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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: Strassens 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.

Strassens 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 Strassens 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 Strassens 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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