Improvement on the crossing number of crossing-critical graphs

The crossing number of a graph $G$ is the minimum number of edge crossings over all drawings of $G$ in the plane. A graph $G$ is $k$-crossing-critical if its crossing number is at least $k$, but if we remove any edge of $G$, its crossing number drops below $k$. There are examples of $k$-crossing-critical graphs that do not have drawings with exactly $k$ crossings. Richter and Thomassen proved in 1993 that if $G$ is $k$-crossing-critical, then its crossing number is at most $2.5k+16$. We improve this bound to $2k+6\sqrt{k}+44$.


Introduction
The crossing number cr(G) of a graph G is the minimum number of edge crossings over all drawings of G in the plane. In the optimal drawing of G, crossings are not necessarily distributed uniformly on the edges. Some edges can be more "responsible" for the crossing number than others. For any positive integer k, there exists a graph G whose crossing number is k, but it has an edge e such that G − e is planar.
On the other hand, Richter and Thomassen [RT93] conjectured that if cr(G) = k, then G contains an edge e such that cr(G − e) ≥ k − c √ k for some constant c. They observed that this bound would be optimal, as shown e.g. by the graph K 3,n . They managed to prove a much weaker bound. Namely, if cr(G) = k, then G contains an edge e such that cr(G − e) ≥ 2k/5 − 8.
A graph G is k-crossing-critical if cr(G) ≥ k, but cr(G − e) < k for any edge e of G.
The result and conjecture of Richter and Thomassen [RT93] can be rephrased as follows. They conjectured that if G is k-crossing-critical, then cr(G) ≤ k + c ′ √ k for some c ′ > 0 and this bound would be optimal. They proved that if G is k-crossing-critical, then cr(G) ≤ 2.5k + 16. This result has been improved in two special cases.
Lomelí and Salazar [LS06] proved that for any k there is an n(k) such that if G is k-crossing-critical and has at least n(k) vertices, then cr(G) ≤ 2k + 23.
Salazar [S00] proved that if G is k-crossing-critical and all vertices of G have degree at least 4, then cr(G) ≤ 2k + 35.
It is an easy consequence of the Crossing Lemma [A19] that if the average degree in a k-crossing-critical graph is large, then its crossing number is close to k [FT08]. More precisely, if G is k-crossing-critical and it has at least cn edges, where c ≥ 7, then cr(G) ≤ kc 2 /(c 2 − 29).
In this note, we obtain a general improvement.
We need a few definitions. Let G be a graph. We call a pair (C, v), where C is a cycle of G and v is a vertex of C, the cycle C with special vertex v. When it is clear from the context, which one is the special vertex, we just write C instead of (C, v).
Suppose C is a cycle with special vertex v. Let x be a vertex of C. An edge, adjacent to x but not in C, is hanging from x in short. Let l(C) be the length of C, that is, the number of its edges. For any vertex x, let d(x) denote the degree of x. Let h(C, v) = h(C) = u∈C,u =v (d(u) − 2), that is, the total number of hanging edges from all non-special vertices of C (with multiplicity).
A set of edges is independent if no two of them have a common endvertex.

The proof of Richter and Thomassen
In [RT93], the most important tool in the proof was the following technical result. In this section we review and analyze its proof. The algorithmic argument finds a cycle C recursively such that h(C) is small.
Theorem 0. [RT93] Let H be a simple graph with minimum degree at least 3. Assume that H has a set E of t edges such that H − E is planar. Then H has a cycle K with special vertex v such that h(K) ≤ t + 36.
The proof is by induction on t. The induction step can be considered as a process, which constructs a graph H * from graph H, and cycle K of H, either directly, or from cycle K * of H * . For convenience, for any planar graph H, define H * = ∅. In the rest of the paper we refer to this as the Richter-Thomassen procedure. The statement of Theorem 0 for t = 0 is the following.
Lemma 0. [RT93] Let H be a simple planar graph with minimum degree at least 3. Then H has a cycle K with special vertex v such that l(K) ≤ 5 and h(K) ≤ 36.
Here we omit the proof of Lemma 0. Suppose now that t > 0 and we have already shown Theorem 0. for smaller values of t. Let H be a simple graph with minimum degree at least 3. Assume that H has a set E of t edges such that H − E is planar, and let e = uw ∈ E. Let H ′ = H − e. We distinguish several cases. Case 1. H ′ has no vertex of degree 2. By the induction hypothesis, H ′ has a cycle K * with a special vertex v such that h(K * ) ≤ t + 35. If e is not a chord of K * , then K = K * with the same special vertex satisfies the conditions for H. Let H * = H ′ .
If e is a chord of K * , then K * + e determines two cycles, and it is easy to see that either one satisfies the conditions. If the new cycle K contains v, then v remains the special vertex. If K does not contain v, then we can choose the special vertex of K arbitrarily. Let H * = H ′ .
Case 2. H ′ has a vertex of degree 2. Clearly, only u and w can have degree 2. Suppress vertices of degree 2. That is, for each vertex of degree 2, remove the vertex and connect its neighbors by an edge. Let H ′′ be the resulting graph. It can have at most two sets of parallel edges.
Clearly, G cannot contain loops, as adding or removing a loop does not change the crossing number. Finally, suppose that e and f are parallel edges, both connecting x and y. Since G is k-crossing-critical, we have cr(G−e) ≤ k − 1. Take a drawing of the graph G − e with at most k − 1 crossings. Add the edge e, drawn very close to f . The obtained drawing of G has at most 2k − 2 crossings.
We assume in the sequel, that G is simple and all vertices have degree at least 3.
We introduce several parameters, for a complete list see the end of the paper. Let k ′ be the smallest integer with the property that we can remove k ′ edges from G so that the remaining graph is planar. Define the function f (x, y) = √ kx + y. Let (t, t ′ ) be the pair of numbers that minimizes the function f (t, t ′ ) = √ kt+t ′ subject to the following property: There exists a set E of t edges such that G − E is planar, and the set E contains at most t ′ independent edges. In the next lemma, part (i) is from [RT93], we repeat it here for completeness.
Lemma 1. The following two statements hold.
Proof of Lemma 1. (i) Since G is k-crossing-critical, G − e can be drawn with at most k − 1 crossings for any edge e. Remove one of the edges from each crossing in such a drawing. We removed at most k edges in total and got a planar graph.
where E contains at most t ′ independent edges, and G − E is planar. Apply the Richter-Thomassen procedure recursively starting with H 0 = G. We obtain a sequence of graphs H 0 , H 1 , . . . , H s , (s ≤ t) such that for 0 ≤ i ≤ s − 1, H * i = H i+1 , and H * s = ∅. The procedure stops with graph H s , where we obtain a cycle C s either directly, in cases 2.2.1, 2.2.2, and 2.3.1, or by Lemma 0, when H s is planar. In all cases, l(C s ) ≤ 5. Following the procedure again, we also obtain cycles C s−1 , . . . , C 0 of H s−1 , . . . , H 0 respectively such that 0 ≤ i ≤ s − 1, C * i = C i+1 . Let C 0 = C with special vertex v. Lemma 2. There is a cycle K of G such that l(K) + h(K)/2 ≤ t + 5 √ k + 45.
Proof of Lemma 2. The cycle K will be either C, or a slightly modified version of C. Let h = h(C), l = l(C). It is clear from the procedure that C does not have a chord in G. In particular, none of e 1 , e 2 , . . . , e t can be a chord of C. Now we partition E into three sets, E = E p ∪ E q ∪ E m , where E p is the subset of edges of E, which have exactly one endvertex on C (these are the hanging edges in E), E q = E ∩ C, E m is the subset of edges of E, which do not have an endvertex on C. Let p = |E p |, q = |E q |, m = |E m |. Let p ′ denote the number of edges of E p hanging from the special vertex v. By definition, and p ≥ p ′ . Consider the moment of the procedure, when we get cycle K from K * . If we get a new hanging edge e, then e ∈ E. Taking into account the initial cases in the procedure, that is, when we apply Lemma 0, or we have Cases 2.2.1, 2.2.2, or 2.3.1, we get the following easy observations. We omit the proofs.
Observation 1. (i) All but 36 edges of G − C adjacent to a non-special vertex of C are in E.
(ii) For all but at most 4 non-special vertices z of C, all edges of G − C adjacent to z, are in E. ✷ Return to the proof of Theorem 1. It follows from Observation 1 (i) that (2) Therefore, h + q + m ≤ p + q + m + 36 = t + 36.
Since all vertices have degree at least 3, and C does not have a chord, Suppose that l > t ′ + 6. Consider t ′ + 5 consecutive vertices on C, none of them being the special vertex v. By Observation 1 (ii), for at least t ′ + 1 of them, all hanging edges are in E p . Consider one of these hanging edges at each of these t ′ + 1 vertices. By the definition of t ′ , these t ′ + 1 edges cannot be independent, at least two of them have a common endvertex, which is not on C. Suppose that x, y ∈ C, z ∈ C, xz, yz ∈ E p . Let the cycle (C ′ , z) be formed by the xy arc of C, which does not contain the special vertex v, together with the path xzy. We have l(C ′ ) ≤ t ′ + 6, h(C ′ ) ≤ h(C). It is not hard to see that all previous statements on C are valid for C ′ as well. We can replace C by C ′ for the rest of the proof. Therefore, we can assume in the sequel that l(C) ≤ t ′ + 6. ( Now, at each vertex x of C, where all hanging edges belong to E p , take one such edge. The set of these edges In G ′′ , each edge of E ′ has an endvertex of degree one. Therefore, we can add all edges of E ′ to G ′ without losing planarity. Consequently, the graph The set F has at most |F | ≤ p + l + m − (l − 5) = p + 5 + m ≤ t + 5 edges by (1). That is Let F ′ ⊆ F be a maximal set of independent edges in F . To estimate |F ′ |, observe that apart from the edges in E m ∩ F ′ , all edges in F ′ are adjacent to a vertex of C. Moreover, at most p − p ′ − (l − 5) + 1 of them have one vertex on C, the rest have two. Therefore, |F ′ | ≤ p − p ′ − (l − 5) + 1 + (l − (p − p ′ − l + 6))/2 + m = (p − p ′ )/2 + 3 + m.
By the choice of the pair (t, t ′ ), Now evoking (2): This concludes the proof of Lemma 2. ✷ Now we can finish the proof of Theorem 1. By Lemma 2, we have a cycle K in G with special vertex v such that h(K)/2 + l(K) ≤ t + 5 √ k + 45. Let e be an edge of K adjacent to v. Since G was k-crossing-critical, the graph G − e can be drawn with at most k − 1 crossings. Let us consider such a drawing D. Let h = h(K), l = l(K).
Suppose the path K − e has cr crossings in D. Remove the edges of K from the drawing, and one edge from each crossing not on K − e. Together with e, we removed at most k + l − cr edges from G to get a planar graph. Therefore, l + k − cr ≥ k ′ . Combining it with Lemma 1 (ii) we have Consider the drawing D of G − e. We can add the missing edge e drawn along the path K − e on either side. The two possibilities together create at most h + 2cr crossings. Choose the one which creates fewer crossings. That makes at most h/2 + cr crossings.

Notations
Here we give a list of the parameters and their definitions, used in the proof.
k: G is k-crossing-critical k ′ : the smallest integer with the property that we can remove k ′ edges from G so that the remaining graph is planar.
(t, t ′ ): the pair of numbers that minimizes the function f (t, t ′ ) = √ kt + t ′ subject to the following property: There exists a set E of t edges such that G − E is planar, and the set E contains at most t ′ independent edges. p = |E p |: the number of edges in E that have exactly one endvertex on C. q = |E q |: the number of edges in E ∩ C. m = |E m |: the number of edges in E that do not have an endvertex on C. p ′ : the number of edges of E p hanging from the special vertex v of C. h = h(C) = h(C, v) = u∈C,u =v (d(u) − 2), the total number of hanging edges from all non-special vertices of C (with multiplicity). l = l(C): the length of C.