文法分类的python实现

#-*-coding:utf-8-*-
G = raw_input("提示输入文法:"); #G为文法
S = G[2] #识别符号S Vn = raw_input("提示输入Vn:"); #Vn为非终结符号集
charactersVn = Vn.split(','); #记录每个非终结符号
VnString = "";
for var in charactersVn:
VnString += var + ",";
vAll = [];
lefties = []; #所有产生式左边的符号
setAllVt = set([]); #含重复元素
setVtAndVn = set([]); #非终结符号集的集合 P = raw_input("提示依次输入产生式规则:"); #P为产生式集
setP = P.split(); #根据空格分
for varP in setP:
n_t = varP.split('::=');
# leftOfP = n_t[0]; #产生式左边
lefties.append(n_t[0]);
rightOfP = n_t[1]; #产生式右边
rightChar = rightOfP.split('|'); #产生式右边的分割
setAllVt = setAllVt | set(rightChar); #所有产生式右边的式子的集合 #只保留式子右边中的单个符号
for ele in list(setAllVt):
# if len(ele) == 1:
# setVtAndVn = setVtAndVn | set(ele);
for i in range(0, len(ele)):
setVtAndVn = setVtAndVn | set(ele[i]); setVt = list(setVtAndVn - set(charactersVn)) #利用集合的差求非终结符号集
VtString = "";
for var in setVt:
VtString += var + ",";
#print setVt; #print len(lefties);
flag = 0; #代表几型文法
length = 0;
#记录并判断产生式左边长度是否都为1
for varLefty in lefties:
length += len(varLefty) if length == len(lefties): #长度都为1
flag = 2;
else:
flag = 1; #在0型文法和1型文法之间进行判断
if flag == 1:
for varP in setP:
n_t = varP.split('::=');
left = n_t[0];
right = n_t[1];
righties = right.split('|');
for righty in righties:
if len(left) > len(righty):
flag = 0; #在2型文法和3型文法中间进行选择
count = 0;
total = 0;
twoOrThird = 0;
if flag == 2:
twoOrThird = 1;
for varP in setP:
n_t = varP.split('::=');
right = n_t[1]; #::=的左边
righties = right.split('|'); #|分割后的每个产生式右边(由于表达形式的原因,要再次区分并计数)
for righty in righties:
total += 1;
if len(righty) == 1 and righty[0] in setVt: #右边为单个终结符合
count += 1;
elif len(righty) == 2 and righty[0] in setVt and righty[1] in charactersVn: #右边为单个终结符号加单个非终结符号
count += 1; if count == total and twoOrThird == 1:
flag = 3; print "文法" + G + "= ({" + VnString[:-1] + "} ,{" + VtString[:-1] + "}, P, " + G[2] + ")\n" ;
print P + "\n"
print "该文法是Chomsky%d型文法" % flag;
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