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Properties of Regular Expressions

The document describes the properties of regular languages and proves that regular languages are closed under various operations including concatenation, union, intersection, complement, reversal, and Kleene star. It provides examples of constructing non-deterministic finite automata (NFAs) and deterministic finite automata (DFAs) to represent languages that are the result of applying these operations to regular languages. Key steps involved in the constructions of NFAs and DFAs for each operation are outlined.

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Costas Busch - LSU 1 Properties of Regular Languages
Costas Busch - LSU 2 1L 2L 21LLConcatenation: * 1LStar: 21 LL Union: Are regular Languages For regular languages and we will prove that: 1L 21 LL  Complement: Intersection: R L1 Reversal:
Costas Busch - LSU 3 We say Regular languages are closed under 21LLConcatenation: Star: 21 LL Union: 1L 21 LL  Complement: Intersection: Reversal: * 1L R L1
Costas Busch - LSU 4 a b b a NFA Equivalent NFA   a b b a A useful transformation: use one accept state 2 accept states 1 accept state
Costas Busch - LSU 5 NFA Equivalent NFA Single accepting state    In General
Costas Busch - LSU 6 NFA without accepting state Add an accepting state without transitions Extreme case
Costas Busch - LSU 7 1LRegular language   11 LML  1M Single accepting state NFA 2M 2L Single accepting state   22 LML  Regular language NFA Take two languages
Costas Busch - LSU 8 }{1 baL n  a b 1M  baL 2 ab 2M 0n Example
Costas Busch - LSU 9 Union NFA for 1M 2M 21 LL    21 LLw  21 or LwLw 
Costas Busch - LSU 10 a b ab   }{1 baL n  }{2 baL  }{}{21 babaLL n NFA for Example
Costas Busch - LSU 11 Concatenation 21LL 1M 2M  21LLw NFA for 221121 and: LwLwwww  change to regular state
Costas Busch - LSU 12 NFA for a b ab }{1 baL n  }{2 baL  }{}}{{21 bbaababaLL nn   Example
Costas Busch - LSU 13 Star Operation * L Lwwwww ik  :21  LML )(    NFA for * Lw or w M
Costas Busch - LSU 14 ** 1 }{ baL n  a b }{1 baL n    Example NFA for
Costas Busch - LSU 15 Reverse 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 )(
Costas Busch - LSU 16 }{1 baL n  a b 1M }{1 nR baL  a b 1M Example
Costas Busch - LSU 17 Complement 1. Take the DFA that accepts L M LML )( M LML )( 2. Make accept states regular and vice-versa
Costas Busch - LSU 18 }{1 baL n  a b 1M ba, ba, }{*},{1 babaL n  a b 1M ba, ba, Example
Costas Busch - LSU 19 NFAs cannot be used for complement Make accept states regular and vice-versa {})( ML M M )(}{)( MLML   ** },{)( baML  NFA NFA it is not the complement
Costas Busch - LSU 20 Same example with DFAs Make accept states regular and vice-versa {})( ML M M )(},{)( * MLbaML  ** },{)( baML  DFA DFA ba, ba, it is the complement
Costas Busch - LSU 21 Intersection 1L regular 2L regular We show 21 LL  regular
Costas Busch - LSU 22 DeMorgan’s Law: 2121 LLLL  21 , LL regular, regular 21 , LL regular, regular 21 LL  regular 21 LL  regular 21 LL  regular
Costas Busch - LSU 23 Example }{1 baL n  },{2 baabL  regular regular }{21 abLL  regular
Costas Busch - LSU 24 1Lfor for 2LDFA 1M DFA 2M Construct a new DFA that accepts Machine Machine M 21 LL  M simulates in parallel and1M 2M Another Proof for Intersection Closure
Costas Busch - LSU 25 States in M ji pq , 1M 2MState in State in
Costas Busch - LSU 26 1M 2M 1q 2qa transition 1p 2pa transition DFA DFA 11, pq a New transition MDFA 22, pq
Costas Busch - LSU 27 0q initial state 0p initial state New initial state 00, pq 1M 2MDFA DFA MDFA
Costas Busch - LSU 28 iq accept state jp accept states New accept states ji pq , kp ki pq , 1M 2MDFA DFA MDFA Both constituents must be accepting states
Costas Busch - LSU 29 Example: }{1 baL n  a b 1M 0n }{2 m abL  b b 2M 0q 1q 0p 1p 0m 2q 2p a a ba, ba, ba,
Costas Busch - LSU 30 00, pq DFA for intersection }{}{}{)( ababbaML mn  10, pqa 21, pq b ab 11, pq 20, pq a 12, pq 22, pq b ba, a b ba, b a M
Costas Busch - LSU 31 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 3 until no new states are added 4. Designate accept states
Costas Busch - LSU 32 00, pq Automaton for intersection }{}{}{ ababbaL mn  initial state
Costas Busch - LSU 33 00, pq Automaton for intersection }{}{}{ ababbaL mn  10, pqa add transition and new state for symbol a 1M 2M 0q 0qa 0p 1pa 00 , pq a M 10 , pq
Costas Busch - LSU 34 00, pq Automaton for intersection }{}{}{ ababbaL mn  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
Costas Busch - LSU 35 Automaton for intersection }{}{}{ ababbaL mn  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
Costas Busch - LSU 36 Automaton for intersection }{}{}{ ababbaL mn  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
Costas Busch - LSU 37 simulates in parallel and1M 2M accepts string w if and only if: accepts string w1M and accepts string w2M )()()( 21 MLMLML  Intersection DFA :M