Finding a knotted cycle in 3D $K_7$ In 1983, Conway and Gordon proved that in each 3D embedding of $K_7$, at least one of the 7-cycles would be knotted. At Intrinsically Knotted Graphs I have 3D points 1-7, no four points in a plane, and a person can chose a cycle.
What are some easier algorithms I can toss 7 3D points at to find the cycles more likely to be knotted?
 A: This does not exactly answer your question, but instead addresses this one: what cycles can I know for sure are knotted?  This is in case you just want to let the user give an embedding then have the program choose a knotted cycle.
The way Conway and Gordon showed this was to define an invariant $\sigma(G)$ of an embedding $G$ of $K_7$ by
$$\sigma(G)=\sum_{C}\alpha(C)\pmod{2},$$
where the sum ranges over all $360$ Hamiltonian cycles of $G$, and where $\alpha(C)$ is the Arf invariant of the knot $C$, an invariant that is defined modulo $2$.  They show $\sigma$ does not change even if the embedding is allowed to pass through itself, and then they find a particular embedding $G'$ of $K_7$ for which $\sigma(G')=1$.
A corollary to the proof, then, is that for any embedding of $K_7$, there is some nontrivially knotted Hamiltonian cycle with Arf invariant equal to $1$.  (There might be nontrivially knotted Hamiltonian cycles with Arf invariant equal to $0$, and of course there might be nontrivially knotted non-Hamiltonian cycles.)
There are a few ways to calculate the Arf invariant.  Maybe the most straightforward way is to use Murasugi's result that $\alpha(C)=0$ iff $\Delta_C(-1)\equiv \pm 1\pmod{8}$, where $\Delta_C(t)$ is the Alexander polynomial and $\lvert\Delta_C(-1)\rvert$ in particular is known as the knot determinant.  Since your embedding is seven points with straight-line paths for edges, it shouldn't be too hard to compute a knot diagram, and then the knot determinant is a determinant of a fairly simple matrix (for example, the Goeritz matrix approach).

It occurs to me that every cycle in $K_7$ has at most $7$ edges, and with a straight-line edge embedding this gives piecewise-linear knots with at most $7$ line segments.  Having stick number at most $7$ really limits what knots you can get.  Using the inequality on the linked Wikipedia page, the crossing number is at most $6$, so according to this table, the the only possible knots you get are the unknot, the trefoil knot, and the figure-eight knot.
Quite a few invariants can detect that the trefoil knot and figure-eight knots are knotted.  Here are a few of the values on KnotInfo.  The "polygon index" is the stick number, for reference.  The unknot has $1$ for its Alexander polynomial, its Jones polynomial, and its determinant.  Its Arf invariant is $0$.
