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.

Undecidable problems for CFGs ALLCFG = {<G>: G is a CFG that generates all strings} ALLCFG is undecidable Reduction ATM = {<M, w>: M is a TM that does not accept w} = {<M, w>: M rejects or loops on input w} ATM is unrecognizable ⇒ ATM is undecidable If ALLCFG is decidable then ATM is decidable. Contradiction.

Undecidable problems for CFGs ALLCFG = {<G>: 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 <G>) { // 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 <M> and w) string U(string <M>, string w){ If M does not accept w, return <G> that generates all strings If M accepts w, return <G> that does not generate all strings If we can simulate H(U(<M>,w)), then U decides ATM and so ATM is decidable.

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

Undecidable problems for CFGs Computation history #q0ab%ab#xq1b%ab# . . . #xx%xx☐qa# Each #xxxxxx# tells you the configuration of the TM in each step … a b ☐ q1 abq1a q1 qacc a/bR … a b ☐ qacc abbqacc

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 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 <G>.

Undecidable problems for CFGs PCP = {<T>: T is a collection of tiles that contains a top-bottom match} PCP is undecidable AMB = {<G>: G is an ambiguous CFG} AMB is undecidable If AMB is decidable, PCP is decidable. Contradiction.

Undecidable problems for CFGs T G (collection of tiles) (CFG) Terminals: a, b, c, 1, 2, 3 Variables: S, T, B 1 2 3 Productions: bab cc c ab a ab S → T | B T → babT1 T → cT2 T → aT3 T → bab1 T → c2 T → a3 B → ccB1 B → abB2 B → abB3 B → cc1 B → ab2 B → ab3

Undecidable problems for CFGs EQCFG = {<G1, G2>: G1 and G2 are two CFGs that describe the same language} Is it decidable?

Undecidable problems for CFGs EQCFG = {<G1, G2>: 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.

Undecidable problems for CFGs EQCFG = {<G1, G2>: 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 <G1>, string <G2>) { 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 <G>) { G1 := a CFG that generates all inputs // hardcode return H(G1, G); Given a CFG <G>, U(<G>) returns accept if G generates all strings, rejects if G does not. Hence ALLCFG is decidable. Contradiction.

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?

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

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?

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.

End Questions?