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Chapter 11: Decidability

Chapter 11: Decidability. Three important questions: Do two regular expressions define the same language? Do two finite automata accept the same language? Does the language defined by a finite automaton have a finite number of words? Infinite? None?. Chapter 11: Decidability.

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Chapter 11: Decidability

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  1. Chapter 11: Decidability Three important questions: • Do two regular expressions define the same language? • Do two finite automata accept the same language? • Does the language defined by a finite automaton have a finite number of words? Infinite? None? Zaguia/Stojmenovic

  2. Chapter 11: Decidability A problem is called a decision problem if its solution is either YES or NO. Definition: a decision problem is decidable, or effectively solvable, if there is an algorithm that can find the answer for any input in a finite number of steps. Zaguia/Stojmenovic

  3. Chapter 11: Decidability • Do two regular expressions define the same language? • Do two finite automata accept the same language? Problems 1 and 2 are equivalent: It is enough to transfer the two regular expressions into equivalent finite automata (it could be done in a finite number of steps) and then resolve Problem 2. Or we can transfer the two finite automata into regular expressions (it could be done in a finite number of steps) and then resolve Problem 1. Zaguia/Stojmenovic

  4. Chapter 11: Decidability Problem 2: Finite algorithm From two finite automata that accept languages L1 and L2, we construct a finite automaton that accepts (L1 L2’) + (L2 L1’) L1 and L2 are equivalent if and only if this languages is empty. Method 1:Convert this finite automaton into a regular expression (using the finite algorithm defined in the proof of Kleene’s theorem). If you end up with a regular expression then the language is not empty, otherwise it is empty. Zaguia/Stojmenovic

  5. r – + a,b a,b + – Chapter 11: Decidability If the algorithm leads to: Then the finite automaton accepts at least one word If the algorithm leads to: OR similar case Then the finite automaton does not accept any word Zaguia/Stojmenovic

  6. Chapter 11: Decidability Method 2: Answer the question: is there any path from – to +? Algorithm: • Mark the start state. (Paint it blue.) • From every blue state, follow every arrow that goes out of the state, and paint the destination state blue. Delete each arrow that was followed. • Repeat step 2 until there are no new blue states. • If there is any final state that is blue, then there are words in the language accepted by the finite automaton. If not, the language is empty. Zaguia/Stojmenovic

  7. a a b a a b b a,b b + – a b a a b a b b a,b + – a b a b a b + – a b Chapter 11: Decidability Zaguia/Stojmenovic

  8. b a b + – a a b a b + – a b Chapter 11: Decidability The language is empty. Zaguia/Stojmenovic

  9. Chapter 11: Decidability Theorem. Let F be a finite automaton with N states. If the language accepted by F is not empty, then it contains at least one word with N or fewer letters. Method 3. Try all words with N and fewer letters. If the finite automaton accepts none of them, then the language accepted by this automaton is empty. Zaguia/Stojmenovic

  10. a a,b b p1+ p2 a a,b b q1– q2+ b s2+ s1– b r2 r1+ a,b a,b a b a b s3 a r3+ a Chapter 11: Decidability a* FA1 FA1’ FA2’ L+aa* FA2 Zaguia/Stojmenovic

  11. a q1 or r1 p1 or s1 a a a — — q1 or r3 p1 or s3 b b b b a,b a,b q2 or r2 p2 or s2 Chapter 11: Decidability (FA2’ + FA1)’ (FA1’ + FA2)’ (FA1’ + FA2)’ + (FA2’ + FA1)’? FA1 and FA2 represent the same language What if one of these two machines has a final state? Zaguia/Stojmenovic

  12. Chapter 11: Decidability Problem 3. Does the language defined by a finite automaton has a finite number of words? Infinite? None? none? Same algorithms, compare FA1 (or r1) with empty L2 or corresponding FA2. finite or infinite language? Solution 1: Does corresponding RE contain * ? Solution 2: Use the following theorem. Zaguia/Stojmenovic

  13. Theorem. Let F be a finite automaton with N states. F accepts a word w such that N  length(w) < 2N if and only if the language accepted by F is infinite. Proof: Apply Pumping lemma.  If w, N  length(w) is accepted then w created loop while following N states  pumping lemma creates infinite number of accepted words  If language accepted by F is infinite then consider any accepted word with length > N. This word creates loop in FA. Shorten the word by cutting one loop, it will be reduced by length up to N. repeat until length falls between N and 2N. Zaguia/Stojmenovic

  14. Chapter 11: Decidability Theorem. There is an algorithm that can decide whether a finite automaton accepts a finite or infinite language. Algorithm: It is enough to check all words of length between N and 2N. There is a finite number of tests to be done, and every check needs a finite number of steps. Zaguia/Stojmenovic

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