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text:
Biografia
Keller ha conseguito il bachelor of arts alla Bucknell University nel 1972. Nel periodo universitario si è avvicinato alla religione cristiana e ha deciso di dedicarsi a studi di teologia. Nel 1975 ha conseguito il Master of Divinity al Gordon-Conwell Theological Seminary e nello stesso anno è stato consacrato pastore della Chiesa presbiteriana. Durante il periodo del seminario ha conosciuto e sposato la moglie Kathy, da cui ha avuto tre figli. Keller si è quindi trasferito con la famiglia in Virginia, dove ha esercitato il suo ministero per nove anni. Durante questo periodo ha anche continuato i suoi studi e nel 1981 ha conseguito il Doctor of Ministry al Westminster Theological Seminary di Filadelfia, dove nel 1984 è stato nominato professore aggiunto di teologia pratica. Nel 1989 è stato inviato a New York per fondare a Manhattan la Redeemer Presbyterian Church. Keller ha fondato inoltre Hope for New York, un'organizzazione senza scopi di lucro che manda volontari e sovvenzioni a ministri e organizzazioni religiose che operano a favore dei poveri della città. Keller è inoltre molto attivo nel campo della diffusione e della difesa della fede cristiana e ha scritto più di venti libri.

Libri pubblicati
 Resources for Deacons: Love Expressed through Mercy Ministries (Christian Education and Publications, 1985)
 Ministries of Mercy: The Call of the Jericho Road (P&R Publishing, 1997)
 Church Planter Manual (Redeemer Presbyterian Church, 2002)
 The Reason for God: Belief in an Age of Skepticism (Dutton Adult, February 2008)
 The Prodigal God: Recovering the Heart of the Christian Faith (Dutton Adult, November 2008)
 Counterfeit Gods: The Empty Promises of Money, Sex, and Power, and the Only Hope that Matters (Dutton Adult, October 2009)
Generous Justice: How God's Grace Makes Us Just (Dutton Adult, November 2010)
King's Cross: The Story of the World in the Life of Jesus (Dutton Adult, February 2011)
The Meaning of Marriage: Facing the Complexities of Commitment with the Wisdom of God (Dutton Adult, November 2011)
The Freedom of Self Forgetfulness: The Path to True Christian Joy (10Publishing, March 2012)
Center Church: Doing Balanced, Gospel-Centered Ministry in Your City (Zondervan, September 2012)
Every Good Endeavor: Connecting Your Work to God's Work (Dutton, November 2012)
Galatians For You (The Good Book Company, February 2013)
Judges For You (The Good Book Company, August 2013)
Walking with God through Pain and Suffering (Dutton, October 2013)
Encounters with Jesus: Unexpected Answers to Life's Biggest Questions (Dutton, 2013)
Romans 1-7 For You (The Good Book Company, February 2014)
Prayer: Experiencing Awe and Intimacy with God (Dutton, 2014)
Romans 8-16 For You (The Good Book Company, February 2015)
Preaching: Communicating Faith in an Age of Skepticism (Viking, June 2015)
The Songs of Jesus: A Year of Daily Devotionals in the Psalms (Viking, November 2015)
Making Sense of GOD: An Invitation to the Skeptical (Viking, 2016)
Hidden Christmas: The Surprising Truth Behind the Birth of Christ (Viking, Nov 2016)
God's Wisdom for Navigating Life: A Year of Daily Devotions in the Book of Proverbs (Viking, Nov 2017)
The Prodigal Prophet: Jonah and the Mystery of God's Mercy (Viking, Oct 2018)

Note

Collegamenti esterni

Pastori protestanti statunitensi
pastori presbiteriani

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The Thrizino-Buragaon, Bomdila, Bameng, Chayang Tajo, Seppa East Assembly constituencies in Arunachal Pradesh will go to poll in the first phase on April 11. Arunachal Pradesh will witness a tough fight between the incumbent NPP, the BJP and Congress Party will be witnessed. Arunachal Pradesh Assembly Elections 2019 Schedule: Voting to be Held on April 11, Result on May 23.


text:
Ursus arctos horribilis és una subespècie de l'os bru (Ursus arctos). A Nord-amèrica se'ls coneix com a grizzlies.

Descripció
 Els seus colors poden ser des del beix fins al marró groguenc, el marró fosc o el negre, sovint en funció de la ubicació geogràfica.
 Pot arribar a pesar entre 180 i 770 kg.
 Presenta dimorfisme sexual: els mascles poden ser 1,5-2 vegades la mida de la femella.
 Té una gepa distintiva entre les espatlles.

Reproducció
La femella madura sexualment al voltant dels 3-5 anys i, depenent del clima, es troba en zel tres setmanes durant el maig i el juliol. Encara que l'aparellament es produeixi en aquest temps, els òvuls fertilitzats no s'implantaran a l'úter fins a la tardor per a donar temps a les femelles a acumular una reserva de greix per a ella i els seus cadells de cara a la hibernació, ja que neixen entre el gener i el març. Les cries neixen cegues, sense pèl i pesant menys de 450 g. Deixen el cau a la primavera, quan assoleixen un pes d'uns 9 kg, i romandran amb sa mare durant, almenys, 2 anys.

Alimentació
És omnívor i es nodreix de matèria vegetal (sobretot baies, com ara Vaccinium cyanococcus, Rubus fruticosus, Rubus spectabilis, Vaccinium oxycoccus i Vaccinium parvifolium), carronya, insectes, rosegadors, peixos (salmons), bisons, caribús, ants i, fins i tot, ossos negres americans.

Distribució geogràfica
Es troba a Nord-amèrica.

Longevitat
La seva esperança de vida és de 15-34 anys en estat salvatge i 47 en captivitat.

Estat de conservació
Des de la dècada de 1800, la seva població s'ha reduït de més de 50.000 a 1.000-1.500 exemplars. Només a Alaska, el Canadà i el Parc Nacional de Yellowstone es considera que no es troba en perill.

Referències

Enllaços externs

 BioLib .
 Brown / Grizzly Bear Facts .
 Why are grizzly bears more aggressive than our black bears?  .
 How to Outrun a Grizzly (and other really bad ideas) .
 Khutzeymateen Grizzly Bear Sanctuary .
 Yellowstone Grizzly Bears - Ursus Arctos .

Grizzly


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New Delhi [India], Nov. 23 : The Indian contention, led by Rio Olympics silver medallist PV Sindhu, will today begin in the Hong Kong Open Super Series Badminton tournament.
Brimming with confidence after clinching her maiden Super Series Premier title at the China Open on Sunday, the ace Indian shuttler would be eying to come up with another good performance to qualify for next month's Dubai World Super Series Finals.
In the women's singles main draw, Sindhu will open her campaign against Indonesia's Susanto Yosephin, while Saina Nehwal, who recently recovered from a knee injury, will face Thailand's Porntip Buranaprasertsuk.
Fifth seeded Saina was last week stunned by the Thai in the first round of the China Open.


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text:
\section{Introduction\label{intro}}
One of the essential observations in the theory of nonperiodic, but
highly ordered structures is the fact that, in many cases, they can be
generated by a certain projection from a higher dimensional (periodic)
point lattice. This is true for the well-known Penrose tilings (cf.~
\cite{gs}) as well as for a lot of other substitution tilings, where
most known examples are living in $\ensuremath{\mathbb{E}}^1,\ensuremath{\mathbb{E}}^2$ or $\ensuremath{\mathbb{E}}^3$. The
appropriate framework for this arises from the theory of model sets
(\cite{mey}, \cite{moo}). Though the knowledge in this field has grown
considerably in the last decade, in many cases it is still hard to
prove rigorously that a given nonperiodic structure indeed is a model
set.

This paper is organized as follows. Section \ref{sec:subst} briefly
collects some well--known facts about substitution
tilings. These facts will be used later. Section
\ref{sec:modset} contains some basics about model sets, in particular
in connection with substitution tilings. In Section 1.3 we give the
definition of Rauzy fractals. This can be done completely in the
framework of model sets; it is just a question of the terminology.
In Section \ref{sec:ansatz} we introduce a family of one--dimensional
substitutions which are generalizations of the well--known Fibonacci
sequences. Four of these possess a substitution factor which is a
PV--number, so these are candidates for being model sets. Section
\ref{sec:n4} is dedicated to one of these substitutions. Therein, the
corresponding Rauzy fractal is obtained. Section \ref{sec:dual} shows
the construction of the dual substitution tiling, which is a
two--dimensional tiling with some interesting properties. Using the
duality, we will show that this dual tiling is a model set. Hence,
this is one of the few known cases of a two--dimensional model set,
generated by a substitution, where the substitution factor is
not an algebraic number of degree one or degree two (and the proof
being rigorous). Section \ref{sec:remarks} contains some additional
remarks, and further examples of dual substitutions, where the duality
principle can be used in order to prove that these are model sets, too.

Let us fix some notation. Throughout the paper, $\cl(M)$ denotes the
closure of a set $M$, $\inn(M)$ the interior of $M$, $\partial M$ the
boundary of $M$, and $\#M$ the cardinality of $M$. $\ensuremath{\mathbb{E}}^d$ denotes the
$d$--dimensional Euclidean space, i.e., $\ensuremath{\mathbb{R}}^d$ equipped with the
Euclidean metric $\| \cdot \|$. $\ensuremath{\mathbb{B}}^d$ denotes the unit ball $\{ {\bs
  x} \, | \, \|{\bs   x}\|\le 1 \}$. If not stated otherwise, $\mu$
denotes the $d$--dimensional Lebesgue measure, where $d$ will be clear
from the context.
A \emph{tile}  is a nonempty compact set $T \subset \ensuremath{\mathbb{E}}^d$ with the property
$\cl(\inn(T))=T$. A \emph{tiling} of $\ensuremath{\mathbb{E}}^d$ is a collection of tiles
$\{T^{}_n\}^{}_{n \ge 0}$ which covers $\ensuremath{\mathbb{E}}^d$ and contains no
{\em overlapping} tiles, i.e., $\inn(T^{}_k) \cap \inn(T^{}_n) =
\varnothing$ for $k \ne n$. In other words: A tiling is both a
covering and a packing of $\ensuremath{\mathbb{E}}^d$. A tiling $\ensuremath{{\mathcal T}}$ is called
\emph{nonperiodic}, if the only solution of  $\ensuremath{{\mathcal T}}+\bs{x}=\ensuremath{{\mathcal T}}$  is
$\bs{x}=\bs{0}$.

\subsection{Substitution Tilings} \label{sec:subst}
A convenient method to generate nonperiodic tilings is by a
{\em substitution}: Choose a set of {\em prototiles}, that is, a set of
tiles $\ensuremath{{\mathcal F}}:=\{T_1, T_2, \ldots, T_m\}$. Choose a
{\em substitution factor} $\lambda \in \ensuremath{\mathbb{R}}, \lambda > 1$ and a rule how to
dissect $\lambda T_i$ (for every $1 \le i \le m$) into tiles, such that
any of these tiles is congruent to some tile $T_j \in \ensuremath{{\mathcal F}}$. It will be
sufficient for this paper to think of a substitution like in
Fig.~\ref{inflbsp}. For completeness, we give the proper definition
here.

\begin{defi} \label{substdef}
Let $\ensuremath{{\mathcal F}}:=\{T_1, T_2, \ldots, T_m\}$ be a set of tiles, the
{\em  prototiles}, and $\lambda>1$ a real number, the
{\em substitution factor}.  Let $\lambda T_i =  \varphi_1(T_{i_1}) \cup
  \varphi_2(T_{i_2}) \cup \ldots  \cup \varphi_{n(i)}(T_{i_{n(i)}})$,
such that every $\varphi_j$ is an isometry of $\ensuremath{\mathbb{E}}^d$ and the
involved tiles do not overlap. Furthermore, let
$\sigma(\{T_i\})= \{ \varphi_1(T_{i_1}), \varphi_2(T_{i_2}), \ldots,
\varphi_{n(i)}(T_{i_{n(i)}}) \}$. Let ${\bs S}$ be the set of all
configurations of tiles congruent to tiles in $\ensuremath{{\mathcal F}}$. That is, ${\bs S}$ is
the set of all sets of the form  $\{T_{j_i} +{\bs x}_i \, | \, T_{j_i} \in
\ensuremath{{\mathcal F}}, i \in I, {\bs x}_i \in \ensuremath{\mathbb{E}}^d \}$, where $I$ is some
finite or infinite index set. By the requirement
$\sigma(\{T_i+{\bs x}\})=\sigma(\{T_i\})+\lambda {\bs x}$ (${\bs x}
\in \ensuremath{\mathbb{E}}^d$), $\sigma$ extends in a
unique way to a well--defined map from ${\bs S}$ to ${\bs S}$.
Then $\sigma: {\bs S} \to {\bs S}$ is called a {\em substitution}.
Every tiling of $\ensuremath{\mathbb{E}}^d$, where any finite part of it is congruent to a
subset of some $\sigma^k(T_i)$ is called a
{\em substitution tiling (with  substitution $\sigma$)}.
\end{defi}

More general, this definition makes sense if we replace $\lambda$
by some expanding linear map. That is, replace $\lambda$ in the
definition above by a matrix $Q$ such that all eigenvalues of $Q$ are
larger than 1 in modulus. In this full generality, many of the following
results may break down. Anyway, in Section \ref{sec:n4} we will use the
generalization in one certain case, where it will cause no problem.

To a substitution $\sigma$ we assign a {\em substitution matrix}:
$A_{\sigma}:=(a_{ij})_{1 \le i,j \le m}$, where $a_{ij}$ is the number
of tiles $T_j+{\bs x}$ (tiles of type $T_j \in \ensuremath{{\mathcal F}}$) in $\sigma(\{T_i\})$.
In this paper we will consider substitutions with {\em primitive}
substitution matrices only. Recall: A nonnegative matrix $A$ is called
{\em primitive}, if some power $A^k$ is a strictly positive matrix. In
this situation we can apply the following well--known theorem.

\begin{thm}[Perron--Frobenius] \label{perron}
Let $A \in \ensuremath{\mathbb{R}}^{m \times m}$ be a primitive nonnegative matrix.
Then $A$ has an eigenvalue $0< \lambda \in \ensuremath{\mathbb{R}}$, which is simple and
larger in modulus than every other eigenvalue of $A$. This
eigenvalue is called {\em Perron--Frobenius--eigenvalue} or shortly
{\em  PF--eigenvalue}.    \\
The corresponding eigenvector can be chosen such that it is strictly
positive. Such an eigenvector is called {\em PF--eigenvector} of $A$.
No other eigenvalue of $A$ has a strictly positive eigenvector.
\end{thm}

Using this theorem, it is a simple exercise to show the following
facts (cf.~\cite{fogg}, Thm.~1.2.7): We consider the
{\em relative frequency} of the tiles of type $T_i$ in a tiling $\ensuremath{{\mathcal T}}$,
i.e., the ratio of the number of tiles of type $T_i$ in $\ensuremath{{\mathcal T}}$ and
the number of all tiles in $\ensuremath{{\mathcal T}}$. Formally, this is  defined by
$\lim_{r\to \infty} \frac{\# \{ T_i + {\bs x} \, \in \ensuremath{{\mathcal T}} \, | \, {\bs x}
  \in r\ensuremath{\mathbb{B}}^d  \}}{\# \{ T+{\bs x} \, \in \ensuremath{{\mathcal T}} \, | \, {\bs x} \in r\ensuremath{\mathbb{B}}^d,
  \, T   \in \ensuremath{{\mathcal F}} \}}$. Since we
consider primitive substitution tilings only, it is true that this
limit exists. Moreover, this limit exists {\em uniformly}, i.e., for
any translate  $\ensuremath{{\mathcal T}}+{\bs x}$ we will obtain the same limit.
\begin{prop} \label{freqvol}
Let $\sigma$ be a primitive substitution in $\ensuremath{\mathbb{E}}^d$ with substitution
factor $\lambda$ and prototiles $T_1, \ldots, T_m$. Then the
PF--eigenvalue of a $A_{\sigma}$ is
$\lambda^d$. In particular, a substitution factor is always an algebraic
number. \\ The normalized (right) PF--eigenvector ${\bs v}=
(v_1,\ldots,v_m)^T$ of $A_{\sigma}$ contains the relative frequencies
of the tiles of  different types in the tiling in the following sense:
The entry $v_i$ is the relative frequency of $T_i$ in $\ensuremath{{\mathcal T}}$.\\
The left PF--eigenvector
(resp.~the PF--eigenvector of $A^T$) contains the $d$--dimensional
volumes of the different prototiles, up to scaling.
\end{prop}

Some people don't like the term 'left eigenvector'. Those may
replace 'left eigenvector of $A$' by 'eigenvector of $A^T$, written as
a row vector' wherever it occurs.
In the next section, we will associate with a tiling a related point
set: By replacing every prototile $T_i$ with a point in its
interior, a tiling $\ensuremath{{\mathcal T}}$ gives rise to a {\em Delone set} $V_{\ensuremath{{\mathcal T}}}$.
(Note that this defines $V_{\ensuremath{{\mathcal T}}}$ not uniquely. It depends on the
distinct choice of points in the prototiles. But the following is
true for any Delone set $V_{\ensuremath{{\mathcal T}}}$ constructed out of a
primitive substitution tiling $\ensuremath{{\mathcal T}}$ in the described way.)
A Delone set is a point set $V \subset \ensuremath{\mathbb{E}}^d$ which is
{\em uniformly discrete} (there exists $r>0$ such that $\# (V \cap
r\ensuremath{\mathbb{B}}^d + {\bs x}) \le 1$ for all ${\bs x} \in \ensuremath{\mathbb{E}}^d$)
and {\em relatively dense} (there exists $R>0$ such
that $\# (V \cap R\ensuremath{\mathbb{B}}^d + {\bs x}) \ge 1$ for all ${\bs x} \in \ensuremath{\mathbb{E}}^d$).
Since the frequencies of the prototiles tiles (cf. remark after Theorem
\ref{perron}) are well--defined for all tilings $\ensuremath{{\mathcal T}}$ considered in this
paper, it makes sense to define the density of the associated
Delone set $V_{\ensuremath{{\mathcal T}}}$: $\dens(V_{\ensuremath{{\mathcal T}}})=\lim_{r\to \infty} \frac{\# V_{\ensuremath{{\mathcal T}}}
  \cap r\ensuremath{\mathbb{B}}^d} {\mu( r \ensuremath{\mathbb{B}}^d)}$. It is not hard to show the following
identity.

\begin{prop} \label{densfreqvol}
Let $\ensuremath{{\mathcal T}}$ be a substitution tiling with substitution
$\sigma$, let ${\bs v}$ be the right normalized PF--eigenvector of
$A_{\sigma}$ (containing the relative frequencies of the prototiles),
let ${\bs w}^T$ be the left PF--eigenvector of $A_{\sigma}$, such that
${\bs w}^T$ contains the $d$--dimensional volumes of the prototiles, and
$V_{\ensuremath{{\mathcal T}}}$ a Delone set constructed out of $\ensuremath{{\mathcal T}}$ by replacing every
prototile with a point in its interior. Then
\[ \dens(V_{\ensuremath{{\mathcal T}}}) = \left( {\bs w}^T \cdot{\bs v} \right)^{-1} \]
\end{prop}

{\em Example:} The golden triangle substitution. The substitution
rule, together with substitution factor, substitution matrix etc. is
shown below.

\begin{figure}[h]
\epsfig{file=inflbsp.eps} \hspace{10mm}
\begin{minipage}[b]{30mm}
$A_{\sigma}:=\begin{pmatrix} 1 & 1 \\ 1 &  0 \end{pmatrix}$
\vspace*{5mm}
\end{minipage}
\caption{'Golden triangles'. A substitution rule in $\ensuremath{\mathbb{E}}^2$ with
  two prototiles $T_1,T_2$ and substitution factor $\lambda=\sqrt{\tau}$,
  where $\tau=\frac{\sqrt{5}+1}{2}$ is the golden mean.
 \label{inflbsp}}
\end{figure}
This substitution uses two prototiles $T_1,T_2$.
The PF--eigenvalue of the substitution matrix $A_{\sigma}$
is  $\tau=\lambda^2$. The normalized PF--eigenvector is
${\bs v}:=(\tau^{-1}, \tau^{-2})^T$. Thus, the relative
frequency of the tiles of type $T_1$ (resp.~$T_2$) in the tiling is
$\tau^{-1}$ (resp.~$\tau^{-2}$). $A_{\sigma}$ is symmetric, thus
${\bs v}^T$ is a left PF--eigenvector. The areas (2--dim volumes) of the
two prototiles are $\mu(T_1)=\sqrt{\tau}^3, \mu(T_2)=\sqrt{\tau}$. If
we replace every triangle in a corresponding substitution tiling by
a point in its interior, then by Prop.~\ref{densfreqvol} the density
of the point set is $\left( (\sqrt{\tau}^3,\sqrt{\tau})\cdot(
  \tau^{-1}, \tau^{-2})^T \right)^{-1} =
\frac{\tau^{3/2}}{(\tau^2+1)}=0.56886448\ldots$

\subsection{Model Sets} \label{sec:modset}
Model sets are a special kind of Delone sets. They show a high degree
of local and global order. For example, a model set $V$ and its
difference set $V-V := \{ {\bs x} - {\bs y} \, | \, x,y \in V \}$
do not differ too much in the following sense (cf.~\cite{mey}): There
is a finite set $F$ such that
\[ V-V \subseteq V+F. \]
Lattices do have this property with $F=\{0\}$, and indeed lattices are
special cases of model sets.  Unfortunately, there is no reference
known to the author which gives a comprehensive overview about model
sets. A good reference is \cite{moo}. A comprehensive
overview for the case of one--dimensional tilings is contained in
\cite{fogg}, mainly chapter 7 and 8. Therein, many connections to
combinatorics, ergodic theory and number theory are compiled (but the
term 'model set' cannot be found in this book, the term 'geometric
representation' is used instead).

A model set is defined by a collection of spaces and maps (a
so--called cut--and--project scheme) as follows.
In (\ref{cps}), let $\Lambda$ be a lattice of full rank in
$\ensuremath{\mathbb{E}}^{d+e}$, $\pi_1,\pi_2$ projections such that
$\pi_1 |_{\Lambda}$ is injective, and $\pi_2(\Lambda)$ is dense in
$\ensuremath{\mathbb{E}}^e$. Let $W$ be a nonempty compact set --- the so--called
{\em window  set} --- with the properties $\cl(\inn(W))=W$
and $\mu(\partial W)=0$, where $\mu$ is the Lebesgue measure in $\ensuremath{\mathbb{E}}^e$.
\begin{equation} \label{cps}
\begin{array}{ccc}
\ensuremath{\mathbb{E}}^d & \stackrel{\pi_1}{\longleftarrow} \ensuremath{\mathbb{E}}^{d+e}
  \stackrel{\pi_2}{\longrightarrow} & \ensuremath{\mathbb{E}}^e \\
\cup & \cup & \cup \\
V & \Lambda & W
\end{array}
\end{equation}
Then the set $V= \{ \pi_1({\bs x}) \, | \, {\bs x} \in \Lambda,
\pi_2({\bs x}) \in W \}$
is called a {\em model set}\footnote{To be precise, this defines a
  {\em regular} model set. Without the property $\mu(\partial W)=0$,
  $V$ is still a model set, but not a regular one.}.
Obviously, $V$ is a set of points in
$\ensuremath{\mathbb{E}}^d$. A tiling can be obtained by applying certain rules to $V$,
e.g., considering the Voronoi cells of $V$. In the case $d=1$ it is
quite simple to get a tiling: If $V=\{ {\bs x}_i \, | \, i \in \ensuremath{\mathbb{Z}} \}$,
such that ${\bs x}_i<{\bs x}_{i+1}$, then $\ensuremath{{\mathcal T}}= \{ [{\bs x}_i, {\bs
  x}_{i+1} ] \, | \, i \in \ensuremath{\mathbb{Z}} \}$ is a tiling of $\ensuremath{\mathbb{E}}^1$.

Since $V \subset \pi_1(\Lambda)$ and $\pi_1|_{\Lambda}$ is injective,
every ${\bs x} \in V$ corresponds to exactly one point $\pi_1^{-1}({\bs x}) \in
\Lambda$, and thus exactly to one point $\pi_2(\pi_1^{-1}({\bs x})) \in W$.
Therefore, we are able to define the {\em star map}:
\begin{equation} \label{starmap}
  \ast : \pi_1(\Lambda) \to \ensuremath{\mathbb{E}}^e, \qquad {\bs x}^{\ast}  =
\pi_2(\pi_1^{-1}({\bs x}))
\end{equation}
This allows for a convenient notation. E.g., we have
\begin{equation} \label{clvstern}
\cl(V^{\ast})=W,
\end{equation}
where $V^{\ast} = \{ {\bs x}^{\ast} \, | \, {\bs x} \in V \}$.

{\em Remark:} In general, $\ensuremath{\mathbb{E}}^d$ and $\ensuremath{\mathbb{E}}^e$ in (\ref{cps}) can be
replaced by some locally compact abelian groups $G$ and $H$. Then
$\mu$ denotes the Haar measure of $H$. This is necessary to show that,
for example, the chair tiling is a model set (cf~\cite{bms}).
If the substitution matrix $A_{\sigma}$ of a given substitution tiling
is {\em unimodular} (i.e., $\det(A_{\sigma} = \pm 1$), then it
suffices to consider Euclidean spaces only. If not, then one needs to
consider more general groups (cf.~\cite{sing}). In this paper, all
occurring substitution matrices are unimodular, therefore we will
consider Euclidean spaces only.

Given a substitution tiling, one may ask if this tiling can be
obtained as a model set $V$. By the following theorem, this is only
possible if the substitution factor is a PV number or a Salem number.

\begin{defi}
Let $\lambda \in \ensuremath{\mathbb{R}}$ be an algebraic integer. $\lambda$ is called a
{\em  Pisot Vijayaraghavan number} or shortly {\em PV number},
if $|\lambda|>1$, and for all its algebraic conjugates $\lambda_i$
holds: $|\lambda_i|<1$. \\
If for all conjugates $\lambda_i$ of $\lambda$ holds
$|\lambda_i|\le 1$, with equality for at least one $i$,  then
$\lambda$ is called a {\em Salem number}.
\end{defi}

\begin{thm}[Meyer] \label{pvsatz}
If $V$ is a model set, $1 < \lambda \in \ensuremath{\mathbb{R}}$ and if $\lambda V \subseteq V$,
then $\lambda$ is a PV number or a Salem number.
\end{thm}

It is known that every substitution $\sigma$ gives rise to at least
one tiling $\ensuremath{{\mathcal T}}$ with the property $\sigma(\ensuremath{{\mathcal T}})=\ensuremath{{\mathcal T}}$. So,
if $V$ is a point set derived from $\ensuremath{{\mathcal T}}$ (where the substitution factor
is $\lambda$), say, $V$ being the vertex set of this tiling, then by the
definition of a substitution follows $\lambda V \subseteq V$.
Now, if $V$ is also a model set, by the theorem above the substitution
factor must be a PV number or a Salem number.

The remaining things to check in order to show that $V$ is a model set
are the following (cf.~(\ref{cps})):
\begin{itemize}
\item Determine the lattice $\Lambda$,
\item determine the window set $W$,
\item show $\mu(W)>0$,
\item show $\mu(\partial W)=0$,
\item show $\dens(V) = \dens(\{ \pi_1({\bs x}) \, |
  \, {\bs x} \in \Lambda, \pi_2({\bs x}) \in W \})$.
\end{itemize}
There are standard constructions for obtaining
$\Lambda$ and $W$. But the latter three items are hard to decide
rigorously in general, or even in special cases. For an example how
hard it may be to give a proper proof, cf.~\cite{bs}. In the next section,
we will use a duality principle to prove that the three latter items
are fulfilled for the considered tiling, thus proving that it arises
from a model set (i.e., the associated point set is a model set). But
first, we will discuss the above list in more detail.

Starting with a substitution tiling, or better a substitution point
set $V$ in $\ensuremath{\mathbb{E}}^1$ (cf.~the remark after (\ref{cps})), the standard
construction for the lattice is as follows: Let $\lambda$ be the
substitution factor (where $\lambda$ from now on is always assumed to be
a PV number) and $\lambda_2,\ldots,\lambda_m$ its algebraic conjugates.
Then \[ \Lambda= \langle (1,1,\cdots,1)^T,
(\lambda,\lambda_2,\cdots,\lambda_m)^T,
(\lambda^2,{\lambda_2}^2,\ldots,{\lambda_m}^2)^T, \ldots,
(\lambda^{m-1},{\lambda_2}^{m-1},\ldots,{\lambda_m}^{m-1})^T \rangle_{\ensuremath{\mathbb{Z}}}. \]
Now, $\pi_1$ (resp.~$\pi_2$) can be chosen as the canonical projections
from $\ensuremath{\mathbb{E}}^m \to \ensuremath{\mathbb{E}}^1$ (resp.~$\ensuremath{\mathbb{E}}^m \to \ensuremath{\mathbb{E}}^{m-1}$). To be precise, if
${\bs  x}=(x_1,x_2,\ldots,x_m)^T$, then $\pi_1({\bs x})=x_1,
\pi_2({\bs   x})=(x_2,\ldots, x_m)^T$.
At this point, we are already able to compute the window set
$W$ numerically: Since the left PF--eigenvector
of the substitution matrix contains the the lengths of the
prototiles (cf.~remark after Thm.~\ref{perron} above), the
lengths can be chosen such that $V \subset \ensuremath{\mathbb{Z}}[\lambda]$. That is, every
${\bs x} \in V$ can be expressed as ${\bs x}=\alpha_0+\alpha_1 \lambda+
\cdots+ \alpha_{m-1} \lambda^{m-1}$, where $\alpha_i \in \ensuremath{\mathbb{Z}}$.
Then
\[ {\bs x}^{\ast} = (\alpha_0+\alpha_1 \lambda_2+ \cdots+ \alpha_{m-1}
{\lambda_2}^{m-1}, \cdots, \alpha_0+\alpha_1 \lambda_{m-1}+ \cdots+
\alpha_{m-1} {\lambda_{m-1}}^{m-1})^T. \]
More general, we may apply the star map to $V$, obtaining the
(countable) set $V^{\ast}$, which in turn yields the window set
$W=\cl(V^{\ast})$. Actually, some of the pictures in the next section
were created in essentially in this way: Using the substitution to
generate a large finite subset $F \subset V$, where the points are
expressed as ${\bs x}$ above, we plot all points of $F^{\ast}$.
If this set $F^{\ast}$ appears to be bounded, even if the number of
points is increased, then this is an indication that the considered
set $V$ is a model set.

To prove rigorously that $V$ is actually a model set, one proceeds by
showing $\mu(W)>0$ and $\mu(\partial W)=0$. If $W$ is a polyhedron,
then this is an easy task. But in the generic case $W$ is of fractal
nature, and it may be even hard to prove that $W$ is of positive
measure.

The last thing to prove is $V=\{ \pi_1({\bs x}) \, | \, {\bs x}
\in \Lambda, \pi_2({\bs x}) \in W \}$, and not just a too small
subset. More precisely: Let $U:=\{ \pi_1({\bs x}) \, | \, {\bs x}
\in\Lambda,\pi_2({\bs x})\in W\}$. Then it must hold
$\dens(V)=\dens(U)$. In this context the following
theorem turns out to be helpful \cite{sch}.

\begin{thm} \label{densvol}
Let $V$ be a model set given by (\ref{cps}). Then
\[ \dens(V) = \mu(W) / \det \Lambda, \]
where $\det \Lambda$ denotes the determinant of the generator matrix of
$\Lambda$. Equivalently, $\det \Lambda$ is the volume of any measurable
fundamental domain of $\Lambda$.
\end{thm}

\subsection{Rauzy Fractals} \label{sec:rauzy}
As mentioned above, 'Rauzy fractal'
is essentially just another term for the window set $W$. The
term 'Rauzy fractal' occurs in the context of dynamical systems and
number theory, while the terms 'window' or 'window set' arose from the
theory of quasicrystals. Originally, 'Rauzy fractal' denoted one special
fractal of this kind, namely the one arising from the substitution
\[ \sigma(1)=1\;2, \quad  \sigma(2) =   1\; 3, \quad \sigma(3)= 1, \]
which is analyzed in \cite{r}. Therefore, the term 'generalized Rauzy
fractal' is sometimes used to
emphasize the fact that one deals with the more general case.
(Of course, this is just a symbolic substitution, but it is trivial to
formulate the geometric substitution according to Def. \ref{substdef},
cf.~comment after (\ref{substn})).

According to \cite{fogg}, the definition of a Rauzy fractal is
essentially as follows: if $\sigma$ is a substitution for $m$
prototiles in $\ensuremath{\mathbb{E}}^1$, where the factor is a PV number of algebraic
degree $m$, then the Rauzy fractal is just the window set $W$ as in
(\ref{clvstern}), where one does not care about $\mu(W)>0$ or
$\mu(\partial W)=0$ or Theorem \ref{densvol}. Note that therefore it
is much more difficult to show that a given substitution point set is
a model set than it is to compute the corresponding Rauzy fractal.

Before we proceed, let us mention a theorem which is important in this
context and which is necessary in the next section.

\begin{thm}[\cite{hut}] \label{hutch}
Let $(X,d)$ be a complete metric space and $\{f_1,\ldots,f_n\}$ a
finite set of contractive maps (i.e., $\exists c<1: \forall x,y \in X:
d(f(x),f(y)) \le c\, d(x,y)$). Then there is a unique compact set $K
\subset X$ such that $K = \bigcup_{i=1}^n  f_i(K)$.
\end{thm}

This theorem is a consequence of the Banach Fixed Point Theorem, and is
easily generalized to partitions of $K$: Let $\{f_1,\ldots,f_n\}$ be
as above and $I_j \subseteq \{1,\ldots,n\}$ for $1 \le j \le \ell$,
then there is a unique tuple $(K_1,\ldots,K_{\ell})$ of compact sets
$K_j$, such that $K_j = \bigcup_{i \in I_j}  f_i(K_{j_i})$ for all
$1 \le j \le \ell$ (\cite{kli}). In the next section, this theorem will
help to show the compactness of the window set $W$.

\section{Two Rauzy Fractals and the Duality Principle} \label{sec:zwo}
We will describe the duality principle by choosing a certain
one--dimensional substitution, constructing the corresponding
Rauzy fractal and the associated dual tiling, and then we will
use the duality principle to show that the dual tiling is a model set.

\subsection{Some Rauzy Fractals} \label{sec:ansatz}
It is easy to see, that for any nonnegative integer $n\times n$--matrix
$A$ there is a substitution tiling in $\ensuremath{\mathbb{E}}^1$ with substitution matrix
$A$. Actually, from any such matrix one can immediately deduce a
substitution, since there are only few geometric restrictions for
tilings in one dimension. (The same problem for higher
dimensions becomes quite difficult!) Here we consider the following
matrices for $n \ge 2$:
\[ M_n:=(m_{ij})_{1 \le i,j \le n}, \quad \mbox{where} \; m_{ij} = \left\{
    \begin{array}{lcl} 1 & : & i+j >n \\ 0 & : & i+j \le n
\end{array} \right. \]
Obviously all these matrices are primitive.
A substitution with substitution matrix $M_n$ for the prototiles
$1,2,\ldots,n$ is given by
\begin{equation} \label{substn}
\begin{array}{ccr} 1 & \to & n \\ 2 & \to & n\!-\!1 \; n \\ & \vdots &
\vdots \\ n & \to & 1 \; 2 \cdots n\!-\!1 \; n \end{array}
\end{equation}
Note that this is just a symbolic substitution. But in one dimension
the geometric realization is obtained from this scheme in a unique way:
The prototiles are intervals of different lengths, and the lengths are
given by the PF--eigenvector of $M_n$. There is a nice trick to deduce the
substitution factor directly from this substitutions: For fixed
$n \in \ensuremath{\mathbb{Z}}, n \ge 2$, let $s_k:=\sin(\frac{k \pi}{2n+1})$.
Because of the wonderful formula (see \cite{nida})
\begin{equation} \label{wformel}
 \frac{s_k}{s_1} s_i =
\sum_{\nu=0}^{k-1}s_{i+1-k+2\nu}
\end{equation}
it follows $\frac{s_n}{s_1} s_1 = s_n,  \frac{s_n}{s_1} s_2 = s_{n-1}+s_n,
\ldots, \frac{s_n}{s_1} s_n = s_1 + s_2 + \cdots + s_n$. Therefore the
substitution factor (which is the PF--eigenvalue of $M_n$) is
$\frac{s_n}{s_1}$, and the length of tile $i$ is $s_i$.
Now, we determine which of these substitutions
define a Rauzy--fractal. The first condition to be fulfilled is that
the substitution factor must be a PV--number\footnote{Or a Salem
  number, but these don't play a role here, so we don't mention them
  explicitly. }, cf.~Theorem
\ref{pvsatz}. This is the case for $n=2,3,4$ and $7$, and for no
other $n \le 30$. (This has been checked by a Maple program, and we
don't expect any more PV--numbers as PF--eigenvalues of $M_n$ for
larger values of $n$.) The corresponding characteristic polynomials of
$M_n$ are:
\[ x^2-x-1, \; x^3-2x^2-x+1, \; x^4-2x^3-3x^2+x+1, \;
x^7-4x^6-6x^5+10x^4+5x^3-6x^2-x+1. \]
The latter two are reducible over \ensuremath{\mathbb{Z}}. The irreducible polynomials
defining $\frac{s_2}{s_1}\, (n=2),\, \frac{s_3}{s_1}\, (n=3), \,
\frac{s_4}{s_1} \, (n=4)$ and $\frac{s_7}{s_1} \, (n=7)$ are
\begin{equation}\label{charpoly}
 x^2-x-1, \; x^3-2x^2-x+1, \; x^3-3x^2+1, \; x^4-4x^3-4x^2+x+1.
\end{equation}
Therefore, the algebraic degree of these numbers is 2,3,3,4, resp.
Here we will focus mainly on the third case ($n=4$). The first case
($n=2$) leads to the well known Fibonacci sequences (see Section
\ref{sec:remarks}, or \cite{fogg}, Sections 2.6 and 5.4).
The fourth case
($n=7$) will lead to a  three-dimensional Rauzy fractal in the end.
The third case ($n=4$) will show the methods and the typical problems
which may occur, and therefore we discuss this case in the following
in detail. This part may be regarded as a recipe how to compute other
Rauzy fractals explicitly.

\subsection{The case ${\bs n=4}$} \label{sec:n4}
In this case we have got four prototiles. Since we consider one--dimensional
tilings here, the tiles can be chosen to be intervals. We denote the
four prototiles by $L$ ('large'), $M$ ('medium'), $S$ ('small') and
$X$ ('extra small'). Their lengths are already known: They are given
by a left PV--eigenvector of the substitution matrix, or by
(\ref{wformel}), so the values can be chosen as
$\frac{s_4}{s_1}, \frac{s_3}{s_1}, \frac{s_2}{s_1},
\frac{s_1}{s_1}=1$, resp.~  (or any positive multiples of these
values).  In order to construct the Rauzy fractal, we need this values
as expressions in $\lambda=\frac{s_4}{s_1}$ (cf. Section \ref{sec:modset}).
W.l.o.g, we choose $X:=[0,1]$. The substitution now reads
\begin{equation}\label{subst}
 X \to L, \; S \to ML, \; M \to SML, \; L \to XSML.
\end{equation}
Since $\lambda X = L$ it follows $L=[0,\lambda]$. Let $s,m,\ell$ denote
the lengths of $S,M,L$, resp. Since $\ell=\lambda$, we conclude from
(\ref{subst}) the equations
\[ \begin{array}{rcrcl}
\lambda s & = &  m+ \lambda & & \\
\lambda m & = & s+m+ \lambda  & = & s(\lambda+1) \\
\lambda^2 & = & 1+s+m+\lambda & = & s(\lambda+1) +1 = \lambda m +1
\end{array} \]
and therefore $m=\frac{\lambda^2-1}{\lambda}$. From (\ref{charpoly})
follows $\lambda^3=3\lambda^2-1$, hence $\lambda^{-1}=3 \lambda -
\lambda^2$, and we obtain
\[ m = (\lambda^2-1)\cdot(3\lambda-\lambda^2) = \lambda^2 - 2 \lambda. \]
Consequently,
\[ s = \lambda^2-\lambda-m-1 = \lambda - 1. \]
To apply the standard construction of Rauzy fractals, the number of
prototiles must equal the algebraic degree of the substitution
factor. So we need to get rid of one letter. By (\ref{subst}), a
tile $X$ in the tiling is always followed by a tile $S$. Since $x+s=\lambda$,
we can merge every pair $XS$ into one tile $L$. It is easy to check
that the new substitution
\begin{equation} \label{subst2}
S \to ML, \; M \to SML, \; L \to LML
\end{equation}
gives rise to the same tilings as (\ref{subst}) does, up to merging
pairs $XS$ into $L$. We identify these tilings with a point set
$V$, where $V$ contains all (right) endpoints of the tiles. If $0 \in
V$, then $V \subset \ensuremath{\mathbb{Z}}[\lambda]$, and every point $x \in V$ can be
expressed by $x=a + b \lambda + c \lambda^2$ ($a,b,c \in \ensuremath{\mathbb{Z}}$).
The lattice $\Lambda \in \ensuremath{\mathbb{E}}^3$ now is given by
\begin{equation} \label{mlambda}
\Lambda = \langle (1,1,1)^T, (\lambda, \lambda_2, \lambda_3)^T, (\lambda^2,
\lambda_2^2, \lambda_3^2)^T \rangle_{\ensuremath{\mathbb{Z}}},
\end{equation}
where $\lambda_2, \lambda_3$ are the algebraic conjugates of $\lambda$.
From $\lambda \lambda_2 \lambda_3 = -1$ and the fact that the values $s_k$
occur frequently above, one easily deduces
$\lambda_2=-\frac{s_1}{s_2}, \lambda_3=\frac{s_2}{s_4}$.
This allows us to write down the star map now (cf.~(\ref{starmap})).
Let $0 \in V$. If $x \in V$, then $x= a + b \lambda + c \lambda^2$ for some
$a,b,c \in \ensuremath{\mathbb{Z}}$, and
\begin{equation} \label{ast}
x^{\ast} = a \begin{pmatrix} 1 \\ 1 \end{pmatrix}
+ b \begin{pmatrix} \lambda_2 \\ \lambda_3 \end{pmatrix}
+ c \begin{pmatrix} {\lambda_2}^2 \\ {\lambda_3}^2 \end{pmatrix}.
\end{equation}
By (\ref{clvstern}), the window set $W$ is just the closure of $V^{\ast}$.
A short computation (using $\lambda_i^3= 3 \lambda_i^2-1$ for $i=2,3$)
shows
\[ (\lambda x)^{\ast} = \begin{pmatrix} \lambda_2 & 0 \\ 0 & \lambda_3
\end{pmatrix} x^{\ast}  =: Qx^{\ast}. \]

In order to show the compactness of $W$, and possibly to draw further
conclusions about $W$ (which is the Rauzy fractal wanted), we consider
a substitution set $V$ such that
$\sigma(V)=V$. (As mentioned above, the considered substitution $\sigma$
can be easily formulated for the point set $V$ as well as for the
tiling $\ensuremath{{\mathcal T}}$.) For example, let
\begin{equation} \label{sv=v}
V = \cdots SMLLM\underline{L}LMLSMLLMLMLSMLLMLLMLSMLL \cdots
\end{equation}
where the right endpoint of the underlined $L$ is located at zero.
(Note, that the condition $\sigma(V)=V$ together with (\ref{sv=v})
defines $V$ uniquely.) Denote by $V_L$ ($V_M,V_S$, resp.) the set of
all points in $V$ which are right endpoints of intervals $L$ ($M,S$,
resp.). Then obviously
\[ V=V_L \cup V_M \cup V_S. \]
Moreover, from (\ref{subst2}) and $\sigma(V)=V$ follows
\begin{equation} \label{vgleich}
\begin{array}{lcl}
V_L & = & \lambda V \quad \cup \quad \lambda V_L -\lambda^2+\lambda, \\
V_M & = & \lambda V -\lambda, \\
V_S & = & \lambda V_M - \lambda^2+\lambda.
\end{array}
\end{equation}
This can be seen as follows: One tile of type $M$ and one tile
of type $L$ occur in {\em every} substituted tile $T$, not depending
on the type of $T$. If $T=[a,b]$, then $\sigma(T)$ contains tiles
$[\lambda b - \lambda - m, \lambda b - \lambda]$ (of type $M$) and
$[\lambda b -\lambda, \lambda b ]$ (of type $L$). Therefore, from
$b \in V$ follows $\lambda b - \lambda \in V_M$,
yielding the second equation, and $\lambda b \in V_L$, yielding the first
part of the first equation. The other parts of (\ref{vgleich}) are
obtained analogously.

Applying the star map to (\ref{vgleich}), we obtain the following
equations for $V^{\ast}$:
\begin{equation} \label{vstern}
\begin{array}{lcl}
V^{\ast} & = & V_L^{\ast} \quad \cup \quad V_M^{\ast} \quad \cup
\quad V_S^{\ast} \\
& & \\
V_L^{\ast} & = & Q V^{\ast} \quad \cup \quad Q V_L^{\ast}
-  ({\lambda_2}^2,  {\lambda_3}^2)^T + ( \lambda_2 , \lambda_3 )^T \\
V_M^{\ast} & = & Q V^{\ast} - ( \lambda_2 , \lambda_3)^T  \\
V_S^{\ast} & = & Q V_M^{\ast} - ( {\lambda_2}^2 , {\lambda_3}^2 )^T +
(\lambda_2 ,   \lambda_3 )^T
\end{array}
\end{equation}
Obviously, $Q$ defines a contracting linear map, thus all maps
appearing in (\ref{vstern}) are contracting. Therefore, Theorem
\ref{hutch} applies: There is one unique compact $W=W_L \cup W_M \cup
W_S$ which fulfills (\ref{vstern}). This assures the existence of this
set, so we obtained a Rauzy fractal $W$.

\begin{figure}[h]
\epsfig{file=win-kp9.eps}
\epsfig{file=win-sml.eps} \hspace{5mm}
\epsfig{file=win-kp7.eps}
\caption{Left: The Rauzy fractal $W$ arising from the substitution
  (\ref{subst2}). Middle: the same set, partitioned according to the
  distinction of tiles $S$, $M$ and $L$. Right: The Rauzy
  fractal arising from the substitution (\ref{substn}) for $n=3$:  $S
  \to L, \; M \to ML, \, L  \to SML$. \label{wkp9} }
\end{figure}

To visualize $W$ and its fractal
nature, $W$ can be approximated numerically. In fact, we just produce a
finite subset of $V$ and apply the star map to it. The resulting
points are plotted into a diagram, which is shown in Fig.~\ref{wkp9}
(left). If we distinguish the points in the diagram according to their
type ($S,M$ or $L$), the result is shown in Fig.~\ref{wkp9} (middle).
The set $W$ is the union of the three sets $W_L, W_M, W_S$, where
$W_i=\cl(V_i^{\ast})$. From (\ref{vstern}) follows that $W_S$ is
an affine image of $W_M$, but not similar to $W_M$, and that $W_M$ is
an affine image of $W$.

In the same way one can compute the Rauzy fractal for the case $n=3$
in (\ref{substn}). The resulting pattern is shown in
Fig.~\ref{wkp9}, right. Note, that we did not claim that one of the
original substitution sets $V$ is a model set until now. In order to
prove this, one needs to prove $\mu(W)>0$, $\mu(\partial W)=0$ and
$\dens(V)=\dens(\{ \pi_1({\bs x}) \, | \, {\bs x} \in  \Lambda,
\pi_2({\bs x}) \in W \})$.

\subsection{The dual tiling} \label{sec:dual}
Recall that $Q$ in (\ref{vstern}) is a contraction. Therefore $Q':=Q^{-1}$
is an expansion. Applying $Q'$ to (\ref{vstern}) and replacing
$V_i^{\ast}$ with its closure $W_i$ ($i \in \{S,M,L\}$)
yields the following equation system:
\begin{equation} \label{dualexp}
\begin{array}{lcl}
Q' W_L & = & W_L \cup W_M \cup W_S \cup W_L - (\lambda_2,
\lambda_3)^T + (1,1)^T  \\
Q' W_M & = & W_L - (1,1)^T  \cup W_M - (1,1)^T
\cup W_S - (1,1)^T  \\
Q' W_S & = & W_M - (\lambda_2, \lambda_3)^T + (1,1)^T
\end{array}
\end{equation}
This defines the substitution $\sigma'$ in $\ensuremath{\mathbb{E}}^2$ with three prototiles
$W_L,W_M,W_S$:
\begin{equation} \label{dualsubst}
\begin{array}{lcl}
\sigma' (\{W_L\}) & = & \{ W_L , W_M , W_S , W_L - (\lambda_2,
\lambda_3)^T + (1,1)^T \} \\
\sigma' (\{W_M\}) & = & \{ W_L - (1,1)^T  , W_M - (1,1)^T
, W_S - (1,1)^T \} \\
\sigma' (\{W_S\}) & = & \{ W_M - (\lambda_2, \lambda_3)^T + (1,1)^T \}
\end{array}
\end{equation}
In contrary to Def.~\ref{substdef}, the expansion by $Q'$ is not a
similarity, but an affinity. Anyway, these substitution rules will
yield a tiling of $\ensuremath{\mathbb{E}}^2$. A part of it is shown in
Fig.~\ref{fractilg}. (To be precise, we need to show that there
will be no overlaps and no gaps, but this is not essential for the
following, since we will consider the corresponding substitution
point set, or a modified version where all tiles are polygons.)

\begin{figure}[h]
\epsfig{file=fractilg.eps} \epsfig{file=polytilg-3.eps}
\caption{ \label{fractilg}
Left: A patch of the tiling arising from (\ref{dualsubst}). Tiles of
type $W_S$ are black, the other tiles come in two slightly different
shadings, such that the boundary between tiles of the same type
remains visible. Right: A polygonalized version. Tiles and colours
correspond to the tiling on the left.}
\end{figure}

Now, we will prove that the dual tiling is a model set.
In analogy to the construction of $V$ out of the original substitution
tiling arising from (\ref{subst2}), we choose in every prototile a
{\em control  point}. A good choice is (cf.~(\ref{vstern})):
$c_L:=0 \in W_L$, $c_M:= Qc_L- (\lambda_2,\lambda_3)^T =
(\lambda_2,\lambda_3)^T \in W_M$, $c_S:= Qc_M
-({\lambda_2}^2,{\lambda_3}^2)^T + (\lambda_2,\lambda_3)^T
\in W_S$. It is not too hard to check that the set $\{W_L, W_M
-({\lambda_2}^2,  {\lambda_3}^2)^T + 2 (\lambda_2,\lambda_3)^T + (1,1)^T\}$
gives rise to a point set $V'$ such that $\sigma'(V')=V'$, and $V'$ is
a Delone set. To be precise, let $V_0':=\{c_L, c_M-({\lambda_2}^2,
{\lambda_3}^2)^T + 2 (\lambda_2,\lambda_3)^T + (1,1)^T\}$, and
$\sigma'$ be defined for points as well as for tiles in the canonical way.
Then $V_0' \subset \sigma'(V_0')$, and inductively $(\sigma')^k (V_0')
\subset  (\sigma')^{k+1} (V_0')$. Therefore,
\begin{equation} \label{vstrich}
V' = \bigcup_{k \ge 1} (\sigma')^k (V_0')
\end{equation}
is well defined, and it holds $\sigma'(V')=V'$.

{\em Remark:} As a first idea, one would try to start with the set
$\{ c_L \}$ instead of $V_0'$. In this case, the property
$(\sigma')^k(\{ c_L \}) \subset (\sigma')^{k+1}(\{ c_L \})$ is
fulfilled, but the resulting point set would be no Delone set, since
it possesses arbitrary large holes.

By the construction, it holds $V \subset \langle (1,1)^T, (\lambda_2,
\lambda_3)^T , ({\lambda_2}^2,{\lambda_3}^2)^T \rangle_{\ensuremath{\mathbb{Z}}} $. In
particular, this shows a
curious property of $V'$: If one coordinate of some ${\bs z}=(x,y)^T \in
V'$ is known, say, $x=a + b \lambda_2 + c {\lambda_2}^2$ ($a,b,c \in \ensuremath{\mathbb{Z}})$,
then the other coordinate $y$ is known, too: $y = a  + b \lambda_3 + c
{\lambda_3}^2$. Another unusual property is that each prototile occurs
in just one orientation. This is not true for most known examples of
nonperiodic substitution tilings in dimension larger than one.

In a similar way as above
(cf.~(\ref{vgleich}))  we obtain from (\ref{dualsubst}) the following
equations, where $V'=V'_S \cup V'_L \cup V'_L$ and $V'_i$ is the set
of all points of type $c_i$ in $V'$ ($i \in \{S,M,L\}$).
\begin{equation} \label{dualgleich}
\begin{array}{lcl}
V'_L & = & Q' V'_L \cup Q' V'_M \cup Q' V'_L - (\lambda_2,
\lambda_3)^T + (1,1)^T  \\
V'_M & = & Q' V'_L - (\lambda_2,\lambda_3)^T  \cup Q' V'_M - (\lambda_2,
\lambda_3)^T \cup Q' V'_S   \\
V'_S & = & Q' V'_L + (\lambda_2, \lambda_3)^T - 2 ({\lambda_2}^2,
{\lambda_3}^2)^T \cup  Q' V'_M + (\lambda_2, \lambda_3)^T - 2
({\lambda_2}^2, {\lambda_3}^2)^T
\end{array}
\end{equation}
In order to show that $V'$ is a model set, we have to determine the
lattice $\Lambda'$ and the window set $W'$ for $V'$. By the
construction of $V'$ follows $\Lambda'=\Lambda$. Since $0 \in V'$, we
know that every ${\bs x} \in V'$ is of the form
$a(1,1)^T+b(\lambda_2,\lambda_3)^T+c({\lambda_2}^2,{\lambda_3}^2)^T$.
Thus our new star map is just the inverse of the original one in
(\ref{ast}), namely
\[ \star: \pi_2(\Lambda) \to \ensuremath{\mathbb{E}}^1, \quad {\bs x}^{\star}=
\pi_1(\pi_2^{-1}({\bs x}))=a+b \lambda + c \lambda^2.  \]
Applying $\star$ to (\ref{dualgleich}) yields:
\begin{equation} \label{dualglstar}
\begin{array}{lcl}
{V'_L}^{\star} & = & \lambda^{-1} {V'_L}^{\star} \cup \lambda^{-1}
{V'_M}^{\star} \cup \lambda^{-1} {V'_L}^{\star} - \lambda + 1  \\
{V'_M}^{\star} & = & \lambda^{-1} {V'_L}^{\star} - \lambda \cup \lambda^{-1}
{V'_M}^{\star} - \lambda \cup \lambda^{-1} {V'_S}^{\star} \\
{V'_S}^{\star} & = & \lambda^{-1} {V'_L}^{\star} + \lambda - 2 \lambda^2 \cup
\lambda^{-1} {V'_M}^{\star} + \lambda - 2 \lambda^2
\end{array}
\end{equation}
Since all occurring maps are contractions, Theorem \ref{hutch} applies,
and the unique compact solution is
\[ \cl({V_L}^{\star}) = [-\lambda,0] = L, \; \cl({V'_M}^{\star}) =
[\lambda-\lambda^2,-\lambda] = M, \;  \cl({V'_S}^{\star}) = [-2
\lambda^2 + 1, -2 \lambda^2 + \lambda] = S. \]
The multiplication of (\ref{dualglstar}) by $\lambda$ yields a
distinct geometric realization of our original substitution
(\ref{subst2}). So, the dual tiling of the dual tiling is the one we
started with.  In particular, the window set of $V'$ is $W'=S \cup M
\cup L = [-2 \lambda^2 + 1, -2 \lambda^2 + \lambda] \cup
[\lambda-\lambda^2,0]$. Therefore $W'$ is compact,
\begin{equation} \label{volw}
\mu(W')= \lambda^2-1 =  7.290859374\ldots >0
\end{equation}
and $\mu(\partial W')=0$. All we are
left with is to compute $\dens(V')$ and use Theorem \ref{densvol}.
If the tiles $W_i$ of the substitution $\sigma'$ in (\ref{dualsubst})
would be polygons, this would be an easy task, using the remark after
Theorem \ref{perron}. Therefore we replace the fractal tiles
$W_S,W_M,W_L$ by appropriate polygons $P_S,P_M,P_L$. Let
\[ \begin{array}{rclrcl}
a & := & ({\lambda_2}^2 -1 ,  {\lambda_3}^2 -1 )^T , &
b & := & ({\lambda_2}^2, {\lambda_3}^2 )^T,  \\
c & := &  ({\lambda_2}^2 - \lambda_2 ,  {\lambda_3}^2 -\lambda_3 )^T, &
d & := & ({\lambda_2}^2 - \lambda_2 -1,  {\lambda_3}^2 -\lambda_3 -1)^T, \\
e & := & (2{\lambda_2}^2 - \lambda_2 -1,  2{\lambda_3}^2 -\lambda_3 -1)^T, &
f & := & (2{\lambda_2}^2 - 1,  2{\lambda_3}^2 -1)^T,  \\
g & := & (\lambda_2,\lambda_3)^T, &
h & := & ({\lambda_2}^2 + \lambda_2 -1,  {\lambda_3}^2 +\lambda_3 -1)^T.
\end{array} \]

\begin{figure}[h]
\epsfig{file=polytiles.eps} \hspace{5mm}
\begin{minipage}[b]{34mm}
$A_{\sigma'}= \begin{pmatrix} 0 & 1 & 1 \\
  1 & 1 & 1 \\ 0 & 1 & 2 \end{pmatrix}$
\vspace*{5mm}
\end{minipage}
\caption{Left: The construction of the polygonal prototiles. Middle:
A sketch of the substitution for them. Right: The corresponding
substitution matrix $A_{\sigma'}$. Note that $A_{\sigma'} =
(A_{\sigma})^T$. \label{polytiles}}
\end{figure}

Let $\conv({\bs x}_1,\ldots,{\bs x}_n)$ denote the convex hull of
${\bs x}_1,\ldots,{\bs x}_n$. Then (see Fig.~\ref{polytiles})
\begin{equation}
\begin{split}
P_S & := \conv(a-b,a,h,h-b), \\
P_M & := \conv(e-g,e,d,d-g) \, \cup \, \conv(c-g,c,e,e-g), \\
P_L & := \conv(0,e,c,b) \, \cup \, \conv(0,g,h,d,e)
\end{split}
\end{equation}
It is quite simple (but lengthy) to show that by replacing $W_i$ with
$P_i$ in (\ref{dualsubst}), the substitution $\sigma'$ leads to a tiling,
without any gaps or overlaps.
(Essentially, one uses the definition of a substitution tiling in
Definition \ref{substdef} and the primitivity of the substitution
matrix. To show that there are no gaps and no
overlaps it suffices to consider all vertex configurations in the
first $n$ substitutions $\sigma'(T),(\sigma')^2(T), \ldots,
(\sigma')^n(T)$ for appropriate $n$ and some prototile $T$. So, in any
certain case this is a finite
problem. Here $n=4$ and $T=P_L$ will do.) It is also quite simple (but
lengthy again) to compute the areas of the prototiles
$P_S,P_M,P_L$. Unfortunately, we cannot use the fact that the left
eigenvector of the corresponding substitution matrix reflects the
ratio of the areas, since the substitution for the polygonal tiles
does not preserve the shape any longer. E.g., $\sigma'(W_S)$ is congruent
to $Q'W_S$, but $\sigma'(P_S)$ is {\em not} congruent to $Q'P_S$.
Altogether, we obtain
\begin{equation}
\begin{split}
  \mu(P_S) & = |\lambda_2| \lambda_3 (\lambda_3-\lambda_2) = \frac{s_1}{s_4}
  \left(  \frac{s_2}{s_4} + \frac{s_1}{s_2} \right), \\
  \mu(P_M) & = \lambda_3-\lambda_2 = \frac{s_2}{s_4} + \frac{s_1}{s_2}, \\
  \mu(P_L) & = {\lambda_2}^2 - {\lambda_3}^2 + 2 |\lambda_2| + 2 \lambda_3 -
  |\lambda_2| {\lambda_3}^2 - \lambda_3 {\lambda_2}^2 = 1 + \frac{s_4
  s_1}{{s_2}^2}, \\
\end{split}
\end{equation}
where the last equality requires an excessive use of (\ref{wformel}).
The vector $(\mu(P_S),\mu(P_M),\mu(P_L))$ indeed is a left
eigenvector of the substitution matrix $A$ (cf.~Fig.~\ref{polytiles})
corresponding to the PF eigenvalue $\lambda$. A right eigenvector
corresponding to $\lambda$ is $(s_2,s_3,s_4)^T$. Therefore,
the normalized right eigenvector of $A$ --- which entries, by Theorem
\ref{freqvol}, are the
relative frequencies $p_i$ of tile types $P_i$ ($i \in \{S,M,L\}$) in
the tiling --- is $(p_S,p_M,p_L)^T := \frac{1}{s_2 + s_3 + s_4}
  (s_2,s_3,s_4)$. Since every point of $V'$ corresponds to exactly one
tile in the polygonal tiling, by Prop. \ref{densfreqvol} the relative
density of $V'$ is
\begin{equation} \label{densv}
\begin{split}
\dens(V') & =
\left( (\mu(P_S),\mu(P_M),\mu(P_L))\cdot(p_S,p_M,p_L)^T \right)^{-1} \\
& = (s_2+s_3+s_4) \left( \frac{s_1{s_2}^2}{s_4^2} + \frac{s_1^2}{s_4}
+ \frac{s_2 s_3}{s_4}+\frac{s_1 s_3}{s_2}+ s_4
+ \frac{s_1 {s_4}^2}{{s_2}^2} \right)^{-1} \\
& = 0.8100954858 \ldots
\end{split}
\end{equation}
In order to apply Theorem \ref{densvol} we need to compute $\det
\Lambda$ (cf.~(\ref{mlambda})). A rather lengthy computation (or the
usage of a computer algebra system, here: Maple) shows
$\det \Lambda = \frac{27}{64 {s_1}^2 {s_2}^2  {s_4}^2}$. Using
elementary trigonometric formulas yields
\[
\begin{split}
s_1 s_2 s_4 & = \frac{1}{4} (s_3+s_4 - s_2 - s_1 ) \\
& = \frac{1}{4} \left( s_3 + 2 \cos \left( \frac{3 \pi}{9} \right)
  s_1 -s_1 \right) = \frac{1}{4} s_3 = \frac{\sqrt{3}}{8},
\end{split}
\]
and therefore $\det \Lambda = 9$. Putting together this with
(\ref{volw}) and (\ref{densv}) in view of Theorem \ref{densvol} we
obtain \[ \mu(W')/\dens(V') = 9 =\det(\Lambda). \]
The leftmost equality requires again a very lengthy computation, or
the usage of Maple. This completes the proof, and we obtain the
following theorem.
\begin{thm} \label{dersatz}
The set $V'$ in (\ref{vstrich}), arising from the substitution $\sigma'$
in (\ref{dualsubst}), is a model set.
\end{thm}
One may ask why we do not state that the original set $V$ arising from
the substitution (\ref{subst2}) is a model set. The reason is that
there is one gap, namely, the value of $\mu(W)$. For example, one needs
to show that the tiling by the fractal tiles $W_S,W_M,W_L$ is not
overlapping. This seems to be clear from the pictures, but we did not
give a rigorous proof here. Indeed, this will require some more work,
and we refer to further publications. Anyway, there are good reasons
to assume that the areas of the
fractal tiles $W_S,W_M,W_L$ are equal to the areas of the polygonal
tiles $P_S,P_M,P_L$, resp. Proving this will complete the
proof that the original set $V$ is a model set, too.

\section{Further Remarks} \label{sec:remarks}
The question whether a given structure is a model set is motivated for
example by the question for the diffraction property of this structure.
It is known that model sets have a pure point diffraction image,
without any continuous part. For details, cf.~\cite{bmrs}.

We did not emphasize it throughout the text, but all tilings considered
here are nonperiodic. For the one--dimensional tilings this is a
consequence of the fact that the relative frequencies of the tiles are
irrational numbers. If a one--dimensional tiling is periodic, then the
relative frequencies are rational numbers. Anyway, all occurring
two--dimensional tilings are nonperiodic, too. This can be shown with
standard tools from the theory of nonperiodic substitution tilings
(cf.~statement 10.1.1 in \cite{gs}).

An important point in proving Theorem \ref{dersatz} is that we know
the occurring values in terms of trigonometric expressions, thus
it is possible to make exact computations of $\mu(W')$, $\dens(V')$,
$\det(\Lambda)$ etc. In general, the fractal nature of the occurring
window sets may forbid this. But if all occurring sets are polytopes
the things become easier. In the following, some examples are listed
where the duality principle provides short and simple proofs that
these considered substitution tilings are model sets.

{\bf Fibonacci sequences:} The symbolic substitution
$S \to L,\, L \to LS$ gives rise to the widely examined Fibonacci
sequences. (Not to be confused with the Fibonacci numbers
$1$,$1$,$2$,$3$,$5$,$8$,$13$,$\ldots$, sometimes also called Fibonacci
sequence. If one considers the first substitutions of $S$, then the
relation becomes clear: $L$, $LS$, $LLS$, $LSLLS$, $LSLLSLSL$, $\ldots$).

A geometric representation as a substitution tiling
(Fibonacci tiling) is for
example given by $L:=[0,1],S=[1, \tau]$ ($\tau$ the golden ratio)
and \[ \sigma (\{S\}) = \{ L+\tau \}, \; \sigma( \{L\} ) = \{L,S\}. \]
By the standard construction (cf.~Section \ref{sec:modset}), we obtain
for some $V$ which fulfills $\sigma(V)=V$ (namely, $V= \bigcup_{k \ge
  0} \sigma^{2k}(\{ L, L-1\})$, up to one point this will do) the
equations
\[ \begin{split}
V_L^{\ast} & = -\tau^{-1} V_L^{\ast} \cup -\tau^{-1} V_S^{\ast} \\
V_S^{\ast} & = -\tau^{-1} V_L^{\ast} + 1
\end{split}, \]
to be read as equations in $\ensuremath{\mathbb{E}}^1$. The unique compact solution is
\[ \cl(V_L^{\ast}) = W_L = [-1, \tau^{-1}] , \;
\cl(V_S^{\ast}) = W_S = [\tau^{-1}, \tau]. \]
The dual tiling is given by the following substitution:
\[ \begin{split}
\sigma'(\{W_L\}) & = \{W_L,W_S\} \\
\sigma'(\{W_S\}) & = \{W_L - \tau \}
\end{split} \]
Note that the substitution factor is $-\tau$. Nevertheless, the
resulting family of substitution tilings for $\sigma'$ is equal to
these for $\sigma$. That means the dual tilings of Fibonacci tilings
are Fibonacci tilings again; or shortly: Fibonacci tilings are
self--dual. Moreover, a short computation yields that the equation in
Theorem \ref{densvol} is fulfilled, and therefore the Fibonacci
sequences are model sets.

\begin{figure}[h]
\epsfig{file=ammch.eps}
\caption{\label{ammch} Left: The substitution for the Ammann
  chair. Right: A part of a corresponding substitution tiling.}
\end{figure}

{\bf Golden triangles and Ammann chair:}
The substitution for the golden triangles (cf.~Fig.~\ref{inflbsp})
defines a model set, if one replaces every triangle with the vertex
at its right angle. This was essentially shown in \cite{dvo} in
detail\footnote{The authors used {\em all} vertices in the tiling
  instead of one vertex of each triangle, but their result is easily
  transferred to the latter case.}. It turns out that the golden
triangle tilings are dual to the so--called Ammann chair tilings (or
'A2 tilings', cf.~\cite{gs}). A sketch of the substitution of the
latter is shown in Fig.~\ref{ammch}, together with a part of the
tiling. The substitution factor and the substitution matrix are equal
to those of the golden triangle substitution. All occurring tiles, and
all occurring window sets are polygons. Therefore, it is simple to
compute the areas and the densities, thus proving --- with the help of
Theorem \ref{densvol} --- that both substitution tilings (resp.~the
corresponding point sets) are model sets.

The duality of the Ammann chair tilings and the golden triangle
tilings was already indicated in \cite{gel} briefly. But a proof of
this needs some care. For example, one has to pay attention to the
fact, that each prototile occurs in four different orientations, not
just in one, as it is true for the tilings in Fig.~\ref{fractilg}.

{\bf Variants of the tilings in Section \ref{sec:zwo}:}
The tilings considered in the last section are a rich source of other
substitutions, or other Rauzy fractals. If we restrict to the case
$n=4$, then the underlying lattice $\Lambda$ will always be the same
(namely, as in (\ref{mlambda})). One possible modification is to
permute the letters in the substitution. E.g., the substitution
\begin{equation} \label{substperm}
 S \to ML, \; M \to MLS, \; L \to LML
\end{equation}
instead of (\ref{subst2}) yields the Rauzy fractal shown in
Fig.~\ref{mehrrosi} (left). Another variant is to use the
matrix $A_{\sigma'}$ of Fig.~\ref{polytiles} (the transpose of the
original substitution matrix $A_{\sigma}$) as a substitution matrix
for a one--dimensional substitution with substitution factor
$\lambda$. One possibility is the substitution
\begin{equation} \label{substtransp}
S \to M, \; M \to SML, \; L \to SMLL,
\end{equation}
where $S,M,L$ are intervals of length $\frac{s_1 s_2}{s_4}, s_2,s_4$,
resp. This substitution yields the Rauzy fractal shown in
Fig.~\ref{mehrrosi} (right). Note that in this case the set $W_M$ is not
connected, it consists of two parts which are similar to each other.
The substitution tiling arising from this Rauzy fractal (resp.~from
the corresponding equation system, multiplied by $Q'$,
cf.~(\ref{dualexp})) would possess a prototile, namely, $W_M$, which is
disconnected. Note, that we did not exclude disconnected
(proto--)tiles in our definition of (substitution) tilings.

\begin{figure}[h]
\epsfig{file=wkp9-alt1.eps} \hfill \epsfig{file=wkp9-alt2.eps}
\caption{\label{mehrrosi}
Two more Rauzy fractals, the left one arising from the substitution
(\ref{substperm}), the right one arising from the substitution
(\ref{substtransp}) }
\end{figure}

Combining these two modifications leads to even more examples, all of
them based on the lattice $\Lambda$. All the two--dimensional tilings
are model sets, which is easily proven by using the results of Section
\ref{sec:zwo}.
From the case $n=3$ one can also construct different tilings by
permuting the letters. In contrary, all variants of the case $n=2$
will yield Fibonacci sequences. Anyway, in the cases $n=3, n=4$ it may
be interesting to examine the different possibilities with respect to
common properties of the corresponding Rauzy fractals.

redpajama_set_name: RedPajamaArXiv


text:
Regal E-Walk Stadium 13 & RPX
247 W. 42nd Street,
13 screens
27 people favorited this theater
Regal Entertainment Group (Official)
Operated by: Regal Entertainment Group
Previously operated by: Loews Cineplex
Architects: David Rockwell
Firms: David Rockwell/Rockwell Group, JKRP Architects
Previous Names: Loews 42nd Street E-Walk Theatre, Regal E-Walk Stadium 13
Manager: 844.462.7342
Harem Theatre
Loew's American Theatre
Show Palace Theatre
AMC Empire 25
Times Theatre
News About This Theater
Apr 12, 2010 — Regal introducing new premium RPX big screen installations
Dec 23, 2005 — Loews and AMC to Sell 10 Theaters
A symbol of the 'new' Times Square (built roughly on the same site of a definitive symbol of the old Times Square – the former 24-hour porn grindhouse, the Harem Theatre), this 13-screen, all-stadium seating megaplex (part of the larger E-Walk entertainment complex), replete with retro-Art Deco style touches, murals, and enough neon to illuminate 50 pizza parlors, opened its doors on November 12, 1999.
Initially it had little neighborhood competition – the faded Criterion Center seven-screen multiplex at Broadway and 45th; the white elephant Loews State Theatre on the basement floor of the Virgin Megastore, one block north of the Criterion; and the single auditorium Loews Astor Plaza Theatre on 44th Street, west of Broadway – but, by April of 2000, with those aforementioned theatres either closed, facing increasing irrelevancy, or facing attendance drops of several fractions, the Loews 42nd Street E-Walk found itself competing with a even bigger movie (and crowd) magnet across the street: AMC's first foray into the NYC film marketplace, the Empire 25.
Despite a battle to fill their combined 38 screens with product, both theatres have succeeded in making Times Square arguably the strongest filmgoing destination it's ever been.
AMC acquired this theatre, along with the rest of the Loews Cineplex chain, in January 2006. As a condition of approving the merger, the state and federal governments required Loews and AMC to sell ten theatres, including this one.
It became a Regal cinema on September 15, 2006.
Contributed by br91975
Recent comments (view all 274 comments)
ridethectrain on November 8, 2017 at 10:31 pm
The Regal recliners are more comfortable than AMC version. Won't pay $30. for 4DX.
The new capacities are: 1. 102 2. 128 4DX 3. 75 4. 79 5. 79 6. 79 7. 79 8. 83 9. 84 (still closed, but will be open when Stars Wars play. 10. 70 11. 76 12. 204 13. 208 RPX/Atmos and Auro 11.1
The EMPIRE accross the street needs to upgrade or when movies play date, E-Walk business will impact the AMC Empire 25. The only time I'll go to the EMPIRE is for IMAX 12 chanels or Dolby Cinema.
Al Alvarez on February 11, 2018 at 9:05 pm
http://deadline.com/2018/02/cineworlds-3-6b-acquisition-of-regal-cinemas-approved-by-shareholders-1202276987/
zoetmb on March 22, 2018 at 5:41 pm
Personally, I think this strategy of pulling out half to two-thirds of the seats to install lounge seating is nuts. Every time they sell-out a show, they're leaving money on the table, plus it limits the number of ideal seats. It might make sense for Monday-Thursday, but it makes no sense for weekends when popular movies draw big crowds. This theater used to have almost 3300 seats. Now it's about 1345. The RPX screen was 508 and #12 was 490. Now they're 208 and 204. They should have left #12 and #13 alone. Why do we have to lie down to see a movie? Ridiculous, IMO.
rayman29 on March 23, 2018 at 1:21 am
I disagree. There are more theaters and showtimes now than 30 years ago. Some shows sell out but most don't. The smaller capacity is actually more profitable due to higher prices. Why go to a theater if its more comfortable to stay at hone? Its helping ticket sales. Recliners are becoming the norm braces on most cases, they increase profits.
moviebuff82 on August 16, 2018 at 5:06 pm
The vr location is closed.
markp on December 24, 2018 at 1:09 pm
3 years ago today I started running The Hateful 8 in 70MM in Cinema 12
moviebuff82 on April 23, 2019 at 8:23 pm
New logo signage has been at this theater since Cineworld took over Regal back in the fall of last year. AMC needs to step up their a game and put more recliners in their other non premium screens to match up with regal.
Al Alvarez on April 23, 2019 at 9:14 pm
What new logo signage? Nothing has been changed.
Mikeoaklandpark on April 24, 2019 at 4:21 pm
Cineworld took over this theater?
Cineworld bought Regal.


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The Invisible Front
Directed by Jonas Ohman, Vincas Sruoginis
Lithuania • 2014 • 87 mins. • In Lithuanian, English
https://www.youtube.com/watch?v=y8uIPg7pF-A
The Invisible Front was the code name used by the Soviet Union's Interior Forces for the armed resistance in Lithuania, a key occupied territory. The resistance arose with hardly any outside support during World War II, continuing in various forms until the fall of the Soviet Union in 1991. The film tells the story of one of the most significant anti-Soviet resistance movements through the words and experiences of its leaders, Juozas Luksa and his fellow Forest Brothers. Luksa risked his life to escape to Paris in 1948, joined up with Western intelligence agencies, wrote a memoir, and married Nijole, the love of his life. Shortly after their wedding, Luksa was air-dropped back into Soviet Lithuania by the CIA to help liberate his country. In English and Lithuanian with English subtitles.
Genres: Documentary
World Trade Center Theater


text:
The Jeffersons – The Complete Series DVD
I Dream of Jeannie - The Complete Series DVD
Full House - The Complete Series DVD
Shipping: Send from the United States, ETA 5-7 working days.
SKU: U_Jeffersons-26 Categories: 70th, Best DVD Sets Feb 2021, CBS, More Sitcom, Old, Sitcom, The Jeffersons, Top 9 Popular DVDs Tag: The Jeffersons DVD
Buy The Jeffersons DVD Box Set
DVD Details
Number of Disc: 33
Starring: Sherman Hemsley, Isabel Sanford, Marla Gibbs
Studio: Shout! Factory
What's it about
Fans of The Jeffersons, rejoice! All 253 episodes of the classic sitcom developed by Norman Lear (All in the Family, Mary Hartman, Mary Hartman) have finally arrived on DVD in The Jeffersons: The Complete Series.
Originally airing from 1975 to 1985, The Jeffersons documented the side-splitting exploits of an upwardly-mobile family: dry cleaning entrepreneur George (Sherman Hemsley), his wife Louise (Emmy winner Isabel Sanford), and their son Lionel, as they move from the working-class section of Queens to a luxury apartment in Manhattan. George's brash, arrogant personality always seems to get him in hot water. Lucky for him, he's got the level-headed "Weezy", by his side and their wise-cracking maid Florence (Marla Gibbs) to keep him humble. Together, they overcome the comic challenges and culture clashes that their new way of life brings.
Featuring a terrific supporting cast (including Paul Benedict, Roxie Roker, and Franklin Cover), memorable guest appearances (Sammy Davis, Jr., Gladys Knight, Reggie Jackson, Billy Dee Williams and more) and of course, one of the catchiest theme songs in television history, The Jeffersons: The Complete Series is another must-own sitcom classic from your friends at Shout! Factory.


Ecology of SO2 resistance: II. Photosynthetic changes of shrubs in relation to SO2 absorption and stomatal behavior.
In an effort to predict SO2 sensitivity of plants from their morphological and physiological features, the effects of SO2 on photosynthesis were partitioned between stomatal and nonstomatal components for a drought deciduous shrub, Diplacus aurantiacus, and an evergreen shrub, Heteromeles arbutifolia. As predicted, the drought deciduous shrub had the higher gas conductance, and hence SO2 absorptance. However, nonstomatal components also play a role in determining SO2 sensitivity. Apparently a plant with a high intrinsic photosynthetic capacity will be more sensitive to SO2 than one with a lower capacity.


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It was during the second month of the siege that the Confederacy began its real assault on the city. Men carrying large planks of wood rushed forward to build bridges over the moat surrounding the walls while their archers sent volleys against the Centurions watching from above. Still, even as the sky was nearly blocked out by missiles, the Centurions were able to send out volleys of their own, killing many of the archers and the men building the bridges. In the end, however, the enemy's sheer numbers allowed them to place seven bridges over the eight-foot deep moat.
Next, siege towers, thirty foot high structures on wheels, moved slowly up to the walls of the city. The Centurions sent hundreds of oil lit arrows at the towers, but each of them was covered by thick animal hides, soaked to the point of having turned green. With dozens of men pushing the towers from behind, they were unreachable for attack.
boulders off the walls, they blocked the paths of the towers. Four of the towers were stopped this way as the rocks jammed up against their wheels.
Only eight towers successfully made it to the walls, their drawbridges dropping over the parapets. Another siege tower almost made it, just a five-foot gap between the tower and the wall, a boulder stuck in its wheels.
With eight towers against the walls, the Confederacy troops that had been pushing them from behind began climbing them and pouring onto the parapet to meet the long spears and crossbows of the Centurion warriors. Less than a dozen Confederate soldiers ever set foot on the stone walls, as the Centurions quickly overwhelmed them. When the Centurions pinned the attackers in their siege towers, they sent in men carrying heavy barrels of oil. Inside the towers, the Centurions cracked the barrels open and poured the fuel down over the dry wooden interiors. After dropping a torch, each tower burst into flames, black smoke billowing out from its top like a chimney.
During all this fighting, the Confederate generals were not idle. Under the clouds of smoke, more bridges were built and additional towers sent toward the walls. With the Centurions busy fighting on the walls, the boulders blocking the other towers were soon removed. Where there had been eight towers against the walls there were soon twelve, sending hundreds of men to the walls.
In the midst of the chaos of battle, a wagon lead by six horses raced toward the city's gates. It almost reached it before Centurion archers from the gate towers shot the horses down. Still, the heavy battering ram in the back of the wagon was now within a few yards of the gate with Confederate troops nearby. Soon the iron shod ram was beating against the heavy gate, the wood cracking. Arrows, hot oil, and boulders rained down on the men manning the ram but as one fell it seemed two more would take his place.
Three times the Centurions killed enough of the attackers to force the ram to be left unmanned, surrounded by the dead. Each time, however, the Confederate troops surged forward and took up beating against the gates again. As Sirrus watched his men's progress, things were not happening fast enough. Already eighteen thousand men were involved in the attack, mostly from the weaker kingdoms. Neither Huron nor Tal Sith was willing to risk their best men until victory was assured.
once the city walls were breached and the Centurions crammed into the palace. He was so sure, he could already imagine receiving his reward.
With his men engaged with thousands of Confederacy troops, the last thing Marcus had expected was to see the rain of arrows darken the sky again, killing friend and foe alike. This time the Centurions, engaged in combat with the enemy, could not duck and hide behind the stone parapets of the city's walls. With wave after wave of arrows falling on the city, the total number of people standing quickly dropped. Seeing that the city's gates were now buckling, Marcus ordered the gate guards to start moving the large stone blocks in front of it. It had become clear, however, that nothing he could do would prevent the enemy from breaking through.
"I do not understand how this commander thinks!" Marcus shouted with disgust. What sort of general would sacrifice his men in such a reckless manner?
"You have to admit, however, it is effective," Varrus groaned.
"Killing thousands of your own men to kill hundreds of ours?"
"With the numbers they have they can afford this blood sacrifice. You have to remember; we are not fighting against a Centurion… moral standards do not apply."
"Well, if I ever get my hands on this leader I will see he is turned into a pin cushion…" Marcus had hoped to hold the enemy back for weeks before retreating behind the palace walls. It was due to their sheer numbers, and total disregard for their own lives, that the Confederacy was overwhelming his Centurions.
Seeing that holding the walls was a hopeless cause, Marcus ordered his men to retreat. The first to fall back were the archers, a few of whom would be left in the city to act as snipers, looking for any chance to take out an important officer and lower morale. Once the archers were in position, Marcus began to call back the infantry using the few cavalry he had kept in the city to hold the city streets. The last to be called back were the gate guards. The city gate was now halfway blocked by stone. Even if the gate fell, it would have to be ripped out before men could pass through it.
After all his men were safe, Marcus called in all his commanders to receive the casualty reports. Of the twenty thousand men who had been guarding the walls, three thousand had been killed and eight thousand were wounded… Not as bad as Marcus had feared. With the Centurion talent for quick healing those eight thousand would soon be ready for battle once more.
"Where were your men?" the general from Forstine demanded, bursting into Sirrus' command tent.
Not for the first time, the Count wished he could place a locked door to the entryway. "My men were occupied elsewhere."
"On some grand field of battle, no less," the general said with scorn.
"They were serving their purpose… just as your men served theirs. Thanks to them we took the city in a day."
"But your archers… how could you have given the order. Hundreds of my soldiers died."
"Hundreds you say… my count is over a thousand." Sirrus smirked as he made his way to a chest, opening it to show it full of silver coins. "Tell me… do you value your men's lives so much that you would have been willing to trade yours for one of theirs?"
"No… that would be foolish. You do not cut of the head to save the tail."
"Of course… how silly of me. How about ten men for your life… a hundred, maybe a thousand? Is any number of your soldiers worth giving up your own life for?"
"…" The general remained silent.
"How about this… I give you ten of these pretty silver coins for every man who was killed in battle today? That is ten months' wages," Sirrus said as he took up a fist full of coins only to let them slip from his fingers.
The general's anger was suddenly replaced with greed. "That would be quite a sum…"
"I think this chest would cover your loss," Sirrus replied as he slammed the chest closed with a foot. Behind that chest were half a dozen others, just waiting to be claimed by other mourners.


Geniuses at Sweet Seeds live by one principle – to recreate and capture the genetics of glorious classic strains that have contributed to hundreds of marijuana strains today. Thanks to a special breeding program involving several generations of none other than Jack 47 Auto, breeders at Sweet Seeds created Jack 47 XL Auto.
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As a Sativa-dominant strain with almost 80 percent Sativa and 20 percent Indica, there are no hints of Indica sedation. What you get instead is a mind-blowing trip to heaven, like you've attained nirvana. High concentrations of THC ranging from 18 to 21 percent make her nothing short of a beast. Waves of euphoria will slam into you relentlessly as you do your best to hang on. You'll not only be stimulated creatively but the urge to be productive will also gain momentum as the high wears on.
Jack 47 is famous as a heavy yielder, and this XL Auto carries the tradition forward with aplomb. Her buds tend to be covered with so much resin that it's easy to make hash using just your fingers. Despite being an autoflowering strain, she generates about 475-600 grams/m2 when grown indoors. Outdoors, the production increases quite a bit with at least 60 to 200 grams of dense buds per plant.
Autoflowers are low-maintenance plants that grow with no interference, and Jack 47 XL stays true to traits even in this department. As long as you provide loads of light and enough nutrients to satisfy her hunger, she will treat you with outstanding connoisseur-quality buds. Don't overdo it, though, because less is more with autos.
She's super sticky, but it's the aroma that usually lures you to her. Like her famous parents, her fragrance is strong with light incense notes. In fact, the grow room won't need a room freshener because she smells like fresh lemonade with floral notes!


After receiving a strong worded letter from IIM-A, IITs may now remove Flipkart from the Day One list of the campus placement programs.
The All-IITs Placement Committee which comprises of all the placement cell heads across all IITs, will meet soon and reach a consensus regarding the Flipkart issue.
The companies usually prefer the day 1 slot so as to pick the best out of the lot. Flipkart was so far given the day 1 slot in almost all the IITs, except a few.
After the letter from IIM-A, IIT-B is also planning to send out a similar email to Flipkart, asking them to reduce the deferment period or else increase the compensation amount.
However, Flipkart has assured that the trainees will be onboard by December 2016, after the entire restructuring process is completed. With the intensifying pressure on the e-commerce player to reduce its cash burn, the skepticisim and the concerns are inevitable.


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Jared Olar: How do you cure PDS?
Jared Olar
The folks at the New York Times, the Washington Post and Mother Jones magazine must be feeling pretty dumb now, or at least they should be.
Last Friday they started rooting through a heap of more than 24,000 emails from Sarah Palin's time as governor of Alaska, even "crowdsourcing" the excavation by enlisting public assistance in reading the emails and flagging all the really juicy ones.
And after all that work, here is what they dug up: "."
Yes, with the release of these emails, we now know that Sarah Palin was the governor of Alaska and did governor-type stuff, was the 2008 GOP vice presidential candidate and did VP candidate-type stuff, is a politician and did politician-type stuff, is a wife and did wife-type stuff, is a mother and did mom-type stuff — but didn't say anything about those things in her emails that is worth mentioning.
I doubt the Old Gray Lady, the WaPo and MJ will win any Pulitzers for their Sarah Palin email coverage. Woodward and Bernstein stuff this sure ain't.
Why did they go to so much trouble? Palin is no longer an elected official, and hasn't sought public office since 2008.
Part of the explanation — but only a very small part — is that the leftist journal Mother Jones initially sought these emails during the 2008 presidential campaign, naturally seeking the inside scoop on Sen. John McCain's choice of runningmate.
It took so long to obtain these emails that there isn't as much relevance in investigating Palin today than there was three years ago. Still, once they got them, it only made sense to see if they held anything of interest.
But probably the primary cause of the Palin email non-event is psychological rather than professional.
Clearly they had hoped that by casting their net so wide and so deep, they would catch something slimy and smelly. But you have to ask why.
The news media generally have not devoted anywhere near the amount of resources to investigating other public figures or candidates, and have shown far less interest in the character, qualifications and lack of paper trail of President Barack Obama.
Yes, Palin has established herself as a permanent fixture and a power broker on the Republican scene who knows how to generate publicity (giving the lie to the belief that she's stupid), and she might possibly run for president (though I doubt she will and don't think she should).
But that can hardly warrant the intensity of interest the media have shown in her — nor the depth of the fear and loathing of Palin that many on the left regularly display.
The email anticlimax followed close on the heels of the preceding week's caravan of journalists trailing her "One Nation" bus tour. Both are just the latest signs of an unhealthy interest in, even an obsession with, the former Alaska governor turned political celebrity — maybe even a full-blown case of PDS, "Palin Derangement Syndrome."
I really can't say why Palin elicits such a strong reaction from so many people on the political left. It can't just be her political stances, which do not meaningfully differ from any other American conservative, nor her occasional verbal gaffes and fact errors, which are as interesting and as newsworthy as the content of her gubernatorial emails.
It's probably futile to seek reasons for the unusual interest that journalists and leftists have in Palin. Maybe it's something about her personality that pushes their buttons (and I suspect she knows how to push their buttons and enjoys doing it).
Or maybe it has to do with how unhealthy our political culture has become, that we are prone to demonize and dehumanize political leaders.
Perhaps it's that we are ruled by our passions rather than our reason, that leads so many of us to be duped by paranoid and incredibly foolish conspiracy theories — Bill and Hillary Clinton and Vince Foster's suicide, George W. Bush and 9/11 Truthism, Obama and Birtherism, Palin and Triggerism (the claim that her son Trig is really her grandson, despite all available evidence including the word of the doctor who delivered him).
Whatever the origins of PDS, we'd be closer to a cure if we stopped obsessing on her in the news.
Jared Olar may be reached at jolar@pekintimes.com.


YES! You too can sit out and DO Free Advice for Free! These Sign images can be copied (as is, with credit due to The Free Advice Man) and printed on card-stock. Just remember that you must not charge money for Free Advice, you must not mislead anyone into thinking that this is your own original idea, you must not provide medical or legal advice (unless you actually are a doctor or lawyer and can afford to), and you must not alter these signs!
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subhana allah!! He alone succeeds in this trial who in the midst of his riches does not forget about Allah's Status and his own status. Abu Hurairah (May Allah be pleased with him) said that: He heard the Prophet (PBUH) said: "There were three men among the Banu Israel, one leper, one bald and one blind.
Allah wanted to test them. He therefore, sent to them an angel who came to the leper and asked him what he would like best. He replied: "A good color, a good skin and to be rid of what makes me loathsome to people ".
He (the angel) rubbed him and his loathsomeness vanished and he was given a good color and a good skin. He then asked him what type of property he would like best. The leper replied that he would like camels - [or perhaps he said cattle, for Ishaq (one of the subnarrator of the Hadith) was uncertain, either said: 'Camels,' or: 'Cattle']. He was given a pregnant she-camel.
The angel invoked for Allah's Blessing on it. The angel then went to the bald man and asked him what he would like best and he replied: "Good hair and to be rid of what makes me loathsome to people".
The angel ran his hand over him and he was given good hair. He then asked him what property he would like best. He replied that he would like cattle, so he was given a pregnant cow. The angel invoked Allah's Blessing on it.
The angel then went to the blind man and asked him what he would like best, and he replied: "I wish that Allah restore my sight to me so that I may see people.'' Thereupon the angel ran his hand over him and Allah restored his sight. The angel then asked what property he would like best. He replied that he would like sheep, so he was given a pregnant ewe. Flocks and herds were produced for the three men, the first having a valley full of camels, the second one, a valley full of cows and the third one full of sheep.
Then the angel came in the form of a leper, to the one who had been a leper, and said: "I am a poor man and my resources have been exhausted in my journey, and my only means of reaching my destination are dependent on Allah and then on you, so I ask you by Him Who gave you the good color, the good skin and the property, for a camel by which I may get to my destination". He replied: "I have many dues to pay.'' The angel then said: "I think I recognize you. Were you not a leper whom people found loathsome and a poor man to whom Allah gave property?'' He replied: "I inherited this property through generations". The angel said: "If you are telling a lie, may Allah return you to your former condition".
The angel went in the form of a bald man to the one who had been bald, and said the same as he had said to the former and received a similar reply. So he said: "If you are telling a lie, may Allah return you to your former condition". The angel then went to the one who had been blind and said: "I am a poor traveler and my resources have been exhausted in my journey.
My only means of reaching my destination are dependent on Allah and then on you, so I ask you by Him Who restored your eyesight for a sheep by which I may get to the end of my journey". He replied: "Yes, I was blind. Allah restored my eyesight, so take what you wish and leave what you wish.
I swear by Allah that I shall not argue with you today to return anything you take, as I give it for Allah's sake". The angel said: "Keep your property. You have all simply been put to a test, and Allah is pleased with you and displeased with both of your companions".
do not tell any hadeth untill you make sure of its truth.