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							@@ -19,7 +19,6 @@
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\end{frame}
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					\end{frame}
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\begin{frame}[plain]{Plan}
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					\begin{frame}[plain]{Plan}
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    \mybox[oliver!10]{\scriptsize The order of the slides is basically the order in which I think about this problem.}
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    \tableofcontents
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					    \tableofcontents
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\end{frame}
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					\end{frame}
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@@ -33,13 +32,13 @@
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    \begin{subfigure}{.5\textwidth}
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					    \begin{subfigure}{.5\textwidth}
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      \centering
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					      \centering
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      \includegraphics[width=.4\linewidth]{images/Piecewise_linear_function.svg.png}
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					      \includegraphics[width=.4\linewidth]{images/Piecewise_linear_function.svg.png}
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      \caption{A 1D piecewise linear function with 4 line segments and 3 breakpoints}
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					      \caption{A 1D pwl function with 4 line segments and 3 breakpoints}
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      \label{fig:sub1}
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					      \label{fig:sub1}
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    \end{subfigure}%
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					    \end{subfigure}%
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    \begin{subfigure}{.5\textwidth}
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					    \begin{subfigure}{.5\textwidth}
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      \centering
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					      \centering
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      \includegraphics[width=.4\linewidth]{images/Piecewise_linear_function2D.png}
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					      \includegraphics[width=.4\linewidth]{images/Piecewise_linear_function2D.png}
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      \caption{A 2D piecewise concave function}
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					      \caption{A 2D pwl concave function}
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      \label{fig:sub2}
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					      \label{fig:sub2}
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    \end{subfigure}
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					    \end{subfigure}
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    % \caption{A figure with two subfigures}
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					    % \caption{A figure with two subfigures}
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@@ -93,9 +92,11 @@ However, observe that in our problem the piecewise linear convex function is not
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\end{frame}
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					\end{frame}
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\section{LP in Low Dimensions}
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					\section{LP in Low Dimensions}
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\begin{frame}[allowframebreaks]{Megiddo's algorithm}
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					\begin{frame}[allowframebreaks]{Megiddo's algorithm}%
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    {\tiny \url{https://people.inf.ethz.ch/gaertner/subdir/texts/own_work/chap50-fin.pdf}}
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					    \mybox[oliver!20]{
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					        \tiny
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					        \url{https://people.inf.ethz.ch/gaertner/subdir/texts/own_work/chap50-fin.pdf}
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					    }
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    The dimension $d$ (in our problem, the dimension of $x$) is small while the number of constraints are huge. We need only $d$ linearly independent tight constraints to identify the optimal solution $x^*$.
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					    The dimension $d$ (in our problem, the dimension of $x$) is small while the number of constraints are huge. We need only $d$ linearly independent tight constraints to identify the optimal solution $x^*$.
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    Thus most of the constraints are useless.
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					    Thus most of the constraints are useless.
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    \BlankLine
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					    \BlankLine
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@@ -167,13 +168,14 @@ However, observe that in our problem the piecewise linear convex function is not
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    with solution $T(n,d)=O(2^{2^d}n)$.
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					    with solution $T(n,d)=O(2^{2^d}n)$.
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\end{frame}
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					\end{frame}
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\begin{frame}{Zemel's conversion}
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					\begin{frame}{Zemel's conversion}
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    Our linear program has \emph{dimension} $n+d$. \href{https://www.sciencedirect.com/science/article/abs/pii/0020019084900140}{Zemel} showed that this kind of problem can be converted to a linear program of dimension $d$.
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    \begin{align*}
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					    \begin{align*}
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        \min &\sum_{i=1}^n f_i\\
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					        \min &\sum_{i=1}^n f_i\\
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        s.t. \quad f_i&\geq \alpha_j(a_i\cdot x -b_i)-\beta_j \quad \forall i\in[n], \forall j\\
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					        s.t. \quad f_i&\geq \alpha_j(a_i\cdot x -b_i)-\beta_j \quad \forall i\in[n], \forall j\\
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    \end{align*}
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					    \end{align*}
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					    Our linear program has \emph{dimension} $n+d$. 
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					    \textbf{\href{https://www.sciencedirect.com/science/article/abs/pii/0020019084900140}{Zemel}} showed that this kind of problem can be solved in linear time.
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    \mybox[oliver!20]{
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					    \mybox[oliver!20]{
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        \scriptsize Here is an intuitive way to understand the conversion. One can think the LP above as a $d$-dimensional search problem with $n+d$ hyperplanes. However, the oracle is quite different. The oracle takes the unknown $x^*$ and a hyperplane $H$ as input, returns the relative position by computing the minimal $f_i$.
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					        This is a \emph{$d$-dimensional search problem} with $n+d$ hyperplanes.
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    }
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					    }
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\end{frame}
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					\end{frame}
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@@ -205,15 +207,32 @@ However, observe that in our problem the piecewise linear convex function is not
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    If a LP problem has low dimension, then its combinatorial dimension is low. \textbf{What about the converse?}
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					    If a LP problem has low dimension, then its combinatorial dimension is low. \textbf{What about the converse?}
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    \newpage
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					    \newpage
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					    \begin{align*}
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					        \min &\sum_{i=1}^n f_i\\
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					        s.t. \quad f_i&\geq \alpha_j(a_i\cdot x -b_i)-\beta_j \quad \forall i\in[n], \forall j\\
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					        &\cdots
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					    \end{align*}%
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    \textbf{Does our LP has low combinatorial dimension?}
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					    \textbf{Does our LP has low combinatorial dimension?}
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    No! A basis contains at least $n$ constraints since otherwise some $f_i$ is unbounded.
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					    No. A basis contains at least $n$ constraints since otherwise some $f_i$ is unbounded.
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					    \begin{problem}
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					        Is it possible to formulate the pwl convex minimization problem as an LP-type problem with low combinatorial dimension?
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					    \end{problem}
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\end{frame}
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					\end{frame}
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\begin{frame}{Aggregate the pwl convex functions}
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					\begin{frame}{Aggregate the pwl convex functions}
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    % blog posts
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					    % blog posts
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					    The sum of pwl convex functions are still pwl convex. 
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					    \newline If we can compute $F=\sum f_i$ in $O(m)$ and the number of line segments on $F$ is also $O(m)$, then the corresponding LP will have low combinatorial dimension.
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					    \begin{align*}
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					        \min \quad F\\
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					        s.t. \quad F&\geq \alpha_j\cdot x -\beta_j \quad \forall j\\
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					        &\cdots
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					    \end{align*}
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					    However, this is not possible for general pwl convex functions in $\R^d$.\footnote{see this \href{https://talldoor.uk/posts/2024-09-16-piecewise-linear.html}{blog post} for detail.}
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\end{frame}
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					\end{frame}
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\end{document}
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					\end{document}
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