Answer: A
To find the asymptotic complexity of T1(n), we solve the recurrence relations step-by-step using the Master Theorem.
Step 1: Solve for T2(n)
The recurrence relation for T2(n) is given by:
T2(n)=5T2(4n)+Θ(log2n)
Comparing this with the standard Master Theorem form T(n)=aT(n/b)+f(n):
- a=5
- b=4
- f(n)=Θ(log2n)
We compare f(n) with nlogba:
nlogba=nlog45≈n1.16
Since f(n)=Θ(log2n)=O(nlog45−ϵ) for any 0<ϵ<0.16, this falls under Case 1 of the Master Theorem. Therefore:
T2(n)=Θ(nlog45)
Step 2: Solve for T1(n)
Now, substitute the asymptotic behavior of T2(n) into the recurrence relation for T1(n):
T1(n)=4T1(2n)+Θ(nlog45)
Comparing this with the Master Theorem form:
- a=4
- b=2
- f(n)=Θ(nlog45)
We compare f(n) with nlogba:
nlogba=nlog24=n2
Since log45≈1.16<2, we have:
f(n)=Θ(nlog45)=O(n2−ϵ)for 0<ϵ<0.84
This also falls under Case 1 of the Master Theorem. Therefore:
T1(n)=Θ(n2)
Thus, the correct option is A.