Regular properties
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This document discusses properties of regular languages and proofs that regular languages are closed under various operations. It begins with an overview of operations like concatenation, union, star, complement, intersection, and reversal. It then provides examples of constructing finite state automata for each operation on sample regular languages. The key points made are that regular languages are closed under these operations and finite state automata can be constructed to recognize the resulting languages.
Regular properties
- 1. © 2020 Wael Badawy 1 Properties of Regular Languages 1 DISCLAIMER: This video is optimized for HD large display using patented and patent-pending “Nile Codec”, the first Egyptian Video Codec for more information, PLEASE check https://NileCodec.com Also available as a PodCast
- 2. © 2020 Wael Badawy Copyright © 2020 Wael Badawy. All rights reserved n This video is subject to copyright owned by Wael Badawy “WB”. Any reproduction or republication of all or part of this video is expressly prohibited unless WB has explicitly granted its prior written consent. All other rights reserved. n This video is intended for education and information only and is offered AS IS, without any warranty of the accuracy or the quality of the content. Any other use is strictly prohibited. The viewer is fully responsible to verify the accuracy of the contents received without any claims of costs or liability arising . n The names, trademarks service marked as logos of WB or the sponsors appearing in this video may not be used in any any product or service, without prior express written permission from WB and the video sponsors n Neither WB nor any party involved in creating, producing or delivering information and material via this video shall be liable for any direction, incidental, consequential, indirect of punitive damages arising out of access to, use or inability to use this content or any errors or omissions in the content thereof. n If you will continue to watch this video, you agree to the terms above and other terms that may be available on http://nu.edu.eg & https://caiwave.net 2
- 3. © 2020 Wael Badawy
- 4. © 2020 Wael Badawy 4 1L 2L 21LLConcatenation: * 1LStar: 21 LL ÈUnion: Are regular Languages For regular languages and we will prove that: 1L 21 LL Ç Complement: Intersection: R L1 Reversal:
- 5. © 2020 Wael Badawy 5 We say Regular languages are closed under 21LLConcatenation: Star: 21 LL ÈUnion: 1L 21 LL Ç Complement: Intersection: Reversal: * 1L R L1
- 6. © 2020 Wael Badawy 6 a b b a NFA Equivalent NFA e e a b b a A useful transformation: use one accept state 2 accept states 1 accept state
- 7. © 2020 Wael Badawy 7 NFA Equivalent NFA Single accepting state e e e In General
- 8. © 2020 Wael Badawy 8 NFA without accepting state Add an accepting state without transitions Extreme case
- 9. © 2020 Wael Badawy 9 1LRegular language ( ) 11 LML = 1M Single accepting state NFA 2M 2L Single accepting state ( ) 22 LML = Regular language NFA Take two languages
- 10. © 2020 Wael Badawy 10 }{1 baL n = a b 1M { }baL =2 ab 2M 0³n Example
- 11. © 2020 Wael Badawy Union NFA for 11 1M 2M 21 LL È e 21 LLw ÈÎ 21 or LwLw ÎÎ e
- 12. © 2020 Wael Badawy 12 a b ab }{1 baL n = }{2 baL = }{}{21 babaLL n È=ÈNFA for Example e e
- 13. © 2020 Wael Badawy Concatenation 13 21LL 1M 2M e 21LLwÎ NFA for 221121 and: LwLwwww ÎÎ= change to regular state
- 14. © 2020 Wael Badawy NFA for 14 a b ab }{1 baL n = }{2 baL = }{}}{{21 bbaababaLL nn == e Example
- 15. © 2020 Wael Badawy Star Operation 15 * L Lwwwww ik Î= :21 ! LML =)( e e NFA for * LwÎ or e=w M change to regular state
- 16. © 2020 Wael Badawy 16 ** 1 }{ baL n = a b }{1 baL n = e e Example NFA for
- 17. © 2020 Wael Badawy Reverse 17 R L M NFA for M¢ 1. Reverse all transitions 2. Make the initial state accept state and the accept state initial state LML =)( R LML =¢)(
- 18. © 2020 Wael Badawy 18 }{1 baL n = a b 1M }{1 nR baL = a b 1M¢ Example
- 19. © 2020 Wael Badawy 19 Complement 1. Take the DFA that accepts L M LML =)( M¢ LML =¢)( 2. Make accept states regular and vice-versa
- 20. © 2020 Wael Badawy 20 }{1 baL n = a b 1M ba, ba, }{},{ * 1 babaL n -= a b 1M¢ ba, ba, Example
- 21. © 2020 Wael Badawy 21 NFAs cannot be used for complement Make accept states regular and vice-versa {})( =ML M M¢ )(}{)( MLML ¹=¢ e ** },{)( baML =S= NFA NFA it is not the complement
- 22. © 2020 Wael Badawy 22 Same example with DFAs Make accept states regular and vice-versa {})( =ML M M¢ )(},{)( * MLbaML ==¢ ** },{)( baML =S= DFA DFA ba, ba, it is the complement
- 23. © 2020 Wael Badawy Intersection 23 1L regular 2L regular we show 21 LL Ç regular
- 24. © 2020 Wael Badawy 24 DeMorgan’s Law: 2121 LLLL È=Ç 21 , LL regular, regular 21 , LL regular, regular 21 LL È regular 21 LL È regular 21 LL Ç regular
- 25. © 2020 Wael Badawy 25 Example }{1 baL n = },{2 baabL = regular regular }{21 abLL =Ç regular
- 26. © 2020 Wael Badawy 26 1Lfor for 2LDFA 1M DFA 2M Construct a new DFA that accepts Machine Machine M 21 LL Ç M simulates in parallel and1M 2M Another Proof for Intersection Closure
- 27. © 2020 Wael Badawy 27 States in M ji pq , 1M 2MState in State in
- 28. © 2020 Wael Badawy 28 1M 2M 1q 2qa transition 1p 2pa transition DFA DFA 11, pq a New transition MDFA 22, pq
- 29. © 2020 Wael Badawy 29 0q initial state 0p initial state New initial state 00, pq 1M 2MDFA DFA MDFA
- 30. © 2020 Wael Badawy 30 iq accept state jp accept states New accept states ji pq , kp ki pq , 1M 2MDFA DFA MDFA Both constituents must be accepting states
- 31. © 2020 Wael Badawy 31 Example: }{1 baL n = a b 1M 0³n }{2 m abL = b b 2M 0q 1q 0p 1p 0³m 2q 2p a a ba, ba, ba,
- 32. © 2020 Wael Badawy 32 00, pq DFA for intersection }{}{}{)( ababbaML mn =Ç= 10, pqa 21, pq b ab 11, pq 20, pq a 12, pq 22, pq b ba, a b ba, b a M
- 33. © 2020 Wael Badawy 33 Construction procedure for intersection 1. Build Initial State 2. For each new state and for each symbol add transition to either an existing state or create a new state and point to it 3. Repeat step 2 until no new states are added 4. Designate accept states
- 34. © 2020 Wael Badawy 34 00, pq Automaton for intersection }{}{}{ ababbaL mn =Ç= initial state
- 35. © 2020 Wael Badawy 35 00, pq Automaton for intersection }{}{}{ ababbaL mn =Ç= 10, pqa add transition and new state for symbol a 1M 2M 0q 0qa 0p 1pa 00, pq a M 10 , pq
- 36. © 2020 Wael Badawy 36 00, pq Automaton for intersection }{}{}{ ababbaL mn =Ç= 10, pqa 21, pq b add transition and new state for symbol b 1M 2M 0q 1qb 0p 2p 00, pq b M 21, pq b
- 37. © 2020 Wael Badawy 37 Automaton for intersection }{}{}{ ababbaL mn =Ç= 00, pq 10, pqa 21, pq b ab 11, pq 20, pq a 12, pq 22, pq b ba, a b ba, b a Repeat until no new states can be added
- 38. © 2020 Wael Badawy 38 Automaton for intersection }{}{}{ ababbaL mn =Ç= add Accept state 00, pq 10, pqa 21, pq b ab 11, pq 20, pq a 12, pq 22, pq b ba, a b ba, b a accept state for accept state for 1q 1M 1p 2M
- 39. © 2020 Wael Badawy 39 simulates in parallel and1M 2M accepts string w if and only if: accepts string w1M and accepts string w2M )()()( 21 MLMLML Ç= Intersection DFA :M