3. This problem is concerned with the string encodings of DFAs. Fix the alphabet =...
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3. This problem is concerned with the string encodings of DFAs. Fix the alphabet Σ = {0, 1}; then one way to encode the DFA (Q, Σ, 8, qo, F) is by the string 10(0,0))((0, 1)) ... ((q|Q|, 1))0(F), where we encode (qi, σ) =q; by 12, and the set F by its characteristic vector: the i-th bit of (F) is 1 if q; E F and 0 otherwise. (thus 10 000 encodes the DFA with one state, go, with 8(90, 0) = 8(90, 1) = qo, and F = {}, while 10001 encodes the DFA with the same transition function and F = {0}.)¹ Once we fix an encoding, we consider the language DDFA = {(M) | M is a DFA that rejects (M)}. So for example, 10000 € DDFA, while 10001 € DDFA. (a) Show that DDFA is decidable: e.g. describe a TM to decide, given the string (M), whether the DFA M will reject (M). (you do NOT need to give the state diagram for this TM.) (b) Prove that DDFA is not regular. 3. This problem is concerned with the string encodings of DFAs. Fix the alphabet Σ = {0, 1}; then one way to encode the DFA (Q, Σ, 8, qo, F) is by the string 10(0,0))((0, 1)) ... ((q|Q|, 1))0(F), where we encode (qi, σ) =q; by 12, and the set F by its characteristic vector: the i-th bit of (F) is 1 if q; E F and 0 otherwise. (thus 10 000 encodes the DFA with one state, go, with 8(90, 0) = 8(90, 1) = qo, and F = {}, while 10001 encodes the DFA with the same transition function and F = {0}.)¹ Once we fix an encoding, we consider the language DDFA = {(M) | M is a DFA that rejects (M)}. So for example, 10000 € DDFA, while 10001 € DDFA. (a) Show that DDFA is decidable: e.g. describe a TM to decide, given the string (M), whether the DFA M will reject (M). (you do NOT need to give the state diagram for this TM.) (b) Prove that DDFA is not regular.
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a To show that DDFA is decidable we need to describe a Turing machine TM that can decide whether a given string M represents a DFA M that rejects itse... View the full answer
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