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# 数学代写|组合学代写Combinatorics代考|A Theorem of Ramsey

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## 数学代写|组合学代写Combinatorics代考|A Theorem of Ramsey

We now discuss without proof a profound and important generalization of the pigeonhole principle which is called Ramsey’s theorem ${ }^8$ after the English logician Frank Ramsey. ${ }^9$

The following is the most popular and easily understood instance of Ramsey’s theorem:
Of six (or more) people, either there are three, each pair of whom are acquainted, or there are three, each pair of whom are unacquainted.
One way to prove this result is to examine all the different ways in which 6 people can be acquainted and unacquainted. This is a tedious task, but nonetheless one that can be accomplished with a little fortitude. There is, however, a simple and elegant proof that avoids consideration of cases. Before giving this proof, we formulate the result more abstractly as
$$K_6 \rightarrow K_3, K_3 \quad\left(\operatorname{read} K_6 \text { arrows } K_3, K_3\right) .$$
What does this mean? First, by $K_6$ we mean a set of 6 objects (e.g., people) and all of the 15 (unordered) pairs of these objects. We can picture $K_6$ by choosing 6 points in the plane, no 3 of which are collinear, and then drawing the edge or line segment connecting each pair of points (the edges now represent the pairs). In general, we mean by $K_n$ a set of $n$ objects and all of the pairs of these objects. ${ }^{10}$ Illustrations for $K_n(n=1,2,3,4,5)$ are given in Figure 2.1. Notice that the picture of $K_3$ is that of a triangle, and we often refer to $K_3$ as a triangle.

## 数学代写|组合学代写Combinatorics代考|Four Basic Counting Principles

Consider an ordinary chessboard which is divided into 64 squares in 8 rows and 8 columns. Suppose there is available a supply of identically shaped dominoes, pieces which cover exactly two adjacent squares of the chessboard. Is it possible to arrange 32 dominoes on the chessboard so that no 2 dominoes overlap, every domino covers 2 squares, and all the squares of the chessboard are covered? We call such an arrangement a perfect cover of the chessboard by dominoes. This is an easy arrangement problem, and one quickly can construct many different perfect covers. It is difficult but nonetheless possible to count the number of different perfect covers. This number was found by Fischer $^1$ in 1961 to be $12,988,816=2^4 \times(901)^2$. The ordinary chessboard can be replaced by a more general chessboard divided into $m n$ squares lying in $m$ rows and $n$ columns. A perfect cover need not exist now. Indeed, there is no perfect cover for the 3-by-3 board. For which values of $m$ and $n$ does the $m$-by- $n$ chessboard have a perfect cover? It is not difficult to see that an $m$-by- $n$ chessboard will have a perfect cover if and only if at least one of $m$ and $n$ is even or, equivalently, if and only if the number of squares of the chessboard is even. Fischer has derived general formulae involving trigonometric functions for the number of different perfect covers for the $m$-by- $n$ chessboard. This problem is equivalent to a famous problem in molecular physics known as the dimer problem. It originated in the investigation of the absorption of diatomic atoms (dimers) on surfaces. The squares of the chessboard correspond to molecules, while the dominoes correspond to the dimers.

Consider once again the 8 -by- 8 chessboard and, with a pair of scissors, cut out two diagonally opposite corner squares, leaving a total of 62 squares. Is it possible to arrange 31 dominoes to obtain a perfect cover of this “pruned” board? Although the pruned board is very close to being the 8 -by- 8 chessboard, which has over twelve million perfect covers, it has no perfect cover. The proof of this is an example of simple but clever combinatorial reasoning. In an ordinary 8 -by- 8 chessboard the squares are alternately colored black and white, with 32 of the squares white and 32 of the squares black. If we cut out two diagonally opposite corner squares, we have removed two squares of the same color, say white. This leaves 32 black and 30 white squares. But each domino covers one black and one white square, so that 31 nonoverlapping dominoes on the board cover 31 black and 31 white squares. Therefore the pruned board has no perfect cover, and the reasoning above can be summarized by
$$31 \mathrm{~B} \mid \mathrm{W} \neq 32 \mathrm{~B}+30 \mathrm{~W} .$$

## 数学代写|组合学代写Combinatorics代考|A Theorem of Ramsey

$$K_6 \rightarrow K_3, K_3 \quad\left(\operatorname{read} K_6 \text { arrows } K_3, K_3\right) .$$

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