a ^ nb ^ m的上下文无关文法

Question

I'm trying to make a CFG for the language:

I did this:

S -> a S b b 
S -> epsilon

It fulfills the requirement, but is it correct?

Answer 1

What if I wanted abbb ? n is 1, therefore 2n is 2 and 4n is 4, m is 3, so 2n <= m < 4n . If you mean "does it create all the words in the language" as "is it correct", then no, it does not. Also, you are creating epsilon (zero a and zero b ), but it's not correct that 2*0 <= 0 < 4*0 because zero is not less than zero (second part of the comparison).

Your grammar is creating only words where there is twice more b than a . However, the language contains also other words - you need to have between 2 times more b (including) and less than 4 times more b than a . Eg for n = 2 your m can be 4, 5, 6 or 7.