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    \hbox to 16cm { {\bf CS602 Applied Algorithms
                        \hfill 2019-20 Sem II} }
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       \hbox to 16cm { {\Large \hfill Lecture #1: #2  \hfill} }
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\begin{document}

%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{3}{January 21}{Rohit Gurjar}{Saurav Yadav}
\section*{Corners/Vertices of a Polyhedron}
\begin{lemma}
    For any face $F$ of $P$, $\exists w \in \mathbb{R}^n$ such that $F$ is exactly the set of points maximizing $w^Tx$ over $P$.
    \label{lemma:2}
\end{lemma}
\begin{proof}
    Let polytope $P \subseteq \mathbb{R}^n$ be
    \begin{eqnarray*}
        a_1^Tx \le b_1\\
        \vdotswithin{\displaystyle\sum}\\
        a_m^Tx \le b_m
    \end{eqnarray*}
    and face $F$ be the set of constraints
    \begin{eqnarray*}
        a_1^Tx = b_1\\
        a_2^Tx = b_2\\
        a_3^Tx \le b_3\\
        \vdotswithin{\displaystyle\sum}\\
        a_m^Tx \le b_m
    \end{eqnarray*}Then $w = a_1 + a_2$ is trivially the required $w$ (As $a_1^Tx + a_2^Tx \le b_1+b_2\; \forall x \in P$ and $a_1^Tx + a_2^Tx = b_1+b_2\; \text{only for}\; x \in F$). This can be extended for any face similarly. Hence, proved.
\end{proof}
\begin{corollary}
    For $w^Tx$, there is a corner of $P$ which attains maximum value.
\end{corollary}
\begin{definition}[Corner/Vertex]
    For polytope $P \subseteq \mathbb{R}^n$
    \begin{eqnarray*}
        a_1^Tx \le b_1\\
        a_2^Tx \le b_2\\
        \vdotswithin{\displaystyle\sum}\\
        a_m^Tx \le b_m
    \end{eqnarray*}
    $z$ is a vertex if $z \in P$ and if there is a subset of $n$ linearly independent constraints which are tight for $z$.
    \label{corner:1}
\end{definition}
\begin{definition}[Corner/Vertex]
    $z$ is a corner of $P$ if $z \in P$ and $\forall y \in \mathbb{R}^n \setminus \{0\}, \; z+y \in P \Rightarrow z - y \notin P$.
    \label{corner:2}
\end{definition}
\begin{definition}[Corner/Vertex]
    $z$ is a corner of $P$ if $z \in P$ and $\exists w \in \mathbb{R}^n$ such that $z$ is the UNIQUE point maximizing $w^Tx$ over $P$.
    \label{corner:3}
\end{definition}
\begin{claim}
    All 3 definitions of Corner/Vertex are equivalent
\end{claim}
\begin{proof}
    \begin{proof}[$\ref{corner:3}\Rightarrow\ref{corner:2}$]
        $\exists w : w^Tz = \alpha^*$\\
        For contradiction, suppose $y \ne 0$ is such that $z+y \in P$ and $z-y\in P$.\\
        $\Rightarrow w^T(z+y) \le \alpha^*$ , $w^T(z-y) \le \alpha^*$\\
        If one of the above is strictly less than $\alpha^*$ then other one would be greater, thus, they must be same as $\alpha^*$. Hence, we get a contradiction to the fact that $z$ is the unique maximizing point.
    \end{proof}
    \begin{proof}[$\ref{corner:1}\Rightarrow\ref{corner:3}$]
        Using \textit{Lemma} \ref{lemma:2}, $\exists w \in \mathbb{R}^n$ such that $z$ if exactly the set of points mazimizing $w^Tx$, but by \textit{Definition} \ref{corner:1}, $z$ is a UNIQUE point ($n$ independent tight constraints in $\mathbb{R}^n$ correspond to a single point). Hence, this implies the condition in \textit{Definition} \ref{corner:3}.
    \end{proof}
        \begin{proof}[$\ref{corner:2}\Rightarrow\ref{corner:1}$]
We want to show the condition in Definition~\ref{corner:1}. 
For the sake of contradiction, let us assume that the maximum number of constraints which are tight for $z$ is $k$.
Without loss of generality, say $a_i^T z = b_i$ for $1\leq i \leq k$. 
If the rank of $(a_1, a_2, \dots, a_k)$ is less than $n$
then we know from linear algebra that there must be common orthogal vector to all of them. That is, there exists $\delta \in \mathbb{R}^n$ such that 
\[a_i^T \delta = 0 \text{ for } 1 \leq i \leq k.
\label{eq:delta} \]
Now, choose two new points $z+\epsilon \delta$ and $z-\epsilon \delta$ for some small $\epsilon >0$.
The constraints which were tight for $z$ will also be tight for these two points (from (\ref{eq:delta})). 
Moreover, the constraints which were not tight for $z$ can still be kept non-tight for the other two points by choosing a small enough $\epsilon$. Thus, the two points are feasible.
And, we get a contradiction to the condition that at least one of $z+y$ and $z-y$ should be outside the polyhedron.
\end{proof}

\end{proof}
% \begin{definition}[Convex Hull]
%     For any set of point $S \in \mathbb{R}^n$, \[S = \{P_1, \dots, P_k\}\]
%     Convex Hull(S) = $\{\lambda_1P_1+\lambda_2P_2+\dots\lambda_kP_k \colon \Sigma\lambda_i=1, 0 \le \lambda_i{}\}$
% \end{definition}
% \begin{claim}
%     A polytope is a Convex Hull of its finitely many vertices
% \end{claim}
% \begin{proof}
%     Let $P = \{p_1, \dots, p_n\}, n \in \mathbb{N}$, such that without loss of generality $p_1, \dots, p_k$ are the vertices of $P := conv(P)$. We prove by induction on $n$ that $\text{conv}(p_1,\dots, p_n) \subseteq \text{conv}(p_1,\dots, p_k)$.For $n = k$ the statement is trivial.\\
% For $n > k$, $p_n$ is not a vertex of $P$ and hence $p_n$ can be expressed as a convex combination $p_n = \Sigma_{i=1}^{n-1}\lambda_ip_i$. Thus for any $x \in P$ we can write $x = \Sigma_{i=1}^{n}\mu_ip_i = \Sigma_{i=1}^{n-1}\mu_ip_i + \mu_n\Sigma_{i=1}^{n}\lambda_ip_i$. As $\Sigma_{i=1}^{n-1}(\mu_i+\mu_n\lambda_i) = 1$, we conclude inductively that $x \in \text{conv}(p_1,\dots , p_{n−1}) \subseteq \text{conv}(p_1,\dots, p_k)$.
% \end{proof}
% \begin{theorem}[Caratheodory's theorem]
%     For a polytope $P \subseteq \mathbb{R}^n$, any point $z \in P$ can be written as a convex combination of atmost $n+1$ vertices
% \end{theorem}
% \begin{proof}
%     Take arbitrary vertex $v$, join $v$ and $z$ by a line and extend till it hits a face $F$. Let that point be $y \in F$. \[z = {\lambda}v + (1-\lambda)y\]. Now $F$ is a $(n-1)$ dimentional polytope\\$\Rightarrow$ it is a convex combination of $(n-1)$ vertices.\\
%     $\Rightarrow z = {\lambda}v+(1+\lambda)[\lambda_1v_1+\lambda_2v_2+\dots+\lambda_{n-1}v_{n-1}]$\\
%     Hence, proved by induction
% \end{proof}
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