Planar Tur\'{a}n Numbers of Cycles: A Counterexample

The planar Turan number $\textrm{ex}_{\mathcal{P}}(C_{\ell},n)$ is the largest number of edges in an $n$-vertex planar graph with no $\ell$-cycle. For $\ell\in \{3,4,5,6\}$, upper bounds on $\textrm{ex}_{\mathcal{P}}(C_{\ell},n)$ are known that hold with equality infinitely often. Ghosh, Gy\"{o}ri, Martin, Paulo, and Xiao [arxiv:2004.14094] conjectured an upper bound on $\textrm{ex}_{\mathcal{P}}(C_{\ell},n)$ for every $\ell\ge 7$ and $n$ sufficiently large. We disprove this conjecture for every $\ell\ge 11$. We also propose two revised versions of the conjecture.


Introduction
The Turán number ex(n, H) for a graph H is the maximum number of edges in an n-vertex graph with no copy of H as a subgraph. Turán famously showed that ex(n, K ℓ ) ≤ (1− 1 ℓ−1 ) n 2 2 ; for example, see [1,Chapter 32]. The Erdős-Stone Theorem [8,Exercise 10.38] generalizes this result, by asymptotically determining ex(n, H) for every non-bipartite graph H: ex(n, H) = (1 − 1 χ(H)−1 ) n 2 2 + o(n 2 ); here χ(H) is the chromatic number of H. Dowden [3] considered the problem when restricting to n-vertex graphs that are planar. The planar Turán number ex P (n, H) ex P (n, H) for a graph H is the maximum number of edges in an n-vertex planar graph with no copy of H as a subgraph (not necessarily induced). This parameter has been investigated for various graphs H in [6] and [4]; but here we focus mainly on cycles. It is well-known that if G is an n-vertex planar graph with no triangle, then G has at most 2n − 4 edges; further, this bound is achieved by every planar graph with each face of length 4. Thus, ex P (n, C 3 ) = 2n − 4 for all n ≥ 4. Dowden [3] proved that ex P (n, C 4 ) ≤ 15(n−2) 7 for all n ≥ 4 and ex P (n, C 5 ) ≤ 12n−33 5 for all n ≥ 11. He also gave constructions showing that both of these bounds are sharp infinitely often.
We prove the first statement of Theorem 1 in Section 2, and sketch a proof of the second statement in Section 3. Our constructions modify that outlined by Ghosh et al. [5]. The main building blocks, which we call gadgets, are triangulations, in which every cycle has length less than ℓ. Clearly, a set of vertex-disjoint gadgets will have no C ℓ . To increase the average degree, we can identify vertices on the outer faces of these gadgets as long as we avoid creating cycles. We can also allow ourselves to create cycles among the gadgets as long as each created cycle has length more than ℓ. So we must find the way to do this most efficiently.
Our notation is standard, but for completeness we record a few things here. We let e(G) and n(G) denote the numbers of edges and vertices in a graph G. We write C ℓ for a cycle of length ℓ.

Disproving the Conjecture: a First Construction
To disprove Conjecture 1, we start with a planar graph in which each face has length ℓ + 1 (and each cycle has length at least ℓ + 1), and then we "substitute" a gadget for each vertex. As a first step, we construct the densest planar graphs with a given girth g, for each fixed g ≥ 6. We will also need that our dense graphs have maximum degree 3, as we require in the following definition. Definition 1. If G is a planar graph of girth g with each vertex of degree 2 or 3, and e(G) = g g−2 (n − 2), then G is a dense graph of girth g dense graph .
An easy counting argument shows that if G is an n-vertex dense graph of girth g, where n = (5g − 10) k 2 − g + 4 (for some positive even integer k), then G has 10k − 8 vertices of degree 3 and all other vertices of degree 2.
Lemma 2. Fix an integer g ≥ 3. If G is an n-vertex planar graph with girth g, then e(G) ≤ g g−2 (n − 2). For each g ≥ 6, this bound holds with equality infinitely often; specifically, it holds with equality if k is a positive even integer and n = (5g − 10) k 2 − g + 4. In fact, for each such k and n, there exists a graph G that attains this bound and that has every vertex of degree 2 or 3.
Proof of Lemma 2. Let G be an n-vertex planar graph with girth g. Denote by n, e, and f the numbers of vertices, edges, and faces in G. Every face boundary contains a cycle, 1 so every face boundary has length at least g. Thus, 2e ≥ gf . Substituting into Euler's formula and simplifying gives the desired bound: e ≤ g g−2 (n − 2). Now we construct graphs for which the bound holds with equality. Before giving our full construction, we sketch a simpler construction which has the desired properties except that it has maximum degree 6 (rather than each degree being 2 or 3, as we require). Begin with a 4-connected n-vertex planar triangulation with maximum degree 6. We will find a set M of edges such that every triangular face contains exactly one edge in M . To see that such a set exists, we consider the planar dual G * . Since G is a triangulation and 2-connected, G * is 3-regular. By Tutte's Theorem, G * contains a perfect matching M * (in fact, this was proved earlier by Petersen). The set M of edges in G corresponding to the edges of M * in G * has the desired property: each triangle of G contains exactly one edge of M .
To get the desired graph G ′ with each face of length g, we replace each edge of G not in M with a path of length ⌊(g + 1)/3⌋ and replace each edge of G in M with a path of length g − 2⌊(g + 1)/3⌋. Now each face of G ′ has length 2⌊(g + 1)/3⌋ + (g − 2⌊(g + 1)/3⌋) = g. Thus, for G ′ the inequality 2e(G ′ ) ≥ gf (G ′ ) in the initial paragraph holds with equality. So e(G ′ ) = g g−2 (n(G ′ ) − 2). Since each non-facial cycle of G has length at least 4, each non-facial cycle of G ′ has length at least g. Now we show how to also guarantee that each vertex of G ′ has degree 2 or 3. The construction is similar, except that it starts from a particular planar graph G with every face of length 6 and every vertex of degree 2 or 3. Again, we find a subset M of edges such that each face of G contains exactly one edge of M . To form G ′ from G, we replace each edge not in M with a path of length ⌊(g + 1)/6⌋ and we replace each edge in M with a path of length g − 5⌊(g + 1)/6⌋. Thus, each face of G ′ has length exactly 5⌊(g + 1)/6⌋ + (g − 5⌊(g + 1)/6⌋) = g.
It will turn out that each non-facial cycle of G has either (i) length at least 10 or (ii) length at least 8 and at least one edge in M . The corresponding non-facial cycle in G ′ thus has length at least g. In Case (ii) this follows from the calculation in the previous paragraph. In Case (i), when g ≥ 10 this holds because 10⌊(g + 1)/6⌋ ≥ 10(g − 4)/6 ≥ g. So consider Case (i) when g ≤ 9. Since each path in G ′ replacing an edge in G has length at least 1, each non-facial cycle in G ′ has length at least 10, which is at least g since g ≤ 9. Thus, what remains is to construct our graph G, specify the set of edges M , and check that each non-facial cycle in G either has length at least 10 or has length 8 and includes an edge in M .
We construct an infinite family of planar graphs G k on 10k−2 vertices, with 5k−2 faces (each of length 6), and with all vertices of degree 2 or 3; here k is an arbitrary positive even integer. Figure 1 shows G k . (By Euler's formula, each G k has 6 vertices of degree 2 and 10k − 8 vertices of degree 3.) Each of k "diagonal columns" contains 10 vertices, except for the first and last, which each contain one vertex fewer. We write v i,j to denote the jth vertex down from the top in column i, except that we start column 1 with , (k, 10)}}. The edge set consists of the boundary cycles of 4(k − 1) 6-faces in the hexagonal grid, k − 1 "curved edges" v i,1 v i−1,10 , when 2 ≤ i ≤ k, as well two "end edges" v 1,2 v 1,7 and v k,4 v k,9 . The matching M 9 . It is easy to check that the only vertices with degree 2 are We now show that every non-facial cycle has either (i) length at least 10 or (ii) length at least 8 and at least one edge in M . We denote by C 2 , C 3 , . . . , C 5k−5 the facial cycles that do not use any end-edge. Informally, C 2 is the "top left" of these (containing v 1,2 ), and subscripts increase as we move down the first diagonal and then wrap around toroidally with the facial cycle containing v 1,10 and two curved edges (see Figure 1), and continue on to the facial cycle containing v k,9 . Formally, each of these is C k , where X denotes its vertex set and k := max{j/2 : v i,j ∈ X} + 5 * min{i − 1 : v i,j ∈ X} + (|{i : v i,j ∈ X}| − 2). The facial cycles containing the left end-edge are C 0 and C 1 , and those containing the right end-edge are C 5k−4 and C 5k−3 .   Figure 1: The planar graph G k has 10k − 2 vertices, 15k − 6 edges, and every face of length 6. Every vertex of G k has degree 2 or 3 and every non-facial cycle either (i) has length at least 10 or (ii) has length 8 and includes a blue edge. The set of blue edges intersects every face exactly once.
Note that the edge-set of any non-facial cycle C is the symmetric difference of the edge-sets of the facial cycles "inside" (or "outside") of C. Consider first a non-facial cycle C that does not contain any end-edge. Pick the side of C that does not contain the right end-edge; take the symmetric difference of the edge-sets of the facial cycles on this side incrementally, in order of increasing subscripts. The symmetric difference of the first two facial cycles has size at least 10 and this size never decreases. Now consider the non-facial cycles that contain exactly one end-edge; by (rotational) symmetry, assume it is the left end-edge. For these cycles, take the symmetric difference incrementally as above for the side not containing the right end-edge; the symmetric difference of the first two facial cycles has size at least 8 and again this size never decreases.
Finally, consider a non-facial cycle C that contains both end-edges. Now take the symmetric difference incrementally as above for the side of C that includes C 1 ; the size of the symmetric difference is now initially at least 8, and never decreases until the final facial cycle (C 5k−4 or C 5k−3 ) is added and the symmetric difference is complete. The final facial cycle C ′ may reduce the size of the symmetric difference by at most 4, but the final symmetric difference still has size at least 12 (due to the position of C ′ relative to C 1 , and the fact that k ≥ 2).
To finish the proof, we should verify that |V (G ′ )| = (5g − 10) k 2 − g + 4, as claimed. By construction, each vertex of G ′ has degree 2 or 3. Each vertex with degree 3 in G ′ also has degree 3 in G, and we have exactly 10k − 8 of these. Let n, e, and f denote the numbers of vertices, edges, and faces in G ′ . Now summing degrees gives where the last two equalities hold as at the start of the proof. Thus, n = (5g − 10) k 2 − g + 4.
Definition 2. Let G be a 2-connected plane graph, with every vertex of degree 2 or 3. Let B be a plane graph with 3 vertices specified on its outer face. To substitute B into G substitute B into G we do the following. Subdivide every edge of G. For each vertex v in G, delete v from the subdivided graph and identify d(v) vertices on the outer face of a copy of B with the neighbors of v in the subdivided graph. Now we consider the result of substituting B into G, as in Definition 2.
Lemma 3. Let G be a plane graph; denote by n 2 and n 3 the numbers of vertices with degree 2 and 3 in G. Let B be a plane graph with n B vertices and e B edges, and with 3 vertices specified on its outer face. Form G ′ by substituting B into G. Now e(G ′ ) = (n 2 + n 3 )e B and n(G ′ ) = n 2 (n B − 1) + n 3 (n B − 3/2). Further, if G has no cycle of length ℓ or shorter, and B has no cycle of length ℓ, then G ′ has no cycle of length ℓ.
Proof. Each vertex in G gives rise to an edge-disjoint copy of B in G ′ ; thus e(G ′ ) = (n 2 + n 3 )e B . Each vertex of degree 2 in G contributes n B − 1 vertices to G ′ , since exactly two of its vertices lie in two copies of B in G ′ (and all others vertices lie in one copy of B). Similarly, each vertex of degree 3 in G contributes n B − 3/2 vertices to G ′ . Finally, assume G and B satisfy the hypotheses on the lengths of their cycles. Now consider a cycle C ′ in G ′ . If C ′ is contained entirely in one copy of B, then C ′ has length not equal to ℓ. If C ′ visits two or more copies of B, then C ′ maps to a cycle C in G with length no longer than the length of C ′ . Since each cycle in G has length longer than ℓ, we are done. Now suppose that we plan to substitute some plane graph B into a dense planar graph of girth ℓ + 1. Which B should we choose? Since B must not contain any ℓ-cycle, a natural choice is a triangulation of order ℓ − 1. Indeed, every such B yields a graph that attains the bound in Conjecture 1. This is Corollary 5, which follows from our next lemma. Proof. Let G be a dense graph of girth ℓ + 1 on n vertices, and let n 2 and n 3 denote, respectively, its numbers of vertices with degree 2 and 3. Recall from Lemma 2 (with g = ℓ + 1) that n = (5ℓ − 5) k 2 − ℓ + 3 for some even integer k, that n 3 = 10k − 8, and that n 2 = n − n 3 . Lemma 3 implies that e(G ′ ) = (n 2 + n 3 )e B = ne B and that n(G ′ ) = n 2 (n B − 1) + n 3 (n B − 3/2) = (n − n 3 )(n B − 1) + n 3 (n B − 3/2) = n(n B − 1) − n 3 /2. Now we show that e(G ′ ) = e B (ℓ−1) ℓ−1 ). The final equality comes from substituting for n 3 and simplifying (using that n = (5ℓ − 5) k 2 − ℓ + 3).
Corollary 5. The bound in Conjecture 1 holds with equality for each graph formed by substituting a triangulation on ℓ − 1 vertices into a dense graph of girth ℓ + 1.
Proof. This follows from the above lemma when B is a plane triangulation on ℓ − 1 vertices, so n B = ℓ − 1 and e B = 3(ℓ − 1) − 6 = 3ℓ − 9. We get To beat the bound of Conjecture 1, it will suffice to instead substitute into a dense graph of girth ℓ + 1 any triangulation with order larger than ℓ − 1, as long as it has each cycle of length at most ℓ − 1. This is because the conjectured average degree is less than 6, and is attained by substituting a triangulation of order ℓ − 1, as shown in Corollary 5. However, the average degree of a triangulation tends to 6 (from below) as its order grows. For each ℓ ∈ {3, . . . , 10}, every triangulation on ℓ vertices is Hamiltonian, i.e., it contains an ℓ-cycle. But for each ℓ ≥ 11, there exists a triangulation on ℓ vertices with no ℓ-cycle; this is a consequence of Lemma 6, which we prove next. (In fact, much more is true, as we show in Section 3.) Lemma 6. For every integer t ≥ 5, there exist a plane triangulation with 3t − 4 vertices and each cycle of length at most 2t, and a plane triangulation with 3t − 3 vertices and each cycle of length at most 2t + 1.
Proof. We start with a plane triangulation on t vertices. First we add into the interior of each face a new vertex, making it adjacent to each vertex on the face. Let A denote the set of vertices in the original triangulation, and let B denote the set of added vertices. Since |A| = t and |B| = 2t − 4, the resulting graph G 1 has order 3t − 4. Further, B is an independent set. Thus, on every cycle C, at least half of the vertices must be from A. Hence, C has length at most 2|A| = 2t.
Now we obtain G 2 by adding a single vertex inside some face of G 1 . It is easy to check that G 2 is a (3t − 3)-vertex triangulation with each cycle of length at most 2t + 1.
We have already outlined the proof of our main result. We let B be a plane triangulation with no ℓ-cycle, and with order at least ℓ, as guaranteed by Lemma 6. We simply substitute B into a dense graph of girth ℓ + 1. For completeness, we include more details in the proof of Theorem 7.
When ℓ ≥ 11 and ℓ is odd, let t 1 := ℓ−1 2 and n B 1 := 3t 1 − 4 = 3(ℓ−1) 2 − 4. We have t 1 ≥ 5; so by Lemma 6, there exists a plane triangulation B 1 with n B 1 vertices and with each cycle of length at most 2t 1 = ℓ − 1. By Euler's formula, e B 1 = e(B 1 ) = 3(3t 1 − 4) − 6 = 9t 1 − 18 = 9( ℓ−1 2 − 2). Form G ′ 1 by substituting B 1 into G. Lemma 3 implies that G ′ is a plane graph with no cycle of length ℓ, and that n( Hence, if ℓ ≥ 11 and ℓ is odd, then whenever k is positive and even and n = ((5ℓ − 5) k Now suppose ℓ ≥ 11 and ℓ is even. Let t 2 := ℓ 2 − 1 and n B 2 := 3t 2 − 3 = 3ℓ 2 − 6. Form G ′ 2 by substituting B 2 into G, where B 2 is a plane triangulation with n B 2 vertices and each cycle of B 2 has length at most 2t 2 + 1 = ℓ − 1. (The existence of B 2 is guaranteed by Lemma 6.) By Euler's formula, e B 2 = e(B 2 ) = 9ℓ 2 − 24. Similarly, it follows from Lemma 3 that G ′ 2 is a plane graph with no cycle of length ℓ, and that n( This completes our proof. Now for each ℓ ≥ 11, we extend the construction in Theorem 7 to all sufficiently large n (which will prove the first sentence of Theorem 1). Our general idea is to build a counterexample with order n ′ , larger than n, and delete vertices to get a counterexample of order precisely n. To see that this works, note that we can substitute different gadgets for different vertices in a sparse planar graph of girth ℓ + 1. As long as each gadget has more than ℓ vertices, we will beat the bound in Conjecture 1. In fact, we still beat the bound if a bounded number of gadgets have exactly ℓ vertices, and all other gadgets have more vertices (this is only needed in the case that ℓ ∈ {11, 12}, since that is when the gadget has precisely ℓ vertices). So we follow the construction in Theorem 7, and then repeatedly remove vertices of degree 3 (that lie in B in Lemma 6). We can remove up to t − 4 of these from each gadget. And the increase to the order of G ′ when we increase k in Theorem 7 is less than (5g − 10)(3t − 5). So it suffices that the number of vertices in the sparse planar graph G is greater than ⌈(5g − 10)(3t − 5)/(t − 4)⌉ ≤ 50(g − 2). This proves the first sentence of Theorem 1.

Denser Constructions and a Revised Conjecture
In this short section, we construct counterexamples to Conjecture 1 that are asymptotically much denser than those in the previous section. We also propose two revised versions of Conjecture 1. By iterating the idea in Lemma 6, Moon and Moser [9] constructed planar triangulations where the length of the longest cycle is sublinear in the order. These triangulations will serve as the gadgets in our denser constructions.

Theorem 8 ([9]
). For each positive integer k there exists a 3-connected plane triangulation G k with n(G k ) = 3 k+1 +5 2 and with longest cycle of length less than 7 2 n(G k ) log 3 2 .
Corollary 9. There exists a positive real D 1 such that for all integers ℓ ≥ 6 there exists a plane triangulation G ℓ with n(G ℓ ) ≥ D 1 ℓ lg 2 3 such that G ℓ has no cycle of length at least ℓ.
Chen and Yu [2] showed that Theorem 8 is essentially best possible.
Theorem 10 ( [2]). There exists a positive real D 2 such that every 3-connected n-vertex planar graph contains a cycle of length at least D 2 n log 3 2 .
We briefly sketch the Moon-Moser construction, which proves Theorem 8. For a more detailed analysis, we recommend Section 2 of [2]. Start with a planar drawing of K 4 , which we call T 1 . To form T i+1 from T i , add a new vertex v f inside each face f (other than the outer face), making v f adjacent to each of the three vertices on the boundary of f , see Figure 2. It is each to check that the order of T i is 3 + (1 + 3 + . . . + 3 i−1 ) ≈ 3 i 2 . To bound the length of the longest cycle in T i , we note that the vertices added when forming T j from T j−1 form an independent set, for each j. Thus, for any cycle in T i , at most half of the vertices were added at the final step. Of those added earlier, at most half were added at the penultimate step, etc. So the length of a longest cycle grows roughly by a factor of 2 at each step (while the order of T i grows roughly by a factor of 3).
To prove the second statement of Theorem 1, we substitute into a sparse planar graph of girth ℓ + 1 a gadget with no cycle of length ℓ, as guaranteed by Corollary 9. We suspect this construction is extremal. So we conclude with the following two conjectures, which are each best possible. Conjecture 2. Fix ℓ ≥ 7, let G be a dense graph of girth ℓ + 1, and let B be a n-vertex planar triangulation with no ℓ-cycle, where B is chosen to maximize n. If G ′ is formed by substituting B into G and n ′ := |V (G ′ )|, then ex P (n ′ , C l ) = |E(G ′ )|.
Proving Conjecture 2 seems plausible for some small values of ℓ. But proving it in general seems difficult. So we also pose the following weaker conjecture. Note that Conjecture 3 would be immediately implied by Conjecture 2 (together with Theorem 10).

Conjecture 3.
There exists a constant D such that for all ℓ and for all sufficiently large n we have ex P (n, C ℓ ) ≤ 3(Dℓ lg 2 3 −1) Dℓ lg 2 3 n.