NFA中的缩写,python

时间:2013-02-16 21:27:21

标签: python finite-automata nfa

我正在尝试创建一种方法,因为疏忽从一个点跳到另一个点。

我创建了一个具有当前边缘的NFA

EDGES = [
(0, 'h', 1),
(1,'a',2),
(2,'z', 3),
(3,'a',4),
(4, 'r', 5),
(5, 'd', 6)
)]

我想要完成的例子 nrec("h-rd", nfa, 1)应该返回accept

nrec是处理NFA字符串并检查其接受或拒绝的方法。

def nrec(tape, nfa, trace=0):
"""Recognize in linear time similarly to transform NFA to DFA """
char = "-"
index = 0
states = [nfa.start]
while True:
    if trace > 0: print " Tape:", tape[index:], "   States:", states
    if index == len(tape): # End of input reached
        successtates = [s for s in states
                          if s in nfa.finals]
        # If this is nonempty return True, otherwise False.
        return len(successtates)> 0
    elif len(states) == 0:
        # Not reached end of string, but no states.
        return False
    elif char is tape[index]:

    # the add on method to take in abreviations by sign: -
    else:
        # Calculate the new states.
        states = set([e[2] for e in nfa.edges
                           if e[0] in states and
                              tape[index] == e[1] 
                      ])
        # Move one step in the string
        index += 1 

我需要添加一个将帐户缩减的方法。我不太确定如何从一个州跳到另一个州。 这就是NFA课程中的内容:

def __init__(self,start=None, finals=None, edges=None):
    """Read in an automaton from python shell"""
    self.start = start
    self.edges = edges
    self.finals = finals
    self.abrs = {}

我努力使用abrs,但在尝试定义我自己的abrs时,我常常会出错,例如

nfa = NFA(
start = 0,
finals = [6],
abrs = {0:4, 2:5},
edges=[
(0,'h', 1),
(1,'a', 2),
(2,'z', 3),
(3,'a', 4),
(4,'r', 5),
(5,'d', 6)
])

我发现错误“TypeError: init ()得到了一个意外的关键字参数'abrs'” 为什么我要回忆这个错误?

对于我所做的修改我会做这样的事情

 elif char is tape[index]:
 #get the next char in tape tape[index+1] so
 #for loop this.char with abrs states and then continue from that point.

明智的选择还是更好的解决方案?

1 个答案:

答案 0 :(得分:0)

错误是由__init__不接受定义的abrs关键字参数引起的。

def __init__(self,start=None, finals=None, edges=None):

您需要abrs=None(或其他值)才能使其成为关键字参数或abrs才能使其成为必需参数。