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Unit 2.6 Pigeonhole Principle. The Pigeonhole Principle. The pigeonhole principle, also known as Dirichlet’s box principle, is a distribution principle. It helps to solve many troublesome problems when applied in a suitable but often subtle manner. The principle
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Unit 2.6 Pigeonhole Principle
The Pigeonhole Principle The pigeonhole principle, also known as Dirichlet’s box principle, is a distribution principle. It helps to solve many troublesome problems when applied in a suitable but often subtle manner. The principle For positive integers m and n, if m pigeons occupy n pigeonholes and m > n, then at least one pigeonhole holds two or more pigeons.
The Pigeonhole Principle The validity of the principle can be argued by contradiction: Assume the contrary, i.e. no pigeonhole has two or more pigeons. Then each pigeon hole has one or no pigeons in it, and the number of pigeons in the pigeons ≤ n. This contradicts the given condition that m ≥ n, so our assumption is incorrect and conclude that at least one of the pigeonhole has two or more pigeons. Note that the principle asserts a certain distribution but does not pinpoint which pigeonhole(s) has two or more pigeons, or which two pigeons are in the same pigeonhole.
Example 1 • If there are 367 people in a birthday party, there must be at least two with the same birthday. [Sol’n] There are only 366 possible birthdays (pigeonholes) and there are 367 people (pigeons), so by the pigeonhole principle, there are two or more people with the same birthday. Note • We don’t know the actual distribution of the birthdays of the people (it may be possible that all 367 people have the same birthday). • In addition, we cannot pinpoint which two people will have the same birthday or the actual date of their birthday.
Example 2 • In the morning, John draws socks randomly from his drawer. If there are 12 pairs of socks, each pair a different colour, in the drawer, how many socks does John have to draw at most in order to get a matched pair? [Sol’n] There are 12 different colours (pigeonholes) and in order to ensure at least two socks (pigeons) have the same colour, no. of socks drawn > 12. So at most 12+1 = 13 socks have to be drawn to get a matched pair. Note We only know that at most 13 socks are required, but we do not know which colour would the matched pair be.
Example 3 • Five points are drawn randomly inside a unit square, show that there is at least two points with a distance less than . [Sol’n] Consider a unit square and cut it into 4 equal smaller squares, as shown below on the left. 1/2 1/2 There are 4 equal smaller squares (pigeonholes) and 5 points (pigeons). By the pigeonhole principle, at least two points are in the same smaller square. The distance between these two points < length of the diagonal of the smaller square. Length of the diagonal = So there is at least two points with a distance less than . Note We cannot pinpoint which two points and its actual distance.
Example 4 • Karen and Chris paint a wall in their house in two colours, red and green. Show that there exists two points of the same colour on the wall that are exactly one metre apart. [Sol’n] Construct an equilateral triangle with each side equal to 1 m on the wall. Since their three vertices (pigeons) are painted in two colours (pigeonholes), by the pigeonhole principle, there are at least two vertices with the same colour. So there exists two points on the wall that are of the same colour and are exactly 1 m apart. Note We do not know the colour of these two points nor their exact locations on the wall.
The Generalized Pigeonhole Principle • In many situations, we can strengthen the results provided by the pigeonhole principle. • Floor and Ceiling functions • For any real number x, define its floor as the largest integer that is smaller than or equal to x, i.e. x = n, where n is an integer such that • n ≤ x < (n + 1) , or equivalently • x – 1 < n ≤ x. • Similarly, we define the ceiling of x as the smallest integer that is greater than or equal to x, i.e. x = n, where n is an integer such that • (n 1) < x ≤ n, or equivalently • x ≤ n < x + 1. • Examples • 2.5 = 2; 2.5 = 3 • 6 = 6 = 6 • = 3; = 4
The Generalized Pigeonhole Principle • The Principle If n pigeons are allocated into k pigeonholes, then there is at least one pigeonhole which contains at least pigeons.
Example 5 • Show that if 9 colours are used to paint 100 houses, then at least 12 houses will be of the same colour. [Sol’n] We have 9 colours (pigeonholes) and 100 houses (pigeons). By the generalized pigeonhole principle, there is at least one colour that will be used to paint at least houses. So there is at least 12 houses painted with the same colour.
Example 6 • How many cards must be selected from a deck of 52 cards to make sure that at least 3 cards of the same suit are selected? [Sol’n] We have 4 suits (pigeonholes) and if all of them does not appear with at least 3 cards (pigeons), i.e. each suit has at most 2 cards, then the maximum number of cards that can possibly be drawn is (4)(2) = 8. So the number of cards that must be drawn to ensure at least 3 cards are of the same suit is 8+1 = 9. Alternatively, let there be n cards drawn, then according to the generalized pigeon principle, = 3 n > (4)(2) = 8. So at least 8+1= 9 cards has to be drawn.
