python 编程练习——一种密码方法：逆矩阵的应用

import numpy as np
letters = [' ','A','B','C','D','E','F','G','H','I','J','K','L','M',
           'N','O','P','Q','R','S','T','U','V','W','X','Y','Z']
word_standard = []
save_numbers=[]
read_number=[]
return_word=[]
decode=''
codes=''
word=input("请输入想要加密的明文：")
word_standard.append(word.upper())
for i in (','.join(word_standard)):
    for j in range(len(letters)):
        if i == letters[j] :
            save_numbers.append(j)
row = len(save_numbers) / 3
M = np.mat(np.array(save_numbers).reshape(int(row),3))
A=np.mat([[-1,0,1],
         [1,-1,1],
         [1,1,0]])
MA = M*A
return_number=MA.tolist()
for m in return_number:
    for n in range(len(m)):
        read_number.append(m[n])
for i in read_number:
    codes += letters[i]
print("密文为：" + codes.lower())
print("是否需要解密？（Y/N）")
judge = input()
if judge == 'N':
    exit()
else:
    m = (MA*A.I).tolist()
    for a in m:
        for b in range(len(a)):
            return_word.append(round(a[b]))
for a in return_word:
    decode += letters[a]
print("解密结果为：" + decode.title())

请输入想要加密的明文：Hill on Tuesday
密文为：mcqcolftncnzvxe
是否需要解密？（Y/N）
Y
解密结果为：Hill On Tuesday

import numpy as np

def generate_plaintext():
    '''
    生成明文
    :return: 返回的是明文对应的数字
    '''
    num = []  # 存储明文对应的数字
    while True:
        info = input('请输入你要传递的明文:')
        if len(info) % 3 == 0:
            info = info.upper()
            for alpha in info:
                for key, values in dict_.items():
                    if alpha == values:
                        num.append(key)
            break
        else:
            print('输入的明文长度要为3的整数倍！请重新输入！')
    return num

def encryption(num, encr_matrix):
    '''
    加密
    :param num:明文對應的數字
    :param encr_matrix:加密矩陣
    :return:返回加密后信息和矩陣
    '''
    encry_mess = ''  # 存储加密信息
    row = int(len(num) / 3)
    col = 3
    num_arr = np.array(num).reshape(row, col)
    encr_info = np.matmul(num_arr, encr_matrix)
    encr_info_ = encr_info.flatten()
    for i in list(encr_info_):
        encry_mess += dict_[i]
    return encry_mess, encr_info

def deciphering(encr_info, encr_matrix):
    '''
    解密
    :param encr_info:加密後的矩陣
    :param encr_matrix:加密矩陣
    :return:
    '''
    deci_info = ''  # 存储解密后的信息
    deci_matrix = np.linalg.inv(encr_matrix)  # 產生解密矩陣
    deci_arr = np.matmul(encr_info, deci_matrix)
    deci_arr = np.around(deci_arr)
    # deci_list = deci_arr.tolist()
    deci_list = list(deci_arr.flatten())
    # print(deci_list)
    for i in deci_list:
        deci_info += dict_[i]
    return deci_info

if __name__ == '__main__':
    dict_ = {0: ' ', 1: "A", 2: 'B', 3: "C", 4: 'D', 5: 'E', 6: "F", 7: 'G', 8: 'H',
             9: "I", 10: 'J', 11: 'K', 12: "L", 13: 'M', 14: 'N', 15: "O", 16: 'P', 17: 'Q', 18: "R", 19: 'S',
             20: 'T', 21: "U", 22: 'V', 23: 'W', 24: "X", 25: 'Y', 26: 'Z'}
    num = generate_plaintext()
    encr_matrix = np.array([[-1, 0, 1], [1, -1, 1], [1, 1, 0]])
    # print(num)
    encr_mess, encr_info = encryption(num, encr_matrix)
    print('加密后的信息为:{}'.format(encr_mess))
    info = input('请选择是否要进行解密:(y/n)')
    if info == 'y':
        deci_info = deciphering(encr_info, encr_matrix)
        print('解密后的信息为:{}'.format(deci_info.lower()))

请输入你要传递的明文:hill on tuesday
加密后的信息为:MCQCOLFTNCNZVXE
请选择是否要进行解密:(y/n)y
解密后的信息为:hill on tuesday