2.1. General formulation of layers occupancy

Let $\mathcal{H}(t)=(L_0(t),L_1(t),\ldots,L_k(t))$ be the occupancy vector of the DLA process at step t. Then,

(1) \begin{align}\mathrm{E} (L_i(t+1) \mid {\mathcal H}(t)) &= L_i(t) + \frac{L_{i+1}(t)}{N_{i+1}} \prod_{j=0}^i \left( 1-\frac{L_{j}(t)}{N_{j}} \right)\!,\quad i < k, \\[-12pt] \nonumber \end{align}

(2) \begin{align}\mathrm{E} (L_k(t+1) \mid {\mathcal H}(t)) &= L_k(t) + \prod_{j=0}^k \left( 1-\frac{L_{j}(t)}{N_{j}} \right)\!. \\[9pt] \nonumber \end{align}

Note that (2) follows from (1) on defining $L_{k+1}(t)/N_{k+1}=1$ for all t, and that if $L_0(t)=1$ , the above recurrences give $L_i(t+1)=L_i(t)$ .

The next proposition gives the solution to these recurrences under suitable conditions.

Proposition 1. For $t \ge 0$ let $\mu_k(t)=t$ , and for $1 \le j \le k-1$ , let

(3) \begin{align}\mu_{k-j}(t)= \frac 1{N_k N_{k-1}\cdots N_{k-j+1}} \frac {t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!}.\end{align}

For $j<k$ in the equal layers model, and for $j \le \sqrt{2k+2}-2$ in the growing layers model there are deterministic values $T_k < \cdots < T_{k-\ell} < \cdots < T_{k-j}$ , such that w.h.p. for all $\ell \le j$ and $T_{k-\ell}\le t \le t_f$ we have $L_{k-\ell}(t) \sim \mu_{k-\ell}(t)$ , whereas for $\ell > j$ , and $0 \le t \le T_{k-j}$ , $L_{k-\ell}(t)=0$ .

This proposition summarizes a gap property that when level $k-j$ starts to fill enough for the occupancy to become concentrated, the levels with lower index $i=0,1,\ldots,k-(\kern1.5pt j+1)$ still remain empty. Most of the work is to prove that $T_{k-j}$ can be chosen to be $t_{k-j}(\omega)$ as given by (6) and (7). The proof of Proposition 1 is inductive backwards from level k. The precise growth of the levels, and values of j which make it work are to be determined. The first steps are common to the equal and growing layers models, and we include them together in this section. For the equal layers model we complete the proof of Proposition 1 in Sections 3.1 and 3.2. The analogous proof for the growing layers model is given in Section 4.

2.2. Upper bound on occupancy at step t

The underlying random walk $\rho_t$ from source v to sink z at step t defines a path given by $\nu=u_0u_1\cdots u_ku_{k+1}=z$ , where $u_i$ is a random vertex in level i. Particle t follows this walk until halting at a vertex $u_i$ , where $u_{i+1}$ is the first occupied vertex encountered by $\rho_t$ .

Let $B_i(t)$ denote the occupied (blocked) vertices in level i in the DLA process at the end of step t. We define an upper-blocked process which we use to upper bound $L_i(t)=|B_i(t)|$ . This new process gives rise to sets $\widehat {B_i}(t)\supseteq B_i(t)$ and random variables $\widehat{L_i}(t)=|\widehat{B_i}(t)|\geq L_i(t)$ . For every vertex $u_j$ , $0 \le j \le k$ , on the walk $\rho_{t+1}$ , if $u_{j+1}$ is occupied ( $u_{j+1} \in \widehat B_{j+1}(t)$ ) add a vertex to $\widehat{B_j}(t)$ as follows. If $u_j \not \in \widehat{B_j}(t)$ add $u_j$ to $\widehat{B_j}(t+1)$ . If $u_j \in \widehat{B_j}(t)$ add some other $u'_j \in S_j \! \setminus \! \widehat{B_j}(t)$ to $\widehat{B_j}(t+1)$ . If the layer $S_j$ becomes full we continue with the layer $S_{j-1}$ . As we prove, this contingency is unlikely to occur, as w.h.p. the first layer to become full is the source.

If particle $t+1$ halts at vertex $u_i$ in the DLA process, then either $u_i$ is added to both $B_i(t+1)$ and $\widehat{B_i}(t+1)$ , or $u_i$ is already a member of $\widehat{B_i}(t)$ . In either case $B_i(t) \subseteq \widehat{B_i}(t)$ for all i and $t \ge 0$ . It follows that $L_i(t) \le \widehat{L_i}(t) \le t$ , as $\rho_{t+1}$ can add at most one vertex to $\widehat{B_i}(t)$ . Moreover, $\widehat{L_k}(t)=t$ deterministically (provided $t \le N_k$ ). As a consequence the finish time $t_f(UB)$ of the upper-blocked process satisfies $t_f(UB) \le t_f(DLA)$ .

Let $\widehat{\mathcal{H}}(t)=(\widehat{L_0}(t),\widehat{L_1}(t),\ldots,\widehat{L_k}(t))$ be the occupancy vector of the upper-blocked process at step t. For $0 \le i \le k$ , the expectation $\mathrm{E}\,\widehat{L_i}(t)$ satisfies the recurrence

(4) \begin{align}\mathrm{E}\big(\widehat{L_i}(t+1) \mid \widehat{H}(t)\big)= \widehat{L_i}(t) + \frac{\widehat L_{i+1}(t)}{N_{i+1}}\, {\unicode{x1D7D9}}_{\{{{\mathcal E}(t)}\}}, \end{align}

where ${\mathcal E}(t)$ is the event that $\widehat{L_i}(t) < N_i$ and that $\widehat L_{i+1}(t) < N_{i+1}$ for $i<k$ . Setting ${\unicode{x1D7D9}}_{\{{{\mathcal E}(t)}\}}=1$ always gives an upper bound on $\mathrm{E}\, \widehat{L_i}(t)$ , and as we only propose to analyse the process as long as no level i ( $1 \le i \le k$ ) is full, we assume ${\unicode{x1D7D9}}_{\{{{\mathcal E}(t)}\}}=1$ forthwith. Equation (4) follows because the upper-blocked process increases the size of $\widehat{B_j}(t)$ (if possible) whenever the walk $\rho_t$ contains a vertex of $\widehat B_{j+1}(t)$ , this being true at all levels $j=0,\ldots,k$ .

The evolution of ${\widehat{\mathcal{H}}}(t)=(\widehat{L_0}(t),\widehat{L_1}(t),\ldots,\widehat{L_k}(t))$ is Markovian, and for $t \le t_f$ we henceforth assume for $i \ge 1$ that $\widehat{L_i}(t) < N_i$ in our calculations. If so, referring to (1) and (2), we have

\[\mathrm{E} (L_i(t+1)\mid {\mathcal H}(t)) \le L_i(t) +\frac{L_{i+1}(t)}{N_{i+1}}\le \widehat{L_i}(t) +\frac{\widehat L_{i+1}(t)}{N_{i+1}}=\mathrm{E} \big( \widehat L_{i+1}(t+1) \mid \widehat{\mathcal H}(t)\big).\]

When $\widehat{L_0}(t)=1$ at $t=\widehat{t}_f$ the upper-blocked process stops, and $\widehat{t}_f$ gives a lower bound on the finish time $t_f$ of the corresponding DLA process.

We next give w.h.p. bounds for $\widehat{L_i}(t)$ assuming none of the layers $i=1,\ldots,k$ are full for $t \le \widehat{t}_f$ . The proofs of level occupancy are inductive backwards from level k. For a given level $k-j$ we identify two (not necessarily integer) times, $t_1(k-j)$ and $t_{k-j}(\omega)$ , defined as follows. For level k, as $\mu_k(t)=t$ , $t_1(k)=1$ and for $j\ge 1$ , let $t_1(k-j)$ is the solution to $\mu_{k-j}(t)=1$ in (3). Thus,

(5) \begin{align}t_{1}(k-j)= \big[(\kern1.5pt j+1)! N_k N_{k-1}\cdots N_{k-j+1}\big]^{1/(\kern1.5pt j+1)}.\end{align}

The variable $t_1(k-j)$ is used as a reference point in many of our calculations.

Let $t_k(\omega)=1$ , and for $1\le j \le k-1$ , let $t_{k-j}(\omega)=(4\omega^{3})^{1/(\kern1.5pt j+1)} t_1(k-j)$ . Thus,

(6) \begin{align}t_{k-j}(\omega)= \big( (4\omega^{3})\big[(\kern1.5pt j+1)! N_k N_{k-1}\cdots N_{k-j+1}\big] \big)^{1/(\kern1.5pt j+1)}.\end{align}

Henceforth, we choose

(7) \begin{align} \omega=6 \log n.\end{align}

In Section 2.3 we show that with the value of $\omega$ given in (7), for $t \ge t_{k-j}(\omega)$ , the value of $\widehat L_{k-j}(t)$ is concentrated around $\mu_{k-j}(t)$ w.h.p., thus establishing the first part of the claim in Proposition 1.

The next lemma gives the expected value of $\widehat{L_i}$ and conditions for its concentration at any step $t \le t_f$ . Recall that $t_1(k-j)$ is given by (5), and $ t_{k-j}(\omega)$ is given by (6).

Proof. As the sink is occupied at $t=0$ , $N_{k+1}=1$ , and we define $\widehat L_{k+1}(t)=1$ . This implies that, provided $t \le N_k$ , $\widehat{L_k}(s)=s=\mu_k(s)$ for $0\leq s\leq t$ . Moreover, as at most one vertex can be added to $\widehat{L_i}(t-1)$ at step t, this implies $\widehat{L_i}(t) \le t$ .

Iterating (4) backwards for $0\leq s\leq t$ , and using $\widehat{L_i}(0)=0$ , gives

(8) \begin{align}\mathrm{E}\,\widehat{L_i}(t)= \frac{1}{N_{i+1}} \sum_{s=0}^{t-1} \mathrm{E}\, \widehat{L}_{i+1}(s).\end{align}

We claim for $j \ge 0$ that

(9) \begin{align} {\unicode{x1D7D9}}_{\{{t \ge j}\}} \frac{(t-j)^{\kern1.5pt j+1}}{(\kern1.5pt j+1)!} \le \big(N_k N_{k-1}\cdots N_{k-j+1}\big) \mathrm{E}\, \widehat L_{k-j}(t) \le \frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!}.\end{align}

For given t, the induction is backwards on $k-j$ from $j=0$ . When $j=0$ , (9) becomes $\mathrm{E}\, \widehat{L_k}(t)=t$ which is true, so the first non-trivial case is $j=1$ . From (8) we see that

(10) \begin{align}\mathrm{E}\, \widehat L_{k-1}(t)= \frac{1}{N_k} \sum_{s=0}^{t-1} s,\end{align}

which illustrates how (9) arises from bounding this sum.

For the induction at step $i=k-(\kern1.5pt j+1)$ , let

(11) \begin{align}M_{j-1}=N_k N_{k-1}\cdots N_{k-j+1}.\end{align}

Multiply (8) by $M_{j-1}$ , and insert the bounds on $M_{j-1} \mathrm{E}\, \widehat L_{k-j}(s)$ from (9) (with $i+1=k-j$ ) into this, to give

(12) \begin{align}\frac 1{N_{k-j}} \sum_{s=j}^{t-1} \frac{(s-j)^{\kern1.5pt j+1}}{(\kern1.5pt j+1)!}\le M_{j-1} \mathrm{E}\, \widehat L_{k-(\kern1.5pt j+1)}(t) \le \frac 1{N_{k-j}} \sum_{{s=1}}^{t-1} \frac{s^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!}\end{align}

By comparison of the sum with the related integral we have that

(13) \begin{align}\frac{(t-1)^{m+1}}{m+1} \le 1^m+2^m+\cdots+(t-1)^m \le \frac{t^{m+1}}{m+1}.\end{align}

Use (13) in (12) with $m=j+1$ to give

\[\frac{{\unicode{x1D7D9}}_{\{{t \ge j+1}\}}}{N_{k-j}} \frac{(t-(\kern1.5pt j+1))^{ \kern1.5pt j+2}}{(\kern1.5pt j+2)!}\le M_{j} \mathrm{E}\, \widehat L_{k-(\kern1.5pt j+1)}(t) \le \frac 1{N_{k-j}} \frac{t^{ \kern1.5pt j+2}}{(\kern1.5pt j+2)!},\]

which completes the induction for (9).

Setting the indicator ${\unicode{x1D7D9}}_{\{{{\mathcal E}(t)}\}}=1$ in (4), provides an upper bound on $\mathrm{E}\, \widehat{L_i}(t)$ , and as the upper bound in this induction is $\mu_i(t)$ , this gives $\mathrm{E}\, \widehat L_{i}(t)\le\mu_{i}(t)$ as required.

Furthermore, provided $\widehat L_{k-j}(t) < N_{k-j}$ , $j^2/t=o(1)$ ,

(14) \begin{align}\mathrm{E}\, \widehat L_{k-j}(t) = \frac {1}{N_k N_{k-1}\cdots N_{k-j+1}} \frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!}\big( 1 -O\big(\kern1.5pt j^2/t \big) \big)=\mu_{k-j}(t) \big( 1 -O\big(\kern1.5pt j^2/t \big) \big).\end{align}

This completes the proof of Lemma 1(1).

The next thing to check is that, for the values of k given in Theorem 1, for $t \ge t_1(k-j)$ , $j^2/t=o(1)$ , allowing us to use Lemma 1(1) in the proof of Lemma 1(2).

Proof. For $j \ge 1$ , the value of $t_{k-j}(\omega)$ given in (6) satisfies

\begin{align*}t_{k-j}(\omega) & = (4 \omega^3)^{1/(\kern1.5pt j+1)} t_1(k-j)\ge t_1(k-j)= ((\kern1.5pt j+1)!N_k\cdots N_{k-j+1})^{1/(\kern1.5pt j+1)} \nonumber \\ & = \left( (\kern1.5pt j+1)! M_{j-1} \right)^{1/(\kern1.5pt j+1)},\end{align*}

where $M_{j-1}$ is given in (11). The product $M_{j-1}$ is model specific. For the equal layers model it has value $M_{j-1}(E)=(n/k)^j$ , as $N_i=n/k$ . For the growing layers model the value is $M_{j-1}(G)=d^{kj-j(\kern1.5pt j-1)/2}$ , see (42).

In the first case $M_{j-1}(E)^{1/(\kern1.5pt j+1)} \ge (n/k)^{1/2}$ , and in the second $M_{j-1}(G)\ge d^{k/2}$ , this minimum being achieved at $j=1$ or $j= k$ . Checking $j^2/M_{j-1}^{1/(\kern1.5pt j+1)}$ we see that the condition $j^2/t_1 (k-j)=o(1)$ is satisfied in the equal layers model provided $k=o(n^{1/5})$ and in the growing layers model provided $d \rightarrow \infty$ .

The other assumptions of Lemma 1(1) and (2) are dealt with as follows. Theorem 1 gives a w.h.p. finish time $t_f \approx T_f$ . As $\mathrm{E}\, \widehat{L_i}(t) \le \mu_i(t)$ follows from Lemma 1, checking that $\mu_i(T_f) \ll N_i$ and using the Markov inequality will imply that w.h.p. $\widehat{L_i}(T_f) \ll N_i$ holds as asserted. For the equal layers model, $\mu_i(T_f) \ll N_i$ is proved in Lemma 4(1), and for the growing layers model in (54) and (55). The condition that $t_1(k-(\kern1.5pt j-1)) \ll t_1(k-j)$ is checked in Lemma 4(2) for the equal layers model and in (53) of for the growing layers model.

With this in hand, we can now continue with the proof of Lemma 1(2).

2.3. Concentration of $\widehat{\boldsymbol{L}_\boldsymbol{i}}(t)$ for sufficiently large t

Proof. The proof is inductive and for level $k-j$ it depends on establishing the result for levels $k,k-1,\ldots,k-j+1$ . By definition, $\widehat{L_k}(t)=t=\mu_k(t)$ . Let $t_k(\omega)=1$ , establishing Lemma 1(2) for $j=0$ . For $j\ge 0$ , let $t_1(k-j)$ as given by (5) and note that $\mu_{k-j}(t_1(k-j))=1$ . For $j \ge 1$ , as $t_{k-j}(\omega)=(4\omega^3)^{1/(\kern1.5pt j+1)}t_1(k-j)$ then $\mu_{k-j}(t_{k-j}(\omega))=4\omega^3$ .

The random variable $\widehat L_{k-j}(t)$ is obtained from $\widehat L_{k-j}(t-1)$ by choosing a random vertex $u \in S_{k-j}$ irrespective of the current occupancy of u, and a random neighbour $w \in S_{k-j+1}$ . If w is occupied then $\widehat L_{k-j}(t+1)=\widehat L_{k-j}(t)+1$ . Thus, $\widehat L_{k-j}(t+1)=\widehat L_{k-j}(t)+ Q_{k-j}(t)$ , where $\mathrm{P}( Q_{k-j}(t)=1)=(\widehat L_{k-j+1}(t)/N_{k-j+1})$ independently of any previous outcomes. As $t_1(k-j) \gg j^2$ , Lemma 1(1) implies that $\mathrm{E}\, \widehat L_{k-j}(t)= \mu_{k-j}(t) (1+o(1))$

For $j \ge 1$ , and $ t \ge t_{k-j}(\omega)$ let ${\mathcal A}_j(t)$ denote the event that $\widehat L_{k-j}(t)\in [\mu_{k-j}(t)(1- 1/\omega), \mu_{k-j}(t)(1+ 1/\omega)]$ . As $\widehat L_{k-j}(t)$ is the sum of independent Bernoulli random variables, by Hoeffding’s inequality,

(15) \begin{align} \mathrm{P}( \neg {\mathcal A}_j(t))\leq 2\exp\left\{-\frac{\mu_{k-j}(t)}{3\omega^2}\left( 1-\delta \right)\right\}\le2\exp\left\{-\frac{\mu(t_{k-j}(\omega))}{4\omega^2}\right\} \le 2\mathrm{e}^{-\omega},\end{align}

where $\delta=O(1/{\omega}+j^2/t)$ , includes the correction from (14).

Let ${\mathcal E}_j(t)$ be the event that ${\mathcal A}_j(s)$ holds for all $s \in [t_{k-j}(\omega), t]$ . For $j=0$ , $\mathrm{P}({\mathcal E}_{0}(t))=1$ , and for $j\ge 1$ , given ${\mathcal E}_{j-1}(t)$ , as an inductive assumption, we have that

(16) \begin{align} \mathrm{P} \big(\neg {\mathcal E}_{j}(t)\mid {\mathcal E}_{j-1}(t) \big) \le \sum_{s=t_{k-j}(\omega)}^t \mathrm{P}\big(\neg {\mathcal A}_j(s)\big).\end{align}

Summing over $i \le j$ we will be able to complete the induction, via

(17) \begin{align}\mathrm{P}\big(\neg{\mathcal E}_j \big)\leq \mathrm{P}\big(\neg{\mathcal E}_j\mid {\mathcal E}_{j-1}\big)+\mathrm{P}\big(\neg{\mathcal E}_{j-1}\big)\le\sum_{i\le j} \sum_{t \ge t_{k-i}(\omega)} \mathrm{P}\big(\neg{\mathcal A}_i(t)\big)\le c j \omega N_k \mathrm{e}^{-\omega},\end{align}

provided we can establish the bound on the right-hand side of (17), which we now do. Here c is some absolute constant.

As $\mu_{k-j}(t)$ in (3) is monotone increasing in t and by (15) is bounded above by $2\mathrm{e}^{-\omega}=o(1)$ , we use the Euler–MacLaurin theorem to replace the summation over t in (16) by an integral. Thus,

\[\sum_{t \ge t_{k-j}(\omega)} \mathrm{P}\big(\neg{\mathcal A}_j(t)\big) \le 2\sum_{t \ge t_{k-j}(\omega)}\exp\left\{-\frac{\mu(t_{k-j}(\omega))}{4\omega^2}\right\} \le 3 \int_{t \ge t_0}\exp\left\{-\frac{t^{ \kern1.5pt j+1}}{C}\right\}\, \mathrm{d}t,\]

where, $t_0=t_{k-j}(\omega)$ , $C=[4\omega^2 (\kern1.5pt j+1)! N_k \cdots N_{k-j+1}]$ , and from (5) and (6), $t_{k-j}(\omega)=(4\omega^3 C)^{1/(\kern1.5pt j+1)}$ .

Put $t^{ \kern1.5pt j+1}/C=z^2/2$ , so that $t=(Cz^2/2)^{1/(\kern1.5pt j+1)}$ , and $dt/dz= (2/(\kern1.5pt j+1)) (C/2)^{1/(\kern1.5pt j+1)} z^{2/(\kern1.5pt j+1)-1}$ . Let $z_0=z(t_0)=\sqrt{2 \omega}$ , and note that as $j\ge 1$ , $z^{2/(\kern1.5pt j+1)-1}\le 1$ . Finally $(C/2)^{1/(\kern1.5pt j+1)} = O(\omega k N_k)$ , as $N_k$ is a largest level in both models. Thus,

(18) \begin{align}\int_{t \ge t_0}\exp\left\{-\frac{t^{ \kern1.5pt j+1}}{C}\right\}\, \mathrm{d}t \le \frac{2}{j+1}\left(\frac{C}{2}\right)^{1/(\kern1.5pt j+1)} \int_{z \ge z_0} \mathrm{e}^{-z^2/2}\,\mathrm{d}z\le O(\omega k N_k) \frac{1}{\sqrt \omega} \mathrm{e}^{-\omega} ,\end{align}

by using a standard bound on the tail of the normal distribution, $\mathrm{P}(Z \ge z) \le 1/(\sqrt{2\pi}z) \mathrm{e}^{-z^2/2}$ .

Lemma 3 follows for all $1\le j \le k$ by adding the upper bound on the right-hand side of (17) given in (18) over $j \le k$ , and choosing $\omega=6 \log n$ .

2.4. Lower bound on occupancy at step t

So far we only have an upper bound on $\mathrm{E}\, L_j(t)$ given by $\mathrm{E}\, L_j(t) \le \mathrm{E}\, \widehat L_j(t)$ . We construct a lower bound $\mathrm{E}\, \widetilde L_j(t)$ and prove that for large enough t these bounds converge, thus giving the asymptotic value of $\mathrm{E}\, L_j(t)$ .

For a given level j, a lower bound on $L_j(t)$ can be found as follows. Define a subprocess of DLA which requires that the particle avoids upper-blocked vertices. Thus, to reach level j, a particle must avoid choosing neighbours in $\widehat B_1,\ldots,\widehat{B_j}$ at steps $0,\ldots,j-1$ of its random walk.

Let $L^*_j(t)$ be an upper bound on $\widehat L_j(t)$ that holds w.h.p. This will, for example, be obtained from Lemma 3 if $t \ge t_j(\omega)$ . Referring to (1) and (2), let $\widetilde L_i(t)$ be obtained by replacing L by $L^*$ in the bracketed terms on the right-hand side. This defines a lower bound $\widetilde L_j(t)$ , such that w.h.p. $\widetilde L_j(t)\le L_j(t) \le \widehat L_j(t)$ . Let $\widetilde{\mathcal{H}}(t)=(\widetilde L_0(t),\widetilde L_1(t),\ldots,\widetilde L_k(t))$ be the occupancy vector obtained from this lower bound, then we have the following recurrences:

(19) \begin{align}\mathrm{E} \big(\widetilde L_i(t+1)\mid \widetilde {\mathcal{H}}(t)\big) &= \widetilde L_i(t) + \frac{ \widetilde L_{i+1}(t)}{N_{i+1}} \prod_{j=0}^i \left( 1-\frac{L^*_{j}(t)}{N_{j}} \right) \!, \\[-12pt] \nonumber \end{align}

(20) \begin{align}\mathrm{E} \big(\widetilde L_k(t+1)\mid \widetilde{\mathcal{H}}(t)\big) &= \widetilde L_k(t) + \prod_{j=0}^k \left( 1-\frac{L^*_{j}(t)}{N_{j}} \right)\!. \\[9pt] \nonumber \end{align}

These recurrences are solved in Section 3.2 for the equal layers model and in Section 4 for the growing layers model.

2.5. The gap property

It is clear that $0 \le \widetilde L_{k-j}(t) \le L_{k-j}(t) \le \widehat L_{k-j}(t) \le t$ , and that these values are monotone non-decreasing. We choose times

\[t^-(k-j) < t_1(k-j) <t^+(k-j)=t_{k-j}(\omega),\]

(where $\omega=6 \log n$ ) such that

\[\mu_{k-j}(t^-) =1/\omega^3, \quad \mu_{k-j}(t_1)=1, \quad \mu_{k-j}(t^+)=4\omega^3.\]

Below $t^-(k-j)$ , $\widehat L_{k-j}(t)=0$ w.h.p. for all $j=1,\ldots,k$ . This follows from Lemma 1, and the Markov inequality, as $k=o(\omega)$ in either model. Above $t^+(k-j)$ we will see that $\mathrm{E}\, \widetilde L_{k-j}(t) \sim \mathrm{E}\,\widehat L_{k-j}(t)$ and thus $\mathrm{E}\, L_{k-j}(t) \sim \mu_{k-j}(t)$ . The main content of Sections 3 and 4 is a proof of this via a gap property which implies that

\[t^+(k-j)= t_{k-j}(\omega) \ll t^-(k-(\kern1.5pt j+1)) \ll t_1(k-(\kern1.5pt j+1)),\]

so that w.h.p. concentration for $L_{k-j}$ occurs while $L_{k-(\kern1.5pt j+1)}$ is still zero.

3.1. Evolution of the state vector $\widehat {\boldsymbol{L}}$ in the equal layers model

We prove there is a large gap in the number of steps between the time when $\mathrm{E}\, \widehat L_{k-j}=1$ with all lower values zero, and the time when $\mathrm{E}\, \widehat L_{k-(\kern1.5pt j+1)}=1$ . The gap allows $\widehat L_{k-j}$ to increase and become concentrated around $\mu_{k-j}$ , whilst all values with a lower index $i<k-j$ remain zero. This confirms the inductive assumption stated below (5) in Lemma 3.

Let $t_1(k-j)$ and $t_{k-j}(\omega)$ be as given by (5) and (6). We list various assumptions used in this and the next section:

(24) \begin{align} 1 \le k \le k^*=\frac{\sqrt{\log n}}{\log \log n},\quad \omega =6\log n, \quad \beta= \frac{N_k}{\omega T_f}.\end{align}

Let $\widehat L=(\widehat{L_0}, \widehat{L_1},\ldots, \widehat{L_k})$ be the state vector of the upper-blocked process. The entries in $\widehat L$ are non-negative integers, and if $\widehat{L_i}=0$ then $\widehat L_{i-1}=0$ .

The following argument for $t \le \omega T_f$ proves there is a large enough gap $t''-t$ between $\mu_{k-j}(t)=1$ and $\mu_{k-(\kern1.5pt j+1)}(t'')=1$ for $\widehat L_{k-j}(t'')$ to be concentrated, as assumed in Lemma 3. In particular w.h.p. at $t_{k-j}(\omega)$ , where $t''\gg t_{k-j}(\omega) > t'$ we will have (32), as is to be shown.

From (22),

(25) \begin{align}\beta=\frac{N_k}{\omega T_f}=\frac{n}{ \omega k}\frac 1{\gamma_k n^{k/(k+1)}}=\frac{1}{\gamma_k} \frac 1{\omega k}n^{1/(k+1)} \ge (6\log n)^{4+k/2}.\end{align}

This is true as $\omega= 6 \log n$ and we assumed that $k \leq \sqrt{\log n}/\log \log n$ .

As $N_{k-j}=N_k=n/k$ in the equal layers model, for any $t \le \omega T_f$ ,

(26) \begin{align}\mu_{k-(\kern1.5pt j+1)}(t) = \frac{t}{(\kern1.5pt j+2){ N_{k-j}}}\cdot \mu_{k-j}(t) \le \frac 1{(\kern1.5pt j+1)\beta}\mu_{k-j}(t).\end{align}

For $j+\ell \le k$ , we can iterate this to give

(27) \begin{align} \mu_{k-(\kern1.5pt j+\ell)}(t) \le \mu_{k-j}(t)\frac 1{\beta^\ell} \frac{1}{(\kern1.5pt j+\ell+1)(\kern1.5pt j+\ell)\cdots (\kern1.5pt j+2)}\le \mu_{k-j}(t)\frac 1{((\kern1.5pt j+1)\beta )^\ell}.\end{align}

Consider $\mathrm{E}\,\widehat L_{k-j}(t)$ . At $t = t_1(k-j)$ when $\mu_{k-j}(t)= 1$ , then by Lemmas 1 and 2, $\mathrm{E}\, \widehat L_{k-j}(t) \sim 1$ . The Markov inequality implies that w.h.p. $\widehat L_{k-j}(t) \in I_\omega=[0,1,\ldots,\omega]$ . From (26) and (27),

(28) \begin{align} \mu_{k-(\kern1.5pt j+\ell)}(t) \le \frac {1+o(1)}{((\kern1.5pt j+1)\beta)^\ell}, \quad\text{for } 1\le \ell\leq k-j,\end{align}

and, thus, w.h.p. $\widehat{L_0}=0,\widehat{L_1}=0,\ldots,\widehat L_{k-(\kern1.5pt j+1)}=0$ . Using (26) and (27), we see that

(29) \begin{align} \mu_{k-j+\ell}(t) \ge \mu_{k-j}(t) j(\kern1.5pt j-1)\cdots(\kern1.5pt j-\ell+1) \beta^{\ell}, \quad \text{for } \ell\leq j.\end{align}

Thus, if $\mu_{k-j}(t)= 1$ then for $1\le \ell \leq j$ , $t\ge t_{k-j+\ell}(\omega)$ , and

(30) \begin{align}\mu_{k-j+\ell}(t) \ge \beta^\ell \ge (6\log n)^{4+k/2}.\end{align}

It now follows that Lemma 1(2) holds, and w.h.p. $\widehat L_{k-j+\ell}$ is equal to $(1+o(1))\mathrm{E}\, \widehat L_{k-j+\ell}\sim\mu_{k-j+\ell}(t)$ .

In summary, at time t such that $\mu_{k-j}(t) \sim 1$ (implying that $t\leq T_f$ , see Lemma 4(2)), w.h.p., the state vector $\widehat L$ is such that $\mu_{k-\ell}\rightarrow \infty$ for $\ell \le j-1$ , and

(31) \begin{align} \widehat L(t)=(0,\ldots,0, \widehat L_{k-j}\in I_\omega, (1+o(1))\mu_{k-j+1}, (1+o(1))\mu_{k-j+2},\ldots, \mu_k).\end{align}

Let $t=t_1(k-j)$ , let $t' =t_{k-j}(\omega)=(4\omega^3)^{1/(\kern1.5pt j+1)} t_1(k-j)$ , so $\mu_{k-j}(t') \sim 4\omega^3$ . By Lemma 3 it holds w.h.p. that $\widehat L_{k-j}(t')\sim \mu_{k-j}(t')$ .

Let $t''=t_1(k-(\kern1.5pt j+1))$ . By Proposition 4(2), $t_1(k-(\kern1.5pt j+1))=\Theta(1) (n/k)^{1/((\kern1.5pt j+1)(\kern1.5pt j+2))} t_1(k-j)$ . As $t'=(4\omega^3)^{1/((\kern1.5pt j+1)}t_1(k-j)$ , it can be checked that $t'' \gg t'$ . In addition, by (27)

\[\sum_{\ell \ge 1} \mu_{k-(\kern1.5pt j+\ell)}(t')=O\left(\frac{\mu_{k-j}(t')}{\beta}\right)=O\left(\frac{\omega^3}{\beta}\right)=O\left(\frac{1}{\log^{1+k/2}n}\right).\]

Thus, applying the Markov inequality to the above, at $t'=t_{k-j}(\omega)$ , w.h.p.

(32) \begin{align} \widehat L(t_{k-j}(\omega))=(0,\ldots,0, (1+o(1))\mu_{k-j}, (1+o(1))\mu_{k-j+1},\ldots, \mu_k),\end{align}

so that $\widehat L_{k-j}(t')$ is concentrated around its mean, and all lower levels are unoccupied, and thus the claimed gap exists. This condition persists w.h.p. until around $t_1(k-(\kern1.5pt j+1))=t''\gg t'=t_{k-j}(\omega)$ , when $\mu_{k-(\kern1.5pt j+1)}(t_1(k-(\kern1.5pt j+1)))=1$ at which point $\widehat L(t'')$ resembles (31) and the induction continues.

3.2. Lower bound on occupancy in the equal layers model

We prove that, for $t \ge t_i(\omega)$ , we have $\mathrm{E}\, \widetilde L_i(t) \sim \mathrm{E}\, \widehat{L_i}(t) \sim \mu_i(t)$ . As with the upper bounds, we have that $\mathrm{E}\, \widetilde L_k \gg \mathrm{E}\, \widetilde L_{k-j}$ for $j \ge 1$ . The first step is to draw a line between these quantities.

Let $t^-=t_1(k-1)/\omega$ , so that $\mu_{k-1}(t^-)=1/\omega^2$ . Then (14) and the Markov inequality imply that w.h.p. $\widehat L_{k-1}(t^-)=0$ and as $\widehat L_{j}(t)$ is monotone non-decreasing in j, w.h.p. $\widehat L_j(t)=0$ for all $j \le k-1$ and $t \le t^-$ . As $\widehat{L_k}(t)=L^*_k(t)=t$ deterministically, this simplifies (20) for $t \le t^-$ .

As before let $\beta=N_k/\omega T_f$ where $\omega, k$ satisfy (24). Provided $k \ge 2$ , if $t \ge t^-$ then $t/\beta \gg \omega^3$ . Indeed using (25),

\[t^-=\frac{t_1(k-1)}{\omega} = \frac{1}{\omega} \sqrt{\frac{2n}{k}} \quad \implies \quad \frac{t^-}{\beta} = \frac{\gamma_k \omega k}{n^{1/(k+1)}} \cdot \frac{1}{\omega} \sqrt{\frac{2n}{k}} \ge 2\gamma_k n^{1/6} \gg \omega^3.\]

For $t \ge t^-$ and $1\le i \le k-1$ , we first consider the product term in (19). Recalling that $N_i=N_k=n/k$ , we next prove that

(33) \begin{align}\prod_{j=0}^i \left( 1-\frac{L^*_{j}(t)}{N_{j}} \right) \ge 1- \sum_{j=0}^i \frac{L^*_{j}(t)}{N_{j}}=1-O\left(\frac{t}{\beta N_k}\right)=1-o(1).\end{align}

By (27), $\mu_{k-j} \le \mu_k/\beta^j$ . Assume $t \ge t^-$ and that for some $\ell \le k-1$ , $t_{\ell+1}(\omega) \le t < t_{\ell}(\omega)$ . Lemma 3 gives concentration of $\widehat L_j(t)$ for $j\ge \ell+1$ , which we use along with (31) and (32).

If $i< \ell$ then $L^*_{j}(t)=0$ , for $j \le i$ . Next, if $i=\ell$ , $L^*_j(t)=0$ for $j<i$ and $L^*_i(t)\le 2 \mu_i(t_i(\omega))\le 8\omega^3$ , since $t < t_i(\omega)$ , and again Lemma 3 gives this upper bound at $t_i(\omega)$ . Finally assume $i \ge \ell+1$ . Then (as $t \ge t^-$ ) using (27),

(34) \begin{align}\sum_{j=0}^i \frac{L^*_{j}(t)}{N_{j}}\le \frac{8\omega^3}{N_k}+2\sum_{j=\ell+1}^{i} \frac{t}{\beta^{k-j} N_k} \le \frac{O(1)}{N_k}\left( \omega^3+\frac{t}{\beta} \right)=O\left(\frac{t}{\beta N_k}\right)\!.\end{align}

This confirms (33).

We turn now to the solution of (19) and (20). Consider first $\mathrm{E}\, \widetilde L_k(t)$ . For all $t \ge 0$ , $\widehat{L_k}(t)=t$ , and $\sum_{j < k} \widehat L_j(t)=O(t/\beta)$ follows from (34). Thus, from (20)

\[\mathrm{E}\, \widetilde L_k(t)=t -\sum_{ s=0}^t O\left(\frac{s}{\beta N_k}\right)=t-O\left(\frac{t^2}{\beta N_k}\right)=t\left( 1-O\left(\frac{t}{\beta N_k}\right) \right)\!.\]

For $t\ge t_0$ where $t_0 \rightarrow \infty$ arbitrarily slowly we have $\widetilde L_k(t) \sim t$ w.h.p., initializing an induction for $\mathrm{E}\, \widetilde L_i(t)$ using arguments equivalent to those in Lemma 1 for $\mathrm{E}\,\widehat{L_i}(t)$ .

At step $t+1$ , (33) implies that, for the lower bounds on the process, with probability $(1-o(1))$ particle $t+1$ will arrive at level i. If $i<k$ , it halts at this level with probability $\widetilde L_{i+1}(t)/N_{i+1}$ . Thus,

(35) \begin{align}\mathrm{E}\,\widetilde L_i(t+1) = \mathrm{E}\,\widetilde L_i(t) + \left( 1-O\left(\frac{t}{\beta N_k}\right) \right)\cdot\frac{\mathrm{E}\, \widetilde L_{i+1}(t)}{N_{i+1}}. \end{align}

Arguing as in (8) on the inductive assumption that

\[\mathrm{E}\, \widetilde L_{i+1}(t)=\mu_{i+1}(t) \left( 1-O\left(\frac{t (k-(i+1) +1)}{\beta N_k}\right) \right)\!,\]

which is true when $i+1=k$ , using (35) we find

\begin{align*} \mathrm{E}\, \widetilde L_i(t)& = \frac{1}{N_{i+1}} \sum_{s=0}^{t-1} \mathrm{E}\, \widetilde L_{i+1}(s) \left( 1-O\left(\frac{s}{\beta N_k}\right) \right)\\ & = \frac{1}{N_{i+1}} \sum_{s=0}^{t-1} \mu_{i+1}(s)\left( 1-O\left(\frac{s(k-(i+1)+1)}{\beta N_k}\right) \right) \left( 1-O\left(\frac{s}{\beta N_k}\right) \right)\\ & = \frac{1}{N_{i+1}} \sum_{s=0}^{t-1} \frac{s^{k-i}}{(k-i)! N_k \cdots N_{i+2}} \left( 1-O\left(\frac{s (k-i+1)}{\beta N_k}\right) \right)\\ & = \mu_i(t)-\frac{t \mu_{i}(t)}{N_k} O\left(\frac{k-i+1}{\beta}\right)=\mu_i(t)\left( 1- O\left(\frac{(k-i+1) t}{\beta N_k}\right) \right).\end{align*}

Provided $t \le \omega T_f$ , which is true w.h.p. by Lemma 5, then $kt/\beta N_k \le k(\omega T_f/N_k)^2 =o(1)$ , see (24) and Lemma 4; in this case

(36) \begin{align}\mathrm{E}\, \widetilde L_i(t) \sim \mathrm{E}\, L_i(t) \sim \mathrm{E}\, \widehat{L_i}(t)\sim \mu_i(t)= \frac{t^{i+1}}{(i+1)! N_k \cdots N_{i+1}}\end{align}

as required.

Let $t_i(\omega)$ be given by (6). For those $i \le k$ , such that $t \ge t_i(\omega)$ , then $\mu_i(t) \rightarrow \infty$ suitably fast and the concentration results of Lemma 3 hold. The gaps inherited from the upper-bound argument of Section 3.1 are essentially unaltered.

This completes the proof of Proposition 1 for the equal layers model.

3.3. Finish time of DLA in the equal layers model

A lower bound on the finish time follows from the upper-blocked process, and an upper bound from the lower bound estimates for DLA. The range of k requires us to split the finish time analysis into two cases; k constant, and $k \rightarrow \infty$ with n.

Case k constant. Let $t'=t_1(0)/\omega'$ , where $\omega' \rightarrow \infty$ arbitrarily slowly. Then w.h.p., $\mathrm{E}\, \widehat{L_0}(t')\le \mu_0(t')=O(1/(\omega')^{k+1})$ and, thus, $\mathrm{P}(\widehat{L_0}(t')>0)=O(1/\omega')$ .

We next investigate the concentration of $L_1(T_f)$ . By Lemma 4(1), as $T_f=t_1(0)$ , $\mu_0(T_f)=1$ and, thus,

(37) \begin{align} \mu_1(T_f)=\mu_0(T_f) \frac{N_1 (k+2)}{T_f}\ge \frac{n}{\gamma_k n^{k/(k+1)}}=\Theta(1) n^{1/(k+1)} \gg 4 \omega^3,\end{align}

where $\omega=6 \log n$ as usual. Thus, $T_f \gg t_1(\omega)$ and, hence, Lemma 3 holds for $\widetilde L_1(T_f)$ . Suppose that $L_0(T_f)=0$ at step $T_f$ . By (37) the expected waiting time $\tau$ for a particle to hit $L_1(T_f)$ is

(38) \begin{align}\tau\le (1+o(1)) \frac{N_1}{\mu_1(T_f)}=\Theta(1) \frac n{k n^{1/(k+1)}}=\Theta(1) \frac{n^{k/(k+1)}}{k}=\Theta(T_f/k)=O(T_f).\end{align}

Thus, by step $\omega' T_f$ , w.h.p. $L_0(\omega'T_f)=1$ .

Case $k \rightarrow \infty$ with n. Let $t'=(1-\varepsilon) t_1(0)$ , where $0<\varepsilon<1$ constant, then $t'=C(1-\varepsilon)T_f$ and $\mu_0(t') \sim (1-\varepsilon)^{k+1}$ . Thus, $\mathrm{P}(\widehat{L_0}(t')>0)\le \mathrm{e}^{-\varepsilon k}$ and so w.h.p. $t_f>t'$ , lower bounding the finish time. On the other hand, for the upper bound, the inequality (37) still holds. By (38), the expected waiting time for the source to be occupied satisfies $\tau =\Theta(T_f/k)=o(T_f)$ . Taking $\varepsilon=o(1)$ we see that $t_f \sim CT_f$ for some constant C.

This completes the proof of Lemma 5 and, hence, Theorem 1(1).

3.4. Existence of a unique connecting path component

We now prove Theorem 2 for the equal layers model. We must show w.h.p. that at step $t_f$ , the finish time, the path of occupied vertices connecting the source to level k (and, hence, to the sink), has no off-path neighbours in $C_{t_f}$ .

At a given step t, the edge-induced component $C_t$ is obtained from $C_{t-1}$ by adding a newly occupied vertex which points to the neighbour in $C_{t-1}$ which halted the particle: Thus, a particle halts at vertex u in level i if it chooses an edge uw to an occupied neighbour w in level $i+1$ . We consider this edge uw as being directed from u to w in the component $C_t$ rooted at the sink. An arborescence is a rooted tree with all edges directed towards the root vertex. Thus $C_t$ is an arborescence with root z. On deletion of the z, $D_t=C_t\! \setminus \!\{z\}$ becomes a directed forest of arborescences each rooted at a vertex in level k.

Let $B_i$ be the set of occupied vertices in level i, where $L_i=|B_i|$ . Given that a particle at u chooses a vertex in $B_{i+1}$ as a neighbour, then this neighbour is chosen uniformly at random (u.a.r.) from the set $B_{i+1}$ .

We regard vertices occupied by halted particles as coloured either red or blue, with all occupied vertices in level k coloured blue. If u is the first in-neighbour of w then u is coloured blue. If however w already has an in-neighbour $u'$ then u, $u'$ and all other in-neighbours are (re)coloured red. At any step, the red vertices in a level are those with siblings and the blue vertices are the unique in-neighbour of some vertex in the next level. The choice of w by the particle at u is independent of the colour of w at this step.

The process halts when there is a directed path of occupied vertices $\nu=u_0 u_1\cdots u_kz$ from the source to sink. The source vertex v is blue at $t_f$ as it is the first in-neighbour of $u_1$ .

Proof. As before, let $B_i$ be the set of occupied vertices in level i, and $L_i=|B_i|$ . The subset Out $(B_i)$ of $B_{i+1}$ with at least one in-neighbour in $B_i$ has size at most $L_i$ . This is because $u \in B_i$ has a unique occupied out-neighbour in $C_t$ .

Let ${\unicode{x1D7D9}}_{\{{k-j,s}\}}$ be the indicator that particle s halts in level $k-j$ and is coloured red due to a pre-existing sibling. In this case particle s has chosen an out-neighbour in the existing set Out $(B_{k-j}) \subseteq B_{k-j+1}$ and, thus, as $|\mathrm{Out}(B_{k-j})| \le \widehat L_{k-j}(s)$ ,

\[\mathrm{E}\,{\unicode{x1D7D9}}_{\{{k-j,s}\}} \le \frac{\mathrm{E}\, \widehat L_{k-j}(s)}{(n/k)} \sim \frac{\mu_{k-j}(s)}{(n/k)}.\]

Let $Z_{k-j}(t)$ be the number of red vertices in level $k-j$ . We associate the number of possibly red vertices at each step with a super-process $\widehat Z_{k-j}(t)=Z_{k-j}(t)+Q_{k-j}(t)$ , where $Q_{k-j}(t)$ is Bernoulli with parameter $(\widehat L_{k-j}-|(\mathrm{Out})B_{k-j}|)$ . Thus, $Z_{k-j}(t) \le \widehat Z_{k-j}(t)$ and

\[\mathrm{E}\, \big(\widehat Z_{k-j}(t+1) \mid \widehat{\mathcal{H}}(t)\big)=\widehat Z_{k-j}(t)+\frac{\widehat L_{k-j}(t)}{N_{k-j}}.\]

The recurrence for $\mathrm{E}\, \widehat Z$ mirrors the recurrence (4) for $\widehat L_{i}(t)$ with $k-j$ replacing $i+1$ . The variable $\widehat Z_{k-j}(t)$ occurs in place of $\widehat{L_i}(t)$ , and will grow as $\widehat L_{k-(\kern1.5pt j+1)}(t)$ . Specifically,

(39) \begin{align}\mathrm{E}\, \widehat Z_{k-j}(t) & \sim \frac 1{(n/k)}\sum_{s=1}^t{\mu_{k-j}(s)}= \frac{1}{(n/k)} \sum_{s=1}^t \frac{s^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)! (n/k)^j} \nonumber\\& \sim \frac{t^{ \kern1.5pt j+2}}{(\kern1.5pt j+2)! (n/k)^{\kern1.5pt j+1}}= \mu_{k-(\kern1.5pt j+1)}(t). \end{align}

We note that $\widehat Z_{k-j}(t)$ is the sum of independent indicator variables, and will be concentrated for $t \ge t_{k-(\kern1.5pt j+1)}(\omega)$ , where $\omega=6 \log n$ as before. The total number of red vertices is at most $2Z_{k-j}(t) \le 2 \widehat Z_{k-j}(t)$ , where the factor 2 covers the case where the pre-existing sibling was blue but is recoloured red.

Assume $T_f/\omega' \le t_f \le \omega' T_f$ , where $\omega'$ from Lemma 5 tends to infinity arbitrarily slowly with n. Denote the path connecting $\nu=u_0$ to level k by $u_0 u_1 u_2\cdots u_k$ , where by definition the source $\nu=u_0$ is blue at $t_f$ . For $i \ge 1$ , let $R_i(s)$ be the red vertices in level i at step s. Let $s_i$ denote the step at which $u_i$ became occupied. Consider the colour of $u_1$ at $t_f$ . As $T_f= t_1(0)$ , and $t_f \ge T_f/\omega' \gg t_1(\omega)$ by the gap property, so we have that $L_1(t_f) \sim \mu_1(t_f)$ . Let $t=\omega' T_f$ , be a w.h.p. upper bound on $t_f$ , then by (3) $\mu_0(t)=(\omega')^{k+1}$ . On the other hand, $\widehat Z_1(t_f)$ may not be concentrated, but we can assume $\widehat Z_1(t_f) \le \omega \mu_0(t)$ where $\omega=6 \log n$ will be large enough to use in the Markov inequality for $k=o(\sqrt{\log n})$ levels.

Vertex $u_1$ was chosen u.a.r. from the occupied vertices in level one by the particle halting at $u_0$ , thus,

(40) \begin{align}\mathrm{P}( u_1 \in R_1(t)) \le \frac{2\widehat Z_1(t)}{L_1(t)} \le \frac{2 \omega \mu_0(t)}{\mu_1(t)} \le \frac{2 \omega t}{(k+1)(n/k)}\le \frac{2 \omega t}{n}.\end{align}

Next consider the colour of $u_2$ . If $u_2$ is red at step t then either (i) it became red before step $s_1$ when it was chosen u.a.r. by $u_1$ or (ii) it became red at some later step.

In the first case,

\[\mathrm{P}(u_2 \in R_2(s_1))\le\frac{2\widehat Z_2(s_1)}{L_2(s_1)} \le 2\omega\frac{\mu_1(s_1)}{\mu_2(s_1)}=\frac{2\omega s_1}{k(n/k)}=\frac{2 \omega s_1}{n}.\]

By the gap property, $L_2(s_1)$ is already concentrated, for otherwise $L_1(s_1)$ would be empty, a contradiction. The $\omega$ covers possible lack of concentration of $\widehat Z_2(s_1)$ as explained previously.

In the second case, as $u_3$ is the chosen out-neighbour of $u_2$ , then $u_2$ will turn red if some particle chooses $u_3$ after time $s_1$ , and thus

\[\mathrm{P}(u_2 \in R_2(t) \! \setminus \! R_2(s_1)) \le \frac{1}{(n/k)} \sum_{\tau=s_1+1}^t \mathrm{E}\, \,{\unicode{x1D7D9}}_{\{{u_3 \text{ chosen at }\tau}\}}=\frac{t-s_1}{(n/k)}.\]

Thus

\[\mathrm{P}(u_2 \in R_2(t)) \le \frac{2\omega kt}{n},\]

and similarly for vertices $u_3,\ldots,u_{k-1}$ . Thus

\[\mathrm{P}(\text{path } vu_1\cdots u_k \text{ is blue}) \ge 1-\sum_{i=1}^{k-1} \frac{2\omega kt}{n}=1-O\left(\frac{\omega k^2t}{n}\right).\]

By Lemma 5, $t= \omega' T_f \le \omega' n^{k/(k+1)}$ , where $\omega' \rightarrow \infty$ arbitrarily slowly. As $k \le \sqrt{\log n}/ \log \log n$ , then $\omega k^2 \omega' T_f/n \le \omega^3/n^{1/(k+1)}=o(1)$ , by upper bounding $\omega' \le \log \log n$ . Thus the path from the source to vertex $w=u_k$ level k is blue w.h.p., and thus the arborescence rooted at w is exactly this path.

4.1. An important level

We next show the existence of a level $i\sim k+1-\sqrt{2k+2}$ , such that the first occupancy of this level effectively determines the finish time of the process.

From (43),

(44) \begin{align}\mu_{k-j}(t_1)=\frac{t_1^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)! d^{kj-j(\kern1.5pt j-1)/2}}= 1 \quad \implies \quad t_1(k-j)=[(\kern1.5pt j+1)!]^{1/(\kern1.5pt j+1)} d^{\frac{ 2kj-j(\kern1.5pt j-1)}{2(\kern1.5pt j+1)}}.\end{align}

Before proceeding we note that rounding $t_1(k-j)$ up or down to the nearest step has negligible effect on $\mu_{k-j}(t_1)$ , as asserted in the note under (6). Indeed

(45) \begin{align} \max_{j \ge 1} \frac{j}{t_1(k-j)} \le \frac{k}{t_1(k-1)}=\frac{k}{d^{k/2}}=o(1).\end{align}

We next determine the maximum value of $t_1(k-j)$ . From (3), for any t,

(46) \begin{align}\frac{\mu_{k-j+1}(t)}{\mu_{k-j}(t)}=\frac{(\kern1.5pt j+1)d^{k-j+1}}{t},\end{align}

so setting $\mu_{k-j}(t_1(k-j))= 1$ gives

(47) \begin{align}\mu_{k-j+1}(t_1) & = \frac{j+1}{[(\kern1.5pt j+1)!]^{1/j+1}} d^{k-j+1-\frac{ 2kj-j(\kern1.5pt j-1)}{2(\kern1.5pt j+1)}}\nonumber\\& = \frac{\Theta(1)}{(2\pi/e (\kern1.5pt j+1))^{1/2(\kern1.5pt j+1)}} d^{\frac{2k+2-j(\kern1.5pt j+1)}{2(\kern1.5pt j+1)}}.\end{align}

The leading term on the right-hand side of (47) is bounded, and the exponent of d on the right-hand side is positive provided $2k+2 > j(\kern1.5pt j+1)$ , which ensures that $\mathrm{E}\, \widehat L_{k-j+1}(t)$ is sufficiently large close to $t_1(k-j)$ .

What value of $k-j$ maximizes the value of $t_1=t_1(k-j)$ at which $\mu_{k-j}(t_1)= 1$ ? Write the exponent of d on the right-hand side of (44) as $f(\kern1.5pt j)/2$ where

(48) \begin{align}f(\kern1.5pt j)=\frac{2kj-j(\kern1.5pt j-1)}{(\kern1.5pt j+1)}=(2k+2) -j -\frac{2k+2}{j+1}.\end{align}

The maximum of f(j) occurs at $j^*$ with solution

(49) \begin{align} (\kern1.5pt j^*+1)^2=(2k+2), \quad \text{giving } f(\kern1.5pt j^*)=(\kern1.5pt j^*)^2.\end{align}

The (not necessarily integer) values of $j^*$ , $k-j^*$ and $d^{f(\kern1.5pt j^*)/2}$ are

(50) \begin{align}j^*=\sqrt{2k+2}-1, \quad k-j^*=k+1-\sqrt{2k+2}, \quad d^{f(\kern1.5pt j^*)/2}=d^{k+3/2-\sqrt{2k+2}}.\end{align}

In the case where $j^*$ is not an integer, the effect of the rounding error is addressed in Section 4.4.

Ignoring rounding effects, we evaluate $t_1=t_1(k-j)$ at $j=j^*$ , in which case $f(\kern1.5pt j)=j^2$ , to find

(51) \begin{align}t_1 & = [(\kern1.5pt j+1)!]^{1/(\kern1.5pt j+1)} d^{\,f(\kern1.5pt j)/2}\nonumber\\& \sim \mathrm{e}^{-1}(\sqrt{2\pi})^{1/(\kern1.5pt j+1)} (\kern1.5pt j+1)^{1+1/2(\kern1.5pt j+1)} d^{ \kern1.5pt j^2/2}\nonumber\\& = C_{k-j} \sqrt{2k+2} d^{k+3/2-\sqrt{2k+2}},\end{align}

on inserting the values from (50), and where $ C_{k-j}=\mathrm{e}^{-1}(1+O(1/j))$ . Note that $t_1=\Theta( T_f)$ . From (46),

\[\frac{\mu_{k-(\kern1.5pt j+1)}(t)}{\mu_{k-j}(t)}=\frac{t}{(\kern1.5pt j+2)d^{k-j}}\]

so that at $t_1$ , for some $C,C'=\Theta(1)$ ,

(52) \begin{align}\mu_{k-j^*}(t_1)= 1, \quad \mu_{k-(\kern1.5pt j^*-1)}(t_1)=Cd^{1/2}, \quad \mu_{k-(\kern1.5pt j^*+1)}(t_1)=C'd^{1/2}.\end{align}

At first this seems confusing, as one might expect to have $ \mu_{k-(\kern1.5pt j^*+1)}(t_1)=o(1)$ by analogy with the equal layers model. Assuming $d^{1/2} \rightarrow \infty$ , level $k-j^*+1$ is the last level at which the condition $\mu_{k-j+1} \rightarrow \infty$ is valid in the recurrence from $k-j+1$ to $k-j$ given in (4). Indeed $j^*-1=\sqrt{2k+2}-2$ is the last value of j for which we claim Proposition 1 holds.

Finally, we need to check $t_1(k-j+1) \ll t_1(k-j)$ for all $j\le j^*-1$ as required in Lemma 1(2). Using (44)

(53) \begin{align}\frac{t_1(k-j)}{t_1(k-j+1)} & = \frac{[(\kern1.5pt j+1)!]^{1/(\kern1.5pt j+1)}}{[j!]^{1/j}}d^{\left( \frac{2kj-j(\kern1.5pt j-1)}{2(\kern1.5pt j+1)}-\frac{2k(\kern1.5pt j-1)-(\kern1.5pt j-1)(\kern1.5pt j-2)}{2j} \right)}\nonumber\\& = \Theta(1)d^{\frac{2k-(\kern1.5pt j-1)(\kern1.5pt j+2)}{2j(\kern1.5pt j+1)}}\nonumber\\& \ge \Theta(1) d^{\frac{3\sqrt{2k+2}-2}{2(2k-3\sqrt{2k+2}+4)}}=\Theta(d^{ 3/\sqrt{8 k}}).\end{align}

The last line is evaluated at $j = j^*-1=\sqrt{2k+2}-2$ , and this is the largest value of j for which Proposition 1 is claimed to hold. By Theorem 1(2), $k\le ((\log d)/( \log \log d))^2$ so the right-hand side of (53) is $\Theta((\log d)^{3/\sqrt{8}})$ which tends to infinity with d, thus validating the of concentration condition of Lemma 1(2) for $\widehat L_{(k-j^*-1)}(t)$ .

4.2. Gap property of $\widehat {\boldsymbol{L}}$

Let $\omega= 6 \log n$ as before. From the definition of $t_1=t_1(k-j^*)$ , at $t_1/\omega$ , we have $\mu_{k-j^*}(t_1/\omega)=O(\omega^{-(\kern1.5pt j^*+1)})$ , where $j^* \sim \sqrt{2k} \rightarrow \infty$ . Thus, w.h.p. $\widehat L_{k-j^*}(t_1/\omega)=o(1)$ in the upper process. Consequently all levels $i=k-\ell$ , $\ell \ge j^*$ , have $\widehat{L_i}(t)=0$ , w.h.p., for $t \le t_1/\omega$ .

In what follows we only consider gaps between indices $k-\ell$ where $0 \le \ell \le k-j^*+1$ .

By (50) we have $k-j^*+1=k+2-\sqrt{2k+2}$ . We see from (52) that $\mu_{k-j^*+1}(t_1(k-j^*))=Cd^{1/2}$ . As $d \rightarrow \infty$ , then $\widehat L_{k-j^*+1}$ will be concentrated around $\mu_{k-j^*+1}(t_1)$ , although we cannot expect it to be as strong as in Lemma 3. Fortunately this will not matter as $k-j^*$ is the last level to which we apply a gap argument. For $\ell \le j^*-1$ at $t_1(k-\ell)$ , the value of $\widehat L_{k-\ell+1}$ obeys Lemma 3. In particular, it can be checked that $t_1(k-j^*+1)=\Theta(1) \sqrt k d^{k+5/2-(3/2)\sqrt{2k+2}}$ . Using (46), we obtain $\mu_{k-j^*+2}(t_1(k-j^*+1)) =\Theta(d^{(\sqrt{2k+2}+1)/2})$ , establishing a gap property between $\widehat L_{k-j^*+1}$ and $\widehat L_{k-j^*+2}$ ; the situation being better for smaller $\ell$ .

4.3. Lower bound on L

Turning to the lower bound $\widetilde L$ as given in (19) and (20), we need to prove that (33) holds. This requires that $\sum_{j=0}^iL^*_j(t)/N_j=o(1)$ for $1 \le i \le k-1$ and $t \le t_f$ . We only consider this sum up to the time (in order of magnitude) when level $k-j^*$ first becomes occupied, and thus levels $0 \le i \le k-j^*-1$ are empty.

The main task is to choose a value of $\beta$ for the growing layers model analogous to (25) which we can use with the arguments given in Section 3.2 for the equal layers model. There is not so much room for manoeuvre here if we do not want to restrict the value of d. Let $\varpi$ to be a function tending to infinity more slowly than $d^{1/4}$ , and assume $t_f \le \varpi T_f$ ; a value which satisfies Proposition 2. We choose $\beta=\sqrt d/\varpi$ . As we assume $d \rightarrow \infty$ , then also $\beta \rightarrow \infty$ .

For $1 \le \ell \le j^*-1$ , we use (41), (43), and $N_{k-\ell}=d^{k-\ell}$ . As $d^\ell/(\ell+1)$ is monotone increasing, the worst case is $\ell=j^*-1$ , so, assuming $t_f\le \varpi T_f$

(54) \begin{align}\frac{\mu_{k-\ell}}{N_{k-\ell}} & = \frac{t}{(\ell+1)d^{k-\ell}}\frac{\mu_{k-\ell+1}}{N_{k-\ell+1}} \\[-12pt] \nonumber \end{align}

(55) \begin{align}& \le \frac{\mu_{k-\ell+1}}{N_{k-\ell+1}} \frac{\varpi \sqrt{k}}{ \sqrt{2k+2}-1} \frac{d^{k+3/2-\sqrt{2k+2}}}{d^{k+2-\sqrt{2k+2}}} \nonumber\\& \le \frac{\varpi }{\sqrt{d}} \frac{\mu_{k-\ell+1}}{N_{k-\ell+1}} = \frac 1{\beta}\frac{\mu_{k-\ell+1}}{N_{k-\ell+1}}\le \left(\frac{1}{\beta}\right)^\ell \frac{t}{N_k}=o(1), \\[9pt] \nonumber \end{align}

as $\varpi T_f/d^k=o(1)$ and $\beta= \sqrt{d}/\varpi$ . Equation (34) becomes

\begin{align*}\sum_{\ell=j^*+1}^i \frac{L^*_{k-\ell}}{N_{k-\ell}} & \le \frac{O(\omega^3)}{N_{k-j^*}}+2 \sum_{\ell=j^*+1}^i\left(\frac{1}{\beta}\right)^\ell \frac{t}{N_k}= \frac{O(\omega^3)}{N_{k-j^*}}+O \left(\frac{t}{\beta^i N_k}\right)\\& \le \left(\frac{\log^3 n}{n^{1-o(1)}}\right) + \left(\frac{ \varpi T_f}{ d^k}\right)\\& \le O\left(\frac{ \varpi \sqrt k}{d^{\sqrt{2k +2}-3/2}}\right)=O\left(\frac{\sqrt k}{d^{\sqrt{2k +2}-2}}\right)=o(1).\end{align*}

The $\omega^3$ term comes from using Lemma 3 to bound the size of $L^*_{k-\ell}(t)$ for $t\le t_{k-\ell}(\omega)$ . The last line follows on assuming $k\le (\log d/\log \log d)^2$ .

It now follows from the proof in Section 3.2 that if $k-j^*+1 \le i \le k$ and $t \ge t_i(\omega)$ then $\mu_i(t) \rightarrow \infty$ suitably fast. We hence obtain that $\mathrm{E}\, \widetilde L_i(t) \sim \mu_i(t) \sim \mathrm{E}\, \widehat{L_i}(t)$ , and so we have $L_{k-\ell}(t) \sim \widehat L_{k-\ell}(t)\sim \mu_{k-\ell}(t)$ .

4.4. The finish time in the growing layers model

We now turn to the proof of (41). We first give a bound ignoring rounding errors on the value of $j^*$ , and subsequently give a correction. We show that, w.h.p., a path (of occupied vertices) grows back to the source from the first vertex to be occupied in level $k-j^*$ , thus halting the process; and, moreover, this occurs before a second vertex becomes occupied in level $k-j^*$ .

4.4.1. Proof ignoring rounding

Using (51) we know that $t_1(k-j^*)= C T_f$ for some constant C. Let $t'=(1-\varepsilon)t_1(k-j^*)$ , then arguing as in Proposition 5, we have that at $t'$ , as by assumption $k \rightarrow \infty$ ,

\[\mathrm{E}\, \widehat L_{k-j^*}(t') \le \mu_{k-j^*}(t') \sim (1-\varepsilon)^{k+1} \le \mathrm{e}^{-\varepsilon k}=o(1).\]

Thus, at $t'=C'T_f$ , w.h.p. level $k-j^*$ is unoccupied.

Let $t_0$ be the first step t for which $L_{k-j^*}(t) =1$ , where w.h.p. $t_0 \ge t'$ . Then either $t_0 \le t_1(k-j^*)$ , or if not we upper bound $t_0$ as follows. Consider the probability $\phi$ that a particle halts at level $k-j^*$ . This is given by

\[\phi=\frac{L_{k-(\kern1.5pt j^*-1)}(t_1)}{N_{k-(\kern1.5pt j^*-1)}} \sim\frac{C d^{1/2}}{d^{k-j^*+1}}=\frac{C}{d^{k+3/2-\sqrt{2k+2}}}.\]

Then the probability no particle halts there in a further $t_1(k-j^*)=O(T_f)$ steps is (see (51))

\[(1-\phi)^{t_1} \le \mathrm{e}^{-t_1\phi}= \mathrm{e}^{-C'\sqrt{2k+2}}=o(1),\]

where $C'\sim C/e$ and we assume $k \rightarrow \infty$ .

Let u be the vertex in level $k-j^*$ containing the unique particle halted at $t_0$ . Construct a path back from u to the source as follows. Wait until a particle halts at $w_{k-j^*-1}$ in level $k-j^*-1$ by choosing edge $w_{k-j^*-1}u$ . The expected time for this is $d^{k-j^*}$ . In a further expected time $d^{k-j^*-1}$ , the path will extend backwards, as a particle will halt in level $k-j^*-2$ by choosing an edge to $w_{k-j^*-1}$ , etc. Thus, in a further

\[T=d^{k-j^*}+ d^{k-j^*-1}+ \cdots +d= d^{k-j^*} \left(\frac{1-1/d^{k-j^*}}{1-1/d}\right)= \Theta(d^{k-j^*})=\Theta \big(d^{k+1-\sqrt{2k+2}}\big)\]

expected steps there will be a path $vw_1\cdots w_{k-j^*-1}u$ of halted particles extending from the source v to vertex u, thus stopping the DLA process (if it has not already halted).

The value of T given above is $O(T_f/\sqrt{kd})$ , and we conclude from the Markov inequality that the source will be occupied in at most $t_0+(kd)^{1/4}T= C'' T_f$ steps, with probability $1-O(1/(kd)^{1/4})$ . This completes the proof that $t_f=\Theta(T_f)$ w.h.p.

We next give the proof of Proposition 2(2). The expected number of steps (after $t_0$ ) needed to create another halted particle in level $k-j^*$ is

\[\frac 1\phi= \Theta \big(d^{k+3/2-\sqrt{2k+2}}\big)= \Theta \big(T d^{1/2} \big),\]

whereas, w.h.p., on the assumption that $d \rightarrow \infty$ , it is in at most $(kd)^{1/4}T$ further steps after $t_0$ the process has halted as claimed. This occurs before a second vertex can become occupied in level $k-j^*$ .

4.4.2. The effect of rounding error on $T_f$ in the growing layers model

Recall that $j^*$ is the solution to $(\kern1.5pt j^*+1)^2=(2k+2)$ as given in (49). Integer solutions exist for some values of k. To see this, let a be a positive integer, and put $j^*+1=2a$ so that $k=2a^2-1$ ; this holds for example for $k=1,5,17,31$ .

In the case that $j^*$ is not an integer, we require the value of j given by $\lfloor j^* \rfloor$ or $ \lceil j^* \rceil$ which maximizes $t_1(k-j)$ . Expand f(j) in (48) about $j^*$ , giving

\begin{align*}f(\kern1.5pt j^*+h) &= f(\kern1.5pt j^*)+h f'(\kern1.5pt j^*)+\frac{h^2}{2} f''(\kern1.5pt j^*+\theta h)\\& = (\kern1.5pt j^*)^2 +0 -h^2 \frac{2k+2}{(\kern1.5pt j^*+1+ \theta h)^3}\\& = (\kern1.5pt j^*)^2 -\frac{h^2}{\sqrt{2k+2}}\left( 1+O\left(\frac{h}{(\kern1.5pt j^*)^3}\right) \right).\end{align*}

Let $\varepsilon=\min \left( j^*-\lfloor j^* \rfloor, \lceil j^* \rceil-j^* \right)$ , then by the above $T_f$ is altered to $T_f(\varepsilon)=T_fd^{-\varepsilon^2/\sqrt{2k+2}(1+o(1))}$ , other arguments about $t_f$ being unchanged.

5.1. Trees with large branching factor: proof of Theorem 3(1)

Let $G=G(k,d)$ be a labelled tree with branching factor d and final level k, where $d^k=n$ . The source of the particles is the unique vertex v at level zero. An artificial sink vertex z placed at level $k+1$ is attached to the vertices at the final level k.

We borrow several ideas from the growing layers model. Let $j^*=\sqrt{2k+2}-1$ as in (50), and as in (41), let $T_f$ be given by

(56) \begin{align}T_f= \sqrt{k} d^{k+3/2-\sqrt{2k+2}}.\end{align}

We first ignore rounding on $j^*$ and prove w.h.p. that $t_f=\Theta(T_f)$ . Correcting for rounding then gives Theorem 3(1).

By symmetry, the expected number of particle walks arriving at a given vertex in level $k-j^*$ by step $T_f$ , is $T_f/d^{(k-j^*)}=\sqrt{kd}$ . However, the expected number arriving at a vertex w at level $k-j^*+1$ is

(57) \begin{align}\frac{T_f}{d^{k-j^*+1}}= \sqrt{\frac{k}{d}}=o(1),\end{align}

provided $k \ll d$ , which we assume to be true. In order for w to be occupied, the subtree rooted at w must contain a path from w to level k on which all $j^*$ vertices are occupied. By the Markov inequality, the event that $j^*=\Theta(\sqrt k)$ particles arrive at w by step $T_ f$ , has probability $O(1/\sqrt d)$ . Thus, there should be few vertices in level $k-j^*+1$ with such a path to level k containing $j^*$ halted particles.

Consider t particles percolating downwards from the root of the infinite d-ary tree $G(d,\infty)$ . Particle s starts at step s and each particle transitions one edge at each step, so that they never collide.

We approximate the distribution of the number of particles arriving at a vertex w of layer $k-j$ in t steps, where eventually we assume $j \le j^*-1$ . Let $X_w=X_w(k-j,\ell,t)$ be the indicator for the event that exactly $\ell$ particles arrive at vertex w of layer $k-j$ at or before step t, then

(58) \begin{align}\mathrm{E}\, X_w(k-j,\ell,t)= \left(\begin{array}{l}t \\ \ell\end{array}\right) \left(\frac{1}{d^{k-j}}\right)^\ell \left( 1-\frac{1}{d^{k-j}} \right)^{t-\ell}.\end{align}

Let $X=X(k-j,\ell,t)$ where $X=\sum_{w \in S_{k-j}}X_w$ is the number of vertices at level ${k-j}$ which have exactly $\ell$ particles reaching them. Thus $\mathrm{E}\, X= d^{k-j} \mathrm{E}\, X_w$ , and

\begin{align*}\mathrm{E}\, X(X-1) & = \big(d^{k-j}\big) \big(d^{k-j}-1 \big) \left(\begin{array}{l}t \\ 2\ell\end{array}\right) { \left(\begin{array}{l}2\ell \\ \ell\end{array}\right)} \left(\frac{1}{d^{k-j}}\right)^\ell\left(\frac{1}{d^{k-j}-1}\right)^\ell \left( 1-\frac{2}{d^{k-j}} \right)^{t-2\ell}\\& = \left( 1+O\left(\frac{\ell}{d^{k-j}}\right)- O\left(\frac{t}{d^{2(k-j)}}\right) -O\left(\frac{\ell^2}{t}\right) \right) (\mathrm{E}\, X)^2,\end{align*}

and so

\[\mathrm{Var}\, X=\mathrm{E}\, X+ O\left(\frac{\ell}{d^{k-j}}\right) (\mathrm{E}\, X)^2.\]

Next let $\ell=j+1$ so that $X(k-j,\ell,t)=X(k-j,j+1,t)$ . As $\mathrm{E}\, X= d^{k-j} \mathrm{E}\, X_w$ , using (58),

\[\mathrm{E}\, X = \frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!} \left(\frac{1}{d^{k-j}}\right)^{ \kern1.5pt j}\left( 1-O\left(\frac{j^2}{t}\right) \right)\left( 1-O\left(\frac{t-j}{d^{k-j}}\right) \right).\]

In what follows we choose $j=j^*-1$ and $k^2 \le t \le CT_f$ where $C=o(\sqrt{d/k})$ so that, by (57), $t/d^{k-j^*+1}\le C\sqrt{k/d}=o(1)$ . In addition, $(\kern1.5pt j^*)^2/t=O(1/k)=o(1)$ .

With these choices, it will be the case that

(59) \begin{align}\mathrm{E}\, X \sim \frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!} \left(\frac{1}{d^{k-j}}\right)^{ \kern1.5pt j}.\end{align}

Furthermore, again by (57)

\[\frac j{d^{k-j}}\mathrm{E}\, X \sim \frac 1{j!} \left(\frac{t}{d^{k-j}}\right)^{\kern1.5pt j+1}=o(1),\]

so that

(60) \begin{align}\mathrm{Var}\, X=\mathrm{E}\, X+ O\left(\frac{j}{d^{k-j}}\right) (\mathrm{E}\, X)^2=(1+o(1))\mathrm{E}\, X.\end{align}

This being the case, by the Chebychev inequality we have that w.h.p. $X\sim \mathrm{E}\, X$ .

Continuing our discussion in the infinite d-ary tree $G(d,\infty)$ , let w be a vertex in level $k-j$ of $G(d,\infty)$ . Let u be a vertex at level k contained in the subtree $R_w$ rooted at w, and let H(w, u) be the unique path $w=w_0w_1\cdots w_j=u$ from w in level $k-j$ to u in level k. Let $1 \le s_0\le s_1 \le s_2 \le \cdots \le s_j \le t$ be the labels of particles which have arrived at w at these steps. Assume that on entering the subtree $R_w$ , the particles behave as follows. Particles $s_0$ and $s_1$ transition all edges of H, particle $s_2$ transitions (at least) edges $w_0w_1\cdots w_{j-1}$ ; in general, $s_{i}$ transitions (at least) $w_0w_1\cdots w_{j-i+1}$ , and $s_{j}$ transitions $w_0w_1$ . The probability of this is

(61) \begin{align}P(s_0,s_1,\ldots,s_j)= \frac 1{d^j} \frac 1{dd^2\cdots d^j}.\end{align}

As there are $d^j$ vertices at level k which belong to the subtree $R_w$ rooted at w, then provided exactly $j+1$ particles have arrived at w, the probability $P_w(\kern1.5pt j+1)$ that such a transition ordering exists is

\[P_w(\kern1.5pt j+1)= \frac 1{dd^2\cdots d^j}=\frac{1}{d^{ \kern1.5pt j(\kern1.5pt j+1)/2}}.\]

For distinct vertices w, $w'$ in level $k-j$ which contribute to X (i.e. exactly $j+1$ particles enter the given vertex in the given time period), the subsequent ordered transition events which occur in the subtrees $R_w$ and $R_{w'}$ are independent. Let $Z(k-j,j+1,t)$ be the number of vertices in level $k-j$ of $G(d,\infty)$ for which exactly $j+1$ particles enter the vertex and which satisfy the ordering. Then $\mathrm{E}\, Z= P_w(\kern1.5pt j+1) \mathrm{E}\, X$ and

(62) \begin{align}\mathrm{E}\, Z(k-j,j+1,t)\sim \frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!} \left(\frac{1}{d^{k-j}}\right)^{ \kern1.5pt j}\frac{1}{d^{ \kern1.5pt j(\kern1.5pt j+1)/2}}=\frac{t^{ \kern1.5pt j+1}}{(\kern1.5pt j+1)!} \frac{1}{d^{kj-j(\kern1.5pt j-1)/2}}= \mu_{k-j}(t),\end{align}

where $ \mu_{k-j}(t)$ is given by (43). We choose $j=j^*-1$ in $P_w(\kern1.5pt j+1)$ , and use the values of $\mathrm{E}\, X$ and $\textbf{Var} X$ derived via (59) and (60). As remarked previously, with these values, w.h.p. $X \sim \mathrm{E}\, X$ . Thus, as long as $ \mu_{k-j}(t)$ is sufficiently large, w.h.p. $Z \sim \mu_{k-j}(t)$ .

Return now to the finite Cayley tree G(k, d). For some vertex in level $k-j$ to be occupied at step t there must be some sequence $(s_0,s_1,\ldots,s_j)$ , which satisfies the ordered transition construction given above for $G(d,\infty)$ . Indeed, on contact with the sink z, particle $s_0$ will halt in level k, causing $s_1$ to halt in level $k-1$ and so on, until $s_j$ halts in level $k-j$ .

Say a path of occupied vertices from vertex w to level k is an exact path, if it is the unique occupied path in the subtree $R_w$ rooted at w. Thus, the paths from level $k-j$ to level k counted by Z are exact in G(d, k). Informally, as t tends to $ T_f$ , w.h.p. there will be many exact paths at level $k-j=k-j^*+1$ waiting to extend backwards towards the source.

5.1.1. Lower bound on $t_f$

The above calculations are used to establish an upper bound on the finish time. For a lower bound, a similar but cruder approach can be used. The total number of $j+1$ tuples which percolate a given vertex in level $k-j$ and form an ordered transition event in $G(d,\infty)$ is

(63) \begin{align}d^{k-j} \left(\begin{array}{l}t \\ j+1\end{array}\right) \left(\frac{1}{d^{k-j}}\right)^{\kern1.5pt j+1} \frac 1{d^{ \kern1.5pt j(\kern1.5pt j+1)/2}} \sim \mu_{k-j}(t).\end{align}

Here $\mu_{k-j}(t)$ is as defined in (43); this latter calculation being made without any regard for exactness.

Let $j=j^*$ and $t'=(1-\varepsilon)t_1(k-j^*)$ where $\varepsilon$ is a small positive constant. Using (63), and as by definition $\mu_{k-j^*}(t_1(k-j^*))\sim 1$ , at $t'$ the expected number of paths between level $k-j^*$ and level k consisting of halted particles is upper bounded by

(64) \begin{align} \mu_{k-j^*}(t')={(1-\varepsilon)^{ \kern1.5pt j^*+1}}\mu_{k-j^*}(t_1(k-j^*))\le \mathrm{e}^{-\varepsilon j^*}=o(1).\end{align}

We conclude that at $t'$ no such path exists w.h.p. and levels $0,1,\ldots,k-j^*$ are empty. By (51), $t_1(k-j^*) =C T_f$ , for some constant $C>0$ , so w.h.p. the finish time $t_f \ge t' \ge C' T_f$ , for some constant C’.

Note. This calculation is valid for any $d \ge 2$ , and will be reused in the next section for the case where d is finite.

5.1.2. Upper bound on $t_f$

Let $t_1=t_1(k-j^*)$ . Using (52), $\mu_{k-j^*+1}(t_1)=C_1 \sqrt d$ , for some positive constant $C_1$ . Thus, w.h.p., $Z(k-j^*+1, j^*,t')=\Theta(\sqrt d)$ which is large, provided $d\rightarrow \infty$ .

A path of occupied vertices back from the sink to level i, is a leading path at level i if all vertices at level $i-1$ are unoccupied. If we increase t above $t_1$ , then either some leading path is already at level $i\le k-j^*$ , or level $k-j^*$ is still unoccupied. If the latter, how long does it take in $G(d,\infty)$ for a particle to hit a vertex with an exact path in level $k-j^*+1$ as counted by $Z(k-j^*+1, j^*,t_1)$ ? The hitting probability of such vertices at a given step is $p_h=Z(t_1)/d^{k-j^*+1}$ , so

(65) \begin{align}\mathrm{E}\, T(\text{wait})=\frac 1p_h \sim \frac{d^{k-j^*+1}}{\mathrm{E}\, Z(t_1)} =\Theta\left(\frac{T_f}{\sqrt k}\right)= o(T_f),\end{align}

assuming $k \rightarrow \infty$ . Thus, the probability we wait a further $T_f$ steps is o(1). Once a leading path is at $i \le k-j^*$ the same waiting time argument as used in Section 4.4 applies, the expected waiting times reducing at each step. In conclusion, w.h.p. the source vertex is now occupied in $\Theta(T_f)$ steps after $t_1$ , completing the proof of Theorem 3(1).

Note. We found $t_f$ on the assumption that $j^*$ is an integer. In the case that it is not an integer, we use the rounding correction of (at most) $d^{O(1/\sqrt k)}$ derived previously for the growing layers model. The value of $t_f \le CT_f$ used above with $C=o(\sqrt{d/k})$ is large enough to absorb any rounding errors.

5.2. Trees with constant or slowly growing branching factor: proof of Theorem 3(2)

In the previous section we assumed that $k \ll d$ . In this section the assumption $k \ll d$ may no longer hold. For example, when $d=2$ , $k=\log n/\log 2$ .

We always assume $k \rightarrow \infty$ with n, even if arbitrarily slowly, which places an upper limit on d. However, if $k \ge \log^2 d$ , for example, then the rounding error is $d^{1/\sqrt k}= \Theta (1)$ , so for d constant or growing suitably slowly (say $d \le k^2$ ) we can ignore rounding.

Proposition 3. For $d \ge 2$ , let $G=G(k,d)$ be the Cayley tree with branching factor d and height k, where $d^k=n$ . Let $T_f$ be given by

\[T_f= \sqrt{k} d^{k+3/2-\sqrt{2k+2}}.\]

Let $t_f$ be the finish time of DLA on G. Then w.h.p. $T_f/C \le t_f \le C T_f$ for some constant $C\ge 1$ .

Proof. The lower bound on the finish time $t_f$ follows directly from the lower bound proof in the previous section, Section 5.1. The upper bound in the previous section used a result from (52) that when $\mu_{k-j^*}(t_1)\sim 1$ , then $ \mu_{k-j^*+1}(t_1)=Cd^{1/2}$ ; this latter value being large when d is. However, then d is constant $\mu_{k-j^*}(t) $ and $\mu_{k-j^*+1}(t)$ are of the same order, so we need to compare two levels whose distance apart, h, is larger than one. Thus, for the upper bound, let $j=j^*-h$ where $h= \lceil 1/2+\log_d k \rceil$ . When $k\ll d$ as in Section 5.1, then $h=1$ , which led to the argument based on the occupancy of levels $k-j^*$ and $k-j^*+1$ . The largest value of k satisfying $h=1$ is when $k=d^{1/2}$ ; in which case as $d^k=n$ we have $d \rightarrow \infty$ and, thus, $k \ll d$ , the case of the previous section. We assume henceforth that $h\ge 2$ .

The value of $\mu_{k-j*}$ given by (46), satisfies $\mu_{k-j+1}(t)=[(\kern1.5pt j+1)d^{k-j+1}/t] \cdot \mu_{k-j}(t)$ . Iterate this, and evaluate at $t=t_1(k-j^*)=(1+O(1/j))\mathrm{e}^{-1}\sqrt{2k+2}\,d^{k-j^*+1/2}$ from (51). As $j^*+1=\sqrt{2k+2}$ , then on the assumption that $h^2 \ll j^*$ ,

(66) \begin{align}\frac{\mu_{k-j^*+h}(t)}{\mu_{k-j^*}(t)}= \frac{(\kern1.5pt j^*+1)\cdots(\kern1.5pt j^*+h)}{t^h}d^{k-j^*+1}\cdots d^{k-j^*+h}=\Theta(1) \mathrm{e}^{h}d^{h^2/2}.\end{align}

This ratio is $\omega(1)$ , as either (i) $d \rightarrow \infty$ or (ii) d is constant and as $h= \lceil 1/2+\log_d k \rceil$ and $d^k=n$ then $k=\log_d n$ , this implies that $h=\omega(1)$ .

In either case, the calculations for X, Z leading to (62) in the previous section hold for the cases given here, with $j^*-h$ now replacing $j^*-1$ . At $t_1=t_1(k-j^*)$ , where $t_1=\Theta(T_f)$ then $\mu_{k-j^*+h}(t_1)=\Theta(\mathrm{e}^hd^{h^2/2})\rightarrow \infty$ , by the comments below (66). Thus, w.h.p. $Z(k-j^*+h,j^*, t_1)=(1-o(1))\mu_{k-j^*+h}(t_1)$ , and there are $Z(t_1)=\Theta(\mathrm{e}^hd^{h^2/2})$ exact paths from level $k-j^*+h$ to level k.

Choose a step $t' \ge T_f >t_1$ . Then either some leading path of halted particles already extends to a level i where $i<k-j^*+h$ , or all vertices in these levels are still unoccupied. In the latter case, as in (65), in expectation, it takes at most

\[\mathrm{E}\, T(\text{wait})=\tau=\frac{d^{k-j^*+h}}{Z(t_1)}=\Theta(1) \frac{d^{k-j^*+h}}{\mathrm{e}^hd^{h^2/2}}\]

further steps for one of these paths to extend back to the source.

The proof in the previous section depended on the ratio $T_f/d^{k-j^*+1}=o(1)$ (see (57)). Recall that $j^*=\sqrt{2k+2}-1$ , and the generic level $k-j$ in the proof between (58) and (62) is now $k-j^*+h$ , so that

(67) \begin{align}\rho_h= \frac{T_f}{d^{k-j^*+h}}=\frac{\sqrt k d^{k+3/2-\sqrt{2k+2}}}{d^{k-j^*+h}}= \frac{\sqrt{kd}}{d^{h}}. \end{align}

The final step is now to prove that $\tau=o(T_f)$ . From the expression for $\rho_h$ in (67), we have that

\[d^{k-j^*+h}= \frac{d^h}{\sqrt{kd}} T_f,\]

and as we assumed $h \ge 2$ ,

\[\tau= \Theta(1) \frac{d^{k-j^*+h}}{\mathrm{e}^hd^{h^2/2}}= \Theta(1)\left(\frac{d}{e d^{h/2}}\right)^h \frac {1}{\sqrt {kd}} T_f \le \Theta(1) \frac 1{\sqrt {kd}} T_f=o(T_f),\]

as required.