T (N)=4T(N/3)+1running time on problem of size N
时间: 2024-03-04 22:52:10 浏览: 18
Based on the provided recursion, the time complexity of the algorithm can be determined using the Master Theorem.
The Master Theorem states that for a recursion of the form T(N) = aT(N/b) + f(N), where a >= 1 and b > 1 are constants and f(N) is a function that represents the cost of the work done outside of the recursive calls, the time complexity can be expressed as follows:
- If f(N) = O(N^log_b(a - ε)) for some constant ε > 0, then T(N) = Θ(N^log_b(a)).
- If f(N) = Θ(N^log_b(a)), then T(N) = Θ(N^log_b(a) * log(N)).
- If f(N) = Ω(N^log_b(a + ε)) for some constant ε > 0, and if af(N/b) <= cf(N) for some constant c < 1 and all sufficiently large N, then T(N) = Θ(f(N)).
In this case, a = 4, b = 3, and f(N) = 1. Therefore, we can apply the Master Theorem as follows:
- log_b(a) = log_3(4) ≈ 1.26
- f(N) = O(N^0) = O(1)
Since f(N) = O(N^log_b(a - ε)) for ε = 0.26, we can apply case 1 of the Master Theorem to obtain the time complexity:
T(N) = Θ(N^log_3(4)) ≈ Θ(N^1.26)
Therefore, the time complexity of the algorithm is Θ(N^1.26).