2. The Growth of Functions Rosen 2.2

3. Basic Rules of Logarithms log z (xy) log z (x/y) log z (x y ) If x = y If x < y log z (-|x|) is undefined = log z (x) + log z (y) = log z (x) - log z (y) = ylog z (x) then log z (x) = log z (y) then log z (x) < log z (y) If log z (x) = a, then x = z a

4. Growth If f is a function from Z or R to R, how can we quantify the rate of growth and compare rates of growth of different functions? Possible problem: Whether f(n) or g(n) is larger at any point may depend on value of n. For example: n 2 > 100n if n > 100

5. How to quantify growth as n gets bigger?

6. Big-O Notation Let f and g be functions from the set of integers or the set of real numbers to the set of real numbers. We say that f(x) is O(g(x)) if there are constants C  N and k  R such that |f(x)|  C|g(x)| whenever x > k. We say “f(x) is big-oh of g(x)”. The intuitive meaning is that as x gets large, the values of f(x) are no larger than a constant time the values of g(x), or f(x) is growing no faster than g(x). The supposition is that x gets large, it will approach a simplified limit.

7. Show that 3x 3 +2x 2 +7x+9 is O(x 3 ) Proof: We must show that  constants C  N and k  R such that |3x 3 +2x 2 +7x+9|  C|x 3 | whenever x > k. Choose k = 1 then 3x 3 +2x 2 +7x+9  3x 3 +2x 3 +7x 3 +9x 3 = 21x 3 So let C = 21. Then 3x 3 +2x 2 +7x+9  21 x 3 when x  1.

8. Show that n! is O(n n ) Proof: We must show that  constants C  N and k  R such that |n!|  C|n n | whenever n > k. n! = n(n-1)(n-2)(n-3)…(3)(2)(1)  n(n)(n)(n)…(n)(n)(n) n times =n n So choose k = 0 and C = 1

9. General Rules Multiplication by a constant does not change the rate of growth. If f(n) = kg(n) where k is a constant, then f is O(g) and g is O(f). The above means that there are an infinite number of pairs C,k that satisfy the Big-O definition. Addition of smaller terms does not change the rate of growth. If f(n) = g(n) + smaller order terms, then f is O(g) and g is O(f). Ex.: f(n) = 4n 6 + 3n 5 + 100n 2 + 2 is O(n 6 ).

10. General Rules (cont.) If f 1 (x) is O(g 1 (x)) and f 2 (x) is O(g 2 (x)), then f 1 (x)f 2 (x) is O(g 1 (x)g 2 (x)). Examples: 10xlog 2 x is O(xlog 2 x) n!6n 3 is O(n!n 3 ) =O(n n+3 )

11. Examples f(x) = 10 is O(1) f(x) = x 2 + x + 1 is O(x 2 ) f(x) = 2x 5 + 100 x 3 + xlogx is O(x 5 ) f(x) = 2 n + n 10 is O(2 n ) How would you prove this?

12. Prove that n 10 is O(2 n ) Proof: We must show that  constants C  N and k  R such that |n 10 |  C|2 n | whenever n > k. Take log 2 of both expressions. log 2 2 n = nlog 2 2 =n,log 2 n 10 = 10log 2 n When is 10log 2 n 10? 2/1 = 2,4/2 = 2,8/3  2.67,16/4 = 4 32/5 = 6.2,64/6  10.67 For n = 64, 2 64 > 64 10. So, if we choose then k = 64, C = 1, then |n 10 |  1*|2 n | whenever n > 64.

13. Example: Big-Oh Not Symmetric Order matters in big-oh. Sometimes f is O(g) and g is O(f), but in general big-oh is not symmetric. Consider f(n) = 4n and g(n) = n 2. f is O(g). Can we prove that g is O(f)? Formally,  constants C  N and k  R such that |n 2 |  C|4n| whenever n > k? No. To show this, we must prove that negation is true for all C and k.  C  N,  k  R,  n>k such that n 2 > C|4n|.

14.  C  N,  k  R,  n>k such that n 2 > 4nC. To prove that negation is true, start with arbitrary C and k. Must show/construct an n>k such that n 2 > 4nC Easy to satisfy n > k, then To satisfy n 2 >4nC, divide both sides by n to get n>4C. Pick n = max(4C+1,k+1), which proves the negation.

16. Is 2 n O(n!)? We must show that  constants C  N and k  Z such that |2 n |  C|n!| whenever n > k. 2 n = 2(2)(2)…(2)(2)(2) n times  n(n-1)(n-2)…(3)(2)(1) =n! if n = 4 So let C = 1 and k = 3.

17. Is 2 n O(n!)? Note that we could also choose k = 1 and C = 2 Since |2 0 |  2*|0!| = 2 |2 1 |  2*|1!| = 2 |2 2 |  2*|2!| = 4 |2 3 |  2*|3!|= 12

18. Is f(x)=(x 2 +1)/(x+1) O(x)? We must show that  constants C  N and k  R such that |f(x)|  C|x| whenever x > k. When x > 1 (Why?) Therefore let k=1, C = 1 |(x 2 +1)/(x+1)|  |x| when x > 1

19. Hierarchy of functions 1 log 2 n 3n 3n nn n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

20. Hierarchy of functions 1 log 2 n 3n 3n nn n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

21. Hierarchy of functions 1, log 2 n 3n 3n nn n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

22. Hierarchy of functions 1, log 2 n, 3  n nn n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

23. Hierarchy of functions 1, log 2 n, 3  n,  n n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

24. Hierarchy of functions 1, log 2 n, 3  n,  n, n nlog 2 n nnnn n2n2 n3n3 2n2n n! n

25. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n nnnn n2n2 n3n3 2n2n n! n

26. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n, n  n n2n2 n3n3 2n2n n! n

27. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n, n  n, n 2 n3n3 2n2n n! n

28. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n, n  n, n 2, n 3 2n2n n! n

29. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n, n  n, n 2, n 3, 2 n n! n

30. Hierarchy of functions 1, log 2 n, 3  n,  n, n, nlog 2 n, n  n, n 2, n 3, 2 n, n!, n n Each one is Big-Oh of any function to its right

31. Prove that log 10 x is O(log 2 x) First we will prove the following lemma: Lemma: log 10 x = clog 2 x where c is a constant. Proof: Let y = log 2 x. Then 2 y = x and log 10 2 y = log 10 x. log 10 2 y = ylog 10 2 = log 10 x. But since y = log 2 x, this means that log 2 xlog 10 2 = log 10 x. Therefore c = log 10 2

32. To Prove that log 10 x is O(log 2 x) We must show that  constants CN CN and kR kR such that |log 10 x|  C|log 2 x| whenever x > k. From the lemma log 10 2log 2 x = log 10 x; so choose C = log 10 2, k=0

33. Prove log(n!) is O(nlogn) We must show that  constants C  N and k  R such that |logn!|  C|nlogn| whenever x > k. We know that n!  n n so log(n!)  log(n n )= nlogn So choose k = 1, C = 1

34. Time Complexity We can use Big-O to find the time complexity of algorithms (i.e., how long they take in terms of the number of operations performed). There are two types of complexity normally considered. Worst-case complexity. The largest number of operations needed for a problem of a given size in the worst case. The number of operations that will guarantee a solution. Average-case complexity. The average number of operations used to solve a problem over all inputs of a given size.

35. Complexity of the Linear Search Algorithm Worst-case complexity The algorithm loops over all the values in a list of n values. –At each step, two comparisons are made. One to see whether the end of the loop is reached, and one to compare the search element x with the element in the list. –If x is equal to list element a i, then 2i comparisons are made. –If x is not in the list, then 2n comparisons are made. –The worst-case complexity is thus 2n and is O(n).

36. Complexity of the Linear Search Algorithm Average-case complexity The algorithm loops over all the values in a list of n values. –At each step, two comparisons are again made. –On average, the number of comparisons is 2 + 4 + 6 +….+ (2n) n What’s the numerator? Average case is O(n) 2 ( )

37. Complexity of Pair-wise Correlation Assume that there are n elements to correlate