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# README.md |
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2021/solution/hw11-planarity-coloring-solution/hw-preamble.tex
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% hw-preamble.tex | ||
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% geometry for A4 paper | ||
% See https://tex.stackexchange.com/a/119912/23098 | ||
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{\bfseries 姓名:}\underline{#1}\hspace{2em} | ||
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{\bfseries 评分:}\underline{#3\hspace{3em}}\hspace{2em} | ||
{\bfseries 评阅:}\underline{#4\hspace{3em}} | ||
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% Please ALWAYS Keep This. | ||
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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。 | ||
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% \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}}}} | ||
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2021/solution/hw11-planarity-coloring-solution/hw11-planarity-coloring-solution.pdf
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2021/solution/hw11-planarity-coloring-solution/hw11-planarity-coloring-solution.tex
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% hw11-planarity-coloring-solution.tex | ||
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% !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} | ||
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\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 | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%% | ||
\begin{problem}[\score{4} $\star\star\star$] | ||
假设 $G$ 是顶点数 $\ge 11$ 的简单图, $\overline{G}$ 是 $G$ 的补图~\footnote{ | ||
补图: 顶点集相同, 但是 $e$ 是 $G$ 的边当且仅当 $e$ 不是 $\overline{G}$ 的边。 | ||
}。 | ||
请证明, $G$ 和 $\overline{G}$ 不同为平面图。 | ||
\end{problem} | ||
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\begin{proof} | ||
反设 $G$ 和 $\overline{G}$ 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 | ||
\[ | ||
n \in \mathbb{Z} \land 3 \le n \le 10. | ||
\] | ||
与 $n \ge 11$ 矛盾。 | ||
\end{proof} | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%% | ||
\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} | ||
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\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 $G$ 是 $d$-正则简单图, 该组中必有两顶点相邻。矛盾。 | ||
\end{proof} | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%% | ||
\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 $G$ 是 $4$-可着色的。 | ||
\end{enumerate} | ||
\end{problem} | ||
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\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 对 $G$ 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, $G'$ 是 $4$-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 $G$ 是 $4$-可着色的。 | ||
\end{description} | ||
\end{enumerate} | ||
\end{proof} | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%% | ||
\begin{problem}[\score{4} $\star\star$] | ||
假设图 $G_{1}$ 与 $G_{2}$ 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} | ||
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\begin{proof} | ||
分别考虑``插入''与``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 $G_{2}$ 的``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} | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%% | ||
\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} | ||
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\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} | ||
%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%%%%%%% | ||
% 如果没有需要订正的题目,可以把这部分删掉 | ||
\begincorrection | ||
%%%%%%%%%%%%%%%%%%%% | ||
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%%%%%%%%%%%%%%%%%%%% | ||
% 如果没有反馈,可以把这部分删掉 | ||
\beginfb | ||
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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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