Question

1

Prove the following using the definitions of the notations, or disprove with a specific counterexample: Theta(g(n)) = O(g(n)) Ohm(g(n)) Theta(alpha g(n) = Theta(g(n)), alpha > 0 If f(n) O(g(n)), then g(n) Ohm(f(n)). For any two non-negative functions f(n) and g(n), either f(n) Ohm(g(n)), or f(n) < O(g(n))

Answer 1

Solution Proof: For a given function g(n), O(g(n) can be denoted as 0(g(n)-(f(n: c and c2 denotes positive constants and n0 so that 0s cig(n) s c2g(n) for all n no A function f(n) belongs to 0(g(n)) if ci and c2 are positive constants such that it could be sandwiched in between cig(n) and c2g(n) For a given function g(n), O(g(n)) can be denoted as O(g(n)-(f(n): c,no are positive constants which exists, such that 0 s f(n) s cg(n) for all O-notation gives an upper bound on the function For a given function g(n), 2(g(n)) can be denoted as 2(g(n))-(f(n): c,no are positive constants which exists, such that 0 all n no cg(n) s f(n) for -notation provides a lower bound on the function For any two functions f(n) and g(n), we have f(n)-(g(n)) if and only if f(n) -O(g(n)) and f(n)-Ω(g(n)) . Since O-notation denotes upper bound of function and Ω-notation denotes lower bound of function. So both lower bound and upper bound are needed to create a tight bound, that is e-notation. As an example, we know that an2+bntc e(n2) for any constants a,b and c, which implies that an2+bntc n2) and an2 +bntc O(n2)