CS 161, Spring 2024: Homework 2 Homework 2: NFAs and Regular Expressions 0. (Ungraded exercise) We rushed/didn’t get to the exercises at the end of worksheet 3 (copied below for convenience). Make sure you understand what is wrong with these proofs. (a) Here is a false statement with a bad proof. What is wrong with the proof? Theorem (Not actually true). Every binary language is regular. Proof. Let A be any language. Here is a DFA M: M q0 0,1 Note that any string in A is accepted by this DFA. Thus, this DFA recognizes A, so A is regular. (b) Here is a false statement with a bad proof. What is wrong with the proof? Theorem (Not actually true). The language A = {00, 11} is not regular. Proof. Here is a DFA M: M q0 q1 0 1 1 0 The string 11, which is in A, is not accepted by this DFA. Thus, the DFA M does not recognize A, so A is not regular.
- (10 points) Let L be the language of binary strings with at least two 0s or at least three 1s. (a) (5 points) Draw a state diagram for an NFA that recognizes L. (b) (5 points) Recall that an NFA is a 5-tuple N = (Q, Σ, δ, q0, F) for finite set of states Q, finite set of alphabet代 写CS 161 NFAs and Regular Expressions characters Σ, transition function δ : Q × Σε → P(Q), start state q0 ∈ Q, and accept states F ⊂ Q. Describe your NFA as a 5-tuple.
- (10 points) Prove the following theorem by generalizing the construction from Worksheet 6. Theorem. The set of regular languages are closed under concatenation. (c) Sara Krehbiel, Ray Li 1 CS 161, Spring 2024: Homework 2 That is, prove that, for any two regular languages A and B, the language A ◦ B = {ab : a ∈ A : b ∈ B} is regular.
- (5 points) Consider the NFA N = ({1, 2, 3}, {0, 1}, δ, 1, {3}) with δ as depicted below (this is the same one from Quiz 6). Give a regular expression for the language recognized by this NFA. N 1 2 3 ε 1 0 1 0
- (10 points) Find an NFA that recognizes the language of (0◦1)∗ ◦(0∪1) (the alphabet is Σ = {0, 1}). Include both a state diagram and a formal specification of your automaton as a 5-tuple.
- (10 points) Let A be the language of strings over Σ = {0, 1} from the first day of class: A = {1 a01b01a+b : a, b ≥ 0}. Prove that A is not regular. (An informal interpretation of this result is: DFAs cannot add in unary) Hint: 1
- (15 points) We see in class on 4/15 how to convert any k-state NFA into an equivalent 2 k -state DFA. This problem shows that this exponential blowup in the number of states is necessary. Let A ⊂ {0, 1} ∗ be the set of all strings (of length at least 101) that have a 0 exactly 100 places from the right hand end. That is A = {w : |w| ≥ 101, w|w|−100 = 0}. (1) (a) (5 points) Draw the state diagram for an NFA with 101102 states that recognizes A. (You can use “· · · ” and don’t have to draw all 101102 states, as long as it’s clear what the states/transitions would be in the omitted states) [Ray: Update: I think you need 102 states. If you have 103 or 104 states, that’s fine.] (b) (10 points) Show that no DFA on less than 2100 states can recognize A. Hint:2 1 In this class, we learn several methods for proving a language A is regular: constructing a DFA recognizing A, constructing an NFA recognizing A, finding a regular expression for A. However, we only learn one method for proving a language is not regular. What is it? 2Give a proof by contradiction and assume such a DFA exists. Apply pigeonhole to all 2100 strings of length 100 to get two strings x and y of length 100 that end up at the same state after digesting. Derive a contradiction by considering the strings xz and yz for some carefully chosen string z. (c) Sara Krehbiel, Ray Li 2 WX：codinghelp
