6
I’m having a hard time proving
f(n)=Omega(g(n))
, then g(n) = O(f(n))
.
How would this proof?
With calculation and theory?
6
I’m having a hard time proving
f(n)=Omega(g(n))
, then g(n) = O(f(n))
.
How would this proof?
With calculation and theory?
0
Recital O as upper limit of F(n)
Whereas Omega is the lower limit of G(n)
We have:
F(n) = Omega (G(n)) -> F(n) <= G(n)
G(n) = O(F(n)) -> G(n) >= F(n)
F(n) <= G(n) <=> G(n) >= F(n)
Soon F(n) <= G(n)
amounts to G(n) >= F(n)
Browser other questions tagged
You are not signed in. Login or sign up in order to post.
In this site they talk about the mathematical part, I don’t know if it helps you: https://math.stackexchange.com/questions/213591/prove-that-if-fn-in-ogn-then-gn-in-omegafn
– Edward Ramos
Possible duplicate of How to prove the asymptotic order of an algorithm?
– Jefferson Quesado
Also worth reading: https://answall.com/q/411911/64969
– Jefferson Quesado