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Finite Automata

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Revision -Construction of PDA ... Construct a PDA to accept strings containing equal number of a's and b's. ... V = {S, A, B, Ca, Cb, C1, C2, C3} ... – PowerPoint PPT presentation

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Title: Finite Automata


1
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

2
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

3
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

4
Inverse homomorphism proof
5
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.

6
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

7
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))

8
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

9
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

10
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

11
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)

12
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, ?)

13
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, ?)

14
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

15
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

16
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

17
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

18
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

19
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

20
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 

21
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

22
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

23
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

24
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

25
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)

26
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)

27
Q AContact Email jibiabraham_at_msrit.edu
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