\documentclass{article} \usepackage[margin=1in]{geometry} \usepackage{amsmath,amsthm,mathtools} \usepackage{amsfonts} \newcounter{lecnum} \newcommand{\lecture}[4]{ \newpage \setcounter{lecnum}{#1} \noindent \begin{center} \framebox{ \vbox{\vspace{2mm} \hbox to 16cm { {\bf CS602 Applied Algorithms \hfill 2019-20 Sem II} } \vspace{4mm} \hbox to 16cm { {\Large \hfill Lecture #1: #2 \hfill} } \vspace{2mm} \hbox to 16cm { {\it Scribe: #4 \hfill Lecturer: #3} } \vspace{2mm}} } \end{center} \vspace*{4mm} } \newtheorem{theorem}{Theorem}[lecnum] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{claim}[theorem]{Claim} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \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} \end{document}