CMPSCI 501: Theory of Computation

Solutions to Second Midterm Exam, Spring 2014

David Mix Barrington

27 March 2014

Directions:

• Answer the problems on the exam pages.
• There are nine problems (some with multiple parts) for 120 total points plus 10 extra credit. Actual scale was A = 96, C = 56.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• Many useful definitions are given below.
• The first six 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: 10 points
  Q7: 30 points
  Q8: 20 points
  Q9: 10+10 points
 Total: 120+10 points

Question text is in black, solutions in blue.

The set N of natural numbers is {1, 2, 3,...} as defined in Sipser.

If C is any class of computers, such as DFA's, CFG's, LBA's, TM's, 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^*}
• SEP_C = {<X, Y, Z>: X, Y, and Z are computers in C ∧ (L(X) ⊆ L(Z)) ∧ (L(Y) ⊆ (the complement of L(Z)))}

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

A language is Turing decidable (TD) if it is equal to L(M) for some Turing machine M that halts on every input.

A language X is co-TR if and only if its complement is TR.

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

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

A reversible stack pushdown automaton (RSPDA) is a nondeterministic machine with a stack that stores a string over its stack alphabet. On any step, the machine may push a new first letter onto the stack, pop the current first letter, or reverse the enture stack. It reads input letters and accepts or rejects just like an ordinary PDA.

A marker Turing machine (MTM) is like an ordinary deterministic one-tape Turing machine except that (1) it has a special marker symbol $ that is in its tape alphabet and not in its input alphabet, (2) its initial configuration on input w is q₀w$, (3) it must always maintain the property that there is exactly one $ on its tape, which is the rightmost non-blank symbol (unless it is in the process of moving the $ to another location), and (4) if the initial input is length n, and the head is ever more than 2n tape cells to the left of the $, the machine halts and rejects.

Sipser defines an LBA to be a deterministic one-tape Turing machine that never accesses any tape cells to the right of where the input initially resides. Here we define a nondeterministic linear bounded automaton (NLBA) to be a nondeterministic one-tape Turing machine with the same property.

• Question 1 (10): True or false with justification: Assuming that the function of Question 9b exists, the languages E_NLBA and ALL_NLBA are both undecidable (that is, neither is TD).

  TRUE. E_LBA ≤_m E_NLBA because any LBA is also an NLBA.

  ALL_CFG ≤_m ALL_NLBA by the reduction of Question 9b. Since both those languages were proved in lecture not to be TD, and the TD languages are closed downward under <_m, the given languages are not TD.

• Question 2 (10): True or false with justification: The language SEP_TM is TR.

  FALSE. We can reduce the complement of A_TM to SEP_TM as follows. We define a function f so that f(<M, w>) = <N, M_∅, M_∅>, where M_∅ is a fixed Turing machine that does not accept any strings, and N is a TM that on input X, runs M on w.

  If w ∈ L(M), then L(N) = Σ^*, L(N) is not a subset of L(M_∅), and f(<M, w>) ∉ SEP_TM.

  If w ∉ L(M), then L(N) = ∅, which is a subset of L(M_∅), so f(<M, w>) ∈ SEP_TM.

  Since the complement of A_TM is not TR, neither is SEP_TM.

• Question 3 (10): True or false with justification: The language {<X, Y>: ∃<Z>: <X, Y, Z> ∈ SEP_DFA} is TD.

  TRUE. Such a Z exists if and only if L(X) ∩ L(Y) = ∅, as in that case we can take Z = X and if that is false, it is clearly impossible to choose a suitable Z, whether DFA-computable or not.

  To test whether L(X) ∩ L(Y) = ∅ we can build a DFA I with L(I) = L(X) ∩ L(Y), and test whether <I> ∈ E_DFA, a language proved in lecture to be TD.

• Question 4 (10): True or false with justification: The language E_RSPDA is co-TR but not TD.

  TRUE. The complement of E_RSPDA is the language {<M>: ∃w: M is an RSPDA and w ∈ L(M)}. A string <M< is in this language if and only if an accepting computation history of M on w exists, and whether a string is such a history is TD, even for this strange sort of machine. So the complement is TR, and E_RSPDA is co-TR.

  To see that E_RSPDA is not TD, we reduce the known non-TD language E_TM to it by noting that we can design an RSPDA to simulate an arbitrary one-tape Turing machine. The RSPDA can use its stack to simulate the tape of the TM, using the top of the stack to simulate the portion to the left of the tape head and the bottom of the stack to simulate the portion to the right of the tape head. By reversing the stack as necessary, it can maintain both of these data structures on the same stack.

• Question 5 (10): True or false with justification: If the languages A and B are not both TR, A ≤_m C, and B ⊆ C, then it is possible that C is TR.

  TRUE. We could take A = C = Σ^*, and take B to be any non-TR language, such as the complement of A_TM. Then A ≤_m C and B ⊆ C are clearly true, and C is clearly TR.

• Question 6 (10): True or false with justification: Let PCP_k be the variant of the Post Correspondence Problem where every string on the top or bottom of any domino has at most k letters. Then for every fixed positive integer k, the language PCP_k is TD.

  FALSE. Let the language BLANK_TM be {<M>: ε ∈ L(M)}, sometimes called the "blank tape halting problem". It is clear that A_TM ≤_m BLANK_TM, by the reduction that takes <M, w> to the usual Turing machine N that on input x, runs M on w. Therefore BLANK_TM is not TD.

  Now we reduce BLANK_TM to PCP by the reduction given in Sipser. The dominos created are now all of a size bounded by a constant, since the first domino now has top ## and bottom #q₀_#, where "_" is the blank character. In fact none of the MPCP dominos in that construction now have strings of more than three letters, so the PCP dominos certainly have strings of no more than seven letters. This reduction thus shows that PCP₇ is not TD.

• Question 7 (40): Let f be any function from N × N to {0, 1}. For any number i ∈ N, we define the function f_i from N to {0, 1} by the rule f_i(x) = f(i, x).
  • (a, 5) Show that if f is any such function, computable or not, there exists a function g from N to {0, 1} such that for all i ∈ N, g ≠ f_i. (That is, there exists an x such that g(x) ≠ f_i(x).)

    Let g(x) be defined as 1 - f_x(x), or 1 - f(x, x). Then for any i, g(i) ≠ f_i and thus g ≠ f_i.

  • (b, 5) Explain why, if f is a Turing computable function, we can find such a g that is also a Turing computable function.

    The definition of g in terms of f in part (a) allows us to calculate g(x) with a TM, by computing 1 - f(x, x) using the assumed TM that computes f.

  • (c, 5) Let {S_i: i ∈ N} be any collection of subsets of N. (That is, for any i, S_i ⊆ N.) Prove that there exists a subset T of N such that for all i ∈ N, T ≠ S_i.

    Let T be the set {n: n ∉ S_n}. Then for any i ∈ N, i ∈ T if and only if i ∉ S_i, and T ≠ S_i.

  • (d, 5) Explain why part (c) implies that the power set of N is uncountable, making clear that you understand the definition of "uncountable".

    If the power set P(N) were countable, there would be a bijection of P(N) with N, meaning a listing of all the subsets of N as {S_i: i ∈ N}. But by part (c), such a listing cannot include all the possible subsets, since it does not include the T computed there.

  • (e, 10) Professor Colbert argues as follows: "Suppose each of the sets S_i in part (c) is TD. Then we can apply the reasoning of part (b) to the construction of the set in part (c), so that the set T constructed there has a Turing computable characteristic function and is therefore TD. Then, reasoning as in part (d), there are uncountably many TD sets." What is wrong with this argument?

    He is arguing that there cannot be a bijection {S_i: i ∈ N} of all the TD sets with N, on the ground that we could use the reasoning of part (d) to come up with a TD set that is not one of the S_i's. If there were a Turing computable listing of all the TD sets, then the function f (given by f(x, y) = 1 if x ∈ S_u and f(x, y) = 0 otherwise) would be Turing computable, and we would have a contradiction. So no such Turing computable listing can exist.

    But to show the TD sets to be uncountable, we would need to show that no such bijection could exist, computable or not. Just because every function f_i is Turing computable does not make f computable. And in fact there is a listing, where we list all the possible Turing machines, cross out the ones that don't compute total functions (that's the uncomputable step), and let S_i be the set decided by the i'th TM in the remaining listing.

• Question 8 (20): Marker Turing machines (MTM's) are defined above.
  • (a, 10) Explain how any MTM can be simulated by an ordinary Turing machine. (You may use any constant number of tapes in your simulation.)

    Our first tape will simulate the single tape of the MTM, one step at a time until or unless it either halts or is halted by us for violating the distance condition on its head.

    Our second tape will contain the number 2n, which we will calculate from the input before starting the MTM.

    Our third tape will contain the current distance between the head of the MTM and its marker.

    To simulate one step of the MTM, we perform the step's action on the first tape, update the distance on the third tape, check whether the distance equals the number on the second tape, and abort the computation and reject if it does.

  • (b, 10) Can every ordinary one-tape TM be simulated by an NTM? That is, for every ordinary one-tape TM, does there exist an MTM with the same language? Prove your answer.

    No such simulation is possible, because the language A_MTM is decidable and the simulation would allow us to reduce A_TM to A_MTM.

    To see that A_MTM is TD, note that only the last 2n cells of the MTM's tape before the $ are every relevant to its future. The $ cannot move left, since it must be to the right of every cell that has ever been touched. (I should have said that the MTM may not write a blank symbol -- I believe I said this during the exam.) So the cells more than 2n distance from the $ can never be accessed again. There are thus only O(2nc²ⁿ) possible configurations of the MTM on input of length n. We can carry out the simulation for that many steps, and halt and reject if it is still running at the end of them. Any computation still running by that time must be stuck in a loop, since the MTM is deterministic, and will never halt on its own.

• Question 9 (10 + 10 extra credit): Nondeterministic linear bounded automata (NLBA's) are defined above.
  • (a, 10) Prove that the language A_NLBA is TD.

    Make a directed graph of the configurations of the NLBA M -- there are |Q||Γ|ⁿn of them, where |Q| is the size of the state set and |Γ| is the size of the tape alphabet. Make a node for each configuration, and an edge from node c to node c' if and only if it is possible for the NLBA to go from c to c' in one step.

    Then determine, for example by depth-first search, whether there is a path in this graph from the start configuration (of the machine M on input w) to an accepting configuration. This path exists if and only if w ∈ L(M). It takes a long time to carry out this configuration, but it is guaranteed to halt with the correct answer.

  • (b, 10) Show that there exists a Turing computable function f such that if G is any context-free grammar, then f(<G>) = <N>, where N is an NLBA such that L(G) = L(N).

    Convert G to a Chomsky Normal Form grammar G'. The NLBA N whose language is L(G) works as follows. It begins with w on the tape. It nondeterministically scans the tape to find the right-hand side of a rule of G', and if it finds one it may choose to replace it with the left-hand side. If the tape contents ever become S, it accepts. Since Chomsky Normal Form rules never replace shorter strings with longer ones (except for the rule S --> ε, which we can handle in the state table of the NLBA), this computation can be carried out within the avialable tape space. It is possible for this NLBA to accept if and only if some derivation of w exists in G', which is true if and only if w is in L(G).

  Last modified 10 April 2014