Proof. Apply the aforementioned result [Reference Pokrovskiy and Sudakov25] for $K_{m_1,\ldots ,m_k}$ with $m_i\geq i^{22}$ for each i to $K_{m^{\prime }_1,\ldots , m^{\prime }_k}$ where $m^{\prime }_1=m_1$ and $m^{\prime }_i=m_k$ if $i>1$ . For each $i>1$ , we have $m^{\prime }_i=m_k\geq k^{22}\geq i^{22}$ , and clearly $m^{\prime }_1\geq 1^{22}$ . Thus, if G is a $C_n$ -free graph on $(n - 1)(k - 1) + m_1$ vertices, then $\overline {G}$ contains a copy of $K_{m^{\prime }_1,\ldots , m^{\prime }_k}$ , which in turn contains $K_{m_1,\ldots , m_k}$ .

Proof. We proceed by induction on k, considering two cases. The base case $k=2$ will be covered by the inductive argument since we argue that we may either (i) construct a suitable subgraph directly for $H'=H$ , (ii) reduce to some smaller $H'\sqsubset H$ with at least two parts or (iii) obtain a contradiction; in the base case only, the first and third possibilities arise. Note that, by removing vertices if necessary, we may assume $\lvert G\rvert =Mmk \log k$ for $k\geq 2$ . We write $m_1\leq \cdots \leq m_k$ for the orders of the vertex classes of H.

The first case is if there exists a set $S\subset V(G)$ such that $m \leq \lvert S\rvert \leq \lvert G\rvert /2$ and $\lvert N_G(S)\rvert \leq \lvert S\rvert /(10\log k) + m\Delta \beta $ . Let $T = V(G)\setminus (N_G(S) \cup S)$ . Since $M \geq 10\Delta \beta $ , we have that

$$ \begin{align*} \lvert T\rvert\geq \lvert G\rvert-\lvert N_G(S)\cup S\rvert & \geq \left(1-1/(10\log k)\right)\lvert G\rvert/2 - m\Delta\beta \\ & = m\left(\tfrac{Mk}{2}(\log k-1/10) - \Delta\beta\right) \\ & \geq km. \end{align*} $$

We distinguish two subcases depending on whether $\lvert S\rvert \geq mk$ or $\lvert S\rvert <mk$ .

First, suppose $\lvert S\rvert \geq mk$ . We choose $t \in \mathbb {N}$ as large as possible such that $\lvert T\rvert \geq M\lvert H_t\rvert \log t$ , where $H_t$ is the graph induced by the t smallest parts of H. We then choose $s \in \mathbb {N}$ as large as possible so that $\lvert S\rvert \geq M\lvert H_{s,t}\rvert \log s$ , where $H_{s,t}$ is the graph induced by the s smallest parts of H not contained in $H_t$ ; observe that $s,t\geq 1$ and $s,t\leq k-1$ .

We will quickly be able to resolve our first subcase after proving the following claim.

Proof of claim. The claim is trivial for $k=2$ since $s,t\geq 1$ . For all $k\geq 3$ , we have

$$\begin{align*}\lvert T\rvert \geq \lvert G\rvert/2-\frac{\lvert G\rvert}{20\log k} - \frac{\lvert G\rvert}{10k\log k} \geq M\lvert H_{\lceil k/2\rceil}\rvert\log(\lceil k/2\rceil), \end{align*}$$

which implies $t\geq \lceil k/2\rceil $ ; in particular, this proves the claim when $k=3$ .

Assume for a contradiction that $s + t < k$ , where $k\geq 4$ is fixed; as $t \geq \lceil k/2\rceil $ , we have $s\leq (k-1)/2$ . Maximalities of s and t ensure that

(2) $$ \begin{align} M\lvert H_{s+1, t}\rvert\log(s+1)> \lvert S\rvert \end{align} $$

and

(3) $$ \begin{align} M\lvert H_{t+1}\rvert\log(t+1)> \lvert T\rvert. \end{align} $$

Observe that $H_{t+1}$ and $H_{s+1, t}$ both contain the $(t+1)$ th smallest part of H.

Also, $\lvert S\rvert +\lvert T\rvert = \lvert G\rvert - \lvert N_G(S)\rvert> Mmk\log k - \frac {\lvert S\rvert }{10\log k} - m\Delta \beta $ ; hence by inequality (2), inequality (3) and the fact that $M\geq 10\beta \Delta $ , we have

(4) $$ \begin{align} \lvert H_{t+1}\rvert\log(t+1) + \lvert H_{s+1, t}\rvert\log(s + 1)c(k) \geq mk\log k - \frac{m}{10}, \end{align} $$

where $c(k)=\left (1 + \frac {1}{10\log k}\right )$ . We want to show that inequality (4) is in fact false, which provides the contradiction we need to prove Claim 4.3. To this end, we may assume $s+t = k-1$ . It follows that

$$ \begin{align*} \lvert H_{t+1}\rvert + \lvert H_{s+1, t}\rvert=mk+m_{t+1} \leq mk+\frac{\lvert H_{s+1, t}\rvert}{s+1}; \end{align*} $$

recall that $m_{t+1}$ is the size of the $(t+1)$ th smallest part of H. Consequently, the left-hand side of inequality (4) is at most

(5) $$ \begin{align} c(k)\log(s+1)\lvert H_{s+1, t}\rvert + \log(k-s)\left(mk - \lvert H_{s+1, t}\rvert\frac{s}{s+1}\right). \end{align} $$

Note that $m(s+1)\leq \lvert H_{s+1, t}\rvert \leq mk$ . For fixed s, expression (5) is linear in $|H_{s+1, t}|$ , and consequently, within this range it is maximised either at $\lvert H_{s+1, t}\rvert =m(s+1)$ or at $\lvert H_{s+1, t}\rvert =mk$ .

We first consider the case $\lvert H_{s+1, t}\rvert =m(s+1)$ when expression (5) becomes

$$\begin{align*}m(c(k)(s+1)\log(s+1)+(k-s)\log(k-s)).\end{align*}$$

This is a convex function of s and so for $1\leq s\leq (k-1)/2$ is maximised when $s=1$ or when $s=(k-1)/2$ . When $s=1$ , we have

(6) $$ \begin{align} 2c(k)\log 2+(k-1)\log(k-1)<k\log k-0.1 \end{align} $$

for all $k\geq 4$ . When $s=(k-1)/2$ , we likewise have

(7) $$ \begin{align} (1+c(k))\frac{k+1}{2}\log\left(\frac{k+1}{2}\right)<k\log k-0.1 \end{align} $$

for all $k\geq 4$ .

Finally, we consider the case $\lvert H_{s+1, t}\rvert =mk$ when expression (5) becomes

$$\begin{align*}mk(\log(s+1)c(k)+\log(k-s)/(s+1)). \end{align*}$$

Suppose $2\leq s\leq \log (k-s)$ . Then $\frac {s}{s+1}\log (k-s)\geq \frac {s^2}{s+1}\geq c(k)\log (s+1)+0.1$ , and so expression (5) is decreasing in $\lvert H_{s+1, t}\rvert $ and maximised when $\lvert H_{s+1, t}\rvert =m(s+1)$ . Similarly, if $s=1$ and $k\geq 7$ , expression (5) is decreasing. For $s=1$ and $4\leq k\leq 6$ , by direct calculation expression (5) contradicts inequality (4). Thus, we may assume that $s\geq \log (k-s)$ . Since $s\leq (k-1)/2$ , we have $s+1\leq k-s$ . Thus,

$$\begin{align*}\frac{\mathrm{d}}{\mathrm{d}s}\left(\log(s+1)c(k)+ \frac{\log(k-s)}{s+1}\right)=\frac{c(k)(s+1)-\frac{s+1}{k-s}-\log(k-s)}{(s+1)^2}>0, \end{align*}$$

and so $\log (s+1)c(k)+\log (k-s)/(s+1)$ is increasing in s for $\log (k-s)\leq s\leq (k-1)/2$ . Hence, within this range $mk(\log (s+1)c(k)+\log (k-s)/(s+1))$ is maximised when $s=(k-1)/2$ , and we obtain

(8) $$ \begin{align} k\left(c(k)+\frac{2}{k+1}\right)\log\left(\frac{k+1}{2}\right)<k\log k - 0.1 \end{align} $$

for all $k\geq 4$ . Then inequalities (6), (7) and (8) together show that expression (5) is less than $mk\log k - \frac {m}{10}$ , contradicting inequality (4).

Hence, $s+t= k$ . Now, either $\overline {G[T]}$ is $H_t$ -free or $\overline {G[S]}$ is $H_{s,t}$ -free since otherwise disjointness of T and $N_G(S)\cup S$ ensures we have a copy of H in $\overline {G[S\cup T]}$ . However, since $\lvert S\rvert ,\lvert T\rvert \geq mk$ , $\overline {G[T]}$ is not $H_1$ -free and $\overline {G[S]}$ is not $H_{1,k-1}$ -free. Thus, in the former case we have $2\leq t\leq k-1$ so we may apply the induction hypothesis to $G[T]$ with k replaced by t and H by $H_t$ , and in the latter $2\leq s\leq k-1$ so we may apply the induction hypothesis to $G[S]$ with k replaced by s and H by $H_{s,t}$ . Here, such applications of the induction hypothesis are possible by the definition of s and t.

Secondly, suppose $m\leq \lvert S\rvert <mk$ . Note that, since $M \geq 10\Delta \beta $ and $M \geq 4k$ , in this case

$$\begin{align*}\lvert T\rvert\geq Mmk\log k-2mk-m\Delta\beta \geq M(mk-1)\log (k-1). \end{align*}$$

Indeed, $Mmk\log \frac {k}{k-1} \geq Mm \geq 2mk + m\Delta \beta $ where we have used $\log (1+x) \geq \frac {x}{1+x}$ with $x = \frac {1}{k-1}$ . Thus, $\lvert T\rvert \geq M\lvert H_{k-1, 1}\rvert \log (k-1)$ and $\lvert S\rvert \geq m \geq \lvert H_1\rvert $ . Since $H_1$ is an independent set of size at most m, $\overline {G[S]}$ contains $H_1$ . Thus, $\overline {G[T]}$ must be $H_{k-1, 1}$ -free, as otherwise $\overline {G[S\cup T]}$ contains H, as before. Since $\lvert T\rvert \geq mk$ , it is not $H_{1,1}$ -free and so $k-1\geq 2$ . We may therefore apply the induction hypothesis to $G[T]$ with $k-1$ replacing k and $H_{k-1,1}$ replacing H.

The final case is where we have $\lvert N_G(S)\cup S\rvert \geq \left (1+1/(10\log {k})\right )\lvert S\rvert + m\Delta \beta $ for every set $S \subseteq V(G)$ with $m \leq \lvert S\rvert \leq \lvert G\rvert /2$ . Consider the largest set $X \subseteq V(G)$ with $\lvert X\rvert \leq 2m$ such that $\lvert N_G(X)\rvert \leq \Delta \lvert X\rvert $ . Since $\beta \geq 4\Delta $ , we have that $\lvert N_G(X) \cup X\rvert \leq (\Delta + 1)\lvert X\rvert \leq m\Delta \beta $ . Thus, we must have $\lvert X\rvert < m$ . Now, consider $F = G- X$ . If there is some $Y\subset V(F)$ with $\lvert Y\rvert \leq m$ and $N_F(Y) < \Delta \lvert Y\rvert $ , then

$$\begin{align*}\lvert N_G(X \cup Y) \cup (X \cup Y)\rvert \leq \lvert N_G(X) \cup X\rvert + \lvert N_F(Y) \cup Y\rvert < (\Delta + 1)\lvert X\cup Y\rvert, \end{align*}$$

contradicting our choice of X.

For $\lvert Y\rvert \geq m$ , we have

$$\begin{align*}\lvert N_F(Y) \cup Y\rvert \geq \lvert N_G(Y) \cup Y\rvert - \lvert X\rvert\geq \left(1 + \frac{1}{10\log{k}}\right)\lvert Y\rvert + m\Delta\beta - m. \end{align*}$$

If $\lvert Y\rvert \leq \beta m$ , then utilising $\beta \geq 10\log k$ we have that $\lvert N_F(Y) \cup Y\rvert \geq \lvert Y\rvert +\Delta \lvert Y\rvert $ , whereas if $\beta m \leq \lvert Y\rvert \leq \lvert F\rvert /2$ , then $\lvert N_F(Y) \cup Y\rvert \geq \left (1 + 1/10\log {k}\right )\lvert Y\rvert $ . Thus, F fulfils the requirements of Lemma 4.2 for $H'=H$ .

Proof. Set $A_0=A$ and $A_{i+1}=(N(A_i)\cup A_i)\setminus C$ for each i. Since we have that $\lvert A_{i+1}\rvert \geq 2\lvert A_i\rvert $ for $\lvert A_i\rvert <\beta d$ , and thus $\lvert A_a\rvert \geq \beta d$ , where $a=\log _2(\beta d)$ . For $i\geq a$ , we have $\lvert N(A_i)\rvert \geq \frac {1}{10\log k}\lvert A_i\rvert \geq 2\lvert C\rvert $ , and so $\lvert A_{i+1}\rvert \geq (1+1/20\log k)\lvert A_i\rvert $ . Consequently, $\lvert A_b\rvert \geq \lvert G\rvert /2$ , where $b=a+\log _{1+1/20\log k}(\lvert G\rvert /2\beta d)$ . Applying $\log (1+x)=-\log (1-\frac {x}{1+x})\geq (1+1/x)^{-1}$ for $x=1/20\log k$ and noting that $1+20\log k\leq 22\log k$ , we get

$$\begin{align*}b\leq a+22\log k\log \lvert G\rvert-22\log k\log(2\beta d)\leq22\log k\log \lvert G\rvert. \end{align*}$$

Since the same argument applies to B, we have that $A_b\cap B_b\neq \varnothing $ , and so there is a path of length at most $2b\leq 44\log k \log \lvert G\rvert $ avoiding C, as desired.

Proof. Since G is not bipartite, it contains an odd cycle. Let C be a shortest odd cycle; it follows that if $x,y\in V(C)$ , then $\operatorname {dist}_G(x,y)=\operatorname {dist}_C(x,y)$ . Indeed, if not, choosing $x,y$ to violate this condition with minimal distance, the shortest path between x and y is internally disjoint from the cycle, and adding this path to whichever path around the cycle has the appropriate parity gives a shorter odd cycle. Consequently, $\lvert C\rvert $ is at most twice the diameter of G. Using Lemma 4.4 with $A_{3.4},B_{3.4}$ being any singletons and $C_{3.4}=\varnothing $ implies the desired bound $\lvert C\rvert \leq 88\log k\log \lvert G\rvert $ .

Now, if C is a triangle then any vertex v can have at most three neighbours on C (trivially). Otherwise, v can have at most two neighbours on C; indeed, if v has at least three neighbours on C, then either v has two neighbours $x,y$ with $\operatorname {dist}_C(x,y)\geq 3>\operatorname {dist}_G(x,y)$ , contradicting the earlier observation, or v has two neighbours $x,y$ which are adjacent, giving a shorter odd cycle $vxy$ . Thus, for any set S we have $\lvert N_G(S)\cap (W\setminus V(C))\rvert \geq \lvert N_G(S)\cap W\rvert -3\lvert S\rvert $ , and, since $G \ (\Delta ,\beta ,d,k)$ -expands into W, G also $(\Delta -3,\beta ,d,k)$ -expands into $W\setminus V(C)$ .

Proof. We iteratively build up sets $A_i,B_i$ with $\lvert A_i\rvert =\lvert B_i\rvert =\Delta ^i$ such that any vertex in $A_i$ is connected to x by a path in $A_i$ of length at most i until i is large enough that $\lvert A_i\rvert \geq \beta d/2$ . We start from $A_0=\{x\},B_0=\{y\}$ . Given disjoint sets $A_i,B_i$ with $\lvert A_i\rvert =\lvert B_i\rvert <\beta d/2$ , we aim to choose disjoint sets $A'\subset N_G(A_i)\cap W$ and $B'\subset N_G(B_i)\cap W$ of size at least $\Delta \lvert A_i\rvert $ . Then we can pick $A_{i+1}\subseteq A_i\cup (A'\setminus B_i)$ and $B_{i+1} \subseteq B_i \cup (B'\setminus A_i)$ with the required disjointness, connectivity and size properties.

First, take $A"=(N(A_i) \cap W)\setminus N(B_i)$ and $B"=(N(B_i) \cap W)\setminus N(A_i)$ . If either of these, say $A"$ has size at least $\Delta \lvert A_i\rvert $ , then we may choose $A'\subseteq A"$ and $B'\subseteq N(B_i) \cap W$ both of size $\Delta \lvert A_i\rvert $ (using the expansion property for the latter); by definition of $A"$ these sets are disjoint.

Consequently, we may assume $\lvert A"\rvert ,\lvert B"\rvert <\Delta \lvert A_i\rvert $ . Since $\lvert A_i\cup B_i\rvert <\beta d$ , the expansion property gives $\lvert N(A_i)\cap N(B_i) \cap W\rvert \geq 2\Delta \lvert A_i\rvert -\lvert A"\rvert -\lvert B"\rvert $ , and so we can add disjoint subsets of $\lvert N(A_i)\cap N(B_i) \cap W\rvert $ to $A"$ and $B"$ to form sets $A',B'$ as required.

Proof. First, note that we may assume, at the cost of replacing the upper bound with $\lvert H\rvert \leq 2mk$ , that every part of H has order at least $m'$ . Indeed, we may replace each part of H with order below $m'$ with a part of order $\lceil m'\rceil $ ; this adds at most $sm'$ vertices, and the new graph has average part size at most $2m'$ . In particular, this means that if $H'\sqsubseteq H$ is a-partite of order $ab$ , then $b\geq m'$ and $ab\leq 2m'(s+1)\leq 2mk$ . Clearly, replacing H with a supergraph preserves the property that $\overline {G}$ is H-free.

Set

$$\begin{align*}M=(10^{4}k\log k) \times \frac{2m}{m'}\quad\text{ and }\quad \beta=M/(200)\geq 100. \end{align*}$$

Then, as $1 \leq s \leq k-1$ , we get

$$\begin{align*}\lvert G\rvert\geq 8\times10^{4}smk\log^2 k\geq M(2m'(s+1))\log (s+1). \end{align*}$$

Further, note that $M \geq 4(s+1)$ and $\beta \geq 10 \log k$ . We use Lemma 4.2 with $(M, \beta , \Delta , m, k)_{3.2} = (M, \beta , 10, m', s+1)$ to pass to a subgraph $G'$ , where $\overline {G'}$ is $H'$ -free for some $H'\sqsubseteq H$ and which $(10,\beta ,m",s')$ -expands into $V(G')$ , where $\lvert H'\rvert =s'm"$ and $\chi (H')=s'$ with $2 \leq s' \leq s+1$ , such that

(9) $$ \begin{align} Mm"s'\log s' - m"\leq\lvert G'\rvert\leq Mm"s'\log s'. \end{align} $$

In particular, since $m"\geq m'$ , we have that $G' \ (10,\beta ,m',s')$ -expands.

As $\overline {G'}$ is $H'$ -free, $G'$ has no independent set of order $\lvert H'\rvert = s'm"$ , and it follows that $\chi (G')\geq \lvert G'\rvert /(s'm")>2$ . Note that as $m \leq k^{22}$ , from inequality (9) we have that $\lvert G'\rvert \leq M (2m'(s+1)) \log k \leq 10^{5}mk^2 \log ^2 k \leq 10^{5}k^{26}$ . Thus, by Lemma 4.5 we may find an odd cycle $C_1$ of length at most $88\log k \log \lvert G'\rvert \leq 10^4\log ^2 k$ , such that we retain $(7,\beta ,m',s')$ -expansion into $V(G')\setminus V(C_1)$ . Choosing almost-antipodal points $v_1,w_1$ on $C_1$ (i.e., $\operatorname {dist}_{C_1}(v_1,w_1) = (\lvert C_1\rvert - 1)/2$ ), we use Lemma 4.6 to find sets $A_1$ and $B_1$ of size $\beta m'/2$ with $A_1\cap B_1=\varnothing $ , $A_1\cap V(C_1)=\{v_1\}$ and $B_1\cap V(C_1)=\{w_1\}$ , with each vertex in $A_1$ (resp. $B_1$ ) having a path in $A_1$ to $v_1$ (resp. in $B_1$ to $w_1$ ) of length at most $\log _{7}(\beta m'/2)<40\log ^2 k$ .

Next, we set $X=A_1\cup V(C_1)\cup B_1$ and $Y=V(C_1)$ . Note that, by inequality (9) and that $\lvert X\rvert \leq \beta m' + 10^4 \log ^2 k$ , we have $\lvert G' - X\rvert \geq \frac {1}{2}Mm"s'\log s'$ . We apply Lemma 4.2 with $(M, \beta , \Delta , m, k)_{3.2} = (M/2, \beta , 10, m", s')$ again to find a subgraph $G"$ with $V(G")\subset V(G')\setminus X$ which $(10,\beta , m"',s")$ -expands and has order at most $\frac {1}{2}Mm"'s"\log s"$ , for some $2 \leq s"\leq s'$ and $m"'\geq m'$ . Within $G"$ , we find another similar structure $A_2\cup C_2\cup B_2$ , and since $G' \ (10, \beta , m', s')$ -expands we can use Lemma 4.4, with parameters $(\Delta , \beta , d, k, A, B, C, W)_{3.4} = (10, \beta , m', s', A_2, B_1, V(C_1 \cup C_2), G')$ to find a path $Q_1$ between $A_2$ and $B_1$ that avoids $V(C_1\cup C_2)$ and is of length at most $5,000\log ^2 k$ . We may assume $Q_1$ is disjoint from $A_1\cup B_2$ since otherwise we may find a shorter path between (say) $A_1$ and $A_2$ then relabel appropriately. We then extend $Q_1$ by paths within $A_2$ to $v_2$ and within $B_1$ to $w_1$ to obtain a $w_1$ - $v_2$ path $P_1$ of length at most $10^4\log ^2 k$ .

Now, we update X to $A_1\cup V(C_1\cup P_1\cup C_2)\cup B_2$ (note that the unused parts of $A_2\cup B_1$ are released) and Y to $V(C_1\cup P_1\cup C_2)$ . As before, we choose $G"'$ with $V(G"')\subset V(G')\setminus X$ such that $G"' \ (10,\beta , m^{(4)},s"')$ -expands for some $2 \leq s"'\leq s'$ and $m^{(4)}\geq m'$ and continue as before, updating X and Y. Observe that we can continue this process as long as $\lvert G' - X\rvert \geq \frac {1}{2}Mm"s'\log s'$ . Thus, since $\lvert G'\rvert \geq Mm"s'\log s' - m"$ , we can continue whenever $\lvert X\rvert \leq \frac {1}{4}Mm"s'\log s'$ . Since $mk + \beta m' \leq \frac {1}{4}Mm"s'\log s'$ , we can continue the process until $\lvert Y\rvert>mk$ , as $\lvert A_i\rvert , \lvert B_i\rvert \leq \beta m'/2$ for all i. Moreover, at each stage of this process $\lvert Y\rvert $ increases by at most $2\times 10^4\log ^2 k$ . Thus, as soon as $\lvert Y\rvert>mk$ we can stop the process and ensure $\lvert Y\rvert \leq mk + (2\times 10^4)\log ^2 k$ at this point. Since $mk + 2\times 10^4\log ^2 k \leq \beta m'/20\log (s+1) \leq \beta m'/20\log s'$ , we can then apply Lemma 4.4 with $(\Delta , \beta , d, k, A, B, C, W)_{3.4} = (10, \beta , m', s', A_1, B_r, Y, G')$ to find a path of length at most $10^4\log ^2 k$ connecting the two remaining large sets $A_1,B_r$ avoiding Y, which closes up the structure into an r-adjuster; since each cycle $C_i$ and each path $P_i$ built in this whole process has length at most $10^4\log ^2 k$ , we have $r \geq \frac {mk}{2 \times 10^4\log ^2 k}$ , and the adjuster contains all of Y, giving the required bounds.

Proof. We proceed as in Lemma 5.2, with the following differences. First, set

$$\begin{align*}M=\frac{n}{10m'\log(s+1)}\geq\frac{n}{10m'\log k}\quad\text{and}\quad\beta=M/200.\end{align*}$$

Secondly, at each stage rather than finding a cycle with two large sets, each of size $\beta m'/2$ , attached we find a single edge with two large sets. This ensures that after each joining step we have a path with two large sets attached. Again, we continue until the total size of the path exceeds $\beta m'/40\log k$ and then join the two large sets, using Lemma 4.4, to obtain a desired long cycle. If this cycle is longer than our desired upper bound, we truncate it using that $\overline {G}$ is H-free (choose $mk$ vertices that are nonadjacent on the cycle).

Proof. We may assume each $P_i$ is internally disjoint from $F_1\cup F_2$ since otherwise we can replace it by a shorter path with this property. Write $x_i$ for the endpoint of $P_i$ in $F_1$ , and $y_i$ for the endpoint in $F_2$ .

We will progressively reduce the number of paths to focus our attention on at least $s^{1/2}/4$ paths which relate to one another in a useful way. First, we give each path $P_i$ an ordered pair of labels from $\{+,-\}$ as follows. If $x_i$ lies on one of the $r_1$ short cycles of $F_1$ , choose the first label to be $+$ if it is on the longer section of that cycle (that is, the longer section between the two vertices of that short cycle that have degree $3$ in $F_1$ ) and $-$ if it is on the shorter section. If $x_i$ lies on a path between cycles, choose the first label arbitrarily. Choose the second label similarly with respect to $y_i$ . Now, we may find a set of at least $s/4$ paths with identical label pairs. Note that this ensures there are routes $C_1$ around $F_1$ and $C_2$ around $F_2$ which cover all the $x_i$ and $y_i$ for $P_i$ in this set of paths, with $\lvert C_i\rvert \in [\ell _i-r_i,\ell _i]$ . In what follows, we consider $C_1$ and $C_2$ to be oriented in a particular direction around $F_1$ and $F_2$ , respectively. Renumbering, if necessary, we may assume these paths are $P_1,\ldots ,P_{\lceil s/4\rceil }$ , and that $x_1,\ldots ,x_{\lceil s/4\rceil }$ are in order of their appearance in $C_1$ . By the Erdős–Szekeres theorem (Theorem 3.1), there is a subset of at least $t\geq \left \lceil \sqrt {s/4}\right \rceil \geq 2$ paths with endvertices $y_i$ either in order of their appearance in $C_2$ or in reverse order. Again, by renumbering and reversing $C_2$ if necessary we may assume these are $P_1,\ldots ,P_{t}$ and the $y_i$ appear in reverse order.

Let $C^{\prime }_1$ be the longest route in $F_1$ , and associate each vertex $x_i$ for $i\in [t]$ with a vertex $x^{\prime }_i$ , chosen as follows. If $x_i$ is internal to a shorter side of a short cycle, let $x^{\prime }_i$ be the vertex on the longer side which is the same distance from the start vertex of the cycle (consequently, $x^{\prime }_i$ is one step further from the end of the cycle than $x_i$ ). If $x_i$ is on one of the paths or is internal to the longer side of a short cycle, set $x^{\prime }_i=x_i$ . Define $C^{\prime }_2$ and each $y_i'$ similarly.

Now, we show that some consecutive pair of paths work. Consider making a new adjuster from $F_1$ , $F_2$ , $P_i$ and $P_{i+1}$ for $i\in [t]$ (we take subscripts modulo t) as follows: Starting from $x_i$ , traverse $P_i$ to $y_i$ , then follow a route around $F_2$ to $y_{i+1}$ , including both sections of any short cycle traversed in this route (apart from any short cycle containing $y_i$ or $y_{i+1}$ ), traverse $P_{i+1}$ to $x_{i+1}$ , then follow a route around $F_1$ to $x_i$ similarly. See Figure 2 for an example.

Figure 2 Creating a new adjuster using paths $P_i$ and $P_{i+1}$ with label $(-,+)$ . Dashed sections are not part of the new adjuster, which has five fewer short cycles. The red section shows how much length is lost from $F_1$ and $F_2$ ; in the former case, this includes one extra edge since $x_{i+1}$ is on the short side of a cycle.

Write $S_1$ for the edges of $C^{\prime }_1$ between $x^{\prime }_i$ and $x^{\prime }_{i+1}$ and $S_2$ for the edges of $C^{\prime }_2$ between $y^{\prime }_{i+1}$ and $y^{\prime }_{i}$ . The short cycles of the new adjuster are those of the original two adjusters except for any which intersect $S_1$ or $S_2$ . The longest route through the new adjuster uses all of $C^{\prime }_1$ except for at most $\lvert S_1\rvert + 1$ edges (the $+1$ added if $x^{\prime }_{i+1}\neq x_{i+1}$ ), and all of $C^{\prime }_2$ except for at most $\lvert S_2\rvert + 1$ edges (the $+1$ added if $y^{\prime }_{i+1}\neq y_{i+1}$ ). It also uses all of the two paths $P_i$ and $P_{i+1}$ , which contain at least one edge each, so its length is at least $\ell _1+\ell _2-\lvert S_1\rvert -\lvert S_2\rvert $ . Note that the sets $S_1\cup S_2$ obtained for different choices of $i\in [t]$ are disjoint.

Now, we claim that for some pair $P_i,P_{i+1}$ the adjuster constructed has the required properties. Note that, since each path $P_i$ has length at most t, the upper bound on length is satisfied for all adjusters constructed in this way, so it suffices to prove the lower bounds on $\ell $ and r. Suppose not, then for each i either more than $(r_1+r_2)4s^{-1/2}+4$ short cycles are lost, or $\lvert S_1\rvert +\lvert S_2\rvert>4s^{-1/2}(\ell _1+\ell _2)$ . In the former case, this means more than $(r_1+r_2)4s^{-1/2}$ short cycles lie entirely in the section between $x_i$ and $x_{i+1}$ or the section between $y_{i+1}$ and $y_i$ since only $4$ can lie partially in these sections. Since these sections are disjoint, the former case occurs for fewer than $s^{1/2}/4$ choices of i. Similarly, the latter case occurs for fewer than $s^{1/2}/4$ choices of i. Since there are at least $s^{1/2}/2$ choices for i, some choice gives an adjuster with the required properties.

Finally, note that the section of the adjuster obtained consisting of vertices not in $F_1$ is contiguous and contributes at least $\ell -\ell _1$ to the overall length. If $r_2=0$ , this section has no short cycles, as required.

Proof of Lemma 5.1.

Let $C= 10^{30}$ . We will prove both parts of the statement simultaneously by finding an r-adjuster of suitable length for some $r\geq 9mk/8$ . If $s\geq 2$ , this adjuster can be taken to have length slightly larger than n such that one of the routes has length exactly n. However, for $s=1$ this is not possible since G itself could have order less than n. In this case, we show that the length of the adjuster can be made significantly larger than $n/2$ .

We repeatedly apply Lemma 5.2, with $H_{4.2}=H'$ and $G_{4.2}=G-X_{i-1}$ , where $X_0=\varnothing $ and $X_i=V(F^{\prime }_{i})$ for $i\geq 1$ is the set of vertices used up from G (see the construction of $F_i'$ , for each i, below), in order to find $f \leq 4\times 10^4\log ^2 k$ adjusters $F_i$ for $i\in [f]$ , such that $F_i$ is an $r_i$ -adjuster and $r_{\mathrm {total}}:=\sum ^{f}_{i=1}r_i\geq 2mk$ . Note that, since $k^{22}\geq m\geq C= 10^{30}$ and $k\geq 10\log ^4 k$ for any $k\geq 2$ , each $F_i$ satisfies

(10) $$ \begin{align} \lvert F_i\rvert\geq mk\geq m_2+4.1\times 10^{18}\log^4k. \end{align} $$

Also, the total size of all these adjusters (using the fact that every adjuster found has size at most $mk+3\times 10^4\log ^4 k\leq 1.1mk$ ) is at most $5\times 10^4mk\log ^2k$ .

Set $F^{\prime }_1=F_1$ and $r^{\prime }_1=r_1$ . After finding each $F_i$ (for $i\geq 2$ ), we merge it with the adjuster $F^{\prime }_{i-1}$ using Lemma 5.4. There are at least $q=4.1\times 10^{18}\log ^4k$ vertex-disjoint paths between the two adjusters, by inequality (10) and the hypothesis, and we may choose these to have length at most $\lvert H'\rvert +s\leq mk+k$ since the fact that $\overline G$ is $H'$ -free implies that any longer path contains a shortcut. Thus, we obtain a merged adjuster $F^{\prime }_i$ , which is an $r^{\prime }_i$ -adjuster with

(11) $$ \begin{align} (r^{\prime}_{i-1}+r_i)\geq r^{\prime}_i\geq (r^{\prime}_{i-1}+r_i)(1-4q^{-1/2}) - 4, \end{align} $$

and

(12) $$ \begin{align} \lvert F^{\prime}_i\rvert\leq \lvert F^{\prime}_{i-1}\rvert+\lvert F_i\rvert+2mk+2k. \end{align} $$

It follows from inequality (12) that for each $i \in [f]$ ,

$$\begin{align*}\lvert F^{\prime}_{i}\rvert\leq \sum_{j\leq i}\lvert F_{j}\rvert+(i-1)(2mk+2k)\leq (5\times 10^4)mk\log^2k+(4\times 10^4)\log^2 k(2mk+2k)\leq n/10,\end{align*}$$

and so, setting $X_i=V(F^{\prime }_i)$ , we have $\lvert G-X_{i-1}\rvert \geq (sn - n/10) - n/10 \geq sn/10$ . Consequently, we can indeed repeatedly apply Lemma 5.2 with $G_{4.2} = G-X_{i-1}$ for all $i \in [f]$ .

Using inequality (11), by induction we have $r^{\prime }_{i-1}\leq \sum _{j<i}r_j$ for each $i\geq 2$ , and so

(13) $$ \begin{align} r^{\prime}_i\geq r^{\prime}_{i-1}+r_i-4q^{-1/2}r_{\mathrm{total}} - 4. \end{align} $$

Summing inequality (13), we obtain $r^{\prime }_{f}\geq r_{\mathrm {total}}(1-((16\times 10^4)q^{-1/2}\log ^2 k))-4f\geq 3r_{\mathrm {total}}/4$ .

We repeatedly apply Lemma 5.3 with $H_{4.3} = H'$ and $G_{4.3} = G - Y_j$ , where $Y_1 = V(F_{f}')$ and $Y_j =Y_1\cup V(C_j')$ for $j \geq 2$ is the set of vertices used up from G so far (see the construction of $C_i'$ below), in order to find $c \leq 1.6 \times 10^6\log ^2 k$ cycles $C_i$ each of length at least $n/(8\times 10^5\log ^2k) \geq m_2 + 4.1 \times 10^{18}\log ^4k$ (which we regard as $0$ -adjusters) satisfying the following bound on their total length:

• • if $s\geq 2$ then $3n/2 \geq \sum _{i=1}^c\lvert C_i\rvert \geq 4n/3$ , whereas

• • if $s=1$ , then $3n/4 \geq \sum _{i=1}^c\lvert C_i\rvert \geq 2n/3$ .

Set $C_1' = C_1$ . After finding each $C_i$ (for $i \geq 2$ ), we merge it with the cycle $C_{i-1}'$ using Lemma 5.4. As before, there are at least $q = 4.1 \times 10^{18}\log ^4 k$ vertex-disjoint paths between the two cycles, by the hypothesis, and we may choose these to have length at most $\lvert H'\rvert + s \leq mk + k$ . Thus, we obtain a merged cycle $C_i'$ with

$$ \begin{align*} (\lvert C_{i-1}'\rvert + \lvert C_i\rvert)(1 - 4q^{-1/2})\leq \lvert C_i'\rvert \leq \lvert C_{i-1}'\rvert + \lvert C_i\rvert +2mk + 2k. \end{align*} $$

By induction, and that after each merging of cycles we can truncate the length of the new cycle (just as we did with the paths), we have $\lvert C_{i-1}'\rvert \leq \left (\sum _{j<i}\lvert C_j\rvert \right ) + 2mk+2m$ for each $i \geq 2$ , and so

(14) $$ \begin{align} \lvert C_i'\rvert \geq \lvert C_{i-1}'\rvert + \lvert C_i\rvert - 4q^{-1/2}\left(\left(\sum_{j=1}^{c}\lvert C_j\rvert\right) + 2mk + 2m\right). \end{align} $$

Summing inequality (14), we obtain

$$\begin{align*}\lvert C_c'\rvert \geq \sum_{j=1}^{c}\lvert C_j\rvert(1 - (6.4 \times 10^6)q^{-1/2}\log^2 k) - ((6.4 \times 10^6)q^{-1/2}\log^2 k)(2mk + 2m),\end{align*}$$

hence for $s\geq 2$ , $\lvert C_c'\rvert \geq 5n/4$ and for $s=1$ , $\lvert C_c'\rvert \geq 0.62n$ .

Now, merge $C_c'$ with $F_{f}'$ using Lemma 5.4 to produce our desired r-adjuster. One can see that $r \geq (3/4)^2r_{\mathrm {total}}\geq 9mk/8\geq (m+1)k$ .

Suppose $s\geq 2$ . Note that the final use of Lemma 5.4 merges a adjuster of length less than n with a $0$ -adjuster of length at least $5n/4$ , and thus, our r-adjuster of length $\ell>n$ contains a section of length at least $\ell -n$ in which there are no short cycles. Using the fact that $\overline G$ is $H'$ -free, we can shortcut parts of this section, if necessary, without reducing r, until $0\leq \ell -n\leq mk+k$ . Now, using the fact that it is an r-adjuster for some $r\geq (m+1)k$ , we may find a route of length exactly n.

Finally, if $s=1$ the final adjuster similarly has length $\ell \geq 0.6 n$ and contains a section of length $\ell -0.6n$ with no short cycles. We can shorten this section, if necessary, as before to obtain an r-adjuster of length between $0.6n$ and $0.7n$ .

Proof of Theorem 1.2.

Set $k=\chi (H)$ and $m=\lvert H\rvert /k$ . By adding edges to H, if necessary, without changing $\sigma (H)$ , we may assume it is a complete k-partite graph for $k=\chi (H)$ . If $m>k^{22}$ , then the same is true of the size of the largest part, and so Corollary 3.2 gives the required bound for some constant $C"$ (and in fact the logarithmic term is not needed). If $C'\leq m\leq k^{22}$ , then Theorem 6.1 gives the required bound with constant $10^{60}$ . Finally, if $m<C'$ , then by increasing the size of the largest part we may replace H with a graph $H'$ of order at most $C'\lvert H\rvert $ and $\sigma (H')=\sigma (H)$ which satisfies the conditions of Theorem 6.1, giving that $C_n$ is $H'$ -good (and hence H-good) for $n\geq 10^{60}C'\lvert H\rvert \log ^4\chi (H)$ .

Proof. Let C be a constant given in Lemma 5.1. Suppose not. Start with $A=V(G)$ and $S=\varnothing $ . If there exists a set X of order at most $4.1\times 10^{18}\log ^4k$ such that removing X from $G[A]$ increases the number of components of $G[A]$ of order at least $m_2$ , remove X from A and add it to S. Do this for k iterations or until no such set exists. At the end of the process, we have $\lvert S\rvert \leq 4.1\times 10^{18}k\log ^4k$ .

Note that any set B of order at least $m_2$ satisfies $\lvert B\cup N(B)\rvert \geq n$ , else $G[V(G)\setminus (B\cup N(B))]$ satisfies (i). In particular, if $B\subset A$ with $\lvert B\rvert \geq m_2$ , then

$$\begin{align*}\lvert B\cup N_{G[A]}(B)\rvert\geq \lvert B\cup N(B)\rvert-\lvert S\rvert\geq n-4.1\times 10^{18}k\log^4 k\geq \lvert H\rvert.\end{align*}$$

It follows that the total size of all components of $G[A]$ of size less than $m_2$ is less than $m_2$ , since otherwise there is a union of some of these components of size between $m_2$ and $2m_2$ , contradicting the fact that sets of this size expand well. It also follows that any other component has order at least $n-4.1\times 10^{18}k\log ^4 k$ ; let these components of $G[A]$ be $A_1,\ldots A_t$ , where $t\geq 1$ and

(15) $$ \begin{align}\sum_{i\in[t]}\lvert A_i\rvert\geq\lvert A\rvert-m_2\geq \lvert G\rvert-\lvert S\rvert-m_2\geq(k-1)n-k-4.1\times 10^{18}k\log^4k-m_2. \end{align} $$

We must have $t\leq k-1$ since otherwise $\overline G$ contains a copy of H obtained by choosing a suitable number of vertices from $A_1,\ldots ,A_k$ .

In particular, we added vertices to S fewer than k times, and so the process described above stopped because no more removals were possible. This will give the required connectivity properties. Indeed, if $B_1,B_2$ are disjoint sets in $A_i$ of size at least $m_2+4.1\times 10^{18}\log ^4 k$ , then no set X of order $4.1\times 10^{18}\log ^4 k$ separates $B_1\setminus X$ from $B_2\setminus X$ within $G[A_i]$ (since both these sets have size at least $m_2$ and by the construction of A). Thus, by Menger’s theorem there are at least $4.1\times 10^{18}\log ^4 k$ disjoint paths between them in $G[A_i]$ .

For each $i\leq t$ , we define $k-t+1$ subgraphs $H^{(1)}_i\sqsubset \cdots \sqsubset H^{(k-t+1)}_i\sqsubset H$ as follows. Set $H_i^{(1)}$ to be an independent set of $m_{k+1-i}$ vertices, and for $2\leq j\leq k-t+1$ set $H_i^{(j)}=K_{m_{k+1-i},m_{k-t},m_{k-t-1},\ldots ,m_{k-t-j+2}}$ , that is, $H_i^{(j)}$ is a complete j-partite subgraph consisting of the ith largest class of H together with $j-1$ of the $k-t$ smallest classes taken in decreasing order.

For each $i\leq t$ , let $s_i\geq 1$ be the largest value such that $\overline {G[A_i]}$ contains $H_i^{(s_i)}$ . Suppose that $\sum _{i\in [t]} s_i\geq k$ , and consider the graph induced in $\overline {G}$ by the vertices of a copy of $H_i^{(s_i)}$ in $\overline {G[A_i]}$ for each i. Taking the largest class from each copy creates a copy of $K_{m_k,\ldots ,m_{k-t+1}}$ , and the other classes give a complete $(k-t)$ -partite graph with jth largest part having size at least $m_{k-t+1-j}$ for each $j\in [k-t]$ , which contains $K_{m_{k-t},\ldots ,m_1}$ . Thus, $\overline G\supset H$ , a contradiction.

Consequently, we have $\sum _{i\in [t]} s_i\leq k-1$ , and in particular $s_i\leq k-t$ for each i. By definition of $s_i$ , $\overline {G[A_i]}$ is $H^{(s_i+1)}_i$ -free for each i. If $s_i>1$ , we have $\lvert A_i\rvert <s_in-n/10$ , since otherwise Lemma 5.1, taking $H^{\prime }_{4.1}=H_i^{(s_i+1)}$ , gives a copy of $C_n$ inside $G[A_i]$ . Contrariwise, if $s_i=1$ we have $\lvert A_i\rvert <n+m_{k-t}$ by Corollary 3.3. In either case, for each $i\in [t]$ we have $\lvert A_i\rvert <s_in+m_{k-t}$ . Note that $(t-1)m_{k-t}\leq \lvert H\rvert =mk$ . Thus, if for any i we have $s_i>1$ , then

$$\begin{align*}\sum_{j\in [t]}\lvert A_j\rvert<s_in-n/10+\sum_{\substack{j\in[t]\\ j\neq i}}(s_in+m_{k-t})\leq (k-1)n-n/10+(t-1)m_{k-t}, \end{align*}$$

contradicting inequality (15). We may therefore assume each $s_i=1$ . Similarly, if $\sum _{i\in [t]} s_i<k-1$ , and so $t<k-1$ , we obtain a contradiction to inequality (15). Thus, $t=k-1$ , and each of $A_1,\ldots ,A_{k-1}$ has size at least $n-4.1\times 10^{18}k\log ^4k\geq 0.95n$ . If for some i we have $K_{m_1,m_2}\subset \overline {G[A_i]}$ , then taking the copy of $K_{m_1,m_2}$ together with $m_k$ vertices from each other part gives a copy of $K_{m_k,\ldots ,m_k,m_2,m_1}$ (with k parts in total) in $\overline {G}$ , and so $\overline G\supset H$ , a contradiction. Together with the connectivity properties heretofore established, this completes the proof.

Proof. Let $P_i$ have ends $x_i$ in $F_1$ and $y_i$ in $F_2$ . We claim that there is a path $Q_1$ of length at least $\ell _1-m_1-m_2-1$ in $F_1$ between $x_1$ and $x_2$ . To see this, first note that, unless $x_1$ and $x_2$ are on opposite sides of the same short cycle, there is a route in $F_1$ which includes both $x_1$ and $x_2$ and has length at least $\ell _1-2$ . If $x_1$ and $x_2$ are at distance at most $m_2$ in this route, then by taking the whole route except for a path of length at most $m_2$ between the two vertices, we are done. If not, then the $m_1$ vertices immediately before $x_1$ on the route do not include $x_2$ , and the $m_2$ vertices immediately before $x_2$ do not include $x_1$ . There is an edge $z_1z_2$ between these two sets of vertices in $G[V(F_1)]$ , since $\overline {G[V(F_1)]}$ is $K_{m_{1},m_{2}}$ -free, and so there is a path which starts at $x_1$ , goes around the route to $z_2$ , via the extra edge to $z_1$ and then around the route in the opposite direction to $x_2$ . This path misses out at most $m_1+m_2-1$ vertices from the route so has at least the required length. See Figure 3. Finally, if $x_1$ and $x_2$ are on opposite sides of the same short cycle, there is a path around the adjuster from $x_1$ to $x_2$ using the vertices on that short cycle after $x_1$ and before $x_2$ and the longer side of every other short cycle. Similarly, there is another path which uses the vertices on that short cycle after $x_2$ and before $x_1$ . At least one of these uses at least half of every short cycle, so has length at least $\ell _1$ .

Figure 3 A long $x_1$ - $x_2$ path using the extra edge $z_1z_2$ (shown in bold and highlighted if colour is available). Dashed lines indicate parts of the adjuster not in the relevant route.

Now, taking this path together with a corresponding path for $F_2$ and the two paths $P_1,P_2$ between them gives a cycle C of length between $\ell _1+\ell _2-2(m_1+m_2)$ and $\ell _1+\ell _2+2t$ . We extend this to an adjuster by including all short cycles from $F_1$ and $F_2$ such that one of the routes round that short cycle is entirely contained in C. Since all but at most $m_1+m_2$ edges of a route of $F_i$ are included in the cycle, this means at most $2(m_1+m_2)$ short cycles of $F_1$ or $F_2$ are not included, giving $r\geq r_1+r_2-2(m_1+m_2)$ . Finally, note that the section of the adjuster obtained consisting of vertices not in $F_1$ is contiguous, and contributes at least $\ell -\ell _1$ to the overall length. If $r_2=0$ , this section has no short cycles, as required.

Proof of Theorem 6.1.

It suffices to prove, via induction on k, the conclusion for $2C\frac {k-1}{k}\le m\le k^{22}$ , where C is at least the constants in Lemma 5.1 and Lemma 6.2. Since we have $C\leq 2C\frac {k-1}{k}\leq 2C$ , we can then take $C'=2C$ to conclude.

Suppose it is not true, and take a counterexample, that is, a graph G of order $(k-1)(n-1)+\sigma (H)$ that is $C_n$ -free and with H-free complement. We apply Lemma 6.2. If statement (i) applies, then we have a graph $G'\subset G$ of order $(k-2)(n-1)+\sigma (H)$ , with $\hat H$ -free complement. Note that by definition of $\hat H$ we have $\sigma (H)=\sigma (\hat H)$ . Also, since $m_2\leq \frac {mk}{k-1}$ , we have

$$\begin{align*}m(\hat H)\geq m\frac{k(k-2)}{(k-1)^2}\geq 2C\frac{k-2}{k-1}.\end{align*}$$

Thus, if $\lvert \hat H\rvert \leq (k-1)^{23}$ we have $G'\supset C_n$ by the induction hypothesis, whereas otherwise we have $G'\supset C_n$ by Corollary 3.2. Consequently, we must have statement (ii).

Within each set $A_i$ , we apply Lemma 4.2 with $G_{3.2}=G[A_i]$ , $H_{3.2}=K_{m_1,m_2}$ , $M_{3.2}=\lvert A_i\rvert /((m_1+m_2)\log 2)$ and $\beta _{3.2}=M_{3.2}/100>40$ (note that also $\beta _{3.2} \geq 10 \log k$ ) to find a $(10,M_{3.2}/100,(m_1+m_2)/2,2)$ -expander $H_i$ of order at least $\lvert A_i\rvert -(m_1+m_2)/2\geq \lvert A_i\rvert -m$ .

Suppose that $H_i$ and $H_j$ are linked (in G) by two disjoint paths for some distinct $i,j\in [k-1]$ . Recall that $\lvert A_i\rvert \geq 0.95n$ . Applying Lemma 5.1 to $G[A_i]$ , we find an r-adjuster $F_i$ for $r\geq 9mk/8$ with length between $0.6n$ and $0.7n$ . Similarly, we may find an adjuster in $A_j$ of length between $0.6n$ and $0.7n$ ; however, instead of using it as an adjuster we simply take the longest route to find a cycle $C_j$ of that length.

Note that the size of the adjuster and cycle ensures that they intersect $H_i$ and $H_j$ , respectively, in at least $\lvert H_i\rvert +0.6n-\lvert A_i\rvert \geq 0.5n$ vertices. Suppose that the disjoint paths between $H_i$ and $H_j$ have endpoints $x_1$ and $x_2$ in $H_i$ and endpoints $y_1$ and $y_2$ in $H_j$ . By Lemma 4.6, we may find disjoint sets $B_1\ni x_1$ and $B_2\ni x_2$ in $V(H_i)$ of size $\lvert A_i\rvert /200$ such that every vertex in $B_1$ is connected to $x_1$ by a path and likewise for $B_2$ and $x_2$ . We may therefore use Lemma 4.4 with $A_{3.4}=V(F_i)\cap V(H_i),B_{3.4}=B_1\cup B_2,C_{3.4}=\varnothing $ to give a path from $V(F_i)$ to $B_1$ or $B_2$ , without loss of generality the former. Extend it to a path from $V(F_1)$ to $x_1$ using vertices from $B_1$ , and remove vertices from this path, if necessary, to obtain a shortest path $P_1$ . This ensures that $\lvert V(P_1)\rvert \leq m_1+m_2+1$ since $\overline {G[A_i]}$ is $K_{m_1,m_2}$ -free; note also that $P_1$ is disjoint from $B_2$ . Next, apply Lemma 4.4 with $A_{3.4}=(V(F_i) \cap V(H_i)) \setminus V(P_1),B_{3.4}=B_2,C_{3.4}=V(P_1)$ to give a disjoint $x_{2}$ - $V(F_{i})$ path $P_{2}$ . We can similarly find two disjoint paths from $y_1$ , $y_2$ to $F_j$ in $H_j$ , and the union of these with the existing paths between $H_i$ and $H_j$ give two disjoint paths between the adjuster and the cycle. Since $\overline {G}$ is H-free, each of these paths contains a shortest path of length at most $mk+k$ . We apply Lemma 6.3 to merge the adjuster and cycle using these shortest paths, obtaining an adjuster with at least $9mk/8-2(m_1+m_2)\geq mk+k$ short cycles, of total length $\ell $ between $1.2n-2(m_1+m_2)> 1.19n$ and $1.4n+2(mk+k)<1.41n$ . Recall that $F_i$ has length at most $0.7n$ , so the adjuster obtained from merging $F_i$ and $C_j$ contains a section of length at least $\ell -0.7n\ge 0.49n>\ell -n$ without short cycles (the part corresponding to $C_j$ ). We may reduce this section until the total length is between n and $n+mk+k$ , since $\overline G$ is H-free, and then use the reducing power of the adjuster to obtain a cycle of length exactly n.

Thus, we may assume that no pair $H_i,H_j$ is linked by two disjoint paths. It follows that for some index i there is a vertex v which separates $H_i$ from all other $H_j$ . Let A be the component of $G-v$ containing $H_i-v$ , and let $G'=G[A\cup \{v\}]$ . Note that $\overline {G'}$ is $K_{m_1+1,m_2+1}$ -free, since otherwise we may choose $m_k$ vertices from $H_j-v$ for each $j\neq i$ together with the vertices of a $K_{m_1,m_2}$ in $\overline {G'-v}$ to give a graph containing H in $\overline {G}$ . Recall that statement (i) of Lemma 6.2 does not apply. It follows that any set $B\subset A$ of size $m_2$ has $\lvert N_G(B)\cup B\rvert \geq n$ , since otherwise $\overline {G-(N_G(B)\cup B)}$ is not $\hat {H}$ -free, implying that $\overline G$ is not H-free. Since also $N_{G'}(B)\cup B=N_G(B)\cup B$ , we have $\lvert N_{G'}(B)\cup B\rvert \geq n$ . Thus, for any set $B'\subset V(G')$ of order at least $m_2+1$ , we have $\lvert N_{G'}(B')\cup B'\rvert \geq n$ (since $B'$ contains a set of order $m_2$ not including v). Now, using Lemma 5.3 with $G_{4.3}=H_i$ and $H_{4.3}=K_{m_2,m_2}$ , we may find a cycle of length at least $10m_2$ in $H_i$ . Taking $x,y$ to be two consecutive vertices on this cycle, Lemma 3.4 applies to $G_{2.4}=G'$ with $m_{2.4}=m_2+1$ , and we may find a path of order exactly n between x and y, giving a copy of $C_n$ .