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User:Jon Awbrey/SANDBOX (view source)

Revision as of 22:14, 19 January 2009

105 bytes added ,  22:14, 19 January 2009
→‎Grammar Stuff
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<pre>
 
<pre>
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1.  The parse of the concatenation Conc^k of the k sentences S_j,
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Table 12.  Algorithmic Translation Rules
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    for j = 1 to k, is defined recursively as follows:
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o------------------------o---------o------------------------o
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|                        | Parse |                        |
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    a. Parse(Conc^0)        =  Node^0.
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| Sentence in PARCE      |  -->  | Graph in PARC          |
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o------------------------o---------o------------------------o
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    b.  For k > 0,
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|                        |        |                        |
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| Conc^0                 |  -->  | Node^0                 |
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        Parse(Conc^k_j S_j)  =  Node^k_j Parse(S_j).
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|                        |        |                        |
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| Conc^k_j S_j         |  -->  | Node^k_j Parse(S_j)   |
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2.  The parse of the surcatenation Surc^k of the k sentences S_j,
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|                        |        |                        |
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    for j = 1 to k, is defined recursively as follows:
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| Surc^0                 |  -->  | Lobe^0                 |
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|                        |        |                        |
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    a.  Parse(Surc^0)        =  Lobe^0.
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| Surc^k_j S_j         |  -->  | Lobe^k_j Parse(S_j)   |
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|                        |        |                        |
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    b.  For k > 0,
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o------------------------o---------o------------------------o
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  −
        Parse(Surc^k_j S_j)  =  Lobe^k_j Parse(S_j).
   
</pre>
 
</pre>
   −
<ol style="list-style-type:decimal">
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{| align="center" border="1" cellpadding="8" cellspacing="0" style="text-align:center; width:96%"
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|
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<li>The parse of the concatenation <math>\operatorname{Conc}_{j=1}^k</math> of the <math>k\!</math> sentences <math>(s_j)_{j=1}^k</math> is defined recursively as follows:</li>
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{| align="center" border="0" cellpadding="8" cellspacing="0" style="text-align:center; width:96%"
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| &nbsp;
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<ol style="list-style-type:lower-alpha">
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| From
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| <math>(A)\!</math>
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<li><math>\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.</math>
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| and
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| <math>(\operatorname{d}A)\!</math>
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<li>
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| infer
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<p>For <math>k > 0,\!</math></p>
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| <math>(A)\!</math>
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| next.
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<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
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| &nbsp;
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|-
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</ol>
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| &nbsp;
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| From
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<li>The parse of the surcatenation <math>\operatorname{Surc}_{j=1}^k</math> of the <math>k\!</math> sentences <math>(s_j)_{j=1}^k</math> is defined recursively as follows:</li>
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| <math>(A)\!</math>
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| and
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<ol style="list-style-type:lower-alpha">
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| <math>\operatorname{d}A\!</math>
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| infer
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<li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math>
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| <math>A\!</math>
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| next.
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<li>
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| &nbsp;
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<p>For <math>k > 0,\!</math></p>
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|-
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| &nbsp;
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<p><math>\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
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| From
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| <math>A\!</math>
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</ol></ol>
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| and
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| <math>(\operatorname{d}A)\!</math>
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| infer
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| <math>A\!</math>
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| next.
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| &nbsp;
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|-
 +
| &nbsp;
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| From
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| <math>A\!</math>
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| and
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| <math>\operatorname{d}A\!</math>
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| infer
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| <math>(A)\!</math>
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| next.
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| &nbsp;
 +
|}
 +
|}
    
==Table Stuff==
 
==Table Stuff==
Jon Awbrey
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