COMP 9024, Class notes, 11s2, Class 1

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1 COMP 90, Class notes, 11s, Class 1 John Plaice Sun Jul 31 1::5 EST 011 In this course, you will need to know a bit of mathematics. We cover in today s lecture the basics. Some of this material is covered in Chapter of your textbook. 1 Functions You should be familiar with the following functions: constant: f(n) = c logarithmic: f(n) = log n linear: f(n) = n linearithmic: f(n) = n log n quadratic: f(n) = n cubic: f(n) = n 3 exponential: f(n) = b n You should also know these two functions floor: f(x) = x ceiling: f(x) = x Properties You should know some basic properties of the exponential and logarithmic functions. First the exponential: Proposition 1. For all a, b, c R, b 0, b a b c = b a+c b a b c = ba c ( b a ) c = b ac For a, c N, this can be proven by induction. Extending these results to Z, then Q, then R, requires deeper mathematical results. Now for the logarithm: Proposition. For all a, b, c R, b > 1, a > 0, c > 0, log b ac = log b a + log b c 1

2 b log b ac = ac = b log b a b log b c = b (log b a+log b c) log b ac = log b a + log b c Proposition 3. For all a, b, c R, b > 1, a > 0, c > 0, log b a c = log b a log b c b log b a c = a c = blog b a b log b c = b (log b a log b c) log b a c = log b a log b c Proposition. For all a, b, c R, b > 1, a > 0, c > 0, log b a c = c log b a b log b ac = a c = ( b log a) c b = b c log b a log b a c = c log b a Proposition 5. For all a, b, d R, b > 1, d > 1, a > 0, log d a = log d b log b a a = b log b a = ( d log b) log d b a = d (log d b log b a) log d a = log d b log b a

3 Proposition. For all a, b, d R, b > 1, d > 1, a > 0, b log d a = a log d b b log d a = b (log d b log b a) = b (log b a log d b) = ( b log a) log b d b = a log d b 3 Arithmetic series To prove the complexity of an algorithm, one needs to be able to count. Below are some important identities for arithmetic series. Proposition 7. For all n N, Proof by induction on n. i = n(n + 1) Case n = 0. 0 i = 0 = 0(0 + 1) Case n = N + 1. Then By the induction hypothesis, Suppose for induction that N i = N+1 i = N(N + 1) N i + (N + 1) N(N + 1) (N + 1) = + (N + 1)(N + ) = = (N + 1)( (N + 1) + 1 ) i = n(n + 1) Proposition 8. For all n N, Proof by induction on n. i = n(n + 1)(n + 1) 3

4 Case n = 0. 0 i = 0 = 0(0 + 1)( 0 + 1) Case n = N + 1. Suppose for induction that N i = N(N + 1)(N + 1) Then N+1 By the induction hypothesis, i = N i + (N + 1) N(N + 1)(N + 1) (N + 1) = + = (N + 1)(N + N + N + ) (N + 1)(N + )(N + 3) = = (N + 1)( (N + 1) + 1 )( (N + 1) + 1 ) i = n(n + 1)(n + 1) Proposition 9. For all n N, Proof by induction on n. Case n = 0. ( ) n(n + 1) i 3 = = n (n + 1) 0 i 3 = 0 = 0 (0 + 1) Case n = N + 1. Suppose for induction that N i 3 = N (N + 1)

5 Then By the induction hypothesis, N+1 i 3 = N i 3 + (N + 1) 3 = N (N + 1) + (N + 1)3 = (N + 1) (N + N + ) = (N + 1) (N + ) = (N + 1)( (N + 1) + 1 ) i 3 = n (n + 1) Geometric series Below are some important identities for arithmetic series. Proposition 10. For all a R, a 1, a i = an+1 1 a 1 a n+1 1 = a n+1 + (a n a n ) + + (a a) 1 = a a i 1 a i = (a 1) a i a i = an+1 1 a 1 Proposition 11. For all a R, 0 < a < 1, a i = 1 1 a 5

6 a i = lim a i n = lim n = 0 1 a 1 = 1 1 a a n+1 1 a 1 5 Asymptotic complexity definitions Definition 1. Let f, g : N R. We say f(n) is O(g(n)) iff there exists c R, c > 0, and n 0 N, n 1, such that n n 0, we have f(n) cg(n). Definition. Let f, g : N R. We say f(n) is Ω(g(n)) iff there exists c R, c > 0, and n 0 N, n 1, such that n n 0, we have f(n) cg(n). Definition 3. Let f, g : N R. We say f(n) is Θ(g(n)) iff there exists c, c R, c > 0, c > 0, and n 0 N, n 1, such that n n 0, we have c g(n) f(n) c g(n). We write O(1) for a constant function and O(n O(1) ) for a polynomial function. Asymptotic geometric series Proposition 1. For all c R, c > 0, g(n) = c i is Θ(1), if c < 1 Θ(n), if c = 1 Θ(c n ), if c > 1 Case c < 1. c i < c i = 1 1 c lim n g(n) = 1 1 c, so g(n) is Θ(1). Case c = 1. g(n) = n, so g(n) is Θ(n). c i = n Case c > 1. c i is a polynomial over c g(n) is Θ(c n ), because in a polynomial, the term of highest degree dominates the other terms with respect to asymptotic complexity.

7 7 Counting divide-and-conquer Let a, b, d, n N, a > 0, b > 1. We consider, as last example, a problem of size n which will be solved using the divide-and-conquer technique. At each stage, it will be divided into a subproblems, each of size n/b, and that the work at that stage is O(n d ). At level k of the subdividing, there will be a k subproblems, each of size (n/b) k. The cost for level k is therefore ( ( n ) ) ( ) d n O b k a k d b dk a k ( n d) ) k b d The total number of levels before getting down to subproblems of size 1 is. The total cost is therefore O ( n d) ( n d ) Case a < b d. is Θ(1) the total cost is O(n d ). Case a = b d. is Θ(logb n) the total cost is Case a > b d. O ( n d ) ( n d log n ) is Θ ( b d ) logb n ) the total cost is ( ( O n d a ) ) logb n b d (n d (b d ) log b ( (n n d a ( b ) d ( )) a (n d ) ( n log b a) n d )) )) 7

8 8 Basic sorts We had a brief look at the following sorts. They will be presented in more detail, with code, in future classes. bogosort: Ω(n), O( ). Best case: sorted input. bubble sort: Ω(n), O(n ). Best case: sorted input. Worst case: reverse sorted input. selection sort: Θ(n ). insertion sort: Ω(n), O(n ). Best case: reverse sorted input. Worst case: sorted input. merge sort: Θ(n log n). quicksort: Ω(n log n), O(n ). Worst case: sorted input. Quicksort can be improved by taking as pivot the median of three randomly chosen inputs. In practice, quicksort is used for large data sets, and insertion sort for small data sets. 8