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Comment by scatters | Hacker News Reader

I don't think it's correct to see the n-sphere as poking out of the sides of the cube; the sphere has constant curvature so it has no protrusions. Rather, the n-cube gets more and more spiky so the constraining spheres get (relatively) smaller and further away from the centre, so constrain the central sphere less and less.

The spikiness of n-cubes is apparent when you look at the (solid) angle at each vertex; it starts constant (1/2 pi radian in 2D, 1/2 pi steradian in 3D) but reduces thereafter at an increasing rate (1/8 pi^2, 1/12 pi^2, 1/64 pi^3, etc.).

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The spiky view of n-spheres is due to all of their volume getting concentrate next to the axes - the proportion of volume which is less than ε away from an axis to the entire volume of the sphere tends to 1 exponentially with n, which leads to the spiky visualization as we imagine 3d but with more axes.

I don't think this is the right way to process this intuitively. When you use words like "volume getting concentrated" it sounds like there is some non-uniformity in the sphere, but the non-uniformity is really in our intuition about space.

What's weird isn't the sphere, it's distance, and I think that's easier to process. Going from a (1d) sidewalk to a (2d) football field to a (3d) ocean, it's easier to see our intuitions about distances slowly breaking down.

>it sounds like there is some non-uniformity in the sphere

How I am understanding this is that the non-uniformity is in the cube, and I think it's very helpful in visualizing it.

I can imagine in 3d, the centers of the sides of the cube being pulled in, so that it's kind of hyperbolic looking.

That's not accurate either. Neither the cube nor the sphere "bend" in any way. Irrespective of the number of dimensions, the cube's surfaces remain (hyper)planar, and the sphere retains its convex shape and constant curvature.

The diagram on the web page is just wrong. The enclosed sphere does not poke out with "spikes".

The error in the diagram is that it shows the corner cubes "filled" with spheres, but this is not what happens in the higher-dimensional analogues. If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all.

Think of a 2D slice through the middle of the 3D example: You'd just see a square, no circles!

>If you take a cube-shaped cross section of a 4D example (through the center), you wouldn't see the corner spheres at all.

Wait wait wait, I thought the "cube" had the same number of dimensions as the "spheres".

They're the same. So square + circle, cube + sphere, 4D hypercube + 4D hypersphere, etc...

But cross-sections don't look the way you think they do. So if you draw a 3D intersection through the 4D case, it looks much like the 2D intersection of the 3D case, etc...

In all such "intersection" diagrams you don't see the spheres in the corners of the cubes. The conceptual diagram on that page showing the high-dimensional case shows the corner spheres and the middle sphere somehow "squished out" in a spiky way.

This is not at all what happens.

I'm not certain I know what you mean, but it doesn't seem to work. If you mean to take epsilon neighborhoods in Euclidean norm of the axes, then I'm skeptical that that contains most of the mass of the sphere, since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes.

If instead you mean most of the mass of the ball is in the points which have all but one coordinate within epsilon of zero, then the intuition doesn't follow. It's equally true that you get most of the mass considering just points with all coordinates within epsilon. And for that you get that the mass of the ball is concentrated in an epsilon cube at the origin, which excludes exactly the spikes that you were basing the intuition on. The weird thing in this thought experiment is the epsilon cube not the sphere. For example the epsilon cube contains points much further than epsilon from the origin and so it's maybe not surprising that it contains most of the sphere.

>since simulating random points on a thousand dimensional sphere doesn't seem to give points near these axes.

that is kind of circular argument as "random" really depends on the density measure underlying the chosen sampling distribution.

If only there was some kind of natural density measure to use in an Euclidean space R^D…

It's no more ill defined than volume itself (and less so since it doesn't need an arbitrary scale).

I don't understand your 'less so' comment. Could you please elaborate?

If you want to define volume in D-dimensional space then you need to choose what the unit volume is, for example that a unit hyper cube has unit volume one. (That choice is relevant for these discussions and you could equally have chosen a unit ball to have volume one since you're defining D-dimensional volume you get to choose.) But a probability distribution is always normalized to integrate to one, which means either choice of volume definition will give you the same uniform probability distribution on the ball. So there's an extra arbitrary constant to choose for volume that isn't important in the probability distribution even though you define the probability density in terms of the volume.

OK. Though you could leave your volume definition parameterized, instead of picking a specific scale, couldn't you?

I do agree that volume and probability distribution are essentially the same.

(That's also why at my university measure theory and probability theory were handled in the same course.)

https://news.ycombinator.com/item?id=3995615

Perhaps you can be more specific since I don't see the claim here being discussed there.

The n-cube is spiky. The n-sphere is kinda defined by its non-spikiness, because it is rotationally invariant.

But there is an alternative visualization where the sides of the unit (hyper)cube are deformed because the location of the vertices gets further and further away from the origin. Of course neither is a true representation of the hyperdimensional case.

Love this explanation, visualizing corner vs centre is intuitive in a way higher dimensions usually aren't.

The radius of the sphere, assuming the cube has side length 1, is (sqrt(D)-1)/2. It seems worthy of note that this radius becomes arbitrarily large.

And the distance from a vertex of the unit cube to the center is sqrt(D)/2, which also becomes arbitrarily large.

https://www.math.ucdavis.edu/~strohmer/courses/180BigData/18...

> The result, when projected in two dimensions no longer appears convex, however all hypercubes are convex. This is part of the strangeness of higher dimensions - hypercubes are both convex and “pointy.”

Isn't it more accurate to say that hypercubes are convex and their projections are pointy?

Yes. Hence the quotes, I guess. But hyperspheres are convex and their projections are also convex. So it's not really appropriate either to say that they are "spiky" as in https://news.ycombinator.com/item?id=29969613

> the (solid) angle at each vertex; it starts constant (1/2 pi radian in 2D, 1/2 pi steradian in 3D)

Unstudied curiosity: steradians seem to be defined in terms of the point of a cone. But the corner of a cube is the intersection of several (for a cube, 3) planes, not a cone. How do you do the calculation of steradians?

Cone is one scenario to think about, but what it really comes down to is the fraction of the full sphere that your solid angle includes.

To make a 2D analogy, you can think of 2D angles as a portion of a circle represented as a fraction of 2pi radians. Cut it in half and you have pi radians, cut that in half (so a quarter of the circle) and you have pi/2 radians, or 90 degrees.

That 90 degrees "quarter of a circle" example is looking at it as the "point of a pie slice", but it's the same 90 degree 2D angle as you have in the corner of a square.

You can look at the corner of a cube the same way. The full sphere of solid angle is 4pi, a hemisphere is 2pi. Now take that hemisphere and cut it into quarters (1/8ths of a sphere). Each of those quarters is pi/2 steradians, and the solid angle at the center is the same solid angle represented by 1/8th of the sphere is the same solid angle you have at the corner of a cube.

Or to put it another way, you could pack the corners of 8 cubes around a point and it would leave no empty gaps, so the corner of each cube is occupying 1/8th of a "full" 4pi steradians.

Now that I've thought about it, I don't think we're on solid ground measuring pointiness in n-dimensional radians. A perfectly non-pointy angle in 2D is half the circle, or π radians. A perfectly non-pointy angle in 3D is half the sphere, or 2π steradians. π is less than 2π, but the straight-line 2D angle is not sharper than the flat-plane 3D angle. What's relevant to the pointiness is the proportion of the sphere, not the numeric value of the surface area.

It corresponds to area if and only if you're talking about a unit sphere.

In the same way that an angle of pi radians corresponds to pi length around the perimeter of a unit circle, but if you're measuring on the perimeter of a larger circle you'll have a correspondingly larger length.

The "4pi steradians in a sphere" comes from the area of a unit sphere (4×pi×r^2 = 4×pi). I've never had cause to use 4+ dimensional solid angle for anything, but you'd determine full n-dimensional radians the same way, I guess with a 4-dimensional sphere the "surface area" is a 3D volume on the perimeter of the 4D unit sphere's space?

> In the same way that an angle of pi radians corresponds to pi length around the perimeter of a unit circle, but if you're measuring on the perimeter of a larger circle you'll have a correspondingly larger length.

I understand this. It's not what I'm talking about.

There was a claim upthread that the corner of a square and the corner of a cube are equally sharp because (1) one of them is π/2 radians; (2) the other is π/2 steradians; and (3) π/2 is equal to π/2, hence equally sharp.

I'm saying I don't think this can be right, because when we consider two angles that we know are equally sharp, the 2D flat line and the 3D flat plane, one of them is π and the other is 2π, so unit-free radian measurements cannot be a valid measure of sharpness. The radian measurement needs to be normed against the full sphere in an equal number of dimensions, so that the flat line equals 1/2 (of a circle) and the flat plane equals 1/2 (of a sphere). I want to conclude that these two angles are equally sharp because 1/2 is equal to 1/2. But even if I can't do that, I need to avoid concluding that a flat line is twice as sharp as a flat plane, because I know that's untrue.

But my reasoning would tell us that the corner of a cube is twice as sharp as the corner of a square, 1/8 to 1/4.

Ah gotcha. I'm not even sure it's meaningful to compare the "relative pointiness" of angles between different dimensions. It's like trying to compare whether the area of a circle is bigger or smaller than the volume of a sphere. And then what if I only have 70% of a circle, but 80% of the sphere?

You could make up some metric like "the unit sphere occupies a greater volume fraction of the unit cube compared to the area fraction of a unit circle in a unit square", and compare like that if you only have some fraction of each circle/sphere instead of the whole thing, but is there any value in making that comparison?

> I'm not even sure it's meaningful to compare the "relative pointiness" of angles between different dimensions.

I think this is actually something we can't avoid. The problem with dodging the question this way is that it's really, really easy to translate a low-dimensional angle up into a higher-dimensional space.

Imagine I have two 2D vectors, U and V, with an angle between them. I can think of them as representing an infinite 1D surface consisting of the points (m·u_1 + n·v_1, m·u_2 + n·v_2) for all nonnegative real m and n.

It's trivial to define a 2D surface in 3D space representing exactly the same angle: it is the points (m·u_1 + n·v_1, m·u_2 + n·v_2, a) for all nonnegative real m and n and all real a. When the angle between U and V is θ radians, this surface will always express an angle of 2θ steradians. I don't think it's a stretch to say that the two angles, θ radians and 2θ steradians, must be equally sharp. (And indeed, the flat-line / flat-plane example is a special case of this one.)

I need to correct myself: I've defined my surfaces as linear combinations of the vectors forming the angle, which is wrong. Rather, the 1D surface expressing the 2D angle consists of the points in the set (mU ∪ nV) for all nonnegative m and n, and the 2D surface extending the angle into 3D space consists of the set [(m·u_1, m·u_2, a) ∪ (n·v_1, n·v_2, a)] for all nonnegative m, n and all a.

Thanks, that was helpful in thinking about it.

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