On this page:
1 Counting to Infinity?
2 Not Counting to Infinity?
3 Sipser is Undecidable?
4 Sipser Helps Prove Undecidability?

Homework 11🔗

Last updated: Mon, 24 Nov 2025 11:18:20 -0500

Out: Mon Nov 17, 12:00pm EDT (noon) Due: Mon Nov 24, 12:00pm EDT (noon)

Note: Assignments are not officially "released" until—and are subject to change without notice up to—the indicated "Out" date and time. If an assignment is posted early, students may look ahead but are responsible for ensuring that they are always working with the most recent version of the homework.

This assignment explores the topic of undecidability.

Homework Problems

  1. Counting to Infinity? (4 + 8 = 12 points)

  2. Not Counting to Infinity? (5 + 10 = 15 points)

  3. Sipser is Undecidable? (15 points)

  4. Sipser Helps Prove Undecidability? (15 points)

  5. README (3 point)

Total: 60 points

Submitting

Submit your solution to this assignment in Gradescope hw11. Please assign each page to the correct problem and make sure your solutions are legible.

A submission must also include a README containing the required information.

1 Counting to Infinity?🔗

In class, we learned about how infinite sets can have different sizes. Specifically, they can be either countable or uncountable.

2 Not Counting to Infinity?🔗

3 Sipser is Undecidable?🔗

Prove that the following language is undecidable:

\textit{SIP} = \left\{\left\langle M\right\rangle\mid M\textrm{ is a \textsf{TM} where }\texttt{Sipser}\in L(M)\right\}

Your proof should be a proof by contradiction and should reduce from E_TM A_\textsf{TM}. It should also use the "modify the TM" technique from class. (You may not use Rice’s theorem in this problem.)

Remember, if the proof requires constructing a decider, even for the purposes of contradiction, all the rules for constructing deciders (e.g., termination arguments) still apply. If the proof includes a statement "machine X decides language L", even if just for the purposes of creating a contradiction, then it must be justified with an Examples Table.

4 Sipser Helps Prove Undecidability?🔗

Now let’s go the other way. Re-prove that E_\textsf{TM} is undecidable (so obviously you can’t assume that it’s already proven.)

Your proof must reduce from the undecidable \textit{SIP} language from the Sipser is Undecidable? problem. (You may assume that the \textit{SIP} language is undecidable, even if you were unable to answer the earlier questions.)

Your proof should be a proof by contradiction. It should also use the "modify the TM" technique from class. (You may not use Rice’s theorem in this problem.)

Remember, if the proof requires constructing a decider, even for the purposes of contradiction, all the rules for constructing deciders (e.g., termination arguments) still apply. If the proof includes a statement "machine X decides language L", even if just for the purposes of creating a contradiction, then it must be justified with an Examples Table.