The Pigeonhole Principle Selected Exercises The PigeonHole Principle

The Pigeonhole Principle: Selected Exercises

The Pigeon-Hole Principle Let 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 2

Exercise 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. (x 1, y 1) ( (x 1+ x 2 )/2 , (y 1+ y 2 )/2 ) (x , y ) Copyright © 2 Peter 2 Cappello 2011 3

Exercise 10 Solution 1. (xj + xi)/2 is integer if xj & xi are both odd or both even. 2. (yj + yi)/2 is integer if yj & yi are both odd or both even. 3. Put each ordered pair into 2 x 2 = 4 categories: 1. x odd, y odd 2. x odd, y even 3. x even, y odd 4. x even, y even 4. 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 4

Exercise 30 Show: If there are 100, 000 wage earners in the US who earn < $1, 000, there are 2 with the same income, to the penny. (Assume each wage earner’s income > 0. ) Copyright © Peter Cappello 5

Exercise 30 Solution Show that if there are 100, 000 wage earners in the US who earn < $1, 000, there are 2 with the same income, to the penny. Solution: 1. Assume that each income is > 0. 2. Denominate the incomes in pennies. 3. The smallest possible income is 1¢. 4. The largest possible income is $999, 999. 99 = 99, 999¢. 5. Since there are more wage earners (100, 000) than distinct income values (99, 999), at least 2 must earn the same income. Copyright © Peter Cappello 6

Exercise 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 7

Exercise 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: 1. 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 8

End Copyright © Peter Cappello 9

20 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 10