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\title{
 The best choice problem\\
 for random number of objects\\ 
  with a refusal probability
}

\author{M Horiguchi\inst{1} \and M Yasuda\inst{2}} 
\institute[Kanagawa Uni, Chiba Uni]{\inst{1} Kanagawa University \and \inst{2} Chiba University}

\date{July 2009, APS at Cornell}

\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\section[Outline]{}
%\frame{\tableofcontents}

%%%
\section{Introduction}

%\subsection{Overview of the Beamer Class}
%%%
%%% スライド
%%%
%\begin{frame}[plain]
%  \frametitle{タイトルのタイトル}[option]
%  option:
%   plain, cotainsverbatim,
%   shrink, squeeze, allowframebreaks
%  2枚目のスライドの内容
%  ...
%\end{frame}

\begin{frame}
\frametitle{Abstract:}
The best choice problem by Markov Potential and Differential equation method. 

Here we dare to try a classical analysis to obtain its optimal value 
for the problem on Markov processes.
\begin{itemize}
 \item[(1)] Optimaility equation(Non-linear Differentila equation) 
    with random number object. \pause
 \item[(2)] Infinitesimal approach(scaling limit).\pause
 \item[(3)] In generally case, it is hard to solve analitically
 (Lambert W-function).
\end{itemize}
\end{frame}


\begin{frame}%[plain]
\begin{itemize}
\item Markov Decision Processes
\item One-Look-Ahead Policy
\item Markov Potential Theory
\item Infinitesimal Analysis(Scaling limit)
\item Variations from the classical problem
      Random Number object, Refusal probability,
      Zero-sum Game version, etc.
\end{itemize}
\end{frame}


\begin{frame}
%\frametitle{Many and many papers}
$\bullet$
Chow/Robbins/Siegmund(1971), Siryaev(1973),
Dynkin/Yushkevitch(1969):  monotone stopping problem, the martingale theory, the potential theory, 

$\bullet$
Ross(1983);
One-Look ahead policy, the optimality principle in the mathematical programming,

$\bullet$
Ferguson(2006), Bruss(2003), Tamaki(2002): the sum-the-odds theorem.

\end{frame}




\begin{frame}%[plain]
The last thing one knows when writing a book is what to put first.
Blaise Pascal(1623-1662)

{\it from J.Havil "Gamma" exploring euler's constant
with a foreword by Freeman Dyson}
\end{frame}




\section{Formulation and Notations}


\begin{frame}
 
Let $x_n, n=0,1,2,\cdots$ be a Markov Chain 
with a transition probability $P=P(i,j), i,j \in S$ 
over a state space $S$ in $R^1$.
\begin{equation*}
 v(i) = \sup_{\tau < \infty} v(i,\tau) 
      = \sup_{\tau < \infty} 
     E\left[ r(x_{\tau}) - \sum_{n=0}^{\tau} c(x_n) \,\,\Big| \,\, x_0=i\right]
\end{equation*}
\end{frame}


\begin{frame}
%\frametitle{Optimality equation}
The relation between Markov potential theory and Dynamic 
Programming or Markov decision processes is well known.
The optimality equation* is reduced as follows.
\begin{equation*}
  v(i) = \max \{r(i), \, Pv(i) - c(i) \}, \, i \in S.
\end{equation*}
Consider the set of states for which stopping immediately is at least
as gppd as stopping after exactly one more period. Denote this
set by
\begin{equation*}\label{eq:1.3}
   B = \{ i \in S \,\, | \, \, r(i) \geq Pr(i) -c(i) \}
\end{equation*}
where $Pr(i) = \sum_{j \in S } P(i,j) r(j)$. 
\end{frame}


\begin{frame}
%\frametitle{Optimal Policy}
The policy, defined by hitting time of this set $B$,
is called a {\it One-stage Loog Ahead}(abridged as OLA) policy,
refer to Ross(1983).

It is sufficient that the OLA policy is optimal if
$B$ is closed and eventually hits to the set.
That the set is closed means if
\begin{equation*}
  P(i,j) = 0 \quad \mbox{for} \quad i \in B, j \in \overline{B} 
\end{equation*}
where $\overline{B}$ denotes the complement of the set $B$.

\end{frame}


\begin{frame}
%\frametitle{abc}
Markov stopping theory as the excessive function, 
the martingale theory and so on.

However the general framework does not cover for specific solution
in the exploit structure or interesting features.

The aim of this note is to obtain the explicit optimal
value associated with OLA policy and various problems in the
asymptotic form. 

\end{frame}

\section{Markov Potential}

\begin{frame}
%\frametitle{Markov Potential}
Firstly  the motivated case for the classical 
secreraly choise problem and then for the variation of the
case with the refusal probability. 

Let $P_A$ denote the restriction to a subset $A$ of S
for the transition probability $P$, that is,
\begin{equation*}
  P_A(i,j) = P(i,j) {\bf 1}_A(i)
\end{equation*}
where the indicator of $A$ is 
\[
{\bf 1}_A(i) = \left\{\begin{array}{ll} 
                 1  & (i \in A) \\
                 0  & (i \notin A)  
                \end{array} \right.
\]
\end{frame}



\begin{frame}
%\frametitle{Markov Potential}
A potential $N$ of $P$ for a function $f$: 
\[
   Nf(i) 
   = \lim_n \left[\{I + P + P^2 + \cdots + P^k \} f
            \right](i)
   \quad i \in S.
\]

%\begin{theorem}
 Assume that
 \begin{equation*}
  \begin{array}{l}
   \lim_k \left[(P_{\overline{B}})^k r\right](i) = 0 \\[2.5ex]
   \left[ N_{\overline{B}} (P_B r -c) \right] (i) < \infty \quad 
   \mbox{for} \quad i \notin B. 
  \end{array}
 \end{equation*}
\end{frame}



\begin{frame}
%\frametitle{Markov Potential}

 Then
 \begin{enumerate}
  \item[(i)] the OLA policy $\tau_B$ is optimal,
  \item[(ii)]
   the optimal value $v(i), i \in S$ is given by
   \begin{equation*}\label{eq:2.7}
    v(i)  \begin{array}[t]{l}  = r(i) + N(Pr -r -c)^+ (i) \\[3ex]
         = \left\{ 
       \begin{array}{lll}
         r(i) & \mbox{on} & B  \\
         N_{\overline{B}} (P_B r -c)(i) 
              & \mbox{on} & \overline{B}
       \end{array}
     \right.
        \end{array}
   \end{equation*}
  \end{enumerate}
%\end{theorem}
\end{frame}

\section{The best choice Problem}

\subsection{Classical Problem}

\begin{frame}

 The secretary problem is an optimal stopping problem based on
relative ranks for objects for objects arriving in a random
fashion. 

The objective is to find the stopping rule
that maximizes the probability of stopping at the best object among
the sequence.

The formulation of the problem as Markov processes is given by
Dynkin/Yushkevitch(1969). 

\end{frame}

\begin{frame}
The state space is $S=\{1,2,3,\cdots,n\}$ and the reward, 
its win of probability is $r(i) = i/n, i \in S$.

The transition probability is 
 \[P(i,j) = \dfrac{i}{j(j-1)},\]
on $1 \le i < j \le n$, and $=0, \mbox{otherwise}$.

The optimal value $v(i)$ satisfies, for $i=1,2,\cdots, n-1,$,  
\begin{equation*}
  \begin{array}{l}
  v(i) = \max \left\{\dfrac{i}{n} ,\, 
         \displaystyle\sum_{j = i+1}^n \dfrac{i}{j(j-1)}v(j)\right\},\\
  v(n) =1.
  \end{array}
\end{equation*}

\end{frame}

\begin{frame}
%\frametitle{Scaling limit}

Taking a scale limit of $\dfrac{i}{n} \to x$ as $i,n \to \infty$,
it becomes that
\begin{equation*}\label{eq:3.11}
  \begin{array}{l}
  v(x) = \max \left\{x ,\, 
          x \displaystyle\int_{x}^1 \dfrac{v(y)}{y^2} \, dy\right\},
  \quad  x \in S=[0,1], \\
   v(1) =1.
  \end{array}
\end{equation*}

The solution is well known as
\begin{equation*}
  v(x) = \left\{\begin{array}{lll}
            e^{-1} & on & [0,e^{-1}], \\
            x      & on & [e^{-1} ,1].
        \end{array}\right.
\end{equation*}

\end{frame}

\begin{frame}

By extending the reward $r(x)$ in general, that is,
\begin{equation*}\label{eq:3.11}
  \begin{array}{l}
  v(x) = \max \{r(x) ,\, x \displaystyle\int_{x}^1 \dfrac{v(y)}{y^2} \, dy\},
  \quad  x \in S=[0,1], \\
 v(1) =1.
  \end{array}
\end{equation*}
It is also the OLA policy is
still optimal if the following assumption holds.

\end{frame}

\begin{frame}

\alert{Assumption:}

The function
\[
   h(x) = r(x) - x \int_{x}^1 \dfrac{r(y)}{y^2} \, dy
\]
is finite on $[0,1]$ for a given $r(x)$ and

it changes its sign from $-$ to $+$ only once as $x$
increases $0$ to $1$.

So there exists a unique solution $\alpha$ of 
the equation $h(x) = 0$, that is, 
\[
\alpha = \inf\{ x ; h(x) \ge 0\}.
\]


\end{frame}

\begin{frame}

%\begin{theorem}
\alert{Theorem:}

Using this solution $\alpha$, the assumption 
implies  
\begin{equation*}
  v(x) = \left\{\begin{array}{lll}
            r(\alpha) & on & [0, \alpha], \\
            r(x)      & on & [\alpha ,1].
        \end{array}\right.
\end{equation*}
%\end{theorem}
\end{frame}

\begin{frame}

Because of using Markov potential,
it is calculated as
\[\begin{array}{l}
  P_Br(x) =r(\alpha) \dfrac{x}{\alpha},\\[3ex]
  \bigl(P_{\overline{B}}\bigr)^nP_B r(x) =
    r(\alpha) \dfrac{x}{\alpha} 
    \log^n \left(\dfrac{\alpha}{x}\right) / n!
  \end{array}
\]
and
\[
  N_{\overline{B}}P_B r(x) =
    r(\alpha)
\]
directly. Here $B = [0, \alpha]$.

\end{frame}

\begin{frame}
%\frametitle{Infinitesimal operator:}
The basic idea based
on the infinitesimal operator of Markov processes.

It is equivalent to, 
\[
   -x \dfrac{d\,v(x)}{dx} = \bigl(r(x) - v(x)\bigr)^{+},
    \quad x \in[0,1] , \qquad v(1) =1.
\]

This method is derived by Mucci(1973). 
\end{frame}

 \subsection{Problem with a refusal probability}


\begin{frame}
%\frametitle{Problem with a refusal probability}
A variant for the classical best choice problem
is considered by Smith(1975) inducing a refusal
probability.

A given $p\, (0<p<1)$ means the probability of \alert{not
refused case} and the quality $1-p$ is a probability
of the refusal. When there occues no refusal$(1-p=0)$, 
that is, the usual standard problem corresponds to $p=1$.

\end{frame}

\begin{frame}
The asymptotic form of the transition probability
is given as
\begin{equation*}
 P(x, dy ) = 
   \left\{ 
    \begin{array}{ll}
      \dfrac{p}{y} \Big(\dfrac{x}{y}\Big)^p \, dy, & 0<x<y<1,\\
      0, & otherwise.
    \end{array}
   \right.  
\end{equation*}
\end{frame}

\begin{frame}
Similarly before

%\begin{assm}\label{ass:3.2}
\alert{Assumption:}
The function
\[
   h_p(x) = r(x) - p \int_{x}^1 \dfrac{r(y)}{y} (x/y)^p  \, dy
\]
is finite on $[0,1]$ for a given $r(x)$ and
it changes its sign from $-$ to $+$ only once as $x$
increases $0$ to $1$.
Its unique solution is denoted by $\alpha_p$ of 
the equation $h_p(x) = 0$, i.e.,
\[
\alpha_p = \inf\{x; h_p(x) \ge 0 \}.
\]
%\end{assm}

\end{frame}

\begin{frame}
%\begin{theorem}
\alert{Theorem:}
Using this solution $\alpha_p$, 
the assumption 
implies that 
\begin{equation*}
  v(x) = \left\{\begin{array}{lll}
            r(\alpha_p) & on & [0, \alpha_p], \\
            r(x)      & on & [\alpha_p ,1].
        \end{array}\right.
\end{equation*}
%\end{theorem}
\end{frame}

\begin{frame}
It is also known the technique of using
a differential equation by Mucci(1973).

Let
\[
   V(x) =   p \int_x^1 \dfrac{r(y)}{y} (x/y)^p  \, dy,
\qquad x \in [0,1]
\]
which means a contditional optimal value. 


This 
satisfies
\begin{equation*}
   \left\{ \begin{array}{ll}
    - x \dfrac{dV(x)}{dx} 
   = p \bigl(r(x) - V(x)\bigr)^+,
     & x \in [0,1], \\[2ex]
    V(1) = 0. & \\
   \end{array} \right.
\end{equation*}

\end{frame}

\begin{frame}

The optimal value at the begining 
\[
v^* = V(0) = r(\alpha_p).
\]

Confirming that the case of $r(x)= x$, 
the equation $h_p(x) =0$  solves 
\[
x = \alpha_p = p^{1/(1-p)}
\]
which consist with the result by Smith(1975).

\end{frame}

\section{Random number of objects}
\subsection{Formulation for random number of objects}

\begin{frame}
%\frametitle{Formulation for random number of objects}
The discussion on the stopping choice problem with
a random number of objects is firstly by
Presman/Sonin(1972). Then Rasmussen/Robbins(1975),
Tamaki(1979),
Irle(1980), Petruccelli(1983) etc.

If we restrict ourselves that the support of distribution
for a number of objects is \alert{bounded}, the technique of
the previous scaling limit could be applied.

This assumption links to the technique of 
\alert{the scaling limit}.
It should be noted that Presman/Sonin discussed the number 
of objects is 
Poisson distribution, so we could not apply this technique
in that case. The next is a fundamental assumption of the
discussion in here.
\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
%\begin{assm}
\alert{Assumption:}

A random number of objects $N$ is bounded and let
\[
   n = \inf\{ k \ge 1; P(N>k) = 0\}.
\]  
%\end{assm}

\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
By using the notation of 
$p_i = P(N=i)$ and $\pi_i = P(N \ge i) = \sum_{k \ge i} p_i$,
the transition probability matrix for this case
is as follows:
\begin{equation*}
  \begin{array}{ll}
  p(i,j)= \dfrac{i \, \pi_j}{j(j-1)\, \pi_i}, & 1 \le i < j \le n,\\
  p(i,n)= \displaystyle\sum_{k=i+1}^n \dfrac{i\, p_k}{k\, \pi_i}, & 1 \le i < n, 
      \quad p(n,n) = 1,\\
  r(i) = \displaystyle\sum_{k=i}^n \dfrac{i\, p_k}{k\, \pi_i}, & c(i) =0, \, \forall i.
  \end{array}
\end{equation*}
\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
If we denote the distribution by $\Phi(x), x \in [0,1]$ 
of its random number for objects,
and taking the scale limit of $k,n \to \infty$ 
with $k/n =x$, then $v(i) \to V(x)$ and 
$p_k=P(N=i) \to d \Phi(x)$,
$\pi_k=\sum_{k \ge i} p_i  \to 1- \Phi(x)$, so
\[
 \begin{array}{l}
  p_k /\pi_k  \to d\Phi(x)/(1-\Phi(x)),\\[1.5ex]
  r(k+1) 
   = \displaystyle\sum_{k=i}^n \dfrac{i\, p_k}{k\, \pi_i}
   = \displaystyle \dfrac{i}{\pi_i}\sum_{k=i}^n \dfrac{p_k}{k} \\
  \qquad \qquad
  \to R(x) =\dfrac{x}{1-\Phi(x)}\displaystyle \int_x^1 y^{-1}\, d \Phi(y).
\end{array}
\]
\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
The resulting the differential
equation is given as
\begin{equation*}\label{eq:4.17}
  \left\{
  \begin{array}{l}
   -x \, dV(x) + x\, \dfrac{V(x)}{1-\Phi(x)}\, d \Phi(x) \\[2ex]
    \qquad \qquad =  (R(x) - V(x))^{+}\, dx, \quad x \in [0,1] \\[2ex]
   V(1) = 0
  \end{array}
  \right.
\end{equation*}
where 
\begin{equation*}\label{eq:4.18}
       R(x) = \dfrac{x}{1-\Phi(x)} \int_x^1 y^{-1}\, d \Phi(y).
\end{equation*}
The details refer to Yasuda(1984).
\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
%\begin{assm}
\alert{Assumption:}

We assume that $d\Phi$ satisfies the conditions:
 \begin{enumerate}
  \item 
    \(
    (1-\Phi(x))^{-1} \displaystyle\int_x^1 y^{-1}d\Phi(y)
    \to 1 \quad as \quad x \to 1,
    \)
  \item
    \(
     x \displaystyle\int_x^1 y^{-1}d\Phi(y)
    \to 0 \quad as \quad x \to 0.
    \)
 \end{enumerate}
%\end{assm}
\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
This assumtion implies that
\(
   R(x) \to 0 \, (x \to 0)
\)  
and 
\( 
   \to 1 (x \to 1).
\)

In the usual non-random case, $R(x) = x$, 
these conditions are satisfied since
 $ \dfrac{1}{1-\Phi(x)} \displaystyle\int_x^1 y^{-1}\, d \Phi(y) =1$
for $x \neq 1$.


\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}
\alert{Example:}

Example of a random number $N$: 

Uniformly distributed case on a partial interval
\begin{center}
$\{n-m, n-m +1,\cdots, n-1,n\}$ of 
$\{1,2,\cdots,n\}$
\end{center}

Letting $\theta = m/n$ is fixed for $n,m \to \infty$,
\[
\Phi(x) = \left\{
\begin{array}{ll} 0, & (0 < x <1-\theta) , \\
  (x - 1 +\theta)/\theta, & (1-\theta < x < 1)
\end{array}
\right.
\]

\end{frame}

\begin{frame}
%\frametitle{Formulation for random number of objects}

\begin{center}
\begin{tabular}{lll}
\hline
 case & $\alpha$: threshold & $v^*$:optimal value \\
\hline 
  $1 - \theta$ $\ge$ $e^{-2}$ & $\sqrt{1-\theta}e^{-1}$ 
  & $- \dfrac{\sqrt{1-\theta}}{\theta} \log(1-\theta) e^{-1}$ \\[1.5ex]
  $1 - \theta$ $\le$ $e^{-2}$ & $e^{-2}$ 
  & $2 e^{-2} /\theta$ \\
\hline
\end{tabular}
\end{center}
\end{frame}

\begin{frame}
If $\theta \to 1$ it tend to $2e^{-2}$
as discussed in Presman/Sonin(1972) etc.
Alternatively, if $\theta \to 0$, this reduces to 
the non-random classical problem and it tend to $e^{-1}$ as is 
well known.
However it is interesting that the threshold $\alpha= e^{-2}$ 
does not depent on $\theta$ in case of 
$1-\theta \approx 0$, that is, case of the ambiguity
is being in spread.

\end{frame}


\subsection{Random number of objects with refusal}


\begin{frame}
Here we would like to impose a refusal probability
for the model. 

Instead of $V(x)$ in the optimality equation 
%(\ref{eq:4.17}), 
%let $v(x) = $,
consider 
$dv(x) = dV(x) - \dfrac{V(x)}{1-\Phi(x)}\, d \Phi(x) $.
Then, it holds that
\begin{equation*} \label{(eq:4.19)}
 v(x) = \max\left\{R(x) ,\,\, 
   \dfrac{x}{ 1-\Phi(x)}
   \int_x^1 (1-\Phi(y)) y^{-2} v(y) \, dy \right\}
\end{equation*}
%with $R(x)$ in (\ref{eq:4.18}).

\end{frame}

\begin{frame}
The optimality equation of stopping problem:
\[
  v(x) = \max\{R(x), Pv(x)\} , \, 0 \le x \le 1
\]
have a modification with a refusal probability
$p(x), 0 \le x \le 1$
becomes
\begin{equation*} \label{eq:41}
  v(x) = \max\{p(x)\,R(x) + (1-p(x))\,Pv(x), \,\,Pv(x)\}.
\end{equation*}
\end{frame}

\begin{frame}
By letting 
\begin{equation*}
  \begin{array}{l}
   Pv(x) = \dfrac{x}{ 1-\Phi(x)}
   \displaystyle\int_x^1 (1-\Phi(y)) y^{-2} v(y) \, dy, \\[1.5ex]
   q(x) = \exp\Big( \displaystyle\int_x^1 y^{-1} (1-p(x)) \, dy \Big)\\[1.5ex]
   \mbox{and} \\[1.5ex]
   h(x) = \displaystyle\int_x^1 y^{-1} \, d\Phi(y),
  \end{array}
\end{equation*}

\end{frame}

\begin{frame}
%\begin{assm}\label{assm:3}
\alert{Assumption:}

Assume that
\begin{equation*} \label{eq:42}
   H_p(x) = h(x) - q(x) \int_x^1 \dfrac{h(y)\, p(y)}{y\, q(y)} dy 
\end{equation*}
changes its sign once from $-$ to $+$ as $x$ 
increase.
%\end{assm}

\end{frame}

\begin{frame}

%\begin{theorem}\label{thrm:4}
%If Assumption (\ref{assm:3}) 
%is satisfies,
\alert{Theorem:}

The solution is given as 
the value
\begin{equation*}
  \alpha_p^* = \left\{ \begin{array}{l}
                 \inf\{ x ; H_p(x) \ge 0 \} \\ 
                 1 \quad {if \, empty} 
               \end{array} \right.
\end{equation*}
determine its threshold policy and so the optimal 
value $v(0+)$.
    

%\end{theorem}
\end{frame}

\begin{frame}

However in order to display the above assertion or
explicit form, it is not easy for us.  

Even the simplest case of 
\[p(x) = p \, (0 < p <1)\]
and
\[\Phi(x) = x \,(0 \le x \le 1),\] 
it needs the discussion on
Lambert W function. 

\end{frame}



\end{document}

% \subsection{A game Version of zero sum cases}

We note here that a game version of zero sum cases
are discussed.
There are several versions for zero sum cases. 

Refer to 
E. B. Dynkin, Game variant of a problem of optimal stopping, Soviet Math. Dokl., 10 (1969), pp. 16-19.
Erik Ekstrom and Goran Peskir, Optimal Stopping Games for Markov Processes
  SIAM J. Control Optim. Volume 47, Issue 2, pp. 684-702 (2008).
Ferenstein, Elzbieta Z. Randomized stopping games and Markov market games. Math. Methods Oper. Res. 66 (2007), no. 3, 531--544.


Let $f(x)$ and $g(x)$ be two functions of bounded variation
over $[0,1]$, right continuous and with left-hand limits, then,
by Fubini's theorem,
\[
 d(fg) (x) = f(x)dg(x) + g(x) df(x) -\{f(x) -f(x-)\}dg(x)
           = f(x-)dg(x) + g(x) df(x).
\]  

\end{document}




\section{Ketpic}
\frame
{\frametitle{KETpic}

\alert{KETpic}(Kisarazu Educational Tpic)\\
A macro package of creating tpic form with help
of Computer Algebra System.

Developed for educations in Math \& Stat \\
using CAS:\alert{Maple, Mathematica, Scilab}.\\

Adding several command to draw figures in 2 dim, 3 dim. 
"ridgeline method", "wireframe for Hidden line".\\

http://www.kisarazu.ac.jp/~masa/math/\\

\begin{itemize}
 \item<1-> Maple version --- ketpic.m, ketpicapp6.m
 \item<2-> Mathematica --- ketpic6.m, ketpicppw.m
 \item<3-> Scilab --- lib(ketpiclib)
\end{itemize}
}

\frame{
From now we show some examples and demonstrations.

}



\begin{frame}
%\usepackage{colortbl}              % colortbl パッケージが必要
\rowcolors[]{1}{blue!20}{blue!10}  % 列毎に色を変更
\begin{tabular}{l!{\vrule}c<{\onslide<2->}c<{\onslide<3->}c<{\onslide}c}
1 & A & B & C \\
2 & D & E & F \\
3 & G & H & I 
\end{tabular}
\end{frame}


\end{document}


\end{document}
