Proof of Theorem 1.4. Let $\alpha,\vartheta \gt 0$ and $n,L\in \mathbb{N}$ such that $1/n\ll \vartheta \ll 1/L\ll \alpha$ . Now let $H= ( [n],E )$ be a $3$ -graph satisfying the degree condition $d(i,j)\geq \min\!\left ( i,j,\frac{n}{2}\right ) + \alpha n$ for all $ij \in [n]^{(2)}$ . Lemmas 2.2, 2.4, and Proposition 2.5 provide a reservoir $\mathcal{R}$ , an absorbing path $P_A\subseteq H\setminus \mathcal{R}$ and a long path $Q\subseteq H\setminus P_A$ with $ \vert \mathcal{R}\cap V(Q) \vert \leq \vartheta ^4 n$ . Let $(a,b)$ and $(c,d)$ be the end-pairs of $P_A$ and let $(r,s)$ and $(t,u)$ be the end-pairs of $Q$ (note that they are disjoint since we have $Q\subseteq H\setminus P_A$ ). Since $ \vert \mathcal{R}\cap V(Q) \vert \leq \vartheta ^4 n$ and $P_A\subseteq H\setminus \mathcal{R}$ and by Lemma 2.3, we can choose a path $P_1$ of length $L$ connecting $(t,u)$ and $(a,b)$ with all internal vertices in $\mathcal{R}\setminus (V(Q)\cup V (P_A ) )$ . Further, by the hierarchy of constants, we also find a path $P_2$ of length $L$ connecting $(c,d)$ and $(r,s)$ with all internal vertices in $\mathcal{R}\setminus ( V(Q)\cup V (P_A )\cup V ( P_1 ) )$ . That leaves us with a cycle $C$ in $H$ which satisfies $v(C) \geq ( 1-2\vartheta ^2 ) n$ and $P_A \subseteq C$ . The absorbing property of $P_A$ guarantees that for $X\;:\!=\;[n] \setminus V(C)$ , there exists a path $P'_{\!\!A}$ with $V ( P'_{\!\!A} ) = V ( P_A ) \cup X$ which has the same end-pairs as $P_A$ (which are connected to $Q$ ), and hence there is a Hamiltonian cycle in $H$ .

Proof of Lemma 2.1. Observe that when we show that there exists an $L\in \mathbb{N}$ and a $\vartheta \gt 0$ such that the statement of Lemma 2.1 holds for these, it easily follows that it holds for all $L\in \mathbb{N}$ and $\vartheta \gt 0$ with $1/n\ll \vartheta \ll 1/L\ll \alpha \ll 1$ . Hence, let the hierarchy and $H$ be as described in the lemma and let $(x,y),(w,z)\in [n]^2$ be two disjoint ordered pairs of distinct vertices.

First, we will show that it is possible to “climb up” along the degree sequence in (relative to $n$ ) few steps, starting from the pairs $(x,y)$ and $(w,z)$ and ending with pairs of vertices $\geq \frac{n}{2}$ .

In the second step, we will connect these two pairs by utilising an analogous “climb up” argument in the link graphs of neighbours of a pair and slipping in an additional connective pair. We first look for walks rather than paths and conclude by deducing that many of them will actually be paths.

First step. By induction on $\ell \geq 3$ , we will prove the following statement: There exist at least $\left ( \frac{\alpha }{5} \right ) ^{\ell -2} n^{\ell -2}$ walks $x_1=x, x_2=y, x_3, \dots, x_{\ell }$ such that for $i\geq 3$ we have:

(3.1) \begin{align} x_i \geq \min\!\left ( \frac{\alpha }{4}n(i-2),\frac{n}{2}\right ) +\frac{\alpha }{4}n \end{align}

We will first show the statement for $\ell =3$ and $\ell = 4$ and then deduce it for any $\ell \geq 5$ given that it holds for $\ell -1$ .

$\ell =3:$ By the degree condition on $H$ we have $d(x,y) \geq \min\!\left ( 1,2, \frac{n}{2}\right )+\alpha n$ . Hence, there exist at least $\frac{\alpha }{5}n$ possible vertices $x_3$ such that $x_1,x_2,x_3$ is a walk and $x_3 \geq \frac{\alpha }{4}n+\frac{\alpha }{4}n$ .

$\ell =4:$ Let $x_1,x_2,x_3$ be one of those $\frac{\alpha }{5} n$ walks satisfying the condition (3.1) that we get by the previous case. We then have $d(x_2,x_3) \geq \min\!\left (1,\frac{\alpha }{2}n, \frac{n}{2}\right )+\alpha n$ , so there exist at least $\frac{\alpha }{5}n$ possible vertices $x_4$ such that $x_1,x_2,x_3,x_4$ is a walk and $x_i \geq \frac{\alpha }{4}n(i-2)+\frac{\alpha }{4}n$ for $i=3,4$ .

$\ell \geq 5:$ Let $x_1,x_2,x_3, \dots,x_{\ell -1}$ be one of the $\left ( \frac{\alpha }{5} \right ) ^{\ell -3} n^{\ell -3}$ walks satisfying, for $i\geq 3$ ,

\begin{equation*}x_i \geq \min\!\left ( \frac {\alpha }{4}n(i-2),\frac {n}{2}\right ) +\frac {\alpha }{4}n\end{equation*}

that we get by induction. Then our pair degree condition implies that

\begin{equation*}d(x_{\ell -2},x_{\ell -1}) \geq \min\!\left ( \frac {\alpha }{4}n(\ell -4)+\frac {\alpha }{4}n, \frac {n}{2}\right )+\alpha n\end{equation*}

which in turn gives rise to at least $\frac{\alpha }{5}n$ possible vertices $x_{\ell }$ such that $x_1,x_2,\dots, x_{\ell }$ build a walk, and we have $x_i \geq \min\!\left (\frac{\alpha }{4}n(i-2),\frac{n}{2}\right )+\frac{\alpha }{4}n$ for all $i \in [\ell ], i \geq 3$ .

This leaves us with $\left ( \frac{\alpha }{5}\right ) ^{ \frac{2}{\alpha }} n^{\frac{2}{\alpha }}$ possibilities for walks

\begin{equation*}x_1=x, x_2=y, x_3, \dots, x_{\frac {2}{\alpha } +2}\end{equation*}

with $x_{\frac{2}{\alpha }+1},x_{\frac{2}{\alpha } +2} \geq \frac{n}{2}$ and an analogous argument for $(w,z)$ with just as many possibilities for walks

\begin{equation*}z_1=z, z_2=w, z_3, \dots, z_{ \frac {2}{\alpha }+2}\end{equation*}

with $z_{\frac{2}{\alpha } +1},z_{ \frac{2}{\alpha } +2} \geq \frac{n}{2}$ .

Second step (Figure 2). Let $m$ be the smallest even number $\geq \frac{1}{\alpha } + 1$ . It now suffices to show that for some $\vartheta '\gt 0$ with $1/n\ll \vartheta ' \ll \alpha$ we have the following. For all ordered pairs $(x^{\prime},y^{\prime}),(w^{\prime},z^{\prime})\in [n]^{2}$ for which the vertices within each pair are distinct and $x^{\prime},y^{\prime},w^{\prime},z^{\prime}\geq \frac{n}{2}$ , the number of $(x^{\prime},y^{\prime})$ - $(w^{\prime},z^{\prime})$ walks with $3m+4$ internal vertices is at least $\vartheta ' n^{3m+4}$ .

Figure 2. Idea of the second step, the picture is similar to [Reference Reiher, Rödl, Ruciński, Schacht and Szemerédi16, Fig. 4.1].

Since $d(x^{\prime},y^{\prime})\geq \frac{n}{2} + \alpha n$ , there exists a set $U_{x^{\prime}y^{\prime}}= \{u_1, \dots, u_{\alpha n} \} \subseteq [n]\setminus [n/2]$ such that $x^{\prime}y^{\prime} \in E ( L_{u_i} )$ , for all $i \in [\alpha n]$ (recall that $L_{u_i}$ denotes the link graph of $u_i$ ). Similarly, there exists $U_{w^{\prime}z^{\prime}}= \{v_1, \dots, v_{\alpha n} \} \subseteq [n]\setminus [n/2]$ such that $w^{\prime}z^{\prime} \in E ( L_{v_i} )$ , for all $i \in [ \alpha n]$ .

For $(a,b) \in [n]^2$ , let $I_{ab} = \{i\in [\alpha n ] \;:\; ab \in E (L_{u_i} ) \cap E (L_{v_i} ) \}$ . Since all vertices $\geq \frac{n}{2}$ (apart from $u_i, v_i$ ) have in both $L_{u_i}$ and $L_{v_i}$ at least $\frac{n}{2} + \alpha n$ neighbours, and therefore $2\alpha n$ vertices that they are adjacent to in both $L_{u_i}$ and $L_{v_i}$ , there are at least $\frac{\alpha n^2}{4}$ edges in $L_{v_i}\cap L_{u_i}$ . Thus, by double counting we have

\begin{equation*}\sum _{(a,b) \in [n]^2} \vert I_{ab} \vert \geq \sum _{i \in \left [\alpha n \right ]} \left \vert E\left (L_{v_i}\right ) \cap E\left (L_{u_i}\right ) \right \vert \geq \frac {\alpha n^2}{4} \alpha n\,.\end{equation*}

Next, for fixed $(a,b)\in [n]^2$ , we find a lower bound on the number $L_{ab}$ of $3$ -uniform walks of the form

\begin{equation*}x^{\prime}y^{\prime}u_{i(1)}r_1r_2u_{i(2)}\dots u_{i\left (\frac {m}{2}\right )}r_{m-1}r_m u_{i\left (\frac {m}{2}+1\right )}ab\end{equation*}

where $y^{\prime}r_1r_2\dots r_{m-1}r_ma$ is a $2$ -uniform walk in $L_{u_{i(k)}}$ and $i(k) \in I_{ab}$ , for all $k \in \left [\frac{m}{2}+1\right ]$ .

To this goal, first observe that for all $i\in [ \alpha n]$ , the number of $y^{\prime}a$ -walks of length $m+1$ in $L_{u_i}$ is at least $\left (\frac{\alpha }{3}\right ) ^m n^m$ . Indeed, since $u_i \geq n/2$ , we know that for $j\in [n]$ , we have $d_{L_{u_i}}(j)\geq \min\!\left (j, \frac{n}{2}\right )+\alpha n$ . Therefore, there are at least $\left (\frac{\alpha n}{2}\right ) ^{m-1}$ walks of length $m-1$ starting in $a$ in which each vertex is either at least $ \frac{n}{2} +\frac{\alpha n}{2}$ or at least $\frac{\alpha n}{2}$ larger than the preceding vertex. Since we set $m\geq 1/\alpha +1$ , each of these walks ends in a vertex $\geq \frac{n}{2}$ and for at least $\left (\frac{\alpha n}{3}\right )^{m-1}$ of them the last vertex is distinct from $y^{\prime}$ . For each such walk $T$ with its last vertex $a_{T^{\prime}}\neq y^{\prime}$ , there are $2\alpha n$ possibilities for common neighbours of $y^{\prime}$ and $a_{T^{\prime}}$ (note that the degrees in $L_{u_i}$ of both $y^{\prime}$ and $a_{T^{\prime}}$ are at least $\frac{n}{2} +\alpha n$ ). In total, that gives us at least $\left (\frac{\alpha n}{3}\right ) ^m$ $y^{\prime}a$ -walks of length $m+1$ in $L_{u_i}$ .

Now for $\vec r \in [n]^m$ , we set $D_{ab} (\vec r ) \;:\!=\; \{ i \in I_{ab} \;:\; y^{\prime}\vec r a \text{ is a walk in }L_{u_i} \}$ . Again by double counting and by the previous observation, we infer

\begin{equation*}\sum _{\vec r \in [n]^m} \left \vert D_{ab}\left (\vec r\right )\right \vert =\sum _{i\in I_{ab}}\left \vert \left \{\vec r \in [n]^m \;:\; y^{\prime}\vec r a \text { is a walk in }L_{u(i)}\right \}\right \vert \geq \left \vert I_{ab}\right \vert \left (\frac {\alpha }{3}\right )^mn^m.\end{equation*}

Note that for each $\vec r \in [n]^m$ that is a $y^{\prime}a$ -walk in $L_{u_{i(k)}}$ for every $k\in \left [\frac{m}{2}+1\right ]$ , we have that

\begin{equation*}x^{\prime}y^{\prime}u_{i(1)}r_1r_2u_{i(2)}\dots u_{i\left (\frac {m}{2}\right )}r_{m-1}r_mu_{i\left (\frac {m}{2}+1\right )}ab\end{equation*}

is a $3$ -uniform $(x^{\prime}y^{\prime})$ - $(ab)$ -walk of length $m+\frac{m}{2}+3$ in $H$ . Hence, with Jensen’s inequality we derive:

\begin{equation*}L_{ab} \geq \sum _{\vec {r} \in [n]^m}\left \vert D_{ab}\left (\vec r\right ) \right \vert ^{\frac {m}{2}+1} \geq n^m\left (\sum \frac {1}{n^m}\left \vert D_{ab}\left (\vec r\right )\right \vert \right )^{\frac {m}{2}+1} \geq n^m \left (\left \vert I_{ab} \right \vert \left (\frac {\alpha }{3}\right )^m\right )^{\frac {m}{2}+1}.\end{equation*}

We define $R_{ab}$ analogously as the number of $3$ -uniform walks of the form

\begin{equation*}abv_{j(1)}s_1s_2v_{j(2)}\dots v_{j\left (\frac {m}{2}\right )}s_{m-1}s_m v_{j\left (\frac {m}{2}+1\right )}w^{\prime}z^{\prime}\,,\end{equation*}

where $bs_1s_2\dots s_{m-1}s_mw^{\prime}$ is a $2$ -uniform walk in $L_{v_{j(k)}}$ and $j(k) \in I_{ab}$ , for all $k \in \left [\frac{m}{2}+1\right ]$ , and get the same lower bound by an analogous argument.

At last, let $W$ be the number of $(x^{\prime}y^{\prime})\text{-}(w^{\prime}z^{\prime})\text{-walks of length }3m+6\text{ in }H$ . We apply Jensen’s inequality a second time to obtain:

\begin{align*} W\geq & \sum _{(a,b)\in [n]^2} L_{ab}R_{ab} \\[5pt] \geq & n^{2m}\left (\frac{\alpha }{3}\right )^{m^2+2m} \sum _{(a,b)\in [n]^2} \left \vert I_{ab} \right \vert ^{m+2} \\[5pt] \geq & n^{2m}\left (\frac{\alpha }{3}\right )^{m^2+2m}n^2\left (\frac{1}{n^2}\frac{\alpha ^2 n^3}{4}\right )^{m+2} \\[5pt] \geq & \left (\frac{\alpha }{3}\right ) ^{m^2+2m}\left (\frac{\alpha ^2}{4}\right )^{m+2} n^{3m+4} \\[5pt] \geq & \left (\frac{\alpha ^2}{4}\right ) ^{m^2+3m+2} n^{3m+4}. \end{align*}

In total, putting together the walks connecting $(x,y)$ and $(x^{\prime},y^{\prime})$ , $(x^{\prime},y^{\prime})$ and $(w^{\prime},z^{\prime})$ , and $(w^{\prime},z^{\prime})$ and $(w,z)$ , we get that the number of $(x,y)$ - $(w,z)$ -walks of length $2 \cdot \frac{2}{\alpha } + 3m+6$ in $H$ is at least

\begin{equation*}\left (\left ( \frac {\alpha }{5}\right ) ^{\frac {2}{\alpha } } n^{\frac {2}{\alpha }}\right )^2 \times \left (\frac {\alpha ^2}{4}\right ) ^{m^2+3m+2} n^{3m+4}\geq \alpha ^{m^3}n^{\frac {4}{\alpha } +3m+4}\,.\end{equation*}

Since only $\mathcal{O}\left (n^{\frac{4}{\alpha } +3m+3}\right )$ of these fail to be a path, we are done.

Proof of Lemma 2.2. Let $\alpha$ , $L$ , $\vartheta$ , $n$ , and $H$ be given as in the statement. We choose a random subset $\mathcal{R} \subseteq [n]$ , where we select each vertex independently with probability $p=\left (1-\frac{1}{10L}\right ) \vartheta ^2$ . Since $ \vert \mathcal{R} \vert$ is now binomially distributed, we can apply Chernoff’s inequality (Lemma 4.1) and utilise the hierarchy to obtain

(4.1) \begin{align} \mathbb{P}\left (\left \vert \mathcal{R}\right \vert \lt \vartheta ^2 n/2\right ) \leq \mathbb{P}\left (\left \vert \mathcal{R}\right \vert \lt \frac{2}{3} \mathbb{E}\left (\mathcal{R}\right )\right ) \leq \exp\!\left (-\frac{\left (1/3\right )^2}{2}pn\right ) \lt 1/3 . \end{align}

We also have $\vartheta ^2 n \geq ( 1+c(L) ) \mathbb{E} ( \vert \mathcal{R} \vert )$ for some small $c(L) \in (0,1)$ not depending on $n$ and therefore, again by Chernoff we get for large $n$ :

(4.2) \begin{align} \mathbb{P}\left (\left \vert \mathcal{R}\right \vert \gt \vartheta ^2 n\right ) \leq \mathbb{P}\left (\left \vert \mathcal{R}\right \vert \geq \left (1+c(L)\right ) \mathbb{E}\left (\mathcal{R}\right )\right ) \leq \exp\!\left (-\frac{c(L)^2}{3}pn\right ) \lt 1/3 \end{align}

By Lemma 2.1, we have that for all disjoint ordered pairs of distinct vertices $(x,y)$ and $(w,z)$ , the number of $(x,y)$ - $(w,z)$ -paths of length $L$ in $H$ is at least $\vartheta n^{L-2}$ . Let $X=X ((x,y),(w,z) )$ denote the random variable counting the number of those $(x,y)$ - $(w,z)$ -paths in $H$ that are of length $L$ and have all internal vertices in $\mathcal{R}$ . We then have $\mathbb{E}(X) \geq p^{L-2} \vartheta n^{L-2}$ .

Now we apply the Azuma-Hoeffding inequality (Lemma 4.2) (with $X_1,\dots, X_n$ being the indicator variables for the events “ $1\in \mathcal{R}$ ”,…,“ $n\in \mathcal{R}$ ”) which gives us, since the presence or absence of one particular vertex in $\mathcal{R}$ affects $X$ by at most $ (L-2 ) n^{L-3}$ , that

\begin{align*} \mathbb{P}\left (X \leq \frac{2}{3} \vartheta (pn)^{L-2}\right ) \leq & \mathbb{P}\left ( X\leq \frac{2}{3} \mathbb{E}(X)\right ) \\[5pt] \leq & 2\exp\!\left ( -\frac{2 \left ( p^{L-2}\vartheta n^{L-2}\right ) ^2}{9n\left ( (L-2)n^{L-3}\right ) ^2}\right ) \\[5pt] = &\exp\!\left ( -\Omega (n)\right ) . \end{align*}

By the union bound, also the probability that there is a pairs of pairs for which the respective number of connecting paths with all internal vertices in $\mathcal{R}$ is less than $\frac{2}{3} \vartheta (pn)^{L-2}$ can be bounded from above by

(4.3) \begin{align} \exp\!\left (-\Omega (n)\right ) \times n^4 \lt 1/3 \end{align}

for $n$ large. Moreover, recalling our hierarchy we have

\begin{equation*}\frac {2}{3}\vartheta p^{L-2} n^{L-2} = \left (1-\frac {1}{10 L}\right )^{L-2} \frac {2}{3}\vartheta \left (\vartheta ^2 n\right )^{L-2}\geq \frac {\vartheta }{2}\left (\vartheta ^2 n\right ) ^{L-2}\end{equation*}

which together with (4.2) and (4.3) implies the following: With probability $\gt 1/3$ the chosen set $\mathcal{R}$ satisfies $\vert \mathcal{R}\vert \leq \vartheta ^2 n$ and has the property that for all disjoint ordered pairs of distinct vertices $(x,y)$ and $(w,z)$ there exist at least $\frac{\vartheta }{2}\left \vert \mathcal{R}\right \vert ^{L-2}$ paths of length $L$ in $H$ that connect those pairs and have all their internal vertices in $\mathcal{R}$ . Therefore, combining this with (4.1) ensures that there indeed exists a version of $\mathcal{R}$ that has all the required properties of our reservoir set.

Proof of Lemma 2.3. Let $H, \mathcal{R}, \mathcal{R}'$ be as in the statement of the Lemma. Consider any two disjoint ordered pairs of distinct vertices $(x,y)$ and $(w,z)$ . We have

\begin{equation*}\left \vert \mathcal {R}'\right \vert \leq 2\vartheta ^4 n \leq \vartheta ^{3/2} \frac {\vartheta ^2}{2} n \leq \vartheta ^{3/2} \left \vert \mathcal {R}\right \vert \end{equation*}

by the lower bound we get from Lemma 2.2. Since every particular vertex in $\mathcal{R}'$ is an internal vertex of at most $(L-2) \vert \mathcal{R}\vert ^{L-3}$ of the $(x,y)$ - $(w,z)$ -paths of length $L$ in $H$ with all internal vertices from $\mathcal{R}$ , the Reservoir Lemma tells us that there are at least

\begin{equation*}\frac {\vartheta }{2}\vert \mathcal {R}\vert ^{L-2} - \left \vert \mathcal {R}'\right \vert (L-2)\vert \mathcal {R}\vert ^{L-3} \geq \frac {\vartheta }{2}\vert \mathcal {R}\vert ^{L-2} - \vartheta ^{3/2} (L-2) \vert \mathcal {R}\vert ^{L-2} \gt 0\end{equation*}

such $(x,y)$ - $(w,z)$ -paths with all internal vertices in $\mathcal{R}\setminus \mathcal{R}'$ .

Proof of Lemma 5.2. Let $1/n\ll \vartheta \ll \alpha \ll 1$ , let $H$ be as in the statement, and let $x\in [n]$ . There are at least $\frac{n}{3}$ possibilities to choose a vertex $w_1\in [n]\setminus (\mathcal{R}\cup \{x\})$ with $w_1 \geq \min (x+\frac{\alpha n}{2},\frac{n}{2})$ . Then, there are at least $\frac{\alpha n}{3}$ choices for a vertex $v_1\in N(w_1,x)\setminus \mathcal{R}$ with $v_1\geq \min (x+\frac{\alpha n}{2},\frac{n}{2})$ since $\vert N(w_1,x)\vert \geq \min (w_1,x,\frac{n}{2})+\alpha n$ and $w_1\geq \min (x+\frac{\alpha n}{2},\frac{n}{2})$ . Similarly, there are at least $\frac{\alpha n}{3}$ choices for a vertex $y_1\in N(w_1,x)\setminus (\mathcal{R}\cup \{v_1\})$ with $y_1\geq \min (x+\frac{\alpha n}{2},\frac{n}{2})$ and at least $\frac{\alpha n}{3}$ choices for a vertex $z_1\in N(x,y_1)\setminus (\mathcal{R}\cup \{v_1,w_1\})$ with $z_1\geq \min (x+\frac{\alpha n}{2},\frac{n}{2})$ .

Now assume that for some $i\in [s-2]$ , vertices $v_j$ , $w_j$ , $y_j$ , and $z_j$ have already been selected, for all $j\in [i]$ , in such a way that all edges required by Definition 5.1 are present and $v_j,w_j,y_j,z_j\geq \min (x+j\frac{\alpha n}{2},\frac{n}{2})$ for all $j\in [i]$ , and denote the set containing all these vertices, all vertices from $\mathcal{R}$ , and $x$ by $A_i$ . Note that for all $i\in [s-2]$ , we have $\vert A_i\vert \leq \frac{\alpha n}{7}$ . Therefore, there are at least $\frac{\alpha n}{3}$ choices for a vertex $w_{i+1}\in N(y_i,z_i)\setminus A_i$ with $w_{i+1}\geq \min (x+(i+1)\frac{\alpha n}{2},\frac{n}{2})$ . Further, there are at least $\frac{\alpha n}{3}$ choices for a vertex $v_{i+1}\in N(w_{i+1},y_i)\setminus A_i$ with $v_{i+1}\geq \min (x+(i+1)\frac{\alpha n}{2},\frac{n}{2})$ . Similarly, there are at least $\frac{\alpha n}{3}$ choices for a vertex $y_{i+1}\in N(v_i,w_i)\setminus (A_i\cup \{v_{i+1},w_{i+1}\})$ with $y_{i+1}\geq \min (x+(i+1)\frac{\alpha n}{2},\frac{n}{2})$ and at least $\frac{\alpha n}{3}$ choices for a vertex $z_{i+1}\in N(w_i,y_{i+1})\setminus (A_i\cup \{v_{i+1},w_{i+1}\})$ with $z_{i+1}\geq \min (x+(i+1)\frac{\alpha n}{2},\frac{n}{2})$ .

Assume that $v_j$ , $w_j$ , $y_j$ , and $z_j$ have been selected for all $j\in [s-1]$ such that all edges required by Definition 5.1 are present and $v_j,w_j,y_j,z_j\geq \min (x+j\frac{\alpha n}{2},\frac{n}{2})$ , for all $j\in [s-1]$ , and denote the set containing all these vertices, all vertices from $\mathcal{R}$ , and $x$ by $A_{s-1}$ . Then there are at least $\frac{\alpha n}{3}$ choices for a vertex $w_{s}\in N(y_{s-1},z_{s-1})\setminus A_{s-1}$ with $w_s\geq \min (x+s\frac{\alpha n}{2},\frac{n}{2})$ and at least $\frac{\alpha n}{3}$ choices for a vertex $y_s\in N(v_{s-1},w_{s-1})\setminus (A_{s-1}\cup \{w_s\})$ with $y_s\geq \min (x+s\frac{\alpha n}{2},\frac{n}{2})$ . Note that by the choice of $s$ we have $v_{s-1},w_{s-1},y_{s-1},z_{s-1},w_s,y_s\geq \min ((s-1)\frac{\alpha n}{2},\frac{n}{2})=\frac{n}{2}$ . Thus, we know that

\begin{equation*}\vert N(w_{s},y_{s-1})\cap N(w_s,y_s)\vert \geq \frac {n}{2}+\alpha n+\frac {n}{2}+\alpha n-n\geq 2\alpha n\end{equation*}

and so there are at least $\alpha n$ choices for $v_s\in N(w_{s},y_{s-1})\cap N(w_s,y_s)\setminus A_{s-1}$ and, similarly, we know that there are at least $\alpha n$ choices for $z_s\in N(w_{s-1},y_{s})\cap N(w_s,y_s)\setminus (A_{s-1}\cup \{v_s\})$ .

Observe that if the vertices $v_1,w_1,y_1,z_1,\dots,v_s,w_s,y_s,z_s$ are chosen in the respective neighbourhoods as described above, they form an $(x,\alpha )$ -absorber. Hence, the number of $(x,\alpha )$ -absorbers is indeed at least $(\frac{\alpha }{3}n)^{4s(\alpha )}$ .

Proof of Lemma 2.4. The proof proceeds in two steps. First, we will use probabilistic arguments, showing that with positive probability a randomly chosen set of $4s$ -tuples contains many absorbers for every vertex while being not too large. In the second part we connect all those paths using the Connecting Lemma (Lemma 2.1).

Let $1/n\ll \vartheta \ll \alpha$ , let $L\in \mathbb{N}$ be given by the Connecting Lemma, let $s=s(\alpha )$ , and let $H,\mathcal{R}$ be given as in the statement of Lemma 2.4.

Let $\mathcal{X} \subseteq ([n] \setminus \mathcal{R} )^{4s}$ be a random selection in which each $4s$ -tuple in $ ([n] \setminus \mathcal{R} )^{4s}$ is included independently with probability $p\;:\!=\;\frac{\vartheta ^23^{4s+2}}{\alpha ^{4s}n^{4s-1}}$ . Then $\mathbb{E}\left [\left \vert \mathcal{X}\right \vert \right ] \leq pn^{4s} = \frac{\vartheta ^23^{4s+2}}{\alpha ^{4s}}n$ and by Markov’s inequality we get

(5.1) \begin{align} \mathbb{P}\left (\left \vert \mathcal{X}\right \vert \gt 2 \frac{\vartheta ^23^{4s+2}}{\alpha ^{4s}}n\right ) \leq \frac{1}{2} . \end{align}

Calling two distinct $4s$ -tuples overlapping if they contain a common vertex, we observe that there are at most $(4s)^2n^{8s-1}$ ordered pairs of overlapping $4s$ -tuples. Let us denote the number of overlapping pairs with both of their tuples occurring in $\mathcal{X}$ by $D$ . We then get $\mathbb{E}[D] \leq (4s)^2n^{8s-1}p^2=(4s)^2\big (\frac{\vartheta ^23^{4s+2}}{\alpha ^{4s}}\big )^2 n$ and Markov yields

(5.2) \begin{align} {\mathbb{P}\left [D\gt \vartheta ^2n\right ]\leq \mathbb{P}\left [D\gt 64s^2\left (\frac{\vartheta ^23^{4s+2}}{\alpha ^{4s}}\right ) ^2 n\right ] \leq \frac{1}{4}} \end{align}

since $1/n\ll \vartheta \ll \alpha$ .

Next, we focus on the number of absorbers contained in $\mathcal{X}$ . For $x\in [n]$ , let $A_x$ denote the set of all $(x,\alpha )$ -absorbers. Lemma 5.2 gives that for every $x\in [n]$ ,

\begin{equation*}\mathbb {E}\left [\left \vert A_x \cap \mathcal {X}\right \vert \right ] \geq \left (\frac {\alpha n}{3}\right )^{4s}p=9\vartheta ^2 n .\end{equation*}

Since $ \vert A_x \cap \mathcal{X} \vert$ is binomially distributed, we may apply Chernoff’s inequality to get for every $x\in [n]$ ,

(5.3) \begin{align} \mathbb{P}\left ( \left \vert A_x \cap \mathcal{X} \right \vert \leq 3\vartheta ^2 n\right ) \leq \exp\!\left (-\frac{\left (\frac{2}{3}\right )^2}{2}9\vartheta ^2 n\right ) \lt \frac{1}{5n}\,. \end{align}

Hence, by the union bound and (5.1), (5.2), and (5.3), there exists a selection $\mathcal{F}_* \subseteq ([n] \setminus \mathcal{R} )^{4s}$ with:

• $\left \vert \mathcal{F}_* \right \vert \leq \frac{2\vartheta ^23^{4s+2}}{\alpha ^{4s}}n$ ,

• $\mathcal{F}_*$ contains at most $\vartheta ^2 n$ overlapping pairs,

• $\mathcal{F}_*$ contains at least $3 \vartheta ^2 n$ $x$ -absorbers, for every $x\in [n]$ .

For each overlapping pair, we delete one of its $4s$ -tuples and thus, for every $x\in [n]$ , we lose at most $\vartheta ^2n$ $x$ -absorbers. Furthermore, we delete every $4s$ -tuple $A\in \mathcal{F}_*$ for which there does not exist an $x\in [n]$ such that $A$ is an $x$ -absorber. Note that now every remaining tuple has all edges present as in Definition 5.1 and all its vertices are distinct. This deletion process gives rise to an $\mathcal{F} \subseteq ([n]\setminus \mathcal{R} ) ^{4s}$ satisfying:

• $\left \vert \mathcal{F} \right \vert \leq \frac{2\vartheta ^23^{4s+2}}{\alpha ^{4s}}n$ ,

• for every $x \in [n]$ , there are at least $2\vartheta ^2n$ $x$ -absorbers in $\mathcal{F}$ ,

• the elements of $\mathcal{F}$ (viewed as sets) are pairwise disjoint, and

• for every $4s$ -tuple $A\in \mathcal{F}$ , there is an $x\in [n]$ such that $A$ is an $x$ -absorber. In particular, all the vertices in $A$ are distinct and there are edges present as in Definition 5.1.

Next, we want to connect the elements in $\mathcal{F}$ to a path utilising the Connecting Lemma. Let $\mathcal{G}$ be the set consisting of all the paths $v_iw_iy_{i+1}z_{i+1}$ and $v_{i+1}w_{i+1}y_iz_i$ for odd $i$ and for each $ (v_1,w_1,y_1,z_1,\dots,v_s,w_s,y_s,z_s ) \in \mathcal{F}$ :

\begin{equation*}\mathcal {G} = \bigcup _{\left (v_1,w_1,y_1,z_1,\dots,v_s,w_s,y_s,z_s\right ) \in \mathcal {F}}\big \{v_{i+j}w_{i+j}y_{i+1-j}z_{i+1-j}\;:\;i\in [s]\text { odd},j\in \{0,1\}\big \}\end{equation*}

We then have $\vert \mathcal{G} \vert = 2\vert \mathcal{F}\vert \leq \frac{4\vartheta ^23^{4s+2}}{\alpha ^{4s}}n$ . Let $\mathcal{G}^* \subseteq \mathcal{G}$ be a maximal subset such that there exists a path $P^* \subseteq H-\mathcal{R}$ with:

• $P^*$ contains all paths in $\mathcal{G}^*$ as subpaths

• $V (P^* )\cap \bigcup _{P\in \mathcal{G}\setminus \mathcal{G}^*}V(P) = \emptyset$

• $P^*$ satisfies $v (P^* ) = (L+2 ) ( \vert \mathcal{G}^* \vert -1) +4$ .

First assume $\mathcal{G}^* \subsetneq \mathcal{G}$ , and let $Q^* \in \mathcal{G}\setminus \mathcal{G}^*$ . Notice that recalling $1/n\ll \vartheta \ll \alpha, 1/L\ll 1$ , we have

(5.4) \begin{align} v\left ( P^*\right ) + \Big \vert \bigcup _{P\in \mathcal{G}\setminus \mathcal{G}^*}V(P)\Big \vert + \left \vert \mathcal{R}\right \vert \leq \left ( L+2\right ) \frac{4\vartheta ^23^{4s+2}}{\alpha ^{4s}}n + \vartheta ^2 n \leq \frac{\vartheta n}{2\left (L-2\right ) } . \end{align}

Now Lemma 2.1 tells us that there are at least $\vartheta n^{L-2}$ paths of length $L$ connecting the ending-pair $(a,b)$ of $P^*$ with the starting-pair $(b,c)$ of $Q^*$ (which are disjoint by the choice of $P^*$ ). By (5.4), at least half of those are disjoint from $\mathcal{R}\cup \bigcup _{P\in \mathcal{G}\setminus (\mathcal{G}^*\cup \{Q^*\})}V(P)$ and (apart from the end-pairs) disjoint from $V (P^* )$ and $V ( Q^* )$ . Hence, there exists a path $P^{**}$ starting with $P^*$ and ending with $Q^*$ whose vertex set is disjoint from $\mathcal{R}\cup \bigcup _{P\in \mathcal{G}\setminus (\mathcal{G}^*\cup \{Q^*\})}V(P)$ and for which we further have

\begin{equation*}v\left (P^{**}\right ) = v\left (P^*\right ) + L-2 + v\left (Q^*\right ) = 4+\left ( L+2\right )(\left \vert \mathcal {G}^* \cup \left \{Q^*\right \}\right \vert -1) \,.\end{equation*}

Therefore, $\mathcal{G}^* \cup \{Q^* \}$ contradicts the maximality of $\mathcal{G}^*$ and thus, $\mathcal{G}^* = \mathcal{G}$ . Further, for $P_A\;:\!=\;P^*$ , the hierarchy $1/n\ll \vartheta \ll \alpha, 1/L\ll 1$ gives us the required bound on $v (P_A )$ :

\begin{equation*}v\left (P_A\right ) \leq 4+(L+2)\frac {4\vartheta ^23^{4s+2}}{\alpha ^{4s}}n \leq \vartheta n .\end{equation*}

Lastly, the structure and the number of the absorbers in $P_A$ ensure the absorbing property: Let $X \subseteq [n]$ with $\vert X \vert \leq 2 \vartheta ^2 n$ . For each $x\in X$ , we can choose one $x$ -absorber $(v_1,w_1,y_1,z_1,\dots,v_s,w_s,y_s,z_s)$ from $\mathcal{F}$ such that all chosen absorbers are distinct, since for every $x\in V$ , the number of $x$ -absorbers in $\mathcal{F}$ is at least $2\vartheta ^2 n$ . For every $x\in X$ , we then “open” $P_A$ at the paths $v_{i+j}w_{i+j}y_{i+1-j}z_{i+1-j}$ for $i\in [s]$ odd and $j\in \{0,1\}$ and reconnect it to a path containing $x$ by instead considering the paths $v_{i+j}w_{i+j}y_{i+1-j}z_{i+1-j}$ , for all even $i\in [s]$ and $j\in \{0,1\}$ , and the paths $v_1w_1xy_1z_1$ and $v_sw_sy_sz_s$ . That leaves us with a path $P'$ which satisfies $V ( P' ) = V (P_A ) \cup X$ and has the same end-pairs as $P_A$ .

Proof of Lemma 6.2. For convenience set $c=\frac{d\xi ^3 -\delta }{6}n$ . Let $\mathcal{P}$ be a maximal set of vertex-disjoint paths of length $3 c-2$ in $H$ , where each path takes alternatingly vertices from each partition class, that is, each path is of the form

\begin{equation*}u_1v_1w_1u_2v_2w_2 \dots u_{ c }v_{c }w_{c }\end{equation*}

with $u_i \in U, v_i \in V, w_i \in W$ .

Assume for a contradiction that $\vert V\vert - \vert \bigcup _{P\in \mathcal{P}} V(P) \vert \gt 3\xi n$ . Then the sets

\begin{equation*}U'\;:\!=\;U\setminus \bigcup _{P\in \mathcal {P}} V(P), V'\;:\!=\;V\setminus \bigcup _{P\in \mathcal {P}} V(P), W'\;:\!=\;W\setminus \bigcup _{P\in \mathcal {P}} V(P)\end{equation*}

satisfy $ \vert U' \vert, \vert V' \vert, \vert W' \vert \gt \xi n$ .

Next, we will delete all the edges that contain vertex pairs of small pair degree. With the edges that still remain after this process we can build a path of the required length.

We start with $F_1=H [ U',V',W' ]$ and set $F_{i+1}$ , for $i\geq 1$ , as the hypergraph obtained from $F_{i}$ by deleting all edges containing a vertex pair $xy$ with $d_{F_i}^{\times }(x,y) \leq c$ , where $d_{F_i}^{\times }(x,y) = \vert \{e\in E (F_i )\;:\; x,y\in e, \vert e\cap U' \vert = \vert e\cap V' \vert = \vert e\cap W' \vert = 1 \} \vert$ . This process stops with a hypergraph $F_j$ in which for all $x,y \in V ( F_j )$ , we either have $d_{F_j}^{\times }(x,y)=0$ or $d_{F_j}^{\times }(x,y)\geq c$ . The deletion condition guarantees

\begin{equation*}e^{\times }(F_1)-e^{\times }\!\left ( F_j\right ) \leq 3cn^2 \,,\end{equation*}

where $e^{\times } ( F_i ) = \vert E^{\times } (F_i )\vert$ and $E^{\times } (F_i )= \{e\in E (F_i )\;:\; \vert e\cap U' \vert = \vert e\cap V' \vert = \vert e\cap W' \vert = 1 \}$ , and the quasirandomness of $U,V,W$ gives that $e^{\times }(F_1)=e (U',V',W' ) \geq ( d\xi ^3 - \delta ) n^3$ . Thus, there still exists an edge $u_1v_1w_1$ in $F_j$ with $u_1\in U'$ , $v_1\in V'$ and $w_1 \in W'$ . But this means that there is a path of length $3c -2$ in $F_j$ : Let $P^*=u_1v_1w_1 \dots u_kv_kw_k$ be a maximal path in $F_j$ with $u_i \in U', v_i \in V'$ and $w_i \in W'$ , for all $i \in [k]$ (note that $k\geq 1$ ). Assuming $k\lt c$ for a contradiction, less than $c$ vertices of $U'$ appear in $P^*$ . But since $v_kw_k$ is contained in the edge $u_kv_kw_k \in E^{\times } ( F_j )$ , we actually have that $d_{F_j}^{\times } (v_k,w_k )\geq c$ , whence there is a $u_{k+1} \in U'\setminus V ( P^* )$ such that $P^*u_{k+1}$ is a path in $F_j$ . The same argument applied to $w_ku_{k+1}$ gives a $v_{k+1} \in V'$ such that $P^*u_{k+1}v_{k+1}$ is a path in $F_j$ and finally applying the argument to $u_{k+1}v_{k+1}$ gives rise to a $w_{k+1} \in W'$ such that the path $P^*u_{k+1}v_{k+1}w_{k+1}$ exists in $F_j$ and thus contradicts the maximality of $P^*$ , telling us that $P^*$ actually contains an alternating path of length $3c-2$ . That, on the other hand, gives us another alternating path of length at least $3 c -2$ that is vertex-disjoint to all paths in $\mathcal{P}$ and, therefore, contradicts the maximality of $\mathcal{P}$ . So we indeed have $\vert V\vert - \vert \bigcup _{P\in \mathcal{P}} V(P) \vert \leq 3\xi n$ .

Proof of Lemma 6.3. Without loss of generality let $\alpha \ll 1$ and $\beta \lt 1/3$ and let $H,G_H$ be given as in the statement. For matchings $M_1,M_2\subseteq H$ of maximum size, we write $M_1\prec M_2$ if $[n]\setminus V(M_1)\leq _{\text{lex}}[n]\setminus V(M_2)$ , where $\leq _{\text{lex}}$ is the usual lexicographic order on $\mathscr{P}([n])$ , that is, $A\leq _{\text{lex}} B$ if $\min A\triangle B\in A$ . Now, let $M\subseteq H$ be a matching of maximum size which is (subject to being of maximum size) maximum with respect to $\prec$ . Assuming the statement is false, gives an $A\subseteq [n]\setminus V(M)$ with $\vert A \vert \geq 3\beta n$ . Let us call a pair of vertices in $H$ true if it is not an edge in $G_H$ . Since $\Delta ( G_H ) \leq \beta n$ , we can find $2\beta n$ distinct vertices $v_1, \dots, v_{\beta n}, w_1,\dots, w_{\beta n}\in A$ such that all the pairs $v_iw_i$ are true. Without loss of generality assume that $v_i\lt w_i$ . Notice that all the neighbours of each such pair lie inside $V(M)$ , otherwise adding the respective edge to $M$ would lead to a larger matching. In what follows, we will show two properties and afterwards deduce the statement from them.

Firstly, we have that for each $v_iw_i$ , there are at least $\frac{\alpha n}{3}$ edges in $M$ in which $v_iw_i$ has at least two neighbours: Let us first consider a pair $v_iw_i$ with $v_i \leq \frac{n}{2}$ . For any edge $abc$ of the matching with $a \in N ( v_i,w_i )$ , we have that $\min \{b,c\}\leq v_i$ as otherwise $E(M) \setminus \{abc\} \cup \{av_iw_i\}$ would be the edge set of a matching $M'$ with the same size as $M$ but with $M\prec M'$ , contradicting our choice of $M$ . This means that in each edge of $M$ which contains only one neighbour of $v_iw_i$ there is at least one vertex $\leq v_i$ . Thus, (and since all those edges are disjoint), at most $v_i$ neighbours of $v_iw_i$ can lie in edges that contain no further neighbour of $v_iw_i$ . Hence, recalling $d ( v_i,w_i ) \geq v_i + \alpha n$ , at least $\frac{\alpha n}{3}$ edges in $M$ contain at least two neighbours of $v_iw_i$ . For a pair $v_iw_i$ with $v_i \geq n/2$ , there exist at least $\frac{\alpha n}{3}$ edges in $M$ containing more than one neighbour of $v_iw_i$ as well since $d\left ( v_i,w_i\right ) \geq \frac{n}{2}+\alpha n$ but $e(M) \leq n/3$ .

Secondly, note that any edge of $M$ that contains at least two neighbours of one true pair $v_iw_i$ cannot contain a neighbour of any other true pair $v_jw_j$ : Assume for contradiction there were true pairs $v_iw_i$ and $v_jw_j$ together with an edge $abc \in E(M)$ such that $a\in N ( v_i,w_i )$ and $ \vert \{abc\} \cap N ( v_j,w_j ) \vert \geq 2$ . Then $b$ or $c$ , without loss of generality $b$ , is a neighbour of $v_jw_j$ and $E(M) \setminus \{abc\} \cup \{av_iw_i, bv_jw_j\}$ is the edge set of a matching in $H$ contradicting the maximum size of $M$ .

Summarised, for each of the $\beta n$ true pairs $v_iw_i$ in $[n] \setminus V(M)$ , we get a set of at least $\frac{\alpha n}{3}$ edges in $M$ that contain more than one neighbour of the respective pair, and thus all those sets of edges are pairwise disjoint. Therefore, we have $\frac{\alpha n}{3}\times \beta n$ distinct edges in $M$ which is a contradiction to $1/n\ll \alpha,\beta$ . So $M$ was indeed a matching satisfying $v(M)\geq ( 1-3\beta ) n$ .

Proof of Proposition 2.5. Let $\alpha,\vartheta$ be given as in the Proposition and set $\alpha ' = \alpha - \vartheta - \vartheta ^2$ . Next choose $\xi, \delta, t_0$ such that we have $1/t_0\ll \delta \ll \xi \ll \vartheta \ll \alpha '$ . Applying the Weak Hypergraph Regularity Lemma 6.1 to $\delta$ and $t_0$ gives us a $T_0$ and by the hierarchy in the Proposition, we may assume $1/n \ll 1/T_0$ . Now let $H$ , $\mathcal{R}$ , and $P_A$ be given as in the statement. Notice that $H'=H [[n]\setminus (\mathcal{R}\cup V (P_A ) ) ]$ after a renaming of the vertices can be seen as a $3$ -graph $H'= ([n'],E' )$ with $n'\geq (1-\vartheta ^2 - \vartheta ) n$ and satisfying the usual degree condition: $d(i,j) \geq \min\!\left ( i,j,\frac{n'}{2}\right ) + \alpha ' n'$ for all $ij \in [n']^{(2)}$ .

For $H'$ , the statement of the Weak Hypergraph Regularity Lemma provides an integer $t\in [t_0,T_0]$ and a partition $V = V_0 \dot \cup V_1 \dot \cup V_2 \dot \cup \cdots \dot \cup V_t$ satisfying all three points of Lemma 6.1. Setting $m= \vert V_1 \vert = \dots = \vert V_t \vert$ , we have that $\frac{n'}{t}\geq m \geq \frac{1-\delta }{t}n'$ and recall that $ \vert V_0 \vert \leq \delta n'$ . Note that for $v_i \in V_i$ , we have $v_i \geq i\cdot m -\frac{n'}{t_0}$ . Summarised, we have the following hierarchy:

(6.1) \begin{align} \frac{1}{n'}\ll \frac{1}{T_0},\frac{1}{t},\frac{1}{t_0}\ll \delta \ll \xi \ll \vartheta \ll \alpha '\ll 1 \end{align}

Let us write $e^{\times } (V^{ijk} ) = \vert \{e\in E' \;:\; \vert e\cap V_i \vert = \vert e\cap V_j \vert = \vert e\cap V_k \vert =1 \} \vert$ for the number of crossing edges in $V^{ijk}$ , and we call a triplet $V^{ijk}$ dense, if $e^{\times }\left ( V^{ijk}\right ) \geq \frac{\alpha 'm^3}{2}$ .

Now we will show that we can almost “transfer” the pair degree condition to a reduced hypergraph. We will do this in two steps: First, we show that every pair $V_iV_j$ belongs to many dense triplets $V^{ijk}$ , and second, we show that we can almost keep that up when restricting ourselves to quasirandom triplets.

Proof. Suppose for a contradiction there is a pair $V_iV_j$ , $ij\in [t]^{(2)}$ , belonging to less than $\min\!\left (i,j,\frac{t}{2}\right ) + \frac{\alpha 't}{3}$ dense triplets $V^{ijk}$ . Let $S$ be the set of hyperedges in $H'$ that contain one vertex in $V_i$ , one in $V_j$ and a third vertex outside of $V_i \dot \cup V_j$ . By invoking the pair degree condition of $H'$ and with the hierarchy (6.1), we get that

\begin{align*} \left \vert S \right \vert \geq & m^2 \left [ \min\!\left (i\cdot m -\frac{n'}{t_0},j\cdot m -\frac{n'}{t_0},\frac{n'}{2}\right ) +\alpha 'n' -2m\right ] \\[5pt] \gt &\frac{{n'}^3}{t^2} \left (\min\!\left ( \frac{i}{t},\frac{j}{t}, \frac{1}{2}\right ) +\frac{7}{8} \alpha '\right ) \end{align*}

We will derive a contradiction by finding a smaller upper bound on $\vert S\vert$ . To this aim, we split $S$ into two parts. By $S_1$ let us denote the set of those edges in $S$ that lie in a dense triplet $V^{ijk}$ , for some $k\in [t]\setminus \{i,j\}$ , (we say an edge $e$ lies or is in $V^{ijk}$ if we have $\vert e\cap V_i\vert = \vert e\cap V_j\vert = \vert e\cap V_k\vert = 1$ ). Since in one triplet there are at most $m^3$ edges and by assumption $V_iV_j$ does not belong to many dense triplets, we get

\begin{equation*}\left \vert S_1\right \vert \lt \left (\min\!\left (i,j,\frac {t}{2}\right ) + \frac {\alpha 't}{3}\right ) m^3 \leq \frac {{n'}^3} {t^2}\left (\min\!\left (\frac {i}{t},\frac {j}{t},\frac {1}{2}\right ) +\frac {\alpha '}{3}\right )\end{equation*}

Let $S_2=S\setminus S_1$ be the set of edges in $S$ lying in triplets that are not dense or that contain a vertex in $V_0$ . There are less than $\frac{\alpha '}{2}m^3$ crossing edges in each triplet that is not dense and $V_iV_j$ belongs to at most $t$ triplets. Further, there are at most $\delta \frac{{n'}^3}{t^2}$ edges containing one vertex in each of $V_i$ , $V_j$ , and $V_0$ . Hence,

\begin{equation*} \left \vert S_2\right \vert \lt \frac {\alpha '}{2}m^3 \times t + \delta \frac {{n'}^3} {t^2} \leq \frac {{n'}^3} {t^2} \left (\frac {\alpha '}{2}+\delta \right ).\end{equation*}

Summarised, we have

\begin{align*} \frac{{n'}^3}{t^2} \left (\min\!\left ( \frac{i}{t},\frac{j}{t}, \frac{1}{2}\right ) +\frac{7}{8} \alpha '\right ) \lt \vert S\vert = \left \vert S_1\right \vert + \left \vert S_2\right \vert \lt \frac{{n'}^3}{t^2}\left (\min\!\left (\frac{i}{t},\frac{j}{t},\frac{1}{2}\right ) +\frac{6\alpha '}{7}\right )\,, \end{align*}

which is a contradiction.

From the Weak Hypergraph Regularity Lemma we also get that in total at most $\delta t^3$ triplets $V^{ijk}$ are not $\delta$ -quasirandom.

Let us now complete the “reduction” of the hypergraph and notice that we can find an almost perfect matching in the reduced hypergraph. Denote by $D$ the hypergraph on the vertex set $[t]$ with $ijk$ being an edge if and only if the triplet $V^{ijk}$ is dense. Let, on the other hand, $IR$ be the hypergraph on the vertex set $[t]$ with $ijk$ being an edge if and only if $V^{ijk}$ is not weakly $\delta$ -quasirandom in $H'$ . In the following, we will remove a few vertices in such a way that $D-IR$ induced on the remaining vertices satisfies our pair degree condition for almost all pairs.

We call a pair $ij \in [t]^2$ a malicious pair if it belongs to more than $\sqrt{\delta }t$ edges of $IR$ . Since $e(IR) \leq \delta t^3$ , there are at most $3\sqrt{\delta }t^2$ malicious pairs. Let $B$ be the graph on vertex set $[t]$ in which the edges are given by the malicious pairs. We call a vertex $i$ a malicious vertex if $d_B(i)\gt \delta ^{1/4}t$ , that is, if it belongs to many malicious pairs. The upper bound on the number of malicious pairs implies that there are at most $6\delta ^{1/4}t$ malicious vertices. Now we remove these malicious vertices and set $D' \;:\!=\; D [[t]\setminus \{v\in [t] \;:\; v \text{ is malicious}\} ]$ and $B'=B [[t]\setminus \{v\in [t] \;:\; v \text{ is malicious}\} ]$ .

The reduced hypergraph that we wished to obtain is now given by $K=D'-IR$ , in which edges encode dense, $\delta$ -quasirandom triplets. In $K$ , every pair $ij \in V(K)^{(2)}$ with $ij \notin E(B')$ satisfies

\begin{equation*}d_K(i,j) \geq \min\!\left (i,j,\frac {t}{2}\right ) + \left (\frac {\alpha '}{3} - 6\delta ^{1/4} - \sqrt {\delta }\right ) t\geq \min\!\left ( i,j, \frac {t}{2}\right ) +\frac {\alpha '}{4}t .\end{equation*}

Thus, we have that the graph $G_K$ on vertex set $V(K)$ with $ij$ being an edge if and only if $ij$ does not satisfy the degree condition $d_K(i,j) \geq \min\!\left ( i,j, \frac{v(K)}{2}\right ) +\frac{\alpha '}{4} v(K)$ is a subgraph of $B'$ . Therefore, and since $v(K)\geq (1-6\delta ^{1/4})t$ , we have

\begin{equation*}{\Delta\!\left ( G_K\right ) \leq \Delta \left ( B'\right ) \leq \delta ^{1/4} t}\leq 2\delta ^{1/4} v(K)\end{equation*}

and we can apply Lemma 6.3 to $K$ with $\frac{\alpha '}{4}$ in place of $\alpha$ and $2\delta ^{1/4}$ instead of $\beta$ and obtain a matching $M$ in $K$ covering all but at most $6\delta ^{1/4}t$ vertices of $K$ .

Finally, notice that each triplet $V^{ijk}$ with $ijk$ being an edge in $K$ is $ (\delta,d_{ijk} )$ -quasirandom with $d_{ijk}\geq \frac{\alpha '}{2} - \delta \geq \frac{\alpha '}{3}$ . Hence, we may apply Lemma 6.2 (with $\xi$ as in (6.1), $d_{ijk}\geq \frac{\alpha '}{3}$ in place of $d$ and $\delta$ as $\delta$ ) to each of the triplets $V^{ijk}$ that corresponds to an edge in $M$ . Doing so and recalling the definition of $H'$ , we notice that in $H$ we can cover at least

\begin{equation*}n-\left (\left ( \delta + 6\delta ^{1/4} + 6\delta ^{1/4} +\xi \right ) n' + \vert \mathcal {R}\vert + v\left ( P_A\right )\right ) \geq n-\left ( 2\vartheta ^2 n + v\left ( P_A\right )\right )\end{equation*}

vertices with paths of length at least $\frac{\frac{\alpha^{\prime}}{3} \xi ^3 -\delta }{2}m-2$ that are all disjoint to $\mathcal{R}$ and $V (P_A )$ . We can connect all those at most $\frac{3t}{\frac{\alpha '}{3} \xi ^3 -\delta }$ paths in $H$ through $\mathcal{R}$ to a path $Q$ by Lemma 2.3 since until we connect the last one we have still only used at most

\begin{equation*}(L-2) \cdot \frac {3t}{\frac {\alpha '}{3} \xi ^3 -\delta } \lt \vartheta ^4 n\end{equation*}

vertices from $\mathcal{R}$ (recall the hierarchy (6.1)). In fact, we have that $Q$ has at most a small intersection with $\mathcal{R}$ , that is, $ \vert V(Q) \cap \mathcal{R} \vert \leq \vartheta ^4 n$ and it covers many vertices, that is, $v(Q) \geq ( 1-2\vartheta ^2 ) n -v ( P_A )$ . Hence, $Q$ is a path satisfying the claims in the statement.