master-theorem

Generic form: T(n) = aT(n/b) + f(n) So i must compare n^logb(a) with f(n) if n^logba > f(n) is case 1 and T(n)=Θ(n^logb(a)) if n^logba < f(n) is case 2 and T(n)=Θ((n^logb(a))(logb(a))) Is that correct? Or I misunderstood something? And what about case 3? When its apply? I think you have misunderstood it. if n^logba > f(n) is case 1 and T(n)=Θ(n^logb(a)) Here you should not be worried about f(n) as result you are getting is T(n)=Θ(n^logb(a)). f(n) is part of T(n) ..and if you get the result T(n) then that value will be inclusive of f(n). so, There is no need to consider that part. Let me know

I am refreshing on Master Theorem a bit and I am trying to figure out the running time of an algorithm that solves a problem of size n by recursively solving 2 subproblems of size n-1 and combine solutions in constant time. So the formula is: T(N) = 2T(N - 1) + O(1) But I am not sure how can I formulate the condition of master theorem. I mean we don't have T(N/b) so is b of the Master Theorem formula in this case b=N/(N-1) ? If yes since obviously a > b^k since k=0 and is O(N^z) where z=log2 with base of (N/N-1) how can I make sense out of this? Assuming I am right so far? ah, enough with the

Example T(n)=T(n/3)+T(n/4)+3n is this solvable with iterative master theorem or recursion tree.Can someone solve it analytically to show how it's done ? We can expand T(n) with a binomial summation : (after some steps - can be proven by induction For some depth of expansion / recursion k . Where do we terminate? When the parameters to all instances of f(n) reach a certain threshold C . Thus the maximum depth of expansion: We choose the smallest between a, b because the parameter with only powers of min(a, b) decreases at the slowest rate : Thus the general expression for T(n) is: The existence