CMPSCI 401: Theory of Computation

Solutions to Second Midterm Exam, Spring 2010

David Mix Barrington

2 April 2010

Directions:

• Answer the problems on the exam pages.
• There are seven problems for 125 total points. Actual scale is A=105, C=65.
• If you need extra space use the back of a page.
• No books, notes, calculators, or collaboration.
• The first four questions are true/false, with five points for the correct boolean answer and up to five for a correct justification of your answer -- a proof, counterexample, quotation from the book or from lecture, etc. -- note that there is no reason not to guess if you don't know.

  Q1: 10 points
  Q2: 10 points
  Q3: 10 points
  Q4: 10 points
  Q5: 35 points
  Q6: 30 points
  Q7: 20 points

  Total: 125 points

If X is a computational model, such as DFA's, CFG's, or TM's, we follow Sipser in defining the following languages:

• A_X = {(M, w): M is a computer in X and w ∈ L(M)}
• E_X = {(M): M is a computer in X and L(M) = ∅}
• ALL_X = {(M): M is a computer in X and L(M) = Σ^*}
• EQ_X = {(M, N): M and N are computers in X and L(M) = L(N)}
• (not in Sipser) COMMON_X = {(M, N): M and N are computers in X and L(M) ∩ L(N) ≠ ∅} (that is, L(M) and L(N) have a string in common)

Recall also that the Post Correspondence Problem takes as input a finite set of dominoes {D₁,..., D_n}, such that each domino D_i has a top string t_i and a bottom string b_i, each a nonempty string in &Sigma'^* for some alphabet Σ. A set of dominoes is in the language PCP if there exists a nonempty sequence of dominoes that is a match, meaning that the concatenation of the top strings and the concatenation of the bottom strings form the same word.

• Question 1 (10): True or false with justification: A dedicated tape Turing machine (or DTTM) is a multitape deterministic TM, each of whose tapes is either read-only or write-only. (The input tape is read-only, and the others are all work tapes that start out with all blanks.) That is, the transition function takes as input the current state and the letter seen on each of the read-only tapes, and gives as output the new state, the letter written to each write-only tape, and the head movement of each tape. (It may help to recall that a tape head ordered to move left from the leftmost cell of a tape just stays there, without indicating to the TM that this has happened.) Then the language A_DTTM is undecidable.

  FALSE. Since the TM can never change its behavior based on any information from the write-only tapes, it would behave the same if they did not exist. Similarly, it will always see a blank on every read-only tape except the input tape. In effect, it is a two-way DFA and thus its A-problem is decidable. (You could prove this quickly by noting that a two-way DFA can be simulated by an LBA, which is enough because A_LBA is decidable. Or you can give a direct proof, that you can decide a two-way DFA with k states on an input of size n with a TM, by simulating the two-way DFA until either it halts or has run for kn+1 steps. In the latter case it must be in a loop and your TM can reject.

• Question 2 (10): True or false with justification: A ten-state binary-tape Turing machine (or 10BTTM) is a deterministic single-tape TM with exactly ten states, whose tape alphabet is {0, 1}. (0 is the blank character, so the input alphabet is {1} and the only valid inputs are strings in 1^*.) Then the language E_10BTTM is decidable.

  TRUE. (This was the hardest question on the test -- several people had the right boolean answer but no one had a full-credit explanation.) The δ function of such a machine is from Q × {0,1} to Q × {0,1} × {L,R}, that is, from a set of size 20 to a set of size 40. Up to renaming of the states, then, there are only 40²⁰ such machines, a finite number. There thus exists a finite lookup table answering the emptiness problem for each of these machines, and our TM deciding the language E_10BTTM just has to figure out which machine it has and look up the answer.

  Note that a ten-state machine with a binary alphabet can use memory beyond its input and thus could run for an insanely long time. (Wikipedia claims that there is a six-state machine with a binary alphabet that halts after running for more than 10²⁸⁷⁹ steps.) But although no imaginable computation could produce this lookup table, it exists mathematically and thus E_10BTTM is a decidable language.

• Question 3 (10): True or false with justification: A ten-state Turing machine (or 10TM) is a deterministic single-tape TM with exactly ten states and some finite tape alphabet Γ and some finite input alphabet Σ ⊆ Γ. Then the language A_10TM is decidable.

  FALSE. We can simulate an arbitrary machine with a ten-state machine, if we are allowed to expand the tape alphabet. If M is a single-tape ordinary TM with state set Q and tape alphabet Γ, define Γ' to be a tape alphabet Γ ∪ (Q × Γ), so a new letter is either an old letter or an old letter marked by a state of M. Our new TM M' will have fewer then ten states, q'₀, a few running states q₁, etc., and special states to allow it to accept or reject. It begins by changing the first letter a of its input to (q₀, a), where q₀ is the start state of M, and going into q₁. In state q₁ it starts the process of simulating a step of M. For example, if δ_M(q,a) = (r,b,R), M' changes a letter (q,a) to (r,b) and then uses its other running states to move the state r, bit by bit, to the cell to the right, leaving the b in place. (We might need some more letters in Γ' to record situations with the state partially copied.) This simulation gives us a reduction from A_TM, which is undecidable, to A_10TM.

• Question 4 (10): True or false with justification: Let Σ be a finite alphabet. A codebook over Σ is a finite set of pairs of strings (p_i, c_i), where in each pair p_i is called the plaintext and c_i is called the ciphertext. A string w in Σ^* is called a plaintext string if it is in the regular language (p₁ ∪ p₂ ∪ ... ∪ p_n), and its corresponding ciphertext string is obtained by replacing each string p_i by the corresponding string c_i. (Note that a given plaintext string could have more than one ciphertext string, if there are two pairs with the same p_i or if there is more than one way to break the string into p_i's.)

  The codebook fixed-point problem (or CFPP) is the language consisting of all codebooks for which there is some nonempty plaintext string that has a ciphertext string that is identical to the palintext. Then this language CFPP is undecidable.

  TRUE. We can reduce PCP, which was proven undecidable in Sipser and in lecture, to CFPP. (Note that the opposite reduction only proves that CFPP is TR.) Given a PCP instance, we make a codebook by making each domino into a pair, making the top string the plaintext and the bottom string the ciphertext of the pair. A match in the domino set thus corresponds exactly to a plaintext string with a ciphertext string identical to it, a fixed point. So this transformation proves PCP ≤_m and thus that CFPP is undecidable.

• Question 5 (35): Let M and N be arbitrary Turing machines.

  • (a,10) Argue carefully that the following language is Turing recognizable: L_M,N consists of al triples of strings (w, x, y) such that either (a) is true or both (b) and (c) are true (that is, "a ∨ (b ∧ c)"):
    • (a) w is in L(M) ∩ L(N)
    • (b) x is not an accepting computation history of M on w
    • (c) y is not an accepting computation history of N on w

    The set of triples satisfying (a) is Turing recognizable because we can start M and N on w in parallel and wait until or unless both accept. The sets satisfying (b) and (c) are each decidable -- we showed in lecture that an LBA, which is among other things an always-halting Turing machine, can decide whether an alleged computation history is valid.

    So we can test (b) and (c), accept the triple if both are true, and otherwise start the machine that recognizes the triples that satisfy (a). If (a) is true we accept, and if not we never find out.

  • (b,15) Argue carefully that the pair (M, N) is in the language EQ_TM if and only if the Turing machine recognizing L_M,N is in the language ALL_TM. (Here we assume that every string represents some triple of strings, so to be in ALL_TM means accepting all possible triples.)

    If (M, N) is in EQ_TM, we know that L(M) = L(M), and we must show that every triple is then in L_M,N. If w is in both L(M) and L(N), then (a) is true for any triple (w,x,y) and thus these triples are in the language. If w is in neither L(M) nor L(N), then there are no accepting histories for either machine on w, and thus (b) and (c) will both be true for any strings x and y. Thus any (w,x,y), for such a w, will also be in L_M,N and we have shown that the machine for L_M,N is in ALL_TM.

    Conversely, suppose that L_M,N is not the set of all triples, so that there exists a triple (w,x,y) such that "(a) or ((b) and (c))" is false. By propositional calculus, this means that (a) is false and at least one of (b) and (c) is false. For (a) to be false, this w cannot be in both L(M) and L(N). For (b) to be false, w must be in L(M), and for (c) to be false, w must be in L(N). So we have that w is in one of L(M) or L(N) but not the other, and thus L(M) ≠ L(N) and (M, N) ∉ ALL_TM.

  • (c,10) Rephrase the result of part (b) as a mapping reduction. What does this mapping reduction (and the results from lecture) imply about the language ALL_TM?

    The function mapping the pair (M, N) to the Turing machine recognizing L_M,N is a mapping reduction proving that EQ_TM ≤_m ALL_TM. Since we proved in lecture that EQ_TM is neither TR nor co-TR, and both the class TR and the class co-TR are closed downward under mapping reduction, we can conclude that ALL_TM is neither TR nor co-TR (and thus certainly undecidable).

• Question 6 (30): This problem uses the definition of the language COMMON_X for a computing model X, given at the start of the exam above.

  • (a,10) Indicate how a multitape Turing machine can decide the language COMMON_DFA.

    We saw in Chapter 1 that if M is a DFA with i states and N is a DFA with j states, then we can construct a new DFA P such that L(P) = L(M) ∩ L(N). By the definition, (M, N) is in COMMON_DFA if and only if P is not in E_DFA, and we proved that the latter language is decidable. (We have only to determine whether the directed graph for P has a path from the start state to any accepting state.) A TM can carry out the construction of P and the test for a path.

  • (b,10) Prove that the language COMMON_CFG is Turing recognizable. (I would ask you whether it's decidable, but I'm not sure...)

    We need to determine, for every string w in Σ^*, whether w is in both L(M) and L(N) (where M and N are now grammars). Since A_CFG is decidable, we can carry out these tests sequentially, accepting if we ever find a common w and running forever if we do not. We could also carry out these tests in parallel by dovetailing as in the answer to part (c).

  • (c,10) Prove that the language COMMON_TM is Turing recognizable but not Turing decidable. You may use any method and any facts known from lecture about the other languages defined for TM's at the start of the exam.

    COMMOM_TM is TR: We run both M and N in parallel on every string w by dovetailing. (For example, for each natural number i in order we could run M and N on each of the first i strings for i steps each.) If we ever find a w that is accepted by both M and N we accept. If we do not find such a w we continue to look for one forever.

    COMMON_TM is not TD: My proof was to reduce from the complement of the known undecidable language E_TM, which must also be undecidable. Given a Turing machine M, let N be a fixed machine such that L(N) = Σ^*. Then L(M) is not empty if and only if L(M) and L(N) have a string in common.

    Several people gave the following proof which I think I like better. Reduce A_TM to COMMON_TM as follows. Given M and w, construct a Turing machine N such that L(N) = {w}. ("On input x, if x = w accept; else reject;") Then (M, w) is in A_TM if and only if (M, N) is in COMMON_TM. We have reduced the known undecidable language A_TM to COMMON_TM, proving that the latter is not decidable.

• Question 7 (20): Here are two problems involving variants of the Post Correspondence Problem, which is defined above at the start of the exam:

  • (a,10) Define UNARY-PCP to be the subset of PCP where the instance's alphabet Σ has one letter (and there is a match). Prove that UNARY-PCP is decidable. (This is Problem 5.17 in Sipser, which we didn't do this term -- obviously you can't just quote the fact that it's an exercise as your answer.)

    Given a set of dominoes with a string in a^* on the top and another string in a^* on the bottom, we decide whether this set is in PCP as follows. If there is a domino with the same string on top and on bottom, accept because the sequence consisting of just this domino is a match. If there is domino with longer top string and another domino with longer bottom string, accept because we can build a match with these two. (Let the first domino have b+c a's on top and b a's on bottom, and the second domino have d a's on top and d+e on bottom. Then e copies of the first domino and c copies of the second give up a match because there are be + ce + cd a's in the top string and the bottom string.) Otherwise the dominoes in the set either all have longer top string or all have longer bottom string, and in this case we reject because a match is clearly impossible.

  • (b,10) Let SINGLE-PCP be the subset of PCP where the instance has a match in which no domino is used more than once. Prove that SINGLE-PCP is decidable.

    We only have to check every sequence of dominoes with no domino more than once -- there are 2ⁿ - 1 possible nonempty sets drawn from n dominoes, and each of these sets can be placed in at most n! orders, so the total number to check is finite. It is clear that checking whether a given sequence forms a match is decidable, so an always-halting TM can just check every possible sequence.

Last modified 2 April 2010