Python 有限自动机是如何在代码中实现的？

如何在Python代码中实现这方面的

dfa

或

nfa

在python中有哪些好方法？

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它们是否曾用于实际项目中？

表示DFA的一种简单方法是将其作为字典字典。为每个州创建一个按字母表字母键控的字典，然后创建一个按州键控的全局字典。例如，以下来自

可以用以下词典表示：

dfa = {0:{'0':0, '1':1},
       1:{'0':2, '1':0},
       2:{'0':1, '1':2}}

在指定初始状态和接受值集后，针对从相关字母表中提取的输入字符串“运行”dfa非常简单：

def accepts(transitions,initial,accepting,s):
    state = initial
    for c in s:
        state = transitions[state][c]
    return state in accepting

您从初始状态开始，一个字符一个字符地遍历字符串，每一步只需查找下一个状态。当您完成对字符串的单步处理后，只需检查最终状态是否在接受状态集中

比如说

>>> accepts(dfa,0,{0},'1011101')
True
>>> accepts(dfa,0,{0},'10111011')
False

---

对于NFA，您可以在转换字典中存储可能的状态集而不是单个状态，并使用

随机

模块从可能的状态集中选择下一个状态。

这里我为NFA提供了一个递归解决方案。考虑下面的NFA：

可以使用列表列表表示转换，如下所示：

transition = [[[0,1],[0]], [[4],[2]], [[4],[3]], [[4],[4]],[[4],[4]]]

注意：状态4是一种假设状态。一旦你进入那种状态，你就不能再前进了。当您无法从当前状态读取输入时，它很有用。您直接转到状态4，并说当前进度不接受输入（通过返回检查其他可能性）。e、 g，如果您处于

q1

，且当前输入符号为

'a'

，则转到状态4并停止进一步计算

以下是Python代码：

#nfa simulation for (a|b)*abb
#state 4 is a trap state


import sys

def main():
    transition = [[[0,1],[0]], [[4],[2]], [[4],[3]], [[4],[4]]]
    input = raw_input("enter the string: ")
    input = list(input) #copy the input in list because python strings are immutable and thus can't be changed directly
    for index in range(len(input)): #parse the string of a,b in 0,1 for simplicity
        if input[index]=='a':
            input[index]='0' 
        else:
            input[index]='1'

    final = "3" #set of final states = {3}
    start = 0
    i=0  #counter to remember the number of symbols read

    trans(transition, input, final, start, i)
    print "rejected"



def trans(transition, input, final, state, i):
    for j in range (len(input)):
        for each in transition[state][int(input[j])]: #check for each possibility
            if each < 4:                              #move further only if you are at non-hypothetical state
                state = each
                if j == len(input)-1 and (str(state) in final): #last symbol is read and current state lies in the set of final states
                    print "accepted"
                    sys.exit()
                trans(transition, input[i+1:], final, state, i) #input string for next transition is input[i+1:]
        i = i+1 #increment the counter


main()

---

如果使用递归，则不需要范围（len（input））上的for循环。你把代码复杂化了。这是一个简化的版本

import sys

def main():
    transition = [[[0,1],[0]], [[4],[2]], [[4],[3]], [[4],[4]]]
    input = raw_input("enter the string: ")
    input = list(input) #copy the input in list because python strings are immutable and thus can't be changed directly
    for index in range(len(input)): #parse the string of a,b in 0,1 for simplicity
        if input[index]=='a':
            input[index]='0' 
        else:
            input[index]='1'

    final = "3" #set of final states = {3}
    start = 0

    trans(transition, input, final, start)
    print "rejected"


def trans(transition, input, final, state):
    for each in transition[state][int(input[0])]: #check for each possibility       
        if each < 4:                              #move further only if you are at non-hypothetical state
            state = each
            if len(input)==1:
                if (str(state) in final): #last symbol is read and current state lies in the set of final states
                    print "accepted"
                    sys.exit()
                else:
                    continue
            trans(transition, input[1:], final, state) #input string for next transition is input[i+1:]

main()

导入系统 def main（）： 转换=[[0,1]，[0]]，[[4]，[2]]，[[4]，[3]]，[[4]，[4]] 输入=原始输入（“输入字符串：”） input=list（input）#复制列表中的输入，因为python字符串是不可变的，因此不能直接更改 对于范围内的索引（len（input））：#为了简单起见，在0,1中解析a，b的字符串 如果输入[索引]=“a”： 输入[索引]=“0” 其他： 输入[索引]=“1” final=“3”#最终状态集={3} 开始=0 传输（转换、输入、最终、启动） 打印“已拒绝” def传输（转换、输入、最终、状态）： 对于每个处于过渡状态的[state][int（输入[0]）]：#检查每个可能性 如果每个<4:#仅当您处于非假设状态时才进一步移动 状态=每个 如果len（输入）==1： if（str（state）in final）：#读取最后一个符号，当前状态位于最终状态集中 打印“已接受” sys.exit（） 其他： 持续 trans（转换，输入[1:]，最终，状态）#下一个转换的输入字符串为输入[i+1:] main（）

---

如果您正在寻找更面向对象的dfa实现，这里是我的dfa实现版本。然而，约翰·科尔曼的回答给了我一点启发

class Node:
    def __init__(self, val):
        self.val = val
        self.links = []
    def add_link(self, link):
        self.links.append(link)
    def __str__(self):
        node = "(%s):\n" % self.val
        for link in self.links:
            node += "\t" + link + "\n"
        return node
    def __add__(self, other):
        return str(self) + other
    def __radd__(self, other):
        return other + str(self)
    def equals(self, node):
        ok = (self.val == node.val)
        if len(self.links) == len(node.links):
            for i in range(len(self.links)):
                ok = ok and (self.links[i] == node.links[i])
            return ok
        else:
            return False

class Link:
    def __init__(self, from_node, etiquette, to_node):
        self.from_node = from_node
        self.etiquette = etiquette
        self.to_node = to_node
    def __str__(self):
        return "(%s --%s--> %s)" % (self.from_node.val, self.etiquette, self.to_node.val)
    def __add__(self, other):
        return str(self) + other
    def __radd__(self, other):
        return other + str(self)
    def equals(self, link):
        return (self.from_node == link.from_node) and (self.etiquette == link.etiquette) and (self.to_node == link.to_node)

class Automata:
    def __init__(self, initial_node, nodes, terminal_node):
        self.initial_node = initial_node
        self.nodes = nodes
        self.terminal_node = terminal_node
    def get_next_node(self, current_node, etiquette):
        for link in current_node.links:
            if link.etiquette == etiquette:
                return link.to_node
        return None
    def accepts(self, string):
        node = self.initial_node
        for character in string:
            node = self.get_next_node(node, character)
        return self.terminal_node.equals(node)
    def __str__(self):
        automata = "Initial node: %s\nTerminal node: %s\n" % (self.initial_node.val, self.terminal_node.val)
        for node in self.nodes:
            automata += node
        return automata
    def __add__(self, other):
        return str(self) + other
    def __radd__(self, other):
        return other + str(self)




if __name__ == '__main__':
    pass

    s0 = Node("s0")
    s1 = Node("s1")
    s2 = Node("s2")

    s0_0_s0 = Link(s0, '0', s0)
    s0_1_s1 = Link(s0, '1', s1)
    s1_0_s2 = Link(s1, '0', s2)
    s1_1_s0 = Link(s1, '1', s0)
    s2_0_s1 = Link(s2, '0', s1)
    s2_1_s2 = Link(s2, '1', s2)

    s0.add_link(s0_0_s0)
    s0.add_link(s0_1_s1)
    s1.add_link(s1_0_s2)
    s1.add_link(s1_1_s0)
    s2.add_link(s2_0_s1)
    s2.add_link(s2_1_s2)

    a = Automata(s0, [s0, s1, s2], s0)

    print(a)
    print(a.accepts('1011101')) #True
    print(a.accepts('10111011')) #False

---

我已经实现了适用于任何dfa的dfa。但这不是在面向对象模式中

states=list(map(int,input("Enter States : ").split(" ")))
symbols=list(input("Enter Symbols : ").split(" "))
initial_state=int(input("Enter initial state : "))
final_states=list(map(int,input("Enter final states : ").split(" ")))

transitions=[]
inlists=[]

for i in range(len(symbols)):
    print("Enter transitions for symbol "+symbols[i]+" from all states :")
    for j in range(len(states)):
        inlists.append(int(input()))
    transitions.append(inlists)
    inlists=[]
cur_state=initial_state

while(1):
    cur_state=initial_state
    string=input("Enter string : ")

    if string=='#':
        break

    for symbol in string:
        print("["+str(cur_state)+"]"+"-"+symbol+"->",end="")
        cur_state=transitions[symbols.index(symbol)][cur_state]

    if cur_state in final_states:
        print("["+str(cur_state)+"]")
        print("String is accepted.")
    else:
        print("["+str(cur_state)+"]")
        print("String is not accepted.")

---

接受字符串101*和001*对@John Coleman的修改

#仅接受101+00101001的Dfa

dfa101 = {0:{'1':1},
       1:{'0':2},
       2:{'1':3},
       3:{'0':3, '1':3}}

#Dfa for accepting only 001+00101001
dfa001={0:{'0':1},
       1:{'0':2},
       2:{'1':3},
       3:{'0':3, '1':3}}



def accepts(transitions,initial,accepting,s):
    state = initial
    try:
        for c in s:
            state = transitions[state][c]
        if(state in accepting):
            return 'Accepted'
        else:
            return 'Rejected'
    except:
        return 'Rejected'
print('Dfa of 101+ ',accepts(dfa101,0,{3},'10101111000')) #Accepted

print('Dfa of 001+ ',accepts(dfa001,0,{3},'00101010101')) #Accepted

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这个问题非常广泛。除非你能把它缩小到一个具体的、客观的、可回答的问题，否则它很可能会被关闭。无论如何是的，它们用于实际项目中。“状态机”是一种常见的实现技术。python中有一种可能的方法：真正的正则表达式可以与DFA匹配；事实上，任何正则语言都可以表示为正则表达式或DFA。谢谢，伙计。这是一个很好的回答。

dfa101 = {0:{'1':1},
       1:{'0':2},
       2:{'1':3},
       3:{'0':3, '1':3}}

#Dfa for accepting only 001+00101001
dfa001={0:{'0':1},
       1:{'0':2},
       2:{'1':3},
       3:{'0':3, '1':3}}



def accepts(transitions,initial,accepting,s):
    state = initial
    try:
        for c in s:
            state = transitions[state][c]
        if(state in accepting):
            return 'Accepted'
        else:
            return 'Rejected'
    except:
        return 'Rejected'
print('Dfa of 101+ ',accepts(dfa101,0,{3},'10101111000')) #Accepted

print('Dfa of 001+ ',accepts(dfa001,0,{3},'00101010101')) #Accepted