text:
\section*{Introduction}

	\title[Automorphisms of left Ideal relation graph over full matrix ring]{Automorphisms
of left Ideal relation graph over full matrix ring}
	\author[Jitender Kumar, Barkha Baloda, Sanjeet Malhotra]{$\text{Jitender Kumar}^{^*}$, Barkha Baloda, Sanjeet Malhotra}
	\address{Department of Mathematics, Birla Institute of Technology and Science Pilani, Pilani, India}
	\email{jitenderarora09@gmail.com, barkha0026@gmail.com, thesanjeetmalhotra@gmail.com}

	\begin{abstract}
		 The left-ideal relation graph on a ring $R$, denoted by $\overrightarrow{\Gamma_{l-i}}(R)$, is a directed graph whose vertex set is all the elements of $R$ and there is a directed edge from $x$ to a distinct $y$ if and only if the left ideal generated by $x$, written as $[x]$, is properly contained in the left ideal generated by $y$. In this paper, the automorphisms of $\overrightarrow{\Gamma_{l-i}}(R)$ are characterized, where $R$ is the ring of all $n \times n$ matrices over a finite field $F_q$. The undirected left relation graph, denoted by $\Gamma_{l-i}(M_n(F_q))$, is the simple graph whose vertices are all the elements of $R$ and two distinct vertices $x, y$ are adjacent if and only if either $[x] \subset [y]$ or $[y] \subset [x]$ is considered. Various graph theoretic properties of $\Gamma_{l-i}(M_n(F_q))$  including connectedness, girth, clique number, etc.  are studied.
		 \end{abstract}

	\subjclass[2010]{05C25}

	\keywords{Matrix ring, left ideals, graph automorphism, finite field\\ *  Corresponding author}

	\maketitle

\section{Introduction}
The study of algebraic structures through graph theoretic properties has emerged as a fascinating and important research topic in the past three decades. A lot of graphs associated with rings such as zero divisor graph, co-maximal graph, commuting graph, inclusion ideal graph, and intersection ideal graph  have been studied ( see \cite{akbari2004commuting, akbari2015inclusion, akbari2006zero, akbari2013some, anderson1999zero, beck1988coloring, redmond2002zero, sharma1995note, wang2009co}). Sharma and Bhatwadekar \cite{sharma1995note}, introduced the graph $\Omega(R)$, for a ring $R$, whose vertices are all the elements of $R$ and two distinct vertices $x$ and $y$ are adjacent if and only if $Rx + Ry = R$. They showed that for a commutative ring $R$, the graph $\Omega(R)$ is finitely colorable if and only if $R$ is a finite ring. Miamani \emph{et al.} \cite{maimani2008comaximal}, studied $\Omega_2(R)$ a  subgraph of $\Omega(R)$ whose vertex set is non-unit elements of $R$ and discussed its connectedness and the diameter.
Meng Ye \emph{et al.} \cite{ye2012co} modified the definition of $\Omega(R)$, and instead of using the elements of $R$ as a vertex set, they used vertices to be the ideals which are not contained in the Jacobson radical of $R$, and two vertices $I_1$ and $I_2$ are adjacent if and only if $I_1 + I_2 = R$. It is referred as the co-maximal ideal graph of $R$. It has a diameter less than or equal to three, and the clique number and the chromatic number of $\Omega(R)$ is equal to the number of maximal ideals of the ring $R$. Inspired by the above mentioned work, and for the idea of revealing relationships between ideals of a ring $R$ and elements of $R$, in \cite{ma2016automorphism}, X. Ma \emph{et al.}, defined a
directed graph, the ideal-relation graph of $R$, written as $\overrightarrow{\Gamma_{i}}(R)$, whose vertex set is $R$ and there is a directed edge
from a vertex $x$ to a distinct vertex $y$ if and only if the ideal of $R$ generated by $x$ is properly
contained in the ideal generated by $y$.
In this paper, they discussed the automorphisms of $\overrightarrow{\Gamma_{i}}(R)$, where $R$ is the ring of all $n \times n$ upper triangular matrices over a finite field $F_q$. The symmetries of a graph are described by its
automorphism group. In general, automorphism groups are important for studying sizeable graphs since these symmetries allow
one to simplify and understand the behavior of the graph. Although, the determination of the full automorphism group is a challenging problem in algebraic graph theory. Recently, the automorphisms of the zero-divisor graph over the matrix ring attracted the attention of researchers (see \cite{ma2016automorphisms, ou2020automorphism, wang2016automorphisms, wong2014group, zhou2017automorphisms}). Also, Feng Xu \emph{et al.} \cite{xu2020automorphism}, determined all the automorphisms of the intersection graph of ideals over a matrix ring.  Automorphisms of the total graph over matrix rings are characterized in \cite{wang2020automorphisms, zhou2017automorphism}. Further,  D. Wong \emph{et al.} \cite{wang2017automorphisms}, characterized the automorphisms of the co-maximal ideal graph over matrix ring.

Motivated by the work of \cite{ma2016automorphism, xu2020automorphism}, we consider the left ideal relation graph over full matrix ring. The left-ideal relation graph $\overrightarrow{\Gamma_{l-i}}(R)$ of a ring $R$ is a directed simple graph whose vertex set is all the elements  of $R$ and there is a directed edge from $x$ to a distinct $y$ if and only if the left ideal generated by $x$ is properly contained in the left ideal generated by $y$. The paper is arranged as follows.  In Section 2, we state necessary fundamental notions and fix our notation. Section 3 comprises the results concerning the automorphisms of left-ideal relation of matrix ring  over a finite field $F_q$.  In Section 4, we study various graph invariants of undirected left-ideal relation graph  $\Gamma_{l-i}(R)$ viz.  girth, dominance number, independence number and clique number etc.

\section{Preliminaries}
In this section, we recall basic definitions from \cite{westgraph} and fix our notation which will be used throughout this paper. Let $\Gamma$ be a graph, $V(\Gamma)$ and $E(\Gamma)$ be the vertex set and edge set, of $\Gamma$ respectively. The \emph{distance} between two vertices $u, v$ in  a graph $\Gamma$ is the number of edges in a shortest path connecting them and it is denoted by $d(u, v)$.  For a vertex $v$, the \emph{eccentricity} of $v$ is the maximum of the distance to any vertex in the graph. The \emph{diameter} is the maximum of the eccentricity of any vertex in $\Gamma$ and the \emph{radius} of $\Gamma$ is the minimum eccentricity among all vertices in the graph $\Gamma$.
The \emph{degree} of the vertex $v$ in $\Gamma$ is the number of edges incident to $v$ and it is denoted by $deg(v)$. A \emph{cycle} is a closed walk with distinct vertices except for the initial and end vertex, which are equal and a cycle of length $n$ is denoted by $C_n$. The \emph{girth} of $\Gamma$ is the length of its shortest cycle and is denoted by ${g(\Gamma)}$. A connected graph $\Gamma$ is Eulerian if and only if degree of every vertex is even {\cite[Theorem 1.2.26]{westgraph}}. The \emph{chromatic number} of $\Gamma$, denoted by $\chi(\Gamma)$, is the smallest number of colors needed to color the vertices  of $\Gamma$ so that no two adjacent vertices share the same color. A \emph{clique} in $\Gamma$ is a set of pairwise adjacent vertices. The \emph{clique number} of $\Gamma$ is the size of maximum clique in $\Gamma$ and it is denoted by $\omega(\Gamma)$. It is well known that $\omega(\Gamma) \leq \chi(\Gamma)$ (see \cite{westgraph}). 	A subset $D$ of $V(\Gamma)$ is said to be a dominating set if any vertex in $V(\Gamma) \setminus D$ is adjacent to at least one vertex in $D$. If $D$ contains only one vertex then that vertex is called dominating vertex. The \emph{domination number} $\gamma(\Gamma)$ of $\Gamma$ is the minimum size of a dominating set in $\Gamma$. A graph $\Gamma$ is said to be planar if it can be drawn on a plane without any crossing of its edges. For  vertices $u$ and $v$ in a graph $\Gamma$, we say that $z$ \emph{strongly resolves} $u$ and $v$ if there exists a shortest path from $z$ to $u$ containing $v$, or a shortest path from $z$ to $v$ containing $u$. A subset $U$ of $V(\Gamma)$ is a \emph{strong resolving set} of $\Gamma$ if every pair of vertices of $\Gamma$ is strongly resolved by some vertex of $U$. The least cardinality of a strong resolving set of $\Gamma$ is called the \emph{strong metric dimension} of $\Gamma$ and is denoted by $\operatorname{sdim}(\Gamma)$. For  vertices $u$ and $v$ in a graph $\Gamma$, we write $u\equiv v$ if $N[u] = N[v]$. Notice that that $\equiv$ is an equivalence relation on $V(\Gamma)$.
We denote by $\widehat{v}$ the $\equiv$-class containing a vertex $v$ of $\Gamma$.
 Consider a graph $\widehat{\Gamma}$ whose vertex set is the set of all $\equiv$-classes, and vertices $\widehat{u}$ and  $\widehat{v}$ are adjacent if $u$ and $v$ are adjacent in $\Gamma$. This graph is well-defined because in $\Gamma$, $w \sim v$ for all $w \in \widehat{u}$ if and only if $u \sim v$.  We observe that $\widehat{\Gamma}$ is isomorphic to the subgraph $\mathcal{R}_{\Gamma}$ of $\Gamma$ induced by a set of vertices consisting of exactly one element from each $\equiv$-class. Subsequently, we have the following result of \cite{ma2018strong} with $\omega(\mathcal{R}_{\Gamma})$ replaced by $\omega(\widehat{\Gamma})$.

\begin{theorem}[{\cite[Theorem 2.2]{ma2018strong}}]\label{strong-metric-dim}
For any graph $\Gamma$ with diameter $2$,  $\operatorname{sdim}(\Gamma) = |V(\Gamma)| - \omega(\widehat{\Gamma})$.
\end{theorem}
Let $F_q$ be a finite field, where $q$ is prime power, and let $M_n(F_q)$ be the ring of all $n \times n$ matrices over $F_q$. The set of all invertible matrices over $F_q$ will be denoted by $M_n^*(F_q)$. Let $E$ be an $ n \times n$ identity matrix, $E_{s,t}$ the $n \times n$ matrix with  entry in $s^{\text{th}}$ row, $t^{\text{th}}$ column as $1$ rest all are $0$. $J_r$ the matrix $\sum_{i=0}^{r}E_{i,i}$ in $M_n(F_q)$,
$E(i,j)$ the matrix obtained by interchanging $i^{\text{th}}$ and $j^{\text{th}}$ column of identity matrix $E$, $E(i(a))$ the matrix obtained by multiplying $i^{\text{th}}$ column of identity matrix $E$ by $a \in F_q$. Let $I_X$ be the left ideal generated by $X$, for convenience,  we denote it by $[X]$, where $X \in M_n(F_q)$. By $S_X$, we mean the subspace of $F_q^n$ spanned by row vectors of $X \in M_n(F_q)$. Now let $\Gamma$ be a directed graph, $V(\Gamma)$ be the vertex set. For $u, v \in V(\Gamma)$, we write $u \rightarrow v$ if there is a directed edge from $u$ to $v$. Also, by  $ U \sim V$ we mean $u \rightarrow v$ and $v \rightarrow u$. For $u, v \in V(\Gamma)$, we have  $N_i(v) = \{ u ~|\ u \rightarrow v \}$ and  $N_o(v)) = \{ u ~|\ v \rightarrow u \}$.  The in-degree $d_i(v)$ of $v$ is the number of vertices
in  $N_i(v)$. Analogusly, the out-degree $d_o(v)$ of $v$ can be defined as the number of vertices in $N_o(v)$. An automorphism of a graph $\Gamma$ is a permutation $f$ on $V (\Gamma)$ with the property that, for any vertices $u$ and $v$, we
	have $uf \rightarrow vf$ if and only if $u \rightarrow v$. The set $Aut(\Gamma)$ of all graph automorphisms of a graph $\Gamma$ forms a group with
	respect to composition of mappings.

\section{Automorphisms of the left-ideal relation graph of ideals over $M_n(F_q)$}
In this section, we obtain the automorphisms of the left-ideal relation graph of ideals over $M_n(F_q)$.
Before characterizing all the automorphisms of
$\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$, we introduce two kinds of standard automorphisms for $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.

\begin{lemma}\label{varphi}
 For $P \in M_n^*(F_q)$, the map $\varphi_P$ $:\ M_n(F_q) \rightarrow M_n(F_q)$ defined by $\varphi_P(X) = XP $ for each  $X \in M_n(F_q)$, is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
 \end{lemma}
 \begin{proof}
 Suppose that $X, Y \in M_n(F_q)$ be arbitrary vertices of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$ and let $\varphi_P(X) = \varphi_P(Y)$. Then $XP = YP$. Since $P \in M_n^*(F_q)$ we get $X = Y$. Thus, $\varphi_P$ is one-one. Also, $\varphi_P$ is onto as $M_n(F_q)$ is finite. Therefore, $\varphi_P$ is bijective. For $X \rightarrow Y$, we have $[X] \subset [Y]$. Let $Z \in [XP]$ so there exists atleast one $ W \in M_n(F_q)$ such that $Z = WXP$. Thus,  $ZP^{-1} = WX$  so that $ZP^{-1} \in [Y]$. There exists  $V \in M_n(F_q)$ such that $ZP^{-1} = VY$. Consequently, $Z = VYP$ so that $Z \in [YP]$. Therefore, $[XP] \subseteq [YP]$. Let $L \in [Y]$ and $L \notin [X]$ be arbitrary vertices of $M_n(F_q)$. Then $L = UY$ for some $U \in M_n(F_q)$. Consequently, $LP  = UYP$ implies that $LP \in [YP]$. Now assume that $LP \in [XP]$. Then there exists $N \in M_n(F_q)$ such that  $LP = NXP$. It follows that $L = NX$ and so $L \in [X]$, a contradiction. Thus, $[XP] \subset [YP]$ and $\varphi_P(X) \rightarrow \varphi_P(Y)$. Thus, $\varphi_P$ is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
 \end{proof}
Define $[X]P = \{AP |\ A \in [X]\}$ and for convenience we write as $[X]P = I_{X}P$. As a consequence, $[\varphi_P(X)] = I_{XP}$.

\begin{lemma}
 For $P\in M_n^*(F_q)$, we have $I_{XP} = I_XP$.
 \end{lemma}
 \begin{proof}
 Let $A \in I_{XP}$ be an arbitrary element of $M_n(F_q)$. Then there exists $C \in M_n(F_q)$ such that $A = CXP$. It follows that $AP^{-1} \in I_X$. Consequently, $A \in I_XP$ implies that $I_{XP} \subseteq I_XP$. Now assume that $B \in I_XP$. Then there exists $D \in M_n(F_q)$ such that $B = DXP$. Therefore, $B \in I_{XP}$. Thus, $I_{XP} = I_XP$.
 \end{proof}
 The next two lemmas are useful in the sequel.
\begin{lemma}[{\cite[Lemma 2.2]{wang2017automorphisms}}]\label{leftidealgeneratedbysingle}
 Let $I$ be any left ideal of $M_n(F_q)$. Then there exists $X \in M_n(F_q)$ such that $I = [X]$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{wang2017automorphisms}}]\label{lemma4}
 Let $I$ be any left ideal of $M_n(F_q)$. Then there exists $P \in M_n^*(F_q)$ such that $I = I_{E_rP}$, where $0 \leq r \leq n$ and $E_0 = 0$.
\end{lemma}
\begin{lemma}[{\cite[Lemma 3.3]{xu2020automorphism}}]\label{lemma5}
Let $X, Y \in M_n(F_q)$. Then $S_X = S_Y$ if and only if $I_X = I_Y$.
\end{lemma}

\begin{lemma}
Let $X, Y \in M_n(F_q)$. Then $S_X \subset S_Y$ if and only if $I_X \subset I_Y$.
\end{lemma}
\begin{proof}
First suppose that $S_X \subset S_Y$ and let $A \in M_n(F_q)$ be an arbitrary element. Then $A\begin{pmatrix}
  X_1\\
  X_2\\
   :\\
  X_n
\end{pmatrix} = \begin{pmatrix}
  X_1^{'}\\
  X_2^{'}\\
   :\\
  X_n^{'}
\end{pmatrix}$, where $X_i \in F_q^n$. It follows that $X_i \in S_X$. Since $S_X \subset S_Y$ so that $X_i \in S_Y$. Therefore, there exists $B \in M_n(F_q)$ such that $BY =    \begin{pmatrix}
  X_1^{'}\\
  X_2^{'}\\
   :\\
  X_n^{'}
  \end{pmatrix}$. This implies $BY = AX$. Consequently, $AX \in   I_Y$ follows that
$I_X \subseteq I_Y$. If $I_X = I_Y$, then by Lemma \ref{lemma5} $S_X = S_Y$, a contradiction. Thus, $I_X \subset I_Y$.
Conversely, suppose that $I_X \subset I_Y$ and let $a \in S_X$ such that $a = \sum_{i=1}^{n}c_iX_i$ where $X_i$ is the $i^{th}$ row of matrix $X$ and $c_i \in F_q$. It follows that \begin{center}
$[c_1 c_2 .... c_n]X = a$
\end{center}
Since $I_X \subset I_Y$ so $X \in I_Y$. It implies that $ \begin{pmatrix}
  c\\
  0
\end{pmatrix}X =\begin{pmatrix}
  a\\
  0
\end{pmatrix} \in I_Y$ and for $d \in F_q^n$ we get $\begin{pmatrix}
  d\\
  0
\end{pmatrix}Y = \begin{pmatrix}
  a\\
  0
\end{pmatrix}$. Therefore, $a \in S_Y$.
Thus, $S_X \subseteq S_Y$. If $S_X = S_Y$ then by Lemma \ref{lemma5} $I_X = I_Y$, again a contradiction.
Hence, $S_X \subset S_Y$.
\end{proof}

Let $\Omega$ be the set of $r$ linearly independent $n-\text{dimensional}$ vectors belongs to $F_q^n$ and let $E_\Omega$ be the matrix such that it's first $|\Omega|$ rows are from set $\Omega$ and rest all rows are $0$ vectors. Let us denote $I_\Omega$ as the left ideal generated by $E_\Omega$ and let $I$ be a left ideal of $M_n(F_q)$. By Lemma \ref{leftidealgeneratedbysingle}, we have, $I = I_X$ for some $X \in M_n(F_q)$. Let $\Omega$ be the set of maximal linearly independent row vectors of $X$. Then $S_X = S_{E_\Omega}$ so that $I_X = I_{E_\Omega}$ and  $I = I_\Omega$. From the above discussion, it is easy to observe the following lemma.

\begin{lemma}\label{independentrows}
Let $I$ be a left ideal of $M_n(F_q)$. Then there exists $\Omega \subseteq F_q^n$ such that $\Omega$ is a set of linearly independent row vectors and $I = I_\Omega$.
\end{lemma}

Let $E_{e_i}$ be the matrix such that its first row is $e_i \in F_q^n$ and rest all rows are $0$ for $1 \leq i \leq n$. Let $X \in M_n(F_q)$. Then by Lemma \ref{independentrows}, there exists $\Omega \subseteq F_q^n$ such that $I_X = I_\Omega$. From this we get that $rank(X) = rank(E_\Omega) = |\Omega|$. In the similar manner to $E_{e_i}$, define $E_a$ to be the matrix with its first row as $a \in F_q^n$ and rest all rows are $0$. Clearly  $rank(E_a) = 1$. Let $I$ be a left ideal of $M_n(F_q)$. Then $I$ is a vector space as we can write $aX = (aE)\cdot X \in I$, where $a \in F_q$  and  $X \in I$. Thus, dimension of $I$ is well defined and we claim that $\dim(I) = n\cdot rank(I)$, where  $rank(I) = rank(X)$ such that $I_X = I$. It follows that $I = [E_rP]$, where $P \in M_n^*(F_q)$. Therefore, by Lemma \ref{leftidealgeneratedbysingle}, we have $rank(I) = rank(E_rP) = r$. Notice that  $I_{E_r}$ is a vector space with bases as $\{E_{i,j}\ |\ 1\leq i \leq n\  and \ 1 \leq  j \leq r\}$. Thus, $\dim(I_{E_r}) = nr = n\cdot rank(I)$ and vector space $I \cong I_{E_r}$ as $I = \varphi_P(E_r)$. By Lemma \ref{varphi}, we get that $\varphi_P$ is bijective and one can verify that $\varphi_P$ is a linear transformation. Thus, $\dim(I) = \dim(I_{E_r}) = n\cdot rank(I)$.

\begin{lemma}
 Let $X,Y \in M_n(F_q)$ such that $rank(X) < rank(Y)$. Then $d_i(X) < d_i(Y)$ and $d_o(X) > d_o(Y)$. Moreover
 \begin{enumerate}
    \item[\rm(i)] $d_i(X) = d_i(Y)$ if and only if $rank(X) = rank(Y)$
     \item[\rm(ii)]$d_o(X) = d_o(Y)$ if and only if $rank(X) = rank(Y)$.
 \end{enumerate}
 \end{lemma}
\begin{proof}
\rm(i) If $rank(X) = s$ and $rank(Y) = t$, then $s < t$. Assume that $[X] = [E_sA]$ and $[Y] = [E_tB]$, where $A, B \in M_n^*(F_q)$. It follows that $d_i(X) = d_i(E_sA)$ so that $d_i(X) = d_i(E_s)$ because $\varphi_A(E_s) = E_sA$. Similarly, we have $d_i(Y) = d_i(E_t)$. Next, we need to prove that $d_i(E_s) < d_i(E_t)$. Now  $N_i(E_s) \subset N_i(E_t)$ as $E_s \in N_i(E_t)$ but $E_s \notin N_i(E_s)$. Then $d_i(E_s) < d_i(E_t)$ implies that $d_i(X) < d_i(Y)$.
Further, assume that $d_i(X) = d_i(Y)$. From the above discussion, if we assume that $rank(X) \neq rank(Y)$, then $ d_i(X) \neq d_i(Y)$ which is a contradiction. Thus, $rank(X) = rank(Y)$.
Now let $rank(X) = rank(Y) = r$. This implies that $d_i(X) = d_i(E_r) = d_i(Y)$. Thus the result holds.

\rm(ii) The proof is similar to Part (i).
\end{proof}

\begin{corollary}\label{rank}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ and $X \in M_n(F_q)$. Then $rank(\Psi(X)) = rank(X)$.
 \end{corollary}

If $n = 2$ then $rank(X) = 0, 1, 2$,  where $X \in M_2(F_q)$. Note that
$rank(X) = 0$ if and only if  $[X] = [0]$. Also,
$rank(X) = 2$ if and only if  $[X] = M_2(F_q)$.

Let $R_i = \{X\ |\ X \in M_2(F_q) \ and \ rank(X) = i\}$. Thus for $X,Y \in M_2(F_q)$, it is clear that $X\rightarrow Y$ if and only if $rank(X) < rank(Y)$. The same has been stated in the following lemma.

\begin{lemma}\label{rankm2}
For $X,Y \in M_2(F_q)$, $X \rightarrow Y$ if and only if $rank(X) < rank(Y)$.
\end{lemma}

Define a mapping $\rho$ from $M_2(F_q)$ to itself such that it permute vertices in $R_i$ for each $i = 0, 1, 2$. Observe that $N_i(X) = N_i(Y)$ and $N_o(X) = N_o(Y)$ for each $ X,Y \in R_i$. Thus, we are in the shape to derive the following lemma.

\begin{lemma}\label{lemma10}
Let $\rho$ be a mapping as defined above. Then $\rho \in Aut(\overrightarrow{\Gamma_{l-i}}(M_2(F_q)))$.
\end{lemma}
\begin{proof}
Since $\rho$ permute vertices in $R_i$, thus it is bijective. Let $X \rightarrow Y$, where $X,Y \in M_2(F_q)$. Then by Lemma \ref{rankm2}, we have $rank(X) < rank(Y)$ and  $rank(\rho(X)) = rank(X)$ and $rank(\rho(Y)) = rank(Y)$ from Corollary \ref{rank}. Therefore, by Lemma \ref{rankm2}, we have $rank(\rho(X)) < rank(\rho(Y))$ implies that $\rho(X) \rightarrow \rho(Y)$. Thus, $\rho \in Aut(\overrightarrow{\Gamma_{l-i}}(M_2(F_q)))$.
\end{proof}

Hence, in what follows, we shall assume  $n \geq 3$.
\begin{lemma}\label{lemma11}
For $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$,  there exists $P \ \in M_n^*(F_q)$ such that $[(\varphi_P\cdot \Psi)(E_{e_1})] = [E_{e_1}]$.
\end{lemma}
\begin{proof}
We know that $rank(E_{e_1}) = 1$ then $rank(\Psi(E_{e_1})) = 1$. Assume that $\Psi(E_{e_1}) = X$ for some $X \in M_n(F_q)$ such that $[\Psi(E_{e_1})] = [E_a]$, where $0\neq a \in F_q^n$. Let $a_k$ be the first non-zero element in row vector $a$. Define \begin{center}
$P = (E-\sum_{i\neq k}a_k^{-1}a_iE_{k,i})E(k(a_k^{-1}))E(1,k)$.
\end{center}
Note that $P \in M_n^*(F_q)$. Moreover, $[\Psi(E_{e_1})]P = [E_a]P$ which implies that $[\Psi(E_{e_1})P] = [E_aP]$. Therefore, $[(\varphi_P\cdot \Psi)(E_{e_1})] = \begin{pmatrix}
  e_1\\
  0
\end{pmatrix}$ and hence $[(\varphi_P\cdot \Psi)(E_{e_1})] = [E_{e_1}]$.
\end{proof}

\begin{lemma}\label{lemma12}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[E_{e_k}] = [\Psi(E_{e_k})]$, where $1 \leq k \leq i$. Then $[E_i] = [\Psi(E_i)]$.
\end{lemma}
\begin{proof}
For $i=1$, we have $E_{e_1} = E_1$, we are done. If $i\geq 2$, then  $[E_{e_k}] \subset [E_i]$. It implies that $E_{e_k}\rightarrow E_i$ and so $\Psi(E_{e_k}) \rightarrow \Psi(E_i)$. Thus, $[\Psi(E_{e_k})] \subset [\Psi(E_i)]$ so that $[E_{e_k}] \subset [\Psi(E_i)]$. It follows that $\sum_{k=1}^{i}[E_{e_k}] \subseteq [\Psi(E_i)]$. Therefore, $[E_i] \subseteq [\Psi(E_i)]$. Since $\dim([E_i]) = \dim([\Psi(E_i)]) = n\cdot i$ we get $[E_i] = [\Psi(E_i)]$.
\end{proof}

\begin{lemma}\label{lemma13}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[E_{e_k}] = [\Psi(E_{e_k})]$ and $1 \leq k \leq i-1$. Then there exists $P \in M_n^*(F_q)$ such that $[(\varphi_P\cdot \Psi)(E_{e_s})] = [E_{e_s}]$, where $1 \leq s \leq i$.
\end{lemma}
\begin{proof}
Let $\Psi(E_{e_i}) = X$ for some $X \in M_n(F_q)$ such that $[\Psi(E_{e_i})] = [E_a]$, where $a (\neq 0) \in F_q^n$. Since $[E_{e_i}]$ is not a proper subset of $[E_{i-1}]$ then $[\Psi(E_{e_i})]$ is not a proper subset of $[\Psi(E_{i-1}]$. By Lemma \ref{lemma12}, we have $[E_{i-1}] = [\Psi(E_{i-1})]$. It follows that $[\Psi(E_{e_i})]$ is not a proper subset of $[E_{i-1}]$. Then there exists $ a_l (\neq 0) \in F_q$ such that $i \leq l \leq n$. Now define \begin{center}
$P = (E-\sum_{j\neq l}a_l^{-1}a_jE_{l,j})E(l(a_l^{-1}))E(i,l)$.
\end{center}
Note that $P \in M_n^*(F_q)$. Moreover, $[\Psi(E_{e_i})]P = [E_a]P$ and  $[\Psi(E_{e_i})P] = [E_aP]$. Therefore,  $[(\varphi_P\cdot \Psi)(E_{e_i})] = \begin{pmatrix}
  e_i\\
  0
\end{pmatrix}$ so that $[(\varphi_P\cdot \Psi)(E_{e_i})] = [E_{e_i}]$. Now, $[\Psi(E_{e_k})]P = [E_{e_k}]P$, where $1 \leq k < i$,  implies that $[\Psi(E_{e_k})P] = [E_{e_k}P]$ and therefore, $[(\varphi_P\cdot \Psi)(E_{e_k})] = [\varphi_P(E_{e_k})]$. It is  easy to observe that $[\varphi_P(E_{e_k})] = [E_{e_k}]$. Thus, $[(\varphi_P\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $1 \leq k \leq i-1$. Hence, $[(\varphi_P\cdot \Psi)(E_{e_s})] = [E_{e_s}]$, where $1 \leq s \leq i$.
\end{proof}

\begin{corollary}\label{corollary14}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$. Then there exists  $P \in M_n^*(F_q)$ such that $[(\varphi_P\cdot \Psi)(E_{e_k})] = [E_k]$ and $1 \leq k \leq n$.
\end{corollary}
\begin{proof}
In view of Lemma \ref{lemma11}, there exists $P_1 \in M_n^*(F_q)$ such that $[(\varphi_{P_1}\cdot \Psi)(E_{e_1})] = [E_{e_1}]$. Then by Lemma \ref{lemma13}, we get $P_2 \in M_n^*(F_q)$ such that $[(\varphi_{P_2}\cdot \varphi_{P_1}\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $k = 1, 2$. On the similar lines of proof of Lemma \ref{lemma13}, there exists $P_1, P_2, \cdots,  P_n$ such that $[(\varphi_{P_n}\cdot \cdots \varphi_{P_2}\cdot \varphi_{P_1}\cdot \Psi)(E_{e_k})] = [E_{e_k}]$ where $k = 1, 2, \cdots, n$. Assume that $P = P_1P_2 \cdots P_n$ and  $P \in M_n^*(F_q)$ and observe that $\varphi_{Q_2}\cdot \varphi_{Q_1} = \varphi_{Q_1Q_2}$, where $Q_1, Q_2 \in M_n^*(F_q)$. This implies that $\varphi_{P_n}\cdot \cdots \varphi_{P_2}\cdot \varphi_{P_1} = \varphi_{P_1P_2 \cdots P_n}$. Therefore, $[(\varphi_P\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $k = 1,2, \cdots, n$.
\end{proof}

Let $\Delta = \{e_1, e_2, \cdots,  e_n\}$ be the set of all unit vectors of $F_q^n$.

\begin{lemma}\label{lemma15}
Let $\delta \subseteq \Delta$. If  $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$ and  $k = 1,2, \cdots, n$, then $[\Psi(E_{\delta})] = [E_{\delta}]$.
\end{lemma}
\begin{proof}
If $|\delta| = 1$ then $E_{\delta} = E_{e_k}$. Therefore, we assume that $|\delta|\geq 2$. Let $e_k \in \delta$. Then $[E_{e_k}] \subset [E_{\delta}]$ implies that $[\Psi(E_{e_k})] \subset [\Psi(E_{\delta})]$. Consequently, we obtain $[E_{e_k}] \subset [\Psi(E_{\delta})]$.
Let $I(\delta) = \{i\ |\ e_i \in \delta\}$. Then we have
    $\sum_{k \in I(\delta)}[E_{e_k}] \subseteq [\Psi(E_{\delta})]$. This implies that $[E_{\delta}] \subseteq [\Psi(E_{\delta})]$. Since $\dim([E_{\delta}]) = \dim([\Psi(E_{\delta})]) = n\cdot |\delta|$ we get, $[E_{\delta}] = [\Psi(E_{\delta})]$.
\end{proof}

For any $X \in M_n(F_q)$ such that $rank(X) = 1$ we have $[X] = [E_a]$, where $a \in F_q^n$. Note that such representation of $[X]$ is not unique and there can be a $b \in F_q^n$ such that $[X] = [E_b]$. But if we place a condition that $a$ should be such that its first non-zero element be $1$ then we get a unique $a \in F_q^n$ such that $[X] = [E_a]$. In next few lemmas we will consider rank $1$ matrices of $M_n(F_q)$.

\begin{lemma}\label{lemma16}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$, where $k = 1,2, \cdots, n$. If $X \in M_n(F_q)$ such that $rank(X) = 1$, $[X] = [E_a]$ and $[\Psi(X)] = [E_b]$, where $a = (a_{1}, a_{2}, \cdots, a_{n}),~ b = (b_{1}, b_{2}, \cdots, b_{n}) \in F_q^n$, then $a_l = 0$ if and only if $b_l = 0$, where $1 \leq l \leq n$.
\end{lemma}
\begin{proof}
Let $a_l = 0$. For each $Y \in [E_a]$, we have $l^{th}$ column of $Y = 0$. Let us denote $\Delta\backslash\{e_l\}$ as $\Delta_l$. Then $[E_a] \subset [E_{\Delta_l}]$ implies that $[\Psi(X)] \subset [\Psi(E_{\Delta_l})]$. Consequently, $[E_b] \subset [\Psi(E_{\Delta_l})]$. By Lemma \ref{lemma15}, we obtain $[E_{\Delta_l}] = [\Psi(E_{\Delta_l})]$. Therefore,  $[E_b] \subset [E_{\Delta_l}]$ which implies that $b_l = 0$.

Further suppose that $a_l \neq 0$. Then $[E_a] \not \subset [E_{\Delta_l}]$. It follows that $[\Psi(X)] \not\subset [\Psi(E_{\Delta_l})]$ and so $[E_b] \not \subset [\Psi(E_{\Delta_l})]$. By Lemma \ref{lemma15}, we get $[E_b] \not \subset [E_{\Delta_l}]$. Now, $b_l = 0$  gives $[E_b] \subset [E_{\Delta_l}]$, a contradiction. Thus, $b_l \neq 0$.
\end{proof}

\begin{lemma}\label{lemma17}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$ where $k = 1,2, \cdots, n$. Let $a_{i} = (a_{i1}, a_{i2}, \cdots, a_{in}) \in F_q^n$ such that $[X_i] = [E_{a_i}]$ where $X_i \in M_n(F_q)$ and $i = 1, 2$. Let $b_i = (b_{i1}, b_{i2}, \cdots, b_{in}) \in F_q^n$ such that $[\Psi(X_i)] = [E_{b_i}]$. Then $b_{1s}b_{2k} = b_{1k}b_{2s}$ if and only if $a_{1s}a_{2k} = a_{1k}a_{2s}$, where $ 1 \leq s < k \leq n$.
\end{lemma}
\begin{proof}
 Suppose that $a_{1s}a_{2k} = a_{1k}a_{2s}$. If $a_{1s}a_{2k}a_{1k}a_{2s} = 0$ then, by Lemma \ref{lemma16}, $b_{1s}b_{2k}b_{1k}b_{2s} = 0$. Therefore, $a_{1s}a_{2k} = a_{1k}a_{2s}$ implies $b_{1s}b_{2k} = b_{1k}b_{2s}$. Thus, we assume $a_{1s}a_{2k}a_{1k}a_{2s} \neq 0$. By Lemma \ref{lemma16}, we have  $b_{1s}b_{2k}b_{1k}b_{2s} \neq 0$. Consider the sets
\begin{equation*}
\begin{split}
\alpha = \{e_1, e_2, \cdots, a_{1s}e_s+a_{1k}e_k, \cdots, e_{k-1}, e_{k+1}, \cdots, e_n\}, \text{and}\\
\beta = \{e_1, e_2, \cdots, b_{1s}e_s+b_{1k}e_k,\cdots, e_{k-1}, e_{k+1}, \cdots, e_n\}
\end{split}
\end{equation*}
Note that $[E_{a_1}] \subset [E_\alpha]$. It follows that $[\Psi(X_1)] \subset [\Psi(E_\alpha)]$ and so  $[E_{b_1}] \subset [E_\alpha]$. And, $[E_{\Delta\backslash\{e_s, e_k\}}] \subset [E_\alpha]$ so $ [E_{\Delta\backslash\{e_s, e_k\}}] \subset [\Psi(E_\alpha)]$. Consequently, $[E_{\Delta\backslash\{e_s, e_k\}}] \ + \ [E_{b_1}]\subseteq [\Psi(E_\alpha)]$. Since $\dim([E_{\Delta\backslash\{e_s, e_k\}}] \ + \ [E_{b_1}]) = \dim([\Psi(E_\alpha)]) = n(n-1)$, we have  $[E_{\Delta\backslash\{e_s, e_k\}}] \ + \ [E_{b_1}] = [\Psi(E_\alpha)]$. It is easy to observe that $[E_{\Delta\backslash\{e_s, e_k\}}] \ + \ [E_{b_1}] = [E_\beta]$. Thus, $[E_\beta] = [\Psi(E_{\alpha})]$.

If $a_{1s}a_{2k} = a_{1k}a_{2s}$ then $[E_{a_2}] \subset [E_\alpha]$. It follows that $[X_2] \subset [E_\alpha]$ which implies that $[E_{b_2}] \subset [E_\beta]$. Therefore, $b_{1s}b_{2k} = b_{1k}b_{2s}$.

Now assume that $a_{1s}a_{2k} \neq a_{1k}a_{2s}$. It follows that $[E_{a_2}] \not\subset [E_\alpha]$. Consequently, $[X_2] \not\subset [E_\alpha]$ gives $[E_{b_2}] \not\subset [E_\beta]$. If $b_{1s}b_{2k} = b_{1k}b_{2s}$, then we get $[E_{b_2}] \subset [E_\beta]$ a contradiction. Thus, $b_{1s}b_{2k} \neq b_{1k}b_{2s}$.
\end{proof}

\begin{lemma}\label{lemma18}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$. Then there exists $P \in M_n^*(F_q)$ such that $[(\varphi_P\cdot \Psi)(E_\mathbf{1})] = [E_\mathbf{1}]$
where $[E_\mathbf{1}] = \begin{pmatrix}
\mathbf{1}\\
0
\end{pmatrix}$ and $\mathbf{1} = (1,1,  \cdots, 1) \in F_q^n$ i.e. all-one vector. Also, $[(\varphi_P\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $1 \leq k \leq n$.
\end{lemma}
\begin{proof}
 In view of Corollary \ref{corollary14}, there exists a matrix $P_1 \in M_n^*(F_q)$ such that $[(\varphi_{P_1}\cdot \Psi)(E_{e_k})]=[E_{e_k}]$, where $1 \leq k \leq n$. Let $[(\varphi_{P_1}\cdot \Psi)(E_\mathbf{1})]  = [E_a]$, where $a \in F_q^n$. Then by Lemma \ref{lemma16}, $a_l \neq 0$ for each $1 \leq l \leq n$.
Now, define $P_2 \in M_n^*(F_q)$ such that $P_2 = diag(a_1^{-1}, a_2^{-1}, \cdots, a_n^{-1})$. We obtain $[(\varphi_{P_1}\cdot \Psi)(E_\mathbf{1})]P_2\   =\  [E_a]P_2$. It follows that $[(\varphi_{P_1}\cdot \Psi)(E_\mathbf{1})P_2]  = [E_aP_2]$. It implies that $[(\varphi_{P_2}\cdot \varphi_{P_1}\cdot \Psi)(E_\mathbf{1})]  = [E_\mathbf{1}]$ so that $[(\varphi_{P_1P_2}\cdot \Psi)(E_\mathbf{1})] = [E_\mathbf{1}]$. Let $P = P_1P_2 \in M_n^*(F_q)$. Then $[(\varphi_P\cdot \Psi)(E_\mathbf{1})] = [E_\mathbf{1}]$. And, $[(\varphi_{P_1P_2}\cdot \Psi)(E_{e_k})] = [\varphi_{P_2}(E_{e_k})] = [E_{e_k}]$. Consequently, $[(\varphi_P\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $ 1 \leq k \leq n$. Thus, $\varphi_P\cdot \Psi$ is the required automorphism.
\end{proof}
\begin{lemma}\label{lemma19}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$, where  $\ 1 \leq k \leq n$ and $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$. Suppose $X \in M_n(F_q)$ such that $[X] = [E_a]$ and $[\Psi(X)] = [E_b]$, where $a, b \in F_q^n$. Then $a_s = a_k$ if and only if $b_s = b_k$ and $1 \leq s < k \leq n$.
\end{lemma}
\begin{proof}
Let $a_s = a_k$. Then $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$. It follows that $a_s\cdot 1 = 1\cdot a_k$ and so $b_s\cdot 1 = 1\cdot b_k$.  By Lemma \ref{lemma17}, we have $b_s = b_k$. If $b_s = b_k$, then $b_s\cdot 1 = 1\cdot b_k$ gives  $a_s\cdot 1 = 1\cdot a_k$. By Lemma \ref{lemma17}, we get $a_s = a_k$.
\end{proof}

Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$, where $\ 1 \leq k \leq n$ and $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$. Let $X \in M_n(F_q)$ such that $[X] = [E_\alpha]$ where $\alpha = (e_1\ +\ ae_2)$ and $a \in F_q$. Further assume that $[\Psi(X)] = [E_\beta]$ where $\beta = {e_1 \ + \ a'e_2}$ and $a' \in F_q$. Note that $a'$ depends upon $a$. Hence, we can define a mapping $\Upsilon$ on $F_q$ such that $\Upsilon(a) = a'$. Thus, $\Upsilon$ is one-one because if $\Upsilon(u) = \Upsilon(v)$, where $u,v \in F_q$, then by Lemma \ref{lemma17}, we get $u = v$. As $\Upsilon$ is a one-one mapping over finite field $F_q$. It follows that $\Upsilon$ is onto. Thus, $\Upsilon$ is bijective over $F_q$. Moreover, by Lemma \ref{lemma16} and Lemma \ref{lemma17} we get, $\Upsilon(0) = 0$ and $\Upsilon(1) = 1$.

\begin{lemma}\label{lemma20}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$, where $1 \leq k \leq n$ and $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$. Let $\Upsilon$ be as defined above. Then the following holds:
\begin{enumerate}
\item[{\rm(i)}] If $[X] = [E_{\{e_1+ae_k\}}]$, where $X \in M_n(F_q)$, then $[\Psi(X)] = [E_{\{e_1+\Upsilon(a)e_k\}}]$ for each $2 \leq k \leq n$ and $a \in F_q$.
\item[{\rm(ii)}] If $[X] = [E_{\{e_1 + a_2e_2 + \cdots + a_n e_n\}}]$, where $X \in M_n(F_q)$, then $[\Psi(X)] =$ $[E_{\{e_1+\Upsilon(a_2)e_2+\cdots+\Upsilon(a_n)e_n\}}]$, where $a_i \in F_q$ and $2\leq i \leq n$.
\item[{\rm(iii)}] If $[X] = [E_{\{e_{i}+a_{i+1}e_{i+1}+ \cdots +a_n e_n\}}]$, where $X \in M_n(F_q)$, then $[\Psi(X)] = $ $[E_{\{e_i +\Upsilon(a_{i+1})e_{i+1}+ \cdots +\Upsilon(a_n)e_n\}}]$, when $ a_j \in F_q$ and $i < j \leq n$.
\end{enumerate}
\end{lemma}
\begin{proof}

\begin{enumerate}
\item[{\rm(i)}] For $k = 2$, there is nothing to prove. We need to prove for $k \geq 3$. Let $Y \in M_n(F_q)$ such that $[Y] = [E_{\{e_1+ae_2+ae_k\}}]$ and $[\Psi(Y)] = [E_{\{e_1+a'e_2+a''e_k\}}]$. By Lemma \ref{lemma19}, we get $a' = a''$ so $[\Psi(Y)] = [E_{\{e_1+a'e_2+a'e_k\}}]$. Let $[Z] = [E_{\{e_1+ae_2\}}]$. Then $[\Psi(Z)] = [E_{\{e_1+\Upsilon(a)e_2\}}]$. By Lemma \ref{lemma17}, and using $[Y]$ and $[Z]$, we get $a' = \Upsilon(a)$. Thus, $ [\Psi(Y)] = [E_{\{e_1+\Upsilon(a)e_2+\Upsilon(a)e_k\}}]$. Let $[\Psi(X)] = [E_{\{e_1+de_k\}}]$. By Lemma \ref{lemma17} and using $[X]$ and $[Y]$, we get $d = \Upsilon(a)$.
\item[{\rm(ii)}] Suppose that $[\Psi(X)] = [E_{\{e_1+c_2e_2+\cdots+c_ne_n\}}]$. Let $Y_k \in M_n(F_q)$ such that $[Y_k] = [E_{\{e_1+a_ke_k\}}]$. It follows that $[\Psi(Y_k)] = [E_{\{e_1+\Upsilon(a_k)e_k\}}]$. By Lemma \ref{lemma17} and by using $[Y_k]$ and $[X]$, we get $c_k = \Upsilon(a_k)$ for each $2 \leq k \leq n$.

\item[{\rm(iii)}] Let $Y \in M_n(F_q)$ such that $[Y] =  [E_{\{e_1+a_2e_2+ \cdots +a_n e_n\}}]$. Then $[\Psi(Y)] = [E_{\{e_1+\Upsilon(a_2)e_2+\cdots+\Upsilon(a_n)e_n\}}]$. Suppose that $Z \in M_n(F_q)$ such that $[Z] = [E_{\{e_1+e_{i}+a_{i+1}e_{i+1}+\cdots +a_n e_n\}}]$. It follows that $[\Psi(Z)] = [E_{\{e_1+e_i +\Upsilon(a_{i+1})e_{i+1}+ \cdots+\Upsilon(a_n)e_n\}}]$. Let $[\Psi(X)] = [E_{\{e_i+d_{i+1}e_{i+1}+ \cdots +d_ne_n\}}]$. By Lemma \ref{lemma17} and by using $[X]$ and $[Z]$, we have $d_j = \Upsilon(a_j)$, where $i < j \leq n$.
\end{enumerate}
\end{proof}
\begin{lemma}\label{lemma21}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$ where $1 \leq k \leq n$ and $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$. Let $\Upsilon$ be a map as defined above. Then $\Upsilon$ is field automorphism of $F_q$.
\end{lemma}
\begin{proof}
Since $\Upsilon$ is bijective in $F_q$ so need to prove that
\begin{center}
$\Upsilon(a+b) = \Upsilon(a)+\Upsilon(b)$ and,
    $\Upsilon(a\cdot b) = \Upsilon(a)\cdot \Upsilon(b)$ where $a,b \in F_q$.
\end{center}
If $a\cdot b = 0$ then either $a = 0$ or $b = 0$. It follows that either $\Upsilon(a) = 0$ or $\Upsilon(b) = 0$. Therefore, $\Upsilon(a)\cdot \Upsilon(b) = 0 = \Upsilon(a\cdot b)$. Hence, we assume that $a\cdot b \neq 0$ which implies that $\Upsilon(a)\cdot \Upsilon(b) \neq 0$. First we claim that $\Upsilon(a^{-1}) = \Upsilon(a)^{-1}$. Let $X_1, X_2 \in M_n(F_q)$ such that $[X_1] = [E_{\{e_1+ae_2+e_3\}}]$ and $[X_2] = [E_{\{e_2+a^{-1}e_3\}}]$. Then $[\Psi(X_1)] = [E_{\{e_1+\Upsilon(a)e_2+e_3\}}]$ and $[\Psi(X_2)] = [E_{\{e_2+\Upsilon(a^{-1})e_3\}}]$. Therefore, by Lemma \ref{lemma17} and by using $[X_1]$ and $[X_2]$, we have $\Upsilon(a)\cdot \Upsilon(a^{-1}) = 1$. It follows that $\Upsilon(a^{-1}) = \Upsilon(a)^{-1}$. Now suppose that $X_3, X_4 \in M_n(F_q)$ such that $[X_3] = [E_{\{e_1+abe_2+e_3\}}]$ and $[X_4] = [E_{\{e_1+be_2+a^{-1}e_3\}}]$. It follows that $[\Psi(X_3)] = [E_{\{e_1+\Upsilon(ab)e_2+e_3\}}] $ and $[\Psi(X_4)] = [E_{\{e_1+\Upsilon(b)e_2+\Upsilon(a^{-1})e_3\}}]$. In view of Lemma \ref{lemma17} and by using $[X_3]$ and $[X_4]$, we get $\Upsilon(ab)\cdot \Upsilon(a^{-1}) = \Upsilon(b)$. Consequently, $\Upsilon(ab) = \Upsilon(a)\cdot \Upsilon(b)$.

Let $Y_1, Y_2 \in M_n(F_q)$ such that $[Y_1] = [E_{\{e_1+ae_3\}}]$ and $[Y_2] = [E_{\{e_2+be_3\}}]$. Then $[\Psi(Y_1)] = [E_{\{e_1+\Upsilon(a)e_3\}}]$ and $[\Psi(Y_2)] = [E_{\{e_2+\Upsilon(b)e_3\}}]$. Now, $[Y_1] \subset  [E_{\{e_1+ae_3, e_2+be_3\}}]$ and $[Y_2] \subset  [E_{\{e_1+ae_3, e_2+be_3\}}]$ implies that $[E_{\{e_1+\Upsilon(a)e_3\}}] \subset [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$ and $[E_{\{e_2+\Upsilon(b)e_3\}}] \subset [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$. It follows that $[E_{\{e_1+\Upsilon(a)e_3\}}] + [E_{\{e_2+\Upsilon(b)e_3\}}] \subseteq [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$. Since $\dim([E_{\{e_1+\Upsilon(a)e_3\}}] + [E_{\{e_2+\Upsilon(b)e_3\}}]) = \dim([\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]) = 2n$ so $[E_{\{e_1+\Upsilon(a)e_3\}}] + [E_{\{e_2+\Upsilon(b)e_3\}}] = [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$. Assume that $Y_3\in M_n(F_q)$ such that $[Y_3] = [E_{\{e_1+e_2+(a+b)e_3\}}]$. Then $[Y_3] \subset [E_{\{e_1+ae_3, e_2+be_3\}}]$ implies that $[\Psi(Y_3)] \subset [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$. It follows that $[E_{\{e_1+e_2+\Upsilon(a+b)e_3\}}] \subset [\Psi(E_{\{e_1+ae_3, e_2+be_3\}})]$ so that $[E_{\{e_1+e_2+\Upsilon(a+b)e_3\}}] \subset [E_{\{e_1+\Upsilon(a)e_3\}}] + [E_{\{e_2+\Upsilon(b)e_3\}}]$. Consequently, $\Upsilon(a+b) = \Upsilon(a)+\Upsilon(b)$. Thus, $\Upsilon$ is field automorphism of $F_q$.
\end{proof}

Now, we extend $\Upsilon$ to $M_n(F_q)$ such that $\Upsilon(X) = [\Upsilon(x_{ij})]_{n\times n}$ for any $X  = [x_{ij}]_{n\times n} \in M_n(F_q)$. Then define a mapping on vertex set of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$ and denote it by $\Upsilon$ only. From now onwards, whether we are referring to mapping on $F_q$ or mapping on $M_n(F_q)$ by $\Upsilon$, will be determined by the context.
\begin{lemma}
The map $\Upsilon$ (as defined above) is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
\end{lemma}
\begin{proof}
Let $X,Y \in M_n(F_q)$ such that $\Upsilon(X) = \Upsilon(Y)$. Then $[\Upsilon(x_{ij})]_{n\times n} = [\Upsilon(y_{ij})]_{n\times n}$, where $X = [x_{ij}]_{n\times n}$ and $Y = [y_{ij}]_{n\times n}$. Since $\Upsilon$ is bijective over $F_q$, we have $[x_{ij}]_{n\times n} = [y_{ij}]_{n\times n}$. Therefore, $X = Y$. Thus, $\Upsilon$ is one-one over $M_n(F_q)$ and because $M_n(F_q)$ is finite so $\Upsilon$ is bijective over $M_n(F_q)$.

Let $X \rightarrow Y$. Then $[X] \subset [Y]$. Further assume that $A \in [\Upsilon(X)]$. It follows that $A = C\Upsilon(X)$, where $C \in M_n(F_q)$. Let $U, V$ be arbitrary element of $M_n(F_q)$. Then $\Upsilon(U\cdot V) = [w_{ij}]_{n\times n} = [\Upsilon(\sum_{k=1}^nu_{ik}\cdot v_{kj})]_{n\times n}$,  where $U \cdot V = W \in M_n(F_q)$. Since $\Upsilon$ is field automorphism of $F_q$ we get $\Upsilon(U\cdot V) = [\sum_{k=1}^n\Upsilon(u_{ik})\cdot \Upsilon(v_{kj})]$ and  $[\sum_{k=1}^n\Upsilon(u_{ik})\cdot \Upsilon(v_{kj})] = \Upsilon(U)\cdot \Upsilon(V)$. Therefore, $\Upsilon(U\cdot V) = \Upsilon(U)\cdot \Upsilon(V)$.

Thus, $A = C\Upsilon(X) = \Upsilon(\Upsilon^{-1}(C)\cdot X)$ so that $\Upsilon^{-1}(A) = \Upsilon^{-1}(C)\cdot X$. It follows that $\Upsilon^{-1}(A) \in [X] \subset [Y]$. Therefore, there exists $D \in M_n(F_q)$ such that $\Upsilon^{-1}(A) = DY$ implies that $A = \Upsilon(D)\cdot \Upsilon(Y)$. Consequently, $A \in [\Upsilon(Y)]$ and so $[\Upsilon(X)] \subseteq [\Upsilon(Y)]$. Now, there exists $Z \in [Y]$ such that $Z \notin [X]$. Let $B \in M_n(F_q)$ such that $Z = BY$. Then $\Upsilon(Z) = \Upsilon(B)\cdot \Upsilon(Y)$. It implies that $\Upsilon(Z) \in [\Upsilon(Y)]$. If  $\Upsilon(Z) \in [\Upsilon(X)]$, then there exists $L \in M_n(F_q)$ such that $\Upsilon(Z) = L \cdot \Upsilon(X)$ and this gives $Z = \Upsilon^{-1}(L)\cdot X$ as $\Upsilon$ is bijective over $M_n(F_q)$. Therefore, $Z \in [X]$, a contradiction. Thus, $\Upsilon(Z) \in [\Upsilon(Y)]$ but $\Upsilon(Z) \notin [\Upsilon(X)]$. It follows that $[\Upsilon(X)] \subset [\Upsilon(Y)]$ and so $\Upsilon(X) \rightarrow \Upsilon(Y)$. Thus, $\Upsilon$ is automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
\end{proof}
\begin{lemma}\label{lemma23}
If $[X] = [Y]$ then $[\Upsilon(X)] = [\Upsilon(Y)]$, where $X,Y \in M_n(F_q)$.
\end{lemma}
\begin{proof}
Let $A \in [\Upsilon(X)]$. Then there exists $W \in M_n(F_q)$ such that $A = W \cdot \Upsilon(X)$. It follows that $\Upsilon^{-1}(A) = \Upsilon^{-1}(W)\cdot X$ and so $\Upsilon^{-1}(A) \in [X] = [Y]$. Then there exists $C \in M_n(F_q)$ such that $\Upsilon^{-1}(A) = C \cdot Y$. Therefore, $A = \Upsilon(C) \cdot \Upsilon(Y)$ implies $A \in [\Upsilon(Y)]$. Thus, $[\Upsilon(X)] \subseteq [\Upsilon(Y)]$. Similarly, we can prove that $[\Upsilon(Y)] \subseteq [\Upsilon(X)]$. Hence, $[\Upsilon(X)] = [\Upsilon(Y)]$.
\end{proof}

\begin{lemma}\label{lemma24}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$ such that $[\Psi(E_{e_k})] = [E_{e_k}]$, where $1 \leq k \leq n$ and $[\Psi(E_\mathbf{1})] = [E_\mathbf{1}]$ and let $\Upsilon$ be a map as defined above. Then for some $\Upsilon$, we have $[(\Upsilon^{-1}\cdot \Psi)(X)] = [X]$ and $X \in M_n(F_q)$.
\end{lemma}
\begin{proof}
In view of Corollary \ref{rank}, if $X = 0$ then $\Upsilon^{-1}\cdot \Psi(X) = X$. It follows that $[(\Upsilon^{-1}\cdot \Psi)(X)] = [X]$. Now, assume that $rank(X) = 1$. Then $[X] = [E_a]$ for some $0 \neq a \in F_q^n$, where $a = (e_i+a_{i+1}e_{i+1}+ \cdots +a_ne_n)$ and $1 \leq i \leq n$. By Lemma \ref{lemma20}, we obtain $[\Psi(X)] = [E_{\{e_i+\Upsilon(a_{i+1})e_{i+1}+ \cdots +\Upsilon(a_n)e_n\}}]$. By Lemma \ref{lemma23}, we have $[(\Upsilon^{-1}\cdot \Psi)(X)] = [\Upsilon^{-1}(E_{\{e_i+\Upsilon(a_{i+1})e_{i+1}+ \cdots +\Upsilon(a_n)e_n\}})]$. It follows that $[(\Upsilon^{-1}\cdot \Psi)(X)] = [E_{\{e_i+a_{i+1}e_{i+1}+ \cdots +a_ne_n\}}]$. Therefore, $[(\Upsilon^{-1}\cdot \Psi)(X)] = [X]$.

Let $rank(X) \geq 2$. Then by Lemma \ref{lemma4}, there exists $P \in M_n^*(F_q)$ such that $[X] = [E_rP]$, where $rank(X) = r$. Since $[E_{e_i}P] \subset [E_rP]$, where $1 \leq i \leq r$, we get $[(\Upsilon^{-1}\cdot \Psi)(E_{e_i}P)] \subset [(\Upsilon^{-1}\cdot \Psi)(X)]$. As $rank(E_{e_i}P) = 1$ implies that $[E_{e_i}P] \subset [(\Upsilon^{-1} \cdot \Psi)(X)]$. It follows that $\sum_{i =1}^r[E_{e_i}P] \subseteq [(\Upsilon^{-1}\cdot \Psi)(X)]$ and so $[E_rP] \subseteq [(\Upsilon^{-1}\cdot \Psi)(X)]$. Since, $\dim([E_rP]) = \dim([(\Upsilon^{-1}\cdot \Psi)(X)]) = nr$, we have  $[E_rP] = [(\Upsilon^{-1}\cdot \Psi)(X)]$. Therefore, $[X] = [(\Upsilon^{-1}\cdot \Psi)(X)]$ for $X \in M_n(F_q)$.
\end{proof}

Define a binary relation $\equiv$ on $M_n(F_q)$ such that $X \equiv Y$ if and only if $[X] = [Y]$. It is easy to check that $\equiv$ is an equivalence relation. Further, we define a mapping $\sigma$ over $M_n(F_q)$ such that it permute elements of each equivalence class arbitrarily.

\begin{lemma}\label{lemma25}
The map $\sigma$ is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
\end{lemma}
\begin{proof}
By the definition, $\sigma$ is bijective over $M_n(F_q)$. Let $X, Y \in M_n(F_q)$ such that $X \rightarrow Y$. Then $[X] \subset [Y]$. Since $[\sigma(X)] = [X]$ and $[\sigma(Y)] = [Y]$ implies that $[\sigma(X)] \subset [\sigma(Y)]$. It follows that $\sigma(X) \rightarrow \sigma(Y)$. Thus, $\sigma$ is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$.
\end{proof}
Now we prove our main result of this section.
\begin{theorem}\label{theorem26}
Let $n \geq 3$ and $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_n(F_q)))$. Then there exists $P \in M_n^*(F_q)$, a field automorphism $\Upsilon$ and $\sigma$, as defined above, such that $\Psi = \varphi_{P} \cdot \Upsilon \cdot \sigma$.
\end{theorem}
\begin{proof}
By Lemma \ref{lemma16}, there exists $P \in M_n^*(F_q)$ such that $[(\varphi_{P^{-1}}\cdot \Psi)(E_{e_k})] = [E_{e_k}]$, where $1 \leq k \leq n$ and $[(\varphi_{P^{-1}}\cdot \Psi)(E_\mathbf{1})] = [E_\mathbf{1}]$. In view of Lemma \ref{lemma24}, there exists $\Upsilon$, a field automorphism of $F_q$ such that $[X] = [(\Upsilon^{-1}\cdot \varphi_{P^{-1}}\cdot \Psi)(X)]$ for all $X \in M_n(F_q)$. Thus, $\Upsilon^{-1}\cdot \varphi_{P^{-1}}\cdot \Psi = \sigma$. Hence, $\Psi = \varphi_P\cdot \Upsilon \cdot \sigma$.
\end{proof}
\begin{theorem}
Let $\Psi \in Aut(\overrightarrow{\Gamma_{l-i}}(M_2(F_q)))$. Then there exists $\rho$, as defined above, such that $\rho = \Psi$.
\end{theorem}
\begin{proof}
By Lemma \ref{lemma10}, $\rho$ is an automorphism of $\overrightarrow{\Gamma_{l-i}}(M_n(F_q))$. Since $rank(\Psi(X)) = rank(X)$, define $\rho$ over $M_2(F_q)$ as $\rho(X) = \Psi(X)$. It follows that $\rho^{-1}\cdot \Psi(X) = X$. Therefore, $\rho^{-1}\cdot \Psi = id$, where $id$ is identity mapping over $M_2(F_q)$. Thus, $\Psi = \rho$.
\end{proof}

\section{Graph Theoretic Properties of undirected Left-Ideal Relation Graph ${\Gamma_{l-i}}(M_n(F_q))$}

In this section, we consider the undirected left-ideal relation graph $\Gamma_{l-i}(M_n(F_q))$.  For $X, Y \in M_n(F_q)$ there is an edge between $X$ and $Y$ i.e. $X \sim Y$ if and only if $[X] \subset [Y]$ or $[Y] \subset [X]$. We investigate various graph theoretic properties of $\Gamma_{l-i}(M_n(F_q))$ including planarity, chromatic number, clique number and strong metric dimension.

\begin{theorem}
 The graph $\Gamma_{l-i}(M_n(F_q))$ is non-planar for $n \geq 2$.
 \end{theorem}
 \begin{proof}
 Let $e_{11} = (1, 1, 0, \cdots, 0) \in F_q^n$. Now, consider $V_1 = \{E_{e_1}, E_{e_2}, E_{e_{11}}\}$ and $V_2 = \{E_{\{e_1,e_2\}}, E_{\{e_2, e_1\}}, E_{\{e_{11},e_1\}}\}$. Observe that graph induced by $V_1 \cup V_2$ is isomorphic to $K_{3,3}$. Therefore, by Kurtowski's theorem, we have $\Gamma_{l-i}(M_n(F_q))$ is non-planar for $n \geq 2$.
  \end{proof}

\begin{theorem}\label{theorem4.3}
 For $M_n(F_q)$, we have  $\chi(\Gamma_{l-i}(M_n(F_q))) = \omega(\Gamma_{l-i}(M_n(F_q))) = n+1$.
 \end{theorem}
 \begin{proof}
 Consider $\mathcal{C} = \{E_0, E_1, E_2, \cdots, E_n\}$. Then the graph induced by $\mathcal{C}$ is clique of $\Gamma_{l-i}(M_n(F_q))$. It follows that $\omega(\Gamma_{l-i}(M_n(F_q))) \geq n+1$. Now we need to show that $\chi(\Gamma_{l-i}(M_n(F_q)))  \leq n+1$. Since no two matrices of same rank are adjacent in $\Gamma_{l-i}(M_n(F_q))$. Thus, we can assign one color to matrices of same rank. It follows that $\chi(\Gamma_{l-i}(M_n(F_q))) \leq n+1$. Thus, $\chi(\Gamma_{l-i}(M_n(F_q))) = \omega(\Gamma_{l-i}(M_n(F_q))) = n+1$.
  \end{proof}

\begin{lemma}
The eccentricity of $X \in V(\Gamma_{l-i}(M_n(F_q)))$ is given below:
\begin{equation*}
ecc(X) =
\begin{cases}
1 & if \ X = 0\\
2 & if \ X \neq 0
\end{cases}
\end{equation*}
\end{lemma}
\begin{proof}
If $X = 0$ then $X$ is adjacent to every other vertex of $\Gamma_{l-i}(M_n(F_q))$. Hence, $ecc(X) = 1$ as eccentricity of a vertex is maximum distance of a vertex to any other vertex. Now, if $X \neq 0$ then $rank(X) \geq 1$. For $rank \geq 1$ there are atleast two vertices of that \(rank\). Let $Y \in M_n(F_q)$ such that $rank(X) = rank(Y)$. Then $d(X, Y) = 2$. Thus, the result holds.
\end{proof}
\begin{corollary}
The graph $\Gamma_{l-i}(M_n(F_q))$ is connected and $diam(\Gamma_{l-i}(M_n(F_q))) = 2$ and $r(\Gamma_{l-i}(M_n(F_q)) = 1$.
 \end{corollary}

\begin{corollary}
The dominance number of $\Gamma_{l-i}(M_n(F_q))$ is $1$.
 \end{corollary}
\begin{theorem}
 If the graph $\Gamma_{l-i}(M_n(F_q))$ has a cycle, then $g(\Gamma_{l-i}(M_n(F_q))) = 3$.
 \end{theorem}
 \begin{proof}
 Suppose that $\Gamma_{l-i}(M_n(F_q))$ has a cycle. Then $n \geq 2$. Now
 consider the set $C = \{E_0, E_1, E_n\}$. The graph induced by vertex set $C$ is a cycle of length $3$. Thus, $g(\Gamma_{l-i}(M_n(F_q))) = 3$.
 \end{proof}
\begin{theorem}
 The strong metric dimension of $\Gamma_{l-i}(M_n(F_q))$ is $q^{n^2}-n-1$.
 \end{theorem}
 \begin{proof}
 Since, $diam(\Gamma_{l-i}(M_n(F_q))) = 2$. Then by Theorem \ref{strong-metric-dim}, we have $ sdim(\Gamma_{l-i}(M_n(F_q))) = |M_n(F_q)|-\omega(R_{\Gamma_{l-i}(M_n(F_q))})$, where $R_{\Gamma_{l-i}(M_n(F_q))} $ is the reduced graph of $\Gamma_{l-i}(M_n(F_q))$. It is known that $|M_n(F_q)| = q^{n^2}$. Since  $\omega(\Gamma_{l-i}(M_n(F_q))) \geq \omega(R_{\Gamma_{l-i}(M_n(F_q))})$, by Theorem \ref{theorem4.3}, we have  $\omega(R_{\Gamma_{l-i}(M_n(F_q))}) \leq n+1$. Define a relation $\equiv$ on $M_n(F_q)$ such that $X \equiv Y$ if and only if $N(X) = N(Y)$, where $N(X) = \{Z\ |\ X \sim Z\ \text{and} \ Z\in M_n(F_q)\}\cup\{X\}$. Observe that $\equiv$ is an equivalence relation. Let $U(\Gamma_{l-i}(M_n(F_q)))$ be the complete set of representative elements for the above mentioned equivalence relation, then notice that $U(\Gamma_{l-i}(M_n(F_q)))$ is the vertex set of $R_{\Gamma_{l-i}(M_n(F_q))}$. Now we consider $\mathcal{C} = \{E_0, E_1, E_2, \cdots, E_n\} \subseteq U(\Gamma_{l-i}(M_n(F_q)))$. Note that $E_i$ and $E_j$ belong to distinct equivalence classes for $i\neq j$. As graph induced by $\mathcal{C}$ is a clique of $R_{\Gamma_{l-i}(M_n(F_q))}$. It follows that $\omega(R_{\Gamma_{l-i}(M_n(F_q))}) \geq n+1$. Therefore, $\omega(R_{\Gamma_{l-i}(M_n(F_q))}) = n+1$. Thus, $sdim(\Gamma_{l-i}(M_n(F_q))) = q^{n^2}-n-1$.
  \end{proof}
\begin{theorem}
The graph $\Gamma_{l-i}(M_n(F_q))$ is not Eulerian.
\end{theorem}
\begin{proof}
Let $deg(Y)$ denote degree of matrix $Y \in M_n(F_q)$ in $\Gamma_{l-i}(M_n(F_q))$. As, zero matrix is adjacent to every other matrix of $\Gamma_{l-i}(M_n(F_q))$. It follows that degree of zero matrix is $q^{n^2}-1$. If $q = 2^m$ for some $m \in \mathbb{N}$, then $q^{n^2}-1$ will be odd and hence degree of zero matrix is odd.
If $X \in M_n(F_q)$ such that $rank(X) = n$ then $deg(X) = q^{n^2}-M(n,n,n,q)$ from \cite{abdel2012counting}, where $M(n,n,n,q)$ is number of $n\times n$ matrices of $rank\ n$ over a finite field $F_q$. Also, $M(n,n,n,q) = \prod_{j=0}^{n-1}(q^n-q^j)$. Now if $q = p^m$ for some odd prime $p$ and $m \in \mathbb{N}$, then $q^n-q^j$ will be even for all $0 \leq j \leq n-1$. It follows that $M(n,n,n,q)$ will be even and hence $q^{n^2}-M(n,n,n,q)$ will be odd as $q^{n^2}$ will be odd. Therefore, $deg(X)$ will be odd for $X \in M_n(F_q)\ and \ rank(X)=n$.
Thus, the zero matrix and any full rank matrix $X$ cannot have even degree simultaneously. Hence, $\Gamma_{l-i}(M_n(F_q))$ is not Eulerian.
\end{proof}

\section{Acknowledgement}
The first author wishes to acknowledge the support of MATRICS Grant  (MTR/2018/000779) funded by SERB, India. The second author gratefully acknowledge for providing financial support to CSIR  (09/719(0093)/2019-EMR-I) government of India.

redpajama_set_name: RedPajamaArXiv


text:
ACCA: 13-ku Kansatsu-ka – 09
So far ACCA has proven a success in the school of the classic slow-burn, in which those patient enough to stick around are lushly rewarded, like the high one gets when about to complete a huge, elaborate jigsaw puzzle (don't judge).
Last week laid out the details of the show's central mystery of what's up with Jean and Lotta, and this week deals with the consequences for everyone once the truth is officially out there, and decide on a course of action. It also allows some previous bit players on the margins play significant roles.
Oh yeah, and a shared love of sandwich bread of all things proves absolutely vital in preventing Lotta's assassination by the First Princess' goons. The food isn't just window-dressing.
As soon as Magie hears from the prince that the knives may be out for Lotta, he makes a call to his comrade-in-bread, Rail, who makes Lotta's protection his top priority. Rail has his suspicions, but doesn't know the whole picture, but that doesn't matter, because he's a decent dude, trusts his fellow bread-lover's warning.
As for poor Lotta, no one's told her anything, and with both Jean and Niino away, a part of her already feels vulnerable. So as out-of-the-blue (or blonde) as it seems, she seems happy to have Rail (a sworn ACCA officer) by her side.
Jean is away because he's on a sprawling three-district trip starting in Peshi (the port district) and moving on to Yakkara (the casino district, and another instance of ACCA imitating Sonic The Hedgehog levels). And Jean is no longer oblivious like Lotta.
He knows what the score is, and even understands what all those cigarettes on his past inspection visits were about. It's need to see the change in Jean's overall demeanor. He seems more focused, alert, and suspicious…as he should. Peshi's chiefs drop the pretense and pledge their support for Jean's ascension, unaware that Jean himself has no such plans.
I like how ultimately, it's only a matter of time before Rail, a good kid but not a professional bodyguard, and Lotta finally get surrounded by the ominous goons and shoved into a car. Unfortunately for the goons, the traffic in Badon flares up just when they need to make their getaway.
Also unfortunately for them, pure dumb luck is on Lotta's side, as Chief Owl (whom Jean asked to keep an eye on her) happens to lean on the open window sill of the goons' car, sees Lotta, and secures her and Rail's release.
Like Rail, Owl doesn't have the whole story, and unlike Rail, he isn't a sandwich bread fanatic (though we've seen him indulge in the office treats du jour) but he does have Jean and Lotta's bests interests at heart, and it's gratifying to see how competently (yet without undue violence) Owlmanages to wrest the crazy kids from certain doom.
With Lotta and Rail nicely rescued, Owl suggests they—what else—go to grab a bite with his ACCA staff. Coups and assassination plots be damned—you gotta eat.
With Lotta out of immediate danger and surrounded by friends, we move on, somewhat relieved but still troubled, to the other major storyline of the episode: Grossular coming clean to the other chiefs, which takes such a crazy turn I'd have nearly fell out of my chair, had I not already been sitting on the carpeted floor.
Grossular lays out the plan that's been in motion since the beginning, with the ultimate goal of instigating an ACCA-led coup d'etat to prevent Schwan from becoming King, thus preserving peace, democracy, an, well, ACCA itself.
Grossular has known about the danger of a King Schwan for some time, but gained a powerful barometer (whom he observed through Crow) for the attitudes (be they pro- or anti-coup) of the districts in Jean, which is why his inspection department was suddenly saved from oblivion.
Once it was clear a majority of districts were in favor of a coup, the time grows near for that coup to commence, but a coup led by ACCA, as an extreme expression of their 'protect & serve' credo. The coup will, Grossular promises, "pose no danger" to ordinary people. Allowing Schwan to dissolve ACCA and create an autocracy might.
Grossular asks his four colleagues whether they stand with him or not, and everyone to a man is with him, all thanks to Lilium, who speaks first in response.
Because Lilium and Grossular have never, to the others' knowledge, ever agreed on anything before, it's all the proof they need to know the right course (on top of their pride in their roles as leaders of ACCA, along with their existing awareness that, ya know, Schwan is bad news). This is to be an act of patriotism, not treason.
Later, we learn that Lilium and Grossular's constant disagreements in front of the others masks the fact that Grossular is, in fact, Lilium's servant. Always a fairly inscrutable guy, we finally see a hint of subservience when Lilium grabs him by the hair and promises him in a threatening tone that "he will manage" in his next objective: do something about Director-General Mauve.
It's this huge, sudden, surprising, yet still well-supported (by both plot and character) shift in character dynamics, as well as the timely utilization of Rail and Owl, that propelled this episode into the '10th district.' It's also a interesting episode in that many cards have been played, but many choice ones remain in the show's hand.
It's that 'floating potential', as it were, that makes episode nine feel special. Hopefully it can be properly harnessed in the tenth, which I eagerly await.
Author braveradePosted on Tue, 7 Mar 2017 Sun, 19 Mar 2017 Categories ACCA: 13-ku Kansatsu-ka, Anime Reviews, Winter 2017Tags acca 13, accident, assassins, aware, bread, casino, coup d'etat, districts, dowa, first princess, grossular, information, inspection department, jean otus, king falke, korore, lilium, lotta, madhouse, magie, mastermind, nino, on the run, owl, peshi, plot, prince schwan, rail, royalty, schwan, sister, snowballs, travel, yakkara1 Comment on ACCA: 13-ku Kansatsu-ka – 09
ACCA had been teasing us for a while now about who exactly Jean and Lotta are, and last week finally answered that question: they are a prince and princess.
This week gives us more of what I yearned for—specifics—by taking us back 33 years, to when King Falke's daughter Princess Schnee leaves the royal family and renounces her titles so that she can spread her wings and be, in her view, of more use to the nation, by traveling and learning more about it.
The four people who arrange Schnee's "detatchment" from the family (complete with faked death) are the king, Privy Council President Qualm, Schnee, and her loyal aide Abend, who looks a lot like a young Grossular.
The twist is that Abend isn't the one Schnee falls for: he fades into the shadows and receives reports and photos from Niino's father, which he then relays to Qualm, who relays them to the king.
Everybody seems to win in this arrangement: Schnee gets to live her life (and fall in love with a commoner, resulting in the birth of Jean and Lotta) out in the world, the king never had to "clip her wings", and the president rids the royal family of a member who he deemed might've caused undue, possibly republican (small r) disruption to the crown.
After she's ensconced in Bodan and starts a family, Schnee and her kids are largely on the margins, Abend disappears altogether, and Niino and his dad are front-and-center. Niino's dad is fiercely loyal to Abend, who in turn is fiercely loyal to Qualm and Falke.
He takes his duty to keep an eye on Schnee and her fam very seriously, but because it is essentially his life, he also takes the time to enjoy it, and imbues that sense of duty, and sense of enjoying one's duty, into his trusty son. That, and we see where Niino got his sweet tooth.
The idea is that King Falke was perfectly fine with his daughter leaving the palace, as long as tabs were kept on her. It doesn't seem like he intended for Schnee and her issue to be some kind of "backup plan" in case his younger daughter's issue (Schwan) wasn't up to snuff.
When Jean enters high school, Niino's dad sends him there to befriend him…even though Niino is ten years older (chalk it up to good genes). Niino also joins the photography club and makes use of the camera given to him by "Master" (Abend). It's clear both Abend and Niino's dad are grooming him for the role of Jean and Lotta's next "royal observer."
When tragedy strikes, it isn't just Schnee and Jean and Lotta's father who perish, but Niino's father, who was traveling with them, as well. Just like that, the torch has been passed. The episode then fast-forwards back to the present, where Niino is telling the adult Jean all of this, and Jean is trying to process it.
Jean doesn't care about whether he's eligible to the throne (technically he isn't); he just remembers how Niino cheered him up back in high school after his parents died, and how that helped him be strong for his little sister, who looks more and more like their gorgeous mother by the day.
The end credits play as they always have, but finally in context: the woman isn't Lotta, but Schnee, barefoot, plainly clothed, and free from the isolation of the royal palace. She gave up everything, but gained freedom and the run of the whole nation, both for her and her children.
Now that Jean (though, notably, not Lotta) knows the truth, what will he do now?
Author braveradePosted on Tue, 28 Feb 2017 Categories ACCA: 13-ku Kansatsu-ka, Anime Reviews, Winter 2017Tags abend, acca 13, accident, bread, camera, chocolate, commoner, districts, dowa, faked death, followed, grossular, information, inspection department, intermediary, jean otus, king falke, korore, lilium, lotta, madhouse, maggie, nino, observer, parents, prince schwan, royalty, schnee, schwan, sister, slice of life, snowballs, travel, voyeurism1 Comment on ACCA: 13-ku Kansatsu-ka – 08
If it were one of many other slow-burn shows, I might be itching for something to happen already near the show's midpoint (assuming this only gets 13 eps – I may be wrong), even to the point of starting to hand out 7's for lack of forward momentum.
And yet, ACCA continues to avoid such scrutiny with its unassuming, calm, quiet competence, all but unique this Winter as a show all about stopping to smell the roses…or freshly-baked bread.
One of the most eventful things to happen happens right at the start, where Jean not only goes up to Grossular, but tells him he knows he's being followed, denies any involvement in a coup, and expresses his certainty Gross isn't involved either. Grossular, after all, was the ACCA officer in Rokkusu who made things right when a horrific train accident claimed, among many others, Jean and Lotta's parents.
Jean meets Mauve at the bakery, but claims to have "nothing of note" to report to her. If he's trying to stay in the director-general's good graces, he could have at least told her about the attempted coup in Suitsu. I'd say that was pretty "of note." Mauve immediately starts to doubt Jean's usefulness…and loyalty.
Things largely quiet down from there, as both the show, the nation of Dowa, and most of its inhabitants kick back and enjoy a New Year's shindig in Jean and Lotta's apartment building.
We learn about the businessmen Jean seems to help out during his auditing duties, and Lotta receives a cake from Rail, only for Owl to give her another cake minutes later.
People seem drawn to Lotta, but there's still no indication she's anyone super-special…yet (the flashbacks also seem to eliminate some possibilities in that arena).
Officers note that the start of ACCA's hundredth year in operation isn't all that different from the start of previous years. Mauve has a pretty standard speech at an all-hands, and that's pretty much it.
Something notable does happen: Grossular joins the other four Top 5 officers in their common room, to ask Lilium why he leaked Crow to Jean. Lilium wanted Jean "to do something," and while Gross may be right that such an action was reckless, Lilium does seem to win the argument by calling for a meeting of the five to discuss what Gross knows and how they'll proceed together, no longer unilaterally. He gets that meeting; it should be a good one.
Jean then heads to Hare, the tropical district where the ACCA uniforms are short-sleeved and informal and the district inhabitants live the longest lives in the nation, and live life with gusto accordingly.
After striking out with Mauve last time, Jean redoubles his efforts to get something, anything out of Hare's chiefs. Yet, when he goes to meet Mauve at the bakery, she's not there. Is it too late?
After Hare (one of Jean's shorter audits), Jean heads to Dowa again, this time for an audit. Prince Schwan continues to try to force his grandfather's hand in subtle ways like hanging his portrait in a place he spends lots of sittin' time. Perhaps Jean will get more juicy info in Dowa.
Author braveradePosted on Tue, 14 Feb 2017 Sun, 19 Mar 2017 Categories ACCA: 13-ku Kansatsu-ka, Anime Reviews, Winter 2017Tags acca 13, accident, bread, caught, crow, dining, districts, dowa, five chiefs, followed, grossular, hare, information, inspection department, intermediary, island, isolated, jean otus, lilium, lotta, madhouse, maggie, nino, parents, plot, prince schwan, public safety, rokkusu, schwan, sister, slice of life, surveillance, travel, world-building3 Comments on ACCA: 13-ku Kansatsu-ka – 06
Two of Lotta's stalkers bond over sandwich bread
I continue to be both bemused and delighted by the sheer obsession with food, eating, and dining in ACCA and the culinary minutiae therein. The fact that these characters need their three squares a day, plus the occasional snacks, really anchors them in the land of reality. It also mirrors real life in its lack of traditional "action", but that lack hasn't dimmed my enthusiasm for the show in the slightest.
One of the most important moments of the episode—and of the show as a whole—surprisingly happens not during a meal, but during a car ride (which is also, in its utilitarian practicality, a tether to reality). In that car, Lilium basically lets slip to Jean that he has a tail, and the reason he can't detect him is that he's someone by whom he's used to being watched.
Jean's a sharp guy, so it doesn't take long for him to narrow it down to his friend the photog. And when he arrives in the wintry, hearty Birra district and Crow is already there with his camera, we finally see a look of shock on his face as Jean stops and turns back to look for…him. It's not clear whether he sees him or not (Crow seems pretty far away, though he's wearing black in the snow), but that look back is all Crow needs to know he's been found out.
"Could you two young brats flirt somewhere else? I'm trying to run a bakery here."
Two elements of the episode that are a little more obscure in their intentions were the plotlines of Eidar very unsubtly showing up to see Jean, only to be crushed when he's not there, and Maggie pursuing Lotta, only to learn where she lives, that her parents are deceased, and that she loves sandwich bread. I'm hoping we'll find out soon what Lotta's significance is, if there turns out to be any, of course.
After a rather stern dinner with Birra's ACCA supervisor, Jean walks into the tranquil, snow-covered forest to confront his tail, and the resulting confrontation is wonderfully offbeat. Jean doesn't seem betrayed; on the contrary, he seems like he could care less…though he doesn't wonder if all those times Niino got him drunk he was pumping him for information.
Back home, Jean and Niino share a meal with Lotta like nothing happened, and Jean even makes him promise to stay with Lotta and take her out to eat every day he's off on his next audit, this time of the Western-style Rokkusu District. Niino obeys, likely hoping he doesn't miss anything good.
The final scene during sunset in Rokkusu is full of intrigue, as is the little chat between Lilium and Pastice at the airport. Is Jean simply saying hello to a vaunted superior in his hometown, or is something else going on? Something…coup-y?
For the first time it looked like Jean was doing something counter to the person we'd seen up to that point; that he was finally showing us something he'd been hiding. I may be dead-wrong, but I'm having fun guessing as I enjoy the food-filled ride.
Author braveradePosted on Tue, 7 Feb 2017 Categories ACCA: 13-ku Kansatsu-ka, Anime Reviews, Winter 2017Tags acca 13, birra, bread, caught, cigarettes, crow, dining, districts, dowa, eidar, five chiefs, followed, grossular, inspection department, isolated, jean otus, king, lilium, lived in, lotta, madhouse, maggie, nino, plot, rokkusu, schwan, sister, slice of life, surveillance, travel, warbler, winter, world-building1 Comment on ACCA: 13-ku Kansatsu-ka – 05
Pan de Peace! – 03
Fuyumi-chan's family owns a bakery and the girls meet on the second floor each day to eat bread. Today they are eating red bean bunds and green tea melon bread.
Then the girls make a loaf of bread and agree to meet tomorrow.
…Roll credits.
PdP is vapid and very little happens in each episode. This wasn't a problem when it was introducing characters quickly. The brevity is a problem now that its trying to tell a story.
Either there's not enough time for the slow burn story it wants to tell or PdP has no story at all and can't even muster fluff to fill 3 minutes. Yes, it is produced well enough to sit through but there's only cute here. No humor, no bread-otaku instruction, no drama.
No real reason to watch it unless you absolutely must see all cute girls being cute shows. (dropped)
Author oigakkosanPosted on Sun, 17 Apr 2016 Categories Anime Reviews, Oigakkosan's Spring '16 Shorts, Spring 2016Tags "soft" Yuri, パンでPeace, bland, bread, comedy, cute, mini, pseudo-yuri, school, short, short format, slice of life
Moa-chan, a French baguette-toting warrioress joins the cast this episode. She may be small but she's a freshman too and all the other girls just want to eat-her-up.
Then the girls talk about drinking milk and comparing breast sizes over lunch. Now there were 4 bread friends!
I'm now 90% certain the "Bread Friends" in Pan de Peace! is knowingly making fun of semi-yuri all girl high school anime. Either that, or it's pandering to that genre's audience.
Despite the vapid boob chat and giggles about milk drinking, this episode was harmless. It was cute and made lethargic attempts at 'quirky' humor, failed, and ended in less than 4 minutes. Not much else to say honestly…
Author oigakkosanPosted on Sun, 10 Apr 2016 Wed, 13 Apr 2016 Categories Anime Reviews, Oigakkosan's Spring '16 Shorts, Spring 2016Tags "soft" Yuri, パンでPeace, bland, bread, comedy, cute, mini, pseudo-yuri, school, short, short format, slice of life
Pan de Peace! – 01 (First Impressions)
First Day of High School-chan notices another girl has bread keychain tchotchkes and immediately becomes friends. Then that girls friend joins them and they all agree to be bread friends and promise to teach each other how to make bread and go to the bakery.
There are generally two types of short format shows: a series of micro episodes and a series of small slices that feel like they are cut from a single full length episode. For better or worse, Pan de Peace! (Bread of Peace) already feels like the latter.
I'd jokingly assumed the bread friends was a eye-rolling pun about lesbianism (the girls are all very touchy-feely) but there's so little content here, I have no idea what the point is, if there is a point at all. PnP is harmless either way.
Author oigakkosanPosted on Tue, 5 Apr 2016 Wed, 13 Apr 2016 Categories Anime Reviews, Oigakkosan's Spring '16 Shorts, Spring 2016Tags パンでPeace, bland, bread, comedy, cute, mini, school, short, short format, slice of life1 Comment on Pan de Peace! – 01 (First Impressions)
Ikoku Meiro no Croisée 2
As the tiny Yune darts around learning about cheese, fresh bread, breakfast, spoons, and evil department stores, I couldn't help but conclude than she, Claude and Oscar were having way more fun than I was having watching it. That's not to say it isn't fun; the episode was pleasant enough, but now we're two weeks in and we still have no idea who Yune is and why she accompanied Oscar to France.
Perhaps that's unfair. In a season full of complex narratives and sprawling casts, it can be tempting to try to fit this round peg into a square hole. No such luck; this series is unapologetic as it is diligent in its detailed depiction of ordinary everyday life in this time and place. The nearest thing to a looming conflict seems to be the big-box store, which looks like an elegant destination for shoppers, but a cancer to local specialty store owners who will die off one by one in its price-cutting shadow.
Will that shadow gradually move in and ruin the light and breezy mood of this series? Unlikely. One thing's certain: Claude has warmed up to Yune much quicker than I envisioned. So far, this is about a girl in a strange new land, but knows the language, has made friends, and is discovering the foreign culture. What happens next? Rating: 3
Author rabujoistaffPosted on Tue, 12 Jul 2011 Tue, 12 Jul 2011 Categories Anime Reviews, Ikoku Meiro no Croisée, Summer 2011Tags 2, big box, bread, cheese, claude, department store, france, frommage, Ikoku Meiro no Croisée, japan, La croisée dans un labyrinthe étranger, oscar, paris, spoon, yune

redpajama_set_name: RedPajamaCommonCrawl


Oh, Paris. The city has been immortalized in art, song, film and literature, not to mention in the hearts of millions of style-obsessed girls the world over. What can I say that hasn't already been said? What tips and secrets can I pass on that you haven't read one hundred other places before?
Come on, did you think something else was going to be first? Pshaw. Whenever I'm in Paris I always hit up Le Bon Marché (24 Rue de Sèvres, 75007 Paris), my favourite department store on earth (with the possible exception of London's Liberty). LBM was the first department store in the world, and it's still tops in terms of selection, service and pure, unadulterated glamour.
I also love fighting over cut-price French beauty products at the (in)famous discount drugstore City Pharma (26 Rue du Four, 75006 Paris), and browsing in all the little jewelry shops in the Haute Marais. You'll need a steely resolve to conquer the former, and a fat wallet for the latter!
You're in Paris, of course you're going to treat yourself! A butter croissant here, a pastel-hued macaron there, none of it counts when you're on holiday. On past trips the macaron has been my treat of choice, but as of last month, I'm a convert to the eclair. L'Eclair du Genie (14 Rue Pavée, 75004 Paris) sells beautiful, flavourful creme-filled treats which can only be described as edible works of art. My friend and I shared a salted caramel one as well as a lemon, and both were incredible. I can't wait for my next trip there (three weeks and counting!) to sample some more.
In Paris, you're always discovering new sights and interesting places. One of my all-time favourite discoveries is L'Institute du Monde Arabe (1 Rue des Fossés Saint-Bernard, 75005 Paris). Not only is the building architecturally interesting (check out the aperture windows on the south facade, as well as Zaha Hadid's futuristic pavilion, originally commissioned by Chanel), but the museum inside has some fascinating exhibits. My favourite part? The view from the rooftop terrace, which is spectacular and (shhhh keep this to yourself) 100% free.
The nightlife in this city is something else. While the skinny-jeans-and-facial-hair crowd can be found in all the hipster bars by the Canal St. Martin, my favourite hangout is actually all the way east, in the 20th arrondissement Philipe Stark's boutique-but-budget hotel Mama Shelter (109 Rue de Bagnolet, 75020 Paris) is a great place to stay, but even if you've got digs elsewhere, take my advice and go to the bar. The scene is cool, the music is great, the bartenders are good-looking and friendly (heeey C & F! Stay cool you guys ;)) and the cocktails are super tasty. What more could you ask for?
While exploring a new city on foot is nothing groundbreaking, I tend to take things to the extreme. I've been known to pound the Parisian pavement all day long (with requisite breaks for pastry and wine, of course), and never tire of discovering new areas to wander through. Bizarrely perhaps, I particularly like walking in the famous cemeteries of Paris: they're beautiful, peaceful, and, if you like hunting down the graves of famous people, really fun. On this trip I discovered Père Lachaise (16 Rue du Repos, 75020 Paris), which on a gorgeous Spring morning, croissant in hand, was all of the above and more.
With all the exploring, eating, drinking and dancing you're doing, you'll need to take a breather now and then. If the weather's fine, head to the Jardins du Luxembourg (6th arrondissement, 75006 Paris) with the rest of Paris and try to grab a patch of grass or an iconic green garden chair. People-watching at its finest.
Bonjour, mes amies! How was your weekend? I spent the past two days in Paris, so mine predictably included walking down grand boulevards and through tiny side streets, shopping in beautiful department stores and lively antique markets, lunching in brasseries and cafés, and downing wine and pastries with wild abandon. Despite the weather (grey, chilly and a bit drizzly from time to time), I managed to have un merveilleux séjour. Here's what I got up to.
1 I love the Paris metro more than any other public transport system I've ever used. It's cheap, efficient, quick to understand and your phone still works on the trains. What more to ask?
2 On Saturday morning we ate at my favourite eatery in Paris, Rose Bakery. The friend we stayed with lives right around the corner from the rue des Martyrs, a great location for eating and drinking.
3, 4, 5, 6 Walks around Paris: the Pompidou Centre, Louvre, Opera and Tuileries Gardens.
7, 8 There's always time for a spot of shopping on my Paris itinerary. Two of my favourites are Le Bon Marche (great home floor) and Kusmi (I pick up a new tea every time I visit).
9,10 As can be expected we ate pretty well on this trip; these are two of my favourite restaurant photos. A small crêperie on the rue de Turenne, and peek inside the famous Angelina, where we ordered their signature hot chocolate (basically, a melted bar of chocolate served with a small dish of whipped cream) and caught up with friends on Sunday afternoon.
As you can imagine, I can hardly stomach the Monday-ness of this morning. In an alternate life, I never have to leave Paris for the drudgery of employment, and spend my days floating on a macaron-shaped cloud of contentment. Until then, however, I'll have to settle for these occasional weekends of escape.
PS. Did you know that Instagram now has online profiles? Check out mine to see the rest of my photos from this weekend!


The report added Gaza's water and sanitation crisis is escalating dangerously, with clean water increasingly scarce and almost a third of households not connected to a sanitation system.
Last week's shutdown of Gaza's only functioning power plant creates even more urgency, with the water utility warning that it does not have the fuel to run water and sanitation facilities when the power is off.
Water pollution is among the factors causing a dramatic increase in kidney problems in the Gaza Strip, Dr Abdallah al-Kishawi said, with a 13-14% increase every year in the number of patients admitted with kidney problems to Gaza City's Shifa Hospital.
Israel's blockade of Gaza severely limits materials from entering, making it incredibly difficult to develop water and sanitation infrastructure to meet the needs of a growing population."


text:
Q: Restoring variables if exit and re-entering activity I have a button that increase a variable by one and the result is printed in a TextView. When I rotate the device the variable value is maintained, but if I exit the activity and then re-enter, every value is reset. How can I fix this?
int n=0;

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

    final TextView tv = (TextView)findViewById(R.id.tvProva);
    Button b = (Button)findViewById(R.id.bProva);

    b.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View view) {
            n++;
            tv.setText(""+n);
        }
    });

    if(savedInstanceState!=null){
        n = savedInstanceState.getInt("n");
        tv.setText(""+n);
    }
}

@Override
protected void onSaveInstanceState(Bundle outState) {

    super.onSaveInstanceState(outState);

    outState.putInt("n", n);
}

@Override
protected void onRestoreInstanceState(Bundle savedInstanceState)
{
    super.onRestoreInstanceState(savedInstanceState);

    savedInstanceState.getInt("n");
}

A: There are a couple of potential options.
SharedPreferences, as Tejas said, below is an example of how you might use this;
SharedPreferences pref = getApplicationContext().getSharedPreferences("MyPref", 0);

To save you value;
Editor editor = pref.edit();
editor.putInt("n", n);
editor.apply();

To retrieve your value;
pref.getInt("n", 0);

The above was taken from a previous post; https://stackoverflow.com/a/24099496/4644276
Alternatively you could leverage the Repository pattern. A repository in simplest terms is an object that can persist for the life of the application, therefore it can persist across different activities. It cannot however survive killing the application unless you leverage SharePreferences as previously discussed or using an API or local database.
If you'd like to look in to the repository pattern further than it is discussed in the Guide to app architecture page on the Android Developer portal.
https://developer.android.com/jetpack/docs/guide

redpajama_set_name: RedPajamaStackExchange


Tremendous 28 acre waterfront estate with over 700 feet of low bank water frontage including a very private cove. Panoramic views of the Cascades and shipping lanes. Literally thousands of hours of land sculpting completed to create this work of art. Two drilled wells in place. Carefully placed roads meander the propertys multiple man made ponds, open space, forest and waterfront. Create your dream estate setting. Spend your days relaxing on a private beach, viewing wildlife and exploring.


Is It Time for A Sales Revolution In The Banking & Financial Services Industry?
There's been a lot scrutiny of the sales practices of financial institutions in New Zealand in recent months. The Sedgwick Report and the Royal Commission (into misconduct in banking) in Australia has cast a long shadow into New Zealand, with most of our biggest financial institutions being Australian owned. The Reserve Bank (RBNZ) and The Financial Markets Authority (FMA) are responding with their own review of the financial services culture and conduct in New Zealand.
Currently, a thematic review is being conducted by the FMA on bank sales incentive structures, which is due to be reported on by the end of the year. This has prompted many financial services institutions to proactively change the way they sell and incentive their front-line sales people. ANZ, our largest bank has already announced that their front-line people will no longer have individual sales targets.
Changes in company policy, sales frameworks, incentives and rhetoric will not be enough if there is no real intent.
Transformational change to a culture that has been embedded over decades will require significant effort and investment. Changes in company policy, sales frameworks, incentives and rhetoric won't be enough if there is no real intent. Lets face it, the old practices have been very successful in contributing to strong growth and profitability, especially in the banking sector, for a very long time – so I imagine there will be an unwavering expectation from the executive suites that any change must deliver the same, or better business outcomes.
Customers are more likely to have their genuine needs met, as front-line people aren't encouraged or incentivised to product push. Isn't this what great service is all about?
Helping customers brings genuine happiness to front-line people, which gives them a real sense of purpose.
Great team spirit is built, where team comes before self and everyone's contribution is valued. People thrive when they feel a strong sense of belonging.
A positive culture of trust and mutual benefit is created, rather than one of fear and greed.
People are coached and supported to learn and master the right behaviours. If you get the behaviours right the numbers will take care of themselves.
People act with integrity, rather than trying to game the system or acting dishonestly due to the pressure of meeting sales targets.
Workplace stress, sick leave and absenteeism are all reduced.
Employees and customers become raving advocates for the organisation.
The business outcomes are fair and sustainable.
Make no mistake I'm not suggesting in any way a regression back to the days of front-line people being order-takers, as this doesn't provide customers with good service experiences either. However, what I am suggesting is a revolution, not an evolution – this is more than just making a couple of minor tweaks.
If an organisation is serious about making change in this area it should conduct a deep dive on all sales related practices (including the digital sales channel), make well thought out decisions about the changes that need to be made and implement and embed well with good change management disciplines.


This is our Winter's Chill Beard & Mustache Wax. Our mustache and beard wax provides a firm hold for your beard and mustache.
Our Winter's Chill Beard Care line is packing the scent of Peppermint, Eucalyptus and a very light forest-like touch for a breathtaking scent that reminds us of winter amongst the frosty trees.
This scent is perfect for the cold days and even awesome on those days where you have a stuffy nose and need some scent relief.
Get some – now by adding this to your cart!
Comes in a 1 oz. tin.


THE International Labor Organization (ILO) Global Commission on the Future of Work has called on governments to commit to a set of measures in order to address the challenges caused by unprecedented transformational change in the world of work.
Co-chaired by South African President Cyril Ramaphosa and Swedish Prime Minister Stefan Löfven, the commission outlines a vision for a human-centred agenda that is based on investing in people's capabilities, institutions of work and in decent and sustainable work.
A universal labor guarantee that protects fundamental workers' rights, an adequate living wage, limits on hours of work and safe and healthy workplaces.
ILO images source: ILO Philippines.
The report is the culmination of a 15-month examination by the 27-member commission, which is made up of leading figures from business and labor, think tanks, academia, government and non-governmental organizations.
"The ILO Global Commission Report on the Future of Work is a vital contribution to global understanding of the changes occurring – and that will continue to unfold – in the world of work," said Ramaphosa.
A welder courtesy of Pixabay.
ILO's report highlights the "unique role" the ILO should play in the development and delivery of the "human-centered economic agenda" in the international system and calls on the organization to give urgent attention to the implementation of the report's recommendations.
"The issues highlighted in this report matter to people everywhere and to the planet," commented ILO Director-General Guy Ryder.


Discussion in '5.9L 24 valve Cummins' started by no mufflers, Nov 1, 2018.
i will finally be getting to the HG job on the cummins and figured maybe add in some new fuel parts.
pretty much looking for some guidance or recommendations on the good or bad.
this is my work/plow truck so im not looking to go crazy with. i plan on getting a new edge tuner and a little bigger injectors, injection pump and an adequate lift pump.
i have herd of guys using exergy injectors but for a set is over $2,000 but all others seem to be in the $800 area.


We are proud to re-open Southbank after a major refurbishment – and can now boast the best bar & brasserie in the area. Starting from street level we have totally transformed the outside area to create a little olive grove oasis opposite the Royal Festival Hall. Step inside into our funky new bar where we are serving cocktails, great beers and our fabulous selection of wines – a perfect after work drop-in before the train home from Waterloo. Our food is as good as ever and we have the new Spring a la carte menu alongside set menu lunch (9.95) and pre-theatre suppers (14.00) – all served in a gorgeous new dining area. The new private dining area is perfect for business meetings, dinners and events as well as weddings and parties – with its own private bar the space is already in demand. Pop in soon and see what's new.


To achieve employees' growth with EcoGreen, the Group has established a full range of staff training, human resources development and interactive career planning through varieties of training, improving employees' knowledge, skills and attitudes. We interact with employees through career planning to best combine employees' personal career development with the company's development, maximizing benefits to both.
EcoGreen adheres to talent and people-oriented principle as they are the most valuable assets to EcoGreen. To attract talents, EcoGreen provides a platform with advanced management and positive corporate culture for unleashing talents and realizing values.
As a leading fine chemicals manufacturer, the company aims to be a leader of green fine chemicals one-stop solution provider in the industry. EcoGreen provides board opportunities and platform for employees' development. Efforts, unity and cooperation are the keys to success and gains.
EcoGreen emphasizes on personnel training and team building at all levels, to build a strong team of professional, knowledgeable, passionate and entrepreneurial experts, managers and staffs, to let the company stay ahead and continue moving forward.
EcoGreen adheres to fairness, efficiency, motivation, consistency, and legitimacy as principles to provide employees a reasonable, competitive incentive pay, sharing the fruits of company development and helping employees achieve their personal values .


text:
Q: How to publish an app on Android play store only for those users who already have the app installed? I want to publish a private app on play store. My use case is that we are going to distribute some Android devices with a specific pre-installed application. This app should not be visible to any other Android user (who does not have the app already installed on the device). We are going to install the app initially on Android devices by ourself using the APK file directly. Then onwards, those users should be able to get the update from play store for the app whenever we want to upload the updated version. But other Android users should not be able to see the app. Only those users who have the app installed should be able to see it. I can't use enterprise email id solution for the private app as those users will use their personal Gmail account on the device. Beta testing solution can also not be used due to its limit.

A: I don't think it is that easy, especially if you only install the the APK, the user can uninstall the app and then never get it again, for something like that you have to contact google directly.
EDIT: since, you install an apk, you don't have to follow google-play policy, and program an updater inside the app, that will update itself

A: AFAIK there are no such options provided by Google Play Store.The only possible way for this is using Internal,Alpha or Beta testing only.
For more info refer link here.Other than this it is not possible

A: Note - I don't recommend doing this, and it is so hacky I wouldn't build a business on doing this. But ...
Option 1
Apps that are "unpublished" on the Play store are invisible to any user who has not installed the app before. But updates can still be given to users who already have the app.  See the documentation
So you could publish your app very briefly, install on the devices, then unpublish it. After this you could send updates using Play.
The caveats are that "existing users" is exactly that - users. Not devices. So any user account that had the app installed could install it on another device, not just the device that already has it.
Option 2
If all you were worried about was other users installing it, not other users seeing it, and you have control over the system image of your device then you have another option. You could add into your manifest a uses-feature element like:
<uses-feature
  android:name="a.very.unique.feature.name.that.only.my.devices.have"
  android:required="true"/>

Then make sure that feature is in the available features for the System rom for your device. This is obviously only available if you are a manufacturer and setting the system image for your device. It is also not secure - anyone with a custom rom can fake the features. But it might be good enough for you.
You could also do a combination of both, which is what phone manufacturers often do for apps only available on certain phones.


No more sitting on an uncomfortable rock after you've hiked for miles. Take your seat with you wherever you go with this portable lightweight Aluminum folding Stool. This seat is designed to fit in your hunting pack so you can scout out your prey more comfortably. Let's be honest, how hard is it to aim and be accurate when your more concerned with how bad your backside is hurting. You may be roughing it outdoors, but you might as well enjoy it once your there. This portable folding stool is a must for hiking, hunting, and fishing.
Folded Size: 14" x 3"
Your item will be shipped from our international warehouse. Please allow 7-35 business days for this item to arrive.


You poor girl, I am so glad they found the infection so you can get better. Those sweaters are so cute and wow, your trip sounds amazing!
Thank you Andrea. I am hoping to start feeling better and enjoying 2019.
Glad they were able to find out what was going on! Hope you feel better! Can't wait to hear all about your trips, I haven't been to Belgium and think that'd be cool!
I can't wait for Belgium. It is definitely going to cost me a lot in chocolates and those caramel waffle things. And thank you so much!
Lady you have had a rough go of doctors and medical stuff as of late. Hang in there.
I completely agree and I just want to be healthy.
Oh no - it has been a week for you! I hope the meds kick in stat. Your work trip sounds AWESOME - definitely something to look forward to!
Me too. It has been a year already. I just hope to get healthy before too long. I can't wait for my work trip!!!
Oh my gosh, Danielle. I didn't even know you could get an infection in your uterus. That is so scary. I'm so glad that you finally got some answers, though. Thank goodness you went to the doc! And girl, your trips make me so jealous. You are going to have an amazing time!
I had no idea either and we have no idea where it came from. I am just glad that they caught it.
OH MY WORD. I have never heard of such a thing. Do you have issues normally with uterus issues? Infections there? That's nuts. I'll pray the antibiotics wipe it out - but man, I feel you having to take antibiotics - I'm STILL recovering from taking a few 4 weeks... they mess me up!!!!
I'm a little annoyed on your behalf that the ER couldn't give you antibiotics! You obviously had an infection with the fever and the pain! I hope things are feeling better now. I also had some stomach pain last weekend and went to the doctor Monday. He thinks it's my gallbladder! I have an ultrasound appointment for this coming Monday. I'm excited for you for your work trip! Have a great weekend!
Me too. Especially after how much I had to spend there. And no obvious answers there. OH no. I hope it isn't your gallbladder. That would be awful. I am sending you positive thoughts.
Oh my gosh I missed that on instagram. I am so sorry! I hope that you are feeling better, how does one get an infection like that? Where are you going in France? I went to school there for a semester!
Ugh, what an ordeal! I hope this infection clears up quickly for you -- February can't come soon enough for you! Although it sounds like you have such an exciting trip coming up. Can't wait to live vicariously through you!
No it can not. I just want to be healthy again and feeling better. I can't wait for my trip. That part will be fun!!!
Ughh I'm sorry you had that pain and had to go to the hospital! Something similar happened to me in March and it was terrible - and we never found out what it was! And wow - what a work trip!!
I am not 100% sure we know what mine is. I hope my antibiotics work though. I can't wait for the work trip!
I'm so sorry to hear about your health issues...horrible. I'm glad they figured out what the problem was!
I really like that cardigan. Hope you get to feeling better soon. Hope you're able to have some fun on the work trip.
Thank you so much. I hope I can start feeling better too. The cardigan is amazing!! I wore it this weekend.
I do need suggestions and tips. I have never been before and I can't wait to go. I will take a peak inside. I am staying at a Hilton property, but I can't remember which one. I will definitely email you. Thank you so much!!!
yikes, that infection sounds awful. hope you are feeling better soon!


Ocunova is developing a novel treatment for Dry Eye Syndrome. Unlike most lubricants, rewetting solutions and anti inflammatories, our therapy targets the specific type of Dry Eye the afflicts people with diabetes. Our novel eye drop solution is the subject of extensive research and is patent protected.
Our novel compound (OCU-001) has been studied in multiple animal models and demonstrated safe in a phase 1 study. We are preparing to assess the safety and efficacy of OCU-001 in a human clinical study in the near future.
Copyright © 2018 Ocunova - All Rights Reserved.


text:
void	printstatus(struct fdc_status *fdcsp, int terse);
void	parse_fmt(const char *, enum fd_drivetype,
		     struct fd_type, struct fd_type *);
struct	fd_type *get_fmt(int, enum fd_drivetype);
void	print_fmt(struct fd_type);
int	getnum(const char *, int *);
void	getname(enum fd_drivetype, const char **, const char **);

redpajama_set_name: RedPajamaGithub