User:Jon Awbrey/SANDBOX

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Grammar Stuff

Working from a structural description of the cactus language, or any suitable formal grammar for \(\mathfrak{C} (\mathfrak{P}),\) it is possible to give a recursive definition of the function called \(\operatorname{Parse}\) that maps each sentence in \(\operatorname{PARCE} (\mathfrak{P})\) to the corresponding graph in \(\operatorname{PARC} (\mathfrak{P}).\) One way to do this proceeds as follows:

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 sequence of \(k\!\) sentences \((s_j)_{j=1}^k\) is defined recursively as follows:
    1. \(\operatorname{Parse} (\operatorname{Conc}^0) ~=~ \operatorname{Node}^0.\)
    2. For \(\ell > 1,\!\)

      \(\operatorname{Conc}_{j=1}^\ell s_j \ = \ \operatorname{Conc}_{j=1}^{\ell - 1} s_j \, \cdot \, s_\ell.\)

  2. The surcatenation \(\operatorname{Surc}_{j=1}^k\) of the sequence of \(k\!\) strings \((s_j)_{j=1}^k\) is defined recursively as follows:
    1. \(\operatorname{Surc}_{j=1}^1 s_j \ = \ ^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.\)
    2. For \(\ell > 1,\!\)

      \(\operatorname{Surc}_{j=1}^\ell s_j \ = \ \operatorname{Surc}_{j=1}^{\ell - 1} s_j \, \cdot \, ( \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime} \, )^{-1} \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_\ell \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.\)

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›