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1 CSE 417: Algorithms and Computational Complexity Winter 2001 Lecture 13 Instructor: Paul Beame

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2 All-pairs shortest paths If no negative-weight edges and sparse graphs run Dijsktra from each vertex O(nm log n) What about other cases?

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3 Floyd’s Algorithm Idea Interior vertices in a path Dijsktra’s Algorithm at each step always computed shortest paths that only had interior vertices from a set S at each step Floyd’s Algorithm slowly add interior vertices on fixed schedule rather than depending on the graph v w interior vertices

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4 Floyd’s Algorithm Vertices V={v 1,...,v n } Let D k [i,j] = length of shortest path from v i to v j that only allows {v 1,...,v k } as interior vertices Note D 0 [i,j] = weight of edge (v i,v j ) if (v i,v j ) E if (v i,v j ) E D n [i,j] = length of shortest path from v i to v j

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5 Floyd’s Algorithm vivi vertices from v 1,...,v k vivi vivi vertices from v 1,...,v k-1 vivi Computing D k [i,j] Case 1: v k not used D k [i,j] = D k-1 [i,j]

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6 Floyd’s Algorithm vjvj vkvk vertices from v 1,...,v k-1 Case 2: v k used vertices from v 1,...,v k-1 vivi

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7 Floyd’s Algorithm vjvj vkvk vertices from v 1,...,v k-1 Case 2: v k used vertices from v 1,...,v k-1 vivi not on shortest path since no negative cycles

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8 Floyd’s Algorithm vjvj vkvk vertices from v 1,...,v k-1 Case 2: v k used vivi vertices from v 1,...,v k-1 D k [i,j] = D k-1 [i,k] + D k-1 [k,j]

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9 Floyd’s Algorithm Floyd(G) D 0 weight matrix of G for k=1 to n do for i=1 to n do for i=1 to n do D k [i,j] min{ D k-1 [i,j], D k-1 [i,k]+D k-1 [k,j]} endfor endfor endfor O(n 3 ) time O(n 2 ) storage by maintaining array for last two values of k

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10 Multiplying Faster On the first HW you analyzed our usual algorithm for multiplying numbers (n 2 ) time We can do better! We’ll describe the basic ideas by multiplying polynomials rather than integers Advantage is we don’t get confused by worrying about carries at first

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11 Note on Polynomials These are just formal sequences of coefficients so when we show something multiplied by x k it just means shifted k places to the left

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12 Polynomial Multiplication Given: Degree n-1 polynomials P and Q P = a n-1 x n-1 + a n-2 x n-2 + … + a 2 x 2 + a 1 x + a 0 Q = b n-1 x n-1 + b n-2 x n-2 + … + b 2 x 2 + b 1 x + b 0 Compute: Degree 2n-2 Polynomial P Q P Q = a n-1 b n-1 x 2n-2 + (a n-1 b n-2 + a n-2 b n-1 ) x 2n-3 +... + (a n-1 b i+1 +a n-2 b i+2 +...+a i+1 b n-1 ) x n+i +... + a 0 b 0 Obvious Algorithm: Compute all a i b j and collect terms (n 2 ) time

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13 Naive Divide and Conquer Assume n is a power of 2 P = (a n-1 x n/2-1 + a n-2 x n/2-2 + … + a n/2 ) x n/2 + (a n/2-1 x n/2-1 +... + a 2 x 2 + a 1 x + a 0 ) = P 1 x n/2 + P 0 Q = Q 1 x n/2 + Q 0 P Q = (P 1 x n/2 +P 0 )(Q 1 x n/2 +Q 0 ) = P 1 Q 1 x n +(P 1 Q 0 +P 0 Q 1 )x n/2 +P 0 Q 0 4 sub-problems of size n/2 + plus linear combining T(n)=4T(n/2)+cn Solution T(n) = O(n 2 )

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14 Karatsuba’s Algorithm A better way to compute the terms Compute P 0 Q 0 P 1 Q 1 (P 1 +P 0 )(Q 1 +Q 0 ) which is P 1 Q 1 +P 1 Q 0 +P 0 Q 1 +P 0 Q 0 Then P 0 Q 1 +P 1 Q 0 = (P 1 +P 0 )(Q 1 +Q 0 ) - P 0 Q 0 - P 1 Q 1 3 sub-problems of size n/2 plus O(n) work T(n) = 3 T(n/2) + cn T(n) = O(n ) where = log 2 3 = 1.59...

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