pfedak

Ha, I guess we just see the world differently. What you're describing sounds exactly like the dismissiveness I'm talking about.

I hope your life is full of "personal side quests" pursuing questions that appeal to your particular sensibilities. You should wrestle with that feeling of not understanding something, use it as a signpost pointing towards interesting ideas rather than as a reason to stop.

I'll get you started on the proof: Suppose for n>2 you have n n-sided dice with disjoint sets of numbers and you want each of them to have an equal chance of rolling the highest number. One of them must have the highest overall number, and it will roll that 1/n of the time. What can you say about the other numbers?

The relative ordering of the other dice doesn't depend on this result. They need to each have an equal chance of coming first. How many total outcomes are there?

You might want to check that your proposed solution works at all before suggesting the problem is trivial. What's the probability that the first die goes first with your numbers?

You also can't generally "and so on" constrained combinatorial arrangements like this.

I believe they are mixing it up with The Guns of August (published in 1962, also by Tuchman), which JFK was fond of and supposedly drew on during the Cuban missile crisis.

This is nonsense, at least in part because it's mixing two different ideas. The notion that the image "looks exactly the same as how it originally appeared" is only true when one of your eyes is positioned exactly where the camera sensor would have been, which requires a specific distance away from the screen.

Lines in 3D remaining straight in a photo is unrelated and not actually demonstrated by the image. I'm having trouble imagining why this matters - you're trying to find the intersection of two lines in an image without drawing anything?

Another aspect of the solution that makes it rather abstract is it effectively assumes we know nothing about the distribution of the number of days.

Paying at 1/2 will be optimal if it ends before you buy, very bad (3x optimal) if it ends right after you buy, and slightly better than the solution in the post if it lasts at least twice that long (1.5x optimal vs e/(e-1)).

The metric in the post is just the worst of those ratios. Assuming the unproven statement in the post (that the solution which is a constant factor worse than optimal is best), any solution of the form you suggest is going to have similar tradeoffs. If we had a distribution, we could choose.

If it wasn't clear, their statements are all true when the events follow a poisson distribution/have exponentially distributed waiting times.

The main image is all at the same 90x level, and those buttons just zoom in (more or less) all the way on the points, while the "140x" are separate scan patches at higher magnification (though the real point is they have 3D/height data, too).

That isn't at all what the central limit theorem says. The whole point is it holds independent of the actual shape of distribution of the population. You could use the same argument to say social security numbers are normally distributed.

One way to explain things like height being normally distributed is that there are a bunch of independent factors which contribute, and the central limit theorem applied to those factors would then suggest the observed variable looking normal-ish.

Actually playing around with it the behavior was very different from what I expected - there was much more zooming. Turns out I missed some parts of the zoom code:

Their zoom actually is my "y" rather than a scale factor, so the metric is ds^2 = dy^2 + (C-y)^2 dx^2 where C is a bit more than the maximal zoom level. There is some special handling for cases where their curve would want to zoom out further.

Normalizing to the same cost to pan all the way zoomed out (zoom=1), their cost for panning is basically flat once you are very zoomed in, and more than the hyperbolic model when relatively zoomed out. I think this contributes to short distances feeling like the viewport is moving very fast (very little advantage to zooming out) vs basically zooming out all the way over larger distances (intermediate zoom levels are penalized, so you might as well go almost all the way).

I think you can reasonably think about the flight path by modeling the movement on the hyperbolic upper half plane (x would be the position along the linear path between endpoints, y the side length of the viewport).

I considered two metrics that ended up being equivalent. First, minimizing loaded tiles assuming a hierarchical tiled map. The cost of moving x horizontally is just x/y tiles, using y as the side length of the viewport. Zooming from y_0 to y_1 loads abs(log_2(y_1/y_0)) tiles, which is consistent with ds = dy/y. Together this is just ds^2 = (dx^2 + dy^2)/y^2, exactly the upper-half-plane metric.

Alternatively, you could think of minimizing the "optical flow" of the viewport in some sense. This actually works out to the same metric up to scaling - panning by x without zooming, everything is just displaced by x/y (i.e. the shift as a fraction of the viewport). Zooming by a factor k moves a pixel at (u,v) to (k*u,k*v), a displacement of (u,v)*(k-1). If we go from a side length of y to y+dy, this is (u,v)*dy/y, so depending how exactly we average the displacements this is some constant times dy/y.

Then the geodesics you want are just the horocycles, circles with centers at y=0, although you need to do a little work to compute the motion along the curve. Once you have the arc, from θ_0 to θ_1, the total time should come from integrating dtheta/y = dθ/sin(θ), so to be exact you'd have to invert t = ln(csc(θ)-cot(θ)), so it's probably better to approximate. edit: mathematica is telling me this works out to θ = atan2(1-2*e^(2t), 2*e^t) which is not so bad at all.

Comparing with the "blub space" logic, I think the effective metric there is ds^2 = dz^2 + (z+1)^2 dx^2, polar coordinates where z=1/y is the zoom level, which (using dz=dy/y^2) works out to ds^2 = dy^2/y^4 + dx^2*(1/y^2 + ...). I guess this means the existing implementation spends much more time panning at high zoom levels compared to the hyperbolic model, since zooming from 4x to 2x costs twice as much as 2x to 1x despite being visually the same.