Proof. Let $k', k", c$ be such that $k_1 \gg k' \gg k" \gg c \gg k_2$ . For each $y \in Y$ and $j \in J$ , choose a j-father of y and denote it by $f_j(y)$ . We wish to find ‘large’ subsets $Y' \subseteq Y$ and $J' \subseteq J$ such that, for every fixed j, the fathers $f_j(y)$ are distinct for $y \in Y'$ . To this end, we construct sequences $Y = Y_0 \supseteq \ldots \supseteq Y_\ell $ and $j_1, \ldots , j_\ell \in J$ recursively as follows: given $Y_0, \ldots , Y_t$ and $j_1, \ldots , j_t$ for $t < \ell $ , if there exists $j \in J \setminus \{j_1, \ldots , j_t\}$ and a subset $Y' \subseteq Y_t$ of size at least $\sqrt {|Y_t|}$ such that $f_j(y)$ is the same for all $y \in Y'$ , define $j_{t+1} = j$ and $Y_{t+1} = Y'$ . Otherwise, stop the process.

Suppose first that $j_1, \ldots , j_\ell $ and $Y_\ell $ are well-defined through the recursive process. Then $f_j(y)$ is the same for all $y \in Y_{\ell }$ , for every $j \in \{j_1, \ldots , j_{\ell }\}$ . Then $\{f_j(y) : j \in \{j_1, \ldots , j_{\ell }\}\}$ is a set of size $\ell $ that is fully joined to $Y_{\ell }$ . Since $|Y_{\ell }| \ge |Y|^{2^{-\ell }}=(k_1)^{2^{-\ell }} \ge \ell $ , there is a copy of $K_{\ell , \ell }$ in G, contradicting the assumption that G is $K_{\ell , \ell }$ -free.

Now suppose that the process stops at the t-th step for some $t<\ell $ with the outputs $j_1, \ldots , j_t$ and $Y_t$ . Let $J' := J \setminus \{j_1, \ldots , j_t\}$ . Then for every $j \in J'$ , each element in the multiset $\{f_j(y) : y \in Y_t\}$ repeats at most $\sqrt {|Y_t|}$ times. Since $|Y_t| \ge |Y|^{2^{-\ell }}=(k_1)^{2^{-\ell }} \ge (k")^2$ , for every $j \in J'$ there is a subset $Y_j' \subseteq Y_t$ of size $k"$ such that $f_j(y)$ are distinct for all $y \in Y_j'$ . Let $Y' \subseteq Y_t$ be the most popular choice for $Y_j'$ . By averaging, $Y_j' = Y'$ for at least $|J'| / \binom {|Y_t|}{\sqrt {|Y_t|}} \ge (p - k_1^2 - \ell )/2^{k_1} \ge k'$ indices $j \in J'$ ; let $J" \subseteq J'$ be a set of size $k'$ such that $Y_j' = Y'$ for every $j \in J"$ .

Let $W := \{f_j(y) : y \in Y', j \in J"\}$ . We claim that for every $y \in Y'$ there are at most $k"$ elements $w \in W$ such that w sends at least $\frac {1}{4k_2}|N_{H_{\mathrm {sb}}}(\mathrm {ext}(y))|$ edges to $N_{H_{\mathrm {sb}}}(\mathrm {ext}(y))$ . Suppose to the contrary that this is not the case for $y \in Y'$ . Write $N := N_{H_{\mathrm {sb}}}(\mathrm {ext}(y))$ for brevity and let $W'$ be the set of vertices $w\in W$ such that w sends at least $\frac {1}{4k_2}|N|$ edges to N satisfying $|W'| \ge k"$ . By Theorem 3.3, $G[N, W']$ Footnote ² contains a copy of $K_{\ell , \ell }$ , a contradiction. Consider the set

$$ \begin{align*} \left\{j \in J" : \text{ for all } z,y\in Y', f_j(z) \text{ sends at most } \frac{1}{4k_2}|N_{H_{\mathrm{sb}}}(\mathrm{ext}(y))| \text{ edges to } N_{H_{\mathrm{sb}}}(\mathrm{ext}(y))\right\}. \end{align*} $$

By the above statement, this set has size at least $|J"| - |Y'| \cdot k" \ge k' - (k")^2 \ge k_2$ . Let $J_{\mathrm {final}}$ be a subset of the above set of size $k_2$ .

For every $j \in J_{\mathrm {final}}$ , define a graph $F_j$ on vertices $Y'$ whose edges are pairs $y_1 y_2$ where $y_1, y_2 \in Y'$ are distinct and one of the following holds: $f_j(y_i)$ is adjacent to $y_{3-i}$ for some $i \in [2]$ ; $f_j(y_i)$ is adjacent to $\mathrm {ext}(y_{3-i})$ ; or $f_j(y_1)$ and $f_j(y_2)$ are adjacent. By Ramsey’s theorem, the graph $F = \bigcup _{j \in J_{\mathrm {final}}}F_j$ contains either an independent set of size $2k_2$ or a clique of size c. Assume the latter case and suppose that U is a subset of $Y'$ of size c that induces a clique. Then for some $j \in J_{\mathrm {final}}$ the graph $F_j$ has at least $\frac {1}{k_2}\binom {|U|}{2}$ edges. Let $U' := U \cup \{\mathrm {ext}(y) : y \in U\} \cup \{f_j(y) : y \in U\}$ . Then, by definition of $F_j$ ,

$$ \begin{align*} e(G[U']) \ge e(F_j) \ge \frac{1}{k_2}\binom{|U|}{2} \ge \frac{1}{k_2} \binom{|U'|/3}{2}. \end{align*} $$

By Theorem 3.3, $G[U']$ contains a copy of $K_{\ell , \ell }$ , a contradiction.

It remains to consider the case where F contains an independent set U of size $2k_2$ . Write $U = \{y_1, \ldots , y_{2k_2}\}$ and $J_{\mathrm {final}} = \{j_1, \ldots , j_{k_2}\}$ . Let $Q_i$ be a unimodal path between $f_{j_i}(y_{2i-1})$ and $f_{j_i}(y_{2i})$ , for $i \in [k_2]$ . By construction, one may easily check that $\{Q_1, \ldots , Q_{k_2}\}$ is a $\frac {1}{4k_2}$ -good collection of size at least $k_2$ .

Proof of Theorem 4.1.

Recall that, while assuming Lemma 4.5, the parameters are chosen according to the hierarchy $\chi \gg p \gg \kappa \gg k_1 \gg k_2 \gg k, \ell , \omega $ . By Lemma 4.6, there is a collection $\mathcal {Q}$ that consists of $k_2$ paths that are $\frac {1}{4k_2}$ -good. Then by Lemma 4.7, there is a collection of $k_2$ independent $\frac {1}{2k_2}$ -good paths; however, Lemma 4.8 then finds a $k_2$ -wheel in G, and hence, a cycle with k chords.

Proof. Let X be the set of vertices in H that are neighbours of u in G. We claim that $|X| \le k+1$ . Suppose to the contrary that $x_1, \ldots , x_{k+2}$ are distinct vertices in X. As H is $\kappa $ -connected with $\kappa \gg k$ , Theorem 4.3 guarantees that there exist pairwise internally vertex-disjoint paths $Q_1, \ldots , Q_{k+2}$ such that each $Q_i$ ends at $x_i$ and $x_{i+1}$ , where the addition of indices taken modulo $k+2$ . Let $Q_i'$ be the path in the copy of $H_{\mathrm {sb}}$ in G that corresponds to the 1-subdivision of $Q_i$ . Then by concatenating the paths $Q_1', \ldots , Q_{k+2}'$ , we obtain an induced cycle in G that contains at least $k+2$ neighbours of u. Then there exists a cycle with exactly k chords, which contradicts the assumption.

Let M be a maximum matching in H such that, for each edge e in M, $\mathrm {sub}(e)$ is adjacent to u in G. Our next claim is $|M| \le k+1$ . Suppose to the contrary that $x_1 y_1, \ldots , x_{k+2} y_{k+2}$ are vertex-disjoint edges in M. By using Theorem 4.3 as before, there exist vertex-disjoint paths $Q_1, \ldots , Q_k$ in $H \setminus \{x_1 y_1, \ldots , x_{k+2} y_{k+2}\}$ such that each $Q_i$ ends at $y_i$ and $x_{i+1}$ . Again, let $Q_i'$ be the 1-subdivision of $Q_i$ in the copy of $H_{\mathrm {sb}}$ in G. Then $x_1 y_1 Q_1' \ldots x_{k+2} y_{k+2} Q_{k+2}'x_1$ is an induced cycle in G that contains at least $k+2$ neighbours of u, which again yields a contradiction.

Take $R = X \cup V(M)$ . Then $|R| \le 3(k+1)$ and u has no neighbours in $(H \setminus R)_{\mathrm {sb}}$ , as required.

Proof of Lemma 4.5.

Let $k_3, k_4$ be such that Let $k_1 \gg k_2 \gg k_3 \gg k_4 \gg \omega \gg k$ . Say that an edge $e = xy$ of H is triangulated if $\mathrm {sub}(e)$ and at least one of x and y have a common j-father, for at least $k_1$ indices j. Then every vertex in H is incident with at least $2k_1$ triangulated edges and hence, there is a matching $M_1$ in H that consists of $k_1$ triangulated edges.

For each $e \in M_1$ let $x(e)$ and $y(e)$ be the two ends of e, let $f(e)$ be a $j(e)$ -father of $\mathrm {sub}(e)$ which is adjacent to $x(e)$ or $y(e)$ , so that the indices $j(e)$ are all distinct for $e \in M_1$ . Let U be the set $\{f(e) : e \in M_1\}$ ; then $|U|=k_1$ . Since $k_1 \gg k_2, \omega $ and G is $K_{\omega +1}$ -free, the set U contains an independent set of size $k_2$ . Let $M_2$ be a submatching of $M_1$ of size $k_2$ such that $\{f(e) : e \in M_2\}$ is independent.

Let F be the auxiliary graph whose vertices are the edges of $M_2$ , where $e_1 e_2$ is an edge in F whenever $f(e_{3-i})$ is adjacent to at least one of $\mathrm {sub}(e_i), x(e_i)$ and $y(e_i)$ for some $i \in [2]$ . By Lemma 4.9, the vertex $\mathrm {sub}(e)$ is adjacent to at most $3(k+1)$ vertices in the copy of $H_{\mathrm {sb}}$ in G, for every $e \in M_2$ . In particular, every $e\in M_2$ has neighbours in at most $3(k+1)$ of the sets $\{\mathrm {sub}(e'), x(e'), y(e')\}$ with $e' \in M_2$ . Thus, F has at most $3(k+1) |F|$ edges. Turán’s theorem then implies that F contains an independent set of size at least $|F|/6(k+1) \ge k_3$ . Let $M_3$ be a submatching of $M_2$ of size $k_3$ that forms an independent set in F. That is, for every distinct $e, e'\in M_3$ , $f(e)$ is not adjacent to any $\mathrm {sub}(e')$ , $x(e')$ or $y(e')$ .

For each $e \in M_3$ , let $R(e)$ be a set of vertices in H of size at most $3(k+1)$ such that $x(e'), y(e') \notin R(e)$ for all $e' \in M_3$ and $f(e)$ has no neighbours in $(H \setminus R(e))_{\mathrm {sb}}$ except for possibly $\mathrm {sub}(e)$ , $x(e)$ and $y(e)$ ; the existence of such a set is guaranteed by the choice of $M_3$ and by Lemma 4.9.

We consider three cases. Throughout the case analysis, we abuse notation by writing e for the vertex $\mathrm {sub}(e)$ for simplicity. Whenever we consider indexed edges $e_1,e_2,\ldots $ , we shall denote $x_i := x(e_i)$ , $y_i := y(e_i)$ , $f_i := f(e_i)$ and $R_i := R(e_i)$ . The 1-subdivision of a path $Q_i$ in H is denoted by $Q_i'$ , and is again a path in the copy of $H_{\mathrm {sb}}$ .

Case 1: there are k edges $e\in M_3$ such that $f(e)$ is not adjacent to $y(e)$ .

Let $e_1, \ldots , e_k \in M_3$ be distinct edges such that $f_i$ is not adjacent to $y_i$ for $i \in [k]$ (by choice of $M_3$ this means that $f_i$ is adjacent to $x_i$ for $i \in [k]$ ). Let $R := R_1 \cup \ldots \cup R_k$ . As H is $\kappa $ -connected, $H \setminus R$ is $10k$ -connected and thus, by Theorem 4.3, there exist pairwise vertex-disjoint paths $Q_1, \ldots , Q_k$ in $H \setminus R$ such that $Q_i$ has ends $y_i$ and $x_{i+1}$ for $i \in [k]$ , where the addition of indices is taken modulo k. Let C be the cycle $x_1 f_1 e_1 y_1 Q_1' \ldots x_k f_k e_k Q_k'x_1$ . Then C has exactly k chords, namely: $x_1 e_1$ , …, $x_k e_k$ .

Figure 2

Case 1: $f_i$ not adjacent to $y_i$ .

Case 2: k is even and $f(e)$ is adjacent to $x(e)$ and $y(e)$ for at least $k/2$ values of e in $M_3$ .

Write $k = 2s$ . Let $e_1, \ldots , e_{s} \in M_3$ be distinct edges such that $f_i$ is adjacent to $x_i$ and $y_i$ for $i \in [s]$ . Let $R := R_1 \cup \ldots \cup R_s$ . As before, there exist pairwise vertex-disjoint paths $Q_1, \ldots , Q_{s}$ in $H \setminus R$ such that $Q_i$ has ends $y_i$ and $x_{i+1}$ for $i \in [s]$ , where the addition of indices is taken modulo s. Then $x_1 f_1 e_1 y_1 Q_1' \ldots x_{s} f_{s} e_{s} y_{s} Q_{s}'x_1$ is a cycle in G with exactly k chords: $x_1 e_1, \ldots , x_{s}e_{s}$ and $f_1 y_1, \ldots , f_{s} e_{s}$ .

Figure 3

Case 2: k even $f_i$ adjacent to $x_i$ and $y_i$ .

Case 3: k is odd and $f(e)$ is adjacent to $x(e)$ and $y(e)$ for at least $k_4 + 1$ values of e in $M_3$ .

Note that if $k = 1$ then $f(e)\, x(e)\, e\, y(e)$ is a cycle with exactly one chord, for any $e \in M_3$ where $f(e)$ is adjacent to $x(e)$ and $y(e)$ . We may thus assume that $k \ge 3$ ; write $k = 2s- 1$ with $s>1$ .

Let $M_4$ be a submatching of $M_3$ of size $k_4$ that consists of edges e where $f(e)$ is adjacent to $y(e)$ and $j(e)> 1$ (recall that $M_3$ contains at most one edge e with $j(e) = 1$ ). For each $e \in M_4$ , let $g(e)$ be a $1$ -father of $f(e)$ . Here the $g(e)$ ’s are not necessarily distinct. We claim that each $g(e)$ , with $e\in M_4$ , is adjacent to at most $k+1$ of the vertices $f(e')$ with $e' \in M_4$ . Indeed, suppose that $e_1, \ldots , e_{k+2}$ are distinct edges in $M_4$ such that each $f_j$ , with $j \in [s]$ , is adjacent to $g(e)$ . As usual, we can find paths $Q_1, \ldots , Q_{k+2}$ in $H \setminus (R(e_1) \cup \ldots \cup R(e_{k+2}))$ that are pairwise vertex-disjoint and $Q_i$ has ends $y_i$ and $x_{i+1}$ for $i \in [k+2]$ . Then $x_1 f_1 y_1 Q_1' \ldots x_{k+2} f_{k+2} y_{k+2} Q_{k+2}'x_1$ is an induced cycle in G that contains at least $k+2$ neighbours of $g(e)$ . Thus, a cycle with exactly k chords exists.

Let F be the auxiliary graph on the edges in $M_4$ , where $e_1$ and $e_2$ form an edge if $g(e_{3-i})$ is adjacent to at least one of $e_i, x(e_i), y(e_i)$ and $f(e_i)$ for some $i \in [2]$ . By Lemma 4.9 and the previous paragraph, each $g(e)$ , with $e\in M_4$ , is adjacent to at most $4k+4$ vertices amongst $\{e', x(e'), y(e'), f(e'): e' \in M_4\}$ . Therefore, F has at most $(4k+4)|F|$ edges and thus, it has an independent set of size at least $|F|/(8k+4) \ge k_4/(8k+4) \ge \lceil k/2 \rceil $ . Let $e_1, \ldots , e_{s}$ be distinct edges in $M_1$ that form an independent set in F. Write $g_i:=g(e_i)$ and let $R_i'$ be a set of at most $3(k+1)$ vertices in H such that $g_i$ has no neighbours in $(H \setminus R_i')_{\mathrm {sb}}$ except for possibly $x_i, y_i, e_i$ , and $x_j, y_j, e_j \notin R_i'$ for $j \in [s]$ . Such a set $R_i'$ exists by Lemma 4.9 and the fact that $e_i$ ’s form an independent set in F. Observe that the $g_i$ ’s are distinct (since $g_i$ is adjacent to $f_i$ but not to $f_j$ with $j \in [s] \setminus \{i\}$ ), and $g_i$ sends no edges to $\{x_j, y_j, e_j, f_j\}$ for $j \neq i$ . Let $R = R_1 \cup \ldots \cup R_{s} \cup R_1' \cup \ldots \cup R_{s}'$ . As usual, let $Q_1, \ldots , Q_{s-1}$ be pairwise vertex-disjoint paths in $H \setminus R$ such that each $Q_i$ ends at $y_i$ and $x_{i+1}$ . Then the path $y_1 Q_1' x_2 f_2 e_2 y_2 Q_2' \ldots x_{s-1} f_{s-1} e_{s-1} y_{s-1} Q_{s-1}x_s$ has exactly $2(s-2)=k-3$ chords. Similarly, paths with exactly $k-3$ chords exist between any $u\in \{x_1, y_1\}$ and $v\in \{x_{s}, y_{s}\}$ . To extend one such path to a cycle with exactly k chords, we need the following claim.

Proof of the claim.

Write $x = x_i$ , $y = y_i$ , $e = e_i$ , $f = f_i$ , $g = g_i$ for some $i \in \{1, s\}$ , and let $\sigma $ be the number of neighbours of g in $\{x, y, e\}$ . We aim to show that there are paths P with vertices in $\{x, y, e, f, g\}$ whose ends are g and one of $x, y$ satisfying the following requirements (separately, namely P varies).

1. (0) P has no chords, if $\sigma = 1$ ;

2. (1) P has exactly one chord, if $\sigma \neq 1$ ;

3. (2) P has exactly two chords;

4. (3) P has three chords, if $\sigma = 1$ .

By using this, Claim 4.10 easily follows. Indeed, unless $\sigma = 1$ for both $i = 1$ and s, we can take one of $P_1$ and $P_{s}$ to have one and two chords, respectively, using (2) and (1). Otherwise, we can take one to have no chords and the other to have three, using (0) and (3). To show that paths P that satisfies (0)–(3) exist, we consider the four possible values of $\sigma $ .

Figure 4

Proof of Claim 4.10.

• • $\sigma = 0$ . For (1) take $P = gfey$ , and for (2) take $P = gfxey$ .

• • $\sigma = 1$ . Without loss of generality, g is not adjacent to x. For (0), (2), and (3), take $P = gfx$ , $P = gfey$ , and $P = gfxey$ , respectively.

• • $\sigma = 2$ . Without loss of generality, g is adjacent to x. For (1) and (2), take $P = gxey$ and $P = gfey$ , respectively.

• • $\sigma = 3$ . For (1) and (2), take $P = gey$ and $P = gxey$ , respectively.

Let $P_1$ and $P_{s}$ be as in Claim 4.10. Without loss of generality, we may assume that $y_1$ is an end of $P_1$ and $x_{s}$ is an end of $P_{s}$ . Let P be a unimodal path with ends $g_1$ and $g_{s}$ . Then the interior of P sends no edges to the copy of $H_{\mathrm {sb}}$ or $\{f_1, \ldots f_{s}\}$ , since each $g_i$ is a $1$ -father of $f_i$ ’s, whereas the $f_i$ ’s and the copy of $H_{\mathrm {sb}}$ are in $G_2$ . Finally, augmenting the path $y_1 Q_1' x_2 f_2 e_2 y_2 Q_2' \ldots x_{s-1} f_{s-1} e_{s-1} y_{s-1} Q_{s-1}'x_s$ by adding the path $x_s P_{s}g_s P g_1 P_1 y_1$ gives a cycle with exactly k chords.

Proof. Let x and y be the ends of P with $x\in V(K)$ and let $x^-$ and $y^-$ be the unique neighbours of x and y in P. By unimodality, vertices in K do not send any edges to $P \setminus \{x, x^-, y, y^-\}$ . Let w be a neighbour of u in K distinct from x, which exists by the assumption on u.

Figure 6

Lemma 6.4.

Suppose that u sends more than $8\sqrt {k}$ edges to P. Let Q be a path in K with ends x and w on $2a$ or $2a+1$ vertices, where $a = \lfloor \sqrt {k} - 2 \rfloor $ . Let $k'$ be the number of chords in the path Q. Then $k'$ is between $a^2 - (2a-1) = (a-1)^2$ and $a(a+1) - 2a = a^2 - a$ . In particular,

$$ \begin{align*} k' \ge (a-1)^2 \ge (\sqrt{k} - 4)^2 \ge k - 8\sqrt{k}. \end{align*} $$

Let $k"$ be the number of chords in the path $uwQxx^-$ . Then, as $e_G(\{x^-, u\}, V(Q))\leq 2(2a+1)$ ,

$$ \begin{align*} k" = k'+e_G(\{x^-, u\}, V(Q)) - 2 \leq a^2 + 3a \le (a+2)^2 \le k. \end{align*} $$

Take $b = k - k"$ , so that $0 \le b \le 8\sqrt {k}$ . Let $P'$ be the subpath of P that starts at x and ends at the $(b+1)$ -th neighbour $u_b$ of u in $P \setminus x$ . Then $u w Q x P' u_b u$ is a cycle with exactly k chords.

Proof. Let $x, y$ be the ends of P, where $x \in K$ , and let $x^-,y^-$ be the neighbours of $x,y$ in P. Take z to be any vertex in K other than x and take v to be any vertex in $K'$ on the opposite side to w. Let Q be a unimodal path with ends z and v from a later layer than P and a different layer than u, and let $z^-, v^-$ be the neighbours of $z, v$ in Q, respectively. By choice of Q, there are no edges between $P \setminus \{x, x^-, y, y^-\}$ and Q (see Observation 3.2).

Figure 7

Lemma 6.5.

Suppose that there is no cycle with exactly k chords and that u has more than $30 \sqrt {k}$ neighbours in P. By Lemma 6.4, there are at most $8\sqrt {k}$ edges between $x^-$ and Q and at most $8\sqrt {k}$ edges between u and Q. Let $Q'$ be a path in K with ends x and z on either $2a$ or $2a+1$ vertices, where $a = \lfloor \sqrt {k} - 13 \rfloor $ . Denote by $k'$ the number of chords in the path $R := uw vQz Q' x x^-$ , excluding the edge $u x^-$ if it exists. Then

$$ \begin{align*} k' \ge \,\, & \#\{\text{chords in } Q'\} \ge (a-1)^2 \ge (\sqrt{k} - 15)^2 \ge k - 30 \sqrt{k}. \end{align*} $$

As $x^-, z^-, v^-$ , and u are the only vertices in $V(R)$ that may have neighbours in $Q'$ ,

$$ \begin{align*} k' \leq \,\, & \#\{\text{chords in } Q'\} + e_G(\{x^-, z^-, v^-, u\}, V(Q')) + e_G(\{x^-, u\}, V(Q)) + e_G(\{x^-, z^-, v^-, v, u, w\}) \\ \le \,\, & (a^2 - a) + 4 \cdot (2a + 1) + 16 \sqrt{k} + \binom{5}{2} \le (a + 4)^2 + 16\sqrt{k} \le (\sqrt{k} - 9)^2 + 16\sqrt{k}\\ = \,\, & k - 2 \sqrt{k} + 81 \le k, \end{align*} $$

where the last inequality follows from the assumption that k is large. Take $b = k - k'$ , so that $0 \le b \le 30 \sqrt {k}$ . Let $P'$ be the subpath of P that starts at x and ends at the $(b+1)$ -th neighbour $u_b$ of u in $P \setminus \{x, x^-\}$ . Then, as there are no edges between $P \setminus \{x, x^-, y, y^-\}$ and K, $K'$ or Q, the cycle $uwvQzQ'xP'u_b u$ has exactly k chords.

Proof of Lemma 6.3.

Let $K_0, \ldots , K_{100}$ be a collection of vertex-disjoint induced copies of $K_{\ell , \ell }$ in the p-extraction of G such that there are no edges across distinct copies. Denote the bipartition of $K_i$ by $\{U_{i, 1}, U_{i, 2}\}$ . Let $P_{i, j}$ be vertex-disjoint unimodal paths from $U_{i, j}$ to $U_{0, j}$ , for $i \in [100]$ and $j \in [2]$ . We take all the unimodal paths from different layers that are not the last two. Let $u_{i,j}$ be the first internal vertex on $P_{i,j}$ from $K_i$ , for each $i \in [100]$ and $j \in [2]$ . Denote by $u_{i,j}'$ the unique neighbour of $u_{i,j}$ in $V(P_{i,j})\cap V(K_i)$ . That is, the end vertex of $P_{i,j}$ in $K_i$ . A vertex u is said to be complete (resp. anti-complete) to $U_{i,j}$ if it is adjacent to all (resp. none) of the vertices in $U_{i,j}\setminus \{u_{i,j}'\}$ , for $i \in [100]$ and $j \in [2]$ .

We first claim that, by possibly shrinking $\ell $ to $\ell '=\ell /2^{200}$ , we may assume that each $u_{i,j}$ is either complete or anti-complete to $U_{s,t}$ , for each $i, s \in [100]$ and $j, t \in [2]$ . For each $v\in \bigcup _{s=1}^{100}(U_{s,1}\cup U_{s,2})$ , write a $0$ - $1$ vector $x_v\in \{0,1\}^{200}$ to encode its adjacency to $u_{i,j}$ . That is, the $(i,j)$ -coordinate is 1 if $vu_{i,j}\in E(G)$ and 0 otherwise. Then each $U_{s, t}\setminus \{u_{s, t}\}$ can be partitioned into at most $2^{200}$ subsets according to the value of $x_v$ . Replacing $U_{s, t}$ by the largest amongst these subsets and adding $u_{s, t}$ suffices for our purpose.

Let $v_{i,j}$ be the first internal vertex on $P_{i,j}$ from $K_0$ , and let $v_{i,j}'$ be the unique vertex in $V(P_{i, j}) \cap V(K_0)$ . By using the same argument possibly shrinking $\ell $ even further, we may assume that each $v_{i,j}$ is either complete or anti-complete to $U_{s, t}$ , for $i, s \in [100]$ and $j, t \in [2]$ .

Our plan is to make a cycle in the following way. We will choose integers $a_1,\ldots ,a_{100} \ge 0$ , depending on k and the structure we have just found. Starting from $v_{1,1}' \in U_{0,1}$ , we take $P_{1,1}$ to reach $u_{1,1}'\in U_{1,1}$ , then go through a path of length $2a_1+1$ in $K_1$ to reach $u_{1,2}'\in U_{1,2}$ . The journey goes back to $K_0$ through $P_{1,2}$ . Then we move to $v_{2,1}'$ to iterate. After 100 iterations, we close the cycle by moving from $v_{100,2}'$ to $v_{1,1}'$ by an edge.

Figure 8

Part of the cycle $\mathcal {C}(a_1, \ldots , a_{100})$ .

Let us calculate how many chords exist in such a cycle, which we call $\mathcal {C}(a_1,\ldots ,a_{100})$ (note that the number of chords depends only on $a_1, \ldots , a_{100}$ , and not on the choice of the paths in $K_i$ for $i \in [100]$ , by previous assumptions). Firstly, a $(2a_i+1)$ -edge path in a copy of $K_{a_i+1, a_i+1}$ gives $(a_i + 1)^2 - (2a_i+1) = a_i^2$ chords and the path in $K_0$ has at most $100^2$ chords. Second, each $u_{i,j}$ or $v_{i,j}$ that is complete to $U_{s, t}$ contributes $a_s$ chords (to $U_{s, t} \setminus \{u_{s, t}'\}$ ), and there are at most $100^2$ chords between vertices $u_{i, j}$ or $v_{i, j}$ and the path in $K_0$ . Third, each $u_{i,j}$ or $v_{i,j}$ adjacent to $u^{\prime }_{s,t}$ , with $(i,j)\neq (s,t)$ , contributes one chord. Finally, there are at most $O(\sqrt {k})$ chords between internal vertices of $P_{i,j}$ s (by Lemmas 6.4 and 6.5, because otherwise there is a cycle with exactly k chords and then, we are done). Overall, the cycle $\mathcal {C}(a_1, \ldots , a_{100})$ has the following number of chords

(1) $$ \begin{align} \sum_{s=1}^{100} a_s^2 +\sum_{s=1}^{100} t_s a_{s} + O(\sqrt{k}), \end{align} $$

where $t_s$ is the number of vertices $u_{i, j}$ or $v_{i, j}$ that are complete to $U_{s, 1}$ plus the number of vertices $u_{i, j}$ or $v_{i, j}$ that are complete to $U_{s, 2}$ , and $O(\sqrt {k})$ is a function that only depends on the graph structure that we have found, that is, the 101 induced $K_{\ell ,\ell }$ ’s and the unimodal paths in between.

In (1), the $O(\sqrt {k})$ term and the terms $t_s$ , with $s \in [100]$ , are all fixed parameters, that is, they do not depend on the choice of $a_1, \ldots , a_{100}$ . Also $t_s \le 400$ for each $s \in [100]$ . Thus, one may write (1) as

(2) $$ \begin{align} f_G(k)+\sum_{s=1}^{100}(a_s^2 +t_sa:s), \end{align} $$

where $f_G(k)=O(\sqrt {k})$ is a parameter depending on k and G, but not on $a_1, \ldots , a_{100}$ .

Among $t_1, \ldots , t_{100}$ , at least $20$ values are even, or at least $80$ are odd. If the former happens, we may assume $t_1,\ldots ,t_{20}$ are the even numbers by relabelling the indices. Then, by Lemma 5.1, we can choose $b_s\geq 200$ for $s \in [20]$ , such that

$$ \begin{align*} \sum_{s = 1}^{20} b_s^2 = k - f_G(k) + \sum_{s=1}^{20}\frac{t_s^2}{4}. \end{align*} $$

Indeed, large enough k guarantees the right-hand side is a large enough positive integer that can be expressed as the sum of exactly $20$ squares at least $200^2$ . Now let $a_s = b_s-t_s/2$ for $s \in [20]$ and 0 otherwise (notice that $a_s \ge 200 - t_s/2 \ge 0$ ). Then the cycle $\mathcal {C}(a_1,a_2,\ldots ,a_{100})$ has exactly k chords, since (2) becomes

$$ \begin{align*} f_G(k)+\sum_{s=1}^{100} (a_s^2+t_sa:s) =f_G(k) + \sum_{s = 1}^{20} \left( b_s^2 - \frac{t_s^2}{4}\right) = k. \end{align*} $$

Otherwise, there are at least $80$ odd $t_s$ ’s, say $t_1,\ldots ,t_{80}$ . Let r be the unique integer divisible by $4$ such that

$$ \begin{align*} r-3 \le k-f_G(k)+\sum_{s=1}^{80}\frac{t_s^2-1}{4} \le r. \end{align*} $$

By our choice of $t_1,\dots ,t_{80}$ , $(t_s^2 - 1)/4$ is an integer for each $s\in [80]$ . Lemma 5.2 then shows that, since r is a large enough integer divisible by $4$ , there exist integers $b_s \ge 300$ , for $s \in [80]$ , such that

$$ \begin{align*} r = \sum_{s=1}^{80} b_s(b_s+1). \end{align*} $$

Now take $a_s = b_s - (t_s - 1)/2$ if $s \in [80]$ (so $a_s \ge 300 - (t_s-1)/2 \ge 0$ ), and $0$ otherwise. Then the number $k'$ of chords in $\mathcal {C}(a_1,a_2,\ldots ,a_{100})$ , using (2), is

$$ \begin{align*} k' & = f_G(k) + \sum_{s=1}^{100}(a_s^2 + t_s a_s) \\ & = f_G(k) + \sum_{s=1}^{80}\left(b_s(b_s+1) - \frac{t_s^2-1}{4}\right) = f_G(k) + r - \sum_{s=1}^{80}\frac{t_s^2-1}{4}. \end{align*} $$

By our choice of r, $k \le k' \le k+3$ . Therefore, there is a cycle with exactly $k'$ chords for some $k' \in \{k, k+1, k+2, k+3\}$ , as required.

Proof. Suppose first that (a) is the case. Let $x_1, \ldots , x_{20} \in X$ be distinct. Let x be the number of edges in the induced subgraph $G[\{x_1, \dots , x_{20}\}]$ . For non-negative integers $a_1, \ldots , a_{20}$ , let $Q_i$ be a path in $K_i$ of length $2a_i + 1$ . Then each $x_i$ is adjacent to the ends of $Q_j$ for $i, j \in [20]$ . Let $\mathcal {C}$ to be the cycle $x_1 Q_1 x_2 \ldots x_{20} Q_{20} x_1$ .

We now wish to count the number of chords in $\mathcal {C}$ , which only depends on $a_1,\dots ,a_{20}$ and x. First, by counting the edges induced on $V(Q_i)$ , the edges from $x_i$ to $Q_j$ , and the edges induced on $\{x_1, \ldots , x_{20}\}$ , the number of edges with both ends in $V(\mathcal {C})$ is

$$ \begin{align*} \sum_{i = 1}^{20} (a_i+1)^2 + \sum_{i = 1}^{20} 20 \cdot 2(a_i+1) + x, \end{align*} $$

whereas the length of the cycle is $\sum _{i = 1}^{20}(2a_i + 3)$ . Subtracting the cycle length from the number of edges induced on $V(\mathcal {C})$ , the number of chords equals to

$$ \begin{align*} \sum_{i = 1}^{20} (a_i^2 + 40a_i + 38) +x = \sum_{i = 1}^{20} (a_i + 20)^2 +x- c, \end{align*} $$

where $c=20 \cdot (400 - 38)$ . By using Lemma 5.1 for large enough k, there exist integers $b_1, \ldots , b_{20} \ge 20$ such that $\sum _{i = 1}^{20} b_i^2 = k -x+ c$ . Given such $b_1, \ldots , b_{20}$ , take $a_i = b_i - 20$ for $i \in [20]$ . Then the cycle $\mathcal {C}$ has exactly k chords, as required.

Figure 9

Part of the cycle $\mathcal {C}$ in Case (a).

Suppose now that (b) happens. Let $x_1, \ldots , x_{21} \in X$ be distinct. Let x be the number of edges in the induced subgraph $G[\{x_1, \dots , x_{20}\}]$ . Given integers $a_1, \ldots , a_{20} \ge 0$ , let $Q_i$ be a path in $K_i$ of length $2a_i$ , with both ends in $U_{i,1}$ . Let $\mathcal {C}$ be the cycle $x_1 Q_1 x_2 \ldots x_{21} Q_{21} x_1$ .

Then the number of edges with both ends in the cycle $\mathcal {C}$ is

$$ \begin{align*} \sum_{i = 1}^{21} a_i(a_i + 1) + \sum_{i = 1}^{21} 21 (a_i + 1)+x, \end{align*} $$

where the length of the cycle is $\sum _{i = 1}^{21} (2a_i + 2)$ . Thus, the number of chords in $\mathcal {C}$ , obtained by subtracting the number of the $\mathcal {C}$ -edges from the number of induced edges on $V(\mathcal {C})$ , is

$$ \begin{align*} \sum_{i = 1}^{21} (a_i^2 + 20 a_i + 19) +x = \sum_{i = 1}^{21} (a_i + 10)^2 +x - c, \end{align*} $$

where $c=21 \cdot (100 - 19)$ . Again by Lemma 5.1, there exist $b_1, \ldots , b_{21} \ge 10$ such that $\sum _{i = 1}^{20} b_i^2 = k-x+ c$ . Given such $b_1, \ldots , b_{21}$ , choosing $a_i = b_i - 10$ yields a cycle $\mathcal {C}$ with exactly k chords.

Proof. We shall choose $a,b, \ell '$ with $\ell ,p\gg M \gg a\gg b\gg m\gg \ell '\gg k$ . For each $u\in K_1\cup \dots \cup K_{\ell }$ and a triple $T=\{j_1,j_2,j_3\}$ in $[a]$ (with $j_1> j_2 > j_3$ ), choose $f_i(u;T)$ , for $i\in [3]$ , arbitrarily from those vertices such that $f_1(u;T)$ is a $j_1$ -father of u and $f_i(u;T)$ is a $j_i$ -father of $f_{i-1}(u;T)$ for $i \in \{2, 3\}$ . Let $F(u)$ be the collection of all $f_i(u;T)$ chosen.

Let $\{X_i, Y_i\}$ be the bipartition of $K_{i}$ . We claim that for $i \in [b]$ we can find subsets $X^{\prime }_i\subseteq X_i$ and $Y^{\prime }_i\subseteq Y_i$ of size m so that for all $u\in X^{\prime }_i\cup Y^{\prime }_i$ and $j\neq i$ , each $v\in F(u)$ is either complete or anti-complete to $X^{\prime }_j$ and either complete or anti-complete to $Y^{\prime }_j$ .

This can be done by an analogous technique to the one used in the proof of Lemma 6.3. Namely, for a vertex w and a set Z, let $\mathbf {x}(w,Z)\in \{0,1\}^{Z}$ be the $0$ - $1$ vector that encodes the adjacency between vertices in Z and w. That is, the corresponding coordinate to $v\in Z$ is $1$ if w and v are adjacent and 0 otherwise. Then one can partition $X_i$ (resp. $Y_i$ ) according to the values of $\mathbf {x}(w,Z)$ with $w\in X_i$ (resp. $X_i$ ). By choosing the largest subset in the partition, we have subsets $X_i'\subseteq X_i$ and $Y_i'\subseteq Y_i$ such that $|X_i'|\geq 2^{-|Z|}|X_i|$ and $|Y_i'|\geq 2^{-|Z|}|Y_i|$ , and each $z\in Z$ is either complete or anti-complete to each $w\in X_i'\cup Y_i'$ .

First, choose arbitrary subsets $X_b^1\subseteq X_b$ and $Y_b^1\subseteq Y_b$ with $|X_b^1| = |Y_b^1| = M$ . Then for each $i<b$ , it is possible to choose $X_i^1 \subseteq X_i$ and $Y_i^1\subseteq Y_i$ of size M such that each vertex $v\in F(u)$ , with $u\in X_j^1\cup Y_j^1$ and $j>i$ , is either complete or anti-complete to $X_i^1\cup Y_i^1$ (indeed, the previous paragraph tells us that there exist such subsets of size at least $\ell \cdot 2^{-a^3 b M} \ge M$ ).

Next, we repeat in the reverse order. That is, starting with $i = 1$ , we choose subsets $X_i' \subseteq X_i^1$ and $Y_i'\subseteq Y_i^1$ of size m such that each vertex $v\in F(u)$ , with $u\in X_j'\cup Y_j'$ and $j<i$ , is either complete or anti-complete to $X_i'\cup Y_i'$ (this is possible since we are guaranteed such subsets of size at least $M \cdot 2^{-a^3 b m} \ge m$ ). The sets $X_i'$ and $Y_i'$ satisfy the requirements.

We then collect all ‘fathers’ $v\in F(u)$ for some $u\in X_i'\cup Y_i'$ . That is, we set $F:=\bigcup _{i=1}^{b}\bigcup _{u\in X^{\prime }_i\cup Y^{\prime }_i}F(u)$ . Consider the auxiliary bipartite graph B between F and $[b]$ where $v\in F$ and $j\in [b]$ are adjacent whenever v is complete to $X^{\prime }_j$ or $Y^{\prime }_j$ . Colour each edge $(v,j)$ by red, blue, or green if v is complete to $X^{\prime }_j$ only, $Y^{\prime }_j$ only, or both $X^{\prime }_j\cup Y^{\prime }_j$ , respectively. By Lemma 7.1, if there is no cycle with exactly k chords, then B contains no monochromatic $K_{\ell ', \ell '}$ . Since $m\gg \ell '$ , the bipartite Ramsey Theorem tells us that B contains no copy of $K_{m,m}$ . In particular, there are at most $m\binom bm$ vertices $f\in F$ with $\deg _B(f)\geq m$ . Since $a\gg m\binom bm+m$ , we can choose a set J of m indices so that for $j\in J$ , all j-fathers $f\in F$ have $\deg _B(f)< m$ . For each $u\in \bigcup _{i=1}^bX_i'\cup Y_i'$ , let $F'(u)\subseteq F(u)$ be the subset of fathers $f_i(u;T)$ with $i \in \{1,2,3\}$ and $T\in \binom {J}{3}$ . Note that $|F'(u)|\leq |J|^3\leq m^3$ .

Consider an auxiliary directed graph D on $[b]$ where we join i to j if there is some $u\in X^{\prime }_i\cup Y^{\prime }_i$ and $v\in F'(u)$ with v complete to $X^{\prime }_j$ or $Y^{\prime }_j$ . As $\big |\bigcup _{u\in X^{\prime }_{i}\cup Y^{\prime }_i}F'(u)\big |\leq |X_i'\cup Y_i'|m^3 = 2m^4$ for each i and each $f\in F'(u)$ is adjacent to at most m of the sets $X_j\cup Y_j$ , the digraph D has maximum out-degree at most $2m^5\leq b/(2m+1)$ . Therefore, D has an independent set I of size m. Letting $K^{\prime }_1, \dots , K^{\prime }_m$ be the complete bipartite graphs induced by $X^{\prime }_i\cup Y^{\prime }_i$ for $i\in I$ , we get a set of complete bipartite graphs satisfying the lemma.

Proof. Let $A_0 := \{a \in A: d_{G_2}(a) \le 2d\}$ . Then $e(G_2) \le d|B|$ implies $|A \setminus A_0| \le e(G_2)/2d \le |B|/2 = m/2$ and hence, $|A_0| \ge m/2$ . Let $A' \subseteq A_0$ be a maximal subset such that there exists $B' \subseteq B$ such that each $G_i[A', B']$ , $i=1, 2$ , induces a perfect matching. We claim that every vertex in $A_0 \setminus A'$ has a $G_2$ -neighbour in $N_{G_2}(A')$ . Indeed, if $a \in A_0 \setminus A'$ has no $G_2$ -neighbour in $N_{G_2}(A')$ then it has a $G_1$ -neighbour $b \in B \setminus N_{G_2}(A')$ , which makes each $G_i[A' \cup \{a\}, B' \cup \{b\}]$ , $i=1, 2$ , a perfect matching. This contradicts the maximality of $A'$ . Thus, $A_0 \subseteq N_{G_2}(N_{G_2}(A'))$ , which implies $|A_0| \le d |N_{G_2}(A')| \le 2d^2|A'|$ . As $|A_0| \ge m/2$ , $|B'| = |A'| \ge m/4d^2$ , as desired.