Sketch of Solutions to Exercises

3.6.1 Similar to 2.5.8.

3.6.2 Given an instance (M, x) of the membership problem a finite state automaton Ax can be constructed to accept just x. By the proof of Theorem 3.5.1 a pushdown automaton M x can be constructed to accept L(M )  /~\ (L(Ax). Now M accepts x if and only if L(M x) = Ø. Hence, the decidability of the membership problem for pushdown automata follows from the decidability of the emptiness problem for pushdown automata (Theorem 3.6.1).

3.6.3 From a given finite state transducer M a pushdown automaton A can be constructed such that A accepts an empty language if and only if M has at most one output on each input. The pushdown automaton A on a give input x nondeterministically simulates two computations of M as input x. The pushdown automaton A discards the outputs of M, but keeps track of the difference in their lengths. The pushdown automaton A accepts x if and only if it determines that the simulated computations produce outputs of either differnt lengths or of different character at a given location.

3.6.4 A pushdown automaton A can be constructed to accept an input of the form x#yrev if and only if the given finite state transducer M has output y or input x.