\documentclass[a4paper,11pt]{article}
\usepackage[tikz]{bclogo}
\usepackage{amsmath,amsthm,amssymb}
\usepackage{color}
\usepackage{CJK}
\title{Introduction to Combinatorics Work 1}
\begin{document}
\maketitle
{\color{red}Caution: You need to add your own name and student number at the end of this file BEFORE you print this out.}
\begin{bclogo}[couleur = blue!30,arrondi = 0.1,logo=\bccrayon, ombre = true]{Theorem}
The number of partitions of n - m into exactly k - 1 parts, none exceeding m, equals the number of partitions of n - k into m - 1 parts, none exceeding k.
\end{bclogo}
\begin{proof}
Let us consider the graphical representation of partition n-m into exactly k-1 parts, none exceeding m. We transform the partitions as follows;first we adjoin a new top row of m nodes; then we delete the first column (which now has k nodes); and then we take the conjugate:
%first graph
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$\leq m$};
\draw (-.5, -.5) node {$k-1 $ };
\foreach \x in {0,...,4}
\filldraw (\x*.25, 0) circle (.5mm);
\foreach \x in {0,...,3}
\filldraw (\x*.25, -.25) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -.5) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -.75) circle (.5mm);
\foreach \x in {0,...,1}
\filldraw (\x*.25, -1) circle (.5mm);
\end{tikzpicture}
\end{center}
%downarrow
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$\Downarrow$};
\end{tikzpicture}
\end{center}
%second graph
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$ m$};
\draw (-.5, -.5) node {$k $ };
\foreach \x in {0,...,5}
\filldraw (\x*.25, 0) circle (.5mm);
\foreach \x in {0,...,4}
\filldraw (\x*.25, -.25) circle (.5mm);
\foreach \x in {0,...,3}
\filldraw (\x*.25, -.5) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -.75) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -1) circle (.5mm);
\foreach \x in {0,...,1}
\filldraw (\x*.25, -1.25) circle (.5mm);
\end{tikzpicture}
\end{center}
%downarrow
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$\Downarrow$};
\end{tikzpicture}
\end{center}
%third graph
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$ m-1$};
\draw (-.5, -.5) node {$\leq k $ };
\foreach \x in {0,...,4}
\filldraw (\x*.25, 0) circle (.5mm);
\foreach \x in {0,...,3}
\filldraw (\x*.25, -.25) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -.5) circle (.5mm);
\foreach \x in {0,...,2}
\filldraw (\x*.25, -.75) circle (.5mm);
\foreach \x in {0,...,1}
\filldraw (\x*.25, -1) circle (.5mm);
\foreach \x in {0,...,0}
\filldraw (\x*.25, -1.25) circle (.5mm);
\end{tikzpicture}
\end{center}
%downarrow
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$\Downarrow$};
\end{tikzpicture}
\end{center}
%fourth graph
\begin{center}
\begin{tikzpicture}
\draw (.5,.5) node {$\leq k$};
\draw (-.5, -.5) node {$m-1 $ };
\foreach \x in {0,...,5}
\filldraw (\x*.25, 0) circle (.5mm);
\foreach \x in {0,...,4}
\filldraw (\x*.25, -.25) circle (.5mm);
\foreach \x in {0,...,3}
\filldraw (\x*.25, -.5) circle (.5mm);
\foreach \x in {0,...,1}
\filldraw (\x*.25, -.75) circle (.5mm);
\foreach \x in {0,...,0}
\filldraw (\x*.25, -1) circle (.5mm);
\end{tikzpicture}
\end{center}
We see immediately that this composite tranformation provides a one-to-one correspondence between the two types or partitions considered and consequently the theorem is established.\\
Problem solved.\\
\begin{flushright}
\begin{CJK*}{GBK}{song}
你的名字\\
你的学号
\end{CJK*}
\end{flushright}
\end{proof}
\end{document}
代码如上，东拼西凑加上网查各种格式，总算搞定了一篇还不算难看的homework paper。这一下午，值了。
但是，核心的创新、思想，是没法从网上找到的，要会思考，这就是永远的财富。