正则表达式 – DFA最小化
我有一个关于DFA最小化的问题.所以我使用了众所周知的技术将正则表达式转换为NFA,然后使用goto / closure算法从中构造DFA.现在问题是如何最小化它?我在这里看过有关它的内容:
http://www.youtube.com/watch?v=T9Z66NF5YRk,我仍然无法理解.什么是DFA最小化?这只是合并IDENTICAL状态(状态在相同的字符上进入相同的状态)还是不同的东西?
所以,我开始使用以下语法: %digit = '0'..'9' %letter = 'a'..'z' | 'A'..'Z' %exponent = ("e" | "E") ("+" | "-")? digit+ T_INT = digit+ T_FLOAT = T_INT exponent T_IDENTIFIER = (letter | "$" | "_") (letter | "$" | "_" | digit)* 最终得到以下DFA(表示为JSON): { "START": [{ "type": "range","from": 36,"to": 36,"shift": "1" },{ "type": "range","from": 48,"to": 57,"shift": "2" },"from": 65,"to": 90,"from": 95,"to": 95,"from": 97,"to": 122,"shift": "1" }],"1": [{ "type": "range",{ "shift": ["t_identifier"] }],"2": [{ "type": "range","from": 69,"to": 69,"shift": "3" },"from": 101,"to": 101,{ "shift": ["t_int"] }],"3": [{ "type": "range","from": 43,"to": 43,"shift": "5" },"from": 45,"to": 45,"shift": "4" }],"4": [{ "type": "range","shift": "4" },{ "shift": ["t_float"] }],"5": [{ "type": "range","shift": "4" }] } 那么我该如何最小化呢? 更新: 好的,这是我的算法.鉴于以下DFA: { 0: [{ from: 97,to: 97,shift: 1 }],1: [{ from: 97,shift: 3 },{ from: 98,to: 98,shift: 2 }],2: [{ from: 98,shift: 4 }],3: [{ from: 97,4: [{ from: 98,shift: 4 }] } 这是我做的最小化它: >对于每个状态(在我的示例中编号为0,1,2,3,4)获取标识此类状态的唯一哈希(例如,对于state0,这将是:from = 97,to = 97,shift = 1,for state1这将是:from = 97,shift = 3& from = 98,to = 98,shift = 2等等……) { 0: [{ from: 97,shift: 1 }],1: [{ from: 97,shift: 3 },{ from: 98,shift: 5 }],3: [{ from: 97,5: [{ from: 98,shift: 5 }] } { 0: [{ from: 97,shift: 6 }],6: [{ from: 97,shift: 6 },shift: 5 }] } 第二次更新: 好的,所以给出以下正则表达式:’a'(‘ce’)*(‘d’|’xa’|”AFe’)| ‘b'(‘ce’)*(‘d’|’xa’|”AFe’)’ce’我有以下DFA(START – >开始状态,[“接受”] – >所以到说过渡到接受状态): { "START": [{ "type": "range","from": 98,"to": 98,"shift": "1.2" },"to": 97,"shift": "17.18" }],"1.2": [{ "type": "range","from": 120,"to": 120,"shift": "10" },"from": 100,"to": 100,"shift": "6.7" },"to": 65,"shift": "8" },"from": 99,"to": 99,"10": [{ "type": "range","shift": "6.7" }],"6.7": [{ "type": "range","shift": "15" },"shift": "13" },"shift": "11" }],"15": [{ "type": "range","shift": "14.accept" }],"14.accept": [{ "type": "range","shift": "16" },{ "shift": ["accept"] }],"16": [{ "type": "range","13": [{ "type": "range","11": [{ "type": "range","from": 70,"to": 70,"shift": "12" }],"12": [{ "type": "range","8": [{ "type": "range","shift": "9" }],"9": [{ "type": "range","shift": "2.3" }],"2.3": [{ "type": "range","shift": "5" }],"17.18": [{ "type": "range","shift": "25" },"shift": "22.accept" },"shift": "23" },"shift": "20" }],"25": [{ "type": "range","shift": "22.accept" }],"22.accept": [{ "type": "range","shift": "28" },"shift": "26" },"28": [{ "type": "range","26": [{ "type": "range","shift": "27" }],"27": [{ "type": "range","23": [{ "type": "range","shift": "24" }],"24": [{ "type": "range","20": [{ "type": "range","shift": "18.19" }],"18.19": [{ "type": "range","shift": "21" }],"21": [{ "type": "range","shift": "18.19" }] } 故事是一样的,我该如何最小化它?如果我遵循经典的Hopcroft算法与所有这些表构造,确定不可区分的状态,将它们合并在一起等等,那么我将得到包含15个状态的DFA(使用此工具:http://regexvisualizer.apphb.com/与此正则表达式a(ce)( d | xa | AFe)| b(ce)(d | xa | AFe)ce检查).以下是使用Hopcroft算法进行缩小后DFA的样子: 在我重新思考Hopcroft的算法之后,我提出的算法构建的DFA小于您上面看到的DFA(对于图像质量而言,我不得不一步一步地重新绘制它以了解为什么它更小): 以下是它的工作原理,关于“状态等价”的决定是基于给定状态的哈希函数的结果(例如“START”)构建短字符串,如果我们从该状态开始,可以从DFA构造.给定上面的DFA和START状态,我们可以构造以下字符串:98-> 120,98-> 100,98-> 65,98-> 99,97-> 120,97-> 100,97-> 65,97-> 99因此,它是START状态的散列函数的结果.如果我们为DFA中的每个状态运行此函数,我们将看到对于某些状态,此函数给出了相同的结果(“1.2”,“6.7”,“2.3”和“10”,“13”和“15”,“16”和“11”,“8”,“26”,“23”和“12”,“9”,“4”,“5”,“20”,“21”和“17.18”,“ 18.19“AND”25“,”28“和”27“,”24“)所以我们需要做的就是将这些状态合并在一起. 我发现我在某处错了,但不明白我的算法产生的最小化DFA有什么问题? 解决方法
您提出的算法不会完全最小化,因为它不会检测行为相同的复杂结构.要理解这个DFA(由
JFLAP绘制):
最小化将结合q1和q2,但概述的算法无法管理. 与此相反,Hopcroft的算法最初会像这样分区: {q0,q1,q2},{q3} 然后拆分第一组,因为q0没有转换到q3: {q0},{q1,{q3} 而不是进一步分裂,因为q1和q2表现相同. (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |