如何使用马尔可夫链得到所有的可能性？

Question

我试图实现一个语言密码破解使用马尔可夫链。你知道吗

这背后的想法是从文本中选择n-gram，选择一个起始n-gram（通常是一个位于句首的单词），并使用前n-1个字符将其表示为一个状态。举个例子，对于“the”，我将有“th”。这将有一个字母及其出现的列表，并将表示为一个字典。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马尔可夫链。这意味着，当马尔可夫链的长度与我要查找的密码相同时，我将停止执行并检查马尔可夫链是否与我的密码相同。我试图实现这一点，使用递归函数，但由于某种原因，我得到堆栈溢出。你知道吗

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]

通过这个算法，我达到了最大的递归深度。但是我试图得到所有的马尔可夫链，直到找到密码短语。你知道吗

编辑1：现在，随着更新的代码，密码是找到的，但只是因为我循环所有可能的最后一个字母。你知道吗

例如：“确实”

ind已经在我的起始字母列表中了，我发现的所有三个字母都有“e”作为它们最常见的下一个字母。所以添加了e，然后是下一个e，然后是下一个e，现在密码是“indeee”，但是我将最后一个字母切分，然后再次遍历for，它最终会找到“确实”，这是可以的。 问题是，如果我给出indedd，它将找不到我的密码，因为第二个“d”永远不会循环通过。我怎样才能回到我的迭代中，在所有级别上遍历所有可能的字母？你知道吗

Answer 1

由于给出了答案，我终于成功了。我张贴的工作算法，希望有一天它会帮助别人。你知道吗

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]