CSCI 3130: Formal languages and automata theory Tutorial 9

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CSCI 3130 Formal languagesand automata
theoryTutorial 9

• Chin

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Reminder

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

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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.

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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.

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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.

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Undecidable problems for CFGs

• Computation history
• q0ababxq1bab . . . xxxx?qa
• Each xxxxxx tells you the configuration of the
  TM in each step

abq1a
abbqacc
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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.

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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.

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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
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Undecidable problems for CFGs

• EQCFG ltG1, G2gt G1 and G2 are two CFGs that
  describe the same language
• Is it decidable?

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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.

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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.

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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?

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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
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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?

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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.

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End

• Questions?