1 / 38

CSE 421 Algorithms

CSE 421 Algorithms. Richard Anderson Lecture 13 Divide and Conquer. What you really need to know about recurrences. Work per level changes geometrically with the level Geometrically increasing (x > 1) The bottom level wins Geometrically decreasing (x < 1) The top level wins

thurstons
Download Presentation

CSE 421 Algorithms

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CSE 421Algorithms Richard Anderson Lecture 13 Divide and Conquer

  2. What you really need to know about recurrences • Work per level changes geometrically with the level • Geometrically increasing (x > 1) • The bottom level wins • Geometrically decreasing (x < 1) • The top level wins • Balanced (x = 1) • Equal contribution

  3. T(n) = aT(n/b) + nc • Balanced: a = bc • Increasing: a > bc • Decreasing: a < bc

  4. Classify the following recurrences(Increasing, Decreasing, Balanced) • T(n) = n + 5T(n/8) • T(n) = n + 9T(n/8) • T(n) = n2 + 4T(n/2) • T(n) = n3 + 7T(n/2) • T(n) = n1/2 + 3T(n/4)

  5. Divide and Conquer Algorithms • Split into sub problems • Recursively solve the problem • Combine solutions • Make progress in the split and combine stages • Quicksort – progress made at the split step • Mergesort – progress made at the combine step

  6. Closest Pair Problem • Given a set of points find the pair of points p, q that minimizes dist(p, q)

  7. Divide and conquer • If we solve the problem on two subsets, does it help? (Separate by median x coordinate) d1 d2

  8. Packing Lemma Suppose that the minimum distance between points is at least d, what is the maximum number of points that can be packed in a ball of radius d?

  9. Combining Solutions • Suppose the minimum separation from the sub problems is d • In looking for cross set closest pairs, we only need to consider points with d of the boundary • How many cross border interactions do we need to test?

  10. A packing lemma bounds the number of distances to check d

  11. Details • Preprocessing: sort points by y • Merge step • Select points in boundary zone • For each point in the boundary • Find highest point on the other side that is at most d above • Find lowest point on the other side that is at most d below • Compare with the points in this interval (there are at most 6)

  12. Identify the pairs of points that are compared in the merge step following the recursive calls

  13. Algorithm run time • After preprocessing: • T(n) = cn + 2 T(n/2)

  14. Divide and Conquer Algorithms • Mergesort, Quicksort • Strassen’s Algorithm • Closest Pair Algorithm (2d) • Inversion counting • Integer Multiplication (Karatsuba’s Algorithm) • FFT • Polynomial Multiplication • Convolution

  15. Inversion Problem • Let a1, . . . an be a permutation of 1 . . n • (ai, aj) is an inversion if i < j and ai > aj • Problem: given a permutation, count the number of inversions • This can be done easily in O(n2) time • Can we do better? 4, 6, 1, 7, 3, 2, 5

  16. Counting Inversions Count inversions on lower half Count inversions on upper half Count the inversions between the halves

  17. Count the Inversions 4 2 3 1 8 6 14 10 19 43

  18. Problem – how do we count inversions between sub problems in O(n) time? • Solution – Count inversions while merging Standard merge algorithm – add to inversion count when an element is moved from the upper array to the solution

  19. Use the merge algorithm to count inversions Indicate the number of inversions for each element detected when merging

  20. Inversions • Counting inversions between two sorted lists • O(1) per element to count inversions • Algorithm summary • Satisfies the “Standard recurrence” • T(n) = 2 T(n/2) + cn

  21. Integer Arithmetic 9715480283945084383094856701043643845790217965702956767 + 1242431098234099057329075097179898430928779579277597977 Runtime for standard algorithm to add two n digit numbers: 2095067093034680994318596846868779409766717133476767930 X 5920175091777634709677679342929097012308956679993010921 Runtime for standard algorithm to multiply two n digit numbers:

  22. Recursive Algorithm (First attempt) x = x1 2n/2 + x0 y = y1 2n/2 + y0 xy = (x1 2n/2 + x0) (y1 2n/2 + y0) = x1y1 2n + (x1y0 + x0y1)2n/2 + x0y0 Recurrence: Run time:

  23. Simple algebra x = x1 2n/2 + x0 y = y1 2n/2 + y0 xy = x1y1 2n + (x1y0 + x0y1) 2n/2 + x0y0 p = (x1 + x0)(y1 + y0) = x1y1 + x1y0 + x0y1 + x0y0

  24. Karatsuba’s Algorithm Multiply n-digit integers x and y Let x = x1 2n/2 + x0 and y = y1 2n/2 + y0 Recursively compute a = x1y1 b = x0y0 p = (x1 + x0)(y1 + y0) Return a2n + (p – a – b)2n/2 + b Recurrence: T(n) = 3T(n/2) + cn

  25. FFT, Convolution and Polynomial Multiplication • Preview • FFT - O(n log n) algorithm • Evaluate a polynomial of degree n at n points in O(n log n) time • Computation of Convolution and Polynomial Multiplication (in O(n log n)) time

  26. Complex Analysis • Polar coordinates: reqi • eqi = cos q + i sin q • a is a nth root of unity if an = 1 • Square roots of unity: +1, -1 • Fourth roots of unity: +1, -1, i, -i • Eighth roots of unity: +1, -1, i, -i, b + ib, b - ib, -b + ib, -b - ib where b = sqrt(2)

  27. e2pki/n • e2pi = 1 • epi = -1 • nth roots of unity: e2pki/n for k = 0 …n-1 • Notation: wk,n = e2pki/n • Interesting fact: 1 + wk,n + w2k,n + w3k,n + . . . + wn-1k,n = 0 for k != 0

  28. Convolution • a0, a1, a2, . . ., am-1 • b0, b1, b2, . . ., bn-1 • c0, c1, c2, . . .,cm+n-2 where ck = Si+j=kaibj

  29. Applications of Convolution • Polynomial Multiplication • Signal processing • Gaussian smoothing • Sequence a1, a2, . . ., an • Mask, w-k, w-(k-1), . . ., w-1, w0, w1, . . ., wk-1, wk • Addition of random variables

  30. FFT Overview • Polynomial interpolation • Given n+1 points (xi,yi), there is a unique polynomial P of degree at most n which satisfies P(xi) = yi

  31. Polynomial Multiplication n-1 degree polynomials A(x) = a0 + a1x + a2x2 + … +an-1xn-1, B(x) = b0 + b1x + b2x2 + …+ bn-1xn-1 C(x) = A(x)B(x) C(x)=c0+c1x + c2x2 + … + c2n-2x2n-2 p1, p2, . . ., p2n A(p1), A(p2), . . ., A(p2n) B(p1), B(p2), . . ., B(p2n) C(p1), C(p2), . . ., C(p2n) C(pi) = A(pi)B(pi)

  32. FFT • Polynomial A(x) = a0 + a1x + . . . + an-1xn-1 • Compute A(wj,n) for j = 0, . . ., n-1 • For simplicity, n is a power of 2

  33. Useful trick A(x) = a0 + a1x + a2x2 + a3x3 +. . . + an-1xn-1 Aeven(x) = a0 + a2x + a4x2 + . . . + an-2x(n-2)/2 Aodd(x) = a1+ a3x + a5x2 + …+ an-1x(n-2)/2 Show: A(x) = Aeven(x2) + x Aodd(x2)

  34. Lemma: w2j,2n = wj,n Squares of 2nth roots of unity are nth roots of unity wj,2n = e2pji/2n The detail of j >= n is being ignored – don’t mention unless it comes up.

  35. FFT Algorithm // Evaluate the 2n-1th degree polynomial A at // w0,2n, w1,2n, w2,2n, . . ., w2n-1,2n FFT(A, 2n) Recursively compute FFT(Aeven, n) Recursively compute FFT(Aodd, n) for j = 0 to 2n-1 A(wj,2n) = Aeven(w2j,2n) + wj,2nAodd(w2j,2n)

  36. Polynomial Multiplication • n-1th degree polynomials A and B • Evaluate A and B at w0,2n, w1,2n, . . ., w2n-1,2n • Compute C(wj,2n) for j = 0 to 2n -1 • We know the value of a 2n-2th degree polynomial at 2n points – this determines a unique polynomial, we just need to determine the coefficients

  37. Now the magic happens . . . C(x) = c0 + c1x + c2x2 + … + c2n-1x2n-1 (we want to compute the ci’s) Let dj = C(wj,2n) D(x) = d0 + d1x + d2x2 + … + d2n-1x2n-1 Evaluate D(x) at the 2nth roots of unity D(wj,2n) = [see text for details] = 2nc2n-j

  38. Polynomial Interpolation • Build polynomial from the values of C at the 2nth roots of unity • Evaluate this polynomial at the 2nth roots of unity

More Related