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Chapter 1. Pigeonhole Principle Prof. Tesler Math 184A Winter 2017 Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 1 / 12 Pigeonhole principle https://commons.wikimedia.org/wiki/File:TooManyPigeons.jpg If you put 10 pigeons into 9 holes, at least one hole has at least two pigeons. Pigeonhole Principle If you put n items into k boxes, for integers n > k > 0, then at least one box has 2 or more items. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 2 / 12 Example: Five real numbers in [0, 1] Pick five real numbers between 0 and 1: x1 , . . . , x5 ∈ [0, 1]. We will show that at least two of them differ by at most 1/4. Split [0, 1] into 4 bins of width 1/4: 0 1/4 1/2 3/4 1 Put 5 numbers into 4 bins. At least two numbers are in the same bin, and thus, are within 1/4 of each other. Technicality: We didn’t say how to handle 41 , 12 , and 34 . Is 41 in the 1st or 2nd bin? However, it doesn’t affect the conclusion. Generalization For an integer n > 2, pick n numbers between 0 and 1. 1 Then at least two of them are within n−1 of each other. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 3 / 12 Example: Picking 6 numbers from 1 to 10 Pick 6 different integers from 1 to 10. Claim: Two of them must add up to 11. Example 1, 4, 7, 8, 9, 10 has 1 + 10 = 11 and 4 + 7 = 11. Proof. Pair up the numbers from 1 to 10 as follows: (1, 10) (2, 9) (3, 8) (4, 7) (5, 6) There are 5 pairs, each summing to 11. At least two of the 6 numbers must be in the same pair. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 4 / 12 Example: 5 points on a sphere Select any five points on a sphere. The sphere may be cut in half (both halves include the boundary) so that one of the two hemispheres has at least four of the points. Pick any two of the five points. Form the great circle through those points, splitting it into two hemispheres. Three of the five points remain. One of the two hemispheres must contain at least two of these points, plus the original two points. Thus, it contains at least four of the points. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 5 / 12 Notation Let x be a real number. bxc = floor of x = largest integer that’s 6 x dxe = ceiling of x = smallest integer that’s > x b2.5c = 2 b−2.5c = −3 b2c = d2e = 2 d2.5e = 3 d−2.5e = −2 b−2c = d−2e = −2 You have probably seen [x] used for greatest integer , and you may encounter some differences in definitions for negative numbers. However, we will use the floor and ceiling notation above, and will use square brackets for [n] = {1, 2, 3, . . . , n} (for integers n > 1) and [0] = ∅. This notation is common in Combinatorics. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 6 / 12 Generalized Pigeonhole Principle Put n items into k boxes. Then There is a box with at least dn/ke items. There is a box with at most bn/kc items. Example 150 pigeons are placed into 60 holes. At least one hole has d150/60e = d2.5e = 3 or more pigeons. At least one hole has b150/60c = b2.5c = 2 or fewer pigeons. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 7 / 12 Proof of Generalized Pigeonhole Principle Put n items into k boxes. Then 1 There is a box with at least dn/ke items. 2 There is a box with at most bn/kc items. Let ai be the number of items in box i, for i = 1, . . . , k. Each ai is a nonnegative integer, and a1 + · · · + ak = n. Assume (by way of contradiction) that all boxes have fewer than dn/ke items: ai < dn/ke for all i. Thus, ai 6 dn/ke − 1 for all i. Then the total number of items placed in the boxes is a + · · · + ak 6 k · (dn/ke − 1) . | 1 {z } =n Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 8 / 12 Proof of Generalized Pigeonhole Principle 1. Show there is a box with at least dn/ke items Assume that every box has fewer than dn/ke items. Then the total number of items placed in the boxes is (∗) a + · · · + ak 6 k · (dn/ke − 1) | 1 {z } =n Dividing n by k gives n = kq + r, where the quotient, q, is an integer and the remainder, r, is in the range 0 6 r 6 k − 1. If there is no remainder (r = 0), then dn/ke = n/k = q. Inequality (∗) becomes n 6 k(q − 1) = n − k. But n − k is smaller than n, a contradiction. If there is a remainder (1 6 r 6 k − 1), then dn/ke = q + 1. Inequality (∗) becomes n 6 k((q + 1) − 1) = kq = n − r, so n 6 n − r. But n − r is smaller than n, a contradiction. Thus, our assumption that all ai < dn/ke was incorrect, so some box must have ai > dn/ke. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 9 / 12 Proof of Generalized Pigeonhole Principle 2. Show there is a box with at most bn/kc items Assume that every box has more than bn/kc items. Then the total number of items placed in the boxes is (∗) a + · · · + ak > k · (bn/kc + 1) | 1 {z } =n Dividing n by k gives n = kq + r, where the quotient, q, is an integer and the remainder, r, is in the range 0 6 r 6 k − 1. bn/kc = q, so inequality (∗) becomes n > k(q + 1) = kq + k = n − r + k, so n > n + (k − r). But k − r > 0, so n is smaller than n + (k − r), a contradiction. Thus, our assumption that all ai > bn/kc was incorrect, so some box must have ai 6 bn/kc. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 10 / 12 Example: Ten points in the unit square (0,1) (1,1) (0,0) (1,0) Pick 10 points (x1 , y1 ), . . . , (x10 , y10 ) in [0, 1] × [0, 1]. Split [0, 1] × [0, 1] into a 3 × 3 grid of 1/3 × 1/3 squares. 10 points placed into 9 squares, and 10 > 9, so some square has at least two points. The farthest apart that points can be within the same square is √ p (1/3)2 + (1/3)2 = 3 2 . Thus, at least two points are at most √ 2 3 apart. If √you pick 10 points in [0, 1] × [0, 1], there must be two of them at most 2/3 ≈ 0.4714045 apart. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 11 / 12 Example: Ten points in the unit square (0,1) (1,1) (0,0) (1,0) Pick 10 points (x1 , y1 ), . . . , (x10 , y10 ) in [0, 1] × [0, 1]. Split [0, 1] × [0, 1] into a 2 × 2 grid of 1/2 × 1/2 squares. One of the 4 squares must have at least d10/4e = d2.5e = 3 points. The farthest apart that two points can be within the same square √ p is (1/2)2 + (1/2)2 = 2 2 . √ Thus, there is a disk (filled-in circle) of diameter 2/2 with at least three of the points. If √you pick 10 points in [0, 1] × [0, 1], there is a disk of radius 2/4 ≈ 0.3535534 with at least three of the points. Prof. Tesler Ch. 1. Pigeonhole Principle Math 184A / Winter 2017 12 / 12