Are 2^n and 4^n in the same Big-Θ complexity class?
2^n
is NOT big-theta (Θ) of 4^n
, this is because 2^n
is NOT big-omega (Ω) of 4^n
.
By definition, we have f(x) = Θ(g(x))
if and only if f(x) = O(g(x))
and f(x) = Ω(g(x))
.
Claim
2^n is not Ω(4^n)
Proof
Suppose 2^n = Ω(4^n)
, then by definition of big-omega there exists constants c > 0
and n0
such that:
2^n ≥ c * 4^n
for all n ≥ n0
By rearranging the inequality, we have:
(1/2)^n ≥ c
for all n ≥ n0
But notice that as n → ∞
, the left hand side of the inequality tends to 0
, whereas the right hand side equals c > 0
. Hence this inequality cannot hold for all n ≥ n0
, so we have a contradiction! Therefore our assumption at the beginning must be wrong, therefore 2^n is not Ω(4^n)
.
Update
As mentioned by Ordous, your tutor may refer to the complexity class EXPTIME, in that frame of reference, both 2^n
and 4^n
are in the same class. Also note that we have 2^n = 4^(Θ(n))
, which may also be what your tutor meant.