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# 数学代写|组合学代写Combinatorics代考|MATH3349 The Pigeonhole Principle

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## 数学代写|组合学代写Combinatorics代考|The Pigeonhole Principle

The following simple statement is known as the pigeonhole principle and may be very useful in proving the existence of some combinatorial configurations.
Theorem 10.4.1. Let $\left{A_1, A_2 \ldots A_k\right}$ be a partition of an $(n k+1)$-set $S$ into $k$ blocks. Then, there exists a block of the partition that contains at least $n+1$ elements.

Proof. Suppose, on the contrary, that $\left|A_j\right| \leqslant n$ for any $j \in{1,2, \ldots, k}$. Then, $|S|=\left|A_1\right|+\left|A_2\right|+\cdots+\left|A_k\right| \leqslant k n$, and this contradicts the condition $|S|=n k+1$.

Example 10.4.2. Suppose that 25 points are given in the plane such that for every three points, we can choose two of them that have a distance of less than 1. We shall prove that there is a circle of radius 1 such that at least 13 of the given 25 points are inside this circle.

Let $S$ be the set of given points, $A$ be an arbitrary point from $S$, and let $c_1$ be a circle with center $A$ whose radius is equal to 1 . If all points from $S$ are inside $c_1$, then the statement is obviously proved. Suppose now that there is a point $B \in S$ such that $d(A, B) \geqslant 1$, and let $c_2$ be a circle with center $B$ whose radius is equal to 1 . Every point from $S$ lies inside at least one of the circles $c_1$ and $c_2$. Indeed, if for some $C \in S, d(A, C) \geqslant 1$ and $d(B, C) \geqslant 1$, then the assumption of the problem is not satisfied for the triple of points $(A, B, C)$. Now by the pigeonhole principle it follows that at least 13 points from $S$ lie inside one of the circles $c_1$ and $c_2 . \triangle$

## 数学代写|组合学代写Combinatorics代考|Ramsey’s Theorem

Example 10.5.1. Let $S$ be an $n$-set, and let $S=S_1 \cup S_2$, where $S_1 \cap S_2=\varnothing$. If $n \geqslant q_1+q_2-1$, where $q_1, q_2 \in \mathbb{N}$, then $\left|S_1\right| \geqslant q_1$ or $\left|S_2\right| \geqslant q_2$. Indeed, if $\left|S_1\right| \leqslant q_1-1$ and $\left|S_2\right| \leqslant q_2-1$, then
$$|S|=\left|S_1 \cup S_2\right|=\left|S_1\right|+\left|S_2\right| \leqslant q_1+q_2-2,$$

and this contradicts the fact that $S$ is an $n$-set. Note that for $n<q_1+q_2-1$, the inequalities $\left|S_1\right| \leqslant q_1-1$ and $\left|S_2\right| \leqslant q_2-1$, may both be valid, as it is in the case:
$$S_1=\left{1,2, \ldots, q_1-1\right}, \quad S_2=\left{q_1, q_1+1, \ldots, q_1+q_2-2\right} . \triangle$$
An important generalization of the result of Example 10.5.1 is given by Ramsey’s theorem.

## 数学代写|组合学代写Combinatorics代考|拉姆齐定理

$$|S|=\left|S_1 \cup S_2\right|=\left|S_1\right|+\left|S_2\right| \leqslant q_1+q_2-2,$$

$$S_1=\left{1,2, \ldots, q_1-1\right}, \quad S_2=\left{q_1, q_1+1, \ldots, q_1+q_2-2\right} . \triangle$$

## MATLAB代写

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