2023年7月31日发(作者：)
仿射密码Python实现算法分析1. 仿射密码结合了移位密码和乘数密码的特点，是移位密码和乘数密码的组合。2. 仿射密码的加密算法就是⼀个线性变化，即对明⽂字符x，对应的密⽂字符为y=ax+b(mod26)其中，a, b属于Z26且gcd(a,b)=13. 实现过程：选取a，b两个参数，其中gcd(a, 26)＝1加密变换： c＝ a∗?＋b 26a=1时，移位密码b=1时，乘数密码解密变换： ?＝ (c−b)∗a^(−1) 26算法实现# 暴⼒破解la = [1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25]lb = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]# 最⼤公约数def gcd(a, b): while b != 0: tem = a % b a = b b = tem return a# 加密def encrypt(m, c, a, b): for i in range(len(m)): # 加密成相应的⼤写字母 (chr(((ord(m[i]) - 97) * a + b) % 26 + 65)) d = ''.join(c) print(d)# 求逆元def niyuan(a, b): ny = 1 while (a * ny) % b != 1: ny += 1 return ny# 解密def decrypt(c, k, b): mw = [] for i in range(len(c)): tem = ord(c[i]) - 65 - b if tem < 0: tem += 26 (chr((k * tem) % 26 + 97)) print("k=" + str(k) + ", b=" + str(b) + "时，解密后的明⽂为：") res = ''.join(mw) print(res)#实现if __name__ == "__main__": # 明⽂ # 密⽂ m = 'ifnottothesunforsmilingwarmisstillinthesuntherebutwewilllaughmoreconfidentcalmifturnedtofoundhisownshadowappropriateescapethesunwillbethroughthe # 密⽂ c = [] x, y = input("请输⼊a和b: ").split() a = int(x) b = int(y) while gcd(a, b) != 1: x, y = input("a和b不互素，请重新输⼊a和b: ").split() a = int(x) b = int(y) print("明⽂内容为：") print(m) print("加密后的密⽂为：") encrypt(m, c, a, b) print("知道密钥破解：") k = niyuan(a, 26) decrypt(c, k, b) print("不知道秘钥破解，暴⼒破解如下： ") for i in range(0, 12): for j in range(0, 26): decrypt(c, la[i], lb[j])加密与解密加密：输⼊a = 3, b = 4时，加密结果如图所⽰：解密：知道秘钥k = 9, b = 4 （k为a的逆元）时，解出相应明⽂。正确性由于算法的前提要求gcd(a,26)==1, 从⽽使加密函数c＝ a∗?＋b 26是⼀个单射函数，故其解必然是唯⼀的。即，gcd(a,26)==1保证了仿射加密函数是⼀个双射函数，故算法正确。安全性分析此密码算法安全性较弱。由算法的实现可知，此算法的秘钥空间⼤⼩为12*26 – 1 ==311（去除a = 1, b = 0时的情况）且a =1,3,5,7,9,11,15,17,19,21,23,25 故很容易便能够通过暴⼒破解获得明⽂。还可以通过统计分析破解：代码如下#统计破解仿射密码# 最⼤公约数def gcd(a, b) while b != 0: tem = a % b a = b b = tem return aif __name__ == "__main__": # a = 3， b = 4时的密⽂ # 根据统计⽽得出的实际各字母出现的概率 reality = dict(a=0.082, b=0.015, c=0.028, d=0.043, e=0.127, f=0.022, g=0.02, h=0.061, i=0.07, j=0.002, k=0.008, l=0.04, m=0.024, n=0.067, o=0.075, p=0.019, q=0.001, r=0.06, s=0.063, t=0.091, u=0.028, v=0.01, w=0.023, x=0.001, y=0.02, z=0.001) # 对字典中各字母出现的概率进⾏降序排序 order = dict(sorted((), key = lambda x:x[1], reverse = True)) print("统计中各字母出现的概率从⼩到⼤如下： ") print(order) # 统计密⽂中各字母出现的次数 example = {} for i in m: example[i] = (i) # 对字典中各字母出现的次数进⾏降序排序 result = dict(sorted((), key = lambda x:x[1], reverse = True)) print("计算得的密⽂中个字母的出现的次数从⼤到⼩如下： ") print(result) # #从结果可推测：Q由e加密⽽得，J由t加密⽽得，进⾏验算。 # (a*4+b)%26==16 # (a*19+b)%26==9 # 从⽽计算出a=3,b=4 # (a*K)%26==1,求得k=9 # ⽤k=9,b=4进⾏解密可得出明⽂ print("根据统计分析，加密所⽤的a, b可能为：") for i in range(1,26): for j in range(1,26): if (i*4+j)%26==16 and (i*19+j)%26==9: if gcd(i, j)==1: print("a="+str(i), "b="+str(j)) m = "CTRUJJUJZQGMRTUDGOCLCRWSEDOCGGJCLLCRJZQGMRJZQDQHMJSQSCLLLEMWZOUDQKURTCNQRJKELOCTJMDRQNJUTUMRNZCGUSR运⾏结果为（此处以破解a=3, b=4时得出的密⽂）：如图所⽰，正确解出a, b 再⽤(a*k)%26==1,求得k=9 ⽤k=9,b=4进⾏解密可得出明⽂。还可以通过差分分析进⾏破解。对于仿射密码来说，“差分”是模26减法，那么，在不知道两对明密⽂对（M1，C1）（M2, C2）的情况下，只需要知道M1-M2和C1-C2便可以确定a。因为C1 = aM1 + b(mod26)C2 = aM2 + b(mod26)易得，a = (C1 – C2)/(M1 – M2) (mod26)得到a 后，进⼀步找到b就很容易了。
2023年7月31日发(作者：)
仿射密码Python实现算法分析1. 仿射密码结合了移位密码和乘数密码的特点，是移位密码和乘数密码的组合。2. 仿射密码的加密算法就是⼀个线性变化，即对明⽂字符x，对应的密⽂字符为y=ax+b(mod26)其中，a, b属于Z26且gcd(a,b)=13. 实现过程：选取a，b两个参数，其中gcd(a, 26)＝1加密变换： c＝ a∗?＋b 26a=1时，移位密码b=1时，乘数密码解密变换： ?＝ (c−b)∗a^(−1) 26算法实现# 暴⼒破解la = [1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25]lb = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25]# 最⼤公约数def gcd(a, b): while b != 0: tem = a % b a = b b = tem return a# 加密def encrypt(m, c, a, b): for i in range(len(m)): # 加密成相应的⼤写字母 (chr(((ord(m[i]) - 97) * a + b) % 26 + 65)) d = ''.join(c) print(d)# 求逆元def niyuan(a, b): ny = 1 while (a * ny) % b != 1: ny += 1 return ny# 解密def decrypt(c, k, b): mw = [] for i in range(len(c)): tem = ord(c[i]) - 65 - b if tem < 0: tem += 26 (chr((k * tem) % 26 + 97)) print("k=" + str(k) + ", b=" + str(b) + "时，解密后的明⽂为：") res = ''.join(mw) print(res)#实现if __name__ == "__main__": # 明⽂ # 密⽂ m = 'ifnottothesunforsmilingwarmisstillinthesuntherebutwewilllaughmoreconfidentcalmifturnedtofoundhisownshadowappropriateescapethesunwillbethroughthe # 密⽂ c = [] x, y = input("请输⼊a和b: ").split() a = int(x) b = int(y) while gcd(a, b) != 1: x, y = input("a和b不互素，请重新输⼊a和b: ").split() a = int(x) b = int(y) print("明⽂内容为：") print(m) print("加密后的密⽂为：") encrypt(m, c, a, b) print("知道密钥破解：") k = niyuan(a, 26) decrypt(c, k, b) print("不知道秘钥破解，暴⼒破解如下： ") for i in range(0, 12): for j in range(0, 26): decrypt(c, la[i], lb[j])加密与解密加密：输⼊a = 3, b = 4时，加密结果如图所⽰：解密：知道秘钥k = 9, b = 4 （k为a的逆元）时，解出相应明⽂。正确性由于算法的前提要求gcd(a,26)==1, 从⽽使加密函数c＝ a∗?＋b 26是⼀个单射函数，故其解必然是唯⼀的。即，gcd(a,26)==1保证了仿射加密函数是⼀个双射函数，故算法正确。安全性分析此密码算法安全性较弱。由算法的实现可知，此算法的秘钥空间⼤⼩为12*26 – 1 ==311（去除a = 1, b = 0时的情况）且a =1,3,5,7,9,11,15,17,19,21,23,25 故很容易便能够通过暴⼒破解获得明⽂。还可以通过统计分析破解：代码如下#统计破解仿射密码# 最⼤公约数def gcd(a, b) while b != 0: tem = a % b a = b b = tem return aif __name__ == "__main__": # a = 3， b = 4时的密⽂ # 根据统计⽽得出的实际各字母出现的概率 reality = dict(a=0.082, b=0.015, c=0.028, d=0.043, e=0.127, f=0.022, g=0.02, h=0.061, i=0.07, j=0.002, k=0.008, l=0.04, m=0.024, n=0.067, o=0.075, p=0.019, q=0.001, r=0.06, s=0.063, t=0.091, u=0.028, v=0.01, w=0.023, x=0.001, y=0.02, z=0.001) # 对字典中各字母出现的概率进⾏降序排序 order = dict(sorted((), key = lambda x:x[1], reverse = True)) print("统计中各字母出现的概率从⼩到⼤如下： ") print(order) # 统计密⽂中各字母出现的次数 example = {} for i in m: example[i] = (i) # 对字典中各字母出现的次数进⾏降序排序 result = dict(sorted((), key = lambda x:x[1], reverse = True)) print("计算得的密⽂中个字母的出现的次数从⼤到⼩如下： ") print(result) # #从结果可推测：Q由e加密⽽得，J由t加密⽽得，进⾏验算。 # (a*4+b)%26==16 # (a*19+b)%26==9 # 从⽽计算出a=3,b=4 # (a*K)%26==1,求得k=9 # ⽤k=9,b=4进⾏解密可得出明⽂ print("根据统计分析，加密所⽤的a, b可能为：") for i in range(1,26): for j in range(1,26): if (i*4+j)%26==16 and (i*19+j)%26==9: if gcd(i, j)==1: print("a="+str(i), "b="+str(j)) m = "CTRUJJUJZQGMRTUDGOCLCRWSEDOCGGJCLLCRJZQGMRJZQDQHMJSQSCLLLEMWZOUDQKURTCNQRJKELOCTJMDRQNJUTUMRNZCGUSR运⾏结果为（此处以破解a=3, b=4时得出的密⽂）：如图所⽰，正确解出a, b 再⽤(a*k)%26==1,求得k=9 ⽤k=9,b=4进⾏解密可得出明⽂。还可以通过差分分析进⾏破解。对于仿射密码来说，“差分”是模26减法，那么，在不知道两对明密⽂对（M1，C1）（M2, C2）的情况下，只需要知道M1-M2和C1-C2便可以确定a。因为C1 = aM1 + b(mod26)C2 = aM2 + b(mod26)易得，a = (C1 – C2)/(M1 – M2) (mod26)得到a 后，进⼀步找到b就很容易了。
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