Regular a regular expression I Example 1.68 Consider the following - - PowerPoint PPT Presentation

Regular ⇒ a regular expression I

Example 1.68 Consider the following DFA 1 2 3 a a b b b a

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Regular ⇒ a regular expression II

It is not that easy to directly see what the regular expression is We need a procedure shown below First, add a start and an accept states s 1 2 3 a a a b b b a ǫ ǫ ǫ

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Regular ⇒ a regular expression III

This generates a generalized NFA (GNFA) Our procedure is DFA → GNFA → regular expression Remove state 1

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Regular ⇒ a regular expression IV

s 2 3 a a b ba ∪ a ab aa ∪ b bb ǫ ǫ Example: the link 3 → 2

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Regular ⇒ a regular expression V

3 1 2 a b a Thus ba ∪ a Idea: now 1 is removed. Need to check how we can go from 3 to 2 via state 1 Need to check all pairs of states Remove state 2

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Regular ⇒ a regular expression VI

s 3 a a(aa ∪ b)∗ a(aa ∪ b)∗ab ∪ b (ba ∪ a)(aa ∪ b)∗ab ∪ bb (ba ∪ a)(aa ∪ b)∗ ∪ ǫ Example: s → 3

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Regular ⇒ a regular expression VII

s 2 3 b a ab aa ∪ b Thus a(aa ∪ b)∗ab ∪ b

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Regular ⇒ a regular expression VIII

Here we need to handle 2

aa∪b

− − → 2 Thus in the early example of removing state 1, we actually have

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Regular ⇒ a regular expression IX

3 1 2 a b a ∅ and b∅∗a ∪ a = bǫa ∪ a = ba ∪ a Remove state 3

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Regular ⇒ a regular expression X

s a (a(aa ∪ b)∗ab ∪ b)((ba ∪ a)(aa ∪ b)∗ab ∪ bb)∗ ((ba ∪ a)(aa ∪ b)∗ ∪ ǫ) ∪ a(aa ∪ b)∗ We will formally explain the procedure

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