Extending a transformation from a curve to the whole space

[maze]

Supposing I have a curve, $\vec{f}\left(s,t\right)$ that lives in $\Re^{3}$ and is deforming over time but never intersecting itself (s is the curve parameter and t is time). I would like to extend the deformation of the curve to the entire space around the curve, creating a transformation of the whole space $\vec{g}\left(\vec{x},t\right)$ that is continuous and doesn't overlap itself.

It would be fine to restrict the curve to a series of connected straight line segments, if that would make it easier.

If I can construct such a function, g, I believe I can use it to create a very general method of solving "tanglement puzzles", where the object is to remove a flexible loop of string from a metal contraption.

Here are some pictures that describe what I am taking about.
Here is the function I start with:
http://img381.imageshack.us/img381/1968/pathcurvingsmww1.png

and here is the function I want to construct:
http://img162.imageshack.us/img162/7487/pathcurvingcoordssmcz2.png

And here are a couple examples of "remove the string" tanglement puzzles:

http://www.puzzles.ca/puzzle_data_3/xastroknot_l.jpg

 

[HallsofIvy]

There are an infinite number of ways of doing that. What additional restrictions do you want to add in order to make that problem unique?

 

[maze]

Yes definitely, there are infinitely many legit possibilities for such transformations. Any will do, so long as I can construct it for any curve. I'm interested in a construction, not an existence proof! The simpler or easier to compute, the better.

I was running through some simple ideas such as linear interpolation with tetrahedra, distance weighting, and so forth, but my constructions keep having problems with overlapping for certain curves.

 

[maze]

Ahh, I believe I may have a way to do this.

Restrict the curves in question to those that consist of straight line segments, as that is acceptable for my purposes. Next create a cubic lattice of gridpoints in the space. Then, using the curve and a bounding box as boundary conditions, force each grid point position to be the average of the position of all it's neighbors.

I think this should work, though I am not 100% sure and will need to prove it...

 

[maze]

Or more generally where the curve isn't straight lines, g is given by the vector laplace equation in each coordinate, using the transform on the curve, f, as a boundary condition. Now I will need to review how to calculate greens functions numerically...