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247 points nabla9 | 1 comments | | HN request time: 0s | source
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FabHK ◴[] No.41833144[source]
For high-dimensional spheres, most of the volume is in the "shell", ie near the boundary [0]. This sort of makes sense to me, but I don't know how to square that with the observation in the article that most of the surface area is near the equator. (In particular, by symmetry, it's near any equator; so, one would think, in their intersection. That is near the centre, though, not the shell.)

Anyway. Never buy a high-dimensional orange, it's mostly rind.

[0] https://www.math.wustl.edu/~feres/highdim

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hansvm ◴[] No.41833317[source]
It's basically the same idea in both cases. Power laws warp anything "slightly bigger" into dominating everything else when the power is big enough. There's a bit more stuff near the outside than the inside, so with a high enough dimension the volume is in the rind. Similarly, the equator is a bit bigger than the other slices, so with enough dimensions its surface area dominates.
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WiSaGaN ◴[] No.41834537[source]
Yes, this seems to be the result of the standard Euclidean metric rather than the high dimension itself. I guess most people assuming the metric to be Euclidean, so it's ok.
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1. hansvm ◴[] No.41863896{3}[source]
The conditions you need for that to be true are substantially weaker than being Euclidean (though, when people are talking about "weird" behavior in high-dimensional spaces nowadays, it's in the context of ML and Euclidean stuff anyway). If you have a meaningful notion of dimension (basic properties like <1,0> being different from <0,1> and <1,0> being closer to <1,1> than <0,1>) and and don't have discretized shenanigans (which would collapse the inequality I'm exploiting to a strict equality, with some sort of 1^n=1 behavior) then the natural measure induced by the metric in question will exhibit the described behavior. You can easily verify that for every metric represented in scipy or whatnot, and a proper proof isn't too much more work.