text:
\section{Introduction}

Rotor-router walk is a deterministic analogue of random walk, first studied by Priezzhev et al.\ \cite{PDDK} under the name ``Eulerian walkers.''  At each site in the integer lattice $\Z^2$ is a {\it rotor} pointing north, south, east or west.  A particle starts at the origin; during each time step, the rotor at the particle's current location is rotated clockwise by $90$ degrees, and the particle takes a step in the direction of the newly rotated rotor.  In rotor-router aggregation, introduced by Jim Propp, we start with $n$ particles at the origin; each particle in turn performs rotor-router walk until it reaches a site not occupied by any other particles.   Let $A_n$ denote the resulting region of $n$ occupied sites.  For example, if all rotors initially point north, the sequence will begin $A_1 = \{(0,0)\}$, $A_2 = \{(0,0),(1,0)\}$, $A_3 = \{(0,0),(1,0),(0,-1)\}$.  The region $A_{1,000,000}$ is pictured in Figure~1.  In higher dimensions, the model can be defined analogously by repeatedly cycling the rotors through an ordering of the $2d$ cardinal directions in $\Z^d$.

\begin{figure}
\centering
\includegraphics[scale=.25]{rotor1mcrop.png}
\caption{Rotor-router aggregate of one million particles in $\Z^2$.  Each site is colored according to the direction of its rotor.}
\label{rotor1m}
\end{figure}

Jim Propp observed from simulations in two dimensions that the regions $A_n$ are extraordinarily close to circular, and asked why this was so \cite{Kleber,Propp}.  Despite the impressive empirical evidence for circularity, the best result known until now \cite{LP} says only that if $A_n$ is rescaled to have unit volume, the volume of the symmetric difference of $A_n$ with a ball of unit volume tends to zero as a power of $n$, as $n \uparrow \infty$.  The main outline of the argument is summarized in \cite{intelligencer}.  Fey and Redig  \cite{FR} also show that $A_n$ contains a diamond.  In particular, these results do not rule out the possibility of ``holes'' in $A_n$ far from the boundary or of long tendrils extending far beyond the boundary of the ball, provided the volume of these features is negligible compared to $n$.

Our main result is the following, which rules out the possibility of holes far from the boundary or of long tendrils in the rotor-router shape.  For $r\geq 0$ let
	\[ B_r = \{ x \in \Z^d ~:~ |x| < r \}. \]

\begin{theorem}
\label{rotorcirc}
Let $A_n$ be the region formed by rotor-router aggregation in $\Z^d$ starting from $n$ particles at the origin and any initial rotor state.  There exist constants $c,c'$ depending only on $d$, such that
	\[ B_{r-c\log r} \subset A_n \subset B_{r (1+c'r^{-1/d}\log r)} \]
where $r=(n/\omega_d)^{1/d}$, and $\omega_d$ is the volume of the unit ball in $\R^d$.
\end{theorem}

We remark that the same result holds when the rotors evolve according to stacks of bounded discrepancy; see the remark following Lemma~\ref{odomflow}.

Internal diffusion limited aggregation (``internal DLA'') is an analogous growth model defined using random walks instead of rotor-router walks.  Starting with $n$ particles at the origin, each particle in turn performs simple random walk until it reaches an unoccupied site.  Lawler, Bramson and Griffeath \cite{LBG} showed that for internal DLA in $\Z^d$, the occupied region $A_n$, rescaled by a factor of $n^{1/d}$, converges with probability one to a Euclidean ball in $\R^d$ as $n \rightarrow \infty$.  Lawler \cite{Lawler95} estimated the rate of convergence.
By way of comparison with Theorem~\ref{rotorcirc}, if $I_n$ is the internal DLA region formed from $n$ particles started at the origin, the best known bounds \cite{Lawler95} are (up to logarithmic factors)
	\[ B_{r-r^{1/3}} \subset I_n \subset B_{r+r^{1/3}} \]
for all sufficiently large $n$, with probability one.

We also study another model which is slightly more difficult to define, but much easier to analyze.
In the {\it divisible sandpile}, each site $x \in \Z^d$ starts with a quantity of ``mass'' $\nu_0(x) \in \R_{\geq 0}$.  A site {\it topples} by keeping up to mass~$1$ for itself, and distributing the excess (if any) equally among its neighbors.  Thus if $x$ has mass $m> 1$, then each of the $2d$ neighboring sites gains mass $(m-1)/2d$ when we topple $x$, and $x$ is left with mass $1$; if $m\leq 1$, then no mass moves when we topple $x$.

  Note that individual topplings do not commute; however, the divisible sandpile is ``abelian'' in the following sense.

\begin{prop}
\label{divsandpileabelianproperty}
Let $x_1, x_2, \ldots \in \Z^d$ be a sequence with the property that for any $x \in \Z^d$ there are infinitely many terms $x_k=x$.  Let
	\begin{align*} u_k(x) &= \text{\em total mass emitted by $x$ after toppling $x_1, \ldots, x_k$}; \\
			       \nu_k(x) &= \text{\em amount of mass present at $x$ after toppling $x_1,\ldots, x_k$}.
			       \end{align*}
Then $u_k \uparrow u$ and $\nu_k \to \nu \leq 1$.  Moreover, the limits $u$ and $\nu$ are independent of the sequence $\{x_k\}$.
\end{prop}

\begin{figure}
\centering
\includegraphics[scale=.3333333333]{sandpile1m.png}
\caption{Classical abelian sandpile aggregate of one million particles in $\Z^2$.  Colors represent the number of grains at each site.}
\label{sandpile1m}
\end{figure}

The abelian property can be generalized as follows: after performing some topplings, we can add some additional mass and then continue toppling.  The resulting limits $u$ and $\nu$ will be the same as in the case when all mass was initially present.  For a further generalization, see \cite{scalinglimit}.

The limiting function $u$ in Proposition~\ref{divsandpileabelianproperty} is the \emph{odometer function} for the divisible sandpile.  This function plays a central role in our analysis.  The limit $\nu$ represents the final mass distribution.  Sites $x \in \Z^d$ with $\nu(x)=1$ are called {\it fully occupied}.  Proposition~\ref{divsandpileabelianproperty} is proved in section~3, along with the following.

\begin{theorem}
\label{divsandcircintro}
For $m \geq 0$ let $D_m \subset \Z^d$ be the domain of fully occupied sites for the divisible sandpile formed from a pile of mass $m$ at the origin.  There exist constants $c,c'$ depending only on $d$, such that
	\[ B_{r-c} \subset D_m \subset B_{r+c'}, \]
where $r = (m/\omega_d)^{1/d}$ and $\omega_d$ is the volume of the unit ball in $\R^d$.
\end{theorem}

The divisible sandpile is similar to the ``oil game'' studied by Van den Heuvel \cite{VdH}.  In the terminology of \cite{FR}, it also corresponds to the $h \rightarrow -\infty$ limit of the classical abelian sandpile (defined below), that is, the abelian sandpile started from the initial condition in which every site has a very deep ``hole.''

In the classical abelian sandpile model \cite{BTW}, each site in $\Z^d$ has an integer number of grains of sand; if a site has at least $2d$ grains, it {\it topples}, sending one grain to each neighbor.  If $n$ grains of sand are started at the origin in $\Z^d$, write $S_n$ for the set of sites that are visited during the toppling process; in particular, although a site may be empty in the final state, we include it in $S_n$ if it was occupied at any time during the evolution to the final state.

\begin{figure}
\centering
\includegraphics[scale=.25]{sandpilebounds.png}
\caption{Known bounds on the shape of the classical abelian sandpile in $\Z^2$.  The inner diamond and outer square are due to Le Borgne and Rossin \cite{LBR}; the inner and outer circles are those in Theorem~\ref{sandpilecircintro}.}
\end{figure}

Until now the best known constraints on the shape of $S_n$ in two dimensions were due to Le Borgne and Rossin \cite{LBR}, who proved that
	\[ \{x \in \Z^2 \,|\, x_1+x_2 \leq \sqrt{n/12}-1 \} \subset S_n \subset \{x \in \Z^2 \,|\, x_1,x_2 \leq \sqrt{n}/2 \}. \]
Fey and Redig~\cite{FR} proved analogous bounds in higher dimensions, and extended these bounds to arbitrary values of the height parameter $h$.  This parameter is discussed in section~4.

The methods used to prove the near-perfect circularity of the divisible sandpile shape in Theorem~\ref{divsandcircintro} can be used to give constraints on the shape of the classical abelian sandpile, improving on the bounds of~\cite{FR} and~\cite{LBR}.

\begin{theorem}
\label{sandpilecircintro}
Let $S_n$ be the set of sites that are visited by the classical abelian sandpile model in $\Z^d$, starting from $n$ particles at the origin.  Write $n = \omega_d r^d$.  Then for any $\epsilon>0$ we have
	\[ B_{c_1r - c_2} \subset S_n \subset B_{c'_1r + c'_2} \]
where
	\[ c_1 = (2d-1)^{-1/d}, \qquad c'_1 = (d-\epsilon)^{-1/d}. \]
The constant $c_2$ depends only on $d$, while $c'_2$ depends only on $d$ and $\epsilon$.
\end{theorem}

Note that Theorem~\ref{sandpilecircintro} does not settle the question of the asymptotic shape of $S_n$, and indeed it is not clear from simulations whether the asymptotic shape in two dimensions is a disc or perhaps a polygon (Figure~\ref{sandpile1m}).  To our knowledge even the existence of an asymptotic shape is not known.

The rest of the paper is organized as follows.  In section~2, we derive the basic Green's function estimates that are used in the proofs of Theorems~\ref{rotorcirc}, \ref{divsandcircintro} and~\ref{sandpilecircintro}.  In section~3 we prove Proposition~\ref{divsandpileabelianproperty} and Theorem~\ref{divsandcircintro} for the divisible sandpile.  In section~4 we adapt the methods of the previous section to prove Theorem~\ref{sandpilecircintro} for the classical abelian sandpile model.  Section~5 is devoted to the proof of Theorem~\ref{rotorcirc}.

\section{Basic Estimate}

Write $(X_k)_{k\geq 0}$ for simple random walk in $\Z^d$, and for $d\geq 3$ denote by
	\[ g(x) = \EE_o \# \{ k|X_k=x \} \]
the expected number of visits to $x$ by simple random walk started at the origin.  This is the \emph{discrete harmonic Green's function} in $\Z^d$; it satisfies $\Delta g(x) = 0$ for $x\neq o$, and $\Delta g(o)=-1$, where $\Delta$ is the discrete Laplacian
	\[ \Delta g(x) = \frac{1}{2d} \sum_{y \sim x} g(y) - g(x). \]
The sum is over the $2d$ lattice neighbors $y$ of $x$.  In dimension $d=2$, simple random walk is recurrent, so the expectation defining $g(x)$ is infinite.  Here we define the {\it potential kernel}
	\begin{equation} \label{potentialkerneldef} g(x) = \lim_{n \rightarrow \infty} g_n(x)-g_n(o) \end{equation}
where
	\[ g_n(x) = \EE_o \# \{k \leq n|X_k=x\}. \]
The limit defining $g(x)$ in (\ref{potentialkerneldef}) is finite \cite{Lawler,Spitzer}, and $g(x)$ has Laplacian $\Delta g(x)=0$ for $x\neq o$, and $\Delta g(o)=-1$.  Note that (\ref{potentialkerneldef}) is the negative of the usual definition of the potential kernel; we have chosen this sign convention so that $g$ has the same Laplacian in dimension two as in higher dimensions.

Fix a real number $m>0$ and consider the function on $\Z^d$
	\begin{equation} \label{gammadef} \widetilde{\gamma}_d(x) = |x|^2 + mg(x). \end{equation}
Let $r$ be such that $m = \omega_d r^d$, and let
	\begin{equation} \gamma_d(x) = \widetilde{\gamma}_d(x)-\widetilde{\gamma}_d(\floor{r}e_1) \label{subtractoffconstant} \end{equation}
where $e_1$ is the first standard basis vector in $\Z^d$.  The function $\gamma_d$ plays a central role in our analysis.  To see where it comes from, recall the divisible sandpile \emph{odometer function} of Proposition~\ref{divsandpileabelianproperty}
	\[ u(x) = \text{total mass emitted from $x$}. \]
Let $D_m \subset \Z^d$ be the domain of fully occupied sites for the divisible sandpile formed from a pile of mass $m$ at the origin.  For $x \in D_m$, since each neighbor $y$ of $x$ emits an equal amount of mass to each of its $2d$ neighbors, we have
	\begin{align*} \Delta u(x) &= \frac{1}{2d} \sum_{y \sim x} u(y) - u(x) \\
					        &= \text{mass received by $x$} - \text{mass emitted by $x$} \\
					        &= 1 - m\delta_{ox}. \end{align*}
Moreover, $u=0$ on $\partial D_m$.  By construction, the function $\gamma_d$ obeys the same Laplacian condition: $\Delta \gamma_d = 1 - m\delta_o$; and as we will see shortly, $\gamma_d \approx 0$ on $\partial B_r$.  Since we expect the domain $D_m$ to be close to the ball $B_r$, we should expect that $u \approx \gamma_d$.  In fact, we will first show that $u$ is close to $\gamma_d$, and then use this to conclude that $D_m$ is close to $B_r$.

We will use the following estimates for the Green's function \cite{FU,Uchiyama}; see also \cite[Theorems 1.5.4 and 1.6.2]{Lawler}.
 \begin{equation}  \label{standardgreensestimate}
  g(x) = \begin{cases}  -\frac{2}{\pi} \log |x| + \kappa + O(|x|^{-2}), & d=2 \\
			 a_d |x|^{2-d} + O(|x|^{-d}), & d\geq 3. \end{cases} \end{equation}
Here $a_d = \frac{2}{(d-2)\omega_d}$, where $\omega_d$ is the volume of the unit ball in $\R^d$, and $\kappa$ is a constant whose value we will not need to know.  For $x \in \Z^d$ we use $|x|$ to denote the Euclidean norm of $x$.  Here and throughout the paper, constants in error terms denoted $O(\cdot)$ depend only on $d$.

We will need an estimate for $\gamma_d$ near the boundary of the ball $B_r$.  We first consider dimension $d=2$.  From (\ref{standardgreensestimate}) we have
	\begin{equation} \label{gammadefn} \widetilde{\gamma}_2(x) = \phi(x) - \kappa m + O(m|x|^{-2}), \end{equation}
where
	\[ \phi(x) = |x|^2 - \frac{2m}{\pi} \log |x|. \]
In the Taylor expansion of $\phi$ about $|x|=r$
	\begin{equation} \label{secondordertaylor} \phi(x) = \phi(r) - \phi'(r)(r-|x|) + \frac12 \phi''(t)(r-|x|)^2 \end{equation}
the linear term vanishes, leaving
	\begin{equation} \label{taylordim2}
	\gamma_2(x) =  \left(1+\frac{m}{\pi t^2} \right)(r-|x|)^2 + O(m|x|^{-2})
	\end{equation}
for some $t$ between $|x|$ and $r$.

In dimensions $d \geq 3$, from (\ref{standardgreensestimate}) we have
	\[ \widetilde{\gamma}_d(x) = |x|^2 + a_d m |x|^{2-d} + O(m|x|^{-d}). \]
Setting $\phi(x) = |x|^2 + a_d m |x|^{2-d}$, the linear term in the Taylor expansion (\ref{secondordertaylor}) of $\phi$ about $|x|=r$ again vanishes, yielding
	\[ \gamma_d(x) = \left( 1 + (d-1)(r/t)^d \right) (r-|x|)^2 + O(m|x|^{-d}) \]
for $t$ between $|x|$ and $r$.  Together with (\ref{taylordim2}), this yields the following estimates in all dimensions $d\geq 2$.

\begin{lemma} \label{gammalowerbound}
Let $\gamma_d$ be given by (\ref{subtractoffconstant}).
For all $x \in \Z^d$ we have
	\begin{equation} \label{quadraticgrowth} \gamma_d(x) \geq (r-|x|)^2 + O\left(\frac{r^d}{|x|^d}\right). \end{equation}
\end{lemma}

\begin{lemma} \label{gammaupperbound}
Let $\gamma_d$ be given by (\ref{subtractoffconstant}).  Then uniformly in $r$,
	\[ \gamma_d(x) = O(1), \qquad x \in B_{r+1}-B_{r-1}. \]
\end{lemma}

The following lemma is useful for $x$ near the origin, where the error term in (\ref{quadraticgrowth}) blows up.

\begin{lemma} \label{gammanearorigin}
Let $\gamma_d$ be given by (\ref{subtractoffconstant}).  Then for sufficiently large $r$, we have
	\[ \gamma_d(x) > \frac{r^2}{4}, \qquad \forall x \in B_{r/3}. \]
\end{lemma}

\begin{proof}
Since $\gamma_d(x)-|x|^2$ is superharmonic, it attains its minimum in $B_{r/3}$ at a point $z$ on the boundary.  Thus for any $x \in B_{r/3}$
	\[ \gamma_d(x)-|x|^2 \geq \gamma_d(z) - |z|^2, \]
hence by Lemma~\ref{gammalowerbound}
	\[ \gamma_d(x) \geq (2r/3)^2 - (r/3)^2 + O(1) > \frac{r^2}{4}. \qed \]
\renewcommand{\qedsymbol}{}
\end{proof}

Lemmas~\ref{gammalowerbound} and~\ref{gammanearorigin} together imply the following.

\begin{lemma} \label{gammanonnegativeeverywhere}
Let $\gamma_d$ be given by (\ref{subtractoffconstant}).  There is a constant $a$ depending only on $d$, such that $\gamma_d \geq -a$ everywhere.
\end{lemma}

\section{Divisible Sandpile}

Let $\mu_0$ be a nonnegative function on $\Z^d$ with finite support.
We start with mass $\mu_0(y)$ at each site $y$.  The operation of {\it toppling} a vertex $x$ yields the mass distribution
	\[ T_x \mu_0 = \mu_0 + \alpha(x)\Delta\delta_x \]
where $\alpha(x) = \text{max}(\mu_0(x)-1,0)$ and $\Delta$ is the discrete Laplacian on $\Z^d$.  Thus if $\mu_0(x)\leq 1$ then $T_x \mu_0 = \mu_0$ and no mass topples; if $\mu_0(x)>1$ then the mass in excess of $1$ is distributed equally among the neighbors of $x$.

Let $x_1, x_2, \ldots \in \Z^d$ be a sequence with the property that for any $x \in \Z^d$ there are infinitely many terms $x_k=x$.  Let
 	\[ \mu_k(y) = T_{x_k} \ldots T_{x_1} \mu_0(y). \]
be the amount of mass present at $y$ after toppling the sites $x_1, \ldots, x_k$ in succession.  The total mass emitted from $y$ during this process is
	\begin{equation} \label{finitetimeodom} u_k(y) := \sum_{j\leq k : x_j=y} \mu_{j-1}(y)-\mu_j(y) = \sum_{j\leq k : x_j=y} \alpha_j(y)
	\end{equation}
where $\alpha_j(y) = \text{max}(\mu_j(y)-1,0)$.

\begin{lemma}
\label{existenceoflimits}
As $k\uparrow \infty$ the functions $u_k$ and $\mu_k$ tend to limits $u_k \uparrow u$ and $\mu_k \rightarrow \mu$.  Moreover, these limits satisfy
	\[ \mu = \mu_0 + \Delta u \leq 1. \]
\end{lemma}

\begin{proof}
Write $M = \sum_x \mu_0(x)$ for the total starting mass, and let $B \subset \Z^d$ be a ball centered at the origin containing all points within $L^1$-distance $M$ of the support of $\mu_0$.  Note that if $\mu_k(x)>0$ and $\mu_0(x)=0$, then $x$ must have received its mass from a neighbor, so $\mu_k(y)\geq 1$ for some $y\sim x$.  Since $\sum_x \mu_k(x) = M$, it follows that $\mu_k$ is supported on $B$.  Let $R$ be the radius of $B$, and consider the {\it quadratic weight}
	\[ Q_k = \sum_{x \in \Z^d} \mu_k(x) |x|^2 \leq MR^2. \]
Since $\mu_k(x_k)-\mu_{k-1}(x_k) = -\alpha_k(x_k)$ and for $y\sim x_k$ we have $\mu_k(y)-\mu_{k-1}(y) = \frac{1}{2d} \alpha_k(x_k)$, we obtain
	\[ Q_k - Q_{k-1} = \alpha_k(x_k) \left( \frac{1}{2d} \sum_{y \sim x_k} |y|^2 - |x_k|^2 \right) = \alpha_k(x_k). \]
Summing over $k$ we obtain from (\ref{finitetimeodom})
	\[ Q_k = Q_0 + \sum_{x \in \Z^d} u_k(x). \]
Fixing $x$, the sequence $u_k(x)$ is thus increasing and bounded above, hence convergent.

Given neighboring vertices $x\sim y$, since $y$ emits an equal amount of mass to each of its $2d$ neighbors, it emits mass $u_k(y)/2d$ to $x$ up to time $k$.  Thus $x$ receives a total mass of $\frac{1}{2d} \sum_{y\sim x} u_k(y)$ from its neighbors up to time $k$.  Comparing the amount of mass present at $x$ before and after toppling, we obtain
	\[ \mu_k(x) = \mu_0(x) + \Delta u_k(x). \]
Since $u_k \uparrow u$ we infer that $\mu_k \rightarrow \mu:= \mu_0 + \Delta u$.  Note that if $x_k=x$, then $\mu_k(x) \leq 1$.  Since for each $x \in \Z^d$ this holds for infinitely many values of $k$, the limit satisfies $\mu \leq 1$.
\end{proof}

A function $s$ on $\Z^d$ is {\it superharmonic} if $\Delta s \leq 0$.  Given a function $\gamma$ on $\Z^d$ the {\it least superharmonic majorant} of $\gamma$ is the function
	\[ s(x) = \text{inf} \{ f(x)~|~ f \text{ is superharmonic and } f\geq \gamma \}. \]
The study of the least superharmonic majorant is a classical topic in analysis and PDE; see, for example, \cite{Koosis}.  Note that if $f$ is superharmonic and $f \geq \gamma$ then
	\[ f(x) \geq \frac{1}{2d} \sum_{y \sim x} f(y) \geq \frac{1}{2d} \sum_{y \sim x} s(y). \]
Taking the infimum on the left side we obtain that $s$ is superharmonic.

\begin{lemma}
\label{discretemajorant}
The limit $u$ in Lemma~\ref{existenceoflimits} is given by $u = s + \gamma$, where
	\[ \gamma(x) = |x|^2 + \sum_{y \in \Z^d} g(x-y) \mu_0(y) \]
and $s$ is the least superharmonic majorant of $-\gamma$.
\end{lemma}

\begin{proof}
By Lemma~\ref{existenceoflimits} we have
	\[ \Delta u = \mu - \mu_0 \leq 1-\mu_0. \]
Since $\Delta \gamma = 1-\mu_0$, the difference $u-\gamma$ is superharmonic.  As $u$ is nonnegative, it follows that $u-\gamma \geq s$.  For the reverse inequality, note that $s+\gamma-u$ is superharmonic on the domain $D = \{x~|~\mu(x)=1\}$ of fully occupied sites and is nonnegative outside $D$, hence nonnegative inside $D$ as well.
\end{proof}

As a corollary of Lemmas~\ref{existenceoflimits} and~\ref{discretemajorant}, we obtain the abelian property of the divisible sandpile, Proposition~\ref{divsandpileabelianproperty}, which was stated in the introduction.

We now turn to the case of a point source mass $m$ started at the origin: $\mu_0=m\delta_{o}$.  More general starting distributions are treated in \cite{scalinglimit}, where we identify the scaling limit of the divisible sandpile model and show that it coincides with that of internal DLA and of the rotor-router model.  In the case of a point source of mass $m$, the natural question is to identify the shape of the resulting domain $D_m$ of fully occupied sites, i.e.\ sites $x$ for which $\mu(x)=1$.  According to Theorem~\ref{divsandcirc}, $D_m$ is extremely close to a ball of volume $m$; in fact, the error in the radius is a constant independent of $m$.
As before, for $r\geq 0$ we write
	\[ B_r = \{ x \in \Z^d ~:~ |x| < r \} \]
for the lattice ball of radius $r$ centered at the origin.
\begin{theorem}
\label{divsandcirc}
For $m \geq 0$ let $D_m \subset \Z^d$ be the domain of fully occupied sites for the divisible sandpile formed from a pile of size $m$ at the origin.  There exist constants $c,c'$ depending only on $d$, such that
	\[ B_{r-c} \subset D_m \subset B_{r+c'}, \]
where $r = (m/\omega_d)^{1/d}$ and $\omega_d$ is the volume of the unit ball in $\R^d$.
\end{theorem}

The idea of the proof is to use Lemma~\ref{discretemajorant} along with the basic estimates on $\gamma$, Lemmas~\ref{gammalowerbound} and~\ref{gammaupperbound}, to obtain estimates on the odometer function
	\[ u(x) = \text{total mass emitted from $x$}. \]
We will need the following simple observation.

\begin{lemma}
\label{boundarypath}
For every point $x\in D_m - \{o\}$ there is a path $x=x_0 \sim x_1 \sim \ldots \sim x_k=o$ in $D_m$ with $u(x_{i+1}) \geq u(x_i)+1$.
\end{lemma}
\begin{proof}
If $x_i \in D_m - \{o\}$, let $x_{i+1}$ be a neighbor of $x_i$ maximizing $u(x_{i+1})$.  Then $x_{i+1} \in D_m$ and
	\begin{align*} u(x_{i+1}) &\geq \frac{1}{2d} \sum_{y \sim x_i} u(y) \\
		&= u(x_i) + \Delta u(x_i) \\
		&= u(x_i)+1,  \end{align*}
where in the last step we have used the fact that $x_i \in D_m$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{divsandcirc}]
We first treat the inner estimate.  Let $\gamma_d$ be given by (\ref{subtractoffconstant}).
By Lemma~\ref{discretemajorant} the function $u-\gamma_d$ is superharmonic,  so its minimum in the ball $B_r$ is attained on the boundary.  Since $u\geq 0$, we have by  Lemma~\ref{gammaupperbound}
	\[ u(x)-\gamma_d(x) \geq -C, \qquad x \in \partial B_r \]
for a constant $C$ depending only on $d$.  Hence by Lemma~\ref{gammalowerbound},
	\begin{equation} \label{odomlowerbound} u(x) \geq  (r-|x|)^2 - C'r^d/|x|^d, \qquad x \in B_r. \end{equation}
for a constant $C'$ depending only on $d$.
It follows that there is a constant $c$, depending only on $d$, such that $u(x)>0$ whenever $r/3\leq |x|<r-c$.  Thus $B_{r-c}-B_{r/3} \subset D_m$.  For $x \in B_{r/3}$, by Lemma~\ref{gammanearorigin} we have $u(x) \geq r^2/4-C>0$, hence $B_{r/3} \subset D_m$.

For the outer estimate, note that $u-\gamma_d$ is harmonic on $D_m$.  By Lemma~\ref{gammanonnegativeeverywhere} we have $\gamma_d \geq -a$ everywhere, where $a$ depends only on $d$.  Since $u$ vanishes on $\partial D_m$ it follows that $u-\gamma_d \leq a$ on $D_m$.
Now for any $x \in D_m$ with $r-1<|x|\leq r$, we have by Lemma~\ref{gammaupperbound}
	\[ u(x) \leq \gamma_d(x) + a \leq c' \]
for a constant $c'$ depending only on $d$.  Lemma~\ref{boundarypath} now implies that $D_m \subset B_{r+c'+1}$.
\end{proof}

\section{Classical Sandpile}

We consider a generalization of the classical abelian sandpile, proposed by Fey and Redig \cite{FR}.  Each site in $\Z^d$ begins with a ``hole'' of depth $H$.  Thus, each site absorbs the first $H$ grains it receives, and thereafter functions normally, toppling once for each additional $2d$ grains it receives.  If $H$ is negative, we can interpret this as saying that every site starts with $h=-H$ grains of sand already present.  Aggregation is only well-defined in the regime $h\leq 2d-2$, since for $h=2d-1$ the addition of a single grain already causes every site in $\Z^d$ to topple infinitely often.

Let $S_{n,H}$ be the set of sites that are visited if $n$ particles start at the origin in $\Z^d$.  Fey and Redig \cite[Theorem~4.7]{FR} prove that
	\[ \lim_{H \to \infty} \limsup_{n \to \infty}  \frac{H}{n} \# \left( S_{n,H} \bigtriangleup B_{H^{-1/d}r} \right) = 0, \]
where $n = \omega_d r^d$, and $\bigtriangleup$ denotes symmetric difference.  The following theorem strengthens this result.

\begin{theorem}
\label{sandpilecirc}
Fix an integer $H \geq 2-2d$.  Let $S_n = S_{n,H}$ be the set of sites that are visited by the classical abelian sandpile model in $\Z^d$, starting from $n$ particles at the origin, if every lattice site begins with a hole of depth $H$.  Write $n = \omega_d r^d$.  Then
	\[ B_{c_1r - c_2} \subset S_{n,H} \]
where
	\[ c_1 = (2d-1+H)^{-1/d} \]
and $c_2$ is a constant depending only on $d$.  Moreover if $H \geq 1-d$, then for any $\epsilon>0$ we have
	 \[ S_{n,H} \subset B_{c'_1r + c'_2} \]
where
 	\[ c'_1 = (d-\epsilon+H)^{-1/d} \]
and $c'_2$ is independent of $n$ but may depend on $d$, $H$ and $\epsilon$.
\end{theorem}

Note that the ratio $c_1/c'_1 \uparrow 1$ as $H \uparrow \infty$.  Thus, the classical abelian sandpile run from an initial state in which each lattice site starts with a deep hole yields a shape very close to a ball.  Intuitively, one can think of the classical sandpile with deep holes as approximating the divisible sandpile, whose limiting shape is a ball by Theorem~\ref{divsandcirc}.  Following this intuition, we can adapt the proof of Theorem~\ref{divsandcirc} to prove Theorem~\ref{sandpilecirc}; just one additional averaging trick is needed, which we explain below.

Consider the odometer function for the abelian sandpile
	\[  u(x) = \text{total number of grains emitted from }x. \]
Let $T_n = \{x|u(x)>0\}$ be the set of sites which topple at least once.  Then
	\[ T_n \subset S_n \subset  T_n \cup \partial T_n. \]
In the final state, each site which has toppled retains between $0$ and $2d-1$ grains, in addition to the $H$ that it absorbed.  Hence
	\begin{equation} \label{dumblaplbounds} H \leq \Delta u(x) + n\delta_{ox} \leq 2d-1+H, \qquad x \in T_n. \
	\end{equation}
We can improve the lower bound by averaging over a small box.  For $x \in \Z^d$ let
	\[ Q_k(x) = \{ y\in \Z^d \,:\, ||x-y||_{\infty} \leq k \} \]
be the box of side length $2k+1$ centered at $x$, and let
	\[ u^{(k)}(x) = (2k+1)^{-d} \sum_{y \in Q_k(x)} u(y). \]
Write
	\[ T_n^{(k)} = \{ x \,|\, Q_k(x) \subset T_n \}. \]
Le Borgne and Rossin \cite{LBR} observe that if $T$ is a set of sites all of which topple, the number of grains remaining in $T$ is at least the number of edges internal to $T$: indeed, for each internal edge, the endpoint that topples last sends the other a grain which never moves again.  Since the box $Q_k(x)$ has $2dk(2k+1)^{d-1}$ internal edges, we have
	\begin{equation} \label{averagedinabox} \Delta u^{(k)}(x) \geq \frac{2k}{2k+1}d +H - \frac{n}{(2k+1)^d} \textbf{1}_{Q_k(o)}(x), \qquad x \in T_n^{(k)}. \end{equation}
The following lemma is analogous to Lemma~\ref{boundarypath}.

\begin{lemma}
\label{boundarypathology}
For every point $x\in T_n$ adjacent to $\partial T_n$ there is a path $x=x_0 \sim x_1 \sim \ldots \sim x_m = o$ in $T_n$ with $u(x_{i+1}) \geq u(x_i)+1$.
\end{lemma}

\begin{proof}
By (\ref{dumblaplbounds}) we have
	\[ \frac{1}{2d} \sum_{y \sim x_i} u(y) \geq u(x_i). \]
Since $u(x_{i-1})<u(x_i)$, some term $u(y)$ in the sum above must exceed $u(x_i)$.  Let $x_{i+1}=y$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{sandpilecirc}]
Let
	\[ \widetilde{\xi}_d(x) = (2d-1+H)|x|^2 + ng(x), \]
and let
	\[ \xi_d(x) = \widetilde{\xi}_d(x) - \widetilde{\xi}_d(\floor{c_1r}e_1). \]
Taking $m = n/(2d-1+H)$ in Lemma~\ref{gammaupperbound}, we have
	\begin{equation} \label{minattainedonboundary} u(x)-\xi_d(x) \geq -\xi_d(x) \geq -C(2d-1+H), \qquad x\in\partial B_{c_1r} \end{equation}
for a constant $C$ depending only on $d$.
By (\ref{dumblaplbounds}), $u-\xi_d$ is superharmonic, so $u-\xi_d \geq -C(2d-1+H)$ in all of $B_{c_1r}$.  Hence by Lemma~\ref{gammalowerbound} we have for $x \in B_{c_1 r}$
	\begin{equation} \label{odomlowerboundagain} u(x) \geq (2d-1+H) \left( (c_1r - |x|)^2 - C'(c_1 r)^d/|x|^d \right),
	\end{equation}
where $C'$ depends only on $d$.
It follows that $u$ is positive on $B_{c_1r - c_2}-B_{c_1 r/3}$ for a suitable constant $c_2$ depending only on $d$.  For $x \in B_{c_1 r/3}$, by Lemma~\ref{gammanearorigin} we have $u(x) > (2d-1+H)(c_1^2 r^2 / 4 -C) > 0$.  Thus $B_{c_1 r - c_2} \subset T_n \subset S_n$.

For the outer estimate, let
	\[ \hat{\psi}_d(x) = (d-\epsilon+H)|x|^2 + ng(x). \]
Choose $k$ large enough so that $\frac{2k}{2k+1} d \geq d-\epsilon$, and define
	\[ \widetilde{\psi}_d(x) = (2k+1)^{-d} \sum_{y \in Q_k(x)} \hat{\psi}_d(y). \]
Finally, let
	\[ \psi_d(x) = \widetilde{\psi}_d(x) - \widetilde{\psi}_d(\floor{c'_1r}e_1). \]
By (\ref{averagedinabox}), $u^{(k)}-\psi_d$ is subharmonic on $T_n^{(k)}$.  Taking $m=n/(d-\epsilon+H)$ in Lemma~\ref{gammanonnegativeeverywhere}, there is a constant $a$ depending only on $d$, such that $\psi_d \geq -a(d+H)$ everywhere.  Since $u^{(k)} \leq (2d+H)^{(d+1)k}$ on $\partial T_n^{(k)}$ it follows that $u^{(k)}-\psi_d \leq a(d+H)+(2d+H)^{(d+1)k}$ on $T_n^{(k)}$.  Now for any $x \in S_n$ with $c'_1r-1 < |x| \leq c'_1 r$ we have by Lemma~\ref{gammaupperbound}
	\[ u^{(k)}(x) \leq \psi_d(x) + a(d+H) + (2d+H)^{(d+1)k} \leq \tilde{c}_2 \]
for a constant $\tilde{c}_2$ depending only on $d$, $H$ and $\epsilon$.  Then $u(x) \leq c'_2 := (2k+1)^d \tilde{c}_2$.  Lemma~\ref{boundarypathology} now implies that $T_n \subset B_{c'_1 r + c'_2}$, and hence
	\[ S_n \subset T_n \cup \partial T_n \subset B_{c'_1 r + c'_2+1}. \qed \]
\renewcommand{\qedsymbol}{}
\end{proof}

We remark that the crude bound of $(2d+H)^{(d+1)k}$ used in the proof of the outer estimate can be improved to a bound of order $k^2H$, and the final factor of $(2k+1)^d$ can be replaced by a constant factor independent of $k$ and $H$, using the fact that a nonnegative function on $\Z^d$ with bounded Laplacian cannot grow faster than quadratically; see \cite{scalinglimit}.

\section{Rotor-Router Model}

Given a function $f$ on $\Z^d$, for a directed edge $(x,y)$ write
	\[ \nabla f (x,y) = f(y)-f(x). \]
Given a function $s$ on directed edges in $\Z^d$, write
	\[ {\text{div}\hspace{0.5ex}} s (x) = \frac{1}{2d} \sum_{y \sim x} s(x,y). \]
The discrete Laplacian of $f$ is then given by
	\[ \Delta f(x) = {\text{div}\hspace{0.5ex}} \nabla f = \frac{1}{2d} \sum_{y \sim x} f(y)-f(x). \]

\subsection{Inner Estimate}

Fixing $n \geq 1$, consider the odometer function for rotor-router aggregation
	\[ u(x) = \text{total number exits from $x$ by the first $n$ particles}. \]
We learned the idea of using the odometer function to study the rotor-router shape from Matt Cook \cite{Cook}.

\begin{lemma}
\label{odomflow}
For a directed edge $(x,y)$ in $\Z^d$, denote by $\kappa(x,y)$ the net number of crossings
from $x$ to $y$ performed by the first $n$ particles in rotor-router aggregation.
Then
	\begin{equation} \label{gradodom} \nabla u(x,y) = -2d\kappa(x,y) + R(x,y)
\end{equation}
for some edge function $R$ which satisfies
	\[ |R(x,y)| \leq 4d-2 \]
for all edges $(x,y)$.
\end{lemma}

\begin{remark}
In the more general setting of rotor stacks of bounded discrepancy, the $4d-2$ will be replaced by a different constant here.
\end{remark}

\begin{proof}
Writing $N(x,y)$ for the number of particles routed from $x$ to $y$, we have
	\[ \frac{u(x)-2d+1}{2d} \leq N(x,y) \leq \frac{u(x)+2d-1}{2d} \]
hence
	\begin{align*}  | \nabla u(x,y) + 2d \kappa(x,y) | &= | u(y) - u(x) + 2dN(x,y) - 2dN(y,x) | \\
		&\leq 4d - 2. \qed \end{align*}
\renewcommand{\qedsymbol}{}
\end{proof}

In what follows, $C_0, C_1, \ldots$ denote constants depending only on $d$.

\begin{lemma}
\label{diameterbound}
Let $\Omega \subset \Z^d-\{o\}$ with $2 \leq \#\Omega < \infty$.  Then
	\[ \sum_{y \in \Omega} |y|^{1-d} \leq C_0\, \text{\em Diam}(\Omega). \]
\end{lemma}

\begin{proof}
For each positive integer $k$, let
	\[ \mathcal{S}_k = \{y \in \Z^d \,:\, k\leq |y| < k+1 \}. \]
Then
	\[ \sum_{y \in S_k} |y|^{1-d} \leq k^{1-d} \# \mathcal{S}_k \leq C'_0 \]
for a constant $C'_0$ depending only on $d$.
Since $\Omega$ can intersect at most $\,\mbox{Diam} (\Omega)+1 \leq 2 \,\mbox{Diam}(\Omega)$ distinct sets $\mathcal{S}_k$, taking $C_0 = 2C'_0$ the proof is complete.
\end{proof}

\begin{lemma}
\label{letsgetthispartystarted}
Let $G = G_{B_r}$ be the Green's function for simple random walk in $\Z^d$ stopped on exiting $B_r$.  For any $\rho\geq 1$ and $x \in B_r$,
	\begin{equation} \label{theparty}
	\sum_{\substack{y \in B_r \\ |x-y| \leq \rho}} \sum_{z \sim  y} |G(x,y)-G(x,z)| \leq C_1\rho.
	\end{equation}
\end{lemma}

\begin{proof}
Let $(X_t)_{t\geq 0}$ denote simple random walk in $\Z^d$,  and let $T$ be the first exit time from $B_r$. For fixed $y$, the function
	\begin{equation} \label{stoppedgreen} A(x) = g(x-y) - \EE_x g(X_T-y) \end{equation}
has Laplacian $\Delta A(x) = - \delta_{xy}$ in $B_r$ and vanishes on $\partial B_r$, hence $A(x)=G(x,y)$.

Let $x,y \in B_r$ and $z\sim y$.  From (\ref{standardgreensestimate}) we have
	\[ |g(x-y)-g(x-z)| \leq\frac{C_2}{|x-y|^{d-1}},  \qquad  y,z \neq x. \]
Using the triangle inequality together with (\ref{stoppedgreen}), we obtain
	\begin{align*} |G(x,y)-G(x,z)| &\leq |g(x-y)-g(x-z)| + \EE_x | g(X_T-y)-g(X_T-z)| \\
		&\leq  \frac{C_2}{|x-y|^{d-1}} + \sum_{w\in \partial B_r} H_x(w) \frac{C_2}{|w-y|^{d-1}}, \end{align*}
where $H_x(w) = \PP_x(X_T = w)$.

Write $D = \{y \in B_r \,:\, |x-y|\leq \}$.  Then
	\begin{equation} \label{twointegrals}  \sum_{\substack{y \in D \\ y \neq x}} \sum_{\substack{z \sim  y \\ z\neq x}} |G(x,y)-G(x,z)| \leq C_3 \rho + C_2 \sum_{w \in \partial B_r} H_x(w) \sum_{y \in D} |w-y|^{1-d}.

	\end{equation}
Taking $\Omega = w-D$ in Lemma~\ref{diameterbound}, the inner sum on the right is at most $C_0 $Diam$(D) \leq 2C_0\rho$, so the right side of (\ref{twointegrals}) is bounded above by $C_1 \rho$ for a suitable $C_1$.

Finally, the terms in which $y$ or $z$ coincides with $x$ make a negligible contribution to the sum in (\ref{theparty}), since for $y\sim x \in \Z^d$
	\[ |G(x,x)-G(x,y)| \leq |g(o)-g(x-y)| + \EE_x |g(X_T-x)-g(X_T-y)| \leq C_4. \qed \]
\renewcommand{\qedsymbol}{}
\end{proof}

\begin{figure}
\label{hitthehyperplanefigure}
\centering
\resizebox{3in}{!}{\input{lionel3.pdf_t}}
\caption{Diagram for the Proof of Lemma~\ref{hitthehyperplane}.}
\end{figure}

\begin{lemma}
\label{hitthehyperplane}
Let $H_1, H_2$ be linear half-spaces in $\Z^d$, not necessarily parallel to the coordinate axes.  Let $T_i$ be the first hitting time of $H_i$.  If $x \notin H_1 \cup H_2$, then
	\[ \PP_x(T_1>T_2) \leq \frac52 \frac{h_1+1}{h_2} \left(1+\frac{1}{2h_2}\right)^2 \]
where $h_i$ is the distance from $x$ to $H_i$.
\end{lemma}

\begin{proof}
If one of $H_1, H_2$ contains the other, the result is vacuous.  Otherwise, let $\widetilde{H}_i$ be the half-space shifted parallel to $H_i^c$ by distance $2h_2$ in the direction of $x$, and let $\widetilde{T}_i$ be the first hitting time of $H_i \cup \widetilde{H}_i$.  Let $(X_t)_{t\geq 0}$ denote simple random walk in $\Z^d$, and write $M_t$ for the (signed) distance from $X_t$ to the hyperplane defining the boundary of $H_1$, with $M_0=h_1$.   Then $M_t$ is a martingale with bounded increments.
Since $\EE_x \widetilde{T}_1 < \infty$, we obtain from optional stopping
	\[ h_1 = \EE_x M_{\widetilde{T}_1} \geq 2h_2 \PP_x(X_{\widetilde{T}_1} \in \widetilde{H}_1) - \PP_x(X_{\widetilde{T}_1} \in H_1), \]
hence
	\begin{equation} \label{justgamblersruin} \PP_x(X_{\widetilde{T}_1} \in \widetilde{H}_1) \leq \frac{h_1+1}{2h_2}. \end{equation}
Likewise, $dM_t^2 - t$ is a martingale with bounded increments, giving
	\begin{align} \label{timeupperbound} \EE_x \widetilde{T}_1 &\leq d\EE_x M_{\widetilde{T}_1}^2 \nonumber \\
		&\leq d (2h_2+1)^2 \PP_x(X_{\widetilde{T}_1} \in \widetilde{H}_1) \nonumber \\
		&\leq d (h_1+1) (2h_2+1)\left(1+\frac{1}{2h_2}\right). \end{align}

Let $T = \text{min}(\widetilde{T}_1, \widetilde{T}_2)$.  Denoting by $D_t$ the distance from $X_t$ to the hyperplane defining the boundary of $H_2$, the quantity
	\[ N_t = \frac{d}{2} \big( D_t^2 + (2h_2 - D_t)^2 \big) - t \]
is a martingale.  Writing $p = \PP_x(T=\widetilde{T}_2)$ we have
	\begin{align*} dh_2^2 = \EE N_0 = \EE N_T &\geq p \frac{d}{2}(2h_2)^2  + (1-p)dh_2^2 - \EE_x T \\
			&\geq (1+p)dh_2^2 - \EE_x T
	\end{align*}
hence by (\ref{timeupperbound})
	\[ p \leq \frac{\EE_x T}{dh_2^2} \leq 2 \frac{h_1+1}{h_2} \left(1+\frac{1}{2h_2}\right)^2. \]
Finally by (\ref{justgamblersruin})
	\[ \PP(T_1>T_2) \leq p + \PP(X_{\widetilde{T}_1} \in \widetilde{H}_1) \leq \frac52 \frac{h_1+1}{h_2} \left(1+\frac{1}{2h_2}\right)^2. \qed \]
\renewcommand{\qedsymbol}{}
\end{proof}

\begin{figure}
\centering
\label{geoshellsfig}
\resizebox{3in}{!}{\input{lionel1.pdf_t}}
\caption{Diagram for the proof of Lemma~\ref{geoshells}.}
\end{figure}

\begin{lemma}
\label{geoshells}
Let $x \in B_r$ and let $\rho = r+1-|x|$.  Let
	\begin{equation} \label{shells} \mathcal{S}^*_k = \{ y \in B_r ~:~ 2^k \rho < |x-y| \leq 2^{k+1} \rho \}. \end{equation}
Let $\tau_k$ be the first hitting time of $\mathcal{S}^*_k$, and $T$ the first exit time from $B_r$.  Then
	\[ \PP_x(\tau_k < T) \leq C_2 2^{-k}. \]
\end{lemma}

\begin{proof}
Let $H$ be the outer half-space tangent to $B_r$ at the point $z\in \partial B_r$ closest to $x$.  Let $Q$ be the cube of side length $2^k \rho/\sqrt{d}$ centered at $x$.  Then $Q$ is disjoint from $\mathcal{S}^*_k$, hence
	\[ \PP_x(\tau_k < T) \leq \PP_x(T_{\partial Q}<T) \leq \PP_x(T_{\partial Q} < T_H) \]
where $T_{\partial Q}$ and $T_H$ are the first hitting times of $\partial Q$ and $H$.  Let $H_1, \ldots, H_{2d}$ be the outer half-spaces defining the faces of $Q$, so that $Q = H_1^c \cap \ldots \cap H_{2d}^c$.  By Lemma~\ref{hitthehyperplane} we have
	\begin{align*}  \PP_x(T_{\partial Q} < T_H) &\leq \sum_{i=1}^{2d} \PP_x(T_{H_i} < T_H) \\
			&\leq \frac52 \sum_{i=1}^{2d} \frac{\text{dist}(x,H)+1}{\text{dist}(x,H_i)} \left(1+\frac{1}{2\,\text{dist}(x,H_i)}\right)^2. \end{align*}
Since dist$(x,H) = |x-z| \leq \rho$ and dist$(x,H_i) = 2^{k-1} \rho/\sqrt{d}$, and $\rho \geq 1$, taking $C_2 = 20\,d^{3/2}(1+\sqrt{d})^2$ completes the proof.
\end{proof}

\begin{lemma}
\label{greengradbound}
Let $G=G_{B_r}$ be the Green's function for random walk stopped on exiting $B_r$.  Let $x \in B_r$ and let $\rho = r+1-|x|$.   Then
 	\[ \sum_{y\in B_r} \sum_{z\sim y} |G(x,y)-G(x,z)| \leq C_3 \rho \log \frac{r}{\rho}. \]
\end{lemma}

\begin{proof}
Let $\mathcal{S}^*_k$ be given by (\ref{shells}), and let
	\[ W = \{ w \in \partial( \mathcal{S}^*_k \cup \partial \mathcal{S}^*_k) ~:~ |w-x|<2^k \rho \} \]
be the portion of the boundary of the enlarged spherical shell $\mathcal{S}^*_k \cup \partial \mathcal{S}^*_k$ lying closer to $x$.
Let $\tau_W$ be the first hitting time of $W$, and $T$ the first exit time from $B_r$.  For $w\in W$ let
	\[ H_x(w) = \PP_x(X_{\tau_W \wedge T} = w). \]
For any $y \in \mathcal{S}^*_k$ and $z \sim y$, simple random walk started at $x$ must hit $W$ before hitting either $y$ or $z$, hence
	\[ |G(x,y) - G(x,z)| \leq \sum_{w \in W}  H_x(w)|G(w,y)-G(w,z)|. \]
For any $y \in \mathcal{S}^*_k$ and any $w \in W$ we have
	\[ |y-w| \leq |y-x| + |w-x| \leq 3\cdot 2^k \rho. \]
Lemma~\ref{letsgetthispartystarted} yields
	\[ \sum_{y \in \mathcal{S}^*_k} \sum_{z \sim y} |G(x,y) - G(x,z)| \leq 3 C_1 2^k \rho \sum_{w \in W} H_x(w). \]
By Lemma~\ref{geoshells} we have $\sum_{w \in W} H_x(w) \leq C_2 2^{-k}$, so the above sum is at most  $3 C_1 C_2 \rho$.  Since the union of shells $\mathcal{S}^*_0, \mathcal{S}^*_1, \ldots, \mathcal{S}^*_{\ceil{\log_2(r/\rho)}}$ covers all of $B_r$ except for those points $y$ within distance $\rho$ of $x$, and $\sum_{|y-x|\leq\rho} \sum_{z \sim y} |G(x,y)-G(x,z)| \leq C_1 \rho$  by Lemma~\ref{letsgetthispartystarted}, the result follows.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{rotorcirc}, Inner Estimate]
Let $\kappa$ and $R$ be defined as in Lemma \ref{odomflow}.  Since the net number of particles to enter a site $x\neq o$ is at most one, we have $2d~{\text{div}\hspace{0.5ex}} \kappa (x) \geq -1$.  Likewise $2d~{\text{div}\hspace{0.5ex}} \kappa(o) = n-1$.  Taking the divergence in (\ref{gradodom}), we obtain
	\begin{align} \Delta u(x) &\leq 1 + {\text{div}\hspace{0.5ex}} R(x), \qquad x\neq o;  \label{laplodom}  \\
	\Delta u(o) &= 1-n + {\text{div}\hspace{0.5ex}} R(o). \label{laplodomat0}
\end{align}
Let $T$ be the first exit time from $B_r$, and define
	\[ f(x) = \EE_x u(X_T) - \EE_x T + n \EE_x \# \{j<T | X_j=0\}. \]
Then $\Delta f(x) = 1$ for $x \in B_r-\{o\}$ and $\Delta f(o)=1-n$.  Moreover $f \geq 0$ on $\partial B_r$.  It follows from Lemma~\ref{gammaupperbound} with $m=n$ that $f \geq \gamma - C_4$ on $B_r$ for a suitable constant $C_4$.

We have
	\[ u(x)-\EE_x u(X_T) = \sum_{k \geq 0} \EE_x \Big( u(X_{k\wedge T}) - u(X_{(k+1)\wedge
T}) \Big). \]
Each summand on the right side is zero on the event $\{T \leq k\}$, hence
	\[ \EE_x \Big( u(X_{k\wedge T}) - u(X_{(k+1)\wedge T}) ~|~ \mathcal{F}_{k\wedge
T} \Big) = -\Delta u (X_k) 1_{\{T>k\}}. \]
Taking expectations and using (\ref{laplodom}) and (\ref{laplodomat0}), we obtain
	\begin{align*}
	u(x)-\EE_x u(X_T)
	&\geq \sum_{k\geq 0} \EE_x \Big[ 1_{\{T>k\}} (n 1_{\{X_k=o\}} -1 - {\text{div}\hspace{0.5ex}} R(X_k) ) \Big] \\
	& \hspace{-0.25in} = n\EE_x \# \{k < T | X_k=o \} - \EE_x T -  \sum_{k\geq 0} \EE_x \Big[ 1_{\{T>k\}}  {\text{div}\hspace{0.5ex}} R(X_k) \Big],
	\end{align*}
hence
	\begin{equation} \label{sumofRs}
	u(x) - f(x) \geq - \frac{1}{2d} \sum_{k \geq 0} \EE_x \left[ 1_{\{T>k\}} \sum_{z \sim X_k} R(X_k,z) \right]. \end{equation}
Since random walk exits $B_r$ with probability at least $\frac{1}{2d}$ every time it reaches a site adjacent to the boundary $\partial B_r$, the expected time spent adjacent to the boundary before time $T$ is at most $2d$.  Since $|R|\leq 4d$, the terms in (\ref{sumofRs}) with $z \in \partial B_r$ contribute at most $16d^3$ to the sum.  Thus
	\[ u(x) - f(x) \geq - \frac{1}{2d}  \sum_{k\geq 0} \EE_x  \left[ \sum_{\substack{y,z\in B_r \\ y\sim z}} 1_{\{T>k\} \cap \{X_k=y\}}R(y,z) \right] - 8d^2. \]
For $y \in B_r$ we have $\{X_k = y\} \cap \{T>k\} = \{X_{k\wedge T} = y\}$, hence
	\begin{equation} \label{triplesum} u(x) - f(x) \geq - \frac{1}{2d} \sum_{k\geq 0} \sum_{\substack{y,z \in B_r \\ y\sim z}} \PP_x(X_{k\wedge T}=y)R(y,z) - 8d^2. \end{equation}
Write $p_k(y) = \PP_x(X_{k\wedge T} = y)$.  Note that since $\nabla u$ and $\kappa$ are antisymmetric,~$R$ is antisymmetric.  Thus
	\begin{align*} \sum_{\substack{y,z\in B_r \\ y\sim z}} p_k(y) R(y,z)
	&=  - \sum_{\substack{y,z\in B_r \\ y\sim z}} p_k(z) R(y,z) \\
	&= \sum_{\substack{y,z\in B_r \\ y\sim z}} \frac{p_k(y)-p_k(z)}{2} R(y,z). \end{align*}
Summing over $k$ and using the fact that $|R|\leq 4d$, we conclude from (\ref{triplesum}) that
	\[ u(x) \geq f(x) - \sum_{\substack{y,z\in B_r \\ y\sim z}} |G(x,y) - G(x,z)| - 8d^2, \]
where $G=G_{B_r}$ is the Green's function for simple random walk stopped on exiting $B_r$.     By Lemma~\ref{greengradbound} we obtain
	\[ u(x) \geq f(x) - C_3 (r+1-|x|) \log \frac{r}{r+1-|x|} - 8d^2. \]
Using the fact that $f \geq \gamma - C_4$, we obtain from Lemma~\ref{gammalowerbound}
	\[ u(x) \geq (r-|x|)^2 - C_3 (r+1-|x|) \log \frac{r}{r+1-|x|} + O\left(\frac{r^d}{|x|^d}\right). \]
The right side is positive provided $r/3 \leq |x| < r- C_5 \log r$.  For $x\in B_{r/3}$, by Lemma~\ref{gammanearorigin} we have $u(x)>r^2/4-C_3 r \log \frac32 >0$, hence $B_{r-C_5\log r} \subset A_n$.
\end{proof}

\subsection{Outer Estimate}

The following result is due to Holroyd and Propp (unpublished); we include a proof for the sake of completeness.  Notice that the bound in (\ref{HPbound}) does not depend on the number of particles.

\begin{prop} \label{HP}
Let $\Gamma$ be a finite connected graph, and let $Y\subset Z$ be subsets of the vertex set of $\Gamma$.  Let $s$ be a nonnegative integer-valued function on the vertices of $\Gamma$.  Let $H_w(s,Y)$ be the expected number of particles stopping in $Y$ if $s(x)$ particles start at each vertex $x$ and perform independent simple random walks stopped on first hitting $Z$.  Let $H_r(s,Y)$ be the number of particles stopping in $Y$ if $s(x)$ particles start at each vertex $x$ and perform rotor-router walks stopped on first hitting $Z$.  Let $H(x) = H_w(1_{x},Y)$.  Then
	\begin{equation} \label{HPbound}
	| H_r(s,Y) - H_w(s,Y) | \leq  \sum_{u \notin Z} \sum_{v \sim u} | H(u)-H(v) |
	\end{equation}
independent of $s$ and the initial positions of the rotors.
\end{prop}

\begin{proof}
For each vertex $u \notin Z$, arbitrarily choose a neighbor $\eta(u)$.  Order the neighbors $\eta(u)=v_1, v_2, \ldots, v_d$ of~$u$ so that the rotor at~$u$ points to~$v_{i+1}$ immediately after pointing to $v_i$ (indices mod $d$).  We assign {\it weight} $w(u,\eta(u))=0$ to a rotor pointing from $u$ to $\eta(u)$, and weight $w(u,v_i) = H(u) - H(v_i) + w(u,v_{i-1})$ to a rotor pointing from $u$ to $v_i$.  These assignments are consistent since $H$ is a harmonic function: $\sum_i (H(u) - H(v_i)) = 0$.  We also assign weight $H(u)$ to a particle located at $u$.  The sum of rotor and particle weights in any configuration is invariant under the operation of routing a particle and rotating the corresponding rotor.  Initially, the sum of all particle weights is $H_w(s,Y)$.  After all particles have stopped, the sum of the particle weights is $H_r(s,Y)$.  Their difference is thus at most the change in rotor weights, which is bounded above by the sum in (\ref{HPbound}).
\end{proof}

For $\rho \in \Z$ let
	\begin{equation} \label{shelldefn} \mathcal{S}_\rho = \{x\in \Z^d ~:~ \rho \leq |x| < \rho+1 \}. \end{equation}
Then
	\[ B_\rho = \{x\in \Z^d ~:~ |x| <\rho \} = \mathcal{S}_0 \cup \ldots \cup \mathcal{S}_{\rho-1}. \]
Note that for simple random walk started in $B_\rho$, the first exit time of $B_\rho$ and first hitting time of $\mathcal{S}_\rho$ coincide.  Our next result is a modification of Lemma~5(b) of \cite{LBG}.

\begin{figure}
\centering
\label{inthezonefigure}
\resizebox{3in}{!}{\input{lionel2.pdf_t}}
\caption{Diagram for the proof of Lemma~\ref{inthezone}.}
\end{figure}

\begin{lemma} \label{inthezone}
Fix $\rho \geq 1$ and $y\in \mathcal{S}_{\rho}$.  For $x \in B_\rho$ let $H(x) = \PP_x(X_T=y)$, where $T$ is the first hitting time of $\mathcal{S}_\rho$.  Then
	\begin{equation} \label{spreadingout} H(x) \leq \frac{J}{|x-y|^{d-1}} \end{equation}
for a constant J depending only on $d$.
\end{lemma}

\begin{proof}
We induct on the distance $|x-y|$, assuming the result holds for all $x'$ with $|x'-y| \leq \frac12 |x-y|$; the base case can be made trivial by choosing $J$ sufficiently large.
By Lemma~5(b) of \cite{LBG}, we can choose $J$ large enough so that the result holds provided $|y|-|x| \geq 2^{-d-3} |x-y|$.  Otherwise, let $H_1$ be the outer half-space tangent to $\mathcal{S}_{\rho}$ at the point of $\mathcal{S}_{\rho}$ closest to $x$, and let $H_2$ be the inner half-space tangent to the ball $\widetilde{S}$ of radius $\frac12 |x-y|$ about $y$, at the point of $\widetilde{S}$ closest to $x$.  By Lemma~\ref{hitthehyperplane} applied to these half-spaces, the probability that random walk started at $x$ reaches $\widetilde{S}$ before hitting $\mathcal{S}_\rho$ is at most $2^{1-d}$.   Writing $\widetilde{T}$ for the first hitting time of $\widetilde{S} \cup \mathcal{S}_{\rho}$, we have
	\[ H(x) \leq \sum_{x' \in \widetilde{S}} \PP_x(X_{\widetilde{T}}=x') H(x') \leq 2^{1-d} J \cdot \left(\frac{|x-y|}{2}\right)^{1-d} \]
where we have used the inductive hypothesis to bound $H(x')$.
\end{proof}

The \emph{lazy} random walk in $\Z^d$ stays in place with probability $\frac12$, and moves to each of the $2d$ neighbors with probability $\frac{1}{4d}$.
We will need the following standard result, which can be derived e.g.\ from the estimates in \cite{Lindvall}, section II.12; we include a proof for the sake of completeness.

\begin{lemma}
\label{lazycoupling}
Given $u \sim v \in \Z^d$, lazy random walks started at $u$ and $v$ can be coupled with probability $1-C/R$ before either reaches distance $R$ from $u$, where $C$ depends only on $d$.
\end{lemma}

\begin{proof}
Let $i$ be the coordinate such that $u_i \neq v_i$.  To define a step of the coupling, choose one of the $d$ coordinates uniformly at random.  If the chosen coordinate is different from $i$, let the two walks take the same lazy step so that they still agree in this coordinate.  If the chosen coordinate is $i$, let one walk take a step while the other stays in place.  With probability $\frac12$ the walks will then be coupled.  Otherwise, they are located at points $u',v'$ with $|u'-v'|=2$.  Moreover, $\PP\big(|u-u'|\geq \frac{R}{2\sqrt{d}}\big)<\frac{C'}{R}$ for a constant $C'$ depending only on $d$.  From now on, whenever coordinate $i$ is chosen, let the two walks take lazy steps in opposite directions.

Let
	\[ H_1 = \left\{x\,\Big|\,x_i = \frac{u'_i+v'_i}{2}\right\} \]
be the hyperplane bisecting the segment $[u',v']$.  Since the steps of one walk are reflections in $H_1$ of the steps of the other, the walks couple when they hit $H_1$.  Let $Q$ be the cube of side length $R/\sqrt{d}+2$ centered at $u$, and let $H_2$ be a hyperplane defining one of the faces of $Q$.  By Lemma~\ref{hitthehyperplane} with $h_1=1$ and $h_2=R/4\sqrt{d}$, the probability that one of the walks exits $Q$ before the walks couple is at most $2d \cdot \frac52 \frac{h_1+1}{h_2} \left(1+\frac{1}{2h_2}\right)^2 \leq 40\, d^{3/2} \big(1+2\sqrt{d}\big)^2 /R$.
\end{proof}

\begin{lemma}
\label{harmonicmeasuregrad}
With $H$ defined as in Lemma~{\em \ref{inthezone}}, we have
	\[ \sum_{u \in B_\rho} \sum_{v \sim u} |H(u)-H(v)| \leq J' \log \rho \]
for a constant $J'$ depending only on $d$.
\end{lemma}

\begin{proof}
Given $u \in B_\rho$ and $v \sim u$, by Lemma~\ref{lazycoupling}, lazy random walks started at $u$ and $v$ can be coupled with probability $1-2C/|u-y|$ before either reaches distance $|u-y|/2$ from $u$.  If the walks reach this distance without coupling, by Lemma~\ref{inthezone} each has still has probability at most $J/|u-y|^{d-1}$ of exiting $B_\rho$ at $y$.  By the strong Markov property it follows that
	\[ |H(u) - H(v)| \leq \frac{2CJ}{|u-y|^d}. \]
Summing in spherical shells about $y$, we obtain
	\[ \sum_{u \in B_\rho} \sum_{v \sim u} |H(u)-H(v)| \leq \sum_{t=1}^{\rho} d\omega_d t^{d-1} \frac{2CJ}{t^d} \leq J' \log \rho. \qed \]

\renewcommand{\qedsymbol}{}
\end{proof}

We remark that Lemma~\ref{harmonicmeasuregrad} could also be inferred from Lemma~\ref{inthezone} using \cite[Thm.\ 1.7.1]{Lawler} in a ball of radius $|u-y|/2$ about $u$.

To prove the outer estimate of Theorem~\ref{rotorcirc}, we will make use of the \emph{abelian property} of rotor-router aggregation.  Fix a finite set $\Gamma \subset \Z^d$ containing the origin.  Starting with $n$ particles at the origin, at each time step, choose a site $x \in \Gamma$ with more than one particle, rotate the rotor at $x$, and move one particle from $x$ to the neighbor the rotor points to.  After a finite number of such choices, each site in $\Gamma$ will have at most one particle, and all particles that exited $\Gamma$ will be on the boundary $\partial \Gamma$.  The abelian property says that \emph{the final configuration of particles and the final configuration of rotors do not depend on the choices}.  For a proof, see \cite[Prop.~4.1]{DF}.

In our application, we will fix $\rho \geq r$ and stop each particle in rotor-router aggregation either when it reaches an unoccupied site or when it reaches the spherical shell $\mathcal{S}_\rho$.  Let $N_\rho$ be the number of particles that reach $\mathcal{S}_\rho$ during this process.  Note that at some sites in $\mathcal{S}_\rho$, more than one particle may have stopped.  If we let each of these extra particles in turn continue performing rotor-router walk, stopping either when it reaches an unoccupied site or when it hits the larger shell $\mathcal{S}_{\rho+h}$, then by the abelian property, the number of particles that reach $\mathcal{S}_{\rho+h}$ will be $N_{\rho+h}$.  We will show that when $h$ is order $r^{1-1/d}$, a constant fraction of the particles that reach $\mathcal{S}_\rho$ find unoccupied sites before reaching $\mathcal{S}_{\rho+h}$.

\begin{proof}[Proof of Theorem~\ref{rotorcirc}, Outer Estimate]
Fix integers $\rho \geq r$ and $h\geq 1$.
In the setting of Proposition~\ref{HP}, let $\Gamma$ be the lattice ball $B_{\rho+h+1}$, and let $Z = \mathcal{S}_{\rho+h}$.  Fix $y \in \mathcal{S}_{\rho+h}$ and let $Y=\{y\}$.  For $x \in \mathcal{S}_\rho$, let $s(x)$ be the number of particles stopped at $x$ if each particle in rotor-router aggregation is stopped either when it reaches an unoccupied site or when it reaches $\mathcal{S}_\rho$.  Write
	 \[ H(x) = \PP_x(X_T=y) \]
where $T$ is the first hitting time of $\mathcal{S}_{\rho+h}$.   By Lemma~\ref{inthezone} we have
	\begin{equation} \label{H_wbound} H_w(s,y) = \sum_{x \in \mathcal{S}_\rho} s(x)H(x) \leq \frac{J N_\rho}{h^{d-1}} \end{equation}
where
	 \[ N_\rho = \sum_{x \in \mathcal{S}_\rho} s(x) \]
is the number of particles that ever visit the shell $\mathcal{S}_\rho$.

By Lemma~\ref{harmonicmeasuregrad} the sum in (\ref{HPbound}) is at most $J' \log h$, hence from Propositon~\ref{HP} and (\ref{H_wbound}) we have
	\begin{equation} \label{tinyRRs} H_r(s,y) \leq \frac{J N_\rho}{h^{d-1}} + J' \log h. \end{equation}
Let $\rho(0) = r$, and define $\rho(i)$ inductively by
	\begin{equation} \label{doublemin} \rho(i+1) = \text{min}\left\{\rho(i)+N_{\rho(i)}^{2/(2d-1)},~\text{min} \{\rho> \rho(i) | N_\rho \leq N_{\rho(i)}/2 \}\right\}. \end{equation}
Fixing $h < \rho(i+1)-\rho(i)$, we have
	\[ h^{d-1} \log h \leq N_{\rho(i)}^{\frac{2d-2}{2d-1}} \log N_{\rho(i)} \leq N_{\rho(i)}; \]
so (\ref{tinyRRs}) with $\rho=\rho(i)$ simplifies to
	\begin{equation} \label{eventinierRRs} H_r(s,y) \leq \frac{CN_{\rho(i)}}{h^{d-1}} \end{equation}
where $C = J + J'$.

Since all particles that visit $\mathcal{S}_{\rho(i)+h}$ during rotor-router aggregation must pass through $\mathcal{S}_{\rho(i)}$, we have by the abelian property
	\begin{equation} \label{alltermsaresmall} N_{\rho(i)+h} \leq \sum_{y\in \mathcal{S}_{\rho(i)+h}} H_r(s,y). \end{equation}
Let $M_k = \# (A_n \cap \mathcal{S}_k)$.  There are at most $M_{\rho(i)+h}$ nonzero terms in the sum on the right side of (\ref{alltermsaresmall}), and each term is bounded above by (\ref{eventinierRRs}), hence
	\[ M_{\rho(i)+h} \geq N_{\rho(i)+h} \frac{h^{d-1}}{CN_{\rho(i)}} \geq  \frac{h^{d-1}}{2C} \]
where the second inequality follows from $N_{\rho(i)+h} \geq N_{\rho(i)}/2$.  Summing over $h$, we obtain
	\begin{equation} \label{lotsavolume}  \sum_{\rho=\rho(i)+1}^{\rho(i+1)-1} M_\rho \geq \frac{1}{2dC} (\rho(i+1)-\rho(i)-1)^d. \end{equation}
The left side is at most $N_{\rho(i)}$, hence
	\[ \rho(i+1)-\rho(i) \leq (2dC N_{\rho(i)})^{1/d} \leq N_{\rho(i)}^{2/(2d-1)} \]
provided $N_{\rho(i)} \geq C' := (2dC)^{2d-1}$.
Thus the minimum in (\ref{doublemin}) is not attained by its first argument.  It follows that $N_{\rho(i+1)}\leq N_{\rho(i)}/2$, hence $N_{\rho(a \log r)} < C' $ for a sufficiently large constant $a$.

By the inner estimate, since the ball $B_{r-c\log r}$ is entirely occupied, we have
	\begin{align*} \sum_{\rho \geq r} M_\rho &\leq \omega_d r^d - \omega_d (r-c\log r)^d \\
	&\leq cd\omega_d r^{d-1} \log r. \end{align*}
Write $x_i = \rho(i+1)-\rho(i)-1$; by (\ref{lotsavolume}) we have
	\[ \sum_{i=0}^{a \log r} x_i^d \leq cd\omega_d r^{d-1} \log r, \]
By Jensen's inequality, subject to  this constraint, $\sum x_i$ is maximized when all $x_i$ are equal, in which case $x_i \leq C'' r^{1-1/d}$ and
	\begin{equation} \rho(a \log r) = r+\sum x_i \leq r+C''r^{1-1/d} \log r. \label{outershellbound} \end{equation}
Since $N_{\rho(a \log r)} < C'$ we have $N_{\rho(a \log r)+C'} = 0$; that is, no particles reach the shell $\mathcal{S}_{\rho(a \log r)+C'}$.  Taking $c' = C' + C''$, we obtain from (\ref{outershellbound})
	\[ A_n \subset B_{r (1 + c'r^{-1/d}\log r)}. \qed \]
\renewcommand{\qedsymbol}{}
\end{proof}

\section{Concluding Remarks}

\begin{figure}
\label{invertedrotor}
\centering
\includegraphics{invertedrotor1m15xtiny.png}
\caption{Image of the rotor-router aggregate of one million particles under the map $z \mapsto 1/z^2$.  The colors represent the rotor directions.  The white disc in the center is the image of the complement of the occupied region.}
\end{figure}

A number of intriguing questions remain unanswered.  Although we have shown that the asymptotic shape of the rotor-router model is a ball, the near perfect circularity found in Figure~\ref{rotor1m} remains a mystery.  In particular, we do not know whether an analogue of Theorem~\ref{divsandcircintro} holds for the rotor-router model, with constant error in the radius as the number of particles grows.

Equally mysterious are the patterns in the rotor directions evident in Figure~\ref{rotor1m}.  The rotor directions can be viewed as values of the odometer function mod $2d$, but our control of the odometer is not fine enough to provide useful information about the patterns.  If the rescaled occupied region $\sqrt{\pi/n} \,A_n$ is viewed as a subset of the complex plane, it appears that the monochromatic regions visible in Figure~\ref{rotor1m}, in which all rotors point in the same direction, occur near points of the form $(1+2z)^{-1/2}$, where $z = a+bi$ is a Gaussian integer (i.e.\ $a,b \in \Z$).  We do not even have a heuristic explanation for this phenomenon.  Figure~7 shows the image of $A_{1,000,000}$ under the map $z \mapsto 1/z^2$; the monochromatic patches in the transformed region occur at lattice points.

L\'{a}szl\'{o} Lov\'{a}sz (personal communication) has asked whether the occupied region $A_n$ is simply connected, i.e.\ whether its complement is connected.  While Theorem~\ref{rotorcirc} shows that $A_n$ cannot have any holes far from the boundary, we cannot answer his question at present.

A final question is whether our methods could be adapted to internal DLA to show that if $n=\omega_d r^d$, then with high probability $B_{r-c\log r} \subset I_n$, where $I_n$ is the internal DLA cluster of $n$ particles.  The current best bound is due to Lawler \cite{Lawler95}, who proves that with high probability $B_{r-r^{1/3}(\log r)^2} \subset I_n$.

\section*{Acknowledgments}

The authors thank Jim Propp for bringing the rotor-router model to our attention and for many fruitful discussions.  We also had useful discussions with Oded Schramm, Scott Sheffield and Misha Sodin.  We thank Wilfried Huss for pointing out an error in an earlier draft.  Yelena Shvets helped draw some of the figures.

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What is fatigue? What is energy?
I've been practicing Sivananda yoga for 12 years. Over that period, I did a degree, began my career, became an M&A investment banker, and moved beyond bulge bracket corporate finance, too.
Recently, I went on a yoga retreat at the Sivananda Ashram in Orleans, France. We were doing two extended yoga classes of 1hr45 mins a day. We were also doing two meditations and chanting everyday. Days one and two, I was largely feeling tired and adjusting to the schedule (5.30am wakeup). Day three I was feeling pretty good, but needed an afternoon nap. By day four, my energy had ramped up, I didn't want a nap, and I didn't really feel like sitting down or vegeing. I literally had so much energy. This pattern was almost universal amongst the people at the retreat, and it's been my pattern on previous retreats.
It really made me think – what is fatigue? What is energy? What makes me feel tired or energised?
Sleep is part of it. But at the ashram you don't sleep very much, yet have a lot of energy. Other times, I can be well rested and still feel lethargic. Getting angry makes me feel tired. A great concert gives me a ton of energy. So energy and fatigue are about something to do with the state of mind, the emotions, in addition to being about sleep.
don't really take time, they give you time. And since that's what they were designed by generations of practitioners to do, I guess I shouldn't be so surprised, and yet somehow I continue to be.


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Brittaney Kiefer
British Arrows: Nike and Adam & Eve/DDB are big winners
Nike's 'Nothing beats a Londoner' swept the awards, while A&E/DDB was named Agency of the Year again.
Nike's 'Nothing beats a Londoner': Commercial of the Year
Adam & Eve/DDB was crowned Agency of the Year at the British Arrows for the third year running, while Nike's "Nothing beats a Londoner" was named Commercial of the Year.
Somesuch took the Production Company of the Year title for the second year at the awards ceremony on Wednesday night in Battersea, London. The Agency and Production Company of the Year prizes are awarded based on the number of gold, silver and bronze wins the companies have received across the show.
Adam & Eve/DDB: Agency of the Year
"Nothing beats a Londoner", created by Wieden & Kennedy London, swept the 2019 British Arrows with 14 prizes – six gold Arrows, six gold Crafts and two silver Crafts. Along with Commercial of the Year, it received the honour of Best Crafted Commercial.
Riff Raff's Megaforce, who directed Nike's film, won the Frank Budgen Award for Best Director. Wieden & Kennedy's Tom Bender, Tom Corcoran, Ryan Fisher, Mark Shanley and Paddy Treacy won in the Writing category for their work on "Nothing beats a Londoner".
Nike was also named Advertiser of Excellence for its continued delivery of inspiring advertising worldwide.
Other big winners from the night were Essity's "Viva la vulva", created by Abbott Mead Vickers BBDO and directed by Kim Gehrig through Somesuch, and John Lewis & Partners' last Christmas ad starring Elton John, "The boy and the piano", created by Adam & Eve/DDB and directed by Seb Edwards through Academy. Each received three gold Arrows, two gold Crafts and one silver Arrow.
Academy's Billy Boyd Cape was named Best New Director for his work on Pride in London's "Somewhere over the rainbow" by BMB.
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Fish Species in Marlborough
Short finned eel
Last Modified: 27-5-2021 8:56
Short finned eels (Anguilla australis), are the commonest of the two native eel species found in New Zealand. They are the most resilient of New Zealand's freshwater fish species, capable of surviving in marginal environments where water quality would not support other fish.
Short finned eels are found throughout New Zealand, and Marlborough is no exception, where they can be found in a number of natural and artificial waterways. With much of what is now Blenheim formerly a flax swamp, the area was an important habitat for eels which provided a major food source for Māori.
This example was observed in an artificial drainage channel that forms an extension to Sutherland Stream in Blenheim. They have also been observed at the Taylor Dam, and in the Taylor River.
Image Date: 4/12/2018
Photographer: Christopher Cookson
Location: Latitude: -41.533436 Longitude: 173.965897
Eels - Te Ara


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Home Latest News Facebook might have to pay billions of dollars in fines to FTC...
Facebook might have to pay billions of dollars in fines to FTC for privacy violations
Facebook could soon set another record…
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By Karissa Bell 2019-02-14 17:15:43 -0800
Facebook may soon set a new record, just not the kind it likes to brag about.
The social network is facing a multibillion-dollar fine from the Federal Trade Commission over privacy violations, according to a new report in The Washington Post.
The FTC previously confirmed it had opened an investigation into the social network last March, following the Cambridge Analytica debacle. Last month, The Washington Post reported Facebook's potential fine could be "record-setting," and significantly higher than the $22.5 million Google was fined in 2012 — the current record-holder for biggest FTC fine against a tech company.
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This isn't the first time Facebook has had a run-in with the FTC. The company reached a settlement with the regulator in 2011 over its deceptive privacy policies.
For Facebook, which takes in billions of dollars every year, a massive fine might not be the most detrimental aspect of a settlement with the FTC. A settlement could also come with increased regulatory scrutiny and, possibly, a requirement to change certain "business practices," according to The Post. The company could also opt to fight the FTC in court, in which case we wouldn't see a resolution to the drama for some time.
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Top 10 Fastest Cars In The World — 2012
GizmoCrazed - February 10, 2012
Unlike a decade back when the number of fast cars were limited, you will now find plenty of options to choose from. These vehicles...
Thousands of NASA Apollo Mission Photos Just Went Online
This Fearsome *Titan Games* Event Reveals the Value of Torque
Elon Musk: 'We need a revolt against the fossil fuel industry'


A bluetooth® wireless connection may fail or the sound cut in and out if the bluetooth signal is weak, out of range or has interference. Refer to the information below to improve the quality of the signal and lessen the chances of interference.
The range of the Bluetooth connection is approximately 30 feet (10 meters). However, maximum communication range will vary depending on obstacles (person, metal, wall, etc.) or electromagnetic environment. Communication range will vary according to whether walls or other obstacles are located between the devices, the material used in the intervening walls, the presence of Radio Frequency (RF) interference and other environmental conditions, and the type of software being used, if applicable.


If you've seen The Martian and thought you too would like to colonize Mars, you actually don't need to go as far you'd think. Matt Damon flew out to Wadi Rum, Jordan, which is also the home of the famous Lawrence of Arabia during WWI. There are numerous sites, both historical and geological, but the real jewel is the insight that you gain into the Bedouin culture and lifestyle.


Fresh, homemade Lancaster County goodness.
take home a piece of Lancaster County.
Unique items for the home cook!
Uplift the spirit and fill the senses!
Tempting toys for both the young and young at heart!
The Whoopie Pie: A Lancaster, PA tradition to share!


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LSUShreveport
NAIA Baseball National Championship Preview
Written by Administrator - Thursday, May 20 2021
Schedule for the first two days of the 64th NAIA Baseball National Championship, Friday-Saturday, May 28-June 4, Harris Field, Lewiston, ID (all times PDT):
8:30 AM: #8Keiser(36-16) vs. #9LSU-Shreveport (44-14)
Keiser (West Palm Beach, FL) is making their second trip to the Baseball National Championship. The first one was in 2017, and they went 1-2. Twitter: @KeiserBaseball
Nickname: Seahawks
Rankings: #17 in coaches' poll, #9 in Boyd Nation's ISR. Strength of Schedule: #10
Quality wins: 3-0 vs. South Carolina-Beaufort, 3-3 vs. Saint Thomas, 2-0 vs. Oklahoma City
Top position players: 3B Tim Bouchard (.403, 16 HR, 67 RBI), OF Matthew Catalfo (.389, 7 HR, 62 RBI), OF Ben McConnell (.356, 6 HR, 40 RBI), SS Mason Dineson (.330, 10 HR, 53 RBI), C Zach Mazur (.253, 14 HR, 53 RBI)
Top Pitchers: RHP Pablo Arevalo (9-3, 1.38 ERA, 114 K), RHP James Reynolds (7-3, 3.65 ERA, 57 K), RHP Deykel Reyes (8-1, 5.52 ERA, 72 K). Arevalo's 1.38 ERA is third-best in the NAIA. He has pitched 117 innings, which is the most in the NAIA.
Coach: Brook Fordyce, fourth season
Outlook: The Seahawks are not at the same level as the top four teams in this tournament.
Louisiana State-Shreveport is making their fourth trip to the Baseball National Championship. The most recent one was 2012, and they finished third. Twitter: @LSUS_Baseball
Nickname: Pilots
Rankings: #25 in coaches' poll, #11 in ISR. SoS: #24
Quality wins: Central Methodist, 2-2 vs. Oklahoma City, Mobile, 2-0 vs. Point Park
Top position players: 3B Payton Robertson (.417, 7 HR, 56 RBI), OF Kody Kolb (.408, 45 RBI), DH Taylor Fajardo (.345, 15 HR, 43 RBI)
Top Pitchers: RHP Trent Sholders (6-1, 3.04 ERA, 30 K), LHP Josh Fortenberry (8-2, 3.23 ERA, 63 K), RHP Sebastian Selway (6-3, 3.36 ERA, 86 K), RHP Kevin Miranda (6-3, 5.30 ERA, 70 K), RHP Luisdiego Quintana (4-3, 5.12 ERA, 70 K)
Coach: Brad Neffendorf, second season
Outlook: Either the Pilots or the Seahawks will get a win.
Interview with LSU Shreveport Coach Brent Lavallee
Written by Administrator - Monday, February 18 2019
Brent Lavallee is in his third season as Head Coach of the LSU Shreveport Pilots. The Pilots won the Red River Athletic Conference regular season championship last season. Lavalee was named RRAC Coach of the Year in 2017.
Bob Broughton, courtesyrunner.com: You played four seasons at LSU Shreveport, and went on to become assistant coach and then head coach there. How did you get from North Delta, British Columbia to Shreveport?
Brent Lavallee, LSU Shreveport: Pretty much, I got lucky. It is not easy for a Canadian to get into a good American college program. A former LSUS player (Aaron McRae) was back in town getting ready to go to Spring training with the Detroit Tigers. He had played his Senior season at LSUS in 2003, leading them to a World Series appearance. He was back home training with our high school team and noticed I had some decent defensive skills and he made a call to the Head Coach at LSUS, Rocke Musgraves. Coach Rocke took his word and spoke with my high school coach, Ari Mellios - who had a reputation of making great catchers, and they gave me a little scholarship money and an opportunity to play as a Freshman, and the rest is history. I owe a lot to my parents, my high school coaches and to Aaron McRae and Coach Rocke Musgraves for giving me that opportunity. There was a lot of hard work on my behalf - I was a hockey player who played baseball in the Summer, but there was a much better chance to get an education playing baseball and I am glad I made that decision.
BB: The Pilots have usually had at least one Canadian on the roster, and this year, you have four of them. What is is about LSUS that attracts them?
BL: I think there are many things that attract Canadian student-athletes to LSUS. We have great weather and terrific facilities, especially compared to most Canadian high school programs. The cost of tuition is very affordable, even with the exchange rate on the Canadian dollar being pretty high. Also, the comfort level of having a Canadian on the coaching staff is a bit of an attraction to the families, just knowing that there is someone who knows what the international students are going through and what to do if something were to go wrong. A lot of it is networking. We don't get the chance to see many of the Canadians that end up here; we go off of recommendations from people and we trust them and hope that they know the talent level that we try to bring in. I find myself sometimes being harder on recruiting Canadians because I don't want to bring a kid 3,000 miles to sit the bench. We want them to be contributors, especially if they are coming from so far and paying that amount of money.
Published in Interviews with coaches
Midland 9, LSU-Shreveport 8
Written by Administrator - Friday, February 10 2017
DH Robert Cummins hit a walkoff RBI single with two out in the bottom of the seventh to give the Midland Warriors a 9-8 come-from-behind victory over the LSU-Shreveport Pilots. The game was part of a tournament hosted by Dallas Christian College.
A stiff wind blowing throughout the game made every fly ball an adventure, and Cummins' game-winning hit was no exception. It was a high popup that dropped between the first baseman and right fielder.
The Pilots led 8-6 going into the bottom of the seventh, and Pilots reliever LHP Pat Day struck out the first two batters. LF Buddy Balsino walked to keep the Warriors alive. RF Nathan Virgen doubled, then pinch-hitter Chris Foster hit a two-RBI single to tie the game, and he went to second on a wild throw. He was then batted in by Cummins.
Published in Game accounts
Canada wins second straight Pan American Games gold medal
Written by Administrator - Monday, July 20 2015
Canada won its second straight gold medal in baseball in the 2015 Pan American Games, played in Ajax, Ontario.
In both 2011 and 2015, alumni of NAIA schools played a key role in Canada's win. The 2011 team had four players who played at NAIA schools, and RHP Chris Kissock of Lewis-Clark State played a key role for that team (story).
The 2015 edition included two NAIA alumni, LHP Jeff Francis of British Columbia and RHP Jared Mortensen of LSU-Shreveport. Francis got two saves during the preliminary round, retired all six batters he faced, struck out two of them. He was called upon to start the gold medal game against the USA. He went seven innings, allowed four runs (two of them on a home run by DH Patrick Kivlehan [Mariners organization, Rutgers]) on eight hits, struck out seven, and walked one. He got a no-decision; the game was tied 4-4 when he left it. Canada ultimately won the game 7-6 in ten innings; the tying and winning runs came on a bad pickoff attempt followed by a throwing error.
Recap for February 2
Written by Administrator - Sunday, February 02 2014
RHP Keegan Yuhl, a transfer from Lee, pitched a no-hitter for #9 Concordia-Irvine in a 9-0 non-conference win over Antelope Valley. Yuhl threw 115 pitches, struck out 12, walked three. Story from Concordia-Irvine: Keegan Yuhl Throws First No-Hitter In Concordia History. The story includes three video clips; scroll down to see them. Box score.
The Eagles didn't just get a no-hitter and a three game sweep of Antelope Valley; all three games were shutouts, and the Eagles allowed only four hits for the series.
#12 The Master's won two of three at #22 San Diego Christian in a Golden State Athletic Conference matchup of two teams that were previously undefeated. The scores were 4-2, 8-11, and 3-1. The Mustangs got three runs on the top of the ninth inning to win the first game; 1B Josh Brown hit a two-RBI single to bring in the winning run. Jason Karkenny pitched two perfect innings to get the win in relief. 3B Jeff Lusardi hit a home run, a double, and five RBI for the Hawks in the second game. RHP Tucker Richardson pitched a seven-inning gem for the Mustangs in the third game; three hits, nine strikeouts, one walk. He was one out short of a shutout; Lusardi hit a home run with two outs in the bottom of the seventh. The Mustangs visit Concordia-Irvine Feb. 7-8. The Hawks host #16 Cal State San Marcos Feb. 8.
The split between The Master's and San Diego Christian enabled Vanguard to take over first place in the GSAC. The Lions swept three games against Arizona Christian, 3-2, 4-2, and 2-1. The Lions were held to three hits by Ismael Castro in the first game, but all three Lions runs were unearned. CF Thomas Mobley hit a two-RBI double in the bottom of the eighth to win the second game. Scott Torrey went the seven-inning distance for the Lions in the third game, didn't allow an earned run, allowed four hits, struck out six. The Lions visit Westmont Feb. 7-8. Live video
Top 25 and recap for April 30
Written by Administrator - Tuesday, April 30 2013
The NAIA Coaches' poll for April 30, with Boyd Nation's ISR ranking. Recap below the fold.
Faulkner (AL)
Lee (TN)
Northwood (TX)
Lubbock Christian (TX)
Embry-Riddle (FL)
The Master's (CA)
Bellevue (NE)
Lewis-Clark State (ID)
York (NE)
Cal State San Marcos
Cumberland (TN)
Concordia-Irvine (CA)
Oklahoma Baptist
Missouri Baptist
Texas Wesleyan
Sterling (KS)
Madonna (MI)
Georgetown (KY)
Point Park (PA)
Lyon (AR)
Lindsey Wilson (KY)
William Carey (MS)
Science & Arts (OK)
San Diego Christian (CA)
Arizona Christian
Menlo (CA)
Westmont (CA)
Vanguard (CA)
Biola (CA)
Southern Polytechnic (GA)
Top 25 and recap for April 2
The NAIA Coaches' poll for April 2, with Boyd Nation's ISR ranking. Recap below the fold.
Rogers State (OK)
Freed-Hardeman (TN)
St. Thomas (FL)
Campbellsville (KY)
Bacone (OK)
Georgetown C. (KY)
Science and Arts of Oklahoma
Recap for March 4
Written by Administrator - Monday, March 04 2013
NAIA baseball action during the past week.
#1 Tennessee Wesleyan defeated Georgia Gwinnett 2-1, and won two of three against Saint Andrew's, 2-0, 6-5, and 2-3. The winning run against Georgia Gwinnett came on an RBI single by SS Wes Minton in the bottom of the ninth. Corey Rhoney and Noah Piard combined for the shutout in the first Saint Andrew's game; Rhoney went eight innings, allowed four hits, struck out seven. In the second game, the Knights out-hit the Bulldogs 12-7; the Bulldogs scored all six of their runs in the first inning. CF Travis Burnside hit a two-RBI single, then scored the winning run on a misplayed ground ball. In the third game, the defending champions were held to six hits by Joe McCarty and Evan Tucker. RF Michael Durig batted in all three of the Knights' runs.
The Bulldogs are now 15-5. They visit Belhaven March 5-6, host Milligan March 9-10, and host Ohio Christian March 11.
#2 LSU-Shreveport got an easy 15-5 mid-week win over Division III Louisiana College, but was swept by Texas Wesleyan, 5-1, 12-8, and 6-5. Derek Vaughn got the win for the Rams in the first game; he went six innings, allowed one run on two hits, struck out four. Connor Zamora and Nolan Barbee kept the Pilots off the board the rest of the way, and Zamora struck out three of the six batters he faced. CF Trevor Moon singled in the winning run (story). In the second game, the Rams scored ten runs in the second inning, capped by a three-RBI double by SS David Kiriakos. The Pilots out-hit the Rams 11-7 in the third game. They got within one run in the top of the ninth, and had the go-ahead run on first with nobody out. However, reliever Patrick Stanley got two strikeout and a fly out to end the game and get a save (story).
Northwood (TX), who has won seven straight, has the lead in the Red River Athletic Conference. The Rams are two games back, and the Pilots are three games back. The Pilots visit LSU-Alexandria March 5, and host Huston-Tillotson March 8-9. Streaming audio
Recap for February 24
#1 Tennessee Wesleyan lost a mid-week game to #12 Lindsey Wilson 4-3. The Bulldogs outhit the Blue Raiders 9-7; LF Leandro Andino hit a sacrifice fly to send in the winning run. (story). The defending champions then swept three games against Indiana-Southeast, 12-2, 3-1, and 2-1. In the first game, Four Bulldog pitchers combined to hold the Grenadiers to four hits; Jarrod Jameson got the win. SS Wes Minton scored the winning run in the second game on a passed ball; Corey Rhoney got the win, went 5 2/3 innings, allowed four hits, struck out five. In the third game, the Bulldogs managed only two hits off Grenadiers pitcher Lucas Harry, but RF David Donald was hit by a pitch and scored on a ground out for the first Bulldog run, and Minton hit a solo home run for the winning run. Tyler Wilson got the win; he went 4 2/3 innings, allowed two hits, struck out seven.
The Bulldogs host Georgia Gwinnett February 27, and Saint Andrew's, who is 9-1, March 2-3.
#2 LSU-Shreveport swept Texas College, 14-4, 3-2, and 8-4. Rory Young got the win in the first game. The Pilots needed a two-RBI single in the bottom of the seventh and final inning by C David Dorton to win the second game. CF Ridge Gonsoulin hit a two-run home run in the third game, finished 2-for-4 with three RBI.
The Pilots host Division III Louisiana College February 26, and visit Texas Wesleyan March 1-2.
#1 Tennessee Wesleyan swept Indiana Tech 7-6, 4-1, and 3-1. The first game was tied 5-5 going into the seventh and last inning. The Warriors got a run in the top of the seventh on an RBI triple by RF Bryan Brudi, but the defending champions tied it on an RBI single by RF David Donald, and won it on an RBI single by 1B Trey Griffin. Griffin went 4-for-4, Donald went 3-for-4 with three RBI. Both teams had three hits in the third game. The winning run came on a sacrifice fly by C Brackton Smith. Jarrod Jameson got the win; he went seven innings, allowed no earned runs, struck out seven.
The Bulldogs have won eight straight after losing the first three games of the season. They host #12 Lindsey Wilson on February 20, and Indiana-Southeast February 22-23.
#2 LSU-Shreveport won two of three against Culver-Stockton. Jason Mouton and Cade Bowen combined for a one-hitter in the first game, won by the Pilots 5-0. Mouton went six innings, allowed the only hit in the fifth inning, struck out five. Bowen pitched a perfect inning, struck out one. In the second game, the Pilots committed five errors, walked seven batters, hit seven batters, and lost 14-3 (story). The third game went eleven innings, and the Pilots won 6-5. CF Ridge Gonsoulin hit an RBI single for the winning run. Winning pitcher Ben Suit retired every batter he faced over the last three innings (story).
The Pilots host Division II Southern Arkansas on February 19, and Texas College February 22-23.
#3 Lee swept Mobile 6-3, 3-1, and 12-0. Myles Smith got the win in the first game, and 2B Mark Silva went 3-for-3 with an RBI. RF Corey Davis hit a two-RBI single for the winning run in the second game. Kyle Briner got the win in relief; he struck out five of the seven batters he faced. The Rams got an excellent effort from Ryan Hines, who struck out 11 (story). 3B Roberto Duran went 3-for-3 with a three-run home run and four RBI in the third game. LHP Clint Terry and C.J. Lusted combined for the three-hit shutout; Terry went six innings, struck out five (story).
The Flames are 10-2. They host Rio Grande February 18, visit Union C. (KY) February 19, and host #18 Auburn-Montgomery February 22-23.
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Bible classes for adults are offered on Sunday morning at 10:15 and refreshments are provided. It is led by the pastor. There are also small group Bible classes that are led by a member.
Sunday School is offered for children in preschool through sixth grade on Sunday mornings at 10:15.
Confirmation classes are offered for junior high (age is not firm) students on Wednesday evenings at 6:30. It is a two year program. The first year is the Life of Jesus and the second year is based on Luther's Small Catechism.
Vacation Bible School Day Camp is offered in August of each year in cooperation with Camp Omega of Waterville, Minnesota. It is open to those entering preschool through eighth grade at no cost. The VBS Day Camp runs from 9:00AM to 11:30 for the preschool class and 9:00 – 3:00 for the rest of the grades.


The price is still rising, so bulls are likely going to test 8/8 MM Level in the coming hours. If a pullback from this level happens, there'll be an opportunity to have a bearish wave (i). In this case, we should keep an eye on 6/8 MM Level as an intraday target.
We've just got a new local high. Previously, the price couldn't fixate under 6/8 MM Level. Therefore, wave v of (c) is likely going to be continued towards 8/8 MM Level, which could be a departure point for wave i of (i).


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Q: Can't deserialize an ASP.NET viewstate in Mono Here's the Base64-encoded viewstate I'm trying to deserialize: https://pastebin.com/VQxvyUGS
…and here's the code I'm attempting to deserialize it with:
string test = (content of the linked paste)
var losFormatter = new LosFormatter();
var viewState = losFormatter.Deserialize(test);

My program crashes on the last line with a NullReferenceException and I have no idea why. Yesterday, the same code worked just fine on Windows, using the real .NET framework. This is on macOS 10.12.6 in Xamarin Studio with Mono 5.2.0.
Is this a bug in Mono itself?

redpajama_set_name: RedPajamaStackExchange