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The Growth of Functions

The Growth of Functions. Rosen 2.2. 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). Basic Rules of Logarithms.

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The Growth of Functions

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  1. The Growth of Functions Rosen 2.2

  2. logz (xy) logz (x/y) logz (xy) If x = y If x < y logz (-|x|) is undefined = logz (x) + logz (y) = logz (x) - logz (y) = ylogz (x) then logz (x) = logz (y) then logz (x) < logz (y) Basic Rules of Logarithms If logz (x) = a, then x = za

  3. 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: n2 > 100n if n > 100

  4. How to quantify growth as n gets bigger?

  5. 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.

  6. Show that 3x3+2x2+7x+9 is O(x3) Proof: We must show that  constants CN and kR such that |3x3+2x2+7x+9|  C|x3| whenever x > k. Choose k = 1 then 3x3+2x2+7x+9  3x3+2x3+7x3+9x3 = 21x3 So let C = 21. Then 3x3+2x2+7x+9  21 x3when x  1.

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

  8. 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) = 4n6 + 3n5 + 100n2 + 2 is O(n6).

  9. General Rules (cont.) • If f1(x) is O(g1(x)) and f2(x) is O(g2(x)), then f1(x)f2(x) is O(g1(x)g2(x)). • Examples: 10xlog2x is O(xlog2x) n!6n3 is O(n!n3) =O(nn+3)

  10. Examples • f(x) = 10 is O(1) • f(x) = x2 + x + 1 is O(x2) • f(x) = 2x5 + 100 x3 + xlogx is O(x5) • f(x) = 2n + n10 is O(2n) How would you prove this?

  11. Prove that n10 is O(2n ) Proof: We must show that  constants CN and kR such that |n10|  C|2n| whenever n > k. Take log2 of both expressions. log22n = nlog22 =n, log2n10 = 10log2n When is 10log2n < n? or n/log2n > 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, 264 > 6410. So, if we choose then k = 64, C = 1, then |n10|  1*|2n| whenever n > 64.

  12. 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) = n2. f is O(g). • Can we prove that g is O(f)? Formally,  constants CN and kR such that |n2|  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 n2 > C|4n|.

  13. CN, kR, n>k such that n2 > 4nC. • To prove that negation is true, start with arbitrary C and k. Must show/construct an n>k such that n2 > 4nC • Easy to satisfy n > k, then • To satisfy n2>4nC, divide both sides by n to get n>4C. Pick n = max(4C+1,k+1), which proves the negation.

  14. Is 2nO(n!)? We must show that  constants CN and kZ such that |2n|  C|n!| whenever n > k. 2n = 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.

  15. Is 2nO(n!)? Note that we could also choose k = 1 and C = 2 Since |20|  2*|0!| = 2 |21|  2*|1!| = 2 |22|  2*|2!| = 4 |23|  2*|3!|= 12

  16. Is f(x)=(x2+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 |(x2+1)/(x+1)|  |x| when x > 1

  17. Hierarchy of functions nn n 3n nlog2n n log2n 2n 1 n! n2 nn n3

  18. Hierarchy of functions 1 nn n 3n nlog2n n log2n 2n n! n2 nn n3

  19. Hierarchy of functions 1, log2n nn n 3n nlog2n n 2n n! n2 nn n3

  20. Hierarchy of functions 1, log2n, 3n nn n nlog2n n 2n n! n2 nn n3

  21. Hierarchy of functions 1, log2n, 3n, n nn n nlog2n 2n n! n2 nn n3

  22. Hierarchy of functions 1, log2n, 3n, n, n nn nlog2n 2n n! n2 nn n3

  23. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n nn 2n n! n2 nn n3

  24. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n, nn 2n n! n2 nn n3

  25. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n, nn, n2 2n n! nn n3

  26. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n, nn, n2, n3 2n n! nn

  27. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n, nn, n2, n3, 2n n! nn

  28. Hierarchy of functions 1, log2n, 3n, n, n, nlog2n, nn, n2, n3, 2n, n!, nn Each one is Big-Oh of any function to its right

  29. Prove that log10x is O(log2x) First we will prove the following lemma: Lemma: log10x = clog2x where c is a constant. Proof: Let y = log2x. Then 2y = x and log102y = log10x. log102y = ylog102 = log10x. But since y = log2x, this means that log2xlog102 = log10x. Therefore c = log102

  30. To Prove that log10x is O(log2x) We must show that  constants CN and kR such that |log10x|  C|log2x| whenever x > k. From the lemma log102log2x = log10x; so choose C = log102, k=0

  31. 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!  nn so log(n!)  log(nn)= nlogn So choose k = 1, C = 1

  32. 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.

  33. 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 ai, 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).

  34. 2( ) 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)

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

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