Homework 7

Last updated: Fri, 6 Nov 2020 21:45:50 -0500

Out: Wed Nov 4, 00:00 EST Due: Tues Nov 10, 23:59 EST

This homework covers material from Chapters 4 and 5 of the textbook.

Homework Problems

1. README (2 points)

2. A Diagonalization Proof (6 points)

3. E_TM is not Turing-recognizable (6 points)

4. Intersection of Context-Free and Regular (6 points)

5. Palindromes in Regular Languages? (6 points)

6. EQ_CFG is undecidable (6 points)

Total: 32 points

Submitting

Submit this assignment at Gradescope hw7.

1 README

Create a README file containing the required information, and submit it along with the rest of the homework.

2 A Diagonalization Proof

In class (and in the book), to show that there are more languages than Turing machines, we used the fact that set of all infinite binary sequences—infinite strings of 1s and 0s—is uncountable. That is, there is no bijection between this set and the natural numbers.

Prove that the set of all infinite binary sequences is uncountable. Use diagonalization.

3 E_TM is not Turing-recognizable

E_{TM} = \{\left\langle M\right\rangle\mid M \textrm{ is a TM and } L(M) = \emptyset\}

We showed in class that E_{TM} is undecidable, by reducing it to A_{TM}.

Show that E_{TM} is not even Turing-recognizable!

4 Intersection of Context-Free and Regular

You proved in Homework 1, The Intersection Operation (Optional Bonus), that regular languages are closed under intersection.

But we showed in class that context-free languages are not closed under intersection.

Prove that the intersection of a context-free and a regular language is context-free.

(It may be helpful to review how we proved intersection closure for regular languages.)

5 Palindromes in Regular Languages?

You showed that the language of palindromes is context-free in Homework 4, CFGs.

Prove that the language:

HASPAL_{DFA} = \{\left\langle D\right\rangle\mid D \textrm{ is a DFA that accepts at least one palidrome}\}

is decidable.

As usual, our library of previous theorems (particularly those about CFLs, and maybe the solution from Intersection of Context-Free and Regular) will again be useful here.

6 EQ_CFG is undecidable

We’ve been speculating that:

EQ_{CFG} = \{\left\langle G,H\right\rangle\mid G \textrm{ and }H\textrm{ are CFGs and } L(G) = L(H)\}

is an undecidable language (page 200 of Sipser).

Now that we’ve learned about reducibility, we can finally prove it.

Prove that EQ_{CFG} is undecidable.

You may find it helpful to use the Theorem 5.13 from the book, that ALL_{CFG} = \{\left\langle G\right\rangle\mid G\textrm{ is a CFG and } L(G) = \Sigma^*\} is undecidable.