Difference between revisions of "User:Jon Awbrey/SANDBOX"
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<li> | <li> | ||
− | <p>For <math>k > | + | <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> | ||
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</ol> | </ol> | ||
− | <li>The | + | <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{ | + | <li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math> |
<li> | <li> | ||
− | <p>For <math> | + | <p>For <math>k > 0,\!</math></p> |
− | <p><math>\operatorname{Surc}_{j=1}^ | + | <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).
- The parse of the concatenation \(\operatorname{Conc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
- \(\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.\)
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For \(k > 0,\!\)
\(\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).\)
- The parse of the surcatenation \(\operatorname{Surc}_{j=1}^k\) of the \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
- \(\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.\)
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For \(k > 0,\!\)
\(\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).\)
Table Stuff
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