text:
\section{Introduction}
The study of bootstrap processes on graphs was initiated in 1979 by Chalupa,
Leath and Reich~\cite{ChLR79}, and is motivated by problems arising from statistical physics, such as the Glauber dynamics of
the zero-temperature Ising model, and kinetically constrained spin models of the liquid-glass transition
(see, e.g.,~\cite{FSS02,Morris09,MMT18}, and the recent survey~\cite{Morris17}).
The $r$-neighbour bootstrap process on a locally finite graph $G$ is a monotone cellular automata on the
configuration space $\{0,1\}^{V(G)}$, (we call vertices in state $1$ ``infected"), evolving in discrete time
in the following way: $0$ becomes $1$ when it has at least $r$ neighbours in state $1$, and infected vertices remain infected forever.
Throughout this paper, $A$ denotes the initially infected set, and we write $\langle A\rangle=G$
if the state of each vertex is eventually 1.

We will focus on \emph{anisotropic} bootstrap models, which are $d$-dimensional analogues of a family of
(two-dimensional) processes studied by Duminil-Copin, van Enter and Hulshof \cite{EH07,DCE13,DEH18}.
In these models the graph $G$ has
vertex set $[L]^d$, and the neighbourhood of each vertex consists of the $a_i$ nearest neighbours in the
$-e_i$ and $e_i$-directions for each $i \in [d]$,
where $a_1\le \cdots\le a_d$ and $e_i\in\mathbb Z^d$ denotes the $i$-th canonical unit vector.
In other words, $u,v\in [L]^d$ are neighbours if (see Figure \ref{figanis3d} for $d=3$)
\begin{align}\label{neigh3}
u-v\in N_{a_1,\dots,a_d}:=\{\pm e_1,\dots, \pm a_1e_1\}\cup \cdots \cup \{\pm e_d,\dots, \pm a_de_d\}.
\end{align}
We also call this process the $\mathcal N_r^{a_1,\dots, a_d}$-{\it model}.
Our initially infected set $A$
is chosen according to the Bernoulli product measure $\mathbb P_p=\bigotimes_{v\in [L]^d}$Ber$(p)$,
and we are interested in the so-called {\it critical length for percolation},
for small values of $p$
\begin{equation}\label{criticalL}
 L_c(\mathcal N_r^{a_1,\dots,a_d},p):= \min\{L\in\mathbbm N: \mathbb P_p(\langle A\rangle=G
 )\ge 1/2\}.
 \end{equation}

The analysis of these bootstrap processes for $a_1=\cdots= a_d=1$ was initiated by Aizenman and Lebowitz~\cite{AL88} in 1988,
who determined the magnitude of the critical length
up to a constant factor in the exponent for the $\mathcal N_2^{1,\dots,1}$-model (in other words, they determined the
`metastability threshold' for percolation). In the case $d = 2$, Holroyd~\cite{H03} determined (asymptotically, as $p \to 0$) the constant in the exponent  (this is usually called a sharp metastability threshold), proving that
$$L_c(\mathcal N_2^{1,1},p) = \exp\bigg( \frac{\pi^2/18 + o(1)}{p} \bigg).$$

For the general $\mathcal N_r^{1,\dots,1}$-model with $2\le r\le d$, the threshold was determined by Cerf and Cirillo \cite{CC99} and Cerf
and Manzo \cite{CM02}, and the sharp threshold by Balogh, Bollob\'as and Morris \cite{BBM09}
and Balogh, Bollob\'as, Duminil-Copin and Morris  \cite{BBDM12}: for all $d\ge r\ge 2$ there exists a computable constant
$\lambda(d,r)$ such that, as $p\to 0$,
\begin{equation*}
 L_c(\mathcal N_r^{1,\dots,1},p) = \exp_{(r-1)}\bigg(\frac{\lambda(d,r) + o(1)}{p^{1/(d-r+1)}}\bigg).
\end{equation*}

In dimension $d=2$, we write $a_1=a, a_2=b$, and the $\mathcal N_r^{a,b}$-model is called isotropic when $a=b$ and anisotropic when $a<b$.
Hulshof and van Enter \cite{EH07} determined the threshold for the first interesting anisotropic model given by the family $\mathcal N_{3}^{1,2}$, and the
corresponding sharp threshold was determined by Duminil-Copin and van Enter  \cite{DCE13}: for $b\ge 2$, as $p\to 0$,
\begin{equation*}
L_c\left(\mathcal N_{b+1}^{1,b},p\right)= \exp\left(\left(\frac{(b-1)^2}{4(b+1)} + o(1)\right)\frac{(\log p)^2}{p}\right).
\end{equation*}

The threshold was also determined in the general case $r=a+b$ by van Enter and Fey
 \cite{AA12} and the proof can be extended to all $b+1\le r\le a+b$: as $p\to 0$,
\begin{equation}\label{paso1}
\log L_c\left(\mathcal N_{r}^{a,b},p\right)=
\begin{cases}
\Theta\left(p^{-(r-b)}\right) & \textup{if }b=a,\\
\Theta\left(p^{-(r-b)}(\log p)^2\right) & \textup{if }b>a.
\end{cases}
\end{equation}

\subsection{Anisotropic bootstrap percolation on $[L]^3$}
In this paper we consider the three-dimensional analogue of the anisotropic bootstrap process studied by Duminil-Copin,
van Enter and Hulshof.
In dimension $d=3$, we write $a_1=a, a_2=b$ and $a_3=c$.
\vskip -.2cm
\begin{figure}[ht]
	\centering

	\includegraphics[width=0.35\textwidth]{Nabc}
	\caption{The neighbourhood $N_{a,b,c}$ with $a=1, b=2$ and $c=4$. The $e_1$-axis is towards the reader, the
	$e_2$-axis is vertical, and the $e_3$-axis is horizontal.}
	\label{figanis3d}
\end{figure}
These models were studied by van Enter and Fey \cite{AA12} for $r=a+b+c$;
they determined the following bounds on the critical length,

as $p\to 0$,

\begin{equation}\label{vefey}
		\log \log L_c\left(\mathcal N_{a+b+c}^{a,b,c},p\right)=
	\begin{cases}
	\Theta\left(p^{-a}\right)                    & \textup{if } b=a, \\
	\Theta\left(p^{-a}(\log \frac 1p)^{2}\right) & \textup{if } b>a.
	\end{cases}
	\end{equation}

Note that, by (\ref{vefey}) the critical length is doubly exponential in $p$ when $r=a+b+c$.
It is not difficult to show that the critical length is polynomial in $p$ if $r\le c$.

On the other hand, the critical length is singly exponential in the case $r=c+1$; indeed,
we determine the magnitude of the critical length up to a constant factor in the exponent.

The following is our main result.
\begin{theor}\label{target0}
As $p\to 0$,
\begin{equation}\label{magnitud}
\log L_c\left(\mathcal N_{c+1}^{a,b,c},p\right)=  \begin{cases}
\Theta\left(p^{-1/2}\right)  & \textup{if } c=b=a, \\
\Theta\left(p^{-1/2}(\log \frac 1p)^{1/2}\right) & \textup{if } c=b>a, \\
\Theta\left(p^{-1/2}(\log \frac 1p)^{3/2}\right) & \textup{if } c\in \{b+1,\dots,a+b-1\}, \\
\Theta\left(p^{-1}\right)                                  &   \textup{if } c=a+b,   \\
\Theta\left(p^{-1}(\log \frac 1p)^{2}\right)   & \textup{if } c> a+b.
	\end{cases}
\end{equation}
\end{theor}
The first three cases of this theorem ($c<a+b$) are obtained by adapting standard ideas used for two-dimensional models.
However, to deal with the lower bounds in the last two cases ($c\ge a+b$), it is necessary to introduce a new algorithm
which we call the {\it beams process}, and to develop new tools in {\it subcritical} bootstrap percolation
(see Theorem \ref{expdecIntro}).

\subsection{The BSU model}
The model we study here is a special case of the following extremely general class of $d$-dimensional monotone cellular automata, which were introduced by Bollob\'as, Smith and Uzzell~\cite{BSU15}.

Let $\mathcal U=\{X_1,\dots,X_m\}$ be an arbitrary finite family of finite
subsets of $\mathbb Z^d\setminus \{0\}$. We call $\mathcal U$ the {\it update family},
each $X\in\mathcal U$ an {\it update rule}, and the process itself $\mathcal U${\it-bootstrap percolation}.
Let $\Lambda$ be either $\mathbb Z^d$ or $\mathbb Z^d_L$ (the $d$-dimensional torus of sidelength $L$).
Given a set $A\subset \Lambda$ of initially {\it infected} sites, set $A_0=A$, and define for each $t\ge 0$,
\[A_{t+1}=A_t\cup\{x\in\Lambda: x+X\subset A_t \text{ for some }X\in\mathcal U\}.\]
The set of eventually infected sites is the {\it closure} of $A$, denoted by
$\langle A\rangle_\mathcal U=\bigcup_{t\ge 0}A_t$, and
we say that there is {\it percolation} when $\langle A\rangle_\mathcal U=\Lambda$.

 Let $S^{d-1}$ be the unit $(d-1)$-sphere and denote the discrete half space orthogonal to $u\in S^{d-1}$ as
 $\mathbb H_u^d:=\{x\in\mathbb Z^d:\langle x,u\rangle <0\}$.
The {\it stable set} $\mathcal{S}=\mathcal{S}(\mathcal U)$ is the set of all $u\in S^{d-1}$
such that no rule $X\in\mathcal U$ is contained in $\mathbb H_u^d$.
Let $\mu$ denote the Lebesgue measure on $S^{d-1}$. The following classification of families was proposed in \cite{BSU15} for $d=2$ and extended to all dimensions in \cite{BDMS15}:
A family $\mathcal U$ is

\begin{itemize}
 \item {\it subcritical} if for every hemisphere $\mathcal H \subset S^{d-1}$ we have $\mu(\mathcal H \cap\mathcal{S})>0$.
 \item {\it critical} if there exists a hemisphere $\mathcal H \subset S^{d-1}$ such that $\mu(\mathcal H \cap\mathcal{S})=0$, and
every open hemisphere in $S^{d-1}$ has non-empty intersection with $\mathcal{S}$;
 \item {\it supercritical} otherwise.
 \end{itemize}

Subcritical families exhibit a behavior which resembles models in classical site percolation,
(see e.g. \cite{BBPS16,H19+}).
For a certain class of subcritical models, we have succeeded in proving an exponential decay property
about the cluster size (see Section \ref{SectionexpdecayK}): denote by $\mathcal K$ the connected component containing $0$ in
$\langle A\rangle_\mathcal U$.
\begin{theor}\label{expdecIntro

Assume that $d=2$. Consider
$\mathcal U$-bootstrap percolation with $\mathcal{S}(\mathcal U)=S^1$ and $A\sim \bigotimes_{v\in \mathbb Z^2}\textup{Ber}(p)$. If $p$ is small enough, then
	\begin{equation}
	\mathbb P_p(|\mathcal K|\ge n)\le
     e^{-\Omega(n)},
	\end{equation}
	for every $n\in\mathbb N$.
\end{theor}

For dimension $d=2$, Bollob\'as, Duminil-Copin, Morris and Smith proved a universality result in \cite{BDMS15},
determining the critical length (with $A\sim \bigotimes_{v\in \mathbb Z^2_L}$Ber$(p)$)
\[L_c(\mathcal U,p):= \min\{L\in\mathbbm N: \mathbb P_p(\langle A\rangle_\mathcal U=\mathbb Z^d_L)\ge 1/2\},\]
up to a constant factor in the exponent for all two-dimensional critical families $\mathcal U$, which we can briefly state as follows.
\begin{theor}[Universality]
	Let $\mathcal U$ be a critical two-dimensional family. There exists a computable positive integer
	$\alpha=\alpha(\mathcal U)$ such that, as $p\to 0$, either
	\begin{equation}
	\log L_c(\mathcal U,p) =\Theta(p^{-\alpha}),
	\end{equation}
	or
	\begin{equation}
	\log L_c(\mathcal U,p) =\Theta(p^{-\alpha}(\log \tfrac 1p)^2).
	\end{equation}
\end{theor}

 Proving a universality result of this kind for three (or higher) dimensions is a challenging open problem.
 However, there is a weaker conjecture concerning all critical families and all $d\ge 3$, stated by the
 authors in \cite{BDMS15}; here for simplicity we state only the case $d=3$.
\begin{conj} Let $\mathcal U$ be a critical three-dimensional family.
As $p\to 0$, either
	\begin{equation}\label{2cri}
	 \log L_c(\mathcal U,p)=p^{-\Theta(1)},
	\end{equation}
	or
	\begin{equation}\label{3cri}
	 \log \log L_c(\mathcal U,p)=p^{-\Theta(1)}.
	\end{equation}
\end{conj}
Let us say that $\mathcal U$ is $2$-{\it critical} if it satisfies condition (\ref{2cri}),
and is $3$-{\it critical} if it satisfies condition (\ref{3cri}).
Observe that we can also think of our $\mathcal N_{r}^{a,b,c}$-model as $\mathcal N_{r}^{a,b,c}$-bootstrap percolation,
where $\mathcal N_{r}^{a,b,c}$ is the family consisting of all subsets of size $r$ of the neighbourhood
$N_{a,b,c}$ in (\ref{neigh3}).
It is easy to check that the family $\mathcal N_{r}^{a,b,c}$ is critical if and only if
 \[r\in\{c+1,\dots,a+b+c\}.\]
Moreover, it turns out that the family $\mathcal N_{r}^{a,b,c}$ is 2-critical for all $r\in\{c+1,\dots,c+b\}$
(see Remark \ref{coveredcases}).
On the other hand, the family $\mathcal N_{a+b+c}^{a,b,c}$ is 3-critical by (\ref{vefey});
it is natural to conjecture that this is the case for all $r\in \{c+b+1,\dots,c+b+a\}$.

\subsection{Outline of the proof}
The proofs of all upper bounds are obtained by adapting standard arguments in bootstrap percolation;
the same is true for the lower bounds in the cases $c<a+b$.

We deal with the lower bounds in the cases $c\ge a+b$ by introducing an algorithm
that we call the {\it beams process}, which will allow us to control the size of the components that can be created in the intermediate steps of the bootstrap dynamics, the trick will be to cover such components with beams (a {\it beam} is a finite 3-dimensional set of the form $H\times[w]$, where $H\subset\mathbb Z^2$ is
 connected and $\langle H\rangle_{\mathcal N_{a+b+1}^{a,b}}=H$, see Definition \ref{beam}).
All initially infected sites are beams, and at every step we merge beams that are within some constant distance, to create a bigger one, then repeat this algorithm and stop it at some finite time; each beam created during the process we call {\it covered}. When we observe the induced process along the $e_3$-direction, it looks like subcritical two-dimensional $\mathcal N_{a+b+1}^{a,b}$-bootstrap percolation,
thus, we can couple the original process and apply the exponential decay property (Theorem \ref{expdecIntro})
 to bound the probability of a beam been covered.

Theorem \ref{expdecIntro} provides new machinery in subcritical bootstrap percolation, we prove it in Section \ref{SectionexpdecayK},
and here we summarize the core idea. First, we need to guarantee the existence of {\it inwards stable droplets},
which are, basically, discrete polygons that can not be infected from outside;
it is possible to show the existence of such droplets by considering families $\mathcal U$ such that $\mathcal{S}(\mathcal U)=S^1$. After that, we combine ideas used by Bollob\'as-Riordan
in classical percolation models to prove that, when the density of initially infected sites is small enough, then the size of the cluster containing the origin decays exponentially fast, in distribution.

\section{Upper bounds}\label{SectionUpper12}
To prove upper bounds, it is enough to give one possible way of growing from $A$
step by step until we fill the whole of $[L]^3$.
The case $c>a+b$ will be deduced in the Appendix as a particular case of Proposition
\ref{genuppbound} (see Remark \ref{coveredcases}). On the other hand, the proof of case $c=b=a$
is similar to the proof given in \cite{AL88} and we will omit here.
Hence, we will focus only on the remaining upper bounds in Theorem \ref{target0} in increasing order of technicality.

More precisely, we will give a full proof of the case
$c\in\{b+1,\dots, a+b-1\}$ in Subsection \ref{fullproof}, then we will only sketch the cases
$c=a+b$ and $c=b>a$ in Subsections \ref{sketch1} and \ref{sketch2}, respectively, by pointing out the
small differences between these cases.
\begin{defi}\label{intfilled}
A {\it rectangle} is a set of the form $R=[x]\times[y]\times[w]\subset\mathbb Z^3$.
We say that a rectangle $R$ is {\it internally filled} if $R\subset \langle A\cap R\rangle_{\mathcal N_{r}^{a,b,c}}$,
and denote this event by $I^\bullet(R)$.
\end{defi}

\subsection{Case $c\in\{b+1,\dots, a+b-1\}$}\label{fullproof}
In this section we consider the families
$\mathcal N_{c+1}^{a,b,c},$
with $c\in\{b+1,\dots, a+b-1\}$ (here $a>1$, otherwise this case does not exist).
As usual in bootstrap percolation, we actually prove a stronger proposition.
\begin{prop}\label{upper2}
Fix $c\in\{b+1,\dots, a+b-1\}$ and  consider $\mathcal N_{c+1}^{a,b,c}$-bootstrap percolation.
There exists a constant $\Gamma=\Gamma(c)>0$ such that, if \[L=\exp\left(\Gamma p^{-1/2}(\log \tfrac 1p)^{3/2}\right),\] then
 $\mathbb P_p\left(I^\bullet([L]^3)\right)\to 1,\ as\ p\to 0.$
\end{prop}
When $h,w\ge c$, for simplicity we denote the event
\[I(h,w) := I^\bullet([h]^2\times[w]).\]

\begin{lemma}\label{lemaesencial}
If $p$ is small enough, then
\[\mathbb P_p(I(h,w+1)|I(h,w))\ge 1-e^{-p h^2},\]
under $\mathcal N_{c+1}^{a,b,c}$-bootstrap percolation.
\end{lemma}
\begin{proof}
If $R_1:=[h]^2\times[w]$ is completely infected, we just need to infect the right-most face
$Q:=[h]^2\times\{w+1\}$, and since we have $c$ already infected vertices in $R_1$, then it is enough to find
$1$ infected vertex in $Q$ (see Figure \ref{fig3ladosuper} below). Thus,
\begin{align*}
\mathbb P_p\left(I^\bullet\left([h]^2\times[w+1]\right)\Big|
I^\bullet\left(R_1\right)\right)
  \ge
1-\prod_{v\in Q}\left(1-\mathbb P_p\left(v\in A\right)\right)
\ge 1-e^{-p h^2}.
\end{align*}
\end{proof}
Lemma \ref{lemaesencial} tells us the cost of growing one step along
the (easiest) $e_3$-direction, and we are also interested in computing the cost of
growing along the $e_1$ and $e_2$ (harder) directions. To do so, we will consider general values of $r$: let us first consider the regime
$r\le a+b,$
this implies that given any rectangle $R$, all three induced 2-dimensional processes in the faces of $R$, namely,
$\mathcal N_{r-c}^{a,b}$, $\mathcal N_{r-b}^{a,c}$ and $\mathcal N_{r-a}^{b,c}$, are supercritical.
\begin{lemma}[Supercritical faces]\label{regimer<a+b}
If $r\le a+b$, and $p$ is small enough, then
\[\mathbb P_p(I(h+1,w)|I(h,w)) \ge
	\left(1-e^{-{c\choose 2}^{-1}p^{{c\choose 2}}wh}\right)^{2},\]

under $\mathcal N_{r}^{a,b,c}$-bootstrap percolation.
\end{lemma}
\begin{proof}
For $s=a,b$, let $\Delta_s$ be the discrete right-angled triangle
whose legs are $[r-s]\times\{1\}$ and $\{1\}\times [r-s]$.
Once $R_1=[h]^2\times[w]$ is completely full, to get $R_2=[h+1]^2\times[w]$ internally filled it is enough
to have one copy of $\Delta_a$ in $A\cap(\{h+1\}\times[h]\times[w])$ (front face), and one copy of $\Delta_b$
in $A\cap([h]\times\{h+1\}\times[w])$ (top face, see Figure \ref{fig3ladosuper}).

\begin{figure}[ht
\centering
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=1.05\linewidth]{Section2-3}
  \caption{$r\le a+b$}
  \label{fig3ladosuper}
\end{subfigure}%
\begin{subfigure}{.5\textwidth}
  \centering
  \includegraphics[width=1.05\linewidth]{Section2-4}
  \caption{$c=b>a$}
 \label{fig3ladosuperc=b}
\end{subfigure}
\caption{A single vertex on the right-most side, one copy of $\Delta_a$ on the front side, and one copy of $\Delta_b$ on the top side.}
\end{figure}

Since $|\Delta_s|=(r-s)(r-s+1)/2$ and $a\ge 2$, then
$|\Delta_b|\le |\Delta_a|\le {c\choose 2}$.   Hence, by independence between the front and top faces,
\begin{align*}
\mathbb P_p(I^\bullet(R_2)|I^\bullet(R_1))
& \ge \left(1-e^{-|\Delta_a|^{-1}p^{|\Delta_a|}wh}\right)
\left(1-e^{-|\Delta_b|^{-1}p^{|\Delta_b|}wh}\right)
  \ge \left(1-e^{-{c\choose 2}^{-1}p^{{c\choose 2}}wh}\right)^{2}.
\end{align*}

\end{proof}

The next step is to determine the size of a rectangle (usually called {\it critical droplet}) such that, once it is internally filled, then it can grow until $[L]^3$ with high probability.
\begin{lemma}\label{secondstep}
Let $\Gamma>0$ be a large constant and set
$h=cp^{-\frac{1}{2}}(\log \frac 1p)^{\frac 12}$, $R_1:=[h]^2\times
[c]$ and
\[L=\exp\left(\Gamma p^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 32}\right).\] Conditionally on $I^\bullet(R_1)$,
the probability of $I^\bullet([L]^3)$ goes to $1$, as $p\to 0$.
\end{lemma}
\begin{proof}
	Consider the rectangles $R_2\subset R_3\subset R_4\subset R_5:= [L]^3$ containing $R_1$, defined by
	\[R_2:=[h]^2\times [c^2p^{-{c\choose 2}+\frac{1}{2}}(\log \tfrac 1p)^{\frac 12}],\
\ \ R_3:=[h^2]^2\times [c^2p^{-{c\choose 2}+\frac{1}{2}}(\log \tfrac 1p)^{\frac 12}],\
\ \ R_4:=[h^2]^2\times [L].\]
	Note that
$\mathbb P_p(I^\bullet([L]^3)|I^\bullet(R_1))\ge\prod_{k=1}^{4} \mathbb P_p(I^\bullet(R_{k+1})|I^\bullet(R_k)).$
We apply Lemma \ref{lemaesencial} to deduce
	\begin{align*}
	\mathbb P_p(I^\bullet(R_{2})|I^\bullet(R_1)) & \ge \left(1-e^{-ph^2}\right)^{c^2p^{-{c\choose 2}+\frac{1}{2}}(\log \frac 1p)^{\frac 12}}
 \ge e^{-2p^{\frac{c^{2}}{2}-{c\choose 2}}}\to 1,
	\end{align*}
	and by Lemma \ref{regimer<a+b},
\begin{align*}
\mathbb P_p(I^\bullet(R_{3})|I^\bullet(R_2)) & \ge
\bigg(1-e^{-\Omega\left(p^{{c\choose 2}} p^{-{c\choose 2}+\frac{1}{2}}(\log \frac 1p)^{\frac 12}\cdot h\right)}
\bigg)^{2h^2}
 \ge \exp\left(-4h^2p^{2c}\right)\to 1,
\end{align*}
We apply these lemmas again to get $\mathbb P_p(I^\bullet(R_{4})|I^\bullet(R_3)) \to 1,$
since $ph^4\ge p^{-1}\gg \Gamma p^{-\frac{1}{2}}(\log \frac 1p)^{\frac 32}$, and also
$\mathbb P_p(I^\bullet(R_{5})|I^\bullet(R_4))\to 1$.
We conclude that $\mathbb P_p(I^\bullet([L]^3)|I^\bullet(R_1))\to 1$, as $p\to 0$.
\end{proof}
Now, we are ready to show the upper bound.
\begin{proof}[Proof of Proposition \ref{upper2}]
	Set $L=\exp\left(\Gamma p^{-\frac{1}{2}}(\log \frac 1p)^{\frac 32}\right)$, where $\Gamma>0$ is a large constant to be chosen.
	Consider the rectangle
	\[R:=\left[cp^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 12}\right]^2\times [c]\subset[L]^3,\]
 and the events  $F_L:=\{\exists$ an internally filled copy of $R$ in $[L]^3\}$,
and $G_L:=\{\langle A\cup R\rangle=[L]^3\}$.
It follows that $\mathbb P_p\left(I^\bullet([L]^3)\right)\ge \mathbb P_p(F_L)\mathbb P_p(G_L|I^\bullet(R)),$
and $\mathbb P_p(G_L|I^\bullet(R))\to 1$, as $p\to 0$,
by the previous lemma.
Therefore, it remains to show that $\mathbb P_p(F_L)\to 1$ too.

Indeed, we claim that there exists a constant $C'>0$ such that
	\begin{equation}\label{spcd'}
	\mathbb P_p(I^\bullet(R))\ge \exp\left(-C' p^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 32}\right),
	\end{equation}
	so using the fact that there are roughly $L^3/|R|$ disjoint (therefore independent) copies of $R$
	(which we label $Q_1,\dots, Q_{L^3/|R|}$),
	and $|R|
	\le p^{-3}$, (\ref{spcd'}) immediately gives

\begin{align*}
	\mathbb P_p(F_L^c)& \le \mathbb P_p\left(\bigcap_i
	I^\bullet(Q_i)^c\right)
 \le \left[1-\mathbb P_p(I^\bullet(R))\right]^{L^3/|R|}
 \le \exp\left(-e^{3\log L-3\log(1/p)-C'p^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 32}}\right).
	\end{align*}
Since $\log L= \Gamma p^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 32}$, by taking $\Gamma>C'/3$ we conclude $\mathbb P_p(F_L)\to 1$, as $p\to 0$.
To finish, it is only left to prove inequality (\ref{spcd'}).

In fact, note that a way to make $R$ be internally filled is the following: start with $[c]^3\subset A$, and then grow from $R_k=[k]^2\times[c]$ to $R_{k+1}$, for
$k=c,\dots, m:=cp^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 12}.$
This gives us
\begin{align*}
  \mathbb P_p\left(I^\bullet(R)\right) & \ge
  \mathbb P_p([c]^3\subset A)\prod_{k=c}^m\mathbb P_p\left(I^\bullet(R_{k+1})|I^\bullet(R_k)\right)
 \ge p^{c^3}\prod_{k=c}^m
  \left(1-e^{-{c\choose 2}^{-1}p^{{c\choose 2}}ck}\right)^{2} \\
&\ge p^{c^3+c^2m}
 \ge \exp\left(-C' p^{-\frac{1}{2}}(\log \tfrac 1p)^{\frac 32}\right),
\end{align*}
for $C'>c^3$, as we claimed.
\end{proof}

\subsection{Case $c=a+b$}\label{sketch1}

In this section we consider the families $\mathcal N_{a+b+1}^{a,b,a+b},$
corresponding to the case  $r=a+b+1$.
To do so, we first compute the cost of growing for all cases $a+b<r\le a+c$,
where, the induced $\mathcal N_{r-c}^{a,b}$ process is still supercritical, but the induced processes $\mathcal N_{r-b}^{a,c}$ and $\mathcal N_{r-a}^{b,c}$ are critical.
\begin{lemma}[Critical faces]\label{regimer>a+b}
If $r\in\{a+b+1,\dots, a+c\}$ and $p$ is small, then
\[\mathbb P_p(I(h+1,w)|I(h,w))\ge
	\left(1-e^{-\frac{1}{r-a}p^{r-a}w}\right)^{r}
	\left(1-e^{-\frac{1}{m}p^{m}w}\right)^{2h},\]
under $\mathcal N_r^{a,b,c}$-bootstrap percolation, with $m:=r-(a+b).$
\end{lemma}
\begin{proof}
Once $[h]^2\times[w]$ is completely full, to fill $[h+1]^2\times[w]$
		it is enough to have the occurrence of the events $F_h^{e_1}$ and $F_h^{e_2}$ defined as follows: $F_h^{e_1}$ as
		(growing along the $e_1$-direction) there exist $r-a$ adjacent vertices
		in $A\cap(\{h+1\}\times\{1\}\times[w])$, $r-(a+1)$ adjacent vertices in
		$A\cap(\{h+1\}\times\{2\}\times[w])$,$\dots$, $r-(a+b-1)$ adjacent vertices in
		$A\cap(\{h+1\}\times\{b\}\times[w])$, and for each $i=b+1,\dots,h$, there exist
		$m=r-(a+b)$ adjacent vertices in $A\cap(\{h+1\}\times\{i\}\times[w])$.
\vspace{-.3cm}
\begin{figure}[ht]\label{sec21}
	\centering
	\includegraphics[width=.5\textwidth]{biggestc2.pdf}
	\caption{$m$ vertices in each of the lines along the $e_3$-direction ($m=2$).}
	\label{figcmayora+b}
\end{figure}
\begin{align*}
\mathbb P_p(F_h^{e_1}) &  \ge \prod_{k=m+1}^{r-a}\left(1-(1-p^{k})^{\frac wk}\right)

\prod_{i=1}^{h}\left(1-(1-p^{m})^{\frac wm}\right)

 \ge \left(1-e^{-\frac{1}{r-a}p^{r-a}w}\right)^{b}
		\left(1-e^{-\frac{1}{m}p^{m}w}\right)^{h}.
\end{align*}
$F_h'$ is defined analogously, this time growing along the $e_2$-direction (see Figure \ref{figcmayora+b}), thus
\begin{align*}
\mathbb P_p(F_h^{e_2}) \ge \left(1-e^{-\frac{1}{r-b}p^{r-b}w}\right)^{a}\left(1-e^{-\frac{1}{m}p^{m}w}\right)^{h}.
\end{align*}
Finally,
$p_r(h+,w)\ge \mathbb P_p(F_h^{e_1})\mathbb P_p(F_h^{e_2})$.
\end{proof}
\begin{rema}
In the regime $a+c<r\le a+b+c$,
all three the induced 2-dimensional processes $\mathcal N_{r-c}^{a,b}$, $\mathcal N_{r-b}^{a,c}$ and $\mathcal N_{r-a}^{b,c}$ are critical.
\end{rema}

As before, we need to set the size of a critical droplet.
\begin{lemma}
Fix $\varepsilon>0$ and let $\Gamma$ be a large constant. Set  $h=p^{-1/2-\varepsilon}$, $R_1:=[h]^2\times [h^{2}]$ and \[L=\exp(\Gamma p^{-1}).\]
Conditionally on $I^\bullet(R_1)$,
the probability of $I^\bullet([L]^3)$ goes to $1$, as $p\to 0$.
\end{lemma}
The proof of this lemma is very similar to that of Lemma \ref{secondstep}, thus, we omit it.

Finally, to deduce the upper bound, we proceed in the same way that we used to prove Proposition \ref{upper2},
this time by showing that the critical droplet $R_1$ satisfies
\begin{equation}\label{spcd}
	\mathbb P_p(I^\bullet(R_1))\ge \exp(-C'p^{-1}),
\end{equation}
for some constant $C'>0$, depending on the integral of the function
$f_1:(0,\infty)\to(0,\infty)$, defined by $f_1(z)=-\log(1-e^{-z})$ (see e.g. \cite{AL88} and \cite{H03}).

\subsection{Case $c=b>a$}\label{sketch2}
In this section we sketch the proof of the last case.
Consider the families \[\mathcal N_{c+1}^{a,c,c}.\]

We follow the same steps, taking into account that the way to grow is slightly different:
in this case, to grow along the $e_2$-direction is as easy as
grow along the $e_3$-direction, so that
it is enough to find a single infected vertex on the right-most and top sides,
while to grow along the $e_1$-direction we still need to find one copy of $\Delta_a$ on the front
side (see Figure \ref{fig3ladosuperc=b} above).

\begin{lemma}\label{grow2}
	Fix $l,w\ge c$ and let $I=I^\bullet([l]\times[w]^2)$. If $p$ is small enough, then
	\begin{enumerate}
	\item[(i)] $\mathbb P_p\left(I^\bullet([l]\times[w+1]^2)|I\right)\ge \left(1-e^{-plw}\right)^2$.
	\item[(ii)]  $\mathbb P_p\left(I^\bullet([l+1]\times[w]^2)|I\right)\ge
	1-e^{-\Omega(p^{c(c+1)/2}w^2)}$.
	\end{enumerate}
\end{lemma}
\begin{proof}
Similar to the proof of Lemmas \ref{lemaesencial} and \ref{regimer<a+b}.
\end{proof}
The size of the critical droplet is given by the following lemma, again, we omit the proof.
\begin{lemma}
Let $\Gamma$ be a large constant. Set
 $R_1:=[p^{-1/2} (\log\frac{1}{p})^{-\frac 12}]\times \left[2\Gamma p^{-1}\log\tfrac 1p\right]^2$ and
 \[L=\exp\left(\Gamma p^{-1/2}\sqrt{\log\tfrac{1}{p}}\right).\]
Conditionally on $I^\bullet(R_1)$,
the probability of $I^\bullet([L]^3)$ goes to $1$, as $p\to 0$.
\end{lemma}
Finally, to deduce the upper bound, we proceed as before, this time by showing that
\begin{equation}\label{spcd}
	\mathbb P_p(I^\bullet(R_1))\ge \exp\left(-C'p^{-\frac{1}{2}}\sqrt{\log \tfrac{1}{p}}\right),
\end{equation}
for some constant $C'>0$, depending on $c$ and the function $f_2(z)=-\log(1-e^{-z^2})$.

\section{Lower bounds via components process}\label{SectionLowerComp}
In this section we only prove the lower bounds corresponding to the cases $c< a+b$, since the proof is an application of the {\it components process} (see Definition \ref{compro} below),
a variant of an algorithm introduced Bollob\'as, Duminil-Copin, Morris, and Smith \cite{BDMS15}.
The lower bound in the case $a=b=c=1$ was proved in \cite{AL88}, and the general case $a=b=c$ follows by using the same arguments.
Thus, we will omit this case, and prove the following.
\begin{prop}\label{lower1weaker}
If $c>a$, there is a constant $\gamma=\gamma(c)>0$
such that, for
\[L<\exp\left(\gamma p^{-1/2}(\log\tfrac{1}{p})^{1/2}\right),\]
 $\mathbb P_p(I^{\bullet}([L]^3))\to 0,$ as $p\to 0$,
under $\mathcal N_{c+1}^{a,c,c}$-bootstrap percolation.
\end{prop}
\begin{prop}\label{lower2weaker}
If $c\in\{b+1,\dots,a+b-1\}$, there exists $\gamma=\gamma(c)>0$
such that, for
\[L<\exp\left(\gamma p^{-1/2}(\log\tfrac{1}{p})^{3/2}\right),\]
 $\mathbb P_p(I^{\bullet}([L]^3))\to 0,$ as $p\to 0$,
under $\mathcal N_{c+1}^{a,b,c}$-bootstrap percolation.
\end{prop}

\begin{nota}
Throughout this paper, when $\mathcal U=\mathcal N_{r}^{a,b,c}$ we will omit the subscript in the closure and
simply write $\langle \cdot \rangle$ instead of $\langle \cdot \rangle_{\mathcal N_{r}^{a,b,c}}$.
\end{nota}
Aizenman and Lebowitz \cite{AL88} obtained the matching lower bound for the family $\mathcal N_2^{1,1,1}$ by using the so-called {\it rectangles process}, and they exploited the fact that for this model, the closure $\langle A\rangle$ is a union of rectangles which are separated by distance at least 2.

In our case, the closure $\langle A\rangle$ is more complicated.
Thus, we need to introduce a notion about rectangles
which is an approximation to being internally filled, and
this notion requires a strong concept of connectedness; we define both concepts in the following.
Consider the superset of $N_{a,b,c}$ (see  (\ref{neigh3})) given by
 \[\bar N_{a,b,c}:= \{(u_1,u_2,u_3)\in\mathbb Z^3: |u_1|\le a, |u_2|\le b, |u_3|\le c
 \textup{ and } u_1u_2u_3=0\}.\]
\begin{defi}\label{strconn}
Let $G=(V,E)$ be the graph with vertex set $[L]^3$ and edge set given by
$E=\{(u,v): u-v\in  \bar N_{a,b,c}\}$.
 We say that a set $S\subset [L]^3$ is {\it strongly connected}
 if it is connected in the graph $G$.
\end{defi}

\begin{defi}\label{intspa}

We say that the rectangle $R\subset [L]^3$ is
{\it internally spanned} by $A$, if there exists a strongly connected set $S\subset\langle A\cap R\rangle$
such that $R$ is the smallest rectangle containing $S$. We denote this event by $I^{\times}(R)$.
\end{defi}
Note that when a rectangle is internally filled then it is also internally spanned, therefore,
Propositions \ref{lower1weaker} and \ref{lower2weaker} are consequences of the following results.
\begin{prop}\label{lower1}
If $c>a$, there is a constant $\gamma=\gamma(c)>0$
such that, for
\[L<\exp\left(\gamma p^{-1/2}(\log\tfrac{1}{p})^{1/2}\right),\]
 $\mathbb P_p(I^{\times}([L]^3))\to 0,$ as $p\to 0$,
under $\mathcal N_{c+1}^{a,c,c}$-bootstrap percolation.
\end{prop}

\begin{prop}\label{lower2}
If $c\in\{b+1,\dots,a+b-1\}$, there exists $\gamma=\gamma(c)>0$
such that, for
\[L<\exp\left(\gamma p^{-1/2}(\log\tfrac{1}{p})^{3/2}\right),\]
 $\mathbb P_p(I^{\times}([L]^3))\to 0,$ as $p\to 0$,
under $\mathcal N_{c+1}^{a,b,c}$-bootstrap percolation.
\end{prop}
We will prove them in Sections \ref{subseclower1} and \ref{subseclower2}, respectively.

\subsection{The components process}
The following is an adaptation of the spanning algorithm in \cite[Section 6.2]{BDMS15}.
We will use it to show
an Aizenman-Lebowitz-type lemma,
which says that when a rectangle
is internally spanned, then it contains internally spanned rectangles of all
intermediate sizes
(see Lemmas \ref{ALlema0} and \ref{ALlema00} below).
\begin{defi}[The components process]\label{compro}
 Let $A=\{v_1,\dots,v_{|A|}\}\subset [L]^3$ and fix $r\ge c+1$.
 Set $\mathcal R:=\{S_1,\dots,S_{|A|}\}$, where $S_i=\{v_i\}$ for each $i=1,\dots,|A|$.
 Then repeat the following steps until STOP:
 \begin{enumerate}
  \item If there exist distinct sets $S_{1},S_{2}\in\mathcal R$ such that
  \[ S_{1}\cup S_{2} \]
  is strongly connected, then remove them from $\mathcal R$,
  and replace by $\langle S_{1}\cup S_{2} \rangle$.
  \item If there do not exist such sets in $\mathcal R$, then STOP.
 \end{enumerate}
\end{defi}
\begin{rema}\label{stopfinitetime}
We highlight two properties that are due to the way the algorithm evolves:
\begin{itemize}
 \item At any stage of the component process, any set
$S=\langle S_{1}\cup S_{2} \rangle$ added to the collection
$\mathcal R$ satisfies $S=\langle A\cap S\rangle=\langle S\rangle\subset [L]^3$ (since $r\ge c+1$).
In particular, the smallest rectangle containing $S$ is internally spanned.
 \item Since $G$ is finite, the process stops in finite time; so that we can consider the final
 collection $\mathcal R'$ and set
 $V(\mathcal R')=\bigcup\limits_{S\in\mathcal R'}S$.
\end{itemize}
\end{rema}
\begin{lemma}\label{final=gen}
  $V(\mathcal R')=\langle A\rangle$.
\end{lemma}
\begin{proof}
 Clearly $A\subset V(\mathcal R')\subset \langle A\rangle$, and to prove that $\langle A\rangle\subset V(\mathcal R')$
 we argue by contradiction.
Suppose this is not the case, since $A\subset V(\mathcal R')$, there would exist vertices
 $v\in\langle A\rangle \setminus V(\mathcal R')$ and $v_1,\dots,v_r\in V(\mathcal R')$ such that $v-v_i\in N_{a,b,c}$, for $i=1,\dots,r$.
Let us say that $v_i\in S_i'$ for some sets $S_i'\in\mathcal R'$.

Since $S_1'=\langle S_1'\rangle$, $v_k\notin S_1'$ for some $k\ne 1$,
so that $S_k'\ne S_1'$.  In particular $S_1'\cup S_k'$ is strongly connected via $v_1,v,v_k$ and
 $\langle S_1'\cup S_k' \rangle\notin \mathcal R'$; this contradicts the definition of $\mathcal R'$.
\end{proof}
\begin{nota}
 From now on, we allow some abuse of notation, by denoting as $[x]\times[y]\times[z]$
 any translate of the rectangle $R=[x]\times[y]\times[z]$ located at the origin.
\end{nota}

\subsection{Case $c=b>a$}\label{subseclower1}
The following is a variant of the Aizenman-Lebowitz Lemma in \cite{AL88}.
\begin{lemma}\label{ALlema0}
 Consider $\mathcal N_{r}^{a,b,c}$-bootstrap percolation with $r\ge c+1$.
 If $[L]^3$ is internally spanned then,
 for every \ $h,k\le L$ there exists an internally spanned rectangle
 $[x]\times[y]\times[z]$ inside $[L]^3$ satisfying $(y+z)/2 \le 2ck$, and either
 \begin{enumerate}
  \item[(a)] $x\ge h$, or
  \item[(b)] $x< h$ and $(y+z)/2 \ge k$.
 \end{enumerate}
\end{lemma}
\begin{proof}
Let $S$ be the first set that appears in the components process such that,
the smallest rectangle $Q:=[x]\times[y]\times[z]$ containing $S$ satisfies either $x\ge h$ or
$(y+z)/2\ge k$ (such a set exists since $V(\mathcal R')=\langle A\rangle$ and $[L]^3$ is internally spanned).
Since $Q$ is internally spanned, it only remains to show that
the semi-perimeter $(y+z)/2$ is at most $2ck$.

In fact, we know that $S=\langle S_{1}\cup  S_{2} \rangle$ for some
sets $S_{t}$ such that, for each $t=1,2$, the smallest rectangle
$[{x_t}]\times[{y_t}]\times[{z_t}]$ containing $S_{t}$ satisfies
$({y_t}+{z_t})/2\le k-1/2$.
Since $S$ is strongly connected, the new semi-perimeter is
\begin{align*}\label{maxdiam2}
	\frac{{y}+{z}}{2} & \le 2 \max_{t= 1,2}\left\{\frac{{y_t}+{z_t}}{2}\right\}
	+\frac {b+c}{2}
 \le 2c\left(k-\frac 12\right)+ c
= 2ck.
\end{align*}
\end{proof}
\begin{proof}[Proof of Proposition \ref{lower1}]
Fix a small constant $\delta >0$ and take $L<\exp(\gamma p^{-1/2}(\log\tfrac 1p)^{1/2})$, where $\gamma=\gamma(\delta)>0$
is another small constant to be chosen. Let us show that $\mathbb P_p(I^{\times}([L]^3))$ goes to $0$, as $p\to 0$. Set
\[h=\delta p^{-\frac 12}(\log\tfrac 1p)^{-\frac 12},\ \ \ k=p^{-\frac 12}\sqrt{\log\tfrac 1p}.\]

If $[L]^3$ is internally spanned, by Lemma \ref{ALlema0} the following event
occurs: there exists an internally
spanned rectangle $Q=[x]\times[y]\times[z]\subset [L]^3$ satisfying $(y+z)/2 \le 2ck$, and either
$x\ge h$, or $x< h$ and $(y+z)/2 \ge k$.

Suppose first that $x< h$ and $(y+z)/2 \ge k$, thus, either $y$ or
$z$ is at least $k$, by symmetry ($b=c$), we can assume $z\ge k$.
Since $Q$ is internally spanned, every copy of the slab
$[x]\times[y]\times[2c]$ must contain at least $1$ element
of $A$. Consider only the $z/2c$ disjoint slabs that partition $Q$; since $xy =O (hk)$, if $\delta$ is small, the probability of this event is at most
\begin{align*}
  (O(pxy))^{z/2c} & \le (O(phk))^{k/2c}
= (O(\delta))^{k/2c}
 \le e^{-k}.
\end{align*}

On the other hand, if $x\ge h$ we use the fact that $a\le (c+1)-2$, thus,
since $Q$ is internally spanned,
every copy of the slab $[3a]\times[y]\times[z]$ must contain at least $2$ elements
$u,v\in A$ such that $u-v\in N_{a,b,c}$. Since $x\ge h$, the probability of this event is at most
\begin{align*}
  \left(O(p^{2}yz)\right)^{x/3a} &\le \left(O(p^{2}k^2)\right)^{h/3a}
\le \left(O(p^{2}p^{-1}\log\tfrac 1p)\right)^{h/3a}
 \le e^{-\Omega(\delta k)}.
\end{align*}

Therefore, the probability that $Q$ is internally spanned is at most $e^{-c(\delta)k}$
for some small constant $c(\delta)>0$. Finally, denoting by $\mathcal R_{k}$ the
collection of rectangles $[x]\times[y]\times[z]\subset[L]^3$ satisfying
$y+z\le 4ck$, it follows by union bound that

\begin{align*}
\mathbb P_p(I^{\times}([L]^3))
& \le \sum_{Q\in\mathcal R_{k}}\mathbb P_p(I^{\times}(Q))
 \le |\mathcal R_{k}|e^{-c(\delta)k}
 \le L^7 \exp\left(-c(\delta)p^{- 1/2}(\log\tfrac 1p)^{1/2}\right) \to 0,

\end{align*}
as $p\to 0$, for $7\gamma<c(\delta)$, and we are finished.
\end{proof}

\subsection{Case $c\in\{b+1,\dots,a+b-1\}$}\label{subseclower2}
In this case,
the corresponding analogue of the Aizenman-Lebowitz Lemma is as follows.
\begin{lemma}\label{ALlema00}
Consider $\mathcal N_{r}^{a,b,c}$-bootstrap percolation with $r\ge c+1$.
 If $[L]^3$ is internally spanned then,
 for every \ $h,k\le L$ there exists an internally spanned rectangle
 $[x]\times[y]\times[z]\subset [L]^3$ satisfying $(x+y)/2 \le rh$, and either
 \begin{enumerate}
  \item[(a)] $z\ge k$, or
  \item[(b)] $z< k$ and $(x+y)/2 \ge h$.
\end{enumerate}
\end{lemma}
The proof of this lemma is identical to that of Lemma \ref{ALlema0},
we therefore omit it and proceed to the proof of the lower bound.
\begin{proof}[Proof of Proposition \ref{lower2}]
Take $L< \exp(\gamma p^{-\frac 12}(\log\tfrac 1p)^{\frac 32})$, where $\gamma>0$ is some small constant.
Fix $\delta>0$ and set
\[h=\delta p^{-\frac 12}(\log\tfrac 1p)^{\frac 12},\ \ \ k=p^{-1}.\]
If $[L]^3$ is internally spanned, by Lemma \ref{ALlema00}, there is an internally
spanned rectangle $Q=[x]\times[y]\times[z]$ satisfying
$(x+y)/2 \le rh$, and either $z\ge k$, or
$z< k$ and $(x+y)/2 \ge h$.

In the case that $z\ge k$ we also know that $xy = O(h^2)$.
As before, every copy of the slab
$S:=[x]\times[y]\times[r]$ intersects $A$.
Thus, by considering the $z/r$ disjoint slabs; if $\delta$ is small,
the probability of this event is at most

\begin{align*}
\mathbb P_p\left(S\cap A\ne \emptyset
\right)^{z/r}
&\le \left(1-e^{-\Omega(ph^2)}\right)^{k/r}
 =\left(1-p^{\Omega(\delta^2)} \right)^{p^{-1}/r}
 \le e^{-p^{-3/4}}.
\end{align*}

In the case that $z< k$ and $(x+y)/2 \ge h$, we can assume w.l.o.g. that $y\ge h$ and use the fact that
$b\le c-1=r-2$. This time there is no gap along the $e_2$-direction, so, every copy of the slab
$[x]\times [2r]\times[z]$ must contain at least $2$ elements of $A$ within constant distance. The probability of this event is at most

\begin{align*}
  \left(O(p^{2}xz)\right)^{y/2r} &\le \left(O(p^{2}hk)\right)^{h/2r}
 \le e^{-\Omega(h\log\frac 1p)}.
\end{align*}
Therefore, the probability that $Q$ is internally spanned is at most
$ e^{-c(\delta)p^{-\frac 12}(\log\tfrac 1p)^{\frac 32}},$
for some small constant $c(\delta)>0$.
Denote by $\mathcal R_{h}'$ the
collection of rectangles $[x]\times[y]\times[z]\subset[L]^3$ satisfying
$x+y \le 2rh$, it follows by union bound that

\begin{align*}
\mathbb P_p(I^{\times}([L]^3))
& \le \sum_{Q\in\mathcal R_{h}'}\mathbb P_p(I^{\times}(Q))
 \le |\mathcal R_{h}'|e^{-c(\delta)p^{-\frac 12}(\log\tfrac 1p)^{\frac 32}}

 \to 0,

\end{align*}
as $p\to 0$, if $\gamma>0$ is small.
\end{proof}

\section{Exponential decay for subcritical families}\label{SectionexpdecayK}
In this section, we develop new machinery for
$\mathcal U$-bootstrap percolation in $\mathbb Z^2$ with $\mathcal U$ subcritical.
The first paper studying these families in such generality is \cite{BBPS16},
it turns out that these families exhibit a behavior which resembles models in classical site percolation,
for instance, in \cite{BBPS16} it is proved that $p_c(\mathbb Z^2,\mathcal U)>0$, for every subcritical family $\mathcal U$,
where
\begin{equation*}\label{pcritico}
p_c(\mathbb Z^2,\mathcal U):=\inf\{p:\mathbb P_p(\langle A\rangle_\mathcal U = \mathbb Z^2)= 1\}.
\end{equation*}

We will only deal with subcritical families $\mathcal U$ satisfying $p_c(\mathbb Z^2,\mathcal U)=1$; the authors of \cite{BBPS16}
proved that this condition is equivalent to $\mathcal{S}(\mathcal U)=S^1$.
Our aim is to show that for such families, if we choose the initial infected set $A$ to be
$\varepsilon$-random with $\varepsilon$ small enough, then the size of the cluster in $\langle A\rangle_{\mathcal U}$ containing the origin decays exponentially fast. More precisely
\begin{defi}
We define the {\it component} (or {\it cluster})
of $0\in\mathbb Z^2$ as the connected component containing $0$ in the graph induced by
$\langle A\rangle_\mathcal U$, and we denote it by $\mathcal K=\mathcal K(\mathcal U,A)$.
If $0\notin \langle A\rangle_\mathcal U$, then we set $\mathcal K=\emptyset$.
\end{defi}
The following is the main result in this section. It will be essential to prove the remaining lower bounds (cases $c\ge a+b$) in Section \ref{SectionLowerBeams}.
\begin{theor}\label{expdecIntro2}
Consider
$\mathcal U$-bootstrap percolation with $\mathcal{S}(\mathcal U)=S^1$. If $p$ is small enough, then
	\begin{equation*}
	\mathbb P_p(|\mathcal K|\ge n)\le
     e^{-\Omega(n)},
	\end{equation*}
	for every $n\in\mathbb N$.
\end{theor}
In order to prove this theorem, first we need to guarantee the existence of {\it inwards stable droplets},
which are, basically, discrete polygons that can not be infected from outside,
it is possible to do so by using the condition $\mathcal{S}(\mathcal U)=S^1$. After that, we introduce the {\it dilation radius}, which is a constant depending on $\mathcal U$, used to obtain an extremal lemma that gives us a quantitative measure of the ratio $|\langle A\rangle_\mathcal U|/|A|$. Finally, we combine ideas used by Bollob\'as and Riordan
in classical percolation models to conclude.

\subsection{Inwards stable droplets and the dilation radius}
Given $x,y\in\mathbb R^2$ we denote the usual euclidean distance between $x$ and $y$ by $\|x-y\|$,
and $B_\rho(x)$ is the ball of radius $\rho>0$ centered at $x$:
\begin{equation}
  B_\rho(x):=\{y\in\mathbb R^2: \|x-y\|\le \rho\}.
\end{equation}
For simplicity, we denote $B_\rho:=B_\rho(0)$.
Imagine for a moment that we have a convex set $D$ in the plane
and suppose it is inscribed in $B_\rho$,
then we know that any other ball with radius $\rho$ and center outside
$B_{3\rho}$ is disjoint from $D$.
This simple remark will be important to prove
Theorem \ref{expdecIntro2}
(see Lemma \ref{extremaldeK}).

\begin{defi}\label{tripling}
	Let us define a {\it rounded droplet} $D$ as the intersection of $\mathbb Z^2$ with a bounded convex set in the plane.
	We say that $D\subset\mathbb Z^2$ is {\it inwards stable} for $\mathcal U$ if
	\begin{equation}
	\langle \mathbb Z^2\setminus D\rangle_\mathcal U=\mathbb Z^2\setminus D.
	\end{equation}
\end{defi}
We need to guarantee the existence of inwards stable (rounded) droplets, note that they are finite; this is the only point where we use the hypothesis $\mathcal{S}(\mathcal U)=S^1$.
\begin{lemma}[Existence, \cite{BBPS16}]\label{existeindrop}
	If $\mathcal{S}(\mathcal U)=S^1$ then, there exist an inwards stable droplet $D$ such that
	$0\in D$.
\end{lemma}
The origin $0\in\mathbb Z^2$ has no special role here, it is just a reference point to locate the droplet $D$.
Any translate of $D$ is inwards stable as well.

There are several choices for the shape of inwards stable droplets. The following proof is included in \cite{BBPS16}, and
shows that $D$ could be a polygon or not; this fact justifies the {\it rounded} term in the definition.
\begin{proof}[Sketch of proof of Lemma \ref{existeindrop}]
Suppose that $B_\rho$ is initially healthy.
	If $\rho$ is large enough then every rule $X\in\mathcal U$ can only infect sites in disjoint circular segments
	`cut off' from $B_\rho$ using chords of length at most
\[\nabla(\mathcal U):=\max_{X\in\mathcal U}\max_{x,y\in X}\|x-y\|,\]
\begin{figure}[ht]

	\includegraphics[width=.3\textwidth]{chords}
	\caption{Set of disjoint circular segments cut off from $B_\rho$ using chords perpendicular to directions $\pi/4$, $\pi/2$ and $7\pi/8$.}
	\label{figBEAM}
\end{figure}
and parallel to the sides of $\textup{Hull}(X)$, and these segments are all either disjoint or contained in each other for different
	rules, since $\rho$ is large. No additional infection takes place in $B_\rho$, therefore
	$D=B_\rho\setminus\langle \mathbb Z^2\setminus B_\rho\rangle_\mathcal U$
	is inwards stable.
\end{proof}
Now, given $\rho>0$ we denote the discrete ball as \[B'_{\rho}:=\mathbb Z^2\cap B_{\rho}.\]
An immediate consequence of the above lemma is the fact that every vertex which is eventually
infected should be within some constant distance from an initially infected vertex.
\begin{coro}\label{dilation}
	If $\mathcal{S}(\mathcal U)=S^1$, there exists $\hat \rho>0$ such that, for every $x\in \langle A\rangle_\mathcal U$,
	\begin{equation}
	  A\cap B'_{\hat \rho}(x)\ne \emptyset.
	\end{equation}
\end{coro}
\begin{proof}
	Let $D$ be an inwards stable droplet with $0\in D$, and $\hat \rho>0$ such that
	$D\subset B_{\hat \rho}$.
	Given $x\in \langle A\rangle_\mathcal U$, the translation $x+D$ is also inwards stable and
	$x\in \langle A\rangle_{\mathcal U}\cap (x+D)$. Thus \[A\cap B_{\hat \rho}(x)\supset A\cap  (x+D)\ne\emptyset.\]
\end{proof}

\begin{defi}[Dilation Radius]
	We define the {\it dilation radius} $\beta:=\beta(\mathcal U)$ to be the smallest radius $\hat \rho\ge 1$
	satisfying the conclusion in Corollary \ref{dilation}.

\end{defi}
Note that
\begin{equation}
  |B'_{3\beta}|\le 30\beta^2.
\end{equation}

\subsection{Exponential decay}
We will use a specific collection of finite subtrees of $\mathbb Z^2$.
\begin{defi}
	For $n \ge 0$ we let $\mathcal T_{0,n}$ to be the collection of all trees $T\subset\mathbb Z^2$ containing the origin $0\in\mathbb Z^2$
	and other $n$ vertices (so that $|T|=n+1$). We also define the collection
	of all trees containing $0$ and having at most $n$ vertices ($|T|\le n$) by
	\begin{equation}
	  \mathcal T_{\le n}:= \bigcup_{k=1}^n \mathcal T_{0,k-1}.
	\end{equation}
\end{defi}
A key ingredient to prove the exponential decay theorem is an upper bound for $|\mathcal T_{\le n}|$.
The following proposition is a particular case of a beautiful problem in the book
{\it The art of mathematics: Coffee time in Memphis} (see Problem 45 in \cite{Btaofm}).
\begin{prop}\label{nofsubt}
	For every $n\ge 1$ we have
	$|\mathcal T_{0,n}|\le (3e)^n$. As a consequence, $|\mathcal T_{\le n}|\le (3e)^n$.
\end{prop}
Consider $\mathcal U$-bootstrap percolation with initially infected set $A\subset\mathbb Z^2$,
where $\mathcal{S}(\mathcal U)=S^1$ and let $\beta$ be the dilation radius.
\begin{lemma}[Extremal lemma for $\mathcal K$]\label{extremaldeK}
	If $|\mathcal K|\ge n$
	then, there exists a tree $T\in\mathcal T_{\le n}$ such that
	\begin{equation}
	  |A\cap T|\ge (30\beta^2)^{-1} n. \end{equation}
\end{lemma}
\begin{proof}
In fact, let us suppose that $|\mathcal K|\ge 30\beta^2 n$,
and recursively find $n$ distinct vertices $x_1',\dots,x_n'\in A\cap T$, for
some tree $T\in\mathcal T_{\le 30\beta^2 n}$.

By definition of $\beta$, for $x_1=0\in\langle A\rangle_\mathcal U$ there exists $x_1'\in A\cap B'_\beta(x_1)$.
Then set $K_1= B'_{3\beta}(x_1)$, and since $|K_1|\le 30\beta^2$
we can find a vertex $x_2\in\mathcal K\setminus K_1$,
which is at distance 1 from $K_1$;
now we apply Corollary \ref{dilation} to $x_2\in\langle A\rangle_\mathcal U$ and find a new vertex
$x_2'\in A\cap B'_{\beta}(x_2)$.
Proceed in this way, for $i\le n$, assume we have found
vertex $x_i'\in A\cap B'_{\beta}(x_{i-1})$, then set
\[K_i=B'_{3\beta}(x_{i})\cup K_{i-1}.\]
Since $|K_i|\le 30\beta^2i$, for $i=1,\dots,n-1$ we have
\[|\mathcal K\setminus K_i|\ge 30\beta^2n-30\beta^2i\ge 1,\] so we can find a vertex $x_{i+1}\in\mathcal K\setminus K_i$,
which is at distance 1 from $K_i$.
Observe that at step $n-1$ we still have $|\mathcal K\setminus K_{n-1}|\ge 30\beta^2\ge 1$, so for $x_n\in\mathcal K\setminus K_{n-1}$ we can apply
the corollary one more time to get our last vertex $x_n'\in A$.
For $i=1,\dots,n$, the vertices $x_i'$ are all distinct because all balls
$B'_\beta(x_i)$ are pairwise disjoint by construction.

Finally, consider a spanning tree $T$ of $K_n$, and note that $x_i,x_i'\in T$ for all
$i=1,\dots,n$. In particular, $|A\cap T|\ge n$, and the fact that
$T\in\mathcal T_{\le 30\beta^2 n}$
follows from $0=x_1\in T$ and $|T|\le|K_n|\le 30\beta^2 n$.
\end{proof}

The same proof allows us to deduce another similar extremal lemma.
\begin{lemma}\label{stuck}
	There exists a constant $\lambda\in(0,30\beta^2]$ such that, if $\langle A\rangle_{\mathcal U}$ is connected then,
\begin{equation}
    |\langle A\rangle_{\mathcal U}|\le \lambda |A|.
\end{equation}
\end{lemma}
\begin{proof}
	If $A$ is infinite we have nothing to show. Assume $A$ is finite, then it is contained in a big
	rectangle $R\subset\mathbb Z^2$, since $\pm e_1,\pm e_2\in\mathcal{S}$, so $\langle A\rangle_\mathcal U\subset R$ is also finite.
	Since $\langle A\rangle_{\mathcal U}$ is connected, the above proof shows that $|\langle A\rangle_\mathcal U|>30\beta^2 n$ implies $|A|>n$.
	In other words, $|A|=n$ implies $|\langle A\rangle_\mathcal U|\le 30\beta^2 n= 30\beta^2 |A|$.
\end{proof}

The following is a quantitative reformulation of Theorem \ref{expdecIntro2}, whose proof is inspired by lines through the book {\it Percolation} of Bollob\'as and Riordan (see pp. 70 in \cite{ByR}).

\begin{theor}[Exponential decay for the cluster size]\label{expdecay}
Consider subcritical $\mathcal U$-bootstrap percolation on $\mathbb Z^2$ with $\mathcal{S}(\mathcal U)=S^1$ and let $\beta\ge 1$ be the dilation radius.
If $0<\varepsilon <e^{-150\beta^2}$ and $C=C(\varepsilon):=-\frac{1}{60\beta^2}\log(\varepsilon)$, then
	\begin{equation}
	\mathbb P_\varepsilon(|\mathcal K|\ge n)\le
 \varepsilon^{\frac{1}{60\beta^2}n}=e^{-Cn},
	\end{equation}
	for every $n\in\mathbb N$.
\end{theor}
\begin{proof}

	By Lemma \ref{extremaldeK} and Proposition \ref{nofsubt}, with
	$\delta=(30\beta^2)^{-1}$, we obtain
	\begin{align*}
	\mathbb P_\varepsilon(|\mathcal K|\ge n) &

	\le  \mathbb P_\varepsilon\Bigg(\bigcup_{T\in\mathcal T_{\le n}} \{|A\cap T|\ge \delta n\}\Bigg)
 \le \sum_{T\in\mathcal T_{\le n}}\mathbb P_\varepsilon(|A\cap T|\ge \delta n)
 \le \sum_{T\in\mathcal T_{\le n}}
	{{n}\choose{\delta n}}\varepsilon^{\delta n}    \\
& \le \sum_{T\in\mathcal T_{\le n}}  (e\delta^{-1}\varepsilon)^{\delta n}

\le ([3e][e\delta^{-1} \varepsilon]^{\delta})^{n}
\le e^{-Cn},
	\end{align*}
	and we are done.
\end{proof}

\section{Lower bounds via beams process}\label{SectionLowerBeams}\label{SectionLowerBeam}
To deal with the cases $c\ge a+b$ we introduce a new tool which we
call the {\it beams process}. This time,
instead of covering the infected vertices step by step with components,
we cover them with beams, so that when we observe this induced process along the
$e_3$-direction it looks like subcritical two-dimensional bootstrap percolation.

Consider the family $\mathcal N_{m}^{a,b}$
given by the collection of all subsets of size $m$ of
\begin{equation}\label{Nab0}
 N_{a,b}=\{a'e_1:\pm a'\in[a]\}\cup \{b'e_2:\pm b'\in[b]\}.
\end{equation}
Observe that $\mathcal{S}(\mathcal N_{m}^{a,b})=S^1$   if and only if $m\ge a+b+1$,
in particular, our exponential decay result (Theorem \ref{expdecay}) holds for these families.
From now on we set
\begin{equation}
m:=a+b+1.
\end{equation}
\begin{defi}\label{beam}
A {\it beam} is a finite subset of $\mathbb Z^3$ of the form $H\times[w]$, where $H\subset\mathbb Z^2$ is
 connected and $\langle H\rangle_{\mathcal N_{m}^{a,b}}=H$.
\end{defi}

It will be important for us to have an upper bound on the number of beams of a given size,
which are contained in $[L]^3$.
The following lemma is another consequence of Proposition \ref{nofsubt}.
\begin{lemma}[Counting beams]\label{nofbeams}
	Let $\mathcal B_{n_1,n_2}$ be the collection of all copies of the beam $H\times[w]$ contained in $[L]^3$ satisfying $w\le n_1$ and $|H|\le n_2$.
	Then
	\[|\mathcal B_{n_1,n_2}|\le n_1L^3(3e)^{n_2}.\]
\end{lemma}
\begin{proof}
The number of segments inside $[L]$ 	with at most $n_1$ vertices, is at most $n_1L$.

	Now we give an upper bound for the number of $H$'s. Let $\mathcal H_h$ denote the collection of all
	connected sets $H\subset[L]^2$ such that $|H|=h$,
	so we can write \[h|\mathcal H_h|=\sum_{H\in\mathcal H_h}|H|=\sum_{x\in[L]^2}\sum_{H\in\mathcal H_h}
	\mathds 1\{x\in H\}= \sum_{x\in[L]^2} \textup{cs}(x),\]
	where $\textup{cs}(x)$ is the number of connected subsets of $[L]^2$ with size $h+1$, containing a fixed point $x$.
	To each of such sets we can associate an spanning tree in an injective fashion, so by Proposition
	\ref{nofsubt}, $|\mathcal H_h|\le L^2(3e)^{h-1}$. It follows that the number of $H$'s is at most
	\[\sum_{h=1}^{n_2} |\mathcal H_h|  \le L^2\sum_{h=1}^{n_2} (3e)^{h-1} \le L^2(3e)^{n_2}.\]
\end{proof}

\subsection{The beams process}
\begin{defi}\label{genbeam}
Given finite connected sets $S_1,S_2\subset\mathbb Z^3$, we say that a beam $H\times[w]$ is {\it generated by} $(S_1,S_2)$ if it can be constructed in the following way: by translating $S_1\cup S_2$ if necessary, we can assume that the smallest rectangle containing it is $R\times[w]$,
 then consider the connected sets $H_1,H_2\subset \mathbb Z^2$ given by
 \[H_t:=\{x\in R: (\{x\}\times[w])\cap S_t\ne\emptyset\},\ \ \ t=1,2.\]
If $\langle H_1\cup H_2\rangle_{\mathcal N_{m}^{a,b}}$ is connected then we take $H:=\langle H_1\cup H_2\rangle_{\mathcal N_{m}^{a,b}}$.
Otherwise, we let $P\subset R$ be any path with minimal size connecting $H_1$ to $H_2$ and then set $H:=\langle H_1\cup H_2\cup P\rangle_{\mathcal N_{m}^{a,b}}$.
\end{defi}
In this definition $\langle S_1\cup S_2\rangle\subset H\times[w]$ for each $r\ge m$, and generated beams could depend on the choice of the path $P$. However, such minimal paths are not relevant for our purposes.
\begin{nota}
 We will denote any fixed beam generated by $(S_1,S_2)$ as $B(S_1\cup S_2)$, regardless the choice of $P$.
\end{nota}
We want to track the process of infection by covering all possible infected
sites with beams, we do that step by step in order to get some control over the sizes.
The following algorithm is a variation of the components process.
We will use it to show an Aizenman-Lebowitz-type lemma which says that when $[L]^3$
is internally filled, then it contains {\it covered} beams of all intermediate sizes (see Lemma \ref{ALbeams} below).
\begin{defi}[The beams process]\label{beampro}
 Let $A=\{x_1,\dots,x_{|A|}\}\subset [L]^3$ and fix $r\ge c+1$.
 Set $\mathcal B:=\{S_1,\dots,S_{|A|}\}$, where $S_i=\{x_i\}$ for each $i=1,\dots,|A|$, and
repeat until STOP:
 \begin{enumerate}
  \item If there exist distinct beams $S_{1},S_{2}\in\mathcal B$ such that
  \[S_{1}\cup S_{2}\]
  is strongly connected, then remove it from $\mathcal B$,
  and replace by $B(S_{1}\cup S_{2})$.
  \item If there do not exist such a family of sets in $\mathcal B$, then STOP.
 \end{enumerate}
We call any beam $S=B(S_{1}\cup S_{2})\subset [L]^3$ added to the collection $\mathcal B$
a {\it covered beam}, and denote the event that $S$ is covered by
$I^{\text{\ding{86}}}(S).$
\end{defi}
Again, there are two properties that are due to the way the algorithm evolves:
\begin{itemize}
 \item  Any covered beam $S$ satisfies $\langle A\cap S\rangle\subset \langle S\rangle=S$.
 \item
 The process stops in finite time, thus, we can consider the final collection $\mathcal B'$ and set  $V(\mathcal B'):=\bigcup\limits_{S\in\mathcal B'}S$.
By using the same arguments in the proof of Lemma \ref{final=gen}, it follows that $\langle A\rangle\subset V(\mathcal B')$.
\end{itemize}

\subsection{Case $c=a+b$}\label{sectionbeam}
In this section we prove the following.
\begin{prop}\label{lower3}
Under $\mathcal N_{m}^{a,b,a+b}$-bootstrap percolation, there is a constant $\gamma=\gamma(a,b)>0$
such that, if
\[L<\exp(\gamma p^{-1}),\]
then
$\mathbb P_p[I^{\bullet}([L]^3)]\to 0,\ as\ p\to 0.$
\end{prop}
The beams process and Lemma \ref{stuck} allow us to prove a beams version of the Aizenman-Lebowitz Lemma for this case.
Let $\lambda>0$ be the constant in Lemma \ref{stuck}
associated
to the subcritical two-dimensional family $\mathcal N_{m}^{a,b}$.
\begin{lemma}\label{ALbeams}
	Consider $\mathcal N_{r}^{a,b,c}$-bootstrap percolation with $r\ge c+1$.
 If $[L]^3$ is internally filled, then there is a constant $\kappa\in\mathbb N$ such that
 for every $k=\kappa,\dots, L$, there exists a covered beam $H\times[w]$ satisfying $w,|H|\le 3\lambda k$, and either $w\ge k$ or $|H|\ge k$.
\end{lemma}
\begin{proof}

	Let $S=H\times[w]$ be the first beam that appears in the beam process satisfying either $w\ge k$ or $|H|\ge k$
	(such a set exists since $V(\mathcal B')=[L]^3$).

	Then, it is enough to show that $w\le rk$ and $|H|\le 3\lambda k$.

	We know that $S=B(S_{1}\cup S_{2})$ for some beams $S_{t}=H_t\times[w_t]$ such that $S_{1}\cup S_{2}$ is strongly connected. Moreover,
	by definition of $S$, $w_t\le k-1$ for $t=1,2$, so
	\begin{equation}\label{maxdiam}
	w\le 2 \max_{t=1,2}\{w_t\} +r \le r(k-1)+r\le rk.
	\end{equation}
	Analogously, $|H_t|\le k-1$, and we know that $H= \langle H_1\cup H_2\cup P\rangle_{\mathcal N_{m}^{a,b}}$ for some path $P$ with bounded (or zero) size, $H$ is connected, so by Lemma \ref{stuck},
	\begin{equation}\label{maxdiam}
	|H|\le \lambda\cdot 2 \max_{t=1,2}\{|H_t|\} + \lambda|P|\le 2\lambda(k-1) + O(1)\le 3\lambda k.
	\end{equation}
\end{proof}
Now, let us prove the lower bound in the case $c=a+b$.
\begin{proof}[Proof of Proposition \ref{lower3}]
Take $L<\exp(\gamma p^{-1})$, where $\gamma>0$ is some small constant. Let us show that
$\mathbb P_p(I^{\bullet}([L]^3))$ goes to $0$, as $p\to 0$. Fix $\varepsilon>0$.

If $[L]^3$ is internally filled, by Lemma \ref{ALbeams} there exists a covered beam $S=H\times[w]\subset [L]^3$
satisfying $w,|H|\le\varepsilon/p$, and moreover, either $w\ge\varepsilon/3\lambda p$ or $|H|\ge \varepsilon/3\lambda p$, hence, by union bound, $\mathbb P_p[I^{\bullet}([L]^3)]$ is at most
\begin{align*}
 \sum_{S\in\mathcal B_{\frac{\varepsilon}{p},\frac{\varepsilon}{p}}}\big(\mathbb P_p[I^{\text{\ding{86}}}(S)\cap
\{w\ge\varepsilon/3\lambda p\}]+\mathbb P_p[I^{\text{\ding{86}}}(S)\cap \{|H|\ge \varepsilon/3\lambda p\}]\big).
\end{align*}

To bound the first term, we use the fact that $H\times[w]$ is covered; this implies that there is no gap
of size $r$ along the $e_3$-direction.
Therefore, by considering the $w/r$ disjoint slabs,
if $\varepsilon$ is small, then there exists some $c_1=c_1(\varepsilon,r)>0$ such that
\begin{align*}
\mathbb P_p[I^{\text{\ding{86}}}(H\times[w])\cap
\{w\ge\varepsilon/3\lambda p\}] & \le
\Big(1-e^{-\Omega(p|H|)}\Big)^{w/r}
 =\Big(1-e^{-\Omega(\varepsilon)}\Big)^{\varepsilon/2r\lambda p}
 \le e^{-c_1/p}.
\end{align*}

To bound the second term, for each $S\in\mathcal B_{\frac{\varepsilon}{p},\frac{\varepsilon}{p}}$ consider the set
\[A':=\left\{x\in[L]^2: (\{x\}\times[w])\cap\langle A\cap S\rangle \ne\emptyset
\right\}.\]
In other words, $x\in A'$ if and only if there exists $y_1\in \{x\}\times [w]$ such that
either $y_1\in A$, or $y_1\in S$ got infected by using at least $m$
infected neighbours in $y+N_{a,b}$, where
$N_{a,b}$ is given by (\ref{Nab0}).
Now, by applying Markov's inequality,
\[\mathbb P_p(A\cap(\{x\}\times [w])\ne \emptyset) =
O(wp)\le \varepsilon.\]
Therefore, by monotonicity we can couple the process on $[L]^2\times[w]$ having initial infected set
$A$, with $\mathcal N_{m}^{a,b}$-bootstrap
percolation on $[L]^2\times\{1\}\subset\mathbb Z^2$ where the initial infected set is chosen to be $\varepsilon$-random.

In particular, under $\mathcal N_{m}^{a,b}$-bootstrap percolation there should exist a connected component of size at least
$|H|\ge \varepsilon/3\lambda p$ inside $[L]^2$.
On the other hand, there are at most
$L^2$ possible ways to place the origin in $H$, so if $\mathcal K$ denotes the cluster of 0, Theorem \ref{expdecay} implies

\begin{align*}
\mathbb P_p[I^{\text{\ding{86}}}(S)\cap \{|H|\ge \varepsilon/3\lambda p\}]
& \le \sum_{x\in [L]^2}\mathbb P_\varepsilon(\{|\mathcal K|>\varepsilon/3\lambda p\}\cap \{x=0\})
 \le L^{2}\mathbb P_\varepsilon(|\mathcal K|\ge \varepsilon/3\lambda p) \\
& \le  e^{2\gamma/p}e^{-C\varepsilon/3\lambda p}
 = e^{-(C\varepsilon/3\lambda -2\gamma)/p},
\end{align*}
where $C=-\frac{1}{60\beta^2}\log\varepsilon$ and we choose $\varepsilon>0$ such that $C\varepsilon>0$ and $\gamma<C\varepsilon/6\lambda$ at first.
By Lemma \ref{nofbeams} we conclude that
\begin{align*}
\mathbb P_p[I^{\bullet}([L]^3)] & \le \sum_{S\in\mathcal B_{\frac{\varepsilon}{p},\frac{\varepsilon}{p}}}\big( e^{-c_1/p}+e^{-(C\varepsilon/3\lambda -2\gamma)/p}  \big)
 \le \frac{\varepsilon}{p}L^3(3e)^{\varepsilon/p}e^{-c_2/p}\\
& \le e^{4\gamma/p}e^{\varepsilon\log(3e)/p}e^{-c_2/p}\to 0,
\end{align*}
for $c_2,\gamma>0$ small enough.
\end{proof}

\subsection{The coarse beams process}
  In this section we study the last case $c\ge a+b+1$. The lower bound will be proved by using a coupling with subcritical two-dimensional bootstrap percolation again, as we did in the previous section, however, this time we infect squares instead of single vertices.
The trick now is to consider the following coarser process.
\begin{defi}[Coarse bootstrap percolation]\label{coarse}
 Assume that $b+1$ divides $L$ and we partition $[L]^2$ as $L^2/(b+1)^2$ copies of
 $\boxplus:=[b+1]^2$ in the obvious way, and think of $\boxplus$ as a single vertex in the new scaled grid $[L/(b+1)]^2$.
 Given a two-dimensional family $\mathcal U$, suppose we have some fully infected copies of $\boxplus\in [L/(b+1)]^2$ and denote this initially infected set by $A$, then we define
 {\it coarse} $\mathcal U$-bootstrap percolation to be the result of applying $\mathcal U$-bootstrap percolation
 to the new rescaled vertices. We denote the closure of this process by $\langle A\rangle_{b}$.
\end{defi}
To avoid trivialities, we assume that $b+1$ divides $L$. Set
\[m:=a+b+1< c+1 = r.\]
\begin{defi}\label{coarbeam}
A {\it coarse beam} is a finite set
of the form $H\times[w]$,
 where $H\subset\mathbb Z^2$ is connected and $\langle H\rangle_{b}=H$
 under coarse $\mathcal N_{m}^{a,b}$-bootstrap percolation.
\end{defi}
\begin{nota}\label{constru2}
Given finite connected sets $S_1,S_2\subset [L]^2\times[L]$, we partition $[L]^2$ as in Definition \ref{coarse} and denote by $B_b(S_1\cup S_2)$ the coarse beam generated by $(S_1,S_2)$ which is constructed in the (coarse) analogous way, as we did in Definition \ref{genbeam}, using coarse paths when needed. Note that every coarse beam is a beam in the sense of the previous section.
\end{nota}
The following algorithm is a refinement of that one given in Definition \ref{beampro}.
\begin{defi}[The coarse beams process]
Let $A=\{x_1,\dots,x_{|A|}\}\subset [L]^3$ and fix $r\ge c+1$.
 Set $\mathcal B:=\{S_1,\dots,S_{|A|}\}$, where $S_i=\{x_i\}$ for each $i=1,\dots,|A|$, and
repeat until STOP:
 \begin{enumerate}
  \item If there exist distinct beams $S_{1},S_{2}\in\mathcal B$ such that
  \[S_{1}\cup S_{2}\]
  is strongly connected, and $\langle S_{1}\cup S_{2} \rangle\ne
  S_{1}\cup  S_{2}$,
  then choose a minimal such family, remove it from $\mathcal B$,
  and replace by a coarse beam $B_b(S_{1}\cup S_{2})$.
  \item If there do not exist such a family of sets in $\mathcal B$, then STOP.
 \end{enumerate}
We call any beam $S=B_b(S_{1}\cup  S_{2})$ added to the collection $\mathcal B$ a {\it covered beam}, and denote the event that $S$ is covered by
$I_b^{\text{\ding{86}}}(S).$
 \end{defi}

 The two highlighted usual properties are preserved for this algorithm too:
\begin{itemize}
 \item  Any covered beam $S$ satisfies $\langle A\cap S\rangle\subset \langle S\rangle=S$.
 \item There is a final collection $\mathcal B'$ and we can set
 $V(\mathcal B'):=\bigcup\limits_{S\in\mathcal B'}S$. Then, we also have $\langle A\rangle\subset V(\mathcal B')$.
\end{itemize}
\subsection{Case $c>a+b$}
  In this section we prove the lower bound corresponding to the last case.

\begin{prop}\label{lower4}
	Under $\mathcal N_{c+1}^{a,b,c}$-bootstrap percolation with $c>a+b$, there exists a constant $\gamma=\gamma(c)>0$

	such that, if
	\[L<\exp(\gamma p^{-1}(\log p)^2),\]
then
$\mathbb P_p[I^{\bullet}([L]^3)]\to 0,\ as\ p\to 0.$
\end{prop}
We state an analogue of Lemma \ref{ALbeams} for the coarse beams setting without proof
because the arguments are exactly the same. However, we obtain slightly different constants since the number of vertices of the form $\boxplus$ in a coarse beam $H$ equals
$|H|/(b+1)^2$.

Consider $\mathcal N_{r}^{a,b,c}$-bootstrap percolation with $r\ge c+1$, and let $\kappa_0$ be a large positive integer depending on $a,b,c$ and $r$.
\begin{lemma}\label{ALlema2}
 If $[L]^3$ is internally filled then for every $h,k=\kappa_0,\dots,L$, there exists
 a covered (coarse) beam
 $H\times[w]\subset [L]^3$ satisfying $w\le rk$, $|H|\le 2(b+1)^2\lambda h$, and
 either $w\ge k$ or $|H|\ge h$.
\end{lemma}

Finally, we prove the lower bound for the remaining case.
\begin{proof}[Proof of Propositon \ref{lower4}]
Take $L<\exp(\gamma p^{-1}(\log p)^2)$, where $\gamma>0$ is some small constant. Let us show that
$\mathbb P_p(I^{\bullet}([L]^3))$ goes to $0$, as $p\to 0$. Fix $\delta>0$ and set
\[h=\delta p^{-1}\log \tfrac 1p ,\ \ \ k=p^{-\frac 32}\]
If $[L]^3$ is internally filled, by Lemma \ref{ALlema2} there exists a covered beam
$S=H\times[w]\subset [L]^3$ satisfying $w\le k$, $|H|\le (b+1)^2h$, and
either $w\ge k/2\lambda$ or $|H|\ge h/2\lambda$ (as we said, the cardinality of
$H$ viewing $S$ as a beam equal $(b+1)^2|H|$ viewing $S$ as a coarse beam), hence $\mathbb P_p[I^{\bullet}([L]^3)]$ is at most
\begin{align*}
 \sum_{S\in\mathcal B_{k,(b+1)^2h}}\Big(\mathbb P_p[I_b^{\text{\ding{86}}}(S)\cap
\{w\ge k/2\lambda\}]+\mathbb P_p[I_b^{\text{\ding{86}}}(S)\cap \{|H|\ge h/2\lambda\}]\Big).
\end{align*}
To bound the first term, we use the fact that
$A\cap (H\times\{rk+1,\cdots,rk+r\})\ne \emptyset$ for all $k=0,\dots,w/r-1$,
since $H\times[w]$ is covered. Therefore, for some $c_1>0$,
\begin{align*}
\mathbb P_p[I_b^{\text{\ding{86}}}(H\times[w])\cap  \{w\ge k/2\lambda\}]
&\le \big(1-(1-p)^{rh}\big)^{w/r}
 \le \big(1-e^{-2r\varepsilon\log\frac 1p}\big)^{k/2r\lambda}
 = \big(1-p^{2r\varepsilon}\big)^{k/2r\lambda}\\
& \le e^{-p^{2r\varepsilon-\frac 32}/2r\lambda}
=e^{-c_1 p^{-1}(\log\frac 1p)^2}.
\end{align*}
To bound the second term we use the fact that $r=c+1\ge a+b+2$. More precisely, if $[L]^3$ is internally filled, then every copy of
$[b+1]^2\times [L]$
should contain at least 2 vertices of $A$ within some constant distance,
otherwise, there is no way to infect such a copy.

Then, given $S=H\times[w]\in\mathcal B_{k,(b+1)^2h}$ consider the set
$A'$ consisting of all copies of $\boxplus\subset [L]^2$ (as in Definition \ref{coarse})
such that the rectangle $\boxplus\times [w]\subset S$ contains at least 2 vertices of $A$ within
distance $r$.
By union bound, the probability of finding such vertices is at most
\[\sum_{x\in\boxplus\times [w]}\sum_{0<\|y-x\|\le r}\mathbb P_p(x,y\in A)\le
\tilde{C}wp^2\le p^{\frac 13}.\]
Therefore, by monotonicity we can couple the process in $[L]^2\times[w]$ having initial infected set
$A$, with coarse $\mathcal N_{m}^{a,b}$-bootstrap
percolation on $[L/(b+1)]^2$ and initial infected set $\varepsilon$-random
with $\varepsilon=\varepsilon(p):=p^{1/3}$.

In particular, under $\mathcal N_{m}^{a,b}$ (coarse) there should exist a connected component of size at least
$|H|\ge h/2\lambda$ inside $[L]^2$. Once more, there are at most
$L^2$ possible ways to place the origin in $H$, so if $\mathcal K$ denotes the (coarse) cluster of 0, Theorem \ref{expdecay} implies

\begin{align*}
\mathbb P_p[I_b^{\text{\ding{86}}}(S)\cap \{|H|\ge h/2\lambda\}]
& \le \sum_{\boxplus\subset [L]^2}\mathbb P_\varepsilon(\{|\mathcal K|>h/2\lambda\}\cap \{\boxplus=0\})
 \le L^{2}\mathbb P_\varepsilon(|\mathcal K|\ge h/2\lambda) \\
& \le  e^{2\gamma p^{-1}(\log\frac 1p)^2}e^{-Ch/2\lambda}
 = e^{-(c' -2\gamma) p^{-1}(\log\frac 1p)^2},
\end{align*}
for some constant $c'=c'(\beta,\lambda)>0$ (recall that $C\sim-\log p$ asymptotically, by
Theorem \ref{expdecay}). Take $\gamma<c'/2$ at first;
by Lemma \ref{nofbeams} we conclude that
\begin{align*}
\mathbb P_p[I^{\bullet}([L]^3)] & \le \sum_{S\in\mathcal
	B_{k,(b+1)^2h}}\Big( e^{-c_1 p^{-1}(\log\frac 1p)^2}
+e^{-(c' -2\gamma)p^{-1}(\log\frac 1p)^2}  \Big) \\
&  \le kL^3(3e)^{(b+1)^2h}e^{-c_3p^{-1}(\log\frac 1p)^2}
 \le e^{4\gamma p^{-1}(\log\frac 1p)^2}e^{-c_3p^{-1}(\log\frac 1p)^2}\to 0,
\end{align*}
as $p\to 0$, for $c_3,\gamma>0$ small enough, and we are finished.
\end{proof}

\section{Future work}
All proofs in this paper extend to the case $r=c+2$, and can be used to determine $\log L_c\left(\mathcal N_{c+2}^{a,b,c},p\right)$ up to a constant factor for all triples $(a,b,c)$, except for $c=a+b-1$ which is a new interesting case to be studied separately.

In general, a problem which remains open is the determination of the threshold for other values of $r$.
We believe that the techniques used to prove Theorem~\ref{target0} can be adapted to cover all $c+1<r\le b+c$
(though significant technical obstacles remain); in this case, by Proposition \ref{genuppbound} below, the critical length is singly exponential.
However, to deal with the cases $b+c<r<a+b+c$,
the techniques required are likely to be more similar to those of~\cite{CC99} and~\cite{AA12}, and the critical length should be doubly exponential.

Finally, Theorem \ref{expdecay} can be generalized to all dimensions $d\ge 3$ and all families $\mathcal U$ such that $\mathcal{S}(\mathcal U)= S^{d-1}$.
However, we do not know if this property holds for subcritical families $\mathcal U$ satisfying $\mathcal{S}(\mathcal U)\ne S^{d-1}$.
In order to determine the critical lengths for general critical models, it could be useful to extend this result to a wider class of subcritical families.
\begin{problem}\label{probExDecay}
Characterize the subcritical $d$-dimensional update families $\mathcal U$
such that $\mathcal K$ has the exponential decay property.
\end{problem}

redpajama_set_name: RedPajamaArXiv


text:
Q: Find all elements in array that contain the search key Is there a built in method in Typescript that searches a string array and finds all items that contain the search string. e.g.
var myarray = ["apple", "orange", "app"];
var searchKey = "app"

this search key should return the array ["apple", "app"]

redpajama_set_name: RedPajamaStackExchange


text:
Upstate Life
Arts + Culture Features
Damsel In thisDress
Best. Sentence. Ever.
Pure + Simple
Michael Lang, Woodstock 1969 and Woodstock 50 Producer
Evan Agostini/Invision/AP/REX/Shutterstock
Woodstock 50 is officially canceled
By Lonny Chin The Scene
After three location changes, loss of investors, and loss of an artist lineup, Woodstock 50 is now officially canceled. The festival was set to be held Aug. 16-18 at the Merriweather Post Pavilion in Columbia, Maryland, as a free concert after losing the initial venue at Watkins Glen and the backup venue at Vernon Downs.
Over the past couple of weeks, headliners such as Jay-Z, Dead & Co., John Fogerty, Miley Cyrus, The Raconteurs, Santana and The Lumineers had all pulled out of the festival. When organizers decided to move it to Maryland, they also released all the artists' contracts, so technically no artists were obligated to play the festival. The Merriweather Post Pavilion also had a Smashing Pumpkins and Noel Gallagher's High Flying Birds concert scheduled for Aug. 17, the Saturday during the Woodstock 50 Festival. Scheduling conflicts, loss of artists and untimely planning led to the organizers' decision to cancel the festival.
This statement was released earlier today:
"We are saddened that a series of unforeseen setbacks has made it impossible to put on the Festival we imagined with the great line-up we had booked and the social engagement we were anticipating, said Michael Lang, co-founder of the Woodstock Music and Arts Festival. "When we lost the Glen and then Vernon Downs we looked for a way to do some good rather than just cancel. We formed a collaboration with HeadCount to do a smaller event at the Merriweather Pavilion to raise funds for them to get out the vote and for certain NGOs involved in fighting climate change. We released all the talent so any involvement on their part would be voluntary. Due to conflicting radius issues in the DC area many acts were unable to participate and others passed for their own reasons. I would like to encourage artists and agents, who all have been fully paid, to donate 10% of their fees to HeadCount or causes of their choice in the spirit of peace. Woodstock remains committed to social change and will continue to be active in support of HeadCount's critical mission to get out the vote before the next election. We thank the artists, fans and partners who stood by us even in the face of adversity. My thoughts turn to Bethel and its celebration of our 50th Anniversary to reinforce the values of compassion, human dignity, and the beauty of our differences embraced by Woodstock"
Bethel Woods Center for the Arts, the original site of the 1969 Woodstock in Bethel, New York, will still host daily events from Aug. 15-18 in celebration of the 50th anniversary. It will have a sold-out showing the 1970 rockumentary film of "Woodstock" on the field on Aug. 15. Ringo Starr & His All Star band will perform Aug. 16. Santana and The Doobie Brothers will play Aug. 17, and John Fogerty, The Tedeschi Trucks Band and Grace Potter will perform on Aug. 18, so attendees will still be able to "re-live" the festival at the original site.
In an attempt to re-create an event that changed the world, it is sad that the actual "festival" will not be happening. Woodstock of 1969 was outrageous and it is possible that an event like that should never happen again. I wish it would.
Arts Festival

redpajama_set_name: RedPajamaCommonCrawl


text:
NBA staffers with COVID-19 issues 'huge' problem
While more than 300 NBA players have entered the league's health and safety protocols this season because of COVID-19-related issues, several league general managers told ESPN that a debilitating issue remains the recent surge of team staffers — from coaches to support staff and beyond — entering protocols for the same reason.
Sources told ESPN that this season there have been more than 500 confirmed COVID-19 cases among 2,400 total Tier 1 staff — anyone working within 15 feet of players — across the NBA. The vast majority of cases among Tier 1 staff have also come within the past month, with more than 450 cases occurring from Dec. 7 to Jan. 10, sources said.
Such positions include equipment managers, player development staffers, video coordinators, analytics staffers, security staffers, scouts and those who work in player health positions, including athletic training staffers who double as their team's respective COVID-19 protocols officers.
General managers across the league point out that while players can often be replaced by G-League backups, replacing core staffers who handle unique roles can prove to be much more challenging and impactful on day-to-day operations.
"A huge and real impact," one Western Conference front-office executive said.
Said an Eastern Conference general manager: "The rising caseloads are stretching other team staff [to the limit]."
One NBA athletic training official, speaking on behalf of peers across the league, told ESPN, "It's been a huge issue from our members' standpoint. You're basically taking an assembly of people who help the athletes and taking a few people off the line every few days for a week or more. It has interfered significantly with the regular protocols and people being given responsibilities/duties they don't normally have or are even qualified to do in order to get the job done. It's been the Wild Wild West."
Since the pandemic began, team executives have also been more concerned about positive cases among staff, as many are older and/or have potential underlying issues that could lead to serious issues if they test positive for COVID-19.
The staffing issue has come to the forefront among teams, while the public focus has largely remained on the 328 total players who have entered protocols this season, with 252 entering in December and 69 having entered so far in January. And half of the 30 NBA head coaches have entered protocols this season, with 10 coming in December alone.
A total of 11 games have been postponed this season as some teams haven't been able to field the league-required eight players.
The league has been encouraging all eligible players and staff to take booster shots. Recently, ESPN reported that nearly 70% of eligible NBA players — a number that comprises about 350 players — have received a booster shot.


Stephanie Driggers is an Attorney with United Parcel Service, a global transportation and logistics company with operations in 220 countries and territories. She is responsible for managing complex commercial litigation both internationally and domestically across all business units. Ms. Driggers formerly had responsibility for global privacy, addressing tactical and strategic privacy and cybersecurity matters around the world.
Prior to joining UPS, Ms. Driggers was a partner at an Am Law 100 law firm, where she concentrated her practice on class action defense and data privacy litigation.
Ms. Driggers is a graduate of Vanderbilt University Law School and received her undergraduate degree from Stetson University. She clerked for Judge Thomas A. Wiseman, Jr. of the United Stated District Court in the Middle District of Tennessee. She is a Fellow of Information Privacy (FIP), a Certified Information Privacy Professional (CIPP-US), and a Certified Privacy Manager (CIPM).
Ms. Driggers focuses her pro bono and volunteer work on human trafficking.


Sales Engineer is responsible for generating and developing new business and customers in assigned territory for US customers. In addition, responsible for increasing business within the existing account base. Provides technical support in assigned territory to support growth objectives. Provides input from account base to management for future product development or strategy adjustment. Relies on experience and judgment to plan and accomplish goals and objectives.
Sales – Primary responsibility is to generate revenue and increase market share in the US assigned territory. Activities include program management, account development and establishing and maintaining key relationships for new and existing accounts. Meets assigned revenue and profitability targets.
Account Management – Develops key account plans to align resources to support company growth objectives with major customers/opportunities.
Forecasting – generates and maintains accurate forecasts for revenue and product / volume mix for assigned territory and customers.
Products – Identifies the need for new products and services. Works with management on development of products in support of market requirements.
Marketing – Gives input to the organization and the Marketing Communications group to support the goal of establishing and implementing promotional, advertising, and communication strategies. Supports trade show's as required.
Coordinates activities with other sales team members as well as with service team.
Working closely with product management and engineering team to interpret customer requirements and to offer customized technical solutions.
Work with scientists and engineers on applying government or academic related projects.
· Preferable up to 2-5 years of sales experience in relevant industries such as Photonics, Lasers, Optical Fiber, Semiconductor, Life Science Imaging System, etc.
Bachelor's Degree, MS or PhD preferred. Technical field in Physics, EE, Photonics or related.


A fantastic opportunity has become available for a Business Manager within our client's franchise approved car dealership in Pontypridd.
Reporting to the Brand Manager and working as part of a team, as a Business Manager you will carry out finance and insurance sales for the Sales Department as well as provide support in the oversight of a small team of Car Sales Executives ensuring they are fully compliant with FCA guidelines and work within relevant financial legislation, in particular, the Consumer Credit Act. Responsibilities will also include assisting in the organising, directing and controlling the activities of the sales staff and administrative support to achieve plans and budgets, ensuring optimum use of materials, financial and personnel resource.
In order to be eligible you will ideally live in and around the Pontypridd region or be within a reasonable commuting distance of Pontypridd, To succeed in this role you will need to have strong and stable business management/sales controller experience within a franchise approved dealer environment, be FCA accredited, and have a comprehensive understanding of financial legislation. You will need to have a desire to produce results, be a strong team leader, have a commitment to excellence, be able to build outstanding customer relationships, and have a genuine passion to work for the company and the manufacturer. You will need to be brand focused, capable of working independently and have the energy and commitment to seek every sales opportunity. A UK driving license with minimal points is essential at this stage.
- Starting basic salary of £18,000 per annum (negotiable pending experience and current earnings).
- Realistic and uncapped on-target-earning of between £45-50,000+ per annum.
- Access to a company car.
- 22 days annual holiday allowance plus the bank holidays (any bank holidays worked you shall gain a day off in lieu).
- Full manufacturer-approved training provided.
- Fantastic career prospects with a well-regarded car dealership that is in the top quarter for the manufacturer for customer service performance.
The hours worked for this Business Manager role are Monday to Friday 8:30am-6:00pm (1 day off in the week), Saturday 9:00am-5:00pm, and Sunday 10:00am-4:00pm (1 in 2 required).
At Perfect Placement, we specialise in Automotive Careers, so if this Car Sales Advisor Vacancy interests you or you want to find out more about other Motor Trade Jobs in Pontypridd, please contact Josh Buck today!
Our team of Automotive Recruitment Consultants all share a passion for finding our jobseekers the best jobs in the Automotive Industry. So if you are looking to improve your career and want to hear about more Motor Trade Jobs in your local area, please get in touch with us today.


text:
package com.auth0.android.lock.events;

public class CountryCodeChangeEvent {
}

redpajama_set_name: RedPajamaGithub


text:
Box Office: $150.3M
UPC: 032429240071 03/22/2016 $7.69 $5.48
Price: $7.69 $5.48
Blu-ray + DVD + Digital HD
4K Ultra HD + Blu-ray + Digital HD
In Theaters December 25 2015
On 4K-UHD February 20 2018 Now Available
On Blu-ray March 22 2016 Now Available
On DVD March 22 2016 Now Available
Will Ferrell, Mark Wahlberg, Linda Cardellini, Thomas Haden Church, Hannibal Buress, Paul Scheer, Scarlett Estevez, Bobby Cannavale, Bill Burr, Jamie Denbo, Owen Vaccaro, Mark L. Young, Matthew Paul Martinez
Adam McKay, Chris Henchy, John Morris, Sean Anders, Brian Burns, Etan Cohen
Diana Pokorny, Sean Anders, Jessica Elbaum, Kevin Messick, Will Ferrell, Adam McKay, Chris Henchy, John Morris
"DADDY'S HOME" follows a mild-mannered radio executive (Ferrell) who strives to become the best stepdad to his wife's two children, but complications ensue when their freewheeling and freeloading real father (Wahlberg) arrives, forcing him to compete for the affection of the kids.
$150.3 Million
thematic elements, crude and suggestive content, and for language


text:
Jarrod James Moseley (born 6 October 1972) is an Australian professional golfer.

Career
Moseley was born in Mandurah, Western Australia. He turned professional in 1997. He won the PGA Tour of Australasia Order of Merit in 1998/99, having won the European Tour co-sanctioned Heineken Classic during the season. He also went on to finish 16th on the European Tour Order of Merit that season, a career best.

Moseley finished well inside the top 100 on the European Tour Order of Merit every season until 2004, when he slipped to 117th, missing out on retaining his card by just one place. He regained his playing privileges immediately via the end of season qualifying school, but was again unable to replicate his form from previous years as he slipped further down the money list.

Amateur wins
1996 Riversdale Cup, Malaysian Amateur Championship

Professional wins (2)

European Tour wins (1)

1Co-sanctioned by the PGA Tour of Australasia

PGA Tour of Australasia wins (2)

*Lonard and Moseley agreed to share the 2002 Australian PGA Championship after failing light caused play to halt after one hole of a playoff.
1Co-sanctioned by the European Tour

PGA Tour of Australasia playoff record (0–1–1)

Results in major championships

Note: Moseley only played in The Open Championship.

CUT = missed the half-way cut
"T" = tied

Results in World Golf Championships

1Cancelled due to 9/11

"T" = Tied
NT = No tournament

Team appearances
Amateur
Nomura Cup (representing Australia): 1995
Eisenhower Trophy (representing Australia): 1996 (winners)

References

External links

Australian male golfers
European Tour golfers
PGA Tour of Australasia golfers
People from Mandurah
1972 births
Living people

redpajama_set_name: RedPajamaWikipedia


moreygirl – Tagged "moreygirl photography"
Theresa and Bill's day ended up pretty awesome!
My sister had her baby! If you visited with us at the Spoon River Drive, she's the one that was sitting with me the first weekend.
I'm so in love with my new little niece!
Kayla & Luke ~ Engaged!
It's really a small world because I went to school with Kayla. She's a bit younger than I am but when you go to a really small school, we were around each other quite a bit. Luke is my neighbor's son. His mom Diana, was one of my mentor moms at MOPS. She lead the crafts while she was there until I took it over.
The whole night was a great night to take pics!
Hope the best for the two of them!
You can check out all their pics over on the photography page!


They are used extensively and generate investment in the Suffolk economy through tourism.
Suffolk has 3,500 miles of public rights of way and over 12,000 acres of open access. This provides car-free access throughout Suffolk, from bustling market towns to open heath land and other areas of outstanding natural beauty.
Suffolk's public rights of way website also contains information for rights of way users and land managers. You can find out more about the different types of rights of way and access, and what your rights and responsibilities are when using or managing them.
Discover Suffolk is our countryside website for walks and rides in Suffolk. Our county is best explored on foot, from a bicycle or on horse-back. From the Brecks to the Sandlings and from the Broads to the estuaries.
Suffolk offers a wealth of opportunities to enjoy the great outdoors, with miles of marked paths, two Areas of Outstanding Natural Beauty, long distance footpaths and national cycle routes.
Within Discover Suffolk you will find inspiration on everything from a riverside stroll for all the family to a day long hike along secluded valleys.


Custard apple is most commonly known as sitaphal in India. They are also known as chermoyas and are native to some parts of Asia, the West Indies and South America. The health benefits of custard apple are immense and they will be discussed in this article.
The custard apple has a hard exterior with a soft and chewy interior. The inside flesh of the fruit is white in colour, has a creamy texture with black shiny seeds. The fruit comes in various shapes like spherical, heart-shaped or round.
As custard apple is sweet and sugary, it is beneficial for those who are trying to gain weight. Being a calorie-dense fruit, the calories come mainly from sugar. So, if you are planning to gain weight in a healthy way consume custard apple with a dash of honey to put on weight  .
Custard apple is rich in vitamin B6 which is effective in reducing bronchial inflammation. Vitamin B6 has been shown to decrease the frequency and severity of asthma attacks, according to a study . Another study also showed the potent ability of vitamin B6 in the treatment of asthma  .
One of the many benefits of custard apple is it improves cardiovascular health. These fruits are an excellent source of potassium and magnesium which prevents cardiac diseases, controls blood pressure and relaxes the artery muscles  . In addition, the presence of dietary fibre and vitamin B6 in custard apples have the ability to reduce cholesterol levels and prevent the development of homocysteine which increases the risk of heart disease .
Many diabetics avoid eating custard apples because of the fear of getting their blood sugar levels raised. Though the fruit is high in sugar content, the glycaemic index of custard apples is low which gets digested, absorbed and metabolised slowly in the bloodstream. This results in the slower rise in blood glucose levels  . However, avoid consuming in excess amounts.
Custard apples are loaded with dietary fibre which helps in easing bowel movement, thereby relieving constipation  . Dietary fibre also binds with harmful toxins in the digestive tract and eliminates them out of the body, resulting in better bowel movements, digestion and proper functioning of the intestines. Furthermore, stomach ulcers, gastritis and heartburn are also lowered if you have a custard apple daily.
Custard apples are rich in iron which can help treat anaemia, a health condition wherein your body suffers low iron levels. Iron is a component of haemoglobin found in red blood cells which carries oxygen from your lungs and transports it throughout your body. If your body doesn't have sufficient amounts of iron, it will not be able to make oxygen-carrying red blood cells.
Custard apple contains loads of magnesium which has the potent ability to balance the water distribution in the body. This helps in eliminating the acids from every joint in the body which aids in lowering inflammation and the joint pains associated with arthritis  . Custard apple is also known to reduce the symptoms of rheumatoid arthritis and that's why most doctors recommend this fruit.
Custard apple has been proved beneficial for pregnant women as they help manage pregnancy symptoms like mood swings, numbness and morning sickness. The fruit is rich in iron, an essential mineral required during pregnancy. According to the European Journal of Biomedical And Pharmaceutical Sciences, expectant mothers should consume custard apple daily for the proper growth of the baby's body and the development of the foetus in the womb.
Custard apples are an excellent source of the antioxidant vitamin C which is known for its anti-inflammatory and immune-boosting properties. Consuming this fruit every day will make you resistant to infections and other harmful free radicals. Vitamin C works by scavenging free radicals in the body, thereby preventing illnesses  .
Vitamin B6 in custard apples helps in proper brain development. This vitamin controls the GABA neuron chemical levels in the brain which reduces stress, tension, depression and irritability and also lowers the risk of developing Parkinson's disease, according to the European Journal of Biomedical And Pharmaceutical Sciences.
Vitamin C in custard apple plays a major role in the development of collagen, a protein which makes up the major portion of scalp and hair. It keeps your hair shiny and reduces fine lines and wrinkles, thus improving the elasticity of the skin  . Eating custard apples every day will help in the regeneration of skin cells which gives the skin a younger look.
Choose ripe custard apple as they are easier to eat and avoid overripe ones.
You can consume the fruit as a snack by adding a pinch of rock salt to make it tasty.
You can either make a custard apple smoothie or a sorbet.
Adding the flesh of the fruit to muffins and cakes would make it healthier.
You can also make ice cream out of this fruit by blending it, adding nuts and freezing it.
Note: As the fruit is very cold in nature, avoid consuming in excess amounts and don't eat it while you are sick. The seeds of the custard apple are poisonous, so make sure you don't swallow it.
 Jamkhande, P. G., & Wattamwar, A. S. (2015). Annona reticulata Linn. (Bullock's heart): Plant profile, phytochemistry and pharmacological properties.Journal of traditional and complementary medicine,5(3), 144-52.
 Sur, S., Camara, M., Buchmeier, A., Morgan, S., & Nelson, H. S. (1993). Double-blind trial of pyridoxine (vitamin B6) in the treatment of steroid-dependent asthma.Annals of allergy,70(2), 147-152.
 WALTERS, L. (1988). Vitamin B, Nutritional Status in Asthma: The Effect of Theophylline Therapy on Plasma Pyridoxal-5'-Phosphate and Pyridoxal Levels.
 Rosique-Esteban, N., Guasch-Ferré, M., Hernández-Alonso, P., & Salas-Salvadó, J. (2018). Dietary Magnesium and Cardiovascular Disease: A Review with Emphasis in Epidemiological Studies.Nutrients,10(2), 168.
 Marcus, J., Sarnak, M. J., & Menon, V. (2007). Homocysteine lowering and cardiovascular disease risk: lost in translation.The Canadian journal of cardiology,23(9), 707-10.
 Shirwaikar, A., Rajendran, K., Dinesh Kumar, C., & Bodla, R. (2004). Antidiabetic activity of aqueous leaf extract of Annona squamosa in streptozotocin–nicotinamide type 2 diabetic rats. Journal of Ethnopharmacology, 91(1), 171–175.
 Yang, J., Wang, H. P., Zhou, L., & Xu, C. F. (2012). Effect of dietary fiber on constipation: a meta analysis.World journal of gastroenterology,18(48), 7378-83.
 Suresh, H. M., Shivakumar, B., Hemalatha, K., Heroor, S. S., Hugar, D. S., & Rao, K. R. (2011). In vitro antiproliferativeactivity of Annona reticulata roots on human cancer cell lines.Pharmacognosy research,3(1), 9-12.
 Zeng, C., Li, H., Wei, J., Yang, T., Deng, Z. H., Yang, Y., Zhang, Y., Yang, T. B., … Lei, G. H. (2015). Association between Dietary Magnesium Intake and Radiographic Knee Osteoarthritis.PloS one,10(5), e0127666.
 Pullar, J. M., Carr, A. C., & Vissers, M. (2017). The Roles of Vitamin C in Skin Health.Nutrients,9(8), 866.


text:
package com.coderconsole.example;

import android.Manifest;
import android.content.Context;
import android.content.Intent;
import android.content.IntentFilter;
import android.os.Bundle;
import android.support.annotation.NonNull;
import android.support.design.widget.Snackbar;
import android.support.v4.app.ActivityCompat;
import android.support.v4.app.FragmentManager;
import android.support.v4.app.FragmentTransaction;
import android.support.v7.app.AppCompatActivity;
import android.view.View;
import android.widget.RelativeLayout;

import com.coderconsole.example.fragment.ContactFragment;

public class MainActivity extends AppCompatActivity {

    private static final String TAG = MainActivity.class.getSimpleName();

    Context mContext;

    String[] WALK_THROUGH = new String[]{Manifest.permission.READ_CONTACTS};

    private RelativeLayout container;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        mContext = MainActivity.this;

        initView();

        askPermission();
    }

    private void askPermission() {
        if (RunTimePermissionWrapper.isAllPermissionGranted(this, WALK_THROUGH))
            initContactFragment();
        else
            RunTimePermissionWrapper.handleRunTimePermission(MainActivity.this, RunTimePermissionWrapper.REQUEST_CODE.MULTIPLE_WALKTHROUGH, WALK_THROUGH);

    }

    private void initView() {
        container = (RelativeLayout) findViewById(R.id.main_container_relativeLayout);
    }

    @Override
    public void onRequestPermissionsResult(int requestCode, @NonNull String[] permissions, @NonNull int[] grantResults) {
        super.onRequestPermissionsResult(requestCode, permissions, grantResults);

        if (RunTimePermissionWrapper.isAllPermissionGranted(this, WALK_THROUGH))
            initContactFragment();
        else {
            showSnack(ActivityCompat.shouldShowRequestPermissionRationale(this, WALK_THROUGH[0]));
        }
    }

    private void initContactFragment() {

        if (isFinishing())
            return;

        FragmentManager fragmentManager = getSupportFragmentManager();
        FragmentTransaction fragmentTransaction = fragmentManager.beginTransaction();
        fragmentTransaction.replace(container.getId(), ContactFragment.newInstance(""));
        fragmentTransaction.commitAllowingStateLoss();
    }

    private void showSnack(final boolean isRationale) {
        final Snackbar snackbar = Snackbar.make(getWindow().getDecorView().getRootView(), "Please provide contact permission", Snackbar.LENGTH_INDEFINITE);
        snackbar.setAction(isRationale ? "VIEW" : "Settings", new View.OnClickListener() {
            @Override
            public void onClick(View view) {
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}


GOODYEAR, Ariz. — The Cincinnati Reds' All-Star second baseman Scooter Gennett is expected to be out for 2-3 months after he suffered a groin injury during a training game Friday, according to a tweet from the Reds.
The injury came while Gennett went to field a grounder hit by the Milwakuee Brewers' Yasmani Grandal.
According to The Associated Press, Gennett slid to make the stop and threw to pitcher Robert Stephenson covering first for the out. Gennett went to his knees and was helped off by Reds head athletic trainer Steve Baumann and manager David Bell.
"I don't know exactly if it was when he fell or when he made the throw," Bell said. "He felt something in his groin."
Gennett made his first All-Star team last season, hitting .310 with 23 homers and 92 RBIs.