Proof. We may assume $n\ge \sum _{k=1}^{r-1}\delta _k$ , or else the quantity under consideration is zero. For an r-tuple $(a_1,\dots ,a_r)$ as in the lemma statement, let $f(a_1,\dots ,a_r)=(b_1,\dots ,b_r)$ , where $b_1=a_1$ and $b_i=a_i-\delta _{i-1}$ for $2\leq i \leq r$ . The desired result follows by noting that f is a bijection between the set of tuples under consideration and the set of unordered r-tuples in $\{1,2,\dots ,n-\sum _{k=1}^{r-1}\delta _k\}$ .

Proof of the upper bound in Theorem 1.7

Fix $r \ge 2$ and $n \ge 1$ . We are going to construct an ordered r-matching M of size $n^{2^{r}-1}$ such that any pair of edges form a collectable r-pattern, and such that the largest clique has size n. We will do this by induction on r.

The base case $r=2$ already appears in [Reference Dudek, Grytczuk and Ruciński21, Reference Dudek, Grytczuk and Ruciński22]. Indeed, consider $M:=M_1[M_2]$ , where $M_1$ is an alignment-clique of size n and $M_2$ is the 2-partite ordered 2-matching of size $n^2$ from Theorem 4.2. Then, M has size $n^3$ . A pair of distinct edges $f_1,f_2\in E(M)$ form an alignment if $(f_1,f_2)$ is an $M_1$ -pair, and they form a nesting or a crossing if $(f_1,f_2)$ is an $M_2$ -pair. It is easy to check that there is no clique of size larger than n.

For the inductive step, suppose we have constructed an ordered $(r-1)$ -matching $M_1$ of size $n^{2^{r-1}-1}$ in which every pair of edges form a collectable $(r-1)$ -pattern, such that the largest clique has size n. For each edge $e \in E(M_1)$ , we add a new vertex $v_e$ just to the right of $e[r-1]$ and extend e to an edge $e'$ of uniformity r by adding $v_e$ to e. Let the resulting ordered r-matching be $M_2$ .

Now, we claim that $M_2$ is free of r-partite r-patterns, and its largest clique has size n. To see this, note that if $e,f\in E(M_1)$ form an $(r-1)$ -pattern with block representation $|{\mathfrak {B}}_1|\dots |{\mathfrak {B}}_k|$ , then $e',f'\in E(M_2)$ form the r-pattern whose block representation is $|{\mathfrak {B}}_1|\dots |{\mathfrak {B}}_{k-1}|{\mathfrak {B}}_k'|$ , where ${\mathfrak {B}}_k'$ is obtained by extending the ‘ ${\mathrm {A}}$ -run’ and the ‘ ${\mathrm {B}}$ -run’ in ${\mathfrak {B}}_k$ by one. (For example, if $e,f$ form a pattern with block representation $|{{\mathrm {A}}{\mathrm {A}}}{{\mathrm {B}}{\mathrm {B}}}|{{\mathrm {A}}{\mathrm {B}}}|{{\mathrm {B}}{\mathrm {A}}}|$ , then $e',f'$ form a pattern with block representation $|{{\mathrm {A}}{\mathrm {A}}}{{\mathrm {B}}{\mathrm {B}}}|{{\mathrm {A}}{\mathrm {B}}}|{{\mathrm {B}}{\mathrm {B}}}{{\mathrm {A}}{\mathrm {A}}}|$ ). This latter pattern is never r-partite, because it has at most $r-1$ blocks.

Now, let $M_3$ be the r-partite ordered r-matching given by Theorem 4.2 (with size $n^{2^{r-1}}$ , and whose largest clique has size n). Every pair of edges in $M_3$ form an r-partite r-pattern. Then, we take M to be the blow-up $M_2[M_3]$ , which has size $n^{2^{r-1}-1}\cdot n^{2^{r-1}}=n^{2^r-1}$ . For any distinct $f_1,f_2\in E(M)$ , the edges $f_1,f_2$ form an r-partite r-pattern if and only if $(f_1,f_2)$ is an $M_3$ -pair. This means that for an r-pattern P, if P is r-partite, then a P-clique in M comes from some copy of $M_3$ and from $M_2$ otherwise. So, the largest clique in M has size n, proving the induction step.

Proof. Let $\mathcal {T}\subseteq \mathbb R^{r}$ be a Poisson process with rate 1 in $\mathbb R^{r}$ . Let $L_{m,n}=L(\mathcal {T}\cap [m,n]^{r})$ be the longest chain in $\mathcal {T}\cap [m,n]^{r}$ with respect to $\mathcal {P}(\mathcal A)$ . Noting that $\mathcal {T}\cap [0,m]^{r}\overset {d}{=}\mathcal {T}_{m}$ , it suffices to prove that the conditions of Kingman’s subadditive ergodic theorem (Theorem 2.6) are satisfied, as follows.

• ○ For $0\le m\le n$ , we have $L_{0,n}\ge L_{0,m}+L_{m,n}$ , because if we have a chain in $\mathcal {T}\cap [0,m]^{r}$ and a chain in $\mathcal {T}\cap [m,n]^{r}$ , their union is always a chain in $\mathcal {T}\cap [0,n]^{r}$ .

• ○ For any $k\ge 1$ , $(L_{nk,(n+1)k})_{n=0}^{\infty }$ is a sequence of i.i.d. random variables.

• ○ For any $m\ge 1$ , we have $(L_{0,k})_{k=0}^{\infty }\overset {d}{=}(L_{m,m+k})_{k=0}^{\infty }$ .

• ○ There exists a constant $M> 0$ such that $\mathbb {E}L_{0,m} \le Mm$ for all m. To see this, we first note that if $\mathcal {S}_N$ is a set of N independent uniformly random points in $[0,m]^{r}$ , then we have ${\mathbb E} L_{\mathcal {A}}(\mathcal {S}_N)\le {\mathbb E} L_{\mathcal {BW}}(\mathcal {S}_N)$ , where $\mathcal {BW}$ is the partition of $\{1,\dots ,r\}$ into r singleton sets. Bollobás and Winkler [Reference Bollobás and Winkler7] proved that ${\mathbb E} L_{\mathcal {BW}}(\mathcal {S}_N)\le (1+o(1))eN^{1/r}$ , so

  $$ \begin{align*} {\mathbb E} L_{0,m}&={\mathbb E} L_{\mathcal{A}}(\mathcal{T}_{m})={\mathbb E}\left[{\mathbb E}\left[\vphantom{\sum}L_{\mathcal{A}}(\mathcal{T}_{m})\,\middle|\,|\mathcal{T}_{m}|\right]\right]={\mathbb E}\left[{\mathbb E}\left[\vphantom{\sum}L_{\mathcal{A}}(\mathcal{S}_{|\mathcal{T}_{m}|})\,\middle|\,|\mathcal{T}_{m}|\right]\right]\\ &\le{\mathbb E}\left[\vphantom{\sum}(1+o(1))e\,|\mathcal{T}_{m}|^{1/r}\right]=(1+o(1))em, \end{align*} $$
  where the last equation holds as $\left\vert {\mathcal {T}_m}\right\vert $ has distribution $\operatorname {Poisson}(m^r)$ .

Proof. First, (1) follows directly from Talagrand’s inequality (Theorem 2.5):

• ○ If we change a single point of $\mathcal {S}_n$ , we change L by at most 1.

• ○ Whenever $L\ge r$ then there is a set of r points which certifies that $L\ge r$ .

Then, for (2), recall the random set $\mathcal {T}_{m}$ from Theorem 5.2. Note that $|\mathcal {T}_{m}|$ has a $\operatorname {Poisson}(m^{r})$ distribution, so by Chebyshev’s inequality we can choose $m_{1},m_{2}\in \mathbb N$ , both of the form $(1+o(1))n^{1/r}$ such that $|\mathcal {T}_{m_{1}}|\le n$ whp and $n\le |\mathcal {T}_{m_{2}}|$ whp.

Note that if we condition on $|\mathcal {T}_{m}|$ , then $\mathcal {T}_{m}$ is conditionally a set of that many independent uniformly random points in $[0,m]^{r}$ (which is equivalent to taking random points in $[0,1]^{r}$ and rescaling them by a factor of m). So, there is a coupling of $\mathcal {S}_{n},\mathcal {T}_{m_{1}},\mathcal {T}_{m_{2}}$ for which $L(\mathcal {T}_{m_{1}})\le L\le L(\mathcal {T}_{m_{2}})$ whp. Then, (2) follows from Theorem 5.2(2).

Finally, (3) follows from (1) and (2).

Proof. For (1), we use the algorithmic approach of Justicz, Scheinerman and Winkler [Reference Justicz, Scheinerman and Winkler41]: We can build a longest chain $\vec {x}^{(1)},\vec {x}^{(2)},\dots ,\vec {x}^{(L)}$ by first taking the point $\vec {x}^{(1)}=(x_{1}^{(1)},\dots ,x_{r}^{(1)})\in \mathcal {T}_{m}$ with the smallest value of $\max _{i}x_{i}^{(1)}$ , then throwing out all other points $\vec {y}\in \mathcal {T}_{m}$ which do not satisfy $\vec {x}\preceq \vec {y}$ , and taking the point $\vec {x}^{(2)}=(x_{1}^{(2)},\dots ,x_{r}^{(2)})$ with the next smallest value of $\max _{i}x_{i}^{(2)}$ , and so on.

If we let $\mathcal {T}$ be the set of points corresponding to a Poisson process of rate 1 in the semi-infinite box $[0,\infty )^{r}$ (instead of a finite box $[0,m)^{r}$ ), then we obtain an infinite sequence of points $(\vec {x}^{(k)})_{k=1}^{\infty }$ as above. Then, $(\vec {x}^{(k)})_{k=1}^{\infty }$ has the following two properties. First, for every $j\geq 1$ the sequence of points $\vec {x}^{(1)},\vec {x}^{(2)},\dots ,\vec {x}^{(j)}$ is a longest chain of points in $\mathcal {T} \cap [0,\max _i x_i^{(j)}]^r$ , and second, the increments $\max _{i}x_{i}^{(k+1)}-\max _{i}x_{i}^{(k)}$ are independent and identically distributed (let Z be a random variable with this common distribution). We have $\Pr [Z> z]=\Pr [\mathcal {T}\cap [0,z]^{r}=\emptyset ]=\exp (-z^{r})$ so the density of Z is

$$\begin{align*}\frac{d}{dz}(1-\exp(-z^{r}))=rz^{r-1}\exp(-z^{r}) \end{align*}$$

and

$$\begin{align*}{\mathbb E} Z=\int_{0}^{\infty}z\cdot rz^{r-1}\exp(-z^{r})\,dz=\frac{\Gamma(1/r)}{r}. \end{align*}$$

So, for any $\varepsilon>0$ , taking $L_{1}\le (r/\Gamma (1/r)-\varepsilon )m$ and $L_{2}\ge (r/\Gamma (1/r)+\varepsilon )m$ , we have $\max _{i}x_{i}^{(L_{1})}\le m\le \max _{i}x_{i}^{(L_{2})}$ whp, by the law of large numbers. Since $\varepsilon $ was arbitrary, it follows that whp $L=(r/\Gamma (1/r)-o(1))m$ ; that is, $a=r/\Gamma (1/r)$ .

For (2), we recall the algorithmic lower bound of Bollobás and Winkler [Reference Bollobás and Winkler7]. They build a chain $\vec {x}^{(1)},\vec {x}^{(2)},\dots ,\vec {x}^{(Q)}$ by first taking the point $\vec {x}^{(1)}=(x_{1}^{(1)},\dots ,x_{r}^{(1)})\in \mathcal {T}_{m}$ such that $\sum _{i}x_{i}^{(1)}$ is minimal, then throwing out all other points $\vec {y}\in \mathcal {T}_{m}$ which do not satisfy $\vec {x}\preceq \vec {y}$ and taking the point $\vec {x}^{(2)}=(x_{1}^{(2)},\dots ,x_{r}^{(2)})$ with the next smallest value of $\sum _{i}x_{i}^{(2)}$ , and so on. It is proved in [Reference Bollobás and Winkler7], by a very similar method as above, that this algorithm produces a chain of length $r^{2}/(r!)^{1/r}\Gamma (1/r)$ .

We then just need to observe that this algorithm is not optimal: For example, instead of finding our chain point-by-point, we can build our chain two points at a time, at step k choosing two points $\vec {x}^{(2k-1)},\vec {x}^{(2k)}$ such that $\vec {x}^{(2k-1)}\preceq \vec {x}^{(2k)}$ and such that $\max \left(\sum _{i}x_{i}^{(2k-1)},\sum _{i}x_{i}^{(2k)}\right)$ is minimized. It is not hard to see that the expected ‘distance travelled’ in one such step is strictly less than two times the distance travelled in a step of the Bollobás–Winkler algorithm.

Proof of Theorem 1.10

Suppose P has block partition given by $J_{1}\cup \dots \cup J_{\ell }=\{1,\dots ,2r\}$ . Let $\mathcal {A}$ be the partition of $\{1,\dots ,r\}$ with parts $I_{i}=\{j:2j\in J_{i}\}$ for $1\le i\le \ell $ . Note that one can uniquely recover $I_{1},\dots ,I_{\ell }$ from $J_{1},\dots ,J_{\ell }$ ; we have $J_i=\bigcup _{j\in I_i}\{2j-1,2j\}$ for all $i\le \ell $ .

Observe that for every P-clique C in M, there is some partition $Q_{1}\cup \dots \cup Q_{\ell }$ of $V(M)=\{1,\dots ,rn\}$ into contiguous intervals such that for every edge $e\in C$ and every $i\in \{1,\dots ,\ell \}$ , we have $|e\cap Q_{i}|=|I_{i}|$ . Say that C is consistent with $Q_{1},\dots ,Q_{\ell }$ if this is the case.

Fix a partition $\mathcal {B}$ of $\{1,\dots ,rn\}$ into contiguous intervals $Q_{1},\dots ,Q_{\ell }$ ; we will only consider cliques consistent with $\mathcal {B}$ (at the end we will take a union bound over all partitions).

Let $M_{\mathcal {B}}$ be the submatching of M containing the edges of M which are consistent with $\mathcal {B}$ . Condition on some outcome N for the number of edges in $M_{\mathcal {B}}$ .

We will realise the conditional distribution of $M_{\mathcal {B}}$ via an alternative construction, as follows.

1. 1. Let $V=[0,\ell ]\subseteq \mathbb R$ , and for each $i\in \{1,\dots ,\ell \}$ , let $V_{i}=[i-1,i]$ , so $V_{1}\cup \dots \cup V_{\ell }=[0,\ell ]$ .

2. 2. For each $j\in \{1,\dots ,r\}$ , let $i(j)$ be such that $j\in I_{i(j)}$ .

3. 3. Let $\mathcal {R}$ be a set of N independent uniformly random points in $\prod _{j=1}^{r}V_{i(j)}$ .

4. 4. For each $\vec {v}\in \mathcal {R}$ , let $e(\vec {v})$ be the set of coordinates of $\vec {v}$ (so $|e(\vec {v})\cap V_{i}|=|I_{i}|$ for each i).

5. 5. Let $M^{*}$ be the hypergraph on the vertex set $\bigcup _{\vec {v}\in \mathcal {R}}e(\vec {v})$ whose edges are the sets $e(\vec {v})$ for $\vec {v}\in \mathcal {R}$ .

Note that with probability 1, $M^{*}$ is a matching, and the distribution of (the order-isomorphism class of) $M^{*}$ is the same as the conditional distribution of (the order-isomorphism class of) $M_{\mathcal {B}}$ , by symmetry.

Our next goal is to realise the distribution of our random point set $\mathcal {R}\subseteq \prod _{j=1}^{r}V_{i(j)}$ in terms of a set $\mathcal {S}_{N}$ of N independent uniformly random points in $[0,1]^{r}$ , in such a way that the size of the largest P-clique $L_{P}(M^{*})$ in $M^{*}$ corresponds precisely to the length of the longest chain $L_{\mathcal A}(\mathcal {S}_{N})$ in $\mathcal {S}_{N}$ with respect to the partial order $\preceq {P}_{\mathcal A}$ defined in Definition 5.1.

Recall that in the block representation (Definition 1.5) of P, the incident vertices of one edge are assigned with label ‘ ${\mathrm {A}}$ ’ while those of the other edge are assigned with label ‘ ${\mathrm {B}}$ ’. Now, consider an isometry $\phi :[0,1]^{r}\to \prod _{j=1}^{r}V_{i(j)}$ defined as follows.

1. 1. For each $i\in \{1,\dots ,r\}$ , define the function $\sigma _i:\mathbb R\to \mathbb R$ by taking $\sigma _{i}(x)=x$ if the block $I_{i}$ consists of an ${\mathrm {A}}$ -run followed by a ${\mathrm {B}}$ -run, and taking $\sigma _{i}(x)=1-x$ if the block $I_{i}$ consists of a ${\mathrm {B}}$ -run followed by an ${\mathrm {A}}$ -run.

2. 2. Let $\phi _0:[0,1]^{r}\to [0,1]^{r}$ be the isometry $(x_{1},\dots ,x_{r})\mapsto (\sigma _{1}(x_{1}),\dots ,\sigma _{r}(x_{r}))$ .

3. 3. Let $\vec {t}$ be such that $[0,1]^r+\vec {t}=\prod _{j=1}^{r}V_{i(j)}$ , and let $\phi _{1}:[0,1]^r\to \prod _{j=1}^{r}V_{i(j)}$ be the translation $\vec {x}\mapsto \vec {x}+\vec {t}$ .

4. 4. Let $\phi =\phi _{1}\circ \phi _{0}$ .

It is easy to check that $L_{P}(M^{*})=L_{I_{1},\dots ,I_{\ell }}(\mathcal {S}_{N})$ . So, still conditioning on an outcome $|M_{\mathcal B}|=N$ , by Theorem 5.3, with conditional probability $1-N^{-\omega (1)}$ we have that the random variable $L_{P}(M_{\mathcal B})$ is of the form $(a_{\mathcal {A}}+o(1))N^{1/r}$ .

Also, by the concentration inequality Theorem 2.3 (see Remark 2.4), with probability at least $1-n^{-\omega (1)}$ we have

$$\begin{align*}|M_{\mathcal B}|=\left(f_{P}\left(\frac{|Q_{1}|}{rn},\dots,\frac{|Q_{\ell}|}{rn}\right)+o(1)\right)n, \end{align*}$$

where $f_{P}(q_{1},\dots ,q_{\ell })$ is the probability that for a set of r independent uniformly random points in $[0,1]$ , exactly $I_{i}$ of them fall between $q_{i-1}$ and $q_{i-1}+q_{i}$ (here, we take $q_{0}=0$ for convenience). Note that $f_{P}(q_{1},\dots ,q_{\ell })$ is a homogeneous polynomial of degree r in the variables $q_{1},\dots ,q_{\ell }$ .

Let $\alpha _{P}$ be the maximum value of $f_{P}(q_{1},\dots ,q_{\ell })$ over all $q_{1},\dots ,q_{\ell }\ge 0$ summing to 1. Then, taking a union bound over all possible partitions $\mathcal {B}$ (of which there are at most $n^{\ell -1}$ ), we see that

$$\begin{align*}\frac{L(Q)}{n^{1/r}}\overset{p}{\to}a_{\mathcal{A}}\alpha_{P}^{1/r}. \end{align*}$$

So, the desired result follows with $b_{P}=a_{\mathcal {A}}\alpha _{P}^{1/r}$ .

Proof of Proposition 1.11

Recall the definitions of $f_{P}$ and $\alpha _{P}$ from the proof of Theorem 1.10, and recall the estimates in that proof.

1. 1. Suppose P has type r. As in the proof of Theorem 1.10, let $\mathcal A$ be the trivial partition of $\{1,\dots ,r\}$ into one part. Note that $\alpha _{P}=f_P(1)=1$ , and by Proposition 5.4, $a_{\mathcal A}=r/\Gamma (1/r)=1/\Gamma ((r+1)/r)$ . So, $b_P=a_{\mathcal A}\alpha _P^{1/r}=1/\Gamma ((r+1)/r)$ .

2. 2. Suppose P has type $1+\dots +1$ . As in the proof of Theorem 1.10, let $\mathcal A$ be the partition of $\{1,\dots ,r\}$ into r singleton parts. Note that $\alpha _{P}=f_{P}(1/r,\dots ,1/r)=r!/r^{r}$ , and note that $a_{\mathcal A}$ is precisely the Bollobás–Winkler constant $c_r$ , which by Proposition 5.4 is strictly larger than $\frac {r^{2}}{(r!)^{1/r}\Gamma (1/r)}$ . Therefore, we have $b_P=c_r(r!/r^{r})^{1/r}=c_r(r!)^{1/r}/r>r/\Gamma (1/r)=1/\Gamma ((r+1)/r)$ .

Proof. Write the edges of Q as $e_1,\dots ,e_m$ , with $e_1[1]<\dots <e_m[1]$ . Say that Q splits into submatchings if there is an index $\ell \in \{1,\dots ,m-1\}$ such that $e_i[r]<e_j[1]$ for any $1 \le i \le \ell <j \le m$ (i.e., Q consists of two matchings placed side-by-side). We proceed differently depending on whether Q splits into submatchings.

If Q does not split into submatchings, then given any Q-free ordered r-graph G, we can ‘glue together two copies of G’ to make a larger Q-free ordered r-graph $G'$ . Specifically, we can take two copies of G, and identify the last vertex of the first copy with the first vertex of the second copy (so except for the single identified vertex, all vertices of the first copy come before all vertices of the second copy). Clearly, if the resulting graph $G'$ were to contain some copy of Q, then this copy must completely lie in one of the two copies of G. But G is Q-free, meaning that $G'$ is also Q-free. In addition, if G had exactly $\left\lceil {n/2} \right\rceil $ vertices and $\operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q)$ edges, then $G'$ has $2\left\lceil {n/2} \right\rceil -1\le n$ vertices and $2\operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q)$ edges, proving the desired inequality.

On the other hand, if Q does split into submatchings, then we can consider the r-graph G on the vertex set $\{1,\dots ,n\}$ containing all possible edges $e\in E(K_n^{(r)})$ with $e[1]\le \left\lceil {n/{2}} \right\rceil <e[r]$ . Note that G is Q-free. We deduce

$$\begin{align*}\operatorname{\mathrm{ex}}_<(n,Q)\ge e(G)\ge \binom{n}{r}-\binom{\left\lceil {n/2} \right\rceil}{r}-\binom{\left\lfloor {n/2} \right\rfloor}{r}\ge 2\binom{\left\lceil {n/2} \right\rceil}{r}\ge 2\operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q), \end{align*}$$

as desired.

Proof. For any $n\in \mathbb N$ , let $\mathcal G_n$ be the set of all r-graphs on the vertex set $\{1,\dots ,n\}$ which are Q-free.

First, we describe how to ‘contract pairs of vertices’ to transform an n-vertex ordered graph $G\in \mathcal G_n$ into an $\left\lceil {n/2} \right\rceil $ -vertex ordered graph $\phi (G)\in \mathcal G_{\left\lceil {n/2} \right\rceil }$ . For $G\in \mathcal G_n$ , partition its vertex set $\{1,\dots ,n\}$ into $\left\lceil {n/2} \right\rceil $ contiguous intervals $I_1,\dots ,I_{\left\lceil {n/2} \right\rceil }$ , where each $I_i$ has size $2$ except possibly $I_{\left\lceil {n/2} \right\rceil }$ (which has size $1$ if n is odd). Then, $\phi (G)$ is obtained by ‘contracting’ each $I_i$ to a single vertex. Specifically, we include $e \in E(K_{\left\lceil {n/2} \right\rceil }^{(r)})$ as an edge of $\phi (G)$ if and only if $\{i_1,\dots ,i_r\} \in E(G)$ for some $i_1\in I_{e[1]}, i_2\in I_{e[2]},\dots ,i_r\in I_{e[r]}$ . Note that this contraction operation cannot create copies of Q.

Now, fixing $G' \in \mathcal {G}_{\left\lceil {n/2} \right\rceil }$ , we are going to estimate the number of $G \in \mathcal {G}_n$ such that $\phi (G)=G'$ . For all possible edges $e\in E(K_n^{(r)})$ of G, say that e is contractible if $\left\lceil {e[1]/2} \right\rceil <\dots <\left\lceil {e[r]/2} \right\rceil $ (i.e., if the vertices of e lie in different $I_i$ , in the definition of $\phi (G)$ ).

First, every edge $e\in E(G')$ arose by contracting some edge $f_e\in E(G)$ . There are $2^r$ possibilities for $f_e$ , and at least one must have been present in G, so given that $\phi (G)=G'$ there are at most $(2^{2^r}-1)^{e(G')}<2^{2^r\operatorname {\mathrm {ex}}_<(n',Q)}$ ways to choose the contractible edges in G.

On the other hand, if $e\in E(G)$ is not contractible, then $e[k]+1=e[k+1]$ for some $1 \le k < r$ . By Lemma 2.2 and Proposition 7.2, the number of $e\in E(K_n^{(r)})$ which have this property is

$$\begin{align*}\binom{n}{r}-\binom{(n-r)_+}{r}<2^r\left(\binom{\left\lceil {n/2} \right\rceil}{r}-\binom{(\left\lceil {n/2} \right\rceil-r)_+}{r}\right)\le 2^r \operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q),\end{align*}$$

assuming that n is sufficiently large in terms of r (say, $n\ge n_r$ ). So, the total number of ways to choose the noncontractible edges in G is at most $2^{2^r\operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q)}$ . An r-graph G is specified by its contractible and noncontractible edges, so we have

$$ \begin{align*} \left\vert {\mathcal{G}_n}\right\vert &\le \left\vert {\mathcal{G}_{\left\lceil {n/2} \right\rceil}}\right\vert\cdot 2^{2^r\operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q)}\cdot 2^{2^r\operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q)} \\&\le \left\vert {\mathcal{G}_{\left\lceil {n/2} \right\rceil}}\right\vert\cdot \exp(2^{r+1}\operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q)). \end{align*} $$

Iterating this inequality and using that $\operatorname {\mathrm {ex}}_<(\lceil n/2^k\rceil ,Q) \leq 2^{-(k-1)} \operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q)$ (by Proposition 7.2), we have

$$ \begin{align*} \left\vert {\mathcal{G}_n}\right\vert &\le \left\vert {\mathcal{G}_{n_r}}\right\vert \prod_{k=1}^{\left\lceil {\log_2 (n/n_r)} \right\rceil} \exp( 2^{r+1}\operatorname{\mathrm{ex}}_<(\lceil n/2^k\rceil,Q)) \\ & \le \left\vert {\mathcal{G}_{n_r}}\right\vert \cdot \exp( (2^{r+2}-1)\operatorname{\mathrm{ex}}_<(\left\lceil {n/2} \right\rceil,Q)). \end{align*} $$

When n is sufficiently large in terms of r, Proposition 7.2 implies $\left\vert {\mathcal {G}_{n_r}}\right\vert \le 2^{\binom {n_r}{r}} \le \exp \left( \operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q) \right)$ . Then, $\left\vert {\mathcal {G}_n}\right\vert < \exp \left(2^{r+2}\operatorname {\mathrm {ex}}_<(\left\lceil {n/2} \right\rceil ,Q)\right)$ , as desired.

Proof of Theorem 7.1

Let $\mathcal {M}_n$ be the set of ordered r-matchings on the vertex set $\{1,\dots ,rn\}$ that are Q-free, and for any $n'$ let $\mathcal G_{n'}$ be the set of r-graphs on the vertex set $\{1,\dots ,n'\}$ that are Q-free.

As in the proof of Lemma 7.4, we will define a ‘contraction’ operation (though this time we will contract larger intervals, and we will not need to iterate our operation). For some B, whose value we will specify shortly, let $n'=\lceil rn/B\rceil $ and partition $\{1,\dots ,rn\}$ into $n'$ contiguous intervals $I_1,\dots ,I_{n'}$ , where each $I_i$ has size B except possibly $I_{n'}$ (which is smaller if n is not divisible by B). For $M \in \mathcal {M}_n$ , let $\psi (M)\in \mathcal G_{n'}$ be the graph on the vertex set $\{1,\dots ,n'\}$ obtained by contracting each interval $I_i$ to a single vertex (specifically, we include $e \in E(K_{n'}^{(r)})$ as an edge of $\psi (G)$ if and only if $(i_1,\dots ,i_r)\in E(M)$ for some $i_1 \in I_{e[1]},\dots ,i_r \in I_{e[r]}$ ). Now, we specify B to be the largest integer such that $B^{r-1}\le \alpha r^r n^{(r-2)}$ . Note that $B^{r-1}\ge \alpha r^rn^{r-2}/2$ and $(n')^{r-1}\le 2(rn/B)^{r-1}= 2{r^{r-1}n^{r-1}}/{B^{r-1}}\le 4n/(r\alpha )$ ; and $(n')^{r-1}\ge (rn/B)^{r-1}= {r^{r-1}n^{r-1}}/{B^{r-1}} \ge n/(r\alpha )$ . By Lemma 7.4 we have

$$ \begin{align*} \left\vert {\mathcal{G}_{n'}}\right\vert &\le \exp\left(2^{r+2}\operatorname{\mathrm{ex}}_<(\lceil n'/2\rceil,Q)\right) \le \exp\left(2^{r+3} \alpha(n'/2)^{r-1}\right) \\& \le \exp\left(\vphantom\sum 16\alpha(n')^{r-1}\right)\le e^{64n/r}\le e^{32n}, \end{align*} $$

assuming n is sufficiently large with respect to r. (In this deduction, we assumed that $\alpha \leq 4n^{3/4}/r$ and n is sufficiently large with respect to $n_Q$ so that $ (n/(r\alpha ))^{1/(r-1)}/2\ge n_Q$ , thus $\lceil n'/2\rceil \ge (n/(r\alpha ))^{1/(r-1)}/2\ge n_Q$ . Note that if $\alpha> 4n^{3/4}/r$ then $(n')^{r-1} \leq 4n/r\alpha < n^{1/4}$ , hence $\left\vert {\mathcal {G}_{n'}}\right\vert \le 2^{\binom {n'}{r}} \leq 2^{(n')^r}<e^{32n}$ ). Now, fixing $G\in \mathcal {G}_{n'}$ , let us estimate the number of $M\in \mathcal {M}_n$ with $\psi (M)=G$ . Viewing each edge $e\in E(G)$ as an ordered r-tuple $(e[1],\dots ,e[r])$ , let

$$ \begin{align*} \mathcal{T}&=E(G)\cup\{(t_1,\dots,t_r)\in[n']^r : t_1\le t_2\le\dots\le t_r \text{ and there exists } k \text{ such that } t_k=t_{k+1} \}. \end{align*} $$

The idea is that $\mathcal T$ specifies the ‘possible places where edges of M can lie’: if $\psi (M)=G$ , then for each edge $e\in M$ there must be some tuple $(t_1,\dots ,t_r)\in \mathcal T$ such that $e[1]\in I_{t_1},\dots ,e[r]\in I_{t_r}$ (say that e is consistent with $\vec t$ ). Then,

$$ \begin{align*} \begin{aligned} \left\vert {\mathcal{T}}\right\vert =&\, |E(G)| + \binom{n'+r-1}{r}-\binom{n'}{r} \le \operatorname{\mathrm{ex}}_<(n',Q) + 2\binom{n'}{r}-2\binom{(n'-r)_+}{r} \\ \le&\, 3\operatorname{\mathrm{ex}}_<(n',Q) \le 3\alpha (n')^{r-1} \le \frac{12n}{r}\le 6n. \end{aligned} \end{align*} $$

(In the deduction above, we assumed that $\alpha \le 4 n^{3/4}/r$ and n is sufficiently large with respect to $n_Q$ and r so that the first inequality holds, and $n'\geq (n/(r\alpha ))^{1/(r-1)} \ge n_Q$ so that the third inequality holds. In the second inequality, we used Proposition 7.2. Also, note that the inequality $\left\vert {\mathcal {T}}\right\vert \le 6n$ holds trivially if $\alpha> 4n^{3/4}/r$ as this implies that $n'< n^{1/(4(r-1))}$ and thus $\left\vert {\mathcal T}\right\vert \le (n')^r<n$ .)

Now, for every $M\in \mathcal {M}_n$ with $\psi (M)=G$ , we consider the number $a_{\vec {t}}$ of edges $e\in M$ which are consistent with $\vec t$ . Since $\sum _{\vec {t}\in \mathcal {T}} a_{\vec {t}}=n$ , the number of possibilities for $(a_{\vec {t}}:\vec t\in \mathcal T)$ is at most

$$\begin{align*}\binom{n+\left\vert {\mathcal{T}}\right\vert-1}{n}<\binom{n+6n}{n}\le 2^{7n}\le e^{5n}.\end{align*}$$

Then, given any choice of $(a_{\vec {t}}:\vec t\in \mathcal T)$ , the number of ways to choose the edges of M is at most

$$\begin{align*}\prod_{\vec{t}\in\mathcal{T}} ({B^r})^{a_{\vec{t}}}\le B^{rn}.\end{align*}$$

Thus,

$$\begin{align*}\left\vert {\mathcal{M}_n}\right\vert \le \left\vert {\mathcal{G}_{n'}}\right\vert\cdot e^{5n}\cdot B^{rn} \le e^{37n}\left(\alpha r^r n^{r-2} \right)^{(r/(r-1))n}, \end{align*}$$

as desired.