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)),a,X)
• If d(q, b, X) (p, ?) where b?T 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.
• (q0,h(w),Z0)-P (p,?,?) iff ((q0,?),w,Z0) -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 a,b 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
  , Cb?b

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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