# 数学代写|组合数学代写Combinatorial mathematics代考|COMP418

## 数学代写|组合数学代写Combinatorial mathematics代考|Extremal graph theory

Another way to look at Theorem $4.0 .2$ is as a density version of Ramsey’s Theorem (cf. Szemerédi’s Theorem and the Density Hales-Jewett Theorem mentioned in the previous chapter). We ask how often we can use the color red without creating a red triangle. As such it makes sense to generalize this result to arbitrary complete graphs. To describe the extremal graphs (those to which no more edges can be added) we use the notion of a k-partite graph: a graph whose vertex set is partitioned into sets $V_1, \ldots, V_k$, and no edge has both endpoints inside a set $V_i$. The complete k-partite graphs are those in which all such edges are present. If $\left|V_i\right|=n_i$ for $i \in[k]$, then this complete $k$-partite graph is denoted by $K_{n_1, \ldots, n_k}$. Moreover, $n=n_1+\cdots+n_k$, and if $\left|n_i-n_j\right| \leq 1$ for all $i, j \in[k]$, then we denote the unique (up to vertex labeling) such graph by $T_{n, k}$.

Proof: Suppose it does. We try to modify the graph to improve the number of edges; we distinguish three cases.

I. If $\operatorname{deg}(u)<\operatorname{deg}(v)$, then we construct a new graph, $G^{\prime}$, as follows from $G$. Delete all edges incident with $u$, and create new edges so that the set of neighbors of $u$, denoted by $N(u)$, satisfies $N(u)=N(v)$. Note that $u$ and $v$ are not adjacent in $G^{\prime}$, so if $G^{\prime}$ has a $K_{k+1}$-subgraph, then it uses at most one of $u$ and $v$. But both $G^{\prime}-u$ and $G^{\prime}-v$ are isomorphic to subgraphs of $G$, and therefore have no $K_{k+1}$. Hence $G^{\prime}$ has none; and the number of edges satisfies $$\left|E\left(G^{\prime}\right)\right|=|E(G)|-\operatorname{deg}(u)+\operatorname{deg}(v)>|E(G)|,$$
contradicting our choice of $G$.
II. $\operatorname{deg}(u)<\operatorname{deg}(w)$ follows from (I) by symmetry. III. If $\operatorname{deg}(u) \geq \operatorname{deg}(v)$ and $\operatorname{deg}(u) \geq \operatorname{deg}(w)$, then we construct a new graph, $G^{\prime}$, as follows from $G$. Delete all edges incident with $v$, and all edges incident with $w$. Then make both $v$ and $w$ adjacent to all neighbors of $u$. Again we can see that $G^{\prime}$ has no $K_{k+1}$-subgraph. Moreover, $$\left|E\left(G^{\prime}\right)\right|=|E(G)|-\operatorname{deg}(v)-\operatorname{deg}(w)+1+2 \operatorname{deg}(u)>|E(G)|,$$
where the $+1$ follows from double-counting the edge $v w$.

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

The sets that attained the bounds in the previous section had a pretty nice structure. You might imagine that it is useful to give up a few sets, if a structure like that arises as a reward. In this section we prove a result showing that we can do this.
4.3.1 Definition. A sunflower with $k$ petals and core $Y$ is a family $\mathscr{F}$ of sets, along with a set $Y$, such that $|\mathscr{F}|=k$, and for all $A, B \in \mathscr{F}$ with $A \neq B$ we have $A \cap B=Y$. Moreover, the sets $A \backslash Y$, which we call the petals, are nonempty.
4.3.2 LEMMA (Sunflower Lemma). Let $\mathscr{F}$ be a family of sets, each of size s. If $|\mathscr{F}|>s !(k-1)^s$ then $\mathscr{F}$ contains a sunflower with $k$ petals.

Proof: We prove the result by induction on $s$. If $s=1$ then $|\mathscr{F}|>k-1$, so $\mathscr{F}$ contains at least $k$ size-1 sets. These form a sunflower (with $Y=\emptyset$ ).

Fix $s \geq 2$, and suppose the result holds for all smaller values of $s$. Let $\left{A_1, \ldots, A_t\right}$ be a maximal collection of pairwise disjoint subsets contained in $\mathscr{F}$. If $t \geq k$ then any $k$ of these form a sunflower, so assume $t\frac{s !(k-1)^s}{s(k-1)}=(s-1) !(k-1)^{s-1} $$sets. Consider \mathscr{F}_x:={A \backslash{x}: A \in \mathscr{F}, x \in A}. By induction we find a sunflower with k petals in this family. Adding x to each member gives the desired sunflower of \mathscr{F}. The Sunflower Lemma has found a number of applications in computer science. As we did in the chapter on Ramsey theory, we may wonder what the best possible bound is that guarantees a sunflower with k petals. Denote this number by f(s, k). We have$$ (k-1)^s<f(s, k) \leq s !(k-1)^s+1 . $$The upper bound was just proven; for the lower bound, consider the family \mathscr{F} of SDRs of a collection of s pairwise disjoint size-( (k-1) sets A_1, \ldots, A_s. A sunflower with k petals in \mathscr{F} must contain two petals using the same element x \in A_1. But each element is in either none, exactly one, or all of the members of the sunflower. It follows that all petals use x. Since we are looking at SDRs, no other element of A_1 is used by the sunflower. This can be repeated for all A_i, and we can only conclude that the sunflower has but one petal! # 组合数学代考 ## 数学代写|组合数学代写Combinatorial mathematics代考|Extremal graph theory 另一种看待定理的方式 4.0 .2 是 Ramsey 定理的密度版本 (参见上一章提到的 Szemerédi 定理和密度 Hales-Jewett 定理)。我们问在不创建红色三角形的情况下我们可以多久使用一次红色。因此，将此结 果推广到任意完整图是有意义的。为了描述极值图 (不能添加更多边的图)，我们使用 k 分图的概念: 顶点集被划分为集合的图 V_1, \ldots, V_k ，并且没有一条边的两个端点都在一个集合内 V_i. 完整的 \mathrm{k} 分图是其 中存在所有此类边的图。如果 \left|V_i\right|=n_i 为了 i \in[k], 那么这个就完成了 k-部分图表示为 K_{n_1, \ldots, n_k}. 而 且， n=n_1+\cdots+n_k ， 而如果 \left|n_i-n_j\right| \leq 1 对所有人 i, j \in[k] ，然后我们表示唯一的（直到顶点 标记) 这样的图 T_{n, k}. 证明：假设成立。我们尝试修改图形以提高边数；我们区分三种情况。 一、如果 \operatorname{deg}(u)<\operatorname{deg}(v) ，然后我们构造一个新图， G^{\prime} ，如下来自 G. 删除所有边事件 u ，并创建新 边，以便 u ，表示为 N(u) ，满足 N(u)=N(v). 注意 u 和 v 不相邻 G^{\prime} ， 因此，如果 G^{\prime} 有个 K_{k+1}-子图，那 么它最多使用其中之一 u 和 v. 但两者 G^{\prime}-u 和 G^{\prime}-v 同构于的子图 G ，因此没有 K_{k+1}. 因此 G^{\prime} 没有; 边 数满足$$ \left|E\left(G^{\prime}\right)\right|=|E(G)|-\operatorname{deg}(u)+\operatorname{deg}(v)>|E(G)|, $$与我们的选择相矛盾 G. 二。 \operatorname{deg}(u)<\operatorname{deg}(w) 通过对称性从 (I) 得出。三、如果 \operatorname{deg}(u) \geq \operatorname{deg}(v) 和 \operatorname{deg}(u) \geq \operatorname{deg}(w) ，然后 我们构造一个新图， G^{\prime}, 如下来自 G. 删除所有边事件 v, 并且所有的边与 w. 然后两者都做 v 和 w 与所有邻 居相邻 u. 我们又可以看到 G^{\prime} 没有 K_{k+1}-子图。而且，$$ \left|E\left(G^{\prime}\right)\right|=|E(G)|-\operatorname{deg}(v)-\operatorname{deg}(w)+1+2 \operatorname{deg}(u)>|E(G)|,$$在哪里$+1$重复计筧边缘$v w$. ## 数学代写|组合数学代写Combinatorial mathematics代考|Sunflowers 上一节中达到边界的集合具有非常好的结构。你可能会想，如果出现这样的结构作为奖励，放弃几组是 有用的。在本节中，我们证明了一个结果表明我们可以做到这一点。$A, B \in \mathscr{F}$和$A \neq B$我们有$A \cap B=Y$. 此外，集$A \backslash Y$，我们称之为花艧，是非空的。 4.3.2 引理 (向日菜引理)。让$\mathscr{F}$是一组集合，每个集合的大小为$\mathrm{s}$。如果$|\mathscr{F}|>s !(k-1)^s$然后$\mathscr{F}$包含 证明: 我们通过归纳法证明结果$s$. 如果$s=1$然后$|\mathscr{F}|>k-1 ，$所以$\mathscr{F}$至少包含$k$大小1 夽。这些形 成向日萡$(与 Y=\emptyset)$. 使固定$s \geq 2$，并假设结果适用于所有较小的值$s$. 让 Veft{A_1, VIdots, A_t\right } } \text { 是包含在中的成对不相交子 } 集的最大集合$\mathscr{F}$. 如果$t \geq k$然后任何$k$这些形成向日葵，所以假设$t \frac{s !(k-1)^s}{s(k-1)}=(s-1) !(k-1)^{s-1}\$

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