Question

Prove and discuss the following reductions.

Walk through the proof to show that the problem of proving the language of a Turning Machine is a context-free language is undecidable. (Do not use Rice’s theorem as a black box and note that this is not the same problem as Theorem 5.13 in the textbook.)

Answer 1

Given α1,…,αm,β1,…,βmα1,…,αm,β1,…,βm: Construct the following CFG G=(V,Σ,R,S)G=(V,Σ,R,S): V={S,S1,S2}V={S,S1,S2}, R={S→S1|S2,S1→α1S1σ1|⋯|αmS1σm|α1σ1|⋯|αmσm,S2→β1S2σ1|⋯|βmS2σm|β1σ1|⋯|βmσm}R={S→S1|S2,S1→α1S1σ1|⋯|αmS1σm|α1σ1|⋯|αmσm,S2→β1S2σ1|⋯|βmS2σm|β1σ1|⋯|βmσm} (where σiσi are new characters added to the alphabet, e.g., σi=i–σi=i_).

If the language is ambiguous, then there is a derivation of some string ww in two different ways. Supposing, wlog, that the derivations both start with the rule S→S1S→S1, reading the new characters backwards until they end makes sure there can only be one derivation, so that's not possible. Hence, we see that the only ambiguity can come from one S1S1 and one S2S2 'start'. But then, taking the substring of ww up to the beginning of the new characters, we have a solution to the PCP (since the strings of indices used after those points match).

Similarly, if there is no ambiguity, then the PCP cannot be solved, since a solution would imply an ambiguity that just follows S⇒S1⇒∗ασ~S⇒S1⇒∗ασ~ and S⇒S2⇒∗βσ~S⇒S2⇒∗βσ~, where α=βα=β are strings of matching αα's and ββ's (since the σ~σ~'s match).

Hence, we've reduced to PCP, and since that's undecidable, we're done.