4. Asymptotically N o Negative Functions f(n) is asymptotically no negative if there exist n 0 N such that for every n n 0 , 0 f(n) n 0 f
5. Theta ( g(n) ) = { f : N R * | ( c 1 , c 2 R + ) ( n 0 N) ( n n 0 ) ( 0 c 1 g(n) f(n) c 2 g(n) ) } c 1 lim f(n) c 2 n g(n ) n 0 g c 1 g f c 2 g
8. For all n n 0 , Dividing by n 2 yields We have that 3/n is a decreasing sequence 3, 3/2, 1, 3/4, 3/5, 3/6, 3/7… and then 1/2 - 3/n is increasing sequence -5/2, -1, -1/2, -1/4, -1/10, 0, 1/14 that is upper bounded by 1/2 .
9. The right hand inequality can be made to hold for n 1 by choosing c 2 1/2 . Likewise the left hand inequality can be made to hold for n 7 by choosing c 1 1/14 .
10. Thus by choosing c 1 = 1/14 , c 2 = 1/2 and n 0 =7 then we can verify that
11. Big O O (g(n)) = { f : N R * | ( c R + ) ( n 0 N) ( n n 0 ) ( 0 f(n) cg(n) ) } lim f(n) c n g(n) n 0 g cg f
14. For all n n 0 , Dividing by n yields The inequality can be made to hold for n 1 by choosing c a+b . Thus by choosing c = a+b and n 0 =1 then we can verify that
16. Big Omega ( g(n) ) = { f : N R * | ( c R + ) ( n 0 N) ( n n 0 ) ( 0 cg(n) f(n) ) } c lim f(n) n g(n) n 0 g f cg
17.
18. Little o o (g(n)) = { f : N R * | ( c R + ) ( n 0 N) ( n n 0 ) ( 0 f(n) < cg (n) ) } lim f(n) = 0 n g(n) “ o(g(n)) functions that grow slower than g ” g f
19.
20. n 2 n Examples Lets show that We only have to show
22. Little Omega ( g(n) ) = { f : N R * | ( c R + ) ( n 0 N) ( n n 0 ) ( 0 c g(n) < f(n ) ) } lim f(n) = n g(n) “ (g(n)) functions that grow faster than g ” f g
23.
24. Analogy with the comparison of two real numbers Asymptotic Real Notation numbers f(n) O(g(n)) f g f(n) (g(n)) f g f(n) (g(n)) f = g f(n) o(g(n)) f < g f(n) (g(n)) f > g Trichotomy does not hold
25. Not all functions are asymptotically comparable Example f(n)=n g(n)=n (1+sin n) (1+sin(n)) [0,2]
26.
27.
28.
29.
30.
31.
32.
33. Asymptotic notation two variables O(g(m, n))= { f: N N R * | ( c R + )( m 0 , n 0 N) ( n n 0 ) ( m m 0 ) (f(m, n) c g(m, n)) }
38. Modular arithmetic For every integer a and any possible positive integer n, a mod n is the remainder of ( or residue ) of the quotient a/n a mod n = a - a / n n.
39. congruency or equivalence mod n If ( a mod n) = ( b mod n) we write a b ( mod n) and we say that a is equivalent to b modulo n or that a is congruent to b modulo n. In other words a b ( mod n) if a and b have the same remainder when divided by n . Also a b ( mod n) if and only if n is a divisor of b-a.
40. 0 1 2 3 -4 -3 -2 -1 -8 -7 -6 -5 -12 -11 -10 -9 12 13 14 15 8 9 10 11 4 5 6 7 Example (mod 4) (mod n) defines a equivalence relation in Z and produces a partitioned set called Z n = Z /n = {0,1,2,…, n -1} in which can be defined arithmetic operations a+b (mod n) a*b (mod n) [0] [1] [2] [3] Z /4 = {[0],[1],[2],[3]} = {0,1,2,3} 4+1 ( mod 4) = 1 5*2 ( mod 4) = 2
41. Polynomials Given a no negative integer d, a polynomial in n of degree d is a function p(n) of the form: Where a 1 , a 2 , …, a d are the coefficients and a d 0.
42.
43.
44.
45.
46.
47.
48.
49. A function f(n) is polylogaritmically bounded if f(n)= O ( lg k n) for some constant k. We have the following relation between polynomials and polylogarithms: then lg k n = o ( n k )
53. Functional iteration Given a function f ( n ) the i-th functional iteration of f is defined as : with I the identity function. For a particular n , we have,