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

## 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.

# 离散数学代写

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

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