Outerplanar Turán number of a cycle
Abstract
A graph is outerplanar if it has a planar drawing for which all vertices belong to the outer face of the drawing. Let be a graph. The outerplanar Turán number of , denoted by , is the maximum number of edges in an -vertex outerplanar graph which does not contain as a subgraph. In 2021, L. Fang et al. determined the outerplanar Turán number of cycles and paths. In this paper, we use techniques of dual graph to give a shorter proof for the sharp upperbound of .
Keywords— Turán number, Outerplanar graph, Cycle
1 Introduction and Main Results
In this paper, all graphs considered are outerplanar, undirected, finite and contain neither loops nor multiple edges. More specifically, we study outerplanar plane graphs what are embeddings (drawings) of graphs in the plane such that the edge curves do not cross each other; they may share just endpoints, and that all vertices belong to the outerface. We use to denote the cycle of vertices. We use -face to denote a face with edges. In particular, we use -innerface to denote an innerface of some outerplanar plane graph with at least edges. We denote the vertex and the edge sets of a graph by and respectively. We also denote the number of vertices and edges of by and respectively. The minimum degree of is denoted .
The Turán number for a graph is the maximum number of edges in an -vertex graph with no copy of as a subgraph. The first result on the topic of Turán number was obtained by Mantel and Turán, who proved that the balanced complete -partite graph is the unique extremal graph of edges. The Erdős-Stone-Simonovits theorem [3, 2] then generalized this result and asymptotically determined for all nonbipartite graphs .
In 2016, Dowden et al. [1] initiated the study of Turán-type problems when host graphs are plane graphs, i.e., how many edges can a plane graph on vertices have, without containing a given graph as a subgraph? Let be a set of graphs. The planar Turán number, , is the maximum number of edges in an -vertex planar graph which does not contain any member of as a subgraph. When has only one element, we usually write instead. Dowden et al. [1] obtained the sharp bounds for all , for all , and for all . For , let denote the graph obtained from by adding a chord. Y. Lan et al. [8] showed that for all , for all . The bounds for and are sharp for infinitely many . The infinitely often sharp upper bound for was proved by D. Ghosh et al. [5]. They proved for all . Recently, R. Shi et al. [9] and E. Győri et al. [7] independently proved the sharp bound of for all . E. Győri et al. [6] also asymptotically determined , and .
A graph is outerplanar if it has a planar drawing for which all vertices belong to the outer face of the drawing. Let be a graph. The outerplanar Turán number of , denoted by , is the maximum number of edges in an -vertex outerplanar graph which does not contain as a subgraph. The topics of outerplanar Turán number was initiated by L. Fang et al. in [4] where they determined the outerplanar Turán numbers of cycles and paths.
Theorem 1.
([4]) Let be integers with . If , then . If , let . Then,
In this paper, we use techniques of dual graph to prove the following theorem:
Theorem 2.
Let be integers with , . Then,
Moreover, for any fixed , we show the bound is sharp for infinitely many :
Theorem 3.
Let be an integer. Then, for any integer s.t. , we have
2 Definitions and Preliminaries
Definition 4 (Outerplanar plane graph).
An outerplanar plane graph is an embedding(drawing) of a graph in which the edge curves do not cross each other and all vertices belong to the outerface.
Proposition 5.
Let be an outerplanar plane graph. Then, the following are equivalent:
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is an edge-maximal outerplanar plane graph, i.e., no edge can be added to while preserving outerplanarity;
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is -connected and all of its innerfaces are triangular;
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.
Proposition 6.
Let be an edge-maximal outerplanar plane graph. Let be an edge of on the outerface. Then, for every , there exists a path between and of length . Moreover, for every , there exists a cycle of length .
Definition 7 (Weak dual graph).
Let be an outerplanar plane graph. The weak dual graph of is the graph whose vertices correspond to the innerfaces of and two vertices are connected if and only if their corresponding innerfaces are adjacent. More precisely, the weak dual graph is obtained from the dual graph of by removing the outerface vertex. Because is outerplanar, its weak dual graph has no double edges or self-loops.
Proposition 8.
Let be an outerplanar plane graph. The weak dual graph of is acyclic.
Definition 9 (Triangular block).
Let be an outerplanar plane graph. A triangular block of is a outerplanar plane subgraph of such that
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is an edge-maximal outerplanar plane graph, and;
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there is no subgraph of such that
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is an edge-maximal outerplanar plane graph, and
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is a subgraph of .
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We remark that triangular blocks may have vertices, edge and no innerface; these triangular blocks are called trivial triangular blocks.
Proposition 10.
Let be an outerplanar plane graph. Each edge of is in exactly one triangular block of . Hence, the collection of triangular blocks of is a partition of the edges of .
Definition 11 (Terminal triangular block).
Let be an outerplanar plane graph. A terminal triangular block of is a triangular block of that shares edges with at most one -innerface of .
Lemma 12.
Let be an outerplanar plane graph with at least one -innerface. Then, there exists an innerface of of size such that at least of the triangular blocks containing each of its edges are terminal triangular blocks.
Proof.
Let be the bipartite graph between set of vertices corresponding to -innerfaces of and the set of vertices corresponding to triangular blocks of , where we connect a -innerface vertex and a triangular block vertex if, and only if, the -innerface and the triangular block share an edge.
The graph can be obtained from the weak dual graph of by contracting all edges between triangles and adding vertices corresponding to trivial triangular blocks of which is either a leaf (when it’s on the outerface), or the middle point of an edge in the weak dual (when it’s shared by two innerfaces). Proposition 8 implies that the weak dual graph of is acyclic, hence is also acyclic.
Let be obtained from by deleting all vertices corresponding to terminal triangular blocks, i.e., by deleting all triangular block vertices of degree at most in . Note that the degree of any triangular block vertex in equals its degree in . Therefore, any triangular block vertex in has degree at least and, consequently, it is not a leaf in .
Since is acyclic, is acyclic. Therefore, has a leaf. Since no triangular block vertex of is a leaf, there exists a -innerface vertex that is a leaf in . Therefore, at least of the neighbors of this innerface vertex in correspond to terminal triangular blocks in , as desired. ∎
3 Proof of Theorem 2
Proof.
We use strong induction on . The base case is . Then,
Since , we know so the base case holds.
Let be a -free outerplanar plane graph with vertices and edges. Assume, by induction hypothesis, that any -free outerplanar graph with vertices and edges satisfies
(1) |
Note that has no innerface of size exactly .
Case 1.
If has an innerface of size at least , then
(2) |
Proof.
Assume has an innerface of size , with . Let be the vertices on this face. Let be plane subgraphs such that for all indices , , for all distinct non-consecutive indices , , for all distinct indices , and . Refer to Figure 1.
Let , . Then, we have and . The plane graphs ’s are outerplanar and -free. Hence,
(3) |
Add these inequalities for all , then
(4) | ||||
as desired. ∎
Case 2.
Assume all innerfaces of have size at most . If has a -innerface, then
(5) |
Proof.
Lemma 12 implies that there exists an innerface of size , with , such that at least of its surrounding triangular blocks are terminal triangular blocks.
Denote by the vertices of this innerface and denote by the terminal triangular blocks surrounding this innerface such that for each index . Let be the plane subgraph consisting of the remaining edges of . Refer to Figure 2(a).
Let be the plane subgraph of induced by . The graph is outerplanar. Proposition 6 implies that, for any , the graph contains a cycle of length . Therefore, .
Let plane graph be obtained from by contracting the edge . The plane graph is outerplanar, and . Therefore, has no cycle of length . Refer to Figure 2(b).
Let , , , . Then, we have and . The plane graphs and are outerplanar and -free. Hence,
(6) | |||
(7) |
Add these inequalities, then
(8) | ||||
as desired. ∎
Case 3.
If is not -connected, then
(9) |
Proof.
Assume is not -connected, i.e., there exist plane subgraphs and satisfying , , , .
Let , , , . Then, we have and . The plane graphs are outerplanar and -free. Hence,
(10) | |||
(11) |
Add these inequalities, then
(12) | ||||
as desired. ∎
Case 4.
Assume is -connected and all of its innerfaces have size . Then,
(13) |
Proof.
All possible outerplanar plane graphs are covered in the four cases above. Therefore, by strong induction, the result follows. ∎
4 Proof of Theorem 3: Extremal Graph Construction
Proof.
We will do induction on the construction. Let be a fixed integer. The base case can be any edge-maximal outerplanar plane graph with vertices and edges. Now we construct outerplanar graph in this way: first we draw a face of size , then randomly pick edges from it, and for each of these edges we attach it to a copy of . Let the other edge on that -face be called . In short, has vertices and edges, with exactly triangular blocks of vertices, trivial triangular block and no other triangular blocks. Given , the inductive step from to is the following: draw a copy of and merge with any edge on the boundary of . Let , then
In Figure 3, we show an example of our construction when .
∎
References
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