The time bounds for the parser are the same as those of the recognizer, while the space bound goes up to n 3 in general in order to store the parse trees. In fact it would probably be most efficient to implementthe algorithm to do just a look-ahead of 1. To implement the full look-ahead for any k would be more costly inprogramming effort and less efficient overall since so few programming languages need the extra lookahead. Improving transformations are performed on a target-specific representation of the program. for both existing and proposed architectures, to gather trace information for code improvements are often applied to a high-level intermediate language though it can be applied to the source level also but it can impact the effectiveness of code improvements
I. Statement is true since there are some parsers which take O(n log2n) time for parsing.
II. Completely false, since there is no use of stack which is required for recursion.
III. False
IV. True since both types of optimizations are applied

Correct Option: B

The time bounds for the parser are the same as those of the recognizer, while the space bound goes up to n 3 in general in order to store the parse trees. In fact it would probably be most efficient to implementthe algorithm to do just a look-ahead of 1. To implement the full look-ahead for any k would be more costly inprogramming effort and less efficient overall since so few programming languages need the extra lookahead. Improving transformations are performed on a target-specific representation of the program. for both existing and proposed architectures, to gather trace information for code improvements are often applied to a high-level intermediate language though it can be applied to the source level also but it can impact the effectiveness of code improvements
I. Statement is true since there are some parsers which take O(n log2n) time for parsing.
II. Completely false, since there is no use of stack which is required for recursion.
III. False
IV. True since both types of optimizations are applied

It is a production p that may be used for reduction in afuture step along with a position in the sentential form where, the right hand side of the production may befound. A “handle” of a string is a substring that matches the RHS of a production and whose reduction to the nonterminal (on the LHS of the production) represents one step along the reverse of a rightmost derivation toward reducing to the start symbol. If S → * αAw → * αbw, then A → β in the position following a is a handle of αβw. In such a case, it is suffice to say that the substring β is a handle of αβw, if the position of β and the corresponding production are clear. Consider the following grammar: E → E + E | E * E | (E) | id and a right-most derivation is as follows: E → E + E → E + E * E → E + E * id3 → E + id2 * id3 → id1 + id2 * id3

Correct Option: D

It is a production p that may be used for reduction in afuture step along with a position in the sentential form where, the right hand side of the production may befound. A “handle” of a string is a substring that matches the RHS of a production and whose reduction to the nonterminal (on the LHS of the production) represents one step along the reverse of a rightmost derivation toward reducing to the start symbol. If S → * αAw → * αbw, then A → β in the position following a is a handle of αβw. In such a case, it is suffice to say that the substring β is a handle of αβw, if the position of β and the corresponding production are clear. Consider the following grammar: E → E + E | E * E | (E) | id and a right-most derivation is as follows: E → E + E → E + E * E → E + E * id3 → E + id2 * id3 → id1 + id2 * id3

LALR parser is reduced form of CLR or LR(1) parser, LALR parser uses the LR(1) items of CLR parser & of any shift reduce conflicts are there then it is due to LR(1) parser.

Correct Option: B

LALR parser is reduced form of CLR or LR(1) parser, LALR parser uses the LR(1) items of CLR parser & of any shift reduce conflicts are there then it is due to LR(1) parser.

It is very clear from the previous solution that the no. of steps required depend upon the no. of a' s & b ' s which ever is higher & exceeds by 2 due to S " AC CB & C"!
So max(l, m) + 2
Hence (a) is correct option.

Correct Option: A

It is very clear from the previous solution that the no. of steps required depend upon the no. of a' s & b ' s which ever is higher & exceeds by 2 due to S " AC CB & C"!
So max(l, m) + 2
Hence (a) is correct option.

In the language language L = {aⁱ b^j | i ≠ j} The expression i ≠ j indicates that language L is a language in which number of a ≠ number of b. Here, we need to search for a language in which number of a ≠ number of b.
Considering the options:
Options (a)
S → AC | CB
C → aCb | a| b
A → aA | ∈
B → Bb | ∈
(a), S → (C) B → a(b) → aBb → ab
Here, the number of a = number of b, hence, it is not the required grammar.
Option (b)
S → as | Sb | a | b
S → as → abB
Here, number of a = number of b, hence it is not the required grammar.
Option (c)
S → AC | CB
C → aC b| ∈
A → aA | ∈
B → Bb | ∈
S(C)B → acbB → ab
Here, number of a = number of b, hence it is not the required grammar.
Option (d)
S → AC | CB
C → aC b| ∈
A → aA | a
B → Bb | b
Consider the production
S → AC | CB
Applying, C → ∈
We get S → A|B
Now, applying A or B, S generates a^n.b^n. So, L = {aⁱ b^j | i ≠ j}
Now, consider the production, S → AC | CB. If the production C → acb is applied then, it generates a^n,b^n.
Now, applying either A or B it will give that number of a ≠ number of b.

Correct Option: D

In the language language L = {aⁱ b^j | i ≠ j} The expression i ≠ j indicates that language L is a language in which number of a ≠ number of b. Here, we need to search for a language in which number of a ≠ number of b.
Considering the options:
Options (a)
S → AC | CB
C → aCb | a| b
A → aA | ∈
B → Bb | ∈
(a), S → (C) B → a(b) → aBb → ab
Here, the number of a = number of b, hence, it is not the required grammar.
Option (b)
S → as | Sb | a | b
S → as → abB
Here, number of a = number of b, hence it is not the required grammar.
Option (c)
S → AC | CB
C → aC b| ∈
A → aA | ∈
B → Bb | ∈
S(C)B → acbB → ab
Here, number of a = number of b, hence it is not the required grammar.
Option (d)
S → AC | CB
C → aC b| ∈
A → aA | a
B → Bb | b
Consider the production
S → AC | CB
Applying, C → ∈
We get S → A|B
Now, applying A or B, S generates a^n.b^n. So, L = {aⁱ b^j | i ≠ j}
Now, consider the production, S → AC | CB. If the production C → acb is applied then, it generates a^n,b^n.
Now, applying either A or B it will give that number of a ≠ number of b.