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Periodic double tilings of the plane

T0 review · 3 major / 4 minor · reviewed 2026-08-08 · deepseek-v4-flash

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Pith's one-line read The least-perimeter periodic tiling of the plane by two unequal-area tiles is exactly one of three shapes, with an explicit interface-length formula.

desk verdict A genuine classification result with an explicit isoperimetric profile, but the proof of the connectedness property rests on an inadmissible competitor for the fixed-lattice problem. read the letter →

arxiv 2502.08396 v3 pith:DU2LGOPA submitted 2025-02-12 math.MG math.APmath.DG

Francesco Nobili , Matteo Novaga , Emanuele Paolini This is my paper · ORCID
classification math.MGmath.APmath.DG MSC 49Q0552C2058E12
keywords periodictilingsKelvinproblemisoperimetricprofileunequalcellareasperimeterminimizationReuleauxtrianglecurvilinearpolygonshoneycombtiling
verification ladder T0 review T1 audit T2 compute T3 formal

The pith

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

The reading

This paper determines the least possible interface length for a periodic tiling of the plane made of two kinds of tiles with prescribed unequal areas, and it writes the answer as explicit formulas. It proves that only three shapes can ever be optimal relative to any fixed lattice: two flat-edged hexagons, a curvilinear square paired with a chipped square, or a Reuleaux triangle paired with a nine-sided chipped hexagon. Which shape wins depends only on the ratio of the two tile areas, with two sharp transitions at approximately $x=0.062$ and $x=0.317$. Because the resulting isoperimetric profile is explicit, any candidate unequal-cell partition can be checked against a closed-form benchmark rather than by numerical search alone.

What carries the argument

The load-bearing mechanism is the pressure-vector description of minimizers, imported from the regularity theory of earlier work and stated as Theorem 2.1. It says that every interface is a finite union of circular arcs and straight segments that meet only at triple points with equal 120-degree angles, that the signed curvature of an arc equals the difference of two pressures $p_1=-p_2$, and that tiles are connected and simply connected. With periodicity and the two-tile assumption, Proposition 2.2 and Proposition 2.3 turn this into a short list of possible edge counts, and the planar-graph bound in Lemma A.5 (each vertex has degree at most 6) cuts the list to the three claimed configurations. The explicit formulas then come from elementary trigonometric identities: the turning-angle relation $\sum_k \alpha_k = (6-n)60^\circ$ (Lemma A.1), the area and perimeter of a Reuleaux triangle (Lemma A.2), the Steiner-tripod length formula (Lemma A.3), and the quadrangular curvilinear polygon computation (Lemma A.4), together with Lagrange-multiplier optimizations over lattice parameters in Section 3.

What would settle it

For a fixed square lattice and area ratio $x = 0.1$, solve the two-tile perimeter-minimization problem numerically with no shape restrictions: the theorem predicts the minimizer is the $(4;8)$ curvilinear square with a chipped square, so finding any admissible tiling with smaller interface length, or one whose interface contains a vertex where three edges do not meet at 120 degrees, would refute the classification.

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Extended reading notes

Core claim

The central claim is that the isoperimetric problem for periodic two-tile tilings of the plane is completely solvable. Theorem 1.1 states that, for any fixed lattice $G$ and any prescribed positive areas of the two generators, a perimeter-minimizing tiling must be one of three configurations: two hexagons with straight edges $(6;6)$, a strictly convex curvilinear quadrangle with four circular arcs paired with an octagonal 'chipped parallelogram' $(4;8)$, or a Reuleaux triangle paired with a nine-sided chipped hexagon $(3;9)$, with all edges meeting at 120 degrees and curvatures governed by a pressure difference. Theorem 1.2 combines the three perimeter formulas with the optimal lattice for each regime and gives the isoperimetric profile $I(x)$ explicitly: $\sqrt{2}\sqrt{\pi-\sqrt{3}}\sqrt{x}+\sqrt[4]{12}$ on $[0,x_1)$, $2\sqrt{\pi/3+1-\sqrt{3}}\sqrt{x}+2$ on $[x_1,x_2)$, $2\sqrt[4]{3}$ on $(x_2,1/2]$, and $I(x)=I(1-x)$ on $(1/2,1]$, with $x_1\approx 0.062$ and $x_2\approx 0.317$.

Load-bearing premise

The classification presupposes that, for a fixed lattice, every area-minimizing tiling has interfaces made of finitely many circular arcs and straight segments that meet only in threes at 120 degrees, with connected and simply connected tiles; if minimizers could have more complicated or disconnected interfaces, the three-configuration list could miss real minimizers.

Editorial extensions

If this is right

  • For $0 < x < x_1$, the unique optimal tiling is a Reuleaux triangle of area $x$ together with a nine-sided chipped regular hexagon on a honeycomb lattice, with interface length $\sqrt{2}\sqrt{\pi-\sqrt{3}}\sqrt{x}+\sqrt[4]{12}$ per unit fundamental-domain area.
  • For $x_1 < x < x_2$, the unique optimum is a curvilinear square with four circular arcs plus a chipped square, on a square lattice, with interface length $2\sqrt{\pi/3+1-\sqrt{3}}\sqrt{x}+2$.
  • For $x_2 < x \le 1/2$, two adjacent regular hexagons sharing an edge are optimal, with constant interface length $2\sqrt[4]{3}$; at $x = 1/2$ this is the honeycomb partition into equal hexagons.
  • At the two transition points $x_1 \approx 0.062$ and $x_2 \approx 0.317$ the adjacent configurations tie, so minimizers are not unique exactly there; for every other area ratio the optimal tiling is unique up to isometry.
  • The squared shifted profile $(I(x)-I(0))^2$ is concave on $[0,1]$, a structural property that may be useful in comparison arguments.

Reading between the lines

Editorial extensions of the paper, not claims the author makes directly.

  • One consequence the authors leave implicit is that the fixed-lattice problem reduces to a finite check: for a given lattice $G$, only the admissibility intervals of the $(3;9)$, $(4;8)$, and $(6;6)$ shapes need to be determined, so their Problem 4.1 could be settled by explicit inequalities rather than by new regularity theory.
  • The same counting argument suggests that for $N$-tilings with one large cell and $N-1$ small cells, the optimal shapes should be a large polygon with small Reuleaux-triangle caps on some vertices, at least when the small cells are sufficiently small; this is exactly the candidate the authors mention in Section 4.2.
  • Because the profile is explicit and $(I(x)-I(0))^2$ is concave, one can bound the perimeter of any periodic partition with many area ratios by a weighted average of these three formulas; an interested reader could test whether such a Jensen-type bound is sharp.
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Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, and a circularity audit.

Referee Report

3 major / 4 minor

Summary. The paper studies periodic tilings of the plane by two tiles with prescribed areas, minimizing interface length either for a fixed lattice or with the lattice free. It claims a classification into three curvilinear configurations, labelled (3;9), (4;8), and (6;6), and an explicit isoperimetric profile with two transition points. The proofs combine regularity theory imported from the authors' earlier work [30] with planar graph arguments and trigonometric computations. The main statements are Theorem 1.1, the fixed-lattice classification, and Theorem 1.2, the explicit profile for the optimal lattice.

Significance. If the regularity and connectivity assumptions are fully justified, this is a substantial explicit solution of a natural unequal-cell variant of the planar Kelvin problem. The formulas are parameter-free, the transition points x1 and x2 are computed rather than fitted, and the paper gives uniqueness statements, a stationary non-minimizing configuration, and clearly formulated open problems. The main caveat is that the classification rests on a regularity/connectivity theorem whose proof is partly sketched and on an imported extension from a companion paper; these points are load-bearing and need to be completed.

major comments (3)
  1. [Section 2, Theorem 2.1(4)] The proof of connectivity of the interface and simple connectivity of the tiles is only a sketch. It says to take a connected component of ∂T together with all enclosed components and their G-translates and move them by a translation, but it does not construct the new generators for the fixed-lattice problem (2.1), does not prove that the new partition has the same areas and perimeter, and does not explain how the moved components eventually produce an illegal vertex. The simple-connectivity assertion is not proved at all. Since Proposition 2.2, Proposition 2.3, and the dichotomy in Theorems 2.4–2.6 all use property (4), this is load-bearing. I do not think the issue is simply that a translation by t∉G breaks G-periodicity, because translating the entire G-orbit of a block can still yield a G-periodic configuration; rather, the argument as written is not a complete proof.
  2. [Section 2, paragraph after (2.1)] The extension of [30, Theorem 5.2] to the fixed-lattice variational problem is asserted with 'readily checked' and a diameter bound. Because Theorem 1.1 and the later profile depend on fixed-lattice regularity, this step needs a real proof: the authors should explain how [30, Lemma 4.2 and Proposition 5.1] adapt when the lattice is fixed and why the diameter bound is compatible with the cell area constraints. As written, the regularity foundation of the classification is an unverified import.
  3. [Section 3, proof of Theorem 3.4, Step 2] The statement that 'the smaller signed distance between E and H is obtained by taking the points on the axis y=x' is used to conclude a=b and u=v, and hence to derive the square-lattice formula (3.7). No proof is supplied for this comparison between a level ellipse of r(x,y) and a level hyperbola of xy. Since the middle branch of I(x) in Theorem 3.5 depends on (3.7), this is another load-bearing point that should be justified.
minor comments (4)
  1. [Theorem 3.5, item (5)] The text says I(x)=q2 on the sixth interval; from the definitions and Theorem 1.2 the constant should be q3.
  2. [Throughout] There are repeated typos: 'preassure' should be 'pressure', and 'Reauleaux' should be 'Reuleaux'.
  3. [Theorem 3.4 proof] The phrase 'the curvilinear rectangle, E1 is a curvilinear square' is missing a verb or comma; it should say that the tile E1 is a curvilinear square.
  4. [Notation] The plain-text rendering '4√12' is ambiguous; the typeset fourth-root notation should be used consistently so that q1 and q3 are not confused with multiples of square roots.

Circularity Check

0 steps flagged · score 1.0 of 10

No circularity: the explicit profile is derived from a classification whose regularity input is an independent prior theorem; the main caveats are a proof gap and same-author citation, not circular reductions.

full rationale

The paper's central claims are Theorem 1.1 (classification) and Theorem 1.2 (explicit isoperimetric profile). Theorem 1.1 follows from Theorems 2.4 and 2.6, which use the regularity structure of Theorem 2.1. Properties (1)-(3) of Theorem 2.1 are imported from [30, Theorem 5.2], a published theorem by the first two authors; this is a self-citation, but it is independent support because [30] was not derived from the present classification or profile and its assumptions do not contain the target result. Property (4) is argued directly in the text, not simply assumed. The computations in Section 3 are self-contained geometric calculations: e.g., Theorem 3.3 derives Per(T) = sqrt(2) sqrt(pi - sqrt(3)) sqrt(|E1|) + fourth-root(12) sqrt(|E1|+|E2|) from explicit area and perimeter formulas for the Reuleaux triangle, and Theorem 3.5 defines x1 and x2 as intersections of the explicit functions m1 sqrt(x)+q1 = m2 sqrt(x)+q2 and m2 sqrt(x)+q2 = q3. No parameter is fitted to data, and no quantity is defined in terms of the result it is used to prove. The only notable weaknesses are non-circular: the proof of Theorem 2.1(4) is a sketch whose 'move these components with a translation' step is not obviously admissible for a fixed lattice, since a translation not lying in G would destroy periodicity, and the regularity extension from [30] to the fixed-lattice problem is asserted with a short argument rather than a full proof. These are correctness risks in load-bearing steps, not circular reductions, because neither the classification nor the profile is assumed as an input. Therefore no significant circularity is present; the score of 1 reflects the minor same-author citation and the sketchy proof of property (4), not a circular derivation.

Assumptions & free parameters 0 free parameters · 4 assumptions · 0 invented entities

The central claim rests on cited regularity and existence theory, standard theorems such as Hales' honeycomb theorem, and explicit trigonometric computations. There are no free parameters fitted to data and no invented physical or mathematical entities; the Reuleaux triangle and chipped polygons are classical shapes used in the classification.

assumptions (4)
  • domain assumption Regularity of isoperimetric periodic tilings: interfaces are circular arcs or straight segments meeting at 120 degrees, with well-defined pressures, and connected components are simply connected. Theorem 2.1, imported from [30, Thm 5.2] and extended to the fixed-lattice problem.
    The whole classification assumes this regularity. The extension to the fixed-lattice variational problem (2.1) is asserted with a sketch, not a full proof, and is load-bearing for Theorem 1.1.
  • domain assumption Existence of minimizers for the periodic fixed-lattice and free-lattice variational problems, taken from [30].
    Needed so that the classification can enumerate all minimizers. The paper does not reproduce the existence proof.
  • standard math Hales honeycomb theorem: among periodic one-tile tilings, the regular hexagon minimizes perimeter for given area. Used in Theorem 3.1 for the degenerate x=0 case.
    Used as the baseline for the degenerate case and as a comparison in optimal-lattice arguments. It is a published theorem, not proved in this paper.
  • standard math Planar graph Euler formula and the lemma that a Z^2-invariant planar graph has vertex degree at most 6, proved in Appendix A.5.
    Used in Theorem 2.6 to bound the number of flat edges of the larger tile and thereby reduce the possible configurations.

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Pith. "Pith review of Periodic double tilings of the plane." pith.science (2026). https://pith.science/paper/DU2LGOPA

@misc{pith2026250208396,
  author       = {Pith},
  title        = {Pith review of: Periodic double tilings of the plane},
  year         = {2026},
  howpublished = {\url{https://pith.science/paper/DU2LGOPA}},
  note         = {Machine review of arXiv:2502.08396}
}
read the original abstract

We study tilings of the plane composed of two repeating tiles of different assigned areas relative to an arbitrary periodic lattice. We classify isoperimetric configurations (i.e., configurations with minimal length of the interfaces) both in the case of a fixed lattice or for an arbitrary periodic lattice. We find three different configurations depending on the ratio between the assigned areas of the two tiles and compute the isoperimetric profile. The three different configurations are composed of tiles with a different number of circular edges, moreover, different configurations exhibit a different optimal lattice. Finally, we raise some open problems related to our investigation.

Figures

Figures reproduced from arXiv: 2502.08396 by the authors.

Figure 1
Figure 1. The three possible configurations of isoperimetric 2-tilings of the plane relative to a given periodic lattice. See the interactive web page: https://paolini.github.io/double-tiling/. A natural variant of this problem is the study of periodic partitions of the Euclidean space minimizing the surface area of the interfaces but with possibly unequal cells. Configurations of this kind have been for instance commented in… view at source ↗
Figure 2
Figure 2. Plot of the isoperimetric profile I(x) where x ∈ [0, 1 2 ] is the area of the smaller tile and 1 − x is the area of the larger tile. On the y-axis, we have y0 := I(0) = √4 12, and yi := I(xi) for i = 1, 2 (see Theorem 1.2 and Theorem 3.5). The dashed lines represent stationary but not optimal tilings. The shape of the optimal tiles is depicted for the values x = 0 (single hexagon), x = x1 (configurations (3; 9) and … view at source ↗
Figure 3
Figure 3. The three possible configurations for an isoperimetric 2-tiling relative to a fixed lattice. The first one is configuration (6;6) where the two tiles are two hexagons with straight edges (equal pressures); the second one is configuration (4;8) where the tile with larger pressure is a curved rectangle and the tile with lower pressure is a chipped parallelogram i.e., an octagon with alternating flat and concave edges;… view at source ↗
Figures from the paper (6 more)
Figure 4
Figure 4. Figure 4: The construction of the hexagons in the proof of Theorem 2.4, Step 3. with the previous and next one. By Proposition 2.2 this means that they are all different translations of the same component D of E2. So they can be written as D + gk with distinct g1, . . . , g6 ∈ G…
Figure 5
Figure 5. Figure 5: The two possible vertices for the tilings in the assumptions of Proposition 2.5. The first case is a vertex of type 1-2-2, and the second is a vertex of type 2-2-2. direction g whose interfaces are zig-zags following the vectors v and w. We can swap the second and thir…
Figure 6
Figure 6. Figure 6: Notation used in configuration (4;8). the curvilinear rectangle, E1 is a curvilinear square and the chipped parallelogram E2 is a chipped square. Moreover, we have 0 < |E1| |E1| + |E2| < (2 + √ 3) 1 − √ 3 + π 3  8 (3.6) , Per(T) = 2r π 3 + 1 − √ 3 p |E1| + 2p (3.7) |E…
Figure 7
Figure 7. Figure 7: Left: the isoperimetric profile I(x) of the periodic 2-tiling, compared with the asymptotic profile J (x), see (4.1), of a non-periodic tiling composed of hexagons with the same areas. Right: the difference I(x) − J (x), highlighting the intervals where I(x) < J (x). P…
Figure 8
Figure 8. Figure 8: A quadrangular curvilinear polygon with circular edges. rotation of 360 degrees. Hence we have Xn k=1 (αk + 60◦ ) = 360◦ and (A.1) follows. □ Lemma A.2 (area and length of Reuleaux triangle). Let E ⊂ R 2 be a Reuleaux triangle i.e., a curvilinear polygon composed of th…
Figure 9
Figure 9. Figure 9: A planar periodic graph can have vertices with a maximum order of 6. centred in these two points and passing through the point A have the same radius and define an angle of 120 degrees in A. It is not difficult to show that the points P, Q are uniquely defined by these…

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