找到小于n的最大素数,n = ~10 ^ 230

Question

我的解决方案是否有问题找到最大的素数少于n什么时候n可以达到~10 ^ 230？对于更好的方法有什么建议吗？

这是我的尝试,在Python中使用以下版本的Miller-Rabin素性测试:

from random import randrange

small_primes = [
    2,  3,  5,  7, 11, 13, 17, 19, 23, 29,
    31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
    73, 79, 83, 89, 97,101,103,107,109,113,
    127,131,137,139,149,151,157,163,167,173,
    179,181,191,193,197,199,211,223,227,229,
    233,239,241,251,257,263,269,271,277,281,
    283,293,307,311,313,317,331,337,347,349,
    353,359,367,373,379,383,389,397,401,409,
    419,421,431,433,439,443,449,457,461,463,
    467,479,487,491,499,503,509,521,523,541,
    547,557,563,569,571,577,587,593,599,601,
    607,613,617,619,631,641,643,647,653,659,
    661,673,677,683,691,701,709,719,727,733,
    739,743,751,757,761,769,773,787,797,809,
    811,821,823,827,829,839,853,857,859,863,
    877,881,883,887,907,911,919,929,937,941,
    947,953,967,971,977,983,991,997
]

def probably_prime(n, k):
    """Return True if n passes k rounds of the Miller-Rabin primality
    test (and is probably prime). Return False if n is proved to be
    composite.

    """
    if n < 2: return False
    for p in small_primes:
        if n < p * p: return True
        if n % p == 0: return False
    r, s = 0, n - 1
    while s % 2 == 0:
        r += 1
        s //= 2
    for _ in range(k):
        a = randrange(2, n - 1)
        x = pow(a, s, n)
        if x == 1 or x == n - 1:
            continue
        for _ in range(r - 1):
            x = pow(x, 2, n)
            if x == n - 1:
                break
            else:
                return False
    return True

我开始测试probably_prime(n)我减少的位置并测试每个值,n直到我得到一个"可能是素数"的数字.当我在n= ~10 ^ 230的值上测试时,我发现质数大约相差20-30个数字.在阅读了关于素数差距的更多信息后,我的结果似乎不太可能,因为我不应该如此频繁地找到素数.我测试k了高达50,000的值,我得到了相同的答案.我做错了什么,有什么建议可以找到更好的解决方案吗？

Answer 1

你是对的,你的代码一旦超出small_primes表格似乎就会遇到困难.仔细观察,这里有一个错误:

    for _ in range(r - 1):
        x = pow(x, 2, n)
        if x == n - 1:
            break
        else:
            return False

你想要返回False(即复合),如果你从未找到x == n-1(或者你可以短路并返回,False如果x == 1,我认为:见这里).这可以通过更改缩进来完成:

    for _ in range(r - 1):
        x = pow(x, 2, n)
        if x == n - 1:
            break
    else:
        return False

(for/else真的是组合for/if-not-break.)

完成此更改后,我得到:

>>> sum(orig(p, 20) for p in range(10**6, 2*10**6))
54745
>>> sum(fixed(p, 20) for p in range(10**6, 2*10**6))
70435
>>> sum(orig(p, 20) for p in range(10**230, 10**230+10**3))
40
>>> sum(fixed(p, 20) for p in range(10**230, 10**230+10**3))
2

哪个是对的.