# cse 421 algorithms richard anderson lecture 13 divide and conquer

Post on 22-Dec-2015

212 views

Category:

## Documents

0 download

Tags:

• #### solution count inversions

Embed Size (px)

TRANSCRIPT

• Slide 1
• CSE 421 Algorithms Richard Anderson Lecture 13 Divide and Conquer
• Slide 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
• Slide 3
• T(n) = aT(n/b) + n c Balanced: a = b c Increasing: a > b c Decreasing: a < b c
• Slide 4
• Classify the following recurrences (Increasing, Decreasing, Balanced) T(n) = n + 5T(n/8) T(n) = n + 9T(n/8) T(n) = n 2 + 4T(n/2) T(n) = n 3 + 7T(n/2) T(n) = n 1/2 + 3T(n/4)
• Slide 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
• Slide 6
• Closest Pair Problem Given a set of points find the pair of points p, q that minimizes dist(p, q)
• Slide 7
• Divide and conquer If we solve the problem on two subsets, does it help? (Separate by median x coordinate) 11 22
• Slide 8
• Packing Lemma Suppose that the minimum distance between points is at least , what is the maximum number of points that can be packed in a ball of radius ?
• Slide 9
• Combining Solutions Suppose the minimum separation from the sub problems is In looking for cross set closest pairs, we only need to consider points with of the boundary How many cross border interactions do we need to test?
• Slide 10
• A packing lemma bounds the number of distances to check
• Slide 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 above Find lowest point on the other side that is at most below Compare with the points in this interval (there are at most 6)
• Slide 12
• Identify the pairs of points that are compared in the merge step following the recursive calls
• Slide 13
• Algorithm run time After preprocessing: T(n) = cn + 2 T(n/2)
• Slide 14
• Divide and Conquer Algorithms Mergesort, Quicksort Strassens Algorithm Closest Pair Algorithm (2d) Inversion counting Integer Multiplication (Karatsubas Algorithm) FFT Polynomial Multiplication Convolution
• Slide 15
• Inversion Problem Let a 1,... a n be a permutation of 1.. n (a i, a j ) is an inversion if i a j Problem: given a permutation, count the number of inversions This can be done easily in O(n 2 ) time Can we do better? 4, 6, 1, 7, 3, 2, 5
• Slide 16
• Counting Inversions 11124172315951686131014 Count inversions on lower half Count inversions on upper half Count the inversions between the halves
• Slide 17
• 111241 72315 11124172315 951686131014 951686131014 Count the Inversions 11124172315951686131014 4123 10 19 86 43
• Slide 18
• Problem how do we count inversions between sub problems in O(n) time? Solution Count inversions while merging 12347111215 568910131416 Standard merge algorithm add to inversion count when an element is moved from the upper array to the solution
• Slide 19
• Use the merge algorithm to count inversions 141112 23715 589166101314 Indicate the number of inversions for each element detected when merging
• Slide 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 x x x x x x x x y y y y y y y y z z z z z z z z z z z z z z z z
• Slide 21
• Integer Arithmetic 9715480283945084383094856701043643845790217965702956767 + 1242431098234099057329075097179898430928779579277597977 2095067093034680994318596846868779409766717133476767930 X 5920175091777634709677679342929097012308956679993010921 Runtime for standard algorithm to add two n digit numbers: Runtime for standard algorithm to multiply two n digit numbers:
• Slide 22
• Recursive Algorithm (First attempt) x = x 1 2 n/2 + x 0 y = y 1 2 n/2 + y 0 xy = (x 1 2 n/2 + x 0 ) (y 1 2 n/2 + y 0 ) = x 1 y 1 2 n + (x 1 y 0 + x 0 y 1 )2 n/2 + x 0 y 0 Recurrence: Run time:
• Slide 23
• Simple algebra x = x 1 2 n/2 + x 0 y = y 1 2 n/2 + y 0 xy = x 1 y 1 2 n + (x 1 y 0 + x 0 y 1 ) 2 n/2 + x 0 y 0 p = (x 1 + x 0 )(y 1 + y 0 ) = x 1 y 1 + x 1 y 0 + x 0 y 1 + x 0 y 0
• Slide 24
• Karatsubas Algorithm Multiply n-digit integers x and y Let x = x 1 2 n/2 + x 0 and y = y 1 2 n/2 + y 0 Recursively compute a = x 1 y 1 b = x 0 y 0 p = (x 1 + x 0 )(y 1 + y 0 ) Return a2 n + (p a b)2 n/2 + b Recurrence: T(n) = 3T(n/2) + cn
• Slide 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
• Slide 26
• Complex Analysis Polar coordinates: re i e i = cos + i sin a is a n th root of unity if a n = 1 Square roots of unity: +1, -1 Fourth roots of unity: +1, -1, i, -i Eighth roots of unity: +1, -1, i, -i, + i , - i , - + i , - - i where = sqrt(2)
• Slide 27
• e 2 ki/n e 2 i = 1 e i = -1 n th roots of unity: e 2 ki/n for k = 0 n-1 Notation: k,n = e 2 ki/n Interesting fact: 1 + k,n + 2 k,n + 3 k,n +... + n-1 k,n = 0 for k != 0
• Slide 28
• Convolution a 0, a 1, a 2,..., a m-1 b 0, b 1, b 2,..., b n-1 c 0, c 1, c 2,...,c m+n-2 where c k = i+j=k a i b j
• Slide 29
• Applications of Convolution Polynomial Multiplication Signal processing Gaussian smoothing Sequence a 1, a 2,..., a n Mask, w -k, w -(k-1),..., w -1, w 0, w 1,..., w k-1, w k Addition of random variables
• Slide 30
• FFT Overview Polynomial interpolation Given n+1 points (x i,y i ), there is a unique polynomial P of degree at most n which satisfies P(x i ) = y i
• Slide 31
• Polynomial Multiplication n-1 degree polynomials A(x) = a 0 + a 1 x + a 2 x 2 + +a n-1 x n-1, B(x) = b 0 + b 1 x + b 2 x 2 + + b n-1 x n-1 C(x) = A(x)B(x) C(x)=c 0 +c 1 x + c 2 x 2 + + c 2n-2 x 2n-2 p 1, p 2,..., p 2n A(p 1 ), A(p 2 ),..., A(p 2n ) B(p 1 ), B(p 2 ),..., B(p 2n ) C(p i ) = A(p i )B(p i ) C(p 1 ), C(p 2 ),..., C(p 2n )
• Slide 32
• FFT Polynomial A(x) = a 0 + a 1 x +... + a n-1 x n-1 Compute A( j,n ) for j = 0,..., n-1 For simplicity, n is a power of 2
• Slide 33
• Useful trick A(x) = a 0 + a 1 x + a 2 x 2 + a 3 x 3 +... + a n-1 x n-1 A even (x) = a 0 + a 2 x + a 4 x 2 +... + a n-2 x (n-2)/2 A odd (x) = a 1 + a 3 x + a 5 x 2 + + a n-1 x (n-2)/2 Show: A(x) = A even (x 2 ) + x A odd (x 2 )
• Slide 34
• Lemma: 2 j,2n = j,n Squares of 2n th roots of unity are n th roots of unity
• Slide 35
• FFT Algorithm // Evaluate the 2n-1 th degree polynomial A at // 0,2n, 1,2n, 2,2n,..., 2n-1,2n FFT(A, 2n) Recursively compute FFT(A even, n) Recursively compute FFT(A odd, n) for j = 0 to 2n-1 A( j,2n ) = A even ( 2 j,2n ) + j,2n A odd ( 2 j,2n )
• Slide 36
• Polynomial Multiplication n-1 th degree polynomials A and B Evaluate A and B at 0,2n, 1,2n,..., 2n-1,2n Compute C( j,2n ) for j = 0 to 2n -1 We know the value of a 2n-2 th degree polynomial at 2n points this determines a unique polynomial, we just need to determine the coefficients
• Slide 37
• Now the magic happens... C(x) = c 0 + c 1 x + c 2 x 2 + + c 2n-1 x 2n-1 (we want to compute the c i s) Let d j = C( j,2n ) D(x) = d 0 + d 1 x + d 2 x 2 + + d 2n-1 x 2n-1 Evaluate D(x) at the 2n th roots of unity D( j,2n ) = [see text for details] = 2nc 2n-j
• Slide 38
• Polynomial Interpolation Build polynomial from the values of C at the 2n th roots of unity Evaluate this polynomial at the 2n th roots of unity

Recommended