Proof. Use $\max _{v\in V}{\mathbb{E}}_{\pi } [\tau _{v}] \leq 20 m \sqrt{t_{\mathsf{rel}}+1}/d_{\mathsf{min}}$ from [Reference Oliveira, Peres, Mishna and Munro51, Theorem 1] in Theorem 3.2.

Proof. For Item (i), by independence of the walks and the union bound we have

\begin{align*} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)} \geq t \right ] & \leq \sum _{v\in V} \prod _{i=1}^k\left (1-{\mathbb{P}}_{\pi }\!\left [N_v^{(i)}(t)\geq 1\right ]\right )\leq \sum _{v\in V}\exp\!\left (-k{\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1\right ] \right ). \end{align*}

For Item (ii), since $N_v(t)$ is a non-negative integer, we have

\begin{equation*} {\mathbb{E}}_{\pi } \!\left [N_v(t)\right ] = {\mathbb{E}}_{\pi } \!\left [N_v(t)\mid N_v(t)\geq 1\right ]\cdot {\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1 \right ]. \end{equation*}

For the two inequalities first observe that ${\mathbb{E}}_{\pi } [N_v(t)]=(t+1)\pi (v)$ . Now, conditional on a walk first hitting $v$ at time $s$ we have $N_v(t)= \sum _{i=0}^{t-s}P_{v,v}^{t}$ . The first inequality in Item (ii) then follows since ${\mathbb{E}}_{\pi } [N_v(t)\mid N_v(t)\geq 1]\leq \sum _{i=0}^t P_{v,v}^{i}$ . For the last inequality in Item (ii), observe that, by reversibility, for any $t\geq 1$

\begin{align*} {\mathbb{P}}_{\pi }\!\left [N_v\left (\left \lceil \tfrac{t-1}{2}\right \rceil \right )\geq 1,\; N_v( t) \geq 1\right ] &= \sum _{v_0v_1\cdots v_t : v\in \bigcup _{i=0}^{\lceil \frac{t-1}{2}\rceil }\{v_i\} } \pi (v_0)P_{v_0,v_1}\cdots P_{v_{t-1},v_t} \\[5pt] &= \sum _{v_0v_1\cdots v_t : v\in \bigcup _{i=0}^{\lceil \frac{t-1}{2}\rceil }\{v_i\} } \pi (v_t)P_{v_t,v_{t-1}}\cdots P_{v_{1},v_0}\\[5pt] &= {\mathbb{P}}_{\pi }\!\left [N_v(t)- N_v\left (\left \lfloor \tfrac{t+1}{2}\right \rfloor -1\right )\geq 1,\; N_v( t) \geq 1\right ]. \end{align*}

Since $\{N_v( t) \geq 1 \}=\{N_v\left (\left \lceil \tfrac{t-1}{2}\right \rceil \right )\geq 1 \}\cup \{N_v(t)- N_v\left (\left \lfloor \tfrac{t+1}{2}\right \rfloor -1\right )\geq 1\}$ we have

\begin{equation*}{\mathbb{P}}_{\pi }\!\left [N_v\left (\left \lceil \tfrac {t-1}{2}\right \rceil \right ) \geq 1\mid N_v( t) \geq 1\right ] \geq 1/2.\end{equation*}

Thus ${\mathbb{E}}_{\pi } \!\left [N_v(t) \mid N_v(t)\geq 1\right ] \geq (1/2)\cdot \sum _{i=0}^{\lceil \frac{t-1}{2}\rceil }P_{v,v}^{i}$ as claimed.

Finally, for Item (iii), the proof of [Reference Oliveira, Peres, Mishna and Munro51, Lemma 1] shows that

\begin{equation*}\sum _{i=0}^t P_{u,u}^{i} - (t+1)\pi (u)\leq \frac {e}{e-1}\left (\sum _{i=0}^{\lceil t_{\mathsf {rel}} \rceil -1}P_{u,u}^{i} -\lceil t_{\mathsf {rel}} \rceil \pi (u)\right ),\end{equation*}

rearranging gives the result.

Proof of Theorem 3.1. Recall that we aim to prove $t_{\mathsf{cov}}^{(k)}(\pi ) = \mathcal{O} \big ( \big (\frac{m}{kd_{\mathsf{min}}}\big )^2\log ^2 n \big )$ . To begin observe that if $k\geq 10 (m/d_{\mathsf{min}} )\log n$ then the probability any vertex $u$ is unoccupied at time $0$ is $(1-\pi (u))^k\leq e^{-\pi (u)k}\leq e^{-d_{\mathsf{min}} k/(2m)}$ . For any graph $t_{\mathsf{cov}} \leq 16mn/d_{\mathsf{min}} \leq 16n^3$ by [Reference Kahn, Linial, Nisan and Saks34, Theorem 2], thus we have

\begin{equation*}{\mathbb{E}}_{\pi ^k} \left [\tau _{\mathsf {cov}}^{(k)}\right ]\leq 16n^3 \cdot n e^{-d_{\mathsf {min}} k/(2m)} = o\Big (\left ( m/(kd_{\mathsf {min}})\right )^2\log ^2 n\Big ).\end{equation*}

It follows that we can assume $k\leq 10 (m/d_{\mathsf{min}} )\log n$ for the remainder of the proof.

We will apply Lemma 3.10 to bound ${\mathbb{P}}_{\pi ^k}\left[\tau _{\mathsf{cov}}^{(k)} \geq t \right]$ for $t\geq 1$ . To begin, by (2) we have

\begin{align*} \sum _{i=0}^{t} P_{u,u}^{i} &\leq \frac{10d(u)}{d_{\mathsf{min}}}\sum _{i=0}^{t} \frac{1}{\sqrt{i+1}} + (t+1) \pi (u) \leq \frac{10 d(u)}{d_{\mathsf{min}}}\left (\int _{1}^{t} \frac{1}{\sqrt{x}}\, \textrm{d}x +1\right ) + (t+1)\pi (u). \end{align*}

Now, since $d_{\mathsf{min}}\leq d(u)$ , $\pi (u)\leq 1$ and $t\geq 1$ , we have

\begin{align*} \sum _{i=0}^{t} P_{u,u}^{i} &\leq \frac{10d(u)}{d_{\mathsf{min}}}\left (\left [2\sqrt{t} - 2\sqrt{1} \right ] + 1 \right )+ t\pi (u) +1 \leq \frac{20d(u)}{d_{\mathsf{min}}}\sqrt{ t} +t\pi (u). \end{align*}

Thus, by Lemma 3.10 (ii) and dividing each term by a factor of $\pi (u)=d(u)/2m$ we have

\begin{equation*}{\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1 \right ] \geq \frac {t}{(40m/d_{\mathsf {min}})\sqrt { t} +t}.\end{equation*}

We now define $t^* =\left (\frac{300m\log n}{kd_{\mathsf{min}}}\right )^2$ . Firstly if $k\geq 10\log n$ , then for any $t\geq t^*$ we have

\begin{equation*}k{\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1 \right ]\geq \frac {k\left (\frac {300m\log n}{kd_{\mathsf {min}}}\right )^2}{(40m/d_{\mathsf {min}})\cdot \frac {300m\log n}{kd_{\mathsf {min}}} + \left (\frac {300m\log n}{kd_{\mathsf {min}}}\right )^2} \geq \frac {300\log n }{40+300/10}\gt 4\log n. \end{equation*}

Now, since $t_{\mathsf{cov}}\leq 16n^3$ , Lemma 3.10(i) gives ${\mathbb{E}}_{\pi ^k} \left [\tau _{\mathsf{cov}}^{(k)}\right ]\leq t^* +o\!\left (n^{-3}\right )\cdot 16n^3 = \mathcal{O}\big (\big (\frac{m}{kd_{\mathsf{min}}}\big )^2\log ^2 n\big )$ .

Finally, if $k\leq 10\log n$ then $ t^*\geq (300m/d_{\mathsf{min}})^2/100 =900(m/d_{\mathsf{min}})^2$ . However, $t_{\mathsf{cov}} \leq 16mn/d_{\mathsf{min}} \leq 32m^2/d_{\mathsf{min}}^2$ by [Reference Kahn, Linial, Nisan and Saks34, Theorem 2]. Thus, ${\mathbb{E}}_{\pi ^k} \!\left[t_{\mathsf{cov}}^{(k)}\right]\leq t^*$ holds as claimed.

Proof of Theorem 3.2. We consider first the case $k \lt 8\log _2 n$ . To begin, for any pair $u,v$ it holds that ${\mathbb{E}}_{u} [\tau _{v}] \leq 2\max _{w\in V}{\mathbb{E}}_{\pi } [\tau _{w}]$ by [Reference Levin, Peres and Wilmer42, Lemma 10.2]. Thus by Markov’s inequality

\begin{align*} {\mathbb{P}}_{u}\!\left [\tau _{v}\geq 4\left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil \right ] \leq \frac{{\mathbb{E}}_{u} \!\left [\tau _v\right ]}{4\left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil }\leq \frac{1}{2}, \end{align*}

for any $v\in V$ . Then, the Markov property yields

\begin{align*} {\mathbb{P}}_{u}\!\left [\tau _{v}\geq 20\left \lceil \frac{\log _2 n}{k}\right \rceil \cdot \left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil \right ] \leq \left (\frac{1}{2}\right )^{5(\!\log _2 n)/k} = \frac{1}{n^{5/k}}. \end{align*}

Thus, by independence of the $k$ walks

\begin{align*} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{v}^{(k)}\geq 20\left \lceil \frac{\log _2 n}{k}\right \rceil \cdot \left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil \right ] \leq \frac{1}{n^5}, \end{align*}

and finally by the union bound,

\begin{align*} {\mathbb{P}}_{\pi ^k}\!\left [t_{\mathsf{cov}}^{(k)}\geq 20\left \lceil \frac{\log _2 n}{k}\right \rceil \cdot \left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil \right ] \leq \frac{1}{n^4}. \end{align*}

Therefore, since the cover time of a single walk satisfies $t_{\mathsf{cov}} =\mathcal{O}(n^3 )$ , we have

\begin{align*} {\mathbb{E}}_{\pi ^k} \left [t_{\mathsf{cov}}^{(k)}\right ] \leq 20\left \lceil \frac{\log _2 n}{k}\right \rceil \cdot \left \lceil \max _{w\in V}{\mathbb{E}}_{\pi } \!\left [\tau _{w}\right ]\right \rceil + o\!\left (1\right ). \end{align*}

We now cover the case $ 8\log _2 n\leq k \leq 100\max _{v\in V}{\mathbb{E}}_{\pi } [\tau _v] \cdot \log n$ , where we apply Items (i) & (ii) of Lemma 3.10 to bound ${\mathbb{P}}_{\pi ^k}\!\left[\tau _{\mathsf{cov}}^{(k)} \geq t \right]$ . Observe that

\begin{align*} \sum _{i=0}^{t} P_{v,v}^i = \sum _{i=0}^{t} (P_{v,v}^i-\pi (v)) + (t+1)\pi (v). \end{align*}

Since the walk is lazy $(P_{v,v}^i-\pi (v))$ is non-negative and non-increasing in $i$ [Reference Levin, Peres and Wilmer42, Exercise 12.5]. Thus

\begin{align*} \sum _{i=0}^{t} P_{v,v}^i \leq \sum _{i=0}^{\infty } (P_{v,v}^i-\pi (v)) + (t+1)\pi (v) = \pi (v){\mathbb{E}}_{\pi } \!\left [\tau _v\right ]+ (t+1)\pi (v), \end{align*}

as $\pi (v){\mathbb{E}}_{\pi } [\tau _v]=\sum _{i=0}^{\infty } (P_{v,v}^i-\pi (v))$ by [Reference Aldous and Fill2, Lemma 2.11]. Now, by Lemma 3.10 (ii) we have

\begin{align*} {\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1 \right ] \geq \frac{(t+1)\pi (v)}{\pi (v){\mathbb{E}}_{\pi } \!\left [\tau _v\right ]+ (t+1)\pi (v)} =\frac{(t+1)}{{\mathbb{E}}_{\pi } \!\left [\tau _v\right ]+ (t+1)}. \end{align*}

Choosing $t = 8\left \lceil \frac{\log n}{k}\max _{v\in V}{\mathbb{E}}_{\pi } \!\left [\tau _v\right ] \right \rceil$ yields

\begin{align*} {\mathbb{P}}_{\pi }\!\left [N_v(t)\geq 1 \right ] \geq \frac{8\log n}{k+8\log n} \geq \frac{4\log n}{k}, \end{align*}

since $k\geq 8\log n$ . Then an application of Lemma 3.10 (i) gives

\begin{align*} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)} \geq t \right ] \leq n\exp\!\left (-4\log n\right ) = \frac{1}{n^3}. \end{align*}

Again, as $t_{\mathsf{cov}} = \mathcal O(n^3)$ , we conclude ${\mathbb{E}}_{\pi ^k} \left [t_{\mathsf{cov}}^{(k)}\right ] \leq 8\left \lceil \frac{\log n}{k}\max _{v\in V}{\mathbb{E}}_{\pi } \!\left [\tau _v\right ] \right \rceil +\mathcal O(1)$ .

Finally assume that $k\geq 100\max _{v\in V}{\mathbb{E}}_{\pi } [\tau _v] \cdot \log n$ . Recall that $ \max _v{\mathbb{E}}_{\pi } [\tau _v] \geq (1/2) \cdot \max _{x,y}{\mathbb{E}}_{x} [\tau _y]$ by [Reference Aldous and Fill2, Lemma 3.15] and $\max _{x,y}{\mathbb{E}}_{x} [\tau _y] \geq \max _{x,y} m R(x,y)$ by the commute time identity [Reference Levin, Peres and Wilmer42, Proposition 10.6], where $R(x,y)$ is the effective resistance between $x$ and $y$ (see [Reference Levin, Peres and Wilmer42, Section 9.4]). Thus we have

(3) \begin{equation} \max _v{\mathbb{E}}_{\pi } \!\left [\tau _v\right ] \geq \frac{1}{2} \cdot m\cdot \max _v \frac{1}{d(v)} \geq \frac{1}{2}\cdot \max _v \frac{1}{\pi (v)}, \end{equation}

as $R(x,y)\geq \max \{1/d(x),1/d(y)\}$ by the definition of the effective resistance. The probability that any vertex $u$ is unoccupied at time $0$ is $(1-\pi (u))^k\leq e^{-\pi (u)k}$ . For any graph $t_{\mathsf{cov}} \leq 16n^3$ by [Reference Kahn, Linial, Nisan and Saks34, Theorem 2] and thus, one can check that by (3), we have

\begin{equation*}{\mathbb{E}}_{\pi ^k} \left [\tau _{\mathsf {cov}}^{(k)}\right ]\leq \max _{v\in V} 16n^3 \cdot n e^{-\pi (v)k}= o\Big (\frac {\max _{v\in V}{\mathbb{E}}_{\pi } \!\left [\tau _v\right ] \cdot \log n}{k}\Big ),\end{equation*}

as claimed.

Proof of Lemma 3.4. The lower bound will follow from the general lower bound we prove later in Theorem 3.6. For the upper bound notice that, by Lemma 3.10 (iii) and our hypothesis that $\sum _{i=0}^{t} P_{vv}^i =\mathcal{O}(1 + t\pi (v) )$ for any $t\leq t_{\mathsf{rel}}$ , we have that for any $T\geq 0$ ,

\begin{equation*} \sum _{t=0}^T P_{v,v}^t \leq \frac {e}{e-1}\left (\sum _{i=0}^{\lceil t_{\mathsf {rel}} \rceil -1}P_{v,v}^{i} -\lceil t_{\mathsf {rel}} \rceil \pi (v)\right )+ (T+1)\pi (v)\leq C(1 + T\pi (v)), \end{equation*}

for some constant $C\lt \infty$ . Now, by Lemma 3.10 (ii) and as $\pi _{\mathsf{min}} =\Omega (1/n)$ holds by hypothesis, for any $v$ and $T= \mathcal{O}(n )$ there exists a constant $C'\lt \infty$ such that

(4) \begin{equation} {\mathbb{P}}_{\pi }\!\left [N_v(T)\geq 1\right ] \geq \frac{T\pi (v)}{C(1 + T\pi (v))}\geq \frac{T}{C'n}. \end{equation}

First consider $k=\omega (\!\log n)$ , and let $T= \lceil 4C'(n/k)\log n \rceil =\mathcal{O}(n )$ . Lemma 3.10 (i) and (4) give

\begin{align*} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)} \geq T\right ] \leq n \exp\!\left (-k{\mathbb{P}}_{\pi }\!\left [N_v(T)\geq 1\right ] \right )=n\exp\!\left (-4\log n\right )=\frac{1}{n^3} . \end{align*}

Otherwise, if $k = \mathcal{O}(\!\log n )$ , then we consider consecutive periods of length $5t_{\mathsf{mix}}= \mathcal{O}(n )$ . For any $u,v\in V$ , (4) gives

\begin{equation*} {\mathbb{P}}_{u}\!\left [N_v(5t_{\mathsf {mix}})\geq 1\right ]\geq \frac {1}{4}{\mathbb{P}}_{\pi }\!\left [N_v(t_{\mathsf {mix}})\geq 1\right ] \geq \frac {t_{\mathsf {mix}}}{4C'n}.\end{equation*}

The probability that a vertex $v$ is not hit in a period, starting from any vector in $V^k$ is

\begin{align*} (1-\min _{u\in V}{\mathbb{P}}_{u}\!\left [N_v(5t_{\mathsf{mix}})\geq 1\right ])^k \leq e^{-kt_{\mathsf{mix}}/(4C'n)} \end{align*}

and thus after $20C'\lceil \frac{n}{k t_{\mathsf{mix}}}\cdot \log n\rceil$ periods the probability $v$ has not been hit is at most $e^{-5 \log n}= n^{-5}$ . By the union bound, and since worst-case cover time of a single walk on any graph is $\mathcal{O}(n^3 )$ , the cover time is $\mathcal{O}\!\left (t_{\mathsf{mix}} \cdot \frac{n}{k t_{\mathsf{mix}}}\cdot \log n \right )= \mathcal{O}\!\left (\frac{n}{k}\cdot \log n\right )$ .

Proof of Lemma 3.5. From Lemma 3.10 (iii) we deduce that $\sum _{i=0}^{t} P_{v,v}^i = \mathcal{O}(1+ t/n+\log t )$ for any $t\leq n(\!\log n)^2$ . Thus we can apply Lemma 3.10 (ii) which, for any $2\leq t \leq n(\!\log n)^2$ , gives

(5) \begin{equation} k{\mathbb{P}}_{\pi }\!\left [\tau _v\leq t\right ] \geq \frac{ckt}{n\log t + t }, \end{equation}

for some fixed constant $c\gt 0$ . Let $t^*=C \frac{n\log n}{k}\log \left (\frac{n\log n}{k}\right )$ and $C\log n \leq k \leq (n\log n )/3$ for some constant $1\lt C\;:\!=\;C(c)\lt \infty$ to be determined later then,

(6) \begin{equation} n\log t^* = n \log \left (C \frac{n\log n}{k}\log \left (\frac{n\log n}{k}\right )\right )\geq n\log \left ( \frac{n\log n}{k}\right )\geq t^* . \end{equation}

In addition we have

(7) \begin{equation} \log t^* =\log \left (C \frac{n\log n}{k}\log \left (\frac{n\log n}{k}\right )\right )\leq (2 + \log C)\log \left ( \frac{n\log n}{k}\right ) . \end{equation}

Thus inserting (6) into (5) then applying (7) yields the following for any $v\in V$ ,

\begin{equation*}k{\mathbb{P}}_{\pi }\!\left [\tau _v\leq t\right ] \geq \frac {c}{2} \cdot \frac {kt^*}{n\log t^*} \geq \frac {c }{2 }\cdot \frac {(Cn\log n)\log \left (\frac {n\log n}{k}\right ) }{(2 + \log C)n\log \left (\frac {n\log n}{k}\right )} = \frac {Cc \log n }{4 + 2\log C } .\end{equation*}

We can assume w.l.o.g. that $c\lt 1$ and thus taking $C= 100/c^2$ yields $\frac{Cc }{4 + 2\log C } \gt \frac{100/c}{14 + 4\log\!(1/c)} \gt 5$ . So by independence of the walks we have

\begin{equation*}{\mathbb{P}}_{\pi ^k}\!\left [\tau _v^{(k)} \gt t^*\right ] \leq \left ( 1-{\mathbb{P}}_{\pi }\!\left [\tau _v\leq t^* \right ]\right )^k\leq \exp\!\left (-k{\mathbb{P}}_{\pi }\!\left [\tau _v\leq t^* \right ]\right ) \leq n^{-5} .\end{equation*}

Thus, by the union bound ${\mathbb{P}}_{\pi ^k}\!\left[\tau _\mathsf{cov}^{(k)} \gt t^* \right]\leq n^{-4}$ . So, since the worst-case expected cover time by a single walk on any graph is $\mathcal{O}(n^3 )$ , we have $t_{\mathsf{cov}}^{(k)}(\mathcal{T}_n,\pi )= \mathcal{O}(t^* )$ , as claimed.

The case $k\leq (100\log n )/c^2$ remains. For any $k = \mathcal{O}(\!\log n )$ , consider periods of length $5t_{\mathsf{mix}}=\mathcal{O}(n )$ . Then by Lemma 3.10 (ii), for any pair of vertices $v,w$ and some $C'\lt \infty$ , we have

\begin{equation*} {\mathbb{P}}_{w}\!\left [N_v(5t_{\mathsf {mix}})\geq 1\right ]\geq \frac {1}{4}{\mathbb{P}}_{\pi }\!\left [N_v(t_{\mathsf {mix}})\geq 1\right ] \geq \frac {t_{\mathsf {mix}}}{C'n\log n}.\end{equation*}

Thus the probability $v\in V$ is not hit in a period, starting from any configuration in $ V^k$ is

\begin{align*} (1-{\mathbb{P}}_{w}\!\left [N_v(5t_{\mathsf{mix}})\geq 1\right ])^k \leq e^{-kt_{\mathsf{mix}}/(C'n\log n)}. \end{align*}

Thus after $5C'\lceil (n\log ^2 n)/(k t_{\mathsf{mix}})\rceil$ periods the probability $v$ has not been hit is at most $e^{-5 \log n}= n^{-5}$ , so (similarly) the cover time is $\mathcal{O}\!\left (t_{\mathsf{mix}} \cdot \frac{n\log ^2 n}{k t_{\mathsf{mix}}}\right )= \mathcal{O}\!\left (\frac{n}{k}\cdot \log ^2 n\right )$ as claimed.

Proof of Lemma 3.8. Let $x = Ck/T$ , and recall that in each step that the random walk on $\widehat{G}(x)$ is not at the vertex $z$ , it moves to $z$ with probability $x$ . For $i\geq 1$ , we refer to the portion of the walk between the $i$ ^th and $(i+1)$ ^th visits to $z$ as the $i$ ^th sub-walk. For $i\geq 0$ let $X_i\sim \operatorname{Geo}\!(x )$ be i.i.d. geometric random variables with mean $1/x$ , then it follows that the $i$ ^th sub-walk has length $X_i+1$ and takes $X_i-1$ steps inside $G$ (the steps leaving and entering $z$ are not in $G$ ) and it takes at most $X_0$ steps to first visit $z$ .

If we let $X=\sum _{i=0}^{Ck/3}X_i$ then we see that $X + Ck/3$ stochastically dominates the time taken for the first $Ck/3$ sub-walks to occur within a walk on $\widehat{G}(x)$ . Observe that ${\mathbb{E}} [X] = T/3$ and so a Chernoff bound for sums of geometric r.v.’s [Reference Janson32, Theorem 2.1] gives

\begin{equation*}{\mathbb{P}}\left [X \gt 2{\mathbb{E}} \left [X\right ] \right ] \lt e^{-x{\mathbb{E}} \left [X\right ](2- 1 -\ln (2))}\leq e^{-\frac {Ck}{10}}.\end{equation*}

Note that $2{\mathbb{E}} [X]+ Ck/3 \leq T$ . Thus, if $\mathcal{E}_1$ is the event that there are at least $Ck/3$ sub-walks in a walk of length $T$ , then ${\mathbb{P}}\left [\mathcal{E}_1^c\right ]\leq e^{-\frac{Ck}{10}}$ .

For each $i\geq 0$ , we have ${\mathbb{P}} [X_i\geq 1/(2x)]= (1-x)^{1/(2x)}\geq 1/2$ by Bernoulli’s inequality. Let $\mathcal{E}_2$ be the event that $\{X_i\geq 1/(2x)\}$ holds for more than $k$ values $1\leq i\leq Ck/3$ . Recall that $C\gt 30$ and, as the $X_i$ ’s are independent, a Chernoff bound [Reference Mitzenmacher and Upfal49, Theorem 4.5] gives

\begin{equation*}{\mathbb{P}}\left [\mathcal {E}_2^c\right ]\leq {\mathbb{P}}\left [\operatorname {Bin}\!\left ( \frac {Ck}{3},\frac {1}{2} \right )\leq k\right ]\leq \exp\!\left (- \frac {1}{2}\cdot \left (1-\frac {6}{C}\right )^2\cdot \frac {Ck}{6} \right ) \leq e^{-\frac {Ck}{20}} .\end{equation*}

Observe that conditional on $\mathcal{E}_1\cap \mathcal{E}_2$ there are at least $k$ sub-walks which take at least $s$ many steps within $G$ where $s= \frac{1}{2x} -1 = \frac{T}{2Ck}-1\geq \frac{T}{10Ck}$ , since $T\geq 5Ck$ . Recall that at each visit to $z$ the walk on $\widehat{G}(x)$ moves to a vertex $v\in V(G)$ proportional to $\pi (v)$ . It follows that, conditional on $\mathcal{E}_1\cap \mathcal{E}_2$ , we can couple $k$ stationary random walks of length $s$ on $G$ to a single random walk of length $T$ on $\widehat{G}$ such that if the latter walk has not covered $\widehat{G}$ then the former walks have not covered $G$ . Thus

\begin{equation*}{\mathbb{P}}_{\pi ^k,G}\!\left [\tau _{\mathsf {cov}}^{(k)} \gt \frac {T}{10Ck}\right ]\geq {\mathbb{P}}_{\widehat {\pi },\widehat {G}(x)}\!\left [\tau _{\mathsf {cov}} \gt T, \mathcal {E}_1\cap \mathcal {E}_2\right ]\geq {\mathbb{P}}_{\widehat {\pi },\widehat {G}(x)}\!\left [\tau _{\mathsf {cov}} \gt T\right ] - {\mathbb{P}}\left [ \mathcal {E}_1^c\cup \mathcal {E}_2^c\right ], \end{equation*}

and the result follows since $ {\mathbb{P}}\left [ \mathcal{E}_1^c\cup \mathcal{E}_2^c\right ] \leq e^{-\frac{Ck}{10}} + e^{-\frac{Ck}{20}}\lt e^{-\frac{Ck}{50}}$ .

Proof of Lemma 3.11. We first show $t_{\mathsf{cov}}\leq 2t_{\mathsf{cov}}(\pi )$ . The Random Target Lemma [Reference Levin, Peres and Wilmer42, Lemma 10.1] states that ${\mathbb{E}}_{x} [\tau _\pi ]=\sum _{v\in V}{\mathbb{E}}_{x} [\tau _v]\pi (v)$ does not depend on $x\in V$ . Thus, for any $u\in V$ ,

\begin{equation*} {\mathbb{E}}_{u} \!\left [\tau _\pi \right ] = {\mathbb{E}}_{\pi } \!\left [\tau _\pi \right ]\leq \max \limits _{v\in V} {\mathbb{E}}_{\pi } \!\left [\tau _v\right ]. \end{equation*}

Then, since choosing a vertex $v$ independently from $\pi$ and waiting for the random walk to hit $v$ is a strong stationary time for the random walk, we have

\begin{align*} {\mathbb{E}}_{u} \!\left [\tau _{\mathsf{cov}}\right ] &\leq {\mathbb{E}}_{u} \!\left [\tau _\pi \right ] + {\mathbb{E}}_{\pi } \!\left [\tau _{\mathsf{cov}}\right ] \leq \max \limits _{v\in V}{\mathbb{E}}_{\pi } \!\left [\tau _v\right ] + {\mathbb{E}}_{\pi } \!\left [\tau _{\mathsf{cov}}\right ] \leq 2 {\mathbb{E}}_{\pi } \!\left [\tau _{\mathsf{cov}}\right ]. \end{align*}

Observe that, by the Markov Property, for any $0\lt b\lt 1$ we have

\begin{align*} t_{\mathsf{cov}}(\pi ) &\leq b t_{\mathsf{cov}}(\pi )\cdot {\mathbb{P}}_{\pi }\!\left [\tau _{\mathsf{cov}}\leq bt_{\mathsf{cov}}(\pi ) \right ]+(b t_{\mathsf{cov}}(\pi ) + t_{\mathsf{cov}})\cdot {\mathbb{P}}_{\pi }\!\left [\tau _{\mathsf{cov}}\gt bt_{\mathsf{cov}}(\pi ) \right ]\\[5pt] &= b t_{\mathsf{cov}}(\pi ) + t_{\mathsf{cov}}\cdot {\mathbb{P}}_{\pi }\!\left [\tau _{\mathsf{cov}}\gt b t_{\mathsf{cov}}(\pi ) \right ]\\[5pt] &\leq b t_{\mathsf{cov}}(\pi ) + 2t_{\mathsf{cov}}(\pi )\cdot {\mathbb{P}}_{\pi }\!\left [\tau _{\mathsf{cov}}\gt b t_{\mathsf{cov}}(\pi ) \right ], \end{align*}

as $t_{\mathsf{cov}}\leq 2t_{\mathsf{cov}}(\pi )$ . Rearranging gives ${\mathbb{P}}_{\pi }\!\left [\tau _{\mathsf{cov}}\gt b\cdot t_{\mathsf{cov}}(\pi ) \right ] \geq \frac{1-b}{2}$ .

Proof of Theorem 3.6. We wish to apply Theorem 3.12 to the geometric reset graph $\widehat{G}(x)$ for some suitable $x\;:\!=\;x(n,k)$ . The following claim is based on [Reference Aldous3, Proposition 2].

Proof of Claim. Let $\pi$ and $\widehat{\pi }$ denote the stationary distributions of $G$ and $\widehat{G}$ , respectively. Note that for any $x\lt 1$ the walk on the weighted graph $\widehat{G}(x)$ satisfies $P_{i,i}=0$ and is reversible. These are items (a) and (c) from Theorem 3.12 and we now prove item (b) also holds for $\theta =1/2$ .

Partition $V(\widehat{G})= V(G)\cup \{z\}$ into sets $I_1$ , $I_2$ , $I_3$ as follows. Let $I_1$ be the set of leaves of $G$ , that is vertices of degree $1$ in $G$ (and degree $2$ in $\widehat{G}$ as all vertices are connected to $z$ ). Let $I_2$ be the set of vertices of $V(G)$ which are adjacent to at least one leaf of $G$ . Let $I_3$ be the remaining vertices (in particular $z \in I_3$ ). Let $J = I_1 \cup I_3$ and fix $j \in J$ . For the case $j=z$ , by the definition of $\widehat{G}$ and by hypothesis, we have $P_{v,z} =x\leq 1/2$ for any $v\in V$ . Otherwise, if $i$ is a neighbour of $j\neq z$ then $i$ cannot be a leaf, so $P_{i,j} = \frac{1-x}{d(i)(1-x)+xd(i)} = \frac{1-x}{d(i)} \leq 1/2$ . Thus it suffices to verify

(8) \begin{equation} \sum _{j\in J}\exp\!\left ( -t^{*} \cdot \widehat{\pi }(j) \right )\geq 1/2 . \end{equation}

Consider an edge $ik\in E(\widehat{G})$ where $i\in I_1$ , then we must have $k\in I_2$ . Since $\widehat{\pi }(i)\lt \widehat{\pi }(k)$ we have $\exp\!(\! -t^{*} \cdot \pi (i) )\geq \exp\!(\! -t^{*} \cdot \pi (k) )$ and also $|I_1|\geq |I_2|$ . Thus, summing over all leaves $i$ gives

(9) \begin{equation} \sum _{i\in I_1}\exp\!\left ( -t^{*} \cdot \widehat{\pi }(i) \right )\geq \sum _{k\in I_2}\exp\!\left ( -t^{*} \cdot \widehat{\pi }(k) \right ). \end{equation}

However by the definition of $t^*$ , the sum over all $i\in I$ is equal to $1$ , thus so (9) implies that the sum over $I_2$ is at most $1/2$ , and so (8) follows.

Now since $t^*\geq n\log n$ by [Reference Aldous3, (1.2)], it follows from Claim 3.13. and Theorem 3.12 that there exists some universal constant $0\lt c\lt \infty$ such that for any $0\lt x\leq 1/2$ we have

\begin{equation*}{\mathbb{E}}_{\widehat {\pi },\widehat {G}(x)} \left [\tau _{\mathsf {cov}}\right ] \geq cn\log n. \end{equation*}

Thus by Lemma 3.11 we have ${\mathbb{P}}_{\widehat{\pi },\widehat{G}(x)}\!\left [\tau _{\mathsf{cov}} \gt \frac{c}{2}\cdot n\log n\right ] \geq \frac{1}{4}$ . We seek to apply Lemma 3.8 with $C=100$ and $T= \frac{c}{2}\cdot n\log n$ . Firstly, we see that the condition $T\geq 5kC$ forces the restriction $k\leq c'\cdot n \log n$ where $c' = \frac{c}{2\cdot 5\cdot C}= \frac{c}{1000}$ for $0\lt c\lt \infty$ universal. Secondly the condition $1/2\geq x = Ck/T$ forces the condition $k\leq c''n\log n$ where $c'' = \frac{c}{2\cdot C}\geq c'$ .

Thus, provided $1\leq k\leq c'\cdot n \log n$ holds, all other assumptions are satisfied and we have

\begin{equation*}{\mathbb{P}}_{\pi ^k,G}\!\left [\tau _{\mathsf {cov}}^{(k)} \gt \frac {(c/2)\cdot n\log n}{ 10\cdot 100\cdot k}\right ]\gt \frac {1}{4} -\exp\!\left (-\frac {100k}{50}\right ) \geq \frac {1}{4} - e^{-2}\gt \frac {1}{10} \end{equation*}

by Lemma 3.8. The result follows by taking the constant in the statement to be $c/2000\leq c'$ , since $c\gt 0$ is universal.

Proof of Lemma 3.9. Let $X_v= \textbf{1}(\tau _v \gt t)$ be the indicator that $v\in V$ has not been visited up-to time $t$ and $X = \sum _{v\in S} X_v$ . Let $p_v = {\mathbb{P}}_{\pi }[\tau _v\leq t ]$ , $p=\max _{v\in S}p_v$ , and observe that ${\mathbb{E}}_{\pi ^k} [X_v] = (1-p_v)^k$ . Thus, we have

(10) \begin{equation} {\mathbb{E}}_{\pi ^k} \left [X\right ]=\sum _{v\in S}(1-p_v)^k\geq \sum _{v\in S} e^{-kp_v}(1-p_v^2 k) \geq \sum _{v\in S} e^{-kp_v}(1-p^2 k)\gt 0, \end{equation}

where the first inequality is by (1) since $0\leq p\leq 1$ and the last is by the assumption $p^2k \leq 2p^2k \lt 1$ . Let $R(t)$ be the number of vertices in $S$ that are visited by a single $t$ -step random walk. Observe that ${\mathbb{E}}_{\pi } [R(t)] = \sum _{v \in S} p_v$ . Let $r(v,w) = {\mathbb{P}}_{\pi }[\tau _v\leq t,\tau _w\leq t ]$ , then for any $v,w\in V$ ,

\begin{align*} {\mathbb{E}}_{\pi ^k} \left [X_v X_w\right ] &= \left (1-{\mathbb{P}}_{\pi }\!\left [\{\tau _v\leq t\} \cup \{\tau _w\leq t\}\right ]\right )^k\nonumber = (1-p_v-p_w +r(v,w))^k. \end{align*}

Recall the identity $a^k-b^k= (a-b)\sum _{i=0}^{k-1}a^{k-1-i}b^i$ for $k\geq 2$ . Thus,

\begin{align*} {\mathbb{E}}_{\pi ^k}[X^2] &= \sum _{v\in S} \sum _{w\in S} (1-p_v-p_w+r(v,w))^k\notag \\[5pt] &=\sum _{v\in S} \sum _{w\in S} \left ( (1-p_v-p_w)^k + ((1-p_v-p_w+r(v,w))^k-(1-p_v-p_w)^k)\right )\notag \\[5pt] &= \sum _{v\in S} \sum _{w\in S} \left ( (1-p_v-p_w)^k +r(v,w)\sum _{i=0}^{k-1}(1-p_v-p_w)^i(1-p_v-p_w+r(v,w))^{k-1-i}\right ). \end{align*}

Since $2p^2k \lt 1$ and $k\geq 2$ , we have $1- p_v -p_w \geq 1-2p \gt 0$ . Additionally we have $ r(v,w)\leq \min (p_v,p_w)$ and so $1- p_v -p_w+r(v,w)\leq 1- \max \{p_v,p_w\} \leq 1$ . Thus

(11) \begin{align} {\mathbb{E}}_{\pi ^k}[X^2] &\leq \sum _{v\in S} \sum _{w\in S} \left ( (1-p_v-p_w)^k +r(v,w) \cdot k\right )\notag \\[5pt] &\leq \sum _{v\in S} \sum _{w\in S} e^{-kp_v-kp_w} +\left (\sum _{v\in S} \sum _{w\in S} r(v,w)\right ) \cdot k\notag \\[5pt] &= \left (\sum _{v\in S}e^{-kp_v} \right )^2+ {\mathbb{E}}_{\pi }[R(t)^2]\cdot k. \end{align}

Recall the Paley-Zygmund inequality: for any non-negative random variable $X$ , ${\mathbb{P}}_{\pi ^k}[X\gt 0 ] \geq {\mathbb{E}}_{\pi ^k} [X]^2/{\mathbb{E}}_{\pi ^k} [X^2]$ . Inserting (10) and (11) into the (inverted) fraction $\frac{{\mathbb{E}}_{\pi ^k} \left [X^2\right ]}{{\mathbb{E}}_{\pi ^k} \left [X\right ]^2}$ gives

\begin{equation*} \frac {{\mathbb{E}}_{\pi ^k} \left [X^2\right ]}{{\mathbb{E}}_{\pi ^k} \left [X\right ]^2} \leq \frac {1}{(1-p^2k)^2}+ \frac {{\mathbb{E}}_{\pi }{[R(t)^2]}k }{|S|^2e^{-2pk}(1-p^2k)^2} =\frac {1}{(1-p^2k)^2}+ \frac {ke^{2pk}{\mathbb{E}}_{\pi }[R(t)^2] } {|S|^2(1-p^2k)^2}. \end{equation*}

Finally, we claim that ${\mathbb{E}}_{\pi } [R(t)^2] \leq 2p^2 |S|(\!\min _{v \in S} \pi (v))^{-1}$ , therefore

\begin{equation*} \frac {{\mathbb{E}}_{\pi ^k} \left [X^2\right ]}{{\mathbb{E}}_{\pi ^k} \left [X\right ]^2} \leq \frac {1}{(1-p^2k)^2}+ \frac {2kp^2e^{2kp}|S|/(\!\min _{v \in S} \pi (v))}{|S|^2(1-p^2k)^2} \leq \frac {1+ \frac {2kp^2e^{2kp}}{|S|\min _{v \in S} \pi (v)}}{(1-p^2k)^2} \end{equation*}

Now since ${\mathbb{P}}_{\pi ^k}\!\left[ \tau _{\mathsf{cov}}^{(k)} \leq t \right] = 1 - {\mathbb{P}}_{\pi ^k}\!\left[ \tau _{\mathsf{cov}}^{(k)} \gt t \right] \leq 1 - {\mathbb{E}}_{\pi ^k} [X]^2/{\mathbb{E}}_{\pi ^k} [X^2]$ we have

\begin{equation*}{\mathbb{P}}_{\pi ^k}\!\left [ \tau _{\mathsf {cov}}^{(k)} \leq t\right ] \leq 1 - \frac {(1-p^2k)^2}{1+ \frac {2kp^2e^{2kp}}{|S|\min _{v \in S} \pi (v)}}\leq 1 - (1-2p^2k)\left (1 - \frac {2kp^2e^{2kp}}{|S|\min _{v \in S} \pi (v)} \right ), \end{equation*}

since $\frac{1}{1+x}\geq 1-x$ for any $x\geq 0$ . Now, as $|S|\min _{v \in S} \pi (v)\leq 1$ we have

\begin{equation*}{\mathbb{P}}_{\pi ^k}\!\left [ \tau _{\mathsf {cov}}^{(k)} \leq t\right ] \leq 2kp^2 + \frac {2kp^2e^{2kp}}{|S|\min _{v \in S} \pi (v)} \leq \frac {4kp^2e^{2kp}}{|S|\min _{v \in S} \pi (v)}, \end{equation*}

which concludes the main proof. To prove the claim, first observe that

(12) \begin{equation} {\mathbb{E}}_{\pi } \!\left [R(t)^2\right ] = {\mathbb{E}}_{\pi } \!\left [\sum _{v\in S}\sum _{w\in S} \textbf{1}(\tau _v\leq t )\textbf{1}(\tau _w\leq t )\right ] = \sum _{v\in S}\sum _{w\in S} {\mathbb{P}}_{\pi }\!\left [\tau _v\leq t, \tau _w\leq t\right ]. \end{equation}

Now, by the Markov property, for any pair $(v,w)\in V^2$ we have

(13) \begin{equation} \begin{aligned} {\mathbb{P}}_{\pi }\!\left [\tau _v\leq t, \tau _w\leq t\right ] &\leq {\mathbb{P}}_{\pi }\!\left [\tau _v\leq t\right ]{\mathbb{P}}_{v}\!\left [\tau _w\leq t\right ] + {\mathbb{P}}_{\pi }\!\left [\tau _w\leq t\right ]{\mathbb{P}}_{w}\!\left [\tau _v\leq t\right ]\\[5pt] &\leq p{\mathbb{P}}_{v}\!\left [\tau _w\leq t\right ] + p{\mathbb{P}}_{w}\!\left [\tau _v\leq t\right ]. \end{aligned} \end{equation}

Thus by inserting the bound from (13) into (12) we have

\begin{align*} {\mathbb{E}}_{\pi } \!\left [R(t)^2\right ] &\leq p\sum _{v\in S}\sum _{w\in S}{\mathbb{P}}_{v}\!\left [\tau _w\leq t\right ] + p\sum _{w\in S}\sum _{v\in S}{\mathbb{P}}_{w}\!\left [\tau _v\leq t\right ] \\[5pt] &\leq p\sum _{v\in S} {\mathbb{E}}_{v} \left [R(t)\right ]+p\sum _{w\in S}{\mathbb{E}}_{w} \left [R(t)\right ]\nonumber \\[5pt] &\leq p\sum _{v\in S} {\mathbb{E}}_{v} \left [R(t)\right ]\cdot \pi (v)\cdot \left(\!\min _{v \in S} \pi (v)\right)^{-1}+p\sum _{w\in S}{\mathbb{E}}_{w} \left [R(t)\right ]\cdot \pi (w)\cdot \left(\!\min _{v \in S} \pi (v)\right)^{-1}\nonumber \\[5pt] &\leq 2p{\mathbb{E}}_{\pi } \!\left [R(t)\right ](\!\min _{v \in S} \pi (v))^{-1}\\[5pt] &\leq 2p^2 |S|(\!\min _{v \in S} \pi (v))^{-1}, \end{align*}

as claimed.

Proof. This follows since $d_{\textrm{TV}}(t)\leq s(t)$ holds for any $t\geq 0$ by [Reference Levin, Peres and Wilmer42, Lemma 6.3].

Proof of Lemma 4.3. We start with Item (i). Let $\bar d_{\textrm{TV}}(t) = \max _{x,y \in V} \|P_{x,\cdot }^t -P_{y,\cdot }^t\|_{\textrm{TV}}$ . Then

\begin{equation*} d_{\textrm{TV}}(t)\leq \bar d_{\textrm{TV}}(t) \leq 2 d_{\textrm{TV}}(t), \qquad \bar d_{\textrm{TV}}(\ell t)\leq \bar d_{\textrm{TV}}(t)^\ell \qquad \text {and}\qquad s(2t)\leq 1-(1-\bar d_{\textrm{TV}}(t))^2, \end{equation*}

hold for any integers $t,\ell \geq 0$ by Lemmas 4.11, 4.12 and 19.3, respectively, of [Reference Levin, Peres and Wilmer42].

Thus if we take $\ell = \lceil \log _{2} (2k/(k-\tilde{k}) ) \rceil$ and $t = t_{\mathsf{mix}}(1/4)$ then we have

\begin{equation*} s(2\ell t) \leq 1-(1-\bar d_{\textrm{TV}}(\ell t))^2 \leq 2\bar d_{\textrm{TV}}(\ell t)\leq 2\bar d_{\textrm{TV}}( t)^\ell \leq 2\cdot ( 2 d_{\textrm{TV}}( t))^\ell \leq 2\cdot \left (\frac {1}{2}\right )^{\log _{2}\left (\frac {2k}{k-\tilde {k}}\right )} \leq 1- \frac {\tilde {k}}{k}, \end{equation*}

it follows that $ t_{\mathsf{mix}}^{(\tilde{k},k)}\leq 2\ell t \leq 2 t_{\mathsf{mix}} \left\lceil \log _{2} (2k/(k-\tilde{k}) ) \right\rceil$ .

For Item (ii), by the Markov property for any non-negative integers $\ell$ and $t$ , we have

\begin{equation*}\max _{x\in V, S\subseteq V, \pi (S)\geq 1/4}{\mathbb{P}}_{x}\!\left [\tau _S\gt \ell t\right ]\leq \left (\max _{x\in V, S\subseteq V, \pi (S)\geq 1/4}{\mathbb{P}}_{x}\!\left [\tau _S\gt t\right ]\right )^\ell . \end{equation*}

By Markov’s inequality we have $\max _{x\in V, S\subseteq V, \pi (S)\geq 1/4}{\mathbb{P}}_{x}[\tau _S\gt 2t_{\mathsf{H}} ]\leq 1/2$ and by Theorem 4.2 there exists some $C$ such that $t_{\mathsf{H}}\leq Ct_{\mathsf{mix}}$ . Thus, if we take $T = 2Ct_{\mathsf{mix}} \cdot \lceil \log _2 (k/(k-\tilde{k}) ) \rceil$ , then

\begin{equation*}\max _{x\in V, S\subseteq V, \pi (S)\geq 1/4}{\mathbb{P}}_{x}\!\left [\tau _S\gt T\right ]\leq \left (\max _{x\in V, S\subseteq V, \pi (S)\geq 1/4}{\mathbb{P}}_{x}\!\left [\tau _S\gt 2t_{\mathsf {H} }\right ]\right )^{\log _2\left (k/(k-\tilde {k})\right )} \leq 1 -\frac {\tilde {k}}{k}. \end{equation*}

It follows that $t_{\mathsf{large-hit}}^{(\tilde{k},k)} \leq T = C't_{\mathsf{mix}} \lceil \log\!(k/(k-\tilde{k}) ) \rceil$ for some $C'\lt \infty$ .

Proof of Lemma 4.5. For Item (i), if we let $\ell = t_{\mathsf{mix}}^{(\tilde{k},k)}+1$ then the separation distances satisfies $s(\ell )\leq 1-\tilde{k}/k$ . Thus, by the definition of separation distance, for any pair of vertices $x,y\in V$ we have $P_{x,y}^\ell \geq (\tilde{k}/k) \cdot \pi (y)$ . Thus for any $x\in V$ and set $S\subseteq V$ satisfying $\pi (S)\geq 1/4$ we have

(17) \begin{equation} {\mathbb{P}}_{x}\!\left [\tau _S \leq \ell \right ]\geq \sum _{y\in S} P_{x,y}^\ell \geq \sum _{y\in S} \frac{\tilde{k}}{k} \cdot \pi (y) = \frac{\tilde{k}}{k}\sum _{y\in S} \pi (y) = \frac{\tilde{k}}{k}\pi (S)\geq \frac{\tilde{k}}{4k}. \end{equation}

Since (17) holds for all $x\in V$ and $S\subseteq V$ where $\pi (S)\geq 1/4$ , we have $ t_{\mathsf{H}} \leq \ell \cdot (4k/\tilde{k})$ , as $\tau _S$ is stochastically dominated by $\ell$ times the number of phases of length $\ell$ before $S$ is hit. By Theorem 4.2 there exists a universal constant $ C\lt \infty$ such that for any graph $t_{\mathsf{mix}} \leq C\cdot t_{\mathsf{H}}$ , thus

\begin{equation*}t_{\mathsf {mix}} \leq C\cdot t_{\mathsf {H}} \leq C\cdot \ell \cdot (4k/\tilde {k}) =\frac {4C k}{\tilde {k}} \left ( t_{\mathsf {mix}}^{(\tilde {k},k)}+1\right ) \leq C'\cdot \frac {k}{\tilde {k}} \cdot t_{\mathsf {mix}}^{(\tilde {k},k)}, \end{equation*}

for some universal constant $C'\lt \infty$ as $ t_{\mathsf{mix}}^{(\tilde{k},k)}\geq 1$ .

For Item (ii), observe that if instead we set $ \ell = t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ then (17) still holds (in fact the stronger bound ${\mathbb{P}}_{x}[\tau _S \leq \ell ] \geq \tilde{k}/k$ holds) and the rest of the proof goes through unchanged.

Proof of Lemma 4.6. If we let $\ell = t_{\mathsf{mix}}^{(4\tilde{k},k)}+1$ then $s(\ell )\leq 1-4\tilde{k}/k$ . So, $P_{x,y}^\ell \geq (4\tilde{k}/k) \cdot \pi (y)$ for any pair of vertices $x,y\in V$ . Thus for any $x\in V$ and set $S\subseteq V$ satisfying $\pi (S)\geq 1/4$ we have

\begin{equation*} {\mathbb{P}}_{x}\!\left [\tau _S \leq \ell \right ]\geq \sum _{y\in S} P_{x,y}^\ell \geq \sum _{y\in S} \frac {4\tilde {k}}{k} \cdot \pi (y) = \frac {4\tilde {k}}{k}\sum _{y\in S} \pi (y) = \frac {4\tilde {k}}{k}\pi (S)\geq \frac {\tilde {k}}{k}. \end{equation*}

Consequently, we have $ t_{\mathsf{large-hit}}^{(\tilde{k},k)}\leq \ell = t_{\mathsf{mix}}^{(4\tilde{k},k)} +1 \leq 2t_{\mathsf{mix}}^{(4\tilde{k},k)},$ as claimed since $t_{\mathsf{mix}}^{(4\tilde{k},k)}\geq 1$ .

Proof of Theorem 4.7. Fix any $1 \leq \tilde{k} \lt k$ . It suffices to prove that with $k$ walks starting from arbitrary positions running for

\begin{equation*} t= t_{\mathsf {mix}}^{(\tilde {k},k)} + 2 t_{\mathsf {cov}}^{(\tilde {k})}(\pi ) \leq 3 \cdot \max \left ( t_{\mathsf {mix}}^{(\tilde {k},k)}, t_{\mathsf {cov}}^{(\tilde {k})}(\pi ) \right ) \end{equation*}

steps, we cover $G$ with probability at least $1/4$ . Consider a single walk $X_1$ on $G$ starting from $v$ . From the definition of $s_v(t)$ we have that at time $T= t_{\mathsf{mix}}^{(\tilde k, k)}$ there exists a probability measure $\nu _v$ on $V$ such that,

\begin{equation*}P_{v,w}^T = (1-s_v(T))\pi (w)+s_v(T)\nu _v(w).\end{equation*}

Now, note that (14) yields $(1-s_v(T))\geq \tilde k/k$ , therefore, we can generate $X_1(T)$ as follows: with probability $1-s_v(T)\geq \tilde k/k$ we sample from $\pi$ , otherwise we sample from $\nu _v$ . If we now consider $k$ independent walks, the number of walks that are sampled at time $T$ from $\pi$ has a binomial distribution $\text{Bin}(k,\tilde{k}/k)$ with $k$ trials and probability $\tilde k/k$ , whose expectation is $\tilde k$ . Since the expectation $\tilde{k}$ is an integer it is equal to the median. Thus, with probability at least $1/2$ , at least $\tilde k$ walks are sampled from the stationary distribution. Now, consider only the $\tilde k$ independent walks starting from $\pi$ . After $2t_{\mathsf{cov}}^{(\tilde k)}(\pi )$ steps, these walks will cover $G$ with probability at least $1/2$ , due to Markov’s inequality.

We conclude that in $t$ time steps, from any starting configuration of the $k$ walks, the probability we cover the graph is at least $1/4$ . Hence in expectation, after (at most) 4 periods of length $t$ we cover the graph.

Proof of Theorem 4.8. By the definition of $ t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ , there exists a vertex $u$ and $S\subseteq V$ such that ${\mathbb{P}}_{u}\!\left[\tau _{S}\leq t_{\mathsf{large-hit}}^{(\tilde{k},k)}-1 \right] \lt \tilde k/k$ . For such a vertex $u$ , we consider $k$ walks, all started from $u$ , which run for $t_{\mathsf{large-hit}}^{(\tilde{k},k)}-1$ steps. It follows that the number of walks that hit $S$ before time $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ is dominated by a binomial distribution with parameters $k$ and $p = \tilde k/k$ , whose expected value and median is $\tilde k$ . We conclude that with probability at least $1/2$ , at most $\tilde k$ walks hit $S$ . Note that $|S| \geq \pi (S)/\pi _{\max } \geq 1/(4 \pi _{\max })$ . Hence even if all $\tilde{k}$ walks that reached $S$ before time $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ were allowed to run for exactly $1/(8 \tilde{k} \pi _{\max })-1$ steps (if a walk exits $S$ , we can completely ignore the steps until it returns to $S$ ), then the total number of covered vertices in $S$ would be at most

\begin{equation*} \tilde {k} \cdot \left (\frac {1}{8 \tilde {k} \pi _{\max }} \right ) \lt 1/(4 \pi _{\max }) \leq |S|, \end{equation*}

which concludes the proof of the first bound.

For the second bound, we follow the first part of the proof before and consider again at most $ \tilde{k}$ walks that reach the set $S$ . Let us denote by $\kappa$ the induced distribution over $S$ upon hitting $S$ from $u$ for the first time. Now, each of the $\tilde{k}$ walks continues for another $\ell =\lfloor \epsilon (n/\tilde{k}) \log\!(n) \rfloor$ steps, where $0\lt \epsilon =\epsilon (\gamma ) \lt 1$ is a sufficiently small constant fixed later. We now define, for any $v \in S$ , the probability $ p_{v} = {\mathbb{P}}_{\kappa }[ \tau _v \leq \epsilon (n/\tilde{k}) \log n ].$ Observe that since a walk of length $\ell$ can cover at most $\ell$ vertices, we have $\sum _{v \in S} p_v \leq \ell$ . Further, define the set $\widetilde{S}= \{ v \in S \,:\, p_v \lt 8 \ell/n \}$ .

Since for every $v$ with $p_v\geq 8 \ell/n$ we have that $np_v/(8 \ell )\geq 1$ and

\begin{align*} |S \setminus \tilde{S}| = \left |\left \{v \in S\;:\; p_v \geq 8\ell/n\right \} \right | \leq \sum _{v \in S} \frac{np_v}{8\ell } \leq \frac{n}{8} \leq \frac{|S|}{2}, \end{align*}

where the last inequality holds since $G$ is regular. Thus $\tilde{S} \geq |S|/2$ . Now, let $Z$ be the number of unvisited vertices in $\widetilde{S}$ after we run $\tilde{k}$ random walks starting from $\kappa$ . If $p_*=\max _{v\in \tilde{S}}p_v\lt 8\ell/n$ then, by definition and recalling that $\ell =\lfloor \epsilon (n/\tilde{k}) \log\!(n) \rfloor$ where $0\lt \varepsilon \lt 1$ , we have

\begin{align*} {\mathbb{E}} \left [Z\right ] \geq |\widetilde{S}| \cdot \left (1 - p_* \right )^{\tilde{k}} \geq (n/2) \cdot e^{-\tilde{k}\cdot p_* } \cdot \left (1 - \tilde{k} p_{*}^2 \right ) \geq n/2 \cdot e^{- 8 \epsilon \log\!(n) } \cdot \left (1 - \frac{(8 \log n)^2}{\tilde{k}} \right ) \geq n^{1-9 \varepsilon }, \end{align*}

where the second inequality holds by (1) and the last since $\tilde{k}\geq n^{\gamma }$ where $\gamma \gt 0$ fixed.

Finally, since each of these $\tilde{k}$ random walks can change $Z$ by at most $\ell$ vertices, by the method of bounded differences [Reference Dubhashi and Panconesi18, Theorem 5.3],

\begin{equation*}{\mathbb{P}}\left [Z\lt {\mathbb{E}} \left [Z\right ]/2\right ]\leq \exp\!\left (-2\frac {(Z/2 )^2 }{\tilde {k}\ell ^2 } \right ) \leq \exp\!\left (- \frac { n^{2-18\epsilon } \tilde {k} }{8 \varepsilon ^2 n^2 \log ^2 n } \right ) \lt 1/2, \end{equation*}

provided we have $\epsilon \lt \gamma/18$ as $\tilde{k} \geq n^{\gamma }$ . This implies that ${\mathbb{P}} [Z \geq 1] \geq 1/2$ .

Proof of Theorem 4.9. Since we are bounding the worst-case cover time from below we can assume that all walks start from a single vertex $u$ . First, consider the $k$ walks running for $ t_{\mathsf{large-hit}}^{(\tilde{k},k)} -1$ steps. Then, by the definition of $ t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ , there exists a vertex $u$ and $S\subseteq V$ such that ${\mathbb{P}}_{u}[\tau _{S}\leq t ] \lt \tilde k/k$ , therefore, the number of walks that, starting from $u$ , hit $S$ before time $t$ is dominated by a binomial distribution with parameters $k$ and $p = \tilde k/k$ , whose expected value and median is $\tilde k$ . We conclude that, if $\mathcal{E}$ is the event that at most $\tilde k$ walks hit $S$ by time $ t_{\mathsf{large-hit}}^{(\tilde{k},k)}-1$ , then ${\mathbb{P}} [\mathcal{E}]\geq 1/2$ .

Although, conditional on $\mathcal{E}$ , we know at most $\tilde{k}$ walks hit $S$ by time $ t_{\mathsf{large-hit}}^{(\tilde{k},k)} -1$ , we do not know when they arrived or which vertices of $S$ they hit first. For a lower bound we assume these $\tilde{k}$ walks arrived at time $0$ and then take the minimum over all sets $S$ such that $\pi (S)\geq 1/4$ and starting distributions $\mu$ on $\partial S$ , the vertex boundary of $S$ (note all particles started from $u$ , so they have the same distribution when they enter $S$ for first time). It follows that, conditional on $\mathcal{E}$ , the expected time for $\tilde{k}$ walks which hit $S$ before time $ t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ to cover $S$ is at least $t_{\mathsf{large-cov}}^{(\tilde k)}$ and so the result follows.

Proof of Lemma 4.10. Let $S \subseteq V$ be a set such that $\Phi (G)=\Phi (S)$ and $\pi (V \setminus S) \geq 1/2$ (such a set exists by symmetry of $\Phi (S)=\Phi (V \setminus S)$ ). Let $\pi _{S}$ be the stationary distribution restricted to $S$ , that is $\pi _{S}(s)= \pi (s)/\pi (S)$ for $s\in S$ and $\pi _{S}(x)=0$ for $x\notin S$ . As shown in [Reference Gharan and Trevisan24, Proposition 8], the probability that a (single) random walk $X_t$ remains in a set $S$ when starting from a vertex in $S$ sampled from $\pi _{S}$ , within $t$ steps is at least $ (1-\Phi (S)/2)^{t} \geq 1 - \Phi (S) t/2$ . Let $t\lt \frac{2\tilde k}{k\Phi (S)}$ , then ${\mathbb{P}}_{\pi _S}[\tau _{S^c} \leq t ] \lt \tilde k/k$ , and thus by taking $u$ as the vertex that minimises the escape probability from $S$ , we conclude that $t_{\mathsf{large-hit}}^{(\tilde{k},k)} \geq \frac{\tilde{k}}{k} \cdot \frac{2}{\Phi (S)}$ .

Proof of Lemma 4.13. Consider a random walk for $t$ steps on a $d$ -dimensional torus (or cycle), where $d\geq 1$ . Let $Z_1,Z_2,\ldots,Z_{t} \in \{-1,0,+1\}$ be the transitions along the first dimension, and let $S_i = Z_1+\ldots + Z_i$ . Note that $S_i$ is a zero-mean martingale with respect to the $Z_i$ . Define $\tau =\min \{i\;:\; |S_i|\geq D/d\}\wedge t$ , which is a bounded stopping time, and thus $S_{\tau \wedge i}$ is another martingale with increments bounded by 1. Then, by Azuma’s inequality [Reference Mitzenmacher and Upfal49, Theorem 13.4],

\begin{align*} {\mathbb{P}}\left [\tau \leq t\right ] = {\mathbb{P}}\left [|S_{\tau \wedge t}| \geq D/d\right ] \leq 2\exp\!\left ( - \frac{(D/d)^2 }{ 2t } \right ) \end{align*}

Now, for the random walk, in order to overcome a distance $D$ during $t$ steps, the above event must occur for at least one of the $d$ dimensions, so by the union bound

\begin{align*} {\mathbb{P}}\left [ \max _{1\leq s \leq t}\textrm{dist}(X_0,X_s) \geq D \right ] \leq 2d \cdot \exp\!\left ( - \frac{D^2}{2t d^2 } \right ), \end{align*}

as claimed.

Proof of Lemma 4.11. Recall from the definition of $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ that there must exist a vertex $u$ and set $S$ with $\pi (S)\geq 1/4$ such that the probability a random walk of length $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ started from $u$ has hit $S$ is at most $\tilde k/ k$ . Notice that since the $d$ -dimensional torus is regular any set $S$ of size at least $n/4$ satisfies $\pi (S)\geq 1/4$ . In order for a random walk to hit the set $S$ within $t$ steps, it must reach a distance $D=\textrm{dist}(u,S)$ from $u$ at least once during $t$ steps. Let $t$ be given by

\begin{equation*}t = \left \lfloor \frac {D^2}{2d^2\log \left (2dk/\tilde k\right )}\right \rfloor .\end{equation*}

Then by Lemma 4.13,

\begin{align*} {\mathbb{P}}_{u}\!\left [ \tau _{S} \leq t \right ] &\leq {\mathbb{P}}\left [ \max _{1 \leq s \leq t} \textrm{dist}(u,X_t) \geq D \right ]\leq 2d \cdot \exp\!\left ( -\frac{D^2}{2td^2} \right ) \leq \frac{\tilde{k}}{k}, \end{align*}

and the result follows.

Proof of Lemma 4.12. In the first part of the proof we will work with random walks whose lengths are independent samples from $\operatorname{Geo}(1/t )$ . That is, we consider walks $(X_0,X_1,\ldots, X_{L-1})$ where $L\geq 1$ is a geometric random variable with mean $t$ , which is independent of the trajectory $(X_s)_{s\geq 0 }$ . Let $\tilde P_{w,u}^s = {\mathbb{P}}_{w}[X_s = u, s\lt L ]$ . We call the above a geometric random walk of expected length $t$ . We consider a collection of $k$ independent geometric random walks of length $t$ , in particular the lengths of these walks are also independent of each other.

Let us lower bound the expected number of unvisited vertices in $S$ by $k$ independent geometric random walks of expected length $t$ . Define a subset $S' \subseteq S$ as

\begin{equation*} S' = \left \{ u \in S \;\colon \sum _{s=0}^{\infty } \sum _{w \in V} \mu (w) \tilde {P}_{w,u}^s \leq 8 t \cdot \pi (u) \right \}. \end{equation*}

Since the geometric random walk has an expected length of $t$ , it visits at most $t$ vertices in expectation and thus

\begin{equation*} \sum _{u \in S} \sum _{s=0}^{\infty } \sum _{w \in V} \mu (w) \tilde {P}_{w,u}^s \leq t. \end{equation*}

It follows by definition of $S'$ that

\begin{equation*} t \geq \sum _{u \in V \setminus S'} \sum _{s=0}^{\infty } \sum _{w \in V} \mu (w) \tilde {P}_{w,u}^s \geq \sum _{u \in V \setminus S'} 8 t \cdot \pi (u), \end{equation*}

and thus $\sum _{u \in V \setminus S'} \pi (u) \leq 1/8$ . Hence $\sum _{u \in S'} \pi (u) \geq \pi (S)-1/8 \geq 1/4-1/8 = 1/8$ .

Let $Z_t=Z_t(u)$ denote the number of visits to $u$ by a geometric random walk of expected length $t$ . The probability a single walk starting from $\mu$ visits a vertex $u \in S'$ before being killed is at most

\begin{align*} {\mathbb{P}}_{\mu }\!\left [ Z_t \geq 1\right ] &= \frac{ {\mathbb{E}}_{\mu } \left [Z_t\right ]}{{\mathbb{E}}_{\mu } \left [Z_t \, \mid \, Z_t \geq 1\right ]} = \frac{ \sum _{s=0}^{\infty } \sum _{w \in V} \mu (w) \tilde{P}_{w,u}^s }{ \sum _{s=0}^{\infty } \tilde{P}_{u,u}^s }, \end{align*}

since conditional on the walk having reached a vertex $u$ , the expected remaining returns before getting killed is equal to $\sum _{s=0}^{\infty } \tilde{P}_{u,u}^s$ . Now observe that $\tilde{P}^{s}_{u,u} \geq \frac{1}{4} \cdot P_{u,u}^{s}$ , which follows since ${\mathbb{P}}\left [ \operatorname{Geo}\!\left (1/t\right ) \gt s\right ] = \left (1 - \frac{1}{t} \right )^{s} \geq \left (1 - \frac{1}{t} \right )^{t}\geq 1/4$ , for any $t\geq 2$ and any $s\leq t$ . Thus we have

\begin{align*} {\mathbb{P}}_{\mu }\!\left [ Z_t \geq 1\right ] &\leq \frac{ \sum _{s=0}^{\infty } \sum _{w \in V} \mu (w) \tilde{P}_{w,u}^s }{ \frac{1}{4}\cdot \sum _{s=0}^{t} P_{u,u}^s } \leq \frac{8 t \cdot \pi (u)}{\frac{1}{4} \cdot ( 32 \cdot t\cdot \pi (u) \cdot k)} =\frac{1}{k}, \end{align*}

by hypothesis. Let $Y$ the stationary mass of the unvisited vertices in $S'$ , then

\begin{equation*} {\mathbb{E}} \left [ Y \right ] \geq \sum _{u \in S'} \pi (u) \cdot \left (1 - \frac {1}{k} \right )^{k} \geq \frac {1}{4} \cdot \pi (S'). \end{equation*}

Hence with probability at least $1/4$ , at least one vertex in $S'$ remains unvisited by the $k$ random walks whose length is sampled from $\operatorname{Geo}(1/t )$ . Finally, since each walk is independent the number of walks which run for more than $t$ steps is binomially distributed with parameters $k$ and $p= (1 - 1/t)^t \geq 1/4$ . Thus by a Chernoff bound the probability that less than $1/8$ of the $k$ random walks run for more than $t$ steps is at most $\exp\!\left (-\frac{(k/8)^2}{(2k/4)} \right )= e^{-k/32}$ . Hence, by coupling, $k/8$ random walks of length $t$ do not visit all vertices in $S'$ with probability at least $1/4-e^{-k/32} \geq 1/5$ , provided $k\geq 100$ .

Proof of Theorem 5.1. We split the analysis for $k\geq 2$ into two cases.

Case (i) [ $2\lt k\leq n^{1/20}$ ]: The lower bound is covered by [Reference Klasing, Kosowski, Pajak, Sauerwald, Fatourou and Taubenfeld37, Lemma 18] so we just prove the upper bound. For a single walk $t_{\mathsf{cov}}(\pi ) = \mathcal{O}(n^2)$ [Reference Aldous and Fill2, Proposition 6.7] and so since $k$ walks take at most as long to cover as a single walk, we can assume that $k\geq 10000$ when proving the upper bound.

To begin, divide the cycle as evenly as possible into $k$ disjoint intervals $\mathcal{I}_1,\dots, \mathcal{I}_k$ of consecutive vertices, each of size $\lfloor n/k\rfloor$ or $\lceil n/k\rceil$ . For $2 \leq c \leq k/\log k$ let $t^*(c)=t^* = \lceil (cn/k)^2\log ^2 k\rceil$ . Now, for each interval $\mathcal{I}_i$ we let $\mathcal{J}_i(c)$ be an interval of length $\ell =\lfloor \sqrt{t^*(c)} \rfloor \leq n$ centred around $\mathcal{I}_i$ . Note that since $c\geq 2$ and $n$ is large we have $\mathcal{I}_i\subset \mathcal{J}_i$ for each $1\leq i\leq k$ .

Proof of Claim 5.3. Let $\mathcal{N}(0,1)$ be the standard normal distribution, then by [Reference Gordon26] for any $x\gt 0$ :

(18) \begin{equation} {\mathbb{P}}\left [\mathcal{N}(0,1)\gt x\right ] \geq \frac{x}{x^2+1} \cdot \frac{1}{\sqrt{2\pi }}e^{-x^2/2} . \end{equation}

Note that all vertices of $\mathcal I_i$ will have been covered if the walk has travelled from one endpoint of $\mathcal J_i$ to the other via a path contained within $\mathcal J_i$ . Let $S_j$ be the distance of a random walk at time $j$ from its start point. Then, for large $n$ , by the Central Limit Theorem and (18) we have

(19) \begin{equation} {\mathbb{P}}\left [\frac{S_{ t^*/2}}{\sqrt{t^*/2}} \gt \frac{\ell }{\sqrt{t^*/2}} \right ] \geq \left (1-o(1)\right ) {\mathbb{P}}\left [\mathcal{N}(0,1)\gt \sqrt{2}\right ]\geq \left (1-o\!\left (1\right )\right )\frac{\sqrt{2}}{3} \cdot \frac{ e^{-1}}{\sqrt{2\pi }} \gt \frac{1}{15}. \end{equation}

Now, by symmetry and (19), regardless of its start point within the interval $\mathcal J_i$ , with probability at least $1/15$ the walk will have hit the ‘left’ end of $\mathcal J_i$ within at most $t^*/2$ steps. Once at the ‘left’ end of $\mathcal J_i$ , then again by (19) with probability at least $1/15$ , it will have reached the right end via a path though $\mathcal J_i$ within at most $t^*/2$ additional steps. Thus, for suitably large $n$ , with probability at least $(1/15)^2 \gt 1/250$ a walk of length $t^*$ starting in $\mathcal J_i$ will cover $\mathcal I_i$ .

Let $w_i$ be the number of walks which start in $\mathcal{J}_i$ . By Chernoff’s bound [Reference Mitzenmacher and Upfal49, Theorem 4.5]:

(20) \begin{equation} {\mathbb{P}}_{\pi ^k}\!\left [w_i\lt \frac{3}{4}\cdot c\log k \right ]={\mathbb{P}}\left [\operatorname{Bin}\!\left ( k,\frac{c\log k }{k} \right )\lt \frac{3}{4}\cdot c\log k \right ] \leq e^{-\frac{(1/4)^2}{2}\cdot c\log k }= k^{-c/32}. \end{equation}

By Claim 5.3, conditional on $w_i$ , none of the walks in $\mathcal{J}_i$ cover $\mathcal{I}_i$ w.p. at most $\left(\frac{249}{250}\right)^{w_i}$ when $n$ is large. Hence, by (20) a fixed interval $\mathcal{I}_i$ is not covered w.p. at most $(249/250)^{(3/4)\cdot c\log k} + k^{-c/32}$ . As $(3/4)\ln (249/250)\lt -3/1000$ and $k\geq 10000$ , for any $1000\leq c \leq k/\log k$ , we have

(21) \begin{equation} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t^*(c)\right ]\leq k\left (k^{-3c/1000} + k^{-c/32}\right ) \leq 2k^{1-3c/1000}\leq k^{-c/1000}, \end{equation}

by the union bound. Now, as $t^*(k/\log k) = n^2$ , observe that we have

(22) \begin{equation} {\mathbb{E}}_{\pi ^k} \left [\tau _{\mathsf{cov}}^{(k)} \right ] \leq t^*(1000) +\sum _{i=1000}^{\lfloor \frac{k}{\log k }\rfloor } [t^*(i+1)-t^*(i)]\cdot {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t^*(i)\right ] + \sum _{t=n^2}^\infty {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t\right ]. \end{equation}

Using ${\mathbb{P}}_{\pi ^k}[\tau _{\mathsf{cov}}^{(k)}\gt t^*(i) ] \leq k^{-i/1000}$ for $1000\leq i \leq k/\log k$ by (21), gives

(23) \begin{equation} \begin{aligned}\sum _{i=1000}^{\frac{k}{\log k }} [t^*(i+1)-t^*(i)]\cdot {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t^*(i)\right ] &\leq \left (\left (\frac{n}{k}\right )^2\log ^2 k + 1 \right ) \sum _{i=0}^{\infty } (2i+1) \cdot k^{-i/1000} \\[5pt] &= \mathcal{O}\!\left (\left (\frac{n}{k}\right )^2\log ^2 k\right ).\end{aligned} \end{equation}

Recall that ${\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t^*(k/\log k) \right ] \leq k^{- \frac{1}{1000}\cdot k/\log k}=e^{-k/1000}$ by (21). If $\textbf{v}$ is the worst-case start position vector for $k$ walks to cover a cycle then for any $t= i\cdot n^2$ and $i\geq 1$ ,

\begin{equation*} {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf {cov}}^{(k)}\gt t \right ]\leq {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf {cov}}^{(k)}\gt t^*(k/\log k) \right ]\cdot {\mathbb{P}}_{\textbf {v}}\!\left [\tau _{\mathsf {cov}}^{(k)}\gt t- n^2 \right ] \leq e^{-k/1000}\cdot {\mathbb{P}}_{\textbf {v}}\!\left [\tau _{\mathsf {cov}}^{(k)}\gt (i-1)n^2 \right ],\end{equation*}

by the Markov property. Note that ${\mathbb{P}}_{\textbf{v}}[\tau _{\mathsf{cov}}^{(k)}\gt (i-1)n^2 ]\leq (1/2)^{i-1}$ by Markov’s inequality since $t_{\mathsf{cov}} = n(n-1)/2$ . Thus the second sum on the RHS of (22) satisfies

(24) \begin{equation} \sum _{t=n^2}^\infty {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt t\right ] \leq n^2\cdot \sum _{i=1}^\infty {\mathbb{P}}_{\pi ^k}\!\left [\tau _{\mathsf{cov}}^{(k)}\gt i\cdot n^2\right ] \leq n^2\cdot \sum _{i=1}^\infty e^{-k/1000}2^{-i+1} = 2e^{-k/1000}n^2 . \end{equation}

Case ( $i$ ) then follows by inserting (23) and (24) into (22).

Case (ii) [ $k\gt n^{1/20}$ ]: The upper bound follows from Theorem 3.1 since in this case $\log k=\Omega (\!\log n )$ . By Lemma A.2 we have $\sum _{i=0}^{\lfloor t/2\rfloor }P_{v,v}^{2i} =\Omega (\sqrt{t} )$ so, by Lemma 3.10 (ii),

(25) \begin{equation} {\mathbb{P}}_{\pi }\!\left [\tau _v\leq t\right ]=\mathcal{O}\!\left (\sqrt{t}/n\right ). \end{equation}

We shall now apply Lemma 3.9 with $S=V$ and so if we set $t= (cn/k)^2\log ^2 k$ for a suitably small constant $c\gt 0$ then $p=\max _{v\in S}{\mathbb{P}}_{\pi }\!\left [\tau _v\leq t\right ] \leq \frac{\log n}{100k},$ by (25). Observe that $|S|\min _{v\in S}\pi (v) = 1$ and also $2p^2k \lt 1$ since $k\geq n^{1/20}$ and $n$ is large. Thus Lemma 3.9 gives

\begin{equation*}{\mathbb{P}}_{\pi ^k}\!\left [ \tau _{\mathsf {cov}}^{(k)} \leq t\right ] \leq \frac {4kp^2e^{2kp}}{|S|\min _{v\in S}\pi (v)} \leq 4k\left (\frac {\log n}{100k}\right )^2\cdot e^{2 k \cdot \frac {\log n}{100k} } = \frac {4(\!\log n)^2 }{100^2 k }\cdot n^{1/50} = o\left (1\right ), \end{equation*}

as $k\geq n^{1/20}$ , which completes Case ( $ii$ ) and finishes the proof.

Proof. Let $Q$ and $P$ be the transition matrices of the lazy walk on the (infinite) $d$ -dimensional integer lattice $\mathbb{Z}^d$ and the (finite) $d$ -dimensional $n$ -vertex torus ${\mathbb{T}}_d$ , respectively. By [Reference Hebisch and Saloff-Coste28, Theorem 5.1 (15)], for each $d\geq 1$ there exist constants $C,C',C''\gt 0$ such that for any $t\geq 1$ and $u,v \in \mathbb{Z}^d$ satisfying $||u -v||_{2} \leq t/C''$ we have

(26) \begin{equation} Q_{u,v}^t \geq \left (\frac{C}{t}\right )^{d/2}\exp\!\left (-C'\cdot \frac{ ||u-v||_2^2}{t} \right ) . \end{equation}

For any $t\geq 0$ and $u,v\in V({\mathbb{T}}_d)$ we have $ P_{u,v}^t \geq Q_{u,v}^t$ and $\| u - v \|_{\infty } \leq n^{1/d}/2$ . Therefore,

(27) \begin{equation} \max _{u,v \in V({\mathbb{T}}_d)} || u-v||_2^2 \leq d\cdot (n^{1/d}/2)^2 = dn^{2/d}/4, \end{equation}

and so if we set

(28) \begin{equation} t= \left \lceil \frac{ dC' \cdot n^{2/d}}{4\log\!(k/\tilde{k})}\right \rceil \leq \frac{dC'\cdot n^{2/d}}{2 \log\!(k/\tilde{k})}. \end{equation}

then, for large $n$ and any $u,v \in V({\mathbb{T}}_d)$ , we have $||u -v||_{2} \leq t/C''$ and thus, by (26) and (27),

\begin{equation*}P_{u,v}^t \geq \left (\frac {C}{t}\right )^{d/2}\cdot \exp\!\left (-C'\cdot \frac {dn^{2/d}/4}{t} \right ) \geq \frac {1}{n} \left (\frac {C2\log\!(k/\tilde {k})}{dC'}\right )^{d/2}\cdot \exp\!\left (- \log \left (\frac {k}{\tilde {k}}\right ) \right )\geq c'\cdot \frac {\tilde {k}}{n k} \end{equation*}

for some $c'\;:\!=\; c'(d)$ , as $ \log\!( k/\tilde{k}) \geq \log 2$ , which holds by hypothesis. It follows from the definition of separation distance that $s(t)\leq 1 - c'\cdot \frac{\tilde{k}}{n k}$ for $t$ given by (28). Note that if $c'\geq 1$ then the statement of the lemma follows by (14), the definition of $t_{\mathsf{mix}}^{(\tilde k,k)}$ . Otherwise, assuming $c' \lt 1$ , if we take $t' = \lceil 2/c'\rceil \cdot t$ then as separation distance is sub-multiplicative [Reference Levin, Peres and Wilmer42, Ex. 6.4] we have

\begin{equation*}s(t') \leq s(t)^{\left\lceil 2/c^{\prime}\right\rceil }\leq \left (1- \frac {c'\tilde {k}}{n k}\right )^{\lceil 2/c'\rceil }\leq \frac {1}{1 +\lceil 2/c'\rceil \cdot \frac {c'\tilde {k}}{n k} } = 1 - \frac { \lceil 2/c'\rceil \cdot \frac {c'\tilde {k}}{n k} }{1 +\lceil 2/c'\rceil \cdot \frac {c'\tilde {k}}{n k} }\leq 1- \frac {\tilde {k}}{n k}, \end{equation*}

for suitably large $n$ where in the second to last inequality we have used the fact that $(1+x)^r \leq \frac{1}{1-rx}$ for any $r\geq 0$ and $x\in [\!-\!1, 1/r)$ .

Proof of Theorem 5.2. We can assume that $k\geq C$ for a large fixed constant $C$ (in particular one satisfying $\log C \geq 1$ ), as otherwise the result holds since $t_{\mathsf{cov}} = \Theta (n^2)$ , and by [Reference Efremenko, Reingold, Dinur, Jansen, Naor and Rolim19, Theorem 4.2] the speed-up of the cover time on any graph is $\mathcal{O}(k^2 )$ .

For $k\geq C$ , define $\tilde k = \lfloor \log\!(k) \rfloor \geq 1$ . Recall that $t_{\mathsf{mix}}^{(\tilde k,k)} = \mathcal{O}(n^2/\log\!(k/\log k) ) = \mathcal{O}(n^2/\log k )$ by Lemma 5.4, and $ t_{\mathsf{cov}}^{\tilde k}(\pi ) =\mathcal{O}\!\left (\frac{n^2}{(\!\log k)^2} (\!\log \log k)^2\right )$ by Theorem 5.1. Thus, Theorem 4.7 yields $t_{\mathsf{cov}}^{(k)}= \mathcal{O}\!\left ( \frac{n^2}{\log k}\right ).$

We will use Theorem 4.9 to prove the lower bound, and for such, we need lower bounds for $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ , and $t_{\mathsf{large-cov}}^{(\tilde{k})}$ for an appropriate choice of $\tilde k$ . We will indeed prove that if we choose $\tilde k$ as a constant, then $t_{\mathsf{large-hit}}^{(\tilde{k},k)} = \Omega ( n^2/\log k )$ and $t_{\mathsf{large-cov}}^{(\tilde{k})} = \Omega (n^2)$ , leading to the desired result as $t_{\mathsf{cov}}^{(k)} = \Omega (\!\min (t_{\mathsf{large-hit}}^{(\tilde{k},k)},t_{\mathsf{large-cov}}^{(\tilde{k})}))$ .

For $t_{\mathsf{large-hit}}^{(\tilde{k},k)}$ we note that for any vertex $u$ of the cycle we can find a set of vertices of size at least $n/2$ with minimum distance at least $ \lfloor n/4\rfloor$ from $ u$ . Thus by Lemma 4.11 for any $\tilde{k} \leq k/2$ we have

\begin{equation*} t_{\mathsf {large-hit}}^{(\tilde {k},k)} = \Omega \left ( \frac {n^2}{\log\!( k/\tilde {k}) } \right ). \end{equation*}

To find a lower bound for $t_{\mathsf{large-cov}}^{(\tilde{k})}$ , by Lemma A.2 there exists some constant $c\gt 0$ such that for any $1\leq t\leq n^2$ and $u\in V$ the return probabilities in a cycle satisfy $\sum _{s=0}^t P_{u,u}^{s} \geq c\sqrt{t}$ . Thus if we take $\tilde{k} \geq \min (100,c/8)$ and let $ t= \lfloor (cn/(256\tilde{k}) )^2 \rfloor$ then $c\sqrt{t}\geq 32\cdot t \cdot \pi (u) \cdot (8\tilde{k})$ is satisfied. Thus by Lemma 4.12 (with $k=8\tilde{k}$ ) we have $t_{\mathsf{large-cov}}^{(\tilde{k})} \geq t/5 = \Omega (n^2)$ .