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Identify the FALSE statement about following CFG: S -> SB|AB A -> CC B -> b C -> a
just 1 mint remainig
CFG generates NULL string
CFG is not in CNF
CFG has 8 Nonterminals
All of the given options
Between the two consecutive joints on a path
One character can be pushed and one character can be popped
Any no. of characters can be pushed and one character can be popped
One character can be pushed and any no. of characters can be popped
Any no. of characters can be pushed and any no. of characters can be popped
In new format of FA (discussed in lecture 37), This state is like initial state of an FA
Consider the following CFG: (NOTE: ^ means NULL) S->a|Xb|aYa X->Y|^ Y->b|X Which Nonterminals are nullable
S and X
X and Y
Y and S
S,X and Y
which path sequence follows the rules of "conversion form" of "PDA"
READ -> POP -> POP
POP -> POP -> POP
READ -> POP -> PUSH a
In the null production N --> ^, N is a
Consider the Following CFG: (NOTE: ^ means NULL) S->Xa X->aX|bX|^ above given CFG can be represented by RE
Before the CFG corresponding to the given PDA is determined, the PDA is converted into the standard form which is called.
Chomsky Normal Form (CNF)
none of given
Tape and Stack alphabets
may be different
must be different
must be same
thank you nayab sis
CFG no 2 only