text:
\section*{1. Introduction}

The {\it vector partition function } that we are interested in is
in the form of
\[
t({\bf b}|M)=\#\{{\bf x}\in {\bf Z}_+^n|{\bf Mx}={\bf b}\},
\]
where, ${\bf Z}_+$ denotes the nonnegative integers, ${\bf M}$ is
a fixed $s\times n$ integer matrix with columns $m_1,\cdots,m_n\in
{\bf Z}^s$ and ${\bf b }$ is a variable vector in ${\bf Z}^s.$ To
guarantee $t({\bf b}|M)$ is finite, we require
$[\{m_1,\cdots,m_n\}]$ does not contain the origin, where $[A]$
denotes the convex hull of a given set $A$. The vector partition
function $t({\bf b}|M),$ which is also called a {\it discrete
truncated power}, has many applications in different mathematical
areas including Algebraic Geometry \cite{pommersheim},
Representation Theory\cite{repre}, Number Theory \cite{number} ,
Statistics\cite{Astati} and Randomized Algorithm \cite{rando} etc.
.

When $s=1,$ an explicit formulation for  $t({\bf b}|M)$, which
counts the integer solutions for the linear Diophantine equation,
is presented in \cite{BeckFrobenius}. In particular, when
$M=(a,b)$ where $a$ and $b$ are relatively prime, Popoviciu gave a
beautiful and surprising formulation for $t(n|(a,b))$
(\cite{popoviciu}).

 For the general matrix $M$, the nature of $t({\bf b}|M)$ is investigated and
the piecewise structure of $t({\bf b}|M)$ is given in
\cite{dahmen2} and \cite{sturmfels}. Moreover, one is also
interested in the  explicit formulation of $t({\bf b}|M)$. For the
general matrix $M$, a powerful method for obtaining $t({\bf b}|M)$
is described  in \cite{brion,vergne1}. Another interesting
algorithm for computing $t({\bf b}|M)$ as a function of ${\bf b}$
is also introduced in \cite{dgbeck}.  When $M$ is unimodular, in
which every nonsingular square submatrix has determinant $\pm 1$,
two algebraic algorithms for generating the explicit formulation
for $t({\bf b}|M)$ is presented in \cite{deloera}. But all these
methods depend on the complex computation.  In \cite{xu}, based on
multivariate truncated power functions $T({\bf x}|M)$,  an
explicit formulation for $t({\bf b}|M)$ is presented. But the
formulation involves multivariate truncated power functions
$T({\bf x}|M)$, which is not explicit form, and high-dimensional
Fourier-Dedekind sums, so we have to give an explicit form for
$T({\bf x}|M)$ and simplify high-dimensional Fourier-Dedekind
sums, in order to predigest the explicit formulation for $t({\bf
b}|M).$

The rest of the paper is organized as follows.  To help make this
paper self-contained we shall first introduce some notations and
definitions  in Section 2. In Section 3, we recall some results
about  vector partition functions $t({\bf b} |M)$. Section 4
generalize the Popoviciu's formulation. In Section 5, the
generalized Frobenius problem is investigated.  Finally, Section 6
give an explicit formulation for multivariate truncated powers in
the case where $s=2$ and show the high-dimensional
Fourier-Dedekind sum can be converted to one-dimensional
Fourier-Dedekind sum, which is convenient for computing. And
hence,  a simplified  explicit formulation for two-dimension
vector partition functions is given.

\section*{2.Preliminaries}

 To describe the nature of $t({\bf b}|{\bf M}),$ we
introduce several notations and  definitions in which the common
terminology of multiset theory is adopted. Intuitively, a {\it
multiset} is a set with possible repeated elements; for instance
$\{2,2,2,3,4,4\}$. Let $Y$ be an $s\times n$ matrix. $Y$ can be
considered as a multiset of elements of ${\bf R}^s.$
 The cone spanned by $Y,$ denoted
by $cone(Y)$, is the set
$$
\{ \sum_{y\in Y}a_y y:a_y\geq 0 \mbox{\, for all\, } y \}.
$$
Denote by $cone^\circ (Y)$ the relative interior of $cone(Y).$ Let
${\mathcal Y}(M)$ denote the set consisting of those submultisets
$Y$ of $M$ for which $M \backslash Y$ does not span ${\bf R}^s.$
Let the set $c(M)$ be the union of cone$(M\setminus Y)$ where $Y$
runs over ${\mathcal Y}(M).$ A connected component of $cone^\circ
(M)\setminus c(M),$ is called a {\it fundamental $M-$cone}.
 For the fundamental $M-$cone $\Omega,$ we
set $v(\Omega|M):=\Omega-[[M)).$ Here,  $
[[M)):=\{\sum_{j=1}^na_jm_j:0\leq a_j< 1, \forall j\},
\Omega-[[M))$ is the set of all elements of the form $a-b,$ where
$a\in \Omega$ and $b\in [[M)).$

We shall use the standard multiindex notation. Specifically, an
element $\alpha\in {\bf N}^m$ is called an {\it $m-$index}, and
$|\alpha |$ is called the length of $\alpha.$ Define $ z^\alpha
:=z_1^{\alpha_1}\cdots z_m^{\alpha_m}$ for $z=(z_1,\cdots,z_m)\in
{\bf C}^m$ and $\alpha=(\alpha_1,\cdots,\alpha_m)\in {\bf N}^m.$
 For $y=(y_1,\cdots, y_s) \in {\bf R}^s$ and a function
$f$ defined on ${\bf R}^s$ , we denote by $D_yf$ the directional
derivative of $f$ in the direction $y$,i.e. $
D_y=\sum_{j=1}^sy_jD_j, $ where, $D_j$ denote the partial
derivative with respect to the $j$th coordinate. For
$v:=(v_1,\cdots,v_m)\in {\bf N}^m,$ we let $D^v=D_1^{v_1}\cdots
D_m^{v_m}$ and $v!=\prod _i v_i!$ Moreover, we  let
$e:=(1,1,\cdots,1)\in {\bf Z}^s.$

Let ${\mathcal S}_k(M)=\{ Y\subseteq M:\#Y=s+k, span(Y)={\bf
R}^s\}$ and ${\mathcal B}(Y)=\{ X\subseteq Y:\#X=s, span(X)={\bf
R}^s\}.$ If for any $Y\in {\mathcal S}_k(M),$ $gcd \{
|det(X)|,X\in {\mathcal B}(Y) \}=1$ , then $M$ is called a {\it
$k-$ prime matrix}. In particular, when $M$ is an $1-$prime
matrix, $M$ is also called a {\it pairwise relative prime matrix}.
When $s=1,$ $k-$prime matrix means that no $k$ of the integers
$m_1,m_2,\cdots,m_n$ have a common factor, where
$m_i,i=1,\cdots,n$ are the elements in $M.$ Moreover, we denote
$e^{\frac{2\pi i}{k}}$ by $W_k$.

 The multivariate truncated power
$T(\cdot |M)$ associated with $M,$  which was introduced by
W.Dahmen \cite{dahemenpower} firstly, is the distribution given by
the rule
\begin{equation}\label{powerdefinition}
T(\cdot |M):\phi \mapsto \int _{{\bf R}_+^n}\phi (Mu)du , \phi \in
{\mathcal D}({\bf R}^s),
\end{equation}
where ${\mathcal D}({\bf R}^s)$ is the space of test functions on
${\bf R}^s$,i.e. the space of all compactly supported and
infinitely differentiable functions on ${\bf R}^s$. In fact,
$T(\cdot |M)$ agrees with some homogeneous polynomial of degree
$n-s$  on each fundamental $M-$cone. When $M$ is an $s\times s$
invertible matrix, $T(\cdot |M)$ agrees with the function on ${\bf
R}^s$ which takes value $\frac{1}{|det(M)|}$ on $cone(M)$ and $0$
elsewhere.

In \cite{numeircalT}, an efficient method for computing the
multivariate truncated power is presented.

\begin{theorem}(\cite{numeircalT})\label{Th:numT}
Let $M$ be an $s\times n$  matrix with columns  $m_1,\cdots,m_n\in
{\bf Z}^s\setminus \{ 0\}$  such that the origin does not contain
in $conv(M).$ For any $\lambda_1,\cdots,\lambda_n\in {\bf R}$, and
${\bf x}=\sum_{j=1}^n\lambda_j m_j,$
\begin{equation}\label{eq:recT}
T({\bf x}|M)=\frac{1}{n-s}\sum_{j=1}^n\lambda_jT(x|M \setminus
m_j).
\end{equation}
\end{theorem}
 For more detailed information about the function, the
reader is referred to \cite{deboor2},\cite{dahemenpower}.

A multivariate Box spline $B(\cdot |M)$ associated with $M$ was
introduced in \cite{deboor1} and \cite{deboor2}, which is the
distribution given by the rule
\begin{equation}\label{Eq:boxspline}
 B(\cdot
|M):\phi \mapsto \int _{{[0,1)}^n}\phi (Mu)du , \phi \in {\mathcal
D}({\bf R}^s).
\end{equation}
By taking $\phi=exp(-iy\cdot )$ in (\ref{Eq:boxspline}), we obtain
the Fourier transform of $B(\cdot |M)$ as
$$
\widehat{B}(\zeta
|M)=\prod_{j=1}^n\frac{1-exp(-i\zeta^Tm_j)}{i\zeta^Tm_j},\zeta \in
{\bf C}^s.
$$
For more detail information about Box splines,the reader is
referred to \cite{deboorbook} .

\begin{remark}
The definition of fundamental M-cone is slightly different with
the one presented in \cite{dahmen2}. In \cite{dahmen2}, a
fundamental M-cone is defined as a connected component of
$cone^\circ (M)\setminus c(M),$ where $c(M)$ is the union of
$span(M\setminus Y)$ and $Y$ runs over ${\mathcal Y}(M).$ In fact,
the fundamental $M-$cone defined in this paper may be larger  than
the one defined in \cite{dahmen2}. But all the conclusions in
\cite{dahmen2} hold for the larger fundamental M-cone. In a
private communication, Prof. M. Vergne introduce the new
definition about the fundamental M-cone.
\end{remark}

\section*{3. Vector partition functions}

To describe  the nature of $t({\bf b}|M),$ we let $ M_{\theta}:=\{
y\in M:\theta^y=1\}$ and let $ A(M):=\{ \theta\in ({\bf
C}\setminus \{ 0\})^s:span(M_{\theta})={\bf R}^s \}. $ Recall
$e=(1,1,\cdots,1)\in {\bf Z}^s.$ Obviously, $e\in A(M).$

The following qualitative result about $t(\cdot |M)$ is presented
in \cite{dahmen2}.

\begin{theorem}(\cite{dahmen2})\label{Theorem:DaMi,truncted power's structure}
Let $M=\{m_1,\cdots,m_n\}$ be a multiset of integer vectors in
${\bf R}^s$ such that $M$ spans ${\bf R}^s$ and the convex hull of
$M$ does not contain the origin. Then for any fundamental $M-$cone
$\Omega,$ there exists a unique element $f_{\Omega}(\alpha
|M)=\sum\limits_{\theta\in
A(M)}\theta^{\alpha}p_{\theta,\Omega}(\alpha)$ such that
$f_{\Omega}(\alpha |M)$ agrees with $t(\alpha |M)$ on $v(\Omega
|M),$ where $p_{\theta,\Omega}(\cdot )$ is a polynomial with
degree less than $\# M_{\theta}-s.$
\end{theorem}

An explicit formulation for $p_{e,\Omega}(\alpha)$, which is the
polynomial part of $t(\alpha|M),$ is presented in the following
theorem.
\begin{theorem}(\cite{xu})\label{Th:FormP}
Under the condition of Theorem \ref{Theorem:DaMi,truncted power's
structure}, $p_{e,\Omega}(x)=\sum_{k=0}^{n-s}p_{k,\Omega}(x),$
where $p_{k,\Omega}(x)$ is homogeneous polynomial of degree
$n-s-k,$ defined inductively by
$$p_{0,\Omega}(x)=T({\bf x}|M),
p_{k,\Omega}(x)=-\sum_{j=0}^{k-1}(\sum_{|v|=k-j}D^vp_{j,\Omega}(x)
(-i)^{|v|}D^v\widehat{B}(0|M)/v!),k\geq 1,$$ where, $x\in \Omega
.$
\end{theorem}

More generally, an explicit formulation for $p_{\theta,\Omega}$ is
also given as follows.
\begin{theorem}(\cite{xu})\label{Th:FormPtheta}
Given $\theta_0\in A(M)\setminus e,$ under the condition of
Theorem \ref{Theorem:DaMi,truncted power's structure},
$p_{\theta_0,\Omega}(x)=\sum_{\mu=0}^{n-s-\kappa}p^{\theta_0}_{\mu,\Omega}(x)$,
where $\kappa=\# (M\setminus
M_{\theta_0}),p^{\theta_0}_{\mu,\Omega}(x)$ is homogeneous
polynomial of degree $n-s-\kappa-\mu,$ defined inductively by
$$p^{\theta_0}_{0,\Omega}(x)=q_{0,r}^{\theta_0}(x),
p^{\theta_0}_{\mu,\Omega}(x)=q_{\mu,r}^{\theta_0}(x)
-\sum_{j=0}^{\mu-1}(\sum_{|v|=\mu-j}D^vp^{\theta_0}_{\mu,\Omega}(x)
(-i)^{|v|}D^v\widehat{B}(0|\widehat{M}_{r})/v!),\mu\geq 1.$$ Here,
$q_{\mu,r}^{\theta_0}(x)$ is a polynomial which is determined by
the following conditions: when $x\in \Omega, $ $ q_{ \mu
,r}^{\theta_0}(x)=\sum\limits_{ j_1+\cdots+j_\kappa =\mu
}\prod\limits_{i=1}\limits^\kappa
\frac{s_{1+j_i}({\theta_0}^{-m_i})}{(j_i+1)!}\frac{1}{r^\kappa}D_{m_1}^{j_1}\cdots
D_{m_\kappa}^{j_\kappa}T(x|M_{{\theta_0}}), $ where
$s_0(x)=\frac{x-x^r}{x-1},s_j(x)=xs_{j-1}'(x),j\in {\bf Z}_+.$
\end{theorem}

In particular, when $M$ is a 1-prime matrix, a simple formulation
for $t(\cdot|M)$ is shown in the following theorem.

\begin{theorem}\cite{xu}\label{Th:form}
Under the condition of Theorem \ref{Theorem:DaMi,truncted power's
structure},when $M$ is a $1-$prime matrix,
$$
f_{\Omega}(\alpha |M)=p_{e,\Omega}(\alpha |M)+\sum_{{\theta} \in
A(M)\setminus e}{\theta} ^{\alpha
}\frac{1}{|det(M_{{\theta}})|}\prod_{w\in M\setminus
M_{{\theta}}}\frac{1}{1-{\theta}^{-w}}
1_{cone(M_{{\theta}})}(\Omega),
$$
where $p_{e,\Omega}(\alpha |M)$ is given in Theorem
\ref{Th:FormP}.
\end{theorem}

For the convenience of description, throughout the rest of the
paper, we suppose $M$ is a 1-prime matrix without further declaration.
 According to Theorem
\ref{Th:form}, to give a simple explicit formulation for $t({\bf
b}|M),$ we have to present  an explicit formulation for $T({\bf
x}|M)$. Moreover, to calculate the elements in $A(M)$ is a
non-trivial problem, hence, we have to predigest the
non-polynomial part in $t({\bf b}|M)$.

\section*{4. The generalized Popoviciu's formulation}
In this section, we are interested in $t({\bf n}|M)$, where
  $M=\begin{pmatrix}
  x_{1} & x_{2} & x_{3} \\
  y_{1} & y_{2} & y_{3}
\end{pmatrix}\in {\bf Z}^{2\times 3},$ ${\bf n}=(n_1,n_2)^T\in {\bf Z}_+^2.$
Without loss of generality, we suppose $
\frac{y_1}{x_1}<\frac{y_2}{x_2}< \frac{y_3}{x_3}$.  Obviously, for
the  matrix $M$, there exit two fundamental $M-$ cones, i.e.
$\Omega_1=\{(x,y)^T|(x,y)^T\in cone(M),\frac{y_1}{x_1}<
\frac{y}{x}< \frac{y_2}{x_2}\}$ and $\Omega_2=\{(x,y)^T|(x,y)^T\in
cone(M),\frac{y_2}{x_2}<\frac{y}{x}< \frac{y_3}{x_3}\}$ (See
Fig.1).

\begin{center}
\end{center}
\begin{center}
Fig.1.The fundamental M-cones.
\end{center}

To describe conveniently, we let $M_{ij}=\begin{pmatrix}
  x_{i} & x_{j}  \\
  y_{i} & y_{j}
\end{pmatrix}$, and let $Y_{ij}=det(M_{ij})$, where $i<j.$
To describe the explicit formulation for $t({\bf n}|M)$, we need
to define the fractional part function $\{x\}$ which denotes the
fractional part of $x$, i.e. $\{x\}=x-\lfloor x\rfloor$.

 In this section, our goal is to generalize the following beautiful formula due
to Popoviciu:
\begin{theorem}\label{th:popoviciu}\cite{popoviciu}
If $a$ and $b$ are relatively prime,
$$
t(n|(a,b))=\frac{n}{ab}-\{\frac{b^{-1}n}{a}\}-\{\frac{a^{-1}n}{b}\}+1,
$$
where $b^{-1}b\equiv 1\,\,\, mod \,\,\, a$, and $a^{-1}a\equiv
1\,\,\, mod \,\,\, b$, $n\in {\bf Z}_+$.
\end{theorem}

In order to generalize Theorem \ref{th:popoviciu}, we firstly
consider the explicit formulation for $T({\bf x}|M)$.

\begin{lemma}
Suppose the matrix
  $M=\begin{pmatrix}
  x_{1} & x_{2} & x_{3} \\
  y_{1} & y_{2} & y_{3}
\end{pmatrix}\in {\bf Z}^{2\times 3}.$
When ${\bf x}=(x,y)^T\in \overline{\Omega}_1, T({\bf x}|M
)=\frac{yx_1-xy_1}{(x_1y_2-y_1x_2)(x_1y_3-y_1x_3)}$; when ${\bf
x}=(x,y)^T\in \overline{\Omega}_2, T({\bf x}|M
)=\frac{xy_3-yx_3}{(x_2y_3-y_2x_3)(x_1y_3-y_1x_3)}.$
\end{lemma}
{\it proof:} Based on Theorem \ref{Th:numT} and $T({\bf
x}|M_{ij})=\frac{1}{det(M_{ij})}, {\bf x}\in cone(M_{ij}),i<j$,
the Lemma can be proved easily after a brief calculation. $\Box$

Hence, we obtain the conclusion as follows.

\begin{theorem}\label{Th:thfor}
Suppose the 1-prime matrix $M=\begin{pmatrix}
  x_{1} & x_{2} & x_{3} \\
  y_{1} & y_{2} & y_{3}
\end{pmatrix}$. When ${\bf n}=(n_1,n_2)^T\in
\overline{\Omega}_1\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)&=&\frac{n_2x_1-n_1y_1}{Y_{12}Y_{13}}-\{\frac{
(f_{12}Y_{13}+g_{12}Y_{23})^{-1}(n_2(f_{12}x_1+g_{12}x_2)-n_1(f_{12}y_1+g_{12}y_2))}{Y_{12}}\}\\
&-&\{\frac{
(f_{13}Y_{12}+g_{13}Y_{23})^{-1}(n_2(f_{13}x_1+g_{13}x_3)-n_1(f_{13}y_1+g_{13}y_3))}{Y_{13}}\}+1;
\end{eqnarray*}

when ${\bf n}=(n_1,n_2)^T\in \overline{\Omega}_2\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)&=&\frac{n_1y_3-n_2y_3}{Y_{23}Y_{13}}-\{\frac{
(f_{23}Y_{13}+g_{23}Y_{12})^{-1}(n_1(f_{23}x_3+g_{23}x_2)-n_2(f_{23}y_3+g_{23}y_2))}{Y_{23}}\}\\
&-&\{\frac{
(f_{13}Y_{12}+g_{13}Y_{23})^{-1}(n_1(f_{13}x_1+g_{13}x_3)-n_2(f_{13}y_1+g_{13}y_3))}{Y_{13}}\}+1,
\end{eqnarray*}
where, $f_{12},g_{12},f_{13}$,  $g_{13}$, $f_{23}$ and $g_{23}\in
{\bf Z}$ satisfy $gcd(f_{12}Y_{13}+g_{12}Y_{23},Y_{12})=1$
$gcd(f_{13}Y_{12}+g_{13}Y_{23},Y_{13})=1$ and
$gcd(f_{23}Y_{13}+g_{23}Y_{12},Y_{23})=1,$ moreover,
$(f_{12}Y_{13}+g_{12}Y_{23})^{-1}(f_{12}Y_{13}+g_{12}Y_{23})\equiv
1\,\,\, mod\,\,\, Y_{12},
(f_{13}Y_{12}+g_{13}Y_{23})^{-1}(f_{13}Y_{12}+g_{13}Y_{23})\equiv
1\,\,\, mod\,\,\, Y_{13},
(f_{23}Y_{13}+g_{23}Y_{12})^{-1}(f_{23}Y_{13}+g_{23}Y_{12})\equiv
1\,\,\, mod\,\,\, Y_{23}. $
\end{theorem}
{\it proof:} We only prove the case where $(n_1,n_2)^T\in
\overline{\Omega}_1\cap {\bf Z}^2.$ Based on Theorem
\ref{Th:FormP}, $p_{e,\Omega_1}({\bf x })$, which is the
polynomial part of $t(\cdot |M)$ on $\Omega_1$, is in the form of
$p_{0,\Omega_1}({\bf x})+p_{1,\Omega_1}({\bf x}),$ where for ${\bf
x}\in \Omega_1$, $p_{0,\Omega_1}({\bf x})=T({\bf x}|M),
p_{1,\Omega_1}({\bf x})=-(\sum_{|v|=1}D^vp_{0,\Omega_1}({\bf
x})(-i)D^v\widehat{B}(0 |M ))$. By the explicit formulation for
$T({\bf x}|M),$ we have $p_{0,\Omega_1}({\bf
x})=\frac{y_1x-x_1y}{(x_2y_1-y_2x_1)(x_1y_3-y_1x_3)}.$ After a
brief calculation, we have $p_{1,\Omega}({\bf
x})=\frac{1}{2}(\frac{1}{Y_{13}}+\frac{1}{Y_{12}}).$ Hence, the
polynomial part of $t({\bf n} |M)$ is
$\frac{n_2x_1-n_1y_1}{Y_{12}Y_{13}}+
\frac{1}{2}(\frac{1}{Y_{13}}+\frac{1}{Y_{12}})$. According to
Theorem \ref{Th:form}, we only need to consider  the sum
\begin{eqnarray*}
& &\sum_{\theta\in A(M)\setminus e}\frac{1}{|det(M_{\theta})|}
\prod_{w\in M\setminus M_{\theta}
}\frac{\theta^n}{1-\theta^{-w}}1_{cone(M_{\theta})}(\Omega_1)\\
& =& \frac{1}{Y_{12}}\sum_{\theta\in A(M)\setminus e\atop
M_\theta=M_{12}}\frac{\theta^n}{1-\theta^{-(x_3,y_3)}} +
\frac{1}{Y_{13}}\sum_{ { \theta \in A(M)\setminus e}\atop
{M_\theta=M_{13}} }\frac{\theta^n}{1-\theta^{-(x_2,y_2)}}.
\end{eqnarray*}

Recall $e^{\frac{2\pi i}{k}}$ is denoted by $W_k$. As pointed out
in \cite{DaMi on the solution}, the elements in the set
$\{\theta|\theta\in A(M),M_{\theta}=M_{12}\}$ have the form
$(W_{Y_{12}}^{\alpha_1^j},W_{Y_{12}}^{\alpha_2^j}),$ where
$(\alpha_1^j,\alpha_2^j)\in {\bf Z}^2,1\leq j\leq Y_{12}$.

 Consider firstly
\begin{eqnarray}\label{eq:formsum}
\frac{1}{Y_{12}}\sum_{\theta\in A(M)\setminus e
\atop M_\theta=M_{12}}\frac{\theta^{\bf n}}{1-\theta^{-(x_3,y_3)}}=
\frac{1}{Y_{12}}\sum_{j=1}^{Y_{12}-1}
\frac{W_{Y_{12}}^{n_1\alpha_1^j+n_2\alpha_2^j}}{1-W_{Y_{12}}^{-(x_3\alpha_1^j+y_3\alpha_2^j)}}.
\end{eqnarray}

We set $x_3\alpha_1^j+y_3\alpha_2^j\equiv k$  $mod$ $Y_{12}$.
Since $M$ is a 1-prime matrix,
$x_3\alpha_1^j+y_3\alpha_2^j\not\equiv
x_3\alpha_1^m+y_3\alpha_2^m$ $mod$ $Y_{12}$ when $j\neq m$. Hence,
$k$ runs over $[1,Y_{12}-1]\cap {\bf Z}.$

For $\theta\in \{\theta| \theta\in A(M),M_{\theta}=M_{12}\},$ we
have $\theta^{(x_1,y_1)}=\theta^{(x_2,y_2)}=1.$ Hence,
\begin{eqnarray}
x_1\alpha_1^j+y_1\alpha_2^j&\equiv &0 \,\,\, mod\,\,\, Y_{12}\label{eq:1},\\
x_2\alpha_1^j+y_2\alpha_2^j&\equiv &0 \,\,\, mod\,\,\, Y_{12}\label{eq:2},\\
x_3\alpha_1^j+y_3\alpha_2^j&\equiv & k \,\,\, mod\,\,\,
Y_{12}\label{eq:3}.
\end{eqnarray}

By $x_1$ on both sides of $(\ref{eq:3}),$ we have

\begin{eqnarray}\label{eq:eq1}
x_1x_3\alpha_1^j+x_1y_3\alpha_2^j\equiv x_1k \,\,\, mod\,\,\,
Y_{12}.
\end{eqnarray}
According to (\ref{eq:1}), we obtain
\begin{eqnarray}\label{eq:eq2}
x_1\alpha_1^j\equiv -y_1\alpha_2^j\,\,\, mod\,\,\, Y_{12}.
\end{eqnarray}

Substituting (\ref{eq:eq2}) into (\ref{eq:eq1}), we have
\begin{equation}\label{eq:diop1}
\alpha_2^jY_{13}\equiv x_1k \,\,\, mod\,\,\, Y_{12}.
\end{equation}
By using similar method,
\begin{equation}\label{eq:diop2}
\alpha_2^jY_{23}\equiv x_2k \,\,\, mod\,\,\, Y_{12}.
\end{equation}
Since $M$ is  a 1-prime matrix, there exits $f_{12},g_{12}\in {\bf
Z}$ such that  $gcd(f_{12}Y_{13}+g_{12}Y_{23},Y_{12})=1.$
Combining (\ref{eq:diop1}) and (\ref{eq:diop2}), we have
$$
\alpha_2^j(f_{12}Y_{13}+g_{12}Y_{23})\equiv
(f_{12}x_1+g_{12}x_2)k\,\,\, mod\,\,\, Y_{12}.
$$
Hence, $\alpha_2^j\equiv
(f_{12}Y_{13}+g_{12}Y_{23})^{-1}(f_{12}x_1+g_{12}x_2)k\,\,\, mod \,\,\, Y_{12}.$

Similarly,  $\alpha_1^j\equiv
-(f_{12}Y_{13}+g_{12}Y_{23})^{-1}(f_{12}y_1+g_{12}y_2)k\,\,\, mod
\,\,\, Y_{12}.$ Hence,(\ref{eq:formsum}) is reduced to
\begin{eqnarray}\label{eq:sums}
\frac{1}{Y_{12}}\sum_{k=1}^{Y_{12}-1}
\frac{W_{Y_{12}}^{(n_2(f_{12}x_1+g_{12}x_2)-n_1(f_{12}y_1+g_{12}y_2))(f_{12}Y_{13}+g_{12}Y_{23})^{-1}k}}
{1-W_{Y_{12
}}^{-k}}.
\end{eqnarray}
According to  discrete Fourier transforms,
\begin{equation}\label{eq:dis}
-\{\frac{t}{a}\}=\frac{1-a}{2a}+\frac{1}{a}\sum_{k=1}^{a-1}\frac{W_a^k}{1-W_a^{-k}},
\end{equation}

(\ref{eq:sums}) can be reduced to
\begin{eqnarray*}
\{\frac{(n_2(f_{12}x_1+g_{12}x_2)-n_1(f_{12}y_1+g_{12}y_2))(f_{12}Y_{13}
+g_{12}Y_{23})^{-1}}{Y_{12}}\} +\frac{1}{2}-\frac{1}{2Y_{12}}.
\end{eqnarray*}

Hence
\begin{eqnarray*}
 & &\frac{1}{Y_{12}}\sum_{\theta\in
A(M)\atop M_\theta=Y_{12}}\frac{\theta^n}{1-\theta^{-(x_3,y_3)}}\\
&=& -
\{\frac{(n_2(f_{12}x_1+g_{12}x_2)-n_1(f_{12}y_1+g_{12}y_2))(f_{12}Y_{13}+g_{12}Y_{23})^{-1}}{Y_{12}}\}
+\frac{1}{2}-\frac{1}{2Y_{12}}.
\end{eqnarray*}
By using similar method, we have
\begin{eqnarray*}
& &\frac{1}{Y_{13}}\sum_{\theta\in
A(M)\setminus e\atop M_\theta=Y_{13}}\frac{\theta^{\bf n}}{1-\theta^{-(x_2,y_2)}}\\
&=&-\{\frac{(n_2(f_{13}x_1+g_{13}x_3)-n_1(f_{13}y_1+g_{13}y_3))(f_{13}Y_{12}+g_{13}Y_{23})^{-1}}
{Y_{13}}\} +\frac{1}{2}-\frac{1}{2Y_{13}}.
\end{eqnarray*}

Hence, when $(n_1,n_2)^T\in v(\Omega_1|M)\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)&=&\frac{n_2x_1-n_1y_1}{Y_{12}Y_{13}}-\{\frac{
(f_{12}Y_{13}+g_{12}Y_{23})^{-1}(n_2(f_{12}x_1+g_{12}x_2)-n_1(f_{12}y_1+g_{12}y_2))}{Y_{12}}\}\\
&-&\{\frac{
(f_{13}Y_{12}+g_{13}Y_{23})^{-1}(n_2(f_{13}x_1+g_{13}x_3)-n_1(f_{13}y_1+g_{13}y_3))}{Y_{13}}\}+1.
\end{eqnarray*}
Note that $\overline{\Omega}_1\subset v(\Omega_1|M).$ Hence, when
${\bf n}\in \overline{\Omega}_1\cap {\bf Z}^2,$ the theorem holds.
 $\Box$

\begin{remark}
 If $f_{12}, g_{12},f_{13},g_{13},f_{23}$ and $g_{23}$ satisfy
$f_{12}Y_{23}+g_{12}Y_{13}=gcd(Y_{23},Y_{13}),f_{13}Y_{12}+g_{13}Y_{23}=gcd(Y_{12},Y_{23}),
f_{23}Y_{13}+g_{23}Y_{12}=gcd(Y_{13},Y_{12})$, then
$gcd(f_{12}Y_{13}+g_{12}Y_{23},Y_{12})=1,$
$gcd(f_{13}Y_{12}+g_{13}Y_{23},Y_{13})=1$ and
$gcd(f_{23}Y_{13}+g_{23}Y_{12},Y_{23})=1.$ Hence,one can determine
$f_{12}, g_{12},f_{13},g_{13},f_{23}$ and $g_{23}$ by Euclidean
algorithm. But in some special cases, such as $Y_{12},Y_{13}$ and
$Y_{23}$ are pairwise relative prime, there exits the simpler
method for obtaining them.
\end{remark}
\begin{corollary}\label{co:for}
Suppose $Y_{12},Y_{13}$ and $Y_{23}$ are pairwise relative prime.
When ${\bf n}=(n_1,n_2)^T\in \overline{\Omega}_1\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)=\frac{n_2x_1-n_1y_1}{Y_{12}Y_{13}}-\{\frac{
Y_{13}^{-1}(n_2x_1-n_1y_1)}{Y_{12}}\}-\{\frac{
Y_{12}^{-1}(n_2x_1-n_1y_1)}{Y_{13}}\}+1,
\end{eqnarray*}
where $Y_{13}^{-1}Y_{13}\equiv 1\,\,\, mod\,\,\, Y_{12}$ and
$Y_{12}^{-1}Y_{12}\equiv 1\,\,\, mod\,\,\, Y_{13}$. When ${\bf
n}=(n_1,n_2)^T\in \overline{\Omega}_2\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)=\frac{n_1y_3-n_2x_3}{Y_{23}Y_{13}}-\{\frac{
Y_{13}^{-1}(n_1x_3-n_2y_3)}{Y_{23}}\}-\{\frac{
Y_{23}^{-1}(n_1x_3-n_2y_3)}{Y_{13}}\}+1,
\end{eqnarray*}
where $Y_{13}^{-1}Y_{13}\equiv 1\,\,\, mod\,\,\, Y_{23}$ and
$Y_{23}^{-1}Y_{23}\equiv 1\,\,\, mod\,\,\, Y_{13}$.
\end{corollary}
{\it proof:} We firstly consider the case where ${\bf n}\in
\overline{\Omega}_1\cap {\bf Z}^2.$ Since $gcd(Y_{12},Y_{13})=1$,
$M$ is a 1-prime matrix. In Theorem \ref{Th:thfor}, we may set
$f_{12}=1,g_{12}=0,f_{13}=1,$ and $g_{13}=0.$ Hence, When ${\bf
n}=(n_1,n_2)^T\in \overline{\Omega}_1\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)=\frac{n_2x_1-n_1y_1}{Y_{12}Y_{13}}-\{\frac{
Y_{13}^{-1}(n_2x_1-n_1y_1)}{Y_{12}}\}-\{\frac{
Y_{12}^{-1}(n_2x_1-n_1y_1)}{Y_{13}}\}+1.
\end{eqnarray*}

Using similar method, when ${\bf n}=(n_1,n_2)^T\in
\overline{\Omega}_2\cap {\bf Z}^2$,
\begin{eqnarray*}
t({\bf n}|M)=\frac{n_1y_3-n_2x_3}{Y_{23}Y_{13}}-\{\frac{
Y_{13}^{-1}(n_1x_3-n_2y_3)}{Y_{23}}\}-\{\frac{
Y_{23}^{-1}(n_1x_3-n_2y_3)}{Y_{13}}\}+1.
\end{eqnarray*}
 $\Box$

\begin{remark}
An interesting observation is that the formulation presented in
Corollary \ref{co:co1} is remarkably similar with Popoviciu's
formulation.
\end{remark}

We now turn to consider the special  case where
$\frac{y_1}{x_1}=\frac{y_2}{x_2}$. Without loss of generality, we
suppose $M=\begin{pmatrix}
  kx_{1} & lx_{1} & x_{3} \\
  ky_{1} & ly_{1} & y_{3}
\end{pmatrix}$.
In this case, there exits only one fundamental M-cone, which is
denoted as $\Omega$. Moreover, since $M$ is a 1-prime matrix, we
have $gcd(k,l)=1,x_1y_3-y_1x_3=1.$
 Then we have
\begin{theorem}\label{th:spM}
Suppose $\frac{y_1}{x_1}<\frac{y_3}{x_3}.$ When $M=\begin{pmatrix}
  kx_{1} & lx_{1} & x_{3} \\
  ky_{1} & ly_{1} & y_{3}
\end{pmatrix},$
$t({\bf n
}|M)=\frac{x_3n_2-y_3n_1}{kl}-\{\frac{l^{-1}}{k}(n_1y_3-n_2x_3)\}-
\{\frac{k^{-1}}{l}(n_1y_3-n_2x_3)\}+1,$ where ${\bf
n}=(n_1,n_2)^T\in \overline{\Omega}\cap {\bf Z}^2.$
\end{theorem}
{\it proof:} By using the recurrence formulation for $T({\bf x}|M)$, we have
 $T({\bf x}|M)=\frac{x_3y-y_3x}{kl}.$ Hence, the polynomial
part of $t(\cdot|M)$ is $\frac{x_3y-y_3x}{kl}+\frac{1}{2}(\frac{1}{k}+
\frac{1}{l}).$ We
now only need to consider the sums
$$
\frac{1}{k}\sum_{\theta:M_{\theta}=Y_k}\frac{\theta^n}{1-\theta^{-(lx_1,ly_1)}},
\frac{1}{l}\sum_{\theta:M_{\theta}=Y_l}\frac{\theta^n}{1-\theta^{-(kx_1,ky_1)}}.
$$
By using the similar method with the one presented in the proof of
Theorem \ref{Th:thfor}, we have $
\frac{1}{k}\sum\limits_{\theta\in A(M) \setminus e \atop
M_{\theta}=Y_k}\frac{\theta^n}{1-\theta^{-(lx_1,ly_1)}}
=-\{l^{-1}\frac{n_1y_3-n_2x_3}{k}\}+\frac{1}{2}-\frac{1}{2k},$
$\frac{1}{l}\sum\limits_{\theta\in A(M)\atop
M_{\theta}=Y_l}\frac{\theta^n}{1-\theta^{-(kx_1,ky_1)}}
=-\{k^{-1}\frac{n_1y_3-n_2x_3}{k}\}+\frac{1}{2}-\frac{1}{2l}.$
Note that $\overline{\Omega}\subset v(\Omega|M)$. By Theorem
\ref{Th:form}, when ${\bf n}=(n_1,n_2)^T\in \overline{\Omega},$
\begin{eqnarray*}
& &t({\bf n }|M)=\frac{x_3y-y_3x}{kl}+\frac{1}{2}(\frac{1}{k}+
\frac{1}{l})+
\frac{1}{k}\sum\limits_{\theta:M_{\theta}=Y_k}\frac{\theta^n}{1-\theta^{-(lx_1,ly_1)}}
+\frac{1}{l}\sum\limits_{\theta:M_{\theta}=Y_l}\frac{\theta^n}{1-\theta^{-(kx_1,ky_1)}}\\
&=&\frac{x_3n_2-y_3n_1}{kl}-\{\frac{l^{-1}}{k}(n_1y_3-n_2x_3)\}-
\{\frac{k^{-1}}{l}(n_1y_3-n_2x_3)\}+1.
\end{eqnarray*}

$\Box$

\begin{remark}
When the matrix $M$ is of the form $\begin{pmatrix}
  x_{1} & kx_{2} & lx_{2} \\
  y_{1} & ky_{2} & ly_{2}
\end{pmatrix},$ a similar result can be obtained using the same method
with the one presented in Theorem \ref{th:spM}.
\end{remark}

\section*{5.Linear Diophantine problem of Frobenius}

Consider the linear Diophantine equation
\begin{equation}\label{eq:dio}
x_1a_1+\cdots x_na_n=N,
\end{equation}
where, $a_i\in {\bf Z}_+, gcd(a_1,\cdots,a_n)=1.$

It is well known that for all sufficiently large $N$ the equation
has solutions. The Frobenius problems asks us to find the largest
integer for which no solution exists. We call the largest integer
the Frobenius number and denote it by $f(a_1,\cdots,a_n).$ For
$n=2$ the largest $N$ for which no solution exists can be
explicitly written as $a_1a_2-a_1-a_2,$ i.e.
$f(a_1,a_2)=a_1a_2-a_1-a_2.$ But no such formula exists for $n\geq
3.$

As pointed out in \cite{xu}, when $gcd\{|Y|:Y\in {\mathcal
B}(M)\}=1,$ for all sufficiently large $N$ the linear Diophantine
equations $Mx=N{\bf n}$ has solution, where ${\bf n}\in cone(M).$
Naturally, we hope to find the largest integer $N$ for which no
solution exits, which is denoted as $f(M,{\bf n}).$ In particular,
we are interested in the linear Diophantine equations $M_0x=N{\bf
n},$ where $ M_0=\begin{pmatrix}
  x_{1} & x_{2} & x_{3} \\
  y_{1} & y_{2} & y_{3}
\end{pmatrix}, {\bf n}\in cone(M_0).
$ In fact, the generalized Frobenius number $f(M_0,{\bf n})$ is a
generalization of $f(a_1,a_2).$

Recall $M_{ij}=\begin{pmatrix}
  x_{i} & x_{j}  \\
  y_{i} & y_{j}
\end{pmatrix}$ and  $Y_{ij}=det(M_{ij}).$
In the following theorem, we shall present an upper  boundary for
$f(M_0,{\bf n}).$
\begin{theorem}\label{th:genf}
 Suppose $Y_{12},Y_{13}$ and $Y_{23}$ are pairwise relative prime. For ${\bf n}\in \overline{\Omega}_1\cap {\bf Z}^2,$
 $f(M_0,{\bf n})< \frac{Y_{12}Y_{13}-Y_{12}-Y_{13}+1}{n_2x_1-n_1y_1}.$
 For ${\bf n}\in \overline{\Omega}_2\cap {\bf Z}^2,$
 $f(M_0,{\bf n})< \frac{Y_{23}Y_{13}-Y_{23}-Y_{13}+1}{n_1y_3-n_2x_3}.$
\end{theorem}
{\it proof:} We only prove the case where ${\bf n}\in
\overline{\Omega}_1\cap {\bf Z}^2.$ Note $t(N{\bf
n}|M)=\frac{N(n_2x_1-n_1y_1)}{Y_{12}Y_{13}}-\{\frac{
(Y_{13})^{-1}(N(n_2x_1-n_1y_1))}{Y_{12}}\}-\{\frac{
(Y_{12})^{-1}(N(n_2x_1-n_1y_1))}{Y_{13}}\}+1=t(N(n_2x_1-n_1y_1)|(Y_{12},Y_{13})).$
Since when $N(n_2x_1-n_1y_1)\geq Y_{12}Y_{13}-Y_{12}-Y_{13}+1,$ $
t(N(n_2x_1-n_1y_1)|Y_{12},Y_{13})=t(N{\bf n}|M)>0.$ Hence, when
$N\geq \frac{Y_{12}Y_{13}-Y_{12}-Y_{13}+1}{(n_2x_1-n_1y_1)},$
$t(N{\bf n}|M)>0.$ So, $f(M,{\bf n})<
\frac{Y_{12}Y_{13}-Y_{12}-Y_{13}+1}{n_2x_1-n_1y_1}.$ $\Box$

\begin{remark}
Theorem \ref{th:genf} only gives an upper boundary for $f(M_0,{\bf
n}).$ According to the proof of Theorem \ref{th:genf}, giving the
exact value of $f(M_0,{\bf n})$ is equivalent for any given
$b_0\in {\bf Z }$ determining the largest integer $N$ for which
the Diophantine equation $x_1a_1+x_2a_2=Nb_0$ no solution exits.
\end{remark}

\section*{6 Two-dimension vector partition functions}

We now turn to the general case. Let $M=\begin{pmatrix}
  x_{1} & x_{2} & \cdots &x_n \\
  y_{1} & y_{2} & \cdots & y_n
\end{pmatrix}$ be  a $2\times n$ integer matrix.
and $\frac{y_{i-1}}{x_{i-1}}<\frac{y_i}{x_i},i=2,\cdots,n$.

For the matrix $M$, there exist $n-1$ fundamental $M-$ cones.
Denote them as $\Omega_i:=\{(x,y)^T|(x,y)^T\in
cone(M),\frac{y_i}{x_i}<\frac{y}{x}<\frac{y_{i+1}}{x_{i+1}}\},i=1,\cdots,n-1$
respectively. In this section, we shall discuss the explicit
formulation for $t({\bf b}|M)$. First, we present an explicit
formulation for $T({\bf x}|M)$.
\begin{theorem}\label{Th:2dT} For ${\bf x}=(x,y)^T\in {\bf R}^2,$
$$
T({\bf x
}|M)=\frac{1}{(n-2)!}\sum_{i=1}^n\frac{(y_ix-x_iy)_+^{n-2}}{\prod_{j\neq
i}(y_ix_j-y_jx_i) },
$$
where, $(y_ix-x_iy)_+=\begin{cases}
y_ix-x_iy , & y_ix-x_iy\geq 0,\\
0 , & otherwise.
\end{cases}
$
\end{theorem}
{\it proof:} According to the definition of $(y_ix-x_iy)_+$ we
only need to prove that when ${\bf x}\in \Omega_k$,
$T({\bf x}|M)=\frac{1}{(n-2)!}\sum_{i=k+1}^n\frac{(y_ix-x_iy)^{n-2}}{\prod_{j\neq
i}(y_ix_j-y_jx_i) }.$

 We argue by induction on $n$.
Initially, when $n=2,3$ the theorem certainly holds. In the
inductive step,  we assume that when $n=n_0$ the theorem holds and
we consider the case when $n=n_0+1$.

 According to the definition of $(y_ix-x_iy)_+$ we
only need to prove that for ${\bf x}\in \Omega_k,T({\bf x
}|M)=\frac{1}{(n_0-1)!}\sum\limits_{i=k+1}^{n_0+1}\frac{(y_ix-x_iy)^{n_0-1}}{\prod_{j\neq
i}(y_ix_j-y_jx_i) },$
 where $M$ is a $2\times (n_0+1)$ matrix.

 After a brief calculation,  it is easy for
obtaining $ {\bf x
}=\frac{xy_{k+1}-x_{k+1}y}{y_{k+1}x_k-y_kx_{k+1}}(x_k,y_k)^T+
\frac{xy_{k}-x_{k}y}{y_{k+1}x_k-y_kx_{k+1}}(x_{k+1},y_{k+1})^T.$
Based on the recurrence formulation of $T(\cdot |M)$ , we have
$$T({\bf x}|M)=
\frac{1}{n_0-1}(\frac{xy_{k+1}-x_{k+1}y}{y_{k+1}x_k-y_kx_{k+1}}
T({\bf x}|M\setminus (x_k,y_k)^T)+
\frac{xy_{k}-x_{k}y}{y_{k+1}x_k-y_kx_{k+1}} T({\bf x}|M\setminus
(x_{k+1},y_{k+1})^T)).$$ By the inductive hypothesis, $ T({\bf
x}|M\setminus
(x_k,y_k)^T)=\frac{1}{(n_0-2)!}\sum\limits_{i=k+1}^{n_0+1}\frac{(y_ix-x_iy)^{n_0-2}}{\prod\limits_{j\neq
i,j\neq k}(y_ix_j-y_jx_i) },$ $T({\bf x}|M\setminus
(x_{k+1},y_{k+1})^T)=\frac{1}{(n_0-2)!}\sum\limits_{i=k+2}^{n_0+1}\frac{(y_ix-x_iy)^{n_0-2}}{\prod\limits_{j\neq
i}(y_ix_j-y_jx_i) }.$ Then we obtain
\begin{eqnarray*}
& &T({\bf
x}|M)=\frac{1}{(n_0-1)!}(\frac{xy_{k+1}-x_{k+1}y}{y_{k+1}x_k-y_kx_{k+1}}\sum_{i=k+1}^{n_0+1}
\frac{(y_ix-x_iy)^{n_0-2}(x_ky_i-y_kx_i)}{\prod\limits_{j\neq
i}(y_ix_j-y_jx_i)} \\
&+&\frac{xy_{k}-x_{k}y}{y_{k+1}x_k-y_kx_{k+1}}\sum_{i=k+2}^{n_0+1}
\frac{(y_ix-x_iy)^{n_0-2}(x_{k+1}y_i-y_{k+1}x_i)}{\prod\limits_{j\neq
i}(y_ix_j-y_jx_i)} )\\
&=&\frac{1}{(n_0-1)!}(\frac{xy_{k+1}-yx_{k+1}}{y_{k+1}x_k-y_kx_{k+1}}\frac{(y_{k+1}x-x_{k+1}y)^{n_0-2}
(x_ky_{k+1}-y_kx_{k+1})}{\prod\limits_{j\neq i}(y_ix_j-y_jx_i)}
+\frac{1}{y_{k+1}x_k-y_kx_{k+1}}\\
& &\sum_{i=k+2}^{n_0+1}
(y_ix-x_iy)^{n_0-2}(\frac{(xy_{k+1}-x_{k+1}y)(x_ky_i-y_kx_i)-
(xy_{k}-x_{k}y)(x_{k+1}y_i-y_{k+1}x_i)}{\prod\limits_{j\neq
i}(y_ix_j-y_jx_i)}))\\
&=&\frac{1}{(n_0-1)!}\sum_{i=k+1}^{n_0+1}\frac{(y_ix-x_iy)^{n_0-1}}{\prod\limits_{j\neq
i}(y_ix_j-y_jx_i) }.
\end{eqnarray*}
Thus, when $n=n_0+1$ the theorem holds also, which completes the
inductive step and the proof. $\Box$

The following statements follow from Theorem \ref{Th:2dT}.

\begin{corollary}\label{co:co1}
$$
D^{v_1,v_2}T({\bf
x}|M)=\frac{1}{(n-2-v_1-v_2)!}\sum_{i=1}^n\frac{(y_ix-x_iy)_+^{n-2-v_1-v_2}}{\prod\limits_{j\neq
i }(y_ix_j-y_jx_i)}y_i^{v_1}(-x_i)^{v_2}.
$$
\end{corollary}

We now turn to non-polynomial part in $t(\cdot |M).$

In \cite{BeckFrobenius}, the Fourier-Dedekind sum is defined as
$\sigma_t(C;n)=\frac{1}{n}\sum\limits_{\lambda^n=1\neq\lambda}\frac{\lambda^t}{\prod_{c\in
C }(\lambda^c-1)}$, where $C$ is an integer multiset and $n$ is an
integer. To simplify the non-polynomial part in $t(\cdot |M),$ we
naturally arrived at the sums
\begin{equation}\label{eq:dedekind}
\frac{1}{Y_{ij}}\sum_{\theta^{M_{ij}}=1,\theta\neq e
}\theta^n\prod_{\omega\in M\setminus
M_{ij}}\frac{1}{1-\theta^{-\omega}},
\end{equation}
 which is considered as a generalized Fourier-Dedekind sum.
Here, $\theta ^{M_{ij}}=1$ means $\theta^{m}=1$ for any $m\in
M_{ij}.$
 In fact, it is a non-trivial problem
for computing all complex vectors satisfying $\theta^{M_{ij}}=1.$
In the following Lemma, we shows the generalized Fourier-Dedekind
sums (\ref{eq:dedekind}) can be converted into the 1-dimensional
Fourier-Dedekind sums.
\begin{lemma}\label{Le:Dedekind}
When $M$ is a 1-prime matrix, for any given integer $m,$ $1\leq
m\leq n, m\neq i,j,$
$$
\frac{1}{Y_{ij}}\sum_{\theta^{M_{ij}}=1\atop \theta\neq e
}\theta^n\prod_{\omega\in M\setminus
M_{ij}}\frac{1}{1-\theta^{-\omega}}
=\sigma_{t_{ij}}(C_{ij};Y_{ij}),
$$
where $C_{ij}=\cup_{1\leq h\leq n,h\neq i,h\neq j
}\{(fY_{im}+gY_{jm})^{-1}(-(fy_i+gy_j)x_h+(fx_i+gx_j)y_h)
\},t_{ij}=(fY_{im}+gY_{jm})^{-1}(-(fy_i+gy_j)n_1+(fx_i+gx_j)n_2)+\sum_{c\in
C_{ij}}c, $ where $f,g\in {\bf Z}$ satisfy
$gcd(fY_{im}+gY_{jm},Y_{ij})=1,$  moreover,
$(fY_{im}+gY_{jm})^{-1}(fY_{im}+gY_{jm})\equiv 1, mod \,\,\,
Y_{ij}.$
\end{lemma}
{\it proof:}
As pointed out in \cite{DaMi on the solution}, the elements in the
set $\{\theta|\theta\in A(M),M_{\theta}=M_{ij}\}$ have the form
$(W_{Y_{ij}}^{\alpha_1^l},W_{Y_{12}}^{\alpha_2^l}),$ where
$(\alpha_1^l,\alpha_2^l)\in {\bf Z}^2,1\leq l\leq Y_{ij}$.

Hence,
\begin{eqnarray}\label{eq:sum1}
\frac{1}{Y_{ij}}\sum_{\theta^{M_{ij}}=1\atop \theta\neq e
}\theta^n\prod_{\omega\in M\setminus
M_{ij}}\frac{1}{1-\theta^{-\omega}}=
\frac{1}{Y_{ij}}\sum_{l=1}^{Y_{ij}-1}
\frac{W_{Y_{ij}}^{n_1\alpha_1^l+n_2\alpha_2^l}}{\prod\limits_{h\neq
i,h\neq j} (1-W_{Y_{ij}}^{-(x_h\alpha_1^l+y_h\alpha_2^l)})}.
\end{eqnarray}

Noting $m\neq i,m\neq j,$ we set
$x_m\alpha_1^l+y_m\alpha_2^l\equiv k$ $mod$ $Y_{ij}$. Since $M$ is
a 1-prime matrix, $k$ runs over $[1,Y_{ij}-1]\cap {\bf Z}.$ Using
the similar method with the one in the proof of Theorem
\ref{Th:thfor}, we have
\begin{eqnarray*}
\alpha_1^l&\equiv &-(f_{ij}Y_{im}+g_{ij}Y_{jm})^{-1}(f_{ij}y_i+g_{ij}y_j)k \,\,\, mod\,\,\, Y_{ij},\\
\alpha_2^l&\equiv
&(f_{ij}Y_{im}+g_{ij}Y_{jm})^{-1}(f_{ij}x_i+g_{ij}x_j)k \,\,\,
mod\,\,\, Y_{ij}.
\end{eqnarray*}

Hence, (\ref{eq:sum1}) is converted into
\begin{eqnarray*}
& &\frac{1}{Y_{ij}}\sum_{k=1}^{Y_{ij}-1}
\frac{W_{Y_{ij}}^{(n_2(f_{ij}x_i+g_{ij}x_j)-n_1(f_{ij}y_i+g_{12}y_j))(f_{ij}Y_{im}+g_{ij}Y_{jm})^{-1}k}}
{\prod\limits_{h\neq i,h\neq j}(1-W_{Y_{ij
}}^{-(f_{ij}Y_{im}+g_{ij}Y_{jm})^{-1}(-x_h(f_{ij}y_i+g_{ij}y_j)+y_h(f_{ij}x_i+g_{ij}x_j))k})}\\
&=&\sigma_{t_{ij}}(C_{ij};Y_{ij}).
\end{eqnarray*}
$\Box$

\begin{remark}
When $|Y_{ij}|=1$,  since $\{\theta:\theta^{M_{ij}}\}=\{e\},$ the
terms in $\sigma_{t_{ij}}(C_{ij}:Y_{ij})$ disappear.
\end{remark}

Combining Theorem \ref{Th:FormP}, Theorem \ref{Th:form}, Theorem
\ref{Th:2dT} and Lemma \ref{Le:Dedekind}, we can present a
simplified  formulation for $t(\cdot |M)$.
\begin{theorem}\label{th:2dvfor}
Suppose $M=\begin{pmatrix}
  x_{1} & x_{2} & \cdots &x_n \\
  y_{1} & y_{2} & \cdots & y_n
\end{pmatrix}$ is  a $2\times n$ integer 1-prime matrix and
$\frac{y_i}{x_i}<\frac{y_{i+1}}{x_{i+1}}$. When ${\bf
n}=(n_1,n_2)^T\in {\Omega}_k\cap {\bf Z}^2$,
$$
 t({\bf n}|M)=p_{e,\Omega_k}({\bf n})+\sum_{(i,j)\in \{(i,j): i\leq k<j \} }\sigma_{t_{ij}}(C_{ij};Y_{ij}),
$$
where,
$p_{e,\Omega_k}({\bf x})=\sum_{j=0}^{n-2}p_{j,\Omega_k}({\bf x}),
p_{0,\Omega_k}({\bf x})=\frac{1}{n-2}
\sum_{l=k+1}^n\frac{(y_lx-x_ly)^{n-2}}{\prod_{j\neq
l}(y_lx_j-y_jx_l)},
p_{j,\Omega_k}({\bf x})=-\sum_{l=0}^{j-1}(\sum_{|v|=j-l}D^vp_{l,\Omega_k}({\bf x})
(-i)^{|v|}\frac{D^v\widehat{B}(0|M)}{v!}),$ $t_{ij}$ and $C_{ij}$  are defined in Lemma
\ref{Le:Dedekind}.
\end{theorem}
{\it proof:} Based on Theorem \ref{Th:form}, when ${\bf n}\in \Omega_k\cap {\bf Z}^2,$
$$
t({\bf n}|M)=p_{e,\Omega_k}({\bf n})+\sum_{{\theta} \in
A(M)\setminus e}{\theta} ^{\bf n
}\frac{1}{|det(M_{{\theta}})|}\prod_{w\in M\setminus
M_{{\theta}}}\frac{1}{1-{\theta}^{-w}}
1_{cone(M_{{\theta}})}(\Omega_k),
$$
where, the $p_{e,\Omega_k}$ can be determined easily. Since $M$ is a 1-prime matrix,
\begin{eqnarray*}
& &\sum_{{\theta} \in A(M)\setminus e}{\theta} ^{\bf n
}\frac{1}{|det(M_{{\theta}})|}\prod_{w\in M\setminus
M_{{\theta}}}\frac{1}{1-{\theta}^{-w}}
1_{cone(M_{{\theta}})}(\Omega_k)\\
&=&\sum_{i<j} \frac{1}{Y_{ij}}\sum_{\theta^{M_{ij}}=1\atop
\theta\neq e }\theta^{\bf n}\prod_{\omega\in M\setminus
M_{ij}}\frac{1}{1-\theta^{-\omega}}1_{cone(M_{ij})}(\Omega_k).
\end{eqnarray*}
Based on Lemma \ref{Le:Dedekind}, the above sum becomes as follows:
\begin{equation}\label{eq:sum}
\sum_{i<j}\sigma_{t_{ij}}(C_{ij}:Y_{ij})1_{cone(M_{ij})}(\Omega_k).
\end{equation}
Since when $k\geq j$ or $k< i$, $cone(M_{ij})\cap \Omega_k=\emptyset$. Hence, (\ref{eq:sum})
is converted into
\begin{equation}
\sum_{(i,j)\in \{(i,j): i\leq k<j \}
}\sigma_{t_{ij}}(C_{ij}:Y_{ij}).
\end{equation}
The theorem holds.
$\Box$

The explicit formulation presented in Theorem \ref{th:2dvfor}
contains $D^v\widehat{B}(0|M).$ Note $$ \widehat{B}(\zeta
|M)=\prod_{j=1}^n\frac{1-exp(-i\zeta^Tm_j)}{i\zeta^Tm_j},\zeta \in
{\bf C}^s.
$$

The following assertion is obvious.
$$
D^{v_1,v_2}\widehat{B}(0|M)=(-i)^{v_1+v_2}\sum_{k_1+\cdots+k_n=v_1}\sum_{l_1+\cdots+l_n=v_2}\frac{v_1!}
{k_1!\cdots k_n!}\frac{v_2!}{l_1!\cdots
l_n!}\prod_{j=1}^n\frac{x_j^{k_j}y_j^{l_j}}{k_j+l_j+1}.
$$

 Using  Theorem \ref{th:2dvfor}, we shall present an explicit formulation for an
actual vector partition function, which is the same with the one
presented in \cite{dgbeck}. By using Theorem \ref{th:2dvfor}, it
is indeed easier for obtaining the explicit formulation for the
actual vector partition function.

\begin{example}\label{ex:ex1}
Let  $A=\begin{pmatrix}
1 & 2 & 1 &0 \\
0 & 1 & 1 &1
\end{pmatrix}.$ We denote by $A_{ij}$ the square matrix containing
the $i$th and the $j$th columns in $A$.

For the matrix $A$, there exit three fundamental cones, which are
denoted  as $\Omega_1,\Omega_2$ and $\Omega_3$ respectively. We
shall discuss the explicit formulation for $t({\bf n}|A).$ After a
brief calculation, we have

$
 T({\bf x}|A)=
\begin{cases}
\frac{y^2}{2} ,& {\bf x}\in \Omega_1,\cr
\frac{1}{4}(-x^2+4xy-2y^2),&{\bf x}\in \Omega_2,\cr
\frac{x^2}{4},&{\bf x}\in \Omega_3.
\end{cases}
$

 Hence,
$p_{0,\Omega_1}=\frac{y^2}{2}$. According to Theorem \ref{Th:2dT},
$p_{1,\Omega_1}=3/2y$ and $p_{2,\Omega_1}=1$ respectively. Since
for any $1<j\leq 3,|det(Y_{1j})|=1$, the terms in Fourier-Dedekind
sum shall not appear when ${\bf n}\in \Omega_1\cap {\bf Z}^2$.
 Based on Theorem \ref{Th:2dT}, we have when
${\bf n}\in \Omega_1\cap {\bf Z}^2 ,t({\bf n|A
})=\frac{n_2^2}{2}+\frac{3n_2}{2}+1.$

Similarly, $p_{0,\Omega_2}=\frac{1}{4}(-x^2+4xy-2y^2),
p_{1,\Omega_2}=\frac{x+y}{2},p_{2,\Omega_2}=\frac{7}{8}$. Based on
Lemma \ref{Le:Dedekind}, the non-polynomial part is
$\frac{1}{Y_{23}}\sum\limits_{\theta^{A_{23}}=1,\theta\neq e
}\theta^n\prod\limits_{\omega\in A\setminus
A_{ij}}\frac{1}{1-\theta^{-\omega}}=(-1)^{n_1}$. Hence, when ${\bf
n}\in \Omega_2\cap {\bf Z}^2 , t({\bf n}|A)=
n_1n_2-\frac{n_1^2}{4}-\frac{n_2^2}{2}+\frac{n_1+n_2}{2}+\frac{7}{8}+\frac{(-1)^{n_1}}{8}.$

Using the same method with the above,  we obtain
$p_{0,\Omega_3}=\frac{x^2}{4},p_{1,\Omega_3}=x,p_{2,\Omega_3}=\frac{7}{8}.$

Hence,  $ t({\bf n}|A)=
\begin{cases}
\frac{n_2^2}{2}+\frac{3n_2}{2}+1,& {\bf n}\in \Omega_1\cap {\bf
Z}^2 \\
n_1n_2-\frac{n_1^2}{4}-\frac{n_2^2}{2}+\frac{n_1+n_2}{2}+\frac{7}{8}+\frac{(-1)^{n_1}}{8},&{\bf n}\in \Omega_2\cap {\bf Z}^2\\
\frac{n_1^2}{4}+n_1+\frac{7}{8}+\frac{(-1)^{n_1}}{8},&{\bf n}\in
\Omega_3\cap {\bf Z}^2.
\end{cases}
$
\end{example}

\begin{remark}
In Theorem  \ref{th:2dvfor}, when the case of
$\frac{y_i}{x_i}=\frac{y_j}{x_j}$ happens, the explicit
formulation for $T({\bf x}|M)$ can be obtained by taking the
limit. Using similar method with the one in the proof of Theorem
\ref{th:spM}, an explicit formulation for $t({\bf n}|M)$ can be
given also.
\end{remark}

\begin{remark}
 To simplify any-dimensional vector partition functions, we have to
give an explicit formulation for multivariate truncated power
functions $T({\bf x}|M)$ and compute the chamber complex
consisting of the fundamental $M-$cones, which are indeed
challenging problems.
\end{remark}

redpajama_set_name: RedPajamaArXiv


If the reservation has not been cancelled within the terms, Kokkola-Camping has the right to charge the full payment. Cancellations on arriving day at 6 PM at the latest.
Fees include linen (not towels) and end cleaning.
Prices include electricity. If there is 2 or more children under 5 years old, there will be extra fee for them.
If electricity is needed, tent will be placed to the caravan site with caravan prices.


Description: It's the final push. The Avengers unite their full might to face down the devious sorcerous Morgan Le Fay. But even they may night have enough power to defeat this enemy alone. Luckily, for them a fallen comrade returns from the grave to lend an ionically powered hand.


It's time for Halloween and what is Halloween with out some trick or treaters!! Help this group of trick or treating friends make their way through the maze and get to the candy! Click on image to view and print full size.
Parents & teachers – feel free to print this activity out and include it in your Halloween goody bags and party activities!
Or download the Trick or Treat printable maze for kids as a PDF file.
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Does this Trick or Treat, Halloween word search have the word "pirate" in it? We can't find it.
A frustrated teacher and her 30 students.
Hi! Love all your ideas and pages for the holidays. Can you email me the answers to the Halloween costume unscramble the words sheet. I want to make sure I have the correct answers before doing them with my class. Thank you so much!
Hello, Where do I find the answers to the Halloween Costume Parade unscramble?
Is the word pirate in this wordsearch. My students as well ,can not find it either.
This question is the same as one that was asked on the 17th from a frustrated teacher and there is no response.
Cheryl – the old word search was missing pirate. I have redone it to include the word pirate, but you will need to download the new one.


Contacts that clicked links in your email newsletter are included in the Analytics Explained: EngagedEngaged statistic in your email analytics. To find The Engaged tab, click on Stats under your campaign thumbnail, on the Campaigns page.
This will take you to a detailed view of the campaign's most recent mailing statistics. From here you can click on any of the main stat types (bounced, accepted, unsubscribed, etc) to view more details.
Click the Engaged tab, then select total or unique clicks.
Total Clicks: displays the total number of clicks, including multiple clicks by the same contact. Note the number next to each contact email in the list. This number represents the number of times that individual contact clicked.
Unique Clicks: displays the list of contacts that clicked. Multiple clicks are not represented in this view.
Hover over any URL listed to view the full link name.
Click any URL to highlight that one in blue, and display clicks for that URL in the statistics.
Note: Only the contacts that clicked links highlighted in blue are displayed in the email analytics list below.
You can combine click lists for multiple URLs, by clicking and highlighting additional links in blue. Click any blue URL a second time to deselect the link and remove the clicks from the list below.
Keep in mind! If no URLs are highlighted in blue, the full clicks list is displayed.
Now that you know how to find your contacts that clicked, use the options on the right to create a list of clicks, or export the clicks list, as you wish.
How Do I Find My Click-Through Rate?
Can I Share My Email Analytics?


text:
Canada added 154,000 jobs final month, pushing jobless price all the way down to pandemic low of 6%
Posted by News Express — December 3, 2021 in Business comments off 5
Canada's financial system added 154,000 new jobs final month, surging previous expectations and sufficient to maneuver the jobless price down to only six per cent.
Statistics Canada reported Friday that the jobless price fell by 0.7 proportion factors, to 6 per cent. That's the bottom jobless price for the reason that pandemic started. Prior to COVID-19, in February 2020, Canada had a jobless price of 5.7 per cent. It topped out at 13.7 per cent in May of that 12 months, earlier than sliding steadily decrease.
The knowledge company calculates that greater than 19.3 million people in Canada had a job final month. That's 183,000 greater than had one pre-pandemic.
The jobs surge means the entire variety of hours labored has lastly additionally risen again to where it was earlier than the pandemic.
Hours labored and wages up, too
There was excellent news on the wage entrance, too, as the info company calculates that wages through the month have been 7.7 per cent increased than they have been two years in the past.
The common hourly wage was $30.40 in November. That's up by $2.18 for the reason that identical interval two years in the past.
While increased wages are good for employees, they arrive with a double-edged sword that the price of dwelling goes up shortly, too. Those larger paycheques are tempered by the very fact that Statistics Canada knowledge reveals costs have elevated by 5.3 per cent in comparison with what they have been two years in the past.
Tanya Gullison, chief income officer with employment agency LHH, stated people are heading again to the workforce in droves as a result of they want the cash to pay for the upper price of the whole lot.
"Inflationary pressure [is] bringing workers back into the labour force in search of increased income," she stated. "Employees are also navigating this with wage increases, mirroring those in south of the border."
It's paying off to alter jobs, too, within the mixture. Statistics Canada says common wage positive aspects are growing at a quicker price for new hires than they're for present employees.
"Over the next fiscal year, bonuses and other perks are also likely to trickle over as a means of drawing new talent and retaining existing staff," Gullison stated.

redpajama_set_name: RedPajamaCommonCrawl


text:
TAKINGS. In GreeneLand last year, according to Department of Environmental Conservation figures (http://dec.ny.gov), 55 black bears were taken, along with 1291 white-tailed deer. Eight of those bears were taken by archers. Most of the bears were taken at, or from, mountain sites: Lexington (8 takings), Jewett (7), Windham, Hunter, Halcott, Ashland, Prattsville. The 2008 takings exceed the 2007 takings, of 31 bears and 1244 deer(s). They contributed to State-wide takings, of 1295 bears and 222,979 deer, exceeding the 2007 takings, or harvests, of 1117 and 219,141. Taken, of course, means killed by hunters. So does harvested. The growth in takings, DEC heads calculate, attests to growth in the target populations. TAKEN: jobs. In GreeneLand, according to State Government figures (www.labor.state.ny.us), the rate of unemployment rose in January to 8.5 per cent of the labor force. That is a jump from 6.2% in January 2008, and it is the second-highest figure in the mid-Hudson Valley (the highest being Sullivan County's 10.2%). The State-wide figure, meanwhile, was 7% in January, a jump from 4.7 in Januaryof 2008. TAKEN: expectations. At least three GreeneLand non-profit organizations were notified recently that State dollars they had been expecting, dollars they had more or less been promised, won't be forthcoming. The notices went "with much regret" from our State Senator, James Seward, to the Athens Cultural Center, the Catskill Chamber of Commerce, and the Greene County Historical Society. They involve sums of as much as $10,000. Senator Seward traced the disappointing news to the deficit reduction measure which Gov. David Paterson had proposed and the majority of legislators (not including Seward) had adopted. That measure "diverted or 'swept'" money from "funding projects set aside to pay legislative initiatives"--also known as member items--into the State Government's general fund. There is a chance, the senator stipulated, that the expected money will become available later. (We wonder, however, what leverage Mr Seward can exert, now that he is a member of the minority party in the Senate). HUDSON IN HUDSON ON HUDSON. At the Hudson Opera House on Sunday afternoon, author-illustrator Hudson Talbott will front a program devoted to his new book, River of Dreams: The Story of the Hudson River; to the current state of that majestic tidal estuary (expounded by Scenic Hudson's Ned Sullivan); and to the incipient musical show based on historic Hudsonian highlights (with composer Frank Cuthbert leading cast members through a couple of his new songs). From 2pm. http://riverofdreamsproject.blogspot.com WHEN? WHERE? On behalf of Jim Tedisco, the Republican candidate in the race to succeed Kirsten Gillibrand as U.S. Representative, publicist Joshua Fitzpatrick put out a news release yesterday that looked ahead and behind. Mr Tedisco, said Mr Fitzpatrick, "tonight conducted" a "Tele-Town Hall" "speaking with and hearing from, thousands of voters" in the 20th congressional district. Mr Fitzpatrick did not say where, or by what channel, that bit of "grassroots outreach" would occur or, alternatively, had occurred. NIMBY NEWS. A new housing development whose environmentally friendly features include "roof insulation, double glazing, lovers and wells" is not coming to GreeneLand. Neither is jazz musician Wayne Tisdale, who, according to Sporting News Today, "will make his first musical appearance since having a portion of his right leg amputated at halftime of the Sooners basketball game...next month." Greene County judges, however, have been active in particularly newsworthy ways. One of them established a record when, for Chi Quang Du, who was suspected of stabbing his girlfriend and had been captured after being at large for 11 years, he set bond at $55 million. Another set a date for the trial of one Randy Sloan, who is charged with felony standing nude in his front yard and whistling at passing school children. Those judges occupy benches in Greene County, Ohio and Greene County, Misssouri. They accordingly were not present at Chatham House in Columbia County (NY) the other day for the annual Crystal Apple award celebration, where Assemblyman Marc Molinaro, according to reporter Seasame Campbell, gave the keynote speech "after hors'devours."
Posted by Dick May at 11:23 AM
Dawn Wallant said...
Relative to lost or taken dollars, and the deleterious affect this will have on area youth, Common Ground is at risk of losing as much as 10,000 in allocations from both Greene and Columbia county youth bureaus as a result of the governor's proposed "Youth Block Grant," which will siphon "diversion" monies into mandated services. This is an example of, to quote a cliche, on "ounce of prevention is worth a pound of cure."
Why Murphy?
Murphy Will Win
Greene Gloom
The Village Elections
Greeneward March


Antonella graduated from the University of Salerno in 1993. Following work experience in New York and London and Milan, she joined Noesis, MTI's partner in Italy, in 1997. In Noesis, she heads the Corporate, Finance and Health Care divisions. In the last few years she has specialized in crisis communications, advising clients on acquisitions, reorganizations, media attacks, judicial enquiries. The most recent crisis she managed was that of Asso Ventidue, the Italian ship occupied by the Libyan military. Antonella is qualified as a professional journalist. She speaks Italian, French and English.


text:
La vallée de l'Oberaar est une vallée des Alpes située en Suisse, dans le canton de Berne.

Géographie

La vallée de l'Oberaar est l'une des deux vallées dont les eaux donnent naissance à l'Aar, avec la vallée de l'Unteraar qui se trouve au nord. Dans cette vallée coule le glacier de l'Oberaar.

D'est en ouest la vallée de l'Oberaar est séparée de la vallée de l'Unteraar par :
 le Vordre Zinggenstock  ;
 le Hindre Zinggenstock  ;
 le Vordre Tierberg  ;
 le Hindre Tierberg  ;
 le Scheuchzerjoch  ;
 la cime nord-est du Scheuchzerhorn  ;
 le Scheuchzerhorn  ;
 le Grunerhorn  ;
 l'Oberaarhorn .

Du nord au sud la vallée est séparée de la vallée de la Wysswasser par :
 l'Oberaarhorn ;
 l'Oberaarjoch  ;
 le Nollen  ;
 l'Oberaarrotjoch  ;
 l'Oberaarrothorn .

D'ouest en est la vallée de l'Oberaar est séparée de la vallée de Conches par :
 l'Oberaarrothorn ;
 le Münsterjoch  ;
 le Talschienpass  ;
 le Talschien  ;
 le Löffelhorn  ;
 le Geschinerstock  ;
 l'Uelistock ;
 le Grosses Sidelhron  ;
 le Sidelhorn .

Annexes

Source
 Carte topographique suisse

Liens externes

Vallée dans le canton de Berne
Vallée des Alpes suisses

redpajama_set_name: RedPajamaWikipedia


text:
Songbirds Phonics Pack: Stage 1+
The UK's most successful reading scheme, Oxford Reading Tree, teams up with star author Julia Donaldson in this brilliant phonics series, full of wonderful stories which practise the basic sounds that make up words. Songbirds builds on children's phonic knowledge by letting them segment, blend and match common letter sounds and letter blends in brilliant stories written with the author's hallmark energy and wit. All the stories support the school curriculum for literacy, providing fully decodeable texts that build on children's phonic knowledge as they work through a total of 36 books spread over six stages.
This set of six Stage One books is intended for children just starting out with phonics. It matches a sound to one letter, and teaches the short vowel sounds and all consonants except x and q. It also includes comprehensive teachers' notes.
"A pedigree scheme worth every penny for helping children get to grips with words, sounds and spelling patterns… Julia Donaldson has described Songbirds as her 'proudest achievement' and it is easy to see why." John Dabell, Teach Primary
Julia Donaldson is one of the UK's best loved authors, known for brilliant picture books like The Gruffalo, Room on the Broom and Stick Man. She was also the Children's Laureate 2011-2013.
Julia has won many major book prizes, including the Smarties and Blue Peter Awards for The Gruffalo, the Blue Peter and Sheffield Children's Book Prizes for Room on the Broom, the British Book Awards Children's Book of the Year for The Gruffalo's Child and the Red House Award for Super Worm. Julia was also the Children's Laureate 2011-2013.
Book Band Pink
Letters and Sounds


David Hartman has experience working as a cartographer, but his claim to fame is helping the state of Minnesota map out its charter school law in 2009.
At head of the state's charter school center, Hartman implemented systems to approve and oversee authorizers in response to the law, which put the state in charge of the authorizing process.
Since coming to SchoolWorks, Hartman has focused on charter school reviews and site renewal visits for the School District of Philadelphia and the National Heritage Academy, among others. He has also overseen SchoolWorks' efforts to assist Chicago Public Schools in overhauling its charter application and renewal processes.
" I've had the opportunity to work on many things in the forefront of the evolution of charter school oversight. There's a lot of room for growth, so our touch at this specific time is pretty significant.""
Running, cross-country skiing, live music.


text:
require_relative "dogids/base"
require_relative "dogids/cache"
require_relative "dogids/config"
require_relative "dogids/deploy"
require_relative "dogids/reload"
require_relative "dogids/run"
require_relative "dogids/ssh"
require_relative "dogids/version"

redpajama_set_name: RedPajamaGithub


text:
Q: SharePoint randomly displays web part error I am noticing that on all pages sometimes SharePoint doesn't load web part and displays the following message:

Unable to display this Web Part. To troubleshoot the problem, open
  this Web page in a Microsoft SharePoint Foundation-compatible HTML
  editor such as Microsoft SharePoint Designer. If the problem persists,
  contact your Web server administrator.

Refreshing the page solves this problem.
And another weird behaviour I noticed today. Let's say I have a page in which I have placed Content Editor web part.
I open the page and do not turn of my computer and go home. Next day when I check that page again (I don't refresh it, nothing, it's just there already opened which I did yesterday) it shows the above web part error message.
Why is this happening?

A: This is quite a generic error message. And a generic answer would be - it is happening as it takes to long to load a query.
In order to see the limit of a query timeout you can run powershell command:
$farm = Get-SPFarm
$farm.XSLTTransformTimeout

If it is 1 sec, you may want to change it to 5 (or whatever you think is reasonable). Use the powershell command:
$farm = Get-SPFarm
$farm.XSLTTransformTimeout = 5
$farm.Update()

If that won't help you have to try to find if there is anything specific that causes the timeout error. In my case it was a SharePoint bug - if you backup/restore a site collection,  you get this error when displaying All Items views. I created a view loading 20 items at a time instead of all and it fixed the issue.

redpajama_set_name: RedPajamaStackExchange


"You're the future," Joe Biden told a rapt audience Sunday afternoon at the South by Southwest Conference here, appealing to the brightest minds in tech and media to join him in his commitment to fighting cancer. "You could make a gigantic impact. We need your ingenuity. You could have a profound impact on cancer."
Biden was a late addition to an uncharacteristically politically loaded SXSW speaker lineup. On Friday, New Jersey Sen. Cory Booker opened the conference — and took a few shots at President Trump — with a packed panel of his own. By early Sunday, Biden's event had become such a hot ticket that wristbands ran out immediately and a nearby theater had to be opened to accommodate the overflow.
Tan and relaxed, Biden dropped a staggering array of cancer research statistics, but he grew emotional when speaking about his son Beau, who died of brain cancer in 2015 at age 46.
The Bidens are now continuing the work of the initiative with their nonprofit Biden Foundation "to spare other families what our family, and so many other families, have gone through," he said.
He told how he first came to spearhead the program after making an offhand comment to then-President Obama during a stroll through the Rose Garden. "I would have loved to have been the president who presides over the end of cancer as we know it," Biden told his then-boss.
Obama then handed Biden the reins to lead the program intended to accelerate cancer research in America.
Biden had come to SXSW to recruit talent. Addressing an audience that included techies and innovators, he pointed out how easily websites and apps had made it to buy movie tickets or cash checks at the swipe of a finger on a smartphone, then wondered why cancer patients like his son couldn't just as easily send test results from one hospital to another.
"Many of you are developing technologies and innovations for purposes large and small, fun and serious, entertaining and lifesaving, that have nothing to do with cancer — but you could make a gigantic impact," he said, his voice booming. "We need you to help us reach people who need to change their behavior and avoid cancer. You're doing it to help them figure out how to buy a product… We need to reach people."
He did talk a little politics, of course, chronicling the successful bipartisan campaign he led to secure $6.3 billion in funding for the 21st Century Cures Act, which includes funding for the cancer initiative.
"Guess what? The only bipartisan thing left in America is the fight against cancer," he said. "I assure you there's still a lot of really decent people left in the Congress in both parties."
Though private philanthropy is important to the funding of cancer research, "the bulk of the money — billions and billions — comes from taxpayers," he said. "Your government that many of you don't like is the vehicle through which this gets funded."
Biden couldn't resist taking a few shots at the current administration, calling out unnamed figures "in the new outfit" who deny that global warming is real. "It frustrates me," he said with a laugh.
But he declared that he would work with Trump's White House to continue seeking a cure.
"It is my hope that this new administration, once it gets organized… will be able to focus on and be as committed and enthusiastic as we were with the goal of ending cancer as we know it," Biden said. "I pledge before all of you and all the world that I will do everything in my power to work with the new administration."
He concluded by referencing John F. Kennedy's famous 1962 moon speech. "He talked about the effort to go to the moon as a commitment the American people had made and that they were quote, 'unwilling to postpone,' " said Biden. "I am unwilling to postpone for one day longer the things we can do now to extend people's lives — and so should you be."


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First Day of School Class Photos!


Abstract. Algorithms that detect silhouettes, creases, and other edge based features often perform per-edge and per-face mesh computations using global adjacency information. These are unsuitable for hardware-pipeline implementation, where programmability is at the vertex and pixel level and only local information is available. Card and Mitchell and Gooch have suggested that adjacency information could be packed into a vertex data structure; we describe the details of converting global/per-edge computations into local/per-vertex computations on a related .edge mesh.. Using this trick, we describe a feature-edge detection algorithm that runs entirely in hardware, and show how to use it to create thick screen-space contours with end-caps that join adjacent thick line segments. The end-cap technique favors speed over quality and produces artifacts for some meshes.
We present two parameterizations for mapping stroke textures onto these thick lines.a tessellation-independent screen space method that is better suited to still images, and an object space method better suited to animation. As additional applications, we show how to create fins for fur rendering and how to extrude contours in world-space to create the sides of a shadow volume directly on the GPU.
The edge mesh is about nine times larger than the original mesh when stored at 16-bit precision and is constructed through a linear time pre-processing step. As long as topology remains fixed, the edge mesh can be animated as if it were a vertex mesh.
NVIDIA has now shipped NV40 and NV47 hardware ("GeForce 6 and 7"). These cards have the ability to perform a texture lookup in the vertex stage. As mentioned at the end of the paper, this allows the edge mesh to store vertex and normal indices instead of 3D vectors, leading to a 6x reduction in the size of our data structure.
Marc ten Bosch has an extension to our technique called Partial Silhouette Detection that helps maintain coherence under animation with very attractive results.
Jorn Lovisach's Stylized Haloed Outlines on the GPU adds more line styles for sketchy strokes and halos for conveying occlusion.
In the paper, we mention that artists have long used Raskar's 1999 Image Precision Silhouette Edges method, which they apparently discovered independently. Ramesh tells us that he is often asked about this, and suggested we provide an example so it can be cited in the future. Here are three: Conor O'Kane's "Honey" model (2000, http://www.planetquake.com/polycount/info/quake3/honey/honey.shtml) uses a 1D texture for 2-tone shading (ala Gooch) and an inverted, black, enlarged version of the mesh to provide cartoon outlines. Seth Galbraith wrote the Cartoon Outline Generator tool (2001, http://www.planetquake.com/gg/tutorial/toon/) to automate the process of producing the second mesh. The Anim8or site offers the following artist submitted tutorial (date unknown): http://www.anim8or.com/tutorials/from_users/toon_notes/toon_rendering_hints.jpg. Note that Quake 3's realistic rendering engine itself is not modified by these techniques and these models work with any Quake 3 renderer; the changes are purely to the geometry of a model and are performed in a modelling tool.
Ramesh has also pointed out to us that his 1999 method can handle thick silhouettes, and that the 2001 paper shows examples of textured strokes. Figure 12 in his 2001 paper shows a thickened stroke exhibiting stroke-cracks; this is where the thick-stroke-cracks problem is first described, as we mentioned in the text.
These are corrected in the PDF on this page.


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Home » 2013 » December » 05
Holiday Thoughts from the President!
Well it's that time of year again and the Holiday Season is upon us! Racers and race fans across the country are hard at work making their Christmas list of items for Santa to bring. A new chassis, new engine, new body and season race tickets to their favorite race tracks. I hope everyone gets all they ask for on their lists. During this time of year we need to stop and remember those that we lost this season and to give thanks to those who have given their life to protect us so we can enjoy our beloved sport of auto racing. So when you see those men & women you see in the military take the time and say Thank You for all they do. "May the Lord Bless & Keep Us All" throughout this holiday season! Bill Barnhart MARFC President
Santa Big Hit at MARFC Children's Christmas Party!
Once again kid's young and old enjoyed at great time at the Annual MARFC Children's Christmas Party on December 1st. First thing "Bubbles the Clown" was on hand to make custom made balloon sculptures for all and after a great lunch that was provided Jimmy Gullett and Pizza from Fran Russo Bubbles entertained the group with a magic show. After that the big man himself "Santa" arrived and everyone got to tell him what they wanted for Christmas and received a gift from him to tide them over until Christmas Morning. A raffle was held for all the children and each went home with a raffle prize and a bag of other goodies. Many thanks to all who donated items for the party your support is greatly appreciated. "Merry Christmas to All"
MARFC February 2020 General Meeting New Location!
MARFC 2020 March Bowling Event Reschduled!
March 14 @ 12:30 pm - 6:00 pm
April 2020 General Meeting!
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Note: AF 92453. Seed from Karel Knize, Peru, sown 1992. First flowering January 2008. Series of photos: anw01451 floral parts. Compare to my casus #41 (identified as P. ferruginea) and casus 97 currently being thought to be P. ferruginea. On my plant petals turn red at anthesis.


Teachable Machine is a A.I. Experiment created as a collaboration between Google Creative Lab, danish duo Støj and the PAIR group at Google that has created deeplearn.js.
The experiment is exciting, because it's the first time we really have been exploring running and training machine learning models right in a browser in a way that everyone can use it. Read more about the project on the site.


Let's see: AvisBudget now realizes 15 months after touting all of the envisioned fleet-sharing synergies with Zipcar that there are costs involved in shuttling the vehicles back and forth between locations? Very curious. One solution is allowing Zipcar to introduce one-way rides, which is a very attractive option for customers regardless.
Zipcar is currently running a beta for one-way rentals, which are expected to be introduced in the third or fourth quarter, but the reason behind the new service isn't merely for passenger convenience.
In fact, the Avis Budget Group found out the hard way that achieving all of those much-touted fleet-sharing synergies between AvisBudget and Zipcar — sending Avis and Budget vehicles to Zipcar during peak-demand periods on weekends and dispatching Zipcars to Avis and Budget locations when road warriors need them on weekdays — was a lot more complicated than initially envisioned.
Speaking at the J.P. Morgan Gaming, Lodging, Restaurant & Leisure Management Access Forum in Las Vegas earlier this week, AvisBudget CEO Ron Nelson said fleet-sharing has turned out to be more complicated than the company thought because there's a cost tied to moving the vehicles from one location to another.
"The cost of shuttling can really impact the incremental yield you expect to get from delivering that car to pick up the incremental demand," Nelson said.
While Avis and Budget do one-way rentals, they were prohibited at Zipcar until the beta began. The plan is to introduce the one-way Zipcar rentals in the third and fourth quarters of 2014 so the much-hyped fleet-sharing synergies can begin to be achieved toward the end of 2014 and into 2015.
In this way, customers will pay to bring the vehicle from the airport to a downtown area "so it is a revenue item and not a cost item," Nelson said.
AvisBudget and Zipcar have already been doing some fleet-sharing, but the profits weren't optimal, Nelson said. Once Zipcar begins its one-way rides "then we'll be able to really optimize the fleet-sharing paradigm," he said.
It's unclear how these complications will impact the $50 million to $70 million in overall annual synergies that AvisBudget estimated would take place when the company announced its plans in January 2013 to acquire Zipcar for $500 million.
The bulk of the synergy-making has already been pushed back toward 2015. As part of a February 2014 AvisBudget investor presentation, Zipcar projected $20 million to $25 million in fleet-sharing synergies, which have not yet been accomplished.
Zipcar, which is currently offering service in the U.S., Canada, the UK and Spain, plans to expand into Australia in 2014.
Nelson said in 2013 AvisBudget cut its inventory in online travel agencies' opaque services, particularly Hotwire, by 40% to 50% in a climate of rising pricing, particularly in the leisure market.
Nelson said AvisBudget expects to take a relatively tough stance when negotiating smaller corporate accounts this year as negotiating large, corporate accounts remains very competitive.
While AvisBudget saw four quarters of "positive pricing in 2013," Hertz has been "erratic," participating in only one-third to one-half of the price increases, and Enterprise, which traditionally has been a "spoiler" in not matching price increases, has "now flipped roles and become the fast follower on most price increases," Nelson said.


The unique conic shaped beater allows maximum incorporation of air that contributes fluffy texture and smooth cake batter. The powerful 450W motor will make light work of the toughest of doughs.
The bowl was designed to seamlessly match the conic shaped beaters for best mixing results. The automatic driven bowl rotates to allow for better and more thorough mixing.
Beaters have been designed with a different pattern on one end to be intuitively assembled for a quick setup. The suction feet keeps your stand and bowl mixer stable & grounded to the surface.


Crown prince, 1972; succeeded to throne, 1972. Coronated King of Bhutan, head of state, 1974. Abdicated throne to his son Jigme Khesar Namgyel Wangchuck, 2006. Chairman Planning Commission Bhutan, since 1972, Council of Ministers, 1972-1998. Also commander-in-chief armed forces.
His Majesty Jigme Singye Wangchuck, former King of Bhutan. Named one of 100 Most Influential People, Time Magazine, 2006.
Wangchuck, His Majesty Jigme Singye was born on November 11, 1955 in Thimphu, Bhutan. Son of Druk Gyalpo Jigme Dorji Wangchuk and Queen Ashi Kesang.
Student, North Point, Darjeeling. Student, Wangchuk Academy, Paro, Bhutan, 1972.
Crown prince, 1972; succeeded to throne, 1972. Coronated King of Bhutan, head of state, 1974. Abdicated throne to his son Jigme Khesar Namgyel Wangchuck, 2006.
Chairman Planning Commission Bhutan, since 1972, Council of Ministers, 1972-1998. Also commander-in-chief armed forces.
Married Dorji Wangmo Wangchuck. Children: Jigyel Ugyen, Sonam Dechen. Married Tshering Pem Wangchuck.
Children: Chimi Yangzom, Kesang Choden, Ugyen Jigme. Married Tshering Yangdön Wanchuck. Children: Jigme Khesar Namgyal, Dechen Yangzom, Jigme Dorji.
Married Sangay Choden Wangchuck. Children: Euphelma Choden, Khamsum Singhye.
Named one of 100 Most Influential People, Time Named one of 100 Most Influential People, Time Magazine, 2006.


Su-Mi Lee is assistant professor in the Political Science Department at the University of Hawai`i at Hilo.
Good Mediator: Relational Characteristics of Effective Mediators is a noteworthy contribution to the conflict management field that deepens our understanding of the forces that drive successful international mediation. The book advances knowledge in the field by nicely unpacking, in ways that have been overlooked by previous research, the roles that mediator impartiality and interest play in shaping the outcome of mediation. Dr. Lee combines both a compelling theoretical narrative and rigorous empirical analysis of mediation, making the book an important resource for both students and practitioners of international mediation.
Lee brings interesting new light to old issues in regard to mediation and in the process provides characteristics of an effective mediator in well-defined circumstances, using both big data and small cases. Her interacting variables are interest and bias, with resources, weight, and conflict stages added as context variables. This book surveys a large number of instances and uses statistical analysis to bring out clear conclusions. Lee then tests her ideas on two cases--Borneo and Yemen--to see how they land on the ground. The study will not lay to rest the debates over interest and bias as a help or hindrance to effective mediation but it will enrich.
This book examines how mediators' relational characteristics can alter the outcome of mediation in international militarized disputes. Rather than focusing on neutrality and bias that are rigorously discussed in the mediation literature, this book redefines the sources of effective mediators, impartiality and interest, and shows how such seemingly opposing characteristics of a mediator can lead to successful mediation through different channels, complementing each other. This book also uncovers the two dimensions of mediator trust, fairness and capacity. For mediation to be successful, disputants must trust that mediators are fair and capable of leading to successful mediation. The identification of these two sources is crucial as each of the two relational characteristics of mediators' discussed above improves each dimension of the mediator trust. Moreover, this book explores how impartial and interested mediators are more or less effective in particular conflicts characterized by the nature of dispute, the nature of disputants' relationships, and the nature of mediators. The book's empirical analysis as well as case studies on the mediation cases led by the Philippines and Syria in the Borneo confrontation and Yemen dispute, respectively, demonstrate how the levels of mediator impartiality and mediator interest counterbalance each other in leading to mediation success.


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Defines behavior when the user swipes horizontally on a row in a table.
Gets or sets the background color of the action button.
Gets or sets the vidual effect that will be used for the action button background.
Gets a value that indicates whether the button for the action will have a default appearance, the appearance for a destructive action, or the appearance for a nondestructive (normal) action..
Gets or sets the text for the action button.
Creates and returns a table view row action with the specified style, title, and handler.


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Review: Bedlam: The Game By Christopher Brookmyre - Work in progress.
Christopher Brookmyre himself was awesome enough to send me a review copy of his brilliant game, Bedlam.
If someone asked me right now, "Hey, Jesper! What game sums the 80's and 90's of online gaming up the most?", I'd definitely answer with "Bedlam, for sure".
Bedlam is a game that is written by Christopher Brookmyre, and developed by RedBedlam. If you ever feel that there is a lack of retro online gaming and 80's & 90's in your life, Bedlam is the game for you.
It all starts off with you, the character, being in some sort of weird test room, that later on takes you to a simulation, sort of. The simulation, is you being an action hero in a world of today's 80's/90's of online gaming. It has the graphics and visuals that you'd expect it to have: good looking in terms of 80's/90's gaming but falls short in today's standards. Obviously, that's what they wanted it to be like. It has the features of today's standards, though. It seems to be running in native 1080p at 60 fps, too. Expected, of course! The game has a very simple and basic menu, with a very cool live background of what I think is the main character. There's also a very interesting feature in the settings, that can change how long the person you've just brutally killed, should lay on the ground, the person can even lay on the ground forever. Pretty unique!
As I walk around in the retro wasteland, I see and enter a building that leads me to this guy:
Talking to him leads to him wanting me to do a salute in front of him (of course). The joke's on him, however: the main character accidentally ("accidentally"?) stabs him to death. Great way to start the game!
No retro game is a retro game without a diverse arsenal of retro gaming inspired weapons! As of now, I've encountered the pistol, double-barrel shotgun and crossbow. The game features the mechanic that we've all seen a lot, the "pick up and throw away" mechanic. Also known as when you're able to pick an object up, as well as either put it down or throw it away. I've, so far, been able to pick up bodies and crates. Now, the game also features ragdoll physics and limb dismemberment, which is very awesome!
So far, the game is very fun to play and it gives you a very welcome sense of nostalgia! The main character needs to hold out as the action hero as long as possible for her to be able to get out of the simulation. When I think about it, however, it seems like the simulation isn't just a simulation, it's more real than what we might think. The protagonist hints at this, when she feels the pain of getting hurt.
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Review: Far Cry 6
Review: 12 Minutes
Review: Knockout City


Posted By: HorrO 0 Comment horror, Press Release, promote horror, Spread the Horror, Watch "Pantheon Woods" Today!
TWO SISTERS WITH THE HELP OF A HUNTER VENTURE DEEP INTO THE WOODS TO INVESTIGATE A SMALL TOWN LEGEND.
Riley Rey feeds her interest in cryptozoology with a steady stream of video uploads and the help of her older sister, Rebecca, who plans to be the next hit cinematographer. With a You- Tube channel featuring the usual ghost haun ngs and UFO videos, Riley decides it's me to up the ante. In an e ort to increase viewership, the two sisters hire the help of Greg, a seasoned tracker with a murky past. He guides the girls on an adventure to inves gate Chanoc a small town legend. Through adversity and con ict, they discover the secrets of the Panteon Woods, but never return.


This entry was posted in Photography and tagged Cariboo Mountains, clearwater, Helmcken Falls, lake panoramas, majestic river views, mountain vistas, old building, old homesteads, photography, roadside photographer, tourists from all over the world, traveling across Canada, waterfalls, Wells Gray Provincial Park. Bookmark the permalink.
I like your photos. I've been through there many times, and it is always a treat. It brings me back in time, as if it were the nineteenth century. Your photos have a fresh, clean look.
Thanks for your kind words John. I am so pleased my blogs have been helpful to you. As to my camera settings, I mostly used Manual. It is only on rare occasions like car races, rodeos and the like that I move my selector to Shutter priority. I might choose Aperture priority if depth of field is really important. The last shot that has the building with white grass in the foreground is from my infrared camera. I'll go thru my RAW metadata from this and send you what I can my friend.
One the additional pleasures of Wells Gray Park is in the winter. The road is kept ploughed, throughout the winter, as far as the Helmcken Falls view point. At other seasons, one can also follow the trail, which leaves from near the Dawson Falls parking area, to the south rim of the Falls. I first did that hike in the summer of 1961 when I was on a survey crew laying out a potential damsite on the Clearwater River just below Clearwater Lake. Thankfully, the dam never proceeded!
Great article. I just got out to Wells Grey Mr.Enman, it was a real adventure. Living in Kamloops, I had thought my days of trudging through snow to get a good shot were done for the year. I had to take the treck through Pyramid Campground road (snow prevented driving in) to get to Dawson Falls, the roadside viewing platform of the falls being a bit too saturated by photographers for my taste. Can you tell me where your third photo in was taken? I really enjoy the look you achieved of these falls. I predict this season will find me getting to know this park through my lens on a whole new level. I like your photo of the dilapidated old house as well.
Glad you liked my article James. I really enjoy that park. I used to take my college photo classes up there for an overnighter twice a year.
The third shot is of my favourite location, Bailey's Shute. I know you are going to have fun photo'ing Wells Grey.


McGregors Corporate, who are an independent firm of Chartered Accountants, Finance Directors and Business Advisers have started to operate in Leicester, with an office in New Walk.
McGregors, currently operate from offices in Birmingham, Nottingham, Mansfield and Lincoln, have a mixed partnership made up of senior directors and partners from large national and smaller mid- tier accountancy firms, together with partners with industry experience as finance directors. They also contribute and are members of Probiz Tax Solutions, who both sponsor and have Leicestershire cricket captain, Mathew Hoggard, as their local ambassador and promoter.
Steven Mugglestone, the partner responsible for the development of the Leicester practise, who had worked in Leicester for many years and still lives with his family in the city, said, "I am very keen and excited about working within the Leicester business community again and opening a presence in Leicester.


text:
Q: Elasticsearch top_hits aggregation vs latest document I am trying to get a list of users who has as their last activity "connect". Ideally, I want this as a metric viz or a data table in Kibana showing the number of users that connected last and the list of them, respectively. I have, however, given up being able to do this in Kibana. I can get something similar directly from Elasticsearch using a terms aggregation followed by top_hits as below. But the problem is, even though I am sorting the top_hits by @timestamp, the resulting document in NOT the most recent.
{
"size" : 0,
"sort": { "@timestamp": {"order": "desc"} },
"aggs" : {
    "by_user" : {
    "terms" : {
            "field" : "fields.username.keyword",
            "size" : 1
        },
        "aggs": {
            "last_message": {
                "top_hits": {
                    "sort": [
                        {
                            "@timestamp": {
                                "order": "desc"
                            }
                        }
                    ],
                    "_source": {
                        "includes": ["fields.username.keyword", "@timestamp", "status"]
                    },
                    "size": 1
                }
            }
        }
        }
}
}

*

*Is there a way to do this directly in Kibana?

*How can I make sure top_hits gives me the latest results, rather than the "most relevant"?

A: I think what you want is field collapsing, which is faster than an aggregation.
Something like this should work for your use case:
GET my-index/_search {
    "query": {
        "match_all": { }
    },
    "collapse" : {
        "field" : "fields.username.keyword"
    },
    "sort": [ {
        "@timestamp": {
            "order": "desc"
         }
    } ] }

I might be missing something, but I don't think Kibana supports this at the moment.


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"A Bronx Tale" Musical Earns Mixed Reviews
By Jennifer R Jones | Posted on December 09, 2016 4:15 PM
A Bronx Tale opened to mixed reviews, receiving a good review in the New York Times but generally earning mediocre reviews.
Robert De Niro and Jerry Zaks Direct the Musical of Chazz Palminteri's Story
This past week, A Bronx Tale opened at the Longacre Theatre. The show had been running in previews since November 3, 2016, and is presently scheduled to continue for an open ended run. This new musical has a book by Chazz Palminteri, who also wrote the play of the same name, in which he starred as a one-man show, first Off-Broadway in 1989, and then on Broadway in 2007 to 2008. When Robert De Niro first saw this play, he offered to buy the film rights, which Palminteri granted as long as he could pen the screenplay and star in the film. De Niro agreed, and also co-starred in and directed the film, which was released in 1993. Therefore, the path was laid long ago for A Bronx Tale the Musical, which is now open and running on Broadway. With lyrics by Glenn Slater (School of Rock, Leap of Faith, Sister Act, The Little Mermaid) and a score by Alan Menken (Aladdin, Leap of Faith, Newsies, Sister Act, The Little Mermaid), the show stars a cast that does not include Palminteri or De Niro. Rather, Bobby Conte Thornton makes his Broadway debut in the lead part of Calogero Anello, originally played by Chazz Palminteri in his one-man show based off his life growing up as an Italian-American in the Bronx.
Mixed Reviews from Critics, Praised by the New York Times
Upon opening, the majority of critics were not entirely blown away by this new musical. Only Charles Isherwood of the New York Times offered it a rave review, but even then he admitted that it doesn't "break ground" or "dazzle." He compared it to comfort food, which in this case is presumably food that is comfortable for committed musical theatre fans only, as he alludes to its "reliable pleasures" and "polished professionalism." Without saying the name of the show, he allows that it is no Hamilton, but nonetheless he enjoyed it. Other reviewers set the bar higher in terms of originality. Frank Scheck of the Hollywood Reporter felt the tale has worn its welcome, after the film and series of plays from the subject matter. Matt Windman from AM New York felt the story was derivative of better musicals such as West Side Story and Jersey Boys, and also that the tone was excessively solemn and sentimental. Joe Dziemianowicz from NY Daily News states that the directors did not squander their gifts with this show, but they haven't dug deeply into their imagination, either. Chris Jones of the Chicago Tribune remarks that the "Rich Dad, Poor Dad" narrative is not new, but that this show represents how Broadway can fulfill the national demand for the same old fare.
Demonstrating Some Promise at the Box Office
Despite these less than stellar reviews, A Bronx Tale is managing to sell fairly well at the box office. While it is by no means selling out, the show has brought in over 75% of its gross potential in every week since it began previews. In the first partial week of 3 performances, the show actually brought in 91.58% of its gross potential, showing strong promise right out of the gate. However, more recent weeks have seen more modest success, with the last reported week of box office figures, the week ending December 4, 2016, bringing in a weekly gross of $723,675, or 76.24% of its gross potential over eight performances. This past week was in fact the first full week of eight performances. Another high performing week was the week before that, which was also the week of the Thanksgiving holiday, which brought in a gross of $717,860 over 7 performances, and which represents 87.09% of its gross potential. Nevertheless, with such questionable quality from these reviews, and with limited brand recognition despite the feature film and the stardom of the co-director De Niro, this show may not break through to hit status at any point in its run.


text:
Fares Sabawi, Digital Journalist
Published: October 9, 2019, 4:35 pm
Tags: Texas, Local, News, San Antonio, Crime
Suspect accused of stealing Converse police car identified
Xavier Deshawn Johnson arrested on 4 criminal charges
CONVERSE – Converse police identified the man accused of stealing a police car while fleeing from police Saturday.
Xavier Deshawn Johnson, 25, was charged with aggravated assault against a public servant, theft of a vehicle, evading arrest and being a felon in possession of a firearm in connection with the incident.
READ MORE: Converse police car stolen, 1 suspect in custody
On Saturday, Converse police spotted a red sedan traveling in the wrong way in the 8000 block of FM Road 78, Chief Fidel Villegas said.
An officer eventually pulled Johnson over in the 300 block of Center Street, police said. Johnson tried to flee the scene in the car, but he and a passenger bailed out of the car by Anteris Park, Villegas said.
The passenger ran on foot, but Johnson pushed a Converse police officer and stole the police car. The driver reached speeds of 80 mph before the car rolled over near I-35 South and Loop 410.
The second suspect has not yet been identified.
Court records show Johnson was out on bail for two pending robbery charges at the time of his arrest.
Fares Sabawi
Fares Sabawi has been a journalist in San Antonio for three years. He has covered several topics, but focuses on crime, courts and record requests the most. He also has a recurring appearance on KSAT News at 9's Trending Now segment.


Brakes took their goal tally to 14 in two games as they dished out a second straight hammering at the New Windmill on Saturday.
Striker Paul Eden scoured four as the Doc Marten's League second string side was dumped out of the Coventry Evening Telegraph Cup. The ex-Racing Club marksman struck four times in the second half, once from the penalty spot.
Jason Cadden and Barry Proctor's side made all the early running and went in front on 19 minutes through top-scorer Josh Blake. Three minutes later Bedworth were level, a deflected free kick fooling keeper Richard Morris.
The impressive Jon Burgess helped put his side back in front. Darren Timms headed home a Steve Thompson corner won by the dimunitive wing back.
Any nerves from Leamington were settled within three minutes of the re-start, when Eden was flattened just inside the box. He converted the penalty himself, tucking the ball just inside a post.
With Thompson, Blake and Burgess all outpacing an increasingly nervy Bedworth back line, there was more to come. Eden grabbed his second on the rebound after 68 minutes.
Bedworth made it 4-2 with ten minutes left, exploiting a period of reorganisation as substitutes Kevin Ariss, Ian Billington and Neale Perry settled in. Eden restored order with two goals inside five minutes. The hat-trick was completed with a powerful drive from a through ball and number four was a header from a pin-point Brian Agar cross.
Timms, who had been superb all afternoon, was denied a second goal when his brilliant lob was deflected off Agar amid defensive panic.
Co-manager Cadden said "We knew what we had to do and we did the job. We played some good football."