On the Halin Turán number of short cycles

: A Halin graph is a graph constructed by embedding a tree with no vertex of degree two in the plane and then adding a cycle to join the tree’s leaves. The Halin Turán number of a graph F , denoted as ex H ( n, F ) , is the maximum number of edges in an n -vertex Halin graph. In this paper, we give the exact value of ex H ( n, C 4 ) , where C 4 is a cycle of length 4. We also pose a conjecture for the Halin Turán number of longer cycles.


Introduction
Let F be a fixed graph.A graph G is called F -free if it contains no isomorphic copy of F as a subgraph.For the graph F and a positive integer n, the Turán number of F , denoted by ex(n, F ), is the maximum number of edges in an n-vertex F -free graph, i.e., ex(n, F ) = max{e(G) : G is an n-vertex F -free graph}.
One of the classical results in extremal graph theory is the Turán's Theorem [1] , which gives the exact value ex(n, K r ), where K r is an r-vertex complete graph.This result is the generalization of the Mantel's Theorem [2] for the case of K 3 .A major breakthrough in the study of Turán number of graphs came in 1966, with the proof of the famous theorem by Erdős, Stone and Simonovits [3,4] .They determined an asymptotic value of the Turán number of any fixed non-bipartite graph F .In particular, they proved ex(n, F ) = + o(n 2 ), where χ(F ) is the chromatic number of F .Since these results, researchers have been interested in working on the Turán number of class of bipartite (degenerate) graphs and extremal problems in a particular family of graphs.In 2016, Dowden [5] initiated the study of Turán-type problems in the family of planar graphs.Definition 1.Let F be a fixed graph and n be a positive integer.The planar Turán number of F , denoted by ex P (n, F ), is the maximum number of edges an n-vertex F -free planar graph contains, i.e., ex P (n, F ) = max{e(G) : G is an n-vertex F -free planar graph}.
Extending the Dowden's result, Lan, Shi and Song [6] obtained an upper bound for ex P (n, C 6 ), and later Ghosh, Győri, Martin, Paulos and Xiao [7] , improved the bound and gave a sharp upper bound with some interesting constructions realizing their bound.However ex P (n, C k ) is still open for general k.We refer [8][9][10][11] for a quick survey and conjectures on planar Turán numbers of graphs.
Theorem 1. [7] For all n ≥ 18, Recently, Fang and Zhai [12] initiated the study of Turán numbers in the family of outerplanar numbers.For a positive integer n and fixed graph F , the outerplanar Turán number of F , denoted by ex OP (n, F ), is the maximum number of edges in an n-vertex outerplanar graph containing no isomorphic copy of F as a subgraph.They completely determined the outerplanar Turán numbers of cycles and paths.
In this paper, we initiate the study of Turán number of cycles in the family of Halin graphs.A Halin graph H is constructed as follows: Start with a tree T in which each non-leaf has degree at least 3, i.e., every non-leaf of T is with degree at least 3. Embed the tree in the plane in a planar fashion and then add new edges to form a cycle C containing all the leaves of T in such a way that the resulting graph H is planar.We write H = T ∪ C, and we call T and C respectively as characteristic tree and outer cycle of the Halin graph H.
Halin graphs were studied by Halin [13] .A Halin graph has at least four vertices.The wheel graph, W n , is an example of Halin graph with the characteristic tree being a star on n leaves.Halin graphs are, edge-minimal and 3-connected [14] .Every edge of a Halin graph is part of some Hamiltonian cycle [15] .

Definition 2.
Let n be a positive integer and F be a fixed graph.The Halin Turán number of F , denoted by ex H (n, F ), is the maximum number of edges in an n-vertex F -free Halin graph, i.e., ex H (n, F ) = max{e(H) : H is an n-vertex F -free Halin graph}.
Bondy and Lovasz [16] have shown that Halin graphs are almost pancyclic.More precisely, they showed that if a Halin graph H on n vertices does not have any vertex of degree three in its characteristic tree, then it has all cycles of length ℓ, where, 3 ≤ ℓ ≤ n.If the characteristic tree contains a vertex of degree three, then cycles of all lengths will still be there with a possible exception of an even-length cycle.In a different study, He and Liu explored the maximum count of short paths in a Halin graph, as discussed in [17] .
The almost pancyclic property of Halin graphs makes them interesting from a theoretical perspective, as it implies that these graphs are highly connected and can be used to model a wide variety of complex systems and phenomena.As a result, much research in this area focuses on developing efficient algorithms and techniques for analyzing the structure and properties of Halin graphs.
Concerning cycles, it is still interesting to study and distinguish the extremal graph structures and the Halin Turán number of cycles of even length.In this paper, we determine the exact value of the Halin Turán number of the 4-cycle, and later we pose our conjecture for longer cycles.The following theorem states our main result.
The following notations and terminologies are needed.Let G be a graph.We denote the vertex and the edge sets of G by V (G) and E(G) respectively.The number of vertices and edges in G respectively are denoted by v(G) and e(G).For a vertex v in G, the degree of v is denoted by d G (v).We may omit the subscript if the underlying graph is clear.The set of all vertices in G which are adjacent to v is denoted as N G (v) or simply N (v) when the underlying graph is clear.For the sake of simplicity, we use the terms k-cycle and k-path to mean a cycle of length k and a path of length k respectively.We denote a k-cycle with vertices we may describe v 1 and v k−1 as semi-pendant vertices of the path.For a plane graph G, the length of a cycle Let H be a Halin graph and T be its characteristic tree that is neither an interior nor a branching vertex.Sometimes we may call a leaf in T a pendant vertex.

Proof of Theorem 3
The following lemmas and observations are important to complete the proof of the theorem.

Lemma 1.
Let H be a C 4 -free Halin graph and T be its characteristic tree.For a longest path L in T , each semi-pendant vertex of L is a branching vertex and is adjacent to only two leaves.
, and hence a contradiction.

Lemma 2. Let H = T ∪ C be a Halin graph, and u 1 and u 2 be leaves in
since H is a Halin graph and the degree of each vertex is at least three.For each vertex u ∈ R, there is a unique leaf u ′ in T such that we have a (u, u ′ )-path with the set of interior vertices disjoint from R. We may call u ′ s as child-pendant vertices of u.Any (u 1 , u 2 )-path other than the edge u 1 u 2 must contain either u or some child-pendant vertex u ′ for each u ∈ R.This implies |C| ≥ |F |.Proof.Since u and v are non-leaf and H is a Halin graph, then d T (u), d T (v) ≥ 3. Therefore, we have vertices  Proof.We give extremal constructions to verify the bounds.First, we give constructions of the characteristic tree of the Halin graph, when n = 16, 17, and n = 18.For the sake of simplicity, we may call the trees as base-tree and denote them by T 16 , T  2, is an important component in describing the constructions.For simplicity reasons, we call it star.Notice the dark-spotted vertices in both the base-trees and the star.
For n ≡ 0 ( mod 3), the Halin graph H n 18 is obtained by having n−18 3 copies of the star and identifying any of the dark-spotted vertices of T 18 and the dark-spotted vertex of the star.Similarly, when n ≡ 1 ( mod 3) and n ≡ 2 ( mod 3), we respectively get H n 16 and H n 17 by having n−16 3 and n−17 3 copies of the star and identifying the dark-spotted vertices of the corresponding base-trees and the star.

It is easy to see that the
From Lemma 1, both v 1 and v k−1 are branching vertices, and each of them is adjacent to two leaves.Denote the leaf, other than v 0 , adjacent to v 1 by v ′ 0 .Denote also the leaf, other than v k , adjacent to v k−1 by v ′ k .Since each non-leaf in T is with a degree at least 3, there must be a vertex, say u, adjacent to v 2 such that either both v 0 v 1 and v 2 u or both v 1 v ′ 0 and v 2 u are incidents to the same bounded face in H. Without loss of generality assume the latter case holds.It can be seen that u can not be a leaf in T .Otherwise, (v 1 , v 2 , u, v ′ 0 , v 1 ) is a 4-cycle in H, which is a contradiction.Hence u is non-leaf in the characteristic tree.Therefore, d T (u) ≥ 3. From the assumption that L is of maximum length in T , u is adjacent to exactly two leaves, and say u 1 and u 2 .For a similar argument, v k−2 is adjacent to a non-leaf w, which is adjacent to two leaves w 1 and w 2 , see Figure 3.  Proof.Our proof relies on induction on the number of vertices.The base cases are shown in the upcoming section.Let L = (v 0 , v 1 , v 2 , . . ., v k ) be a longest path in T .It is easy to check that k ≥ 4. Next, we prove the following sequence of lemmas as part of the proof.
. Moreover, v is not a leaf in T .Indeed, suppose for contradiction v is a leaf.Since L is the longest path, then the two faces incident to the edge v 2 v are with size either 3 or 4. The latter, can not happen as H is C 4 -free.Thus we may assume both faces are with size three.Hence we get two triangles sharing the same edge v 2 v.However, this also results in a 4-cycle, which is a contradiction.Hence, each vertex in T adjacent to v 2 is a non-leaf.Since L is a longest path, each vertex adjacent to v 2 is a branching vertex.That means the vertex is adjacent to two pendant vertices.Therefore, H is obtained by identifying the dark-spotted vertex of n−7 3 copies of stars with v 2 .It can be checked that e(H) = 5 3 (n − 1), and this completes the proof of Claim 1.

Claim 2. If
We verify that each vertex v ∈ V (T )\S incident to v 2 or v 3 is a branching vertex.Indeed, without loss of generality assume v ∈ N (v 2 ).
Since L is a longest path in T , the faces incident to v 2 v and located on its left side must be either a 3-face or a 4-face.The latter can not happen, as H is a C 4 -free graph.Thus we may assume the face is a 3-face and let the leaf forming the 3-face be v ′ , i.e., the 3-face is (v 2 , v, v ′ , v 2 ).For the same reason, we have a leaf , which is a contradiction.Therefore, each vertex in V (T )\S adjacent to v 2 or v 3 is a branching vertex.This implies, H is obtained by identifying the black-spotted vertex of n−8 3 copies of the star to either v 2 or v 3 .It can be checked that e(H) = 5 3 (n − 2) + 1.This completes the proof of Claim 2. Claim 3.For k ≥ 6, then H meets either the conditions of Lemma 4 or the conditions of Lemma 5 or the conditions of Lemma 6.

Proof. Consider the longest path
As L is a longest path, v 1 is a branching vertex and hence it is adjacent to two leaves where v 0 is one of the two vertices.Let the other vertex be v ′ 0 .From the degree condition of Halin graph, d T (v 2 ) ≥ 3.Moreover, every vertex adjacent to v 2 is not a leaf.Since again L is a longest path, each vertex adjacent to v 2 must be a branching vertex.If v 3 is a semi-branching vertex of degree at least 4, then H satisfies the condition of Lemma 4 and we are done.So we may assume that v 3 is not a semi-branching vertex or a semi-branching vertex with d T (v 3 ) = 3.In the former case, the edge v 2 v 3 is an edge with the property that its end vertices are non-leaf and the two faces incident to the edge are with size at least 6, and hence H satisfies the conditions of Lemma 6.In the latter case, since L is a longest path in T , v 2 is not a semi-branching vertex.Hence, the path (v 2 , v 3 , v 4 ) is with a size of at least 6, and hence H meets conditions of Lemma 5.This completes the proof of Claim 3.

Conjectures and concluding remarks
As mentioned earlier in the beginning, Bondy and Lovász proved that a Halin graph is pancyclic if every non-leaf in its characteristic tree is of degree at least 4. It is also remarked that, if the characteristic tree contains a vertex of degree three, cycles of all lengths will still be in the graph with a possible exception of an evenlength cycle.The following is our conjecture concerning the sharp upper bound of the Halin Turán number of the 6-cycle.

Lemma 3 .
Let H be a Halin graph with a characteristic tree T .Let e = uv ∈ E(T ) such that both u and v are non-leaf in T .If F 1 and F 2 are the two bounded faces incident to e, then for a cycle C in T containing e, then |C| ≥ min{|F 1 |, |F 2 |}.

Lemma 4 . 2 . 5 .Lemma 6 .Lemma 7 . 5 3
Let H be an n-vertex C 4 -free Halin graph with the characteristic tree T .If T contains a semibranching vertex of degree at least 4, then there is an (n − 1)-vertex C 4 -free Halin graph H ′ such that e(H) = e(H ′ ) + 2.Proof.Let C be the outer cycle of H. Let v ∈ V (T ), with d H (v) ≥ 4, be a semi-branching vertex and u ∈ N T (v) be a leaf.Let the path (u 1 , u, u 2 ) be the portion of C in the clockwise direction and denote F 1 and F 2 as faces in H incident to the paths (v, u, u 1 ) and (v, u, u 2 ) respectively.It can be seen that either |F 1 | and |F 2 | is at least 5. Indeed, since H is a C 4 -free graph, no face is of size 4. On the other hand if|F 1 | = |F 2 | = 3, then (v, u 1 , u, u 2 , v)is in H and this contradicts the C 4 -free assumption of H. Now obtain the graph H ′ by deleting u and joining the vertices u 1 and u 2 with an edge.H ′ is a Halin graph with characteristic treeT ′ = T − u, as d T ′ (v) ≥ 3 and d T ′ (w) = d T (w) for every w ∈ V (T )\{v}.Let C ′ bethe characteristic tree and the outer cycle of H ′ .H ′ is C 4 -free as the bounded face, say F , incident to the edge u 1 u 2 is of size at least 5, and by Lemma 2 the boundary cycle of F is the smallest cycle containing u 1 u Lemma Let H be an n-vertex C 4 -free Halin graph with characteristic tree T .Let (u, v, w) be a path in T such that v is a semi-branching vertex with d T (v) = 3.If the bounded face incident to the path is with size at least 6, then there is an (n − 2)-vertex C 4 -free Halin graph H ′ such that e(H) = e(H ′ ) + 3. Proof.Let v ′ ∈ N (v) and F 1 , F 2 and F 3 as the faces incident to the paths (u, v, w), (u, v, v ′ ) and (v ′ , v, w) respectively.By assumption |F 1 | ≥ 6.Since H is C 4 -free and v is a semi-branching vertex, then |F 2 |, |F 3 | ≥ 5. Denote u ′ and w ′ as the leaves in T such that v ′ u ′ is incident to F 2 and v ′ w ′ is incident to the face F 3 .Let H ′ be a graph obtained from H by deleting v and adding the edges uw and u ′ w ′ .It can be checked that H ′ is an (n − 2)-vertex Halin graphs, with the two faces incident to uw with size at least 5 and at least 6, and hence by Lemma 2 H ′ contains no 4-cycle and e(H) = e(H ′ ) + 3. Let H be an n-vertex C 4 -free Halin graph with characteristic tree T .Let e ∈ E(T ) such that its end vertices are non-leaf in T .If the two faces incident to e are with size at least 6, then there is an (n − 1)-vertex C 4 -free Halin graph, H ′ , such that e(H) = e(H ′ ) + 1. Proof.Denote H = T ∪ C, where C is the outer cycle of H. Let e = vu and F 1 and F 2 be the two bounded faces in H incident to e. u and v by assumption are non-leaf, and hence d T (u) ≥ 3 and d T (v) ≥ 3. Let T ′ be the graph obtained after contracting e in T .Clearly, T ′ an (n − 1)-vertex tree and a leaf in T ′ is a leaf in T .Moreover, for each non-leaf vertex w ∈ V (T ′ ), d T ′ (w) ≥ 3. Therefore by contracting e in H we get a Halin graph H ′ = T ′ ∪ C. Since |F 1 |, |F 2 | ≥ 6, then by Lemma 3, for each cycle C containing e we have |C| ≥ 6.Hence, by contracting e, every cycle in H ′ is with no 4-cycle.Therefor, H ′ is an (n − 1)-vertex C 4 -free Halin graph.This completes the proof of Lemma 6.For n ≥ 16, we have ex H (n, C 4 ) (n − 3) + 3, 3|(n − 3).

Figure 3 .
Figure 3. Distribution of vertices on a longest path of a Halin graph.
are paths incident to F 1 and F 2 respectively.Moreover, we have vertices u ′ 1 , v ′ 1 and u ′ 2 , v ′ 2 , which are leaves in T such that e 1 = u ′ 1 v ′ 1 and e 2 = u ′ 2 v ′ 2 are edges incident to F 1 and F 2 respectively.Notice that, u ′ 1 can be u 1 and v ′ 1 can be v 1 , and similarly for u ′ 2 and v ′ 2 with u 2 and v 2 .Clearly, C contains either e 1 or e 2 , but not both.If C contains e 1 , then by Lemma 2, |C| ≥ |F 1 |.Moreover if C contains e 2 , the |C| ≥ |F 2 |.Therefore, |C| ≥ min{|F 1 |, |F 2 |}.
17and T 18 .Denote also the corresponding Halin graphs by H 16 , H 17 and H 18 respectively.It is easy to see the Halin graphs are C 4 -free (see Figure1).17 and T 18 as follows.The star K 1,3 , which is shown in Figure Figure 1.Characteristic trees of Halin graphs on 16, 17, and 18 vertices.Now let n ≥ 19.We define an n-vertex Halin graphs H n 16 , H n 17 and H n 18 based on the base-trees T 16 , T Therefore, for n ≡ 0 ( mod 3), n ≡ 1 ( mod 3) and n ≡ 2 ( mod 3), we have ex H (n, C 4 ) ≥ e(H n 18 ), ex H (n, C 4 ) ≥ e(H n 16 ), and ex H (n, C 4 ) ≥ e(H n 17 ) respectively.
Figure 2. Star K 1,3 .Observation 1.Let H be an n-vertex C 4 -fee Halin graph with a characteristic tree