Pyjama Set Covering

The Quantitative Pyjama Covering Problem

Definitions and mathematical statement

Shared notation.

Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

For $0<\varepsilon<1/2$, let $P_\varepsilon=\{(x,y):\min_{m\in{\mathbb{Z}}}|x-m|\leq\varepsilon\}$. A permitted motion is a rotation $R_\theta$ about the origin, with no translation. Let $m(\varepsilon)$ be the minimum size of a finite family of angles for which $\bigcup_j R_{\theta_j}P_\varepsilon={\mathbb{R}}^2$, taking $m=\infty$ if none exists. Determine this number or its sharp growth as $\varepsilon\downarrow0$. [S1] [S2]

Statement recorded in the supplied source.

How many rotated (about the origin) copies of the "pyjama set" $\{(x, y) \in \mathbb{R}^2 : \operatorname{dist}(x, \mathbb{Z}) \leq \varepsilon\}$ are needed to cover $\mathbb{R}^2$? [S1]

Residual task reported by the archive.

The embedded review $2026-08-17$ records: Obtain substantially smaller upper bounds, such as epsilon^{-C}, and nontrivial matching lower bounds for the minimum covering number. [S1]

Short English statement

How many differently rotated families of narrow periodic strips are needed to cover the whole plane?

Research attempt

Provenance and scope.

This individual attempt was generated by GPT-6 Astra Ultra on 1 October 2026. The method was to derive explicit partial results and test the first proposed extension adversarially. The original statement and sources below are retained. No claim of mathematical novelty or complete solution is made.

Formal target and primary-source update.

For $0<\varepsilon<1/2$, a rotated pyjama set has the form $P_u=\{x\in{\mathbb{R}}^2:\operatorname{dist}(x\cdot u,{\mathbb{Z}})\leq\varepsilon\}$, where $u$ is a unit normal. Rotations are about the origin, with no independent translations. The inspected Kravitz--Leng paper proves a quantitative triple-exponential upper bound for the covering number. We do not reproduce that theorem. Instead we derive a density lower bound, solve a substantial large-width regime exactly, and explain why generic random normals cannot give a finite cover.

A uniform density lower bound.

Inside a disk of radius $R$, every fixed stripe family occupies fraction $2\varepsilon+O(1/R)$. One direct justification is to tile the plane by unit squares aligned with its normal and tangent. On each full unit square the occupied area is exactly $2\varepsilon$, because the normal coordinate runs through one period. Only $O(R)$ squares meet the disk boundary, and their total area is $O(R)$. Dividing by disk area proves the assertion, uniformly for each fixed rotation.

If $M$ such families cover the plane, their union covers every disk. The union bound for area therefore gives, after letting $R\to\infty$, \[ 1\leq2\varepsilon M,\qquad M\geq\frac1{2\varepsilon}. \] Overlaps can only strengthen this lower bound. It depends on global density, not on the behaviour near the origin, which belongs to every rotated family.

Exactly three rotations suffice for widths at least one third.

Choose three unit normals at mutual angles $120$ degrees, so $u_1+u_2+u_3=0$. Suppose a point avoids all three stripe sets at $\varepsilon=1/3$. Then each fractional part $t_i=\{x\cdot u_i\}$ lies strictly between one third and two thirds. Their sum is an integer, because the three actual projections sum to zero. But the sum lies strictly between one and two, containing no integer. This contradiction proves coverage. Wider stripes preserve coverage.

Two rotations never suffice when $\varepsilon<1/2$. If their normals are parallel, they define the same stripe family up to sign and leave uncovered points. If they are independent, the linear map $x\mapsto(x\cdot u_1,x\cdot u_2)$ is onto ${\mathbb{R}}^2$, so choose both projections equal to one half. That point lies outside both families. Hence the exact minimum is three for $1/3\leq\varepsilon<1/2$. For $\varepsilon\geq1/2$, one stripe family already covers the plane.

The chosen three-direction cover fails below one third. Prescribe $x\cdot u_1=x\cdot u_2=1/3$; independence gives such a point, and the third projection is $-2/3$. All three distances to integers are one third, so the point is uncovered for smaller widths.

A structural obstruction to generic choices.

Suppose the normals $u_1,\ldots,u_M$ have no nonzero integer relation $\sum k_ju_j=0$. Then the map $x\mapsto(x\cdot u_j\pmod1)_j$ has dense image in the $M$-torus. Here is the relevant harmonic argument. Choose a vector $v$ not orthogonal to any nonzero integer combination of the normals; the excluded set is a countable union of lines. Along $x=tv$, the average of every nonconstant torus character tends to zero, since it is the average of $e^{2\pi it\beta}$ with $\beta\ne0$. Approximation of continuous functions by trigonometric polynomials then implies that this trajectory meets every nonempty open torus box.

In particular it meets the box $(\varepsilon,1-\varepsilon)^M$, producing a point outside every stripe family. Thus normals without an integer relation cannot cover the plane for any width below one half. Generic randomly chosen normals have no such relation, so a random-direction strategy needs additional arithmetic structure. The successful three-direction example uses exactly the relation $u_1+u_2+u_3=0$.

Verification and remaining gap.

The open complement intervals are essential at the boundary width one third; equality belongs to the stripe sets and causes no loophole. The torus argument uses continuous averaging over the entire line, with no unjustified independence of projected coordinates. The density bound and the exact three-rotation regime are complete partial results. They leave the small-$\varepsilon$ growth largely unresolved: one needs structured families of many normals that cover with substantially fewer rotations than the known quantitative upper bound, or a stronger lower obstruction exploiting their mandatory integer relations.

Sources

• S1 ULAM.AI, UnsolvedMath, supplied problems.json, record 171 (GREEN-083), statement and background. The embedded literature-review date is 2026-08-17; that is the archive’s review date, not a fresh certification. Dataset landing page: https://huggingface.co/datasets/ulamai/UnsolvedMath.

• S2 F. Manners, A solution to the pyjama problem, Invent. Math. 202 $2015$, 239-270,

  \nolinkurl{doi:10.1007/s00222-014-0571-7.}

  https://arxiv.org/abs/1305.1514. Bibliographic information from the supplied record.

• S3 N. Kravitz, J. Leng, Quantitative pyjama, arXiv:2510.17744 $2025$. https://arxiv.org/abs/2510.17744. Bibliographic information from the supplied record.

• S4 B. Green, 100 Open Problems, Problem 41, version. https://people.maths.ox.ac.uk/greenbj/papers/open-problems.pdf. The author-hosted document was inspected on 27 September 2026; the archive’s local code is not a problem-number locator in this paper.

• E1 N. Kravitz and J. Leng, Quantitative pyjama, 2025. https://arxiv.org/abs/2510.17744. Inspected 1 October 2026.