Relations Over Asymptotic Notations

Overview of Previous Lecture Although Estimation but Useful It is not always possible to determine behaviour of an algorithm using Θ-notation. For example, given a problem with n inputs, we may have an algorithm to solve it in a.n 2 time when n is even and c.n time when n is odd. OR We may prove that an algorithm never uses more than e.n 2 time and never less than f.n time. In either case we can neither claim  (n) nor  (n 2 ) to be the order of the time usage of the algorithm. Big O and  notation will allow us to give at least partial information Mashhood's Web Family mashhoood.webs.com 2

Definition: Let X be a non-empty set and R is a relation over X then R is said to be reflexive if (a, a)  R,  a  X, Example 1: Let G be a graph. Let us define a relation R over G as if node x is connected to y then (x, y)  G. Reflexivity is satisfied over G if for every node there is a self loop. Reflexive Relation Mashhood's Web Family mashhoood.webs.com 3

Example 1 Since, 0  f(n)  cf(n)  n  n 0 = 1,if c = 1 Hence f(n) = O(f(n)) Example 2 Since, 0  cf(n)  f(n)  n  n 0 = 1, if c = 1 Hence f(n) =  (f(n)) Example 3 Since, 0  c 1 f(n)  f(n)  c 2 f(n)  n  n 0 = 1,if c 1 = c 2 = 1 Hence f(n) =  (f(n)) Note: All the relations, , , O, are reflexive Reflexivity Relations over , , O Mashhood's Web Family mashhoood.webs.com 4

Example As we can not prove that f(n) 0 Therefore 1. f(n)  o(f(n)) and 2.f(n)   (f(n)) Note : Hence small o and small omega are not reflexive relations Little o and  are not Reflexivity Relations Mashhood's Web Family mashhoood.webs.com 5

Definition: Let X be a non-empty set and R is a relation over X then R is said to be symmetric if  a, b  X, (a, b)  R  (b, a)  R Example 1: Let P be a set of all persons, and B be a relation over P such that if (x, y)  B then x is brother of y. This relation is not symmetric because (Anwer, Sadia)  B  (Saida, Brother)  B Symmetry Mashhood's Web Family mashhoood.webs.com 6

Property : prove that f(n) =  (g(n))  g(n) =  (f(n)) Proof Since f(n) =  (g(n)) i.e. f(n)   (g(n))   constants c 1, c 2 > 0 and n 0  N such that 0  c 1 g(n)  f(n)  c 2 g(n)  n  n 0 (1) (1)  0  c 1 g(n)  f(n)  c 2 g(n)  0  f(n)  c 2 g(n)  0  (1/c 2 )f(n)  g(n)(2) (1)  0  c 1 g(n)  f(n)  c 2 g(n)  0  c 1 g(n)  f(n)  0  g(n)  (1/c 1 )f(n)(3) Symmetry over  Mashhood's Web Family mashhoood.webs.com 7

From (2),(3): 0  (1/c 2 )f(n)  g(n)  0  g(n)  (1/c 1 )f(n)  0  (1/c 2 )f(n)  g(n)  (1/c 1 )f(n) Suppose that 1/c 2 = c 3, and 1/c 1 = c 4, Now the above equation implies that 0  c 3 f(n)  g(n)  c 4 f(n),  n  n 0  g(n) =  (f(n)),  n  n 0 Hence it proves that, f(n) =  (g(n))  g(n) =  (f(n)) Symmetry over  Mashhood's Web Family mashhoood.webs.com 8

Definition: Let X be a non-empty set and R is a relation over X then R is said to be transitive if  a, b, c  X, (a, b)  R  (b, c)  R  (a, c)  R Example 1: Let P be a set of all persons, and B be a relation over P such that if (x, y)  B then x is brother of y. This relation is transitive this is because (x, y)  B  (y, z)  B  (x, z)  B Example 2: Let P be a set of all persons, and F be a relation over P such that if (x, y)  F then x is father of y. Of course this relation is not a transitive because if (x, y)  F  (y, z)  F  (x, z)  F Transitivity Mashhood's Web Family mashhoood.webs.com 9

Prove the following 1.f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) 2.f(n) = O(g(n)) & g(n) = O(h(n))  f(n) = O(h(n)) 3.f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) 4.f(n) = o (g(n)) & g(n) = o (h(n))  f(n) = o (h(n)) 5.f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) Note It is to be noted that all these algorithms complexity measuring notations are in fact relations which satisfy the transitive property. Transitivity Relation over , , O, o and  Mashhood's Web Family mashhoood.webs.com 10

Property 1 f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) Proof Since f(n) =  (g(n)) i.e. f(n)   (g(n))   constants c 1, c 2 > 0 and n 01  N such that 0  c 1 g(n)  f(n)  c 2 g(n)  n  n 01 (1) 2.Now since g(n) =  (h(n)) i.e. g(n)   (h(n))   constants c 3, c 4 > 0 and n 02  N such that 0  c 3 h(n)  g(n)  c 4 h(n)  n  n 02 (2) 3.Now let us suppose that n 0 = max (n 01, n 02 ) Transitivity Relation over  Mashhood's Web Family mashhoood.webs.com 11

4.Now we have to show that f(n) =  (h(n)) i.e. we have to prove that  constants c 5, c 6 > 0 and n 0  N such that 0  c 5 h(n)  f(n)  c 6 h(n)? (2)  0  c 3 h(n)  g(n)  c 4 h(n)  0  c 3 h(n)  g(n)(3) (1)  0  c 1 g(n)  f(n)  c 2 g(n)  0  c 1 g(n)  f(n)  0  g(n)  (1/c 1 )f(n)(4) From (3) and (4),0  c 3 h(n)  g(n)  (1/c 1 )f(n)  0  c 1 c 3 h(n)  f(n) (5) Transitivity Relation over  Mashhood's Web Family mashhoood.webs.com 12

(1)  0  c 1 g(n)  f(n)  c 2 g(n)  0  f(n)  c 2 g(n)  0  (1/c 2 )f(n)  g(n)(6) (2)  0  c 3 h(n)  g(n)  c 4 h(n)  0  g(n)  c 4 h(n)(7) From (6) and (7),0  (1/c 2 )f(n)  g(n)  (c 4 )h(n)  0  (1/c 2 )f(n)  (c 4 )h(n)  0  f(n)  c 2 c 4 h(n) (8) From (5), (8), 0  c 1 c 3 h(n)  f(n)  0  f(n)  c 2 c 4 h(n) 0  c 1 c 3 h(n)  f(n)  c 2 c 4 h(n) 0  c 5 h(n)  f(n)  c 6 h(n) And hence f(n) =  (h(n))  n  n 0 Transitivity Relation over  Mashhood's Web Family mashhoood.webs.com 13

Property 2 f(n) = O(g(n)) & g(n) = O(h(n))  f(n) = O(h(n)) Proof Since f(n) = O(g(n)) i.e. f(n)  O(g(n))   constants c 1 > 0 and n 01  N such that 0  f(n)  c 1 g(n)  n  n 01 (1) 2.Now since g(n) = O(h(n)) i.e. g(n)  O(h(n))   constants c 2 > 0 and n 02  N such that 0  g(n)  c 2 h(n)  n  n 02 (2) 3.Now let us suppose that n 0 = max (n 01, n 02 ) Transitivity Relation over Big O Mashhood's Web Family mashhoood.webs.com 14

Now we have to two equations 0  f(n)  c 1 g(n)  n  n 01 (1) 0  g(n)  c 2 h(n)  n  n 02 (2) (2)  0  c 1 g(n)  c 1 c 2 h(n)  n  n 02 (3) From (1) and (3) 0  f(n)  c 1 g(n)  c 1 c 2 h(n) Now suppose that c 3 = c 1 c 2 0  f(n)  c 1 c 2 h(n) And hence f(n) = O(h(n))  n  n 0 Transitivity Relation over Big O Mashhood's Web Family mashhoood.webs.com 15

Assignment Prove that 1.f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) 2.f(n) = o (g(n)) & g(n) = o (h(n))  f(n) = o (h(n)) 3.f(n) =  (g(n)) & g(n) =  (h(n))  f(n) =  (h(n)) To be submitted within the first lecture after Mid Term Mashhood's Web Family mashhoood.webs.com 16