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Decidability 02 Thumrongsak Kosiyatrakul [email protected] Thumrongsak Kosiyatrakul [email protected] Decidability 02
Decidable Languages E DFA Let E DFA = {h A i | A is a DFA and L ( A ) = ∅} . Show that E DFA is decidable. How do we know a DFA accepts no string? This machine has no accept state ( F = ) or This machine has some accept states but are not reachable from the start state To show that E DFA is decidable, we need a Turing machine T such that If h A i ∈ E DFA , T accepts h A i If h A i 6∈ E DFA , T rejects h A i Thumrongsak Kosiyatrakul [email protected] Decidability 02
Not a Decider Example Consider this TM S : S = “On input h A i where A is a DFA: 1 For all strings w : 2 Run TM M on input h A, w i 3 If M accepts h A, w i , reject Do you see the problem of TM S ? TM S is not a decider Given h A i where A is a DFA and L ( A ) = , S will run indefinitely For this problem: We may have to play around with its state diagram Create a Turing machine that analyze a DFA by marking all states reachable from the start state. If no accept state is marked, the DFA accepts no string Thumrongsak Kosiyatrakul [email protected] Decidability 02
Decidable Languages Consider the following TM T : T = “On input h A i , 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 .” We should not need to go down to the state diagram of a DFA unless it is necessary Rely on graph theory To prove, it must be based on graph theory Thumrongsak Kosiyatrakul [email protected] Decidability 02
Decidable Languages EQ DFA Let EQ DFA = {h A, B i | A and B are DFAs and L ( A ) = L ( B ) } . Show that EQ DFA is decidable. Given two DFAs A and B , is there any algorithm to check whether L ( A ) = L ( B ) ?

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