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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

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

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  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)

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