Master Theorem

Notes maitre dhotel Theorem Slides by Christopher M. Bourke Instructor: Berthe Y. Choueiry Spring 2006 Computer comprehension & engineer 235 Introduction to Discrete Mathematics cse235@cse.unl.edu hold in Theorem I Notes When analyzing algorithms, rally that we b atomic number 18ly cargon about the asymptotic behavior. Recursive algorithms are no di?erent. Rather than solve exactly the recurrence congenator associated with the hail of an algorithm, it is enough to give an asymptotic characterization. The main son of a bitch for doing this is the rule theorem. Master Theorem II Notes Theorem (Master Theorem) allow T (n) be a monotonically change magnitude function that satis?es T (n) = aT ( n ) + f (n) b T (1) = c where a ? 1, b ? 2, c > 0. If f (n) ? ?(nd ) ? if ? ?(nd ) ?(nd logarithm n) if T (n) = ? ?(nlogb a ) if where d ? 0, accordingly a < bd a = bd a > bd Master Theorem Pitfalls Notes You cannot use the Master Theorem if T (n) is not monotone, ex: T (n) = sin n f (n) is not a polynomial, ex: T (n) = 2T ( n ) + 2n 2 ? b cannot be expressed as a constant, ex: T (n) = T ( n) Note here, that the Master Theorem does not solve a recurrence relation. Does the base fictitious character bear a concern? Master Theorem Example 1 Notes permit T (n) = T n 2 + 1 n2 + n. What are the parameters? 2 a = 1 b = 2 d = 2 Therefore which ramification down? Since 1 < 22 , causal agent 1 applies.

hence we finish that T (n) ? ?(nd ) = ?(n2 ) Master Theorem Example 2 Notes ? Let T (n) = 2T n 4 + n + 42. What are the parameters? a = 2 b = 4 d = 1 2 Therefore which condition? Since 2 = 4 2 , case 2 applies. Thus we conclude that ?! T (n) ? ?(nd log n) = ?( n log n) 1 Master Theorem Example 3 Notes Let T (n) = 3T n 2 + 3 n + 1. What are the parameters? 4 a = 3 b = 2 d = 1 Therefore which condition? Since 3 > 21 , case 3 applies. Thus we conclude that T (n) ? ?(nlogb a ) = ?(nlog2 3 ) Note that log2 3 ? 1.5849 . . .. undersurface we say that T (n) ? ?(n1.5849 ) ? quaternary Condition...If you want to get a full essay, order it on our website: OrderEssay.net

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