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Replies to This Discussion

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

ACCEPT
REJECT
START
READ

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

Terminal
Non terminal
Word
Semi word

Consider the Following CFG: (NOTE: ^ means NULL) S->Xa X->aX|bX|^ above given CFG can be represented by RE
A*B*
a*b*a
(a+b)*a
a(a+b)*a

Before the CFG corresponding to the given PDA is determined, the PDA is converted into the standard form which is called.
Finite Automaton
Chomsky Normal Form (CNF)
Conversion form
none of given

Tape and Stack alphabets
are same
may be different

must be different
must be same

Attachments:

thank you nayab sis 

S

CFG no 2 only 

4th 

2nd 

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