For parameters a and b, both of which are ω(1), T(n)=T(n1/a)+1, and T(b)=1. Then T(n) is
(A) Θ(logalogbn)
(B) Θ(logabn)
(C) Θ(logblogan)
(D) Θ(log2log2n)
Answer: (A)
Explanation: Given,
T(n) = T(n1/a)+1, T(b) = 1
Now, using iterative method,
= T(n) = [T(n1/a2)+1] + 1 = [T(n1/a3)+1] + 2 = [T(n1/a4)+1] + 3 . . . = [T(n1/ak)+1] + (k-1) = T(n1/ak) + k
Let,
→ n1/ak = b → log(n1/ak) = log(b) → ak = log(n) / log (b) = logb(n) → k = logalogb(n)
Therefore,
= T(n1/ak) + k = T(b) + logalogb(n) = 1 + logalogb(n) = Θ(logalogb(n))
Option (A) is correct.
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