鴿巢原理 容斥原理

1、2006級2008. 11. 9鴿巢原理與容斥原理1Contents鴿巢原理：簡單形式鴿巢原理：加強形式容斥原理2Pigeonhole PrincipleIf 7 pigeons are to live in 6 boxes (holes), then there is at least one box containing two or more pigeons.3Dirichlet (狄里克雷)“鴿巢原理”，最先是由19世紀的德國數(shù)學家狄里克雷應用于解決問題，后人們?yōu)榱思o念他從這么平凡的事情中發(fā)現(xiàn)的規(guī)律，就把這個規(guī)律用他的名字命名，叫“狄里克雷原理”，又把它叫做“抽屜原理”。4Pigeon

2、hole Principle If we put n+1 balls into n boxes, then at least one box must contain two or more balls. 將n+1個球放入n個盒子內(nèi), 最小有一個盒子藏有2個或以上的球。5ProofWe want to prove that at least one box must contain two or more balls.反證法Assume that the statement above is wrong then all the n boxes contains 1 or 0 ball. Th

3、erefore the total number of balls is less than or equal to n. This is a contradiction (矛盾). The contradiction is due to our assumption that the statement is wrong.We conclude that the statement must be true.6Examples For any 368 people, at least two of them must have the same birthday.There are at l

4、east two people in the world having same number of hair. At least two of you in this class (assuming the class size 12) born in the same month. 7Exercise There are 10 married couples. How many of the 20 people must be selected in order to guarantee that one has selected a married couple ?Answer: 118

5、Pigeonhole Principle : Strong Form (鴿巢原理加強形式)If we put (k*n+1) balls in n boxes, then at least one box must contain k+1 or more balls.將 (k*n+1) 個球放入n個盒子內(nèi), 最小有一個盒子藏有k+1個或以上的球。9ProofWe want to prove that at least one box must contain k+1 or more balls.Assume that the statement above is wrong then all

6、the n boxes contains k or less balls. Therefore the total number of balls is less than or equal to n*k. This is a contradiction (矛盾). The contradiction is due to the assumption that the statement is wrong.We conclude that the statement must be true.10Exercise There are 90 people in a hall. Some of t

7、hem know each other, some are not. Prove that there are at least two persons in the hall who know the same number of people in the hall.What should be the holes and pigeons? How many holes are there?11Solution If there is a person in the hall who does not know any other people, then each of the othe

8、r persons in the hall may know either 0, or 1, or 2, or 3, . , or 88 people. Therefore we have 89 holes numbered 0, 1, 2, 3, . , 88, and have to distribute among 90 people.Next, assume that every person in the hall know at least one other person. Again, we have 89 holes 1, 2, 3, . , 89 and 90 people

9、.In either case, we can apply the Pigeonhole Principle to get the conclusion.12Pigeonhole Principle : Strong Form (鴿巢原理加強形式)令q1, q2, , qn為正整數(shù)。 如果將 q1 + q2 + qn n + 1個物體放入n個盒子內(nèi)，那么， 或者第一個盒子至少含有q1個物體，或者第二個盒子至少含有q2個物體，, 或者第n個盒子至少含有qn個物體。13Pigeonhole Principle : Strong Form (鴿巢原理加強形式)證明：（反證法）設將q1 + q2 + qn n + 1個物體分到n個盒子中。 如果對于每個i=1, 2, , n, 第i個盒子含有少于qi個物體， 那么所有盒子中的物體總數(shù)不超過(q1-1) + (q2-1) + (qn-1) = q1 + q2 + qn n該數(shù)比所分發(fā)的物體總數(shù)少1， 因此我們斷言， 對于某一個i=1, 2, , n, 第i個盒子至少包含 qi個物體。14Pigeonhole Principle : Strong Form (鴿巢原理加強形式)將m個物體放入n個盒子內(nèi)