Re: languages

quote:

Try starting with this production. This generates the language
L={a^(2n): n >= 0}

S->aSa | lambda


So from this I know the langage is equal number of a's. Therefore
aa,aaaa,etc will be accepted. So from here is where I need to add the
b* so that it will accept the first langauge with addition of b* in the
middle.

From that I believe the correct grammar is of the form:

S-> aSa | B | lambda
B-> bB | lambda

.

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