Re: languages

"jason_box" <cppisfun@xxxxxxxxx> writes:

Here is the grammar I came up with.

S-> aBa
B-> b | BB | S | lambda

for the gramma L = {a^nb*a^n : a >= 0 }

That is definitely wrong, as you can derive, e.g.:

S => aBa => aBBa => aSBa => aSSa => aaBaSa => aaBaaBaa
=> aabaaBaa => aabaabaa

which is not in the language.

The problem is the production B->S, so drop this and change the
production(s) for S to allow repeated a's.

Torben
.

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