Stability in the Erd\H{o}s--Gallai Theorem on cycles and paths, II

TL;DR

This paper strengthens Kopylov's stability theorem for the Erd ext{o}s--Gallai cycle length problem, characterizing the structure of near-extremal 2-connected graphs with no long cycles.

Contribution

It completes a stability theorem for the Erd ext{o}s--Gallai problem, identifying the structure of graphs close to extremal edge counts for odd cycle length constraints.

Findings

Characterizes 2-connected graphs near extremal edge counts

Identifies extremal graphs with maximum edges under cycle constraints

Provides tight bounds for the number of edges in such graphs

Abstract

The Erd\H{o}s--Gallai Theorem states that for $k \geq 3$, any $n$-vertex graph with no cycle of length at least $k$ has at most $\frac{1}{2}(k-1)(n-1)$ edges. A stronger version of the Erd\H{o}s--Gallai Theorem was given by Kopylov: If $G$ is a 2-connected $n$-vertex graph with no cycle of length at least $k$, then $e(G) \leq \max\{h(n,k,2),h(n,k,\lfloor \frac{k-1}{2}\rfloor)\}$, where $h(n,k,a) := {k - a \choose 2} + a(n - k + a)$. Furthermore, Kopylov presented the two possible extremal graphs, one with $h(n,k,2)$ edges and one with $h(n,k,\lfloor \frac{k-1}{2}\rfloor)$ edges. In this paper, we complete a stability theorem which strengthens Kopylov's result. In particular, we show that for $k \geq 3$ odd and all $n \geq k$, every $n$-vertex $2$-connected graph $G$ with no cycle of length at least $k$ is a subgraph of one of the two extremal graphs or $e(G) \leq \max\{h(n,k,3),h(n,k,\frac{k-3}{2})\}$. The upper bound for $e(G)$ here is tight.

Equations

$$h(n,k,a):=e(H_{n,k,a})=\left(\genfrac{}{}{0ex}{}{k-a}{2}\right)+a(n-k+a).$$

$$e(G)\le \max \{h(n,k,2),h(n,k,t)\}$$

$$h(n,k,t)-h(n,k,t-1)=\left\{\begin{array}[]{ll}n-t-3&\mbox{ if }k=2t+1,\\ n-t-5&\mbox{ if }k=2t+2.\end{array}\right.$$

$$\ell_{n,d}=\max\left\{\binom{n-d}{2}+d^{2},\binom{\lceil\frac{n+1}{2}\rceil}{2}+{\Big{\lfloor}\frac{n-1}{2}\Big{\rfloor}}^{2}\right\}.$$

$$d_{H/uv}(w)\ge d_H(w)-1$$

$$T(H/uv)\ge T(H)-1$$

$$x\in V_{\le h+1}(G^{\prime }).$$

$$v\in V_{\le h}(G)-x\ast y$$

$$e(H_{k,k,t})-e(F)\le h(k,k,t)-h(k,k,t-1)-1=t-4,$$

$$m=k.$$

$$e(G_k)\le r^2+\left(\genfrac{}{}{0ex}{}{k-r}{2}\right).$$

$$k=2t+1$$

$$\delta (G_j)\le j-k+r-1\text{mbox}and|V_{\le j-k+r}(G_j)|\ge j-k+r.$$

$$\delta (G_j)\le t-1\text{mbox}forallk\le j\le n,$$

$$\delta (G_j)\le j-k+r-1\le (n-k)+r-1\le \frac{t-1}{2}+\frac{t+6}{3}-1=\frac{5t+3}{6}<t.$$

$$e(G)\le r+r+t(2t-1-t)+\left(\genfrac{}{}{0ex}{}{t}{2}\right),$$

$$h(k,k,t-1)-e(G)\ge \left(\genfrac{}{}{0ex}{}{2t+1-(t-1)}{2}\right)+(t-1{)}^2-\left[r+r+t(2t-1-t)+\left(\genfrac{}{}{0ex}{}{t}{2}\right)\right]=t+2-2r,$$

$$e(G)\le r+r+t(2t-t)+\left(\genfrac{}{}{0ex}{}{t}{2}\right),\text{and}$$

$$h(k,k,t-1)-e(G)\ge \left(\genfrac{}{}{0ex}{}{2t+2-(t-1)}{2}\right)+(t-1{)}^2-[r+r+t(2t-t)+\left(\genfrac{}{}{0ex}{}{t}{2}\right)]=t+4-2r,$$

$$e(G)$$

$$\displaystyle\leq\big{[}(t-1)+(t-1)+\ldots+(t-1)\big{]}+h(k,k,t-1)$$

$$=(t-1)(n-k)+h(k,k,t-1)$$

$$=h(n,k,t-1),$$

$$e(G)$$

$$\le (n-k)(t-1)+h(k,k,t-1)$$

$$=h(n,k,t-1).$$

$$H$$

$$t+2\le \ell \le k-2$$

$$e(G^{\ast })\le \left(\genfrac{}{}{0ex}{}{\ell }{2}\right)+(n-\ell )(k-\ell )=h(n,k,k-\ell ).$$

$$e(G)>\max \{h(n,k,t-1),h(n,k,3)\}$$

$$e(G^{-})>\max \{h(|V(G^{-})|,k,t-1),h(|V(G^{-})|,k,3)\}.$$

$$e(G^{-})\ge e(G)-(t-1)>h(n-1,k,t-1)\ge h(n-1,k,3),$$

Full text

In this paper, we complete a stability theorem which strengthens Kopylov’s result. In particular, we show that for $k\geq 3$ odd and all $n\geq k$, every $n$-vertex $2$-connected graph $G$ with no cycle of length at least $k$ is a subgraph of one of the two extremal graphs or $e(G)\leq\max\{h(n,k,3),h(n,k,\frac{k-3}{2})\}$. The upper bound for $e(G)$ here is tight.

Mathematics Subject Classification: 05C35, 05C38.

Keywords: Turán problem, cycles, paths.

1 Introduction

One of the basic Turán-type problems is to determine the maximum number of edges in an $n$-vertex graph with no $k$-vertex path. Erdős and Gallai [3] in 1959 proved the following fundamental result on this problem.

Theorem 1.1** (Erdős and Gallai [3]).**

*Fix $n,k\geq 2$. If $G$ is an $n$-vertex graph that does not contain a path with $k$ vertices, then $e(G)\leq\frac{1}{2}(k-2)n$. *

When $n$ is divisible by $k-1$, the bound is best possible. Indeed, the $n$-vertex graph whose every component is the complete graph $K_{k-1}$ has $\frac{1}{2}(k-2)n$ edges and no $k$-vertex paths. Also, if $H$ is an $n$-vertex graph without a $k$-vertex path $P_{k}$, then by adding to $H$ a new vertex $v$ adjacent to all vertices of $H$ we obtain an $(n+1)$-vertex graph $H^{\prime}$ with $e(H)+n$ edges that contains no cycle of length $k+1$ or longer. Then Theorem 1.1 follows from another theorem of Erdős and Gallai:

Theorem 1.2** (Erdős and Gallai [3]).**

*Fix $n,k\geq 3$. If $G$ is an $n$-vertex graph that does not contain a cycle of length at least $k$, then $e(G)\leq\frac{1}{2}(k-1)(n-1)$. *

The bound of this theorem is best possible for $n-1$ divisible by $k-2$. Indeed, any connected $n$-vertex graph in which every block is a $K_{k-1}$ has $\frac{1}{2}(k-1)(n-1)$ edges and no cycles of length at least $k$. In the 1970’s, some refinements and new proofs of Theorems 1.1 and 1.2 were obtained by Faudree and Schelp [4, 5], Lewin [9], Woodall [10], and Kopylov [8] – see [7] for more details. The strongest version was proved by Kopylov [8]. His result is stated in terms of the following graphs. Let $n\geq k$ and $1\leq a<\frac{1}{2}k$. The $n$-vertex graph $H_{n,k,a}$ is as follows. The vertex set of $H_{n,k,a}$ is the union of three disjoint sets $A,B,$ and $C$ such that $|A|=a$, $|B|=n-k+a$ and $|C|=k-2a$, and the edge set of $H_{n,k,a}$ consists of all edges between $A$ and $B$ together with all edges in $A\cup C$ (Fig. 1 shows $H_{14,11,3}$). Let

[TABLE]

For a graph $G$, let $c(G)$ denote the length of a longest cycle in $G$. Observe that $c(H_{n,k,a})<k$: Since $|A\cup C|=k-a$, any cycle $D$ of at length at least $k$ has at least $a$ vertices in $B$. But as $B$ is independent and $2a<k$, $D$ also has to contain at least $k+1$ neighbors of the vertices in $B$, while only $a$ vertices in $A$ have neighbors in $A$. Kopylov [8] showed that the extremal $2$-connected $n$-vertex graphs with no cycles of length at least $k$ are $G=H_{n,k,2}$ and $G=H_{n,k,t}$: the first has more edges for small $n$, and the second — for large $n$.

Theorem 1.3** (Kopylov [8]).**

Let $n\geq k\geq 5$ and $t=\lfloor\frac{1}{2}(k-1)\rfloor$. If $G$ is an $n$-vertex $2$-connected graph with $c(G)<k$, then

[TABLE]

with equality only if $G=H_{n,k,2}$ or $G=H_{n,k,t}$.

2 Main results

2.1 A previous result

Recently, three of the present authors proved in [6] a stability version of Theorems 1.2 and 1.3 for $n$-vertex $2$-connected graphs with $n\geq 3k/2$, but the problem remained open for $n<3k/2$ when $k\geq 9$. The main result of [6] was the following:

Theorem 2.1** (Füredi, Kostochka, Verstraëte [6]).**

Let $t\geq 2$ and $n\geq 3t$ and $k\in\{2t+1,2t+2\}$. Let $G$ be a $2$-connected $n$-vertex graph $c(G)<k$. Then $e(G)\leq h(n,k,t-1)$ unless

[TABLE]

Note that

[TABLE]

The paper [6] also describes the $2$-connected $n$-vertex graphs with $c(G)<k\leq 8$ for all $n\geq k$.

2.2 The essence of the main result

Together with [6], this paper gives a full description of the 2-connected $n$-vertex graphs with $c(G)<k$ and ‘many’ edges for all $k$ and $n$. Our main result is:

Theorem 2.2**.**

Let $t\geq 4$ and $k\in\{2t+1,2t+2\}$, so that $k\geq 9$. If $G$ is a $2$-connected graph on $n\geq k$ vertices and $c(G)<k$, then either $e(G)\leq\max\{h(n,k,t-1),h(n,k,3)\}$ or

[TABLE]

Note that the case $n<k$ is trivial and the case $k\leq 8$ was fully resolved in [6].

2.3 A more detailed form of the main result

In order to prove Theorem 2.2, we need a more detailed description of graphs satisfying (b) in the theorem that do not contain ‘long’ cycles.

Let ${\mathcal{G}}_{1}(n,k)=\{H_{n,k,t},H_{n,k,2}\}$. Each $G\in{\mathcal{G}}_{2}(n,k)$ is defined by a partition $V(G)=A\cup B\cup C$ and two vertices $a_{1}\in A$, $b_{1}\in B$ such that $|A|=t$, $G[A]=K_{t}$, $G[B]$ is the empty graph, $G(A,B)$ is a complete bipartite graph, and $N(c)=\{a_{1},b_{1}\}$ for every $c\in C$. Every member of $G\in{\mathcal{G}}_{3}(n,k)$ is defined by a partition $V(G)=A\cup B\cup J$ such that $|A|=t$, $G[A]=K_{t}$, $G(A,B)$ is a complete bipartite graph, and

    — $G[J]$ has more than one component,

    — all components of $G[J]$ are stars with at least two vertices each,

    — there is a $2$-element subset $A^{\prime}$ of $A$ such that $N(J)\cap(A\cup B)=A^{\prime}$,

    — for every component $S$ of $G[J]$ with at least $3$ vertices, all leaves of $S$ have degree 2 in $G$ and are adjacent to the same vertex $a(S)$ in $A^{\prime}$.

The class $\mathcal{G}_{4}(n,k)$ is empty unless $k=10$. Each graph $H\in{\mathcal{G}}_{4}(n,10)$ has a $3$-vertex set $A$ such that $H[A]=K_{3}$ and $H-A$ is a star forest such that if a component $S$ of $H-A$ has more than two vertices then all its leaves have degree 2 in $H$ and are adjacent to the same vertex $a(S)$ in $A$. These classes are illustrated below:

We can refine Theorem 2.2 in terms of the classes $\mathcal{G}_{i}(n,k)$ as follows:

Theorem 2.3**.**

(Main Theorem)* Let $k\geq 9$, $n\geq k$ and $t=\left\lfloor\frac{1}{2}(k-1)\right\rfloor$. Let $G$ be an $n$-vertex $2$-connected graph with no cycle of length at least $k$. Then $e(G)\leq\max\{h(n,k,t-1),h(n,k,3)\}$ or $G$ is a subgraph of a graph in $\mathcal{G}(n,k)$, where*

[TABLE]

Since every graph in $\mathcal{G}_{2}(n,k)\cup\mathcal{G}_{3}(n,k)$ and many graphs in $\mathcal{G}_{4}(n,k)$ have a separating set of size $2$ (see Fig. 3), the theorem implies the following simpler statement for $3$-connected graphs:

Corollary 2.4**.**

Let $k\in\{2t+1,2t+2\}$ where $k\geq 9$. If $G$ is a $3$-connected graph on $n\geq k$ vertices and $c(G)<k$, then either $e(G)\leq\max\{h(n,k,t-1),h(n,k,3)\}$ or $G\subseteq H_{n,k,t}$ or $k=10$ and $G$ is a subgraph of some graph $H\in\mathcal{G}_{4}(n,10)$ such that each component of $H-A$ has at most 2 vertices.

3 The proof idea

3.1 Small dense subgraphs

First we define some more graph classes. For a graph $F$ and a nonnegative integer $s$, we denote by $\mathcal{K}^{-s}(F)$ the family of graphs obtained from $F$ by deleting at most $s$ edges.

Let $F_{0}=F_{0}(t)$ denote the complete bipartite graph $K_{t,t+1}$ with partite sets $A$ and $B$ where $|A|=t$ and $|B|=t+1$. Let $\mathcal{F}_{0}=\mathcal{K}^{-t+3}(F_{0})$, i.e., the family of subgraphs of $K_{t,t+1}$ with at least $t(t+1)-t+3$ edges.

Let $F_{1}=F_{1}(t)$ denote the complete bipartite graph $K_{t,t+2}$ with partite sets $A$ and $B$ where $|A|=t$ and $|B|=t+2$. Let $\mathcal{F}_{1}=\mathcal{K}^{-t+4}(F_{1})$, i.e., the family of subgraphs of $K_{t,t+2}$ with at least $t(t+2)-t+4$ edges.

Let $\mathcal{F}_{2}$ denote the family of graphs obtained from a graph in $\mathcal{K}^{-t+4}(F_{1})$ by subdividing an edge $a_{1}b_{1}$ with a new vertex $c_{1}$, where $a_{1}\in A$ and $b_{1}\in B$. Note that any member $H\in\mathcal{F}_{2}$ has at least $|A||B|-(t-3)$ edges between $A$ and $B$ and the pair $a_{1}b_{1}$ is not an edge.

Let $F_{3}=F_{3}(t,t^{\prime})$ denote the complete bipartite graph $K_{t,t^{\prime}}$ with partite sets $A$ and $B$ where $|A|=t$ and $|B|=t^{\prime}$. Take a graph from $\mathcal{K}^{-t+4}(F_{3})$, select two non-empty subsets $A_{1}$, $A_{2}\subseteq A$ with $|A_{1}\cup A_{2}|\geq 3$ such that $A_{1}\cap A_{2}=\emptyset$ if $\min\{|A_{1}|,|A_{2}|\}=1$, add two vertices $c_{1}$ and $c_{2}$, join them to each other and add the edges from $c_{i}$ to the elements of $A_{i}$, ($i=1,2$). The class of obtained graphs is denoted by ${\mathcal{F}}(A,B,A_{1},A_{2})$. The family $\mathcal{F}_{3}$ consists of these graphs when $|A|=|B|=t$, $|A_{1}|=|A_{2}|=2$ and $A_{1}\cap A_{2}=\emptyset$. In particular, $\mathcal{F}_{3}(4)$ consists of exactly one graph, call it $F_{3}(4)$.

Graph $F_{4}$ has vertex set $A\cup B$, where $A=\{a_{1},a_{2},a_{3}\}$ and $B:=\{b_{1},b_{2},\dots,b_{6}\}$ are disjoint. Its edges are the edges of the complete bipartite graph $K(A,B)$ and three extra edges $b_{1}b_{2}$, $b_{3}b_{4}$, and $b_{5}b_{6}$ (see Fig. 4 below). Define $F_{4}^{\prime}$ as the (only) member of ${\mathcal{F}}(A,B,A_{1},A_{2})$ such that $|A|=|B|=t=4$, $A_{1}=A_{2}$, and $|A_{i}|=3$. Let $\mathcal{F}_{4}:=\{F_{4},F^{\prime}_{4}\}$, which is defined only for $t=4$.

Define \mathcal{F}(k):=\left\{\begin{array}[]{ll}\mathcal{F}_{0},&\mbox{if kis odd},\\ \mathcal{F}_{1}\cup\dots\cup\mathcal{F}_{4},&\mbox{ifk is even.}\end{array}\right.

3.2 Proof idea

For our proof, it will be easier to use the stronger induction assumption that the graphs in question contain certain dense graphs from $\mathcal{F}(k)$. We will prove the following slightly stronger version of Theorem 2.3 which also implies Theorem 2.2.

Theorem 2.3′ Let $t\geq 4$, $k\in\{2t+1,2t+2\}$, and $n\geq k$. Let $G$ be an $n$-vertex $2$-connected graph with no cycle of length at least $k$. Then $e(G)\leq\max\{h(n,k,t-1),h(n,k,3)\}$ or

[TABLE]

The method of the proof is a variation of that of [6]. Also, when $n$ is close to $k$, we use Kopylov’s disintegration method. We take an $n$-vertex graph $G$ satisfying the hypothesis of Theorem 2.3′, and iteratively contract edges in a certain way so that each intermediate graph still satisfies the hypothesis. We consider the final graph of this process $G_{m}$ on $m$ vertices and show that $G_{m}$ satisfies Theorem 2.3′. Two results from [6] will be instrumental. The first is:

Lemma 3.1** (Main lemma on contraction [6]).**

Let $k\geq 9$ and suppose $F$ and $F^{\prime}$ are $2$-connected graphs such that $F=F^{\prime}/xy$ and $c(F^{\prime})<k$. If $F$ contains a subgraph $H\in\mathcal{F}(k)$, then $F^{\prime}$ also contains a subgraph $H^{\prime}\in\mathcal{F}(k)$.   $\Box$

This lemma shows that if $G_{m}$ contains a subgraph $H\in\mathcal{F}(k)$, then the original graph $G$ also contains a subgraph in $\mathcal{F}(k)$. The second result (proved in Subsection 4.5 of [6]) is:

Lemma 3.2** ([6]).**

Let $k\geq 9$, and let $G$ be a $2$-connected graph with $c(G)<k$ and $e(G)>h(n,k,t-1)$. If $G$ contains a subgraph $H\in\mathcal{F}(k)$, then $G$ is a subgraph of a graph in $\mathcal{G}(n,k)-\{H_{n,k,2}\}$.   $\Box$

We will split the proof into the cases of small $n$ and large $n$. The following observations can be obtained by simple calculations (for $t\geq 4$):

[TABLE]

In the case of large $n$ we will contract an edge such that the new graph still has more than $h(n-1,k,t-1)$ edges. In order to apply induction, we also need the number of edges to be greater than $h(n-1,k,3)$. To guarantee this, we pick the cutoffs for the two cases $n\leq k+(t-1)/2$ and $n>k+(t-1)/2$ (therefore $n-1>k+(t-3)/2$).

4 Tools

4.1 Classical theorems

Theorem 4.1** (Erdős [2]).**

Let $d\geq 1$ and $n>2d$ be integers, and

[TABLE]

Then every $n$-vertex graph $G$ with $\delta(G)\geq d$ and $e(G)>\ell_{n,d}$ is hamiltonian.   $\Box$

Theorem 4.2** (Chvátal [1]).**

Let $n\geq 3$ and $G$ be an $n$-vertex graph with vertex degrees $d_{1}\leq d_{2}\leq\ldots\leq d_{n}$. If $G$ is not hamiltonian, then there is some $i<n/2$ such that $d_{i}\leq i$ and $d_{n-i}<n-i$.   $\Box$

Theorem 4.3** (Kopylov [8]).**

If $G$ is $2$-connected and $P$ is an $x,y$-path of $\ell$ vertices, then $c(G)\geq\min\{\ell,d(x,P)+d(y,P)\}$.   $\Box$

4.2 Claims on contractions

A helpful tool will be the following lemma from [6] on contraction.

Lemma 4.4** ([6]).**

Let $n\geq 4$ and let $G$ be an $n$-vertex $2$-connected graph. For every $v\in V(G)$, there exists $w\in N(v)$ such that $G/vw$ is $2$-connected.   $\Box$

For an edge $xy$ in a graph $H$, let $T_{H}(xy)$ denote the number of triangles containing $xy$. Let $T(H)=\min\{T_{H}(xy)\,:\,xy\in E(H)\}$. When we contract an edge $uv$ in a graph $H$, the degree of every $x\in V(H)-u-v$ either does not change or decreases by $1$. Also the degree of $u*v$ in $H/uv$ is at least $\max\{d_{H}(u),d_{H}(v)\}-1$. Thus

[TABLE]

Similarly,

[TABLE]

We will use the following analog of Lemma 3.3 in [6].

Lemma 4.5**.**

*Let $h$ be a positive integer. Suppose a $2$-connected graph $G$ is obtained from a $2$-connected graph $G^{\prime}$ by contracting edge $xy$ into $x*y$ chosen using the following rules:

$(i)$ one of $x,y$, say $x$ is a vertex of the minimum degree in $G^{\prime}$;

$(ii)$ $T_{G^{\prime}}(xy)$ is the minimum among the edges $xu$ incident with $x$ such that $G^{\prime}/xu$ is $2$-connected. (Such edges exist by Lemma 4.4). If $G$ has at least $h$ vertices of degree at most $h$, then either $G^{\prime}=K_{h+2}$ or

$(a)$ $G^{\prime}$ also has a vertex of degree at most $h$, and

$(b)$ $G^{\prime}$ has at least $h+1$ vertices of degree at most $h+1$.*

Proof. Since $G$ is $2$-connected, $h\geq 2$. Let $V_{\leq s}(H)$ denote the set of vertices of degree at most $s$ in $H$. Then by (2), each $v\in V_{\leq h}(G)-x*y$ is also in $V_{\leq h+1}(G^{\prime})$. Moreover, then by (i),

[TABLE]

Thus if $x*y\notin V_{\leq h}(G)$, then (b) follows. But if $x*y\in V_{\leq h}(G)$, then by (2), also $y\in V_{\leq h+1}(G^{\prime})$. So, again (b) holds.

If $V_{\leq h-1}(G)\neq\emptyset$, then (a) holds by (2). So, if (a) does not hold, then

[TABLE]

Case 1: $|V_{\leq h}(G)-x*y|\geq h$. Then by (4), $d_{G^{\prime}}(x)=h+1$. This in turn yields $N_{G^{\prime}}(x)=V_{\leq h}(G)+y$. Since $G^{\prime}$ is $2$-connected, each $v\in V_{\leq h}(G)-x*y$ is not a cut vertex. Furthermore, $\{x,v\}$ is not a cut set. If it was, because $y$ is a common neighbor of all neighbors of $x$, all neighbors of $x$ must be in the same component as $y$ in $G^{\prime}-x-v$. It follows that

[TABLE]

If $uv\notin E(G)$ for some $u,v\in V_{\leq h}(G)$, then by (6) and (i), we would contract the edge $xu$ rather than $xy$. Thus $G^{\prime}[V_{\leq h}(G)\cup\{x,y\}]=K_{h+2}$ and so either $G^{\prime}=K_{h+2}$ or $y$ is a cut vertex in $G^{\prime}$, as claimed.

Case 2: $|V_{\leq h}(G)-x*y|=h-1$. Then $x*y\in V_{\leq h}(G)$. This means $d_{G^{\prime}}(x)=d_{G^{\prime}}(y)=h+1$ and $N_{G^{\prime}}[x]=N_{G^{\prime}}[y]$. So by (5), there is $z\in V(G)$ such that $N_{G^{\prime}}[x]=N_{G^{\prime}}[y]=V_{\leq h}(G)\cup\{x,y,z\}$. Again (6) holds (for the same reason that $N_{G^{\prime}}[x]\subseteq N_{G^{\prime}}[y]$). Thus similarly $vu\in E(G^{\prime})$ for every $v\in V_{\leq h}(G)-x*y$ and every $u\in V_{\leq h}(G)+z$. Hence $G^{\prime}[V_{\leq h}(G)\cup\{x,y,z\}]=K_{h+2}$ and either $G^{\prime}=K_{h+2}$ or $z$ is a cut vertex in $G^{\prime}$, as claimed.   $\Box$

4.3 A property of graphs in $\mathcal{F}(k)$

A useful feature of graphs in $\mathcal{F}(k)$ is the following.

Lemma 4.6**.**

Let $k\geq 9$ and $n\geq k$. Let $F$ be an $n$-vertex graph contained in $H_{n,k,t}$ with $e(F)>h(n,k,t-1)$. Then $F$ contains a graph in $\mathcal{F}(k)$.

Proof.

Assume the sets $A,B,C$ to be as in the definition of $H_{n,k,t}$. We will use induction on $n$.

Case 1: $k=2t+1$. If $n=k$, then $F\in\mathcal{K}^{-t+3}(H_{k,k,t})$ because $h(k,k,t)-h(k,k,t-1)-1=t-3$. Thus, since $H_{k,k,t}\supseteq F_{0}(t)$, $F$ contains a subgraph in $\mathcal{F}_{0}$. Suppose now the lemma holds for all $k\leq n^{\prime}<n$. If $\delta(F)\geq t$, then each $v\in V(F)-A$ is adjacent to every $u\in A$. Hence $F$ contains $K_{t,n-t}$. If $\delta(F)<t$, then since $A$ is dominating and $n>2t$, there is $v\in V(F)-A$ with $d_{F}(v)\leq t-1$. Then $F-v\subseteq H_{n-1,k,t}$, and we are done by induction.

Case 2: $k=2t+2$. Let $C=\{c_{1},c_{2}\}$. If $n=k$ then as in Case 1,

[TABLE]

i.e., $F\in\mathcal{K}^{-t+4}(H_{k,k,t})$. Since $F_{1}(t)\subseteq H_{k,k,t}$, $F$ contains a subgraph in $\mathcal{F}_{1}$. Suppose now the lemma holds for all $k\leq n^{\prime}<n$. If $\delta(F)<t$, then there is $v\in V(F)-A$ with $d_{F}(v)\leq t-1$. Then $F-v\subseteq H_{n-1,k,t}$, and we are done by induction.

Finally, suppose $\delta(F)\geq t$. So, each $v\in B$ is adjacent to every $u\in A$ and each of $c_{1},c_{2}$ has at least $t-1$ neighbors in $A$. Since $|B\cup\{c_{1}\}|\geq n-t-1\geq t+2$, $F$ contains a member of $\mathcal{K}^{-1}(F_{1}(t))$. Thus $F$ contains a member of $\mathcal{F}_{1}$ unless $t=4$, $n=2t+3$ and $c_{1}$ has a nonneighbor $x\in A$. But then $c_{1}c_{2}\in E(F)$, and so $F$ contains either $F_{3}(4)$ or $F^{\prime}_{4}$.   $\Box$

5 Proof of Theorem 2.3′

5.1 Contraction procedure

If $n>k$, we iteratively construct a sequence of graphs $G_{n},G_{n-1},...G_{m}$ where $|V(G_{j})|=j$. In [6], the following Basic Procedure (BP) was used:

At the beginning of each round, for some $j:k\leq j\leq n$, we have a $j$-vertex $2$-connected graph $G_{j}$ with $e(G_{j})>h(j,k,t-1)$.

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Remark 5.1. By definition, (BP3) applies only when $j\geq k+t-1$. As observed in [6], if $j\leq 3t-2$, then a $j$-vertex graph $G_{j}$ with a $2$-vertex set $\{x,y\}$ separating the graph into at least $3$ components cannot have $T_{G_{j}}(uv)\geq t-1$ for every edge $uv$. It also was calculated there that if $3t-1\leq j\leq 3t$, then any $j$-vertex graph $G^{\prime}$ with such $2$-vertex set $\{x,y\}$ and $T_{G^{\prime}}(uv)\geq t-1$ for every edge $uv$ has at most $h(j,k,t-1)$ edges and so cannot be $G_{j}$.

In this paper, we also use a quite similar Modified Basic Procedure (MBP): start with a $2$-connected, $n$-vertex graph $G=G_{n}$ with $e(G)>h(n,k,t-1)$ and $c(G)<k$. Then

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5.2 Proof of Theorem 2.3′ for the case $n\leq k+(t-1)/2$

Apply to $G$ the Modified Basic Procedure (MBP) starting from $G_{n}=G$. By Remark 1, (BP3) was never applied, since $k+(t-1)/2<k+t-1$. Therefore at every step, we only contracted an edge. Denote by $G_{m}$ the terminating graph of (MBP). Then $G_{j}$ is $2$-connected and $c(G_{j})\leq c(G)<k$ for each $m\leq j\leq n$. By construction, after each contraction, we lose at most $t-1$ edges. It follows that $e(G_{m})>h(m,k,t-1)$.

If $m>k$, then the same argument as in [6] gives us the following structural result:

Lemma 5.1** ([6]).**

Let $m>k\geq 9$ and $n\geq k$.

• •

If $k\neq 10$, then $G_{m}\subseteq H_{m,k,t}$.

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*If $k=10$, then $G_{m}\subseteq H_{m,k,t}$ or $G_{m}\supseteq F_{4}$.   *$\Box$

If $k=10$ and $G_{m}\supseteq F_{4}$, then $G_{m}$ contains a subgraph in $\mathcal{F}(k)$. Otherwise, by Lemma 4.6, again $G_{m}$ has a subgraph in $\mathcal{F}(k)$. Next, undo the contractions that were used in (MBP). At every step for $m\leq j\leq n$, by Lemma 3.1, $G_{j}$ contains some subgraph $H^{\prime}\in\mathcal{F}(k)$. In particular, $G=G_{n}$ contains such a subgraph. Thus by Lemma 3.2, we get our result. So, below we assume

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Since $c(G_{k})<k$, $G_{k}$ does not have a hamiltonian cycle. Denote the vertex degrees of $G_{k}$ $d_{1}\leq d_{2}\leq...\leq d_{k}$. By Theorem 4.2, there exists some $2\leq i\leq t$ such that $d_{i}\leq i$ and $d_{k-i}<k-i$. Let $r=r(G_{k})$ be the smallest such $i$.

Because $G_{k}$ has $r$ vertices of degree at most $r$, similarly to [2],

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For $k=2t+1$, $r^{2}+{k-r\choose 2}>h(n,k,t-1)$ only when $r=t$ or $r<(t+4)/3$, and for $k=2t+2$, when $r=t$ or $r<(t+6)/3$. If $r=t$, then repeating the argument in [6] yields:

Lemma 5.2** ([6]).**

If $r(G_{k})=t$ then $G_{k}\subseteq H_{k,k,t}$.

Then by Lemma 4.6, Lemma 3.1, and Lemma 3.2, $G\subseteq H_{n,k,t}$ and contains some subgraph in $\mathcal{F}(k)$. So we may assume that

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Our next goal is to show that $G$ contains a large “core”, i.e., a subgraph with large minimum degree. For this, we recall the notion of disintegration used by Kopylov [8].

Definition: For a natural number $\alpha$ and a graph $G$, the $\alpha$-disintegration of a graph $G$ is the process of iteratively removing from $G$ the vertices with degree at most $\alpha$ until the resulting graph has minimum degree at least $\alpha+1$. This resulting subgraph $H=H(G,\alpha)$ will be called the $\alpha$-core of $G$. It is well known that $H(G,\alpha)$ is unique and does not depend on the order of vertex deletion.

Claim 5.3**.**

The $t$-core $H(G,t)$ of $G$ is not empty.

Proof of Claim 5.3: We may assume that for all $m\leq j<n$, graph $G_{j}$ was obtained from $G_{j+1}$ by contracting edge $x_{j}y_{j}$, where $d_{G_{j+1}}(x_{j})\leq d_{G_{j+1}}(y_{j})$. By Rule (MBP2), $d_{G_{j+1}}(x_{j})=\delta({G_{j+1}})$, provided that $\delta({G_{j+1}})\leq t-1$.

By definition, $|V_{\leq r}(G_{k})|\geq r$. So by Lemma 4.5 (applied several times), for each $k+1\leq j\leq k+t-r$, because each $G_{j}$ is not a complete graph (otherwise it would have a hamiltonian cycle),

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To show that

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by (9) and (8), it is enough to observe that

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We will apply a version of $t$-disintegration in which we first manually remove a sequence of vertices and count the number of edges they cover. By (10) and (MBP2), $d_{G_{n}}(x_{n-1})=\delta({G_{n}})\leq n-k+r-1$. Let $v_{n}:=x_{n-1}$. Then $G-v_{n}$ is a subgraph of $G_{n-1}$. If $x_{n-2}\neq x_{n-1}*y_{n-1}$ in $G_{n-1}$, then let $v_{n-1}:=x_{n-2}$, otherwise let $v_{n-1}:=y_{n-1}$. In both cases, $d_{G-v_{n}}(v_{n-1})\leq n-k+r-2$. We continue in this way until $j=k$: each time we delete from $G-v_{n}-\ldots-v_{j+1}$ the unique survived vertex $v_{j}$ that was in the preimage of $x_{j-1}$ when we obtained $G_{j-1}$ from $G_{j}$. Graph $G-v_{n}-...-v_{k+1}$ has $r\geq 2$ vertices of degree at most $r$. We additionally delete 2 such vertices $v_{k}$ and $v_{k-1}$. Altogether, we have lost at most $(r+n-k-1)+(r+n-k-2)+...+r+2r$ edges in the deletions.

Finally, apply $t$-disintegration to the remaining graph on $k-2\in\{2t-1,2t\}$ vertices. Suppose that the resulting graph is empty.

Case 1: $n=k$. Then

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where $r+r$ edges are from $v_{k}$ and $v_{k-1}$, and after deleting $v_{k}$ and $v_{k-1}$, every vertex deleted removes at most $t$ edges, until we reach the final $t$ vertices which altogether span at most ${t\choose 2}$ edges.

For $k=2t+1$,

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which is nonnegative for $r<(t+3)/3$. Therefore $e(G)\leq h(k,k,t-1)$, a contradiction.

Similarly, if $k=2t+2$,

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which is nonnegative when $r<(t+6)/3$.

Case 2: $k<n\leq k+(t-1)/2$. Then for $k=2t+1$,

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where the last inequality holds because $r+n-k-1\leq t-1$.

Similarly, for $k=2t+2$,

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This contradiction completes the proof of Claim 5.3.   $\Box$

For the rest of the proof of Theorem 2.3′, we will follow the method of Kopylov in [8] to show that $G\subseteq H_{n,k,2}$. Let $G^{*}$ be the $k$-closure of $G$. That is, add edges to $G$ until adding any additional edges creates a cycle of length at least $k$. In particular, for any non-edge $xy$ of $G^{*}$, there is an $(x,y)$-path in $G^{*}$ with at least $k-1$ edges.

Because $G$ has a nonempty $t$-core, and $G^{*}$ contains $G$ as a subgraph, $G^{*}$ also has a nonempty $t$-core (which contains the $t$-core of $G$). Let $H=H(G^{*},t)$ denote the $t$-core of $G^{*}$. We will show that

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Indeed, suppose there exists a nonedge in $H$. Choose a longest path $P$ of $G^{*}$ whose terminal vertices $x\in V(H)$ and $y\in V(H)$ are nonadjacent. By the maximality of $P$, every neighbor of $x$ in $H$ is in $P$, similar for $y$. Hence $d_{P}(x)+d_{P}(y)=d_{H}(x)+d_{H}(y)\geq 2(t+1)\geq k$, and also $|P|=k-1$ (edges). By Kopylov’ Theorem 4.3, $G^{*}$ must have a cycle of length at least $k$, a contradiction.

Therefore $H$ is a complete subgraph of $G^{*}$. Let $\ell=|V(H)|$. Because every vertex in $H$ has degree at least $t+1$, $\ell\geq t+2$. Furthermore, if $\ell\geq k-1$, then $G^{*}$ has a clique $K$ of size at least $k-1$. Because $G^{*}$ is $2$-connected, we can extend a $(k-1)$-cycle of $K$ to include at least one vertex in $G^{*}-H^{\prime}$, giving us a cycle of length at least $k$. It follows that

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and therefore $k-\ell\leq t$. Apply a weaker $(k-\ell)$-disintegration to $G^{*}$, and denote by $H^{\prime}$ the resulting graph. By construction, $H\subseteq H^{\prime}$.

Case 1: There exists $v\in V(H^{\prime})-V(H)$. Since $v\notin V(H)$, there exists a nonedge between a vertex in $H$ and a vertex in $H^{\prime}-H$. Pick a longest path $P$ with terminal vertices $x\in V(H^{\prime})$ and $y\in V(H)$. Then $d_{P}(x)+d_{P}(y)\geq(k-\ell+1)+(\ell-1)=k$, and therefore $c(G^{*})\geq k$.

Case 2: $H=H^{\prime}$. Then

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If $3\leq(k-\ell)\leq t-1$, then $e(G)\leq\max\{h(n,k,3),h(n,k,t-1)\}$, so by (12), $k-\ell=2$, and $H$ is the complete graph with $k-2$ vertices. Let $D=V(G^{*})-V(H)$. If there is an edge $xy$ in $G^{*}[D]$, then because $G^{*}$ is $2$-connected, there exist two vertex-disjoint paths, $P_{1}$ and $P_{2}$, from $\{x,y\}$ to $H$ such that $P_{1}$ and $P_{2}$ only intersect $\{x,y\}\cup H$ at the beginning and end of the paths. Let $a$ and $b$ be the terminal vertices of $P_{1}$ and $P_{2}$ respectively that lie in $H$. Let $P$ be any $(a,b)$-hamiltonian path of $H$. Then $P_{1}\cup P\cup P_{2}+xy$ is a cycle of length at least $k$ in $G^{*}$, a contradiction.

Therefore $D$ is an independent set, and since $G^{*}$ is $2$-connected, each vertex of $D$ has degree 2. Suppose there exists $u,v\in D$ where $N(u)\neq N(v)$. Let $N(u)=\{a,b\},N(v)=\{c,d\}$ where it is possible that $b=c$. Then we can find a cycle $C$ of $H$ that covers $V(H)$ which contains edges $ab$ and $cd$. Then $C-ab-cd+ua+ub+vc+vd$ is a cycle of length $k$ in $G^{*}$. Thus for every $v\in D$, $N(v)=\{a,b\}$ for some $a,b\in H$. I.e., $G^{*}=H_{n,k,2}$, and thus $G\subseteq H_{n,k,2}$.   $\Box$

5.3 Proof of Theorem 2.3′ for all $n$

We use induction on $n$ with the base case $n\leq k+(t-1)/2$. Suppose $n\geq k+t/2$ and for all $k\leq n^{\prime}<n$, Theorem 2.3′ holds. Let $G$ be a $2$-connected graph $G$ with $n$ vertices such that

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Apply one step of (BP). If (BP4) was applied (so neither (BP2) nor (BP3) applies to $G$), then $G_{m}=G$ (with $G_{m}$ defined as in the previous case). By Lemmas 5.1, 4.6, and 3.2, the theorem holds.

Therefore we may assume that either (BP2) or (BP3) was applied. Let $G^{-}$ be the resulting graph. Then $c(G^{-})<k$, and $G^{-}$ is $2$-connected.

Claim 5.4**.**

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Proof.

If (BP2) was applied, i.e., $G^{-}=G/uv$ for some edge $uv$, then

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so (14) holds. Therefore we may assume that (BP3) was applied to obtain $G^{-}$. Then $n\geq k+t-1$ and $e(G)-e(G^{-})={t+1\choose 2}-1$. So by (13),

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The right hand side of (15) equals $h(n-(t-1),k,t-1)+t^{2}/2-5t/2+2$ which is at least $h(n-(t-1),k,t-1)$ for $t\geq 4$, proving the first part of (14).

We now show that also $e(G^{-})>h(n-(t-1),k,3)$. Indeed, for $k=2t+1$,

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Similarly, for $k=2t+2$,

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Thus if $n\geq 3t+1$, then (14) is proved. But if $n\in\{3t-1,3t\}$ then by Remark 5.1, no graph to which (BP3) applied may have more than $h(n,k,t-1)$ edges.   $\Box$

By (14), we may apply induction to $G^{-}$. So $G^{-}$ satisfies either (a) $G^{-}\subseteq H_{|V(G^{-})|,n,2}$, or (b) $G^{-}$ is contained in a graph in $\mathcal{G}(n,k)-H_{|V(G^{-})|,k,2}$ and contains a subgraph $H\in\mathcal{F}(k)$. Suppose first that $G^{-}$ satisfies (b). If (BP3) was applied to obtain $G^{-}$ from $G$, then because $G^{-}$ contains a subgraph $H\in\mathcal{F}(k)$ and $G^{-}\subseteq G$, $G$ also contains $H$. If (BP2) was applied, then by Lemma 3.1, $G$ contains a subgraph $H^{\prime}\in\mathcal{F}(k)$. In either case, Lemma 3.2 implies that $G$ is a subgraph of a graph in $\mathcal{G}(n,k)-H_{n,k,2}$.

So we may assume that (a) holds, that is, $G^{-}$ is a subgraph of $H_{|V(G^{-})|,n,2}$. Because $\delta(G^{-})\leq 2$, $\delta(G)\leq 3$, and so $G$ has edges in at most $2\leq t-2$ triangles. Therefore (BP2) was applied to obtain $G^{-}$, where $G/uv=G^{-}$. Let $D$ be an independent set of vertices of $G^{-}$ of size $(n-1)-(k-2)$ with $N(D)=\{a,b\}$ for some $a,b\in V(G^{-})$. Since $T_{G^{-}}(xa),T_{G^{-}}(xb)\leq 1$ for every $x\in D$, we have that $T_{G}(uv)\leq 2$ with equality only if $T(G)=2$ where $T(G)=\min_{xy\in E(G)}T_{G}(xy)$.

We want to show that $T_{G}(uv)\leq 1$. If not, suppose first that $u*v\in D\subseteq V(G^{-})$. Then there exists $x\in D-u*v$, and $x$ and $u*v$ are not adjacent in $G^{-}$. Therefore $x$ was not in a triangle with $u$ and $v$ in $G$, and hence $T_{G}(xa)=T_{G^{-}}(xa)\leq 1$, so the edge $xa$ should have been contracted instead. Otherwise if $u*v\notin D$, at least one of $\{a,b\}$, say $a$, is not $u*v$. If $T(G)=2$, then for every $x\in D\subseteq V(G)$, $T_{G}(xa)=2$, therefore each such edge $xa$ was in a triangle with $uv$ in $G$. Then $T_{G}(uv)\geq|D|=(n-1)-(k-2)\geq k+t/2-1-k+2\geq 3$, a contradiction.

Thus $T_{G}(uv)\leq 1$ and $e(G)\leq 2+e(G^{-})\leq 2+h(n-1,k,2)=h(n,k,2)$. But for $n\geq k+t/2$, we have $h(n,k,t-1)\geq h(n,k,2)$, a contradiction.   $\Box$

Acknowledgment. The authors thank Zoltán Király for helpful comments.