What is Non-Immediate Left Recursion in compiler design?

A Grammar G (V, T, P, S) is left recursive if it has a production in the form.

A → A α |β.

The above Grammar is left recursive because the left of production is occurring at a first position on the right side of production. It can eliminate left recursion by replacing a pair of production with

A → βA′

A → αA′|ϵ

The general form for left recursion is

A → Aα₁|Aα₂| … . |Aα_m|β₁|β₂| … . . β_n

can be replaced by

A → β₁A′|β₂A′| … . . | … . . |β_nA′

A → α₁A′|α₂A′| … . . |α_mA′|ε

In the following grammar, it does not have the immediate or direct left recursion. But here, it can have non-immediate left recursion.

S → Ab

A → Sc|d

Here, S is left Recursive, because S ⇒ A b ⇒ S c b

Algorithm to Remove Left-Recursion

Input − Grammar G with no cycles or ε − productions.

Output − Equivalent Grammar with no Left Recursion.

Method

• Order the non-terminals as A₁, A₂, … … … . . A_n

for i = 1 to n
{
   for j = 1 to i – 1
   {
      replace each production Ai → Ajγ
      by productions Ai → δ1γ| δ2γ| … … . . |δKγ
      where Aj → δ1|δ2| … … . |δK
   }
Remove immediate left Recursion among Ai productions.
}

Example1 − Remove or Eliminate Left Recursion from the following Grammar.

S → Ab

A → Sc | d

Solution

These productions don’t have immediate left Recursion. So it cannot be eliminated directly. We can use the algorithm here to remove Left Recursion.

Step1 − Order the non-terminals S, A as A₁, A₂i.e., Mark S =A₁, A = A₂

Substituting values

∴   A₁ → A₂b                             …………… (1)

     A₂ → A₁c|d                           …………… (2)

Step2 − Put values of (1) in (2)

∴  A₁ → A₂b

    A₂ → A₁bc|d

Step3 − Again substitutes A₁ = S and A₂ = A

∴  S → A b                              ……………. (3)

    A → A b c | d                        ……………. (4)

Step4 − Remove Immediate Left recursion in A → A b c | d

So,

Example2 − Remove left recursion from the following Grammar.

S → Aa |b

A → Ac | Sd|e

Solution

As we don’t have immediate left recursion so, we have to apply the algorithm here.

Step1 − Rename S, A as A₁, A₂ respectively.

A₁ → A₂ a | b                        …………… (1)

A₂ → A₂c | A1d|e                  ……………. (2)

Step2 − Substitute value of A₁ of the statement (1) in R.H.S of the statement (2).

A₁ → A₂ a | b

A₂ → A₂ c |A₂ a d|bd|e.

Step3 − Again, substitute A₁ = S, A₂ = A

S → Aa |b                               …………….. (3)

A → Ac |A a d|bd|e                   ……………… (4)

Step4 − Statement (4) has an immediate left recursion. Removing immediate left recursion we get