大概步骤
LL(1)文法的判定
准备工作
数据结构:本次实验选用python语言进行编程,主要用到的数据结构为字典。因其键值对一一对应较为方便,适合模拟LL(1)文法产生式左部和右部的对应关系。对于同一个非终结符多次出现在产生式左部的情况,本次实验采取的策略是用列表作为该键的值,这样便可以实现“一对多”的关系。
任务一 计算First
求解每个符号的First集:遍历所有符号,当该符号为终结符时,其First集仅有它自己这一个元素。否则如果是非终结符,先判断其是否能推出ε,如果能则将ε加入该符号的First集。接着,对以该终结符为左部的产生式的右部串进行扫描,判断该串的第一个符号,如果符号为终结符,则将此终结符加入First集。至此,我们可以得到所有能推出ε和直接推出终结符的First集。
接下来处理不能直接推出的情况。对每个非终结符,遍历对应产生式右部的每个符号,将每个符号的First集并入此非终结符的First集直到遇到第一个不能推出空串ε的非终结符为止。这里对空串ε的处理方法是:开始遍历每个非终结符时,设置一个标记all_null,并初始化为1,仅当遇到第一个不能推出空串ε的非终结符时,将all_null置为0,并不再判断该串后面的符号。值得注意的是,每次合并First集合时,并不把空串合并进去,当一个右部串判断结束后再判断all_null的值,如果all_null为1,则把空串ε加入First集。后面将多次用到all_null标记,思想较为相似,不再赘述。
这里还有一个问题,在进行取并集时,由于数据结构用的是列表,会有重复元素,因此要进行消重。这在后面也多次用到。
任务二 计算每个产生式右部的First集
求解每个产生式右部的First集:这里求解思路和求每个符号的First差不多,主要需要注意的地方当遇到第一个不能推出空串得非终结符时,停止后续符号的判断。这里也用到all_null进行控制,并且同样需要消重。
任务二 计算Follow
6. 求解每个非终结符的Follow集:第一步,需要把“#”放入开始符号“S”的Follow集中。接下来主要思想是找到每一非终结符所在串的下一个符号的First集,这里有几种特殊情况:(1)终结符为位于串尾,此时要将左部的Follow集并入该终结符的Follow集;(2)下一个符号推出空串ε,此时需要把下一个的后一个非终结符的Follow并入。注意:Follow集中不能出现空串ε。在代码实现上,可以把该非终结符后面所有的非终结符的Follow都并入Follow中,直到遇到串中的一个非终结符不能推出空串ε。这种合并…直到…的思想前面已经多次用到,可以是代码更简洁。最后再把所有的空串ε移除Follow集。消重不再赘述。
任务三 计算Select
求解每个产生式的Select集:求解出所有的First集和Follow后,可以直接利用公式得到Select。
任务四 判断
判断是否为LL(1)文法:最后一步进行判断,设置标记LL_1初始化为1,当交集不为空时LL_1置为0,得到最终判断结果。
总之,本次实验编码过程并不难,只需要注意一些细节和技巧,更为重要的是理解LL(1)文法判定的每一步思想,要真正地理解非终结符的First集和Follow集的意义是当词法分析遇到该非终结符时,下一步能够推得的终结符是哪些,进一步理解寻找Select交集的原因是遇到同一个终结符时,选择产生式会不会有多个产生式可选的冲突。如果要进一步探索本实验,可以增加第一步判断空串ε,还可以实现输入任意的上下文无关文法,判断其是否为LL(1)文法。
源代码
// An highlighted block
G = {'S': ['AB', 'bC'], 'A': ['ε', 'b'], 'B': ['ε', 'aD'], 'C': ['AD', 'b'], 'D': ['aS', 'c']} # G[S]
non_terminal = ['S', 'A', 'B', 'C', 'D'] # non_terminal
terminal = ['a', 'b', 'c'] # terminal
null = 'ε' # null string
First = {}
right_First = {}
print("G[S]:")
for left in G:
for i in range(len(G[left])):
print(left, '->', G[left][i])
# ============================== First Sets ========================================
for char in non_terminal+terminal:
First[char] = []
if char in terminal: # terminal char itself belongs to First(char)
First[char].append(char)
for left in G:
if left == char:
if null in G[left]: # X-> ε
First[char].append(null)
for i in range(len(G[left])):
if G[left][i][0] in terminal: # X-> a...
First[char].append(G[left][i][0])
for char in non_terminal:
for right_null in G[char]:
right = [r for r in right_null if r != null]
all_null = 1
for j in right:
if null in First[j]:
First[j].remove(null)
First[char] += First[j]
First[j].append(null)
else:
First[char] += First[j] # the first char that cannot deduce ε
all_null = 0 # all_null flag becomes 0
break
if all_null: # if every char can deduce ε, then append ε in First[char]
First[char].append(null)
redundancy = []
for x in First[char]: # eliminate redundancy
if x not in redundancy:
redundancy.append(x)
First[char] = redundancy
print("==================First sets======================")
for key in First:
print('First', '(', key, ')', '=', set(First[key]))
# ============================== right_First Sets ========================================
for left in G:
for right in G[left]:
right_First[right] = []
all_null = 1 # assume that all characters in right section can deduce ε
if right == null: # right section α = ε
right_First[right].append(null)
else:
if null not in First[right[0]]:
right_First[right] = First[right[0]]
else:
for i in range(len(right)):
right_First[right] += First[right[i]]
if null not in First[right[i]]: # when find one character that cannot deduce ε, set all_null = 0
all_null = 0
break
redundancy = []
for x in right_First[right]: # eliminate redundancy
if x not in redundancy:
redundancy.append(x)
right_First[right] = redundancy
if not all_null: # if not all_null, remove 'ε'
right_First[right].remove(null)
print("==================right_First sets======================")
for right in right_First:
print('First', '(', right, ')', '=', set(right_First[right]))
# ============================== Follow Sets ========================================
Follow = {}
for char in non_terminal:
Follow[char] = []
if char == 'S': # S is the start character
Follow[char].append('#')
else:
for left in G:
for right in G[left]:
all_null = 1 # assume that all characters in right section can deduce ε
if char == right[-1]:
Follow[char] += Follow[left]
else:
if char in right:
for i in range(right.index(char) + 1, len(right)): # check the remainder after char
Follow[char] += First[right[i]]
if null not in First[right[i]]: # when find one character cannot deduce ε
all_null = 0 # set all_null = 0
break
if all_null: # if all_null, add Follow[leftSection]
Follow[char] += Follow[left]
redundancy = []
for x in Follow[char]: # eliminate redundancy
if x not in redundancy:
redundancy.append(x)
Follow[char] = [f for f in redundancy if f != null]
print("==================Follow sets======================")
for left in Follow:
print('Follow', '(', left, ')', '=', set(Follow[left]))
# ============================== Select Sets ========================================
Select = {}
for left in G:
for right in G[left]:
key = left + '->' + right
Select[key] = []
if null not in right_First[right]:
Select[key] = right_First[right]
else:
Select[key] = [s for s in right_First[right] if s != null] + Follow[left]
print("==================Select sets======================")
for generator in Select:
print('Select', '(', generator, ')', '=', set(Select[generator]))
# ============================== Judgement of LL(1) ========================================
intersection = {}
for left in G:
intersection[left] = []
union = []
for right in G[left]:
generator = left + '->' + right
for s in Select[generator]:
if s not in union:
union.append(s)
else:
intersection[left].append(s)
LL_1 = 1
for left in intersection:
if len(intersection[left]) == 0:
intersection[left] = 'Φ'
else:
intersection[left] = set(intersection[left])
LL_1 = 0
print("==================Judgement of LL(1)======================")
for left in G:
generator_1 = left + '->' + G[left][0]
generator_2 = left + '->' + G[left][1]
print('Select', '(', generator_1, ')', '∩', 'Select', '(', generator_2, ')', '=', intersection[left])
if LL_1:
print("The intersection is Φ, so the G[s] belongs to LL(1) grammar!")
else:
print("The intersection is not Φ, so the G[s] does not belong to LL(1) grammar!")
运行结果
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