# Eliminating Epsilon Production for Left Recursion Elimination

Im following the algorithm for left recursion elimination from a grammar.It says remove the epsilon production if there is any

I have the following grammer

S-->Aa/b

A-->Ac/Sd/∈


I can see after removing the epsilon productions the grammer becomes

  1) S-->Aa/a/b

2)A-->Ac/Sd/c/d


Im confused where the a/b comes in 1) and c/d comes in 2) Can someone explain this?

A is nullable, that is, A has a production which is epsilon.

Given that A is nullable, you have to revisit any productions containing one or more instances of A and replace them with the set of productions formed from every possible replacement of A with null, including no instances replaced and with all instances replaced.

So looking at the production S->Aa we need to replace it with the production where A is not replaced (the same production) and the production where A is replaced with null (which is s->a). Now there's three total productions for S.

S-> Aa/a/b

When we look at the production A-> Ac we have to do the same thing, so Ac is replaced with Ac/c and we get...

A->Ac/c/Sd

You could rearrange it as necessary.

Sd does NOT become replaced with Sd/d because S is not nullable. There is no way to produce epsilon from S, so we don't replace S in productions.