text:
\section*{Introduction}

There is a significant amount of research devoted to finding Gray
codes for classes of permutations, where two permutations are considered
adjacent if they differ by an involution of some special kind. For a
survey of some key results we refer the reader to Savage's
paper~\cite[Section 11]{Savage}. The simplest and most elegant result
in this area is the Johnson-Trotter algorithm~\cite{Johnson, Trotter},
providing an adjacent transposition Gray code for all permutations of a
finite set.

The present work is motivated by King's recent paper~\cite{King}, providing a
transposition Gray code for the set of all indecomposable permutations
of a finite set. The combinatorial interest in these permutations is
long-standing, both Comtet~\cite[p.\ 261]{Comtet-AC} and
Stanley~\cite[Ch.\ 1, Exercise 32]{Stanley-EC1} discuss them in their
textbooks. As shown by Ossona de Mendez and
Rosenstiehl~\cite{OM-Rosenstiehl-conn}, they are bijectively equivalent to
rooted hypermaps. Refining Dixon's famous result~\cite{Dixon}, stating
that a random pair of permutations generates almost always a transitive
group is also related to enumerating indecomposable permutations, see
Cori's work~\cite{Cori-indec}.

Our main result is finding an adjacent transposition Gray code for a
signed variant of the indecomposable permutations, which we call {\em
  sign-connected permutations of the set $\{\pm 1,\pm 2,\ldots,
  \pm n\}$}. The importance of this signed variant is highlighted by the
fact that our sign-connected permutations encode the shelling types of
the boundary of the hypercube. Here we consider two shellings to be of
the same type, if they may be transformed into each other by an isometry
of the underlying hypercube. Adjacent transpositions correspond to
swapping adjacent entries in a list of facets.

We achieve our goal in two steps. First, in Section~\ref{sec:gray}, we provide a
very simple adjacent transposition Gray code for a larger class of
permutations, which we call {\em standard permutations of the set
  $\{\pm 1,\pm 2,\ldots, \pm n\}$}. The simplicity of this code is
comparable to the simplicity of the output of the Johnson-Trotter
algorithm. The second step, contained in Section~\ref{sec:conn},  is to
show the following, surprisingly simple statement: an adjacent
transposition Gray code for the class of sign-connected permutations of
$\{\pm 1,\pm 2,\ldots, \pm n\}$ may be obtained from the Gray code of
all standard permutations by simply deleting those standard permutations
of $\{\pm 1,\pm 2,\ldots, \pm n\}$ which are not
  sign-connected. Although the statement is very simple, its proof
  requires a careful look at a special way of coding standard
  permutations of $\{\pm 1,\pm 2,\ldots, \pm n\}$. This coding is
    inspired by the place-based inversion tables, introduced for
    ordinary permutations in~\cite{Hetyei}. Our result justifies the
    hope that the
    same coding applied to ordinary permutations may help us find an
    adjacent transposition Gray code for indecomposable permutations,
if such a Gray code exists.

Our standard permutations are bijectively equivalent to
{\em standard double-occurrence words} with $n$ symbols. As shown by
Ossona de Mendez and Rosenstiehl~\cite{OM-Rosenstiehl}, these words are
bijectively equivalent to rooted maps with $n-1$ edges, this result was
later refined by Drake~\cite{Drake}. Standard double-occurrence words are
also bijectively equivalent to fixed point free involutions, as defined by
Cori~\cite{Cori}. Under these bijections, sign-connected standard
permutations correspond to connected rooted maps and indecomposable
fixed-point free involutions. Our Gray codes may be useful in the study
of these related objects.

Our paper is structured as follows. In the Preliminaries we gather a few
basic facts on permutation Gray codes and shellings of the boundary of a
hypercube that we will need. We also define sign-connected permutations
of $\{\pm 1,\pm 2,\ldots, \pm n\}$. In Section~\ref{sec:hyp} we
introduce standard permutations of $\{\pm 1,\pm 2,\ldots, \pm n\}$ and show
that the sign-connected standard permutations are bijectively equivalent
to the shelling types of the boundary of an $n$-dimensional
hypercube. The fact that standard sign-connected permutations correspond
to standard double-occurrence words is shown in Section~\ref{sec:arc}
where we also introduce arc diagrams that help visualize standard
permutations of $\{\pm 1,\pm 2,\ldots, \pm n\}$. Essentially the same
visualization was used by
Drake~\cite{Drake}. Inspired by the place-based noninversion tables
introduced in~\cite{Hetyei}, we introduce a new encoding of our arc
diagrams. It is this encoding that makes the definition of the simple
Gray code for all standard permutations in Section~\ref{sec:gray}
truly easy. Essential properties of this Gray
code are explored in Section~\ref{sec:prop}. Finally, our main result
may be found in Section~\ref{sec:conn}.

Our result highlights a connection between the study of
hypermaps and the theory of shellings which may be worth exploring
further in the future. It also raises the hope that, by using a similar
encoding to the one introduced in~\cite{Hetyei}, one could find an
adjacent transposition Gray code for indecomposable
permutations. Finally, although a large amount of literature exists on shelling
and shellability, very little has been done to explore the set of all
shellings of the same object. Our paper is among the first in this
new direction.

\section{Preliminaries}

\subsection{Gray codes for sets of permutations}
Given a finite set $X$ with $n$ elements, we define a {\em permutation} of $X$ as a word $\pi=\pi(1)\cdots \pi(n)$ in which every element of $X$ appears exactly once. We denote the set of all permutations of $X$ by ${\mathcal
  S}_X$. Given any subset ${\mathcal V}$ of ${\mathcal S}_X$, we consider the following graph on the vertex set ${\mathcal V}$: the unordered pair $\{\pi,\rho\}\subseteq {\mathcal V}$ is an edge if $\pi$ and $\rho$
differ by an {\em adjacent transposition}, i.e., there is an $i\in \{1,\ldots,n-1\}$ such that $\rho(i+1)=\pi(i)$, $\rho(i)=\pi(i+1)$, and $\rho(j)=\pi(j)$ for $j\in\{1,\ldots,n\}\setminus \{i,i+1\}$. An {\em adjacent transposition Gray code for ${\mathcal V}$} is a Hamiltonian path in the resulting graph. Obviously not all sets of permutations have
such a Gray code.

The Johnson-Trotter algorithm generates an adjacent transposition Gray
code on the set ${\mathcal S}_X$ of all permutations of the set $X$.  When $X=\{1,2,\dots,n\}$, this algorithm recursively defines the Gray code as follows: set the initial permutation as $12...n$; replace each permutation in the Gray code for permutations of length $n-1$ with $n$ new permutations of length $n$; for each of the length $n-1$ permutations, insert $n$ into every position $1$ through $n$ to get the new permutations of length $n$.  For odd-indexed, length $n-1$ permutations, insert $n$ from right to left, and left to right for even-indexed permutations~\cite{Johnson, Trotter}.  See Table~\ref{tab:JT} for the Gray code when $n=2$ and $n=3$.

\begin{table}[h]
\caption{The Gray code produced by the Johnson-Trotter algorithm for $n=2$ and $n=3$ (read down)}
\label{tab:JT}
\begin{tabular}{cccc}
\hline
$n=2$ &      &  $n=3$ &     \\
\hline
12  &       & 123 & 321  \\
21  &       & 132 & 231  \\
      &       & 312 & 213  \\
\hline
\end{tabular}
\end{table}

A permutation $\pi$ of ${\mathcal S}_n$ is {\em connected} if there is no $m < n$ such that $\pi$ sends ${\{1,2,\ldots ,m\}}$ into ${\{1,2,\ldots ,m\}}$.  King found a {\em transposition Gray code} for such connected permutations~\cite{King}, i.e., a Gray code for the graph whose vertices are connected permutations and whose edges connect permutations that differ by a (not necessarily adjacent) transposition. It is still open, whether there is an adjacent transposition Gray code for connected permutations.

In this paper, we be interested in the following signed variant of connected permutations.

\begin{definition}
\label{def:sign-conn}
We call a permutation $\pi=\pi(1) \cdots \pi(2n)$ of $\{\pm 1, \pm 2, \ldots , \pm n\}$ {\em sign-connected} if and only if for all $1\leq m<2n$ there is at least one $j\in \{1,2,\ldots,n\}$ such that $|\{\pi(1),\ldots,\pi(m)\}\cap
\{-j,j\}|=1$. If a permutation of $\{\pm 1, \pm 2, \ldots , \pm n\}$ is not sign-connected then we call it {\em sign-disconnected}.
\end{definition}

As we will see in Section~\ref{sec:hyp}, a subset of these permutations is identifiable with the shelling types of the boundary of a hypercube. We will show in Section~\ref{sec:conn} that there is an adjacent transposition Gray code for this subset.

\subsection{Shellings of the boundary of the hypercube}
\label{sec:shell}

We define the standard $n$-dimensional hypercube ($n$-cube) to be $[-1,1]^n \subset \mathbb{R}^n$.  As observed by Metropolis and Rota~\cite{Metropolis-Rota}, each non-empty face of the standard hypercube may be denoted by a vector $(u_1,\ldots,u_n)\in \{-1,*,1\}^n$, where $u_i=-1$ or $u_i=1$ indicates that all points in the face have the $i^{th}$ coordinate equal to $u_i$; whereas, $u_i=*$ indicates the $i^{th}$ coordinate of the points in the face range over the entire set $[-1,1]$. Using this notation, the code for each facet of the boundary is of the form $(u_1,\ldots,u_n)\in \{-1,*,1\}^n$ such that exactly one $u_i$ belongs to $\{-1,1\}$. Thus, we can encode each facet by a single number $-k$ or $k$ where $k\in\{1,\ldots,n\}$ is the unique index $k$ such that $u_k=\pm 1$.  We write $-k$ or $k$ when $u_k=-1$ or $u_k=1$, respectively. Using this notation, we will identify each enumeration of the facets of the boundary of the $n$-cube with a permutation of the set $\{\pm 1, \pm2, \ldots, \pm n\}$. This correspondence is a bijection.

A {\em shelling} is a particular way of listing the facets of the boundary of a polytope, see \cite[Definition 8.1]{Ziegler} for a definition.  The fact that the boundary complex of any polytope is shellable was shown by Bruggeser and Mani~\cite{Bruggeser-Mani}. A polytope is {\em cubical} if each of its proper faces is combinatorially equivalent to a cube. The boundary complex of a cubical polytope is a cubical complex; see Chan's paper~\cite{Chan} for a definition of a cubical complex as the geometric realization of a cubical poset. The boundary complex of a cubical polytope is {\em pure}, i.e., all facets have the same dimension. Following Chan~\cite{Chan}, we define a shelling of a $(d-1)$-dimensional cubical complex as an enumeration $(F_1,\ldots,F_r)$ of its facets in such a way that for all $m>1$ the intersection $F_m\cap (F_1\cup\cdots\cup F_{m-1})$ is a union of $(d-2)$-faces homeomorphic to the ball or sphere. Such an intersection is a {\em cubical shelling component}, whose {\em type} is $(i,j)$ if it is the union of $i$ antipodally unpaired $(d-2)$-faces and $j$ pairs of antipodal $(d-2)$-faces. The ``only if part'' of the following statement is obvious and used in Chan's work~\cite{Chan} without proof.  Both implications are shown by Ehrenborg and Hetyei~\cite[Lemma 3.3]{Ehrenborg-Hetyei}.

\begin{lemma}[Ehrenborg-Hetyei]
\label{l:EH}
The ordered pair $(i,j)$ is the type of a shelling component in a shelling of a cubical $(d-1)$-complex if and only if one of the following holds:
\begin{itemize}
\item[(i)] $i=0$ and $j=d-1$; or
\item[(ii)] $0<i<d-1$ and $0\leq j\leq d-1-i$.
\end{itemize}
Furthermore, in case (i), the shelling component is homeomorphic to a $(d-2)$-sphere, and in case (ii), the shelling component is homeomorphic to a $(d-2)$-ball.
\end{lemma}

As a consequence of Lemma~\ref{l:EH}, we may describe the shellings of the boundary complex of an $n$-cube in the following way.

\begin{lemma}
An enumeration $(F_1,\ldots,F_{2n})$ of the facets of the boundary complex of the $n$-cube is a shelling if and only if for each $m<2n$, the set $\{F_1,\ldots, F_m\}$ contains at least one antipodally unpaired facet.
\end{lemma}
\begin{proof}
On the one hand, the cubical complex $F_1\cup\cdots\cup F_m$ is shellable and $(n-1)$-dimensional, and as such it is homeomorphic to a $(n-1)$-ball or an $(n-1)$-sphere. Since the boundary complex $F_1\cup \cdots F_{2n}$ is an $(n-1)$-sphere, the proper subcomplex $F_1\cup\cdots\cup F_m$ can only be an $(n-1)$-ball. By \cite[Lemma 3.3]{Ehrenborg-Hetyei} (see part (i) of Lemma~\ref{l:EH} above), $F_1\cup\cdots\cup F_m$ must contain at least one antipodally unpaired facet.

Conversely, assume that $F_1\cup\cdots\cup F_m$ contains at least one antipodally unpaired facet for each $m<2n$. In other words, each $F_1\cup\cdots\cup F_m$ could be used as a shelling component in the shelling of an $n$-dimensional cubical complex, and its type $(i_m,j_m)$ satisfies $i_m>0$. Adding $F_m$ to $F_1\cup\cdots\cup F_{m-1}$ results either in introducing a new antipodally unpaired facet or $F_m$ is the antipodal pair of a previously listed facet. In the first case, we have $i_m=i_{m-1}+1$ and $j_m=j_{m-1}$, so $F_m\cap (F_1\cup\cdots\cup F_{m-1})$ is a shelling component of type $(i_m,j_m)$. In the second case, we have $i_m=i_{m-1}-1$ and $j_m=j_{m-1}+1$, so $F_m\cap (F_1\cup\cdots\cup F_{m-1})$ is a shelling component of type $(i_m,j_m)$. Finally, for $m=2n$, $F_{2n}\cap (F_1\cup\cdots\cup F_{2n-1})$ is a shelling component of type $(0,n-1)$.
\end{proof}

For each initial segment of an enumeration of the facets of a shelling
of the boundary of the $n$-cube, there is at least one antipodal pair of
facets such that exactly one of the two facets belongs to the shelling
component, i.e., the facet $k$ is listed in the first $i$ facets of the
shelling but facet $-k$ facet is not~\cite{Ehrenborg-Hetyei}.

\begin{corollary}
\label{cor:conn}
A permutation of $\{\pm 1, \pm 2, \ldots , \pm n\}$ is sign-connected if and only if it represents a shelling of the facets of the hypercube.  Similarly, a sign-disconnected permutation represents an enumeration of the facets which is not a shelling.
\end{corollary}

As a result of Corollary~\ref{cor:conn}, the number of sign-connected permutations equals the number of shellings of the boundary of the $n$-cube.  Given any $n$, this number can be found by the recursive formula $a_n = (2n-1)!!-\sum_{k=0}^{n-1} (2k-1)!! \cdot a_{n-k}$.  The sequence $\{a_n\}$ is sequence A000698 in the On-Line Encyclopedia of Integer Sequences~\cite{A000698}.

\section{Equivalence Classes of the Enumerations of the Facets of the $n$-Cube}
\label{sec:hyp}

The isometries of the $n$-cube permute its facets, inducing a $B_n$-action on the enumerations of all facets of the boundary of the $n$-cube.  This action is free, i.e., any nontrivial isometry takes each enumeration into a different enumeration.

\begin{definition}
We consider two enumerations of the facets of the $n$-cube equivalent if they can be transformed into each other by an isometry of the $n$-cube.
\end{definition}

As noted in Section~\ref{sec:shell}, every enumeration of the facets of
the boundary of the $n$-cube can be identified with a signed
permutation.  The induced action of $B_n$ on these signed permutations
is generated by the following operations:
\begin{enumerate}
\item For each $k\in\{1,\ldots,n\}$ there is a reflection
  $\varepsilon_k$  interchanging $k$ with $-k$ and leaving all other
  entries unchanged;
\item For each $\{i,j\}\subset\{1,\ldots,n\}$ there is a
  reflection $\rho_{i,j}$ interchanging $i$ with $j$ and $-i$ with $-j$,
  leaving all other entries unchanged.
\end{enumerate}
It is worth noting that we may also identify the elements of $B_n$ with
signed permutations the usual way, the action of $B_n$ we consider is
then the action of $B_n$ on itself, via conjugation. There exist $2^n
\cdot n!$ symmetries of the $n$-cube~\cite{Humphreys}.  Since the $B_n$
action is free, all equivalence classes have the same cardinality,
giving a total of $\dfrac{(2n)!}{2^n \cdot n!}=(2n-1)!!$ equivalence
classes.

\begin{definition}
\label{def:std}
A standard permutation is defined to be an permutation of $\{\pm1,\ldots ,\pm n\}$ such that the following two properties hold:
\begin{enumerate}
\item for all  $i$, $i$ occurs before $-i$ in the list, and
\item the negative numbers in the list appear in the following order:
  $-1, -2, \ldots , -n$.
\end{enumerate}
\end{definition}

Since there are $2^n$ ways to assign the negative sign to the pair $k$ and $-k$ and $n!$ ways to order $\{-1,\ldots ,-n\}$, we have a total of $\dfrac{(2n)!}{2^n \cdot n!}=(2n-1)!!$ standard permutations.

\begin{lemma}
Each equivalence class of the enumerations of the facets of an $n$-cube corresponds to exactly one standard permutation.
\end{lemma}
\begin{proof}
Since the number of equivalence classes and the number of standard permutations are both $(2n-1)!!$, we need only to show that every $\pi \in {\mathcal S}_{\{\pm 1,\ldots ,\pm n\}}$ is equivalent to a standard permutation.

Pick any $\pi \in {\mathcal S}_{\{\pm 1,\ldots ,\pm n\}}$, and suppose
$\Pi$ is the equivalence class which contains $\pi$.
Applying only reflections $\varepsilon_k\in B_n$ we may replace $\pi$
with a $\pi'\in \Pi$ which satisfies condition (1) in
Definition~\ref{def:std}. Applying only reflections $\rho_{i,j}\in
B_n$ we may replace $\pi'$ with a $\pi''\in \Pi$ that also satisfies
condition (2) in Definition~\ref{def:std}. Note that the application of
an operator $\rho_{i,j}$ leaves the validity of condition (1) unchanged.
\end{proof}

The set of sign-connected permutations is closed under the action of
$B_n$; thus, we can think of equivalence classes of shellings as types
of shellings. The following characterization of standard sign-connected
permutations is an immediate consequence of
Definitions~\ref{def:sign-conn} and \ref{def:std}.
\begin{lemma}
\label{l:std-conn}
A standard permutation $\pi=\pi(1) \ldots  \pi(2n)$ of $\{\pm 1, \pm 2, \ldots , \pm n\}$ is sign-disconnected if and only if there exists an $i$ such that $|\pi(j)| \le i$ for all $j \le 2i$ (and $|\pi(j)| \ge i+1$ for all $j > 2i$).
\end{lemma}

Applying the definition of a standard permutation to our previous notion of a sign-connected permutation, we get the following characterization.

\begin{lemma}
\label{l:ineq}
A standard permutation $\pi=\pi(1) \cdots \pi(2n)$ is sign-connected if
and only if for all $m<2n$, the sum $\pi(1)+\cdots+\pi(m)> 0$.
\end{lemma}
\begin{proof}
By definition of a standard permutation, each $j$ appears before $-j$,
thus we have $\pi(1)+\cdots+\pi(m)\geq 0$ for every $m\leq 2n$. Equality
occurs exactly when the set $\{\pi(1),\ldots,\pi(m)\}$ is the union of
pairs of the form $\{j,-j\}$. The existence of an $m<2n$ satisfying
$\pi(1)+\cdots+\pi(m)=0$ is thus equivalent to the standard permutation
being sign-disconnected.
\end{proof}

\begin{remark}
\label{rem:ineq}
Note that the proof of the ``if'' part of Lemma~\ref{l:ineq} may be
restated in the following, stronger from: $\pi=\pi(1) \cdots \pi(2n)$ is
sign-connected if for all $m<2n$ we have $\pi(1)+\cdots+\pi(m)\neq 0$.
\end{remark}

The standard permutation in Example~\ref{ex:discon} is sign-disconnected.

\section{Arc Diagrams}
\label{sec:arc}

A standard permutation $\pi \in {\mathcal S}_{\{\pm 1,\ldots, \pm n\}}$
of the facets of the boundary of the $n$-cube can be visually represented by an arc
diagram.  The arc diagram is constructed as follows: put $2n$ vertices
in a row; label them left to right with $\pi(1),\ldots, \pi(2n)$;
then for each $k\in\{1,\ldots,n\}$, create an arc connecting the
vertices labeled $-k$ and $k$.  See Fig.~\ref{fig:arc01} for an example
when $n=3$. Hence, in the associated arc diagram of a standard permutation,
the order of the labels of the vertices will match the order of the standard
permutation.

Note that the labels of the vertices in the arc diagram are uniquely
determined by the underlying complete matching.  This matching is
represented by the arcs where the right endpoints of the arcs must be
labeled left to right by $-1,-2,\ldots, -n$, in this order, and the left
endpoint of the arc whose right end is labeled $-k$ must be be labeled
$k$. Hence, using these rules, we can uniquely reconstruct the
associated standard permutation from its the arc diagram. Thus, the arc diagram
representation provides a bijection
between standard permutations in ${\mathcal S}_{\{\pm 1,\ldots, \pm n\}}$
and complete matchings of a $2n$ element set.

\begin{figure}[h]
\input{arc01.pspdftex}
\caption{The arc diagram associated to $(3,1,-1,2,-2,-3)$}
\label{fig:arc01}
\end{figure}

Almost the same bijection appears in the work of
Drake~\cite{Drake}, the only difference being that Drake encodes
complete matchings with {\em standard double occurrence words} in the letters
$1,2,\ldots,n$. A double occurrence word is a word
in which each letter occurs exactly twice. Ossona de Mendez and
Rosenstiehl~\cite{OM-Rosenstiehl} call a double occurrence word
{\em standard} if the first occurrence of the letters happens in
increasing order. (Note that Drake~\cite{Drake} omits the adjective
standard, but he adopts his terminology from~\cite{OM-Rosenstiehl}, and
the words he uses to encode complete matchings are the standard
double occurrence words.)

Taking the {\em reverse complement} of such a double occurrence word
and changing the second occurrence of each $k$ to $-k$ results in a
standard permutation. This correspondence is a bijection. (As usual,
the complement of a word in the letters $1,2,\ldots,n$ is the word
obtained by replacing each letter $i$ with $n+1-i$, and the reverse of a
word $w_1 w_2\cdots w_{2n}$ is the word $w_{2n}w_{2n-1}\cdots w_1$.) For
example, the standard permutation $(3,1,-1,2,-2,-3)$ corresponds to the
same matching as the double occurrence word $122331$ in Drake's
work~\cite{Drake}.

\begin{remark}
Let us associate to each standard permutation $\pi$ of $\{\pm 1, \pm 2,
\ldots, \pm n\}$
the {\em fixed point free involution $\widehat{\pi}$} of $\{\pm 1, \pm
2, \ldots , \pm n\}$
that exchanges the elements $\pi^{-1}(-i)$ and $\pi^{-1}(i)$ for
$i=1,2,\ldots,n$.
It is easy to see that this correspondence is a bijection. Under this
bijection, sign-connected standard permutations correspond to
indecomposable fixed point free involutions, as defined by Cori~\cite{Cori}.
\end{remark}

For each $k\in\{1,\ldots,n\}$, the definition of a standard permutation
forces the vertex labeled $k$ to be to the left of the vertex labeled
$-k$ in the arc diagram.  Any arc diagram associated to a standard
permutation can be represented by a word $a_1 a_2 \ldots  a_n$,
recursively constructed as follows. Let $a_n$ be the position of the
rightmost arc's left endpoint, i.e., $a_n$ is the position of the
vertex labeled $n$.  Remove the rightmost arc from the diagram, and repeat
the process until all arcs are removed.  As a result of how we defined
this representation, $1 \le a_i \le 2i-1$ for each $a_i$ in the word
$a_1 a_2 \ldots  a_n$, and each word of this form corresponds to exactly
one standard permutation.  Conversely, every such word $a_1\ldots a_n$
will encode some arc diagram.
\begin{example}
\label{ex:con}
$(3,1,-1,2,-2,-3) \cong 131$.
\end{example}

\begin{definition}
We will denote the standard permutation which the word $a_1\cdots a_n$ encodes by $\pi(a_1\cdots a_n)$.
\end{definition}

\begin{lemma}
\label{lem:discon}
A standard permutation $\pi(a_1 a_2 \ldots  a_n)$ is sign-disconnected if and only if there exists a $k \ge 2$ such that $a_k=2k-1$ and $a_j \ge a_k = 2k-1$ for all $j > k$.
\end{lemma}
\begin{proof}
Suppose $a_1\ldots a_n$ encodes a sign-disconnected standard permutation $\pi(1)\ldots \pi(2n)$.  Then by Lemma~\ref{l:std-conn}, there exists an $i$ such that $|\pi(j)| \ge i+1$ for all $j > 2i$, meaning that in the associated arc diagram both ends of the $(i+1)^{st}$ through $n^{th}$ arcs are located at vertex positions $2i+1$ or greater.  Hence, $a_{i+1}=2i+1$ and $a_j \ge 2i+1$ for $j > i$.

Conversely, let $a_1 \ldots  a_n$ encode a standard permutation $\pi(1) \ldots  \pi(2n)$.
Suppose $k$ is the smallest integer such that $k\geq 2$, $a_k=2k-1$ and
$a_j \ge a_k = 2k-1$ for all $j > k$.
Then in the associated arc diagram, the right endpoint of the $(k-1)^{st}$ arc is located at vertex position $2k-2$.  Thus, $\pi(2k-2)=-(k-1)$ since the right endpoint of any arc is labeled with the negative number of the antipodal pair.
Because $\pi(1) \ldots  \pi(2n)$ is a standard permutation, we have $|\pi(j)| \le k-1$ for all $j \le 2k-2$.
Also, $a_j \ge 2k-1$ for all $j \ge k$ in the word encoding $\pi$'s associated arc diagram means that the left endpoints of the $j^{th}$ arcs are all at vertex position $2k-1$ or greater.
Hence, $|\pi(j)| \ge k$ for all $j \ge 2k-1$.
\end{proof}

Clearly, a standard permutation is sign-disconnected if and only if
a vertical line can be drawn between the first and last vertices of the
associated arc diagram which does not intersect any of the arcs
and which separates two adjacent vertices. In other words, the
arc diagram is comprised of more than one component of overlapping arcs.
Drake~\cite{Drake} noted that this occurs if there exists a vertex $k$,
$1<k<2n$, which is not nested under any arc in the diagram.
We will refer to the arc diagram of a sign-connected standard permutation as
connected and to the arc diagram of a sign-disconnected standard
permutation as disconnected. An arc diagram is connected
exactly when it represents a connected matching as defined by
Drake~\cite{Drake}. In terms of standard double occurrence words, connected arc
diagrams correspond to connected standard double occurrence words in the
work of Ossona de Mendez and Rosenstiehl~\cite{OM-Rosenstiehl}.

We will call a word $a_1 \cdots a_n$ {\it arc-connected} if and only if
the arc diagram associated to $\pi(a_1 \cdots a_n)$ is connected,
otherwise we call the word {\it arc-disconnected}.

\begin{example}
\label{ex:discon}
The arc diagram associated to $(1,2,-1,-2,3,-3) \cong 125$, shown in
Fig.~\ref{fig:arc02}, is disconnected.

\begin{figure}[h]
\input{arc02.pspdftex}
\caption{The arc diagram associated to $(1,2,-1,-2,3,-3)$}
\label{fig:arc02}
\end{figure}
\end{example}

\begin{corollary}
\label{cor:arc-conn}
A word is arc-disconnected if and only if there exists a $k$ such that $a_k=2k-1$ and $a_j \ge a_k = 2k-1$ for all $j > k$.  Similarly, a word is arc-connected if and only if no such $k$ exists.
\end{corollary}

\begin{definition}
A minimal arc will be defined to be a connected component of an arc diagram such that the component consists of exactly one arc.
\end{definition}

For example, in the arc diagram of Example~\ref{ex:discon} (see Fig.~\ref{fig:arc02}), there is one minimal arc located at the rightmost end of the diagram.  However, in the arc diagram of Example~\ref{ex:con} (see Fig.~\ref{fig:arc01}), there is no minimal arc since the third arc stretches over the first two arcs in the diagram, i.e., $a_3=1$.

The effect of an adjacent transposition of the facets of the boundary of the $n$-cube, i.e., in the standard permutation,  on the associated arc diagram is exactly one swap of adjacent ends of two distinct arcs.  Namely, an interchange of two left ends, or an interchange of one left end and one right end.

If $\pi(a_1\ldots a_n)$ and $\pi(b_1\ldots b_n)$ differ by an adjacent
transposition, then there exists a unique $i$ such that $b_i=a_i \pm 1$
and $b_j=a_j$ for all $j \neq i$.  The converse is not true.  For
example, the word $122$ is obtained from the word $112$ by changing the
second letter by $1$, but the standard permutations $\pi(112)$ and
$\pi(122)$ differ by more than just a single adjacent transposition, see
Figure~\ref{fig:arct01}.

\begin{figure}[h]
\input{arct01.pspdftex}
\caption{The arc diagrams associated to $\pi(112)$ and $\pi(122)$}
\label{fig:arct01}
\end{figure}

\section{A Gray Code for all Standard Permutations}
\label{sec:gray}

In this section, we define a Gray code for the standard
permutations of the set $\{\pm 1,\ldots,\pm n\}$ for each $n\geq 1$. We
will write our permutations in the form $\pi(a_1\ldots a_n)$, using the encoding
introduced in Section~\ref{sec:arc}. To simplify our notation, we omit
the operator $\pi$ and write $a_1\cdots a_n$ instead of $\pi(a_1\ldots
a_n)$ in our lists.  We will refer to this simplified way of writing
our Gray code as {\em a Gray code for words}.

We define our Gray code for words recursively as follows. For $n=1$, we
have only one word to list, namely $1$. To write the Gray code for words
of length $n$, start with $11\ldots 1$.
Increase $a_n$ by $1$ to get the next word: $11\ldots 12$.  Continue to
increase $a_n$ by $1$ to get a list of codes.  Stop increasing $a_n$
once $a_n=2n-1$ (recall $1\le a_i \le 2i-1$ for any $i$). Now replace
$a_1 \ldots  a_{n-1}$ with the next word in the Gray code for words of
length $n-1$ to get $11\ldots 12(2n-1)$.  Decrease $a_n$ by $1$ to
get the next string of codes. When $a_n=1$, replace $a_1 \ldots
a_{n-1}$ with the next word in the Gray code of length $n-1$.  Continue
in this fashion until the Gray code terminates with $135\ldots (2n-1)$.
See Table~\ref{tab:GC} for the Gray code when $n=2$ and $n=3$.

\begin{table}[h]
\caption{The Gray Code (read down)}
\label{tab:GC}
\begin{tabular}{ccccc}
\hline
$n=2$ &      &  $n=3$ &   &  \\
\hline
11  &       & 111 & 125 & 131 \\
12  &       & 112 & 124 & 132 \\
13  &       & 113 & 123 & 133 \\
      &       & 114 & 122 & 134 \\
      &       & 115 & 121 & 135 \\
\hline
\end{tabular}
\end{table}

\begin{theorem}
\label{thm:fullgc}
The enumeration defined above is an adjacent transposition Gray code for
the standard permutations of the set $\{\pm1,\ldots,\pm n\}$.
\end{theorem}
\begin{proof}
We proceed by induction on $n$. Comparing two consecutive words
$a_1\ldots a_n$ and $b_1\ldots b_n$, they
differ in exactly one letter. We distinguish two cases, depending on
whether this letter is the last letter of the word or some other letter.

{\noindent\it Case 1:} $a_n\neq b_n$. In this case, the associated standard
permutations differ by an adjacent transposition since the first $n-1$
arcs in the two associated arc diagrams remain stationary, and the
$n^{th}$ arc moves to the right or to the left by one vertex position.
This move switches the left end of the $n^{th}$ arc with an adjacent end
of a different arc. As mentioned at the end of Section~\ref{sec:arc},
switching adjacent ends of two arcs corresponds to an adjacent
transposition on the associated  signed permutation.

{\noindent\it Case 2:} $a_n=b_n$. This case occurs only if $a_n=1$,
corresponding
to the $n^{th}$ arc stretching over the first $n-1$ arcs; or if
$a_n=2n-1$, which means the $n^{th}$ arc is a minimal arc.  Thus, in
either situation, the $n^{th}$ arc will not affect whether or not the
move produces an adjacent transposition in the associated standard
permutations.  By the recursive definition of the Gray code, we replace
$a_1 \ldots  a_{n-1}$ with the next word in the Gray code for words of
length $n-1$. By our induction hypothesis, consecutive words in the Gray
code for words of length $n-1$ correspond to standard permutations on
$\{\pm 1,\ldots ,\pm(n-1)\}$ which differ by an adjacent transposition.
\end{proof}

\section{Properties of the full Gray Code}
\label{sec:prop}

The Gray code for the equivalence classes of the enumerations of the
facets of the boundary of the $n$-cube, referred to by their standard
permutations, will be referred to as the {\em full Gray code}.  As seen
in Section~\ref{sec:gray}, this Gray code may be defined in terms of a
list of standard permutations in the form $\pi(a_1 \ldots a_n)$ for each
word $\underline{a}=a_1\ldots a_n$.  We introduce the notation
$\tau(\underline{a})=a_1\ldots a_{n-1}$ to represent the truncated word,
obtained by removing the last letter of $\underline{a}$.

\begin{definition}
The collection of all words $\underline{a}$ such that $\tau (\underline{a})=a_1\cdots a_{n-1}$ is fixed will be referred to as a run in the Gray code for words.
\end{definition}

All words in a run form a sublist of codes corresponding to standard permutations differing by an adjacent transposition.  In the run, only $a_n$ changes, either increasing from $1$ to $2n-1$ or decreasing from $2n-1$ to $1$.
A run is increasing if $a_n$ increases to get each subsequent word in the run.  If an increasing run is the $k^{th}$ run in the Gray code, then $k \equiv 1$ mod 2.  Hence, increasing runs will also be referred to as odd runs.
A run is decreasing if $a_n$ decreases to get each subsequent code in the run.  If a decreasing run is the $k^{th}$ run in the Gray code, then $k \equiv 0$ mod 2.  Decreasing runs will also be referred to as even runs.
Note, odd and even refer to the count of the run and not whether $a_{n-1}$ is odd or even.  For example, $1111$ through $1117$ is the first run in the Gray code for words of length $4$, and $a_3=1$.  However, $12561$ through $12569$ is the $37^{th}$ run in the Gray code for words of length $5$, and $a_4=6$.

Several properties of the Gray code follow directly from its definition:
\begin{enumerate}
\item  Suppose $\pi(\underline{b})=\pi(b_1\ldots b_n)$ immediately
  follows $\pi(\underline{a})=\pi(a_1\ldots a_n)$ in the Gray code and
  $\underline{a}$ and $\underline{b}$ are in different runs. If
  $\underline{a}$ is in an odd (increasing) run, then $\underline{b}$ is
  in an even (decreasing) run and $a_n=b_n=2n-1$.  If $\underline{a}$ is
  in an even (decreasing) run, then $\underline{b}$ is in an odd
  (increasing) run and $a_n=b_n=1$.
\item  In every run, there is at least one arc-disconnected word (when $a_n=2n-1$) and at least two arc-connected words (when $a_n=1$ or $2$).
\item  There are $(2n-3)!!$ runs in the Gray code where each word that encodes a standard permutation has length $n$.
\end{enumerate}

\begin{lemma}
\label{fact-m}
Suppose $\pi(\underline{a})=\pi(a_1 \ldots a_n)$ is the $m^{th}$ standard permutation in the Gray code.
Then $(n-1)+\sum_{i=1}^{n} a_i \equiv m$ mod 2.
\end{lemma}
\begin{proof}
We will use induction on $m$. For $m=1$, $a_1=a_2=\cdots=a_{n}=1$ and
$(n-1)+\sum_{i=1}^{n} a_i =2n-1 \equiv 1 =m$ mod 2. Assume the statement is true for the $m^{th}$ standard permutation $\pi(\underline{a})$, and let $\underline{b}$ be the word which encodes the $(m+1)^{st}$ standard permutation in the Gray code. As noted above, there exists a unique $i$ such that $b_i=a_i\pm 1$ and $b_j=a_j$ for all $j\neq i$. Hence the parity of $(n-1)+\sum_{i=1}^{n} b_i$ is the opposite of the parity of $(n-1)+\sum_{i=1}^{n} a_i$, that is $(n-1)+\sum_{i=1}^{n} b_i \equiv m+1$ mod 2.
\end{proof}

\begin{corollary}
\label{run-parity}
If $\underline{a}$ is in an increasing run, then $a_1+a_2+\ldots +a_{n-1}+(n-2)\equiv 1$ mod 2, and if $\underline{a}$ is in a decreasing run, then $a_1+a_2+\ldots +a_{n-1}+(n-2)\equiv 0$ mod 2.
\end{corollary}

\begin{lemma}
\label{fact6}
Suppose that $\pi(\underline{a})=\pi(a_1\ldots a_n)$ is immediately
followed by $\pi(\underline{b})=\pi(b_1\ldots b_n)$ in the full Gray
code. Let $i$ be the unique index such that $b_i=a_i \pm 1$
and $b_j=a_j$ for all $j \neq i$.  Then if $i<n$, either $a_k=1$ for all
$k>i$ or $a_k=2k-1$ for all $k>i$.
\end{lemma}
\begin{proof}
By the recursive nature of the Gray code, for
each $k>i$ we must have either $a_k=1$ or $a_k=2k-1$ .  In particular,
$a_n=1$ or $2n-1$ if $i<n$.  We would like to show that in a single word
we cannot have both $a_k=1$ and $a_j=2j-1$ for some $k,j>i$. Assume by
way of contradiction, that there is such a change, then there is a least
$k>i$ such that exactly one of $a_k$ and $a_{k+1}$ is equal to one. By
the recursive nature of our Gray code we may assume $k=n-1$, since
$a_j=b_j$ for $j\geq k+2$ implies that $b_1\ldots b_{k+1}$ immediately
follows $a_1\dots a_{k+1}$ in the full Gray code of words.
We will show by way of contradiction the impossibility of the case when
$a_n=2n-1$ and $a_{n-1}=1$. The case when $a_{n-1}=2n-3$ and $a_n=1$ is
completely analogous.

Since $a_n=b_n=2n-1$, $\underline{a}$ is in an increasing run.  By
Corollary~\ref{run-parity}, $a_1+\ldots +a_{n-1}+(n-2) \equiv 1$ mod 2.
Thus, $a_1+\ldots +a_{n-2}+(n-3) \equiv 1$ mod 2, which means
$\tau(\underline{a})$ is in an increasing run in the Gray code for words
of length $n-1$.  Since $a_{n-1}=1$, we must have $b_{n-1}=2$, in
contradiction with $i<n-1$.
\end{proof}

\begin{corollary}
\label{parity}
If $\underline{a}=a_1\ldots a_k 1\ldots 1$ and the next word in the Gray
code for words is $a_1\ldots a'_k 1\ldots 1$, then $a'_k=a_k+1$ implies
that $a_k$ is even, and $a'_k=a_k-1$ implies that $a_k$ is odd.
Similarly, if $\underline{a}=a_1\ldots a_k (2k+1)\ldots (2n-1)$ and the
next word in the Gray code for words is $a_1\ldots a'_k (2k+1)\ldots
(2n-1)$, then $a'_k=a_k+1$ implies that $a_k$ is odd, and $a'_k=a_k-1$
implies that $a_k$ is even.
\end{corollary}
Lemma~\ref{fact-m} has the following consequence.
\begin{corollary}
\label{parity-truncated}
Under the conditions of Lemma~\ref{fact6}, if $\underline{a}=a_1\ldots
a_k 1\ldots 1$, then $a_1+\ldots +a_k+(k-1)=0$ mod 2;
and if $\underline{a}=a_1\ldots a_k (2k+1)\ldots (2n-1)$, then
$a_1+\ldots +a_k+(k-1)=1$ mod 2.
\end{corollary}

\begin{lemma}
\label{fact8:prelim}
In each run in the Gray code for words of length $n$, there exists a $k$
such that if $a_n \le 2k-2$, $\underline{a}$ is arc-connected; and if
$a_n \ge 2k-1$, $\underline{a}$ is arc-disconnected. This $k$ is the
least index $k'\leq n-1$ such that $a_j\geq 2k'-1$ holds for all $j\in \{k',
k'+1, \ldots,n-1\}$, if such an index exists; otherwise $k=n$.
\end{lemma}
\begin{proof}
Consider a run in the Gray code for words of length $n$. All the words in
the run have the same truncated word $\tau(\underline{a})=a_1\ldots a_{n-1}$.

{\noindent \it Case 1}: $\tau(\underline{a})$ is arc-connected.
Applying Corollary~\ref{cor:arc-conn} to $\tau(\underline{a})$ we see
that there is no $k'\leq n-1$ such that
$a_j\geq 2k'-1$ holds for all $j\in \{k',k'+1, \ldots,n-1\}$. By the
same Corollary~\ref{cor:arc-conn}, $\underline{a}$ is arc-disconnected if
and only if there exists a $k$ such that $a_k=2k-1$ and $a_j \ge
a_k=2k-1$ for all $j>k$.  Since $a_1 \cdots a_{n-1}$ is arc-connected,
such a $k$ must satisfy  $k=n$.  Thus, $a_1 \cdots a_n$ is
arc-disconnected if and only if $a_k=a_n=2n-1$.\\

{\noindent\it Case 2}: $\tau(\underline{a})$ is arc-disconnected.
Applying Corollary~\ref{cor:arc-conn} to $\tau(\underline{a})$ yields
that there is a smallest $k'\leq n-1$
such that $a_{k'}=2k'-1$ and $a_j \ge 2k'-1$ for $k'<j<n$.  Clearly, if $a_n \ge
2k'-1$ then $\underline{a}$ is arc-disconnected. We are left to show
that $\underline{a}$ is arc-connected whenever $a_n \le 2k'-2$.
Assume, by way of contradiction, that $\underline{a}$ is
arc-disconnected for some $a_n \le 2k'-2$. By
Corollary~\ref{cor:arc-conn}, there is a $k''$ such that $a_j\geq
2k''-1$ holds for all $j\geq k''$. By the minimality of $k'$ we must
have $k''\geq k'$. On the other hand $a_n\geq 2k''-1$ and $a_n\leq
2k'-2$ imply $k''<k'$, a contradiction.
\end{proof}

\begin{lemma}
\label{fact8}
Suppose $k$ is defined as in Lemma~\ref{fact8:prelim}. Then
$\pi(a_1\ldots a_{k-1})$ is the first sign-connected component of the
standard permutation $\pi(a_1\ldots a_{n-1})$.
\end{lemma}
\begin{proof}
Assume first there is a least index $k'\leq n-1$ such that $a_j\geq
2k'-1$ holds for all $j\in \{k', k'+1, \ldots,n-1\}$.
The we have $k=k'$ and, by Corollary~\ref{cor:arc-conn}, the
standard permutation $\pi(a_1\cdots a_{k-1})$ is sign-connected.
Since $a_j\geq 2k-1$ holds for $j=k,\ldots,n-1$, the left endpoints of
the corresponding arcs are to the right of the arcs associated to
$\pi(a_1\cdots a_{k-1})$. Therefore $\pi(a_1\ldots a_{k-1})$ is the
first sign-connected component of $\pi(\tau(\underline{a}))$.

Assume now that there is no $k'\leq n-1$ satisfying $a_j\geq
2k'-1$ for all $j\in \{k', k'+1, \ldots,n-1\}$. In this case $k=n$ and,
by Corollary~\ref{cor:arc-conn}, $\pi(\tau(\underline{a}))$ is
arc-connected.
\end{proof}

\begin{lemma}
\label{fact5}
Suppose $\underline{a}$ is immediately followed by
$\underline{b}$ in the full Gray code for words and let $k$ be the
unique index such that $b_k\neq a_k$.
If $b_k=a_k+1$ and $\underline{a}$ is arc-disconnected,
then $\underline{b}$ is arc-disconnected; and
if $b_k=a_k-1$ and $\underline{a}$ is arc-connected,
then $\underline{b}$ is arc-connected.
\end{lemma}
\begin{proof}
Consider first the case when $k=n$. In this case $\underline{a}$ and
$\underline{b}$ belong to the same run and the statement follows from
Lemma~\ref{fact8:prelim}. We are left to consider the case when $k \neq
n$. In this case Lemma~\ref{fact6} tells us that either $a_i=1$ holds
for all $i > k$  or $a_i=2i-1$ holds for all $i > k$. By
Lemma~\ref{fact8:prelim}, $\underline{a}$ is arc-connected exactly when
$a_i=1$ holds for all $i>k$ and $\underline{a}$ is arc-disconnected exactly when
$a_i=2i-1$ holds for all $i>k$. The same characterization also applies
to $b$ since $a_i=b_i$ for all $i>k$.
\end{proof}

\section{Restricting the Gray Code to the Shelling Types of the $n$-Cube}
\label{sec:conn}

Our goal is to define a Gray code for the facet enumerations of the
boundary of the $n$-cube, restricted to shelling types.  This is
equivalent to finding a Gray code for the arc-connected words $a_1
\ldots  a_n$ since arc-connected words encode sign-connected standard
permutations, and we know the sign-connected standard permutations
represent shellings of the boundary of the $n$-cube.

In this section, we will show that the sublist obtained by removing all
arc-disconnected words from the full Gray code of words from
Section~\ref{sec:gray} yields a Gray code for the sign-connected standard
permutations. This sublist of the standard permutations will be referred
to as the connected Gray code. Our main result is the following.

\begin{theorem}
\label{prop1}
Suppose $\underline{a}$ and $\underline{b}$ are arc-connected, but every
code listed between these two words in the full Gray code of words is
arc-disconnected.  Then $\pi(\underline{a})$ and $\pi(\underline{b})$
differ by an adjacent transposition.
\end{theorem}

\begin{proof}
Without loss of generality we may assume that $\underline{b}$ follows
$\underline{a}$ in the full Gray code of words.
By Lemma~\ref{fact8:prelim}, $\underline{a}$ cannot be at the end of a
run and, Lemma~\ref{fact5}, $\underline{a}$ must be in an increasing
run. So $a_1\ldots a_{n-1}a_n$ is arc-connected, and the next consecutive
word in the Gray code of words is $a_1\ldots a_{n-1}a'_n$, where
$a'_n=a_n+1$.  This word is disconnected as are all the remaining codes
in the run up to and including $a_1\ldots a_{n-1}(2n-1)$.

The next run in the Gray code of words starts with $c_1\ldots
c_{n-1}(2n-1)$ which is arc-disconnected.  This run decreases down to
$c_1\ldots c_{n-1}1$, an arc-connected code.  We must have $c_1\ldots
c_{n-1}=b_1\ldots b_{n-1}$ since $b_1\ldots b_n$ is the first
arc-connected code following $\underline{a}$.  Thus, the next run in the
Gray code of words actually starts with $b_1\ldots b_{n-1}(2n-1)$, and
the subsequent codes in the run down to $b_1\ldots b_{n-1}(b_n+1)$ are
arc-disconnected.

By the construction of the Gray code, $\tau(\underline{a})$ and
$\tau(\underline{b})$ are consecutive words in the full Gray code for
words of length $n-1$; thus, there is exactly one $k<n$ such that
$b_k=a_k \pm 1$.  By Lemma~\ref{fact6}, either we have $k=n-1$ or we have
$k<n-1$ and $\tau(\underline{a})$ is either of the form
$\tau(\underline{a})=a_1\ldots a_k 1 \ldots  1$, or of the form
$\tau(\underline{a})=a_1\ldots a_k (2k+1) \ldots  (2n-3)$.

{\noindent\it Case 1:} $\tau(\underline{a})=a_1\ldots a_k 1 \ldots  1$
where $a_k\neq1$. By our assumption, $b_k=a_k \pm 1$.  Thus, both
$\tau(\underline{a})$ and $\tau(\underline{b})$ are arc-connected; by
Lemma~\ref{fact8}, we must have $a_n=2n-2=b_n$.  In the arc diagram,
$a_j=b_j=1$ for $k < j < n$ means the first $k$ arcs are under $n-k-1$
nested arcs.  An arc stretching over all previous arcs will not change
how any two ends of the previous arcs are interchanged since that move
occurs completely under the arcs.  The $n^{th}$ arc ($a_n=2n-2$) will
only overlap the $(n-1)^{st}$ arc, meaning this last arc will not affect
how the first $k$ arcs change when we make the move associated to
changing from $\underline{a}$ to $\underline{b}$.
By the recursive definition of the full Gray code, we know that
$\pi(a_1\ldots a_k)$ and $\pi(b_1\ldots b_k)$ differ by an adjacent
transposition.  This means the arc diagrams encoded by $a_1\ldots a_k$
and $b_1\ldots b_k$ differ by exactly two adjacent ends of two distinct
arcs swapping positions; and since we already know that the $(k+1)^{st}$
through $n^{th}$ arcs will not affect that swap, we get that
$\underline{a}=a_1\ldots a_k 1 \ldots  1 (2n-2)$ and
$\underline{b}=b_1\ldots b_k 1 \ldots  1 (2n-2)$ encode two standard
permutations which differ by an adjacent transposition.

{\noindent\it Case 2:} $\tau(\underline{a})=a_1\ldots a_k (2k+1) \ldots
(2n-3)$ where $a_k \neq 2k-1$. By Corollary~\ref{cor:arc-conn},
$\tau(\underline{a})$ is arc-disconnected, so there must exist a $j \le
k+1$ such that $a_j=2j-1$ and $a_i \ge 2j-1$ for $j<i<n$.  Let us choose
$j$ to be the smallest index with this property. By
Lemma~\ref{fact8:prelim}, the word
$a_1\ldots a_k(2k+1)\ldots (2n-3)(2j-2)$ is arc-connected, but the rest
of the codes in the run, $a_1\ldots a_k(2k+1)\ldots (2n-3)(2j-1)$
through $a_1\ldots a_k(2k+1)\ldots (2n-3)(2n-1)$, are arc-disconnected.
Thus, we must have $a_n=2j-2$.

Since $\underline{b}$ is in a decreasing run
immediately following the run in which $\underline{a}$ is contained, the
run of $\underline{b}$ starts with $b_1\ldots b_k(2k+1)\ldots
(2n-3)(2n-1)$ where $b_k=a_k \pm 1$.  By Corollary~\ref{parity}, if
$b_k=a_k+1$, then $a_k$ is odd, and if $b_k=a_k-1$, then $a_k$ is even.
Since $j \neq k$ and $j\leq k+1$, we have two subcases:

{\noindent\it Case 2a:} $j=k+1$. In this case, $a_1\ldots a_k$ is
arc-connected and $a_n=2j-2=2k$. We claim that $b_1\ldots b_k$ is also
arc-connected. This is an immediate consequence of Lemma~\ref{fact5}
when $b_k=a_k-1$. If $b_k=a_k+1$, then $b_k$ is even and $b_1\ldots b_k$
is arc-connected because, by Lemma~\ref{fact8:prelim}, the first
arc-disconnected code in an increasing run ends with an odd letter.
Thus, both $\underline{a}=a_1\ldots a_k (2k+1) \ldots  (2n-3) (2k)$ and
$b_1\ldots b_k (2k+1) \ldots  (2n-3) (2k)$ are
arc-connected codes, and there are only arc-disconnected codes between
these two words in the full Gray code of words. Therefore $b_n=2k$.

In the arc diagram of $\pi(\underline{a})=\pi(a_1\ldots a_k (2k+1)
\ldots  (2n-3)(2k))$, the arcs of $\pi(a_1\ldots a_k)$ form the first
connected component of $\pi(\tau(\underline{a}))$, which is followed by
$n-k-1$ minimal arcs (see Section~\ref{sec:arc} for a definition).  Then
the $n^{th}$ arc stretches over the minimal arcs to intersect only the
$k^{th}$ arc of the first connected component of
$\pi(\tau(\underline{a}))$.  Thus, the $(k+1)^{st}$ through $n^{th}$
arcs will not affect any moves among the first $k$ arcs.
Since $\pi(a_1\ldots a_k)$ and $\pi(b_1\ldots b_k)$ must differ by an
adjacent transposition, the recursive construction of the full Gray code
guarantees that $\pi(\underline{a})$ and $\pi(\underline{b})$ will also
differ by an adjacent transposition.

{\noindent\it Case 2b:} $j < k$.
In this case, $a_1\ldots a_{j-1}$ is arc-connected and $2j-1 \le
a_k<2k-1$.  By Corollary~\ref{cor:arc-conn}, we have that $a_1\ldots
a_k$ is arc-disconnected and, by Lemma~\ref{fact8}, the first arc-connected
component of $\pi(a_1\ldots a_k)$ is $\pi(a_1\ldots a_{j-1})$.
We claim that $b_1\ldots b_k$ is also arc-disconnected. This is an
immediate consequence of Lemma~\ref{fact5} when $b_k=a_k+1$.
If $b_k=a_k-1$ then $a_k$ is even and $b_1\ldots b_k$
is arc-disconnected because $a_k$ is strictly greater than
$2j-1$, implying $b_k \ge 2j-1$.

Since $b_1\ldots b_k$ is arc-disconnected and the same holds for
$b_1\ldots b_m=a_1\ldots a_m$ for any $m$ strictly between $j$ and $k$,
$\pi(b_1\ldots b_{j-1})$ is the first connected component of
$\pi(b_1\ldots b_k)$.
Hence, both $\underline{a}=a_1\ldots a_j\ldots a_k (2k+1) \ldots  (2n-3)
(2j-2)$ and $b_1\ldots b_j\ldots b_k (2k+1) \ldots  (2n-3)
(2j-2)$ are arc-connected codes such that there are only
arc-disconnected words between them in the full Gray code of words. Thus
we have $b_n=2j-2$.

In the arc diagram of $\pi(\underline{a})=\pi(a_1\ldots a_k (2k+1)
\ldots  (2n-3)(2j-2))$, the arcs of $\pi(a_1\ldots a_{j-1})$ form the
first connected component of $\pi(\tau(\underline{a}))$; the
$(k+1)^{st}$ through $(n-1)^{st}$ arcs are minimal arcs located at the
right end of the diagram; and the $n^{th}$ arc stretches over the second
through last connected components of the arc diagram, intersecting only
the $(j-1)^{st}$ arc of the first connected component.  Thus, the
$(k+1)^{st}$ through $n^{th}$ arcs will not affect any changes occurring
in the first $k$ arcs.  The recursive construction of the full
Gray code ensures that $\pi(\underline{a})$ and $\pi(\underline{b})$
will differ by an adjacent transposition since $\pi(a_1\ldots a_k)$ and
$\pi(b_1\ldots b_k)$ differ by an adjacent transposition.

{\noindent\it Case 3:} $k=n-1$ (namely, $b_{n-1}=a_{n-1} \pm 1$).
By Lemma~\ref{fact8:prelim}, there exists a $j$ such that $a_1\ldots a_{n-1}(2j-2)$ is arc-connected but that $a_1\ldots a_{n-1}(2j-1)$ is arc-disconnected.
If $\tau(\underline{a})=a_1\ldots a_{n-1}$ and
$\tau(\underline{b})=b_1\ldots b_{n-1}$ are both arc-connected, this is
a degenerate case of case $2a$, where the number of minimal arcs to the
right of the first connected component of the arc diagram of
$\pi(\tau(\underline{a}))$ zero (since $\tau(\underline{a})$
is arc-connected). This does not change the conclusion of the argument.
Similarly, if $\tau(\underline{a})=a_1\ldots a_{n-1}$ and
$\tau(\underline{b})=b_1\ldots b_{n-1}$ are both arc-disconnected, we
have a degenerate case of case $2b$. We are left to consider the case
when $\tau(\underline{a})$ is
arc-connected but $\tau(\underline{b})$ is arc-disconnected, and the
case when $\tau(\underline{a})$ is
arc-disconnected but $\tau(\underline{b})$ is arc-connected.
We will show by way of contradiction that neither of these
cases can occur.

Assume first that $\tau(\underline{a})$ is
arc-connected, but is immediately
followed by the arc-disconnected word $\tau(\underline{b})$ in the Gray
code for words of length $n-1$.  Then by Lemma~\ref{fact8:prelim},
$\tau(\underline{a})=a_1\ldots a_{n-2}(2j-2)$ for some $j$.  Since
$\underline{a}$ is in an increasing run, Corollary~\ref{run-parity}
gives $a_1+\ldots +a_{n-2}+(2j-2)+(n-2) \equiv 1$ mod 2.  Thus,
$a_1+\ldots +a_{n-2}+(n-3) \equiv 0$ mod 2, so $\tau(\underline{a})$ is
in a decreasing run in the Gray code of length $n-1$.  Hence,
$b_{n-1}=a_{n-1}-1=2j-3$.  By Lemma~\ref{fact5}, we get that
$\tau(\underline{b})$ is arc-connected, a contradiction.

Assume finally that $\tau(\underline{a})$ is arc-disconnected, but is
immediately followed by the arc-connected word in $\tau(\underline{b})$
the Gray code for words of length $n-1$.  Then by
Lemma~\ref{fact8:prelim}, $\tau(\underline{a})=a_1\ldots a_{n-2}(2j-1)$
for some $j$.  Since $\underline{a}$ is in an increasing run,
Corollary~\ref{run-parity} gives $a_1+\ldots +a_{n-2}+(2j-1)+(n-2)
\equiv 1$ mod 2.  Thus, $a_1+\ldots +a_{n-2}+(n-3) \equiv 1$ mod 2, so
$\tau(\underline{a})$ is in an increasing run in the Gray code for words
of length $n-1$.  Hence, $b_{n-1}=a_{n-1}+1=2j$.  By Lemma~\ref{fact5}
we get that $\tau(\underline{b})$ is arc-disconnected, a contradiction.
\end{proof}

In the proof of Theorem~\ref{prop1} we have also shown the following
statement.
\begin{proposition}
\label{prop2}
Suppose $\underline{a}$ and $\underline{b}$ as in Theorem~\ref{prop1}.  Then
$a_n=b_n=2i-2$ for some $i\in \{2, \ldots, n\}$, satisfying $a_i=b_i=2i-1$ and
$a_j,b_j \ge 2i-1$ for all $j\in \{i,\ldots, n-1\}$.
\end{proposition}

If we look at the list of words encoding the standard permutations of
the connected Gray code, then each run starts with $a_n=1$ and ends with
$a_n=2i-2$ for some $i$ (or vice versa).  The proof of
Theorem~\ref{prop1} allows us to describe the relationship between the
any arc-connected $\underline{a}$ and the arc-connected $\underline{b}$
immediately following it in the connected Gray code. We see that in the
case when $\underline{a}$ and $\underline{b}$ are in different runs of
the full Gray code, $\underline{a}$ and $\underline{b}$ have the
following properties:
\begin{enumerate}
\item $\pi(\underline{a})$ and $\pi(\underline{b})$ differ by an adjacent transposition,
\item $a_n=b_n$,
\item if $a_n=b_n=2i-1$ for some $i>1$ then $\tau(\underline{a})$ and $\tau(\underline{b})$ are either both connected or both disconnected,
\item if $a_n=b_n=2i-1$ for some $i>1$, $\tau(\underline{a})$ and
  $\tau(\underline{b})$ are both arc-disconnected and $\pi(a_1\ldots a_k)$
  is the first connected component of $\pi(\tau(\underline{a}))$, then
  the first connected component of $\pi(\tau(\underline{b}))$ is
  $\pi(b_1\ldots b_k)$.
\end{enumerate}

As an immediate consequence of Theorem~\ref{prop1}, the sublist of the
full Gray code obtained by simply removing all of the sign-disconnected
standard permutations is also a Gray code.  Hence, by working with the
words which encode standard permutations, we have found a Gray code,
namely the connected Gray code, for the standard permutations which
represent the shelling types of the of the facets of the boundary of the
$n$-cube.

redpajama_set_name: RedPajamaArXiv


Granite is crystalline in structure, so it always has tiny pits or spaces between the various mineral crystals. They are not visible prior to polishing, and usually remain unobtrusive on finished pieces once the surface is highly polished. Granite also contains natural fissures that may appear to be cracks, but they are not structural defects and will not impair the function or durability of the material. They occur naturally and are considered to be part of the beauty of stone.
Granite is not recommended to customers who prefer surfaces that are perfectly uniform in color and pattern, those that are totally free of blemishes. For those individuals, engineered stone would be the preferred choice.
Although granite is very durable when it's installed properly, it's not unbreakable. It can be chipped or cracked if it's struck a sharp blow by a heavy object. It can also break if it's dropped during installation. It is not flexible, and will crack if it is forced to twist or bend. Therefore, granite should only be handled by professionals and must always be adequately supported by proper framing or cabinetry.
Granite is the least susceptible of all natural products to scratches. If not abused, it will hold its luster forever. However, harsh chemicals and abrasive cleaners will dull the surface over time.
Granite will not scorch or burn through ordinary use. It's also resistance to stains. However, a few varieties may absorb some moisture with prolonged contact. Usually, no evidence remains when the liquid is removed and the granite dries, but this could be a problem with dark pigmented liquids or oils. A stone sealer should always be applied to its surface after installation.
Granite is a primordial stone with naturally occurring variations in color, tone, granularity, pattern, etc. These variations, referred to as 'movement', should be expected and are the source of its natural beauty.
Also, keep in mind that the veining in the granite can effect color perception. Sometimes two different grain sizes occurring in the same slab will appear to be of a different color. Mineral concentrations may cause patches that appear darker or lighter.
Consumers who are less acquainted with the material may expect the granite ordered to be identical to the sample they were shown. While the samples are intended to represent the quarry's product, each slab may differ slightly in color and veining. Indeed, even a single granite slab will have color variations from one end to the other. This lack of predictability gives the product its unique character and adds an element of nature into human-designed spaces. Indeed, each specimen is an original artwork.
As with all surfaces, most of the wear on a granite countertop or a marble or slate floor is caused by dirt and grit that tend to scratch the finish over an extended period. So, the more you keep the stone tile flooring or granite countertop or granite vanity top free of these elements, the longer the marble or granite will retain its original sheen. It may also mean less polishing for you to do! Using a damp mop or cloth, and taking care not to over-wet the stone tile floor, granite countertop, or granite vanity top will be enough to make sure that the marble or granite is free of the kinds of material that promote scratching and wear.


Romance and adventure go hand-in-hand when a set of identical twins with powerful psychic abilities work for the military. Sergeant Jinn Dennison goes AWOL when her sister, Julie, is shot and nearly drowned in a Mexican rainforest. When Jinn's mistaken for Julie by her captain, she soon finds herself branded a traitor.
Tab, (Toxic) McTavish never would have believed Julie would ever betray the team, yet something was strange about her all of a sudden. Toxic Tango Troop's mission was to take down a drug dealer by the name of Iceman, and get out of Mexico fast, but a traitor in their midst has given them away. While on the covert mission, he tries to find his kidnapped sister, Tara, but Jinn wants to rescue her own sister, whose life is in jeopardy.
After a long, uneventful day, Jinn was exhausted and tired of waiting for Tab to follow through on his threats. With no chance to get past his defenses and the team never far away, she remained captive.
There'd been no contact from Julie, but the mild pain in Jinn's chest assured her that her sister was alive. They'd found few signs of humans passing, and the trail was getting cold. Many of the teammates were getting frustrated.
Jane Burgess, one of their female soldiers who they called Spitfire, was a prisoner of the enemy and that boded ill for her. From what Jinn had seen with Angel, the rebels were brutal with women.
Angel still hadn't spoken more than a few words since her release. She'd withdrawn into herself as though to block out the world. Tab and Gonzales watched over her. Jinn studied Tab as he fed the woman and realized by the tender care as he wiped some meat juice off her chin with his thumb that he had feelings for her. A twinge of jealousy reared its green head.
Jennings sat down beside Jinn and studied her for a few moments before she spoke, the first of the soldiers to even acknowledge her since she'd been unveiled by Tab.
A flush turned Jennings' cheeks red as she stood.
"As long as your job doesn't jeopardize the rest of us, I'm good with that." Jennings stood and moved off to sit with the others, leaving Jinn to her thoughts.
The hairs on the back of her neck stood up, and she realized that someone was watching her, she looked up and met Tab's brooding gaze. Her lips tingled at the memory of his mouth on hers, his body mastering her on the ground earlier. She licked her lips unconsciously and nearly grinned when his eyes dropped to watch the movement. He might pretend to be impartial, but by the growing bulge in his pants, he was anything but.
Tab turned around and disappeared into the forest. Satisfaction was a sweet tool.


© comes out as '?' when rendered with -O ascii.
2. Look at the copyright notice.
going to use alternative ASCII characters for such symbols then.
enhancement because there is workaround.
No, that doesn't work.  Tried with w3m-0.5.1-4.
US-ASCII in mozilla, it's screwed up a bit.


text:
Escource is een gemeente in het Franse departement Landes (regio Nouvelle-Aquitaine) en telt 606 inwoners (1999). De plaats maakt deel uit van het arrondissement Mont-de-Marsan.

Geografie
De oppervlakte van Escource bedraagt 102,0 km², de bevolkingsdichtheid is 5,9 inwoners per km².

Demografie
Onderstaande figuur toont het verloop van het inwonertal (bron: INSEE-tellingen).

Externe links

Gemeente in Landes

redpajama_set_name: RedPajamaWikipedia


The South Hadley Bike-Walk Committee (BWC) invites the public on a guided walking trek to explore the golf course grounds. The golfing season has closed for the winter season affording an opportunity for safe, non-vehicular passive recreation, including being mindful of the various tees, fairways and greens to protect them during their winter sleep while walking, snowshoeing and cross-country skiing. Overall, the trek is intended as a fun, orientation walk on selected paths that are relatively flat with a few minor-moderate ascents and descents. Total trek time is approximately 90 minutes. Post-trek, please join BWC members in the parking lot for light refreshments and socializing! Be aware that the club facilities are closed for the season, including restrooms. Non-enjoyable or hazardous inclement weather/conditions will force a cancellation; if that should appear necessary, please check the Town's website (www.southhadley.org) for a Community News notification.


Discover ways to release blockages, heighten intuition, clear programmed patterns and limited beliefs, so that you can ultimately live the life you were destined to enjoy. Working within a heart-centred group enhances the dynamic shifting process, where the transformation is amplified.
Want amazing results unleash success in 2019?
Lack confidence and trust in decision making?
Frustrated with inability to follow through, and commit to completion of ideas- projects-goals?
BREAKTHROUGH the top 3 blocks, that have created resistance in 2018, discover the key strengths that will help to unleash ultimate success in 2019.
Discover HOW to unlock, attract and accelerate your aligned path of prosperity, by releasing the patterns that spin our wheels, and unfulfilled intentions from the beginning of 2018.
Discover HOW to unlock, attract and accelerate your aligned path of prosperity.
Discover the 4 keys to being a radiant woman... Move through doubt, not being seen, and not being understood.
Feel understood, respected and honoured for YOU, your gifts, talents, skills and experience.
As we release the past conditioning, where our mothers, grandmothers, and other female ancestors were held back, through culture, society and conditioning.
When you let go of doubt and fear … you are free to embrace Trust.
Want amazing results in 2018?
Mariangela is a true healer. I had seen so many other therapists,but she was the only one that made a difference.


text:
Prophecy: What the Future Holds for You
By Sylvia Browne , with Lindsay Harrison
Read by Jeanie Hackett
Sylvia Browne Highbridge Audio, HighBridge
Category: Nonfiction/Body, Mind & Spirit
Publisher: Highbridge Audio Publisher: Highbridge Audio Publisher: Highbridge Audio
In Prophecy, Sylvia Browne turns her psychic wisdom to the puzzling, often contradictory predictions proposed by major historical and contemporary figures, ranging from biblical prophets and Nostradamus to George Washington and NASA scientists.
Examining the most famous—and notorious—prophetic voices throughout the ages, Browne offers a clear and fascinating vision of the world as it will be from five years into the future to five hundred, with a startling and revealing look into the future of our nation and our world—and a beacon of hope in these uncertain times.
Author Bio: Sylvia Browne
Sylvia Browne is a New York Times bestselling author and popular psychic. She has appeared regularly on Larry King Live and the Montel Williams Show and has made countless other media and public appearances. She is also the founder of a church, the Society of Novus Spiritus, which celebrated its twenty-fifth anniversary in 2011.

redpajama_set_name: RedPajamaCommonCrawl


The 3M HB900 is a sturdy dispenser made of rugged, high impact-resistant black plastic that stands up to the toughest work environments and looks good doing it. It easily handles film tapes up to 1 inch wide x 72 yards on a 3 inch core. The non-slip weighted base keeps the dispenser in place, permitting one hand to pull and tear the tape while freeing the other hand to hold the work.


text:
100+ new creatures coming soon to Pokémon GO
Christian Zibreg on November 28, 2016
After analyzing new code in a Pokémon GO update coming next month, The Silph Road has discovered that developer Niantic Labs will be adding more than a hundred new creatures to the popular augmented-reality game, including normal and legendary ones.
In addition to the new species, the code confirms other new features like Pokémon trading, player vs. player battles, new ways to raise/train your monsters and more.
The most recent Pokémon GO update added a Nearby Pokémon feature, the purple shape-shifting Pokémon, called Ditto, and other perks.
So-called "moveset data" for the hundred new species has yet to appear on the servers. A client-side update will be required to pave the way for a whole bunch of new monsters to begin appearing in-game.
Niantic did not officially confirm the new creatures nor did it say when they might appear in the game. For what it's worth, some sources have called for a major update to the game next month, ahead of Christmas.
Pokémon GO is a free download from the App Store.
Source: The Silph Road
Lara Croft GO gains new levels and iMessage sticker pack via Mirror of Spirits update
Pokémon GO celebrates Halloween with first in-game event
Spotify to bring a block feature for muting artists you don't want to hear from


Knowing the different types of ladders is the first step to prevent injury. Each different type of ladder has its own ability, so knowing the abilities and functions will help prevent injury to yourself in the long run. Before using a ladder, make sure all the components, i.e. bolts, screws, steps, are attached properly, etc.
Always face the ladder when walking up and down the steps, to lower the chances of slipping and falling. When setting up your ladder, make sure it is set up on a secure foundation, so that as you climb up, the ladder doesn't move side to side and have that potential of falling over with you on it.
If you are on a roof, be aware of where the ladder is at all times. If it does fall, and you don't know it, you could unwittingly go to step back on it with nothing being there, thus leading to a fall. Finally, wearing the right kind of safety gear is vital as well. Gear like hard hats, eye protection, heavy duty boots, and gloves will help keep you steady on the ladder and will prevent injury to your head if you do fall.
Keeping these safety tips in mind will help you to remain injury free, and complete your projects in a reasonable amount of time, and with the best outcome.
Diminish your exposure to the EMFs you find. Determine how far you must stay away from the EMF emitters in your home and work environment to achieve less than 2.5 mG of exposure — the microwave oven, the alarm clock, the computer, and so on. Rearrange your furniture (especially the beds, desks, and couches where you spend the most time) away from heaters, wiring, fluorescent lights, electric doorbells, and other EMF "hot spots." Where practical, replace electrical appliances with non-electric devices. Have an electrician correct faulty high EMF wiring and help you eliminate dangerous stray ground currents.
If you have a pool and young children as well, what is a danger that can fall upon these young ones? Drowning. What can you do to prevent this from happening? Well, in this article we will talk a little bit about a specific device to detect surface waves in your pool, whether it be little children unsupervised or the family pet.
Basically the device is called a pool alarm. There is one contact that sits on top of the water, and one below it. When both contacts detect surface waves in your pool, the alarm goes off. The nice thing about it is that you can also decrease or increase the sensitivity of the alarm. With this in mind, you can more accurately tell who or what is swimming in your pool, unsupervised or not. A drawback to the alarm however is that you may forget to turn it on, so you may not hear what is going on in your pool until it's too late.
It should be noted that this device should NOT be used as a stand alone safeguard for your children or pets who go into your pool unsupervised. Using other strategies is not only a good idea, but it will benefit your household in the long run, and keep everyone safe.


Amazon has discounted the eBook version of my novel PREDATORY INSTINCT by 32%! If you've been sitting on the fence, now's the time to take the plunge.
This entry was posted on March 7, 2012 by evilanche. It was filed under Uncategorized .


text:
EntertainmentMovies, Celebs and TVArt & DesignArt and The ArtsLiteratureBooks, Genres, ReviewsMusicInstruments, Musicians & GenresVideo GamesPC & Console Gaming
Movie Review: The Right Stuff (1983)
by StevenHelmer
A review of the 1983 docudrama starring Sam Shepard and Scott Glenn.
Synopsis: Six years after the sound barrier is broken, the United States competes with the Soviet Union to be the first nation to send a man into space. Already falling behind, NASA recruits seven test pilots and begins training them to be the nation's first astronauts in the experimental Mercury space program.
Who's In It?
The movie stars Sam Shepard, Ed Harris, Scott Glenn, Dennis Quaid and Fred Ward.
My wife had to work Friday night and, as always, I looked for a movie to watch while waiting up for her. I had not seen "The Right Stuff" in forever and, knowing it was a movie nobody else in my family would watch with me (mostly because of it being more than 3 hours long), decided to put it on.
I was about 10 years old the last time I had seen this particular film and didn't remember much about it as a result. However, it turned out to be a pretty decent choice.
I think the thing I really like about this movie is the way it presents history through the eyes of the men who made it. A movie about the space race would be somewhat entertaining to begin with. But, presenting that aspect as more of a background story and, instead, focusing on the astronauts and their families, really made me feel like I was there, despite the fact it was about events that predated me by about two decades.
I found the part about the astronauts' wives particularly intriguing because, while there was still some pride, there was quite a bit of negative to go with that role, especially with the media pounding on their doors all the time. I found it easy to feel sorry for them, especially since you rarely hear anything about them in history books.
Probably my only real complaint about this movie was the extra emphasis on Chuck Yeager's (Shepard) career. While his historical contributions were rightfully not overlooked, the movie does seem to go out of its way to make sure he gets the last laugh, even after the film shifts its focus from setting speed records to the space race he wasn't invited to join. My only thought with that is maybe there's something in the book this movie is based off of that explains that better and I may have to look for it the next time I'm at the library as a result.
Final Opinion
As I said, this movie is over three hours long and you really do need to have the time to watch it from start to finish. However, if you have the time to watch it, it's worth doing so.
More By This Writer
Murder By Chili (Maxwell V. Diamond Book 1)
Lester Greene was the favorite to win the Emerald Springs' annual chili cook off and potentially sell his recipe to the Ma Kettle Canned Chili Company. However, when he suddenly...
Journey to the Bubble Gum Planet
The Honest Company Truly Calming Lavender Bubble Bath Tear Free Kids Bubble Bath Naturally Derive...
Sunstar 3614 GUM Go-Betweens Proxabrush Cleaners Wide, 10 Count
Updated: 04/18/2016, StevenHelmer
Movie Review: Nancy Drew and the Hidden Staircase (1939)
A review of the 1939 movie starring actress Bonita Granville as detective Nan...
Movie Review: Halloween (2007)
A review of the 2007 remake starring Malcolm McDowell and Scout Taylor-Compton.
Movie Review: The Return of the Vampire (1943)
A review of the 1943 horror movie starring Bela Lugosi as the vampire Armand ...
StevenHelmer
1065 Pages, 84%
Movie Review: Exiled (1998)on 01/23/2021 Law & Order: Special Victims Unit Episode Review: The Long Arm of the Witnesson 01/22/2021 Review of Walker Pilot Episodeon 01/22/2021 Review of Luige's Western Omelette Pizzaon 01/21/2021
23% 0 4 StevenHelmer
A review of the 1939 movie starring actress Bonita Granville as detective Nancy Drew.
A review of the 1943 horror movie starring Bela Lugosi as the vampire Armand Tesla


Spotted an error on the main site? Let us know!
A forum for all kinds of forum games. Your post count is not increased here. This doesn't allow spamming, however.
Split-personality mystery! Can YOU discover who the werewolves are?
Page created in 0.144 seconds with 15 queries.


The original Bad Girls of R&B aka Total experienced a resurgence after being a part of the Bad Boy reunion tour in 2016 and 2017. Original members Pam and Kima, who had been touring on their own as Total prior to that, even got Keisha to finally join them for the comeback.
One thing we haven't got from the group is new music.
However, Pam had put out singles over the years, and Kima even gave us the single "Distraction" featuring Vita a couple of years ago.
Kima has now just released the visual for her latest single "Luv Me Back" which she put out earlier this month.
Stay tuned as more details emerge about a potential project.


text:
Brian S. Rothman
Contact Attorney
Call : 305-662-8005 Email : br@marcosandrothman.com
Mr. Rothman received a Bachelor of Arts Degree in Criminal Justice from the University of Florida (1985). Brian received his Juris Doctorate Degree from Stetson University College of Law in 1987
Brian is the managing partner of the Firm and practices workers compensation defense in the South Miami office. He received a Bachelor of Arts Degree in Criminal Justice from the University of Florida in 1985 and his Juris Doctorate Degree from Stetson University College of Law in 1987 where he served on the Moot Court Board. Prior to starting the firm in 1993, Brian worked at an insurance defense firm in both Tampa and Miami practicing exclusively workers compensation.
Brian's litigation expertise results from thirty years of practice handling complex medical/legal issues and successfully defending insurance carriers and self-insured companies through every stage of the litigation process. As a result of his outstanding negotiation skills, he has established an exceptional track record effectively diminishing his clients' exposure.
Brian has lectured on both fundamental and novel issues affecting workers' compensations claims and construction and trucking litigation. His seminars include: Litigation Issues for Managers in the Workplace; Expiration of Temporary Benefits and PTD Under the New Law; Litigation Roundtable: Truck Accident Investigation/Preservation of Evidence; and Legislative Changes to the Workers' Compensation Law.
Brian is responsible for oversight of the firm's business and financial operations, including human resources, recruiting and professional development.
Brian is a supporter of local charitable organizations and he enjoys cycling, traveling, watching any sporting event involving the Gators and spending time with his family. He is a native Floridian and he attended every home game of the Dolphins' 1973 perfect season when he was 10 years old.
The law firm of Marcos Rothman Valdes & Goldstein, P.L. is an insurance defense firm practicing primarily in Miami-Dade, Broward, Palm Beach, Martin and Monroe counties. Our firm provides statewide defense litigation coverage to select clients at their request. Specifically, we provide insurance defense coverage for Personal Injury and Workers' Compensation.
© 2017 Marcos Rothman Valdes Goldstein, P.L.


The final temptation of power was too much. He hadn't considered the cost.
Little does he know, his every move is being recorded. Every misstep is being judged. As he gets ever closer to winning over his demon, heavenly eyes watch from above. Some root for his success while others hope he'll fail.
Will Joe win out? Will Michael be able to save Joe's soul?
The image of the Soul and Belath fades into a cloud of smoke as Prophecy breaks the psychic link to that moment in time, called an event. It takes few minutes for me to gather my wits. The only thing worse is the Soul Memory Wipe. At the beginning of a Judgment Council, every instance of the Soul on trial is removed from the Defense and Prosecution Ministers' memory.
The Memory Wipe is to remove any possible bias, including family affiliation; some Souls have high-ranking Angelic and Demonic parents. Many millennia ago, before our current Grand Master, an Angel convinced the Prosecution Minister to lose a case, allowing for an easier Paradise ruling for his son. When that Grand Master became aware of it, nothing could be done as all rulings are final. To ensure this never recurred, he declared all memories of the Soul would be removed before trial and would be returned after the final ruling.
Removing memories of the Soul also forces the ruling to be made only on the events brought before the Judgment Council. Actions that can equally be argued for good or evil are brought to the attention of the Defense and Prosecution Ministers. It is their responsibility to argue these events to their own favor. The Defense highlights benevolent portions of the event and attempts to show the actions are justified by the end result. The Prosecution, on the other hand, spotlights even the slightest of grievances in an attempt to force the Soul to The Abyss. Grand Master is aware of actions that are primarily benevolent, as well as those primarily malevolent, and includes those when making his final ruling. After the arguments have been made on the event, the memory is removed from each Minister. This means each event brought to the Judgment Council is argued as if it is the only thing separating the Soul from Paradise and The Abyss.
Sahiva is much more experienced and always beats me to the punch with initial arguments. I have grown so accustomed to this I tend to let her take the first strike even when I have a great case. My father taught me, "He who strikes before gaining his footing is most often felled by a patient sniper," so I have learned to let her arrogance be my greatest ally. As expected, she goes for an easy target.
"This Soul killed three men whom he knew he had outmatched. He even tells them 'this is a fight you cannot win,' only to massacre them in a matter of moments. He didn't even break a sweat when speaking to the last man; he just walked up and cleaned his knife on the guy's pants. There is no doubt this is malicious," Sahiva proclaims in her standard, near grandstand manner.
With this opening, I am confident I can argue this event in my favor, but I must not let it show on my face. I wait patiently as Sahiva flails her arms, yelling in her over-dramatic style, and finally she takes her place in the Prosecution Box. No objection is needed; she has opened the door for my win. I only have to tell the story right. I calmly walk to the center of the hall and take my place in the Speaker's Box.
"I am glad you emphasized the fact he killed three men, Sahiva. I would also like to thank you for pointing out he did not kill the fourth man after he was no longer a threat," I say, nodding to Sahiva in feigned gratitude. Turning back to Grand Master, I continue.
"This proves this Soul was only defending himself. It can further be argued he wasn't defending only himself, but also defending the man he was with, Belath." I pause to take a breath; I can't let the feeling of victory show on my face. I take a brief second to compose myself and turn to those in the Colosseum seating in an attempt to hide my face from both Sahiva and Grand Master.
"As you can see by the event, Belath is an older gentleman, possibly incapable of defending himself in situations such as these. Even if he is able to defend himself, remember the Soul stated, 'I don't want to know you.' This tells us he does not know the man, thus showing he is not aware of any defense or fighting abilities he may or may not possess. I submit he was only defending the weak, thus making this event benevolent." I give my prescribed bow and return to my seat.
Grand Master seems to have been moved by my position because he gives me an applauding nod before turning back to Sahiva.
"Does Prosecution have further cross arguments?" He asks in a tone that makes me feel he doesn't think cross is necessary.
"But of course, Grand Master, thank you for allowing it," she says, giving a sarcastic curtsy, knowing cross is allowed, if not required, for all events. She takes her place in the Speaker's Box and continues.
"Objection! Those men will have their own Judgment Council," I protest more childishly than I had hoped.
"Grand Master, if the motives of these men cannot be determined as malicious, we cannot simply assume the Soul was acting in righteous anger." Sahiva is always quick on her feet.
"I agree with the Defense Minister. Sahiva, please keep your arguments based on the actions of the Soul in question," Grand Master says in his judicial voice while again giving me an approving nod.
"As you wish. I have nothing further, Grand Master," she says as she almost stomps back to the Prosecution Box.
"Defense Minister, do you have anything further?" Grand Master asks in a tone that tells me I have nearly won this event's argument.
"I have only one more thing to mention, Grand Master. I feel I must point out the Soul stayed with Belath after the attack. I argue this was to protect him against possible attacks in the future. Please take this into consideration when making your decision." I finish my arguments and hope for the best.
"I most certainly will take that into account, Defense Minister. With that final argument, I rule this event as benevolent. We will take a short recess, during which you are to eat and go through a Soul Memory Wipe for this event. We will return in two hours. You are dismissed. Prophecy, be sure the next event is ready when we return." And with that, Grand Master and Council retire to the antechamber.
Josh has had a knack for writing from a young age; mostly poetry.��The Demon Within�is his debut novel and, according to Josh, nearly wrote itself.� He is a US Army veteran and has been deployed to the Balkans and Middle East.� He has been an IT professional for about a decade.
Many of the events in�The Demon Within�were taken from Josh's life and "put through the fiction blender" as he puts it.� When pressed for more details, he said he couldn't give specifics for fear of "giving spoilers," but, he did say the book includes fictional spins on things ranging from childhood bullying to being placed in the Las Vegas foster care system.
He currently lives in Columbus OH with his family and is working on book two of his 'The Last War' series.
This giveaway is for one print copy of the book and a $50 gift card to a U.S. winner.� The ebook with $50 gift card is open worldwide.� This giveaway ends on April 27, 2018.� Entries are accepted via Rafflecopter only.
Congrats on the tour and thank you for the excerpt and giveaway. I appreciate the opportunity to win.
This sounds like a book I'd really enjoy reading, thank you for sharing!


early 14c., in a theological sense, "religious truth via divine revelation, hidden spiritual significance, mystical truth," from Anglo-French *misterie, Old French mistere "secret, mystery, hidden meaning" (Modern French mystère), from Latin mysterium "secret rite, secret worship; a secret thing," from Greek mysterion (usually in plural mysteria) "secret rite or doctrine," from mystes "one who has been initiated," from myein "to close, shut" (see mute (adj.)); perhaps referring to the lips (in secrecy) or to the eyes (only initiates were allowed to see the sacred rites).
The Greek word was used in Septuagint for "secret counsel of God," translated in Vulgate as sacramentum. Non-theological use in English, "a hidden or secret thing," is from late 14c. In reference to the ancient rites of Greece, Egypt, etc. it is attested from 1640s. Meaning "detective story" first recorded in English 1908.
"Marvellous, indeed, is the mystery of our being," exclaimed Anaxagoras.
From that day the fate of Leichardt and his companions has been involved in mystery.
It is the old land of mystery and wonder which the Greeks called Mesopotamia.
The difference too is radical; it goes to the heart of the mystery.
Will Nature teach them the mystery of a plate of turtle-soup?
Like a sentinel on that solitary plain it overwhelms me with a sense of mystery.
What is mystery for one age is not of necessity mystery for another.
But the mystery was to remain unsolved, since Gilder now entered the office.
She felt that this mystery held in it something sinister to herself.
He breakfasted at Mrs. McKee's, and was initiated into the mystery of the ticket punch.


text:
A werewolf can't get a second date in "Mike Tyson Mysteries"
Did you know that you can watch Mike Tyson solve mysteries and punch monsters in an Adult Swim series animated in the style of Hanna-Barbera classics like Scooby Doo? Well, thanks to a kind and benevolent universe, you can, and the latest episode is about a werewolf, leopard print boots, and pie.
"Mike Tyson Mysteries" really does feature the voice of Mike Tyson, along with Norm Macdonald, Rachel Ramras, and the incomparable Jim Rash. Episode 9, "Night Moves", airs tomorrow (February 1st) at 11:30 PM (Eastern and Pacific) on Adult Swim, but if you want to catch it a few hours early, it's available right now on Amazon. Here's the synopsis, via Toon Zone:
When Mike and the gang try to solve the mystery of why a likeable guy named Andrew can never get a second date, Yung Hee discovers the hard way that it's probably because he's a werewolf. She makes it home alive, but will she stay that way once that the gang becomes convinced that she might now be a werewolf herself? And why does Mike Tyson have a drawer full of silver bullets? Run, Yung, run!
I watched it and I laughed a lot, particularly at Tyson's non-sequiturs and malapropisms, which are delivered to his incredulous colleagues via mumbly, awkward and utterly charming line readings. The werewolf design and vocalizations were great (bonus points awarded for his name), and now I really really want to eat pie.
Angela Quinton	Jan. 31st 2015
In Film, Television & Music, Pop Culture
#animation #Humour #romance


text:
Communication with traffic police
By	Swetalana
How to apply for the examination, the traffic police through public Services
The most convenient way of writing the exam or re-sit in the traffic police - through the national portal of public Services. You can sign up for the exam for obtaining both Russian and international law.
How to register for exams in the SAI
How to pass the practical exam in the traffic police
How to register on the portal of public services?
- access to the portal of public Services;
- medical certificate;
- a document confirming the registration of candidates for the drivers;
- the document proving the identity;
- application for issuance of driver's licenses;
- the document about passing of training;
receipt about payment of state duty.
Log in to the portal Gosuslugi and go to the most popular services. Select the tab "licensing" and navigate to the page of the description of services.
Then click the right blue button "Obtaining services". You will be asked to read the rules of submission of the application electronically. Here you need to tick that you are aware of the proposed order, and click the button "Next".
In the window that appears, choose your region and the type of service (receipt of Russian or international rights). Check category for which you have been trained and are planning to get right. Click "Next".
The next step is to complete and verify your personal information. It is also necessary to specify information about the driving school and a medical facility that issued the certificate. Required fields include: name of school; date and number of certificate of completion of a driving school; the name of the medical facility; the license number of the medical institution; a medical certificate; the date of passing the examination; has category.
Then specify the date and convenient for visiting address and examination divisions of traffic police. To visit update on available appointment times.
After completing all the steps save the electronic ticket on the service receipt and a completed application form. You can also pay a state fee or to print a receipt in the dashboard, in the masonry of "My accounts".
Users of the portal of public Services there is also the possibility of registration or replacement driver's license, opening a new driving category and getting help on the driver's licence.
The procedure of registration of the vehicle in the traffic police via the Internet
How to pass the exam in the traffic police of the city
How is the exam in the SAI
How to obtain a license in another city
How to pass your driving exam
How to get a driver's license in Moscow
How to register a new car
How to restore the card to the driver
Where to sell an old car
Selling an old car can be quite time-consuming, difficult work. In fact, if you disassemble...
Indicators yellow car dashboard
Useful tips on how to save money on gasoline
How to choose a loan on the car for 1 million rubles
How to get rid of sleepiness at the wheel


Crans, John H., factory owner — Portland — born in Christine, North Dakota, July 15, 1877. Came to S. Bend, Washington, in 1890 and stayed there for a few years. Later he stayed in Portland and in Southern Oregon. Since 1906 he has lived in Portland, where he is half owner of Portland Top Co., 306 E. Morrison St., which manufactures roofs et cetera for carriages. Member of the Independent Order of Odd Fellows. In 1900 he married Minnie Hart; they have three children.


If there is a best way to travel around Jeju, I would have to say that it is by bicycle. I have done it twice now, just to make sure. First I did it with one friend, stopping to drink some beers along the way, then with a group of friends, stopping to swim and jump off fences. Having a circumference of a couple of hundred kilometres, it is a pleasant two day ride, but can be done in one if you are a killer cyclist, or more than two if you like to trundle. On both my circumnavigations, I cycled half of the island each day.
This is a little taste of cycling around Jeju. Don't think about it too much, don't plan it, just get a bicycle and do it. Stick to the coast roads when you can and you will have a great time. Happy cycling.
I am planning a trip there in early aug this year. Heard it may not be the best time to cycle there because of the hot and humid weather, as well as being typhoon season. But I wanna go during that time so the water will be warm enough for swimming. SO shld I risk it?
The summer is super warm and humid, but if you're there, go, enjoy it, and the water should be lovely.
Great idea! I think this is how I will spend one of my long weekends off while teaching in Korea. Did you rent bikes or bring your own?
I bought a decent second hand bike in Seogwipo – there is a little shop with a friendly man who fixes up old bikes and sells them on. People also sell a lot of bikes through the FB groups. Have fun!
Thanks for your post. Love the up in the air shots. How can you guys jump so high!
Btw, what are the name of those facebook groups?
I don't remember anymore, but if you search around on FB I am sure you will come across them.


Found 217 tutors in West Grove, PA 19390 – page 1 of 22.
I am a certified K-12 teacher and have over seventeen years of experience in education. I taught high school French and English as a Second Language high school classes for seven years in Delaware and Texas.
I've been fascinated by foreign languages and foreign cultures all my life. I majored in German in college, and then went on to get a PhD in Russian. I was a professor of Russian at the University of Missouri for 31 years.


Can I clear a title overview?
I want clear my title overview for a new phase (instead of create new title or delete all player), Is there a way to do that?
No, we don't provide a way to clear the metrics that drive the overview dashboard. If it's important that you start from a blank slate, the best alternative is to create a new title under your studio. We have heard feedback that it would be great if the backend settings from one title were easily portable to another title. It sounds like such a feature would be beneficial in your case as well.


My wife and I love to minister to people the love of Jesus, His grace and His power to heal. We were pastors of a church which we helped plant for about 12 years.
Bill is a teacher at a local college, but also loves to teach followers of Jesus from the scripture how to minister to one another.


The Holiday-Inn-turned-Days-Inn-turned-Heaven-on-Earth-Hotel at 801 St. Joseph Pkwy. has now been posted to CoStar by commercial broker CBRE following a 2-week period of higher-than-usual on-site cleanup activity. Crews removed debris from its garage and pool late last week; and just yesterday, 2 workers scaled nearly all 31 stories of its western face to take down the semicircular light box of the Days Inn sign that once hung near the top of it (see photo above).
The seller, SFK Development, bought the property in 2012 — becoming the building's third owner in its post-Maharishi era. (The building — which for a time was home to a Vedic school run by the former Beatles spiritual advisor — was shut down by the city in 1998.) Since then, it has been fertile ground for numerous urban explorers, as well as the imagination of several would-be redevelopers.
How about a rendering of a huge pile of rubble? From what I've read in this venue, the building is not salvageable.
that building has been an eye sore since I moved here in the 90's. I cannot fathom how or why the city put up with it for over two decades. Houston is a thriving and growing city, not Detroit.
The city should look into converting this into a residential tower for middle class workers that either work downtown, or would like to, but can't afford the high cost of most of the new residential projects that have gone up in the last three years.
That would be an awesome implosion.
Just in time for the fracking boom and the big $15k per unit downtown living initiative. Good thing they are on this before anyone else starts building residential housing downtown. Oh, wait . . .
I will never understand commercial real estate.
I don't know who would be so gullible to buy this turkey from SFK Development (the current owner). Even after that quick cleanup, the place still looks like a fresh hot mess. Well, not so fresh.
I'd pay $5 towards an implosion. Somebody should start a GoFundMe campaign for it.
@Bob Russell Please, no. The taxpayers are already in debt for another 100 million to turn the Astrodome into a historically significant parking lot.
Raise your hand if you're old enough to remember when this was a functioning Holiday Inn. (Raising hand.) Would love to see it repurposed…ya kinda get used to things after awhile.
Very tired of this building; I just feel like it's in the wrong spot.
Repurpose site to house new greyhound/megabus terminal with hotel/hostel and retail. Great job for METRO to tackle. The transit agency could buy and redevelop site, with METRO owning it giving MPD jurisdiction. This would end the vice associated with current location.
My Dad was the Innkeeper when the Holiday Inn opened, and we lived in the building for a couple of years. It was quite a place, with the Orbit Lounge, one of A. J. Foyt's Indy cars in the lobby, and so on. But its day has passed.
@Sinplysid: That is brilliant! How can we make this happen? I'm in.
METRO doesn't have enough funds to enact its own plans, let alone relocated and take over a private bus operators' facilities over which the agency has zero authority. Where do folks come up with this craziness?
'Reposition' the folks living under 59 over by BBVA and midtown Sears into rehabbed units. Two birds, one stone.
Patrick, you need to write your story – the screenplay will rival Urban Cowboy!
Last night prior to the rain storm there was some strong winds that blew glass and metal debris over the block south of it. The portion of St. Joseph Pkwy. in front of the bldg was covered in broken safety glass with crumpled mullions littering the sidewalk in front of the Lonestar bar's parking lot.