E → E - T |
T → T + F | F
F → ( E ) | id
Using the rule for removal of left recursion is
A → Aα / β
A → βA '
A ' → α A ' / ∈
Then, the given grammar is written as :-
E' → −TE' / ∈
E → +TE '
T ' → + FT ' / ∈
T → FT '
F → (E) | id
Now by putting E ' as X and T ' as Y, then
X → −TX / ∈
E → TX
Y → +FY / ∈
T → FY
F → (E) | id
Hence option (c) is correct.

Correct Option: C

E → E - T |
T → T + F | F
F → ( E ) | id
Using the rule for removal of left recursion is
A → Aα / β
A → βA '
A ' → α A ' / ∈
Then, the given grammar is written as :-
E' → −TE' / ∈
E → +TE '
T ' → + FT ' / ∈
T → FT '
F → (E) | id
Now by putting E ' as X and T ' as Y, then
X → −TX / ∈
E → TX
Y → +FY / ∈
T → FY
F → (E) | id
Hence option (c) is correct.

The number of control Paths is depends on number of if statement that is 2ⁿ, where, n is the number. of if statements For (2 “if statements”) 2² = 4 control flow Paths are possible :

These four control flow are :
(1) e₁ → e₂
(2) e₁ → e₃
(3) (e₂ if e₄ ) → e₅
(4) (e₃ if e₄ ) → e₅
So, for (10 “if statements”), 2¹⁰ = 1024.
Control flow path are possible. Hence, answer is 1024.

Correct Option: A

The number of control Paths is depends on number of if statement that is 2ⁿ, where, n is the number. of if statements For (2 “if statements”) 2² = 4 control flow Paths are possible :

These four control flow are :
(1) e₁ → e₂
(2) e₁ → e₃
(3) (e₂ if e₄ ) → e₅
(4) (e₃ if e₄ ) → e₅
So, for (10 “if statements”), 2¹⁰ = 1024.
Control flow path are possible. Hence, answer is 1024.

The context free grammar given over alphabets
∑ = {a, b, c}, with S and T are nonterminals.
Given, G₁ : S → aSb | T , T → cT | ∈
G₂ : S → bSa | T , T → cT | ∈
Lets L(G₁) is the language for Grammar (G₁) and L(G₂) is the language for Grammar (G₂).
where L(G₁) = { aⁿ c^m bⁿ | m , n ≥ 0 }
L(G₂) = { bⁿ c^m aⁿ | m , n ≥ 0 }
then L(G₁) ∩ L(G₂) = { aⁿ c^m bⁿ } ∩ { bⁿ c^m aⁿ }
= { c^m | m ≥ 0 } = C^*
Since the only common strings will be those strings with only ‘C’, so the intersection is not finite but regular.

Correct Option: B

The context free grammar given over alphabets
∑ = {a, b, c}, with S and T are nonterminals.
Given, G₁ : S → aSb | T , T → cT | ∈
G₂ : S → bSa | T , T → cT | ∈
Lets L(G₁) is the language for Grammar (G₁) and L(G₂) is the language for Grammar (G₂).
where L(G₁) = { aⁿ c^m bⁿ | m , n ≥ 0 }
L(G₂) = { bⁿ c^m aⁿ | m , n ≥ 0 }
then L(G₁) ∩ L(G₂) = { aⁿ c^m bⁿ } ∩ { bⁿ c^m aⁿ }
= { c^m | m ≥ 0 } = C^*
Since the only common strings will be those strings with only ‘C’, so the intersection is not finite but regular.

The given grammar with productions,
S → SaS | aSb | bSa | SS | ∈
Now consider,
S → aSb | bSa | SS | ∈
This grammar generates all strings with equal number of ‘a’ and ‘b’.
Now, S → SaS can only generate strings where ‘a’ is more than ‘b’. Since on left and right of ‘a’ in SaS, S will have only strings with n_a = n_b or n_a > n_b.
Now consider each options
1. S → SS → aSbS → abS → abaSb → abab ,when, S → ∈
2. S → aSb → aSaSb → aaaSb → aaab , when S → ∈
3. S → SS → aSbS → abS → abbSa → abbSaSa → abbaa, when ( S → ∈ )
4. “babba” which is a string with nb > na is not possible to generate by the given grammar.
Hence, option (d) is correct.

Correct Option: D

The given grammar with productions,
S → SaS | aSb | bSa | SS | ∈
Now consider,
S → aSb | bSa | SS | ∈
This grammar generates all strings with equal number of ‘a’ and ‘b’.
Now, S → SaS can only generate strings where ‘a’ is more than ‘b’. Since on left and right of ‘a’ in SaS, S will have only strings with n_a = n_b or n_a > n_b.
Now consider each options
1. S → SS → aSbS → abS → abaSb → abab ,when, S → ∈
2. S → aSb → aSaSb → aaaSb → aaab , when S → ∈
3. S → SS → aSbS → abS → abbSa → abbSaSa → abbaa, when ( S → ∈ )
4. “babba” which is a string with nb > na is not possible to generate by the given grammar.
Hence, option (d) is correct.

Follow (Q) ?
Follow (Q) is First (R) Hence First (R) = {W, E}.
We add ‘W’ in Follow (Q) and for ∑.
We calculate :
First (S) = {y}
So, follow (Q) = First (R S)
= {{W} – ∑} ∪ First (S)
Follow (Q) = {W, y} ( ∴ First (S) = {y} )
Hence option (c) is correct.

Correct Option: C

Follow (Q) ?
Follow (Q) is First (R) Hence First (R) = {W, E}.
We add ‘W’ in Follow (Q) and for ∑.
We calculate :
First (S) = {y}
So, follow (Q) = First (R S)
= {{W} – ∑} ∪ First (S)
Follow (Q) = {W, y} ( ∴ First (S) = {y} )
Hence option (c) is correct.