1 / 16

CSC401 – Analysis of Algorithms Lecture Notes 11 Divide-and-Conquer

CSC401 – Analysis of Algorithms Lecture Notes 11 Divide-and-Conquer. Objectives: Introduce the Divide-and-conquer paradigm Review the Merge-sort algorithm Solve recurrence equations Iterative substitution Recursion trees Guess-and-test The master method

dillan
Download Presentation

CSC401 – Analysis of Algorithms Lecture Notes 11 Divide-and-Conquer

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. CSC401 – Analysis of Algorithms Lecture Notes 11Divide-and-Conquer • Objectives: • Introduce the Divide-and-conquer paradigm • Review the Merge-sort algorithm • Solve recurrence equations • Iterative substitution • Recursion trees • Guess-and-test • The master method • Discuss integer and matrix multiplications

  2. 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

  3. 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 7 2  9 4 2 4 7 9 7  2 2 7 9  4 4 9 7 7 2 2 9 9 4 4 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)

  4. 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.

  5. 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).

  6. 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)

  7. 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) < cnlogn. • Wrong: we cannot make this last line be less than cn log n

  8. 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.

  9. Master Method • Many divide-and-conquer recurrence equations have the form: • The Master Theorem:

  10. Master Method, Examples • The form: • The Master Theorem: • Example 1: • Solution: a=4, b=2, logba=2, f(n)=n, so case 1 says T(n) is O(n2). • Example 2: • Solution: a=b=2, logba=1, f(n)=nlogn, so case 2 says T(n) is O(n log2 n).

  11. Master Method, Examples • The form: • The Master Theorem: • Example 3: • Solution: a=1, b=3, logba=0, f(n)=nlogn, so case 3 says T(n) is O(nlogn). • Example 4: • Solution: a=2, b=8, logba=3, f(n)=n2, so case 1 says T(n) is O(n3).

  12. Master Method, Examples • The form: • The Master Theorem: • Example 5: • Solution: a=9, b=3, logba=2, f(n)=n3, so case 3 says T(n) is O(n3). • Example 6: (binary search) • Solution: a=1, b=2, logba=0, f(n)=1, so case 2 says T(n) is O(log n). • Example 7: (heap construction) • Solution: a=2, b=2, logba=1, f(n)=logn, so case 1 says T(n) is O(n).

  13. 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

  14. 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.

  15. 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).

  16. Matrix Multiplication Algorithm • Problem: Given n×n matrices X and Y, find their product Z=X×Y, • General algorithm: O(n3) • Strassen’s Algorithm: • Rewrite the matrix product as: • Where I=AE+BG, J=AF+BH K=CE+DG, L=CF+DH • Define 7 submatrix products: • S1=A(F-H), S2=(A+B)H, S3=(C+D)E, S4=D(G-E) • S5=(A+D)(E+H), S6=(B-D)(G+H), S7=(A-C)(E+F) • Construct: • I=S4+S5+S6-S2, J=S1+S2, K=S3+S4, L=S1-S7-S3-S5 • Running time: T(n) = 7T(n/2)+bn2 for some constant b, which gives: O(nlog7), better than O(n3)

More Related