We think you have liked this presentation. If you wish to download it, please recommend it to your friends in any social system. Share buttons are a little bit lower. Thank you!
Presentation is loading. Please wait.
Published byJaylin Bibby
Modified over 2 years ago
The Pigeonhole Principle: Selected Exercises
The Pigeon-Hole PrincipleLet k be a positive integer. If k + 1 or more objects are placed in k boxes, Then at least 1 box contains 2 or more objects. Copyright © Peter Cappello
Copyright © Peter Cappello 2011Exercise 10 Let ( xi, yi ), i = 1, 2, 3, 4, 5, be a set of 5 distinct points with integer coordinates in the xy plane. Show that the midpoint of the line joining at least 1 pair of these points has integer coordinates. (x1, y1) ( (x1+ x2 )/2 , (y1+ y2 )/2 ) (x2, y2) Copyright © Peter Cappello 2011
Copyright © Peter CappelloExercise 10 Solution (xj + xi)/2 is integer if xj & xi are both odd or both even. (yj + yi)/2 is integer if yj & yi are both odd or both even. Put each ordered pair into 2 x 2 = 4 categories: x odd, y odd x odd, y even x even, y odd x even, y even With 5 distinct ordered pairs, at least 1 category has ≥ 2 points. A line connecting 2 points in the same category has a midpoint with integer coordinates. Copyright © Peter Cappello
Copyright © Peter CappelloExercise 30 Show: If there are 100,000,000 wage earners in the US who earn < $1,000,000, there are 2 with the same income, to the penny. (Assume each wage earner’s income > 0.) Copyright © Peter Cappello
Copyright © Peter CappelloExercise 30 Solution Show that if there are 100,000,000 wage earners in the US who earn < $1,000,000, there are 2 with the same income, to the penny. Solution: Assume that each income is > 0. Denominate the incomes in pennies. The smallest possible income is 1¢. The largest possible income is $999, = 99,999,999¢. Since there are more wage earners (100,000,000) than distinct income values (99,999,999), at least 2 must earn the same income. Copyright © Peter Cappello
Copyright © Peter CappelloExercise 40 There are 51 houses on a street. Each house has an integer address between 1000 and 1099, inclusive. Show that at least 2 houses have consecutive addresses. Copyright © Peter Cappello
Copyright © Peter CappelloExercise 40 Solution There are 51 houses on a street. Each house has an integer address between 1000 and 1099, inclusive. Show that at least 2 houses have consecutive addresses. Solution: Partition the address space into 50 intervals: [1000, 1001], [1002, 1003], …, [1098, 1099]. Associate each of the 51 addresses with an interval. There are more addresses than intervals. At least 1 interval is associated with 2 distinct addresses. Copyright © Peter Cappello
Copyright © Peter CappelloEnd Copyright © Peter Cappello
Copyright © Peter Cappello20 Find an increasing subsequence of maximal length and a decreasing subsequence of maximal length in the sequence 22, 5, 7, 2, 23, 10, 15, 21, 3, 17. Solution: Brute force. 5, 7, 10, 15, 21 is an increasing subsequence of length 5. 22, 10, 3 is a decreasing subsequence of length 3 (not unique). Copyright © Peter Cappello
Recursive Definitions & Structural Induction: Selected Exercises
The Growth of Functions: Selected Exercises
Recurrence Relations: Selected Exercises
Generalized Permutations & Combinations: Selected Exercises.
The Basics of Counting: Selected Exercises. Copyright © Peter Cappello2 Sum Rule Example There are 3 sizes of pink shirts & 7 sizes of blue shirts. How.
The Pigeonhole Principle
Inclusion-Exclusion Selected Exercises
1 Let’s Recapitulate. 2 Regular Languages DFAs NFAs Regular Expressions Regular Grammars.
The Pigeon hole problem
Who Wants To Be A Millionaire? Decimal Edition Question 1.
Half Life. The half-life of a quantity whose value decreases with time is the interval required for the quantity to decay to half of its initial value.
1 The Pigeonhole Principle CS 202 Epp section 7.3.
1 The Pigeonhole Principle CS 202 Epp section ??? Aaron Bloomfield.
Pigeonhole Principle. If n pigeons fly into m pigeonholes and n > m, then at least one hole must contain two or more pigeons A function from one finite.
1 Chapter 4 Greedy Algorithms Slides by Kevin Wayne. Copyright © 2005 Pearson-Addison Wesley. All rights reserved.
Lecture 4 4.1,4.2 Counting. 4.1 Counting Two Important Principles: Product Rule and Sum Rule. Product Rule: Assume we need to perform procedure 1 AND.
Consecutive Integers Consecutive Integers are integers that follow one after another. Example 1: are consecutive integers.
CMPS 2433 Chapter 8 Counting Techniques Midwestern State University Dr. Ranette Halverson.
Powerpoint Jeopardy Category 1Category 2Category 3Category 4Category
Multiplying Up. Category 1 4 10 4 5 4 4 56 ÷ 4.
Discrete Mathematics Math 6A Homework 3 Solution.
The Pigeonhole Principle Alan Kaylor Cline. The Pigeonhole Principle Statement Children’s Version: “If k > n, you can’t stuff k pigeons in n holes without.
22C:19 Discrete Math Counting Fall 2011 Sukumar Ghosh.
LO: Count up to 100 objects by grouping them and counting in 5s 10s and 2s. Mrs Criddle: Westfield Middle School.
The Pigeonhole Principle. Pigeonhole principle The pigeonhole principle : If k is a positive integer and k+1 or more objects are placed into k boxes,
Fraction X Adding Unlike Denominators
C) between 18 and 27. D) between 27 and 50.
Tutorial 2: First Order Logic and Methods of Proofs
Graphing y = nx2 Lesson
a*(variable)2 + b*(variable) + c
COMP 482: Design and Analysis of Algorithms
1 Copyright © 2010, Elsevier Inc. All rights Reserved Fig 2.1 Chapter 2.
Counting. Counting = Determining the number of elements of a finite set.
Least Common Multiples and Greatest Common Factors
1 Functions and Applications
11.2 Arithmetic Sequences & Series
Applications of Inclusion-Exclusion: Selected Exercises.
25 seconds left…...
Copyright © 2014 by McGraw-Hill Higher Education. All rights reserved.
DIVIDING INTEGERS 1. IF THE SIGNS ARE THE SAME THE ANSWER IS POSITIVE 2. IF THE SIGNS ARE DIFFERENT THE ANSWER IS NEGATIVE.
Fraction XII Subtracting Unlike Denominators
Chapter 5 Relations and Functions
Business Transaction Management Software for Application Coordination 1 Business Processes and Coordination.
1 Decidability continued…. 2 Theorem: For a recursively enumerable language it is undecidable to determine whether is finite Proof: We will reduce the.
Year 3 Objectives: Number NUMBER AND PLACE VALUE Objective 1: Read and write numbers up to 1000 in numerals and words Read and write all numbers to 100.
What is the best way to start? 1.Plug in n = 1. 2.Factor 6n 2 + 5n Let n be an integer. 4.Let n be an odd integer. 5.Let 6n 2 + 5n + 4 be an odd.
© 2017 SlidePlayer.com Inc. All rights reserved.