. Let G ′ = G − ( X ∪ Y ). Since ( X ∪ Y ) ∩ B ( G ) = ∅ and G [ X ∪ Y ] is connected,any special cycle of G ′ is also a special cycle of G . So τ ( G ′ ) = 0 and ∂ ( G ) ≤ ∂ ( G ′ ) + ( | X ∪ Y | ) − s . As X ∪ Y = ∅ and ( G ; A ) is extreme, f ( G ′ ; A ) ≥ ∂ ( G ′ ). Thus f ( G ′ ; A ) + | Y | ≤ f ( G ; A ) < ∂ ( G ) ≤ ∂ ( G ′ ) + 34 | X ∪ Y | − s ≤ f ( G ′ ; A ) + 34 | X ∪ Y | − s . This implies that s + | Y | < | X | . Lemma 4.10. G is -connected and | A | = 2 . roof. Suppose G is not 2-connected. If | V ( G ) | ≤
3, then G is (3 , A )-degenerate, so f ( G ; A ) = ∂ ( G ) and we are done. Else | V ( G ) | >
3. As G is connected, it has a cut-vertex x . Let G , G be subgraphs of G such that G = G ∪ G , V ( G ) ∩ V ( G ) = { x } ,and | V ( G ) ∩ A | ≤ | V ( G ) ∩ A | . Observe that if x / ∈ A , then A ∩ V ( G ) = ∅ by thechoice of G because A is usable in G .Let A = V ( G ) ∩ A if x ∈ A and A = { x } otherwise. Let A = V ( G ) ∩ A . Notethat for each i = 1 , A i is usable in G i . For i = 1 ,
2, let X i be a maximum A i -goodset in G i .Let X := ( X ∪ X − { x } ) ∪ ( X ∩ X ). We claim that X is A -good in G . If x ∈ A ,then use the collection order X − A , X − A , A ∩ X . If x / ∈ A and x ∈ X ∩ X , thenuse the collection order X − { x } , X . If x / ∈ A and x / ∈ X ∩ X , then use the collectionorder X − { x } , X − { x } . This proves the claim that X is A -good in G .As ( G ; A ) is extreme, f ( G i ; A i ) ≥ ∂ ( G i ) for i = 1 , x ∈ B then B ( G i ) = B ( G ) ∩ V ( G i ) for i = 1 ,
2. Note that any special cycleof G i is a special cycle of G and so τ ( G i ) = 0 for i = 1 , ∂ ( G ) = ∂ ( G ) + ∂ ( G ) − x / ∈ B , then we may assume V ( G ) ∩ B ( G ) = ∅ . Hence B ( G ) = B ( G ). Sinceonly one inner face of G contains vertices of G , τ ( G ) ≤ ∂ ( G ) = ∂ ( G ) + ∂ ( G ) −
34 + 14 τ ( G ) −
14 ( | B ( G ) | − τ ( G )) . Since τ ( G ) ≤ | B ( G ) | −
2, we have ∂ ( G ) ≤ ∂ ( G ) + ∂ ( G ) − f ( G ; A ) ≥ | X | + | X | − f ( G ; A ) + f ( G ; A ) − ≥ ∂ ( G ) + ∂ ( G ) − ≥ ∂ ( G ) . Thus G is 2-connected, and hence | A | ≤
2. As ( G ; A ) is extreme, we have | A | = 2.In the following, set A = { a, a ′ } . Lemma 4.11.
The boundary cycle B has no chord.Proof. Assume B has a chord e := xy . Let P , P be the two paths from x to y in B such that A ⊆ V ( P ). Since e is a chord, both P and P have length at least two.Set G = int[ P + e ] and G = int[ P + e ]. As τ ( G ) = 0 by Lemma 4.8, we knowthat τ ( G ) = τ ( G ) = 0. Hence ∂ ( G ) = ∂ ( G ) + ∂ ( G ) −
2. We may assume that A ⊆ V ( G ). Let A = { x, y } and A = A .For i = 1 ,
2, let X i be a maximum A i -good set in G i . Then X = ( X ∪ X − { x, y } ) ∪ ( X ∩ X ) is an A -good set in G : use the collection order X − { x, y } , ( X − { x, y } ) ∪ ( X ∩ X ). Thus f ( G ; A ) ≥ f ( G ; A ) + f ( G ; A ) − ≥ ∂ ( G ) + ∂ ( G ) − ∂ ( G ) , contrary to the choice of G . Lemma 4.12. G is a near plane triangulation. roof. By Lemma 4.10, every face boundary of G is a cycle of G . Assume to the contrarythat G has an interior face F which is not a triangle. Then V ( F ) has a pair of verticesnon-adjacent in G because G is a plane graph. Let e / ∈ E ( G ) be an edge drawn on F joining them. Then G ′ = G + e is a plane graph with B ( G ′ ) = B ( G ). As G is extreme, G ′ is not a counterexample. As f ( G ′ ; A ) ≤ f ( G ; A ), we conclude τ ( G ′ ) > τ ( G ), and hence G ′ has an exposed special cycle C and e is an edge of G ′ C . By (d) of Observation 3.2,there is a vertex v ∈ X C such that after deleting v , we can collect all the three verticesof T C . In G − ( V ( T C ) ∪ { v } ), all vertices in ( V C ∪ V ( F )) − ( V ( T C ) ∪ { v } ) are exposed.By Lemma 4.9, none of these vertices can be an interior vertex of G , because otherwise | B ( G − ( V ( T C ) ∪ { v } ) | ≥ | B ( G ) | . So all these vertices are boundary vertices of G . ByLemmas 4.10 and 4.11, G is 2-connected, | A | = 2, and B ( G ) has no chord, so G has noother vertices and int( B ( G )) = T C , as v ∈ X C is also a boundary vertex of G .By Lemma 4.4, k u, V ( T C ) k ≥ u ∈ B ( G ) − A , and k w, B ( G ) k ≥ w ∈ V ( T C ). On the other hand, the number of vertices u ∈ B ( G ) with k u, V ( T C ) k ≥ | B ( G ) | ≤ | B ( G ) | = 3, then G is triangulated. Suppose | B ( G ) | = 4. If k u, V ( T C ) k ≥ u ∈ B ( G ), then G is isomorphic to Q ; else G is isomorphic to Q .Both are contradictions. If | B ( G ) | = 5, then G is isomorphic to Q or Q +4 , again acontradiction. In a plane graph G , a cycle C is called separating if both V (int( C )) and V (ext( C )) arenonempty. In this section we will discuss properties of separating cycles in G . Lemma 5.1.
Suppose T is a separating triangle of G and let I = int( T ) . Then(a) k V ( T ) , V ( I ) k ≥ ,(b) | I | ≥ ,(c) k x, V ( I ) k ≥ for all x ∈ V ( T ) , and(d) for all distinct x , y in V ( T ) , | N ( { x, y } ) ∩ V ( I ) | ≥ .Proof. If | I | ≤
2, then I contains a vertex v with d G ( v ) ≤
3, contrary to Lemma 4.4.Thus | I | ≥ I + := int[ T ] is triangulated and therefore k I + k = 3 | I + | − k I k ≤ | I | −
6. Thus k V ( T ) , V ( I ) k = k I + k − k T k − k I k ≥ | I | ) − − − (3 | I | −
6) = 6 . Thus (a) holds. As I + is triangulated and T is separating, every edge of T is containedin a triangle of I + other than T ; so (c) holds.If | ( N ( x ) ∪ N ( y )) ∩ V ( I ) | ≤
1, then | I | = 1 because G is a near plane triangulation.This contradicts (b). So (d) holds. Lemma 5.2.
Let C be a separating cycle in G such that V ( C ) ∩ A = ∅ . Assume X , Y are disjoint subsets of G such that X ∪ Y = ∅ , Y is collectable in G − X , and G [ X ∪ Y ]14 s connected. Let G = int[ C ] − ( X ∪ Y ) , G = ext( C ) − ( X ∪ Y ) , B = B ( G ) , B = B ( G ) , G ′ = ext[ C ] − ( X ∪ Y ) , A ′ = V ( C ) − ( X ∪ Y ) . If A ′ is usable in G andcollectable in G ′ , then | Y | + | B | + | B | < | X | + | B | + τ ( G ) ≤ | X | + | B | + 1 . In particular, | Y | < ( | X | + | B | − | B | − | B | if ( X ∪ Y ) ∩ B = ∅ , | X | + τ ( G ) − | B | otherwise.Proof. Since A ′ is usable, ( X ∪ Y ) ∩ V ( C ) = ∅ and so X ∪ Y lies in the infinite faceof G . Thus any special cycle of G is also a special cycle of G . Thus by Lemma 4.8, τ ( G ) = τ ( G ) = 0. By Lemma 4.2, in an optimal special cycle packing of G , at mostone cycle is type-c and there are no type-a or type-b cycles. Therefore τ ( G ) ≤ A ′ is collectable in G ′ , we have f ( G ; A ) ≥ f ( G ; A ′ ) + f ( G ; A ) + | Y | . On the other hand, ∂ ( G ) = ∂ ( G ) + ∂ ( G ) + 34 ( | X | + | Y | ) −
14 ( | B | + | B | − | B | − τ ( G )) . As f ( G ; A ′ ) ≥ ∂ ( G ) and f ( G ; A ) ≥ ∂ ( G ), we have ∂ ( G ) −
34 ( | X | + | Y | )+ 14 ( | B | + | B |−| B |− τ ( G )) ≤ f ( G ; A ′ )+ f ( G ; A ) ≤ f ( G ; A ) −| Y | . As f ( G ; A ) < ∂ ( G ), it follows that | Y | + | B | + | B | < | X | + | B | + τ ( G ) ≤ | X | + | B | + 1 . Note that if ( X ∪ Y ) ∩ B = ∅ , then τ ( G ) = 0. In this case, we have | Y | + | B | + | B | < | X | + | B | . If ( X ∪ Y ) ∩ B = ∅ , then B = B . In this case, we have | Y | + | B | < | X | + τ ( G ). Lemma 5.3.
Let C be a separating triangle of G . If C has no vertex in B ( G ) , theneither k v, V (ext( C )) k ≥ for all vertices v ∈ V ( C ) or k v, V (ext( C )) k ≥ for twovertices v ∈ V ( C ) .Proof. Suppose not. Let C = xyzx be a counterexample with the minimal area. Wemay assume that k x, V (ext( C )) k ≤ k y, V (ext( C )) k ≤
3. By Lemma 5.1(c), z has a neighbour w in I := int( C ). If w is the only neighbour of z in I , then byLemma 5.1(b), C ′ := xwyx is a separating triangle. However, w has only 1 neighbourin ext( C ′ ) and x has at most 3 neighbours in ext( C ′ ), contradicting the choice of C .Thus k z, V ( I ) k ≥ C , X = { z } and Y = ∅ . Then A ′ := { x, y } is usable in G := int[ C ] − z , A ′ is collectable in G ′ := ext[ C ] − z and B := B ( G ) ⊇ { x, y }∪· N I ( z ).So | B | ≥
4, and this contradicts Lemma 5.2.15 emma 5.4.
Let C be a separating induced cycle of length in G having no vertex in B ( G ) . Then exactly one of the following holds.(a) | B (int( C )) | ≥ .(b) | V (int( C )) | ≤ and every vertex in int( C ) has degree in G .Proof. Suppose that | B (int( C )) | ≤
3. By Euler’s formula, we have k int[ C ] k = 3 | V (int[ C ]) | − | V (int( C )) | + 5as G is a near plane triangulation. Then since C is induced, by Lemma 4.4,0 ≤ X v ∈ V (int( C )) ( d ( v ) − k int[ C ] k − k C k + k int( C ) k − | V (int( C )) | = (3 | V (int( C )) | + 5) − k int( C ) k − | V (int( C )) | = 1 − | V (int( C )) | + k int( C ) k . (5.1)Suppose that int( C ) has a cycle. Since | B (int( C )) | ≤
3, we deduce that B (int( C )) = xyzx is a triangle. By Euler’s formula applied on G [ V ( C ) ∪ B (int( C ))], we have k V ( C ) , B (int( C )) k = (3 · − − − , hence B (int( C )) is a facial triangle by Lemma 5.3. Therefore, x, y, z have degree 4, 4,5 in G by (5.1) and Lemma 4.4. Let w, w ′ ∈ V ( C ) be consecutive neighbours of x in V ( C ). From G , we can delete w and collect x, y, z . Let G ′ = G − { w, x, y, z } . If G ′ has an exposed special cycle, then the face of G ′ containing w has length at most 5,implying that k w, V (ext( C )) k ≤ C − w is a subpath of an exposed specialcycle of G ′ , as C is induced. Then we can delete w ′ and collect x, y, z, w , contradictingLemma 4.5. Therefore G ′ has no exposed special cycles. Then ∂ ( G ) = ∂ ( G ′ ) + 3 and f ( G ; A ) ≥ f ( G ′ ; A ) + 3 ≥ ∂ ( G ′ ) + 3 = ∂ ( G ), a contradiction.Therefore int( C ) has no cycles. Then k int( C ) k ≤ | V (int( C )) | −
1, and so in (5.1)the equality must hold. This means int( C ) is a tree and every vertex in int( C ) hasdegree 4 in G by Lemma 4.4. If int( C ) has at least 3 vertices, then let w be a vertex in V ( C ) adjacent to some vertex in int( C ). By deleting w , we can collect all the verticesin int( C ). Similarly we can choose w so that G ′ = G − w − V (int( C )) contains nospecial cycle, and that leads to the same contradiction. Thus we deduce (b). Lemma 6.1.
Each vertex in B has degree at most .Proof. Assume to the contrary that x ∈ B has d ( x ) ≥
6. Then deleting x exposes atleast 4 interior vertices. Apply Lemma 4.9 with X = { x } , Y = ∅ and s = 3, we obtaina contradiction. 16 x ′ xu vyy ′ Figure 4: The case 1 in the proof of Lemma 6.2. The dashed line may have othervertices and the gray region has other edges but no interior vertices.
Lemma 6.2.
Each vertex in B − A has degree .Proof. Suppose that there is a vertex x ∈ B − A with d ( x ) <
5. By Lemma 4.4, d ( x ) = 4. Consider two cases. Case 1: x has a neighbour y ∈ B − A . As | A | = 2, we have | B | ≥
4. As G is anear plane triangulation, there is a vertex z ∈ N ( x ) ∩ N ( y ) such that xyzx is a facialtriangle. As B has no chords by Lemma 4.11, ( N ( x ) ∩ N ( y )) ∩ B ( G ) = ∅ .Suppose there is z ′ ∈ N ( x ) ∩ N ( y ) − { z } . Since d ( x ) = 4 and G is a near planetriangulation, xzz ′ x is a facial triangle. Since d ( z ) ≥ T := yzz ′ y is a separating triangle. As d ( y ) ≤ y has a unique neighbour y ′ ∈ V (int( T )) and therefore both yy ′ zy and yy ′ z ′ y are facial triangles. By Lemma 5.1(b),int( T ) contains at least three vertices and so T ′ := zz ′ y ′ z is a separating triangle with k z, V (ext( T ′ )) k = 2 and k y ′ , V (ext( T ′ )) k = 1, contrary to Lemma 5.3. So N ( x ) ∩ N ( y ) = { z } .If d ( y ) = 5, then deleting z and collecting x and y exposes three vertices in ( N ◦ ( x ) ∪ N ◦ ( y )) − { z } , the resulting graph G ′ = G − { x, y, z } has | B ( G ′ ) | ≥ | B | + 1. ApplyLemma 4.9 with X = { z } , Y = { x, y } , and s = 1, we obtain a contradiction.Hence d ( y ) = 4. By repeating the same argument, we deduce that for all edges vv ′ ∈ B − A , we have (i) d ( v ) = 4 = d ( v ′ ) and (ii) | N ( v ) ∩ N ( v ′ ) | = 1.Let x ′ , y ′ be vertices such that N ◦ ( x ) = { x ′ , z } and N ◦ ( y ) = { y ′ , z } . As G is anear plane triangulation and B is chordless, G − B is connected. Let J = { x ′ , z, y ′ } .If V − B = J , then there exist b ∈ J and t ∈ ( V − B ) − J such that b and t areadjacent. Then deleting b and collecting x , y exposes vertices in ( J − { b } ) ∪ { t } . Let G ′ = G − { x, y, b } . Then | B ( G ′ ) | ≥ | B | + 1. With X = { b } , Y = { x, y } , and s = 1,this is in contrary to Lemma 4.9.Hence V − B = J .Let u , v be vertices in B so that uxyv is a path in B . Since G is a near planetriangulation, x ′ is adjacent to u and z , and y ′ is adjacent to v and z , see Figure 4.Then ux ′ zy, xzy ′ v are paths in G . As | A | = 2, u = v . If A = { u, v } , then B is a 4-cycleand as d ( x ′ ) , d ( y ′ ) ≥
4, we must have x ′ y ′ ∈ E ( G ), which implies that G is isomorphicto Q +2 and B is a special cycle, contrary to Lemma 4.8. Therefore A = { u, v } and bysymmetry, we may assume v / ∈ A . This implies d ( v ) = 4. Then v has another neighbourin J , and by the observation that x ′ is non-adjacent to y ′ , we deduce that v = z . Thus v is adjacent to x ′ , and x ′ is adjacent to y ′ .17 ′ ∈ Azya ∈ A x
Figure 5: The case 2 in the proof of Lemma 6.2. The gray region may have othervertices.Furthermore every vertex in B − { u, x, y, v } has degree at most 3, because B hasno chords and x ′ is the only possible interior neighbor. By Lemma 4.4, every vertex in B − { u, x, y, v } is in A . Then G is isomorphic to Q ++4 and B is a special cycle, contraryto Lemma 4.8. Case 2: N G ( x ) ∩ B ⊆ A . Then B = xaa ′ x . (Note that A = { a, a ′ } .) Since G is a nearplane triangulation and d ( x ) = 4, the neighbours of x form a path of length 3 from a to a ′ , say ayza ′ where a , y , z , a ′ are the neighbours of x . (See Figure 5.)If | N ◦ ( y ) | ≥
3, then deleting y and collecting x exposes at least three vertices in N ◦ ( y ). Let G ′ = G − { x, y } . Then | B ( G ′ ) | ≥ | B | + 2. With X = { y } , Y = { x } , and s = 2, this is in contrary to Lemma 4.9.Thus | N ◦ ( y ) | ≤ d ( y ) ≤
5. (Note that y may be adjacent to a ′ .) Bysymmetry, | N ◦ ( z ) | ≤ d ( z ) ≤ y is adjacent to a ′ , then z is non-adjacent to a and so d ( z ) = 4 by Lemma 4.4.Then T := yza ′ y is a separating triangle, as int( T ) contains a neighbour of z . Since d ( y ) ≤ d ( z ) = 4, we have | N ( { y, z } ) ∩ V (int( T )) | = 1, contrary to Lemma 5.1(d).So y is non-adjacent to a ′ . By symmetry, z is non-adjacent to a . As | N ◦ ( y ) | , | N ◦ ( z ) | ≤ d ( y ) , d ( z ) ≥ y and z have a unique common neighbour w and d ( y ) = d ( z ) = 4.Since G is a near plane triangulation, w is adjacent to both a and a ′ .If d ( w ) >
4, then deleting w and collecting y , z , x exposes at least one vertex andso | B ( G − { x, y, z, w } ) | ≥ | B | . With X = { w } , Y = { x, y, z } , and s = 0, this is incontrary to Lemma 4.9. This implies d ( w ) = 4, hence B ( G ) is a special cycle, contraryto Lemma 4.8. In this section we prove that | B | = 3. Lemma 7.1. If xy ∈ E ( B − A ) , then the following hold:(a) There are S := { x , x , u, y , y } ⊆ V − B and x ∗ , y ∗ ∈ B such that x ∗ x x uy is apath in G [ N ( x )] and xuy y y ∗ is a path in G [ N ( y )] .(b) d ( x ) , d ( u ) , d ( y ) ≥ .(c) The vertices x , x , u, y , y are all distinct.(d) | N ◦ ( { x , u } ) − S | ≤ and | N ◦ ( { y , u } ) − S | ≤ .(e) x y , x y , x y , ux , uy / ∈ E . x = y x x yy Figure 6: When x = y in the proof of Lemma 7.1(c). Gray regions may have othervertices. (f ) There is w ∈ ( N ( { x , u, y } ) ∩ B ) − { x, y } ; in particular G [ S ] is an induced path.(g) x , u / ∈ N ( x ∗ ) and y , u / ∈ N ( y ∗ ) .(h) w / ∈ { x ∗ , y ∗ } .Proof. (a) By Lemma 6.2, d ( x ) = 5 = d ( y ). By Lemmas 4.10 and 4.11, there are x ∗ , y ∗ ∈ B with N ( x ) ∩ B = { x ∗ , y } and N ( y ) ∩ B = { x, y ∗ } . As G is a near planetriangulation, there is u ∈ N ( x ) ∩ N ( y ). So (a) holds.(b) Assume d ( x ) = 4. Note that | N ◦ ( x ) ∩ N ◦ ( y ) | ≤ d ( v ) ≥ v ∈ V ( G ) byLemma 4.4. If x / ∈ { y , y } , then deleting u and collecting x , x , y exposes y , y (notethat it is possible that x ∈ { y , y } , so we do not count it as exposed). If x ∈ { y , y } ,say, x = y , then x / ∈ { y , y } and deleting u and collecting x , x , y exposes x , y .In both cases, we have | B ( G − { u, x , x, y } ) | ≥ | B | . With X = { u } , Y = { x , x, y } ,and s = 0, this in contrary to Lemma 4.9. Thus d ( x ) ≥ d ( u ) ≥ x = y . By symmetry, d ( y ) ≥
5. If d ( u ) = 4, then we candelete x , collect u, x, y , and expose y , y . This contradicts Lemma 4.9 applied with X = { x } , Y = { u, x, y } , and s = 0. So (b) holds.(c) Since d ( u ) ≥
5, we deduce x = y , and if x = y , then T := x x ux is aseparating triangle (See Figure 6), since d ( x ) ≥
5. As k x , V (ext( T )) k = 1 and k u, V (ext( T )) k = 2, this contradicts Lemma 5.3. So x = y . By symmetry, x = y .It remains to show that x = y . Suppose not. By (b), d ( x ) ≥
5, so C := x x uy x is a separating 4-cycle (See Figure 7). We first prove the following.For all u ′ ∈ V ( C ) − { u } , | N ( { u, u ′ } ) ∩ V (int( C )) | ≤
3. (7.1)Suppose not. Then deleting u , u ′ and collecting x , y exposes two vertices in V ( C ) −{ u, u ′ } and at least 4 vertices in int( C ). So | B ( G − { u, u ′ , x, y } ) | ≥ | B | − X = { u, u ′ } , Y = { x, y } , and s = 4. This proves (7.1).If u is adjacent to x , then C := x x ux and C := x uy x are both separating tri-angles by (b). Then | N ( { u, x } ) ∩ V (int( C i )) | ≥ i ∈ { , } by Lemma 5.1(d).Thus | N ( { u, x } ) ∩ V (int( C )) | ≥
4, contrary to (7.1). So u is non-adjacent to x .If x is adjacent to y , then C := ux y u is a separating triangle by (b). Then | N ( { u, x } ) ∩ V (int( C )) | ≥ k u, V (ext( C )) k = 2, Lemma 5.319 x = y x x yy Figure 7: When x = y in Lemma 7.1(c). Gray regions may have other vertices. ux x xx ∗ y ∗ yy y Figure 8: The situation in the proof of Lemma 7.1(d); x , x , u, y , y are all distinct.The gray region has other vertices.implies that k x , V (ext( C )) k ≥
4, hence | N ( { u, x } ) ∩ V (int( x x y x )) | ≥
2. Thus | N ( { u, x } ) ∩ V (int( C )) | ≥
4, contrary to (7.1). So C has no chord.By (b), C is a separating induced cycle of length 4 in G . By Lemma 5.4, either | B (int( C )) | ≥ | V (int( C )) | ≤ C ) has degree 4 in G .By (7.1), d ( x ) , d ( y ) ≤
6. If | B (int( C )) | ≥
4, then deleting u , x and collecting x , y , x , y exposes at least 4 vertices and therefore | B ( G − { u, x , x, y, x , y } ) | ≥ | B | + 2.This contradicts Lemma 4.9 applied with X = { u, x } , Y = { x, y, x , y } , and s = 2.Therefore we may assume 1 ≤ | V (int( C )) | ≤ C ) has degree4 in G . As x is non-adjacent to y , x has at least one neighbour in int( C ) and thereforeafter deleting x , we can collect all vertices in V (int( C )) and then collect x , y and u ,this contradicts Lemma 4.5. So (c) holds.(d) (See Figure 8.) If | N ◦ ( { x , u } ) − S | ≥
3, then deleting x , u and collecting x , y exposes x , y , y , and three other vertices and so | B ( G − { x , u, x, y } ) | ≥ | B | − X = { x , u } , Y = { x, y } , and s = 4, we obtain acontradiction. So we deduce that | N ◦ ( { x , u } ) − S | ≤
2. By symmetry, | N ◦ ( { y , u } ) − S | ≤ x is adjacent to u . By (b) and (d), d ( x ) = 5. Thus T := ux x u is aseparating triangle. Let w , w be the two neighbours of x other than x , x , u so that x w w u is a path in G . Such a choice exists because G is a near plane triangulation. As d ( w ) ≥ u has no neighbours in int( ux w w u ) by (d), x is adja-cent to w . As d ( w ) ≥ T ′ := x w w x is a separating triangle. Note that x x w x , x w w x , x w ux , and ux w u are facial triangles. Thus k w , V (ext( T ′ )) k = 1 and k w , V (ext( T ′ )) k = 2, contrary to Lemma 5.3. So x is non-adjacent to u . By symmetry, y is non-adjacent to u . 20uppose that x is adjacent to y . Let T ′′ := ux y u . By (b), d ( u ) ≥
5, so T ′′ is aseparating triangle. By (d), k z, V (int( T ′′ )) k ≤ z ∈ V ( T ′′ ). By Lemma 5.1, X z ∈ V ( T ′′ ) k z, V (int( T ′′ )) k = k V ( T ′′ ) , V (int( T ′′ )) k ≥ k z, V (int( T ′′ )) k = 2 for all z ∈ V ( T ′′ ). By (d), N ( u ) ∩ V (int( T ′′ )) = N ( x ) ∩ V (int( T ′′ )) = N ( y ) ∩ V (int( T ′′ )). Then u , x , y , and their neighbours inint( T ′′ ) induce a K subgraph, contradicting our assumption on G . Thus x is non-adjacent to y .Suppose that x is adjacent to y . Since x is non-adjacent to y , (b) and (d) implythat d ( y ) = 5. Let w , w be the two neighbours of y other than u , y , y such that uw w y is a path in G . By (d), N ◦ ( u ) − S ⊆ { w , w } . If u is adjacent to both w and w , then uw w u , uy w u , y w w y are facial triangles, implying that w has degree3, contradicting Lemma 4.4. Thus, as d ( u ) ≥ d ( u ) = 5. Since G is a near plane triangulation, x is adjacent to w and ux w u , uw y u are facialtriangles. If x w w y x is a separating cycle, then deleting w , w and collecting y , u , y , x exposes at least 4 vertices and so | B ( G − { w , w , y , u, y, x } ) | ≥ | B | − X = { w , w } , Y = { y , u, y, x } , and s = 2, we obtain acontradiction. So x w w y x is not a separating cycle. By Lemma 4.4, d ( w ) ≥ w is adjacent to x and d ( w ) = 4 = d ( w ). Then, deleting y and collecting w , w , u , y , x exposes 3 vertices and | B ( G − { y , w , w , u, y, x } ) | = | B | − X = { y } , Y = { w , w , u, y, x } , and s = 1, we obtain acontradiction. So x is non-adjacent to y . By symmetry, x is non-adjacent to y .(f) Suppose that none of x , u , y has neighbours in B − { x, y } . By (b), (d), and (e), d ( x ) = 5 = d ( y ). If | N ◦ ( { x , y } ) − { u }| ≥ u , x and collecting x , y , y exposes all vertices in N ◦ ( { x , y } ) − { u } andso | B ( G − { u, x , x, y, y } ) | ≥ | B | − X = { u, x } , Y = { x, y, y } , and s = 3, we obtain a contradiction. Thus | N ◦ ( { x , y } ) − { u }| ≤ x , y have the same set of neighbours in V ( G ) − ( B ∪ S ) by (c) and (e).Let w , w ′ be the neighbours of x (and also of y ) such that w ∈ V (int( uy w ′ x u )).Then w is the unique common neighbour of x , u , and y . By (d) and Lemma 4.4, w is adjacent to w ′ . Thus d ( w ) = 4. Deleting u and collecting w , x , y , x , y exposesat least 3 vertices including w ′ and so | B ( G − { u, w, x , y , x, y } ) | ≥ | B | − X = { u } , Y = { w, x , y , x, y } , and s = 1.Thus at least one vertex of x , u , and y is adjacent to a vertex in B − { x, y } . Then x is non-adjacent to y . By (e), G [ S ] is an induced path and (f) holds.(g) Suppose that x ∗ is adjacent to x . As d ( x ) ≥ T := x ∗ x x x ∗ is a separating triangle. Since d ( x ∗ ) ≤ x ∗ has a unique neighbour w ∈ V (int( T )). So w is adjacent to both x and x . As d ( w ) ≥ T ′ := wx x w is a separating triangle with k w, V (ext( T ′ )) k = 1 and k x , V (ext( T ′ )) k = 2,21ontrary to Lemma 5.3. So x ∗ is non-adjacent to x . By symmetry, y ∗ is non-adjacentto y .Suppose u is adjacent to x ∗ . As d ( x ) ≥ d ( x ∗ ) ≤ x ∗ has a unique neighbour w ∈ V (int( x ∗ x x ux ∗ )) adjacent to both x and u .By (b) and (d), w is adjacent to x . If uwx u is a separating triangle, then byLemma 5.1(d), | N ( { x , u } ) ∩ V (int( uwx u )) | ≥
2, hence | N ◦ ( { x , u } ) − S | ≥
3, contraryto (d). So uwx u is facial. As d ( x ∗ ) ≤ wx ∗ x w and wx ∗ uw are facial triangles. As d ( x ) ≥ T ′ := wx x w is a separating triangle. So k x , V (ext( T ′ )) k = 2and k x , V (ext( T ′ )) k = 2, contrary to Lemma 5.3. Thus u is non-adjacent to x ∗ . Bysymmetry, u is non-adjacent to y ∗ . So (g) holds.(h) Suppose that w = y ∗ . By (g), y ∗ is adjacent to x . Let C := y ∗ x uy y y ∗ and C ′ be the cycle formed by the path from x ∗ to y ∗ in B ( G ) − x − y together with thepath y ∗ x x x ∗ . Since G is a near plane triangulation and d ( y ) ≥
4, by (f) there is w ∈ N ( y ∗ ) ∩ N ( y ) ∩ V (int( C )). By Lemma 6.1, d ( y ∗ ) = 5, and therefore x is adjacentto w and x wy ∗ x is a facial triangle. Let y ∗∗ ∈ B be the neighbour of y ∗ other than y . Then x y ∗∗ y ∗ x is also a facial triangle in G . Because x is non-adjacent to x ∗ by(g), y ∗∗ = x ∗ . By (f) applied to yy ∗ , we have y ∗ ∈ A because uy y wx is not aninduced path in G . Thus x ∗ / ∈ A because | A | = 2. By Lemma 4.11, B ( G ) is chordless.Therefore by Lemma 6.2, d ( x ∗ ) = 5 and so k x ∗ , V (int( C ′ )) k = 2. By (b), (d), and (e),we have | N ◦ ( y ) − S | = 2. Deleting x , u and collecting x , x ∗ , y , y exposes at least 6vertices, including two neighbours of x ∗ in int( C ′ ) and two neighbours of y in int( C ).So | B ( G − { x , u, x, x ∗ , y, y } ) | ≥ | B | − X = { x , u } , Y = { x, x ∗ , y, y } , and s = 3, we obtain a contradiction. So w = y ∗ . By symmetry, w = x ∗ . Thus (h) holds. Lemma 7.2. | B | = 3 .Proof. For an edge e = xy ∈ E ( B − A ), let x ∗ , x , x , u, y , y , y ∗ be as in Lemma 7.1.By Lemma 7.1(h), | B | ≥ N ◦ ( u ) = { x , y } . Suppose not. By Lemma 4.10, | A | = 2, so at leastone vertex of { x ∗ , y ∗ } , say, y ∗ is not in A . By Lemma 7.1(f) applied to yy ∗ , we deducethat u is non-adjacent to vertices in N ◦ ( y ∗ ). Thus deleting u , y and collecting y , x , y ∗ exposes at least 6 vertices and so | B ( G − { u, y , y, x, y ∗ } ) | ≥ | B | − X = { u, y } , Y = { y, x, y ∗ } , and s = 3, we obtain a contradiction. So N ◦ ( u ) = { x , y } .Since d ( u ) ≥ u has at least one boundary neighbour z = x, y .Let B ( x, z ) be the boundary path from x to z not containing y , and B ( y, z ) be theboundary path from y to z not containing x . So B ( x, z ) and B ( y, z ) has only onevertex in common, namely z . One of B ( x, z ), B ( y, z ) has no internal vertex in A . Wedenote this path by P ( e, z ). We choose e = xy and z so that P ( e, z ) is shortest. Assume P ( e, z ) = B ( y, z ). Let e ′ = yy ∗ . Then e ′ ∈ E ( B − A ). Let y be the common neighbourof y and y ∗ and let z ′ = y, y ∗ be a boundary neighbour of y . Then P ( e ′ , z ′ ) is a propersubpath of P ( e, z ), and hence is shorter. This is in contrary to our choice of e and z . 22 ′ va x y zw CFigure 9: An illustration of Lemma 8.1(c).
In this section we complete the proof of Theorem 3.2. First we prove a lemma.
Lemma 8.1.
Suppose B = { a, a ′ , v } and axyza ′ is a path in G [ N ( v )] .(a) x is non-adjacent to z .(b) y is adjacent to neither a nor a ′ .(c) z is non-adjacent to a and x is non-adjacent to a ′ .(d) d ( x ) , d ( y ) , d ( z ) ≥ .(e) | N ◦ ( { x, y, z } ) | ≤ .(f ) x and z have a common neighbour w / ∈ { y, v } .(g) N ( x ) ∩ N ( z ) = { v, w, y } .Proof. (a) Suppose x is adjacent to z . As K * G , x is non-adjacent to a ′ or z is non-adjacent to a ; by symmetry, assume x is non-adjacent to a ′ . Since d ( y ) ≥ T := xyzx is a separating triangle. By Lemma 5.1, | V (int( T )) | ≥
3. Since k y, V (ext( T )) k = 1,Lemma 5.3 implies k x, V (ext( T )) k ≥
4, and so | N ◦ ( x ) ∩ V (ext( T )) | ≥ d ( y ) ≤
6, then deleting x , z and collecting v , y exposes at least 5 vertices from B (int( T )) and N ◦ ( x ) ∩ V (ext( T )) and so | B ( G − { x, z, v, y } ) | ≥ | B | − X = { x, z } , Y = { v, y } , and s = 4, we obtain a contradiction.Therefore, d ( y ) ≥
7. Then | N ◦ ( y ) ∩ V (int( T )) | ≥ x , y and collecting v exposes at least 7 vertices, and | B ( G − { x, y, v } ) | ≥ | B | − X = { x, y } , Y = { v } , and s = 6, we obtain a contradiction. So (a) holds.(b) Suppose y is adjacent to a . Then T := axya is a separating triangle, because d ( x ) ≥ x are facial. As d ( a ) ≤ a has a unique neighbour w in int( T ). As d ( w ) ≥ T ′ := xwyx is a separating triangle.Now k w, V (ext( T ′ )) k = 1, and k x, V (ext( T ′ )) k = 2, contrary to Lemma 5.3. Thus y isnon-adjacent to a . By symmetry, y is non-adjacent to a ′ . So (b) holds.(c) Suppose that z is adjacent to a . By (a), z is non-adjacent to x . As d ( x ) ≥ d ( a ) ≤ w ∈ ( N ( a ) ∩ N ( x ) ∩ N ( z )) − { v } , and xawx , wazw , aza ′ a are all facial triangles. (See Figure 9.) By (b), y = w . Since d ( y ) ≥ C := xyzwx is a separating cycle of length 4. Let I = int( C ). Then V = B ∪· V ( C ) ∪· V ( I ), (i) k x, V (ext( C )) k = 2, (ii) k y, V (ext( C )) k = 1, (iii) k z, V (ext( C )) k = 3,and (iv) k w, V (ext( C )) k = 1. 23f w is adjacent to y , then we apply Lemma 5.2 with C , X = { w } , and Y = ∅ . As y is adjacent to w , A ′ := { x, y, z } is usable in G := int[ C ] − w , and by (i–iii), A ′ iscollectable in G ′ := ext[ C ] − w . As | V ( G ) | = | B | = 3, τ ( G ) = 0. This is in contraryto Lemma 5.2.So using (a), C is chordless and x has at least one neighbour in int( C ).By Lemma 5.4, either | B ( I ) | ≥ | V ( I ) | ≤ I has degree 4 in G . If | V ( I ) | ≤ I has degree 4 in G , then V − { x } is A -good as wecan collect V ( I ) , y, w, z, v, a ′ , a . Then f ( G ; A ) ≥ | V ( G ) | − ≥ ∂ ( G ), a contradiction.Therefore | B ( I ) | ≥ uu ′ ∈ E ( C ) with | N ( { u, u ′ } ) ∩ V ( I ) | ≥
4, then we apply Lemma 5.2with C , X = { u, u ′ } and Y = ∅ . Now A ′ := V ( C ) − { u, u ′ } is usable in G :=int[ C ] − { u, u ′ } , A ′ is collectable in G ′ := ext[ C ] − { u, u ′ } , | B | ≥
6, and B = B . As G = B , τ ( G ) = 0. This is in contrary to Lemma 5.2. So | N ( { u, u ′ } ) ∩ V ( I ) | ≤ uu ′ ∈ E ( C ) and in particular, k u, V ( I ) k ≤ u ∈ V ( C ). This implies d ( y ) ≤ | N ( { x, y, z } ) ∩ V ( I ) | ≥
4, then we apply Lemma 5.2 with C , X = { x, z } and Y = { v, y } . Then Y is collectable in G − X , A ′ := { w } is usable in G := int[ C ] −{ x, y, z } , A ′ is collectable in G ′ := ext[ C ] −{ x, y, z } , | B | ≥
5, and B = B −{ v } . As ( X ∪ Y ) ∩ B = ∅ ,this is in contrary to Lemma 5.2.Therefore | N ( { x, y, z } ) ∩ V ( I ) | ≤
3. Since | B ( I ) | ≥
4, there exists a vertex u in B ( I ) − N ( { x, y, z } ). Then w is the only neighbour of u in C .Because G is a plane triangulation and d ( u ) ≥ w is adjacent to u . Since u is non-adjacent to x , y , z , we deduce that B ( I ) ∩ N ( w ) contains u and at least twoof the neighbours of u . Since k w, V ( I ) k ≤
3, we deduce that k w, V ( I ) k = 3. Since | N ( { x, w } ) ∩ V ( I ) | ≤
3, all neighbours of x in I are adjacent to w . Similarly allneighbours of z in I are adjacent to w . Since | B ( I ) | ≥
4, there is a vertex t in B ( I )non-adjacent to w . Then t is non-adjacent to x and z . Therefore t is adjacent to y . Bythe same argument, k y, V ( I ) k = 3 and every neighbour of x or z in I is adjacent to y .Thus, every vertex in N ( { x, z } ) ∩ V ( I ) is adjacent to both y and w .If k x, V ( I ) k ≥
2, then x , y , w , and their common neighbours in I together with a are the branch vertices of a K , -subdivision, using the path avy . So G is nonplanar,a contradiction. Thus, k x, V ( I ) k ≤ k z, V ( I ) k ≤
1. This means that d ( x ) ≤ B (int[ C ] − { x, y, w } ) = B ( I ) ∪ { z } .We apply Lemma 5.2 with C , X = { w, y } and Y = { x } . Then Y is collectablein G − X and A ′ = { z } is usable in G := int[ C ] − { w, x, y } , A ′ is collectable in G ′ := ext[ C ] − { w, x, y } , | B | = | B ( I ) ∪ { z }| ≥ B = B , and G = B . Thus τ ( G ) = 0 and this is in contrary to Lemma 5.2. Hence z is non-adjacent to a . Bysymmetry, x is non-adjacent to a ′ . Thus (c) holds.(d) Suppose d ( u ) ≤ u ∈ { x, y, z } . By Lemma 4.4, d ( u ) = 4. Let u ′ := y if u = y , u ′ := x otherwise. Then, deleting u ′ and collecting u , v exposes at least 2vertices in N ◦ ( { u, u ′ } ) by (a) and (c) and so | B ( G − { u, u ′ , v } ) | ≥ | B | − X = { u ′ } , Y = { u, v } , and s = 1, we obtain a contradiction.24 a a ′ x y zx x z z C ∗ Figure 10: Proof of Lemma 8.1(f). There are no vertices in int( C ∗ ).So (d) holds.(e) Suppose | N ◦ ( { x, y, z } ) | ≥
5. If d ( y ) ≤
6, then deleting x , z and collecting v , y exposes at least 5 vertices and so | B ( G − { x, z, v, y } ) | ≥ | B | − X = { x, z } , Y = { v, y } , and s = 4, we obtain a contradiction. Thus d ( y ) ≥
7. Then either | N ◦ ( { x, y } ) −{ z }| ≥ | N ◦ ( { z, y } ) −{ x }| ≥
5. We may assumeby symmetry that | N ◦ ( { x, y } ) − { z }| ≥
5. Then deleting x , y and collecting v exposesat least 6 vertices and so | B ( G − { x, y, v } ) | ≥ | B | − X = { x, y } , Y = { v } , and s = 5, we obtain a contradiction. So (e) holds.(f) Suppose N ( x ) ∩ N ( z ) = { y, v } . By (d), d ( x ) , d ( z ) ≥
5. By (e), | N ◦ ( { x, z } ) −{ y }| ≤ z is non-adjacent to a and x is non-adjacent to a ′ and by (a), x is non-adjacent to z . So each of x and z have exactly two neighbours in int( axyza ′ a ) and d ( x ) = d ( z ) = 5.Let x , x be those neighbours of x and z , z be those two neighbours of z . We mayassume that x x yz z is a path in G by swapping labels of x and x and swappinglabels of z and z if necessary. By (e), we have N ◦ ( y ) − { x, z } ⊆ { x , x , z , z } . As d ( x ) ≥ y is not adjacent to x because otherwise x x yx is a separating triangle,that will make a new interior neighbour of y by Lemma 5.1(c), contrary to (e). Bysymmetry, y is not adjacent to z . So x is adjacent to z as G is a plane triangulation.Therefore d ( y ) = 5.Let C ∗ := ax x z z a ′ a . Suppose that w ∈ N ( { x , x , z , z } ) ∩ V (int( C ∗ )). Thenby symmetry, we may assume w is adjacent to x or x . Deleting x , x and collecting x , y , v , z exposes w , z , z and so | B ( G − { x , x , x, y, v, z } ) | ≥ | B | − X = { x , x } , Y = { x, y, v, z } , and s = 2, we obtain a contradiction.Thus N ( { x , x , z , z } ) ∩ V (int( C ∗ )) = ∅ and therefore | G | = 10. See Figure 10.By Observation 3.1 applied to int[ C ∗ ], there is a vertex w ∈ { x , x , y , y } havingdegree at most 2 in int[ C ∗ ]. By symmetry, we may assume that w = x i for some i ∈ { , } . Since d ( x i ) ≤
4, after deleting x − i , we can collect x i , x , y , v , z , resulting anouterplanar graph, which can be collected by Observation 3.1. So, f ( G ; A ) ≥ ≥ ∂ ( G ),a contradiction. So (f) holds.(g) Suppose there is w ′ ∈ N ( x ) ∩ N ( z ) − { v, w, y } . Let C := xyzwx . We may assumethat w is chosen to maximize | V (int( C )) | . So w ′ is in V (int( C )) and together with ( a ),25e deduce that C is an induced cycle.We claim that y is non-adjacent to w ′ . Suppose not. As d ( y ) ≥ xw ′ yx or zw ′ yz is a separating triangle. By symmetry, we may assume xw ′ yx is a separatingtriangle. Thus | N ( { x, y } ) ∩ V (int( xw ′ yx )) | ≥ G is aplane triangulation, by (e), w is adjacent to w ′ and xww ′ x , zww ′ z , and yzw ′ y are facialtriangles. Thus k y, V (ext( xw ′ yx )) k = k w ′ , V (ext( xw ′ yx )) k = 2, contrary to Lemma 5.3.This proves the claim that y is non-adjacent to w ′ .Therefore k y, V (int( xyzw ′ x )) k = 2 by (d) and (e). Let y , y be two neighbours of y in int( xyzw ′ x ) such that xy y z is a path in G . Because G is a plane triangulation,by (e), w ′ is adjacent to both y and y and int( xw ′ zwx ) has no vertex. Then C is aseparating induced cycle of length 4 and | B (int( C )) | = 3, contrary to Lemma 5.4. So(g) holds. Proof of Theorem 3.2.
Let ( G ; A ) be an extreme counterexample. Then G is a nearplane triangulation. Let B = B ( G ) and B = B ( G ). By Lemmas 4.10 and 7.2, | B | = 3and | A | = 2. Let A = { a, a ′ } and v ∈ B − A . By Lemma 6.2, d ( v ) = 5. Let x , y , z bethe neighbours of v other than a , a ′ . As G is a plane triangulation, axyza ′ is a path.By Lemma 8.1(g), x and z have exactly one common neighbour w in G − v − y . Then C := xyzwx is a cycle of length 4. By symmetry and Lemma 8.1(d), we may assumethat d ( x ) ≥ d ( z ) ≥
5. By Lemma 8.1(e),( d ( x ) −
3) + ( d ( z ) − − ≤ | N ◦ ( { x, y, z } ) | ≤ . Therefore d ( z ) = 5 and d ( x ) = 5 or 6.We claim that y is non-adjacent to w . Suppose not. By Lemma 8.1(d), d ( y ) ≥ xywx and yzwy is a separating triangle. If both of them are separatingtriangles, then | N ( { x, y } ) ∩ V (int( xywx )) | ≥ | N ( { y, z } ) ∩ V (int( yzwy )) | ≥
2, byLemma 5.1(d). Therefore | N ◦ ( { x, y, z } ) | ≥ xywx and yzwy is a separating triangle.Suppose yzwy is a separating triangle. Then xywx is a facial triangle, and z hasa neighbour in int( yzwy ). As d ( z ) = 5, z has no neighbour in int( axwza ′ a ). There-fore, w is adjacent to a ′ , and wza ′ w is a facial triangle. Thus k y, V (ext( yzwy )) k = k z, V (ext( yzwy )) k = 2, contrary to Lemma 5.3. So yzwy is not a separating triangle.Therefore xywx is a separating triangle. By Lemma 5.1(d), int( xywx ) has at leasttwo vertices in N ◦ ( { x, y, z } ). By Lemma 8.1(d), z has a neighbour in int( axwza ′ a ).Then already we found four vertices in N ◦ ( { x, y, z } ). This means that x has no neigh-bours in int( axwza ′ a ) by Lemma 8.1(f). Hence k y, V (ext( xywx )) k = k x, V (ext( xywx )) k =2, contrary to Lemma 5.3. This completes the proof of the claim that y is non-adjacentto w .Therefore C is chordless by Lemma 8.1(a). By Lemma 8.1(d), d ( y ) ≥
5. Thus C isa separating induced cycle of length 4. By Lemma 5.4, either | B (int( C )) | ≥ | V (int( C )) | ≤ C ) has degree 4 in G .If | B (int( C )) | ≥
4, then deleting w , y and collecting z , v , x exposes at least 4vertices and so | B ( G − { w, y, z, v, x } ) | = | B | − X = { w, y } , Y = { z, v, x } , and s = 3, we obtain a contradiction.26herefore 1 ≤ | V (int( C )) | ≤ C ) has degree 4 in G .Deleting y and collecting all vertices in int( C ) and z , v , x exposes w and so B ( G − ( { y, z, v, x } ∪ V (int( C )))) ≥ | B | − X = { y } , Y = V (int( C )) ∪ { z, v, x } , and s = 0, we obtain a contradiction. References [1] Jin Akiyama and Mamoru Watanabe,
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