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CSCI 3130: Formal languages and automata theory Tutorial 9

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CSCI 3130: Formal languages and automata theory Tutorial 9 Chin Reminder Homework 5 is due at next Tuesday! No tutorial next week! :D This is the last one. – PowerPoint PPT presentation

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Title: CSCI 3130: Formal languages and automata theory Tutorial 9


1
CSCI 3130 Formal languages and automata
theory Tutorial 9
  • Chin

2
Reminder
  • Homework 5 is due at next Tuesday!
  • No tutorial next week! D
  • This is the last one.

3
Undecidable problems for CFGs
  • ALLCFG ltGgt G is a CFG that generates all
    strings
  • ALLCFG is undecidable
  • Reduction
  • ATM ltM, wgt M is a TM that does not accept w
  • ltM, wgt M rejects or loops on input w
  • ATM is unrecognizable ? ATM is undecidable
  • If ALLCFG is decidable then ATM is decidable.
  • Contradiction.

4
Undecidable problems for CFGs
  • ALLCFG ltGgt G is a CFG that generates all
    strings
  • ALLCFG is undecidable
  • Suppose it is decidable, then there exists a
    universal TM H
  • string H(string ltGgt)
  • // G is a CFG
  • if G generates all strings, return accept
  • if G does not generates all strings, return
    reject
  • Consider the following TM U (dependent on ltMgt and
    w)
  • string U(string ltMgt, string w)
  • If M does not accept w, return ltGgt that
    generates all strings
  • If M accepts w, return ltGgt that does not
    generate all strings
  • If we can simulate H(U(ltMgt,w)), then U decides
    ATM and so ATM is decidable.

5
Undecidable problems for CFGs
  • ALLCFG ltGgt G is a CFG that generates all
    strings
  • ALLCFG is undecidable
  • Consider the following TM U (dependent on ltM, wgt)
  • string U(string ltMgt, string w)
  • If M does not accept w, return ltGgt that
    generates all strings
  • If M accepts w, return ltGgt that does not
    generate all strings
  • We construct U as follows instead
  • string U(string ltMgt, string w)
  • If M does not accept w, construct ltGgt that
    generates all strings
  • If M accepts w, construct ltGgt that generates
    all strings except
  • the computation history of M(w)
  • return H(ltGgt)
  • How to construct ltGgt? We need to construct ltGgt so
    that even if M loops on w, ltGgt generates all
    strings.

6
Undecidable problems for CFGs
  • Computation history
  • q0ababxq1bab . . . xxxx?qa
  • Each xxxxxx tells you the configuration of the
    TM in each step

abq1a
abbqacc
7
Undecidable problems for CFGs
  • Computation history
  • Suppose the current configuration is
  • xbq1xab
  • What does the next configuration look like?
  • The head must be one symbol left or one symbol
    right
  • xbq1xab
  • Only these 3 symbols can change
  • Use this fact to design a PDA that generates the
    computation history (see lecture notes).
  • Convert the PDA to ltGgt.

8
Undecidable problems for CFGs
  • PCP ltTgt T is a collection of tiles
    that contains a top-bottom match
  • PCP is undecidable
  • AMB ltGgt G is an ambiguous CFG
  • AMB is undecidable
  • If AMB is decidable, PCP is decidable.
    Contradiction.

9
Undecidable problems for CFGs
T
G
(CFG)
(collection of tiles)
Terminals
a, b, c, 1, 2, 3
Variables
S, T, B
1
2
3
Productions
bab cc
c ab
a ab
T ? cT2
T ? aT3
T ? babT1
S ? T B
T ? c2
T ? a3
T ? bab1
B ? abB2
B ? abB3
B ? ccB1
B ? ab2
B ? ab3
B ? cc1
10
Undecidable problems for CFGs
  • EQCFG ltG1, G2gt G1 and G2 are two CFGs that
    describe the same language
  • Is it decidable?

11
Undecidable problems for CFGs
  • EQCFG ltG1, G2gt G1 and G2 are two CFGs that
    descript the same language
  • EQCFG is undecidable.
  • Same idea as EQTM
  • If EQCFG is decidable, ALLCFG is also decidable.
  • Let G1 to be a CFG that generates all inputs.

12
Undecidable problems for CFGs
  • EQCFG ltG1, G2gt G1 and G2 are two CFGs that
    descript the same language
  • If EQCFG is decidable, there exists a TM H such
    that
  • string H(string ltG1gt, string ltG2gt)
  • if G1 and G2 are equal, return accept
  • if G1 and G2 are not equal, return reject
  • Let G1 to be a CFG that generates all inputs. We
    construct U as follows
  • string U(string ltGgt)
  • G1 a CFG that generates all inputs //
    hardcode
  • return H(G1, G)
  • Given a CFG ltGgt, U(ltGgt) returns accept if G
    generates all strings, rejects if G does not.
  • Hence ALLCFG is decidable. Contradiction.

13
Post Correspondence Problem
  • PCP1 PCP over the alphabet S 1.
  • The alphabet of a tile is the set of symbols that
    appear on the tiles.
  • Is PCP1 decidable?

14
Post Correspondence Problem
  • PCP1 PCP over the alphabet S 1.
  • The alphabet of a tile is the set of symbols that
    appear on the tiles.
  • PCP1 is decidable.
  • 3 1 2
  • 2 5 -3
  • Find x, y such that 2x 3y 0. Easy.
  • If exists two tiles like above, accept. Otherwise
    reject.

111 1
11 11111
15
Post Correspondence Problem
  • PCP2 PCP over the alphabet S 0, 1.
  • The alphabet of a tile is the set of symbols that
    appear on the tiles.
  • Is PCP2 decidable?

16
Post Correspondence Problem
  • PCP2 PCP over the alphabet S 0, 1.
  • The alphabet of a tile is the set of symbols that
    appear on the tiles.
  • PCP2 is undecidable.
  • If PCP2 is decidable, then PCP is also decidable.
  • Represent the alphabet in PCP in binary form.

17
End
  • Questions?
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