Chapter 4 Divide-and-Conquer Copyright © 2007 Pearson Addison-Wesley. All rights reserved.

Tromino puzzle  Exercise A tromino is an L-shape tile formed by 1- by-1 adjacent squares. The problem is to cover any 2 n -by-2 n chessboard with one missing square (anywhere on the board) with trominos. Trominos should cover all the squares except the missing one with on overlaps  Design a divide-and-conquer algorithm for this problem

Divide-and-Conquer The most-well known algorithm design strategy: 1. Divide instance of problem into two or more smaller instances 2. Solve smaller instances (recursively) 3. Obtain solution to original (larger) instance by combining these solutions Pergunta: sempre melhor que força bruta? contra-exemplo?

Divide-and-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 Como ficaria o uso em computadores paralelos?

Divide-and-Conquer Examples  Sorting: mergesort and quicksort  Binary tree traversals  Binary search (?)  Multiplication of large integers  Matrix multiplication: Strassen’s algorithm  Closest-pair and convex-hull algorithms

Casos: divisão por 2 e geral  Caso típico: divide por 2  Caso geral: Instância de tamanho n é dividida em b instâncias de tamanho b/n das quais a devem ser resolvidas (a e b constantes, a>=1; b>1)Instância de tamanho n é dividida em b instâncias de tamanho b/n das quais a devem ser resolvidas (a e b constantes, a>=1; b>1) Assumindo que n é potência de b (para simplificar análise) temosAssumindo que n é potência de b (para simplificar análise) temos T(n) = a T (n/b) + f(n) Onde f(n) responde pelo tempo que se leva para dividir o problema em menores e/ou para combinar as soluçõesOnde f(n) responde pelo tempo que se leva para dividir o problema em menores e/ou para combinar as soluções Ex: somar série com Divide and Conquer …Ex: somar série com Divide and Conquer …

General Divide-and-Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n)   (n d ), d  0 Master Theorem: If a < b d, T(n)   (n d ) If a = b d, T(n)   (n d log n) If a = b d, T(n)   (n d log n) If a > b d, T(n)   ( n log b a ) If a > b d, T(n)   ( n log 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) + n 2  T(n)  ? T(n) = 4T(n/2) + n 2  T(n)  ? T(n) = 4T(n/2) + n 3  T(n)  ? T(n) = 4T(n/2) + n 3  T(n)  ? Ver Apêndice B para prova do teorema

General Divide-and-Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n)   (n d ), d  0 Master Theorem: If a < b d, T(n)   (n d ) If a = b d, T(n)   (n d log n) If a = b d, T(n)   (n d log n) If a > b d, T(n)   ( n log b a ) If a > b d, T(n)   ( n log b a )Examples: T(n) = 2T(n/2) + 1  T(n)  ? (a=2, b=2, f(n)=1, d=0) T(n) = 2T(n/2) + n  T(n)  ? (a=2, b=2, f(n)=n, d=1) T(n) = 2T(n/2) + n 2  T(n)  ? (a=2, b=2, f(n)=n 2, d =2) T(n) = 2T(n/2) + n 3  T(n)  ? (a=2, b=2, f(n)=n 3, d=3)

General Divide-and-Conquer Recurrence T(n) = aT(n/b) + f (n) where f(n)   (n d ), d  0 Master Theorem: If a < b d, T(n)   (n d ) If a = b d, T(n)   (n d log n) If a = b d, T(n)   (n d log n) If a > b d, T(n)   ( n log b a ) If a > b d, T(n)   ( n log b a )Examples: T(n) = 2T(n/2) + 1  T(n)  ? (a=2, b=2, f(n)=1, d=0) n T(n) = 2T(n/2) + n  T(n)  ? (a=2, b=2, f(n)=n, d=1) n log n T(n) = 2T(n/2) + n 2  T(n)  ? (a=2, b=2, f(n)=n 2, d =2) n 2 T(n) = 2T(n/2) + n 3  T(n)  ? (a=2, b=2, f(n)=n 3, d=3) n 3

Mergesort  Split array A[0..n-1] in two about equal halves and make copies of each half in arrays B and C  Sort arrays B and C recursively  Merge sorted arrays B and C into array A as follows: Repeat the following until no elements remain in one of the arrays (total de n elementos):Repeat the following until no elements remain in one of the arrays (total de n elementos): –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.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

Mergesort Example

Analysis of Mergesort  All cases have same efficiency: Θ (n log n)  Number of comparisons in the worst case is close to theoretical minimum for comparison-based sorting:  log 2 n!  ≈ n log 2 n n  log 2 n!  ≈ n log 2 n n  Space requirement: Θ (n) (not in-place)  Can be implemented without recursion (bottom-up)  Estável ???

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 and all the elements in the remaining n-s positions are larger than or equal to the pivot (see next slide for 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 p A[i]  p A[i]  p

Partitioning Algorithm < =

Quicksort Examples Estável ???

Analysis of Quicksort  Best case: split in the middle — Θ (n log n) n+1 comparações se indices cruzam, n se coincidemn+1 comparações se indices cruzam, n se coincidem C best (n) = 2 C best (n/2) +n p/ n>1, C best (1)=0 (Master Theorem)C best (n) = 2 C best (n/2) +n p/ n>1, C best (1)=0 (Master Theorem)  Worst case: sorted array! — Θ (n 2 ) C worst (n) = (n+1) + n + … + 3 = (n+1)(n+2)/2 – 3C worst (n) = (n+1) + n + … + 3 = (n+1)(n+2)/2 – 3  Average case: random arrays — Θ (n log n) C avg (n) = 1/n ∑ s=0 n=1 [(n+1) + C avg (s) + C avg (n-1-s)] p/ n>1, C avg (0)=0, C avg (1)=0C avg (n) = 1/n ∑ s=0 n=1 [(n+1) + C avg (s) + C avg (n-1-s)] p/ n>1, C avg (0)=0, C avg (1)=0  Improvements (combine to 20-25% ): better pivot selection: median of three partitioningbetter pivot selection: median of three partitioning switch to insertion sort on small subfilesswitch to insertion sort on small subfiles elimination of recursionelimination of recursion  Considered the method of choice for internal sorting of large files (n ≥ 10000)

Binary Search Very efficient algorithm for searching in sorted array: K vs 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] and 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 if K = A[m] return m else if K < A[m] r  m-1 else if K < A[m] r  m-1 else l  m+1 else l  m+1 return -1

Analysis of Binary Search  Time efficiency worst-case recurrence: C w (n) = 1 + C w (  n/2  ), C w (1) = 1 solution: C w (n) =  log 2 (n+1)  This is VERY fast: e.g., C w (10 6 ) = 20worst-case recurrence: C w (n) = 1 + C w (  n/2  ), C w (1) = 1 solution: C w (n) =  log 2 (n+1)  This is VERY fast: e.g., C w (10 6 ) = 20  Optimal for searching a sorted array  Limitations: must be a sorted array (not linked list)  Bad (degenerate) example of divide-and-conquer  Has a continuous counterpart called bisection method for solving equations in one unknown f(x) = 0 (see Sec. 12.4)

Binary Tree Algorithms Binary tree is a divide-and-conquer ready structure! Ex. 1: Classic traversals (preorder, inorder, postorder) Algorithm Inorder(T) if T   a a Inorder(T left ) b c b c Inorder(T left ) b c b c print(root of T) d e d e print(root of T) d e d e Inorder(T right ) Inorder(T right ) Efficiency: Θ (n)

Binary Tree Algorithms (cont.) Ex. 2: Computing the height of a binary tree h(T) = max{h(T L ), h(T R )} + 1 if T  and h(  ) = -1 h(T) = max{h(T L ), h(T R )} + 1 if T   and h(  ) = -1 Efficiency: Θ(n)

Multiplication of Large Integers Consider the problem of multiplying two (large) n-digit integers represented by arrays of their digits such as: A = B = The grade-school algorithm: a 1 a 2 … a n b 1 b 2 … b n (d 10 ) d 11 d 12 … d 1n (d 20 ) d 21 d 22 … d 2n (d 20 ) d 21 d 22 … d 2n … … … … … … … … … … … … … … (d n0 ) d n1 d n2 … d nn Efficiency: n 2 one-digit multiplications Efficiency: n 2 one-digit multiplications

First Divide-and-Conquer Algorithm A small example: A  B where A = 2135 and B = 4014 A = (21· ), B = (40 · ) So, A  B = (21 · )  (40 · ) = 21  40 · (21   40) ·  14 = 21  40 · (21   40) ·  14 In general, if A = A 1 A 2 and B = B 1 B 2 (where A and B are n-digit, A 1, A 2, B 1, B 2 are n/2-digit numbers), A  B = A 1  B 1 ·10 n + (A 1  B 2 + A 2  B 1 ) ·10 n/2 + A 2  B 2 Recurrence for the number of one-digit multiplications M(n): M(n) = 4M(n/2), M(1) = 1 Solution: M(n) = n 2 M(n) = 4M(n/2), M(1) = 1 Solution: M(n) = n 2

Second Divide-and-Conquer Algorithm A  B = A 1  B 1 ·10 n + (A 1  B 2 + A 2  B 1 ) ·10 n/2 + A 2  B 2 The idea is to decrease the number of multiplications from 4 to 3: (A 1 + A 2 )  (B 1 + B 2 ) = A 1  B 1 + (A 1  B 2 + A 2  B 1 ) + A 2  B 2, I.e., (A 1  B 2 + A 2  B 1 ) = (A 1 + A 2 )  (B 1 + B 2 ) - A 1  B 1 - A 2  B 2, which requires only 3 multiplications at the expense of (4-1) extra add/sub. (A 1 + A 2 )  (B 1 + B 2 ) = A 1  B 1 + (A 1  B 2 + A 2  B 1 ) + A 2  B 2, I.e., (A 1  B 2 + A 2  B 1 ) = (A 1 + A 2 )  (B 1 + B 2 ) - A 1  B 1 - A 2  B 2, which requires only 3 multiplications at the expense of (4-1) extra add/sub. Recurrence for the number of multiplications M(n): M(n) = 3M(n/2), M(1) = 1 Solution: (backward substitutions, a log b c = c log b a ) M(n) = 3 log 2 n = n log 2 3 ≈ n 1.585

Example of Large-Integer Multiplication 2135  4014

Strassen’s Matrix Multiplication Strassen [1969] observed that the product of two matrices can be computed as follows: Strassen [1969] observed that the product of two matrices can be computed as follows: C 00 C 01 A 00 A 01 B 00 B 01 = * = * C 10 C 11 A 10 A 11 B 10 B 11 M 1 + M 4 - M 5 + M 7 M 3 + M 5 M 1 + M 4 - M 5 + M 7 M 3 + M 5 = M 2 + M 4 M 1 + M 3 - M 2 + M 6 M 2 + M 4 M 1 + M 3 - M 2 + M 6  Requer 7 multiplicações enquanto força bruta requer 8

Formulas for Strassen’s Algorithm M 1 = (A 00 + A 11 )  (B 00 + B 11 ) M 2 = (A 10 + A 11 )  B 00 M 3 = A 00  (B 01 - B 11 ) M 4 = A 11  (B 10 - B 00 ) M 5 = (A 00 + A 01 )  B 11 M 6 = (A 10 - A 00 )  (B 00 + B 01 ) M 7 = (A 01 - A 11 )  (B 10 + B 11 )

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 M(n) = 7M(n/2), M(1) = 1 Solution: M(n) = 7 log 2 n = n log 2 7 ≈ n vs. n 3 of brute-force alg. Algorithms with better asymptotic efficiency are known but they are even more complex.

Closest-Pair Problem by Divide-and-Conquer Step 1 Divide the points given into two subsets S 1 and S 2 by a vertical line x = c so that half the points lie to the left or on the line and half the points lie to the right or on the line.  We can assume that the points are ordered on x coordinates (may use mergesort, O(nlogn)).  Can use as c the median of x coordinates

Closest Pair by Divide-and-Conquer (cont.) Step 2 Find recursively the closest pairs for the left and right subsets. Step 3 Set d = min{d 1, d 2 } We can limit our attention to the points in the symmetric vertical strip of width 2d as possible closest pair. Let C 1 and C 2 be the subsets of points in the left subset S 1 and of the right subset S 2, respectively, that lie in this vertical strip. The points in C 1 and C 2 are stored in increasing order of their y coordinates, which is maintained by merging during the execution of the next step. We can limit our attention to the points in the symmetric vertical strip of width 2d as possible closest pair. Let C 1 and C 2 be the subsets of points in the left subset S 1 and of the right subset S 2, respectively, that lie in this vertical strip. The points in C 1 and C 2 are stored in increasing order of their y coordinates, which is maintained by merging during the execution of the next step. Step 4 For every point P(x,y) in C 1, we inspect points in C 2 that may be closer to P than d. There can be no more than 6 such points (because d ≤ d 2 )!

Closest Pair by Divide-and-Conquer: Worst Case The worst case scenario is depicted below:

Efficiency of the Closest-Pair Algorithm Running time of the algorithm is described by (the time M(n) for merging solutions is O(n)) T(n) = 2T(n/2) + M(n), where M(n)  O(n) T(n) = 2T(n/2) + M(n), where M(n)  O(n) By the Master Theorem (with a = 2, b = 2, d = 1) T(n)  O(n log n) T(n)  O(n log n)

Brute force Convex hull P1P1 P3P3 P2P2 P6P6 P8P8 P4P4 P5P5 P7P7

Simple but inefficient:  Dois pontos P1 e P2 fazem parte da borda se e só se todos os demais pontos estão de um único lado do segmento P1P2 (assumindo que não há 3 pontos na mesma linha)  Linha (x1, x2), (x2, y2) definida por ax+by=c onde a=y2-y1, b=x1-x2, c=x1x2-y1y2  Essa linha divide o plano de 2 half-planes: para todos os pontos de un deles, ax+by > c, para o outro ax+by c, para o outro ax+by<c  O(n 3 ) P1P1 P3P3 P2P2 P6P6 P8P8 P4P4 P5P5 P7P7

Quickhull Algorithm Convex hull: smallest convex set that includes given points  Assume points are sorted by x-coordinate values  Identify extreme points P 1 and P 2 (leftmost and rightmost)  Compute upper hull recursively: find point P max that is farthest away from line P 1 P 2find point P max that is farthest away from line P 1 P 2 compute the upper hull of the points to the left of line P 1 P maxcompute the upper hull of the points to the left of line P 1 P max compute the upper hull of the points to the left of line P max P 2compute the upper hull of the points to the left of line P max P 2  Compute lower hull in a similar manner P1P1 P2P2 P max

Efficiency of Quickhull Algorithm  Finding point farthest away from line P 1 P 2 can be done in linear time  Time efficiency: worst case: Θ (n 2 ) (as quicksort)worst case: Θ (n 2 ) (as quicksort) average case: Θ (????) (under reasonable assumptions about distribution of points given)average case: Θ (????) (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 for convex hull are known