Finite Automata

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Finite Automata Formal Languages (FAFL)Push
Down Automata Properties of CFG - Revision

• Jibi AbrahamAsst. Professor Dept of CSEM S
  Ramaiah Institute of TechnologyBangalore. Email
  jibiabraham_at_msrit.edu

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Set Difference

• L1 and L2 are CFLs. L1 - L2 is not necessarily a
  CFL
• Proof
• L1 - L
• is regular and is also CFL
• But - L LC
• If CFLs were closed under set difference then
  - L LC would always be a CFL.
• But CFLs are not closed under complementation

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Inverse homomorphism

• To recall If h is a homomorphism and L is any
  language then h-1(L) called an inverse
  homomorphism is the set of all strings w such
  that h(w)L
• The CFLs are closed under inverse homomorphism.
• Theorem If L is a CFL and h is a homomorphism
  then h-1(L) is a CFL

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Inverse homomorphism proof
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Proof Contd...

• After input a is read h(a) is placed in a
  buffer.
• Symbols of h(a) are used one at a time and fed to
  PDA being simulated.
• Only when the buffer is empty does the PDA read
  another of its input symbol and apply
  homomorphism to it.

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Proof Contd...

• Suppose h applies to symbols of alphabet S and
  produces strings in T.
• Let PDA P (Q T G d q0 Z0 F) that accept
  CFL L by final state.
• Construct a new PDA P (Q S G d (q0 )
  Z0 F X ) to simulate language of h-1(L)
  where
• Q is the set of pairs (q x) such that
• q is a state in Q
• x is a suffix of some string h(a) for some input
  string a in S

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Proof Contd...

• d is defined by
• d((q ) a X) ((q h(a))aX)
• If d(q b X) (p ) where bT or b then
  d((q bx) X) ((p x) )
• The start state of P is (q0 )
• The accepting state of P is (q ) where q is
  an accepting state of P.
• (q0h(w)Z0)-P (p) iff ((q0)wZ0) -P
  ((p ) )
• P accepts h(w) if and only if P accepts w
  because of the way the accepting states of P are
  defined.  
• Thus L(P)h-1(L(P))

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Revision

• Based on previous question papers
• Construction of PDA
• Convert CFG to PDA
• Simplification of CFG
• Conversion to CNF
• Pumping lemma of CFL
• Closure properties of CFL

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Revision -Construction of PDA

• To accept the language L aibjck i j k i
  0 j 0 (Feb/Mar 2004 model question paper)
• Solution PDA M (q0 q1 q2 q3 a b c
  a b Z0 d q0 Z0 q3) where d is defined
  by following rules
•   d(q0 a Z0) (q0 aZ0) d(q0 a a)
  (q0 aa)
• d(q0 b a) (q1 ba) d(q0 b Z0)
  (q1 bZ0)
• d(q0 c a) (q2 ) d(q1 b b)
  (q1 bb)
• d(q1 c b) (q2 ) d(q2 c b)
  (q2 )
• d(q2 c a) (q2 ) d(q2 Z0)
  (q3 )
• d(q0 Z0) (q3 )
• d(q x Y) for all other possibilities

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Revision -Construction of PDA Contd

• To accept the language L anb2n a b S n
  1 by final State (Jul/Aug 2004 Jan/Feb 2005
  new scheme)
• Solution PDA M (q0 q1 q2 a b a
  Z0 d q0 Z0 q2) where d is defined by
  following rules
• d(q0 a Z0) (q0 aaZ0) d(q0 a a) (q0
  aaa)
• d(q0 b a) (q1 ) d(q1 b a) (q1 )
• d(q1 Z0) (q2 )
• d(q x Y) for all other possibilities

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Revision -Construction of PDA Contd

• Construct a PDA to accept strings containing
  equal number of as and bs. (Jan/Feb 2005 old
  scheme)
• Design PDAs to accept by final scheme and by
  empty stack 0n1n n 1(July/Aug 2005)

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Revision - Convert CFG to PDA

• July/Aug 2004
• I a b S aA A aABC bB a B b C c 
• Solution PDA accepted by empty stack M (q
  a b c a b c S A B C I d q S)
  where transition functions d are given below
• d(q I) (q a) (q b)
• d(q S) (q aA)
• d(q A) (q aABC) (q bB) (q a)
• d(q B) (q b)
  d(q C) (q c)
• d(q a a) (q )
  d(q b b) (q )
• d(q c c) (q )

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Revision - Convert CFG to PDA Contd

• Jan/Feb 2005 old scheme
• S aBB B bS c
• Solution PDA accepted by empty stack M (q
  a b c a b c S B d q S) where
  transition functions d are given below 
• d(q S) (q aBB)
• d(q B) (q bS) (q c)
• d(q a a) (q ) d(q b b) (q )
• d(q c c) (q )

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Revision simplification of CFG

• Eliminate unit production from the grammar below
  (July/Aug 2004) 
• S Aa B B A bb A a bc B
• Solution
• Unit productions are S B B A and A B
• A B and S are derivable
• Eliminating B in the A production gives A a
  bc bb.
• Eliminating A in the B production gives B a
  bc bb.
• Eliminating B in the S production gives S Aa
  a bc bb

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Revisionsimplification of CFG Contd

• The final set of productions after eliminating
  unit productions are given below 
• S Aa a bc bb
• B a bc bb
• A a bc bb

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Revisionsimplification of CFG Contd

• Eliminate useless symbols and productions from
  the following grammar (Jul/Aug 2004 old scheme) 
• S ABa BC A aC BCC C a B bcc
• D E E d F e
• Solution
•  Step 1 Eliminate non-generating symbols All
  variables are generating 
• Step 2 Elimination of non-reachable variables
   Draw the dependency graph

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Revisionsimplification of CFG Contd

•  
• D E and F are non reachable from S
• After Removing useless symbols
• S ABa BC
• A aC BCC
• C a
• B bcc

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Revisionsimplification of CFG Contd

• Eliminate useless symbols and productions from
  the following grammar G (V T P S) where V
  S A B C T ab and productions P given
  below (July/Aug 2005)
• S ab A C A a B ac C aCb

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Revision Conversion to CNF

• Feb / Mar 2004 model question paper
• S ASB A aAS a B SbS A bb
• Solution 
• Step 1 Simplify the grammar with productions P
• Step 1a Eliminate - productions to obtain P1
• P1 S ASB AB A aAS a aA B SbS
  A bb Sb bS b
• Step 1b Eliminate unit productions to obtain P2
• P2 S ASB AB A aAS a aA B SbS
  bb Sb bS b aAS a aA

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Revision Conversion to CNF Contd

• Step 1cEliminate useless symbols to obtain P3
• All variables are generating and all are
  reachable.
• Simplified grammar is G (S A B a b
  S ASB AB A aAS a aA B SbS bb
  Sb bS b aAS a aA S)
• Step 2 Convert G to CNF
•  Step 2a Add productions of the form A BC A a
• P S AB A a B b a 

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Revision Conversion to CNF Contd

• Step 2b Eliminate terminals from RHS of the
  other productions
• A aAS to A CaAS and Ca a
• A aA to A CaA
• B SbS to B SCbS and Cb b
• B bb to B CbCb
• B bS to B CbS
• B Sb to B SCb
• B aAS to B CaAS
• B aA to B CaA
• Add productions in the CNF form to P S
  AB A a CaA B CbCb CbSSCbCaA b a Ca a
  Cbb

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Revision Conversion to CNF Contd

• Step 2c Reduce the RHS of the productions with
  more than 2 variables to the form of A BC
• A CaAS to A CaC1 and C1 AS
• B SCbS to B SC2 and C2 CbS
• B CaAS to B Ca C3 and C3 AS
• Adding these productions to P S AB A
  a CaA CaC1 B CbCb CbS SCb CaA
  SC2 CaC3 b a Ca a Cb b C1 AS C2
  CbS C3 AS

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Revision Conversion to CNF Contd

• The grammar in CNF form G (V a b P
  S)
• V S A B Ca Cb C1 C2 C3
• P S AB A a CaA CaC1 B CbCb
  CbS SCb CaA SC2 CaC3 b a Ca a Cb
  b C1 AS C2 CbS C3 AS

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Revision Conversion to CNF Contd

• Jul / Aug 2004 old scheme
•   S abAB A bAB B BAa A
• Jan / Feb 2005 old scheme
• S S T T R T T a b R R a
• Jul / Aug 2005 old scheme
•   S ABa A aab B AC
• Jan / Feb 2005 new scheme
•   S ASB A aAS a B AbA A bb

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Revision Pumping Lemma of CFL

• State and prove pumping lemma for CFL (Jul / Aug
  2004 old scheme and new scheme) (Jan/Feb 2005
  old scheme)
• Show that L an bn cn n 1 is not a CFL (Jul
  / Aug 2004 old scheme) (Jan/Feb 2005 old scheme)

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Revision Closure Properties of CFL

• Show that the family of CFL is not closed under
  intersection and complementation (Jul / Aug 2005)
• Show that the family of CFL is closed under
  union concatenation and star closure (Jul / Aug
  2004 old scheme) (Jan/Feb 2005 new scheme)
  (Jan/Feb 2005 old scheme)

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Q AContact Email jibiabraham_at_msrit.edu