Outline 1 PCP. 2 Decision problems about CFGs. M.Mitra (ISI) Post s Correspondence Problem 1 / 10

Transcription

1 Outline 1 PCP 2 Decision problems about CFGs M.Mitra (ISI) Post s Correspondence Problem 1 / 10

2 PCP reduction Given: M, w in encoded form To construct: an instance (A, B) of MPCP such that M accepts w if and only if (A, B) has a solution. Preliminaries: Assume w.l.o.g. that M never prints a blank; M never moves left from the first tape square (i.e. never hangs). Starting configuration: (q 0, #w) Representation of configuration: αqβ (If head is on a blank, i.e. at the end of the written portion of the tape, β = ε.) M.Mitra (ISI) Post s Correspondence Problem 2 / 10

3 PCP instance List A List B Group I q o w Group II a a for any a Σ Group IIIA qa pb if b Σ when δ(q, a) = (p, b) qa ap if b = R cqa pca if b = L Group IIIB q pb if b Σ when δ(q, #) = (p, b) q # p if b = R cq pc if b = L M.Mitra (ISI) Post s Correspondence Problem 3 / 10

4 PCP instance List A List B Group IV ahb h ah h hb h Group V h M.Mitra (ISI) Post s Correspondence Problem 4 / 10

5 PCP is undecidable Lemma: M accepts w if and only if the MPCP instance (A, B) above has a solution. Proof sketch: ( ) 1 Start with pair from group I. 2 Use pairs from group III to simulate one move in the region around the head position; use pairs from group II to copy rest of the tape contents. 3 If M halts, use pairs from group IV to progressively erase the tape in the B string. A string:... a 1... a i ha i+1... B string:... a 1... a i ha i+1... a 1... a i 1 ha i Terminate with pair from group V. M.Mitra (ISI) Post s Correspondence Problem 5 / 10

6 PCP is undecidable Proof sketch (contd.): ( ) Solution must begin with: A string = B string = q o w. Partial solutions have the form: A string = x B string = xy where x = sequence of configurations separated by, last one (w C, say) may be incomplete y = computation upto and including w C followed by prefix of a configuration that follows w C Partial solution can be completed only if M halts; otherwise, B string is always longer. M.Mitra (ISI) Post s Correspondence Problem 6 / 10

7 Outline 1 PCP 2 Decision problems about CFGs M.Mitra (ISI) Post s Correspondence Problem 7 / 10

8 Ambiguity of CFGs Given: A = w 1, w 2,..., w k, B = x 1, x 2,..., x k. Reduction: Let G A : A w 1 Aa 1... w k Aa k w 1 a 1... w k a k and G B : B x 1 Ba 1... x k Ba k x 1 a 1... x k a k Let G AB : S A B... Then G AB is ambiguous if and only if (A, B) has a solution. M.Mitra (ISI) Post s Correspondence Problem 8 / 10

9 Ambiguity of CFGs Given: A = w 1, w 2,..., w k, B = x 1, x 2,..., x k. Reduction: Let G A : A w 1 Aa 1... w k Aa k w 1 a 1... w k a k and G B : B x 1 Ba 1... x k Ba k x 1 a 1... x k a k Let G AB : S A B... Then G AB is ambiguous if and only if (A, B) has a solution.. List languages. Ḷ(G A ) and L(G B ) are called list languages. M.Mitra (ISI) Post s Correspondence Problem 8 / 10

10 List languages Lemma: If L A is a list language, then L A is a CFL. Proof: Let P be a (D)PDA that works as follows. 1 On symbols in Σ, P pushes the input symbol on stack. 2 On symbols a 1,..., a k, it pops the stack to see if the top of the stack contains w R i. 3 If input and stack are both empty, reject. 4 In all other cases, accept. M.Mitra (ISI) Post s Correspondence Problem 9 / 10

11 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? M.Mitra (ISI) Post s Correspondence Problem 10 / 10

12 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? Let L(G 1 ) = L A, L(G 2 ) = L B. M.Mitra (ISI) Post s Correspondence Problem 10 / 10

13 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? Let L(G 1 ) = L A, L(G 2 ) = L B. L(G 1 ) = L(G 2 )? M.Mitra (ISI) Post s Correspondence Problem 10 / 10

14 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). M.Mitra (ISI) Post s Correspondence Problem 10 / 10

15 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). L(G 1 ) = L(R)? M.Mitra (ISI) Post s Correspondence Problem 10 / 10

16 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? L(G 1 ) = L(R)? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). Let L(G 1 ) = L A L B, R = (Σ I). M.Mitra (ISI) Post s Correspondence Problem 10 / 10

17 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? L(G 1 ) = L(R)? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). Let L(G 1 ) = L A L B, R = (Σ I). L(G 1 ) L(G 2 )? M.Mitra (ISI) Post s Correspondence Problem 10 / 10

18 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? L(G 1 ) = L(R)? L(G 1 ) L(G 2 )? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). Let L(G 1 ) = L A L B, R = (Σ I). Let L(G 2 ) = L A L B, L(G 1 ) = (Σ I). M.Mitra (ISI) Post s Correspondence Problem 10 / 10

19 CFG related problems Theorem: Let G 1 and G 2 be CFGs, and R a regular expression. Then the following are undecidable. L(G 1 ) L(G 2 ) =? L(G 1 ) = L(G 2 )? L(G 1 ) = L(R)? L(G 1 ) L(G 2 )? Let L(G 1 ) = L A, L(G 2 ) = L B. Let L(G 1 ) = L A L B, L(G 2 ) = (Σ I). Let L(G 1 ) = L A L B, R = (Σ I). Let L(G 2 ) = L A L B, L(G 1 ) = (Σ I). L(R) L(G 1 )? M.Mitra (ISI) Post s Correspondence Problem 10 / 10