## 数学代写|组合数学代写Combinatorial mathematics代考|MATH418

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## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

9.1. (a) We can prove by induction on $n$ that there is a party of $n$ people such that every integer $0,1, \ldots, n-2$ appears as the number of acquaintances of a party member.

Two non-acquainted people deliver an example of such a party for $n=2$.

Assume that there is a party $P$ of $n$ people such that every integer $0,1, \ldots, n-2$ appears as the number of acquaintances. In fact, by the Pigeonhole Principle, one of these numbers, say $k(0 \leq k \leq n-2)$, must appear twice.

Now we have to construct a party $P^{\prime}$ of $n+1$ people such that every integer $0,1, \ldots, n-1$ appears as the number of acquaintances. We start with the party $P$ of $n$ people and add one more person, who is acquainted with exactly one of two people with $k$ acquaintances, and with everyone having more than $k$ acquaintances. You can easily verify that $P^{\prime}$ satisfies the required condition.
(b) There is such a party. The construction is similar to the one in Exercise 9.1(a).
9.2. Due to Theorem 9.1, the total number $T$ of acquaintances is even, on the other hand, if we assume that there exists a party of 21 such that everyone has exactly seven acquaintances, we have $T=21 \times 7$, an odd number. This contradiction proves that such a party does not exist.
We would like to mention here that similar reasoning proves the following result: there is no polyhedron with an odd number of oddsided faces.

Graphs appear in our discussion as diagrams of acquaintances. Thus, the only thing that matters when we represent a graph in the plane is the set of vertices (but not their positions on the plane) and which vertices are adjacent (but not the shapes of the edges, which we presume have no points in common except the vertices of the graph). In fact, think of a graph as a set of pins, some of which are connected by rubber bands. A graph remains the same if we reposition the pins and stretch the rubber bands.

Two graphs are called isomorphic if “pins” of one of them can be repositioned and its “rubber bands” stretched so that the two become identical.

More formally, two graphs $G$ and $G_1$ are said to be isomorphic if there is a one-to-one correspondence $f: V \rightarrow V_1$ of their vertex sets that preserves adjacency, i.e., vertices $v_1$ and $v_2$ of $G$ are adjacent if and only if $f\left(v_1\right)$ and $f\left(v_2\right)$ of $G_1$ are adjacent.
We denote the isomorphism of graphs $G$ and $G_1$ by $G \cong G_1$.

Solution. Let us manipulate “pins” and “rubber bands” of $G$ : first we flip $v_2 v_5$, then stretch it (Figure 10.2).

Lo and behold, we end up with a graph identical to $G_1$. $G$ and $G_1$ are isomorphic.

Needless to say, two isomorphic graphs must have equal numbers of vertices and edges; therefore, the numbers of vertices and edges are what we call invariants of graphs (invariants are characteristics shared by all isomorphic graphs). The equality of these two invariants, however, is not sufficient to prove the isomorphism of two graphs.
Example 10.2. Are the graphs in Figure $10.3$ isomorphic?
Solution. Even though $G$ and $G_1$ have equal numbers of vertices and edges, they are not isomorphic. Under no one-to-one correspondence of vertices can adjacency be preserved, because $G_1$ contains a vertex $v$ of degree one. This vertex must correspond under isomorphic correspondence to a vertex of degree one in $G$. But $G$ has no such vertex!

While solving Example 10.2, we discovered a new invariant of graphs: the degrees of its vertices. Do we have enough invariants to guarantee isomorphism of two graphs? In other words, given $G$ and $G_1$ with equal numbers of vertices and edges and equal sequences of vertex degrees, do $G$ and $G_1$ have to be isomorphic?

## 数学代写|组合数学代写Combinatorial mathematics代考|Solutions to Exercises

9.1. (a) 我们可以通过对n的归纳来证明n，有一个n人的聚会，使得每个整数0,1,…,n−2都表现为一个聚会成员的熟人数目。

GG1G≅G1

GG1G1v一级的。该顶点必须在同构对应下对应于中的一阶顶点。但是没有这样的顶点！GG

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