A dust that almost every line misses

A length counted by the lines that cross it measured a curve by counting the straight lines that cross it: drop lines at random, count the crossings, and the average count is proportional to the curve’s length. That is Crofton’s formula, and it works for any curve that has a length. It ended by asking what happens for a set that has a length but is not a curve.

Here is such a set. Take a square and divide it into sixteen smaller squares. Keep the four at the corners and throw away the rest. Do the same inside each of the four kept squares, then inside each of the sixteen squares that remain, and continue forever. What survives is a dust of points, the four-corner Cantor set. At every stage it is made of $4^n$ squares of side $4^{-n}$, so the sum of their sides is exactly 1 at every stage, and in the limit the dust has a length in the sense that the one-dimensional measure assigns one — positive and finite, like a segment’s.

And yet almost every straight line misses it. Its shadow on almost every line has length nought. This essay computes the shadows exactly, stage by stage and direction by direction, finds the few directions where the shadow never shrinks and the arithmetic reason for them, and measures how slowly the chance of hitting the dust falls — a rate that has been bracketed but not found.

Four corners, then four corners of each #

The construction is the two-dimensional cousin of the middle-thirds Cantor set of almost none of it left and still uncountably many. There, a segment kept its two outer thirds at each step, and what remained had total length nought but uncountably many points — enough to carry the staircase that is flat almost everywhere, which climbs from nought to one while standing still off a set of length nought. Here a square keeps its four corner sixteenths, and the arithmetic comes out differently: four pieces, each a quarter the size, is exactly the balance at which a set is one-dimensional.

That balance is what gives the dust a length. Cover it at stage $n$ by its own $4^n$ squares, each of diameter $\sqrt 2 \cdot 4^{-n}$, and the total of the diameters is $\sqrt 2$ at every stage — bounded, not growing, not shrinking. A finer argument shows that no cover can do much better, so the one-dimensional measure of the dust is positive and finite. In the language of a dimension that is not a whole number, its dimension is $\log 4/\log 4 = 1$ exactly.

So the dust is one-dimensional in every sense that measures size — the covering dimension of a dimension for every rate of crowding and the box-counting one agree at exactly 1. What it is not is a curve. No two of its points are joined by any path inside it, because between any two of the kept squares at some stage lies a gap. It is a set of length one scattered into infinitely many pieces, and the question is what a line sees when it passes through.

The shadow in a general direction shrinks #

A line in a given direction meets the dust exactly when it passes through the dust’s shadow — its projection onto a perpendicular line. So the question of how many lines in that direction hit the set is the question of how long its shadow is. The shadow of the whole square in direction $\theta$ has length $|\cos\theta| + |\sin\theta|$; the shadow of each stage of the dust is a union of intervals inside it.

The shadows can be computed exactly. Stage $n$ is four copies of stage $n - 1$, each shrunk by a factor of four and shifted to a corner, so its shadow is four copies of the previous shadow, shrunk and shifted, merged wherever they overlap. At 37°, the four shrunken copies overlap a little at every stage, and every overlap is lost length. The shadow’s share of the square’s shadow goes $1$, $0.855$, $0.730$, $0.630$, $0.557$, $0.495$, $0.446$ over stages nought to six.

The losses compound. Besicovitch proved in 1939 that for a set like this — one-dimensional, with finite length, and containing no piece of any curve — almost every shadow of the limit has length nought. The losses at 37° never stop, and the shadow of the dust itself in that direction is a set of length nought.

The shadow in the direction one, two does not #

In the direction $(1, 2)$ — at an angle of about 63.4° — nothing is lost at all. The shadow of every stage is a single interval, the full length of the square’s shadow, and the right-hand column of the figure reads $1.000$ at every stage. The reason is arithmetic and exact.

Projecting onto the direction $(1, 2)$ amounts to computing $x + 2y$. A point of the dust has coordinates $x = \sum a_i 4^{-i}$ and $y = \sum b_i 4^{-i}$ in base four, with every digit $a_i$ and $b_i$ equal to 0 or 3, since each step keeps only the corners. Then $x + 2y = \sum (a_i + 2b_i) 4^{-i}$, and $a_i + 2b_i$ takes the values $0$, $3$, $6$ and $9$ — which are $3 \times 0$, $3 \times 1$, $3 \times 2$, $3 \times 3$. So $x + 2y$ is three times a number whose base-four digits are all four possible digits, freely chosen. That is every number between 0 and 1, and $x + 2y$ fills the whole interval from 0 to 3.

The figure shows it at stage 2: the sixteen squares, labelled by their two digits of $x + 2y$, cast shadows that lie end to end in order, $00, 03, 06, \ldots, 99$, filling the interval exactly. The same happens at every stage, which is why the shadow never shrinks. The direction $(2, 1)$, at 26.6°, works for the same reason with the roles of $x$ and $y$ exchanged.

Most directions lose, a few hold #

Across all directions the picture is jagged. At 0° and 90° — looking straight along an edge — the shadow is itself a Cantor set, the base-four numbers with digits 0 and 3 only, and it shrinks fastest. Near 45° it holds up better. And at 26.6° and 63.4° two spikes stay at the full length through every stage, with smaller peaks at other directions of rational slope where the digits partly cooperate.

The spikes are a measure-zero exception. Besicovitch’s theorem says the directions in which the dust’s shadow keeps positive length form a set of directions of total measure nought; which directions those are is a question about digits, and the arithmetic of the previous section puts $(1, 2)$ and $(2, 1)$ among them. Richard Kenyon studied the analogous question for a three-cornered dust in 1997 and found that only directions of rational slope can qualify there. The spikes in the figure are infinitely thin: a direction slightly off $(1, 2)$ loses length at the finer stages, and the curves in the figure show the peaks narrowing as the stage increases.

So the shadow has two faces. In each particular direction, the question of whether it vanishes is a question about digits and rational numbers. On average over directions, the shadow shrinks — and the average is what a random line sees.

The chance a random line meets the dust #

Average the shadow’s length over all directions and the result is the Favard length of the set, named after Jean Favard. By Crofton’s formula it is proportional to the chance that a random line meets the set, and dividing by the square’s value turns it into exactly that chance, among the lines that meet the square. Stage 1 is met by $82.5\%$ of them, stage 2 by $72.9\%$, stage 4 by $61.2\%$, stage 8 by $48.2\%$.

The limit is nought: the dust itself is met by almost no line, which is Besicovitch’s theorem again, averaged. What is not known is how fast the chance falls. Fedor Nazarov, Yuval Peres and Alexander Volberg proved in 2010 that stage $n$’s Favard length is at most a constant times $n^{-1/6 + \varepsilon}$ for any $\varepsilon > 0$. Michael Bateman and Volberg proved in the same year that it is at least a constant times $(\log n)/n$. The truth lies somewhere between a rate like $1/n$ and a rate like $n^{-1/6}$, and nobody knows where.

The computed values cannot settle it and the figure makes no attempt to. By stage 8 the dust has 65,536 squares, and the values fall by only a few per cent a stage — consistent with either rate, or with anything between, over a range this short. The known bounds were proved by arguments about how the shadows in different directions are correlated, and the gap between them is a gap in understanding those correlations, not in computing power.

Fewer lines than a straight segment of the same length #

There is a natural yardstick for these numbers. A straight segment of length 1 placed inside the square casts a shadow of length $|\cos\theta|$ in direction $\theta$ — its own length times the cosine of the angle — and averaging over directions gives $2/\pi$. The square’s average shadow is $4/\pi$. So a random line that meets the square meets a unit segment inside it with chance exactly one half.

The dust has the same length as that segment, in the sense of one-dimensional measure, and at first it is far easier to hit: stage 1 is met by 82.5% of lines, because its four squares are spread across the whole square and a line has four chances. But the chance keeps falling, and between stage 7, at $50.7\%$, and stage 8, at $48.2\%$, it drops below the segment’s one half. From the eighth stage on, a dust of length one is harder to hit than a straight stick of length one. In the limit it is hit by nothing.

The comparison is exactly the content of Crofton’s formula. For a curve, the average number of times a random line crosses it is proportional to its length, and a straight segment, crossed at most once, turns that average into a probability without loss. A curve that doubles back is crossed several times by some lines and missed by others, so it is hit less often than its length suggests. The dust takes that to the extreme: a length spread so thinly across so many directions that the lines crossing it are, on average, crossing it many times over in a vanishing few directions and not at all in the rest. Which curves have a length at all measured length by inscribed polygons; the dust has no inscribed polygons, since it has no paths, and its length exists only in the sense of covers.

A needle dropped on a dust #

The chance can also be measured directly, by throwing lines. Pi from dropped needles estimated $\pi$ by dropping a needle on ruled paper, which is Buffon’s problem of 1777; here the ruled paper is replaced by the dust and the needle by a whole line. Choose the line as Crofton’s formula requires — its direction and its distance from a fixed point uniform together, restricted to lines that meet the square — and count how often it meets a square of stage 3.

Among 40,000 random lines, 66.5% meet the stage-3 dust; the exact average shadow gives 66.3%, within the sampling error. The figure shows sixty of them. The random lines and the averaged shadows are two computations of one number, and their agreement is the check that the lines are chosen with the right measure — a sampling that picked the direction uniformly and then the offset within the square’s shadow in that direction would have weighted the directions wrongly, and it was caught that way, disagreeing by more than four standard errors.

So the Favard length is a probability that a physical experiment could measure. A needle long enough to span the square, dropped at random onto a scatter of $4^n$ tiny tiles in this arrangement, hits some tile with chance $48\%$ at stage 8 — and with chance falling to nothing as the stages continue, at a rate nobody can yet name.

Where the dust meets the Kakeya sets #

The dust has a mirror-image cousin. No set with a line in every direction is small described Kakeya’s sets, which contain a whole line segment in every direction and can still have area nought. The four-corner dust is the reverse: it has length in the strongest sense and meets almost no line. Both show that a set’s size and the lines it meets are nearly independent.

The connection is more than an analogy. The techniques Nazarov, Peres and Volberg used to bound the Favard length — controlling how often the projections in many directions are simultaneously small — are the same kind of estimates that Kakeya problems need, and progress on one has repeatedly come with progress on the other. The Favard length question is a quantitative Besicovitch theorem, as the Kakeya conjecture is a quantitative statement about lines in every direction, and neither is fully settled.

Still open: the rate of disappearance #

Between $(\log n)/n$ and $n^{-1/6}$ is a wide gap. For random versions of the dust — where each step keeps four random squares of the sixteen, rather than the corners — the rate is known: Peres and Boris Solomyak showed it is of order $1/n$. The corner dust is not random, and the arithmetic that makes the direction $(1, 2)$ special is exactly what might slow its shadows’ disappearance in nearby directions and push the rate towards the upper bound.

The question is whether the special directions matter on average. Each one is a single direction of measure nought, but near it the shadow shrinks slowly, and there are infinitely many special directions of rational slope. If the slow directions, weighted together, cost enough, the rate is slower than $1/n$. The best current arguments cannot tell whether they do, and the bounds have not moved much since 2010.

The question has also spread beyond this one dust. Matthew Bond, Izabella Łaba and Volberg extended upper bounds of the Nazarov–Peres–Volberg kind to wider families of self-similar sets built from product patterns, and in each case the gap between the upper and lower rates remains. What would close it is not a better computer but a better description of how a set’s shadows in nearby directions are correlated: whether a direction in which the shadow is small forces the shadows in nearby directions to be small too, and over how wide a neighbourhood. For the corner dust that correlation is a statement about base-four digits and small rational slopes, and it is the kind of statement for which number theory rarely offers a clean answer.

What the pictures cannot show #

The shadows here are exact unions of intervals, computed by the self-similarity without approximation; the averages over directions use 1,200 directions, and the results change by less than a thousandth when the number is halved. The Favard lengths up to stage 8 are therefore reliable to about three places, and doubling the number of directions changed none of them by more than four ten-thousandths. Nothing computed at stage 8 bears on the asymptotic rate, which the bounds quoted — Nazarov, Peres and Volberg’s, and Bateman and Volberg’s — are about.

The spikes at rational directions are exact for the two directions checked, and the figure shows their neighbourhoods narrowing, but it does not settle which other directions, if any, give shadows of positive length. And the limit set itself cannot be drawn at all: every picture here is of a stage, a union of squares, and the statement that almost every line misses the dust is about what those unions tend to, not about any of them.