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CSI 3104 Winter 2006: Introduction to Formal Languages Chapter 17: ContextFree Languages

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X bXa / ba. A aA / a. L1 L2 = anbnan (is not a context-free language ) Zaguia/Stojmenovic ... PDA is deterministic with other properties, we could use for the ... – PowerPoint PPT presentation

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Title: CSI 3104 Winter 2006: Introduction to Formal Languages Chapter 17: ContextFree Languages


1
CSI 3104 /Winter 2006 Introduction to Formal
Languages Chapter 17 Context-Free Languages
  • Chapter 17 Context-Free Languages
  • I. Theory of Automata
  • ? II. Theory of Formal Languages
  • III. Theory of Turing Machines

2
Chapter 17 Context-Free Languages
  • Theorem. The set of context-free languages is
    closed under union, concatenation, and Kleene
    closure.
  • Union. L1L2
  • L1 and L2 are generated by two context-free
    grammars G1 and G2. We replace each nonterminal X
    in G1 by X1, and each nonterminal X in G2 by X2.
    We add the productions
  • S ? S1 S ? S2
  • L1L2 is the language generated by this new CFG.

3
Chapter 17 Context-Free Languages
  • Example L1 PALINDROME S ? aSa bSb a
    b ?
  • L2 anbn S ? aSb ?
  • L1L2 S ? S1 S2
  • S1 ? aS1a bS1b a b
    ?
  • S2 ? aS2b ?
  • Example L1 aa,bb S ? aA bB A ? a
    B ? b
  • L2 ? S ? ?
  • L1L2 S ? S1 S2
  • S1 ? aA1 bB1 A1 ?
    a B1 ? b
  • S2 ? ?

4
Chapter 17 Context-Free Languages
Proof by machines
5
Chapter 17 Context-Free Languages
(begin with a)
(contain aa)
6
Chapter 17 Context-Free Languages
L1 L2
7
Chapter 17 Context-Free Languages
  • Concatenation. L1L2
  • Similar to union except we add
  • S ? S1S2
  • Example
  • L1 palindrome S ? aSa bSb a b ?
  • L2 anbn S ? aSb ?
  • L1L2 S ? S1S2
  • S1 ? aS1a bS1b a b ?
  • S2 ? aS2b ? proof by PDAs?

8
L
  • Kleene star. L
  • We replace S by S1 and add
  • S ? S1S ?
  • S ? S1S ? S1S1S ? S1S1S1S ?
  • Example L S ? aSa bSb a b ?
  • L S ? S1S ?
  • S1 ? aS1a bS1b a b ?

9
Chapter 17 Context-Free Languages
  • Theorem. The intersection of two context-free
    languages may or may not be context-free.
  • L1 anbnam
  • S ? XA
  • X ? aXb / ab
  • A ? aA / a
  • L2 anbmam
  • S ? AX
  • X ? bXa / ba
  • A ? aA / a
  • L1 n L2 anbnan (is not a context-free
    language )

10
Chapter 17 Context-Free Languages
  • Theorem. The intersection of a context-free
    language and a regular language is always
    context-free.
  • Proof By constructive algorithm using pushdown
    automata, similar to the intersection algorithm
    for two finite automata.
  • (details not covered)

11
Example use the theorem
  • L doubleword ww abbabb,
  • Assume L is CFL
  • L n abab anbmanbm
  • is then context free (n regular lang.)
  • But we have proven that anbmanbm is not CFL
  • ? contradiction.

12
Chapter 17 Context-Free Languages
  • Theorem. The complement of a context-free
    language may or may not be context-free.
  • When the PDA is deterministic with other
    properties, we could use for the complement a
    similar technique as for FA

13
Chapter 17 Context-Free Languages
PALINDROME-X
14
Chapter 17 Context-Free Languages
For the complement We interchange the states
ACCEPT and REJECT. It does not work all the
time.
15
Complement of a CFL may not be a CFL
  • Proof by indirect arguments
  • Suppose complement of every CFL is a CFL.
  • L1 and L2 CFLs ? L1 and L2 CFLs ? L1L2 CFLs
    ? (L1L2) L1 ? L2 is CFL
  • Contradicts intersection theorem.

16
Nondeterminism
  • L accepts w if some paths lead to accept (some
    paths may lead to reject)
  • L rejects w if all paths lead to reject
  • Exchange accept and reject states
  • L rejects w if some paths lead to reject ??
  • L accepts w if all paths lead to accept ??
  • Nondeterminism does not support complement

17
Example
  • Lanbnan is not CFL but L is CFL
  • LMpqMqpMprMrpMqrMrqM
  • Mpqapbqar pgtq aanbna CFL
  • Mqpapbqar qgtp anbnba CFL
  • Mprapbqar pgtr aanban CFL
  • Mqr, Mrq, Mrp similarly CFL
  • Mabc complement of a regular language

18
Deterministic PDA lt Nondeterministic PDA
  • Proof Complement of anbncn is CFL and cannot
    have deterministic PDA since otherwise its
    complement has deterministic PDA and would be CFL
    (accept ?reject )
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