1 / 42

Chapter 4 Divide-and-Conquer

Chapter 4 Divide-and-Conquer. About this lecture. Recall the divide-and-conquer paradigm, which we used for merge sort: Divide the problem into a number of sub-problems that are smaller instances of the same problem. Conquer the sub-problems by solving them recursively.

paxton
Download Presentation

Chapter 4 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. Chapter 4 Divide-and-Conquer

  2. About this lecture • Recall the divide-and-conquer paradigm, which we used for merge sort: • Divide the problem into a number of sub-problems that are smaller instances of the same problem. • Conquer the sub-problems by solving them recursively. • Base case: If the sub-problems are small enough, just solve them by brute force. • Combine the sub-problem solutions to give a solution to the original problem. • We look at two more algorithms based on divide-and-conquer.

  3. About this lecture • Analyzing divide-and-conquer algorithms • Introduce some ways of solving recurrences • Substitution Method (If we know the answer) • Recursion Tree Method (Very useful !) • Master Theorem (Save our effort)

  4. Maximum-subarray problem • Input: an array A[1..n] of n numbers • Assume that some of the numbers are negative, because this problem is trivial when all numbers are nonnegative • Output: a nonempty subarray A[i..j]having the largest sum S[i, j] = ai + ai+1 +... + aj 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 A maximum subarray

  5. A brute-force solution • Examine all possible S[i, j] • Two implementations: • compute each S[i, j] in O(n) time O(n3) time • compute each S[i, j+1] from S[i, j] in O(1) time • (S[i, i] = A[i] and S[i, j+1] = S[i, j] + A[j+1]) • O(n2) time Ex: i 1 2 3 4 5 6 A[i] 13 -3 -25 20 -3 -16 S[1, j] = 13 10 -15 5 2 -14 S[2, j] = -3 -28 -8 -11 -27

  6. A divide-and-conquer solution • Possible locations of a maximum subarray A[i..j] of A[low..high], where mid = (low+high)/2 • entirely in A[low..mid] (lowi j mid) • entirely in A[mid+1..high] (mid < i j high) • crossing the midpoint (lowi mid < j high) crossing the midpoint low mid high mid +1 entirely in A[low..mid] entirely in A[mid+1..high] Possible locations of subarrays of A[low..high]

  7. A[mid+1..j] i low mid high mid +1 j A[i..mid] A[i..j] comprises two subarrays A[i..mid] and A[mid+1..j]

  8. FIND-MAX-CROSSING-SUBARRAY (A,low, mid, high) left-sum = - 􀀀// Find a maximum subarray of the form A[i..mid] sum = 0 fori = mid downto low sum = sum + A[i ] if sum > left-sum left-sum = sum max-left = i right-sum = - 􀀀// Find a maximum subarray of the form A[mid + 1 .. j ] sum =0 for j = mid +1 to high sum = sum + A[j] ifsum > right-sum right-sum = sum max-right = j // Return the indices and the sum of the two subarrays Return (max-left, max-right, left-sum + right-sum)

  9. Example: mid =5 A S[5 .. 5] = -3 S[4 .. 5] = 17 (max-left = 4) S[3 .. 5] = -8 S[2 .. 5] = -11 S[1 .. 5] = 2 mid =5 A S[6 .. 6] = -16 S[6 .. 7] = -39 S[6 .. 8] = -21 S[6 .. 9] = (max-right = 9)  -1 S[6..10] = -8 maximum subarray crossing mid is S[4..9] = 16

  10. FIND-MAXIMUM-SUBARRAY (A, low, high) if high == low Return (low, high, A[low])// base case: only one element else mid = (left-low, left-high, left-sum) = FIND-MAXIMUM-SUBARRAY(A, low, mid) (right-low, right-high, right-sum)= FIND-MAXIMUM-SUBARRAY(A, mid + 1, high) (cross-low, cross-high, cross-sum) = FIND-MAX-CROSSING-SUBARRAY(A, low, mid, high) if left-sum ≧right-sum and left-sum ≧ cross-sum return (left-low, left-high, left-sum) elseif right-sum ≧ left-sum and right-sum ≧ cross-sum return (right-low, right-high, right-sum) else return (cross-low, cross-high, cross-sum) Initial call: FIND-MAXIMUM-SUBARRAY (A, 1, n)

  11. Analyzing time complexity • FIND-MAX-CROSSING-SUBARRAY : (n), where n = highlow + 1 • FIND-MAXIMUM-SUBARRAY T(n) = 2T(n/2) + (n) (with T(1) = (1)) = (nlg n) (similar to merge-sort)

  12. Matrix multiplication • Input: two nn matrices A and B • Output: C= AB, An O(n3) time naive algorithm SQUARE-MATRIX-MULTIPLY(A, B) n A.rows let C be an nn matrix fori 1 ton forj 1 ton cij 0 fork 1 ton cijcij + aikbkj returnC

  13. Divide-and-Conquer Algorithm • Assume that n is an exact power of 2 (4.1)

  14. Divide-and-Conquer Algorithm • A straightforward divide-and-conquer algorithm • T(n) = 8T(n/2) + (n2) • = (n3) Computing A+B O(n2)

  15. Strassen’s method (4.2)

  16. Strassen’s method (4.4) (4.3)

  17. Strassen’s divide-and-conquer algorithm • Step 1: Divide each of A, B, and C into four sub-matrices as in (4.1) • Step 2: Create 10 matrices S1, S2, …, S10 as in (4.2) • Steep 3: Recursively, compute P1, P2, …, P7 as in (4.3) • Step 4: Compute according to (4.4) 

  18. Time complexity T(n) = 7T(n/2) + (n2) = (nlg 7 ) (why?) = (n2.81)

  19. Discussion • Strassen’s method is largely of theoretical interest for n 45 • Strassen’s method is based on the fact that we can multiply two 2  2 matrices using only 7 multiplications (instead of 8). • It was shown that it is impossible to multiply two 2  2 matrices using less than 7 multiplications.

  20. Discussion • We can improve Strassen’s algorithm by finding an efficient way to multiply two k k matrices using a smaller number q of multiplications, where k > 2. The time is T(n) = qT(n/k) + θ(n2). • A trivial lower bound for matrix multiplication is (n2). The current best upper bound known is O(n2.376). • Open problems: • Can the upper bound O(n2.376) be improved? • Can the lower bound (n2) be improved? 

  21. Substitution Method(if we know the answer) How to solve this? T(n) = 2T( ) + n, with T(1) = 1 • Make a guess e.g., T(n) = O(n log n) 2. Show it by induction • e.g., to show upper bound, we find constants c and n0 such that T(n)  cf(n) for n = n0, n0+1, n0+2, …

  22. Substitution Method(if we know the answer) How to solve this? T(n) = 2T( ) + n, with T(1) = 1 • Make a guess e.g., T(n) = O(n log n) • Show it by induction • Firstly, T(2) = 4, T(3) = 5.  We want to have T(n) cn lg n  Let c = 2  T(2) and T(3) okay • Other Cases ?

  23. Substitution Method(if we know the answer) • Induction Case: Assume the guess is true for all n = 2, 3,…, k For n = k+1, we have: T(n) = 2T( ) + n = cn lg n – cn + ncn log n Induction case is true

  24. Substitution Method(if we know the answer) Q. How did we know the value of c and n0 ? • If induction works, the induction case must be correct c ≥1 Then, we find that by setting c= 2, our guess is correct as soon as n0 = 2 Alternatively, we can also use c= 1.5 Then, we just need a larger n0 = 4 (What will be the new base cases? Why?)

  25. Substitution Method(New Challenge) How to solve this? 1. Make a guess (T(n) = O(n)), and • Show T(n) ≤ cn by induction • What will happen in induction case?

  26. Substitution Method(New Challenge) Induction Case: (assume guess is true for some base cases) This term is not what we want …

  27. Substitution Method(New Challenge) • The 1st attempt was not working because our guess for T(n) was a bit “loose” Recall: Induction may become easier if we prove a “stronger” statement 2nd Attempt: Refine our statement Try to show T(n) ≤ cn- b instead

  28. Substitution Method(New Challenge) Induction Case: We get the desired term (when b  1) It remains to find c and n0, and prove the base case(s), which is relatively easy Can you prove T(n) ≤ cn2 by induction?

  29. Avoiding Pitfalls • For T(n) = 2T( ) + n, we can falsely prove T(n) = O(n) by guessing T(n)  cn and then arguing Wrong!!

  30. Substitution Method(New Challenge 2) How to solve this? T(n) = 2T( ) + lg n ? Hint: Change variable: Set m = lg n

  31. Substitution Method(New Challenge 2) Set m = lg n , we get T(2m) = 2T(2m/2) + m Next, set S(m) = T(2m) = T(n) S(m) = 2S(m/2) + m We solve S(m) = O(m lg m) T(n) = O(lg n lg lg n)

  32. Recursion Tree Method( Nothing Special… Very Useful ! ) How to solve this? T(n) = 2T(n/2) + n2, with T(1) = 1

  33. Recursion Tree Method( Nothing Special… Very Useful ! ) Expanding the terms, we get: T(n) = n2 + 2T(n/2) = n2 + 2n2/4 + 4T(n/4) = n2 + 2n2/4 + 4n2/16 + 8T(n/8) = . . . = = Q(n2) + Q(n) = Q(n2)

  34. Recursion Tree Method( Recursion Tree View ) We can express the previous recurrence by:

  35. This term is from T(n/2) Further expressing gives us:

  36. Recursion Tree Method( New Challenge ) How to solve this? T(n) = T(n/3) + T(2n/3) + n, with T(1) = 1 What will be the recursion tree view?

  37. The corresponding recursion tree view is: The depth of the tree is log3/2n. Why?

  38. Master Method( Save our effort ) When the recurrence is in a special form, we can apply the Master Theorem to solve the recurrence immediately The Master Theorem has 3 cases …

  39. Master Theorem LetT(n) = aT(n/b) + f(n) with a  1 and b  1 are constants, where we interpret n/b to mean either n/b or n/b. Theorem: (Case 1) If f(n) = O(nlogb a- e) for some constant e  0 then T(n) = Q(nlogb a)

  40. Theorem: (Case 2) If f(n) = Q(nlogb a), then T(n) = Q(nlogb a lg n) Theorem: (Case 3) If f(n) = W(nlogb a + e) for some constant e  0, and if af(n/b)  c f(n) for some constant c 1, andall sufficiently large n, then T(n) = Q(f(n))

  41. Master Theorem • Solve T(n) = 9T(n/3) + n (case 1) • Solve T(n) = T(2n/3) + 1 (case 2) • Solve T(n) = 3T(n/4) + nlgn (case 3) • How about this? • T(n) = 2T(n/2) + n lg n ? • 5. T(n) = 8T(n/2) + n2 , T(n) = 8T(n/2) + n • 6. T(n) = 7T(n/2) + n2 ,T(n) = 7T(n/2) + 1

  42. Homework • Exercise: 4.1-3 (Programming) (due: Oct. 17) • Practice at home: 4.1-5, 4.2-1, 4.2-7 • Exercise: 4.3-6, 4.4-6, 4.5-1 (due: Oct. 19) • Problem: 4.1 (a, g) (due: Oct.19) • Practice at home: 4.3-7, 4.4-8, 4.4-7, 4.5-4

More Related