MIT6_042JF10_rec14_sol

MIT6_042JF10_rec14_sol - 6.042/18.062J Mathematics for...

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Unformatted text preview: 6.042/18.062J Mathematics for Computer Science October 24, 2006 Tom Leighton and Marten van Dijk Notes for Recitation 14 1 The Akra-Bazzi Theorem Theorem 1 (Akra-Bazzi, strong form) . Suppose that: is defined for x x T ( x ) = k a i T ( b i x + h i ( x )) + g ( x ) for x > x i =1 where: a 1 ,...,a k are positive constants b 1 ,...,b k are constants between and 1 x is large enough in a technical sense we leave unspecified | g ( x ) | = O ( x c ) for some c N | h i ( x ) | = O ( x/ log 2 x ) Then: x T ( x ) = x p 1 + u g ( p u +1 ) du 1 k where p satisfies the equation i =1 a i b p i = 1 . The only difference between the strong and weak forms of Akra-Bazzi is the appearance of this h i ( x ) term in the recurrence, where h i ( x ) represents a small change in the size of the subproblems ( O ( x/ log 2 x )). Notice that, despite the change in the recurrence, the solution T ( x ) remains the same in both the strong and weak forms, with no dependence on h i ( x )! In algorithmic terms, this means that small changes in the size of subproblems have no impact on the asymptotic running time. Example: Lets compare the bounds for the following divide-and-conquer recurrences. n n T a ( n ) = 3 T + n T b ( n ) = 3 T + n 3 3 Recitation 14 2 For T a ( n ) we have a 1 = 3, b 1 = 1 / 3, g ( n ) = n , p = 1. For T b ( n ) we have the same parameters as for T a ( n ), plus h 1 ( n ) = n/ 3 n/ 3. Using the strong Akra-Bazzi form, the h 1 ( n ) falls out of the equation: n u T a ( n ) = T b ( n ) = ( n (1 + 2 du )) = ( n log n ) ....
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MIT6_042JF10_rec14_sol - 6.042/18.062J Mathematics for...

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