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Fix a knot type $K \subset S^3$, and consider the set $$Y_K = \{ \mbox{Diagrams of }K \} / \mbox{planar isotopy}.$$ We can turn $Y_K$ into a metric space by considering the distance induced by Reidemeister moves: $d(D_0,D_1) = $ minimum length of a sequence of Reidemeister moves (each move is possibly followed by a planar isotopy), and $D_0,D_1$ are two diagrams of $K$.

We can build a graph $\mathcal{G}_K$ as follows: the vertices are given by the points of $Y_K$, while the edges are given by the pairs $(D_0,D_1)$ such that $d(D_0,D_1)=1$.

This construction parallels Hirasawa and Yoshiaki's in "The Gordian Complex of Knots" for the Gordian metric space of knots, but unlike the Gordian complex, each vertex of $\mathcal{G}_K$ has only finitely many edges (the number of edges at a vertex $D$ corresponds to the possible inequivalent diagrams - up to planar isotopy - that can be reached from $D$ by a single Reidemeister move).

For each knot $K$ we can thus define a function $$\varphi_K : Y_K \longrightarrow \mathbb{N}$$ called the simplicity, as $\varphi_K (D) = \#\{$edges of $\mathcal{G}_K$ that contain $D$ in their boundary$\}$.

As an example the "standard" diagram of a trefoil trefoil-1 has a lower simplicity than trefoil-2 (you need to take into account that the former is 3-symmetric).

Taking the minimum of the simplicity over all diagrams of a knot yields an invariant $\mathcal{S}(K)$.

Is the invariant $\mathcal{S}(K)$, or any possible application/connection, known?

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    $\begingroup$ Possibly stupid question: in the definition of simplicity, what does it mean for an edge to contain a vertex in its boundary? (What is the boundary of an edge?) At first I thought you just meant the two vertices on which the edge is incident, but then simplicity would be the same as degree. $\endgroup$ – Nate Eldredge Feb 11 '14 at 23:35
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    $\begingroup$ I didn't understand your example- the diagrams of trefoil-1 and of trefoil-2 are the same up to reparametrization. $\endgroup$ – Daniel Moskovich Feb 12 '14 at 2:03
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    $\begingroup$ Other than by knowing a complete set of diagrams for K, is there any way to compute S(K)? $\endgroup$ – Theo Johnson-Freyd Feb 12 '14 at 8:16
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    $\begingroup$ @NateEldredge you are right, it is the degree, i just didn't know it was called that way. $\endgroup$ – Daniele Celoria Feb 12 '14 at 9:13
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    $\begingroup$ @DanieleCeloria Yeah, I didn't read carefully enough and was imagining we were talking about the graph where there would be multiple edges for multiple Reidemeister links between the same ambient-isotopic diagrams. $\endgroup$ – Gabriel C. Drummond-Cole Feb 13 '14 at 16:26
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I think the minimal degree of "Reidemeister complex" has some meaning.

I know one paper about this topic.

A distance for diagrams of a knot

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  • $\begingroup$ Thanks, this paper is really useful, but it actually focuses on prime Reidemeister moves, and doesn't seem to care for planar isotopies, so the invariants defined should be different from $\mahtcal{S}$. $\endgroup$ – Daniele Celoria Feb 13 '14 at 15:01

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