测试数字是否为素数的最快方法 [英] Fastest way of testing if a number is prime with Python

问题描述

我正在尝试一种快速的方法来确定使用Python是否为质数.

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

我有两个功能可以做到这一点.两者都返回True或False.

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

函数isPrime1返回的速度非常快False是数字不是素数.例如，数量很大.但是对于大质数测试True的速度很慢.

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.

函数isPrime2更快地返回质数的True.但是，如果一个数字很大而又不是素数，则返回值花费的时间太长.第一个功能可以更好地工作.

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.

我如何提出一个解决方案，对于一个不是素数的大数，可以快速返回False，而对于一个素数的大数，可以快速地工作呢?

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`

推荐答案

穷尽除法，直到平方根达到您能想到的最简单的平方根为止.最坏的情况是素数，因为必须执行所有除法.无论如何，直到十亿，实际上没有可测量的时间(对于 1000000007 而言，约为1.2毫秒).

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

请注意，此版本返回最小除数或 0 而不是布尔值.

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

可以进行一些微优化(例如使用增量表)，但是我认为它们不会带来很大的收益.

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

有很多更先进，更快速的方法可用，但我不确定它们是否值得这么小的 n 大惊小怪.

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

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