1. The Pigeonhole Principle: Selected Exercises

2. 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.
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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.
(x1, y1)
( (x1+ x2 )/2 , (y1+ y2 )/2 )
(x2, y2)
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Exercise 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.
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Exercise 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.)
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Exercise 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.
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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.
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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:
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.
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End
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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).
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