Chapter 5.2
Chapter 5.2The Pigeonhole Principle1The pigeonhole
principleSuppose a flock of pigeons fly into a set of pigeonholes
to roostIf there are more pigeons than pigeonholes, then there must
be at least 1 pigeonhole that has more than one pigeon in itIf k+1
or more objects are placed into k boxes, then there is at least one
box containing two or more of the objects
2Pigeonhole principle examplesIn a group of 367 people, there
must be two people with the same birthdayAs there are 366 possible
birthdays
In a group of 27 English words, at least two words must start
with the same letterAs there are only 26 letters
3Pigeonhole principle examplesShow that for every integer n
there is a multiple of n that has only Os and I s in its decimal
expansion.Solution: Let n be a positive integer. Consider the n + 1
integers 1, 11, 111, ... , 11 ... 1 (where the last integer in this
list is the integer with n + 1 I s in its decimal expansion). There
are n possible remainders when an integer is divided by n. Because
there are n + 1 integers in this list, by the pigeonhole principle
there must be two with the same remainder when divided by n. The
larger of these integers less the smaller one is a multiple of n,
which has a decimal expansion consisting entirely of Os and 1
s.Generalized pigeonhole principleIf N objects are placed into k
boxes, then there is at least one box containing N/k objectsProof
by contradiction: Suppose that none of the boxes contains more than
N/k-1 objects. Then, the total number of objects is at most:
*
5Generalized pigeonhole principleAmong 100 people, there are at
least 100/12 = 9 born on the same month
How many students in a class must there be to ensure that 6
students get the same grade (one of A, B, C, D, or F)?The boxes are
the grades. Thus, k = 5Thus, we set N/5 = 6Lowest possible value
for N is 266ExampleA bowl contains 10 red and 10 yellow ballsHow
many balls must be selected to ensure 3 balls of the same color?One
solution: consider the worst caseWe can select all yellow balls
before we select a single red ballThe next two balls will be red,
we may need to select 13 ballsVia generalized pigeonhole
principleHow many balls are required if there are 2 colors, and one
color must have 3 balls?How many pigeons are required if there are
2 pigeon holes, and one must have 3 pigeons?number of boxes: k = 2,
We want N/k = 3What is the minimum N?N = 57More elegant
applicationsShow that among any n+1 positive integers not exceeding
2n, there must be an integer that divides one of the other
integers.Let a1, a2, , an+1 be such integers (increasing).For any
j, aj could be written as 2kjqj where qj is odd.The integers q1,
q2, , qn+1 are all odd positive integers less than 2n.Since there
are only n odd numbers less than 2n, qi = qj by the Pigeonhole
principle.Let q be qi = qj. Then ai = 2kiq and aj = 2kjq. If ki
