CMPSCI 601: Theory of Computation

Offered through the PEEAS distance learning program

Solutions to Homework Assignment #1

David Mix Barrington

Solutions posted Wed 25 June 2003

Problem text in black, solution text in blue.

• Question 1 (15): Let A be the set of strings over alphabet {a,b} such that every a in the string has a b immediately before it and a b immediately after it. (Note that a string with no a's at all satisfies this condition and is thus in A.) Design a DFA that recognizes the language A. Determine the four Myhill-Nerode equivalence classes of A and give a regular expression for each.

  I'll describe the four-state DFA in words:

  • State 1, the start state, is final, with a-arrow to 4 and b-arrow to 2
  • State 2 is final, with a-arrow to 3 and b-arrow to itself
  • State 3 is non-final, with a-arrow to 4 and b-arrow to 2
  • State 4 ("the death state") is non-final with a-arrow and b-arrow to itself

  The four Myhill-Nerode classes are the sets of strings that take this DFA from the start state to each of the states:

  • Class 1 consists of the empty string only, the set (emptyset)^*
  • Class 2 can be written b(b+ab)^*, using "+" to mean "union"
  • Class 3 can be written b(b+ab)^*a
  • Class 4 can be written (a+b(b+ab)^*aa)(a+b)^*

  Many other correct solutions are possible.

• Question 2 (30): Let M be a one-tape Turing machine that never writes on its tape. (Note that its one tape is the input tape -- it has no work tapes.) Show that L(M) is a regular language.

  Hint: M is in essence a two-way DFA, like an ordinary DFA except that it can move in either direction based on its state and the letter it sees. That is, its transition function is from (Q times Σ) to (Q times {L,R}) where Q is its state set and Σ its alphabet.

  To show that L(M) is regular, we define a set of questions about a string w and show that two strings with the same answers to all these questions are Myhill-Nerode equivalent with respect to L(M). Here are the questions:

  • If M is started on the left end of w in the start state, does it ever leave w to the right, and if so in what state does it do so?
  • If M is started on the right end of w in state s, does it ever leave w to the right and if so in what state does it do so?

  If you choose to follow this Hint, your job is to show (a) that if strings v and w have the same answers for each of these questions, then they are Myhill-Nerode equivalent for L(M), (b) there are only finitely many different possible answers for these questions, and (c) it follows that L(M) is regular.

  • (a) Suppose that v and w have the same answers for all these questions and that x is an arbitrary string. We must show that M has the same behavior on vx and ux -- either it fails to leave the right end of either (and thus neither is in L(M)) or it leaves the right end of each in the same state (so both are in L(M) if this state is final and neither is in L(M) otherwise).

    Consider these two behaviors of M from the beginning, with M started on the left end of each string. If M never leaves v when running on vx, it also never leaves w running on wx (because the first question has the same answer) and so neither string is accepted. If it does enter x on vx, it enters x on wx in the same state, because the first question has the same answer. Running on x starting from the same state, the two machines have the same behavior because they see the same letters of x. If they both leave the right end of x, they do so in the same state, and we are done.

    The remaining case is when the two machines leave x to the left, doing so in the same state. Now because v and w have the same answers to each of the second class of questions, the machines either both fail to return or both return to x in the same state. In the former case we are done, in the latter the two machines again behave identically on x, either leaving to the right in the same state or returning once more to v and w.

    By induction on the number of times the machines enter x, we see that they always do so in the same state given our assumption about the answers to the questions for v and w. Thus if one machine ever leaves x to the right, the other also does so after the same number of visits to x, and in the same state.

  • (b) There are k+1 questions, the first one and k in the second class (one for each state of M). Each question has k+1 possible answers, one for each state plus the special answer "doesn't come back". So there are (k+1)^k+1 possible sets of answers, a number that does not depend on the length of v or w but only on the number of states in n.

  • (c) Since any string must have one of the (k+1)^k+1 possible sets of answers and any two strings with the same answers are Myhill-Nerode equivalent for L(M), there are only at most this many equivalence classes. By the Myhill-Nerode theorem, then, there is a DFA for L(M) with at most this many states and L(M) is then regular by Kleene's Theorem.

• Question 3 (15): Using the result of Question 2 even if you did not prove it yourself, argue that the complexity class DSPACE(1) is equal to the class of regular languages. Recall that a DSPACE(1) machine has a read-only input tape and a read-write worktape, and the size of the latter is bounded by a number that does not depend on the input size.

  No one explained this correctly. Several people noted that if there was no worktape at all in the DSPACE(1) machine, it would be a two-way DFA and its language would be regular by Question 2. But of course there is a worktape, and we need to explain why the worktape doesn't allow the DSPACE(1) machine to accept a non-regular language.

  The key point is that the size of the worktape is bounded by a constant that does not depend on the input size. Let a be the number of worktape cells and b the size of the worktape alphabet. There are only b^a possible strings to be on the worktape, and only a possible head positions, so the worktape is always in one of at most ab^a states, a number that does not depend on the input size.

  Let q be the number of states in the DSPACE(1) machine. By treating the state of the worktape as part of the machine state, we can build a two-way DFA with qab^a states, one state for each possible combination of machine state and worktape state. Since the DSPACE(1) machine's behavior and next worktape state depend only on the current state, current worktape state, and input letter it sees, this can be modeled as a two-way DFA. Then by Question 2 we may conclude that L(M) is regular, though with a potentially huge number of equivalence classes: (qab^a+1)^(qaba+1).

• Question 4 (25):Define a two-stack PDA to be like an ordinary PDA except that it has two stacks. Thus each transition includes a push or pop from each stack as well as reading of an input character and a change of state. Prove that a language A is recursively enumerable iff there exists some two-stack PDA M such that A = L(M). (Hint: Use the two stacks to simulate the tape of a Turing machine. You may omit details if your general argument is convincing.)

  We must prove that every r.e. language is the language of some 2SDPA and that the language of every 2SPDA is r.e.:

  • Suppose that A is an r.e. language. By definition, there must be some one-tape TM M such that A = L(M). We simulate M with a 2SPDA P as follows. At any point of M's computation, it is looking at a letter of its tape. We have one stack of P represent the part of M's tape to the left of this letter, with the left end of the tape at the bottom of the stack. We have the other stack of P represent the part of M's tape to the right of the current letter, with the right end of the tape at the bottom of the stack. We can maintain this relationship between the tape and the stacks through each step of the calculation, because the tape changes only locally around the current letter and we can make the corresponding changes at the top of each stack of P.

    We can define the acceptance condition of P so that L(P) = L(M) = A, and thus we prove that A is the language of a 2SPDA as desired. Many students described the simulation convincingly, but did not explain how this simulation's existence proved the desired result.

  • The other half of the problem is to prove that the language of a 2SPDA must be r.e., that is, must be the language of some TM. Let P be an arbitrary 2SPDA. We can easily simulate P with a nondeterministic three-tape TM, using one worktape to represent each stack. Then by known results this three-tape TM can be simulated by a deterministic one-tape TM, so its language is r.e. as desired. We must establish the accepting conditions for the TM so that it accepts a string iff P does. Note that L(P) need not be recursive -- since P might be simulating an arbitary TM, we have no idea whether P will eventually halt or not on an arbitary input.

• Question 5 (15): Let A be a recursively enumerable language. Prove that there is a Turing machine M such that if M is started with input consisting of two strings u and v, and exactly one of u and v is in A, then M halts and its output determines which of the strings is in A. Don't worry about what M does if both input strings are in A or neither input string is in A.

  Because A is recursively enumerable, there exists some TM M_A such that A = L(M_A), that is, such that M_A halts on an input string w in an accepting state iff w is in A.

  We design M as follows. Its input must be two strings u and v, separated by a space. It runs M_A on both u and v in parallel, alternating between running a step on u and running a step on v. It does this on two sections of its single tape, moving the v-tape part to the right whenever necessary to create enough room for the u-tape part.

  We are given that exactly one of u and v is in A, and thus that exactly one of these two computations will eventually halt and accept. We design M to halt when this happens, and indicate with its output which one accepted. By the assumption, we are guaranteed that M will eventually halt.

  Several students had the right idea for this problem but did not indicate what machine they were running in the two parallel computations.

Last modified 25 June 2003