Divide-and-Conquer. Combine solutions to sub-problems into overall solution. Break up problem of size n into two equal parts of size!n.

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1 Chapter 5 Divide and Conquer Slides by Kevin Wayne. Copyright 25 Pearson-Addon Wesley. All rights reserved. Divide-and-Conquer Divide-and-conquer. Break up problem into several parts. Solve each part recursively. Combine solutions to sub-problems into overall solution. Most common usage. Break up problem of size n into two equal parts of size n. Solve two parts recursively. Combine two solutions into overall solution in linear time. Consequence. Brute force: n 2. Divide-and-conquer: n log n. Divide et impera. Veni, vidi, vici. - Julius Caesar 2

2 5. Mergesort Sorting Sorting. Given n elements, rearrange in ascending order. Obvious sorting applications. Lt files in a directory. Organize an MP3 library. Lt names in a phone book. Dplay Google PageRank results. Problems become easier once sorted. Find the median. Find the closest pair. Binary search in a database. Identify stattical outliers. Find duplicates in a mailing lt. Non-obvious sorting applications. Data compression. Computer graphics. Interval scheduling. Computational biology. Minimum spanning tree. Supply chain management. Simulate a system of particles. Book recommendations on Amazon. Load balancing on a parallel computer.... 4

3 Mergesort Mergesort. Divide array into two halves. Recursively sort each half. Merge two halves to make sorted whole. Jon von Neumann (945) A L G O R I T H M S A L G O R I T H M S divide O() A G L O R H I M S T sort 2T(n/2) A G H I L M O R S T merge O(n) 5 Merging Merging. Combine two pre-sorted lts into a sorted whole. How to merge efficiently? Linear number of comparons. Use temporary array. A G L O R H I M S T A G H I Challenge for the bored. In-place merge. [Kronrud, 969] using only a constant amount of extra storage 6

4 A Useful Recurrence Relation Def. T(n) = number of comparons to mergesort an input of size n. Mergesort recurrence. Solution. T(n) = O(n log 2 n). Assorted proofs. We describe several ways to prove th recurrence. Initially we assume n a power of 2 and replace with =. 7 Proof by Recursion Tree T(n) n T(n/2) T(n/2) 2(n/2) T(n/4) T(n/4) T(n/4) T(n/4) log 2 n 4(n/4)... T(n / 2 k ) 2 k (n / 2 k )... T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) n/2 (2) n log 2 n 8

5 Proof by Telescoping Claim. If T(n) satfies th recurrence, then T(n) = n log 2 n. assumes n a power of 2 Pf. For n > : T(n) n = 2T(n /2) n + = T(n/2) n/2 + = L = T(n/4) n/4 T(n/n) n/n = log 2 n L log 2 n 9 Proof by Induction Claim. If T(n) satfies th recurrence, then T(n) = n log 2 n. assumes n a power of 2 Pf. (by induction on n) Base case: n =. Inductive hypothes: T(n) = n log 2 n. Goal: show that T(2n) = 2n log 2 (2n).

6 Analys of Mergesort Recurrence Claim. If T(n) satfies the following recurrence, then T(n) n lg n. log 2 n Pf. (by induction on n) Base case: n =. T() = = lg. Define n = n / 2, n 2 = n / 2. Induction step: Let n"2, assume true for, 2,..., n. 5.4 Closest Pair of Points

7 Closest Pair of Points Closest pair. Given n points in the plane, find a pair with smallest Euclidean dtance between them. Fundamental geometric primitive. Graphics, computer vion, geographic information systems, molecular modeling, air traffic control. Special case of nearest neighbor, Euclidean MST, Voronoi. fast closest pair inspired fast algorithms for these problems Brute force. Check all pairs of points p and q with Θ(n 2 ) comparons. -D version. O(n log n) easy if points are on a line. Assumption. No two points have same x coordinate. to make presentation cleaner 3 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. L 4

8 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. Obstacle. Impossible to ensure n/4 points in each piece. L 5 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly n points on each side. L 6

9 Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly n points on each side. Conquer: find closest pair in each side recursively. L Closest Pair of Points Algorithm. Divide: draw vertical line L so that roughly n points on each side. Conquer: find closest pair in each side recursively. Combine: find closest pair with one point in each side. Return best of 3 solutions. seems like Θ(n 2 ) L

10 Closest Pair of Points Find closest pair with one point in each side, assuming that dtance < δ. L 2 2 δ = min(2, 2) 9 Closest Pair of Points Find closest pair with one point in each side, assuming that dtance < δ. Observation: only need to consider points within δ of line L. L 2 2 δ = min(2, 2) δ 2

11 Closest Pair of Points Find closest pair with one point in each side, assuming that dtance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate. 7 L δ = min(2, 2) 2 δ 2 Closest Pair of Points Find closest pair with one point in each side, assuming that dtance < δ. Observation: only need to consider points within δ of line L. Sort points in 2δ-strip by their y coordinate. Only check dtances of those within positions in sorted lt 7 L δ = min(2, 2) 2 δ 22

12 Closest Pair of Points Def. Let s i be the point in the 2δ-strip, with the i th smallest y-coordinate. Claim. If i j 2, then the dtance between s i and s j at least δ. Pf j No two points lie in same δ-by-δ box. Two points at least 2 rows apart have dtance 2(δ). 2 rows 29 3 δ δ Fact. Still true if we replace 2 with 7. i δ Scan points in y-order and compare dtance between each point and next neighbours. If any of these dtances less than δ, update δ. δ δ 23 Closest Pair Algorithm Closest-Pair(p,, p n ) { Compute separation line L such that half the points are on one side and half on the other side. O(n log n) δ = Closest-Pair(left half) δ 2 = Closest-Pair(right half) δ = min(δ, δ 2 ) 2T(n / 2) Delete all points further than δ from separation line L Sort remaining points by y-coordinate. O(n) O(n log n) Scan points in y-order and compare dtance between each point and next neighbours. If any of these dtances less than δ, update δ. O(n) } return δ. 24

13 Closest Pair of Points: Analys Running time. Q. Can we achieve O(n log n)? A. Yes. Don't sort points in strip from scratch each time. Each recursion returns two lts: all points sorted by y coordinate, and all points sorted by x coordinate. Sort by merging two pre-sorted lts Integer Multiplication

14 # Multiply Multiply. Given two n-digit integers a and b, compute a # b. Brute force solution: Θ(n 2 ) bit operations. 27 Integer Arithmetic Add. Given two n-digit integers a and b, compute a + b. O(n) bit operations. + Add 28 To multiply two n-digit integers: Multiply four n-digit integers. Add two n-digit integers, and shift to obtain result. Divide-and-Conquer Multiplication: Warmup assumes n a power of 2

15 Proof by Telescoping Claim. assumes n a power of 2 Pf. For n > : T(n)/n = 4T(n/2)/n + C = 2 T(n/2)/(n/2) + C = 2 [ 2 T(n/4)/(n/4) + C ] + C = 4 T(n/4)/(n/4) + 2C + C = 4 [ 2 T(n/8)/(n/8) + C ] + 2C + C = 8 T(n/8)/(n/8) + 4C + 2C +... = n T()/() + n/2 C + n/4 C C + 2C + = C (n/2+n/ ) = C(n-). 29 Karatsuba Multiplication To multiply two n-digit integers: Add two n digit integers. Multiply three n-digit integers. Add, subtract, and shift n-digit integers to obtain result. Theorem. [Karatsuba-Ofman, 962] Can multiply two n-digit integers in O(n.585 ) bit operations. 3

16 Karatsuba: Recursion Tree T(n) n T(n/2) T(n/2) T(n/2) 3(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) T(n/4) 9(n/4) T(n / 2 k ) 3 k (n / 2 k ) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) 3 lg n (2) 3 Matrix Multiplication

17 Matrix Multiplication Matrix multiplication. Given two n-by-n matrices A and B, compute C = AB. Brute force. Θ(n 3 ) arithmetic operations. Fundamental question. Can we improve upon brute force? 33 Matrix Multiplication: Warmup Divide-and-conquer. Divide: partition A and B into n-by-n blocks. Conquer: multiply 8 n-by-n recursively. Combine: add appropriate products using 4 matrix additions. 34

18 Matrix Multiplication: Key Idea Key idea. multiply 2-by-2 block matrices with only 7 multiplications. 7 multiplications. 8 = + 8 additions (or subtractions). 35 Fast Matrix Multiplication Fast matrix multiplication. (Strassen, 969) Divide: partition A and B into n-by-n blocks. Compute: 4 n-by-n matrices via matrix additions. Conquer: multiply 7 n-by-n matrices recursively. Combine: 7 products into 4 terms using 8 matrix additions. Analys. Assume n a power of 2. T(n) = # arithmetic operations. 36

19 Fast Matrix Multiplication in Practice Implementation sues. Sparsity. Caching effects. Numerical stability. Odd matrix dimensions. Crossover to classical algorithm around n = 28. Common mperception: "Strassen only a theoretical curiosity." Advanced Computation Group at Apple Computer reports 8x speedup on G4 Velocity Engine when n ~ 2,5. Range of instances where it's useful a subject of controversy. Remark. Can "Strassenize" Ax=b, determinant, eigenvalues, and other matrix ops. 37 Fast Matrix Multiplication in Theory Q. Multiply two 2-by-2 matrices with only 7 scalar multiplications? A. Yes [Strassen, 969] Θ(n log 2 7 ) = O(n 2.8 ) Q. Multiply two 2-by-2 matrices with only 6 scalar multiplications? (n log 2 A. Impossible. [Hopcroft and Kerr, 97] Θ 6 ) = O(n 2.59 ) Q. Two 3-by-3 matrices with only 2 scalar multiplications? (n log 3 A. Also impossible. Θ 2 ) = O(n 2.77 ) Q. Two 7-by-7 matrices with only 43,64 scalar multiplications? A. Yes [Pan, 98] (n log 7 Θ 4364 ) = O(n 2.8 ) Decimal wars. December, 979: O(n ). January, 98: O(n ). 38

20 Fast Matrix Multiplication in Theory Best known. O(n ) [Coppersmith-Winograd, 987.] Conjecture. O(n 2+ε ) for any ε >. Caveat. Theoretical improvements to Strassen are progressively less practical. 39 CLRS 4.3 Master Theorem

21 Master Theorem from CLRS Proof by Recursion Tree T(n) f(n) T(n/b) T(n/b) T(n/b) af(n/b) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) T(n/b 2 ) a 2 f(n/b 2 )... log b n... T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k ) T(n / b k )... a k f(n/b k )... T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) a logb n T(2) = T(2) n logb a T(n) = a k f(n/b k ) 42

22 Master Theorem from CLRS Master Theorem from CLRS

23 Master Theorem from CLRS Master Theorem from CLRS

24 Master Theorem from CLRS Master Theorem from CLRS

25 Master Theorem from CLRS Master Theorem from CLRS 4.3 5

26 Extra Slides 5.3 Counting Inversions

27 Counting Inversions Music site tries to match your song preferences with others. You rank n songs. Music site consults database to find people with similar tastes. Similarity metric: number of inversions between two rankings. My rank:, 2,, n. Your rank: a, a 2,, a n. Songs i and j inverted if i < j, but a i > a j. Songs Me You A B C D E Inversions 3-2, 4-2 Brute force: check all Θ(n 2 ) pairs i and j. 53 Applications Applications. Voting theory. Collaborative filtering. Measuring the "sortedness" of an array. Sensitivity analys of Google's ranking function. Rank aggregation for meta-searching on the Web. Nonparametric stattics (e.g., Kendall's Tau dtance). 54

28 Counting Inversions: Divide-and-Conquer Divide-and-conquer Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate lt into two pieces Divide: O()

29 Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate lt into two pieces. Conquer: recursively count inversions in each half Divide: O() Conquer: 2T(n / 2) 5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, , 9-3, 9-7, 2-3, 2-7, 2-, -3, Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide: separate lt into two pieces. Conquer: recursively count inversions in each half. Combine: count inversions where a i and a j are in different halves, and return sum of three quantities Divide: O() Conquer: 2T(n / 2) 5 blue-blue inversions 8 green-green inversions 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, -6, -9, -3, -7 Combine:??? Total = =

30 Counting Inversions: Combine Combine: count blue-green inversions Assume each half sorted. Count inversions where a i and a j are in different halves. Merge two sorted halves into sorted whole. to maintain sorted invariant blue-green inversions: Count: O(n) Merge: O(n) 59 Counting Inversions: Implementation Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] L sorted. Sort-and-Count(L) { if lt L has one element return and the lt L Divide the lt into two halves A and B (r A, A) Sort-and-Count(A) (r B, B) Sort-and-Count(B) (r B, L) Merge-and-Count(A, B) } return r = r A + r B + r and the sorted lt L 6