穷举法：在密钥空间较小的情况下，采用暴力破解方式攻击方法。
频率统计：在密文长度足够长的时候，可使用词频分析。
爬山法：选择性地尝试不同解密密钥，然后给每一个解密出来的明文标记上一个适应度。若解密出来的明文越接近我们的日常用的英语，它的适应度就越高；若解密出来的明文越难读懂或者越不像我们日常用的英语，则其适应度越低。
# -*- coding: UTF-8 -*-
import random
from ngram_score import ngram_score
#参数初始化
ciphertext = 'JGRMQOYGHMVBJWRWQFPWHGFFDQGFPFZRKBEEBJIZQQOCIBZKLFAFGQVFZFWWEOGWOPFGFHWOLPHLRLOLFDMFGQWBLWBWQOLKFWBYLBLYLFSFLJGRMQBOLWJVFPFWQVHQWFFPQOQVFPQOCFPOGFWFJIGFQVHLHLROQVFGWJVFPFOLFHGQVQVFIEOGQILHQFQGIQVVOSFAFGBWQVHQWIJVWJVFPFWHGFIWIHZZRQGBABHZQOCGFHX'
parentkey = list('ABCDEFGHIJKLMNOPQRSTUVWXYZ')
#只是用来声明key是个字典
key = {'A':'A'}
#读取quadgram statistics
fitness = ngram_score('quadgrams.txt')
parentscore = -99e9
maxscore = -99e9
j = 0
print('---------------------------start---------------------------')
while 1:
j = j+1
#随机打乱key中的元素
random.shuffle(parentkey)
#将密钥做成字典
for i in range(len(parentkey)):
key[parentkey[i]] = chr(ord('A')+i)
#用字典一一映射解密
decipher = ciphertext
for i in range(len(decipher)):
decipher = decipher[:i] + key[decipher[i]] + decipher[i+1:]
parentscore = fitness.score(decipher)#计算适应度
#在当前密钥下随机交换两个密钥的元素从而寻找是否有更优的解
count = 0
while count < 1000:
a = random.randint(0,25)
b = random.randint(0,25)
#随机交换父密钥中的两个元素生成子密钥，并用其进行解密
child = parentkey[:]
child[a],child[b] = child[b],child[a]
childkey = {'A':'A'}
for i in range(len(child)):
childkey[child[i]] = chr(ord('A')+i)
decipher = ciphertext
for i in range(len(decipher)):
decipher = decipher[:i] + childkey[decipher[i]] + decipher[i+1:]
score = fitness.score(decipher)
#此子密钥代替其对应的父密钥，提高明文适应度
if score > parentscore:
parentscore = score
parentkey = child[:]
count = 0
count = count+1
#输出该key和明文
if parentscore > maxscore:
maxscore = parentscore
maxkey = parentkey[:]
for i in range(len(maxkey)):
key[maxkey[i]] = chr(ord('A')+i)
decipher = ciphertext
for i in range(len(decipher)):
decipher = decipher[:i] + key[decipher[i]] + decipher[i+1:]
print ('Currrent key: '+''.join(maxkey))
print ('Iteration total:', j)
print ('Plaintext: ', decipher)
原文：https://blog.csdn.net/qq_35964497/article/details/78985332
