Algorithms and Discrete Mathematics 20062007

1
Algorithms and Discrete Mathematics 2006/2007

• Lecture 7
• Fundamentals of Discrete Mathematics
• Asymptotic Notation II

Ioannis Ivrissimtzis
17-Nov-2006
2
Overview of the lecture

• Review of the Big-O notation
• Examples

3
The Big-O notation

• By O(g(n)) ( read Big-O of g of n ) we denote
  the set of functions dominated by g(n).
• We commonly write
  instead of to
  denote that f(n) is a member of O(g(n)).

4
The Big-O notation

• Some orders that often occur in practice are

5
Overview of the lecture

• Review of the Big-O notation
• Examples

6
Examples

• Example 7.1 Let
• f(n)5n, g(n)n2
• where n is a positive integer.
• The table gives f(n) and g(n) for small values of
  n.

7
Examples

• We notice that f(n) g(n) for n 1,2,3,4.
• But,
• f(n) g(n) for n 5.
• To see this notice that
• n 5 n2 5n
• So, with c 1 and n0 5, we have that for all n
  n0 , f(n) cg(n).
• Therefore, g dominates f and thus f O(n2).

8
Examples

• Remark
• Other values of c,n0 would also work. For example
  
• c1, n010
• c2, n05
• c5, n01
• That is, we can increase c or n0, or decease the
  one and increase
• the other without destroying the proof. Thus, c
  and n0 are not unique.
• However, they must be fixed for the proof to be
  valid.

9
Examples

• Remark
• We proved that 5n O(n2).
• That means that as n increases, the value of f
  will be (up to a constant)
• at most n2.
• However, we can further improve the result.
• Exercise 7.1 Prove that 5n O(n).
• Hint Use the definition with c 5, n0 1.

10
Examples

• Example 7.2 Let
• f(n)4n2 17n 5 , g(n)n2
• with n a positive integer.
• We have
• f(n) 4n2 17n 5 4n2 17n2 5n2 26n2
  26g(n)
• Because the above equation holds for every
  positive integer n, if we
• choose n01 and c26 we have
• f(n) cg(n)
• Giving, f O(g), that is, f O(n2).

11
Examples

• In the same example we have,
• g(n) n2 4n2 17n 5 f(n)
• for all n 1. Using any c 1, we get
• g O(f).
• Thus, we proved that O(f) O(g).
• Remark
• The trick of replacing all the terms of the
  polynomial 4n2 17n 5 with
• n2 is quite common in such proofs. In fact, it is
  quite common in proving
• inequalities.

12
Examples

• Example 7.3 Let
• f(n)3n3 7n2 - 4n 2 , g(n)n3
• with n a positive integer.
• For all n1, we have,
• So, with n0 1, c 16 we have f(n)
  cg(n), giving, f O(g) O(n3).

13
Examples

• Remark
• In this example we had to use the absolute values
  because of the
• negative term -4n in function f.
• The inequality
  
• can be obtained with repetitive use of the
  triangle inequality
• with a,b real numbers.

14
Examples

• Remark
• You can prove the triangle inequality by
  considering separately all the
• possible cases for the signs of a,b.
• Exercise 7.2
• Using the triangle inequality show that
• where a,b,c,d are real numbers.
• Can you further generalize the result ?

15
Examples

• Example 7.4 Let f(n) be a polynomial of degree
  k, that is,
• with n a positive integer, ak, ak-1, ,a2, a1,
  a0 real numbers and ak?0.
• We have

16
Examples

• With
  and n01 we
• have f(n) cnk, giving, f O(nk).
• In words, any polynomial function is big-O its
  leading degree without
• the leading constant.

17
Examples

• Remark
• In Example 7.4, we first used a generalization of
  Exercise 7.2.
• Then, we used the property of the absolute values
  
• ab ab, for b0.
• Then, as usual, we replaced all the terms of the
  polynomial with nk.
• Remark
• If ak 0 then the degree of the polynomial is
  less than k. Therefore we
• use the constraint ak? 0 to make sure that the
  degree is indeed k.
• It is a formality commonly used for the sake of
  accuracy.

18
Examples

• Example 7.5
• Let,
• f(n) n2, g(n)5n
• where n is a positive integer. We want to show
  that g O(f).
• Notice that in all the previous examples we
  proved statements of the
• form g O(f). To do that, it was sufficient to
  find a pair c, n0 such that
• the inequality f(n) cg(n), holds for every
  n n0.
• To prove that g is not a member of O(f) we have
  to show that such pair
• c, n0 does not exist. In other words, for every
  pair c, n0 , there is a nn0
• such that f(n) gt cg(n).

19
Examples

• In this particular example we have to show that
  for every pair c, n0 there
• exist nn0 such that n2gtc5n, or equivalently
  ngtc5.
• We notice that any n gt max n0 , 5c will do,
  because it satisfies both
• nn0 and ngtc5, as required.
• Therefore, n2 O(5n).