Presentation is loading. Please wait.

Presentation is loading. Please wait.

Fall 2006Costas Busch - RPI1 Properties of Regular Languages.

Similar presentations


Presentation on theme: "Fall 2006Costas Busch - RPI1 Properties of Regular Languages."— Presentation transcript:

1 Fall 2006Costas Busch - RPI1 Properties of Regular Languages

2 Fall 2006Costas Busch - RPI2 Concatenation: Star: Union: Are regular Languages For regular languages and we will prove that: Complement: Intersection: Reversal:

3 Fall 2006Costas Busch - RPI3 We say: Regular languages are closed under Concatenation: Star: Union: Complement: Intersection: Reversal:

4 Fall 2006Costas Busch - RPI4 NFA Equivalent NFA A useful transformation: use one accept state 2 accept states 1 accept state

5 Fall 2006Costas Busch - RPI5 NFA Equivalent NFA Single accepting state In General

6 Fall 2006Costas Busch - RPI6 NFA without accepting state Add an accepting state without transitions Extreme case

7 Fall 2006Costas Busch - RPI7 Regular language Single accepting state NFA Single accepting state Regular language NFA Take two languages

8 Fall 2006Costas Busch - RPI8 Example

9 Fall 2006Costas Busch - RPI9 Union NFA for

10 Fall 2006Costas Busch - RPI10 NFA for Example

11 Fall 2006Costas Busch - RPI11 Concatenation NFA for

12 Fall 2006Costas Busch - RPI12 NFA for Example

13 Fall 2006Costas Busch - RPI13 Star Operation NFA for

14 Fall 2006Costas Busch - RPI14 NFA for Example

15 Fall 2006Costas Busch - RPI15 Reverse NFA for 1. Reverse all transitions 2. Make initial state accepting state and vice versa

16 Fall 2006Costas Busch - RPI16 Example

17 Fall 2006Costas Busch - RPI17 Complement 1. Take the DFA that accepts 2. Make accepting states non-final, and vice-versa

18 Fall 2006Costas Busch - RPI18 Example

19 Fall 2006Costas Busch - RPI19 Intersection regular We show regular

20 Fall 2006Costas Busch - RPI20 DeMorgan’s Law: regular

21 Fall 2006Costas Busch - RPI21 Example regular

22 Fall 2006Costas Busch - RPI22 for DFA Construct a new DFA that accepts Machine simulates in parallel and Another Proof for Intersection Closure

23 Fall 2006Costas Busch - RPI23 States in State in

24 Fall 2006Costas Busch - RPI24 transition DFA New transition DFA

25 Fall 2006Costas Busch - RPI25 initial state New initial state DFA

26 Fall 2006Costas Busch - RPI26 accept state accept states New accept states DFA Both constituents must be accepting states

27 Fall 2006Costas Busch - RPI27 Example:

28 Fall 2006Costas Busch - RPI28 Automaton for intersection

29 Fall 2006Costas Busch - RPI29 simulates in parallel and accepts stringif and only if: accepts string and accepts string


Download ppt "Fall 2006Costas Busch - RPI1 Properties of Regular Languages."

Similar presentations


Ads by Google