! Break up problem into several parts. ! Solve each part recursively. ! Combine solutions to sub-problems into overall solution.

Transcription

1 Divide-and-Conquer Chapter 5 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. Slides by Kevin Wayne. Copyright 25 Pearson-Addison Wesley. All rights reserved.! Divide-and-conquer: n log n. Divide et impera. Veni, vidi, vici. - Julius Caesar 2 Sorting 5. Mergesort Sorting. Given n elements, rearrange in ascending order. Obvious sorting applications.! ist files in a directory.! Organize an MP3 library.! ist names in a phone book.! Display Google PageRank results. Non-obvious sorting applications.! Data compression.! Computer graphics.! Interval scheduling.! Computational biology.! Minimum spanning tree. Problems become easier once sorted.! Find the median.! Find the closest pair.! Binary search in a database.! Identify statistical outliers.! Supply chain management.! Simulate a system of particles.! Book recommendations on Amazon.! oad balancing on a parallel computer.! Find duplicates in a mailing list. 4

2 Mergesort Merging Mergesort.! Divide array into two halves.! Recursively sort each half.! Merge two halves to make sorted whole. Merging. Combine two pre-sorted lists into a sorted whole. How to merge efficiently?! inear number of comparisons.! Use temporary array. Jon von Neumann (945) A G O R I T H M S A G O R H I M S T A G O R I T H M S divide O() A G H I A G O R H I M S T sort 2T(n/2) A G H I M O R S T merge O(n) Challenge for the bored. In-place merge. [Kronrud, 969] using only a constant amount of extra storage 5 6 A Useful Recurrence Relation Proof by Recursion Tree Def. T(n) = number of comparisons to mergesort an input of size n. Mergesort recurrence. " if n = T(n) = # 2T(n /2) + n { otherwise sorting both halves merging T(n) " if n = ) ( T( # n/2 ) + T( n/2& ) { n otherwise ) * solve left half solve right half merging T(n) n T(n/2) T(n/2) 2(n/2) Solution. T(n) = O(n log 2 n). log 2 n 4(n/4) Assorted proofs. We describe several ways to prove this recurrence. Initially we assume n is a power of 2 and replace! with =. 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) 7 n log 2 n 8

3 Proof by Telescoping Proof by Induction Claim. If T(n) satisfies this recurrence, then T(n) = n log 2 n. Claim. If T(n) satisfies this recurrence, then T(n) = n log 2 n. assumes n is a power of 2 assumes n is a power of 2 " if n = T(n) = # 2T(n/2) + n { otherwise sorting both halves merging " if n = T(n) = # 2T(n /2) + n { otherwise sorting both halves merging Pf. For n > : T(n) n = = = = 2T(n/2) n T(n /2) n/2 T(n / 4) n/4 T(n /n) n/n = log 2 n log 2 n Pf. (by induction on n)! Base case: n =.! Inductive hypothesis: T(n) = n log 2 n.! Goal: show that T(2n) = 2n log 2 (2n). T(2n) = 2T(n) + 2n = 2nlog 2 n + 2n = 2n( log 2 (2n) ") + 2n = 2nlog 2 (2n) 9 Analysis of Mergesort Recurrence Claim. If T(n) satisfies the following recurrence, then T(n)! n "lg n#. if n = log 2 n ) T(n) " ( T( # n /2 ) + T( n/2& ) + n { otherwise ) * solve left half solve right half merging 5.3 Counting Inversions Pf. (by induction on n)! Base case: n =.! Define n = n / 2, n 2 = "n / 2#.! Induction step: assume true for, 2,..., n. T(n) " T(n ) + T(n 2 ) + n " n # lgn + n 2 # lg n 2 + n " n # lgn 2 + n 2 # lg n 2 + n = n # lgn 2 + n " n( # lgn ) + n = n# lgn n 2 = " n/2# " # 2 " lg n # / 2 = 2 " lg n # / 2 lgn 2 " lg n# &

4 Counting Inversions Applications 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. Applications.! Voting theory.! Collaborative filtering.! Measuring the "sortedness" of an array.! Sensitivity analysis of Googles ranking function.! Rank aggregation for meta-searching on the Web.! Nonparametric statistics (e.g., Kendalls Tau distance). Me You Songs A B C D E Inversions 3-2, 4-2 Brute force: check all &(n 2 ) pairs i and j. 3 4 Counting Inversions: Divide-and-Conquer Counting Inversions: Divide-and-Conquer Divide-and-conquer. Divide-and-conquer.! Divide: separate list into two pieces Divide: O()

5 Counting Inversions: Divide-and-Conquer Counting Inversions: Divide-and-Conquer Divide-and-conquer.! Divide: separate list into two pieces.! Conquer: recursively count inversions in each half. Divide-and-conquer.! Divide: separate list 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() Divide: O() Conquer: 2T(n / 2) Conquer: 2T(n / 2) 5 blue-blue inversions 8 green-green inversions 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, -7 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, -6, -9, -3, -7 Combine:??? Total = = Counting Inversions: Combine Counting Inversions: Implementation Combine: count blue-green inversions! Assume each half is sorted.! Count inversions where a i and a j are in different halves.! Merge two sorted halves into sorted whole blue-green inversions: to maintain sorted invariant Count: O(n) Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] is sorted. Sort-and-Count() { if list has one element return and the list Divide the list into two halves A and B (r A, A) Sort-and-Count(A) (r B, B) Sort-and-Count(B) (r B, ) Merge-and-Count(A, B) Merge: O(n) } return r = r A + r B + r and the sorted list T(n) " T( # n/2 ) + T( n/2& ) + O(n) T(n) = O(nlog n) 9 2

6 Closest Pair of Points 5.4 Closest Pair of Points Closest pair. Given n points in the plane, find a pair with smallest Euclidean distance between them. Fundamental geometric primitive.! Graphics, computer vision, 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 &(n 2 ) comparisons. -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 22 Closest Pair of Points: First Attempt Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. Divide. Sub-divide region into 4 quadrants. Obstacle. Impossible to ensure n/4 points in each piece

7 Closest Pair of Points Closest Pair of Points Algorithm.! Divide: draw vertical line so that roughly!n points on each side. Algorithm.! Divide: draw vertical line so that roughly!n points on each side.! Conquer: find closest pair in each side recursively Closest Pair of Points Closest Pair of Points Algorithm.! Divide: draw vertical line 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 &(n 2 ) Find closest pair with one point in each side, assuming that distance < ( ( = min(2, 2) 27 28

8 Closest Pair of Points Closest Pair of Points Find closest pair with one point in each side, assuming that distance < (.! Observation: only need to consider points within ( of line. Find closest pair with one point in each side, assuming that distance < (.! Observation: only need to consider points within ( of line.! Sort points in 2(-strip by their y coordinate ( = min(2, 2) 2 3 ( = min(2, 2) 2 ( 29 ( 3 Closest Pair of Points Closest Pair of Points Find closest pair with one point in each side, assuming that distance < (.! Observation: only need to consider points within ( of line. Def. et s i be the point in the 2(-strip, with the i th smallest y-coordinate.! Sort points in 2(-strip by their y coordinate.! Only check distances of those within positions in sorted list! Claim. If i j ) 2, then the distance between s i and s j is at least (. Pf j! No two points lie in same!(-by-!( box. 7 6! Two points at least 2 rows apart have distance ) 2(!().! 2 rows 29 3!(!( Fact. Still true if we replace 2 with 7. i 27 28!( 2 3 ( = min(2, 2) ( 3 ( ( 32

9 Closest Pair Algorithm Closest Pair of Points: Analysis Closest-Pair(p,, p n ) { Compute separation line such that half the points are on one side and half on the other side. ( = Closest-Pair(left half) ( 2 = Closest-Pair(right half) ( = min((, ( 2 ) Delete all points further than ( from separation line Sort remaining points by y-coordinate. Scan points in y-order and compare distance between each point and next neighbors. If any of these distances is less than (, update (. O(n log n) 2T(n / 2) O(n) O(n log n) O(n) Running time. T(n) " 2T( n/2) + O(n log n) # T(n) = O(n log 2 n) Q. Can we achieve O(n log n)? A. Yes. Dont sort points in strip from scratch each time.! Each recursive returns two lists: all points sorted by y coordinate, and all points sorted by x coordinate.! Sort by merging two pre-sorted lists. } return (. T(n) " 2T( n/2) + O(n) # T(n) = O(n log n) Integer Arithmetic 5.5 Integer Multiplication Add. Given two n-digit integers a and b, compute a + b.! O(n) bit operations. Multiply. Given two n-digit integers a and b, compute a " b.! Brute force solution: &(n 2 ) bit operations. * Multiply + Add 36

10 Divide-and-Conquer Multiplication: Warmup Karatsuba Multiplication To multiply two n-digit integers:! Multiply four!n-digit integers.! Add two!n-digit integers, and shift to obtain result. 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. x = 2 n /2 " x + x y = 2 n /2 " y + y ( ) ( 2 n /2 " y + y ) = 2 n " x y + 2 n /2 " x y + x y xy = 2 n /2 " x + x ( ) T(n) = 4T n / "(n) 23 # T(n) = "(n2 ) recursive calls add, shift assumes n is a power of 2 ( ) + x y x = 2 n /2 " x + x y = 2 n /2 " y + y xy = 2 n " x y + 2 n /2 " x y + x y ( ) + x y ( ) + x y = 2 n " x y + 2 n /2 " (x + x )(y + y ) # x y # x y A B A C C Theorem. [Karatsuba-Ofman, 962] Can multiply two n-digit integers in O(n.585 ) bit operations. ( ) + T( n/2 ) + T( + n /2 ) T(n) " T # n/2 & & + (n) recursive calls ( T(n) = O(n log 2 3 ) = O(n.585 ) add, subtract, shift Karatsuba: Recursion Tree " if n = T(n) = # 3T(n/2) + n otherwise log 2 n T(n) = " n 2 3 = k= ( ) k 3 2 ( ) +log 2 n # 3 2 # = 3n log2 3 # 2 Matrix Multiplication T(n) n T(n/2) T(n/2) T(n/2) 3(n/2) 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) 39

11 Matrix Multiplication Matrix Multiplication: Warmup Matrix multiplication. Given two n-by-n matrices A and B, compute C = AB. n c ij = " a ik b kj k= " c c 2 c n c 2 c 22 c 2n = M M O M # c n c n2 c nn & Brute force. &(n 3 ) arithmetic operations. " a a 2 a n a 2 a 22 a 2n ( M M O M # a n a n2 a nn & " b b b 2 n b 2 b 22 b 2n M M O M # b n b n2 b nn & 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. " C C 2 " = A A 2 " ( B B 2 # C 2 C 22 & # A 2 A 22 & # B 2 B 22 & ( ) + ( A 2 " B 2 ) ( ) + ( A 2 " B 22 ) ( ) + ( A 22 " B 2 ) ( ) + ( A 22 " B 22 ) C = A " B C 2 = A " B 2 C 2 = A 2 " B C 22 = A 2 " B 2 Fundamental question. Can we improve upon brute force? ( ) T(n) = 8T n/ "(n2 ) recursive calls add, form submatrices # T(n) = "(n 3 ) 4 42 Matrix Multiplication: Key Idea Fast Matrix Multiplication Key idea. multiply 2-by-2 block matrices with only 7 multiplications. " C C 2 " A = A 2 " ( B B 2 P = A " (B 2 # B 22 ) # C 2 C 22 & # A 2 A 22 & # B 2 B 22 & P 2 = ( A + A 2 ) " B 22 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. C = P 5 + P 4 " P 2 + P 6 C 2 = P + P 2 C 2 = P 3 + P 4 C 22 = P 5 + P " P 3 " P 7! 7 multiplications.! 8 = + 8 additions (or subtractions). P 3 = ( A 2 + A 22 ) " B P 4 = A 22 " (B 2 # B ) P 5 = ( A + A 22 ) " (B + B 22 ) P 6 = ( A 2 # A 22 ) " (B 2 + B 22 ) P 7 = ( A # A 2 ) " (B + B 2 ) Analysis.! Assume n is a power of 2.! T(n) = # arithmetic operations. ( ) T(n) = 7T n/ "(n2 ) # T(n) = "(n log 2 7 ) = O(n 2.8 ) recursive calls add, subtract 43 44

12 Fast Matrix Multiplication in Practice Fast Matrix Multiplication in Theory Implementation issues.! Sparsity.! Caching effects.! Numerical stability.! Odd matrix dimensions.! Crossover to classical algorithm around n = 28. Common misperception: "Strassen is 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 its useful is a subject of controversy. Remark. Can "Strassenize" Ax=b, determinant, eigenvalues, and other matrix ops. Q. Multiply two 2-by-2 matrices with only 7 scalar multiplications? A. Yes! [Strassen, 969] "(n log 2 7 ) = O(n 2.8 ) Q. Multiply two 2-by-2 matrices with only 6 scalar multiplications? A. Impossible. [Hopcroft and Kerr, 97] "(n log 2 6 ) = O(n 2.59 ) Q. Two 3-by-3 matrices with only 2 scalar multiplications? A. Also impossible. "(n log 3 2 ) = O(n 2.77 ) Q. Two 7-by-7 matrices with only 43,64 scalar multiplications? A. Yes! [Pan, 98] "(n log ) = O(n 2.8 ) Decimal wars.! December, 979: O(n ).! January, 98: O(n ) Fast Matrix Multiplication in Theory Best known. O(n ) [Coppersmith-Winograd, 987.] Conjecture. O(n 2+* ) for any * >. Caveat. Theoretical improvements to Strassen are progressively less practical. 47