Proof of Theorem 2. Let $\delta = \epsilon ^{24/\epsilon ^{12}}$ . Run the DFS algorithm described in Section 2 on $G$ . We claim that after $\epsilon n$ steps in the algorithm, whp, $|U|\geq \frac {\epsilon ^2n}{3d}$ , and thus, as $U$ forms an induced path in $G_p$ , the statement follows.

Indeed, after $\epsilon n$ steps, $|U\cup S_1\cup S_2|\sim Bin(\epsilon n,p)$ . Thus, by Lemma2.1, with probability at least $1-2\exp \left \{-\frac {(n/d)^{4/3}}{4\epsilon n /d}\right \}$ we have that $\bigg ||U_1\cup S_1\cup S_2|-\frac {(1+\epsilon )\epsilon n}{d}\bigg |\le (n/d)^{2/3}$ . Further, by Lemma2.7 together with our assumption on $\delta$ , we have that whp $|S_2|=O(\epsilon ^3)n/d$ . Assume towards contradiction that $|U|\lt \frac {\epsilon ^2n}{3d}$ . Then, whp,

\begin{align*} |S_1|\ge \frac {(1+\epsilon )\epsilon n}{d}-n^{2/3}-O(\epsilon ^3)\frac {n}{d}-\frac {\epsilon ^2 n}{3d}\ge \frac {(\epsilon +3\epsilon ^2/5)n}{d}. \end{align*}

On the other hand, whp,

\begin{equation*}|S_1|\le |U\cup S_1\cup S_2|\le 4\epsilon n/(3d)\lt n/(3d),\end{equation*}

and thus by Lemma2.3 applied with $\alpha =\epsilon ^{12}$ (which is possible by our assumption on $\delta$ ), we have that whp

\begin{align*} |N_G(S_1)|\ge (1-\epsilon ^{12})\left (\epsilon +3\epsilon ^2/5-\frac {1}{2}(\epsilon +3\epsilon ^2/5)^2\right )n\gt \epsilon n, \end{align*}

a contradiction, since $N_G(S_1)\subseteq S_2\cup U\cup W$ and $|S_2\cup U\cup W|\le |S_1\cup S_2\cup U\cup W|\le \epsilon n$ .

Proof of Theorem 1. Let $\epsilon ,\delta (\epsilon )\gt 0$ be as in the statement of Theorem2. We may assume that $\delta _1\le \delta (\epsilon )$ . Let $p=\frac {1+\epsilon }{d}$ . Then, by Theorem2, whp $G_p$ contains an induced path of length at least $\frac {\epsilon ^2n}{3d}$ . Thus, deterministically, $G$ contains an induced path $P$ on $k=\frac {\epsilon ^2n}{3d}$ vertices (note that for the latter to hold, we merely needed the first statement to hold with positive probability).

Let $P=\{v_1,\ldots ,v_{k}\}$ . Let $P_1=\{v_1,\ldots , v_{k/3}\}$ be the first $\frac {\epsilon ^2n}{9d}$ vertices of $P$ , and let $P_2=\{v_{2k/3+1},\ldots ,v_{k}\}$ be the last $\frac {\epsilon ^2n}{9d}$ vertices of $P$ . Let $P^{\prime}=\{v_{k/2-k/20+1},\ldots ,v_{k/2+k/20}\}$ be the set of $\frac {\epsilon ^2n}{30d}$ vertices at the middle of $P$ . By Lemma2.3, $|N_G(P_1)|, |N_G(P_2)|\geq \frac {\epsilon ^2n}{10}$ . Since $G$ is $d$ -regular, $|N_G(P^{\prime})|\leq d|P^{\prime}|=\frac {\epsilon ^2n}{30}$ . Set $N_1=N_G(P_1)\setminus \left (P\cup N_G(P^{\prime})\right )$ and similarly $N_2=N_G(P_2)\setminus \left (P\cup N_G(P^{\prime})\right )$ . Note that $|N_1|,|N_2|\geq \frac {\epsilon ^2n}{30}$ . In particular, by Lemma2.2, $e(N_1,N_2)\ge \epsilon ^5 dn\gt 0$ .

Let $u_1\in N_1$ , $u_2\in N_2$ be such that $u_1u_2\in E(G)$ . Consider $L=\{v_1,\ldots ,v_{k/2-k/20}\}$ and $R=\{v_{k/2+k/20+1},\ldots ,v_k\}$ . If $u_1$ has a neighbour in both $L$ and $R$ , we let $w_1$ be the neighbour of $u_1$ from $L$ closest to $P^{\prime}$ , that is, to $v_{k/2-k/20}$ . Similarly, let $w_2$ be the neighbour of $u_1$ from $R$ closest to $P^{\prime}$ . Recall that $u_1$ has no neighbours in $P^{\prime}$ , hence $u_1$ , together with the subpath of $P$ starting at $w_1$ and ending at $w_2$ forms an induced cycle in $G$ whose length is at least $|P^{\prime}|=\frac {\epsilon ^2n}{30d}$ . This completes the proof, with $\delta _2=\epsilon ^2/30$ . A similar argument applies when $u_2$ has a neighbour in both $L$ and $R$ .

Now, since $u_1\in N_1\subseteq L$ and $u_2\in N_2\subseteq R$ , it remains to consider the case where $u_1$ has a neighbour in $L$ but not in $R$ and $u_2$ has a neighbour in $R$ but not in $L$ . Let $w_1$ as the neighbour of $u_1$ from $L$ closest to $P^{\prime}$ , and similarly, define $w_2$ as the neighbour of $u_2$ from $R$ closest to $P^{\prime}$ . $u_1u_2$ together with the subpath of $P$ starting at $w_1$ and ending at $w_2$ forms an induced cycle in $G$ whose length is at least $|P^{\prime}|=\frac {\epsilon ^2n}{30d}$ which again completes the proof, with $\delta _2=\epsilon ^2/30$ .

Proof of Theorem 4. Let $\epsilon ,\delta (\epsilon )\gt 0$ be as in the statement of Theorem2. We may assume that $\delta _1\le \delta (\epsilon )$ , and in particular, $\delta _1\le \epsilon ^{8/\epsilon ^4}$ . Let $p=\frac {1+\epsilon }{d}$ . Then, by Theorem2, whp $G_p$ contains an induced path $P$ of length exactly $\frac {\epsilon ^2n}{3d}$ . Furthermore, by Lemma2.3 and by our assumption on $\delta _1$ , whp $|N_G(P)|\ge (1-\epsilon ^4)\left (\frac {\epsilon ^2n}{3}-\frac {\epsilon ^4n}{18}\right )\ge \frac {\epsilon ^2n}{3}-\epsilon ^4n$ . Thus, there exists (deterministically) an induced path $P=\{v_1,v_2,\ldots ,v_k\}$ in $G$ on $k=\frac {\epsilon ^2n}{3d}$ vertices, such that $|N_G(P)|\ge \frac {\epsilon ^2n}{3}-\epsilon ^4n$ .

Now, since $G$ is $d$ -regular, we have that $\sum _{v\in N_G(P)}d(v,P)\le \sum _{v\in P}d(v)=\epsilon ^2n/3$ . Suppose towards contradiction that there are more than $2\epsilon ^4n$ vertices in $N_G(P)$ each having at least two neighbours in $P$ . Then,

\begin{align*} \sum _{v\in N_G(P)}d(v,P)\ge |N_G(P)|+2\epsilon ^4n\gt \epsilon ^2n/3, \end{align*}

which is a contradiction. Hence, there exists a set $U\subseteq N_G(P)$ of size at least $\frac {\epsilon ^2n}{3}-3\epsilon ^4n$ , such that every $u\in U$ has exactly one neighbour in $P$ . In particular, this implies that there are at least $\frac {\epsilon ^2n}{10d}$ vertices in $P$ which have at least $\frac {d}{10}$ neighbours in $U$ . Let us denote the set of these vertices in $P$ by $W$ . Let us then choose a subset $W^{\prime}\subseteq W$ , such that $v_1,v_2,v_{k-1},v_k\notin W^{\prime}$ , and the distance in $P$ between any $u,u^{\prime}\in W^{\prime}$ is at least $4$ . Note that we can choose such $W^{\prime}$ with $|W^{\prime}|\ge |W|/5\ge \frac {\epsilon ^2n}{50d}$ . Crucially, observe that every $u\in N_G(W^{\prime})\cap U\,:\!=\, N_{W^{\prime}}$ has a unique neighbour $v\in W^{\prime}$ on $P$ .

Now, let $H\,:\!=\, G[V(P)\cup N_{W^{\prime}}]$ . Let $A\neq A^{\prime}$ be subsets of $N_{W^{\prime}}$ , such that $G[V(P)\cup A]\cong G[V(P)\cup A^{\prime}]$ . Let $\psi :V(P)\cup A\to V(P)\cup A^{\prime}$ be an isomorphism between these two graphs. We claim that $\psi \mid _{V(P)}$ is either the trivial automorphism of $P$ , or the non-trivial involution automorphism of $P$ . First, note that the only two vertices in $H$ which are of degree one and have a neighbour of degree two are $v_1$ and $v_k$ . Indeed, $v_2,v_{k-1}\notin W^{\prime}$ and thus have degree two, and any other vertex of degree one is in $N_{W^{\prime}}$ , and thus has a neighbour on $V(P)$ of degree at least three. It thus suffices to show that $\psi$ sends vertices of degree two in $V(P)$ to (possibly other) vertices of degree two in the same set $V(P)$ .

Suppose towards contradiction that there exists $u\in V(P)$ such that $d_H(u)=2=d_{G[V(P)\cup A]}(u)$ and $\psi (u)\in A^{\prime}$ . Since every vertex in $A^{\prime}\subseteq N_{W^{\prime}}$ has a unique neighbour in $P$ and since $\psi (u)$ has degree two in $G[V(P)\cup A^{\prime}]$ , we have that $\psi (u)$ has a neighbour in $A^{\prime}$ , which we denote by $v$ . Let $x_1$ be the unique neighbour of $\psi (u)$ in $P$ , and let $x_2$ be the unique neighbour of $v$ in $P$ , noting that both $x_1$ and $x_2$ have degree at least three in $G[V(P)\cup A^{\prime}]$ . We then have that $\psi (u)$ belongs to the path $x_1\psi (u)vx_2$ in $G[V(P)\cup A^{\prime}]\subseteq H$ . However, by construction of $H$ , since $d_H(u)=2$ , the two closest vertices to $u$ which have degree at least $3$ in $G[V(P)\cup A]$ (and therefore, in $H$ ) lie on $P$ , and thus are at distance at least four from each other – contradiction.

Therefore, any automorphism $\psi$ has that $\psi \mid _{V(P)}$ is one of the two involution automorphisms. Hence, for a fixed automorphism of $P$ , any two subsets $A,A^{\prime}\in N_{W^{\prime}}$ having $|A^{\prime}\cap N(v)\cap N_{W^{\prime}}|\neq |A^{\prime}\cap N(v)\cap N_{W^{\prime}}|$ for some $v\in W^{\prime}$ cannot span (together with $P$ ) isomorphic subgraphs. We thus conclude that, for every tuple $(0\leq s_v\leq d/10)_{v\in W^{\prime}}$ , there exists at most one other tuple $(0\leq s^{\prime}_v\leq d/10)_{v\in W^{\prime}}$ satisfying the following: there exist two subsets $A,A^{\prime}\subset N_{W^{\prime}}$ such that $|A\cap N(v)\cap N_{W^{\prime}}|=s_v$ and $|A^{\prime}\cap N(v)\cap N_{W^{\prime}}|=s^{\prime}_v$ for every $v\in W^{\prime}$ and $G[V(P)\cup A],G[V(P)\cup A^{\prime}]$ are isomorphic. Therefore,

\begin{equation*} \mu (G)\geq \frac {1}{2}(d/10)^{|W^{\prime}|}\geq \frac {1}{2}\left (\frac {d}{10}\right )^{\frac {\epsilon ^2n}{50d}}\ge \exp \left \{\frac {\delta _2n\log d}{d}\right \}, \end{equation*}

for small enough $\delta _2$ , as required.

Proof of Theorem 5. Let $\delta \gt 0$ be a constant. Let $d\in \mathbb{N}$ be sufficiently large, and let $n\in \mathbb{N}$ be such that $d,n$ satisfy the parity assumptions which are implicit below (in particular, $n$ is divisible by $d$ ). Let $d_0\,:\!=\, d_0(\delta )$ be the smallest integer satisfying $d_0\ge 3$ and $\sqrt {d_0}\ge \frac {3}{\delta }$ .

Assume first, for the sake of clarity of presentation, that $d$ is divisible by $d_0$ . Let $n_0\,:\!=\, n\cdot d_0/d$ , and let $H_1$ be an $(n_0, d_0, \lambda _0)$ -graph with $\lambda _0\le 3\sqrt {d_0}$ — indeed, such a graph exists, since a random $d_0$ -regular graph on $n_0$ vertices typically satisfies this (see, e.g., [Reference Krivelevich and Sudakov23]). We note that we assume here that $n_0$ is sufficiently large with respect to $d_0$ , and in turn, $n$ is sufficiently large with respect to $d$ . Let $H_2$ be the empty graph on $d/d_0$ vertices. Set $G=\mathrm{lex}(H_1,H_2)$ . By Lemmas5.1 and 5.2 we have that $G$ is an $(n,d,\lambda )$ -graph with $\lambda =\lambda _0\cdot \frac {d}{d_0}$ . Hence, by our choice of $d_0$ ,

\begin{align*} \frac {\lambda }{d}=\frac {\lambda _0}{d_0}\le \frac {3}{\sqrt {d_0}}\le \delta , \end{align*}

and thus satisfies the requirement of Theorem5.

Let $V(H_1)=\{v_1,\ldots , v_{n_0}\}$ . Let $V_1,\ldots , V_{n_0}$ be subsets of $V(G)$ , where $V_i$ is the blow-up of $v_i$ . Note that, by construction, given two graphs $G_1,G_2\subseteq G$ , if for every $i\in [n_0]$ , $|V(G_1)\cap V_i|=|V(G_2)\cap V_i|$ , then $G_1\cong G_2$ . We thus conclude that the number of non-isomorphic induced subgraphs of $G$ is bounded from above by the number of different tuples $(0\leq s_i\leq d/d_0)_{i=1}^{n_0}$ . Hence,

\begin{align*} \mu (G)\le \left (\frac {d}{d_0}+1\right )^{n_0}=\exp \left \{\frac {nd_0}{d}\log \!(d/d_0+1)\right \}\le \exp \left \{\frac {Cn\log d}{d}\right \}, \end{align*}

as required.

In the general case, when $d$ is not divisible by $d_0$ , let us write $d=qd_0+r$ where $q\in \mathbb{N}$ and $1 \le r \le d_0-1$ . Suppose first that $qr$ is even. We then set $H_2$ to be a graph on $q$ vertices, formed by taking a disjoint union of $\lfloor \frac {q}{r+1}\rfloor -1$ cliques of size $r+1$ , and on the remaining $q-(r+1)\left (\lfloor \frac {q}{r+1}\rfloor -1\right )$ vertices we draw an arbitrary $r$ -regular graph (indeed, such a graph exists since $qr$ and $r(r+1)$ are both even). If $qr$ is odd, then both $q$ and $r$ are odd, and thus $q-1$ is even. Let us write $d=(q-1)d_0+(d_0+r)$ . We then set $H_2$ to be a graph on $q-1$ vertices, formed by taking a disjoint union $\lfloor \frac {q-1}{d_0+r+1}\rfloor -1$ cliques of size $d_0+r+1$ , and on the remaining $(q-1)-(d_0+r+1)\left (\lfloor \frac {q-1}{d_0+r+1}\rfloor -1\right )$ vertices we draw an arbitrary $(d_0+r)$ -regular graph (indeed, such a graph exists since $q-1$ and $(d_0+r)(d_0+r+1)$ are both even). The rest of the proof, for both cases (both choices of $H_2$ ), is quite similar to the case where $d$ is divisible by $d_0$ .

Proof of Theorem 3. Let $\delta \gt 0$ be a constant. Let $d\in \mathbb{N}$ be sufficiently large, and let $n\in \mathbb{N}$ be such that $d,n$ satisfy the parity assumptions which are implicit below (in particular, $n$ is divisible by $d$ ). Let $d_0$ be the smallest integer satisfying $d_0\ge 3$ and $\sqrt {d_0}\ge \frac {4}{\delta }$ .

Assume first, for the sake of clarity of presentation, that $d+1$ is divisible by $d_0+1$ . Let $k\,:\!=\, \frac {d+1}{d_0+1}$ , and let $n_0=n/k$ . Let $H$ be an $(n_0,d_0,\lambda _0)$ -graph satisfying that $\lambda _0\le 3\sqrt {d_0}$ (indeed, a random $d_0$ -regular graph on $n_0$ vertices typically satisfies this). In particular, here too we assume that $n_0$ is sufficiently large with respect to $d_0$ , and in turn, $n$ is sufficiently large with respect to $d$ . Set $G=\mathrm{lex}(H,K_k)$ . Noting that $kd_0+k-1=d$ , we then have that $G$ is a $d$ -regular graph on $kn_0=n$ vertices. Furthermore, by Lemmas5.1 and 5.2 we have that the second largest eigenvalue of $G$ , $\lambda$ , satisfies that

\begin{align*} \frac {\lambda }{d}=\frac {k\lambda _0+k-1}{kd_0+k-1}\le \frac {4}{\sqrt {d_0}} \le \delta , \end{align*}

and thus $G$ satisfies the assumptions of Theorem3.

By the construction of $G$ , we have that a largest independent set of $G$ is of size at most $n_0=(d_0+1)n/d$ . Therefore, a longest induced path in $G$ is of length at most $2(d_0+1)n/d$ . Setting $C\,:\!=\, C(\delta )=2(d_0+1)$ completes the proof.

In the general case, where $d+1$ is not divisible by $d_0+1$ , let $r$ be the residue of $d+1$ modulo $d_0+1$ . Let $k\,:\!=\, \frac {d+1-r}{d_0+1}$ , and let $H_2$ be a copy of $K_k$ with an $r$ -regular graph removed. We then take $G=\mathrm{lex}(H,H_2)$ . The rest of the proof is quite similar to the case where $d+1$ is divisible by $d_0+1$ .