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Let the alphabet = {a, b}. Part (a): Give a Context Free Grammar for the language {a^nb^m  n > 2m}.
I have NO clue at all.
 one year ago
 one year ago
Let the alphabet = {a, b}. Part (a): Give a Context Free Grammar for the language {a^nb^m  n > 2m}. I have NO clue at all.
 one year ago
 one year ago

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Tomas.ABest ResponseYou've already chosen the best response.4
if zero b's then there must be at least one a if one b then there must be at least three a's if two b's then tehre must be at least five a's and so on
 one year ago

Tomas.ABest ResponseYou've already chosen the best response.4
also first should come a's and only then b's
 one year ago

Tomas.ABest ResponseYou've already chosen the best response.4
so we start with S > aQT (we must have at least one a) Q > aQ  ε (we can generate any number of a's before b) T > aaTb  ε (if we create at least one b we must put two a's before b)
 one year ago

seanwalsh1984Best ResponseYou've already chosen the best response.0
Wow! Thanks! Here is what I came up with this morning toying around. \[S_0 \rightarrow aaS_1b  aaS_1b  aS_1\] \[S_1 \rightarrow S_0\epsilon  a\] Is this equivalent?
 one year ago

Tomas.ABest ResponseYou've already chosen the best response.4
no, for example you can't get just a, but according to rules you should be able to however maybe you made mistake by typing because if it will be S_1 > S_0ϵa i think it would be ok and aaS_1b is same as aaS_1b so why you mention it 2 times? :D
 one year ago

seanwalsh1984Best ResponseYou've already chosen the best response.0
Yes you are right, I mistyped, and my computer was acting up. I did have S1>S0  epsilon  a. Thanks for the help!
 one year ago
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