问题
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?
回答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.
来源：https://stackoverflow.com/questions/21312669/eliminating-epsilon-production-for-left-recursion-elimination