=============================================================== Lect 14 - November 13, 2008 - ECS 20 - Fall 2008 - Phil Rogaway =============================================================== Today: o Pigeonhole principle o Induction, recursion and recurrence relations Announcements o Tomorrow discussion section will be a lecture, given by me. --------------------------- 1. The Pigeonhole Principle --------------------------- If N pigeons roost in n holes, N>n, then some two pigeons share a hole. Restated: if f: A -> B where A and B are finite sets, |A|>|B|, the f is NOT injective. Ex 0. Any room with 3 or more people has some two of the same gender. Ex 1. 20 people at a party, some two have the same number of friends. number of friends proof: 0..18 or 1..19 Ex 2: Given five points inside the square whose side is of length 2, prove that two are within \sqrt{2} of each other. Soln: divide square into four 1 x 1 cells. Diameter of each cell = \sqrt{2} Ex 3: In any list of 10 numbers, a_1, ..., a_10, there's a string of consecutive numbers whose sum is divisible by 10. Consider a_1 a_1 + a_2 ... a_1 + a_2 + ... + a_10 If any of these divisible by 10: done. Otherwise, each is congruent to 1,..., 9 mod 10, so two are congruent to the same thing, eg, a_1 + a_2 + a_3 = 6 (mod 5) a_1 + a_2 + a_3 + a_4 + a_5 = 6 (mod 5) But then a+4 + a_5 = 0 (mod 10) Ex 4, In any room of 6 people, there are 3 mutual friends or 3 mutual strangers (Ramsey theorem) Remove person 1 5 people left. Put into two pots: friends with 1, non-friends with 1. One has at least three people. If three friends: Case 1: some two know each other: DONE Case 2: no two know each other: DONE If three non-friends: ...o Difficult Puzzle: What is the minimum number of people that must assemble in a room such that there will be at least n friends or n non-friends: R(n,n) R(4,4) = 18 (1955) R(5,5) = ?? open!!! known to be between 43 (1989) and 49 (1995) R(10,10) =?? open and not tightly determined at all: range 798 (1986)- 23,556 (2002) The above example uses the ---------------------------------------------- "Strong form of Pigeonhole principle: If f: A -> B, A and B finites sets, then some point in B will have at least \lceil |A|/|B| \rceil preimages under f ---------------------------------------------- -------------------------- 2. Induction and recursion -------------------------- EXAMPLE 1. Sam's Dept Store sells enveloped in packages of 5 and 12. Prove that, for any n\ge 44, the store can sell you exactly n envelopes. [GP, p.147] Try it: 44 = 2(12) + 4(5) 45 = 9(5) 46 = 3(12) + 2(5) ?...? SUPPOSE: it is possible to buy k envelopes for some k\ge44. SHOW: it is possible to buy k+1 envelopes If purchasing at least seven packets of 5, trade them for three packets of 12: 7(5) -> 3(12) 35 36 If <7 packets of 5, ie <=6 fewer packets of 5, so at most 30 of the envelopes are in packets of 5; so there are >= 44-30 = 14 envelopes being bought in packets of 12, so >= 2 two packets of twelve. So take two of the packets of 12 and trade them for 5 packets of 5: 2(12) -> 5(5) 24 25 ----------------------------------- Principle of mathematical induction ----------------------------------- To prove a proposition P(n) for all integers n\ge n_0: 1) Prove P(n_0) (Basis) 2) Prove that P(k) --> P(k+1) for all k > n_0 (inductive step) (Inductive hypothesis) (initially make n_0=1) And that's as far as we got today!!