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## Chapter 4 Divide- in addition to -Conquer Copyright © 2007 Pearson Addison-Wesley. All rights

Vernon, Connor, Meteorologist has reference to this Academic Journal, PHwiki organized this Journal Chapter 4 Divide- in addition to -Conquer Copyright © 2007 Pearson Addison-Wesley. All rights reserved. Divide- in addition to -Conquer The most-well known algorithm design strategy: Divide instance of problem into two or more smaller instances Solve smaller instances recursively Obtain solution to original (larger) instance by combining these solutions Divide- in addition to -Conquer Technique (cont.) subproblem 2 of size n/2 subproblem 1 of size n/2 a solution to subproblem 1 a solution to the original problem a solution to subproblem 2 a problem of size n (instance) It general leads to a recursive algorithm!

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Divide- in addition to -Conquer Examples Sorting: mergesort in addition to quicksort Binary tree traversals Binary search () Multiplication of large integers Matrix multiplication: Strassens algorithm Closest-pair in addition to convex-hull algorithms General Divide- in addition to -Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n) (nd), d 0 Master Theorem: If a < bd, T(n) (nd) If a = bd, T(n) (nd log n) If a > bd, T(n) (nlog b a ) Note: The same results hold with O instead of . Examples: T(n) = 4T(n/2) + n T(n) T(n) = 4T(n/2) + n2 T(n) T(n) = 4T(n/2) + n3 T(n) (n^2) (n^2log n) (n^3) Mergesort Split array A[0 n-1] into about equal halves in addition to make copies of each half in arrays B in addition to C Sort arrays B in addition to C recursively Merge sorted arrays B in addition to C into array A as follows: Repeat the following until no elements remain in one of the arrays: compare the first elements in the remaining unprocessed portions of the arrays copy the smaller of the two into A, while incrementing the index indicating the unprocessed portion of that array Once all elements in one of the arrays are processed, copy the remaining unprocessed elements from the other array into A.

Pseudocode of Mergesort Pseudocode of Merge Time complexity: (p+q) = (n) comparisons Mergesort Example The non-recursive version of Mergesort starts from merging single elements into sorted pairs.

Analysis of Mergesort All cases have same efficiency: (n log n) Number of comparisons in the worst case is close to theoretical minimum as long as comparison-based sorting: log2 n! n log2 n – 1.44n Space requirement: (n) (not in-place) Can be implemented without recursion (bottom-up) T(n) = 2T(n/2) + (n), T(1) = 0 Quicksort Select a pivot (partitioning element) here, the first element Rearrange the list so that all the elements in the first s positions are smaller than or equal to the pivot in addition to all the elements in the remaining n-s positions are larger than or equal to the pivot (see next slide as long as an algorithm) Exchange the pivot with the last element in the first (i.e., ) subarray the pivot is now in its final position Sort the two subarrays recursively Partitioning Algorithm Time complexity: (r-l) comparisons or i > r or j = l < Quicksort Example 5 3 1 9 8 2 4 7 2 3 1 4 5 8 9 7 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 1 2 3 4 5 7 8 9 Analysis of Quicksort Best case: split in the middle (n log n) Worst case: sorted array! (n2) Average case: r in addition to om arrays (n log n) Improvements: better pivot selection: median of three partitioning switch to insertion sort on small subfiles elimination of recursion These combine to 20-25% improvement Considered the method of choice as long as internal sorting of large files (n 10000) T(n) = T(n-1) + (n) Binary Search Very efficient algorithm as long as searching in sorted array: K vs A[0] A[m] A[n-1] If K = A[m], stop (successful search); otherwise, continue searching by the same method in A[0 m-1] if K < A[m] in addition to in A[m+1 n-1] if K > A[m] l 0; r n-1 while l r do m (l+r)/2 if K = A[m] return m else if K < A[m] r m-1 else l m+1 return -1 Analysis of Binary Search Time efficiency worst-case recurrence: Cw (n) = 1 + Cw( n/2 ), Cw (1) = 1 solution: Cw(n) = log2(n+1) This is VERY fast: e.g., Cw(106) = 20 Optimal as long as searching a sorted array Limitations: must be a sorted array (not linked list) Bad (degenerate) example of divide- in addition to -conquer because only one of the sub-instances is solved Has a continuous counterpart called bisection method as long as solving equations in one unknown f(x) = 0 (see Sec. 12.4) Binary Tree Algorithms Binary tree is a divide- in addition to -conquer ready structure! Ex. 1: Classic traversals (preorder, inorder, postorder) Algorithm Inorder(T) if T a a Inorder(Tleft) b c b c print(root of T) d e d e Inorder(Tright) Efficiency: (n). Why Each node is visited/printed once. Binary Tree Algorithms (cont.) Ex. 2: Computing the height of a binary tree h(T) = max{h(TL), h(TR)} + 1 if T in addition to h() = -1 Efficiency: (n). Why Multiplication of Large Integers Consider the problem of multiplying two (large) n-digit integers represented by arrays of their digits such as: A = 12345678901357986429 B = 87654321284820912836 The grade-school algorithm: a1 a2 an b1 b2 bn (d10) d11d12 d1n (d20) d21d22 d2n (dn0) dn1dn2 dnn Efficiency: (n2) single-digit multiplications First Divide- in addition to -Conquer Algorithm A small example: A B where A = 2135 in addition to B = 4014 A = (21·102 + 35), B = (40 ·102 + 14) So, A B = (21 ·102 + 35) (40 ·102 + 14) = 21 40 ·104 + (21 14 + 35 40) ·102 + 35 14 In general, if A = A1A2 in addition to B = B1B2 (where A in addition to B are n-digit, A1, A2, B1, B2 are n/2-digit numbers), A B = A1 B1·10n + (A1 B2 + A2 B1) ·10n/2 + A2 B2 Recurrence as long as the number of one-digit multiplications M(n): M(n) = 4M(n/2), M(1) = 1 Solution: M(n) = n2 Second Divide- in addition to -Conquer Algorithm A B = A1 B1·10n + (A1 B2 + A2 B1) ·10n/2 + A2 B2 The idea is to decrease the number of multiplications from 4 to 3: (A1 + A2 ) (B1 + B2 ) = A1 B1 + (A1 B2 + A2 B1) + A2 B2, I.e., (A1 B2 + A2 B1) = (A1 + A2 ) (B1 + B2 ) - A1 B1 - A2 B2, which requires only 3 multiplications at the expense of (4-1) extra add/sub. Recurrence as long as the number of multiplications M(n): M(n) = 3M(n/2), M(1) = 1 Solution: M(n) = 3log 2n = nlog 23 n1.585 What if we count both multiplications in addition to additions Example of Large-Integer Multiplication 2135 4014 = (2110^2 + 35) (4010^2 + 14) = (2140)10^4 + c110^2 + 3514 where c1 = (21+35)(40+14) - 2140 - 3514, in addition to 2140 = (210 + 1) (410 + 0) = (24)10^2 + c210 + 10 where c2 = (2+1)(4+0) - 24 - 10, etc. This process requires 9 digit multiplications as opposed to 16. Conventional Matrix Multiplication Brute- as long as ce algorithm c00 c01 a00 a01 b00 b01 = c10 c11 a10 a11 b10 b11 a00 b00 + a01 b10 a00 b01 + a01 b11 = a10 b00 + a11 b10 a10 b01 + a11 b11 8 multiplications 4 additions Efficiency class in general: (n3) Strassens Matrix Multiplication Strassens algorithm as long as two 2x2 matrices (1969): c00 c01 a00 a01 b00 b01 = c10 c11 a10 a11 b10 b11 m1 + m4 - m5 + m7 m3 + m5 = m2 + m4 m1 + m3 - m2 + m6 m1 = (a00 + a11) (b00 + b11) m2 = (a10 + a11) b00 m3 = a00 (b01 - b11) m4 = a11 (b10 - b00) m5 = (a00 + a01) b11 m6 = (a10 - a00) (b00 + b01) m7 = (a01 - a11) (b10 + b11) 7 multiplications 18 additions

Strassens Matrix Multiplication Strassen observed [1969] that the product of two matrices can be computed in general as follows: C00 C01 A00 A01 B00 B01 = C10 C11 A10 A11 B10 B11 M1 + M4 – M5 + M7 M3 + M5 = M2 + M4 M1 + M3 – M2 + M6 Formulas as long as Strassens Algorithm M1 = (A00 + A11) (B00 + B11) M2 = (A10 + A11) B00 M3 = A00 (B01 – B11) M4 = A11 (B10 – B00) M5 = (A00 + A01) B11 M6 = (A10 – A00) (B00 + B01) M7 = (A01 – A11) (B10 + B11) Analysis of Strassens Algorithm If n is not a power of 2, matrices can be padded with zeros. Number of multiplications: M(n) = 7M(n/2), M(1) = 1 Solution: M(n) = 7log 2n = nlog 27 n2.807 vs. n3 of brute- as long as ce alg. Algorithms with better asymptotic efficiency are known but they are even more complex in addition to not used in practice. What if we count both multiplications in addition to additions

Closest-Pair Problem by Divide- in addition to -Conquer Step 0 Sort the points by x (list one) in addition to then by y (list two). Step 1 Divide the points given into two subsets S1 in addition to S2 by a vertical line x = c so that half the points lie to the left or on the line in addition to half the points lie to the right or on the line. Closest Pair by Divide- in addition to -Conquer (cont.) Step 2 Find recursively the closest pairs as long as the left in addition to right subsets. Step 3 Set d = min{d1, d2} We can limit our attention to the points in the symmetric vertical strip of width 2d as possible closest pair. Let C1 in addition to C2 be the subsets of points in the left subset S1 in addition to of the right subset S2, respectively, that lie in this vertical strip. The points in C1 in addition to C2 are stored in increasing order of their y coordinates, taken from the second list. Step 4 For every point P(x,y) in C1, we inspect points in C2 that may be closer to P than d. There can be no more than 6 such points (because d d2)! Closest Pair by Divide- in addition to -Conquer: Worst Case The worst case scenario is depicted below:

Efficiency of the Closest-Pair Algorithm Running time of the algorithm (without sorting) is: T(n) = 2T(n/2) + M(n), where M(n) (n) By the Master Theorem (with a = 2, b = 2, d = 1) T(n) (n log n) So the total time is (n log n). Quickhull Algorithm Convex hull: smallest convex set that includes given points. An O(n^3) brute as long as ce time is given in Levitin, Ch 3. Assume points are sorted by x-coordinate values Identify extreme points P1 in addition to P2 (leftmost in addition to rightmost) Compute upper hull recursively: find point Pmax that is farthest away from line P1P2 compute the upper hull of the points to the left of line P1Pmax compute the upper hull of the points to the left of line PmaxP2 Compute lower hull in a similar manner P1 P2 Pmax Efficiency of Quickhull Algorithm Finding point farthest away from line P1P2 can be done in linear time Time efficiency: T(n) = T(x) + T(y) + T(z) + T(v) + O(n), where x + y + z +v <= n. worst case: (n2) T(n) = T(n-1) + O(n) average case: (n) (under reasonable assumptions about distribution of points given) If points are not initially sorted by x-coordinate value, this can be accomplished in O(n log n) time Several O(n log n) algorithms as long as convex hull are known

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