質數篩選算法

判斷方法

1.直接計算
2.埃拉托色尼篩選法
3.拉賓米勒算法

直接計算

import math

def isPrime(num):
    # 判斷數是不是質數
    if num < 2:
        return False
    # 目標數字的平方根內的整數是否能整除，如果能整除說明不是質數
    for i in range(2, int(math.sqrt(num)) + 1):
        if num % i == 0:
            return False
    return True

埃拉托色尼篩選法

1.假設一組表格(知道其大小)裏裝着每個數，都標記爲"質數"
2.把1標記爲"非質數"，然後把2的倍數(除了2之外)的標記爲"非質數",知道這個表格最大值
3.重複第2步過程，將3(除3之外)的倍數標記爲"非質數"，對4…對5…
4.直到表格大小的平方根爲止，因爲另外一半已經被標記完了
5.然後再進行篩選出表格裏還是"質數"的數

def primeSieve(sieveSize):
    # 通過埃拉托色尼篩選算法計算出指定範圍內的質數列表

    sieve = [True] * sieveSize
    sieve[0] = False # zero and one are not prime numbers
    sieve[1] = False

    # create the sieve
    for i in range(2, int(math.sqrt(sieveSize)) + 1):
        pointer = i * 2
        while pointer < sieveSize:
            sieve[pointer] = False
            pointer += i

    # compile the list of primes
    primes = []
    for i in range(sieveSize):
        if sieve[i] == True:
            primes.append(i)

    return primes

拉賓米勒算法

對於很大的質數，直接計算會太慢，而埃拉托色尼算法又太消耗內存。
判斷大數字是不是質數可以通過拉賓米勒質數測試來檢驗。
拉賓米勒算法數學理論太複雜，但能判斷數百個數位長的數字是不是質數。

import random


def rabinMiller(num):
    # Returns True if num is a prime number.

    s = num - 1
    t = 0
    while s % 2 == 0:
        # keep halving s while it is even (and use t
        # to count how many times we halve s)
        s = s // 2
        t += 1

    for trials in range(5): # try to falsify num's primality 5 times
        a = random.randrange(2, num - 1)
        v = pow(a, s, num)
        if v != 1: # this test does not apply if v is 1.
            i = 0
            while v != (num - 1):
                if i == t - 1:
                    return False
                else:
                    i = i + 1
                    v = (v ** 2) % num
    return True


def isPrime(num):
    # Return True if num is a prime number. This function does a quicker
    # prime number check before calling rabinMiller().

    if (num < 2):
        return False # 0, 1, and negative numbers are not prime

    lowPrimes = [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]

    if num in lowPrimes:
        return True

    # See if any of the low prime numbers can divide num
    for prime in lowPrimes:
        if (num % prime == 0):
            return False

    # If all else fails, call rabinMiller() to determine if num is a prime.
    return rabinMiller(num)


def generateLargePrime(keysize=1024):
    # Return a random prime number of keysize bits in size.
    while True:
        num = random.randrange(2**(keysize-1), 2**(keysize))
        if isPrime(num):
            return num