Trefoil Gauss traversal and rotation reconstruction

The visits of the given word are, in order, under at crossing $1$, over at $3$, under at $2$, over at $1$, under at $3$, and over at $2$. The source code in [F6], started at its fourth visit, is $u1+,o2+,u3+,o1+,u2+,o3+$; exchanging labels $2$ and $3$ gives exactly the stated word and the same positive sign at each crossing. Thus the word is the stated based traversal of the right trefoil.

Let $x_i$ be the directed edge from visit $i$ to visit $i+1$, with indices modulo $6$. Then $x_1$ joins $1:U_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $3:O_{\mathrm{i}\mathrm{n}}$, $x_2$ joins $3:O_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $2:U_{\mathrm{i}\mathrm{n}}$, $x_3$ joins $2:U_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $1:O_{\mathrm{i}\mathrm{n}}$, $x_4$ joins $1:O_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $3:U_{\mathrm{i}\mathrm{n}}$, $x_5$ joins $3:U_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $2:O_{\mathrm{i}\mathrm{n}}$, and $x_6$ joins $2:O_{\mathrm{o}\mathrm{u}\mathrm{t}}$ to $1:U_{\mathrm{i}\mathrm{n}}$. At each crossing the sign is positive by the given vector and [F2]. Reading each crossing clockwise from its outgoing over-dart therefore gives $X_1=(x_4,x_6,x_3,x_1)$, $X_2=(x_6,x_2,x_5,x_3)$, and $X_3=(x_2,x_4,x_1,x_5)$. Scanning crossing $1$ then $2$ then $3$ assigns canonical edge names $e_1=x_4,e_2=x_6,e_3=x_3,e_4=x_1,e_5=x_2,e_6=x_5$, yielding the three normalized records in the statement.

Write $j:e_k$ for the dart labelled by edge $e_k$ at crossing $j$. Applying $\varphi =\alpha \circ \sigma$ gives $\begin{array}{rl}{\displaystyle 1:e_1}& {\displaystyle \mapsto 2:e_2\mapsto 3:e_5\mapsto 1:e_1,}\\ {\displaystyle 1:e_2}& {\displaystyle \mapsto 2:e_3\mapsto 1:e_2,}\\ {\displaystyle 1:e_3}& {\displaystyle \mapsto 3:e_4\mapsto 2:e_6\mapsto 1:e_3,}\\ {\displaystyle 1:e_4}& {\displaystyle \mapsto 3:e_1\mapsto 1:e_4,}\\ {\displaystyle 2:e_5}& {\displaystyle \mapsto 3:e_6\mapsto 2:e_5\mathrm{.}}\end{array}$ For example, $1:e_1$ advances clockwise to $1:e_2$, whose paired dart is $2:e_2$. The five displayed cycles are disjoint and contain all twelve darts, so they are exactly the face cycles. Hence $V=3$, $E=6$, $F=5$, and $V-E+F=2$.

Each edge occurs at its two ends once incoming and once outgoing by the six directed segments in step 2.1. Each crossing record has its two opposite oriented strands, the over/under designation, and the positive cyclic order from [F1]–[F2]. Traversing from $1:U_{\mathrm{o}\mathrm{u}\mathrm{t}}$ follows $x_1$ to $3:O_{\mathrm{i}\mathrm{n}}$, then the opposite over-dart follows $x_2$ to $2:U_{\mathrm{i}\mathrm{n}}$, then follows $x_3$ to $1:O_{\mathrm{i}\mathrm{n}}$, then $x_4$ to $3:U_{\mathrm{i}\mathrm{n}}$, then $x_5$ to $2:O_{\mathrm{i}\mathrm{n}}$, and finally $x_6$ to $1:U_{\mathrm{i}\mathrm{n}}$; the opposite under-dart there is the starting $1:U_{\mathrm{o}\mathrm{u}\mathrm{t}}$. This single cycle covers all six visits, and the edge graph connects crossings $1,2,3$. By [F3], the Euler count gives the spherical realization; choosing $F_0$ yields the selected plane model.

The code therefore passes the validator's ordered checks, and [F4] returns its diagram with these face cycles and outer face. Its normalized crossing signs, edge pairing, and selected cycle occur in the enumerator at $c=3$; the one-cycle component trace passes the knot filter. By [F5], the reconstructed candidate is included at bound $3$.