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# 数学代写|组合学代写Combinatorics代考|COMP418 Paths and Connected Graphs

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## 数学代写|组合学代写Combinatorics代考|Paths and Connected Graphs

Let $G$ be a graph with the set of vertices $V$, and the set of edges $E$. A walk in $\operatorname{graph} G$ is a finite sequence
$$W=v_{0} e_{1} v_{1} e_{2} \ldots v_{n-1} e_{n} v_{n},$$
where $v_{0}, v_{1}, \ldots, v_{n} \in V$, and $e_{1}, e_{2}, \ldots, e_{n} \in E$, and for any $i \in{1,2, \ldots, n}$, the vertices $v_{i-1}$ and $v_{i}$ are incident to the edge $e_{i}$. The vertices $v_{0}$ and $v_{k}$ are called, respectively, the origin, and the terminus of walk $W$, while the vertices $v_{1}, v_{2}, \ldots, v_{k-1}$ are called internal vertices of the walk.

The number of edges in the walk is called the length of the walk. Origin $v_{0}$ and terminus $v_{n}$ of walk $W=v_{0} e_{1} v_{1} e_{2} \ldots v_{n-1} e_{n} v_{n}$, need not be distinct. If $v_{0}=v_{n}$, then we say that walk $W$ is closed. If $v_{0} \neq v_{n}$, then walk $W$ is open. A vertex $v_{0}$ can be considered as a trivial walk.

A trail is a walk $W=v_{0} e_{1} v_{1} e_{2} \ldots v_{n-1} e_{n} v_{n}$ without repetition in the sequence of its edges $e_{1}, e_{2}, \ldots, e_{n}$.

If the vertices $v_{0}, v_{1}, \ldots, v_{n}$ of walk $W=v_{0} e_{1} v_{1} e_{2} \ldots v_{n-1} e_{n} v_{n}$ are distinct, then $W$ is called a path from $v_{0}$ to $v_{n}$.

## 数学代写|组合学代写Combinatorics代考|Isomorphic Graphs

Let $G$ be a graph with the set of vertices $V$, and the set of edges $E$, and let $G^{\prime}$ be a graph with the set of vertices $V^{\prime}$, and the set of edges $E^{\prime}$. The graphs $G$ and $G^{\prime}$ are isomorphic if there exist two bijective functions
$$f: V \rightarrow V^{\prime}, \quad g: E \rightarrow E^{\prime}$$

such that for each $v \in V$, and each $e \in E$, if $v$ is an endpoint of the edge $e$, then $f(v)$ is andpoint of the edge $g(e)$.

Example 8.5.1. Let $V={1,2,3,4,5}$ be the set of vertices, and ${i, j}$ be the edge with incident vertices $i$ and $j$, for any $i, j \in V$. Let us define $G_{1}$ and $G_{2}$ to be graphs with the same set of vertices $V$, and with the sets of edges $E_{1}=\left{e_{1}, e_{2}, e_{3}, e_{4}, e_{5}\right}$, and $E_{2}=\left{\varepsilon_{1}, \varepsilon_{2}, \varepsilon_{3}, \varepsilon_{4}, \varepsilon_{5}\right}$, respectively, where:
\begin{aligned} &e_{1}={1,3}, \quad e_{2}={2,4}, \quad e_{3}={3,5}, \quad e_{4}={4,1}, \quad e_{5}={5,2}, \ &\varepsilon_{1}={1,2}, \quad \varepsilon_{2}={2,3}, \quad \varepsilon_{3}={3,4}, \quad \varepsilon_{4}={4,5}, \quad \varepsilon_{5}={5,1} . \end{aligned}
Graphs $G_{1}$ nd $G_{2}$ are given in Figures $8.5 .1$ and 8.5.2, where every edge ${i, j}$ is presented as a segment with endpoints $i$ and $j$.

It is obvious that graphs $G_{1}$ and $G_{2}$ are complements to each other. We shall prove that these two graphs are also isomorphic.

## 数学代写|组合学代写Combinatorics代考|Paths and Connected Graphs

$$W=v_{0} e_{1} v_{1} e_{2} \ldots v_{n-1} e_{n} v_{n},$$

## 数学代写|组合学代写Combinatorics代考|lsomorphic Graphs

$$f: V \rightarrow V^{\prime}, \quad g: E \rightarrow E^{\prime}$$

$\backslash$ left 的分隔符缺失或无法识别，，分别，其中:
$e_{1}=1,3, \quad e_{2}=2,4, \quad e_{3}=3,5, \quad e_{4}=4,1, \quad e_{5}=5,2, \quad \varepsilon_{1}=1,2, \quad \varepsilon_{2}=2,3, \quad \varepsilon_{3}=3,4, \quad \varepsilon_{4}=4,5, \quad \varepsilon_{5}=5,1 .$

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