Question

Let S = {0,1}. Show that the problem of determining whether a CFG generates some string in 1* is decidable. In other words, show that { <G>G is a CFG over {0,1} and 1* n L(G) != 0 }

Answer 1

*We have to show that the problem of determining whether a CFG generates some string 1* is decidable

Given a CFG over (0,1) that generates some string in 1. The language generated by this CFG, denoted by Language A can be rephrased in the following way: A={<G>G is a CFG over (0,1) and n L(G)=0} • To test decidability, the fact that intersection of a Regular Language (RL) & Context Free Language (CFL) is a CFL shall be harnessed.

. Note that 1 is a Regular Language & L(G)is a CFL (since G is a CFG), therefore, in L(G) is clearly a CFL • A Turing Machine (TM) has the task to test whether the problem at hand is decidable or undecidable. • If for an input, TM culminates in either ACCEPT or REJECT state, the problem is referred to as decidable.

Let H be the TM that decides A. Using the Theorem 4.8, the decidability is employed by the following algorithm: H="on input <G>, where G is a CFG": Construct a CFGB, such that L(B)=1*nL(G) (remember that 1*nL(G) is CFL so the statement is valid). 1. Run the TM R that decides Ecro on <B>. It may be elaborated here that Ecro denotes the problem of determining whether a CFG (here B) generates any strings at all is decidable. Formally, Ecrc = {<B>| B is a CFG and L(B)=0} 2. If R accepts, reject. It means that the language generated by the CFG B is empty. Therefore, TM H culminates in REJECT state. 3. If R rejects, accept. In other words, the problem at hand is Decidable since language generated by the CFG B is NOT empty. Elaboration: If TM Raccepts, it would imply that, L(B)=1*nL(G) That, is for some string wel, wę LG), hence R should reject.