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I have a few questions about nonplanar "stick circuits" (or hexagons and higher $n$-gons) that you might be able to help with:

(I know that $n=6$ is the minimum number of points to form a stick knot.)

  1. Given $n=6$ points in $\mathbb{R}^3$ in general position connected by a specific "stick circuit" (nonplanar hexagon), what test can be done to see if it forms a stick knot vs. an unknot?

  2. Given $n=6$ points in $\mathbb{R}^3$ in general position, there are 60 different stick circuits connecting them. True or false, at least one forms a knot?

  3. Given $n=6$ points in $\mathbb{R}^3$ in general position, does the minimum-length stick circuit on these $n$ points ever form a knot? ("knotted 6-point traveling salesman problem with return)

All these can be generalized to $n > 6$.

These questions occurred to me over the last few days. I suspect (1) has a known answer but I have no idea about (2) or (3).

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For (2), no. Take the six points to be approximately the vertices of a regular octahedron. – Tom Goodwillie Jun 10 '12 at 2:57
Actually, a regular octahedron (or close to it) is a configuration where I would always expect a stick knot. A pentagonal pyramid (or close to it) would be more challenging. Gerhard "Ask Me About System Design" Paseman, 2012.06.09 – Gerhard Paseman Jun 10 '12 at 3:13
For (1) I know no simple test. However, all my mental images of 6-stick nots involve a stick with two acute angles at either end. I imagine this is a necessary condition. Gerhard "Or My Imagination Is Limited" Paseman, 2012.06.09 – Gerhard Paseman Jun 10 '12 at 3:22
If the knot consists entirely of edges of the octahedron (or any other convex polytope) then it is a Jordan curve in that convex surface, therefore unknotted. The other possibilities can easily be handled case by case. – Tom Goodwillie Jun 10 '12 at 4:04
(3) was asked a few days ago on,… – Gerry Myerson Jun 10 '12 at 6:19

The answer to question (2) is

  • no for $n=6$,
  • yes for $n=7$.

For $n=6$, take, for example, the following six points as vertices of a straight-line (stick) embedding of $K_6$:

$A = (-2,-2,1), B= (2,-2,0), C= (0,2,0), D= (-1,-1,0), E= (1,-1,1), F= (0,1,2)$

The projection onto the $xy$ plane has crossing number $3$.

a projection of K_6 with crossing number 3

Moreover, the crossings are between disjoint pairs of edges. Therefore, since every nontrivial knot has at least three crossings, there is at most one possible cycle that could form form a nontrivial knot; that is, the cycle $AECDBF$ formed by the six edges participating in the crossings. But by the above-below relations at the crossings, this cycle clearly forms an unknot.

For $n=7$, Conway and Gordon proved that every embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a nontrivial knot, using the parity of the sum of the quadratic terms of the Conway polynomials of the Hamiltonian cycles as an invariant.

Edit: See also J. L. Ramirez Alfonsin, Spatial Graphs and Oriented Matroids: the Trefoil, Discrete and Computational Geometry 22:149--158 (1999) for the following stronger result:

Every stick embedding of $K_7$ in $\mathbb{R}^3$ contains a Hamiltonian cycle forming a (left-handed or a right-handed) trefoil.

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A partial test for (1) is provided by the Fary-Milnor theorem. See also this question.

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There is a problem about matching red and blue dots in the plane in pairs by straight line segments, with a length minimal matching involving no crossings. I imagine (3) could be answered negatively by similar reasoning.

Gerhard "Ask Me About System Design" Paseman, 2012.06.09

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A paper by Gerard Venema and Tom Clark classified stick knots with 6 segments (using the lengths of the segments); they are using chains for their knots.

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I realize now that you're looking specifically for circuits - my apologies. – John Engbers Jun 12 '12 at 1:10

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