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Lect 14 - November 13, 2008 - ECS 20 - Fall 2008 - Phil Rogaway
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Today: o Pigeonhole principle
o Induction, recursion and recurrence relations
Announcements
o Tomorrow discussion section will be a lecture, given by me.
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1. The Pigeonhole Principle
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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
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"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
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2. Induction and recursion
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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
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Principle of mathematical induction
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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!!