nfa

Model Checking : Bad Prefixes using NFA

由 回眸只為那壹抹淺笑 提交于 2020-04-16 05:12:33
问题 We use NFA to model BadPrefixes for the safety property.I want to understand for a given Safety property , how to model the NFA. The following images are for reference. For instance, for safety property P2 ,Can someone explain how to know how many states are required(solution has 4) and which logic to use on the edges, how in Fig.,3 and Fig.4 , the edges are selected to satisfy the badprefixes P1 and P2.Thanks. 回答1: We have several definitions and notations here, let's go through these first:

正则基础 之 NFA引擎匹配原理

由 折月煮酒 提交于 2020-02-29 14:09:16
NFA 引擎匹配原理 1 为什么要了解引擎匹配原理 一个个音符杂乱无章的组合在一起，弹奏出的或许就是噪音，同样的音符经过作曲家的手，就可以谱出非常动听的乐曲，一个演奏者同样可以照着乐谱奏出动听的乐曲，但他 / 她或许不知道该如何去改变音符的组合，使得乐曲更动听。 作为正则的使用者也一样，不懂正则引擎原理的情况下，同样可以写出满足需求的正则，但是不知道原理，却很难写出高效且没有隐患的正则。所以对于经常使用正则，或是有兴趣深入学习正则的人，还是有必要了解一下正则引擎的匹配原理的。 2 正则表达式引擎 正则引擎大体上可分为不同的两类： DFA 和 NFA ，而 NFA 又基本上可以分为传统型 NFA 和 POSIX NFA 。 DFA Deterministic finite automaton 确定型有穷自动机 NFA Non-deterministic finite automaton 非确定型有穷自动机 Traditional NFA POSIX NFA DFA 引擎因为不需要回溯，所以匹配快速，但不支持捕获组，所以也就不支持反向引用和 $number 这种引用方式，目前使用 DFA 引擎的语言和工具主要有 awk 、 egrep 和 lex 。 POSIX NFA 主要指符合 POSIX 标准的 NFA 引擎，它的特点主要是提供 longest-leftmost 匹配

I am trying to implement NFA in Python to recognize words but my code doesn't work,

由 拥有回忆 提交于 2020-01-24 23:57:09
问题 I am trying to implement a method that recognizes words. I have written the following code and have tried to follow my code on paper and execute it step by step with example inputs, but I can't find the reason why my code is not doing what I want him to do. Does anyone see the flaw? I can't see it and I am confused on why it doesn't work. from collections import defaultdict class NFA: def __init__(self, initial, trns, final): self.initial = initial self.final = set(final) self.trns =

time complexity trade offs of nfa vs dfa

由 不想你离开。 提交于 2020-01-14 12:10:44
问题 I am looking for a discussion on which is better used and in what circumstances in a compiler an nfa or dfa. what are the time complexity trade-offs of simulating an nfa vs dfa and which one is more suitable during what circumstances in a compiler?? 回答1: The construction time for a DFA from an NFA is O(2^m) where m is the number of nodes. The running time of a DFA is O(n) where n is the length of the input string. This is because there is only 1 path through the DFA for a given string. The

time complexity trade offs of nfa vs dfa

由 假如想象 提交于 2020-01-14 12:10:13
问题 I am looking for a discussion on which is better used and in what circumstances in a compiler an nfa or dfa. what are the time complexity trade-offs of simulating an nfa vs dfa and which one is more suitable during what circumstances in a compiler?? 回答1: The construction time for a DFA from an NFA is O(2^m) where m is the number of nodes. The running time of a DFA is O(n) where n is the length of the input string. This is because there is only 1 path through the DFA for a given string. The

time complexity trade offs of nfa vs dfa

由 拟墨画扇 提交于 2020-01-14 12:09:30
问题 I am looking for a discussion on which is better used and in what circumstances in a compiler an nfa or dfa. what are the time complexity trade-offs of simulating an nfa vs dfa and which one is more suitable during what circumstances in a compiler?? 回答1: The construction time for a DFA from an NFA is O(2^m) where m is the number of nodes. The running time of a DFA is O(n) where n is the length of the input string. This is because there is only 1 path through the DFA for a given string. The

time complexity trade offs of nfa vs dfa

由 て烟熏妆下的殇ゞ 提交于 2020-01-14 12:09:02
问题 I am looking for a discussion on which is better used and in what circumstances in a compiler an nfa or dfa. what are the time complexity trade-offs of simulating an nfa vs dfa and which one is more suitable during what circumstances in a compiler?? 回答1: The construction time for a DFA from an NFA is O(2^m) where m is the number of nodes. The running time of a DFA is O(n) where n is the length of the input string. This is because there is only 1 path through the DFA for a given string. The

Convert a NFA to Regular Expression

由 心已入冬 提交于 2020-01-13 11:46:30
问题 I found a same question on this website, and the answer was a PDF describing how to convert an NFA to a regex. But this is not working because this method has some conditions: There are transitions going from the initial state to all other states, and there are no transitions into the initial state. There is a single accept state that has only transitions coming into it (and no outgoing transitions). The accept state is distinct from the initial state. Except for the initial and accepting

Efficient matching of text messages against thousands of regular expressions

由 三世轮回 提交于 2019-12-24 01:08:48
问题 I am solving a problem where I have text message to match with thousands of regular expressions of the form <some string> {0 or 300 chars} <some string> {0 or 300 chars} e.g. "on"[ \t\r]*(.){0,300}"."[ \t\r]*(.){0,300}"from" or a real example can be "Dear"[ \t\r]*"Customer,"[ \t\r]*"Your"[ \t\r]*"package"[ \t\r]*(.){0,80}[ \t\r]*"is"[ \t\r]*"out"[ \t\r]*"for"[ \t\r]*"delivery"[ \t\r]*"via"(.){0,80}[ \t\r]*"Courier,"[ \t\r]*(.){0,80}[ \t\r]*"on"(.){0,80}"."[ \t\r]*"Delivery"[ \t\r]*"will"[ \t

Converting NFA to DFA

由 本秂侑毒 提交于 2019-12-21 23:27:24
问题 I'm having trouble understanding how to convert. If 2 gets an input of 'a' would it become (1,4) or (1,2,4) because of the empty string? Thanks! 回答1: If state Q2 gets an input of 'a' next states may be either Q1 , Q2 , 0r Q4 . In your NFA your get final state Q4 Its equivalent DFA is as below: a- || ▼| --►(Q0)---a---►((Q1))---b----►((Qf)) ▲-----a--------| Where Q1 and Q2 are final state. And its Regular Expression is: a (a + ba)* (b + ε ) Where ε is null symbol (epsilon) 回答2: We begin to
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