Proof. As long as there is a $(k-1)$ -set $S$ with degree between $1$ and $t$ , delete all edges containing $S$ . Obviously, every $(k-1)$ -set is considered at most once during this process, and when it is considered, we delete at most $t$ edges.

Proof of Lemma 4.1. Let $k \in \mathbb{N}$ with $k \ge 3$ and ${\varepsilon },\rho \gt 0$ be given, and observe that we can assume $\varepsilon$ to be small enough for Lemma 3.1 to hold. Let

\begin{equation*}1/n \ll \zeta \ll \eta \ll {\varepsilon }, \rho, 1/k\, .\end{equation*}

Let $G$ be a $2$ -edge-coloured (with red and blue) $n$ -vertex $k$ -graph on $V$ such that all but at most $\zeta n^{k-1}$ of the $(k-1)$ -subsets of $V$ satisfy $d_G(S) \ge (1/2+{\varepsilon })n$ . We apply Proposition 4.2 with parameter $t\,:\!=\,\eta {\varepsilon } n$ twice, once to the subhypergraph induced by the red edges and once to the subhypergraph induced by the blue edges. This gives a subhypergraph $G'$ of $G$ with $e(G') \geq e(G) - 2\eta {\varepsilon } n^k$ such that each $(k-1)$ -set $S$ satisfies the following condition for each of the colours: either $S$ is not contained in any edge of $G'$ of this colour, or it is contained in at least $\eta {\varepsilon } n$ edges of $G'$ of this colour.

Let $H$ be the $(k-1)$ -uniform shadow of $G'$ , equipped with the following $2$ -colouring of its edges: Colour an edge of $H$ red (resp. blue) if it is contained in at least one red (resp. blue) edge of $G'$ (and hence at least $\eta {\varepsilon } n$ such edges), noting that we allow edges of $H$ to receive both colours. For the purpose of finding coloured structures in $H$ , an edge that has both colours can be used either way.

Proof of Claim 1. Suppose $H$ contains an alternating $(k-1)$ -grid. Then there exists $W\,:\!=\,\{x_{ij}\,{:}\,1 \le i,j \le k-1\} \subseteq V$ such that $x_{i1} \cdots x_{i(k-1)}$ is a red edge of $H$ for each $i \in [k-1]$ and $x_{1j} \cdots x_{(k-1)j}$ is a blue edge of $H$ for each $j \in [k-1]$ . Let $\mathcal{R}$ be the set of ordered tuples $(x_{1k},x_{2k},\ldots, x_{(k-1)k})$ such that $x_{1k},x_{2k},\ldots, x_{(k-1)k} \in V \setminus W$ are pairwise distinct, $\{x_{i1} \cdots x_{ik}\}$ is a red edge of $G$ for each $i \in [k-1]$ , and $d_G(\{x_{1k},x_{2k},\ldots, x_{(k-1)k}\}) \ge (1/2+{\varepsilon })n$ . Owing to the fact that a red edge of $H$ can be extended to a red edge of $G'$ (and hence $G$ ) in at least $\eta {\varepsilon } n$ ways and that for all but $\zeta n^{k-1}$ of the $(k-1)$ -subsets $S$ of $V$ it holds that $d_G(S) \ge (1/2+{\varepsilon })n$ , we have $|{\mathcal{R}}| \ge (\eta {\varepsilon } n)^{k-1} - n^{k-2} - (k-1)! \cdot \zeta n^{k-1} \ge (\eta {\varepsilon } n)^{k-1}/2$ , where we used $\zeta \ll \eta, {\varepsilon },1/k$ (and the $n^{k-2}$ term accounts for the choices of $x_{1k},\ldots, x_{(k-1)k}$ which are not pairwise distinct). Similarly, let $\mathcal{B}$ be the set of ordered tuples $(x_{k1},x_{k2},\ldots, x_{k(k-1)})$ such that $x_{k1},x_{k2},\ldots, x_{k(k-1)} \in V \setminus W$ are pairwise distinct, $\{x_{1j} \cdots x_{kj}\}$ is a blue edge of $G$ for each $j \in [k-1]$ , and $d_G(\{x_{k1},x_{k2},\ldots, x_{k(k-1)}\}) \ge (1/2+{\varepsilon })n$ . As above for $\mathcal{R}$ , we have $|{\mathcal{B}}| \ge (\eta {\varepsilon } n)^{k-1}/2$ . Therefore, there exist vertex-disjoint $R \in {\mathcal{R}}$ and $B \in {\mathcal{B}}$ . As $d_G(R),d_G(B) \ge (1/2+{\varepsilon })n$ , we have $|N_G(R) \cap N_G(B)|\ge 2(1/2+{\varepsilon })n - n = 2{\varepsilon } n$ . Hence, there exists $x_{kk} \in (N_G(R) \cap N_G(B)) \setminus W$ , i.e. a vertex $x_{kk}$ which forms an edge of $G$ with both $R$ and $B$ (although we have no control of the colours of these edges). This gives a near-alternating $k$ -grid in $G$ , as desired.

Proof of Claim 2. We begin by bounding the number of $(k-1)$ -sets $S$ with $d_{G'}(S) \lt (1/2+{\varepsilon }/2)n$ . Observe that if $d_{G'}(S) \lt (1/2+{\varepsilon }/2)n$ , then either $d_G(S) \lt (1/2+{\varepsilon })n$ (i.e. $S$ has small degree already in $G$ ), or we removed at least ${\varepsilon } n/2$ edges of $G$ containing $S$ during the cleaning process (i.e. when obtaining $G'$ from $G$ ). The former case holds for at most $\zeta n^{k-1}$ sets $S$ , by assumption. Let us now bound the number of $S$ in the latter case. Since removing an edge of $G$ decreases (by one) the degree of precisely $k$ $(k-1)$ -sets, and as the total number of removed edges is at most $2\eta {\varepsilon } n^k$ , the number of such sets $S$ is at most $\frac {k \cdot 2 \eta {\varepsilon } n^k}{{\varepsilon } n/2} = 4 k \eta n^{k-1}$ .

Next, we bound the number of double-coloured edges of $H$ . Let $\tilde {H}$ be the subhypergraph of $H$ on $V$ induced by the double-coloured edges. Then $e(\tilde {H}) \le \eta n^{k-1}$ . Indeed, otherwise $\tilde {H}$ contains a copy of the $(k-1)$ -grid by Theorem4.3, since the $(k-1)$ -grid is a $(k-1)$ -partite $(k-1)$ -graph (with each vertex class of size $k-1$ ). However, as each edge of $\tilde {H}$ receives both colours, this gives an alternating $(k-1)$ -grid, which is a contradiction.

Summarising, the number of bad sets is at most $(\zeta + 4 k \eta +\eta )n^{k-1} \le 5k\eta n^{k-1}$ , where we used $\zeta \ll \eta$ .

Proof of Claim 3. For each $0 \le j \le k-1$ , we bound the number of $(k-1)$ -sets $S$ for which the number of bad sets $T$ with $|S \cap T| = j$ is more than $\eta ^{1/2} n^{k-1-j}$ . If $j=0$ , this cannot happen by Claim2, so we can assume $0 \lt j \le k-1$ . Given a bad set $T$ , the number of $(k-1)$ -sets $S$ intersecting $T$ in $j$ vertices is at most $\binom {k-1}{j}n^{k-1-j}$ . Therefore, using the bound in Claim2, the number of $(k-1)$ -subsets of $V$ which are not clean is at most

\begin{equation*} \sum _{j=1}^{k-1} \frac {\binom {k-1}{j} n^{k-1-j} \cdot 5 k \eta n^{k-1}}{\eta ^{1/2} n^{k-1-j}} \le \eta ^{1/3} n^{k-1}\, . \end{equation*}

Proof of Claim 4. Consider two arbitrary edges $T_1, T_2 \in E(H')$ . We will show that they have the same colour. Fix $C,\beta$ with $\eta \ll 1/C \ll \beta \ll {\varepsilon }, \rho, 1/k$ . We claim that there exists a set $R\,{\subseteq}\, V$ such that

1. (i) $|R| \ge C/2$ ;

2. (ii) for $i=1,2$ , for every $(k-1)$ -set $S$ which is contained in $R \cup T_i$ and is not bad, it holds that $|N_{G'}(S) \cap R|\ge (1/2+{\varepsilon }/8)|R \cup T_i|$ ;

3. (iii) no $(k-1)$ -subset contained in $R \cup T_1$ or $R \cup T_2$ is bad.

Let $R \subseteq V$ be obtained by independently including each vertex of $V$ with probability $C/n$ . We show that such $R$ satisfies (i), (ii) and (iii) with positive probability.

Since $\mathbb{E}[|R|]=C$ and $1/C \ll \beta$ , an easy application of Chernoff’s inequality (Lemma 3.4) shows that $(1-\beta ) C \le |R| \le (1+\beta ) C$ with probability at least $0.9$ .

Fix $i \in \{1,2\}$ , let $S \subseteq V$ be a $(k-1)$ -set which is not bad, and define the events $A_S\,:\!=\, \{S \subseteq R \cup T_i\}$ and $B_S\,:\!=\,\{X_S \lt (1/2+{\varepsilon }/4)C\}$ , where $X_S \,:\!=\, |N_{G'}(S) \cap R|$ . Furthermore, let $j=j_S\,:\!=\,|S \cap T_i|$ , so $0 \le j \le k-1$ . We are going to show that with probability at least $0.9$ , the event $A_S \cap B_S$ does not hold for any such $S$ . Note that $A_S$ and $B_S$ are independent, and ${\mathbb{P}}[A_S]=\left (\frac {C}{n}\right )^{k-1-j}$ .

Since $S$ is not bad, we have $\mathbb{E}[X_S] = |N_{G'}(S)| \cdot \frac {C}{n} \ge (1/2 + {\varepsilon }/2) C$ and, using Chernoff’s inequality (Lemma 3.4), we conclude that ${\mathbb{P}}[B_S] \le 2 \exp \left (-\frac {(\varepsilon /4)^2}{3} \mathbb{E}[X_S]\right )=\exp (-\Omega _{{\varepsilon }}(C))$ . Therefore, ${\mathbb{P}}[A_S \cap B_S] \le \left (\frac {C}{n}\right )^{k-1-j} \cdot \exp (-\Omega _{{\varepsilon }}(C))$ . By taking the union bound over the at most $2^{k-1}n^{k-1-j}$ $(k-1)$ -sets $S \subseteq V$ with $j_S=j$ , we see that the probability that $A_S \cap B_S$ holds for some such $S$ is at most $\left (\frac {C}{n}\right )^{k-1-j} \cdot \exp (-\Omega _{{\varepsilon }}(C)) \cdot 2^{k-1}n^{k-1-j} \le 0.1 k^{-1}$ , where we used that $1/C \ll {\varepsilon }, 1/k$ . The conclusion now follows by taking another union bound over the $k$ choices for $j$ .

Next, let $Y$ be the random variable counting the number of bad sets contained in $R \cup T_i$ , and observe that $Y= \sum _{j=0}^{k-1} Y_j$ , where $Y_j$ counts the number of bad sets $T \subseteq R \cup T_i$ with $|T \cap T_i|=j$ . Since $T_i$ is a clean set, the number of bad sets $T$ with $|T \cap T_i|=j$ is at most $\eta ^{1/2} n^{k-1-j}$ . Therefore, $\mathbb{E}[Y_j] \le \eta ^{1/2} n^{k-1-j} \cdot \left ( \frac {C}{n} \right )^{k-1-j} \leq \eta ^{1/2}C^{k-1}$ and hence $\mathbb{E}[Y] \le k \eta ^{1/2} C^{k-1}\le 0.1$ , using $\eta \ll 1/C$ . By Markov’s inequality, we have that $Y=0$ with probability at least $0.9$ .

Therefore, with positive probability we have that $(1-\beta )C \le |R| \le (1+\beta )C$ , the event $A_S \cap B_S$ does not hold for any $(k-1)$ -set which is not bad (for both $i=1,2$ ), and no $(k-1)$ -subset contained in $R \cup T_1$ or $R \cup T_2$ is bad. These in turn imply properties (i)-(iii). Indeed, (i) and (iii) follow directly from the above, and for (ii) it is enough to observe that if $S$ is not bad and is contained in $R \cup T_i$ (i.e. $A_S$ holds), then $B_S$ cannot hold and thus $X_S \ge (1/2+{\varepsilon }/4)C \ge (1/2+{\varepsilon }/8)|R \cup T_i|$ , where the first inequality follows from the definition of $B_S$ , and the second inequality uses $|R| \le (1+\beta )C$ , $|T_i| = k-1$ , and $1/C, \beta \ll {\varepsilon },1/k$ .

We conclude that a set $R$ satisfying (i), (ii) and (iii) does indeed exist. Note that (ii)–(iii) imply that $\delta (G'[R \cup T_i]) \geq (1/2 + \varepsilon /8)|R \cup T_i|$ . Moreover, by (i), $|R|$ is large enough to apply Lemma 3.1 to $G'[R \cup T_1]$ and $G'[R \cup T_2]$ . Now, fix two arbitrary orderings $\overrightarrow {T_1}$ and $\overrightarrow {T_2}$ of $T_1, T_2$ , respectively, and let $\overrightarrow {T}$ be an arbitrary ordering of a $(k-1)$ -set $T \subseteq R \setminus (T_1 \cup T_2)$ . Using Lemma 3.1 twice, we find a tight path $P_1$ connecting $\overrightarrow {T_1}$ and $\overrightarrow {T}$ in $G'[R \cup T_1]$ and a tight path $P_2$ connecting $\overrightarrow {T_2}$ and $\overrightarrow {T}$ in $G'[R \cup T_2]$ . Owing to (iii), we know that every $k-1$ consecutive vertices in $P_1$ form a set which is not bad, which means that all edges of $G'$ containing this set have the same colour. By the definition of the colouring of $H$ , every such $(k-1)$ -set has the same colour and, in particular, this holds for $T_1$ and $T$ . By repeating the same argument for $P_2$ , this holds for $T_2$ and $T$ . We conclude that $T_1$ and $T_2$ have the same colour, as desired.

This concludes the proof of Lemma 4.1.

Proof. Let $v\in V$ . For $1 \le i \le k-1$ , observe the following identity where the first sum runs over all $\overrightarrow {S} \in (V)_{k-1}$ such that $v$ is the $i$ -th element of $\overrightarrow {S}$ , and the second sum runs over all $S \in \binom {V}{k-1}$ with $v \in S$ :

\begin{equation*} \sum _{\substack {\overrightarrow {S} = (v_1,\ldots, v_{k-1}) \\ v_i = v}} \mathbf{x}(S) = (k-2)! \cdot \sum _{S \in \binom {V}{k-1} \,{:}\, v \in S} \mathbf{x}(S) = (k-1)! \cdot \sum _{e\,:\,e \ni v} \mathbf{x}(e) = (k-1)! \,. \end{equation*}

Together with (5.2), we get ${\mathbf{Pr}}[Y_i=v] = 1/n$ .

Since $\pi$ as defined in (5.1) is the stationary distribution of $\mathcal{Z}$ , it holds that ${\mathbf{Pr}}[\overrightarrow {Z_i}=\overrightarrow {S}] = \pi (\overrightarrow {S})$ for each $i \ge k-1$ . Let $i \ge k$ . Then by the law of total probability we have

where we used (5.2), that for an edge $e$ with $e \ni v$ there are $(k-1)!$ choices for $\overrightarrow {S} \in (V)_{k-1}$ such that $S \cup \{v\}=e$ , and that $\mathbf{x}$ is a perfect fractional matching.

For the ‘moreover’-part of Fact 5.1, let $k \le i \le t$ and $e \in E(G)$ . Then

where we used (5.2) and that there are $k!$ ways to choose $\overrightarrow {S} \in (V)_{k-1}$ such that $S{\subseteq } e$ .

Proof. Let

\begin{equation*}1/n \ll \mu \ll \eta \ll \zeta \ll \rho \ll \mu _0 \ll {\varepsilon }, 1/r, 1/k\,, \end{equation*}

where $\mu _0$ is small enough for Lemma 3.3 to hold on input $k$ and $\varepsilon$ , and $\zeta$ is small enough for Lemma 4.1 to hold on input $k,{\varepsilon }/2$ and $\rho /2$ . Let $\mathbf{x}_0$ be a $\mu _0$ -normal perfect fractional matching as given by Lemma 3.3.

Let $c_1, \ldots, c_r$ be the $r$ colours used in the edge-colouring of $G$ . Since Lemma 4.1 is only stated for $2$ -edge-coloured graphs, we will now consider the following $2$ -edge-colouring of $G$ , obtained by identifying $r-1$ of the colours to a unique colour: Colour by ‘red’ any edge coloured by $c_1$ , and by ‘blue’ every other edge.

We claim that either $G$ contains $\eta n^k$ edge-disjoint gadgets or one of the colour classes of $G$ has size at most $\rho n^k$ . This follows by applying Lemma 4.1 iteratively. Suppose indeed that a maximal collection of edge-disjoint gadgets has size $\ell \lt \eta n^k$ , and let $G'$ be the subhypergraph of $G$ obtained by removing all the edges of such gadgets. Let $\mathcal{S}$ be the collection of the $(k-1)$ -subsets $S$ of $V$ with $d_{G'}(S) \lt (1/2+{\varepsilon }/2)n$ . We now bound the size of $\mathcal{S}$ . The total number of removed edges is $2k \ell \lt 2k\eta n^k$ . In order to have $S \in {\mathcal{S}}$ , we must have removed at least ${\varepsilon } n/2$ edges containing $S$ , and each removed edge decreases the degree of $k$ $(k-1)$ -sets by one. Therefore $|{\mathcal{S}}| \le \frac {2k^2 \eta n^k}{{\varepsilon } n/2} \le \zeta n^{k-1}$ , where we used $\eta \ll \zeta, {\varepsilon }$ for the last inequality. Since the collection was maximal, invoking Lemma 4.1, we get that one of the colour classes of $G'$ has size at most $\rho n^k/2$ . Therefore one of the colour classes of $G$ has size at most $\rho n^k/2 + 2k \cdot \ell \lt \rho n^k$ , where we used $\ell \lt \eta n^k$ and $\eta \ll \rho$ .

If one of the colour classes of $G$ , say $\mathcal{C}$ , has size at most $\rho n^k$ , then, since $\mathbf{x}_0$ is $\mu _0$ -normal, this colour class gets a total weight of at most $\sum _{e \in {\mathcal{C}}} \mathbf{x}_0(e) \le |{\mathcal{C}}| \mu _0^{-1} n^{-k+1} \le \mu _0^{-1} \rho n$ . Then the other colour class gets a total weight of at least $(1/k- \mu _0^{-1} \rho )n$ . By averaging, $\mathbf{x}_0$ assigns to one of the colour classes of the original $r$ -edge-colouring of $G$ a total weight of at least $\frac {(1/k-\mu _0^{-1}\rho )n}{r-1} \ge (1+\mu ) \frac {n}{rk}$ , where we used $\rho \ll \mu _0, 1/r, 1/k$ . In particular, $\mathbf{x}_0$ is already a desired $\mu$ -normal perfect fractional matching with high discrepancy.

We are left with the case where there exists a collection $\mathcal{L}\,:\!=\,\{L_i\,{:}\,i \in [\eta n^k]\}$ of $\eta n^k$ edge-disjoint gadgets. Now, either the total weight assigned by $\mathbf{x}_0$ to the red edges is at least $\frac {n}{rk}$ , or the total weight assigned by $\mathbf{x}_0$ to the blue edges is at least $\frac {(r-1)n}{rk}$ .

Suppose we are in the first case. We will modify $\mathbf{x}_0$ on the edges of each gadget in $\mathcal{L}$ to obtain a $\mu$ -normal perfect fractional matching $\mathbf{x}$ with high discrepancy in colour $c_1$ . For $L \in \mathcal{L}$ , let $e_1^L,\ldots, e_k^L$ (resp. $f_1^L,\ldots, f_k^L)$ be the horizontal (resp. vertical) edges of $L$ , and recall these are pairwise distinct. By the definition of a near-alternating $k$ -grid, all edges $e_1^L,\ldots, e_k^L$ but at most one are red, and all edges $f_1^L,\ldots, f_k^L$ but at most one are blue. Define $\mathbf{x} \colon E(G) \to [0,1]$ as follows: $\mathbf{x}(e_i^L)=\mathbf{x}_0(e_i^L)+ \mu _0 n^{-k+1}/2$ and $\mathbf{x}(f_i^L)=\mathbf{x}_0(f_i^L)- \mu _0 n^{-k+1}/2$ for each $i \in [k]$ and $L \in \mathcal{L}$ . Moreover, set $\mathbf{x}(e)=\mathbf{x}_0(e)$ for any other edge $e \in E(G)$ . In other words, $\mathbf{x}$ is obtained from $\mathbf{x}_0$ by decreasing the weight of each vertical edge by $\mu _0 n^{-k+1}/2$ and increasing the weight of each horizontal edge by the same quantity, in each of the gadgets in $\mathcal{L}$ . Observe that if a vertex is contained in a gadget, then it belongs to precisely one horizontal edge and one vertical edge of this gadget. Therefore for every vertex $v \in V(G)$ we have that $\sum _{e \ni v} \mathbf{x}(e) = \sum _{e \ni v} \mathbf{x}_0(e) = 1$ . Also, for every $e \in E(G)$ we have $\mu n^{-k+1} \le \mathbf{x}_0(e)-\mu _0 n^{-k+1}/2 \le \mathbf{x}(e) \le \mathbf{x}_0(e)+\mu _0 n^{-k+1}/2 \le \mu ^{-1} n^{-k+1}$ , where we used that $\mu _0 n^{-k+1} \le \mathbf{x}_0(e) \le \mu _0^{-1} n^{-k+1}$ and $\mu \ll \mu _0$ . This shows that $\mathbf{x}$ is a $\mu$ -normal perfect fractional matching. Moreover, for each $L \in \mathcal{L}$ , the weight given by $\mathbf{x}$ to the red edges in $L$ is bigger by at least $\mu _0 n^{-k+1}/2$ than the weight given by $\mathbf{x}_0$ to these edges. This is because we increased (by $\mu _0 n^{-k+1}/2$ ) the weight of at least $k-1 \geq 2$ red edges, and decreased (by the same amount) the weight of at most one such edge. As the gadgets in $\mathcal{L}$ are edge-disjoint and $|\mathcal{L}| = \eta n^k$ , we conclude that the total weight assigned by $\mathbf{x}$ to the red edges is at least $\frac {n}{rk}+\eta n^k \mu _0 n^{-k+1} /2\ge (1+\mu )\frac {n}{rk}$ , using that $\mu \ll \eta, \mu _0$ . Therefore, the total weight assigned by $\mathbf{x}$ to the colour class of $c_1$ in the original edge-colouring of $G$ is at least $(1+\mu )\frac {n}{rk}$ .

Suppose now that we are in the second case, namely, that the total weight assigned by $\mathbf{x}_0$ to the blue edges is at least $\frac {(r-1)n}{rk}$ . We then use the same argument as above to find a $\mu$ -normal perfect fractional matching $\mathbf{x}$ which assigns a total weight of at least $(r-1)(1+\mu )\frac {n}{rk}$ to the blue edges. Recall that the blue edges are precisely those coloured by $c_2,\ldots, c_r$ in the original colouring of $G$ , and thus, by averaging, there is $2 \le i \le r$ such that the total weight assigned by $\mathbf{x}$ to the colour class of $c_i$ in the original colouring of $G$ is at least $(1+\mu )\frac {n}{rk}$ .

Proof. Let $\mathbf{x}$ be a $\mu$ -normal perfect fractional matching of $G$ such that the total weight received by some colour class, say ‘red’, is at least $(1+\mu )\frac {n}{rk}$ . This exists by Lemma 5.2. Let $\mathcal{Y}=(Y_1,Y_2,\ldots )$ be the random walk defined via $\mathbf{x}$ , with notation as introduced at the beginning of this section. (In particular, we use $\mathbf{Pr}$ as the probability measure corresponding to the random walk, whereas $\mathbb{P}$ will denote the desired probability measure on $\Omega$ .) Here, we will only be interested in the first $t$ vertices of $\mathcal{Y}$ and we note that those form a tight walk of order $t$ . Now, sample elements of $\Omega$ according to the first $t$ vertices of $\mathcal{Y}$ , conditioned on $\mathcal{Y}$ being self-avoiding up to $Y_t$ . More precisely, let $\mathcal{B}$ be the event that $Y_1, \ldots, Y_t$ are pairwise distinct, and denote by $Q$ any tight path of order $t$ and by $q_1, \ldots, q_t$ the vertices of $Q$ (appearing in this order). Then take the distribution on $\Omega$ where a randomly chosen element $P \in \Omega$ satisfies

\begin{equation*} {\mathbb{P}}[P=Q] = {\mathbf{Pr}}\Big [\{Y_1=q_1,\ldots, Y_t=q_t\} \cup \{Y_1=q_t,\ldots, Y_t=q_1\}\; |\; {\mathcal{B}}\Big ]\,, \end{equation*}

where we considered both orders of $Q$ since the elements of $\Omega$ are unordered. We claim that this distribution on $\Omega$ has the desired properties. Before proving that this is the case, we need some preliminary observations.

Since $\mathbf{x}$ is $\mu$ -normal, the probability that the walk starts with $\overrightarrow {S}=(q_1,\ldots, q_{k-1})$ is

\begin{equation*} \pi (\overrightarrow {S}) = \frac {\mathbf{x}(S)}{\sum _{\overrightarrow {S'} \in (V)_{k-1}} \mathbf{x}(S')} \le \frac {n \cdot \mu ^{-1} n^{-k+1}}{(k-1)! \cdot n} = O_{\mu }(n^{-k+1})\,, \end{equation*}

where we used (5.2). By using in addition that $\delta (G) \ge (1/2+{\varepsilon })n$ , we can show that the transition probabilities of $\mathcal{Y}$ are $O_{\mu }(n^{-1})$ . Indeed, with $\overrightarrow {Z_i}=(Y_{i-(k-2)},\ldots, Y_i)$ being the ordered set of the last $k-1$ vertices of $\mathcal{Y}$ , for $v \in V(G) \setminus Z_i$ we have

\begin{equation*} {\mathbf{Pr}}[Y_{i+1}=v|Y_{i-(k-2)},\ldots, Y_{i}] = \frac {\mathbf{x}(Z_{i}\cup \{v\})}{\mathbf{x}(Z_i)} \le \frac {\mu ^{-1} n^{-k+1}}{1/2 \cdot n \cdot \mu n^{-k+1}} = O_{\mu }(n^{-1})\,, \end{equation*}

while for $v \in Z_i$ we have ${\mathbf{Pr}}[Y_{i+1}=v|Y_{i-(k-2)},\ldots, Y_{i}]=0$ . Therefore, by applying the chain rule, ${\mathbf{Pr}}[Y_1=q_1,\ldots, Y_t=q_t]=O_{t,\mu }(n^{-t})$ . Moreover, the number of walks of order $t$ which are not self-avoiding is $O_t(n^{t-1})$ and thus ${\mathbf{Pr}}[{\mathcal{B}}^c]=O_t(n^{t-1}) \cdot O_{t,\mu }(n^{-t})=O_{t,\mu }(n^{-1})$ .

Now Item (1) follows easily as we have

\begin{equation*}{\mathbb{P}}[P=Q]\le \frac {{\mathbf{Pr}}[Y_1=q_1,\ldots, Y_t=q_t]+{\mathbf{Pr}}[Y_1=q_t,\ldots, Y_t=q_1]}{{\mathbf{Pr}}[{\mathcal{B}}]} =O_{t,\mu }(n^{-t}), \end{equation*}

where we used that ${\mathbf{Pr}}[{\mathcal{B}}] = 1-O_{t,\mu }(n^{-1})\ge 1/2$ .

Fix $v \in V(G)$ and $1 \le i \le t$ . The number of walks of order $t$ which are not self-avoiding and whose $i$ -th vertex is $v$ is $O_t(n^{t-2})$ and thus ${\mathbf{Pr}}\left [\{Y_i=v\} \cap {\mathcal{B}}^c\right ] = O_t(n^{t-2}) \cdot O_{t,\mu }(n^{-t})=O_{t,\mu }(n^{-2})$ . Therefore,

\begin{equation*} {\mathbb{P}}[v \in V(P)] = {\mathbf{Pr}}\left [\bigcup _{i \in [t]} \{Y_i=v\} \; | \; {\mathcal{B}} \right ] = \sum _{i \in [t]} \frac {{\mathbf{Pr}}\big [\{Y_i=v\} \cap {\mathcal{B}} \big ]}{{\mathbf{Pr}}[{\mathcal{B}}]} \ge tn^{-1}-O_{t,\mu }(n^{-2}) \,, \end{equation*}

where we used that the events $\{Y_i=v\} \cap {\mathcal{B}}$ with $i \in [t]$ are pairwise disjoint, that ${\mathbf{Pr}}[\{Y_i=v\}]=n^{-1}$ by Fact 5.1, that

\begin{equation*} {\mathbf{Pr}}\big [\{Y_i=v\} \cap {\mathcal{B}} \big ] = {\mathbf{Pr}}\big [Y_i=v\big ] - {\mathbf{Pr}}\big [\{Y_i=v\} \cap {\mathcal{B}}^c\big ] = n^{-1} - O_{t,\mu }(n^{-2}), \end{equation*}

and that ${\mathbf{Pr}}[{\mathcal{B}}]\le 1$ . Similarly, by using ${\mathbf{Pr}}[{\mathcal{B}}]=1-O_{t,\mu }(n^{-1})$ , we get that ${\mathbb{P}}[v \in V(P)] \leq tn^{-1}+O_{t,\mu }(n^{-2})$ . This proves Item (2).

For a given $k \le i \le t$ , denote by $e_i\,:\!=\,\{Y_{i-k+1},\ldots, Y_i\}$ the $(i-k+1)$ -st edge of $\mathcal{Y}$ . Let $e \in E(G)$ . Similarly as above, ${\mathbf{Pr}}[\{e_i=e\} \cap {\mathcal{B}}^c]=O_{t,\mu }(n^{-k-1})$ because there are $O_t(n^{t-k-1})$ tight walks which are not self-avoiding and satisfy $e_i = e$ , and each such walk has probability $O_{t,\mu }(n^{-t})$ . Thus,

\begin{equation*} {\mathbf{Pr}}\Big [e_i=e \; | \; {\mathcal{B}}\Big ]=\frac {{\mathbf{Pr}}\big [e_i=e\big ]-{\mathbf{Pr}}\big [\{e_i=e\} \cap {\mathcal{B}}^c\big ]}{{\mathbf{Pr}}[{\mathcal{B}}]} \ge \frac {k}{n}\cdot \mathbf{x}(e) - O_{t,\mu }(n^{-k-1})\,, \end{equation*}

where we used Fact 5.1 and ${\mathbf{Pr}}[{\mathcal{B}}]\le 1$ . Let $\mathcal{R}$ be the set of edges of $G$ which are coloured red. Then

\begin{equation*} {\mathbf{Pr}}\Big [e_i\in {\mathcal{R}} \; | \; {\mathcal{B}}\Big ] \ge \sum _{e \in {\mathcal{R}}} \left (\frac {k}{n}\cdot \mathbf{x}(e) - O_{t,\mu }(n^{-k-1})\right ) \ge \frac {1+\mu }{r}-O_{t,\mu }(n^{-1})\,, \end{equation*}

where we used that $|{\mathcal{R}}| \le n^k$ and $\sum _{e \in {\mathcal{R}}} \mathbf{x}(e) \ge (1+\mu )\frac {n}{rk}$ . Since $P$ is a path of order $t$ , it has $t-k+1$ edges, and thus Item (3) follows by linearity of expectation.

Proof. Let

\begin{equation*} 1/n \ll \delta \ll 1/t \ll \beta \ll \mu \ll {\varepsilon }, 1/k, 1/r\,, \end{equation*}

where $2\mu$ is given by Lemma 5.3 on input ${\varepsilon }, r, k$ , and set $\gamma \,:\!=\,\delta /(50t^2)$ and $V\,:\!=\,V(G)$ .

Set $N\,:\!=\,n^{t-1/2}$ and let $\mathcal{P}\,:\!=\,\{P_i:i \in [N]\}$ be a collection of $N$ tight paths of order $t$ independently sampled according to the distribution given by Lemma 5.3. For a given $v \in V$ , define $X_v\,:\!=\,\{i \in [N]\,{:}\,v \in V(P_i)\}$ . Using Item (2) in Lemma 5.3, we have $\mathbb{E}[|X_v|]=N \cdot t n^{-1} \cdot \left [1 \pm O_{t,\mu }(n^{-1})\right ] = t n^{t-3/2} \cdot \left [1 \pm O_{t,\mu }(n^{-1})\right ]$ . Therefore, by Chernoff’s inequality (Lemma 3.4) and a union bound over $v \in V$ , w.h.p. we have

(6.1) \begin{equation} |X_v| = (1 \pm \beta /3) t n^{t-3/2} \end{equation}

for every $v \in V$ . For a path $P \in \mathcal{P}$ , let $\text{red}(P)$ be the number of red edges of $P$ , and note that $0 \leq \text{red}(P) \leq t-k+1$ . Let $R\,:\!=\, \sum _{i = 1}^N \text{red}(P_i)$ . By Item (3) of Lemma 5.3, $\mathbb{E}[R] \ge N \cdot \left [(t-k+1) \cdot \frac {1+2\mu }{r} - O_{t,\mu }(n^{-1})\right ]$ and by Chernoff’s inequality (Lemma 3.4), we have w.h.p. that

(6.2) \begin{equation} R \ge (1-\beta ) \cdot (t-k+1) \cdot \frac {1+2\mu }{r} \cdot n^{t-1/2}\, . \end{equation}

We now show that we can pass to a large subcollection $\mathcal{P}' \subseteq \mathcal{P}$ such that no two paths in $\mathcal{P}'$ have the same vertex set. Let $Y$ be the set of pairs $1 \leq i\lt j \leq N$ such that $V(P_i) = V(P_j)$ . We now bound $|Y|$ . Given any $t$ -subset $S$ of $V$ , observe that there are $t!/2$ (unordered) paths $P$ with $V(P)=S$ and that, using Item (1) in Lemma 5.3, the probability that $P_i = P$ for a given $1 \leq i \leq N$ is $O_{t,\mu }(n^{-t})$ . Therefore, the probability that a given pair $i,j$ belongs to $Y$ is $O_{t,\mu }(n^{-t})$ , and thus $\mathbb{E}[|Y|] \le N^2 \cdot O_{t,\mu }(n^{-t}) = O_{t,\mu }(n^{t-1})$ . It follows from Markov’s inequality that w.h.p. $|Y| \le n^{t-2/3}$ .

We now fix such a collection of paths that satisfies (6.1), (6.2) and $|Y| \le n^{t-2/3}$ . Let $\mathcal{P}' \subseteq \mathcal{P}$ be obtained from $\mathcal{P}$ by deleting $P_i,P_j$ for every pair $\{i,j\} \in Y$ . Then $|\mathcal{P}'| \ge |\mathcal{P}| - 2|Y| \ge (1-\beta /3) n^{t-1/2}$ . Let $\mathcal{H}$ be the auxiliary $t$ -graph on $V$ with edge set $\{V(P)\,{:}\,P \in \mathcal{P}'\}$ . By the definition of $\mathcal{P}'$ , we have $V(P) \neq V(Q)$ for each $P, Q \in \mathcal{P}'$ , and thus $e(\mathcal{H})=|\mathcal{P}'|$ (i.e., $\mathcal{H}$ has no multiple edges). We now show that $\mathcal{H}$ is suitable for an application of Theorem3.5, by establishing bounds on $\Delta (\mathcal{H})$ and $\Delta ^c(\mathcal{H})$ . Using (6.1), we have $d_{\mathcal{H}}(\{v\}) \le |X_v| \le (1+\beta /3)t n^{t-3/2}$ . Setting $\Delta \,:\!=\,(1 + \beta /3) t n^{t-3/2}$ , we have $\Delta (\mathcal{H}) \le \Delta$ , and it trivially holds that $\Delta ^c(\mathcal{H}) \le n^{t-2} \le \Delta ^{1-\delta }$ , as $\delta \ll 1/t$ .

We would like the matching given by Theorem3.5 to be almost-spanning and with large discrepancy. To this end, we define two weight functions $w_1,w_2\colon E(\mathcal{H}) \rightarrow \mathbb{R}_{\ge 0}$ as follows: $w_1 \equiv 1$ ; and for an edge $e \in E(\mathcal{H})$ corresponding to a path $P \in \mathcal{P}'$ , we define $w_2(e) \,:\!=\, \text{red}(P)$ . We claim that $w_i(E(\mathcal{H})) \ge \max _{e \in E(\mathcal{H})} w(e) \Delta ^{1+\delta }$ for each $i=1,2$ . For $i=1$ , this is obvious as

\begin{equation*} w_1(E(\mathcal{H})) = e(\mathcal{H}) \ge (1-\beta /3) n^{t-1/2} \ge \Delta ^{1+\delta } = \max _{e \in E(\mathcal{H})} w_1(e) \Delta ^{1+\delta }\,, \end{equation*}

where the last inequality uses that $\delta \ll 1/t$ . And for $i=2$ , we have

\begin{align*} w_2(E(\mathcal{H})) &= \sum _{P \in \mathcal{P}'}\text{red}(P) = R - \sum _{P \in \mathcal{P}\setminus \mathcal{P}'}\text{red}(P) \geq R - (t-k+1) \cdot 2n^{t-2/3} \\ &\geq (1-2\beta ) \cdot (t-k+1) \cdot \frac {1+2\mu }{r} \cdot n^{t-1/2}, \end{align*}

where the first inequality uses $|\mathcal{P}|-|\mathcal{P}'| \le 2 |Y| \le 2n^{t-2/3}$ and that each path has at most $t-k+1$ red edges, while the second inequality uses (6.2). Therefore

\begin{equation*} w_2(E(\mathcal{H})) = \Omega (n^{t-1/2}) \ge (t-k+1) \Delta ^{1+\delta } \ge \max _{e \in E(\mathcal{H})} w_2(e) \Delta ^{1+\delta }\,, \end{equation*}

using that $\Delta = O_t(n^{t-3/2})$ and $\delta \ll 1/t$ .

Let $\mathcal{M}$ be the matching in $\mathcal{H}$ given by applying Theorem3.5 with $\mathcal{W} \,:\!=\, \{w_1,w_2\}$ . Using $w_1(E(\mathcal{H})) = e(\mathcal{H}) = |\mathcal{P}'| \ge (1-\beta /3)n^{t-1/2}$ and the guarantees of Theorem3.5, we have

\begin{equation*} |\mathcal{M}|=w_1(\mathcal{M}) \ge \left (1-\Delta ^{-\gamma }\right ) \cdot \frac {w_1(E(\mathcal{H}))}{\Delta } \ge \left (1-\Delta ^{-\gamma }\right ) \cdot \frac {(1-\beta /3)n^{t-1/2}}{(1 + \beta /3) t n^{t-3/2}} \geq (1-\beta ) \frac {n}{t}\,. \end{equation*}

Similarly, for $w_2$ we have

\begin{align*} w_2(\mathcal{M}) \ge \left (1-\Delta ^{-\gamma }\right ) \cdot \frac {w_2(E(\mathcal{H}))}{\Delta } & \ge (1-o(1)) \cdot \frac {(1-2\beta ) \cdot (t-k+1) \cdot \frac {1+2\mu }{r} \cdot n^{t-1/2}}{(1 + \beta /3) t n^{t-3/2}} \\ &= (1 - o(1)) \cdot \frac {1-2\beta }{1+\beta /3} \cdot \frac {t-k+1}{t} \cdot \frac {1+2\mu }{r} \cdot n \\ &\ge (1-3\beta ) \cdot \frac {1+2\mu }{r} \cdot n \ge (1+\mu )\frac {n}{r}\,, \end{align*}

where we used the bound on $w_2(E(\mathcal{H}))$ established above, the definition of $\Delta$ , together with $1/t \ll \beta \ll \mu \ll 1/k$ .

Therefore the matching $\mathcal{M}$ corresponds to a collection of vertex-disjoint paths of $G$ of order $t$ such that their union covers all but at most $\beta n$ vertices and the colour red appears on at least $(1+\mu )\frac {n}{r}$ edges in the paths, as desired.

Proof of Theorem 1.1. Let

\begin{equation*} 1/n \ll 1/t \ll \beta \ll \mu \ll {\varepsilon } \ll 1/k, 1/r\,, \end{equation*}

where we have assumed without loss of generality that $\varepsilon$ is sufficiently small. Let $G$ be an $n$ -vertex $k$ -graph with $\delta (G) \ge (1/2+{\varepsilon })n$ whose edges are $r$ -coloured. By Lemma 3.2, there exists a tight path $P_0$ such that $|V(P_0)| \le \mu n$ and for each $W \subseteq V(G) \setminus V(P_0)$ of size $|W| \le 3 \beta n$ there is a tight path covering $V(P_0) \cup W$ and with the same ends as $P_0$ . Let $R$ be a random subset of $V(G) \setminus V(P_0)$ obtained by including each vertex with probability $\beta$ independently. Then w.h.p. it holds that $\beta n/2 \leq |R| \leq 2\beta n$ and that $|N_G(S) \cap R| \ge (1/2+{\varepsilon }/2)|R|$ for every $(k-1)$ -subset $S \subseteq V(G)$ . Indeed, this can be shown by a standard application of Chernoff’s inequality (Lemma 3.4) and a union bound over $S$ .

Let $G'\,:\!=\,G \setminus (V(P_0) \cup R)$ and observe that $\delta (G') \ge \delta (G) - |V(P_0) \cup R| \ge (1/2+{\varepsilon }/2)n \ge (1/2+{\varepsilon }/2)|V(G')|$ . Then, by Lemma 6.1 (with $4\mu$ playing the role of $\mu$ ), $G'$ contains a collection $\mathcal{P}\,:\!=\,\{P_i\,{:}\,i \in [N]\}$ of vertex-disjoint tight paths of order $t$ such that their union covers all but at most $\beta n$ vertices, and there is a colour, say red, which appears on at least $(1+4\mu )|V(G')|/r \ge (1+4\mu )(1-\mu -2\beta )n/r \ge (1+\mu )n/r$ edges in the paths. Therefore, if we manage to connect the paths in $\mathcal{P}$ into a tight cycle while covering all the vertices then we are done. This connection can be achieved using the standard tools collected in Section 3.1. The details follow.

Denote by $\overrightarrow {S_i}$ and $\overrightarrow {T_i}$ the ends of $P_i$ for each $0 \le i \le N$ . Add the uncovered vertices of $G'$ to $R$ to get a set $R'$ , and observe that $|R| \leq |R'| \leq |R| + \beta n$ . Using multiple applications of the connecting lemma (Lemma 3.1), we can connect the path $P_0$ and the paths in $\mathcal{P}$ into an almost-spanning tight cycle using vertices in $R'$ , as proved by the following claim.

Proof. Suppose we can find vertex-disjoint connecting paths $Q_0, \ldots, Q_{m-1}$ as in the statement of the claim, and let $m$ be as large as possible. If we are not done yet, then $m \le N$ . The union of $Q_0, \ldots, Q_{m-1}$ covers at most $m \cdot 2k/{\varepsilon }^2 \le 2kn/({\varepsilon }^2 t)$ vertices of $R'$ , where we used that $N \le n/t$ . Let $R''$ denote the subset of $R'$ consisting of the uncovered vertices. Then for every $(k-1)$ -subset $S \subseteq V(G)$ we have $|N_G(S) \cap R''| \ge (1/2 + {\varepsilon }/2)|R| - |R' \setminus R''| \geq (1/2+{\varepsilon }/4)|R'' \cup T_{m+1} \cup S_{m+2}|$ , where we used that $|R| \geq |R'| - \beta n$ , $|R' \setminus R''| \leq 2kn/({\varepsilon }^2 t)$ and $1/t \ll \beta \ll {\varepsilon }, 1/k$ . Therefore, we can apply Lemma 3.1 to $G[R'' \cup T_{m+1} \cup S_{m+2}]$ to get a tight path of order at most $2k/{\varepsilon }^2$ connecting $\overleftarrow {T_{m+1}}$ and $\overleftarrow {S_{m+2}}$ , which is vertex-disjoint from $Q_0, \ldots, Q_{m-1}$ . This contradicts the maximality of $m$ .