### 数学代写|图论作业代写Graph Theory代考|Math780

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## 数学代写|图论作业代写Graph Theory代考|Ramsey numbers

Ramsey’s theorem may be rephrased as follows: if $H=K^r$ and $G$ is a graph with sufficiently many vertices, then either $G$ itself or its complement $\bar{G}$ contains a copy of $H$ as a subgraph. Clearly, the same is true for any graph $H$, simply because $H \subseteq K^h$ for $h:=|H|$.

However, if we ask for the least $n$ such that every graph $G$ of order $n$ has the above property – this is the Ramsey number $R(H)$ of $H$-then the above question makes sense: if $H$ has only few edges, it should embed more easily in $G$ or $\bar{G}$, and we would expect $R(H)$ to be smaller than the Ramsey number $R(h)=R\left(K^h\right)$.

A little more generally, let $R\left(H_1, H_2\right)$ denote the least $n \in \mathbb{N}$ such that $H_1 \subseteq G$ or $H_2 \subseteq \bar{G}$ for every graph $G$ of order $n$. For most graphs $H_1, H_2$, only very rough estimates are known for $R\left(H_1, H_2\right)$. Interestingly, lower bounds given by random graphs (as in Theorem 11.1.3) are often sharper than even the best bounds provided by explicit constructions.

The following proposition describes one of the few cases where exact Ramsey numbers are known for a relatively large class of graphs:

Proposition 9.2.1. Let $s, t$ be positive integers, and let $T$ be a tree of order $t$. Then $R\left(T, K^s\right)=(s-1)(t-1)+1$.

Proof. The disjoint union of $s-1$ graphs $K^{t-1}$ contains no copy of $T$, while the complement of this graph, the complete $(s-1)$-partite graph $K_{t-1}^{s-1}$, does not contain $K^s$. This proves $R\left(T, K^s\right) \geqslant(s-1)(t-1)+1$.
Conversely, let $G$ be any graph of order $n=(s-1)(t-1)+1$ whose complement contains no $K^s$. Then $s>1$, and in any vertex colouring of $G$ (in the sense of Chapter 5) at most $s-1$ vertices can have the same colour. Hence, $\chi(G) \geqslant\lceil n /(s-1)\rceil=t$. By Corollary $5.2 .3, G$ has a subgraph $H$ with $\delta(H) \geqslant t-1$, which by Corollary 1.5 .4 contains a copy of $T$.

## 数学代写|图论作业代写Graph Theory代考|Induced Ramsey theorems

Ramsey’s theorem can be rephrased as follows. For every graph $H=K^r$ there exists a graph $G$ such that every 2-colouring of the edges of $G$ yields a monochromatic $H \subseteq G$; as it turns out, this is witnessed by any large enough complete graph as $G$. Let us now change the problem slightly and ask for a graph $G$ in which every 2-edge-colouring yields a monochromatic induced $H \subseteq G$, where $H$ is now an arbitrary given graph.

This slight modification changes the character of the problem dramatically. What is needed now is no longer a simple proof that $G$ is ‘big enough’ (as for Theorem 9.1.1), but a careful construction: the construction of a graph that, however we bipartition its edges, contains an induced copy of $H$ with all edges in one partition class. We shall call such a graph a Ramsey graph for $H$.

The fact that such a Ramsey graph exists for every choice of $H$ is one of the fundamental results of graph Ramsey theory. It was proved around 1973 , independently by Deuber, by Erdős, Hajnal \& Pósa, and by Rödl.

Theorem 9.3.1. Every graph has a Ramsey graph. In other words, for every graph $H$ there exists a graph $G$ that, for every partition $\left{E_1, E_2\right}$ of $E(G)$, has an induced subgraph $H$ with $E(H) \subseteq E_1$ or $E(H) \subseteq E_2$.

We give two proofs. Each of these is highly individual, yet each offers a glimpse of true Ramsey theory: the graphs involved are used as hardly more than bricks in the construction, but the edifice is impressive.

# 图论代考

## Matlab代写

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