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

MyWikiBiz, Author Your Legacy — Sunday November 10, 2024
Jump to navigationJump to search
Line 30: Line 30:
  
 
<li>
 
<li>
<p>For <math>k > 1,\!</math></p>
+
<p>For <math>k > 0,\!</math></p>
  
 
<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
 
<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
Line 36: Line 36:
 
</ol>
 
</ol>
  
<li>The ''surcatenation'' <math>\operatorname{Surc}_{j=1}^k</math> of the sequence of <math>k\!</math> strings <math>(s_j)_{j=1}^k</math> is defined recursively as follows:</li>
+
<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>
  
 
<ol style="list-style-type:lower-alpha">
 
<ol style="list-style-type:lower-alpha">
  
<li><math>\operatorname{Surc}_{j=1}^1 s_j \ = \ ^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></li>
+
<li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math>
  
 
<li>
 
<li>
<p>For <math>\ell > 1,\!</math></p>
+
<p>For <math>k > 0,\!</math></p>
  
<p><math>\operatorname{Surc}_{j=1}^\ell s_j \ = \ \operatorname{Surc}_{j=1}^{\ell - 1} s_j \, \cdot \, ( \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime} \, )^{-1} \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_\ell \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></p></li>
+
<p><math>\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
  
 
</ol></ol>
 
</ol></ol>

Revision as of 20:30, 19 January 2009

Grammar Stuff

1.  The parse of the concatenation Conc^k of the k sentences S_j,
    for j = 1 to k, is defined recursively as follows:

    a.  Parse(Conc^0)        =  Node^0.

    b.  For k > 0,

        Parse(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:

    a.  Parse(Surc^0)        =  Lobe^0.

    b.  For k > 0,

        Parse(Surc^k_j S_j)  =  Lobe^k_j Parse(S_j).
  1. The parse of the concatenation \(\operatorname{Conc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
    1. \(\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.\)
    2. For \(k > 0,\!\)

      \(\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).\)

  2. The parse of the surcatenation \(\operatorname{Surc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
    1. \(\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.\)
    2. For \(k > 0,\!\)

      \(\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).\)

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›