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.
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.
The nice thing about deciding on a distance metric is that it gives you both a path (geodesics) and the speed, and if you trust your distance metric it should be perceptually constant velocity. I agree it's non-euclidean, I think the hyperbolic geometry description works pretty well (and has the advantage of well-studied geodesics).
I did finally find the duration logic when I was trying to recreate the path, I made this shader to try to compare:
https://www.shadertoy.com/view/l3KBRd
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.
i'm not clear from that how many trials were run for each test condition, but the percentage is average speed reduction, not a chance for binary hit/not hit. edit: the paper pdf says up to three trials each.
Wikipedia (citing a book on archive.org but no page number) [1] claims the largest order of an element of the Rubik's cube group is only 1260, so the simple repetition strategy would have to have a repeated unit way too long to be practical. (It sounds like you could potentially get a longer sequence if you allow "rotating the cube" as a move vs just turning the sides, though probably not enough to matter)
The linked circuit does have a repeating pattern like this at its core, and visits entire cosets of a subgroup at a time (presumably the same way each time) so it seems like there's already a bit more structure than a random 200MB file.
The example is poorly chosen in terms of practicality for this reason, but otherwise, no, this is a poor summary that misses something interesting.
The memory layout isn't changing in the faster versions, and there are no additional cache misses. It's easy to convince yourself that the only difference between the naive linked list and assuming linear layout is the extra pointer load - but TFA shows this is false! The execution pipeline incurs extra costs, and you can influence it.
Restated, if abelian G acts transitively on a set X, X and G have the same size. There's a tacit assumption, then, that you want as many possible states as possible, which the group action result immediately belies.
I'm not sure the author of TFA really thought through the implications of the "block" stuff, all of the conclusions feel pretty uninspiring. The elliptic curve solution is just taking G to be cyclic with prime order (smaller than 2^n). This avoids some pathological behavior that power-of-two abelian groups give you for the multi-set use case - collision probabilities are sort of bunched up around power-of-two multiples, with some unlucky hashes having extremely low order and e.g. adding two of an element doubling the number of potential collisions.
The index of refraction (including its dependence on frequency) fully describes the physics, there's nothing else to look up. Group velocity is a derived property taking into account the mathematics of waves, and in the typical textbook example of a very-nearly-monochromatic wave can be computed from the derivative of the index of refraction with respect to frequency.