从 {a, b}* 的 {w 元素构造 CFG : 2#a(w)=3#b(w)}

Question

If i have following language { x is element of {a,b}*, where 2#a(x)=3#b(x), then the cfg of that language is :

S=>SaSaSaSbSbS |SaSaSbSaSbS|SaSaSbSbSaS | SaSbSaSaSbS| SaSbSaSbSaS | SaSbSbSaSaS |SbSaSaSaSbS |SbSaSaSbSaS |SbSaSbSaSaS |SbSbSaSaSaS | epsilon/lambda

Is this correct? If this isnt correct/there's another more simple form, can you tell it? I have no clue on another form other than that.

Answer 1

At a glance it looks like this probably works:

1. your base case is good; the empty string is in the language

2. you cover all your inductive cases: you only add 2 a and 3 b and you cover all arrangements

I am not seeing a fundamentally simpler solution than this, although you might be able to remove either the leading or the trailing S from the right-hand side of all productions; then, by choosing a production you'd be committing to that first or last terminal symbol, but I think that still works out. Possibly even removing both leading and trailing S so you commit to both the first and the last. Any other simplification seems like it would increase the number of productions or the number of nonterminals, or both, which while possibly reducing the total number of symbols needed to encode the grammar, arguably doesn't make the grammar any simpler (indeed, more nonterminals and productions is typically seen as more complicated, not less). If you want to experiment with adding productions or nonterminals, consider eg T => Sa and R => Sb, just to cut down on repetition.