MTAT Introduction to Theoretical Computer Science

Transcription

1 MTAT.5.25 Introduction to Theoretical Computer Science Autumn 25 Vitaly Skachek Estonian version by Reimo Palm English version by Yauhen Yakimenka Lecture 2. Decidable languages. Decidable languages We can express different computational problems as languages. For example, testing whether a particular DFA accepts the given string: L DFA = { A,w A is a DFA, Here, A,w represents a pair: that accepts the input string w}. encoding of the DFA A (list of five ingredients: Q,Σ,δ,q,F); input string w. The task of deciding whether DFA A accepts a string w is equivalent to checking if the pair A,w is in the language L DFA. Theorem. L DFA is a decidable language. Proof. We design a TM M that decides the language L DFA. On the input A,w, the machine M will simulate the automaton A on w, and accept/reject according to the automaton s decision. First, M scans the input and determines if the input properly represents a DFA (which we denote as A) and a string (which we denote as w). If not, M rejects.

2 Second, M simulatesa. It keeps track ofa s current state anda s current position in the input w by writing the information directly on the tape. In the beginning, the input ofm isw, and the head position is the leftmost symbol of w. The states and the positions are updated according to the transition function δ. When M is finishing processing the last symbol of w, it goes to accept/reject state depending on whether A is in the accept/reject state. Similarly define L NFA = { A,w A is an NFA Theorem. L NFA is a decidable language. that accepts the input string w}. Proof. We present a TM M that decides L NFA : on the input A,w, M does the following:. Converts A into equivalent DFA A, by using the procedure that was studied in the course. 2. Run the machine M from the previous theorem on the input A,w. 3. If M accepts accepts, otherwise rejects. One more example. Let L = { A A is a DFA and L(A) = }. I.e. all DFAs that do not accept anything. Theorem. L is a decidable language. Proof. A DFA A accepts some string if and only if reaching one of the accept states by travelling along the arrows of the DFA is possible. Therefore, a TM M will test if there exists such a path. For example, in the automaton q q q 2 q 3 q 4 2

3 there is a path q q q 3 q 4. This correspond to the input. Therefore L(A) as L(A). TM M works as follows.. Mark the start state of A. 2. Repeat until no new states are marked: Mark any unmarked state that has an incoming arrow from any state that was marked already. 3. If no accept state is marked accept, otherwise reject. For the example above, M will mark the states in the following order: q q q 2 q 3 q q 4 is marked, so M rejects (A accepts at least one string). Undecidable languages Define: L TM = { M,w M is a TM and M accepts the input w}. Theorem. L TM is undecidable. Note. We present a proof based on a technique called diagonalisation. Proof. We prove by contradiction. Assume, that there exists a Turing machine H, where { accepts, if M accepts w, H( M,w ) = rejects, if M does not accept w (either rejects or loops). 3

4 Now we construct a new machine D, which uses H as a subroutine. On input M, D does the following:. Runs H on input M, M. 2. Outputs the opposite of what H outputs. That is, if H accepts D rejects; if H rejects D accepts. In summary, D( M ) = { accepts, if M does not accept M, rejects, ifi M accepts M. Question: what happens when we run D with its own encoding D as an input? In this case { accepts, if D does not accept D, D( D ) = rejects, if D accepts D. No matter what D is supposed to do, it does the opposite. Contradiction. Therefore such H does not exist. Practise session. Define the language L REX = { R,w R is a regular expression that generates the string w}. Show that L REX is a decidable language. Solution. We construct a TM M that on the input R,w does the following:. Converts R into an equivalent NFA A by using the procedure for conversion that we studied. 2. Gives an input A,w to the TM that decides L NFA. 3. If A,w L NFA accepts, otherwise rejects. 2. Definte the language: L DFAEQ = { A,B A and B are DFA Prove that L DFAEQ is a decidable language. 4 and L(A) = L(B)}.

5 Solution. We construct a new DFA C, which accepts strings that are accepted by either A or B, but not by both. Then L(C) = ( L(A) L(B) ) ( L(A) L(B) ). L(A) L(B) L(C) If L(A) = L(B), then L(A) L(B) = and L(A) L(B) =, hence L(C) =. If L(A) L(B), then there exists w L(A), w / L(B) (or vice versa). Then w L(A) L(B) (or, respectively, w L(A) L(B)) and therefore w L(C) and L(C). So L(A) = L(B) if and only if L(C) =. We construct a TM M as follows. On the input A,B it does the following:. Constructs C as described. 2. Runs TM that decides the language L on C. 3. If C L accepts. If C / L rejects. 3. Define the language L = { A A is a DFA that accepts at least one string of the form }. Prove that L is decidable. Solution. We construct TM M that decides L. On the input A, M does the following:. Constructs a DFA B that accepts exactly language described by. Such an automaton is easy to build: it runs A and B in parallel and accepts if and only if exactly one of A and B accepts 5

6 2. Constructs a DFA C, such that L(C) = L(A) L(B). 3. Checks if C L. If no accepts, if yes rejects. Let us justify the construction. If C L then L(C) = and so L(A) L(B) =. This means that for each w L(A), it holds that w / L(B) and therefore w does not have the form. If C / L, then L(C) and L(A) L(B). Thus there exists w, such that w L(A) and w L(B). This means that w has the form and w L(A). Correct. 4. Definte the language L k-str = { A,k A is a DFA Prove that L k-str is decidable. and L(A) consists of exactly k strings, k N}. Proof. We construct a TM M, which decides L k-str. On the input A,k, M does the following.. Checks the number of states of A. Denote this number by p. 2. Constructs a DFA B, that accepts all strings of length p or longer. Also constructs a DFA C, such that L(C) = L(A) L(B). 3. Generates all strings of length p and tests whether each string is accepted by A. Counts the number of such strings, denote this number by c A. 4. Tests whether L(C) =. 5. If L(C) = and c A = k accepts, otherwise rejects. Let us show that M does what we want. First, note that due to the pumping lemma, if A accepts any string of length p, then it accepts infinitely many strings. This condition is tested by testing if L(C) =. Provided A does not accept any strings of length p, c A is exactly the cardinality of L(A). Thus M accepts if and only if L(A) = k. 6