text:
\section{Introduction}
\label{intro}
The \emph{module isomorphism problem} (MIP) can be formulated as follows: design a deterministic algorithm that, given a ring $R$ and two left $R$-modules $M$ and $N$, decides in polynomial time whether they are isomorphic, and if yes, exhibits an isomorphism.

This problem is as fundamental as it is easily stated and has been studied extensively, mainly due to its broad range of applications. In particular, polynomial time algorithms were given in \cite{BL,CIK} for the case where $R$ is a finite dimensional algebra over a field and $M,N$ are finite dimensional modules over that field. For our purposes, $R$ will be a finite ring (not necessarily containing a field) and $M,N$ will be finite $R$-modules. We give an algorithm as described by the following theorem:

\begin{thm}\label{mth1}
There exists a deterministic polynomial time algorithm that, given a finite ring $R$ and two finite $R$-modules $M$ and $N$, outputs maximal length direct summands of $M$ and $N$ that are isomorphic. Moreover, the algorithm produces a map $f\in\Hom_R(M,N)$ inducing such an isomorphism.
\end{thm}

We establish the result in Theorem \ref{mth1} by a direct generalisation of the methods given in \cite{BL}, where the rings considered were finitely generated algebras over a field and the modules were finite dimensional over that field. This approach relies on the ability of finding non-nilpotent elements in non-nilpotent ideals of the endomorphism ring $\End_{R}(M)$ (cf. Proposition \ref{find}). As a direct consequence of Theorem \ref{mth1} we have that:

\begin{cor}
There exists a deterministic polynomial time algorithm that, given a finite ring $R$ and two finite $R$-modules $M$ and $N$, decides whether $M$ and $N$ are isomorphic, and if they are, exhibits an isomorphism.
\end{cor}

In Section 3, we give a second algorithm that solves the module isomorphism problem over finite rings. As a by-product, we get:

\begin{thm}\label{sideexit}
There exists a deterministic polynomial time algorithm that, given a finite ring $R$ and a finite $R$-module $M$, determines the minimum number of $R$-generators of $M$ and outputs a list of such generators.
\end{thm}

In particular, we can determine if a module is cyclic or free (by comparing cardinalities).\\

It is important to note that the algorithms given here work for any finite ring, do not distinguish between large and small characteristic and avoid computation of the Jacobson radical, thus constituting an improvement to known results.
Moreover, the algorithm in Theorem \ref{sideexit} is an interesting object \emph{per se}, due to its structure and the techniques it employs.
A common approach to this type of problems is to reduce to the semisimple case and then ``lift" (e.g. \cite{CIK,IKS}). In our algorithm, we work \emph{as if} the ring were semisimple and we have a list, $S_1,\ldots ,S_t$, of candidates for the isomorphism classes of simple modules composing it. During the running of the algorithm, we allow ourselves to be contradicted in our assumption about the simplicity of the $S_i$, in which case we update our list, quotient the ring by an appropriate two-sided nilpotent ideal and start again. If we are not contradicted, we may still draw conclusions. In this way, there is always a side-exit available and what forces an output in polynomial time is that we cannot take the side-exit too many times.\\

Rings are always assumed to contain a unit element, but are not necessarily commutative. Modules are always left unital, unless otherwise specified. We aim to develop the theory in as great a generality as possible, before restricting to finite rings for algorithmic purposes.

In the absence of an underlying field, a finite ring $R$ is assumed to be given to the algorithm via its group structure. Let $R^{+}$ be the underlying additive group of $R$. Then
\begin{equation}
R^{+}\cong \bigoplus_{i=1}^t \mathbb{Z}/n_i\mathbb{Z},
\end{equation}
for some $t\in\mathbb{Z}_{\geq 0}$, $n_1,\ldots ,n_t \in \mathbb{Z}_{> 0}$. So, in order to specify $R^{+}$, we give the algorithm a sequence of integers $n_1,\ldots ,n_t$ . To make it into a ring, we also give a bilinear map
\begin{equation}
R^{+}\times R^{+}  \to R^{+},\quad (e_i,e_j)  \mapsto e_ie_j,
\end{equation}
where $\lbrace e_i\rbrace_{i=1}^t$ are the additive generators of $R^{+}$ and we express $e_ie_j$ linearly in terms of the $e_k$'s. This amounts to giving $t^3$ numbers.

To input a module, we give a finite abelian group $(M,+)$, as before, and a bilinear map
$R\times M\to M$,
where, for every additive generator of $R$ and $M$, we express the image in terms of the additive generators of $M$.

\section{MIP via non-nilpotent endomorphisms}
\label{sec:1}

\subsection{Finding non-nilpotent elements}
\label{sec:2}

An ideal $I$ of a ring $R$ is said to be \emph{nil} if all its elements are nilpotent. The following is a well known fact:

\begin{prop}
If $R$ is a left-Artinian ring, then a left (or right) ideal $I$ of $R$ is nil if and only if it is nilpotent.
\end{prop}
\begin{proof}
Clearly if a left (or right) ideal is nilpotent, then it is nil. For the other direction, note that any nil left (or right) ideal of $R$ is contained in the Jacobson radical $\J(R)$ (see \cite{lam2001first}, Lemma 4.11). So it is enough to show that $J:=\J(R)$ is nilpotent.

Suppose $J$ is not nilpotent. Since $R$ is left-Artinian, the sequence of consecutive powers of $J$ must stabilize, i.e. $\exists n\in \mathbb{Z}_{> 0} \text{ such that } J^n=J^{2n}\neq 0$. Then there exists a left ideal $L$ s.t. $J^nL\neq 0$. Suppose $L$ is minimal with this property. Since $J^nL\neq 0$, there exists $a\in L$ such that $J^na\neq 0$. Now $Ra\subseteq L$ and $J^n a\subseteq L$. Moreover, $0\neq J^na=J^{n}J^{n}a$ and $0\neq J^n a\subseteq J^n Ra$, so by minimality of $L$, we have that $L=Ra=J^na$. Hence $a=xa$, for some $x\in J^n$. It follows that $a=0$ (otherwise, $1-x$ would be both a unit and a left zero-divisor, which is not possible). This gives the desired contradiction.
\end{proof}

If $R$ is finite, the proof above can be translated into an algorithm for finding a non-nilpotent element in a non-nilpotent left ideal of $R$:

\begin{prop}\label{find}
There exists a deterministic polynomial time algorithm that, given a finite ring $R$ and a left ideal $I$, determines whether or not $I$ is nilpotent and if it is not, produces a non-nilpotent element lying inside it.
\end{prop}
\begin{proof}
Suppose $R$ is a $k$-algebra, where $k$ is a commutative ring ($R$ will certainly be an algebra over its centre or its prime subring) and let $I$ be a non-nilpotent left ideal of $R$.  Since $R$ is finite, we can find $n\in\mathbb{Z}_{>0}$ such that $I^n=I^{2n}\neq 0$. Suppose $I^n$ is generated over $k$ by a set $A$ and over $R$ by a set $B$, i.e.
\begin{equation}
I^n=\sum_{\alpha\in A} k\alpha=\sum_{\beta\in B} R\beta.
\end{equation}
Then $0\neq I^{2n}=\sum_{\alpha\in A \atop \beta\in B} R\beta\alpha$, so there exists $b\in B$ and $a\in A$ such that $b\cdot a\neq 0$.
Consider $I^n a\subseteq Ra$ (note $I^na\neq 0$ since it contains $ba$). If equality holds, then we can write $a=xa$, for some $x\in I^n$ and since $a\neq 0$, it must be the case that $x$ is non-nilpotent, as required (otherwise $1-x$ would be a unit and a left zero divisor at the same time, which is impossible). Suppose now that the inclusion is strict. Then there exists $c\in I^na$ such that $I^n c\neq 0$ (otherwise $I^nI^na=I^na=0$, which is a contradiction, since $0\neq ba\in I^{n}a$).
Moreover, we will be able to find such a $c$ among the $k$-generators of $I^n a$.
We have now produced a smaller ideal $Ra\supsetneq I^na\supset Rc$ such that $I^nc\neq 0$. We replace $a$ by $c$ and keep going. This process must terminate after a finite number of steps, since $R$ is finite.
We also note that $\length({_R}R)\leq \log_2{\vert R \vert}$ and $I$ is nilpotent if and only if $I^{\length({_R}R)}=0$, so it is possible to test if $I$  is nilpotent.
\end{proof}

\subsection{Splitters}
Let $R$ be an algebra over a commutative Artinian ring $k$ such that $R$ is finitely generated as a $k$-module. Let $M_1$ and $M_2$ be two $R$-modules of finite length over $k$. Note that $\Hom_R(M_1,M_2)$ is a left $k$-module and a right $\End_{R}(M_1)$-module. Moreover, $\End_{R}(M_1)$ is left (and right) Artinian.
\\

Following \cite{BL}, we make the following definition:
\begin{definition}
Let $f\in\Hom_{R}(M_1,M_2)$.
A decomposition $M_1=N_1\oplus K_1$, for $N_1,K_1\leq M_1$, is called an \emph{$f$-decomposition} if $N_1\neq 0$, $\ker(f)\leq K_1$ and the image of $N_1$ under $f$, which we denote by $fN_1$, is a direct summand of $M_2$.
If $M_1$ has an $f$-decomposition, we say that $f$ is a \emph{splitter}.
\end{definition}

The following proposition and its proof, together with Proposition \ref{find}, allow us to algorithmically decide if a given homomorphism $f$ is a splitter, and if it is, to produce an $f$-decomposition:

\begin{prop}\label{splitternilp}
Let $f\in\Hom_{R}(M_1,M_2)$. Then $f$ is a splitter if and only if there exists $g\in \Hom_{R}(M_2,M_1)$ such that $gf$ is not nilpotent.
\end{prop}
\begin{proof}
The proof of this proposition is the same as the one given in \cite{BL}, Lemma 3.3, which treats the case when $R$ is a finitely generated algebra over a field and $M_1,M_2$ are finite dimensional over that field. For completeness we include the proof of the ``if" statement here.

Suppose $g\in\Hom_R(M_2,M_1)$ is such that $gf$ is not nilpotent. Let $s=gf$ and $t=fg$. Since $M_1$ and $M_2$ have finite length over $k$ (so are both Artinian and Noetherian over R), we can apply Fitting's Lemma (see \cite{lang}, Chapter X, Proposition 7.3) to say that $M_1=\ker(s^d)\oplus \im(s^d)$, $M_2=\ker(t^d)\oplus \im(t^d)$, for $d=\max\lbrace \length_R(M_1),\length_R(M_2)\rbrace$ and the restriction of $t$ to $\im(t^d)$ is an automorphism. We have that $s^dM_1\neq 0$ since $s$ is not nilpotent, $\ker(f)\leq \ker(s^d)$ by definition of $s$ and
\begin{equation*}
f(s^dM_1)=t^d(fM_1)\subseteq t^dM_2 = fg(t^dM_2)=fs^d(gM_2 )\subseteq f(s^dM_1),
\end{equation*}
so $f\im(s^d)=f(s^dM_1) = t^dM_2 =\im(t^d)$, which is what is required for $f$ to be a splitter.
\end{proof}

Suppose now that we are given $f\in \Hom_R(M_1,M_2)$ and we wish to know whether it is a splitter. To do this, we first compute a set $C$ of $k$-generators of $\Hom_{R}(M_2,M_1)$. Now consider the left ideal $I$ of $\End_R(M_1)$ generated by the set $Cf=\lbrace cf \mid c\in C\rbrace$. If $I$ is nilpotent (which we can determine by Proposition \ref{find}), then since $\Hom_{R}(M_2,M_1)f=\spans _k (Cf)= I$, we will not be able to find a non-nilpotent element of the form $gf$, so $f$ cannot be a splitter. Otherwise the algorithm of Proposition \ref{find} will produce some $g\in\Hom_R(M_2,M_1)$ witnessing that $f$ is a splitter and we can produce an $f$-decomposition by Proposition \ref{splitternilp}.\\

We now have a way of identifying splitters. But we would not like to have to look for them over all homomorphisms. The following proposition tells us that we can restrict our attention to a considerably smaller set:
\begin{prop}\label{lookfor}
If a splitter exists, then there exists a splitter in any set of $k$-module generators of $\Hom_{R}(M_1,M_2)$ and in any set of $\End_{R}(M_1)$-generators of $\Hom_{R}(M_1,M_2)$.
\end{prop}
\begin{proof}
To see that a set of $\End_R(M_1)$-module generators is enough, note that
\begin{equation*}
f \text{ is not a splitter } \iff \Hom_R(M_2,M_1)f\subseteq \J(\End_R(M_1)).
\end{equation*}
Let $B$ be a set of $\End_R(M_1)$-module generators of $\Hom_R(M_1,M_2)$. Suppose that $B$ does not contain any splitters. Then
\begin{align*}
\Hom_R(M_2,M_1)\Hom_R(M_1,M_2) & =\sum_{b\in B}\Hom_R(M_2,M_1)b\End_R(M_1) \\
& \subseteq \J(\End_R(M_1)),
\end{align*}
and therefore $\Hom_R(M_1,M_2)$ cannot contain a splitter.

Finally, note that any set of left-$k$-module generators is also a set of right-$\End_R(M_1)$-module generators.
\end{proof}

Putting these results together, we can construct an algorithm satisfying the requirements of Theorem \ref{mth1} as follows. We view $R$ as an algebra over its prime subring $k$. Let $M$ and $f$ be two auxiliary variables that at the end of the algorithm will become equal to the desired maximal length direct summand of $M_1$ that has an isomorphic copy as a direct summand of $M_2$, and the isomorphism-inducing $f\in\Hom_R(M_1,M_2)$.
At the beginning of the algorithm, we put $M$ and $f$ equal to zero. We compute $B$, a set of $k$-generators of $\Hom_{R}(M_1,M_2)$ (or a set of $\End_{R}(M_1)$-module generators thereof). For each element of $B$, we test if it is a splitter: by Proposition \ref{lookfor}, if a splitter exists, we will find one inside $B$.
Finding a splitter $b\in B$ also gives us a decomposition $M_1=N_1\oplus K_1$ and $M_2=N_2\oplus K_2$ with $N_1\cong N_2$ via $b$. We make the following replacements: $M:=M\oplus N_1$ and $f:=f\oplus (\text{the restriction of $b$ to $N_1$})$, $M_1:=K_1$, $M_2:=K_2$, $B:=\text{set of $k$-module generators of the new $\Hom_{R}(M_1,M_2$})$ and we start again. Note that we are all the time assuming the Krull-Remak-Schmidt Theorem (\cite{lang}, Chapter X, Theorem 7.5), which ensures existence and uniqueness up to isomorphism of the direct summands of $M_1$ and $M_2$.

\section{MIP via an approximation of the Jacobson radical}

Let $R$ be a finite ring. We observe that determining whether two finite $R$-modules $M_1$ and $M_2$ are isomorphic reduces to determining if a module is free of rank one. Let $E:=\End_{R}(M_1)$ and $K:=\Hom_{R}(M_2,M_1)$. Note that $K$ is a left $E$-module.

\begin{prop}\label{tfae}
Let $R,M_1,M_2,E,K$ be as above. The following are equivalent:

(i) $M_1\cong M_2$ as $R$-modules.

(ii) $E\cong K$ as $E$-modules and the image of $1_E$ in $K$ under any such isomorphism is an isomorphism between $M_1$ and $M_2$.

(iii) There exists an $E$-module isomorphism $\phi: E\rightarrow K$ such that $\phi(1_E)$ is an isomorphism between $M_1$ and $M_2$.
\end{prop}
\begin{proof}
The implications $(ii\Rightarrow iii)$ and $(iii\Rightarrow i)$  are immediate.
For $(i\Rightarrow ii)$, suppose $f:M_1\xrightarrow{\sim} M_2$. Then $E\cong K$, so let $\phi:E\xrightarrow{\sim}K$ be any such isomorphism. Let $\lambda:=\phi(1)$. Then there exists a unique $\epsilon\in E$ such that
\begin{equation*}
f=\phi(\epsilon)=\phi(\epsilon \cdot 1)=\epsilon \phi(1)=\epsilon \lambda,
\end{equation*}
where the third equality follows by $E$-linearity of $\phi$.
Since $f$ is injective, so must $\lambda$ be. Moreover, since $M_1$ and $M_2$ have the same length, $\lambda$ must also be surjective.
\end{proof}

Hence an algorithm that can establish freeness of rank one of a module provides a solution to the module isomorphism problem.

\subsection{Computing minimum number of generators}
Let $R$ be a left-Artinian ring. Suppose, along with $R$, we are given a collection $S_1,\ldots , S_t$ of nonzero left $R$-modules of finite length. Ideally, we would like this collection to be a set of representatives for the isomorphism classes of simple $R$-modules. However, for now, we only require that each simple $R$-module occurs in at least one of the $S_i$, i.e. it occurs as a quotient in its composition series. We can take, for example, $t=1$ and $S_1={_R}R$.

Let
\begin{equation}
\mathfrak{a}=\bigcap_{i=1}^t \ann_R(S_i),
\end{equation}
where $\ann_R(S_i)=\ker(R\to\End_{\mathbb{Z}}(S_i))$. Again, ideally we would like $\mathfrak{a}$ to be the Jacobson radical, $\J(R)$. In reality, we only have one inclusion, namely $\mathfrak{a}\subseteq \J(R)$, since an element of $\mathfrak{a}$ will kill all the $S_i$'s and so will kill all submodules, quotients and submodules of quotients of the $S_i$. Since every simple $R$-module occurs in at least one of the $S_i$, we have that $\mathfrak{a}$ kills all simple $R$-modules, which is equivalent to being inside the Jacobson radical. Furthermore, since $R$ is left-Artinian, $\J(R)$ is nilpotent and hence $\mathfrak{a}$ is nilpotent.

We will construct an algorithm that, given $R$ and a collection of $S_i$, either ``improves" the sequence of $S_i$ or computes $\mathfrak{a}$ and an isomorphism of $R$-modules
\begin{equation}\label{iss}
R/\mathfrak{a}\xrightarrow{\sim} \bigoplus_{i=1}^t S_i^{a_i},
\end{equation}
for suitable $a_i\in\mathbb{Z}_{\geq 0}$.
Similarly, we construct an algorithm that, when also given a finitely generated $R$-module $M$, either ``improves" the sequence of $S_i$ or computes an isomorphism of $R$-modules
\begin{equation}
M/\mathfrak{a}M\xrightarrow{\sim} \bigoplus_{i=1}^t S_i^{c_i},
\end{equation}
for suitable $c_i\in\mathbb{Z}_{\geq 0}$.
Consider the quantity $l{(S_i)_{i=1}^t}=\sum_{i=1}^t(2\length(S_i)-1)\in\mathbb{Z}_{\geq 0}$. An ``improvement" in the sequence is measured by a decrease in $l{(S_i)_{i=1}^t}$ and occurs either when we remove one of the $S_i$ from the list (either because the list already contains an isomorphic copy of it or because it is not needed) or when we discover a nonzero proper submodule $T$ of one of the $S_i$ which witnesses nonsimplicity of $S_i$ and which we use to replace $S_i$ by $S_i/T$ and $T$, that now stand a better chance of being simple. Note that the factor 2 in the expression of $l{(S_i)_{i=1}^t}$ ensures it decreases even when we remove an $S_i$ from the list. \\

Let us first write out the details of the routine that finds an isomorphism as in (\ref{iss}). We will call this routine UPDATE and within the main algorithm we will call it whenever we have improved on our sequence of $S_i$'s and we want to update our $\mathfrak{a}$, the sequence of $a_i$'s and the isomorphism $R/\mathfrak{a}\xrightarrow{\sim} \bigoplus_{i=1}^t S_i^{a_i}$.

Let $\mathfrak{S}=\lbrace (S_i)_{i=1}^t \mid t\in \mathbb{Z}_{\geq 0}$, each $S_i$ is a nonzero finite length $R$-module and each simple $R$-module occurs as a factor in the composition series of at least one $S_i\rbrace$.

\begin{prop}\label{aa1}
There exists a deterministic polynomial time algorithm that takes as input a finite ring $R$ and a collection of modules $(S_i)_{i=1}^t\in\mathfrak{S}$ and outputs a sequence of integers $a_1,\ldots ,a_t\in\mathbb{Z}_{>0}$, a two-sided nilpotent ideal $\mathfrak{a}'$ of $R$ and an isomorphism $\varphi:R/\mathfrak{a}'\xrightarrow{\sim} \bigoplus_{i=1}^{t'} (S_i')^{a_i}$, where $t' \in\mathbb{Z}_{\geq 0}$, $\mathfrak{a}'=\cap_{i=1}^{t'}\ann_R(S_i')$ and $(S_i')_{i=1}^{t'}\in\mathfrak{S}$ is such that $l{(S_i')_{i=1}^{t'}}\leq l{(S_i)_{i=1}^t}$.
\end{prop}
\begin{proof}
Let $\mathfrak{b}$ be a left ideal of $R$ which we will use as an intermediate variable that at the end of the algorithm will become equal to the desired $\mathfrak{a}'$.
We start off by setting $\mathfrak{b}=R$, $t'=t$, $a_i=0$ and $S_i'=S_i$ for $1\leq i\leq t$, and $\varphi=0$. Throughout the algorithm $\varphi:R/\mathfrak{b}\xrightarrow{\sim} \bigoplus_{i=1}^{t'} (S_i')^{a_i}$ and $\cap_{i=1}^{t'}\ann_R(S_i')\subseteq \mathfrak{b}$ will be invariant. If for all $i$ we have $\mathfrak{b}S_i'=0$, then we are done. Otherwise, we choose $1\leq h\leq t'$ and $s\in S_h'$ such that $\mathfrak{b}s\neq 0$. We define
\begin{equation}
\psi:R\to S_h'\oplus \bigoplus_{i=1}^{t'} (S_i')^{a_i}, \quad \psi(r)=(rs,\varphi(r)).
\end{equation}
Then $\ker(\psi)= \ann_R(s)\cap\mathfrak{b}\subsetneq\mathfrak{b}$, since $\mathfrak{b}$ did not annihilate $s$. Let $\overline{\psi}$ be the map induced by $\psi$ on $R/\ker(\psi)$. Then
\begin{equation*}
\overline{\psi} \text{ isomorphism } \iff \psi \text{ surjective }
\iff S_h' \oplus\lbrace 0 \rbrace  \subseteq \im(\psi)
\iff \mathfrak{b}s=S_h'.
\end{equation*}
This comes as a confirmation of our intuition: we are treating the $S_i'$ as if they were simple, so if $0\neq \mathfrak{b}s\leq S_h'$, then we would want $\mathfrak{b}s$ to be the whole of $S_h'$.

If $\psi$ is surjective, we make the following replacements:
$a_h:=a_{h}+1$, $\varphi:=\psi$ and $\mathfrak{b}:=\ker{\psi}$. Note that now $\mathfrak{b}$ has smaller length than before.

If $\psi$ is not surjective, we replace $S_h'$ by $S_h'/\mathfrak{b}s$, $t'$ by $t'+1$ and put $S_{t'+1}':=\mathfrak{b}s$ and $a_{t'+1}:=0$. In addition, we replace $\varphi$ by the composition $\pi\circ \varphi$, where $\pi$ is the canonical map $\pi:\bigoplus_{i=1}^{t'} [\old (S_i')^{\old a_i}]\to \bigoplus_{i=1}^{t'+1} [\new (S_i')^{\new a_i}]$ and we replace $\mathfrak{b}$ by the kernel of our new $\varphi$. Note that the new $\mathfrak{b}$ will contain the old $\mathfrak{b}$, but the improvement is now sitting in the new $S_i'$.

Consider the quantity $l{(S_i')_{i=1}^{t'}}=\sum_{i=1}^{t'}(2\length(S_i')-1)$. Then $0\leq t'\leq l{(S_i')_{i=1}^{t'}}$, since each $S_i'$ has length at least 1. Also, $l{(S_i')_{i=1}^{t'}}$ is bounded above by its initial value. At each iteration of the algorithm, we either see a decrease in the value of $l{(S_i')_{i=1}^{t'}}$, when we improve our sequence, or we see a decrease in the length of $\mathfrak{b}$, whose length is initially equal to $\length(R)$. Since $\length(R)\leq \log_2(\vert R \vert)$ (recall that $R$ was finite), the algorithm runs in polynomial time.
\end{proof}

We now turn to the main algorithm:

\begin{thm}
There exists a deterministic polynomial time algorithm that, given a finite ring $R$ and a finite $R$-module $M$, determines the minimum number of $R$-generators of $M$ and outputs a list of such generators.
\end{thm}
\begin{proof}
We begin by running the UPDATE algorithm presented in Proposition \ref{aa1} with input $t=1$ and $S_1={_R}R$.
Let $X=\lbrace x_1,\ldots, x_d\rbrace$ be a set of $R$-generators of $M$. Then $\overline{X}=\lbrace x_i+\mathfrak{a}M\mid 1\leq i\leq d\rbrace$ generates $M/\mathfrak{a}M$ over $R/\mathfrak{a}$. This gives a surjective
homomorphism $(R/\mathfrak{a})^d\cong \bigoplus_{i=1}^t S_i^{a_i d} \twoheadrightarrow M/\mathfrak{a}M$. Relabel the $S_i$ to get a map $\bigoplus_{i\in I} S_i\twoheadrightarrow M/\mathfrak{a}M$. We would like to find a subset $J\subseteq I$ for which this map becomes an isomorphism. In this process, the standard proof of its existence in the case where the $S_i$ are simple (see \cite{lang}, Chapter XVII, Lemma 2.1) may produce a witness of nonsimplicity of some $S_i$, in which case we refine the sequence and call the UPDATE subroutine to start over.
In the end, we will have produced an isomorphism $\bigoplus_{i=1}^t S_i^{c_i} \xrightarrow{\sim}M/\mathfrak{a}M $, for some $c_i\in\mathbb{Z}_{>0}$ (note that the ideal $\mathfrak{a}$ may now be different from the one we started with).

Let $n=\max_i\lbrace \lceil \frac{c_i}{a_i}\rceil \rbrace$.
If for all $i\neq h$ we have $\Hom_R(S_i,S_h)=0$, then there is a surjective map $(R/\mathfrak{a})^n\twoheadrightarrow M/\mathfrak{a}M$ and $n$ is minimal with this property. Since $\mathfrak{a}\subseteq \J(R)$, we can lift this to a map $R^n\twoheadrightarrow M$ and produce $n$ generators of $M$.
If, however, for some $i\neq h$ we have that $\Hom_R(S_i,S_h)$ contains a nonzero element $f$, then we can once again refine our sequence (either using $\ker(f)\neq 0$ or $\im(f)\neq S_h$ or removing one of $S_i$ or $S_h$ from the list if they are isomorphic) and start over by calling the UPDATE routine on our newly improved sequence.

To establish the running time, consider again the quantity $l{(S_i)_{i=1}^t}=\sum_{i=1}^t (2\length(S_i)-1)$. At each iteration of the algorithm, we either produce an output or we improve on our sequence, which results in a decrease in $l{(S_i)_{i=1}^t}$.
\end{proof}

In particular, we may use this algorithm to determine if a module is cyclic or free of rank 1 (by comparing cardinalities), thus  settling the module isomorphism problem yet again (see Proposition \ref{tfae}).

\begin{note}
As we have described the algorithm, after we have improved our sequence, calling the UPDATE subroutine overwrites our so far acquired knowledge about $\mathfrak{a}$.
There is possibly a way of saving some of this information by running a modified version of the UPDATE routine, which would thus make the main algorithm slightly more efficient. However, being more precise about this here would obscure the idea of the algorithm.
\end{note}

\begin{acknowledgements}
I would like to thank my PhD supervisor, Professor Hendrik Lenstra, for his generous guidance and his unfailing patience, particularly in the production of the side-exit algorithm.
\end{acknowledgements}

redpajama_set_name: RedPajamaArXiv


text:
Q: Prove that $\mathbb{N}$ with cofinite topology is not path-connected space. $\mathbb{N}$ is the set of natural numbers.
Let $U_{\alpha \in A} \subset \mathbb{N}$ be the subset such that its complement $\mathbb{N}$ \ $U_\alpha$ is a finite subset.
Then $T= \{\emptyset, \mathbb{N}, U_\alpha |\alpha \in A \}$ is a topology on $\mathbb{N}$.
Prove that $\mathbb{N}$ is not path-connected space with the following definitions below
Definition: A topological space $X$ is path-connected space if for every points $a,b \in X$, there is a continuous map $f:[0,1]\rightarrow X$ such that $f(0)=a$ and $f(1)=b$.
Definition: Let $X$ and $Y$ be two topological spaces. A mapping $f:X \rightarrow Y$ is continuous at a point $a \in X$ if and only if for any open set $W$ containing $f(a)$ in $Y$, there exists an open set $G$ containing $a$ in $X$ such that $f(G) \subseteq W$.
Following statements are equivalent about continuity of map $f$ on two topological spaces $X$ and $Y$:

*

*"$f$ is continuous on $X$" is equivalent to "for every open set $W$ in $Y$, the inverse image $f^{-1}(W)$ is open in $X$.

A: U can easily show that:
$[0,1]\subset \mathbb R$ can not be written as countable union of closed and disjoint sets $\implies $ $\mathbb N$ with the cofinite topology is not patch-connected
Prove: If $\mathbb N$ where patch-connected then there is a continuous map $\gamma :[0,1] \rightarrow \mathbb N$ such that $\gamma(0)=x$ and $\gamma(1)=y$ for all x,y $\in \mathbb N$. Then $[0,1]=\gamma^{-1}(\mathbb N)=\gamma^{-1}(\bigcup_{i=1}^{\infty}\{i\})=\bigcup_{i=1}^{\infty}\gamma^{-1}\{i\}$ Since finite elements in the confinite topology are closed and $\gamma$ is continuous we have a countable union of closed disjoints sets. A contradiction.
Now it is enough to prove that there does not exist such an union.
Here is the basic idea: (Again by contradiction)
1) Suppose that $\bigcup_{i=1}^{\infty}B_n=[0,1]$ with closed and disjoint $B_n$
2) Construct a decreasing sequence of closed intervals $...I_4 \subset I_3 \subset I_2 \subset I_1 \subset [0,1]$ such that $B_n \cap I_n =\emptyset$
3) The set $\bigcap_{n=1}^{\infty}I_n$ is not empty. Let $x$ be an element of the set. Then $x$ is element of every $I_n$ and of exactly one $B_n$, hence $B_j \cap I_j=\{x\}$ for one $j\in \mathbb N$. A contradiction. The statement follows.
I left the construction of the intervals $I_n$ by the reader :)

redpajama_set_name: RedPajamaStackExchange


1206-1222 Fleeing the fall of Constantinople, Theodore, related by marriage to the former emperor Alexius III, took control of Nicaea where he soon began the arduous task of forming a government in exile. To add credibility to his authority he reclaimed as much of the imperial staff as he could find and restored to them their former salaries and titles. The new Latin emperor based out of Constantinople, Baldwin, tried defeating Theodore before he became too powerful. He thus departed Constantinople at the head of an army, met Theodore's forces and defeated them but Theodore himself managed to escape and Baldwin could not afford a protracted pursuit. Once he quit the battle scene Theodore just began consolidating his domains all over again. He went on to repel an incursion by the Seljuks on the eastern front and then to be formally recognized by the Latins who by now realized they were not going to be able to get rid of him. A few more prosperous years passed until his peaceful death in 1222 at which time the imperial title was passed on to his son-in-law John.
SB 2064, DOC IV 2 Electrum Trachy Obv: Jesus Christ seated, facing, raising hand and holding Gospels.
Rev: Theodore I and St. Theodore standing, facing, each holding a sword and together a staff with star atop. (Magnesia).
SB 2061, DOC IV 5a Billon Trachy Obv: The Virgin Mary seated facing, holding baby Jesus.
Rev: Theodore and St. Theodore standing facing together holding patriarchal cross. (Nicaea).
SB 2071, DOC IV 11 Billon Trachy Obv: St. Theodore standing facing, holding spear and shield.
Rev: Theodore standing facing, holding scepter and globus cruciger. (Magnesia). $190 4/1/01.


"Makasib contest allows 10 participants per day to win 5000 points by answering a daily question about the loyalty program. Costumer can participate in the contest by sending (M) as a text message to 80588 and answer the question which will give the chance of entering the ruffle draw and earn the points." He further added.
"A special corner at our festival outlets in Amerat and Naseem parks have been set up to provide the public with more information about Makasib Loyalty program" Usama Al Rawas concluded.
The rewarding loyalty initiative' the Makasib' program was launched last November in line with the company's strategy to enhance customer experience by offering customers the opportunity to benefit from a diverse range of rewards and benefits through earning points with usage of any Omantel mobile, fixed line, and internet services. These points can then be exchanged for rewards and benefits that best suit each customer's needs individually.
How to enjoy Makasib ….
All Omantel costumers are invited to join Makasib by dialing the short code *224# and following the simple instructions or sending an SMS with the word "Makasib" or "M" to 91224 . After registration, customers will start earning Makasib and Status points for every 100 Baisa they spend on any Omantel service. The status points, which are valid for one year, determine the membership status: Sliver, Gold, or Platinum. Upon joining the program, customers will be automatically enrolled to Silver Status and will start earning one Makasib and one Status point for every 100 Baisa they spend. Once a customer has accumulated 3,000 Status points, the customer will be upgraded to Gold. Here, the customers will receive 50% bonus Makasib points for every 100 Baisa they spend. When a customer has collected 12,000 Status points, the customer membership will be upgraded to Platinum and will start receiving 100% bonus Makasib points for every 100 Baisa they spend.
Omantel is the leading provider of broadband services in the Sultanate of Oman connecting individuals and corporate customers in different regions through advanced fixed, mobile and Internet network. By offering innovative products and unmatched after-sales service, Omantel has the largest customer base in the Sultanate with more than 3 million customers. Omantel has been recently selected as the winner of the "Platinum Award for Quality and Best Trade Name" by European "Other Ways" Association for the quality of its network and efforts made to improve the customer experience.


Czech Republic: On October 25th, pupils and teachers from our school prepared a project day called Let´s Help Each Other. Pupils tried out helping little children at the local nursery schools to create Halloween toys or decorations. Other children took part in a marathon or some different activities during which they practiced cooperation, support and help.


A factor VIII activity blood test lets doctors evaluate the functioning of a protein that helps blood to clot. A clot is a lump of blood that the body produces to prevent excessive bleeding by sealing leaks from blood vessels caused by wounds, cuts, scratches, or other conditions.
Blood clotting is a process involving platelets (also called thrombocytes) and proteins called clotting factors. Platelets are oval-shaped cells made in the bone marrow. Most clotting factors are made in the liver. When a blood vessel breaks, platelets are first to the area to help seal the leak and temporarily stop or slow bleeding. But for the clot to become strong and stable, the action of clotting factors is required.
Factor VIII, with factor IX, is involved in the last step of the clotting process — the creation of a "net" that closes a torn blood vessel. When an abnormal gene causes a child to be deficient in factor VIII, the result is a bleeding disorder known as hemophilia A. A factor IX deficiency is known as hemophilia B. Both conditions are usually hereditary, but also can occur spontaneously.


The summer is off to a bang with three new and very exciting mobile slot games from Microgaming. All three games are available at your favorite Microgaming mobile casino, and they're already gaining a huge following thanks to the fun gameplay, huge payouts, and multiple bonus features.
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text:
kogami
No-Header (SNES)
Action > Shooter
Readme File
Star Fox 2 is an action shoot 'em up game for the Super Nintendo. The game's release on the Super Nintendo was cancelled in 1996, however in 2017 the game got an official release on the Super NES Classic Edition (SNES Mini).
Dialogues…….. 100 %
Objets………… 100 %
Graphismes….. 100 %
Crédits……….. 100 %
Everything is translated into French in the game. The patch is in IPS format. Additional optional patches are included that fix some bugs in the game. See the readme for details.
Star Fox 2 (J) (Beta).smc (1 048 576 bytes)
CRC 79F3433D
kogami Hacking Project leader, translated credits
Palpatine Translation
Skeud Hacking Decompression/recompression
Happexamendios Script Editing/Revision

redpajama_set_name: RedPajamaCommonCrawl


text:
Browsing stories tagged GENERATIONS
Oricon Jpop charts June 27th - July 3rd 2016 GENERATIONS from EXILE and The Idolmaster top
GENERATIONS from EXILE tops the singles chart with Namida (涙) selling over 93,000 copies. Kuwata Keisuke from Southern All Stars lands at #2 with solo single Yoshiko-san. SawanoHiroyuki[nZk] (aka Sawano Hiroyuki) lands at #6 with Into the Sky EP...
Ayumi Hamasaki, Choshinsei, EXILE, GENERATIONS, Idolmaster, MY FIRST STORY, Oricon, Oricon Weekly, SawanoHiroyuki[nZk], SawanoHiroyuki[nZk] Into the Sky
Oricon Jpop charts June 13th - 19th 2016 BTOB and HiGH & LOW top
Korean unit BTOB tops the singles chart with Japanese single L.U.V. selling over 76,000 copies. Its their 5th Japanese single and first #1 charting in Japan. Idol group AKB48 slips to #3 with 44th single Tsubasa wa Iranai (翼はいらない) with a three week total of over 1.55...
2PM, BTOB, EXILE, Galileo Galilei, Galileo Galilei Sharin no Jiku, GENERATIONS, GENESIS OF 2PM, Haruka Tomatsu, Haruka Tomatsu BEST SELECTION starlight, Haruka Tomatsu BEST SELECTION sunshine, J Soul Brothers, kobukuro, Oricon, Oricon Weekly
Oricon Jpop charts February 29th - March 6th 2016 KAT-TUN and GENERATIONS from EXILE TRIBE top
KAT-TUN tops the singles chart with 26th single UNLOCK selling over 194,000 copies. The title track has tie-up as theme song for TV drama series Kaito Yamaneko staring member Kamenashi Kazuya. Landing at #2 is fictional idol group μ's (Muse) from Love Live! School Idol ...
Afilia Saga, Arashi, Arashi Fukkatsu LOVE, ClariS, ClariS SPRING TRACKS - Haru no Uta, DOES, DOES KNOW KNOW KNOW, Eir Aoi, Eir Aoi Accentier, GENERATIONS, Idolmaster, KAT-TUN, KAT-TUN UNLOCK, Love Live!, Oricon, Oricon Weekly, SCANDAL, SCANDAL YELLOW, μ's (Muse)
Oricon Jpop charts January 25th - 31st 2016 GLAY and Yamapi top
GLAY tops the singles chart with single G4 IV selling over 39,000 copies. It's the 4th volume of their G4 series and includes Sora ga Aozara de Arutami ni (空が青空であるために) featured as opening theme for anime Ace of the Diamond Second Season. Landing at #7 is...
Acid Black Cherry, EXILE, FUNKY MONKEY BABYS, GENERATIONS, GLAY, Kishidan, Mamoru Miyano, Mamoru Miyano HOW CLOSE YOU ARE, Oricon, Oricon Weekly, SAO, Sword Art Online, Yamapi, Yamashita Tomohisa, Yuki Kajiura
Oricon Jpop charts September 14th - 20th 2015 GENERATIONS and AAA top
GENERATIONS (from EXILE TRIBE) tops the singles chart with 10th single ALL FOR YOU selling over 153,000 copies. Idol unit DENPA GUMI.inc climbs to #2 with Ashita Chikyuu ga Konagona ni Natte mo and Korean band FTISLAND climb to #2 with 15th Japanese single ...
AAA, AAA 10th Anniversary, Crystal Kay, DEMPA GUMI.inc, FTISLAND, FTISLAND PUPPY, GENERATIONS, Kalafina, Kalafina far on the water, Mamoru Miyano, Mamoru Miyano FRONTIER, Namie Amuro, Oricon, Oricon Weekly
Oricon charts February 16th - 22nd 2015 SMAP and Generations
SMAP tops the singles charts with double a-side single Karei naru Gyakushu / Humor Shichau yo selling over 160,000 copies. Their 54th single is featured as theme song for TV drama Zeni no Senso (銭の戦争) staring member Kusanagi Tsuyoshi and Oshima Yuko. ...
Aoi Eir, Aoi Eir GENESIS, GENERATIONS, J Soul Brothers, Milky Holmes, Milky Holmes Milky A GO GO, Nitta Emi, Nitta Emi NEXT PHASE, Nitta Emi Tankyu Dreaming, Oricon, Oricon Weekly, Shoko Nakagawa, Shoko Nakagawa DoriDori, SMAP, SMAP Humor Shichau yo, SMAP Karei naru Gyakushu, SUPER GiRLS
Oricon charts January 26th - February 1st 2015, 2PM and J Soul Brothers top
2PM tops the singles charts with 9th Japanese single Guilty Love selling 98,000 copies. The Momoiro Clover Z vs KISS collaboration single Yume no Ukiuyo ni Saitemina (夢の浮世に咲いてみな) climbs into #2. And the boys of GENERATIONS from EXILE TRIBE hit #3...
2PM, 2PM Guilty Love, AAA, AAA I'll be there, AKB48, AKB48 Koko ga Rhodes da Koko de Tobe!, GENERATIONS, GRANRODEO, GRANRODEO Punky Funky Love, J Soul Brothers, Maaya Sakamoto, Maaya Sakamoto Shiawase ni Tsuite Watashi ga Shitteiru Itsutsu no Hoho, Maaya Sakamoto Shikisai, miwa, miwa fighting-φ-girls, Momoclo, Momoclo x KISS, Momoclo Yume no Ukiuyo ni Saitemina, Momoiro Clover Z, nano, nano Rock on., NMB48, NMB48 Rashikunai, Oricon, Oricon Weekly, SEKAI NO OWARI, Sonar Pocket
Oricon charts for April 21st - 27th 2014, Johnny's WEST and Not yet top
Johnny's WEST, a new unit from Johnny's Entertainment, debuts at #1 with their first single Ei ja nai ka selling over 262,000 copies. In #2 is NEVER LET YOU GO the 5th single from GENERATIONS from EXILE TRIBE selling 86,000 copies. A...
CNBLUE, CNBLUE Truth, EXILE, Fukuyama Masaharu, GENERATIONS, HKT48, HKT48 Sakura Minna de Tabeta, Johnny's West, Love Live!, May J., Nana Mizuki, Nana Mizuki SUPERNAL LIBERTY, Not yet, Not yet already, Oricon, Oricon Weekly, Oshima Yuko, SCANDAL, SCANDAL Departure, Sore ha Bokutachi no Kiseki, Wagakki Band, μ's (Muse)
Oricon charts for November 11th - 17th 2013, Kis-My-Ft2 and Generations from EXILE TRIBE Top
On top of the singles charts are the boys of Kis-My-Ft2 with their 9th #1 and 9th single SNOW DOME no Yakusoku / Luv Sick selling over 268,000 copies. Climbing into the #4 spot is idol group Afilia Saga with single S.M.L featured as the OP for...
Afilia Saga, Aoi Eir, Aoi Eir Sirius, Arashi, Arashi LOVE, EXILE, GENERATIONS, IDOLING!!!, Kimura Kaela, Kis-My-Ft2, Koda Kumi, Koda Kumi Dreaming Now, NEGOTO, NEGOTO Synchromanica, NMB48, NMB48 Kamonegikkusu, OLDCODEX, OLDCODEX Walk, Oricon, Oricon Weekly, Shiina Ringo, Yui Horie, Yui Horie Golden Time, Yuzu
Oricon charts for October 7th - 13th 2013, Sexy Zone and Superfly top
Topping the singles charts is male idol group Sexy Zone with double-A side single Bye Bye Du Bye - See You Again / A My Girl Friend selling over 144,000 copies. Its their 5th single and 5th #1. In the #2 spot another male idol group GENERATIONS from ...
BIGBANG, EXILE, GENERATIONS, Kana Nishino, Kana Nishino Love Collection ~ mint ~, Kana Nishino Love Collection ~ pink ~, NMB48, NMB48 Kamonegikkusu, Perfume, Perfume LEVEL 3, Sexy Zone, Superfly, T.M.R., T.M.Revolution
Oricon charts for May 13th - 19th 2013, KAT-TUN and Eikichi Yazawa top
Topping the singles charts is popular vocal group KAT-TUN with their 21st single "FACE to Face" featured as theme for movie "Ore Ore" that stars group leader Kazuya Kamenashi. The group keeps their record is perfect with 21 #1 charting singles. Close...
EXILE, Fudanjuku, FUNKY MONKEY BABYS, GENERATIONS, KAT-TUN, KAT-TUN 21st single, KAT-TUN FACE to Face, Nana Mizuki, Oricon, Oricon Weekly, Sonar Pocket, T.M.R., T.M.Revolution x Nana Mizuki, T.M.Revolution x Nana Mizuki Preserved Roses, Uta no Prince-sama
Oricon weekly charts for Jan 28th - Feb 3rd 2013, SKE48 and Mika Nakashima top
On top of the single charts this week is SKE48 with their 11th single "Choco no dorei" selling big with over 539,000 units sold. It's the AKB48 sister group's best selling opening week single and 7th #1 in a row. Based out of Nagoya they continue to...
Aoi Eir, Aoi Eir BLAU, Eir Aoi, GENERATIONS, Kyary Pamyu Pamyu, LinQ, Mika Nakashima, Oricon, Oricon Weekly, SKE48
Oricon single and album charts for November 19th-25th, Mayuyu and Yumi Matsutoya top
Topping this week's singles charts is popular AKB48 idol Mayu Watanabe (Mayuyu) and her 3rd solo single "Hikaru mono tachi / Sayonara no Hashi" selling over 91,000 copies. The single's coupling track "Sayonara no Hashi" is featured as the...
2PM, 2PM Masquerade, Eir Aoi, Eir Aoi INNOCENCE, EXILE, GENERATIONS, Girls Generation, Girls Generation FLOWER POWER, JUJU, Kis-My-Ft2, Kis-My-Ft2 Ai no Beat, Mayu Watanabe, Mayu Watanabe Hikaru mono tachi, Mayu Watanabe Sayonara no Hashi, Mayuyu, Miliyah Kato, Momoiro Clover Z, Oricon, Oricon Weekly, SID, Yumi Matsutoya, Yumi Yumi
August 29, 2016T.M.Revolution single RAIMEI updated details


Happy Thanksgiving! I love this time of year & the chance we get to really think about all we are thankful for & how many blessing we really do have! Today I have many, many things to be thankful for!
I'm thankful for the Gospel of Jesus Christ and all the blessings it brings to my life.
I'm thankful for the knowledge that families can be together forever, and we will see our loved ones again.
I'm thankful for my 10 year old daughter who keeps my up to date on all the 5th grade drama. :) I love our "girl talks". :) I'm thankful for her kind & helpful personality.
I'm thankful for my 4 year old little lady who would rather hang out with me than be anywhere else in the world. :) I'm also thankful for her sassy attitude!
I'm thankful for my 2 year old tornado who can make a mess so fast but he's also just as quick to give a hug & a kiss. Oh how I love this little boy!
I'm thankful for family. We are so blessed on both sides of our family. Our parents are the greatest. We wouldn't be where we are today with out all of their help, advice & support.
I'm thankful for my sisters, brother, for my sister-in-laws and brother-in-laws. You guys are the best!
I'm thankful for a warm home, good food to eat & comfortable surroundings.
I'm thankful for this country and all the freedoms we enjoy.
I'm thankful for this little blog!
I'm thankful to each of YOU & for all of your support & friendship you have given me!
I hope each one of YOU has a very Happy Thanksgiving!


text:
Parnaenus cyanidens är en spindelart som först beskrevs av Koch C.L. 1846.  Parnaenus cyanidens ingår i släktet Parnaenus och familjen hoppspindlar. Inga underarter finns listade i Catalogue of Life.

Källor

Hoppspindlar
cyanidens

redpajama_set_name: RedPajamaWikipedia


text:
County: Los Angeles County
Dana Point is a city located in southern Orange County, California. The population was 33,351 at the 2010 census. It has one of the few harbors along the Orange County coast, and with ready access via State Route 1, it is a popular local destination for surfing.
Woodland Hills is a neighborhood bordering the Santa Monica Mountains in the San Fernando Valley region of the city of Los Angeles, California.
Winnetka is a neighborhood in the west-central San Fernando Valley in the city of Los Angeles. It is a highly ethnically diverse area, both for the city and for Los Angeles County, with a relatively large percentage of Hispanic and Asian people. Winnetka was founded in 1922 as a small farming community. As of 2018, there are six public and four private […]
Wilmington is a neighborhood in the Los Angeles Harbor Region area of Los Angeles, California, covering 9.14 square miles. Featuring a heavy concentration of industry and the third-largest oil field in the continental United States, this neighborhood has a high percentage of Latino and foreign-born residents.
Whittier is a city in Southern California located within Los Angeles County, California. As of the 2010 United States Census, the city had a population of 85,331, reflecting an increase of 1,631 from the 83,680 counted in the 2000 Census, and encompasses 14.7 square miles (38.0 km2). Like nearby Montebello, the city constitutes part of the Gateway Cities. Whittier was incorporated in […]
West Hollywood, commonly referred to as WeHo , is a city in Los Angeles County, California, United States. Incorporated in 1984, it is home to the Sunset Strip. As of the 2010 U.S. Census, its population was 34,399. It is considered one of the most prominent gay villages in the United States.
West Covina is a city in Los Angeles County, California, located 19 miles (31 km) east of Downtown Los Angeles in the eastern San Gabriel Valley and is part of Greater Los Angeles. The population for the city was 106,098 at the 2010 census.
Walnut is a city in the eastern part of Los Angeles County in the U.S. state of California. Money's Best Places to Live ranked Walnut #70 in 2009 and #57 in 2011, the highest ranking for a California city in both years. As of the 2010 census, the city had a population of 29,172, and in 2018 the population was estimated at 30,006.
The Verdugo Mountains are a small, rugged mountain range of the Transverse Ranges system, located just south of the western San Gabriel Mountains in Los Angeles County, Southern California. The range is sometimes known as the Verdugo Hills or simply the Verdugos.
Venice is a residential, commercial, and recreational beachfront neighborhood within Los Angeles, California. It is located within the urban region of western Los Angeles County known as the Westside.


Christel Bartelse sets the tone for her show by crawling out from under a gynaecological examination bed as a hideous, giant foetus. You can almost hear her biological clock ticking as she guides you through one of life's biggest decision-making processes: to breed or not to breed. From infertility, to the possibility of having ungrateful snot-nosed brats, to environmental collapse, Bartelse humorously explores all the anxieties that come with parenthood. Dynamic, dramatic and thoughtfully observant, this is a show that will find more uses for an umbilical cord than you could have ever dreamed possible. Squeamish people: watch out for some seriously graphic descriptions of vaginal-related problems and examinations, which wouldn't feel out of place in a Gothic thriller.


text:
David Wilmot
Irish Film
Sebastian Lelio's The Wonder kicks off principal photography in Ireland
Principal photography has begun on Sebastian Lelio's The Wonder, adapted by Emma Donoghue, Lelio and Alice Birch from Donoghue's novel of the same name, and produced
CompetitionFeatured
Competition: Win a copy of Niall Heery's Gold on DVD
Win a copy of Niall Heery's Irish drama Gold on DVD, with thanks to Wildcard Distribution. Out now to buy or rent.
A heart wrenching tale of a young British soldier's night on the streets of Belfast.
Movie NewsVideos and Clips
Two clips from Yann Demange's Northern Irish set '71
Two clips from Yann Demange's Northern Irish set '71, starring Jack O'Connell, David Wilmot, and Richard Dormer. Out October 10th.
Irish Abroad: Gold and The Survivalist represent Ireland at the 26th Dinard British Film Festival
Gold and The Survivalist represent Ireland at the 26th Dinard British Film Festival, September 30th to October 4th.
Competition: 12 Days of Irish Film Christmas – Day 11
12 Days of Irish Film Christmas Scannain Giveaway: On the Eleventh day of Irish Film Christmas Scannain sent to me...John Michael McDonagh's Calvary.
Watch: New clip from Yann Demange's thriller '71
New clip from Yann Demange's Northern Irish set '71, starring Jack O'Connell, David Wilmot, and Richard Dormer. Out October 10th.
ImagesMovie News
First image from John Michael McDonagh's War On Everyone
First image from John Michael McDonagh's War On Everyone, starring Michael Peña, Alexander Skarsgård, Theo James, and David Wilmot.
Box OfficeMovie News
Yann Demange's powerful drama '71 bucks box-office trend
Yann Demange's powerful drama '71 bucks box-office trend, increasing by 47% at the Irish box-office in its second week.
New clip from Yann Demange's Northern Irish set '71
New clip from Yann Demange's Northern Irish set '71, starring Jack O'Connell, David Wilmot, and Richard Dormer. Out October 10th.


My client has an exciting opportunity for a Delivery Technician (Healthcare) to join their growing team in Scarborough.
This role is a driving based role covering the North Yorkshire region, therefore it is essential that you hold a full UK driving licence.
It is ideal if you live in either of the following postcode areas: DN15, DN16, DN17.
My client is an independent specialist provider of healthcare services to patients with long term conditions, delivering a high-quality, cost-effective solutions with care closer to home.
* Completion of, installation, servicing, and removals of concentrators, cylinders and LOX equipment with some problem solving in line with our Service Level Agreement with the NHS.
* Completion of necessary documentation and processes to ensure activities are completed in full.


The Preparatory Commission for the Marist International Mission Assembly met at Rome from the 27th to the 30th July 2006. All the members of the Commission were able to attend the meeting, including Mr Colin Quine, an expert facilitator in the running of many international meetings.
The meeting room was decorated with many things produced by different Provinces, including the Assembly's logo with the text translated into about twenty of the languages used in the Marist world.
The first part of our meeting was consecrated to giving information on what had been done in the various geographical zones in distributing the material for the preparation of the Assembly and in motivating the Provincial delegates. A copy of a summary of information received from the said delegates was also handed out. The general impression is that the initiative has been well received and the material has been very much appreciated. Many groups are in favour of continuing this work in the future. There are difficulties, especially due to the geographical distances and extent of some Provinces.
From this moment, our facilitator took over the guiding of the meeting.
After a time of personal reflection, analysing our hopes and our concerns, we shared our ideas on the current situation of the preparation of the Assembly as well as the results that we hope to obtain from the Marist mission as fruit of this process.
A few places have not yet started the preparation process. We believe that we must respect the local realities, without imposing anything, but that it is worth communicating with them the advantages and benefits that are being experienced by those who have started the process.
Shortly, we will send a letter from the Commission to all those who have participated in the process, to appreciate their collaboration and to remind them that the preparation of the Assembly passes through Provincial, inter-Provincial and international phases.
We consecrated a good amount of time in the meeting to anticipating the development of the International Assembly itself. This work will finish at the next Commission meeting, which will take place in the same house as the Assembly at Mendés, Rio de Janeiro (Brazil), from the 12th to the 15th April 2007. Some members of the Commission have been entrusted with the task of collecting practical information (voyage, layout of the house, possibilities of technical support, budgets) in relation with this theme.
Finally, we discussed the preparation that we are going to try to give the representatives who are chosen in the Provinces. We have anticipated a series of activities, suggestions, information and documents that we will send them. We also need to collect and systematise the information that we will receive from the Provincial phases.
We also evaluated the work already realised. The general impression is one of satisfaction. The help given by Colin as facilitator was stressed and for that reason we have decided to invite him to participate also in the next meeting of the Commission and during the Assembly itself at Mendes.


text:
John Joseph Abercrombie (March 4, 1798 – January 3, 1877) was a career United States Army officer who served in numerous wars, finally reaching the rank of brigadier general during the American Civil War.

Early life and career
Abercrombie was born and baptized in Baltimore, Maryland, although some accounts suggest he was a native of Tennessee, born in 1802. The son of John Joseph Abercrombie Sr. and Sarah DeNormandie, their family was living in Nashville, Tennessee when the younger John entered the United States Military Academy in 1818. Graduating 37th of 40 from the United States Military Academy in 1822, Abercrombie's class included future Union Army generals Joseph K. Mansfield, David Hunter, and George A. McCall, as well as future Confederate Army general Isaac R. Trimble.

Abercrombie began his long military career with garrison duty at Baton Rouge, Louisiana, until 1825. He became the adjutant of the 1st U.S. Infantry and was assigned administrative duty at regimental headquarters. He was promoted to first lieutenant in September 1828 and served in the Black Hawk War against the Sauk Indians in 1832. He was then on garrison duty in Illinois and Wisconsin.

Seeing additional combat action as a captain during the Seminole Wars, Abercrombie was brevetted a major for gallantry at the Battle of Lake Okeechobee. Serving on frontier duty throughout the 1830s and 40s within the South and Northwest, Abercrombie fought again during the Mexican–American War, where he gained distinction at the Battle of Monterrey and was promoted to lieutenant colonel.

Following the war, he married Mary Engle Patterson, the daughter of General Robert Patterson, a wealthy Philadelphia merchant. They would have nine children, including Frank Abercrombie, who briefly played professional baseball for the Troy Haymakers of the National Association.

He founded Fort Abercrombie in what is now North Dakota in 1857, selecting the site and supervising construction of the wooden fort. He then commanded the garrison until promoted and reassigned to duty in Minnesota.

Civil War service

By the start of the Civil War, at the age of 63, Abercrombie was one of the oldest field officers in both the Union and Confederacy. He served as the commanding officer of the 7th U.S. Infantry in Minnesota with the full rank of colonel in the Regular Army.

Promoted to brigadier general of Volunteers in the Union Army on August 31, 1861, Abercrombie commanded troops under his father-in-law Robert Patterson in the Shenandoah Valley and at the Battle of Hoke's Run. He was reassigned as commander of the IV Corps' 2nd Brigade, attached to the Army of the Potomac. He led the brigade throughout the Peninsula Campaign in 1862, during which he was wounded at the Battle of Seven Pines. During the Battle of Malvern Hill, his command assisted in repulsing the Confederate attack against Union forces. His unit later participated in several skirmishes during the Federal retreat to Harrison's Landing.

After the conclusion of the Peninsula Campaign, Abercrombie was replaced as commander in favor of younger officers and spent the next year involved in the defense of Washington, D.C., and, in 1864, headed various Union supply depots in Virginia during the Overland Campaign. Abercrombie commanded the troops that defended a depot in Fredericksburg, Virginia, in June 1864 against an attack by Hampton's Legion.

On March 13, 1865, Abercrombie was brevetted a brigadier general in the Regular Army in recognition of his long service. Abercrombie retired from active duty on June 12 after commanding Fort Schuyler in New York.

Postbellum
In his retirement, Abercrombie stayed associated with the U.S. Army. He served on court-martial duty for the next 3 years.

Abercrombie died at Roslyn, Long Island, New York on January 3, 1877. He was buried in The Woodlands Cemetery in Philadelphia, Pennsylvania.

See also

List of American Civil War generals (Union)

References

Sources
 Eicher, John H., and Eicher, David J., Civil War High Commands, Stanford University Press, 2001, .

 Linedecker, Clifford L., ed. Civil War, A-Z: The Complete Handbook of America's Bloodiest Conflict. New York: Ballantine Books, 2002.

External links

Appleton's Cyclopedia of American Biography, edited by James Grant Wilson, John Fiske and Stanley L. Klos. Six volumes, New York: D. Appleton and Company, 1887-1889
Abercrombie genealogy and biography

1798 births
1877 deaths
Military personnel from Baltimore
United States Military Academy alumni
American people of the Seminole Wars
American military personnel of the Mexican–American War
Members of the Aztec Club of 1847
Union Army generals
Burials at The Woodlands Cemetery


Millions of engaged couples turn to WeddingWire as their trusted wedding resource, with the largest directory of local wedding vendors in Wyoming. Easily find the best wedding services in your area with detailed vendor...Read more reviews, prices and availability for your wedding day.
Explore the Cowboy State's incredible natural wonders-the Grand Tetons, Yellowstone National Park, Devils Tower and more-and create memories that will last a lifetime as you begin your life together at your Wyoming wedding.
Cheyenne, is a western town through and through, with ranches, rodeos, and cowgirls & boys all over the city. The state capital exemplifies Wyoming and the Old West. For those interested in Cheyenne history, the Historic Governor's' Mansion is available for tours as well as events. The recently restored and renovated mansion is a unique venue for your Wyoming wedding. The Cheyenne Depot Museum, a cultural focus in the heart of Cheyenne, celebrates the history of "America's Railroad Capital" and the surrounding railroad. The stunning lobby is also available for events and would be a fitting space for a vintage Wyoming wedding.
Laramie, on the plains of the Laramie River Valley, is a popular destination for lovers of the outdoors. At the base of the Snowy Range Mountains, visitors can climb, hike, ski, or snowmobile, depending on the season. The nearby Medicine Bow National Forest contains hundreds of miles of hikeable paths, as well as lakes and granite peaks to explore. The Vedauwoo Recreation Area is here, where you can camp, sightsee, or get married! The incredible rock formations make for a picturesque backdrop for your Wyoming wedding.
Casper, in Central Wyoming, features both gorgeous landscapes and big city culture. At the foot of Casper Mountain and within the North Platte River Valley, Casper is beautiful and green. Nature makes its way into the city here, so you can enjoy the best of all Wyoming has to offer. Fly fishing and other outdoor activities are popular here, but the city has plenty of urban sophistication to enjoy as well. You can visit the National Historic Trails Interpretive Center to learn about the pioneers who journeyed through the area, or any of the 20 museums in the city to learn about art, history, and the unique landscape of Wyoming. Many of the museums offer spaces for events, but for those who prefer a more metropolitan wedding, the Parkway Plaza is the ideal Wyoming wedding venue. With multiple spaces available for events, the Parkway Plaza is a chic locale for your Wyoming wedding.
The weather varies quite a bit depending on your location in Wyoming, but in general you can expect winters to be cold and windy, while summers are fairly mild with cool evenings. Even in the summer, it's a good idea to keep a jacket or warm wrap nearby. When planning décor for your wedding, consider the season and location. A fall wedding can celebrate the beauty of Wyoming's scenery by incorporating aspects of the couple's favorite outdoor activity. Couples who fish can place fishing rods on walls as decoration and use the vibrant flies used for fly fishing on place cards, bouquets, and boutonnieres. Use local wild flowers for a beautiful local touch. A summer wedding is likely to be temperate and lovely, so plan a wedding outdoors with any of the Wyoming mountains as your backdrop. For a truly western wedding, wear cowboy boots and ride in and out on a grand horse. Create a casual, homespun atmosphere with burlap runners and low bouquets on long picnic-style tables for your Wyoming wedding.
Wyoming is the home where the buffalo roam, and in cowboy country there is nothing more cowboy than jerky. Buffalo jerky has been popular for over 100 years in these parts, and you can find it nearly everywhere, along with beef and elk jerky. You'll also find bison burgers in many restaurants, which are delicious and a bit healthier than regular beef burgers. Whether or not you want to serve bison at your Wyoming wedding, a Wyoming wedding caterer can help you choose the best menu for your event.
A wedding in Wyoming costs about $10,000 less than the national average, so as long as you pay attention to how you spend your money, you won't break the bank. Wyoming also does not have income tax, so if you're a Wyoming resident, this will work to your budget's benefit as well.
A marriage license in Wyoming will cost you about $30 in cash. Be sure to bring a government-issued ID, your Social Security number, and your parents' full names and places of birth. There is no waiting period once you've received your marriage license and it will remain valid for one year.


Innovative medical devices have undoubtedly saved lives and improved the field of medicine over the past few decades. From hip joints to defibrillators, medical devices have added years to some patients' lives, while improving the overall quality of life for others. Sometimes, however, these medical devices are implanted incorrectly, the parts are defective or they create problems inside the body that result in injuries and even death.
When a surgical error or a defective medical device leads to further health deterioration, the outcome can be devastating. If you or someone you love has suffered an injury from a defective medical device, there is recourse. Talk to a defective medical device attorney to learn how to move forward with your claim and what your options are for pursuing compensation.
Each year, millions of people use various medical devices such as hip replacements, pacemakers and artificial knee joints without believing they present a serious health risk. Most patients assume these devices will allow them to lead healthy and more fulfilling lives. But this isn't always the case. Defective medical devices can leave users at risk of serious and lifelong injuries.
Oftentimes, medical devices are sold without the proper disclosures, meaning that patients do not have the opportunity to decide whether a device's benefits outweigh its risks.
Although the Food and Drug Administration (FDA) is the organization responsible for scrutinizing medical devices before they are used. However, the Administration is under continual pressure from device manufacturers to approve devices quickly. This has led to premature approval, as devices do not always undergo the necessary clinical tests and trials. As a result, some medical devices intended to help patients actually create more harm than if their condition was left untreated.


Moral Dilemma – two words that do not feature prominently , jointly or severally, in the tax advisor's lexicon.
In the first half of the 20th Century when world wars were all the rage, moral dilemmas evolved from the gritty reality of combat. By the late sixties serious debate was banished to the periphery of existence along with Bras and the Bible.
In 1967 the ubiquitous moral dilemma that had every moral philosopher thinking involved a trolley hurtling out of control along a railway line towards 5 people bound to the tracks. All 5 were bound to die unless the chap with the moral dilemma diverted the trolley to a siding where a single individual would be unavoidably sent to kingdom come. Was the loss of one life preferable to the loss of five where there was an, albeit indirect, contribution to the lone death?
By 1996 the search for sound-bytes, together with advances in the human condition, had led to the addition of a bridge over the railway with a slobbering fat man sitting precariously on the parapet. Our friend with the moral dilemma now had a number of choices. He could pull the lever and be the indirect cause of the unpleasant demise of the single person on the siding; he could let the 5 on the track go west; or he could – very directly – push the fat git off the bridge , thus dispatching him to an oversized grave while stopping the trolley in its tracks. (The chap with the moral dilemma would not sacrifice himself because (1) he was too thin to stop the trolley and (2) moral philosophers are often people with their heads up their own backsides who have difficulty practicing what they preach).
Although the Utilitarian answer to this conundrum would be a toss-up between diverting the trolley or deflating the fat geezer (a 5-1 win for the human race), it was established that the most likely outcome would be to watch in horror as the 5 are trampled. Coming in second would be diversion of the trolley. The Fat Man would survive to spill over both sides of the middle-seat on an economy flight to Australia, as well as to invariably eat the last cake on the plate. Even though they could not stand his guts – normal people would shy away from actively taking the man's life.
Last week I read in The Economist, however, that according to a recent book "Would You Kill the Fat Man?" when the question is posed to respondents in a foreign language, it is more likely that they will elect to kill the bridge-balancing- blubbery- baboon. This has nothing to do with xenophobia. (I must apologize to any rotund gentlement reading this post if I am sounding unintentionally offensive. I am currently on a diet trying to shrink my girth and I figured that, just like anti-Semites never become Jews and the English never become Irish – if I am disparaging enough about fat people, I will have conquered obesity for ever – fat chance).
Nobel Laureate Daniel Kahneman in his book "Thinking, fast and slow' talks about "expert intuition" – a Fire Chief who senses exactly when to leave a burning house before it collapses or a Chess Master who can instinctively advise the next three moves in somebody else's game. It turns out that this comes from enormous practice and experience and not some magical eureka moment – a combination of System 1 (automatic) and System 2 (conscious) thinking. This allows for quick, highly complex, thoughts.
It turns out that when you are thinking foreign, you slide into System 2 thinking most of the time. As a result, while the Natural Born English Speaker instinctively decides he cannot shove the Fat Man over the edge, the foreigner is forced to ruminate more slowly over the question and is more likely to come to a utilitarian decision, which might just explain why Germany started two world wars.
This has consequences for tax consultancy (and just about every other verbal interaction for that matter). Last week I met with a gentleman from Southern Europe who, for the purposes of this post, we shall refer to as Umberto. Now, because Umberto was not called Frank, Henry or Michael, as we sat over some pretty complex tax issues I realized that, even putting my mouth into low gear and applying the hand brake, he was left hanging on for life to the rear bumper. Until I read the Economist article I assumed it was just a matter of language but now I realize that our thought processes were probably working on two different planes. The upshot is that, given that there are far more of them than us, English tax advisors should be forced to attend a course in English as a Foreign Language in which they are brainwashed into thinking in Pidgin.
What would he have made of it all?
It is parochial and naive to think that English has conquered the world. It is the world that has conquered English.
I think I am going to start speaking like Borat. It's a very nice.


Storm bringer (25 points): Obtain 20,000 points in the Singapore mission in the Campaign.
A one-man riot (25 points): Obtain 11,000 points in the Shanghai mission in the Campaign.
Terror of the deep (25 points): Obtain 11,000 points in the South China Sea mission in the Campaign.
Gladiator (25 points): Obtain 12,000 points in the Kunlun Mountains mission in the Campaign.
Above and beyond the call (25 points): Complete all Assignments in the Campaign.
Demolition man (25 points): Obtain 15,000 points in the Tashgar mission in the Campaign.
Guardian of the fleet (25 points): Obtain 6,000 points in the Suez mission in the Campaign.
Full arsenal (65 points): Unlock all assignments and collectibles in the Campaign.
Fish (20 points): Complete the Campaign on Easy.
Dunn's pride (25 points): Obtain 7,000 points in the Baku mission in the Campaign.
Braving the storm (20 points): Complete the Singapore mission in the Campaign.
Wolves in sheep's clothing (20 points): Complete the Shanghai mission in the Campaign.
The fall of a Titan (20 points): Complete the South China Sea mission in the Campaign.
Dead by dawn (20 points): Complete the Kunlun Mountains mission in the Campaign.
Guns at dawn (20 points): Complete the Suez mission in the Campaign.
Fishing in Baku (20 points): Complete the Baku mission in the Campaign.
Antediluvian (20 points): Complete the Tashgar mission in the Campaign.
It was on the way... (20 points): Find 6 collectibles in the Campaign.
Took a casual look around (20 points): Find 9 collectibles in the Campaign.
Methodical search (20 points): Find 15 collectibles in the Campaign.
Done some searching (20 points): Find 12 collectibles in the Campaign.
No stone left unturned (20 points): Find 18 collectibles in the Campaign.
Every nook and cranny (20 points): Find 21 collectibles in the Campaign.
Stumbled over it (20 points): Find 3 collectibles in the Campaign.
Wolf (25 points): Complete the Campaign on Normal.
Tombstone (25 points): Complete the Campaign on Hard.
Recon (65 points): Find 28 collectibles in the Campaign.
Well placed (20 points): Get 10 kills with C4 in the Baku mission in the Campaign.
Wrecker (20 points): Get 10 multi-kills in the Shanghai mission in the Campaign.
Blood wake (20 points): Get 30 headshots in the South China Sea mission in the Campaign.
War turtle (20 points): Get 15 kills with RPG in the Singapore mission in the Campaign.
Infiltrator (20 points): Get 10 adrenaline kills in the Tashgar mission in the Campaign.
Shawshank (20 points): Get 5 kills with Shank in the Kunlun Mountains mission in the Campaign.
For tombstone (20 points): Let the Valkyrie be destroyed in the Campaign.
For the people (20 points): Sacrifice Irish to save the Valkyrie in the Campaign.
For the cause (20 points): Sacrifice Hannah to save the Valkyrie in the Campaign.
Patience is a virtue (20 points): Experience all 3 endings in the Campaign.
Turn around... (10 points): Perform 5 dog tag kills in Multiplayer.
Won them all (10 points): Win a round of each game mode.
.45 old school (25 points): Perform 45 kills with the M1911 Handgun in Multiplayer.
Bomb squad (25 points): Deliver 5 bombs in Obliteration.
Call me "Sir" (50 points): Reach rank 25.
Fledgling (20 points): Play a round of Air Superiority.
2 Wheels (20 points): Get a kill with the Dirtbike.
Mini Kamikaze (30 points): Kill an enemy with the SUAV.
Death From Above (20 points): Get a kill with the bomber.
New Superpower (30 points): Complete all China Rising Assignments.
Link Repeater (20 points): Win 2 rounds of Chainlink.
Bulletproof. Sort Of... (20 points): Spend 5 minutes using the Ballistic Shield.
The Metropolitan (20 points): Play a round on all Dragon's Teeth maps.
RC Assassin (30 points): Get 5 kills in a round with the RAWR.
Street Fighter (30 points): Complete all Dragon's Teeth Assignments.
The Big Leagues (20 points): Play a round of Carrier Assault.
Killing Me Softly (20 points): Get a kill in the ACV.
Spotted (20 points): Climb the highest point in Naval Strike.
Fly Swatter (30 points): Kill an enemy with the AA Mine.
No Parley (30 points): Get a kill with the old cannon.
Blind Bomber (30 points): In a round get 5 kills with air vehicles in Gulf of Oman.
Dirty Job (20 points): (Find the hidden tunnel in Caspian Border.
Falling Down (20 points): Kill an enemy with a ceiling collapse in Operation Metro.
Risky Business (30 points): (Kill an enemy flag holder while carrying your team's flag in CTF.
Torched (20 points): Ignite a brush fire using the repair tool in Operation Metro or Caspian Border.


text:
<?php

namespace Symfony\Component\DependencyInjection\Loader\Configurator\Traits;

trait ArgumentTrait
{
    /**
     * Sets the arguments to pass to the service constructor/factory method.
     *
     * @param array $arguments An array of arguments
     *
     * @return $this
     */
    final public function args(array $arguments)
    {
        $this->definition->setArguments(static::processValue($arguments, true));

        return $this;
    }

    /**
     * Sets one argument to pass to the service constructor/factory method.
     *
     * @param string|int $key
     * @param mixed      $value
     *
     * @return $this
     */
    final public function arg($key, $value)
    {
        $this->definition->setArgument($key, static::processValue($value, true));

        return $this;
    }
}

redpajama_set_name: RedPajamaGithub


text:
Q: How can I manage the state 'call in progress' in an outgoing call? I'm italian and my english could be bed, i'm sorry for it.
I want manage the state 'call in progress' in all outgoing call, to explain better I want do some things  (like play music) during the "TUUU TUUU" noise you hear before the recipient answers (sorry but I don't know what you call it).
It's possible?
Thank you.

A: If any change is detected in the call state the of the device,(I assume that you know how to register a listener to listen to call states) call back function - public void onCallStateChanged(int state, String incomingNumber) is triggered, with the state and incoming number.
Various states are - CALL_STATE_IDLE, CALL_STATE_OFFHOOK, and CALL_STATE_RINGING
1) IDLE             – when there is no incoming call
2) OFFHOOK   – when the line is busy
3) RINGING      – when call is incoming
The "TUUUU TUUUUU" sound you are talking about, is the functionality provided by the network provider and you CAN'T do anything related to that.


American cartoonists and writers may not have invented the comic strip, but some argue that the comics, as we know them today, are an American creation. Clearly, the United States has played an outsize role in the development of this underappreciated art form.
1.11.1900: Hal Rasmusson, the creator of Aggie Mack, is born in Crookston, Minnesota. His teen strip, which starred a blonde named Aggie Mack, launched in 1946 and ran until the early 1970s.
1.11.1958: Frank Willard, the creator of Moon Mullins, dies at 64. After his death, Ferd Johnson continued the strip until its demise in 1991. It focused on a group of lowlifes living in a boardinghouse.
Most of the information listed here from one day to the next comes from two online sites -- Wikipedia, and Don Markstein's Toonopedia -- as well as 100 Years of American Newspaper Comics, edited by Maurice Horn. Note that my focus is on American newspaper comic strips (and the occasional foreign strip that gained popularity in the United States). Thus, comic books and exclusively online comics are not included here.


LLNL's unique combination of capabilities, expertise, and relationships provides leadership and collaboration opportunities in flagship neutrino physics experiments. LLNL either leads or plays an important role in the experiments described below.
LLNL is the founder and lead institution for AIT-WATCHMAN, an underground, 1000-ton water Cherenkov detector doped with the neutron-capture agent gadolinium. In addition to its reactor monitoring and technology development applications, WATCHMAN will be one of the world's largest supernova detectors, with a unique ability to distinguish between the neutrino and antineutrino components of the supernova burst.
Unique expertise in time-projection chamber technology has enabled LLNL to assume a leading role in the nEXO collaboration, which plans to field a ton-scale time-projection chamber detector wherein xenon-136 serves both as the decay source and detector. When nEXO is ready to deploy, it will be the largest collection of xenon-136 atoms on earth and will be more sensitive to neutrinoless double beta decay than any other experiment.
LLNL is a founding member of the PROSPECT collaboration and the lead laboratory for the deployment and operation of the PROSPECT detector at the Oak Ridge High Flux Reactor. This experiment seeks to determine whether a new type of neutrino exists and to measure the flux of reactor antineutrinos with unprecedented accuracy.


text:
Q: c# what is the best way to record live frames as Video I have been checking around to convert live frames into video. And I found (NReco.VideoConverter) ffmpeg lib to convert live frames to Video, but the problem is it is taking time to write each frame to ConvertLiveMediaTask (async live media task conversion).
I have an event that provides (raw) frames (1920x1080) (25fps) from IpCamera. Whenever I get frame I am doing the following
//Image availbale event fired
//...

//...
// Record video is true
if(record)
{
//////////////############# Time taking part ##############//////////////////////
var bd = frameBmp.LockBits(new Rectangle(0, 0, frameBmp.Width, frameBmp.Height), ImageLockMode.ReadOnly, PixelFormat.Format24bppRgb);
  var buf = new byte[bd.Stride * frameBmp.Height];
  Marshal.Copy(bd.Scan0, buf, 0, buf.Length);
  // write to ConvertLiveMediaTask
  convertLiveMediaTask.Write(buf, 0, buf.Length); // ffMpegTask
  frameBmp.UnlockBits(bd);
//////////////////////////////////////////////////////////////////////////////////
}

As the above part is taking much time, I am loosing the frames.
//Stop recording
convertLiveMediaTask.Stop(); //ffMpegTask

Stop recording, for this part I have used BackgroundWorker, because this takes too smuch time to save the media to file.
My question is how can I write the frame to ConvertLiveMediaTask in faster way? are there any possibilites to write it in background?
Please give me suggestions.

A: I'm sure that most time takes encoding and compressing raw bitmaps (if you encode them with h264 or something like that) by FFMpeg because of FullHD resolution (NReco.VideoConverter is a wrapper to FFMpeg). You must know that real-time encoding of FullHD is VERY CPU consuming task; if your computer is not able to do that, you may try to play with FFMPeg encoding parameters (decrease video quality / compression ratio etc) or use encoder that requires less CPU resources.

A: If You need to record some limited time live stream, You can split video capturing and compressing/saving into two threads.
Use for example ConcurrentQueue to buffer live frames (En-queue) on one thread without delay, and other thread could save those frames at a pace it can (De-queue). This way you will not loose frames.
Obviously You will have strain on RAM and also after stopping Live video You will have a delay while saving thread finishes.


Over the years, The Dark Mod has accumulated a number of Fan Missions that form a larger interconnected series or an overarching/continuing narrative. Many series are connected by their main characters, some by other FM inter-related elements.
This article serves as an overview of the known existing FM series.
The FMs of each series are sorted in chronological order.
Beware of potential (minor) spoilers behind the character links.
While all TDM missions are technically Fan Missions, the ones belonging to these official campaigns are as close as TDM will ever get to its own official in-game missions, or in-game story campaigns. These core missions for the base game are in slow and patient continuous development (due to TDM being a hobbyist project of a fan community), and have been gradually added to the base game over the course of several years.
For additional series by TDM authors, please see the next section.
The missions of this campaign serve as a gradual introduction into the setting and gameplay of The Dark Mod. The missions are shorter and fairly easy, and gradually put the player's newly acquired skills to the test.
Currently, this is the only series that is bundled directly as part of the base game. By installing TDM, the first three missions available to you will be part of this official mini-campaign.
Series' status: In development, but currently on hiatus. 3 missions currently available, including the (generally plotless) training mission.
TDM dev team 2009-10-17 (original version) Unique. The official tutorial mission of TDM, with open-ended gameplay.
The majority of missions created for The Dark Mod. For the sake of easier searching, these FM series are arranged in alphabetical order, based on their titles. In terms of release, a Single-mission (S) series consists of several single-mission FM releases over time (the commonest sort), while a Campaign (C) series is an FM containing a storyline consisting of several missions.
Series' status: Currently on hold.
Description: Though originally planned as a major campaign for future releases of TDM, development progress has been halted.
Release: Released as Single-mission (S), possibly would have been a Campaign (C) if finished.
Series' status: Development halted, campaign apparently cancelled. Only 1 mission has been completed and made available.
Horror Winner of the 2014 Halloween Speed-build Contest.
Series' status: Completed and released.
Series' status: Currently on hiatus.
Release: Single-mission (S) for the first mission and Campaign (C) for the second mission.
Series' status: Ongoing, currently on hiatus.
Lost Civilizations Winner of the 2014 official Thief Mod Competition by Square Enix & Amazon.com.
Series' status: Ongoing, currently in gradual development.
Melan (original version), (version 2.0) 2010-01-10 (original version), 2015-01-03 (2.0), 2015-01-03 (current version, 3.0) City Missions Winner of the 2009 Grand Christmas Contest.
Note: Sir Talbot's Collateral by Bikerdude and Baal is actually an unrelated mission, just with a similarly named character.
Description: The de facto oldest FM series in The Dark Mod. Follow the skillful thief Farrell as he navigates the criminal underworld of Bridgeport and the surrounding world.
Horror Winner of the 2010 Summer Vertical Contest.
Release: Single-mission (S) for the first two published missions, Campaign (C) for the final release that also includes a new first mission.
Series' status: Ongoing, though the wait for the fourth mission was longer. A fifth mission is in the works.
For a key to the descriptions of each Mission Type, please see this overview.
Fan Missions for The Dark Mod - Main article on TDM fan missions.


I cant seem to get xsendkey to respond. I wonder if I'm missing something? I believe I have it configured correctly. I needed to send ALT:DOWN, but when the event is executed, nothing happens. Any help would be appreciated.


text:
Download: iOS 12.1.3 Beta 3 Released
By Paul Morris | January 7th, 2019
You can now download iOS 12.1.3 beta 3 IPSW links and OTA update for your iPhone, iPad and iPod touch devices. Here are the details on it.
Apple is back from the App Store Connect Christmas shutdown and its release cycle is back in full effect. In evidence of that, the Cupertino-based company has just released iOS 12.1.3 beta 3 and made it available to all developers with immediate effect.
Apple's previous iOS 12.1.3 release arrive just before the holiday shutdown period on December 20th. That release came during a particularly troubling time for Apple and it's iOS device users who had been served up a public release of iOS 12.1.2 which seemed to break a lot more than it fixed for a lot of device owners. In particular, a growing number of iPhone users found that iOS 12.1.2 caused them a lot of cellular data troubles. Hopefully, the Christmas shutdown has given Apple the space it needed to regain composure and push ahead into 2019 with releases of this nature which are bulletproof.
Time will tell on that front, but, for now, this latest release is a simple progression of the second beta which was released into the development community on December 20th. As the versioning suggests, it is a relatively minor bug fix release which looks to eradicate known issues and fix any niggles which occurred due to the release of iOS 12.1.2. As we already know, that release not only caused users some problems with cellular data but also contained a number of amends to comply with court rulings in China relating to Apple's infringement of Qualcomm Inc. patents.
There is no such excitement held within iOS 12.1.3 but it remains an important update, nevertheless. It will, of course, appeal to developers who are not only committed to ensuring that the Apple ecosystem is the best that it can be but also ensuring that their own apps and games are functioning as expected on this latest iOS release.
If you are one of those registered developers who has a beta configuration profile installed on your device, then you can grab iOS 12.1.3 beta 3 OTA update directly from the Software Update mechanism found in Settings > General > Software Update section. For those who want a little more control in the form of a fresh installation on a test device, then you can grab the IPSW directly from Apple and go through a manual iTunes installation.
Download iOS 12.1.2 16C104 IPSW Links, OTA For iPhone XS, Max, XR, X And More
How To Downgrade iOS 12.1.2 To iOS 12.1 / iOS 12.1.1
iOS 12.1.2 Release Notes, Changelog And Other Changes As Highlighted By Apple
Download iOS 12.1.2 Final IPSW Links, OTA Update For iPhone XS, XS Max, XR, X, And More
Jailbreak iOS 12.1.2 Status Update: Here's What You Need To Know


text:
Montgomery County Schools Consider Legal Action Over County Budget Cuts
Les Knapp
County Budgets / Education
A May 20 Gazette.net article discusses the ongoing threat by the Montgomery County Public School System to sue the County over $45 million in cuts to the school system's proposed budget. The Montgomery County Council's final operating budget provides for $4.4 billion, a 2.2 percent increase over FY 2011. School aid also increased over its FY 2011 total but includes a $45 million reduction supported by the Council and County Executive Isiah "Ike" Leggett, who believe school employees and teachers should contribute more towards their health care and retirement costs. The Council and County Executive do not believe the reduction to the increase in aid will affect student performance.
Of the county's fiscal 2012 budget, $1.37 billion in county tax dollars will go to schools.
Including state funding, total tax dollars spent on schools in fiscal 2012 will be $1.95 billion — 55.9 percent of all tax-supported funding in the county.
The county spent $1.415 billion on Montgomery County Public Schools in fiscal 2011. …
Montgomery School Board President Christopher S. Barclay (Dist. 4) called the decision unacceptable and said the council's claims that the cuts will not affect students are false.
"Every school will feel the effects of these cuts," he said.
The school board has argued that cuts could harm the academic progress of students.
Barclay said the school board is awaiting a ruling from the Maryland State Board of Education on a petition the local school board filed earlier this year. The petition asked the state board to rule that the County Council does not have the authority to cut its budget.
The state board could announce its decision Wednesday — the day before the council formally adopts its budget.
In response to the reduction, Superintendent of Schools Jerry Weast has released a new school budget that proposes the elimination of 155 non-teaching postions. Further information can be obtained from this May 23 Gazette.net article.
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Register Today for the 2015 Governor's Grants Training Conference – November 9
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Baltimore County to Consider 'Learning Hub Centers' for Students