# Gauss Codes that produce classical knots as opposed to virtual knots

I have been doing some research in Gauss codes and have been reading Kauffman's paper Virtual Knot Theory. In section 3.3, Theorem 2, he states that

If $K$ is a virtual knot whose underlying Gauss code is planar and whose sign sequence is standard, then $K$ is equivalent to a classical knot.

He then goes on to say that "The fundamental problem in Gauss codes is to give an algorithm for determining whether a given code can be realized by a planar shadow."

One problem I have is that I can't figure out what exactly the word standard means in the theorem - he says he will define it later and never does. And does he actually solve this "fundamental problem" in the rest of the section 3.3. If anyone knows a technical definition of standard, I would be very appreciative.

The point is to figure out if the the follow question is open:

Given a Gauss Code, is there an (explicit) algorithm for detecting if it can be realized as a classical knot?

So if you know that it is or is not open, I would love to know too. In the case that it is known, a reference would be wonderful too. Thanks.

In principle, there exists an algorithm to tell if a virtual knot is classical following from an observation of Kuperberg, that the minimal genus of a virtual knot is well-defined, and equal to $0$ if and only if it is classical. There is a standard way to embed a virtual knot in a thickened surface, and using algorithms from normal surface theory (due to Jaco and Tollefson), one may determine the JSJ decomposition as described in Kuperberg's paper, and therefore the genus.
I believe what Kauffman means by standard is that the sign sequence corresponds to a particular way in which $g^*$ is dually paired. In the proof at the end of section 3.3, this is what allows Kauffman to assert that there is a planar embedding with the same local orientations. A quasi-restatement of this theorem is that if the underlying combinatorial map of a virtual knot diagram is planar (with the classical crossings thought of as degree-$4$ vertices), then that diagram is virtually equivalent to one with no virtual crossings, and in particular by using only moves of type (B) and (C) in Figure 2. More generally, any two virtual knot diagrams that have the same underlying combinatorial map (annotated with over/under at the vertices) are virtually equivalent.