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 Post subject: Contex free grammar
PostPosted: Thu, 2 Jun 2011 08:08:43 UTC 
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Construct Contex free grammar that recognize the language :
L= { a^{n} b^{m} c^{2k} | m = 2n , k >= 1 }
Can someone help me :)


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 Post subject: Re: Contex free grammar
PostPosted: Thu, 2 Jun 2011 08:21:12 UTC 
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counter wrote:
Construct Contex free grammar that recognize the language :
L= { a^{n} b^{m} c^{2k} | m = 2n , k >= 1 }
Can someone help me :)


Can you construct a rule for a^n b^m? Now stick the c^{2k} at the end.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject:
PostPosted: Thu, 2 Jun 2011 08:43:54 UTC 
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Something like this - > {a^{n}b^{2n}|| S -> E , S -> aSb^{2}}
////{c^{2k}|k >= 1}


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 Post subject:
PostPosted: Thu, 2 Jun 2011 09:37:04 UTC 
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counter wrote:
Something like this - > {a^{n}b^{2n}|| S -> E , S -> aSb^{2}}
////{c^{2k}|k >= 1}


That isn't exactly what we want.

OK, so S\to aSb^2\mid\epsilon generates all a^nb^m that we want. Unfortunately that isn't the whole story, so we can't really use the letter S. Switch that to T, say: T\to aTb^2\mid \epsilon

Now can you cook up a rule for c^{2k}, k\geqslant 1? Say that is the language U.

Finally, glue these pieces together, which forms the entire language we want: S\to TU.

So the end result is three rules: one for how to get T, one for how to get U, and one rule for S.

(By the way, it is context-free grammar)

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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