Chapter 4: Divide and Conquer

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Chapter 4: Divide and Conquer The Design and Analysis of Algorithms

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The Design and Analysis of Algorithms. Chapter 4: Divide and Conquer. Chapter 4. Divide and Conquer Algorithms. Basic Idea Merge Sort Quick Sort Binary Search Binary Tree Algorithms Closest Pair and Quickhull Conclusion. Basic Idea. - PowerPoint PPT Presentation

Transcript of Chapter 4: Divide and Conquer

Page 1: Chapter 4: Divide and Conquer

Chapter 4: Divide and Conquer

The Design and Analysis of Algorithms

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Chapter 4. Chapter 4. Divide and Conquer Algorithms Basic Idea Merge Sort Quick Sort Binary Search Binary Tree Algorithms Closest Pair and Quickhull Conclusion

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Basic Idea Divide instance of problem into two or

more smaller instances

Solve smaller instances recursively

Obtain solution to original (larger) instance by combining these solutions

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Basic Idea

subproblem 2 of size n/2

subproblem 1 of size n/2

a solution to subproblem 1

a solution tothe original problem

a solution to subproblem 2

a problem of size n

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General Divide-and-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 ) Examples: T(n) = T(n/2) + n T(n) (n )

Here a = 1, b = 2, d = 1, a < bd

T(n) = 2T(n/2) + 1 T(n) (n log 2

2 ) = (n )Here a = 2, b = 2, d = 0, a > bd

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Examples

T(n) = T(n/2) + 1 T(n) (log(n) )Here a = 1, b = 2, d = 0, a = bd

T(n) = 4T(n/2) + n T(n) (n log 2

4 ) = (n 2 )Here a = 4, b = 2, d = 1, a > bd

T(n) = 4T(n/2) + n2 T(n) (n2 log n)Here a = 4, b = 2, d = 2, a = bd

T(n) = 4T(n/2) + n3 T(n) (n3) Here a = 4, b = 2, d = 3, a < bd

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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

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Mergesort

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Mergesort

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Mergesort8 3 2 9 7 1 5 4

8 3 2 9 7 1 5 4

8 3 2 9 7 1 5 4

8 3 2 9 7 1 5 4

3 8 2 9 1 7 4 5

2 3 8 9 1 4 5 7

1 2 3 4 5 7 8 9

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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:

log2 n! ≈ n log2 n - 1.44n

Space requirement: Θ(n) (not in-place)

Can be implemented without recursion (bottom-up)

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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., ) sub-array — the pivot is now in its final position

Sort the two sub-arrays recursively

p

A[i]p

A[i]p

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Analysis of Quicksort Best case: split in the middle — Θ(n log n) Worst case: sorted array — Θ(n2) Average case: random arrays — Θ(n log n)

Improvements: better pivot selection: median of three partitioning switch to insertion sort on small subfiles elimination of recursion

Quicksort is considered the method of choice for internal sorting of large files (n ≥ 10000)

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Binary Search Very efficient algorithm for searching in sorted array: K

vsA[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]

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Analysis of Binary Search

Time efficiencyWorst-case recurrence: Cw (n) = 1 + Cw( n/2 ), Cw (1) = 1

solution: Cw(n) = log2(n+1)

Optimal for searching a sorted array Limitations: must be a sorted array (not linked list)

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Binary Tree Algorithms Binary tree is a divide-and-conquer ready structure

Ex. 1: Classic traversals (preorder, inorder, postorder)

Algorithm Inorder(T) if T Inorder(Tleft) print(root of T) Inorder(Tright) Efficiency: Θ(n)

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Binary Tree Algorithms

Ex. 2: Computing the height of a binary tree

T TL R

h(T) = max{ h(TL), h(TR) } + 1 if T and h() = -1

Efficiency: Θ(n)

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Closest Pair Problem

Step 1: Divide the points given into two subsets S1 and S2 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.

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Closest Pair Problem Step 2 Find recursively the closest pairs for the left and

right subsets. Step 3 Set d = min {d1, d2} Let C1 and C2 be the subsets of points in the left subset

S1 and of the right subset S2, respectively, that lie in this vertical strip.

The points in C1 and C2 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 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)

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Closest Pair Problem

The worst case scenario is depicted below

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Efficiency of the Closest Pair Problem

Running time of the algorithm

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)

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Quickhull AlgorithmConvex hull: smallest convex set that includes given points Assume points are sorted by x- coordinate values Identify extreme points P1 and P2 (leftmost and

rightmost) Compute upper hull recursively:

find point P max 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

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Quickhull Algorithm

P1

P2

Pmax

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Efficiency of Quickhull Algorithm

Finding point farthest away from line P1P2 can be done in linear time

Time efficiency: worst case: Θ(n2) (as quicksort)average case: Θ(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

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Conclusion Very efficient solutions. It is the choice when the following is true:

the problem can be described recursively - in terms of smaller

instances of the same problem.

the problem-solving process can be described recursively, i.e.

we can combine the solutions of the smaller instances to

obtain the solution of the original problem

Note that we assume that the problem is split in at least two parts of equal size, not simply decreased by a constant factor.