This simple isprime(number) function checks if the given integer number is a prime number and returns True or False. The function makes sure that the number is a positive integer, and that 1 is not considered a prime number.

To find out if an integer n is odd one can use n & 1, to check for even one can then use not n & 1 or the more traditional n % 2 == 0 which is about 30% slower.

Edited 1 Year Ago by vegaseat: odd even

# prime numbers are only divisible by unity and themselves
# (1 is not considered a prime number by convention)

def isprime(n):
    '''check if integer n is a prime'''
    # make sure n is a positive integer
    n = abs(int(n))
    # 0 and 1 are not primes
    if n < 2:
        return False
    # 2 is the only even prime number
    if n == 2: 
        return True    
    # all other even numbers are not primes
    if not n & 1: 
        return False
    # range starts with 3 and only needs to go up the squareroot of n
    # for all odd numbers
    for x in range(3, int(n**0.5)+1, 2):
        if n % x == 0:
            return False
    return True

# test ...
print isprime(1)       # False
print isprime(2)       # True
print isprime(3)       # True
print isprime(29)      # True
print isprime(345)     # False
print isprime(999979)  # True
print isprime(999981)  # False

# extra test ...
print isprime(49)      # False
print isprime(95)      # False

Hi,

The code listed above does not seem accurate as numbers like 49 and 95 pass the test and yet they are not prime numbers!

Perhaps a flaw in the algorithm used?

Editor's note: as you can see, 49 and 95 do not pass the test, not sure what's wrong with you?

Edited 5 Years Ago by vegaseat: 49 55

I think the mistake is that you converted the square root to an int, because the code I wrote a couple of years ago works fine, and its almost the same

def isprime(n):
    if n == 2:
        return 1
    if n % 2 == 0:
        return 0

    max = n**0.5+1
    i = 3

    while i <= max:
        if n % i == 0:
            return 0
        i+=2

    return 1

Editors note: the range function in vegaseat's code needs integers, so it's correct.

Edited 3 Years Ago by vegaseat: range() needs integers

Comments
could you please explain line 7 m a newbie

Simple RSA primtest. Does not work on universal pseudo primes.

def isprime(n,PROB):
    '''returns if the number is prime. Failure rate: 1/4**PROB '''
    if n==2: return '1'
    if n==1 or n&1==0:return '0'
    s=0
    d=n-1
    while 1&d==0:
        s+=1
        d>>=1
    for i in range(PROB):
        a=random.randint(2,n-1)
        composit=True
        if pow(a,d,n)==1:
            composit=False
        if composit:
            for r in xrange(0,s):
                if pow(a,d*2**r,n)==n-1:
                    composit=False
                    break
        if composit: return False
    return True

this works fine returning true if prime and false with the smallest number that divides it if it is false.

def isprime(x):
	x = abs(int(x))
	if x < 2:
		return "Less 2", False
	elif x == 2:
		return True
	elif x % 2 == 0:
		return False	
	else:
		for n in range(3, int(x**0.5)+2, 2):
			if x % n == 0:
				return n, False
		return True

Edited 5 Years Ago by kungfujam: Previous incorrect

Pardon my previous post. The indentation was not right.This one works fine.

for n in range(2,10):
  for x in range(2,n):
      if n%x == 0:
        print n,'equal',x,'*',n/x
        break
  else:
       print n,'is a prime number'

Edited 4 Years Ago by pyTony: Code block by TAB

Attachments
for n in range(2,10):
  for x in range(2,n):
      if n%x == 0:
        print n,'equal',x,'*',n/x
        break
  else:
       print n,'is a prime number'

Hey thank you for this epic post!!! I used it as a base for checking if a number was prime in an algorithm I made for Project Euler #3.
Thanks again!!!!

how avout this one?:

def is_prime(n):
    i = 2
    if n < 2:
        return False
    while i < n:        
        if n % i == 0:
            return False
        else:
            i += 1
    return True

Editor's note: too slow for large n

Edited 3 Years Ago by vegaseat: slow algorithm

gmorcan,
using Python module timeit and a rather small prime number like 999979, your function is 2700 times slower than vegaseat's function.

Edited 3 Years Ago by vegaseat: spell

OK, here is a really optimized version for this classic problem. This is not originally mine (but have done very similar).
It utilizes the fact that primes > 4 are 1(mod 6) or 5(mod 6). Actually I editted it a lot now before posting to be more according to format standards of PEP8.

def is_prime(n):
    if n == 2 or n == 3: 
        return True
    elif n < 2 or n % 2 == 0: 
        return False
    elif n < 9:
        return True
    elif n % 3 == 0: 
        return False

    r = int(sqrt(n))
    f = 5

    while f <= r:
        if n % f == 0 or n % (f + 2) == 0: 
            return False
        else:
            f += 6

    return True

Edited 4 Years Ago by pyTony

How do you feel witth this?

import re 
def is_prime(num): return not re.match(r"^1?$|^(11+?)\1+$", "1" * num)
def pri(num): return is_prime(num)

Real hack, and surely stupendously inefficient. Best as joke or concept. Do not try to test (2**13-1) with it.

Super fast for very large numbers:

def miller_rabin_isprime(a, i, n):
    """
    Miller-Rabin primality test
    returns a 1 if n is a prime
    usually i = n - 1
    does not test prime 2
    """
    if i == 0:
        return 1
    x = miller_rabin_isprime(a, i // 2, n)
    if x == 0:
        return 0
    y = (x * x) % n
    if ((y == 1) and (x != 1) and (x != (n - 1))):
        return 0
    if (i % 2) != 0:
        y = (a * y) % n
    return y
i = 0
while(i <100):
    j = 2
    while (j<=(i/j)):
        if not (i%j):break
        j=j+1
    if (j > i/j):print i ," is prime number"
    i = i+1

Editor's note:
Let's say you want to know if 999979 is a prime number. How long would that take?

Edited 3 Years Ago by vegaseat: timing

You have made the classic mistake of including even numbers = testing twice as many numbers. Start out with i=3 and add two on each pass.

Edited 3 Years Ago by woooee

I just got an automated email from Daniweb suggesting that I read this thread based on my recent activity. The code posts in this thread are fine, but most of the comment posts are curmudgeonly, rude, standoffish, exclusivist, and socially inept. People seem to think it's a violation of some natural law if a beginner programmer joins a discussion and tries to make a positive contribution by posting code. I'll definitely be putting some of you on my ignore list.

People seem to think it's a violation of some natural law if a beginner programmer joins a discussion and tries to make a positive contribution by posting code.

On the contrary, we answer beginner programmer's questions every day in the python forum and try to help them as much as we can. Correct errors and inefficiencies in code is not rude or socially inept: we all learn by our own mistakes.

I would advise beginner programmers to start their own thread with well described issues to solve. They will probably get better answers than by adding to a code snippet thread.

They may also find many ideas in Vegaseat's thread projects-for-the-beginner.

Edited 3 Years Ago by Gribouillis

Hi, Grib,

Thanks for the link to that thread. I'll definitely be checking it out. What I'm really hungry for is some real-world problems to try to provide solutions for, but I realize it might be a bit too early for me to do that. Doing those kinds of projects could be a good alternative.

If there are similar threads available for html and javascript, I'd love to see links for them as well.

Eyetee

I don't understand what's going on with line 15 in vegaseat's code. What does not n & 1 do?

What does not n & 1 do?

& is the integer bitwise and operator. When n is an integer, n & 1 is the value of its first bit in base 2. Therefore, n & 1 has value 0 if n is even and 1 if n is odd. not n & 1 has the same meaning as n is even (which is less cryptic but is not python). Another way to say it is n % 2 == 0, but vegaseat thinks he gains some speed.

Edited 3 Years Ago by Gribouillis

`

def is_prime(x):
    prime = False
    if x > 1:
        for i in range(2, x+1):
            if x % i == 0 and i != x:
                prime = False
                break
            else:
                prime = True
    return prime

`

Simpler than above examples, and creates a list of primes (to 10,000 in this example). Also, you only have to test x up to a half of n, as anything over a half of the number is not worth testing. (not worth testing if 17 can be dividied by 9, 10, 11, 12 etc.!) Also only tests odd numbers with the range (1,n,2):

primes = []
for n in range(1,10000,2):
  for x in range(2,n/2):
      if n%x == 0:
        break
  else:
       primes.append(n)
print primes

@Gu1tar1st: As the square root of 17 is 4.123 you don't even have to test for divisibility from 5,6,7,8,9....

The article starter has earned a lot of community kudos, and such articles offer a bounty for quality replies.