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数学代写|离散数学作业代写discrete mathematics代考|Hamiltonian Paths

A Hamiltonian path ${ }^3$ in a graph $\mathrm{G}=(\mathrm{V}, \mathrm{E})$ is a path that visits every vertex once and once only. Another words, the length of a Hamiltonian path is $|\mathrm{V}|-1$. A graph is Hamiltonian-connected if for every pair of vertices there is a Hamiltonian path between the two vertices.

Hamiltonian paths are applicable to the travelling salesman problem, where a salesman ${ }^4$ wishes to travel to $k$ cities in the country without visiting any city more than once. In principle, this problem may be solved by looking at all of the possible routes between the various cities, and choosing the route with the minimal distance.
For example, Fig. $9.7$ shows five cities and the connections (including distance) between them. Then, a travelling salesman starting at A would visit the cities in the order AEDCBA (or in reverse order ABCDEA) covering a total distance of 14.
However, the problem becomes much more difficult to solve as the number of cities increases, and there is no general algorithm for its solution. For example, for the case of ten cities, the total number of possible routes is given by $9 !=362,880$, and an exhaustive search by a computer is feasible and the solution may be determined quite quickly. However, for 20 cities, the total number of routes is given by $19 !=1.2 \times 10^{17}$, and in this case it is no longer feasible to do an exhaustive search by a computer.

There are several sufficient conditions for the existence of a Hamiltonian path, and Theorem $9.4$ describes a condition that is sufficient for its existence.

Theorem 9.4 Let $G=(V, E)$ be a graph with $|V|=n$ and such that deg $v+$ deg $w \geq n-1$ for all non-adjacent vertices $v$ and $w$. Then $G$ possesses a Hamiltonian path.
Proof The first part of the proof involves showing that $\mathrm{G}$ is connected, and the second part involves considering the largest path in $\mathrm{G}$ of length $k-1$ and assuming that $k<n$. A contradiction is then derived and it is deduced that $k=n$.

We assume that $\mathrm{G}^{\prime}=\left(\mathrm{V}^{\prime}, \mathrm{E}^{\prime}\right)$ and $\mathrm{G}^{\prime \prime}=\left(\mathrm{V}^{\prime \prime}, \mathrm{E}^{\prime \prime}\right)$ are two connected components of $\mathrm{G}$, then $\left|\mathrm{V}^{\prime}\right|+\left|\mathrm{V}^{\prime \prime}\right| \leq n$ and so if $v \in \mathrm{V}^{\prime}$ and $w \in V^{\prime \prime}$ then $n-1 \leq v+{ }^v w \leq\left|\mathrm{V}^{\prime}\right|-1+\left|\mathrm{V}^{\prime \prime}\right|-1=\left|\mathrm{V}^{\prime}\right|+\left|\mathrm{V}^{\prime \prime}\right|-2 \leq n-2$ which is a contradiction, and so $\mathrm{G}$ must be connected.

Let $\mathrm{P}=v_1, v_2, \ldots, v_k$ be the largest path in $\mathrm{G}$ and suppose $k<n$. From this a contradiction is derived, and the details for are in [1].

数学代写|离散数学作业代写discrete mathematics代考|Trees

An acylic graph is tcrmed a forest and a conncctcd forcst is tcrmed a tree. A graph $\mathrm{G}$ is a tree if and only if for each pair of vertices in $\mathrm{G}$ there exists a unique path in $\mathrm{G}$ joining these vertices. This is since $\mathrm{G}$ is connected and acyclic, with the connected property giving the existence of at least one path and the acylic property giving uniqueness.
A spanning tree $\mathrm{T}=\left(\mathrm{V}, \mathrm{E}^{\prime}\right)$ for the connected graph $\mathrm{G}=(\mathrm{V}, \mathrm{E})$ is a tree with the same vertex set $\mathrm{V}$. It is formed from the graph by removing edges from it until it is acyclic (while ensuring that the graph remains connected).

Theorem 9.5 Let $G=(V, E)$ be a tree and let $e \in E$ then $G^{\prime}=(V, E$ Vell is disconnected and has two components.
Proof Let $e=u v$ then since $\mathrm{G}$ is connected and acyclic $u v$ is the unique path from $u$ to $v$, and thus $\mathrm{G}^{\prime}$ is disconnected since there is no path from $u$ to $v$ in $\mathrm{G}^{\prime}$.

It is thus clear that there are at least two components in $\mathrm{G}^{\prime}$ with $u$ and $v$ in different components. We show that any other vertex $w$ is connected to $u$ or to $v$ in $\mathrm{G}^{\prime}$.

Since $\mathrm{G}$ is connected there is a path from $w$ to $u$ in $\mathrm{G}$, and if this path does not use $e$ then it is in $\mathrm{G}^{\prime}$ as well, and therefore $u$ and $w$ are in the same component of $\mathrm{G}^{\prime}$.
If it does use $e$ then $e$ is the last edge of the graph since $u$ cannot appear twice in the path, and so the path is of the form $w, \ldots, v, u$ in $\mathrm{G}$. Therefore, there is a path from $w$ to $v$ in $\mathrm{G}^{\prime}$, and so $w$ and $v$ are in the same component in $\mathrm{G}^{\prime}$. Therefore, there are only two components in $\mathrm{G}^{\prime}$.
Theorem $9.6$ Any connected graph $G=(V, E)$ possesses a spanning tree. Proof This result is proved by considering all connected subgraphs of $(\mathrm{G}=\mathrm{V}, \mathrm{E})$ and choosing a subgraph T with $\left|E^{\prime}\right|$ as small as possible. The final step is to show that $\mathrm{T}$ is the desired spanning tree, and this involves showing that $\mathrm{T}$ is acyclic. The details of the proof are left to the reader.

数学代写|离散数学作业代写discrete mathematics代考|MATH300

离散数学代写

数学代写|离散数学作业代写离散数学代考|哈密顿路径


图$\mathrm{G}=(\mathrm{V}, \mathrm{E})$中的哈密顿路径${ }^3$是一条访问每个顶点一次且仅一次的路径。换句话说,哈密顿路径的长度是$|\mathrm{V}|-1$。如果每一对顶点之间都有一条哈密顿路径,那么这个图就是哈密顿连通图

哈密顿路径适用于旅行推销员问题,其中推销员${ }^4$希望旅行到国内的$k$个城市,但不访问任何城市超过一次。原则上,这个问题可以通过查看各个城市之间所有可能的路线,并选择距离最小的路线来解决。例如,图$9.7$显示了五个城市和它们之间的联系(包括距离)。然后,从a开始的旅行推销员将按AEDCBA顺序(或倒序ABCDEA)访问城市,总距离为14。
然而,随着城市数量的增加,这个问题变得更加难以解决,并且没有通用的算法来解决它。例如,对于十个城市的情况,可能的路线总数由$9 !=362,880$给出,用计算机进行详尽的搜索是可行的,解决方案可能会很快确定。但是,对于20个城市,总路线数是由$19 !=1.2 \times 10^{17}$给出的,在这种情况下,用计算机进行穷尽搜索已不再可行

哈密顿路径存在几个充分条件,定理$9.4$描述了其存在的充分条件

定理9.4设$G=(V, E)$是一个包含$|V|=n$的图,并且对于所有非相邻顶点$v$和$w$, deg $v+$ deg $w \geq n-1$。那么$G$拥有一条哈密顿路径。证明的第一部分涉及到证明$\mathrm{G}$是连通的,第二部分涉及到考虑$\mathrm{G}$中长度为$k-1$的最大路径,并假设$k<n$。然后推导出一个矛盾,推导出$k=n$ .

我们假设$\mathrm{G}^{\prime}=\left(\mathrm{V}^{\prime}, \mathrm{E}^{\prime}\right)$和$\mathrm{G}^{\prime \prime}=\left(\mathrm{V}^{\prime \prime}, \mathrm{E}^{\prime \prime}\right)$是$\mathrm{G}$的两个相互连接的组成部分,那么$\left|\mathrm{V}^{\prime}\right|+\left|\mathrm{V}^{\prime \prime}\right| \leq n$,那么如果$v \in \mathrm{V}^{\prime}$和$w \in V^{\prime \prime}$那么$n-1 \leq v+{ }^v w \leq\left|\mathrm{V}^{\prime}\right|-1+\left|\mathrm{V}^{\prime \prime}\right|-1=\left|\mathrm{V}^{\prime}\right|+\left|\mathrm{V}^{\prime \prime}\right|-2 \leq n-2$,这是一个矛盾,所以$\mathrm{G}$必须被连接

设$\mathrm{P}=v_1, v_2, \ldots, v_k$为$\mathrm{G}$中的最大路径,并设$k<n$。由此衍生出一个矛盾,而关于的细节在[1].

数学代写|离散数学作业代写discrete mathematics代考|树


一幅曲线图代表一片森林,而一幅连环画代表一棵树。当且仅当$\mathrm{G}$中的每对顶点在$\mathrm{G}$中存在连接这些顶点的唯一路径时,图$\mathrm{G}$就是树。这是因为$\mathrm{G}$是连接的和非循环的,其中connected属性给出了至少一条路径的存在性,acylic属性给出了唯一性。
连通图$\mathrm{G}=(\mathrm{V}, \mathrm{E})$的生成树$\mathrm{T}=\left(\mathrm{V}, \mathrm{E}^{\prime}\right)$是具有相同顶点集$\mathrm{V}$的树。它是由图的边移除,直到它是无环的(同时确保图保持连接)

定理9.5设$G=(V, E)$为树,设$e \in E$则$G^{\prime}=(V, E$ Vell是断开的,有两个组件。
证明设$e=u v$,那么由于$\mathrm{G}$是连接的,而非环的$u v$是从$u$到$v$的唯一路径,因此$\mathrm{G}^{\prime}$是断开的,因为在$\mathrm{G}^{\prime}$中没有从$u$到$v$的路径


可见,$\mathrm{G}^{\prime}$中至少有两个组件,$u$和$v$在不同的组件中。我们表明,任何其他顶点$w$都连接到$u$或$\mathrm{G}^{\prime}$中的$v$

自从 $\mathrm{G}$ 是否有连接的路径 $w$ 到 $u$ 在 $\mathrm{G}$,如果此路径不使用 $e$ 然后它就在 $\mathrm{G}^{\prime}$ 也是,因此 $u$ 和 $w$ 都在相同的组分中 $\mathrm{G}^{\prime}$.
如果它使用 $e$ 然后 $e$ 图的最后一条边是自 $u$ 不能在路径中出现两次,因此路径是该形式的 $w, \ldots, v, u$ 在 $\mathrm{G}$。因此,有一个路径从 $w$ 到 $v$ 在 $\mathrm{G}^{\prime}$,所以 $w$ 和 $v$ 都在同一组分中 $\mathrm{G}^{\prime}$。因此,只有两个成分 $\mathrm{G}^{\prime}$.
定理 $9.6$ 任意连通图 $G=(V, E)$ 拥有生成树。的所有连通子图证明了该结果 $(\mathrm{G}=\mathrm{V}, \mathrm{E})$ 选择子图T $\left|E^{\prime}\right|$ 尽可能的小。最后一步是展示这一点 $\mathrm{T}$ 这是我们想要的生成树吗 $\mathrm{T}$ 是无环的。证明的细节留给读者。

数学代写|离散数学作业代写discrete mathematics代考

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