Question

I have solved a recurrence relation that has a running time of Θ(2^n), exponential
time.
How do I find Ω, and O for the same recurrence relation.

I guess if it is Θ(2^n), it should also be O(2^n), am I right?
How do I find the Ω, the lower bound?

I tried solving the recurrence relation:

T(n) = 2T(n-1) + C.

Answer 1

If it’s homework, you could try to draw it as a recursion tree, where nodes represent the complexity of operations required by the function calls.

EDIT: About the lower bound, The Omega is defined as a lower bound. If you have Theta (the exact behavior), you have also Omicron and Omega… they are just less precise.

So

Theta(n) <=> O(n) AND Omega(n)

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SPOILER

If I remember correctly, this is how you interpret it…

you have a tree, at its root you have only C (the work to marge the solutions), and you have to spit in two branches (again with C as work), the nodes get branched n times

     C
    /|
   C C
  /| |\
 C C C C
/| ......

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Complete solution

because the tree has a depth of n, at the bottom you have 2^n nodes all with complexity of C, then you have n-1 levels with the complexity C, 2C, 4C....(2(n-1)*C), they should sum up to 2^n*c

So the final complexity should be 2*(2^n)*C which is theta(2^n)