Transcription of Master Theorem: Practice Problems and Solutions
1 Master Theorem: Practice Problems and SolutionsMaster TheoremThe Master Theorem applies to recurrences of the following form:T(n) =aT(n/b) +f(n)wherea 1 andb >1 are constants andf(n) is an asymptotically positive are 3 cases:1. Iff(n) =O(nlogba ) for some constant >0, thenT(n) = (nlogba).2. Iff(n) = (nlogbalogkn) with1k 0, thenT(n) = (nlogbalogk+1n).3. Iff(n) = (nlogba+ ) with >0, andf(n) satisfies the regularity condition, thenT(n) = (f(n)).Regularity condition:af(n/b) cf(n) for some constantc <1 and all sufficiently ProblemsFor each of the following recurrences, give an expression for the runtimeT(n) if the recurrence can besolved with the Master Theorem.
2 Otherwise, indicate that the Master Theorem does not (n) = 3T(n/2) + (n) = 4T(n/2) + (n) =T(n/2) + (n) = 2nT(n/2) + (n) = 16T(n/4) + (n) = 2T(n/2) +nlogn1most of the time,k= (n) = 2T(n/2) + (n) = 2T(n/4) + (n) = (n/2) + 1 (n) = 16T(n/4) +n! (n) = 2T(n/2) + (n) = 3T(n/2) + (n) = 3T(n/3) + (n) = 4T(n/2) + (n) = 3T(n/4) + (n) = 3T(n/3) + (n) = 6T(n/3) + (n) = 4T(n/2) + (n) = 64T(n/8) (n) = 7T(n/3) + (n) = 4T(n/2) + (n) =T(n/2) +n(2 cosn) (n) = 3T(n/2) +n2= T(n) = (n2) (Case 3) (n) = 4T(n/2) +n2= T(n) = (n2logn) (Case 2) (n) =T(n/2) + 2n= (2n) (Case 3) (n) = 2nT(n/2) +nn= Does not apply (ais not constant) (n) = 16T(n/4) +n= T(n) = (n2) (Case 1) (n) = 2T(n/2) +nlogn= T(n) =nlog2n(Case 2) (n) = 2T(n/2)
3 +n/logn= Does not apply (non-polynomial difference betweenf(n) andnlogba) (n) = 2T(n/4) + T(n) = ( ) (Case 3) (n) = (n/2) + 1/n= Does not apply (a <1) (n) = 16T(n/4) +n! = T(n) = (n!) (Case 3) (n) = 2T(n/2) + logn= T(n) = ( n) (Case 1) (n) = 3T(n/2) +n= T(n) = (nlg3) (Case 1) (n) = 3T(n/3) + n= T(n) = (n) (Case 1) (n) = 4T(n/2) +cn= T(n) = (n2) (Case 1) (n) = 3T(n/4) +nlogn= T(n) = (nlogn) (Case 3) (n) = 3T(n/3) +n/2 = T(n) = (nlogn) (Case 2) (n) = 6T(n/3) +n2logn= T(n) = (n2logn) (Case 3) (n) = 4T(n/2) +n/logn= T(n) = (n2) (Case 1) (n) = 64T(n/8) n2logn= Does not apply (f(n) is not positive) (n) = 7T(n/3) +n2= T(n) = (n2) (Case 3) (n) = 4T(n/2) + logn= T(n) = (n2) (Case 1) (n)
4 =T(n/2) +n(2 cosn) = Does not apply. We are in Case 3, but the regularity conditionisviolated. (Considern= 2 k, wherekis odd and arbitrarily large. For any such choice ofn, you canshow thatc 3/2, thereby violating the regularity condition.)3