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Working from a structural description of the cactus language, or any suitable formal grammar for <math>\mathfrak{C} (\mathfrak{P}),</math> it is possible to give a recursive definition of the function called <math>\operatorname{Parse}</math> that maps each sentence in <math>\operatorname{PARCE} (\mathfrak{P})</math> to the corresponding graph in <math>\operatorname{PARC} (\mathfrak{P}).</math>  One way to do this proceeds as follows:
 
Working from a structural description of the cactus language, or any suitable formal grammar for <math>\mathfrak{C} (\mathfrak{P}),</math> it is possible to give a recursive definition of the function called <math>\operatorname{Parse}</math> that maps each sentence in <math>\operatorname{PARCE} (\mathfrak{P})</math> to the corresponding graph in <math>\operatorname{PARC} (\mathfrak{P}).</math>  One way to do this proceeds as follows:
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<pre>
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<ol style="list-style-type:decimal">
1.  The parse of the concatenation Conc^k of the k sentences S_j,
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    for j = 1 to k, is defined recursively as follows:
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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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<ol style="list-style-type:lower-alpha">
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<li><math>\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.</math>
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<li>
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<p>For <math>k > 0,\!</math></p>
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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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    a.  Parse(Conc^0)        =  Node^0.
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</ol>
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    b.  For k > 0,
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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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        Parse(Conc^k_j S_j)  = Node^k_j Parse(S_j).
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<ol style="list-style-type:lower-alpha">
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2.  The parse of the surcatenation Surc^k of the k sentences S_j,
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<li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math>
    for j = 1 to k, is defined recursively as follows:
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    a.  Parse(Surc^0)        =  Lobe^0.
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<li>
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<p>For <math>k > 0,\!</math></p>
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    b. For k > 0,
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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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        Parse(Surc^k_j S_j)  =  Lobe^k_j Parse(S_j).
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</ol></ol>
    
For ease of reference, Table 12 summarizes the mechanics of these parsing rules.
 
For ease of reference, Table 12 summarizes the mechanics of these parsing rules.
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<pre>
 
Table 12.  Algorithmic Translation Rules
 
Table 12.  Algorithmic Translation Rules
 
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