Question

Consider the pushdown automation (PDA) below which runs over the input alphabet (a, b, c). It has the stack alpabet $\{Z_o,X\}$ where $Z_o$ is the bottom-of-stack marker. The set of states of the PDA is (s, t, u, f) where s is the start state and f is the final state. The transitions of the PDA given below are depicted in a standard manner. For example, the transition (s, b, X) $\to$ $(t,X{Z}_o)$ means that if the PDA is in states and the symbol on the top of the stack is X, then it can read b from the input and move to state t after popping the top of stack and pushing the symbols $Z_o$ and X (in that order) on the stack.
$(s,\in ,Z_o\to (f,\in )$
$(s,a,Z_o)\to (S,XX{Z}_o)$
$(s,a,X)\to (S,XXX)$
$(s,b,X)\to (t,\in )$
$(t,b,x)\to (t,\in )$
$(t,c,x)\to (u,\in )$
$(u,c,x)\to (u,\in )$
$(u,\in ,Z_o)\to (f,\in$)
The language accepted by the PDA is

Answer 1

For every a we put two X in stack [at state s] After that for every b we pop out one X[reach to state t (getting b after a)]

After that for every c we pop out one X [reach to state u (getting c after b)]

If all X are popped out then reached to final state f, mean for every b there is a, for every c there is a .

a was followed by b and b was followed by c [ state s to t, t to u, u to f, final]

means sum of no of b's and no of c's == twice of no of a's[ one a for one b , one a for one c ]

i.e. $\{a^l{b}^m{c}^n|2l=m+n\}$
So, option (c) is correct.