Question

我正在尝试使用markov链实现语言密码破解程序。

其背后的想法是从文本中选择n-gram，选择起始n-gram（通常是句子开头的单词），并使用前n-1个字符将其表示为状态。例如，对于“ the”，我将拥有“ th”。这将具有出现字母的列表，并将其表示为字典。 dict["th"] = [('e', 120), ('a', 79)]等。对于这些值中的每一个，我将尝试创建一个满足我的密码或密码长度的markov链。这意味着当markov链与我要查找的密码具有相同的长度时，我将停止执行并检查markov链与我的密码是否相同。我正在尝试使用递归函数来实现这一点，但是由于某种原因，我正在堆栈溢出。

def ceva(myTry, good_all, pwd, guess, level):
        save = myTry
        if len(pwd) == len(guess):
            if pwd == guess:
                return 1
        else:
            if myTry in good_all.keys():
                values = good_all[myTry]
                for i in range(0,len(values)):
                    #print(i, len(values))
                    letter = values[i][0]
                    #print("First",myTry, letter)

                    pwd += letter
                    if i != len(values)-1:

                        if len(pwd) == len(guess):
                            #print("In if", pwd, myTry)
                            if pwd == guess:
                                print("I found:", pwd)
                                return 1
                            else:

                                pwd = pwd[0:len(pwd)-1]
                        else:

                            myTry += letter
                            myTry = myTry[1:]
                            #print("In else: ",pwd, myTry)
                            return ceva(myTry, good_all, pwd, guess, level)
                    else:
                        if len(pwd) == len(guess):
                            #print("In if", pwd, myTry)
                            if pwd == guess:
                                print("I found:", pwd)
                                return 1

                        pwd = pwd[0:len(pwd)-1]


    for key, letterList in starter_follows.items():
        myTry = key.replace("_", "")

        # i will not treat the case when the starting phrase
        # is a single character
        if myTry == "i":
            pass
        else:
            for letter in letterList:

                if letter[0] not in "_.-\"!":
                    myTry += letter[0]
                    pwd = copy.copy(myTry)
                    #print("Starter:", pwd)
                    res=ceva(myTry, good_all, pwd, toGuess, 1)
                    myTry = myTry[0:len(myTry)-1]

使用此算法，我达到了最大递归深度。但是我正在尝试获取所有的马尔可夫链，直到找到密码。

编辑1：现在，使用更新的代码，可以找到密码，但这仅是因为我在循环所有可能的最后一个字母。

例如：“确实”

ind已经在我的入门指南列表中，我发现的所有三字母组都以“ e”作为最常见的下一个字母。因此，先添加e，再添加下一个e，再添加下一个e，现在密码为“独立”，但是我将最后一个字母切成薄片并再次输入for，它最终找到“独立”，这是可以的。问题是，如果我提供indedd，它将找不到我的密码，因为第二个“ d”永远不会循环通过。如何返回迭代并遍历所有级别的所有可能字母？

Answer 1

由于给出的答案，我最终设法实现了目标。我发布了有效的算法，希望有一天能对某人有所帮助。

def ceva(myTry, good_all, pwd, guess, flag):

if len(pwd) > len(guess):
    return 0
if len(pwd) == len(guess):
    if pwd == guess:
        flag = 1
        return 1
    else: 
        return 0
save = copy.copy(myTry)
#print("Start of functionn:", "[", pwd, ",", myTry, "]")

if flag == 1:
    return 1
else:   

    if myTry in good_all.keys():
        # get the list of letters for this specific trigram
        values = good_all[myTry]

        if len(pwd) <= len(guess):
            for i in range(0,len(values)):
                #myTry = copy.copy(save)
                # get the letter
                letter = values[i][0]

                # add the letter to the password
                pwd += letter


                # if i found the password, set flag to 1 and break the loop
                if pwd == guess:
                    flag = 1
                    print("I found:", pwd)
                    return 1

                # add the new letter to the trigram, and the get rid of the first
                # letter, in order to create the new trigram
                myTry += letter
                myTry = myTry[1:]
                #print("Pwd before cutting: [", pwd, "]")
                res = ceva(myTry, good_all, pwd, guess, flag)
                #print(res)
                if res == 0:
                    pwd = pwd[0:len(pwd)-1]
                    myTry = pwd[-3:]
                #   print("This is after stop: [", pwd,",", myTry, "]")
                else:
                    return 1


    if flag == 0:
        return 0
    else:
        return 1

flag = 0

for key, letterList in starter_follows.items():
    myTry = key.replace("_", "")
    # i will not treat the case when the starting phrase
    # is a single character
    if myTry == "i":
        pass

    else:
        #print("Aaaa")
        for letter in letterList:

            if letter[0] not in "_.-\"!":

                # add the letter to create a (n-1)-gram
                myTry += letter[0]
                # create a copy of the starting password
                pwd = copy.copy(myTry)

                # call the recursive function
                res=ceva(myTry, good_all, pwd, toGuess, flag)
                # go back to the begining, to try a new start
                # e.g.: "the" first, "tha" second
                myTry = myTry[0:len(myTry)-1]

如何使用马尔可夫链获得所有可能性？

1 个答案: