I have been out of puzzle world for sometime. Leaving IIT Kanpur meant leaving company of theoretical souls such as Ramprasad, Deepanjan and Sagarmoy. Interactions with them drove me to write most of the articles I write here.

Anyways, I had an interesting chat with Ramprasad and pretty much copying the conversation.

Imagine four unit circles centred around the 4 points (1,1) (1,-1)

(-1,1) (-1,-1). All the four circles of course can be enclosed within

a square of side two (corners at (plusminus 2, plusminus 2)). Now

there is this tiny gap around the origin right? Put the largest

possible circle there that touches all these other circles. What is

the radius of this circle? I won’t insult your intelligence; by

simple geometry it is sqrt(2) – 1.

Now repeat this in 3 dimensions. Unit spheres at all points (plusminus

1, plusminus 1, plusminus 1), and draw the largest sphere centred at

the origin to touch all these sphere. What is its radius now? Again,

easy to see it is sqrt(3) -1.

Now what is the answer to the same question in n-dimensions? It is sqrt(n) – 1.

Okay good. Here is the freaky part. The (higher-dimensional)cube that

encloses all these circles is still of side-length two. The sphere you

drew at the origin has radius sqrt(n) – 1, which is way way larger

than 2 for large n. So the sphere you drew there actually *leaks out*

of the box.

I went insane after first hearing about this. It took me quite a while to get my head around this mathematical fact. Yes, indeed this is a fact, no matter how counter-intuitive it sounds. There is not catch or fallacy here. It’s hard to imagine dimensions beyond 3. But probably here is an explanation why the inner sphere *leaks* out of the box.

If it helps, imagine the following picture. Take a box of side-length

4, and put 4 circles in each of the corners that touches two sides of

the square, but make the radius of the circles something like 1/100.

Now, if you draw a circle centred at the origin but touching these

four 1/100 radius circles, you would see that the circle actually goes

out of the square in the middle of the edge. This I guess is similar

to what happens in higher dimensions. The volume of circles in higher

dimensions is really tiny.

Of course, even in higher dimensions the spheres touch the adjacent sphere which is not the case in the explanation above with circles of radius 1/100. But one can see that the cube side-length remains 4, the main diagonal of the cube is sqrt(16n). Hence, the inner sphere, does not go out of the main diagonal but it very well can *leak* out of the box from one of its face (or surface or facet, whatever you want to call them).

For N=9, it would be the first time when the inner sphere can not be fully contained within the N-dimensional box that we defined earlier.

Let’s talk about volumes The volume of a sphere of radius r in n dimension is. pi^{n/2} r^n /

Gamma(n/2 + 1) [Gamma(x) is roughly like the factorial of x, but

defined even all reals. Gamma(x) = xGamma(x-1), and Gamma(1/2) =

sqrt(pi)]. Please see more about Gamma Functions here.

Now, the thing is that the volume of an N-dimensional sphere starts decreasing after the dimension grows above 5 because the denominator increases faster as compared to the numerator in the equation of the volume!! And as N-tends to infinity, the volume approaches zero!!!

My mind has shaken up enough for the day!! That’s why I am going on a 5-day leave starting tomorrow (thanks to easter holidays).

I knew there must be a good reason why I hate geometry :-)

Tags: geometry, high-dimensional sphere, math

April 9, 2013 at 6:52 am |

i did n’t get the part why you would have a sphere (3-dimension) in the gap created in N-dimension.

April 9, 2013 at 10:20 am |

Well,

I was talking about the N-dimensional space so the dimension of the sphere (manifold) is N-1. For example on a plane (2D) when you draw a circle, the circle itself is a 1D manifold embedded in a 2D space. If you do not know what a manifold is, please refer to wikipedia or wolfram mathworld.

April 22, 2013 at 6:58 am |

For two dimensions we put a circle of maximum possible radius in the gap at the center

Similarly for three dimensions we put a sphere of maximum volume(not area) in the central gap

for the fourth dimension we will maximize some other metric lets say x. (what area is in 2 dimensions and volume in three dimensions, x is the the equivalent in 4 dimensions) not volume.

Your argument based on volume is hence incorrect.

To answer the queston,

note that we use area in 2 dimensions

and volume in 3 dimensions

now imagine if we are unware of the third dimension and live in a world of two dimensions

we would argue similarly that the area(surface area) of higher dimension cubes would soon become too big that it will overshoot the box. But it is not the case because we use volume(which we dont know about as we live in 2 dimensional world) to see if the sphere would fit in the space

April 22, 2013 at 10:09 am |

Well,

Your argument is more philosophical and it seems you do not understand dimensionality much. Since you seems to be too disturbed about the terms area ( which is nothing but volume in 2 dimension ) and volume. Let us define a measure m.

A sphere in N-dimension is defined with (p,r) where p is a point in N-dimensional space and r is the radius. Now, a sphere in N-dimension is nothing but a set of points which are AT MOST r distance from p in any direction. Now, this measure m is nothing but the integration of the (N-1)-dimensional manifold that wraps the sphere in question. Remember that integration gives area under a curve ( measure m under 1D manifold in a 2D plane ). I have given the formula on how to calculate volume or measure of a sphere in N-dimension. I request you to read more about manifolds and higher dimensions before you quibble over terminology such as “area” (volume in 2D), “length/periphery” (volume in 1D) and volume.

Regards,

November 14, 2013 at 11:47 pm |

Hello,

I am a high school student, so perhaps this is just my limited understanding of the subject, but something doesn’t make sense to me here.

For the 2-dimensional model, we had an inner circle with a radius of sqrt(2) – 1 and an outer square area of 2^2.

For the 3-dimensional model, we had an inner circle with a radius of sqrt(3) – 1 and an outer box volume of 2^3.

So, for the N-dimensional model, we had an inner circle with a radius of sqrt(n) – 1, so wouldn’t we have an outer box volume (or whatever the shape would be called in other dimensions) of 2^n.

If that part is true then the example of N=9 (or this model in the 9th dimension) would be:

sqrt(9) -1

3-1

2

area of the circle:

(3.14)(2)^2

12.56

area of the outer box:

2^9

512

Which makes more sense to me than having the inner circle leak out of the box. Of course, I could be wrong (I’ve been wrong once or twice) and if I am please explain it to me.

P.S. No offence intended by showing all of my steps like that. it is sort of habit for me anymore :/

Thanks, Krys

November 15, 2013 at 9:55 am |

Hi Krys,

No offence taken :-).

The calculation you show for the “area” of the circle is the “area” of the circle IF it is projected onto a 2d plane (pi*r^2). For three dimension it would be (4/3 * pi *r ^3). Where as the “area” that you calculate of the outer box is indeed 2^9 since you are calculating it in 9 dimensions. Please refer to the formula about how to calculate “area” or “volume” of a sphere in 9-dimension.

Besides, the easy way to see is that sqrt(9)-1=3-1=2. So the circle has a radius 2 and its diameter is 4. But each side of the box is still of unit 2. It is easy to see now that the sphere will leak out of the box since a sphere with diameter 4 can not fit inside a box where every side is at most 2.

December 26, 2013 at 2:38 pm |

Hope you enjoy the following, which is more of a research problem than a puzzle (but sounds like a puzzle).

What is the sidelength of the largest square centered at 0 that contains no point with all integer coordinates? (You are allowed to rotate the square any way you want). It’s not hard to see the answer is sqrt(2).

How about the side length of the largest n-dimensional cube centered at 0 that contains no point with integer coordinates? My knee-jerk reaction was sqrt(n), but the 2-dimensional case is completely misleading (because in two dimensions a cube and a cross-polytope are the same thing). In fact it’s no more than C*sqrt(log(n)), for a relatively small constant C which is independent of n. An in fact, there is a conjecture that it’s even less: that the side length of a cube that contains no integer coordinates is bounded above by a constant independent of dimension.