使用基于文本的字典暴力破解凯撒密码

Question

#CODE
ALPHABET = "abcdefghijklmnopqrstuvwxyz0123456789!@#$%/."
MESSAGE = "6@ a2/.zb/4 ya a5ya a6!2 y 0#!0ya21 .a/b4492 dy. z26@4 0y//621 #@ d6a5 4/2ya2/ 52ya a5y@ 2c2/ 6@ a52 56452.a 06/092., z2ad22@ a52 $y/a62. #3 /b!fy@a.2c, a52 3/2@05, !y/fy 321#/#c@y, a52 A.y/2c605, y@1 #a52/., 1/#d@21 y. b.by9 zf a52 zbgg6@4 #3 a52 0#b/a 1/#@2.x Zba a52 0y9!, 9beb/6#b. 9632 #3 a2/.zb/4, 0#@02/@21 #@9f yz#ba y@a#!. y@1 /23920a6#@. #3 /2y9 9632, d2@a #@ 6@ 6a. #91 dyf y@1 !y12 6a 5y/1, 2e02$a zf y 4/2ya 233#/a, a# /2y96g2 a52 1y@42/ y@1 a52 163360b9a $#.6a6#@ #3 a52 /b..6y@ #x A52/2 d2/2 a52 .y!2 /202$a6#@. y@1 zy99., a52 .y!2 3/2@05 a52ya2/, a52 .y!2 0#b/a 6@a2/2.a. y@1 .2/c602 6@a2/2.a. y@1 6@a/64b2. y. b.by9x #@9f 6@ a52 c2/f 56452.a 06/092. d2/2 yaa2!$a. !y12 a# 822$ 6@ !6@1 a52 163360b9a62. #3 a52 y0aby9 $#.6a6#@x .a#/62. d2/2 d56./21 #3 5#d 16332/2@a9f a52 ad# 2!$/2..2. z25yc21 6@ a52.2 163360b9a 06/0b!.ay@02.x A52 2!$/2.. !y/fy, 0#@02/@21 3#/ a52 d293y/2 #3 a52 05y/6ayz92 y@1 21b0ya6#@y9 6@.a6aba6#@. b@12/ 52/ $ya/#@y42, 5y1 46c2@ 16/20a6#@. a5ya a52f .5#b91 y99 z2 /2!#c21 a# 8ygy@, y@1 a52 a56@4. z29#@46@4 a# a52.2 6@.a6aba6#@. 5y1 y9/2y1f z22@ $y0821 b$x A52 2!$/2.. 296.yz2a5, 5#d2c2/, d52@ y.821 d5ya 6@.a/b0a6#@. .52 d#b91 z2 y.21 a# 46c2- d6a5 52/ 05y/y0a2/6.a60 /b..6y@ $ya/6#a6.! 5y1 /221 a5ya .52 0#b91 46c2 @# 16/20a6#@. yz#ba .aya2 6@.a6aba6#@. 3#/ a5ya dy. a52 y33y6/ #3 a52 .#c2/264@, zba y. 3y/ y. .52 /.#@y99f dy. 0#@02/@21 .52 d#b91 z2 a52 9y.a a# %b6a a2/.zb/4x"

def importDictionary():
    f = open(r"dictionary.txt")
    english = []
    for word in f.read().split('\n'):
        english.append(word.lower())
    f.close()
    return english

english = importDictionary()     
     

def isEnglish(plaintext):
    """
    Does not work
    checking if plaintext letter is upper case
    for letter in plaintext:
        uppercase = letter.isupper()
    """
    possibleWord = [i.lower() for i in plaintext.split()]
    newWords = []
    match = 0
    totalWords = 0
    
    for word in possibleWord:
        totalWords += 1
        if word in english:
            match += 1
            newWords.append(word)
    
    """
    Does not work
    If the letter was upper case reassign the letter to it's original form
    if uppercase:
        letter = plaintext.upper()
        possibleWord.append(letter)
    """
    if (match/totalWords) * 100 >= 60:
        return ''.join(map(str, plaintext))
    else:
        return ""

def decryptstr(message,key):
    #creating a list for new letters
    newletters = []
    #loop to assign new letter value
    for letter in message:
        uppercase = letter.isupper()
        letter = letter.lower()
    #checking if the letter is upper or lower
        if letter in ALPHABET:
            index = ALPHABET.find(letter)
            newindex = (index - key) % len(ALPHABET)
            letter = ALPHABET[newindex]
            if uppercase:
                letter = letter.upper()
        newletters.append(letter)
    #joining the list
    return ''.join(newletters)


def main():
    x = 0
    while x < len(ALPHABET):
        key = int(x)
        decryptstr(MESSAGE,key)
        newString = decryptstr(MESSAGE,key)
        statment = isEnglish(newString)
        print("The Decryption key is: "+str(x)+"\n"+statment)
        x = x + 1
main()

大家好，代码，据我所知，代码功能齐全。我确实需要有关 print("The Decryption key is: "+str(x)+"\n"+statment) 的帮助，我只希望它在实际返回某些内容时打印，而不是在 returns 和 "" 时打印。我对代码的唯一其他问题是当 MESSAGE 这么长时，它需要很长时间。有没有人对我如何加速这段代码有任何想法 Here is the dictionary

Answer 1

def main():
    x = 0
    while x < len(ALPHABET):
        key = int(x)
        decryptstr(MESSAGE,key)
        newString = decryptstr(MESSAGE,key)
        statment = isEnglish(newString)
        if statment:
            print("The Decryption key is: "+str(x)+"\n"+statment)
        x = x + 1
main()

在 main 中刚刚在 print 语句之前添加了 if statment:。还要进行拼写检查。

使用基于文本的字典暴力破解凯撒密码

Using a text based dictionary to brute force crack Caesar Ciphers

python

python-3.x

caesar-cipher