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Which of the following statements are true?

1. Every left-recursive grammar can be converted to a right-recursive grammar and vice-versa
2. All $\epsilon$-productions can be removed from any context-free grammar by suitable transformations
3. The language generated by a context-free grammar all of whose productions are of the form $X \rightarrow w$ or $X \rightarrow wY$ (where, $w$ is a string of terminals and $Y$ is a non-terminal), is always regular
4. The derivation trees of strings generated by a context-free grammar in Chomsky Normal Form are always binary trees
1. I, II, III and IV
2. II, III and IV only
3. I, III and IV only
4. I, II and IV only

Statement 1 is true: Using GNF we can convert Left recursive grammar to right recursive and by using reversal of CFG and GNF we can convert right recursive to left recursive.

Statement 2 is false: because if $\epsilon$ is in the language then we can't remove  $\epsilon$ production from Start symbol. (For example $L = a^*$)

Statement 3 is true because right linear grammar generates regular set

Statement 4 is true, only two non-terminals are there in each production in CNF. So it always form a binary tree.

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How can we convert using GNF(left recursive grammar to right recursive grammar and vice versa.?
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2nd statement says ϵ  can be removed from Grammar.

Suppose we have a grammar like

S -> AbaC

A -> BC

B -> b/ ϵ

C -> D/ ϵ

D -> d

In this case ϵ  productions can be removed. Correct me if i am wrong..

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@Shamim Ahmed option b says all epsilon can be removed. If language can generate epsilon, how can we remove it? start symbol will always have it to generate epsilon.

1.Every left recursive grammar can be converted to a right recursive grammar and vice versa

yes their is algo.

2. All ∈ production can be removed from any CFG by suitable transformation

NO when language itself contain null then u can't remove null.

3. The language generated by a CFG all whose production are to the form X→ w or X→ wY

yup their is formula for it X-> aT* then regular since right linear form

4.The derivation trees of strings generated by a CFG in CNF are alays binary trees.

yes b/c cfn in form of S->AB or S->a so max two child at a time.

2nd is false

S->∈ if we have this production in our cfg we can remove it if we do so it changes language

All other options are true so ans is c
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but,we have a procedure to remove null production in cfg isnt it???and the qs asked that all epsilon production can be removed or not...
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Yes, we can remove any Null production except in one case if we can derive $\epsilon$ from grammar, we cannot remove that.

eg. $S\rightarrow aS|\epsilon$ generating language $L = a^*$, and if we remove $\epsilon$ from it,
we will get $S\rightarrow aS|a$ generating a different language ,$L = a^+$

Explanation for I.

• Reverse "LLG for L" to get "RLG for LR" by changing A → Ba to A → aB
• Convert "RLG for LR" directly to "FA for LR"
• Reverse "FA for LR" to get "FA for L" by
• Change starting state to final state
• Reverse direction of each transition
• Create a new start state with ϵ transitions to all accepting states
• Convert "FA for L" directly to "RLG for L"

Similar approach can be derived for converting "RLG for L" to "LLG of L".

–1 vote

Statement 1 is true: Using GNF we can convert Left recursive grammar to right recursive and by using reversal of CFG and GNF we can convert right recursive to left recursive.

Statement 3 is true because right linear grammar generates regular set

Statement 4 is true, only two non-terminals are there in each production in CNF. So it always form a binary tree.

Statement 2 is true, we can remove null productions from any context free grammar G producing language L(G) using suitable transformations such that the new grammar G1 will produce language L1(G1).

The new language L1 will produce every string produced by L except for Null string.

But the statement given in question does not mention the new grammar to be equivalent of the original grammar. Therefore statement 2 is also TRUE and the answer is A

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