304 lines
14 KiB
TeX
304 lines
14 KiB
TeX
\chapter{树}
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\section{树的概念}
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\begin{definition}
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\newnoun{树}{tree}是一张无圈连通图。
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在树中，度为1的顶点称为\newnoun{叶}{leaf}，度不小于2的顶点称为\newnoun{内点}{branch node}。
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树的集合称为\newnoun{林}{forest}。
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\newnoun{平凡树}{trivial tree}是只有一个顶点、无边的树。
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\end{definition}
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\begin{theorem}
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对于$T=(V,E)$，下列叙述等价：
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\begin{enumerate}
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\item $T$是一棵树，即其连通且无圈；
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\item $T$连通，但删去任何一条边都会使它变为不连通的图；
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\item $T$无圈，但是添加任意一条边都会使它有圈；
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\item $T$中的任意一对顶点都有唯一的一条路相连；
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\item $T$连通，且$\vert E \vert = \vert v \vert - 1$；
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\item $T$无圈，且$\vert E \vert = \vert v \vert - 1$。
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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对$G$中的一条边$e$，若$c(G-e) > c(G)$则称$e$为是图$G$的一个\newnoun{桥}{bridge}。
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\end{definition}
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\begin{theorem}
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任意连通图都有生成树。
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\end{theorem}
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\begin{remark}
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同一个连通图可能有多个不同的生成树。
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\end{remark}
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\begin{figure}[H]
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\centering
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\subfloat[连通图$G$]{
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\begin{tikzpicture}[scale=1.5]
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\draw node[circle, draw=black] (a) at (1,0) {};
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\draw node[circle, draw=black] (b) at (-1,0) {};
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\draw node[circle, draw=black] (c) at (0,1) {};
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\draw node[circle, draw=black] (d) at (1,-1) {};
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\draw node[circle, draw=black] (e) at (-1,-1) {};
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\draw (a)--(b)--(c)--(a)--(d)--(e)--(b)--(d);
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\end{tikzpicture}
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}
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\hspace{2cm}
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\subfloat[$G$的一种生成树]{
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\begin{tikzpicture}[scale=1.5]
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\draw node[circle, draw=black] (a) at (1,0) {};
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\draw node[circle, draw=black] (b) at (-1,0) {};
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\draw node[circle, draw=black] (c) at (0,1) {};
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\draw node[circle, draw=black] (d) at (1,-1) {};
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\draw node[circle, draw=black] (e) at (-1,-1) {};
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\draw (e)--(b)--(c)--(a)--(d);
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\end{tikzpicture}
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}
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\hspace{2cm}
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\subfloat[$G$的另一种生成树]{
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\begin{tikzpicture}[scale=1.5]
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\draw node[circle, draw=black] (a) at (1,0) {};
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\draw node[circle, draw=black] (b) at (-1,0) {};
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\draw node[circle, draw=black] (c) at (0,1) {};
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\draw node[circle, draw=black] (d) at (1,-1) {};
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\draw node[circle, draw=black] (e) at (-1,-1) {};
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\draw (e)--(b)--(a)--(c);
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\draw (b)--(d);
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\end{tikzpicture}
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}
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\end{figure}
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\section{生成一棵树}
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\begin{theorem}
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任何至少有两顶点的树至少有两个叶子顶点。
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\end{theorem}
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\begin{theorem}
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利用一下方法生成一张图：
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\begin{enumerate}
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\item 开始于一个根节点。
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\item 重复进行一下操作：对已有的图$G$，建立一个新的顶点（子顶点），并将这个顶点通过一条边与$G$中的一个顶点（父顶点）相连。
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\end{enumerate}
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通过上述过程生成的图是一棵树，且任何树都能通过上述过程得到。
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\end{theorem}
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\begin{theorem}
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任何$n$个顶点的树都有$n-1$条边。
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\end{theorem}
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\begin{example}
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一棵树有$n_2$个顶点的度为2，$n_3$个顶点度为3，……，$n_k$个顶点的度为$k$，问它有几片叶？
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\end{example}
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\begin{proof}[解]
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设它有$n_1$片叶，则顶点数为$n_1 + n_2 + \dots + n_k$，那么它有$n_1 + n_2 + \dots + n_k - 1$条边。所有顶点的度的和为$n_1 + 2n_2 + \dots + kn_k$。根据握手定理，
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\[n_1 + 2n_2 + \dots + kn_k = 2(n_1 + n_2 + \dots + n_k - 1)\]
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解得
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\[n_1 = n_3 + 2n_4 + \dots + (k-2)n_k + 2\eqper\qedhere\]
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\end{proof}
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\section{树的计数}
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首先我们需要区分\newnoun{无编号树}{unlabeled tree}与\newnoun{有编号树}{labeled tree}。
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对于无编号树，下面两个图被认为是相同的无编号树：
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\begin{figure}[H]
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\centering
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\subfloat{
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\begin{tikzpicture}
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\draw node[circle, draw=black] (a) at (18:2) {};
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\draw node[circle, draw=black] (b) at (90:2) {};
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\draw node[circle, draw=black] (c) at (162:2) {};
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\draw node[circle, draw=black] (d) at (234:2) {};
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\draw node[circle, draw=black] (e) at (306:2) {};
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\draw (a)--(e)--(d)--(c);
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\draw (b)--(e);
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\end{tikzpicture}
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}
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\hspace{1cm}
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\subfloat{
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\begin{tikzpicture}
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\draw node[circle, draw=black] (a) at (18:2) {};
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\draw node[circle, draw=black] (b) at (90:2) {};
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\draw node[circle, draw=black] (c) at (162:2) {};
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\draw node[circle, draw=black] (d) at (234:2) {};
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\draw node[circle, draw=black] (e) at (306:2) {};
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\draw (a)--(e)--(b)--(c);
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\draw (d)--(b);
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\end{tikzpicture}
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}
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\caption{一组相同的无编号树}
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\end{figure}
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而与之不同的是，下面的两个图被认为是不同的有编号树：
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\begin{figure}[H]
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\centering
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\subfloat{
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\begin{tikzpicture}
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\draw node[circle, draw=black] (a) at (18:2) {1};
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\draw node[circle, draw=black] (b) at (90:2) {2};
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\draw node[circle, draw=black] (c) at (162:2) {3};
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\draw node[circle, draw=black] (d) at (234:2) {4};
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\draw node[circle, draw=black] (e) at (306:2) {5};
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\draw (a)--(e)--(d)--(c);
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\draw (b)--(e);
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\end{tikzpicture}
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}
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\hspace{1cm}
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\subfloat{
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\begin{tikzpicture}
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\draw node[circle, draw=black] (a) at (18:2) {1};
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\draw node[circle, draw=black] (b) at (90:2) {2};
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\draw node[circle, draw=black] (c) at (162:2) {3};
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\draw node[circle, draw=black] (d) at (234:2) {4};
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\draw node[circle, draw=black] (e) at (306:2) {5};
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\draw (a)--(e)--(b)--(c);
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\draw (d)--(b);
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\end{tikzpicture}
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}
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\caption{一组不同的有编号树}
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\end{figure}
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\begin{theorem}[Cayley's Theorem]\label{Cayley's Theorem}
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$n$个顶点的有编号树的个数为$n^{n-2}$。
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\end{theorem}
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\section{如何存储树}
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如果想要证明定理\ref{Cayley's Theorem}，我们需要先来考虑如何储存一棵树。
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在这节中，我们将以下面的树为例考虑如何存储。
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}
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\draw node[circle, draw=black] (0) at (-2,0) {0};
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\draw node[circle, draw=black] (1) at (4,1) {1};
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\draw node[circle, draw=black] (2) at (0,0) {2};
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\draw node[circle, draw=black] (3) at (-4,0) {3};
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\draw node[circle, draw=black] (4) at (0,-2) {4};
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\draw node[circle, draw=black] (5) at (4,-1) {5};
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\draw node[circle, draw=black] (6) at (2,0) {6};
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\draw node[circle, draw=black] (7) at (-1,4) {7};
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\draw node[circle, draw=black] (8) at (1,4) {8};
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\draw node[circle, draw=black] (9) at (0,2) {9};
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\draw (3)--(0)--(2)--(6)--(1);
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\draw (5)--(6);
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\draw (4)--(2)--(9)--(8);
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\draw (9)--(7);
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\end{tikzpicture}
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\caption{一棵有编号树}
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\end{figure}
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最普通的方法是利用它的\newnoun{邻接矩阵}{adjacency matrix}：
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\setcounter{MaxMatrixCols}{11}
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\begin{equation*}
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\begin{bNiceMatrix}[first-row,first-col]
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& 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\
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0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\
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1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
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2 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 1\\
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3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
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4 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\
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5 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0\\
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6 & 0 & 1 & 1 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\
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7 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
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8 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\
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9 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 0\\
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\end{bNiceMatrix}
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\end{equation*}
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\setcounter{MaxMatrixCols}{10}
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然而这明显收入了过多的无用信息（过多的零！）。
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\subsection{The father code}
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一个减少储存信息的自然的想法是：把0看作根节点，对于剩下的每个节点，只要记录下它的父节点是谁就足够描述这个图了。例如：
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\begin{table}[H]
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\centering
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\begin{tabular}{ccccccccc}
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1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\
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6 & 0 & 0 & 2 & 6 & 2 & 9 & 9 & 2
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\end{tabular}
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\end{table}
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\subsection{The Prüfer code}
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上面的father code显然已经比邻接矩阵要高效地多，而我们还有更好的方法。我们每次都从（不是根的）叶子中找到标号最小的那个，并记录下它的父亲，之后删除我们已经记录的这个叶子和边，重复操作直到只剩下根，于是我们得到了扩展的Prüfer code（extended Prüfer code）：
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\begin{table}[H]
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\centering
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\begin{tabular}{ccccccccc}
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1 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 2\\
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6 & 0 & 2 & 6 & 2 & 9 & 9 & 2 & 0
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\end{tabular}
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\end{table}
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首先，我们可以注意到，第二行的最后一个一定是零，因此它是多余的，对于一个$n$个端点的树，我们只需要$n-2$列。
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实际上，在这个表格中，整个第一行也都是多余的：
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\begin{lemma}
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扩展Prüfer code中的第二行可以确定第一行。
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\end{lemma}
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\begin{proof}
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如果我们只有第二行，且我们已经填好了前$n-1$位。现在，我们要考虑第$n$列的第一行的数：首先，第一行前面已经出现过的数不会填在这里，因为它们已经被删掉了；其次第二行中第$n$列及之后的数也不会填在这里，因为它们在之后还是父亲顶点，现在不可能作为叶子。这样我们就已经将范围缩小了；因为我们生成Prüfer code的规则要求我们每次都取编号最小的顶点，因此我们只要取不等于第一行已经出现过和第二行还未考虑到的数中最小的一个填入空格即可。
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\end{proof}
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在这里，我们可以看到$n-2$位的$n$元Prüfer code与$n$个顶点的树一一对应。由此，我们可以证明定理\ref{Cayley's Theorem}的正确性，因为Prüfer code 有$n^{n-2}$种。
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\section{无编号树的数量}
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至今我们还没有办法得知给定端点树的无编号树的数量的具体值。
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\begin{theorem}
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无编号$n$顶点树的数量$T_n$满足
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\[\frac{n^{n-2}}{n} \leq T_n \leq \binom{2n-4}{n-2}\eqper\]
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\end{theorem}
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为了证明这个定理，我们要先引入Planar code：
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首先，我们先将无编号树``展平''，即使它的边互相不交叉。之后，我们从它的某个顶点旁边开始，使自己的右手边始终挨着树走一圈。那么，如果我们将我们第一次经过某个边记作1，第二次经过它记作0，那么我们就会得到一个序列1111100100011011010000。
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\begin{figure}[H]
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\centering
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\begin{tikzpicture}
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\draw node[circle, draw=black, fill=black] (a) at (0,0) {};
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\draw node[circle, draw=black] (b) at (0,2) {};
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\draw node[circle, draw=black] (c) at (-2,2) {};
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\draw node[circle, draw=black] (d) at (-4,2) {};
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\draw node[circle, draw=black] (e) at (-6,2) {};
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\draw node[circle, draw=black] (f) at (-8,2) {};
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\draw node[circle, draw=black] (g) at (-4,4) {};
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\draw node[circle, draw=black] (h) at (2,2) {};
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\draw node[circle, draw=black] (i) at (2,4) {};
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\draw node[circle, draw=black] (j) at (4,2) {};
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\draw node[circle, draw=black] (k) at (4,4) {};
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\draw node[circle, draw=black] (l) at (6,2) {};
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\draw (f)--(e)--(d)--(c)--(b)--(h)--(j)--(l);
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\draw (d)--(g);
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\draw (b)--(a);
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\draw (h)--(i);
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\draw (j)--(k);
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\draw (j)--(l);
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\draw[dashed,-{Stealth[width=5pt]}] (-0.5,0)--(-0.5,1.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-0.5,1.5)--(-8,1.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-8,1.5) arc (-90:-270:0.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-8,2.5)--(-4.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-4.5,2.5)--(-4.5,4);
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\draw[dashed,-{Stealth[width=5pt]}] (-4.5,4) arc (-180:-360:0.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-3.5,4)--(-3.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (-3.5,2.5)--(1.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (1.5,2.5)--(1.5,4);
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\draw[dashed,-{Stealth[width=5pt]}] (1.5,4) arc (-180:-360:0.5);
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\draw[dashed,-{Stealth[width=5pt]}] (2.5,4)--(2.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (2.5,2.5)--(3.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (3.5,2.5)--(3.5,4);
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\draw[dashed,-{Stealth[width=5pt]}] (3.5,4) arc (-180:-360:0.5);
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\draw[dashed,-{Stealth[width=5pt]}] (4.5,4)--(4.5,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (4.5,2.5)--(6,2.5);
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\draw[dashed,-{Stealth[width=5pt]}] (6,2.5) arc (90:-90:0.5);
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\draw[dashed,-{Stealth[width=5pt]}] (6,1.5)--(0.5,1.5);
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\draw[dashed,-{Stealth[width=5pt]}] (0.5,1.5)--(0.5,0);
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\draw[dashed,-{Stealth[width=5pt]}] (0.5,0) arc (0:-180:0.5);
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\end{tikzpicture}
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\end{figure}
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\begin{proof}
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Planar code 的第一位一定是1，最后一位一定是0，剩余$2n-4$位中有$n-2$（一半）是1，共有$\dbinom{2n-4}{n-2}$种，而有的Planar code不合法（例如11000110前五位有两个1却有三个0），因此$T_n < \dbinom{2n-4}{n-2}$，即右侧的不等式成立；
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其次，对于一个无编号树，我们最多有$n!$种方法给他的不同节点编号成为不同的有编号树，因此无编号树的数量多于$\dfrac{n^{n-2}}{n!}$种，即$\dfrac{n^{n-2}}{n!} \leq T_n$。
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\end{proof}
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