CSE 202 Divide-and-conquer algorithms. Fan Chung Graham UC San Diego

Transcription

1 CSE 22 Divide-and-conquer algorithms Fan Chung Graham UC San Diego

2 Announcements Homework due today before the class. About homework, write your own homework, allowing oral discussion with one fixed partner. Fan s office hour will be held at CSE226 this week. Olivia s office hour will be changed.

3 Chapter 4 Greedy Algorithms Slides by Kevin Wayne. Copyright 25 Pearson-Addison Wesley. All rights reserved.

4 4.4 Shortest Paths in a Graph shortest path from Princeton CS department to Einstein's house Shortest path tree in Bay area

5 Coin Changing Greed is good. Greed is right. Greed works. Greed clarifies, cuts through, and captures the essence of the evolutionary spirit. - Gordon Gecko (Michael Douglas)

6 Trees with at most 4 edges 2

7 A useful fact about trees Any tree on n vertices contains a vertex v whose removal separates the remaining graph into two parts, one of which is of sizes at most n/2 and the other is at most 2n/3. 3

8 A useful fact about trees Any tree on n vertices contains a vertex v whose removal separates the remaining graph into two parts, one of which is of sizes at most n/2 and the other is at most 2n/3. Try to write a proof for this! 5

9 A planar graph is a graph that can be drawn in the plane without crossings. 6

10 A planar graph is a graph that can be drawn in the plane without any crossing. Are these planar graphs? 7

11 A planar graph is a graph that can be drawn in the plane without any crossing. Are these planar graphs? 8

12 A useful fact about planar graphs Any planar graph on n vertices contains cn n vertices whose removal separates the remaining graph into two parts, one of which is of sizes at most n/2 and the other is at most 2n/3. Tarjan and Lipton, 977 c = 2 2 9

13 Chapter 5 Divide and Conquer Slides by Kevin Wayne. Copyright 25 Pearson-Addison Wesley. All rights reserved.

14 Sorting Sorting. Given n elements, rearrange in ascending order. 3, 6, 5, 2,, 4, 2, 3, 4, 5, 6 B, U, S, H B, H, S, U Obvious sorting applications. List files in a directory. Organize an MP3 library. List names in a phone book. Display Google PageRank results. 3

15 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) 6

16 Merging Merging. Combine two pre-sorted lists into a sorted whole. How to merge efficiently?! Linear number of comparisons.! 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 7

17 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A auxiliary array

18 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G auxiliary array 2

19 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H auxiliary array 3

20 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H I auxiliary array 4

21 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H I L auxiliary array 5

22 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H I L M auxiliary array 6

23 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H I L M O auxiliary array 7

24 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. smallest smallest A G L O R H I M S T A G H I L M O R auxiliary array 8

25 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. first half exhausted smallest A G L O R H I M S T A G H I L M O R S auxiliary array 9

26 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. first half exhausted smallest A G L O R H I M S T A G H I L M O R S T auxiliary array

27 Merging Merge.! Keep track of smallest element in each sorted half.! Insert smallest of two elements into auxiliary array.! Repeat until done. first half exhausted second half exhausted A G L O R H I M S T A G H I L M O R S T auxiliary array

28 Merging Merging. Combine two pre-sorted lists into a sorted whole. How to merge efficiently?! Linear number of comparisons.! 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 7

29

30 A Useful Recurrence Relation Def. T(n) = number of comparisons to mergesort an input of size n. Mergesort recurrence. T(n)! & ( ' ( ) if n = ( " # ) T n /2!#" # $ solve left half ( $ % ) + T n /2!#" # $ solve right half + % n otherwise merging Solution. T(n) = O(n log 2 n). Proof: First try the recurrence T (n) 2T (n / 2) + n Assorted proofs. We describe several ways to prove this recurrence. Prove by induction: Suppose for k < n, T (k) ck log k. Initially we assume n is a power of 2 and replace! with =. T (n) 2c(n / 2)log(n / 2) + n cn(logn ) + n cnlogn 8

31 Solving recurrences: T (n) = 2T (n / 2) + n T (n) = O(n logn)

32 Solving recurrences: T (n) = 2T (n / 2) + n T (n) = O(n logn) T (n) = 2T (n / 2) + n 2

33 Solving recurrences: T (n) = 2T (n / 2) + n T (n) = O(n logn) T (n) = 2T (n / 2) + n 2?

34 Solving recurrences: T (n) = 2T (n / 2) + n T (n) = O(n logn) T (n) = 2T (n / 2) + n 2? T (n) = 3T (n / 2) + n log n

35 Solving recurrences: T (n) = 2T (n / 2) + n T (n) = O(n logn) T (n) = 2T (n / 2) + n 2? T (n) = 3T (n / 2) + n log n?

36 5.3 Counting Inversions

37 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!(n 2 ) pairs i and j. 24

38 Applications Applications.! Voting theory.! Collaborative filtering.! Measuring the "sortedness" of an array.! Sensitivity analysis of Google's ranking function.! Rank aggregation for meta-searching on the Web.! Nonparametric statistics (e.g., Kendall's Tau distance). 25

39 Counting Inversions: Divide-and-Conquer Divide-and-conquer

40 Counting Inversions: Divide-and-Conquer Divide-and-conquer.! Divide: separate list into two pieces Divide: O()

41 Counting Inversions: Divide-and-Conquer Divide-and-conquer.! Divide: separate list 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, -7 28

42 Counting Inversions: Divide-and-Conquer 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() 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 = =

43 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

44 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves 6 2 auxiliary array Total: 6 2

45 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves 6 2 auxiliary array Total: 6 3

46 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 4

47 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 5

48 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 6

49 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 7

50 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 8

51 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 9

52 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 + 3

53 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: 6 + 3

54 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

55 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

56 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

57 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

58 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

59 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

60 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

61 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

62 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

63 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. first half exhausted i = two sorted halves auxiliary array Total:

64 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

65 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

66 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total:

67 Merge and Count Merge and count step.! Given two sorted halves, count number of inversions where a i and a j are in different halves.! Combine two sorted halves into sorted whole. i = two sorted halves auxiliary array Total: = 3 25

68 Counting Inversions: Combine 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. to maintain sorted invariant blue-green inversions: Count: O(n) Merge: O(n) T(n)! T " n/2# ( ) + T ( $ n/2% ) + O(n) & T(n) = O(nlog n) 3

69 Counting Inversions: Implementation Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] L is sorted. Sort-and-Count(L) { if list L has one element return and the list L 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, L)! Merge-and-Count(A, B) } return r = r A + r B + r and the sorted list L 3

70 Merge-and-Count(A,B) { Initialize Pointer to the front of A. Pointer2 to the front of B. Count = While A and B are nonempty, compare a i at Pointer with b at Pointer2, append the smaller one to output and advance the pointer by one. If b j is smaller, then increase Count by the number of elements still in A. Endwhile j

71 Counting Inversions: Implementation Pre-condition. [Merge-and-Count] A and B are sorted. Post-condition. [Sort-and-Count] L is sorted. Sort-and-Count(L) { if list L has one element return and the list L 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, L)! Merge-and-Count(A, B) } return r = r A + r B + r and the sorted list L 3

72 5.4 Closest Pair of Points

73 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 33

74 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. L 34

75 Closest Pair of Points: First Attempt Divide. Sub-divide region into 4 quadrants. Obstacle. Impossible to ensure n/4 points in each piece. L 35

76 Closest Pair of Points Algorithm.! Divide: draw vertical line L so that roughly!n points on each side. L 36

77 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

78 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. seems like!(n 2 )! Return best of 3 solutions. L

79 Closest Pair of Points Find closest pair with one point in each side, assuming that distance <!. L 2 2! = min(2, 2) 39

80 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 L. L 2 2! = min(2, 2)! 4

81 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 L.! Sort points in 2!-strip by their y coordinate. 7 L ! = min(2, 2) 2! 4

82 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 L.! Sort points in 2!-strip by their y coordinate.! Only check distances of those within positions in sorted list! 7 L ! = min(2, 2) 2! 42

83 Closest Pair of Points Def. Let s i be the point in the 2!-strip, with the i th smallest y-coordinate. 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.! 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!! 26 25!! 43

84 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.! = Closest-Pair(left half)! 2 = Closest-Pair(right half)! = min(!,! 2 ) Delete all points further than! from separation line L 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) } return!. 44

85 Closest Pair of Points: Analysis 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. Don't 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. T(n)! 2T ( n/2) + O(n) " T(n) = O(n log n) 45

86 5.5 Integer Multiplication

87 47 Integer Arithmetic 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. * + Add Multiply

88 Divide-and-Conquer Multiplication: Warmup To multiply two n-digit integers:! Multiply four!n-digit integers.! Add two!n-digit integers, and shift 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 ( ) + x y ( ) T(n) = 4T n /2!# "# $ +!!(n) " $ " T(n) =!(n2 ) recursive calls add, shift assumes n is a power of 2 48

89 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. 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!####### "####### $ recursive calls ' T(n) = O(n log 2 3 ) = O(n.585 ) + &(n)!# "# $ add, subtract, shift 49

90 Karatsuba: Recursion Tree T(n) =! " # if 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 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) 5

91 Matrix Multiplication

92 Matrix Multiplication 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 & # " " # " & # " c n c n2! & c nn % =! a a 2! a n $ # & # a 2 a 22! a 2n & # " " # " & # " a n a n2! & a nn % '! b b 2! b n $ # & # b 2 b 22! b 2n & # " " # " & # " b n b n2! & b nn % Brute force.!(n 3 ) arithmetic operations. Fundamental question. Can we improve upon brute force? 52

93 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.! 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 ( ) T(n) = 8T n/2!# "# $ +!(n2 )!##" ## $ recursive calls add, form submatrices " T(n) =!(n 3 ) 53

94 Matrix Multiplication: Key Idea Key idea. multiply 2-by-2 block matrices with only 7 multiplications.! C C 2 $! # & = A A 2 $! # & ' B B 2 $ # & " C 2 C 22 % " A 2 A 22 % " B 2 B 22 % 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 P = A! (B 2 " B 22 ) P 2 = ( A + A 2 )! B 22 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 )! 7 multiplications.! 8 = + 8 additions (or subtractions). 54

95 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. Analysis.! Assume n is a power of 2.! T(n) = # arithmetic operations. ( ) T(n) = 7T n /2!# "# $ +!(n2 )!#" # $ recursive calls add, subtract " T(n) =!(n log 2 7 ) = O(n 2.8 ) 55

96 Fast Matrix Multiplication in Practice 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 it's useful is a subject of controversy. Remark. Can "Strassenize" Ax=b, determinant, eigenvalues, and other matrix ops. 56

97 Fast Matrix Multiplication in Theory 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] Q. Two 3-by-3 matrices with only 2 scalar multiplications? A. Also impossible.!(n log 2 6 ) = O(n 2.59 )!(n log 3 2 ) = O(n 2.77 ) Q. Two 7-by-7 matrices with only 43,64 scalar multiplications? A. Yes! [Pan, 98] Decimal wars.! December, 979: O(n ).! January, 98: O(n ).!(n log ) = O(n 2.8 ) 57

98 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. 58