4 ms·
Isn't it basically this algorithm + overhead from using strings/regex? def is_prime(n): if n < 2: return False for i in range(2, int(n ** 0.5)
by SushiHippie 2y ago
Isn't it basically this algorithm + overhead from using strings/regex?
def is_prime(n):
if n < 2:
return False
for i in range(2, int(n ** 0.5) + 1):
if n%i==0:
return False
return True
Haven't tested this, just quickly jotted it down in the HackerNews textarea, but theoretically it should be python code, which checks if n is divisible by any number between 2 and sqrt(n).
And wouldn't this algorithm be just O(n), what does the regex engine do differently (except the string overhead)?
- chx 2y agoFine. It's now only 10^10 times the age of the universe. much better.