6. What is the asymptotic solution to the recurrence relation T(n) = 3T(n/2)+n3 log(n)?
please explain
Master theorem: ----------------- T(n) = aT(n/b) + f(n) If f(n) = Θ(n^c) where c < Logb(a) then T(n) = Θ(n^Logb(a)) If f(n) = Θ(n^c) where c = Logb(a) then T(n) = Θ((n^c)(Log(n))) If f(n) = Θ(n^c) where c > Logb(a) then T(n) = Θ(f(n)) Given T(n) = 3T(n/2)+n3log(n) T(n) = 3T(n/2)+Θ(n3log(n)) Where a = 3, b = 2, f(n) = Θ(n3log(n)) then c is greater than 3 Log2(3) = 1.5849625007 < c So, It satisfies rule 3 T(n) = Θ(f(n)) = Θ(n3log(n))
6. What is the asymptotic solution to the recurrence relation T(n) = 3T(n/2)+n3 log(n)? please explain
3. Determine the asymptotic complexity of the function defined by the recurrence relation. Justify your solution using expansion/substitution and upper and/or lower bounds, when necessary. You may not use the Master Theorem as justification of your answer. Simplify and express your answer as O(n*) or O(nk log2 n) whenever possible. If the algorithm is exponential just give exponential lower bounds c) T(n) T(n-4) cn, T(0) c' d) T(n) 3T(n/3) c, T() c' e) T(n) T(n-1)T(n-4)clog2n, T(0) c' 3. Determine the...
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