Induced 2-degenerate Subgraphs of Triangle-free Planar Graphs

09/12/2017 ∙ by Zdeněk Dvořák, et al. ∙ University of Waterloo 0

A graph is k-degenerate if every subgraph has minimum degree at most k. We provide lower bounds on the size of a maximum induced 2-degenerate subgraph in a triangle-free planar graph. We denote the size of a maximum induced 2-degenerate subgraph of a graph G by α_2(G). We prove that if G is a connected triangle-free planar graph with n vertices and m edges, then α_2(G) ≥6n - m - 1/5. By Euler's Formula, this implies α_2(G) ≥4/5n. We also prove that if G is a triangle-free planar graph on n vertices with at most n_3 vertices of degree at most three, then α_2(G) ≥7/8n - 18 n_3.

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1 Introduction

A graph is -degenerate if every nonempty subgraph has a vertex of degree at most . The degeneracy of a graph is the smallest for which it is -degenerate, and it is one less than the coloring number. It is well-known that planar graphs are 5-degenerate and that triangle-free planar graphs are 3-degenerate. The problem of bounding the size of an induced subgraph of smaller degeneracy has attracted a lot of attention. In this paper we are interested in lower bounding the size of maximum induced 2-degenerate subgraphs in triangle-free planar graphs. In particular, we conjecture the following.

Every triangle-free planar graph contains an induced 2-degenerate subgraph on at least of its vertices.

Conjecture 1, if true, would be tight for the cube, which is the unique 3-regular triangle-free planar graph on 8 vertices (see Figure 1). For an infinite class of tight graphs, if is a planar triangle-free graph whose vertex set can be partitioned into parts each inducing a subgraph isomorphic to the cube, then does not contain an induced 2-degenerate subgraph on more than vertices.

Figure 1: The cube.

Towards Conjecture 1, we prove the following weaker bound.

Every triangle-free planar graph contains an induced 2-degenerate subgraph on at least of its vertices.

We believe the argument we use can be strengthened to give a bound , however the technical issues are substantial and since we do not see this as a viable way to prove Conjecture 1 in full, we prefer to present the easier argument giving the bound .

Triangle-free planar graphs have average degree less than , and thus they must contain some vertices of degree at most three. Nevertheless, they may contain only a small number of such vertices—there exist arbitrarily large triangle-free planar graphs of minimum degree three that contain only vertices of degree three. It is natural to believe that -degenerate induced subgraphs are harder to find in graphs with larger vertex degrees, and thus one might wonder whether a counterexample to Conjecture 1 could not be found among planar triangle-free graphs with almost all vertices of degree at least four. This is a false intuition—such graphs are very close to being -regular grids, and their regular structure makes it possible to find large -degenerate induced subgraphs. To support this counterargument, we prove the following approximate form of Conjecture 1 for graphs with small numbers of vertices of degree at most three.

If is a triangle-free planar graph on vertices with vertices of degree at most three, then contains an induced 2-degenerate subgraph on at least vertices.

Theorems 1 and 1 are corollaries of more technical results.

We say a graph is difficult if it is connected, every block is either a vertex, an edge, or isomorphic to the cube, and any two blocks isomorphic to the cube are vertex-disjoint.

We actually prove the following, which easily implies Theorem 1 since, by Euler’s formula, a triangle-free planar graph on at least three vertices satisfies .

If is a triangle-free planar graph on vertices with edges and difficult components, then contains an induced 2-degenerate subgraph on at least

vertices.

The proof of Theorem 1 is the subject of Section 2.

If is a plane graph, we let denote the minimum size of a set of faces such that every vertex in of degree at most three is incident to at least one of them.

We actually prove the following, which easily implies Theorem 1.

If is a triangle-free plane graph on vertices, then either is 2-degenerate or contains an induced 2-degenerate subgraph on at least

vertices.

The proof of Theorem 1 is the subject of Section 3.

Let us discuss some related results. To simplify notation, for a graph we let denote the size of a maximum induced subgraph that is -degenerate. Alon, Kahn, and Seymour [4] proved in 1987 a general bound on based on the degree sequence of . They derive as a corollary that if is a graph on vertices of average degree , then . Since triangle-free planar graphs have average degree at most four, this implies that if is triangle-free and planar then . Our Theorem 1 improves upon this bound.

For the remainder of this section, let be a planar graph on vertices.

Note that a graph is 0-degenerate if and only if it is an independent set. The famous Four Color Theorem, the first proof of which was announced by Appel and Haken [5] in 1976, implies that . In the same year, Albertson [2] proved the weaker result that , which was improved to by Cranston and Rabern [7]; the constant factor is the best known to date without using the Four Color Theorem. The factor is easily seen to be best possible by considering copies of .

If additionally is triangle-free, a classical theorem of Grőtzsch [9] says that is 3-colorable, and therefore . In fact, Steinberg and Tovey [15] proved that , and a construction of Jones [10] implies this is best possible. Dvor̆ák and Mnich [8] proved that there exists such that if has girth at least five, then .

Note that a graph is 1-degenerate if and only if it contains no cycles. In 1979, Albertson and Berman [3] conjectured that every planar graph contains an induced forest on at least half of its vertices, i.e. . The best known bound for for planar graphs is , which follows from a classic result of Borodin [6] that planar graphs are acyclically 5-colorable.

Akiyama and Watanabe [1] conjectured in 1987 that if additionally is bipartite then , and this may also be true if is only triangle-free. The best known bound when is bipartite is , which was proved by Wan, Xie, and Yu [16]. The best known bound when is triangle-free is , which was proved by Le [13] in 2016. Kelly and Liu [11] proved that if has girth at least five, then .

Kierstead, Oum, Qi, and Zhu [12] proved that if is a planar graph on vertices then , but the proof is yet to appear. A bound for of may be possible, which is achieved by both the octahedron and the icosahedron.

In 2015, Lukot’ka, Mazák, and Zhu [14] studied for planar graphs. They proved that . A bound for of may be possible, which is achieved by the icosahedron.

So far, bounds on for planar graphs have not been studied. However, as Lukot’ka, Mazák, and Zhu [14] pointed out, it is easy to see that every planar graph contains an induced outerplanar subgraph on at least half of its vertices. Since outerplanar graphs are 2-degenerate, this implies . Nevertheless, a bound of may be possible, which is achieved by the octahedron. If has girth at least five, may be as large as , which is achieved by the dodecahedron.

2 Proof of Theorem 1

In this section we prove Theorem 1. First we prove some properties of a hypothetical minimal counterexample (i.e., a plane triangle-free graph with the smallest number of vertices such that , where and is the number of difficult components of ).

2.1 Preliminaries

A minimal counterexample to Theorem 1 is connected and has no difficult components.

Proof.

Note that the union of induced 2-degenerate subgraphs from each component of is an induced 2-degenerate subgraph of . Thus if is not connected, then one of its components is a smaller counterexample, a contradiction.

Now suppose for a contradiction that has a difficult component. Since is connected, is difficult. Note that is not a single vertex, or else is not a counterexample. Suppose that contains a vertex of degree ; in this case, note that is a difficult graph. Since is a minimal counterexample, there exists a set that induces a 2-degenerate subgraph of size at least

But then induces a 2-degenerate subgraph in , contradicting that is a minimal counterexample.

Therefore, has minimum degree at least . Note that is not a cube, or else is not a counterexample. Since is difficult, we conclude that is not -connected and any end-block of is a cube. Let be the vertex set of an end-block of , and observe that is a difficult graph. Since is a minimal counterexample, there exists a set that induces a 2-degenerate subgraph of size at least

But then for any , induces a 2-degenerate subgraph in , contradicting that is a minimal counterexample. ∎

We will often make use of the following induction lemma. Let be a minimal counterexample to Theorem 1, and let . If every induced 2-degenerate subgraph of can be extended to one of by adding vertices, then

where is the number of difficult components of .

Proof.

Let induce a maximum 2-degenerate subgraph in . Since is a minimal counterexample,

(1)

Note that has no difficult components by Lemma 2.1. Since can be extended to induce a 2-degenerate subgraph in by adding vertices of ,

(2)

Combining (1) and (2) yields , which gives the desired inequality since both sides are integers. ∎

A minimal counterexample to Theorem 1 has no subgraph isomorphic to the cube that has fewer than six edges leaving.

Proof.

Let induce a cube in where and are 4-cycles and is adjacent to for each , as in Figure 1. Suppose for a contradiction that . Let induce a 2-degenerate subgraph in . First, we claim that there is some vertex such that induces a 2-degenerate subgraph in .

If has at least three neighbors not in , then induces a 2-degenerate subgraph in : Since is -degenerate, it suffices to verify that for every non-empty , there exists a vertex with at most two neighbors in . Since the cube is -edge-connected, there are at least three edges with one end in and the other end in . Since there are at most five edges leaving and at least three of them are incident with , at most two such edges are incident with vertices of . Consequently, , and thus indeed contains a vertex whose degree in is less than three.

By symmetry, we may assume no vertex in has more than two neighbors not in . If has two neighbors not in , an analogous argument using the fact that the only -edge-cuts in the cube are the neighborhoods of vertices shows that induces a 2-degenerate subgraph in , unless each of , , and has a neighbor not in . However, in that case it is easy to verify that induces a 2-degenerate subgraph in .

Hence, we may assume that each vertex of has at most one neighbor not in . Let be a set of size exactly containing all vertices of with a neighbor outside of . If contains all vertices of a face of the cube, then by symmetry we can assume that , and induces a 2-degenerate subgraph in . Otherwise, we have and , and since , by symmetry we can assume that and . However, then induces a 2-degenerate subgraph in .

This confirms that every set inducing a -degenerate subgraph of can be extended to a set inducing a -degenerate subgraph of by the addition of vertices. Let be the number of difficult components of . By Lemma 2.1, . Since is connected, it follows that consists of exactly difficult components, each connected by exactly one edge to the cube induced by . But then is a difficult graph, contradicting Lemma 2.1. ∎

A minimal counterexample to Theorem 1 has minimum degree at least three.

Proof.

Suppose not. Let be a vertex of degree at most two. Note that has degree at least one by Lemma 2.1. Note also that any induced 2-degenerate subgraph of can be extended to one of by adding . By Lemma 2.1, if has difficult components, then . But then is a difficult graph, contradicting Lemma 2.1. ∎

2.2 Reducing vertices of degree three

A cycle in a plane graph is separating if both the interior and the exterior of contain at least one vertex. The main result of this subsection is the following lemma. A minimal counterexample to Theorem 1 contains no vertex of degree three that is not contained in a separating cycle of length four or five.

For the remainder of this subsection, let be a minimal counterexample to Theorem 1, and suppose is a vertex of degree three that is not contained in a separating cycle of length four or five. Recall that a minimal counterexample is a plane graph, so has a fixed embedding.

The vertex has no neighbors of degree at least five.

Proof.

Suppose for a contradiction has a neighbor of degree at least five, and let . Note that any induced 2-degenerate subgraph of can be extended to one of including . By Lemma 2.1, the number of difficult components of is positive.

Let be a difficult component of . First, suppose contains a vertex of degree at most one. By Lemma 2.1, this vertex is adjacent to and , contradicting that is triangle-free. Therefore has an end-block isomorphic to the cube. Since is triangle-free and planar, has at most two neighbors in , and and do not both have two neighbors in . Hence , so has at most four edges leaving, contradicting Lemma 2.1. ∎

The vertex has no neighbors of degree three.

Proof.

Let and be the neighbors of , and suppose for a contradiction that has degree three.

First, let us consider the case has degree at least four (and thus exactly four by Claim 2.2). Note that any induced 2-degenerate subgraph of can be extended to one of including and . By Lemma 2.1, the number of difficult components of is positive.

Let be a difficult component of . Note that each leaf of is adjacent to and and not adjacent to by Lemma 2.1, since is triangle-free. Now if has at least two leaves, then is contained in a separating cycle of length four, a contradiction. Note also that is not an isolated vertex. Hence, contains an end-block isomorphic to the cube. If contains another end-block, then we can choose among the end-blocks isomorphic to the cube so that has at most five edges leaving, contradicting Lemma 2.1. Therefore is isomorphic to the cube. By Lemma 2.1, every neighbor of and is in , contradicting that is planar and triangle-free.

Therefore we may assume and symmetrically have degree three. Note that any induced 2-degenerate subgraph of can be extended to one of including and . By Lemma 2.1, the number of difficult components of is positive.

Let be a difficult component of . First, suppose is a tree. If is an isolated vertex, this vertex is adjacent to and by Lemma 2.1, but then is contained in a separating cycle of length four, a contradiction. Note that is not an edge, or else it is contained in a triangle with one of or , by Lemma 2.1. Similarly, is not a path, or else contains a triangle or a vertex of degree at most two. Therefore has at least three leaves. Since has minimum degree three and has only six edges leaving, is isomorphic to . In this case, is isomorphic to the cube, a contradiction.

Therefore we may assume is not a tree, so contains a block isomorphic to the cube. Let be a block in isomorphic to the cube with the fewest edges leaving. If contains an endblock different from , then at most five edges are leaving , contradicting Lemma 2.1. Therefore is isomorphic to the cube and all six edges leaving end in , contradicting that is planar and triangle-free. ∎

The vertex is not contained in a cycle of length four that contains another vertex of degree three.

Proof.

Suppose for a contradiction that and are neighbors of with a common neighbor of degree three that is distinct from , and let . By Claims 2.2 and 2.2, and have degree four. Note that any induced 2-degenerate subgraph of can be extended to one of including . By Lemma 2.1, if is the number of difficult components of , then .

Let and be difficult components of . Since there are only six edges leaving , we may assume without loss of generality that . Note that is not an isolated vertex by Lemma 2.1 since is triangle-free. If contains a leaf, then it is adjacent to either both and or both and by Lemma 2.1. In either case, is contained in a separating cycle of length four, a contradiction. Therefore contains an end-block isomorphic to the cube, contradicting Lemma 2.1. ∎

Every edge incident with is contained in a cycle of length four.

Proof.

Suppose for a contradiction is a neighbor of such that the edge is not contained in a cycle of length four. Let be the graph obtained from by contracting the edge into a new vertex, say , and observe that is planar and triangle-free.

Let induce a maximum-size induced 2-degenerate subgraph of . We claim that contains an induced 2-degenerate subgraph on at least vertices. If , then induces a 2-degenerate subgraph of on at least vertices, as claimed. Therefore we may assume . It suffices to show induces a 2-degenerate subgraph in . Given , we will show contains a vertex of degree at most two. If , then equals , which contains a vertex of degree at most two, as desired. Therefore we may assume . Note that contains a vertex of degree at most two. If , then since is triangle-free, is not adjacent to both and , and thus has degree at most two in , as desired. So we may assume has degree at most two in . Now at least one of and has degree at most two in , as desired.

Since is a minimal counterexample and , we have

where is the number of difficult components of . Furthermore, contains an induced 2-degenerate subgraph on at least vertices as argued, and thus

It follows that . Since is connected, is difficult. By Lemmas 2.1 and 2.1, cannot have an endblock not containing , and thus is isomorpic to the cube. But then either or has degree at most two in , which is a contradiction. ∎

We can now prove Lemma 2.2.

Proof of Lemma 2.2.

Suppose for a contradiction that contains such a vertex . By Claim 2.2, the vertex has a neighbor such that the edge is contained in two cycles of length four. Let and denote the other neighbors of . Since is contained in two cycles of length four, for , and have a common neighbor that is distinct from . By Claims 2.2 and 2.2, and have degree four. Since is not contained in a separating cycle of length four, , and are not adjacent, and and are not adjacent. By Claim 2.2, and have degree at least four. Let , and note that . Note also that any induced 2-degenerate subgraph of can be extended to one of by adding and . By Lemma 2.1, if is the number of difficult components of , . Let be a difficult component of such that the number of edges between and is minimum. Note that if or , then . Otherwise, .

Since is triangle-free and is not contained in a separating cycle of length at most 5, each vertex of has at most two neighbors in , and if it has two, these neighbors are either or . By Claim 2.2, if is a leaf of , we conclude that is adjacent to and . By planarity, has at most two leaves. Furthermore, if had two leaves, then all edges between and would be incident with and , and by planarity and absence of triangles, we would conclude that contains a vertex of degree two or a cube subgraph with at most four edges leaving, which is a contradiction. Hence, has an end-block isomorphic to the cube. Label the vertices of according to Figure 1. By Lemma 2.1, has at most one end-block isomorphic to the cube. Hence, either , or has precisely two end-blocks, one of which is a leaf and one of which is .

Suppose or . Then there are at most 5 edges between and . By Lemma 2.1, , so has at least two end-blocks. Therefore there are at most edges between and , so there are at most edges leaving , contradicting Lemma 2.1. Hence, .

By planarity, all edges between and are contained in one face of . Since is triangle-free and is not contained in a separating -cycle, there are at most 3 edges between and . If has a leaf, then as we observed before, the leaf is adjacent to and , and by planarity, all edges between and are incident with either or . By Lemma 2.1, the former is not possible, and in the latter case, there are edges between and , both and have a neighbor in , and consists of and the leaf. However, this is not possible, since is triangle-free. Consequently, is isomorphic to the cube.

Let us now consider the case that has a neighbor in . We may assume without loss of generality that is adjacent to . Since is not in a separating cycle of length at most five, and are not adjacent to or . Therefore and each have at most one neighbor in . By Lemma 2.1, one of and has two neighbors in , and we may assume without loss of generality it is . Since is planar and triangle-free, is adjacent to and , and is not adjacent to a vertex in . Therefore and have no neighbors in , so , a contradiction.

Hence, we may assume has no neighbor in . By Lemma 2.1, at least two of the vertices have two neighbors in . Suppose has two neighbors in . Then and have at most one, since does not have a common neighbor with or . Therefore has two neighbors in . Then and have no neighbors in , contradicting Lemma 2.1. Therefore we may assume by symmetry that and have two neighbors in . Then and have no neighbors in , again contradicting Lemma 2.1. ∎

2.3 Discharging

In this section, we use discharging to prove the following.

Every triangle-free plane graph with minimum degree three contains a vertex of degree three that is not contained in a separating cycle of length four or five.

For the remainder of this subsection, suppose is a counterexample to Lemma 2.3. We assume is connected, or else we consider a component of . Since is planar and triangle-free, it contains a vertex of degree at most three, and thus contains a separating cycle of length at most five. We choose a separating cycle of length at most five in so that the interior of contains the minimum number of vertices, and we let be the subgraph of induced by the vertices in and its interior. Note that has no chords since is triangle-free. By the choice of , we have the following.

The only separating cycle of of length at most five belonging to is .

Figure 2: A vertex of degree three.

Now we need the following claim about vertices of degree three in the interior of (see Figure 2). If some vertex has degree three, then , and has precisely one neighbor in and is incident to a face of length five whose boundary intersects in a subpath with three vertices.

Proof.

Suppose has degree three. Since is a counterexample, is contained in a separating cycle in of length four or five. By Claim 2.3, is not contained in , and since is chordless, contains a vertex not in . Since has length at most five, has at least one neighbor in . By Claim 2.3, has at most one neighbor in . Hence has precisely one neighbor in , as desired. Note that is a pair of nonadjacent vertices, or else contains a triangle. If is not incident to a face of length five containing three vertices of , or if , then contains a separating cycle of length at most five containing , contradicting Claim 2.3. ∎

Proof of Lemma 2.3.

For each , let , for each , let , and for each face , let . Note that by Euler’s formula, if denotes the set of faces of ,

Now we redistribute the charges in the following way, and we denote the final charge . For each , if has degree three and is adjacent to , let send one unit of charge to . Note that by Claim 2.3, for each , . Note also that for each , . The sum of charges is unchanged, i.e., it is .

First, suppose , and thus the sum of the charges is . Note that every vertex and face has precisely zero final charge, so every face has length precisely four. By Claim 2.3, every vertex has degree precisely four. Therefore every vertex in has degree precisely two. Since is separating, is not connected, a contradiction.

Therefore we may assume , so the sum of the charges is 2. Note that the outer face has final charge . Since

has an even number of odd-length faces, it follows that

has another face of length and final charge , and all other faces and vertices have zero final charge. In particular, all faces of distinct from and have length and each vertex in is adjacent only to vertices of degree three in . Using Claim 2.3, we conclude there are more than two vertices of with a neighbor in and at least two faces of length at least five in the interior of , a contradiction. ∎

Now the proof of Theorem 1 follows easily from Lemmas 2.1, 2.2, and 2.3.

3 Proof of Theorem 1

For the remainder of this section, let be a counterexample to Theorem 1 such that is minimum, and subject to that, is minimum, and let be a set of faces of such that every vertex in of degree at most three is incident to at least one of them.

3.1 Preliminaries

The graph has minimum degree three.

Proof.

Suppose not. Since is planar and triangle-free, has minimum degree at most three. Therefore we may assume contains a vertex of degree at most two. By assumption, there is a face in incident with . Therefore . Note that is not 2-degenerate or else is. By the minimality of , there exists of size at least such that is 2-degenerate. Now induces a 2-degenerate subgraph of on at least vertices, contradicting that is a counterexample. ∎

If is a triangle-free plane graph of minimum degree at least two such that , then has at least four vertices of degree two.

Proof.

Let be a face of incident to all the vertices in of degree at most three. We use a simple discharging argument. For each vertex , assign initial charge , and for each face , assign initial charge . Now let send one unit of charge to each vertex of degree at most three incident with , and denote the final charge . By Euler’s formula, the sum of the charges is . However, , and every other face has nonnegative final charge. Therefore the vertices have total final charge at most . Every vertex of degree at least three has nonnegative final charge, and every vertex of degree two has final charge . Therefore contains at least four vertices of degree two, as desired. ∎

Lemmas 3.1 and 3.1 imply that . A cylindrical grid is the Cartesian product of a path and a cycle.

If is a triangle-free plane graph such that , then either has minimum degree at most two, or is a cylindrical grid.

Proof.

Let be a triangle-free plane graph of minimum degree three such that . It suffices to show that is a cylindrical grid. Let and be faces of such that every vertex of degree at most three is incident to either or . Again we use a simple discharging argument. For each vertex , assign initial charge , and for each face , assign initial charge . Now for , let send one unit of charge to each vertex incident to , and denote the final charge . By Euler’s formula, the sum of the charges is . However, , and every other face and every vertex has nonnegative final charge. It follows that , and that every other face and every vertex has precisely zero final charge. Therefore the boundaries of and are disjoint, and every vertex incident with either or has degree three. Every other vertex has degree four, and every face that is not or has length four. It is easy to see that the only graphs with these properties are cylindrical grids, as desired. ∎

A triangle-free cylindrical grid on vertices contains an induced 2-degenerate subgraph on at least vertices.

Proof.

Let be a triangle-free cylindrical grid on vertices. The vertices of can be partitioned into sets that induce cycles of equal length such that for each , every vertex in has a unique neighbor in and in . Let be any set of vertices containing precisely one vertex in for each . Note that is an induced 2-degenerate subgraph on at least vertices, as desired. ∎

By Lemmas 3.1 and 3.1, we have .

We say a subset of the plane is -normal if it intersects only in vertices. If and are faces of , we define to be the smallest number of vertices contained in a -normal curve with one end in and the other end in . If is a face of and is a vertex of , we define to be the minimum of over all faces incident with .

Let be a -normal connected subset of the plane that intersects a face in or its boundary. Let be the set of vertices of contained in . Suppose that and are disjoint induced subgraphs of such that . If and , then .

Proof.

Note that there is a face of containing in its interior, and any vertex of of degree at most three that has degree at least four in is incident with this face. Therefore . By the minimality of , for each , there exists of size at least such that is 2-degenerate. But is 2-degenerate, and

Since is a counterexample, , so , as desired. ∎

Note that Lemma 3.1 together with Lemmas 3.1 and 3.1 imply that is connected.

All distinct faces satisfy .

Proof.

Suppose not. Then there is a set of at most vertices such that and are contained in the same face of . Therefore .

Let . Recall that , and thus , as otherwise the empty subgraph satisfies the requirements of Theorem 1. Note that is not 2-degenerate or else is an induced 2-degenerate subgraph on at least vertices, contradicting that is a counterexample. So by the minimality of , there exists of size at least such that is 2-degenerate, contradicting that is a counterexample. ∎

For each and , if , then induces a cycle in . Furthermore, every vertex in has at most one neighbor satisfying .

Proof.

We assume without loss of generality that is the outer face of . We use induction on . In the base case, is the set of vertices incident with . We prove this case as a special case of the inductive step.

By induction, we assume that for each , induces a cycle in and each vertex has at most one neighbor in . Let . Note that is the set of vertices incident with the outer face of . By Lemma 3.1, if then every vertex of has degree at least four in .

First we show that every has at most one neighbor in . Here the base case is trivial, so we may assume . Suppose for a contradiction that a vertex has two neighbors and in . Let and be the two paths in the cycle with ends and . Since is triangle-free, and have length at least two. For , note that the subgraph of drawn in the closure of the interior of the cycle has minimum degree at least two and at most three vertices (, , and ) of degree two. Therefore by Lemma 3.1, it contains a face .

For , there exists a simple -normal curve from to containing exactly one vertex from for each . Let consist of the vertices on and together with , and note that . Let , where is a face of and is a face of —neither nor is incident with a vertex of by Lemma 3.1, and for the same reason the vertices in incident with have degree at least three for . By Lemma 3.1, for we have either or contains vertices of degree at most two. In the latter case, the vertices of degree at most two in are incident with the outer face, and thus . This contradicts Lemma 3.1. Therefore every vertex of has at most one neighbor in , as claimed. Note that this implies every vertex of has degree at least three in .

Now we claim that is connected and does not contain a cut-vertex of . Suppose not. Then contains at least two end-blocks and . Note that and have minimum degree at least two and at most one vertex of degree two. Therefore by Lemma 3.1, . But there is a connected -normal subset of the plane intersecting in a set of vertices containing only one vertex of and at most two vertices from each for such that and are in different components of . Note that . By Lemma 3.1, , contradicting Lemma 3.1. Hence is connected, and does not contain a cut-vertex of , as claimed.

Since does not contain a cut-vertex of , the outer face of is bounded by a cycle, say . Now if does not induce a cycle in , then there is a chord of , say . Let and be paths in with ends at and such that . For , let be the graph induced by on the vertices in and its interior. Since has minimum degree two and at most two vertices of degree two, by Lemma 3.1, . But there is a connected -normal subset of the plane containing , , and intersecting in a set of vertices containing at most two vertices from each for . Note that . By Lemma 3.1, , contradicting Lemma 3.1. ∎

Consider a face and for , let be the cycle induced by according to Lemma 3.1. For and , let denote the number of neighbors of in (note that ) and . Let be the sum of over all faces such that , i.e., the faces between cycles and . Let if and otherwise, and let . Let us also define