Solving recurrence of divide and conquer using master theorem

Question

Can T(n) = 3T(n/2) + c T(1)=0, solved using master theorem? If yes, I am struggling on understand master theorem now, could someone tell me which case it falls to and why.

Answer 1

There are different possible approaches for the master theorem. I like the one proposed by Cormen et al. in their book Introduction to Algorithms .

1. If f(n) = O(n^(log b (ae))) for some constant e>0, then T(n) = Θ(n^(log b (a)))
2. If f(n) = Θ(n^(log b (a))), then T(n) = Θ(n^(log b (a)). lg n)
3. If f(n) = Ω(n^(log b (a+e))) for some constant e>0, and if af(n/b) < c.f(n) for some constant c<1 and all sufficiently large n, then T(n) = Θ(f(n))

Now we need to compare f(n) with n^(log b (a)) .

• if f(n) is polynomially smaller than n^(log b (a)) , it falls under the 1st case
• if n^(log b (a)) is polynomially smaller than f(n) , it falls under the 3rd case
• if f(n) and n^(log b (a)) are the same size, it falls under the 2nd case

Note that the three cases do not cover all the possibilities for f(n). There is a gap between cases 1 and 2 when f(n) is smaller than n^(log b (a)) but not polynomially smaller. Similarly, there is a gap between cases 2 and 3 when f(n) is larger than n^(log b (a)) but not polynomially larger. If the function f(n) falls into one of these gaps, or if the regularity condition in case 3 fails to hold, you cannot use the master method to solve the recurrence.

Now to solve the recurrence in question...

a=3, b=2, f(n) = c = n^0

so we have n^(log2(3)) ≈ n^(1.58) which is polynomially larger than n^0, falling under the 1st case. Then the time complexity is T(n) = Θ(n^(log b (a))) --> T(n) = Θ(n^1.58)