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update font and sty

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Yu Cong 2025-05-04 16:45:57 +08:00
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beamerthemeSimple.sty
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% !TEX program = xelatex % !TEX program = xelatex
% !TEX TS-program = xelatex % !TEX TS-program = xelatex
% fonts % fonts
\RequirePackage[sfdefault]{FiraSans} % \usefonttheme{professionalfonts}
% \RequirePackage[sfdefault]{FiraSans}
\RequirePackage{FiraMono} \RequirePackage{FiraMono}
\renewcommand{\rmfamily}{\sffamily} \renewcommand{\rmfamily}{\sffamily}
% \RequirePackage[fakebold]{firamath-otf} % \RequirePackage[fakebold]{firamath-otf}
\usepackage[mathrm=sym]{unicode-math} \RequirePackage{unicode-math}
\setmathfont{Fira Math} \unimathsetup{math-style=ISO, bold-style=ISO, mathrm=sym}
\setsansfont{FiraGO}[BoldFont=* SemiBold, Numbers=Monospaced]
\setmathfont{Fira Math}[BoldFont=*-SemiBold]
\RequirePackage{xeCJK} \RequirePackage{xeCJK}
\setCJKmainfont{Source Han Sans SC} \setCJKmainfont{Source Han Sans SC}
@ -138,7 +141,7 @@
\setbeamerfont{frametitle}{series=\bfseries\boldmath} \setbeamerfont{frametitle}{series=\bfseries\boldmath}
\setbeamerfont{block title}{series=\bfseries\boldmath} \setbeamerfont{block title}{series=\bfseries\boldmath}
\setbeamerfont{title}{series=\bfseries\boldmath} \setbeamerfont{title}{series=\bfseries\boldmath}
\setbeamertemplate{frametitle}{\vskip2pt\hskip-6pt\underline{\insertframetitle}} % add line under frametitle \setbeamertemplate{frametitle}{\vskip2pt\hskip-6pt\underbar{\insertframetitle}} % add line under frametitle
% theorem env % theorem env
\setbeamertemplate{theorem begin}{% \setbeamertemplate{theorem begin}{%
@ -167,8 +170,9 @@
} }
% more theorem env % more theorem env
\newtheorem{observation}{Observation} \newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}{Question} \newtheorem{observation}[theorem]{Observation}
\newtheorem{question}[theorem]{Question}
% ---------------------------------------------------------------------- % ----------------------------------------------------------------------

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main.tex
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\documentclass{beamer} \documentclass{beamer}
\title[template example]{Minimizing the Sum of Piecewise Linear Convex Functions} \title[template example]{Title}
\date{\today} \date{\today}
\author{丛宇} \author{丛宇}
% \AtBeginSection[]{ % \AtBeginSection[]{
@ -23,6 +23,7 @@
\section{Problems \& Definitions} \section{Problems \& Definitions}
\begin{frame}{$\min \sum f_i(a_i\cdot x-b_i)$} \begin{frame}{$\min \sum f_i(a_i\cdot x-b_i)$}
$A\setminus B$ 测试中文:
\begin{problem} \begin{problem}
Given $n$ piecewise linear convex functions $f_1,...,f_n:\R \to \R$ of total $m$ breakpoints, and $n$ linear functions $a_i\cdot x-b_i:\R^d\to \R$, find $\min_x \sum_i f_i(a_i\cdot x-b_i)$. Given $n$ piecewise linear convex functions $f_1,...,f_n:\R \to \R$ of total $m$ breakpoints, and $n$ linear functions $a_i\cdot x-b_i:\R^d\to \R$, find $\min_x \sum_i f_i(a_i\cdot x-b_i)$.
\end{problem} \end{problem}
@ -78,7 +79,7 @@ However, observe that in our problem the piecewise linear convex function is not
Let $n_i$ be the number of line segments in $f_i$. Note that $\sum_i n_i=m+n$. Let $n_i$ be the number of line segments in $f_i$. Note that $\sum_i n_i=m+n$.
We can formulate the optimization problem as the following linear program, We can formulate the optimization problem as the following linear program,
\newpage \framebreak
\begin{align*} \begin{align*}
\min &\sum_{i=1}^n f_i\\ \min &\sum_{i=1}^n f_i\\
@ -92,10 +93,8 @@ However, observe that in our problem the piecewise linear convex function is not
\section{LP in Low Dimensions} \section{LP in Low Dimensions}
\begin{frame}[allowframebreaks]{Megiddo's algorithm}% \begin{frame}[allowframebreaks]{Megiddo's algorithm}%
\boxfill{ \url{https://people.inf.ethz.ch/gaertner/subdir/texts/own_work/chap50-fin.pdf}
\tiny
\url{https://people.inf.ethz.ch/gaertner/subdir/texts/own_work/chap50-fin.pdf}
}
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 $x^*$. 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 $x^*$.
Thus most of the constraints are useless. Thus most of the constraints are useless.
@ -106,7 +105,7 @@ However, observe that in our problem the piecewise linear convex function is not
Through inquiries. Let $a\cdot x \leq b$ be the constraint. Define 3 hyperplanes, $a\cdot x = c$ where $c\in \set{b,b-\e,b+\e}$. Now solve three $d-1$ dimension linear programming. The largest of the three objective functions tells us where $x^*$ lies with respect to the Through inquiries. Let $a\cdot x \leq b$ be the constraint. Define 3 hyperplanes, $a\cdot x = c$ where $c\in \set{b,b-\e,b+\e}$. Now solve three $d-1$ dimension linear programming. The largest of the three objective functions tells us where $x^*$ lies with respect to the
hyperplane. hyperplane.
\newpage \framebreak
Finding the optimal solution $x^*$ is therefore equivalent to the following problem, Finding the optimal solution $x^*$ is therefore equivalent to the following problem,
\begin{problem}[Multidimensional Search Problem] \begin{problem}[Multidimensional Search Problem]
Suppose that there exists a point $x^*$ which is not known to us, but there is a oracle that can tell the position of $x^*$ relative to any hyperplane in $\R^d$. Given $n$ hyperplanes, we want to know the position of $x^*$ relative to each of them. Suppose that there exists a point $x^*$ which is not known to us, but there is a oracle that can tell the position of $x^*$ relative to any hyperplane in $\R^d$. Given $n$ hyperplanes, we want to know the position of $x^*$ relative to each of them.
@ -114,7 +113,7 @@ However, observe that in our problem the piecewise linear convex function is not
\textbf{What about 1 dimension search?} A fastest way will be using the linear time median algorithm. We can find the median of $n$ numbers and call the oracle to compare the median with $x^*$. Thus with $O(n)$ time median finding and one oracle call, we find the relative position of $n/2$ elements relative to $x^*$. \textbf{What about 1 dimension search?} A fastest way will be using the linear time median algorithm. We can find the median of $n$ numbers and call the oracle to compare the median with $x^*$. Thus with $O(n)$ time median finding and one oracle call, we find the relative position of $n/2$ elements relative to $x^*$.
\newpage \framebreak
If we can do similar things in $\R^d$, i.e., there is a method which makes $A(d)$ oracle calls and determines at least $B(d)$ fraction of relative positions, then we can apply this method $\log_{\frac{1}{1-B(d)}} n$ times to find all relative positions. If we can do similar things in $\R^d$, i.e., there is a method which makes $A(d)$ oracle calls and determines at least $B(d)$ fraction of relative positions, then we can apply this method $\log_{\frac{1}{1-B(d)}} n$ times to find all relative positions.
@ -124,7 +123,7 @@ However, observe that in our problem the piecewise linear convex function is not
\[T(n,d)=n(3T(n-1,d-1)+O(nd))\] \[T(n,d)=n(3T(n-1,d-1)+O(nd))\]
Note that in this setting $A(d)=1$ and $B(d)=1/n$. Note that in this setting $A(d)=1$ and $B(d)=1/n$.
\newpage \framebreak
Megiddo designed a clever method where $A(d)=2^{d-1}$ and $B(d)=2^{-(2^d-1)}$. Megiddo designed a clever method where $A(d)=2^{d-1}$ and $B(d)=2^{-(2^d-1)}$.
\begin{lemma} \begin{lemma}
@ -133,7 +132,7 @@ However, observe that in our problem the piecewise linear convex function is not
\end{figure} \end{figure}
Given two lines through the origin with slopes of opposite sign, knowing which quadrant $x^*$ lies in allows us to locate it with respect to at least one of the lines. Given two lines through the origin with slopes of opposite sign, knowing which quadrant $x^*$ lies in allows us to locate it with respect to at least one of the lines.
\end{lemma} \end{lemma}
\newpage \framebreak
Let $l_H$ be the intersection of hyperplane $H$ and $x_1x_2$ plane. Let $l_H$ be the intersection of hyperplane $H$ and $x_1x_2$ plane.
Compute a partition $S_1\sqcup S_2=\mathcal H$. Compute a partition $S_1\sqcup S_2=\mathcal H$.
$H\in S_1$ iff $l_H$ has positive slope. Otherwise $l_H\in S_2$. We further assume that $|S_1|=|S_2|=n/2$. $H\in S_1$ iff $l_H$ has positive slope. Otherwise $l_H\in S_2$. We further assume that $|S_1|=|S_2|=n/2$.
@ -148,12 +147,8 @@ However, observe that in our problem the piecewise linear convex function is not
Now we have $n/2$ pairs $(H_1,H_2)$, where $H_i\in S_i$. Let $l_i$ be the intersection of $H_i$ and $x_1x_2$ plane. Now we have $n/2$ pairs $(H_1,H_2)$, where $H_i\in S_i$. Let $l_i$ be the intersection of $H_i$ and $x_1x_2$ plane.
Let $H_{x_i}$ be the linear combination of $H_1$ and $H_2$ s.t. $x_i$ is eliminated. Let $H_{x_i}$ be the linear combination of $H_1$ and $H_2$ s.t. $x_i$ is eliminated.
\end{minipage} \end{minipage}
By the previous lemma, calling oracle on $l_{x_1}$ and $l_{x_2}$ locate $x^*$ with respect to at least one of $H_1$ and $H_2$.
{ \framebreak
% Now we have $n/2$ pairs $(H_1,H_2)$, where $H_i\in S_i$. Let $l_i$ be the intersection of $H_i$ and $x_1x_2$ plane.
% Let $H_{x_i}$ be the linear combination of $H_1$ and $H_2$ s.t. $x_i$ is eliminated.
By the previous lemma, calling oracle on $l_{x_1}$ and $l_{x_2}$ locate $x^*$ with respect to at least one of $H_1$ and $H_2$.}
\newpage
Input: $S_1,S_2$ and the pairs. Input: $S_1,S_2$ and the pairs.
\begin{enumerate} \begin{enumerate}
\item recursively locate $x^*$ respect to $B(d-1)n/2$ hyperplanes($H_{x_i}$) with $A(d-1)$ oracle calls in $S_1$. \item recursively locate $x^*$ respect to $B(d-1)n/2$ hyperplanes($H_{x_i}$) with $A(d-1)$ oracle calls in $S_1$.

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readme.md Normal file
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To use this theme:
1. install fonts. [FiraGO](https://github.com/bBoxType/FiraGO), [firamath](https://github.com/firamath/firamath)(build from source) and [Source Han Sans SC](https://github.com/adobe-fonts/source-han-sans/releases).
2. use xelatex. `.latexmkrc` generated by chatgpt
```latexmk
$pdf_mode = 1;
$latex = 'xelatex %O %S';
$pdflatex = 'xelatex %O %S';
$dvi_mode = 0;
$postscript_mode = 0;
```