sneekula, I don't want to sound rude, but your one-sentence-question-posts are annoying. Why don't you try to find the answer alone first? If you're stuck, post what you have tried and we will try to help you.
BTW: The internet is full of isprime functions in any language you can dream of. 2 minutes of googling, that's all you need to do ;)
You are absolutely right, a search should always start in our snippets section :) But, searching for prime python via Google, I found a link to a snippet here (on the second page). You see, Google is not that bad :)
Our snippet examples give you a list of primes. You could get a list of primes in the range of your numbers, and then see if your numbers are in it. For simplicity sake you can use this small function ...
# prime numbers are only divisible by unity and themselves
# (1 is not considered a prime number)
'''check if integer n is a prime'''
# range starts with 2 and only needs to go up the squareroot of n
for x in range(2, int(n**0.5)+1):
if n % x == 0:
# test ...
print isprime(29) # True
print isprime(345) # False
print isprime(8951) # True
If you're looking to test really large numbers, here's an implementation of the Rabin-Miller probabilistic test:
primes(n) --> primes
Return list of primes from 2 up to but not including n. Uses Sieve of Erasth.
if n < 2:
nums = range(2,int(n))
p = 
new_prime = nums
for i in nums[1:]:
if i % new_prime == 0:
def power_mod(a, b, n):
power_mod(a,b,n) --> int
Return (a ** b) % n
if b < 0:
elif b == 0:
elif b % 2 == 0:
return power_mod(a*a, b/2, n) % n
return (a * power_mod(a,b-1,n)) % n
def rabin_miller(n, tries = 7):
rabin_miller(n, tries) --> Bool
Return True if n passes Rabin-Miller strong pseudo-prime test on the
given number of tries, which indicates that n has < 4**(-tries) chance of being composite; return False otherwise.
if n == 2:
if n % 2 == 0 or n < 2:
p = primes(tries**2)
# necessary because of the test below
if n in p:
s = n - 1
r = 0
while s % 2 == 0:
r = r+1
s = s/2
for i in range(tries):
a = p[i]
if power_mod(a,s,n) == 1:
for j in range(0,r):
if power_mod(a,(2**j)*s, n) == n - 1:
I can't guarantee that the implementation is correct, although it has passed testing so far. Also, the RM test only gives a probability of primeness, not absolute certainty. However, with default settings one has a 0% chance of false positives and a 0.006% chance of a false negative.
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