C语言实现NFA转DFA

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了C语言实现NFA转DFA相关的知识,希望对你有一定的参考价值。

请给我一个可以运行的代码,最好附上实验结果截图让我可以跑起来,代码里面最好有注释

参考技术A ε只能出现在NFA中,当然不是为了方便直观,而是连通NFA和DFA的桥梁。编译原理讲授的不是如何绘制NFA或者DFA,二是告诉读者怎样能够自动实现NFA或DFA的构造。在实际应用中ε可以帮助计算机转换NFA为DFA,而在属性文法和语法制导阶段,它也是沟通综合属性与继承属性、执行语义动作不可或缺的一部分。另外ε的使用可以大大简化文法产生式的构造难度。我记得最初使用ε是为了使得文法体系(字母表)更加完善,但是在实际应用中却变得应用广泛(此观点不一定正确)。
最后想说的是,在编译中,ε也带来了不小的麻烦,否则也就不会有诸如“去空产生式”这样的算法了:)

编译原理:NFA转DFA(含数据可视化)

第三方库graphviz的安装方法:https://blog.csdn.net/lizzy05/article/details/88529483
建议直接使用Linux环境,不需要使用环境变量
代码如下

def printlist(l):
    for i in range(len(l)):
        print(l[i], end=" ")


print("----------------------------输入NFA样例-----------------------------")
ab = eval(input("请输入变量的个数"))
print("请输入变量名称(用_(下划线)表示空)")
aabb = []
for i in range(1, ab + 1):
    aabb_eval = input("第" + str(i) + "个变量的名称是: ")
    aabb.append(aabb_eval)
printlist(aabb)
# [a,b,c]
# 输入变量的个数之后将为这个创造字典
l = [{} for i in range(ab)]  # 创建和变量相同个数的字典
n = 0
while n < ab:
    print("开始构造变量" + aabb[n] + "的两节点状态")
    while True:
        status = input("请输入起始状态")
        end_status = input("请输入终止状态")
        if status != '#' or end_status != '#':
            l[n].setdefault(status, end_status)
        else:
            print(aabb[n] + "变量的构造输入已经结束")
            break
    n = n + 1

for i in range(len(l)):
    if aabb[i] == '_':
        print("空变量的节点起始关系为:" + str(l[i]))
    else:
        print("变量" + aabb[i] + "的节点起始关系为:" + str(l[i]))

print(l)
# 空变量的节点起始关系为:{'2': '3', '1': '4', '5': '6'}
# 变量a的节点起始关系为:{'1': '4', '3': '5'}
# 变量b的节点起始关系为:{'5': '2', '3': '5'}

key_list = []
for i in range(len(l)):
    for k in l[i].keys():
        if k not in key_list:
            key_list.append(k)
print("状态节点有:", end=" ")
# {1,2,3,4,5}
printlist(key_list)
print()
start = input('请输入起始节点')
end_list = []
a = ""
while a != "#":
    a = input("请输入终止节点(集),以#表示结束")
    end_list.append(a)
end_list.pop()
print("起始节点是:" + start)
print("终止节点(集)是:", end="")
printlist(end_list)
# 变量a的节点起始关系为:{'1': '2', '3': '4'}
# 变量b的节点起始关系为:{'1': '3', '5': '2'}
# 空变量的节点起始关系为:{'2': '3', '4': '5'}
# 状态节点有: 1 3 5 2 4
# 请输入起始节点1
# 请输入终止节点(集),以#表示结束3
# 请输入终止节点(集),以#表示结束#
# 起始节点是:1
# 终止节点(集)是:3
print()
print("\\n-------------------将状态节点保存到文件中-----------------")
with open('nfa.txt', 'w') as f:
    f.write(start + "\\n")
    for i in range(len(end_list)):
        f.write(end_list[i] + " ")
    f.write("\\n")
    for j in range(len(l)):  # j表示第几个字典
        for k in range(len(l[j])):  # k表示第几个键值对
            for m in l[j].keys():  # m表示第几个键
                f.write(m + " " + aabb[j] + " " + l[j][m] + "\\n")
print("\\n------------------读取文件中的信息----------------------")
with open('nfa.txt', 'r') as r:
    nfa = []
    for line in r.readlines():  # 将文件数据内容保存到nfa列表当中
        line_rstrip = line.rstrip('\\n')
        nfa.append(line_rstrip)
    # print(nfa)
begin_start = nfa[0].split()  # 起始节点
#print(begin_start)
end_end_list = nfa[1].split()  # 终止节点
#print(end_end_list)
# 如果是按照前面的方法创建的,那么以上代码没有必要,直接用start和end_list即可
trans_nfa = nfa[2:]
#print(trans_nfa)
for i in range(len(trans_nfa)):
    trans_nfa[i] = trans_nfa[i].split()
#print(trans_nfa)
# #[['1', 'a', '2'], ['3', 'a', '4'], ['1', 'a', '2'], ['3', 'a', '4'], ['1', 'b', '2'], ['3', 'b',
# '4'], ['1', 'b', '2'], ['3', 'b', '4'], ['1', '_', '2']] 所有的状态已经标识好了
state_to_draw = []  # 状态节点
para_to_draw = []  # 转换变量
for i in range(len(trans_nfa)):
    state_to_draw.append(trans_nfa[i][0])
    state_to_draw.append(trans_nfa[i][2])
    para_to_draw.append(trans_nfa[i][1])

# print(state_to_draw)
# print(para_to_draw)

state0 = list(set(state_to_draw))  # 去除节点重复
state0.sort(key=state_to_draw.index)  # 对节点进行排序
para0 = list(set(para_to_draw))
para0.sort(key=para_to_draw.index)
# print(state0)  # ['0', '1', '2', '4', '3', '6', '5', '7', '8', '9', '10']
# print(para0)  # ['_', 'a', 'b']
if '_' in para0:
    para0.remove('_')
# 去掉空

_closure = dict()  # 表示空串可以到达的集合

print("\\n--------------开始作图-------------------")
from graphviz import Digraph


def draw_nfa():
    g = Digraph('G', filename='nfa.gv', format='png')
    for i in range(len(trans_nfa)):
        g.edge(trans_nfa[i][0], trans_nfa[i][2], label=trans_nfa[i][1])
    for i in range(len(begin_start)):
        g.node(begin_start[i], color='red')
    for i in range(len(end_end_list)):
        g.node(end_end_list[i], shape='doublecircle')
    g.view()


draw_nfa()
print("\\n------------nfa作图完毕------------------")
print("\\n-----------开始进行NFA转DFA---------------")

## 寻找所有空闭包
for i in range(len(state0)):
    res = [state0[i]]  # 第i个节点
    # print(res)
    for j in range(len(trans_nfa)):
        if trans_nfa[j][0] == state0[i] and trans_nfa[j][1] == '_':
            # 如果目前研究的节点里面存在着空
            res.extend(trans_nfa[j][2])
            # 这里用
    _closure[state0[i]] = list(set(res))
    # 到此,一个状态所有的空闭包都会显现出来,接下来只要一查到底即可


# print(_closure)
# {'0': ['1', '0', '7'], '1': ['1', '2', '4'], '2': ['2'], '4': ['4'], '3': ['6', '3'], '6': ['7', '6', '1'], '5': ['5', '6'], '7': ['7'], '8': ['8'], '9': ['9'], '10': ['10']}


# 空闭包的递归查询------
def find_closure(state_input):
    state_now = _closure[state_input]  # 查询输入一个状态下是否有空闭包 ['1', '0', '7']
    state_now_list = []
    for state_now_para in state_now:
        state_now_list = state_now_list + _closure[state_now_para]  # 递归查询
    state_now_list = list(set(state_now_list))
    ## 尝试输出
    #print(state_now_list)  # ['0', '7', '1', '2', '4']
    while set(state_now) != set(state_now_list):  # 多次查询
        state_now = state_now_list
        state_now_list = []
        for state_now_para in state_now:
            state_now_list = state_now_list + _closure[state_now_para]
        state_now_list = list(set(state_now_list))
    return state_now_list


A = find_closure(state0[0])  # 初始状态的空闭包
#print(A)  # ['1', '7', '4', '2', '0']


def find_state(l_state):
    res = dict()  # 创建一个字典,里面的值是状态集合
    for c in para0:  # ['_', 'a', 'b']
        res_two = []
        for i in range(len(l_state)):
            for j in range(
                    len(trans_nfa)):  # [['0', '_', '1'], ['1', '_', '2'], ['1', '_', '4'], ['2', 'a', '3'], ['3', '_', '6'], ['4', 'b', '5'], ['5', '_', '6'], ['6', '_', '7'], ['7', 'a', '8'], ['8', 'b', '9'], ['9', 'b', '10'], ['6', '_', '1'], ['0', '_', '7']]
                if trans_nfa[j][0] == l_state[i] and trans_nfa[j][1] == c:
                    res_two.append(trans_nfa[j][2])
        result = []
        for k in res_two:
            result = result + find_closure(k)
        res[c] = list(set(result))
    return res


number = 0
length = 1
state_list = []
state_list.append(A)  # [['0', '7', '1', '2', '4']]
while number < length:
    A2 = find_state(state_list[number])
    number = number + 1
    for c in para0:
        temp = 1
        for p in range(length):
            if set(A2[c]) == set(state_list[p]):
                temp = 0
        if temp == 1:
            state_list.append(A2[c])
            length = length + 1
#print(state_list)
# [['0', '2', '4', '1', '7'],
# ['2', '3', '6', '4', '8', '1', '7'],
# ['2', '6', '4', '5', '1', '7'],
# ['9', '2', '6', '4', '5', '1', '7'],
# ['2', '6', '4', '10', '5', '1', '7']]

#获取开始节点和结束节点
dfa_begin = []
dfa_end = []
for i in range(len(begin_start)):
    # 查看是否存在空……变量
    dfa_begin.append(find_closure(begin_start[i]))
for j in range(len(end_end_list)):
    for k in range(len(state_list)):
        if end_end_list[j] in state_list[k]:
            #如果nfa的结束节点是在我的某一个状态列表里面
            if state_list[k] not in dfa_end:
                dfa_end.append(state_list[k])

print(dfa_begin)
print(dfa_end)
while [] in state_list:
    state_list.remove([])

print('DFA状态表: ', state_list)
print("DFA起始状态: ", dfa_begin)
print("DFA终止状态:", dfa_end)

# DFA状态表为: [['4', '2', '0', '1', '7'], ['4', '2', '1', '7', '8', '3', '6'], ['4', '2', '1', '7', '6', '5'], ['4', '2', '1', '7', '9', '6', '5'], ['4', '2', '1', '7', '10', '6', '5']]
# DFA开始状态: [['4', '2', '0', '1', '7']]
# DFA终止状态: [['4', '2', '1', '7', '10', '6', '5']]
def draw_dfa(state_list, begin, end):
    g = Digraph('G', filename='dfa.gv', format='png')
    begin_aaa = []
    end_aaa = []
    for i in range(len(begin)):
        begin_aaa.append(" ".join(sorted(begin[i])))
    for j in range(len(end)):
        end_aaa.append(" ".join(sorted(end[j])))

    for i in range(len(state_list)):
        dicttttt = find_state(state_list[i])
        for j in para0:
            if state_list[i] != [] and dicttttt[j] != []:
                g.edge(" ".join(sorted(state_list[i])), " ".join(sorted(dicttttt[j])), label=j)
    for k in range(len(begin_aaa)):
        g.node(begin_aaa[k], color='red', shape='circle')
    for i in range(len(state_list)):
        g.node(" ".join(sorted(state_list[i])), color='blue', shape='circle')
    for j in range(len(end_aaa)):
        if end_aaa[j]!=begin_aaa[0]:
            g.node(end_aaa[j], shape='doublecircle')
        else:
            g.node(end_aaa[j], shape='doublecircle',color='red')
    g.view()


draw_dfa(state_list, dfa_begin, dfa_end)



state_dfa_other_name=[ chr(65+i) for i in range(len(state_list))]
#['A', 'B', 'C', 'D', 'E']
start_char=state_dfa_other_name[ state_list.index(dfa_begin[0])]
start_char=list(start_char)
print(start_char)
end_char=[]
for i in range(len(dfa_end)):

    s=state_dfa_other_name[state_list.index(dfa_end[i])]
    end_char.append(s)
print(end_char)

效果图:

以上是关于C语言实现NFA转DFA的主要内容,如果未能解决你的问题,请参考以下文章

C语言itoa()函数和atoi()函数详解(整数转字符C实现)

C语言游戏玩转扫雷——简单扫雷功能的实现!

c语言如何实现16进制字符串转换为 base64

中缀表达式转后缀表达式c语言实现

求助!C语言用指针函数实现十进制转,十六进制,八进制,二进制

(转)通过汇编语言实现C协程