For anyone still following the random-cross-sections-of-a-cube saga, I should mention that I've finally reopened my copy of Klain and Rota's "Introduction to Geometric Probability" (after a decade-long hiatus) and checked that Crofton's formula (Theorem 9.3.2) confirms that a random nonempty cross-section of a cube has area 2/3 and perimeter pi on average. However, I don't know of any variant of Crofton's formula that would compute the expected number of vertices of the intersection. Klain and Rota don't go into that sort of thing because they're concerned with valuations, and the "count-the-corners" map from convex polygons to integers isn't additive the way area and perimeter are. If any of you know of variants of Crofton's formula and/or the kinematic formula that apply to those more general sorts of questions, please let me know! Jim Propp On Thu, Aug 8, 2019 at 7:41 PM James Propp <jamespropp@gmail.com> wrote:
I. What is the question?
One can describe the measure on Graff(3,2) as follows:
1) The orientations of the planes (viewed as elements of the projective plane)
are uniformly distributed relative to the measure on the projective plane that
you get by taking the uniform measure on the 2-sphere and modding it out by
the antipodal identification.
2) The restriction of the measure to an orientation-equivalence class of
planes looks just like Lebesgue measure on the real line.
This is an infinite measure, but if we restrict this measure to the subset of
Graff(3,2) consisting of those planes that intersect some ball centered at the
origin, we get a finite measure, which can be rescaled to yield a probability
measure. Here's a concrete way to sample from that probability measure:
1) Choose a point P uniformly at random on surface of the ball and draw the
line segment connecting it to its antipode P'.
2) Choose a random point Q uniformly at random from segment PP', and draw
the plane through Q perpendicular to PP'.
More generally, if we want to sample from those planes that intersect some
compact convex body K, we can find a ball B containing K and use rejection
sampling: repeatedly choose a random plane intersecting B until you find one
that intersects K.
There are other ways of describing the measure on Graff(3,2), and other ways
to sample from it, but I like it because the rotational invariance of the
measure is obvious. The same cannot be said for schemes that sample a plane
ax+by+cz=d by choosing a, b, and c independently and then rejecting them
if some inequality fails to hold.
Anyway, this well-defined notion of a random plane that cuts a cube allows
us to ask "What is the expected perimeter of a random cross-section of the
unit cube?"
II. Random cross-sections of a square
It turns out that answering the preceding question in R^3 requires us to
answer a similar question in R^2: "What is the expected length of a random
cross-section of the unit square?" Here we are taking the natural measure
on Graff(2,1) (the set of lines in the plane), restricting it to those
lines that intersect a fixed unit square, and turning it into a probability
measure.
Here I will cite without proof a couple of results relating to mean width
in two dimensions (I won't recite the definition of mean width since you
can get that from Wikipedia). First, the mean width of a convex compact polygon
A in the plane is its perimeter divided by pi; second, the average length of
a random nonempty cross-section of A equals the area of A divided by the mean
width of A. (For the proof of the analogous statement in three dimensions,
see https://mathoverflow.net/questions/335293/mean-cross-sectional-area .)
Since the unit square has perimeter 4, its mean width is 4/pi; and since its
area is 1, the average length of a cross-section of the square is pi/4.
Now imagine that the Euclidean 2-space E^2 that contains our unit square
is actually sitting in a Euclidean 3-space E^3, and consider the subset of
Graff(3,2) consisting of all planes that intersect that square. What is the
expected length of the intersection of such a plane with the square? It takes
a little thought to see that the answer must be the same as it is one dimension
down, but it's true because of the symmetry properties of the measure on
Graff(3,2). Specifically, if you look at any line in E^2 that intersects
the unit square in a particular line segment, then spinning that line around
the line segment gives you all the planes in E^3 that intersect the unit square
in that particular segment, and the measure is rotationally uniform. (This
is the only part of the argument I might have some trouble writing down
formally, but I'm sure it's right.)
The upshot is, if you look at a random plane that intersects a square sitting
in 3-space, the expected length of the intersection is pi/4.
III. Vital digression: the Wall of Fire Theorem
The Wall of Fire Theorem says that the expected number of sides in a random
cross-section of a cube is 4. This tells us that the probability that a random
plane that intersects the cube intersects a particular face of the cube is
4/6. That's because we can write the number of sides of the cross-section as
a sum of six indicator variables, one for each face of the cube, equalling 1
when random plane intersects the face in question and equalling 0 otherwise.
Linearity of expectation, combined with symmetry, tells us that the expected
number of sides of the intersection (which we know equals 4) must be equal to
6 times the probability that the random plane intersects a particular face of
the cube. So that probability must equal 4/6, aka 2/3.
IV. Random cross-sections of a hollow cube
Consider a random plane that intersects a unit cube, and consider a specific
face of the cube. With probability 2/3, the plane intersects the face, and
conditional upon that event, the expected length of the intersection is pi/4.
On the other hand, with probability 1/3, the plane doesn't intersect the face,
and conditional upon that event, the expected length of the intersection is 0.
So the unconditioned expectation of the length of that intersection is equal to
(2/3)(pi/4) + (1/3)(0) = pi/6.
Lastly, if we sum over all the faces of the cube, we find that the expected
length of the intersection between the random plane and the surface of the
cube is (6)(pi/6) = pi. So this is indeed the expected perimeter of a random
cross-section of the unit cube.
V. Random cross-sectional area
To prove that a random cross-section of the unit cube has area 2/3, one can
appeal to the MathOverflow post I cited before, showing that the average
cross-sectional area of a convex compact body in 3-space equals the volume
divided by the mean width, and the fact that a cube of side-length s has
mean width 3/2. I know of a nice way to prove the latter fact by way of
Archimedes' hat-box theorem, but this email has gone for way too long already.)
To recap: a random cross-section of the unit cube has 4 sides, has area 2/3,
and has perimeter pi. I think that's swell. Thanks for helping me figure
this out!
Jim Propp
On Thu, Aug 8, 2019 at 6:27 PM James Propp <jamespropp@gmail.com> wrote:
Cool!
Better still, I think I know how to prove this. That is, I think I can prove that on average a random cross-section of a unit cube has perimeter pi.
More soon,
Jim
On Thu, Aug 8, 2019 at 10:21 AM Tomas Rokicki <rokicki@gmail.com> wrote:
And here are the final lines after an overnight run. Note that accumulating billions of doubles in this fashion loses precision; I should have done my summing more carefully. But at least five of the digits are probably good.
At 8589934592 pts 4.0000236480264 area 2.66669753208071 perim 6.2832343008038
At 17179869184 pts 4.00001316709677 area 2.66668286448053 perim 6.28322083261956
At 34359738368 pts 4.00000967484084 area 2.66667855572835 perim 6.28320124178968