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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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%\lecture{**LECTURE-NUMBER**}{**DATE**}{**LECTURER**}{**SCRIBE**}
\lecture{2}{January 20}{Rohit Gurjar}{Jay Bansal}


\section*{Maximizing set and faces}
\begin{definition}[Face of a Polyhedron]
 Let $P$ be a polyhedron given by a set of linear constraints, i.e., $P = \{ x \in \mathbb{R}^n\ |\ Ax \le b\}$, and let $(A',b')$ be any subset of rows of $(A,b)$. Then if
$F = \{ x \in P\ |\ A'x = b'\}$, and $F$ is not empty, we call $F$ a face of the polyhedron $P$.
We say that the constraints $A'x \leq b'$ are tight for the face $F$ (or every point in $F$).
\end{definition}

\paragraph{Example:} Consider the following linear program. Let $P$ be a polyhedron given by the following set of linear constraints (see Figure~\ref{fig:example}):
\begin{eqnarray} 
x_1 & \ge & 0 \nonumber \\
x_2 & \ge & 0 \nonumber \\
x_1 & \le & 2 \nonumber \\
x_2 & \le & 2 \label{eq:x22}\\
x_1 + x_2 & \le & 3 \label{eq:x123}
\end{eqnarray}

\begin{figure}[!hbt]
    \centering
    \subfloat[A point maximizing function $f_1$]{{\includegraphics[height=6cm]{im1.png} }}%
    \qquad
    \subfloat[A line segment maximizing function $f_2$]{{\includegraphics[width=6cm]{im2.png} }}%
    \caption{A linear function is optimized at a face}
    \label{fig:example}
\end{figure}

Now consider these two optimizing functions, with the above constraints:
\begin{eqnarray*} 
f_1 &=& x_1 + 2x_2 \\
f_2 &=& 2x_1 + 2x_2 
\end{eqnarray*}
It can be seen from the Figure~\ref{fig:example} that $f_1$ is maximized at a point in $P$ which is a face of $P$ (tight constraints are (\ref{eq:x22}) and (\ref{eq:x123})) and $f_2$ is maximized at a line segment in $P$ which is also a face of $P$ (tight constraint is (\ref{eq:x22})). 
In general, maximizing set will be a face of any dimension (could be a corner, a line segment/edge of the polyhedron, or any $k(\leq n)$-dimensional face). 

\begin{definition}[Convex Set] A set $P \subseteq \mathbb{R}^n$ is called convex if for all $a,b \in P$, %\implies (\frac{a+b}{2} \in P)$, or in general, 
any convex combination of $a$ and $b$, that is, $a\lambda+(1-\lambda)b$ for $0\le \lambda \le1$ is also in $P$.
\end{definition}

\begin{lemma}
The feasible region of a set of linear constraints, i.e., $P = \{x \in \mathbb{R}^n\ \mid \ Ax \le b\}$ for some $A \in \R^{m \times n}$, and $b \in \R^m$, is a convex set.
\label{lem:polyconvex}
\end{lemma}
\begin{proof}
Let $a_1$ and $a_2$ be two points in $P$, that is, 
$Aa_1 \le b $ and 
$Aa_2 \le b $. Consider some $0 \leq \lambda \leq 1$.
\[A (\lambda a_1 + (1-\lambda) a_2)
= \lambda Aa_1 + (1-\lambda) A a_2 
\leq \lambda b + (1-\lambda) b = b\]
Thus, by the definition of a convex set, $P$ is convex. 
\end{proof}

\begin{claim}
\label{cla:c1}
Suppose $P = \{x \in  \mathbb{R}^n | Ax \le b\}$ and $z \in P$ is a point maximizing $w^\transpose x$. Let $(A', b')$ be a subset of rows of $(A, b)$ that gives all the tight constraints for $z$, i.e., $A'z = b'$.
Then, any point $y \in P$ satisfying $A'y = b'$ will also maximize $w^\transpose x$ over P.
\end{claim}

\begin{proof}
Let $w^\transpose x = w^* $.
Now, for the sake of contradiction, suppose $w^\transpose y < w^\transpose z = w^* $
This implies
\begin{equation} 
w^\transpose (z-y) > 0 
\label{eq:z-y}
\end{equation}
Now, let us define a point $p = z+(z-y)\epsilon$ for some $\epsilon > 0$.
Then, using Equation~(\ref{eq:z-y}),
\begin{equation} 
w^\transpose p = w^ \transpose z+\epsilon w^\transpose (z-y) > w^\transpose z = w^*
\label{eq:wpw*}
\end{equation}
We argue that $p \in P$. First, we show that the tight constraints for $z$ are also tight for $p$.
\begin{equation} 
A'p = A'z+A'[(z-y)\epsilon] = A'z = b'.
\label{eq:A'pb'}
\end{equation}
Now, consider any non-tight constraint for $z$, i.e., $a_i^\transpose z < b_i$. We have
\begin{equation} 
a_i^\transpose p = a_i^\transpose z + a_i^\transpose [(z-y)\epsilon] < b_i
\label{eq:aipbi}
\end{equation}
which holds because there always exists a small enough $\epsilon$ which makes it true. 
%
(\ref{eq:A'pb'}) and (\ref{eq:aipbi}) prove our claim that $p \in P$. Combining it with (\ref{eq:wpw*}), we get that the function $w^\transpose x$ takes a larger value at $p$ than at $z$, which is contradiction. 
\end{proof}

\begin{claim}
\label{cla:c2}
If $\alpha$ and $\beta$ are two points maximizing $w^\transpose x$ in $P$. Then $\frac{\alpha + \beta}{2}$ will also maximize $w^\transpose x$ in $P$.
\end{claim}
\begin{proof}
By Lemma~\ref{lem:polyconvex}, $\frac{\alpha + \beta}{2} \in P$. 
%Now
%since $\alpha$ and $\beta$ maximize $w^\transpose x$ in $P$, let,
Let \[w^\transpose \alpha = w^\transpose \beta = w^*. \]
Then 
\[w^\transpose \left( \frac{\alpha + \beta}{2} \right) = w^\transpose \alpha/2 + w^\transpose \beta/2 = w^* \]
which makes $\frac{\alpha + \beta}{2}$ maximize $w^\transpose x$ in $P$. 
\end{proof}

\begin{claim}
{For any $w \in \mathbb{R}^n$ and a polyhedron $P$, the set of points $x \in P$ maximizing $w^\transpose x$ forms a face.}
\end{claim}
\begin{proof}
For any two optimal points which have different \textbf{smallest} face containing them, if both points optimize $w^\transpose x$, we need to prove that both of them lie on a face and that entire face  optimizes $w^\transpose x$. If we prove this, we can take all possible such distinct points two at a time and recursively claim that the entire set of points in $P$ maximizing $w^\transpose x$ forms a face of $P$. 

Let $z_1 \in P$ and $z_2 \in P$ be two points that maximize $w^\transpose x$.
By Claim~\ref{cla:c2}, ($\frac{z_1 + z_2}{2}$) is also optimal, and by Claim~\ref{cla:c1}, all the points on the smallest face containing ($\frac{z_1 + z_2}{2}$), say $F$ are also optimal. 
Now, we just need to prove that $z_1$ and $z_2$ lie on $F$ and we are done. 

Let us say $z_1$ doesn't lie on $F$, then among the set of tight constraints for $F$, there must be one,
say $a_i^\transpose x \le b_i$, that is not tight for $z_1$. That is,
$a_i^\transpose z_1 < b_i$.
 Since ($\frac{z_1 + z_2}{2}$) lies on $F$ and satisfies $a_i^\transpose (\frac{z_1 + z_2}{2}) = b_i$, we get that $a_i^\transpose z_2 > b_i$.
This is a contradiction, since $z_2 \in P$ and cannot violate $a_i^\transpose \le b_i$. Thus $z_1$ and $z_2$ lie on $F$.
\end{proof}
\end{document}