Question

Construct a PDA for the context free grammar: S → AxA | By | zC

A → B | a

B → C | b

C → SS | c | ε

Answer 1

Answer:------------

Algorithm to find PDA corresponding to a given CFG:------------
Input − A CFG, G = (V, T, P, S)
Output − Equivalent PDA, P = (Q, ∑, S, δ, q0, I, F)

Step 1 − Convert the productions of the CFG into GNF.

Step 2 − The PDA will have only one state {q}.

Step 3 − The start symbol of CFG will be the start symbol in the PDA.

Step 4 − All non-terminals of the CFG will be the stack symbols of the PDA and all the terminals of the CFG will be the input symbols of the PDA.

Step 5 − For each production in the form A → aX where a is terminal and A, X are combination of terminal and non-terminals, make a transition δ (q, a, A).

Given, Problem:--------------
Construct a PDA from the following CFG.
G = ({S, A,B,C}, {a, b, c, x, y, z}, P, S)
where the productions are −
S → AxA | By | zC

A → B | a

B → C | b

C → SS | c | ε

Solution:--------
Let the equivalent PDA,

P = ({q}, {a, b, c, x, y, z}, {a, b, c, x, y, z, A, B, C, S}, δ, q, S)

where δ −

δ(q, ε , S) = { (q, AxA), (q, By), (q, zC) }

δ(q, ε , A) = { (q, B), (q, a) }

δ(q, ε , B) = { (q, C), (q, b) }

δ(q, ε , C) = { (q,SS), (q, c), (q, ε) }

δ(q, a, a) = { (q, ε ) }

δ(q, b, b) = { (q, ε ) }\

δ(q, c, c) = { (q, ε ) }

δ(q, x, x) = { (q, ε ) }

δ(q, y, y) = { (q, ε ) }

δ(q, z, z) = { (q, ε ) }