Are there any 3D hilbert curves wherein there are no double-length edges?
An example of a "double-length edge" would be the two sections outlined in red below:
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In 3D, there are only three seven-segment polylines with vertices at the vertices of the unit cube and all segments axis-aligned, if we ignore their rotated and/or mirrored copies:
The middle one in black is the basic 3D Hilbert curve.
The left one in blue is similar to the basic 3D Hilbert curve, except that the copy in the third dimension is rotated by 90°. This causes the end point to be diagonally opposite to the initial point, which makes this one difficult to use.
The right one in red has no 2D equivalent; it is a 3D primitive. If closed, it matches the closed Hilbert curve, so in some sense it is a permutation of the 3D Hilbert curve (i.e., with a different open edge).
As far as I understand, none of the above recursively produce a space-filling curve without two consecutive line segments being parallel.
However, if we replace each edge in the middle/black curve with rotated (red) and rotated and mirrored (purple) copies of the right/red block, we get a $4 \times 4 \times 4$ curve, with no consecutive parallel line segments:
I do suspect that this approach does generalize (that is, one could replace each purple or red edge with a rotated and/or mirrored copy of the middle/black block, and get a $8 \times 8 \times 8$ curve with no consecutive parallel line segments; and so on for any $2^n \times 2^n \times 2^n$ lattice), but I have not verified this: I've only examined curves that fill a $4 \times 4 \times 4$ lattice.