Fastest way of testing if a number is prime?

Question

I'm trying to get a fast way to determine if a number is prime using Python.

I have two functions to do this. Both return either True or False.

Function isPrime1 is very fast to return False is a number is not a prime. For example with a big number. But it is slow in testing True for big prime numbers.

Function isPrime2 is faster in returning True for prime numbers. But if a number is big and it is not prime, it takes too long to return a value. First function works better with that.

How can I come up with a solution that could quickly return False for a big number that is not prime and would work fast with a big number that is prime?

def isPrime1(number): #Works well with big numbers that are not prime
    state = True
    if number <= 0:
        state = False
        return state
    else:          
        for i in range(2,number):
            if number % i == 0:
                state = False
                break
        return state

def isPrime2(number): #Works well with big numbers that are prime   
    d = 2
    while d*d <= number:
        while (number % d) == 0:            
            number //= d
        d += 1
    if number > 1:       
        return True
    else:
        return False`

Solution 1

Exhaustive division until the square root is about the simplest you can think of. Its worst case is for primes, as all divisions must be performed. Anyway, until a billion, there is virtually no measurable time (about 1.2 ms for 1000000007).

def Prime(n):
    if n & 1 == 0:
        return 2
    d= 3
    while d * d <= n:
        if n % d == 0:
            return d
        d= d + 2
    return 0

Note that this version returns the smallest divisor or 0 rather than a boolean.

Some micro-optimizations are possible (such as using a table of increments), but I don' think they can yield large gains.

There are much more sophisticated and faster methods available, but I am not sure they are worth the fuss for such small n.

Solution 2

Here is what I came up with

def is_prime(number):
    # if number is equal to or less than 1, return False
    if number <= 1:
        return False

    for x in range(2, number):
        # if number is divisble by x, return False
        if not number % x:
            return False
    return True

Solution 3

Primality tests is a very tricky topic.

Before attempting to speed up your code, try to make sure it works as intended. I suggest you start out with very simple algorithms, then build from there.

Of interest, isPrime2 is flawed. It returns True for 6, 10, 12, ...

lines 3 to 6 are very telling

while d*d <= number:
    while (number % d) == 0:            
        number //= d
    d += 1

When a factor of number d is found, number is updated to number = number // d and at the end of the while loop, if number > 1 you return True

Working through the code with number = 6:

isPrime2(6)
initialise> number := 6
initialise> d := 2
line3> check (2 * 2 < 6)     :True
line4> check (6 % 2 == 0)    :True
line5> update (number := 6//2) -> number = 3
line6> update (d : d + 1) -> d = 3
jump to line3
line3> check (3 * 3 < 3)      :False -> GOTO line7
line7> check(number > 1) -> check(3 > 1) :True
line8> return True -> 6 is prime

Comments

[Ry-]

en.wikipedia.org/wiki/Primality_test

[Mark Ransom]

Use a Bloom filter that is pre-initialized with a list of prime numbers up to the largest you need to consider.

[alvas]

compoasso.free.fr/primelistweb/page/prime/accueil_en.php

[Yves Daoust]

How big ? Please be very specific, because the answer will directly depend on that.

[Yves Daoust]

Your first version continues trying all divisors past the square root of n, plus uses a huge range. It must be banned forever ! Also, trying all even divisors is a pity !

[Rory Daulton]

There are many posts on this site about fast primality testing in Python. Did you look through them, and why do they not meet your needs?

[Mark Dickinson]

@RoryDaulton: Not sure that's a good duplicate. Primality testing is an easier problem than full-blown factorization.

[Mark Dickinson]

Still the wrong duplicate. Prime factorization and primality testing are not the same thing. Checking whether a 1000-digit number is (probably) prime is trivially quick; doing a determistic primality check for such a number is slower, but still feasible. But that's way out of the range of effective factorization.

[numersoz]

isPrime1 was a function that I wrote, and isPrime2 was something that I modified which was actually intended to calculate Prime Factors of a number. The first function is working, let me work on the second one and fix it. Thanks

[Xero Smith]

haufa. The first line of your function doesn't match the comment. The part `` not number % 2`` will return False for all odd numbers which means it will return False for every prime number except 2. Please modify the function accordingly.

[numersoz]

Your function is faster than my first one. isPrime1(1000000007) gave a result in 70 seconds and is_Prime(1000000007) gave a result in 60 seconds. Its better but lets see what other people have that would work faster.

[ephrim]

@XeroSmith I've made the changes. Thanks for the correction

[Yves Daoust]

Testing up to number is a total waste. You can stop at √number. In the case of 1000000007, this is about 30000 times faster.

[Hacker]

Why do you only need to go up to the square root