CMPSCI 401: Theory of Computation

Second Midterm Exam, Spring 2013

David Mix Barrington

28 March 2013

Directions:

• Answer the problems on the exam pages.
• There are eight problems for 120 total points plus 15 extra credit. Actual scale was A = 105, C = 69.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• The first five questions are statements -- in each case say whether the statement is true or false and give a convincing justification of your answer -- a proof, counterexample, quotation from the book or from lecture, etc. You get five points for the correct boolean answer (so there is no reason not to guess if you don't know) and up to five for the justification.

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 10 points
  Q6: 25 points
  Q7: 15 points
  Q8: 30+15 points
 Total: 120+15 points

If C is any class of computers, such as DFA's, CFG's, LBA's TM, strange variant TM's, etc.:

• A_C = {(M, w): M is a computer in C and w ∈ L(M)}
• E_C = {(M): M is a computer in C and L(M) = ∅}
• ALL_C = {(M): M is a computer in C and L(M) = Σ^*}
• EQ_C = {(M, N): M and N are computers in C and L(M) = L(N)}
• REG_C = {(M): M is a computer in C and L(M) is a regular language}
• COMP_C = {(M, N): M and N are computers in C and L(M) is the complement of L(N)}

The language DECIDER is the set {(M): M halts (either accepts or rejects) on every input}.

Recall that if A and B are two languages, A is mapping reducible to B, written A ≤_m B, if there exists a function f: Σ^* → Σ^* such that for any string w, w ∈ A ↔ f(w) ∈ B.

A language A is Turing recognizable (TR) if it equals L(M) for some Turing machine M.

A is Turing decidable if A = L(M) for some Turing machine M that always halts.

A function f is Turing computable if there exists a Turing machine M such that for any string w, M when started on w halts with f(w) on its tape.

A stupid Turing machine (STM) has one tape which "wraps around" -- moving left from its leftmost cell takes you to its rightmost cell, and moving right from its leftmost cell takes you to its leftmost cell. It begins with the input string filling the entire tape.

A very stupid Turing machine (VSTM) is like an STM except that for some of its state-letter pairs it may replace the current cell with two cells, writing to both of them, instead of just writing to the current cell. Thus its transition function is δ (Q × Σ) ↔ (Q × (Σ ∪ (Σ × Σ)) × {L, R}).

The function S from N × N to N is defined as follows. Consider all the k-state Turing machines with tape alphabet {0, 1, blank} and consider all input strings in {0, 1}ⁿ. S(k, n) is the number of steps taken by the longest of any of the halting computations of any of these machines on any of these inputs. (This is similar to the "busy beaver function" but is not exactly the same thing.)

• Question 1 (10): True or false with justification: The language A_STM is Turing decidable.

• Question 2 (10): True or false with justification: The language A_VSTM is Turing decidable.

• Question 3 (10): True or false with justification: Any regular language is the language of some STM.

• Question 4 (10): True or false with justification: It is not true that REG_CFG ≤_m REG_TM.

• Question 5 (10): True or false with justification: The set of all TR languages is countably infinite.

• Question 6 (25): These questions use the definition of the language COMP_C above.

  • (a, 10) Prove that the language COMP_NFA is Turing decidable.

  • (b, 15) Is the language COMP_CFG Turing decidable? Prove your answer.

• Question 7 (15): Prove that the function S from N × N to N, defined above, is not Turing computable. (Hint: Assume that it is, and from this assumption prove the statement "A_TM is Turing decidable", which we know to be false.)

• Question 8 (30+15): These questions use the definitions of the languages DECIDER and COMP_C from above.
  • (a, 10): Prove, by any valid method, that DECIDER is not Turing decidable.

  • (b, 5XC): Prove, using the Recursion Theorem, that DECIDER is not Turing decidable. (Hint: Assume that there is a decider for DECIDER and that you can design a machine to obtain its own description, and derive a contradiction.)

  • (c, 10): Prove that DECIDER is not Turing recognizable. (Hint: You could show that the complement of A_TM is ≤_m DECIDER, but it may be easier to show that ALL_TM ≤_m DECIDER. You may assume without proof that ALL_TM is not TR.

  • (d, 10): Prove that DECIDER ≤_mCOMP_TM. What can you conclude about COMP_TM from this result?

  • (e, 10XC): Recall that a language A is in the class Π₂ if there is some Turing decidable language B such that for any string w, w ∈ A if and only if ∀x:∃y: (w, x, y) ∈ B. Prove that if A is any language in Π₂, then A ≤_m DECIDER.

Last modified 3 April 2013