Discrete Mathematics with Applications MATH236 Dr. Hung P. - PowerPoint PPT Presentation

Discrete Mathematics with Applications MATH236 Dr. Hung P. Tong-Viet School of Mathematics, Statistics and Computer Science University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 Tong-Viet (UKZN) MATH236 Semester 1, 2013 1 / 20

Table of contents Pigeonhole Principle 1 One-to-one functions and permutation 2 One-to-one functions Permutations Tong-Viet (UKZN) MATH236 Semester 1, 2013 2 / 20

Pigeonhole Principle Strong Pigeonhole Principle (cont.) Theorem (Strong Pigeonhole Principle) Let n 1 , n 2 , · · · , n k be positive integers. If n 1 + n 2 + · · · + n k − k + 1 objects are placed into k boxes, then there is an integer i ∈ { 1 , 2 , · · · , k } such that the ith box contains at least n i objects. Tong-Viet (UKZN) MATH236 Semester 1, 2013 3 / 20

Pigeonhole Principle Strong Pigeonhole Principle (cont.) Example Suppose that we choose n 2 + 1 integers from the integers 1 , 2 , · · · , n . Then at least one of the integers 1 , 2 , · · · , n is chosen at least n + 1 times. Place n 2 + 1 integers into n boxes marked from 1 to n depending on its value Since n 2 + 1 = ( n + 1) + ( n + 1) + · · · + ( n + 1) − n + 1 , � �� � n terms the Strong Pigeonhole Principle implies that at least one box contains at least n + 1 objects that is, at least one of the integers 1 , 2 , · · · , n is chosen at least n + 1 times Tong-Viet (UKZN) MATH236 Semester 1, 2013 4 / 20

Pigeonhole Principle Increasing and decreasing sequences Definition Let a 1 , a 2 , · · · , a k be a sequence of real numbers A subsequence is a sequence of the form a i 1 , a i 2 , · · · , a i t where i 1 < i 2 < · · · < i t The sequence a 1 , a 2 , · · · , a k is increasing if a 1 ≤ a 2 ≤ · · · ≤ a k The sequence a 1 , a 2 , · · · , a k is decreasing if a 1 ≥ a 2 ≥ · · · ≥ a k Tong-Viet (UKZN) MATH236 Semester 1, 2013 5 / 20

Pigeonhole Principle Example Example Consider the sequence 7 , 1 , 2 , 8 , 4 , 9 , 5 , 6 , 3 . Then 1 , 2 , 6 , 3 is a subsequence But 1 , 2 , 7 , 8 is NOT a subsequence The sequence 1 , 2 , 8 , 9 is increasing The sequence 8 , 4 , 3 is decreasing Tong-Viet (UKZN) MATH236 Semester 1, 2013 6 / 20

Pigeonhole Principle Erd¨ os-Szekeres Theorem Theorem Every sequence a 1 , a 2 , · · · , a n 2 +1 of n 2 + 1 real numbers contains an increasing subsequence of length n + 1 or a decreasing subsequence of length n + 1 . Tong-Viet (UKZN) MATH236 Semester 1, 2013 7 / 20

Pigeonhole Principle Proof of Erd¨ os-Szekeres Theorem Proof. Suppose that there is no increasing subsequence of length n + 1 We will show that there exists a subsequence of length n + 1 which is decreasing For each k ∈ { 1 , 2 , · · · , n 2 + 1 } , let ℓ k be the length of a longest increasing subsequence beginning with a k By our assumption, each ℓ k is one of the numbers 1 , 2 · · · , n . Tong-Viet (UKZN) MATH236 Semester 1, 2013 8 / 20

Pigeonhole Principle Proof of Erd¨ os-Szekeres Theorem Proof. By the Strong Pigeonhole Principle, we know that n + 1 of these numbers ℓ k are equal Assume these numbers are ℓ k 1 , ℓ k 2 , · · · , ℓ k n +1 We can also assume that k 1 < k 2 < · · · < k n +1 We now claim that a k 1 ≥ a k 2 ≥ · · · ≥ a k n +1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 9 / 20

Pigeonhole Principle Proof of Erd¨ os-Szekeres Theorem Proof. Suppose to the contrary that, there is some integer i ∈ { 1 , 2 , · · · , n } such that a k i < a k i +1 Consider a sequence beginning with a i i and then taking a longest increasing subsequence starting at a k i +1 we then obtain a subsequence beginning at a i i of length greater than ℓ k i +1 this implies that ℓ k i > ℓ k i +1 , a contradiction Hence, a k 1 ≥ a k 2 ≥ · · · ≥ a k n +1 Thus a k 1 , a k 2 , · · · , a k n +1 is a decreasing subsequence of length n + 1 as required. Tong-Viet (UKZN) MATH236 Semester 1, 2013 10 / 20

One-to-one functions and permutation One-to-one functions One-to-one functions Definition Recall that a function is a relation in which each element of the domain is related to exactly one element of the range A function f is called a one-to-one or an injection if no two elements of its domain are related to the same element of its range, that is, x � = y implies that f ( x ) � = f ( y ) If f is one-to-one, then the relation f − 1 = { ( b , a ) : ( a , b ) ∈ f } is also a function Conversely, if f is not one-to-one, then f − 1 is not a function If f is an injection, then f − 1 is called the inverse function of f Tong-Viet (UKZN) MATH236 Semester 1, 2013 11 / 20

One-to-one functions and permutation One-to-one functions Examples Let f 1 = { (1 , a ) , (2 , c ) , (3 , b ) } and f 2 = { (1 , a ) , (2 , b ) , (3 , b ) } Both f 1 and f 2 are functions f 1 is one-to-one but f 2 is NOT one-to-one ( b is related to both 2 and 3) f − 1 = { ( a , 1) , ( c , 2) , ( b , 3) } 1 Tong-Viet (UKZN) MATH236 Semester 1, 2013 12 / 20

One-to-one functions and permutation One-to-one functions Examples The relation f 3 = { ( x , x 4 + 1) : x ∈ R } is a function. However, both (1 , 2) and ( − 1 , 2) are members of f 3 , so f 3 is not one-to-one The relation f 4 = { ( x , 5 x + 2) : x ∈ R } is one-to-one. f − 1 = { ( x , ( x − 2) / 5) : x ∈ R } 4 Tong-Viet (UKZN) MATH236 Semester 1, 2013 13 / 20

One-to-one functions and permutation Permutations Definition of permutations Definition Let S be a nonempty set. A permutation of S is a one-to-one function whose domain and range are S . Example Let S = { 1 , 2 , 3 } Let f be a function on S with f (1) = 1 , f (2) = 3 and f (3) = 2 . Then f is a permutation on S since it is a one-to-one function from S to S This permutation can be written in the form � 1 � 2 3 f = 1 3 2 Tong-Viet (UKZN) MATH236 Semester 1, 2013 14 / 20

One-to-one functions and permutation Permutations Permuations The top row of the matrix is the domain the bottom row is the range In general, the permutation has the form: � � 1 2 3 f = f (1) f (2) f (3) Tong-Viet (UKZN) MATH236 Semester 1, 2013 15 / 20

One-to-one functions and permutation Permutations Permutations There are six permutations of a set of three elements. � 1 � 1 � 1 � � � 2 3 2 3 2 3 f 1 = f 2 = f 3 = 1 2 3 2 3 1 3 1 2 � 1 � 1 � 1 � � � 2 3 2 3 2 3 f 4 = f 5 = f 6 = 1 3 2 3 2 1 2 1 3 Tong-Viet (UKZN) MATH236 Semester 1, 2013 16 / 20