我已经提出了使用Python的加密例程,我很感激任何人都可以看一下.我正在寻找一些信息; 之前是否使用过此方法/变体,如果是这样,它的名称是什么？它有多安全？
我们的想法是通过互联网传输数据,并使用双方都知道的两个密码进行加密.
它使用SHA1哈希对密码进行编码,然后使用哈希中的字符创建偏移查找表.将偏移值添加到普通字符以生成加密字符.它使用一对一方法而不是压缩或添加数据.
字符串'Hello StackOverflow'生成'wPjOew6AdoNOYgjf7y'如果两个SHA1哈希是使用'burger'和'meat'生成的.
这是整个代码,对于超长字典数组抱歉:S
代码运行使用:win32上的Python 2.7.2(默认,2011年6月12日,15:08:59)[MSC v.1500 32位(英特尔)]
import sys
# burger
sha1_pass1 = '7a86b15480e0a870f0b07a4d23a54ef8f9acac44'
# meat
sha1_pass2 = 'bb40f75a9c6038e0da200fc5c3a6f371c1592c66'
# Characters available to encrypt (can be extended)
valid_chars = '0123456789 abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ.,!?' * 2
offset = {'00':0,
'01':1,
'02':2,
'03':3,
'04':4,
'05':5,
'06':6,
'07':7,
'08':8,
'09':9,
'0a':10,
'0b':11,
'0c':12,
'0d':13,
'0e':14,
'0f':15,
'10':16,
'11':17,
'12':18,
'13':19,
'14':20,
'15':21,
'16':22,
'17':23,
'18':24,
'19':25,
'1a':26,
'1b':27,
'1c':28,
'1d':29,
'1e':30,
'1f':31,
'20':32,
'21':33,
'22':34,
'23':35,
'24':36,
'25':37,
'26':38,
'27':39,
'28':40,
'29':41,
'2a':42,
'2b':43,
'2c':44,
'2d':45,
'2e':46,
'2f':47,
'30':48,
'31':49,
'32':50,
'33':51,
'34':52,
'35':53,
'36':54,
'37':55,
'38':56,
'39':57,
'3a':58,
'3b':59,
'3c':60,
'3d':61,
'3e':62,
'3f':63,
'40':64,
'41':65,
'42':66,
'43':0,
'44':1,
'45':2,
'46':3,
'47':4,
'48':5,
'49':6,
'4a':7,
'4b':8,
'4c':9,
'4d':10,
'4e':11,
'4f':12,
'50':13,
'51':14,
'52':15,
'53':16,
'54':17,
'55':18,
'56':19,
'57':20,
'58':21,
'59':22,
'5a':23,
'5b':24,
'5c':25,
'5d':26,
'5e':27,
'5f':28,
'60':29,
'61':30,
'62':31,
'63':32,
'64':33,
'65':34,
'66':35,
'67':36,
'68':37,
'69':38,
'6a':39,
'6b':40,
'6c':41,
'6d':42,
'6e':43,
'6f':44,
'70':45,
'71':46,
'72':47,
'73':48,
'74':49,
'75':50,
'76':51,
'77':52,
'78':53,
'79':54,
'7a':55,
'7b':56,
'7c':57,
'7d':58,
'7e':59,
'7f':60,
'80':61,
'81':62,
'82':63,
'83':64,
'84':65,
'85':66,
'86':0,
'87':1,
'88':2,
'89':3,
'8a':4,
'8b':5,
'8c':6,
'8d':7,
'8e':8,
'8f':9,
'90':10,
'91':11,
'92':12,
'93':13,
'94':14,
'95':15,
'96':16,
'97':17,
'98':18,
'99':19,
'9a':20,
'9b':21,
'9c':22,
'9d':23,
'9e':24,
'9f':25,
'a0':26,
'a1':27,
'a2':28,
'a3':29,
'a4':30,
'a5':31,
'a6':32,
'a7':33,
'a8':34,
'a9':35,
'aa':36,
'ab':37,
'ac':38,
'ad':39,
'ae':40,
'af':41,
'b0':42,
'b1':43,
'b2':44,
'b3':45,
'b4':46,
'b5':47,
'b6':48,
'b7':49,
'b8':50,
'b9':51,
'ba':52,
'bb':53,
'bc':54,
'bd':55,
'be':56,
'bf':57,
'c0':58,
'c1':59,
'c2':60,
'c3':61,
'c4':62,
'c5':63,
'c6':64,
'c7':65,
'c8':66,
'c9':0,
'ca':1,
'cb':2,
'cc':3,
'cd':4,
'ce':5,
'cf':6,
'd0':7,
'd1':8,
'd2':9,
'd3':10,
'd4':11,
'd5':12,
'd6':13,
'd7':14,
'd8':15,
'd9':16,
'da':17,
'db':18,
'dc':19,
'dd':20,
'de':21,
'df':22,
'e0':23,
'e1':24,
'e2':25,
'e3':26,
'e4':27,
'e5':28,
'e6':29,
'e7':30,
'e8':31,
'e9':32,
'ea':33,
'eb':34,
'ec':35,
'ed':36,
'ee':37,
'ef':38,
'f0':39,
'f1':40,
'f2':41,
'f3':42,
'f4':43,
'f5':44,
'f6':45,
'f7':46,
'f8':47,
'f9':48,
'fa':49,
'fb':50,
'fc':51,
'fd':52,
'fe':53,
'ff':54,}
cipher = []
# create the lookup table in cipher
for n in range(40):
sp1 = sha1_pass1[n]
sp2 = sha1_pass2[n]
cipher.append(offset[sp1 + sp2])
# get a user defined string
ask = raw_input('\n\n>>> ')
print ('\n') # make some space
# exit if return
if not ask:
sys.exit(1)
cipher_pos = 0
# progress through the user string
for n in range(len(ask)):
c = ask[n] # character n
# get the position of character in string
p = valid_chars.find(c)
if p == -1: sys.exit(1) # if not found then end
p += cipher[cipher_pos] # add the offset created by the passwords
cipher_pos += 1
if cipher_pos == 40: cipher_pos = 0 # reset lookup table position so it repeats
# get new character
x = valid_chars[p]
sys.stdout.write(x)
sys.stdout.flush()
print('\n')
您的代码是Vigenère密码,密码长度为40个字符.因此,您只需在链接的维基百科页面上查找其弱点即可.
最明显的攻击是:
频率分析
一旦知道了密钥的长度,就可以将密文重写成那么多列,每列对应于密钥的单个字母.每列由明文组成,该明文由单个凯撒密码加密; 凯撒键(shift)只是用于该列的Vigenère键的字母.使用类似于打破凯撒密码的方法,可以发现密文中的字母.
你有点落后于时代,你的计划已于1553年发明.
您的代码的另一个问题是它是没有初始化向量的同步流密码.这些密码具有通用的弱点,即不止一次使用密钥允许攻击者减去两个密文,从而消除密钥.结果就是普通文本的不同之处.在实践中,这通常足以获得两个明文.