1 the pigeonhole principle cs/apma 202 rosen section 4.2 aaron bloomfield

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The Pigeonhole The Pigeonhole PrinciplePrinciple

CS/APMA 202CS/APMA 202

Rosen section 4.2Rosen section 4.2

Aaron BloomfieldAaron Bloomfield

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The pigeonhole principleThe pigeonhole principle

Suppose a flock of pigeons fly into a set of Suppose a flock of pigeons fly into a set of pigeonholes to roostpigeonholes to roost

If there are more pigeons than pigeonholes, then If there are more pigeons than pigeonholes, then there must be at least 1 pigeonhole that has there must be at least 1 pigeonhole that has more than one pigeon in itmore than one pigeon in it

If If kk+1 or more objects are placed into +1 or more objects are placed into kk boxes, boxes, then there is at least one box containing two or then there is at least one box containing two or more of the objectsmore of the objects This is Theorem 1This is Theorem 1

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Pigeonhole principle examplesPigeonhole principle examples

In a group of 367 people, there must be In a group of 367 people, there must be two people with the same birthdaytwo people with the same birthday As there are 366 possible birthdaysAs there are 366 possible birthdays

In a group of 27 English words, at least In a group of 27 English words, at least two words must start with the same lettertwo words must start with the same letter As there are only 26 lettersAs there are only 26 letters

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Generalized pigeonhole principleGeneralized pigeonhole principle

If If NN objects are placed into objects are placed into kk boxes, then boxes, then there is at least one box containing there is at least one box containing NN//kk objectsobjects This is Theorem 2This is Theorem 2

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Generalized pigeonhole principle Generalized pigeonhole principle examplesexamples

Among 100 people, there are at least Among 100 people, there are at least 100/12100/12 = 9 born on the same month = 9 born on the same month

How many students in a class must there How many students in a class must there be to ensure that 6 students get the same be to ensure that 6 students get the same grade (one of A, B, C, D, or F)?grade (one of A, B, C, D, or F)? The “boxes” are the grades. Thus, The “boxes” are the grades. Thus, kk = 5 = 5 Thus, we set Thus, we set NN/5/5 = 6 = 6 Lowest possible value for Lowest possible value for NN is 26 is 26

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Rosen, section 4.2, question 4Rosen, section 4.2, question 4

A bowl contains 10 red and 10 yellow ballsA bowl contains 10 red and 10 yellow ballsa)a) How many balls must be selected to ensure 3 balls of How many balls must be selected to ensure 3 balls of

the same color?the same color? One solution: consider the “worst” caseOne solution: consider the “worst” case

Consider 2 balls of each colorConsider 2 balls of each colorYou can’t take another ball without hitting 3You can’t take another ball without hitting 3Thus, the answer is 5Thus, the answer is 5

Via generalized pigeonhole principleVia generalized pigeonhole principleHow many balls are required if there are 2 colors, and one color How many balls are required if there are 2 colors, and one color must have 3 balls?must have 3 balls?How many pigeons are required if there are 2 pigeon holes, and How many pigeons are required if there are 2 pigeon holes, and one must have 3 pigeons?one must have 3 pigeons?number of boxes: number of boxes: kk = 2 = 2We want We want NN//kk = 3 = 3What is the minimum What is the minimum NN??NN = 5 = 5

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A bowl contains 10 red and 10 yellow A bowl contains 10 red and 10 yellow ballsballs

b)b) How many balls must be selected to How many balls must be selected to ensure 3 yellow balls?ensure 3 yellow balls?

Consider the “worst” caseConsider the “worst” caseConsider 10 red balls and 2 yellow ballsConsider 10 red balls and 2 yellow balls

You can’t take another ball without hitting 3 You can’t take another ball without hitting 3 yellow ballsyellow balls

Thus, the answer is 13Thus, the answer is 13

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6 computers on a network are connected to at least 1 6 computers on a network are connected to at least 1 other computerother computerShow there are at least two computers that are have Show there are at least two computers that are have the same number of connectionsthe same number of connections

The number of boxes, The number of boxes, kk, is the number of computer , is the number of computer connectionsconnections

This can be 1, 2, 3, 4, or 5This can be 1, 2, 3, 4, or 5

The number of pigeons, The number of pigeons, NN, is the number of computers , is the number of computers That’s 6That’s 6

By the generalized pigeonhole principle, at least one By the generalized pigeonhole principle, at least one box must have box must have NN//kk objects objects

6/56/5 = 2 = 2 In other words, at least two computers must have the same In other words, at least two computers must have the same

number of connectionsnumber of connections

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Consider 5 distinct points (Consider 5 distinct points (xxii, , yyii) with integer values, where ) with integer values, where ii = 1, = 1, 2, 3, 4, 52, 3, 4, 5Show that the midpoint of at least one pair of these five points Show that the midpoint of at least one pair of these five points also has integer coordinatesalso has integer coordinates

Thus, we are looking for the midpoint of a segment from (Thus, we are looking for the midpoint of a segment from (aa,,bb) to ) to ((cc,,dd))

The midpoint is ( (The midpoint is ( (aa++cc)/2, ()/2, (bb++dd)/2 ))/2 )Note that the midpoint will be integers if Note that the midpoint will be integers if aa and and cc have the same have the same parity: are either both even or both oddparity: are either both even or both odd

Same for Same for bb and and ddThere are four parity possibilitiesThere are four parity possibilities

(even, even), (even, odd), (odd, even), (odd, odd)(even, even), (even, odd), (odd, even), (odd, odd)Since we have 5 points, by the pigeonhole principle, there must Since we have 5 points, by the pigeonhole principle, there must be two points that have the same parity possibilitybe two points that have the same parity possibility

Thus, the midpoint of those two points will have integer coordinatesThus, the midpoint of those two points will have integer coordinates

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More elegant applicationsMore elegant applications

Not going over…Not going over…

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Quick surveyQuick survey

I felt I understood the material in this I felt I understood the material in this slide set…slide set…

a)a) Very wellVery well

b)b) With some review, I’ll be goodWith some review, I’ll be good

c)c) Not reallyNot really

d)d) Not at allNot at all

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The pace of the lecture for this The pace of the lecture for this slide set was…slide set was…

a)a) FastFast

b)b) About rightAbout right

c)c) A little slowA little slow

d)d) Too slowToo slow

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How interesting was the material in How interesting was the material in this slide set? Be honest!this slide set? Be honest!

a)a) Wow! That was SOOOOOO cool!Wow! That was SOOOOOO cool!

b)b) Somewhat interestingSomewhat interesting

c)c) Rather bortingRather borting

d)d) ZzzzzzzzzzzZzzzzzzzzzz

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Today’s demotivatorsToday’s demotivators

1 the pigeonhole principle cs/apma 202 rosen section 4.2 aaron bloomfield

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