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Brian K. Strickland. 12 03 45 a ba Λ a aa b Λ -NFA for Regular Expression (aab)*(a + aba)*

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Presentation on theme: "Brian K. Strickland. 12 03 45 a ba Λ a aa b Λ -NFA for Regular Expression (aab)*(a + aba)*"— Presentation transcript:

1 Brian K. Strickland

2 12 03 45 a ba Λ a aa b Λ -NFA for Regular Expression (aab)*(a + aba)*

3 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b Computation Tree for Input String: aababa

4 12 03 45 a ba Λ a aa b 0 1 2 0 44 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

5 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

6 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

7 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

8 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

9 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

10 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

11 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

12 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b

13 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b Accepted!

14 12 0 3 45 a ba Λ a aa b 0 1 2 0 4 4 5 4 5 1 Λ Λ a aa a a a a a a 3 3 3 3 3 3 a a 3 b b b The key is the lambda transition from the Kleene star operation (*) which allows the regular expression to transition to the accepting state of the NFA.


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