Consider the following recurrence relation: T(n)=3T(n/3)+n2/3,T(1)=1,T(0)=0 What is the complexity of T(n)? Options : Θ(n) Θ(nlogn) Θ(n^2/3) Θ(n^1.5)
Question
Consider the following recurrence relation:
T(n) = 3T(n/3) + n^{2/3}, T(1) = 1, T(0) = 0
What is the complexity of T(n)?
Options :
- Θ(n)
- Θ(n log n)
- Θ(n^{2/3})
- Θ(n^{1.5})
Solution
The complexity of the given recurrence relation T(n) = 3T(n/3) + n^2/3, T(1) = 1, T(0) = 0 is Θ(n^1.5). This is because the recurrence relation follows the form of the Master Theorem, which is used to determine the time complexity of divide and conquer algorithms. The Master Theorem states that for a recurrence relation of the form T(n) = aT(n/b) + f(n), where a >= 1, b > 1, and f(n) is an asymptotically positive function, the time complexity can be determined based on the comparison of f(n) and n^log_b(a). In this case, a = 3, b = 3, and f(n) = n^2/3. Since 2/3 > log_3(3) = 1, the time complexity is Θ(n^1.5).
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