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Comment by hgsgm | Hacker News Reader

How is that different from the solution on the page?

To give the answer credit, that answer does suggest adding the vectors (the same). It is also much more thorough than what I said, and I like the images. I like the answer and I was attempting to iterate.

I think these two things could be improved from that answer:

* I'm suggesting a general approach to measuring track tension, which is the average of the vectors. I didn't see that idea in the answer.

* I think the answer could be communicated a little more simply. For example, we don't need to think in terms of Q[sqrt(3)]; I see that as a distraction.

> For example, we don't need to think in terms of Q[sqrt(3)]; I see that as a distraction.

Except for replacing Q[sqrt(3)] by a suitable ring extension of Z, I see no possibiliy to simplify the argument. So what kind of simplification do you have in mind?

You're speaking as if floating point representations of numbers have zero utility. One approach here could be numerical. An algorithm could work with vectors represented as floating points. Another approach, which depends on the relative angles and lengths of track pieces, may be to encode each piece as an (angle, distance^2) pair. Many angles can give us an exact distance^2 values.

For starters, the "solution" here doesn't ensure that the ending piece meets the starting piece in the right direction. :p

Ok, a snarky way to phrase things, but it's a good point - in general the vectors could add to zero and the angles may not align. So perhaps a better expression may be

a metric for tension = [ norm(sum(piece_vectors)) + abs(angle_displacement) ] / n

where

angle_displacement = sum(angles) - 360

and the angles are signed accordingly.

Counterexample: a dozen (+1, 0) tiles followed by a dozen (-1, 0) tiles. Vector sum is fine, total angle is fine, but at each end you have an impossible 180 deg connection. I don’t think you can get away from local measures of tension with a global look.

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