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4.1 Decidable languages Decidable problems concerning regular languages: Let ADFA = {B,w | B is a DFA that accepts input string w}. The problem of testing whether a DFA B accepts an input w is the same as the problem of testing whether B,w ADFA. Theorem 4.1 ADFA is a decidable language. M = “On B,w, whether B is a DFA and w is a string: 1. Simulate B on input w. 2. If the simulation ends in an accept state, accept. If it ends in a nonaccepting state, reject.” ANFA = {B,w | B is a NFA that accepts input string w}. Theorem 4.2 ANFA is a decidable language. N = “On input B,w where B is an NFA ,and w is a string: 1. Convert NFA B to an equivalent DFA C using the procedure for this conversion given in Theorem 1.19. 2. Run TM M from Theorem 4.1 on input C,w. 3. If M accepts, accept; otherwise reject.” Let AREX = {R,w | R is a regular expression that generates string w}. Theorem 4.3 AREX is a decidable language. PROOF The following TM P decides AREX. P =“On input R,w where R is a regular expression and w is a string: 1. Convert regular expression R to an equivalent DFA A by using the procedure for this conversion given in Theorem 1.28. 2. Run TM M on input A,w. 3. If M accepts, accept; if M rejects, reject.” EDFA = {A | A is a DFA and L(A)=}. Theorem 4.4 EDFA is a decidable language. T = “On input A where A is a DFA: 1. Mark the start state of A. 2. Repeat until no new states get marked: 3. Mark any state that has a transition coming into it from any state that is already marked. 4. If no accept state is marked, accept; otherwise reject.” EQDFA = {A,B | A and B are DFAs and L(A) = L(B)}. Theorem 4.5 EQDFA is a decidable language. L(C ) L( A) L( B ) L( A) L( B ) F = “On input A, B, where A and B are DFAs: 1. Construct DFA C as described. 2. Run TM T from Theorem 4.4 on input C. 3. If T accepts, accept. If T rejects, reject.” DECIDABLE PROBLEMS CONCERNING CONTEXT-FREE LANGUAGE ACFG = {G,w | G is a CFG that generates string w}. Theorem 4.6 ACFG is a decidable language. Proof The TM S for ACFG follows. S = “On input G,w, where G is a CFG and w is a string: 1. Concert G to an equivalent grammar in Chomsky normal form. 2. List all derivations with 2n-1 steps, where n is the length of w. 3. If any of these derivations generate w, accept; if not, reject.” ECFG = {G | G is a CFG and L(G) = }. Theorem 4.7 ECFG is a decidable language. PROOF R = “On input G, where G is a CFG: 1. Mark all terminal symbols in G. 2. Repeat until no new variables get marked: 3. Mark any variable A where G has a rule AU1U2…Uk and each symbol U1,…,Uk has already been marked. 4. If the start symbol is not marked, accept; otherwise reject.” Theorem 4.8 Every context-free language is decidable. Proof Let G be a CFG for A and design a TM MG that decides A. We build a copy of G into MG. It works as follows. MG = “On input w: 1. Run TM S on input G, w, where S is a Turing machine design in Theorem 4.6. 2. If this machine accepts, accept; if it rejects, reject.” FIGURE 4.2 The relationship among classes of languages. 4.2 The Halting Problem 1. ATM = {M, w | M is a TM and M accepts w} The following Turing machine U recognizes ATM. U =“On input M, w, where M is a TM and w is a string: 1. Simulate M on input w. 2. If M ever enter its accept state, accept; If M ever enters its reject state, reject.” ATM is sometimes called the halting problem. 2. The proof of the undecidability of the halting problem uses a technique called “diagonalization”, discovered by mathematician Georg Cantor in 1873. 3. Assume that we have two sets A and B and a function f: AB – f is one-to-one if f(a)f(b) whenever ab. – f is onto if for every bB there is an aA s.t. f(a)=b. – A and B are the same size if there is a one-to-one, onto function f: AB. – A function that is both one-to-one and onto is called a correspondence. Ex. Let N be the set of natural numbers {1, 2, 3, …}and let be the set of even natural numbers {2, 4, 6, …}. Using Cantor’s definition, N and have the same size. f(n)=2n: 4. A set S is countable if either it is finite or it has the same size as N. Ex. Let Q be the set of positive rational numbers, m i.e. Q { m, n N }, Q is countable. n 5. For some infinite sets no correspondence with N exists, such sets are called uncountable. Theorem 4.14 R is uncountable. 需證明 no correspondence exists between N and R. The proof is by induction: 1. 假設存在一種 correspondence f between N and R (每個 element in N 有唯一之 element in R 與其對應). 2. 我們只要找出一 xR 使得 x 不包含在此對應 關係即可證明. 如何找？以下列範例說明： 永遠可以找到一 x 不存在 f 之中！ Corollary 4.15 Some language are not Turing-recognizable. 1. 觀察一：the set of all string * is countable, for any alphabet . 觀察二：the set of all Turing machines is countable because each Turing machine M has an encoding into a string M. The set of all Turing machines is countable. 2. Let B be the set of all infinite binary sequences. B is uncountable. 3. Let L be the set of all languages over . Let * = {s1, s2, s3, …}. Each language AL has unique sequence A in B: The ith bit of that sequence is a “1” if siA and is a “0” if siA. 例: * = { , 0, 1, 00, 01, 10, 11, 000, 001, …}; A ={ 0, 00, 01, 000, 001, …}; A = 0 1 0 1 1 0 0 1 1 … . The function f: LB, where f(A) equals the characteristic sequence of A, is one-to-one and onto and hence a correspondence. B is uncountableL is uncountable. Some languages are not recognized by any Turing machine. The Halting Problem is Undecidable Now we are ready to prove Theorem 4.9, the undecidability of the language ATM = {M,w| M is a TM and M accepts w}. Proof: Assume ATM is decidable and obtain a contradiction. 1. Suppose that H is a decider for ATM. H(M,w) = accept if M accepts w. reject if M does not accept w. 2. Now we construct a new Turing machine D with H as a subroutine. D = “On input M, where M is a TM: 1. Run H on input M, M. 2. Output the opposite of what H outputs; that is, if H accepts, reject and if H rejects, accept.” 3. D(M) = accept if M does not accept M. reject if M accepts M. D(D) = accept if D does not accept D. reject if D accepts D. Figure 4.4 Entry i,j is accept if Mi accepts Mj Figure 4.5 Entry i,j is the value of H on input Mi , Mj FIGURE 4.6 If D is in the figure, a contradiction occures at “?” Theorem 4.16 A language is decidable if and only if it is both Turing-recognizable and co-Turing- recognizable. In other words, a language is decidable if and only if both it and its complement are Turing- recognizable. Pf: 1. 証 Easy 2. 証 2. 証 If both A and A are Turing-recognizable, we let M1 be the recognizer for A and M2 be the recognizer for A. The following Turing machine M is a decider for A: M = “On input w: 1. Run both M1 and M2 on input w in parallel. 2. If M1 accepts, accept; if M2 accepts, reject.” Corollary 4.17 ATM is not Turing-recognizable. 習題 4.3 4.5 4.8 4.11 4.12

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