Merge pull request #1 from congyu711/firasans

.latexmkrc

@@ -0,0 +1,5 @@
+$pdf_mode = 1;              # 生成PDF
+$latex = 'xelatex %O %S';   # 用 xelatex 替代 latex
+$pdflatex = 'xelatex %O %S'; # 用 xelatex 替代 pdflatex
+$dvi_mode = 0;              # 禁用DVI输出
+$postscript_mode = 0;       # 禁用PS输出

beamerthemeSimple.sty

@@ -1,13 +1,32 @@
-% Copyright 2018 by Zhibo Wang
-%
-% This file may be distributed and/or modified
-% under the LaTeX Project Public License
-\usepackage{libertine}
+% !TEX program = xelatex
+% !TEX TS-program = xelatex
+% fonts
+\RequirePackage{fontspec}
+% \usefonttheme{professionalfonts}
+% \RequirePackage[sfdefault]{FiraSans}
+\RequirePackage{FiraMono} 
+\renewcommand{\rmfamily}{\sffamily}
+% \RequirePackage[fakebold]{firamath-otf}
+\RequirePackage{unicode-math}
+\RequirePackage{amsmath}
+\RequirePackage{amsthm}
+\RequirePackage{amssymb}
+\RequirePackage{inputenc}
+\unimathsetup{math-style=ISO, bold-style=ISO, mathrm=sym}
+\setsansfont{FiraGO}[BoldFont=* SemiBold, Numbers=Monospaced]
+\setmathfont{Fira Math}[BoldFont=*-SemiBold]
+% \setmathfont[range=bb]{XITS Math Bold}
+\RequirePackage{xeCJK}
+\setCJKmainfont{Source Han Sans SC}
+\setCJKsansfont{Source Han Sans SC}
+\setCJKmonofont{Source Han Sans SC}
+
 \RequirePackage[english]{babel}
 \RequirePackage{fancyhdr}        % header footer
-\RequirePackage{xcolor}
+\RequirePackage[dvipsnames]{xcolor}
 \RequirePackage{bookmark}
 \RequirePackage{hyperref}[colorlinks=true,urlcolor=Blue,citecolor=Green,linkcolor=BrickRed,unicode]
+\RequirePackage{natbib}
 \RequirePackage{graphicx} % Allows including images
 \RequirePackage{booktabs} % Allows the use of \toprule, \midrule and \bottomrule in tables
 % \RequirePackage{tikz}
@@ -18,11 +37,8 @@
 % \RequirePackage[ruled,linesnumbered]{algorithm2e}
 % \RequirePackage{adjustbox}
 \RequirePackage{subcaption}
-\RequirePackage{amsmath}
-\RequirePackage{amsthm}
-\RequirePackage[utf8]{inputenc}
-\RequirePackage{CJKutf8}
-\def\zh#1{\begin{CJK}{UTF8}{gbsn}#1\end{CJK}}
+% \RequirePackage{CJKutf8}
+% \def\zh#1{\begin{CJK}{UTF8}{gbsn}#1\end{CJK}}
 \RequirePackage{aliascnt}
 
 % a color box
@@ -125,25 +141,24 @@
 \setbeamercolor{alerted text}{fg=beamer@simple@color}
 \setbeamerfont{block title alerted}{series=\mdseries}
 \setbeamerfont{alerted text}{series=\bfseries\boldmath}
-\hypersetup{colorlinks,linkcolor=,urlcolor=oliver}
+\hypersetup{colorlinks=true,linkcolor=,citecolor=Green,urlcolor=oliver}
 % \usefonttheme[onlymath]{serif}
 \setbeamerfont{frametitle}{series=\bfseries\boldmath}
 \setbeamerfont{block 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
 \setbeamertemplate{theorem begin}{%
 {
-  \vspace{5pt}
+  \vspace{5pt}%
   \usebeamerfont*{block title}%
-  \selectfont
+  \selectfont%
   \usebeamercolor[fg]{block title}%
-  \textbf{
+  \textbf{%
   \inserttheoremname
-  % \inserttheoremnumber
+  \inserttheoremnumber
   \ifx \inserttheoremaddition \empty \else\ [\inserttheoremaddition]\fi
-  \inserttheorempunctuation
   }
 }
 }
@@ -152,7 +167,7 @@
 \setbeamertemplate{proof begin}{%
 {\vspace{5pt}
 \usebeamercolor[fg]{block title}
-\textit{Proof.}}
+\textit{\textbf{Proof:}}}
 }
 \setbeamertemplate{proof end}{
 \qedhere
@@ -160,8 +175,9 @@
 }
 
 % more theorem env
-\newtheorem{observation}{Observation}
-\newtheorem{question}{Question}
+\newtheorem{conjecture}[theorem]{Conjecture}
+\newtheorem{observation}[theorem]{Observation}
+\newtheorem{question}[theorem]{Question}
 
 
 % ----------------------------------------------------------------------
@@ -174,13 +190,13 @@
 %  I give up.  These are in the wrong font, but my kludged versions 
 %  LOOK like kludges, especially \Z, \Q, and \C.
 %
-\def\Real{\mathbb{R}}
-\def\Proj{\mathbb{P}}
-\def\Hyper{\mathbb{H}}
-\def\Integer{\mathbb{Z}}
-\def\Natural{\mathbb{N}}
-\def\Complex{\mathbb{C}}
-\def\Rational{\mathbb{Q}}
+\def\Real{ℝ}
+\def\Proj{ℙ}
+\def\Hyper{ℍ}
+\def\Integer{ℤ}
+\def\Natural{ℕ}
+\def\Complex{ℂ}
+\def\Rational{ℚ}
 
 \let\N\Natural
 \let\Q\Rational

main.tex

@@ -1,8 +1,8 @@
 \documentclass{beamer}
 
-\title[template example]{Minimizing the Sum of Piecewise Linear Convex Functions}
+\title[template example]{Title}
 \date{\today}
-\author{\texorpdfstring{\zh{丛宇}}{Yu Cong}}
+\author{丛宇}
 % \AtBeginSection[]{
 %     \frame{\frametitle{Outline}\tableofcontents[currentsection, 
 %     subsectionstyle=show/show/shaded]}
@@ -23,6 +23,7 @@
 
 \section{Problems \& Definitions}
 \begin{frame}{$\min \sum f_i(a_i\cdot x-b_i)$}
+    $A\setminus B$ 测试中文:
 \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 $\min_x \sum_i f_i(a_i\cdot x-b_i)$.
 \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 $\sum_i n_i=m+n$.
 
     We can formulate the optimization problem as the following linear program,
-    \newpage
+    \framebreak
 
     \begin{align*}
         \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}
 \begin{frame}[allowframebreaks]{Megiddo's algorithm}%
-    \boxfill{
-        \tiny
-        \url{https://people.inf.ethz.ch/gaertner/subdir/texts/own_work/chap50-fin.pdf}
-    }
+    \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 $x^*$.
     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
     hyperplane.
 
-    \newpage
+    \framebreak
     Finding the optimal solution $x^*$ is therefore equivalent to the following 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.
@@ -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 $x^*$.
 
-    \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.
 
@@ -124,7 +123,7 @@ However, observe that in our problem the piecewise linear convex function is not
     \[T(n,d)=n(3T(n-1,d-1)+O(nd))\] 
     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 $B(d)=2^{-(2^d-1)}$.
 
     \begin{lemma}
@@ -133,7 +132,7 @@ However, observe that in our problem the piecewise linear convex function is not
         \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.
     \end{lemma}
-    \newpage
+    \framebreak
     Let $l_H$ be the intersection of hyperplane $H$ and $x_1x_2$ plane.
     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 $|S_1|=|S_2|=n/2$.
@@ -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.
         Let $H_{x_i}$ be the linear combination of $H_1$ and $H_2$ s.t. $x_i$ is eliminated.
     \end{minipage}
-
-    { 
-    % 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
+    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
     Input: $S_1,S_2$ and the pairs.
     \begin{enumerate}
         \item recursively locate $x^*$ respect to $B(d-1)n/2$ hyperplanes($H_{x_i}$) with $A(d-1)$ oracle calls in $S_1$.

readme.md

@@ -0,0 +1,11 @@
+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;
+```