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2 changes: 1 addition & 1 deletion 2021/solution/hw10-trees-solution/hw10-trees-solution.tex
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%%%%%%%%%%%%%%%%%%%%
\title{10. 图论: 树 (10-trees)}
\me{魏恒峰}{hfwei@nju.edu.cn}{}{}
\date{2021年05月13日 发布作业 \\ 2021年06月xx日 发布答案}
\date{2021年05月13日 发布作业 \\ 2021年06月12日 发布答案}
%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
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1 change: 1 addition & 0 deletions 2021/solution/hw11-planarity-coloring-solution/README.md
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# README.md
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198 changes: 198 additions & 0 deletions 2021/solution/hw11-planarity-coloring-solution/hw-preamble.tex
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% hw-preamble.tex

% geometry for A4 paper
% See https://tex.stackexchange.com/a/119912/23098
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left=20.0mm,
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% for Chinese
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% for fonts
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% To fix the ``MakeTextLowerCase'' bug:
% See https://github.com/Tufte-LaTeX/tufte-latex/issues/64#issuecomment-78572017
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linkcolor = teal,
urlcolor = teal,
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anchorcolor = blue}

\newcommand{\me}[4]{
\author{
{\bfseries 姓名:}\underline{#1}\hspace{2em}
{\bfseries 学号:}\underline{#2}\hspace{2em}\\[10pt]
{\bfseries 评分:}\underline{#3\hspace{3em}}\hspace{2em}
{\bfseries 评阅:}\underline{#4\hspace{3em}}
}
}

% Please ALWAYS Keep This.
\newcommand{\noplagiarism}{
\begin{center}
\fbox{\begin{tabular}{@{}c@{}}
请独立完成作业,不得抄袭。\\
若得到他人帮助, 请致谢。\\
若参考了其它资料,请给出引用。\\
鼓励讨论,但需独立书写解题过程。
\end{tabular}}
\end{center}
}

% \newcommand{\goal}[1]{
% \begin{center}{\fcolorbox{blue}{yellow!60}{\parbox{0.50\textwidth}{\large
% \begin{itemize}
% \item 体会``思维的乐趣''
% \item 初步了解递归与数学归纳法
% \item 初步接触算法概念与问题下界概念
% \end{itemize}}}}
% \end{center}
% }

% Each hw consists of four parts:
\newcommand{\beginrequired}{\hspace{5em}\section{作业 (必做部分)}}
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\newtheorem*{remark}{注}

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% for algorithms
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% for footnote/marginnote
% see https://tex.stackexchange.com/a/133265/23098
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% hw11-planarity-coloring-solution.tex

% !TEX program = xelatex
%%%%%%%%%%%%%%%%%%%%
% see http://mirrors.concertpass.com/tex-archive/macros/latex/contrib/tufte-latex/sample-handout.pdf
% for how to use tufte-handout
\documentclass[a4paper, justified]{tufte-handout}

\input{hw-preamble} % feel free to modify this file if you understand LaTeX well
%%%%%%%%%%%%%%%%%%%%
\title{11. 图论: 平面图与图着色 (11-planarity-coloring)}
\me{魏恒峰}{hfwei@nju.edu.cn}{}{}
\date{2021年05月21日 发布作业 \\ 2021年06月14日 发布答案}
%%%%%%%%%%%%%%%%%%%%
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%
\noplagiarism % PLEASE DON'T DELETE THIS LINE!
%%%%%%%%%%%%%%%%%%%%
\begin{abstract}
% \mfigcap{width = 0.85\textwidth}{figs/George-Boole}{George Boole}
% \begin{center}{\fcolorbox{blue}{yellow!60}{\parbox{0.65\textwidth}{\large
% \begin{itemize}
% \item
% \end{itemize}}}}
% \end{center}
\end{abstract}
%%%%%%%%%%%%%%%%%%%%
\beginrequired
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%
\begin{problem}[\score{4} $\star\star\star$]
假设 $G$ 是顶点数 $\ge 11$ 的简单图, $\overline{G}$$G$ 的补图~\footnote{
补图: 顶点集相同, 但是 $e$$G$ 的边当且仅当 $e$ 不是 $\overline{G}$ 的边。
}。
请证明, $G$$\overline{G}$ 不同为平面图。
\end{problem}

\begin{proof}
反设 $G$$\overline{G}$ 都是平面图。因此,
\begin{align}
m \le 3n - 6 \\
\overline{m} \le 3n - 6 \\
m + \overline{m} = \frac{n(n-1)}{2}
\end{align}
解得~\footnote{\url{https://www.wolframalpha.com/input/?i=n\%5E2+-+13n+\%2B+24+\%3C\%3D+0}},
\[
\frac{13 -\sqrt{73}}{2} \le n \le \frac{13 + \sqrt{73}}{2}.
\]
其整数解为
\[
n \in \mathbb{Z} \land 3 \le n \le 10.
\]
$n \ge 11$ 矛盾。
\end{proof}
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%
\begin{problem}[\score{4} $\star\star\star$]
假设 $G$ 是包含 $n$ 个顶点的$d$-正则简单图。
请证明
\[
\chi(G) \ge \frac{n}{n-d}.
\]
\end{problem}

\begin{proof}
反设
\[
\chi(G) < \frac{n}{n-d}.
\]
因此, 将 $G$ 中顶点按照颜色归类, 可以将 $G$ 划分成 $< \frac{n}{n-d}$ 组,
每组内的顶点两两不相邻~\footnote{注, 这种组被称为``独立集''}。
根据鸽笼原理, 至少存在一组含有 $> \frac{n}{\frac{n}{n-d}} = n - d$ 个顶点。
由于 $G$$d$-正则简单图, 该组中必有两顶点相邻。矛盾。
\end{proof}
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%
\begin{problem}[\score{4} $\star\star\star$]
假设 $G$ 是不包含三角形 $\triangle$ 的简单平面图。
\begin{enumerate}[(1)]
\item 请使用Euler公式证明 $G$ 含有度数 $\le 3$ 的顶点。
\item 请使用数学归纳法证明 $G$$4$-可着色的。
\end{enumerate}
\end{problem}

\begin{proof}
\begin{enumerate}[(1)]
\item 反设 $\delta(G) \ge 4$。因此,
\[
\sum_{v} \deg(v) \ge 4n.
\]
由于 $G$ 是不包含$\triangle$的简单平面图,
\[
m \le 2n - 4.
\]
根据握手定理,
\[
\sum_{v} \deg(v) = 2m \le 4n - 8.
\]
矛盾。
\item$G$ 中顶点数 $n$ 作归纳。
\begin{description}
\item[基础步骤:] $n = 1$$G$ 显然是 $4$-可着色的。
\item[归纳假设:] 假设命题对包含 $1 \le k = n-1$ 个顶点的任意符合要求的图均成立。
\item[归纳步骤:] 考虑包含 $n$ 个顶点的图 $G$。根据 (1), $G$ 中包含度数 $\le 3$
的顶点, 记为 $v$。从 $G$ 中删除 $v$, 得到图 $G'$$G'$ 包含 $n-1$ 个顶点,
且是不包含$\triangle$的简单平面图。根据归纳假设, $G'$$4$-可着色的。
考虑 $G'$ 的任意一种 $4$-着色方案, 将 $v$ (及删除的边) 加入 $G'$, 重新得到图 $G$
因为 $\deg(v) \le 3$, 所以 $G$$4$-可着色的。
\end{description}
\end{enumerate}
\end{proof}
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%
\begin{problem}[\score{4} $\star\star$]
假设图 $G_{1}$$G_{2}$ 是 homeomorphic 的。请证明~\footnote{
$m$, $n$ 分别表示边数与点数。
}:
\[
m_{1} - n_{1} = m_{2} - n_{2}.
\]
\end{problem}

\begin{proof}
分别考虑``插入''``收缩''2度顶点的操作:
\begin{itemize}
\item 对于``插入''一个2度顶点, 顶点数 $n$ 加一, 边数 $m$ 加一, 所以 $m-n$ 保持不变。
\item 对于``收缩''一个2度顶点, 顶点数 $n$ 减一, 边数 $m$ 减一, 所以 $m-n$ 保持不变。
\end{itemize}
对从 $G_{1}$ 转化到 $G_{2}$``插入/收缩''操作序列的长度作归纳, 即得证。
% 假设 $G_{1}$ 通过长度为 $n$ 的``插入/收缩''操作序列得到 $G_{2}$:
% \[
% G_{1}^{0} = G_{1} \to G_{1}^{1} \to G_{1}^{2} \to \dots \to G_{1}^{n} = G_{2}.
% \]
% 我们证明一个更强的结论:
% \[
% \forall 0 \le i \le j \le n.\;
% m_{i} - n_{i} = m_{j} - n_{j}.
% \]
% 对序列长度 $n$ 作归纳。
% \begin{description}
% \item[基础步骤:] $n = 0$。$G_{1} = G_{2}$。结论显然成立。
% \item[归纳假设:] 假设结论对 $1 \le k = n - 1$ 成立。
% \item[归纳步骤:]
% \end{description}
\end{proof}
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%
\begin{problem}[\score{4} $\star\star$]
请使用 Kuratowski 定理说明下图不是平面图~\footnote{你不需要制作\textsf{.gif}。}:
\fig{width = 0.50\textwidth}{figs/planarity}
\end{problem}

\begin{proof}
先去掉图内部中间的两个点, 然后收缩剩下的所有2度顶点, 得到一个 $K_{3,3}$
根据 Kuratowski 定理, 该图不是平面图。
\end{proof}
%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%
% 如果没有需要订正的题目,可以把这部分删掉
\begincorrection
%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%
% 如果没有反馈,可以把这部分删掉
\beginfb

你可以写 (也可以发邮件或者使用``教学立方'')
\begin{itemize}
\item 对课程及教师的建议与意见
\item 教材中不理解的内容
\item 希望深入了解的内容
\item $\cdots$
\end{itemize}
%%%%%%%%%%%%%%%%%%%%
\end{document}
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