I'm studying algorithm complexity and I am trying to figure out this one question that runs in my mind- is O(n!) faster than O(2^n) or is it the opposite way around?
CodePudding user response:
O(2^n)
is 2 * 2 * 2 * ...
where O(n!)
is 1 * 2 * 3 * 4 * ...
O(n!)
will quickly grow much larger - so O(2^n)
is faster.
For example: 2^10 = 1024
and 10! = 3628800
CodePudding user response:
You can try working with Stirling's approximation for n!
https://en.wikipedia.org/wiki/Stirling's_approximation
n! = (n / e)^n * sqrt(2 * Pi * n) * (1 o(n))
Now, let's compare
O(n!) <=> O(2^n)
In order to find out the right letter <
, =
or >
let's compute limit
lim (n! / 2^n) =
n -> inf
lim (n / e)^n * sqrt(2 * pi * n) / 2^n >=
n -> inf
lim n^n / (2 * e)^n >= // when n > 4 * e
n -> inf
lim (4 * e)^n / (2 * e)^n =
n -> inf
lim 2^n = inf
n -> inf
So
lim (n! / 2^n) = inf
n -> inf
which means that O(n!) > O(2^)