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

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==Grammar Stuff==
 
==Grammar Stuff==
 
<pre>
 
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
 
</pre>
 
  
 
<br>
 
<br>
  
 
{| align="center" border="1" cellpadding="8" cellspacing="0" style="text-align:center; width:90%"
 
{| align="center" border="1" cellpadding="8" cellspacing="0" style="text-align:center; width:90%"
 +
|+ '''Table 13.  Algorithmic Translation Rules'''
 
|- style="background:whitesmoke"
 
|- style="background:whitesmoke"
 
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Revision as of 11:50, 22 January 2009

Grammar Stuff


Table 13. Algorithmic Translation Rules
\(\text{Sentence in PARCE}\!\) \(\xrightarrow{\operatorname{Parse}}\) \(\text{Graph in PARC}\!\)
\(\operatorname{Conc}^0\) \(\xrightarrow{\operatorname{Parse}}\) \(\operatorname{Node}^0\)
\(\operatorname{Conc}_{j=1}^k s_j\) \(\xrightarrow{\operatorname{Parse}}\) \(\operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j)\)
\(\operatorname{Surc}^0\) \(\xrightarrow{\operatorname{Parse}}\) \(\operatorname{Lobe}^0\)
\(\operatorname{Surc}_{j=1}^k s_j\) \(\xrightarrow{\operatorname{Parse}}\) \(\operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j)\)


Table Stuff


fixy
u =
v =
1 1 0 0
1 0 1 0
= u
= v
fjuv
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›