[lsy]

I'm sure it's nearly an academic distinction, but:

> Basically, for any given region, we find its highest point and assume that there is a perfectly placed sibling peak of the same height that is mutually visible.

Shouldn't you always add 335km to the horizon distance to account for the possibility of Everest (i.e. a taller sibling peak) being on the other side of the horizon?

[tennysont]

You're right, but all lines of sight are mutual, so we will notice this oversight when checking the other peak.

This seems poorly explained, but I think the author was in a hurry to get to the main algorithm, and sped through the intro.

[tombh]

Author here. I really appreciate this question because it's the entire reason I wrote the post. I feel like this is a unique problem and so I'm sure I'm not considering all the possibilities.

I _think_ your suggestion is covered by the fact that I'm basing the size of the tile on the single highest point that it contains. The steps are:

    1. Take any point and calculate the furthest theoretical distance it could see if there were another point at just the perfect distance away for mutual visibility. Note how there could of course be an Everest, but I don't check because step 2 should solve that.
    2. Based on that furthest theoretical distance I create a tile of that exact width around the point and then check to see what the new highest point is within that tile _and_ its surroundings. "Surroundings" here means a border region around the tile of the same width as the tile itself. These surroundings don't get viewsheds calculated for them, they're just auxiliary data.
    3. If a higher point is found, then increase the width based on that new highest point and repeat step 2. If no higher point is found then the tile is ready.