Terminating computations for classical knot invariants — Examples and Counterexamples

With $c=8$, the state-count formula in [F1] gives $2^8=256$ visits.

The stated incidences and [F3] give the equations $2x_1-x_2-x_3=0$, $2x_2-x_3-x_1=0$, and $2x_3-x_1-x_2=0$. In ${\mathbb{F}}_3$, $-1=2$, so all three rows are $(2,2,2)$. Row reduction leaves one nonzero row and two zero rows, hence one pivot and two free variables. By [F4], the algorithm returns $3^2=9$ trefoil colorings.

The crossingless unknot has one arc and no crossing equations, so its matrix has one column and zero rows, rank zero, and one free variable. By [F4], it has $3^1=3$ colorings.

The crossing equation $2x_o=x_i+x_j$ is symmetric in the two underarc colors and independent of strand arrows, so reversing a component orientation does not change its Fox coloring set. If the trefoil and unknot were ambient-isotopic as unoriented knots, choose their orientations so the equivalence preserves orientation. By [A1] and [F6] the oriented diagrams would then be joined by a finite Reidemeister sequence, and [F5] would force equal coloring counts. The computed values $9$ and $3$ contradict this, so the underlying knot types are distinct.

Starting with $U_0$, insert $2N+1$ disjoint type-I kinks, one at a time, to form a diagram $U_N$. Each insertion is one R1 move, so deleting the kinks gives an explicit sequence of $2N+1$ moves from $U_N$ to $U_0$. Both diagrams are regular one-component knot diagrams; their normalized codes are valid and occur among the enumerated candidates at their crossing bounds by [F3] and [F4]. By [A1] and [F9], the finite move sequence makes the represented oriented knots equivalent.

The crossing count difference is $c(U_N)-c(U_0)=2N+1$. By [F1], any single Reidemeister move changes crossing count by at most two. A path of at most $N$ moves could therefore change the crossing count by at most $2N$, so no such path connects these two diagrams. The search through depth $N$ completes its finite layers by [F2] without finding a path, although the diagrams are equivalent by step 1.1. This refutes the statement.

The crossingless unknot has one arc and no crossing equations, so [F5]–[F7] give $3^1=3$ Fox 3-colorings. For the standard trefoil with arcs $x_1,x_2,x_3$, the crossing equations are $2x_1-x_2-x_3=0$, $2x_2-x_3-x_1=0$, and $2x_3-x_1-x_2=0$. In ${\mathbb{F}}_3$, $-1=2$, so all rows are $(2,2,2)$; the matrix has rank one and two free variables, giving $3^2=9$ colorings by [F6] and [F7]. These equations do not use strand arrows, so reversing either orientation leaves the count unchanged. If the underlying unoriented knots were ambient-isotopic, orient them compatibly; [A1] and [F9] give a finite oriented move sequence, and [F8] forces the counts to agree, contradicting $3\mathrm{\ne }9$. Thus the Fox count is an actual distinction, while a finite failed search alone is not one.