Eliminating Epsilon Production for Left Recursion Elimination

Question

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?

Answer 1

lets look at the rule S->Aa , if A->∈ then S->∈a giving just S->a , so together with the previous rules we get S->Aa|a|b

now lets check the rule A->Ac and A->∈c which gives us A->c .

what about A->Sd ? I dont see how you got A->d as a rule. if that is a rule, then the string "da" is accepted by this grammar ( S->Aa & A->d --> "da" ), but try to construct this string with the original grammar - if you start with S and the string finishes with a , it means you must use S->Aa , but then in order to have a "d" you must use A->Sd , which forces us to have another "a" or "b" , meaning we cannot construct this string, and the rule A->d is not correct.