Difference between revisions of "User:Jon Awbrey/SANDBOX"

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<pre>
 
<pre>
−
1.  The parse of the concatenation Conc^k of the k sentences S_j,
+
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          |
−
 
+
o------------------------o---------o------------------------o
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    b.  For k > 0,
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|                        |        |                        |
−
 
+
| Conc^0                 |  -->  | Node^0                 |
−
        Parse(Conc^k_j S_j)  =  Node^k_j Parse(S_j).
+
|                        |        |                        |
−
 
+
| Conc^k_j S_j         |  -->  | Node^k_j Parse(S_j)   |
−
2.  The parse of the surcatenation Surc^k of the k sentences S_j,
+
|                        |        |                        |
−
    for j = 1 to k, is defined recursively as follows:
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| Surc^0                 |  -->  | Lobe^0                 |
−
 
+
|                        |        |                        |
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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)   |
−
 
+
|                        |        |                        |
−
    b.  For k > 0,
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o------------------------o---------o------------------------o
−
 
 
−
        Parse(Surc^k_j S_j)  =  Lobe^k_j Parse(S_j).
 
 
</pre>
 
</pre>
  
−
<ol style="list-style-type:decimal">
+
{| align="center" border="1" cellpadding="8" cellspacing="0" style="text-align:center; width:96%"
−
 
+
|
−
<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>
+
{| align="center" border="0" cellpadding="8" cellspacing="0" style="text-align:center; width:96%"
−
 
+
| &nbsp;
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<ol style="list-style-type:lower-alpha">
+
| From
−
 
+
| <math>(A)\!</math>
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<li><math>\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.</math>
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| and
−
 
+
| <math>(\operatorname{d}A)\!</math>
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<li>
+
| infer
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<p>For <math>k > 0,\!</math></p>
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| <math>(A)\!</math>
−
 
+
| next.
−
<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
+
| &nbsp;
−
 
+
|-
−
</ol>
+
| &nbsp;
−
 
+
| From
−
<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>
+
| <math>(A)\!</math>
−
 
+
| and
−
<ol style="list-style-type:lower-alpha">
+
| <math>\operatorname{d}A\!</math>
−
 
+
| infer
−
<li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math>
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| <math>A\!</math>
−
 
+
| next.
−
<li>
+
| &nbsp;
−
<p>For <math>k > 0,\!</math></p>
+
|-
−
 
+
| &nbsp;
−
<p><math>\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
+
| From
−
 
+
| <math>A\!</math>
−
</ol></ol>
+
| and
 +
| <math>(\operatorname{d}A)\!</math>
 +
| infer
 +
| <math>A\!</math>
 +
| next.
 +
| &nbsp;
 +
|-
 +
| &nbsp;
 +
| From
 +
| <math>A\!</math>
 +
| and
 +
| <math>\operatorname{d}A\!</math>
 +
| infer
 +
| <math>(A)\!</math>
 +
| next.
 +
| &nbsp;
 +
|}
 +
|}
  
 
==Table Stuff==
 
==Table Stuff==

Revision as of 22:14, 19 January 2009

Grammar Stuff

Table 12.  Algorithmic Translation Rules
o------------------------o---------o------------------------o
|                        |  Parse  |                        |
| Sentence in PARCE      |   -->   | Graph in PARC          |
o------------------------o---------o------------------------o
|                        |         |                        |
| Conc^0                 |   -->   | Node^0                 |
|                        |         |                        |
| Conc^k_j  S_j          |   -->   | Node^k_j  Parse(S_j)   |
|                        |         |                        |
| Surc^0                 |   -->   | Lobe^0                 |
|                        |         |                        |
| Surc^k_j  S_j          |   -->   | Lobe^k_j  Parse(S_j)   |
|                        |         |                        |
o------------------------o---------o------------------------o
  From \((A)\!\) and \((\operatorname{d}A)\!\) infer \((A)\!\) next.  
  From \((A)\!\) and \(\operatorname{d}A\!\) infer \(A\!\) next.  
  From \(A\!\) and \((\operatorname{d}A)\!\) infer \(A\!\) next.  
  From \(A\!\) and \(\operatorname{d}A\!\) infer \((A)\!\) next.  

Table Stuff


fi‹x, y›
u =
v =
1 1 0 0
1 0 1 0
= u
= v
fj‹u, v›
x =
y =
1 1 1 0
1 0 0 1
= f‹u, v›
= g‹u, v›


A
u =
v =
1 1 0 0
1 0 1 0
= u
= v
B
x =
y =
1 1 1 0
1 0 0 1
= f‹u, v›
= g‹u, v›


u =
v =
1 1 0 0
1 0 1 0
= u
= v
x =
y =
1 1 1 0
1 0 0 1
= f‹u, v›
= g‹u, v›


u =
v =
x =
y =
1 1 0 0
1 0 1 0
1 1 1 0
1 0 0 1
= u
= v
= f‹u, v›
= g‹u, v›


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