判断方法
1.直接计算
2.埃拉托色尼筛选法
3.拉宾米勒算法
直接计算
import math
def isPrime(num):
# 判断数是不是质数
if num < 2:
return False
# 目标数字的平方根内的整数是否能整除,如果能整除说明不是质数
for i in range(2, int(math.sqrt(num)) + 1):
if num % i == 0:
return False
return True
埃拉托色尼筛选法
1.假设一组表格(知道其大小)里装着每个数,都标记为"质数"
2.把1标记为"非质数",然后把2的倍数(除了2之外)的标记为"非质数",知道这个表格最大值
3.重复第2步过程,将3(除3之外)的倍数标记为"非质数",对4…对5…
4.直到表格大小的平方根为止,因为另外一半已经被标记完了
5.然后再进行筛选出表格里还是"质数"的数
def primeSieve(sieveSize):
# 通过埃拉托色尼筛选算法计算出指定范围内的质数列表
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
拉宾米勒算法
对于很大的质数,直接计算会太慢,而埃拉托色尼算法又太消耗内存。
判断大数字是不是质数可以通过拉宾米勒质数测试来检验。
拉宾米勒算法数学理论太复杂,但能判断数百个数位长的数字是不是质数。
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