1 / 28

Growth Rates

Growth Rates. Growth rates of functions: Linear  n Quadratic  n 2 Cubic  n 3 In a log-log chart, the slope of the line corresponds to the growth rate of the function. Constant Factors. The growth rate is not affected by constant factors or lower-order terms Examples

phoebe
Download Presentation

Growth Rates

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Growth Rates • Growth rates of functions: • Linear  n • Quadratic  n2 • Cubic  n3 • In a log-log chart, the slope of the line corresponds to the growth rate of the function Analysis of Algorithms

  2. Constant Factors • The growth rate is not affected by • constant factors or • lower-order terms • Examples • 102n+105is a linear function • 105n2+ 108nis a quadratic function Analysis of Algorithms

  3. Big-Oh Notation • Given functions f(n) and g(n), we say that f(n) is O(g(n))if there are positive constantsc and n0 such that f(n)cg(n) for n n0 • Example: 2n+10 is O(n) • 2n+10cn • (c 2) n  10 • n  10/(c 2) • Pick c = 3 and n0 = 10 Analysis of Algorithms

  4. Big-Oh Example • Example: the function n2is not O(n) • n2cn • n c • The above inequality cannot be satisfied since c must be a constant Analysis of Algorithms

  5. More Big-Oh Examples • 7n-2 7n-2 is O(n) need c > 0 and n0 1 such that 7n-2  c•n for n  n0 this is true for c = 7 and n0 = 1 • 3n3 + 20n2 + 5 3n3 + 20n2 + 5 is O(n3) need c > 0 and n0 1 such that 3n3 + 20n2 + 5  c•n3 for n  n0 this is true for c = 4 and n0 = 21 • 3 log n + log log n 3 log n + log log n is O(log n) need c > 0 and n0 1 such that 3 log n + log log n  c•log n for n  n0 this is true for c = 4 and n0 = 2 Analysis of Algorithms

  6. Big-Oh and Growth Rate • The big-Oh notation gives an upper bound on the growth rate of a function • The statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n) • We can use the big-Oh notation to rank functions according to their growth rate Analysis of Algorithms

  7. Big-Oh Rules • If is f(n) a polynomial of degree d, then f(n) is O(nd), i.e., • Drop lower-order terms • Drop constant factors • Use the smallest possible class of functions • Say “2n is O(n)”instead of “2n is O(n2)” • Use the simplest expression of the class • Say “3n+5 is O(n)”instead of “3n+5 is O(3n)” Analysis of Algorithms

  8. Relatives of Big-Oh • big-Omega • f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 • big-Theta • f(n) is (g(n)) if there are constants c’ > 0 and c’’ > 0 and an integer constant n0  1 such that c’•g(n)  f(n)  c’’•g(n) for n  n0 • little-oh • f(n) is o(g(n)) if, for any constant c > 0, there is an integer constant n0 > 0 such that f(n) < c•g(n) for n  n0 • little-omega • f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0 > 0 such that f(n) > c•g(n) for n  n0 Analysis of Algorithms

  9. Intuition for Asymptotic Notation Big-Oh • f(n) is O(g(n)) if f(n) is asymptotically less than or equal to g(n) big-Omega • f(n) is (g(n)) if f(n) is asymptotically greater than or equal to g(n) big-Theta • f(n) is (g(n)) if f(n) is asymptotically equal to g(n) little-oh • f(n) is o(g(n)) if f(n) is asymptotically strictly less than g(n) little-omega • f(n) is (g(n)) if is asymptotically strictly greater than g(n) Analysis of Algorithms

  10. Example Uses of the Relatives of Big-Oh • 5n2 is (n2) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 let c = 5 and n0 = 1 • 5n2 is (n) f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1 such that f(n)  c•g(n) for n  n0 let c = 1 and n0 = 1 • 5n2 is (n) f(n) is (g(n)) if, for any constant c > 0, there is an integer constant n0 > 0 such that f(n) > c•g(n) for n  n0 need 5n02 > c•n0  given c, the n0 that satifies this is n0 > c/5 > 0 Analysis of Algorithms

  11. Divide-and conquer is a general algorithm design paradigm: Divide: divide the input data S in two or more disjoint subsets S1, S2, … Recur: solve the subproblems recursively Conquer: combine the solutions for S1,S2, …, into a solution for S The base case for the recursion are subproblems of constant size Analysis can be done using recurrence equations Divide-and-Conquer Analysis of Algorithms

  12. Merge-sort on an input sequence S with n elements consists of three steps: Divide: partition S into two sequences S1and S2 of about n/2 elements each Recur: recursively sort S1and S2 Conquer: merge S1and S2 into a unique sorted sequence Merge-Sort Review AlgorithmmergeSort(S, C) Inputsequence S with n elements, comparator C Outputsequence S sorted • according to C ifS.size() > 1 (S1, S2)partition(S, n/2) mergeSort(S1, C) mergeSort(S2, C) Smerge(S1, S2) Analysis of Algorithms

  13. Recurrence Equation Analysis • The conquer step of merge-sort consists of merging two sorted sequences, each with n/2 elements and implemented by means of a doubly linked list, takes at most bn steps, for some constant b. • Likewise, the basis case (n< 2) will take at b most steps. • Therefore, if we let T(n) denote the running time of merge-sort: • We can therefore analyze the running time of merge-sort by finding a closed form solution to the above equation. • That is, a solution that has T(n) only on the left-hand side. Analysis of Algorithms

  14. Iterative Substitution • In the iterative substitution, or “plug-and-chug,” technique, we iteratively apply the recurrence equation to itself and see if we can find a pattern: • Note that base, T(n)=b, case occurs when 2i=n. That is, i = log n. • So, • Thus, T(n) is O(n log n). Analysis of Algorithms

  15. The Recursion Tree • Draw the recursion tree for the recurrence relation and look for a pattern: Total time = bn + bn log n (last level plus all previous levels) Analysis of Algorithms

  16. Guess-and-Test Method • In the guess-and-test method, we guess a closed form solution and then try to prove it is true by induction: • Guess: T(n) < cn log n. • Wrong: we cannot make this last line be less than cn log n Analysis of Algorithms

  17. Guess-and-Test Method, Part 2 • Recall the recurrence equation: • Guess #2: T(n) < cn log2 n. • if c > b. • So, T(n) is O(n log2 n). • In general, to use this method, you need to have a good guess and you need to be good at induction proofs. Analysis of Algorithms

  18. Master Method • Many divide-and-conquer recurrence equations have the form: • The Master Theorem: Analysis of Algorithms

  19. Master Method, Example 1 • The form: • The Master Theorem: • Example: Solution: logba=2, so case 1 says T(n) is O(n2). Analysis of Algorithms

  20. Master Method, Example 2 • The form: • The Master Theorem: • Example: Solution: logba=1, so case 2 says T(n) is O(n log2 n). Analysis of Algorithms

  21. Master Method, Example 3 • The form: • The Master Theorem: • Example: Solution: logba=0, so case 3 says T(n) is O(n logn). Analysis of Algorithms

  22. Master Method, Example 4 • The form: • The Master Theorem: • Example: Solution: logba=3, so case 1 says T(n) is O(n3). Analysis of Algorithms

  23. Master Method, Example 5 • The form: • The Master Theorem: • Example: Solution: logba=2, so case 3 says T(n) is O(n3). Analysis of Algorithms

  24. Master Method, Example 6 • The form: • The Master Theorem: • Example: (binary search) Solution: logba=0, so case 2 says T(n) is O(log n). Analysis of Algorithms

  25. Master Method, Example 7 • The form: • The Master Theorem: • Example: (heap construction) Solution: logba=1, so case 1 says T(n) is O(n). Analysis of Algorithms

  26. Iterative “Proof” of the Master Theorem • Using iterative substitution, let us see if we can find a pattern: • We then distinguish the three cases as • The first term is dominant • Each part of the summation is equally dominant • The summation is a geometric series Analysis of Algorithms

  27. Integer Multiplication • Algorithm: Multiply two n-bit integers I and J. • Divide step: Split I and J into high-order and low-order bits • We can then define I*J by multiplying the parts and adding: • So, T(n) = 4T(n/2) + n, which implies T(n) is O(n2). • But that is no better than the algorithm we learned in grade school. Analysis of Algorithms

  28. An Improved Integer Multiplication Algorithm • Algorithm: Multiply two n-bit integers I and J. • Divide step: Split I and J into high-order and low-order bits • Observe that there is a different way to multiply parts: • So, T(n) = 3T(n/2) + n, which implies T(n) is O(nlog23), by the Master Theorem. • Thus, T(n) is O(n1.585). Analysis of Algorithms

More Related