有限自动机如何在代码中实现？

Question

如何在Python代码中实现dfa或nfa？

在python中有什么好方法可以做到这一点？ 他们曾经在现实世界的项目中使用过吗？

Answer 1

表示DFA的直接方式是字典词典。对于每个州，创建一个字典，该字典由字母表中的字母键入，然后是由状态键入的全局字典。例如，Wikipedia article on DFAs

中的以下DFA

可以用这样的字典表示：

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

针对从相关字母表中提取的输入字符串“运行”dfa（在指定初始状态和接受值集合之后）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，逐字逐句地逐字逐句，并在每一步只需查找下一个状态。完成单步执行后，只需检查最终状态是否处于接受状态集中。

例如

>>> 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多组可能的状态而不是单个状态，并使用random模块从可能的状态集中选择下一个状态。

Answer 2

好吧，我在这里为NFA提供递归解决方案。

考虑以下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是假设状态。一旦你进入那个州，你就无法继续前进。当您无法从当前状态读取输入时，它会很有用。您直接进入状态4并说当前进度不接受输入（通过返回检查其他可能性）。例如，如果您在q1，并且当前输入符号为'a'，则转到状态4并停止进一步计算。

这是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()

示例输出：（接受以abb结尾的字符串）

enter the string: abb
accepted

enter the string: aaaabbbb
rejected

...

Answer 3

如果您使用递归，则不需要在范围（len（input））上进行for循环。您使代码过于复杂。这是简化版

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()

Answer 4

如果您正在寻找一个更加面向对象的实现，这是我的dfa实现版本。但是，约翰·科尔曼（John Coleman）的回答让我有些受启发。

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

Answer 5

我已经实现了适用于任何dfa的dfa。但这不是面向对象的模式。

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.")

Answer 6

接受@John Coleman 的字符串 101* 和 001* 修改

#Dfa 只接受 101+00101001

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