【題目】
實現維吉尼亞密碼演算法，具體要求：
a. 實現維吉尼亞密碼加密過程，由使用者輸入金鑰，可以對任意輸入的明文進行加密；
b. 根據使用者輸入的金鑰，對密文進行解密；
c. 實現維吉尼亞密碼的唯密文攻擊破解（基於重合互指數方法）
【實現**】
# -*- coding: utf-8 -*-
"""created on wed dec 13 08:17:01 2017
@author: hp
"""#!/usr/bin/env python
"維吉尼亞"
from string import ascii_lowercase as lowercase
#加密def vigenereencrypto (p , key) :
p = get_trim_text(p)
ptlen = len(p)
keylen = len(key)
quotient = ptlen // keylen #商
remainder = ptlen % keylen #餘
out = ""
for i in range (0 , quotient) :
for j in range (0 , keylen) :
c = int((ord(p[i*keylen+j]) - ord('a') + ord(key[j]) - ord('a')) % 26 + ord('a'))
#global output
out += chr (c)
for i in range (0 , remainder) :
c = int((ord(p[quotient*keylen+i]) - ord('a') + ord(key[i]) - ord('a')) % 26 + ord('a'))
#global output
out += chr (c)
return out
#解密def vigeneredecrypto (output , key) :
ptlen = len (output)
keylen = len (key)
quotient = ptlen // keylen
remainder = ptlen % keylen
inp = ""
for i in range (0 , quotient) :
for j in range (0 , keylen) :
c = int((ord(output[i*keylen+j]) - ord('a') - (ord(key[j]) - ord('a'))) % 26 + ord('a'))
#global input
inp += chr (c)
for i in range (0 , remainder) :
c = int((ord(output[quotient*keylen + i]) - ord('a') - (ord(key[i]) - ord('a'))) % 26 + ord('a'))
#global input
inp += chr (c)
return inp
def get_trim_text(text):
text = text.lower()
trim_text = ''
for l in text:
if lowercase.find(l) >= 0:
trim_text += l
return trim_text
#計算重合指數
def get_coincidence_index(text):
text = get_trim_text(text)
length = len(text)
letter_stats =
for l in lowercase:
lt = {}
count = text.count(l)
lt[l] = count
index = 0
for d in letter_stats:
v = list(d.values())[0]
index += (float(v)/length) ** 2
return index
#計算和0.067的差距大小
def get_var(data, mean=0.067):
if not data:
return 0
var_sum = 0
for d in data:
var_sum += (d - mean) ** 2
return float(var_sum) / len(data)
#求秘鑰長度
def get_key_length(text):
# assume text length less than 26
text = get_trim_text(text)
group =
for n in range(1, len(text)+1):
group_str = ['' for i in range(n)]
for i in range(len(text)):
l = text[i]
for j in range(n):
if i % n == j:
group_str[j] += l
var_list =
length = 1
for tex in group:
data =
for t in tex:
index = get_coincidence_index(t)
length += 1
var_list = sorted(var_list, key=lambda x: x[1])
print(var_list)
return [v[0] for v in var_list[:int(n/2)+1]] #var_list[0][0]
# 統計字母頻度
def countlist(lis):
li =
alphabet = [chr(i) for i in range(97,123)]
for c in alphabet:
count = 0
for ch in lis:
if ch == c:
count+=1
return li
# 根據金鑰長度將密文分組
def texttolist(text,length):
text = get_trim_text(text)
textmatrix =
row =
index = 0
for ch in text:
index += 1
if index % length ==0:
row =
return textmatrix
# 獲取金鑰
def getkey(text,length):
text = get_trim_text(text)
key = # 定義空白列表用來存金鑰
alpharate =[0.08167,0.01492,0.02782,0.04253,0.12705,0.02228,0.02015,0.06094,\
0.06996,0.00153,0.00772,0.04025,0.02406,0.06749,0.07507,0.01929,\
0.0009,0.05987,0.06327,0.09056,0.02758,0.00978,0.02360,0.0015,0.01974,0.00074]
matrix = texttolist(text,length)
for i in range(length):
w = [row[i] for row in matrix if len(row) > i] #獲取每組密文
li = countlist(w)
powli = #算乘積
for j in range(26):
sum = 0.0
for k in range(26):
sum += alpharate[k]*li[k]
li = li[1:]+li[:1]#迴圈移位
abs = 100
ch = ''
for j in range(len(powli)):
if abs(powli[j] -0.065546)
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