IS CS-2017S1-01
题目来源:Problem 1 日期:2024-08-09 题目主题: CS-Formal Languages-Regular Languages
解题思路
这个问题涉及有限自动机的构造和正则语言的性质。我们需要构造一个 NFA 以满足特定语言的描述,证明任意有限语言都是正则的,然后构造 DFA 来识别补语言。最后,我们需要为判断 NFA 识别的语言是否是无穷集提供一个决策过程。
Solution
Question 1
Problem: Design an NFA such that , where , and the number of states of is not greater than 4.
Solution:
To solve this, we can create an NFA that detects if any character (either ‘a’ or ‘b’) occurs more than once in a string . The automaton will have 4 states to achieve this.
States:
- : Start state, initial state, no repeated character detected.
- : An ‘a’ has been seen.
- : A ‘b’ has been seen.
- : Accept state, a repeat of either ‘a’ or ‘b’ has been detected.
Transitions:
- From :
- On ‘a’, go to , .
- On ‘b’, go to , .
- From :
- On ‘a’, go to (accept state, since ‘a’ is repeated).
- On ‘b’, remain in .
- From :
- On ‘b’, go to (accept state, since ‘b’ is repeated).
- On ‘a’, remain in .
- From :
- On any input, stay in .
This NFA is illustrated by the following mermaid diagram:
stateDiagram-v2
[*] --> q0
q0 --> q0 : a, b
q0 --> q1 : a
q0 --> q2 : b
q1 --> q3 : a
q1 --> q1 : b
q2 --> q3 : b
q2 --> q2 : a
q3 --> q3 : a, b
q3 --> [*]
Question 2
Problem: Prove that any finite language is regular.
Solution:
A language is regular if there exists a finite automaton that recognizes it. Given a finite language , we can construct a finite automaton (specifically a DFA) that recognizes exactly this set of strings.
Proof:
- Consider each word . Since is finite, we can construct a path in a DFA that accepts .
- Create a unique path from the start state to an accept state for each word . The DFA will move through states corresponding to the characters of and reach a final state when the entire word is read.
- Since is finite, the DFA will have a finite number of paths, each corresponding to one word in .
- Thus, the DFA has a finite number of states and recognizes . Hence, is regular.
Question 3
Problem: Design a DFA such that , where is the language from Question 1. The number of states in should not exceed 5.
Solution:
To accept the complement of , we should focus on strings where no character appears more than once. These strings include (the empty string), “a”, “b”, “ab”, and “ba”. Any other string would contain repeated characters.
Let’s define the DFA where:
- States:
- Alphabet:
- Transition Function: is defined as:
- ,
- ,
- ,
- Start State:
- Accept States:
Here is the breakdown of the DFA’s behavior:
- : The initial state that accepts the empty string .
- : This state is reached after reading “a” and can move to if “b” follows.
- : This state is reached after reading “b” and can move to if “a” follows.
- : This state is reached after reading “ab” or “ba”, and does not accept any further input.
- : Any transition that introduces a repeated character will move the DFA to a trap state where no further transitions lead to an accepting state.
stateDiagram-v2
[*] --> q0
q0 --> [*]
q1 --> [*]
q2 --> [*]
q3 --> [*]
q0 --> q1 : a
q0 --> q2 : b
q1 --> q4 : a
q1 --> q3 : b
q2 --> q4 : b
q2 --> q3 : a
q3 --> q4 : a, b
q4 --> q4 : a, b
Question 4
Given an NFA , where:
- is the set of states
- is the alphabet
- is the transition function
- is the initial state
- is the set of final states
Follow these steps to determine if is infinite:
-
Convert the NFA to an equivalent DFA using the subset construction method.
-
In the resulting DFA , identify all states reachable from the initial state by performing a depth-first search (DFS) or breadth-first search (BFS).
-
Construct a directed graph where:
- Vertices represent the reachable states of
- Edges represent transitions between states
-
Perform a topological sort on . If a cycle is detected during this process, proceed to step 5. Otherwise, go to step 6.
-
If a cycle is found:
- Check if there exists a path from any state in the cycle to a final state in .
- If such a path exists, is infinite. Terminate the procedure.
-
If no cycle is found or no cycle leads to a final state:
- is finite if and only if for each path from the initial state to a final state, the length of the path is less than (the number of states in the original NFA ).
- Check all such paths. If any path has length , is infinite. Otherwise, it is finite.
Correctness:
- If there’s a cycle that can reach a final state, we can pump the cycle arbitrarily many times to generate infinitely many accepted strings.
- If there’s no such cycle, the only way to have infinitely many strings is if there’s a path longer than the number of states, which would imply a cycle by the pigeonhole principle.
知识点
重点词汇
- Finite Automaton (有限自动机)
- Regular Language (正则语言)
- Deterministic Finite Automaton (确定有限自动机)
- Nondeterministic Finite Automaton (非确定有限自动机)
- Language Complement (语言的补集)
- Cycle Detection (环检测)
参考资料
- Michael Sipser, Introduction to the Theory of Computation, Chapter 1.
- John E. Hopcroft, Rajeev Motwani, Jeffrey D. Ullman, Introduction to Automata Theory, Languages, and Computation, Chapter 2.