Hypothesis (language regularity) and algorithm (L-graph to NFA) in TOC

Prerequisite – Finite automata, L-graphs and what they represent
L-graphs can generate context sensitive languages, but it’s much harder to program a context sensitive language over programming a regular one. This is why I’ve came up with a hypothesis about what kind of L-graphs can generate a regular language. But first, I need to introduce you to what I call an iterating nest.

As you can remember a nest is a neutral path $T_1T_2T_3$ , where $T_1$ and $T_3$ are cycles and $T_2$ path is neutral. We will call $T_1T_2T_3$ an iterating nest, if $T_1$ , $T_2$ and $T_3$ paths print the same string of symbols several times, to be more exact $T_1$ prints $\alpha^k$ , $T_2$ prints $\alpha^l$ , $T_3$ prints $\alpha^m$ , $k, \: l, \: m \geqslant 0$ and $\alpha$ is a string of input symbols (better, if at least one of $k, \: l\: and\: m\: is \geqslant 1$ ).
From this definition comes out the next hypothesis.

Hypothesis – If in a context free L-graph G all nests are iterating, then the language defined by this L-graph G, L(G), is regular.
If this hypothesis will be proven in the near future, it can change a lot in programming that will make creating new easy programming languages much easier than it already is. The hypothesis above leads to the next algorithm of converting context free L-graphs with iterating nests to an NFA.

Algorithm – Converting a context free L-graph with iterating complements to a corresponding NFA
Input – Context free L-graph $G=(\Sigma, V, P, \lambda, P_0, F)$ with iterating complements
Output – $G'=(\Sigma', V', \lambda', P'_0, F')\\*$

Step-1: Languages of the L-graph and NFA must be the same, thusly, we won’t need a new alphabet $\Rightarrow \Sigma'' = \Sigma, \: P'' = P$ . (Comment: we build context free L-graph G’’, which is equal to the start graph G’, with no conflicting nests)
Step-2: Build Core(1, 1) for the graph G.
V’’ := {(v, $\varepsilon$ ) | v $\in$ V of $\forall$ canon k $\in$ Core(1, 1), v $\notin$ k}
$\lambda''$ := { arcs $o \in \lambda$ | start and final states $o', o'' \in$ V’’}

For all k $\in$ Core(1, 1):
Step 1’. v := 1st state of canon k. $\eta := \varepsilon$ .
V’’ $\cup= (v, \eta)$
Step 2’. $\lambda'' \cup=$ arc from state $(v, \eta)$ followed this arc into new state defined with following rules:
$\eta := \eta$ , if the input bracket on this arc $= \varepsilon$ ; $\eta'the\: input\: bracket'$ , if the input bracket is an opening one; $\eta 'without\: the\: last\: bracket'$ , if the input bracket is a closing bracket
v := 2nd state of canon k
V’’ $\cup= (v, \eta)$
Step 3’. Repeat Step 2’, while there are still arcs in the canon.
Step-3: Build Core(1, 2).
If the canon has 2 equal arcs in a row: the start state and the final state match; we add the arc from given state into itself, using this arc, to $\lambda''$ .
Add the remaining in $\lambda$ arcs v – u $(\alpha)$ to $\lambda''$ in the form of $(v, \varepsilon) - (u, \varepsilon) (\alpha)$
Step-4: $P''_0 = (P_0, \varepsilon).\: F'' = \{(f, \varepsilon) | f \in F\}$
(Comment: following is an algorithm of converting context free L-graph G’’ into NFA G’)
Step-5: Do the following to every iterating complement $T = T_1T_2T_3$ in G’’:
Add a new state v. Create a path that starts in state $beg(T_3)$ , equal to $T_3$ . From v into $T_3$ create the path, equal to $T_1$ . Delete cycles $T_1$ and $T_3$ .
Step-6: G’ = G’’, where arcs are not loaded with brackets.

So that every step above is clear I’m showing you the next example.
$\textbf{Example:}\\ \textbf{Input:}$ Context free L-graph with iterating complements
$G = ( \{a, b, c\}, \\*\{1, 2, 3\} \\*\{( (, ) ), ( [, ] )\}, \\*\\*\{ (: \{ 1 - a - 1 \}, \\*): \{ 2 - a - 2 \}, \\*\big[: \{ 1 - b - 2 \}, \\*\big]: \{ 2 - c - 3 \}, \\*\varepsilon: \{ 1 - a - 2 \} \}, \\*\\*1, \\*\{2, 3\} \}$ ,
which determines the $language = \{a^(^2^n^+^1^) | n \geqslant 0\} \cup \{bc\}$
Start graph G
$\Sigma'' = \{a, b, c\}\\ V'' = \varnothing\\ \lambda'' = \varnothing$
Core(1, 1) = { 1 – a – 2 ; 1 – a, (1 – 1 – a – 2 – a, )1 – 2 ; 1 – b, (2 – 2 – c, )2 – 3 }
Core(1, 2) = Core(1, 1) $\cup$ { 1 – a, (1 – 1 – a, (1 – 1 – a – 2 – a, )1 – 2 – a, )1 – 2 }
Step 2: Step 1’ – Step 3’
$\Rightarrow\\ V'' = \{(1, \varepsilon), (2, (_2), (3, \varepsilon), (1, (_1), (2, )_1), (2, \varepsilon)\}\\* \lambda'' = \{ \\*(: \{ (1, \varepsilon) - a - (1, (); (1, () - a - (1, () \}, \\*): \{ (2, )) - a - (2, )); (2, )) - a - (2, \varepsilon) \}, \\*\big[: \{ (1, \varepsilon) - b - (2, [) \}, \\*\big]: \{ (2, [) - c - (3, \varepsilon) \}, \\*\varepsilon: \{ (1, \varepsilon) - a - (2, \varepsilon); (1, () - a - (2, )) \} \}\\ P''_0 = (1, \varepsilon)\\ F'' = \{(2, \varepsilon), (3, \varepsilon)\}\\ G'' = (\Sigma'', V'', P'', \lambda'', P''_0, F'')$
Intermediate graph G’’
$\Sigma' = \{a, b, c\}\\ V' = V'' \cup \{4\}\\ P'_0 = P''_0\\ F' = F''\\ \lambda' = \{ \\*(1, \varepsilon) - a - (1, (); \\*(2, )) - a - 4; \\*4 - a - (2, )); \\*(2, )) - a - (2, \varepsilon); \\*(1, \varepsilon) - b - (2, [); \\*(2, [) - c - (3, \varepsilon); \\*(1, \varepsilon) - a - (2, \varepsilon); \\*(1, () - a - (2, )) \}\\ G' = (\Sigma', V', \lambda', P'_0, F')$
NFA G’