[nwjtkjn]

Good point, though I'm not convinced Hilbert really avoids this. For example, look at n=3 here:

http://www.texample.net/media/tikz/examples/PNG/hilbert-curv...

This might lead to say

. . . . . . . .

1 . . . . . . 2

4 . . . . . . 3

. . . . . . . .

when

. . . . . . . .

1 . . . . . . 4

2 . . . . . . 3

. . . . . . . .

is more efficient.

Generally speaking, it seems this is basically TSP.

[Merovius]

I tried describing here https://www.reddit.com/r/programming/comments/6oxra8/using_h... why I think that this is theoretically a TSP problem, I don't consider that necessarily the best way to treat it. tl;dr is, that at least to me, at the time, for this problem, the speedup of an optimal solution wouldn't have outweighed the additional thinking time to get to that optimal solution (and, most importantly, to actually make it useful; after all, I wasn't actually drawing a path, I was outputting a list of names).

You can even see that in the argument I added to the post; while a zig-zag line might've been, in theory, a stupider, easier way to solve the problem, in practice it would've meant spending some time thinking about the right discretization. With a Hilbert-curve, I could just choose some n that is definitely large enough and be done with it and the additional cost of the more complicated curve doesn't really factor in, as I could just copy-paste it anyway.

But yes. The theoretical problem is a TSP and with the right set of tools, I could've added some efficiency to the search by viewing it as such.

[hidal]

I think is TSP only if you start by the first point in the ordered list. If you start any other point, you will be going in only one direction (top-bottom/left-right or bottom-top/left-right) (See how the space is filled)