How is PDA to CFG conversion done? Consider a PDA that accepts by empt | Theory of Computation 2080

Theory of Computation20805 marksNumerical: PDA to CFG conversionConversion of PDA to CFG

How is PDA to CFG conversion done? Consider a PDA that accepts by empty stack, Now construct an equivalent CFG. $P = ((p,q);\{0,1\},\{Z\},\delta,p,Z);$ $\delta(p,0,Z)=(p,0z), \delta(p,0,0)=(p,00), \delta(p,1,0)=(p,\varepsilon), \delta(p,\varepsilon,z)=(q,\varepsilon)$ [5]

How is PDA to CFG conversion done? Consider a PDA that accepts by empty stack, Now construct an equivalent CFG. $P = ((p,q);{0,1},{Z},\delta,p,Z);$ $\delta(p,0,Z)=(p,0z), \delta(p,0,0)=(p,00), \delta(p,1,0)=(p,\varepsilon), \delta(p,\varepsilon,z)=(q,\varepsilon)$ [5]

PDA $P = (\{p, q\}, \{0,1\}, \{Z, 0\}, \delta, p, Z)$ accepting by empty stack. Transitions: - $\delta(p, 0, Z) = (p, 0Z)$ - $\delta(p, 0, 0) = (p, 00)$ - $\delta(p, 1, 0) = (p, \varepsilon)$ - $\delta(p, \varepsilon, Z) = (q, \varepsilon)$ (Stack alphabet includes both $Z$ and $0$ since $0$ is p...

PDA $P = ({p, q}, {0,1}, {Z, 0}, \delta, p, Z)$ accepting by empty stack. Transitions: - $\delta(p, 0, Z) = (p, 0Z)$ - $\delta(p, 0, 0) = (p, 00)$ - $\delta(p, 1, 0) = (p, \varepsilon)$ - $\delta(p, \varepsilon, Z) = (q, \varepsilon)$ (Stack alphabet includes both $Z$ and $0$ since $0$ is p...

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How is PDA to CFG conversion done? Consider a PDA that accepts by empty stack, Now construct an equivalent CFG. P = ((p,q);0,1,Z,delta,p,Z); delta(p,0,Z)=(p,0z), delta(p,0,0)=(p,00), delta(p,1,0)=(p,varepsilon), delta(p,varepsilon,z)=(q,varepsilon)