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Outerplanar Turán number of a cycle

Ervin Győri Rényi Institute, Budapest, Hungary. Research partially supported by the NKFIH Grant 132696. E-mail: [email protected] Guilherme Zeus Dantas e Moura Haverford College, PA 19041, USA. E-mail: [email protected]; [email protected] Runtian Zhou Davidson College, Davidson, North Carolina 28035 E-mail: [email protected]
Abstract

A graph is outerplanar if it has a planar drawing for which all vertices belong to the outer face of the drawing. Let HH be a graph. The outerplanar Turán number of HH, denoted by ex𝒪𝒫(n,H)ex_{\mathcal{OP}}(n,H), is the maximum number of edges in an nn-vertex outerplanar graph which does not contain HH 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 ex𝒪𝒫(n,Ck)(2k5)(knk1)k22k1ex_{\mathcal{OP}}(n,C_{k})\leq\frac{(2k-5)(kn-k-1)}{k^{2}-2k-1}.

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 CkC_{k} to denote the cycle of kk vertices. We use nn-face to denote a face with nn edges. In particular, we use (4+)(4+)-innerface to denote an innerface of some outerplanar plane graph with at least 44 edges. We denote the vertex and the edge sets of a graph GG by V(G)V(G) and E(G)E(G) respectively. We also denote the number of vertices and edges of GG by v(G)v(G) and e(G)e(G) respectively. The minimum degree of GG is denoted δ(G)\delta(G).

The Turán number ex(n,H)ex(n,H) for a graph HH is the maximum number of edges in an nn-vertex graph with no copy of HH as a subgraph. The first result on the topic of Turán number was obtained by Mantel and Turán, who proved that the balanced complete rr-partite graph is the unique extremal graph of ex(n,Kr+1)ex(n,K_{r+1}) edges. The Erdős-Stone-Simonovits theorem [3, 2] then generalized this result and asymptotically determined ex(n,H)ex(n,H) for all nonbipartite graphs H:H: ex(n,H)=(11𝒳(H)1)(n2)+o(n2)ex(n,H)=(1-{1\over\mathcal{X}(H)-1}){n\choose 2}+o(n^{2}).

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 nn vertices have, without containing a given graph as a subgraph? Let \mathcal{H} be a set of graphs. The planar Turán number, ex𝒪𝒫(n,)ex_{\mathcal{OP}}(n,\mathcal{H}), is the maximum number of edges in an nn-vertex planar graph which does not contain any member of \mathcal{H} as a subgraph. When ={H}\mathcal{H}=\{H\} has only one element, we usually write ex𝒪𝒫(n,H)ex_{\mathcal{OP}}(n,H) instead. Dowden et al. [1] obtained the sharp bounds ex𝒫(n,K4)=3n6ex_{\mathcal{P}}(n,K_{4})=3n-6 for all n4n\geq 4, ex𝒫(n,C4)15(n2)7ex_{\mathcal{P}}(n,C_{4})\leq\frac{15(n-2)}{7} for all n4n\geq 4, and ex𝒫(n,C5)12n335ex_{\mathcal{P}}(n,C_{5})\leq\frac{12n-33}{5} for all n11n\geq 11. For k{4,5}k\in\{4,5\}, let Θk\Theta_{k} denote the graph obtained from CkC_{k} by adding a chord. Y. Lan et al. [8] showed that ex𝒪𝒫(n,Θ4)15(n2)5ex_{\mathcal{OP}}(n,\Theta_{4})\leq{15(n-2)\over 5} for all n4n\geq 4, ex𝒪𝒫(n,Θ5)5(n2)2ex_{\mathcal{OP}}(n,\Theta_{5})\leq{5(n-2)\over 2} for all n5n\geq 5. The bounds for ex𝒪𝒫(n,Θ4)ex_{\mathcal{OP}}(n,\Theta_{4}) and ex𝒪𝒫(n,Θ5)ex_{\mathcal{OP}}(n,\Theta_{5}) are sharp for infinitely many nn. The infinitely often sharp upper bound for ex𝒪𝒫(n,C6)ex_{\mathcal{OP}}(n,C_{6}) was proved by D. Ghosh et al. [5]. They proved ex𝒪𝒫(n,C6)5n142ex_{\mathcal{OP}}(n,C_{6})\leq{5n-14\over 2} for all n18n\geq 18. Recently, R. Shi et al. [9] and E. Győri et al. [7] independently proved the sharp bound of ex𝒪𝒫(n,C7)187n487ex_{\mathcal{OP}}(n,C_{7})\leq{18\over 7}n-{48\over 7} for all n60n\geq 60. E. Győri et al. [6] also asymptotically determined ex𝒫(n,{K4,C5})ex_{\mathcal{P}}(n,\{K_{4},C_{5}\}), ex𝒫(n,{K4,C6)ex_{\mathcal{P}}(n,\{K_{4},C_{6}) and ex𝒫(n,{K4,C7})ex_{\mathcal{P}}(n,\{K_{4},C_{7}\}).

A graph is outerplanar if it has a planar drawing for which all vertices belong to the outer face of the drawing. Let HH be a graph. The outerplanar Turán number of HH, denoted by ex𝒪𝒫(n,H)ex_{\mathcal{OP}}(n,H), is the maximum number of edges in an nn-vertex outerplanar graph which does not contain HH 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 n,kn,k be integers with k3k\geq 3. If 2nk12\leq n\leq k-1, then ex𝒪𝒫(n,Ck)=2n3ex_{\mathcal{OP}}(n,C_{k})=2n-3. If nkn\geq k, let λ=kn2k1k22k1+1\lambda=\lfloor{kn-2k-1\over k^{2}-2k-1}\rfloor+1. Then,

ex𝒪𝒫(n,Ck)={2nλ2λk3,if k|λ2nλ2λk2,otherwise.ex_{\mathcal{OP}}(n,C_{k})=\begin{cases}2n-\lambda-2\lfloor{\lambda\over k}\rfloor-3,&\text{if $k|\lambda$}\\ 2n-\lambda-2\lfloor{\lambda\over k}\rfloor-2,&\text{otherwise.}\end{cases}

In this paper, we use techniques of dual graph to prove the following theorem:

Theorem 2.

Let n,kn,k be integers with n2n\geq 2, k3k\geq 3. Then,

ex𝒪𝒫(n,Ck)(2k5)(knk1)k22k1.ex_{\mathcal{OP}}(n,C_{k})\leq\frac{(2k-5)(kn-k-1)}{k^{2}-2k-1}.

Moreover, for any fixed kk, we show the bound is sharp for infinitely many nn:

Theorem 3.

Let k3k\geq 3 be an integer. Then, for any integer nn s.t. nk1modk22k1n\equiv k-1\mod k^{2}-2k-1, we have

ex𝒪𝒫(n,Ck)=(2k5)(knk1)k22k1.ex_{\mathcal{OP}}(n,C_{k})=\frac{(2k-5)(kn-k-1)}{k^{2}-2k-1}.

It turns out that Theorem 3 agrees with Theorem 1.

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 G=(V,E)G=(V,E) be an outerplanar plane graph. Then, the following are equivalent:

  • GG is an edge-maximal outerplanar plane graph, i.e., no edge can be added to GG while preserving outerplanarity;

  • GG is 22-connected and all of its innerfaces are triangular;

  • |E|=2|V|3|E|=2|V|-3.

Proposition 6.

Let G=(V,E)G=(V,E) be an edge-maximal outerplanar plane graph. Let (uv)E(uv)\in E be an edge of GG on the outerface. Then, for every 1i|V|11\leq i\leq|V|-1, there exists a path between uu and vv of length ii. Moreover, for every 3i|V|3\leq i\leq|V|, there exists a cycle of length ii.

Definition 7 (Weak dual graph).

Let GG be an outerplanar plane graph. The weak dual graph of GG is the graph whose vertices correspond to the innerfaces of GG 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 GG by removing the outerface vertex. Because GG is outerplanar, its weak dual graph has no double edges or self-loops.

Proposition 8.

Let GG be an outerplanar plane graph. The weak dual graph of GG is acyclic.

Definition 9 (Triangular block).

Let GG be an outerplanar plane graph. A triangular block of GG is a outerplanar plane subgraph BB of GG such that

  • BB is an edge-maximal outerplanar plane graph, and;

  • there is no subgraph BB^{\prime} of GG such that

    • BB^{\prime} is an edge-maximal outerplanar plane graph, and

    • BB is a subgraph of BB^{\prime}.

We remark that triangular blocks may have 22 vertices, 11 edge and no innerface; these triangular blocks are called trivial triangular blocks.

Proposition 10.

Let GG be an outerplanar plane graph. Each edge of GG is in exactly one triangular block of GG. Hence, the collection of triangular blocks of GG is a partition of the edges of GG.

Definition 11 (Terminal triangular block).

Let GG be an outerplanar plane graph. A terminal triangular block of GG is a triangular block of GG that shares edges with at most one (4+)(4+)-innerface of GG.

Lemma 12.

Let GG be an outerplanar plane graph with at least one (4+)(4+)-innerface. Then, there exists an innerface of GG of size 4\ell\geq 4 such that at least 1\ell-1 of the triangular blocks containing each of its \ell edges are terminal triangular blocks.

Proof.

Let HH be the bipartite graph between set of vertices corresponding to (4+)(4+)-innerfaces of GG and the set of vertices corresponding to triangular blocks of GG, where we connect a (4+)(4+)-innerface vertex and a triangular block vertex if, and only if, the (4+)(4+)-innerface and the triangular block share an edge.

The graph HH can be obtained from the weak dual graph of GG by contracting all edges between triangles and adding vertices corresponding to trivial triangular blocks of GG 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 GG is acyclic, hence HH is also acyclic.

Let HH^{\prime} be obtained from HH by deleting all vertices corresponding to terminal triangular blocks, i.e., by deleting all triangular block vertices of degree at most 11 in HH. Note that the degree of any triangular block vertex in HH^{\prime} equals its degree in HH. Therefore, any triangular block vertex in HH^{\prime} has degree at least 22 and, consequently, it is not a leaf in HH^{\prime}.

Since HH is acyclic, HH^{\prime} is acyclic. Therefore, HH^{\prime} has a leaf. Since no triangular block vertex of HH^{\prime} is a leaf, there exists a (4+)(4+)-innerface vertex that is a leaf in HH^{\prime}. Therefore, at least 1\ell-1 of the \ell neighbors of this innerface vertex in HH correspond to terminal triangular blocks in GG, as desired. ∎

3 Proof of Theorem 2

Proof.

We use strong induction on nn. The base case is n=2n=2. Then,

e1(2k5)(knk1)k22k1=(2k5)(k1)k22k1=1+(k2)(k3)k22k1.e\leq 1\leq\frac{(2k-5)(kn-k-1)}{k^{2}-2k-1}={(2k-5)(k-1)\over k^{2}-2k-1}=1+{(k-2)(k-3)\over k^{2}-2k-1}.

Since k3k\geq 3, we know (k2)(k3)k22k10{(k-2)(k-3)\over k^{2}-2k-1}\geq 0 so the base case holds.

Let G=(V,E)G=(V,E) be a CkC_{k}-free outerplanar plane graph with nn (n2)(n\geq 2) vertices and ee edges. Assume, by induction hypothesis, that any CkC_{k}-free outerplanar graph GG^{\prime} with nn^{\prime} (2n<n)(2\leq n^{\prime}<n) vertices and ee^{\prime} edges satisfies

e(k22k1)(2k5)(knk1).e^{\prime}(k^{2}-2k-1)\leq(2k-5)(kn^{\prime}-k-1). (1)

Note that GG has no innerface of size exactly kk.

Case 1.

If GG has an innerface of size at least k+1k+1, then

e(k22k1)(2k5)(knk1).e(k^{2}-2k-1)\leq(2k-5)(kn-k-1). (2)
Proof.

Assume GG has an innerface of size \ell, with k+1\ell\geq k+1. Let v1,,vv_{1},\dots,v_{\ell} be the vertices on this face. Let G1=(V1,E1),,G=(V,E)G_{1}=(V_{1},E_{1}),\dots,G_{\ell}=(V_{\ell},E_{\ell}) be plane subgraphs such that ViVi+1={vi}V_{i}\cap V_{i+1}=\{v_{i}\} for all indices 1i<1\leq i<\ell, VV1={v}V_{\ell}\cap V_{1}=\{v_{\ell}\}, ViVj=V_{i}\cap V_{j}=\varnothing for all distinct non-consecutive indices i,ji,j, V1V=VV_{1}\cup\dots\cup V_{\ell}=V, EiEj=E_{i}\cap E_{j}=\varnothing for all distinct indices i,ji,j, and E1E=EE_{1}\cup\dots\cup E_{\ell}=E. Refer to Figure 1.

Refer to caption
G1G_{1}
Refer to caption
G2G_{2}
Refer to caption
GG_{\ell}
Refer to caption
Figure 1: Outerplanar plane graph GG and its innerface of size k+1\ell\geq k+1.

Let ni=|Vi|n_{i}=|V_{i}|, ei=|Ei|e_{i}=|E_{i}|. Then, we have n+=n1++nn+\ell=n_{1}+\cdots+n_{\ell} and e=e1++ee=e_{1}+\cdots+e_{\ell}. The plane graphs GiG_{i}’s are outerplanar and CkC_{k}-free. Hence,

ei(k22k1)(2k5)(knik1).e_{i}(k^{2}-2k-1)\leq(2k-5)(kn_{i}-k-1). (3)

Add these inequalities for all ii, then

e(k22k1)\displaystyle e(k^{2}-2k-1) (2k5)(k(n+)k)\displaystyle\leq(2k-5)(k(n+\ell)-k\ell-\ell) (4)
(2k5)(knk1),\displaystyle\leq(2k-5)(kn-k-1),

as desired. ∎

Case 2.

Assume all innerfaces of GG have size at most k1k-1. If GG has a (4+)(4+)-innerface, then

e(k22k1)(2k5)(knk1).e(k^{2}-2k-1)\leq(2k-5)(kn-k-1). (5)
Proof.

Lemma 12 implies that there exists an innerface of size \ell, with 4k14\leq\ell\leq k-1, such that at least 1\ell-1 of its \ell surrounding triangular blocks are terminal triangular blocks.

Denote by v1,v2,,vv_{1},v_{2},\dots,v_{\ell} the vertices of this innerface and denote by B1=(V1,E1),,B1=(V1,E1)B_{1}=(V_{1},E_{1}),\dots,B_{\ell-1}=(V_{\ell-1},E_{\ell-1}) the terminal triangular blocks surrounding this innerface such that vi,vi+1Viv_{i},v_{i+1}\in V_{i} for each index 1i<1\leq i<\ell. Let G=(V,E)G^{\prime}=(V^{\prime},E^{\prime}) be the plane subgraph consisting of the remaining edges of GG. Refer to Figure 2(a).

Refer to caption
B1B_{1}
v1v_{1}
v2v_{2}
v3v_{3}
v1v_{\ell-1}
vv_{\ell}
B2B_{2}
B1B_{\ell-1}
GG^{\prime}
(a) Outerplanar plane graph GG and its innerface for which all but one surrounding triangular blocks are terminal triangular blocks.
Refer to caption
B1B_{1}
Refer to caption
B2B_{2}
Refer to caption
B1B_{\ell-1}
v1vv_{1}\equiv v_{\ell}
v2v_{2}
v3v_{3}
v1v_{\ell-1}
Refer to caption
(b) Outerplanar plane graph HH^{*}.
Figure 2: Graphs considered on the proof of Case 2.

Let HH be the plane subgraph of GG induced by V1V1V_{1}\cup\cdots\cup V_{\ell-1}. The graph HH is outerplanar. Proposition 6 implies that, for any i|H|\ell\leq i\leq|H|, the graph HH contains a cycle of length ii. Therefore, |H|<k|H|<k .

Let plane graph H=(V,E)H^{*}=(V^{*},E^{*}) be obtained from HH by contracting the edge (v1v)(v_{1}v_{\ell}). The plane graph HH^{*} is outerplanar, and |H|=|H|1<k1|H^{*}|=|H|-1<k-1. Therefore, HH^{*} has no cycle of length kk. Refer to Figure 2(b).

Let n=|V|n^{\prime}=|V^{\prime}|, n=|V|n^{*}=|V^{*}|, e=|E|e^{\prime}=|E^{\prime}|, e=|E|e^{*}=|E^{*}|. Then, we have n+1=n+nn+1=n^{\prime}+n^{*} and e=e+ee=e^{\prime}+e^{*}. The plane graphs GG^{\prime} and HH^{*} are outerplanar and CkC_{k}-free. Hence,

e(k22k1)(2k5)(knk1),\displaystyle e^{\prime}(k^{2}-2k-1)\leq(2k-5)(kn^{\prime}-k-1), (6)
e(k22k1)(2k5)(knk1).\displaystyle e^{*}(k^{2}-2k-1)\leq(2k-5)(kn^{*}-k-1). (7)

Add these inequalities, then

e(k22k1)\displaystyle e(k^{2}-2k-1) (2k5)(k(n+1)2k2)\displaystyle\leq(2k-5)(k(n+1)-2k-2) (8)
<(2k5)(knk1)\displaystyle<(2k-5)(kn-k-1)

as desired. ∎

Case 3.

If GG is not 22-connected, then

e(k22k1)(2k5)(knk1).e(k^{2}-2k-1)\leq(2k-5)(kn-k-1). (9)
Proof.

Assume GG is not 22-connected, i.e., there exist plane subgraphs G1=(V1,E1)G_{1}=(V_{1},E_{1}) and G2=(V2,E2)G_{2}=(V_{2},E_{2}) satisfying |V1|>1,|V2|>1|V_{1}|>1,|V_{2}|>1, |V1V2|1|V_{1}\cap V_{2}|\leq 1, V1V2=VV_{1}\cup V_{2}=V, E1E2=EE_{1}\cup E_{2}=E.

Let n1=|V1|n_{1}=|V_{1}|, n2=|V2|n_{2}=|V_{2}|, e1=|E1|e_{1}=|E_{1}|, e2=|E2|e_{2}=|E_{2}|. Then, we have n+1n1+n2n+1\geq n_{1}+n_{2} and e=e1+e2e=e_{1}+e_{2}. The plane graphs G1,G2G_{1},G_{2} are outerplanar and CkC_{k}-free. Hence,

e1(k22k1)(2k5)(kn1k1),\displaystyle e_{1}(k^{2}-2k-1)\leq(2k-5)(kn_{1}-k-1), (10)
e2(k22k1)(2k5)(kn2k1).\displaystyle e_{2}(k^{2}-2k-1)\leq(2k-5)(kn_{2}-k-1). (11)

Add these inequalities, then

e(k22k1)\displaystyle e(k^{2}-2k-1) (2k5)(k(n+1)2k2)\displaystyle\leq(2k-5)(k(n+1)-2k-2) (12)
<(2k5)(knk1),\displaystyle<(2k-5)(kn-k-1),

as desired. ∎

Case 4.

Assume GG is 22-connected and all of its innerfaces have size 33. Then,

e(k22k1)(2k5)(knk1).e(k^{2}-2k-1)\leq(2k-5)(kn-k-1). (13)
Proof.

Proposition 5 implies that GG is edge-maximal outerplanar plane graph, and e=2n3e=2n-3. Proposition 6 implies that, for every 3in3\leq i\leq n, there is a cycle of length ii in GG. Hence, nk1n\leq k-1. Therefore,

e(k22k1)\displaystyle e(k^{2}-2k-1) =(2n3)(k22k1)\displaystyle=(2n-3)(k^{2}-2k-1) (14)
(2k5)(knk1),\displaystyle\leq(2k-5)(kn-k-1),

as desired. ∎

All possible outerplanar plane graphs GG are covered in the four cases above. Therefore, by strong induction, the result follows. ∎

4 Proof of Theorem 3: Extremal Graph Construction

G0G_{0}vvuuHHG1G_{1}G2G_{2}
Figure 3: Example of construction when k=5k=5
Proof.

We will do induction on the construction. Let k3k\geq 3 be a fixed integer. The base case G0G_{0} can be any edge-maximal outerplanar plane graph with k1k-1 vertices and 2k52k-5 edges. Now we construct outerplanar graph HH in this way: first we draw a face of size k+1k+1, then randomly pick kk edges from it, and for each of these edges we attach it to a copy of G0G_{0}. Let the other edge on that (k+1)(k+1)-face be called (uv)(uv). In short, HH has k22k+1k^{2}-2k+1 vertices and 2k25k+12k^{2}-5k+1 edges, with exactly kk triangular blocks of k1k-1 vertices, 11 trivial triangular block and no other triangular blocks. Given GmG_{m}, the inductive step from GmG_{m} to Gm+1G_{m+1} is the following: draw a copy of HH and merge (uv)(uv) with any edge on the boundary of GmG_{m}. Let Gm=(Vm,Em)G_{m}=(V_{m},E_{m}), then

|Vm|\displaystyle|V_{m}| =k1+(m+1)(k22k1)\displaystyle=k-1+(m+1)(k^{2}-2k-1)
|Em|\displaystyle|E_{m}| =(2k5)(k+1+mk)\displaystyle=(2k-5)(k+1+mk)
|Em|\displaystyle|E_{m}| =(2k5)(k|Vm|k1)k22k1.\displaystyle={(2k-5)(k|V_{m}|-k-1)\over k^{2}-2k-1}.

In Figure 3, we show an example of our construction when k=5k=5.

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