A→bdA′∣A′ → Here why did this single A’ come that is the A’ in the second production come

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(b) As it is the case of indirect recursion so let first make it as direct recursion then apply rules of removal of left recursion.

to make it as direct recursion first production remain unchanged while in second production substitute the right hand side of first production wherever it comes.In the question $S$ comes in middle of $A$ so substitute the right hand side of production $S$.Now after substituting it looks like:

- $A \rightarrow Ac\mid Aad \mid bd \mid \epsilon$

Now remove direct recursion from it

For removal of direct recursion rule:

- $A \rightarrow A\alpha_1 \mid \ldots \mid A\alpha_n \mid \beta_1 \mid \ldots \mid \beta_m$

Replace these with two sets of productions, one set for $A:$

- $A \rightarrow \beta_1A^\prime \mid \ldots \mid \beta_mA^\prime$

and another set for the fresh nonterminal $A^{\prime}$

- $A^\prime \rightarrow \alpha_1A^\prime \mid \ldots \mid \alpha_nA^\prime \mid \epsilon$

After applying these rule we'll get:

- $A \rightarrow bdA'\mid A'$
- $A' \rightarrow cA'\mid adA' \mid \epsilon$

Now complete production without left recursion is:

- $S \rightarrow Aa \mid b$
- $A \rightarrow bdA'\mid A'$
- $A' \rightarrow cA'\mid adA' \mid \epsilon$

@svas7246

Yes $\epsilon$ is also being considered as $\beta$ for the conversion… If you see the problem a bit intuitively, then you shall get your answer as to why is it working…

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2 votes

- S →Aa∣b
- A →Ac∣Sd∣ϵ

Remove indirect recursion first:-

A-> Ac | Aad | bd |ϵ

I have replace S production with RHS in above . Now we have got the grammer with direct left recursion.

Now let us remove left recursion

S →Aa∣b

A -> bd A' | A'

A' -> cA' | ad A' | ϵ

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@Gate Fever there is a rule of doing it , we number the productions then replace in an order, you can google how to remove indirect left recursion.

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