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Home , Dihedron, Polyhedron, Prismatoid, Regular dodecahedron, Regular icosahedron

CCCG 2021, Halifax, Canada, August 10–12, 2021

Any Regular Can Transform to Another by O(1) Refoldings

Erik D. Demaine∗ Martin L. Demaine∗ Yevhenii Diomidov∗ Tonan Kamata† Ryuhei Uehara† Hanyu Alice Zhang‡

Abstract al. [9] found an infinite sequence of that can each fold into a and an approaching-regular tetra- We show that several classes of polyhedra are joined hedron. by a sequence of O(1) refolding steps, where each re- More broadly, Demaine et al. [4] showed that any con- folding step unfolds the current polyhedron (allowing vex polyhedron can always be refolded to at least one cuts anywhere on the and allowing overlap) and other convex polyhedron. Xu et al. [11] and Biswas and folds that unfolding into exactly the next polyhedron; Demaine [3] found common unfoldings of more than two in other words, a polyhedron is refoldable into another (specific) polyhedra. On the negative side, Horiyama polyhedron if they share a common unfolding. Specifi- and Uehara [6] proved impossibility of certain refoldings cally, assuming equal , we prove that (1) any when the common unfolding is restricted to cut along two tetramonohedra are refoldable to each other, (2) any the edges of polyhedra. doubly covered triangle is refoldable to a tetramono- In this paper, we consider the connectivity of polyhe- hedron, (3) any (augmented) regular and dra by the transitive closure of refolding, an idea sug- doubly covered regular is refoldable to a tetra- gested by Demaine and O’Rourke [5, Section 25.8.3]. monohedron, (4) any has a 3-step refold- Define a (k-step) refolding sequence from Q to ing sequence to a tetramonohedron, and (5) the regu- Q0 to be a sequence of convex polyhedra Q = lar has a 4-step refolding sequence to a 0 Q0,Q1,...,Qk = Q where each Qi−1 is refoldable to tetramonohedron. In particular, we obtain a ≤ 6-step Qi. We refer to k as the length of the refolding se- refolding sequence between any pair of Platonic solids, quence. To avoid confusion, we use “1-step refoldable” applying (5) for the dodecahedron and (1) and/or (2) to refer to the previous notion of refoldability. for all other Platonic solids. As far as the authors know, this is the first result about common unfolding involving the . Our results. Do all pairs of convex polyhedra of the same surface area (a trivial necessary condition) have a finite-step refolding sequence? If so, how short of a se- 1 Introduction quence suffices? As mentioned in [5, Section 25.8.3], the open problem mentioned above is 0 A polyhedron Q is refoldable to a polyhedron Q if Q equivalent to asking whether 1-step refolding sequences can be unfolded to a planar that folds into ex- exist for all pairs of regular polyhedra. We solve a 0 0 actly the surface of Q , i.e., Q and Q share a common closely related problem, replacing “1” with “O(1)”: for unfolding/development, allowing cuts anywhere on the any pair of regular polyhedra Q and Q0, we give a re- 0 surfaces of Q and Q . (Although it is probably not nec- folding sequence of length at most 6. essary for our refoldings, we also allow the common un- More generally, we give a series of results about O(1)- folding to self-overlap, as in [6].) The idea of refolding step refolding certain pairs of polyhedra of the same was proposed independently by M. Demaine, F. Hur- surface area: tado, and E. Pegg [5, Open Problem 25.6], who specifi- cally asked whether every regular polyhedron (Platonic 1. In Section 3, we show that any two tetramonohedra solid) can be refolded into any other regular polyhe- are 1-step refoldable to each other, where a tetra- dron. In this context, there exist some specific results: monohedron is a tetrahedron that consists of four Araki et al. [2] found two Johnson-Zalgaller solids that congruent acute triangles. are foldable to regular tetrahedra [2], and Shirakawa et This result offers a possible “canonical form” for ∗CSAIL, MIT, USA. {edemaine,mdemaine,diomidov}@mit.edu finite-step refolding sequences between any two †School of Information and Science, Japan Ad- polyhedra: because a refolding from Q to Q0 is also vanced Institute of Science and Technology, Japan. a refolding from Q0 to Q, it suffices to show that {kamata,uehara}@jaist.ac.jp ‡School of Applied and Engineering Physics, Cornell Univer- any polyhedron has a finite-step refolding into some sity, USA. [email protected] tetramonohedron. 33rd Canadian Conference on Computational , 2021

2. In Section 4, we show that every regular prismatoid the polyhedron Q is convex, the following result is cru- and every augmented regular prismatoid are 1-step cial: refoldable to a tetramonohedron. Lemma 1 (Alexandrov’s Theorem [8, 5]) If we In particular, the regular tetrahedron is a tetra- fold a polygon P in a way that satisfies the following monohedron, the regular (cube) is a three Alexandrov’s conditions, then there is a regular prismatoid, and the regular and unique convex polyhedron Q realized by the folding. regular are both augmented regular . Therefore, the regular tetrahedron 1. Every point on the boundary of P is used in the has a 2-step refolding sequence to the regular hexa- gluing. hedron, octahedron, and icosahedron (via an inter- mediate tetramonohedron); and every pair of poly- 2. At any glued point, the summation of interior an- hedra among the regular hexahedron, octahedron, gles (cocurvature) is at most 2π. and icosahedron have a 3-step refolding sequence (via two intermediate tetramonohedra). 3. The obtained surface is homeomorphic to a .

3. In Section 5, we prove that a regular dodecahedron By this result, when we fold a polygon P to a poly- is refoldable to a tetramonohedron by a 4-step re- hedron Q, it is enough to check that the gluing sat- folding sequence. isfies Alexandrov’s conditions. (In this paper, it is easy to check that the conditions are satisfied by our As far as the authors know, there are no previous (re)foldings, so we omit their proof.) explicit refolding results for the regular dodecahe- A polyhedron Q is (1-step) refoldable to a polyhe- dron, except the general results of [4]. dron Q0 if Q can be unfolded to a polygon that folds Combining the results above, any pair of regu- to Q0 (and thus they have the same surface area). A lar polyhedra (Platonic solids) have a refolding se- (k-step) refolding sequence of a polyhedron Q to quence of length at most 6. a polyhedron Q0 is a sequence of convex polyhedra 0 Q = Q0,Q1,...,Qk = Q where Qi−1 is refoldable to 4. In addition, we prove that every doubly covered tri- Qi for each i ∈ {1, . . . , k}. To simplify some arguments (Section 3) and every doubly covered regular that Q is refoldable to Q0, we sometimes only partially polygon (Section 4) are refoldable to a tetramono- unfold Q (cutting less than needed to make the surface hedron, and that every tetrahedron has a 3-step re- unfold flat), and refold to Q0 so that Alexandrov’s con- folding sequence to a tetramonohedron (Section 6). ditions hold. Therefore, every pair of polyhedra among the list We introduce some key polyhedra. A tetrahedron is a above have an O(1)-step refolding sequence. tetramonohedron if its faces are four congruent acute triangles.1 We consider a doubly covered polygon as a special polyhedron with two faces. Precisely, for a 2 Preliminaries given n-gon P , we make a mirror image P 0 of P and glue corresponding edges. Then we obtain a doubly For a polyhedron Q, V (Q) denotes the set of vertices covered n-gon which has 2 faces, n edges, and zero of Q. For v ∈ V (Q), define the cocurvature σ(v) of . v on Q to be the sum of the incident to v on the facets of Q. The κ(v) of v is defined by κ(v) = 2π − σ(v). In particular, if κ(v) = σ(v) = π, we 3 Refoldabilty of Tetramonohedra and Doubly Cov- call v a smooth . We define Πk to be the class of ered Triangles polyhedra Q with exactly k smooth vertices. It is well- known that the total curvature of the vertices of any In this section, we first show that any pair of tetra- convex polyhedron is 4π, by the Gauss–Bonnet Theo- monohedra can be refolded to each other. We note that rem (see [5, Section 21.3]). Thus the number of smooth a doubly covered rectangle is a (degenerate) tetramono- vertices of a convex polyhedron is at most 4. Therefore, hedron, by adding edges along two crossing (one on the front side and one on the back side). It is the classes Π0, Π1, Π2, Π3, Π4 give us a partition of all convex polyhedra. known that a polyhedron is a tetramonohedron if and An unfolding of a polyhedron is a (possibly self- only if it is in Π4 [7]. In other words, Π4 is the set of overlapping) planar polygon obtained by cutting and tetramonohedra. developing the surface of the polyhedron (allowing cuts Theorem 2 For any Q, Q0 ∈ Π , Q is 1-step refoldable anywhere on the surface). Folding a polygon P is an 4 to Q0. operation to obtain a polyhedron Q by choosing crease lines on P and gluing the boundary of P properly. When 1This notion is also called isotetrahedron in some literature. CCCG 2021, Halifax, Canada, August 10–12, 2021

Q Q′′ b′ b p p p q h a a 2a′ a′ q q h 2a Q b Q′ b 2a q′ b′ p′ p′ q′ a   2 ′ 2a 2a′ 2a′

Figure 2: A refolding from a doubly covered triangle to a doubly covered rectangle Figure 1: A refolding between two tetramonohedra

along the crease lines defined by the grid, we can obtain Proof. Let T be any triangular of Q. Let a be a doubly covered rectangle Q00 of size b/2×h (matching the length of the longest of T and b the height the doubled surface area of Q). (Intuitively, this folding of T for the edge of length a. We define T 0, a0, wraps T and T 0 on the surface of the rectangle pp0q0q.) and b0 in the same manner for Q0; refer to Figure 1. Because Q00 is also a tetramonohedron, the second We assume a > a0 without loss of generality. Now we claim follows from Theorem 2. When Q has an edge of have a0 > b0 because a0 is the longest edge of T 0, and the same length as an edge of Q0, as in the third claim, a0b0 = ab because T and T 0 are of the same area. Thus, we can cut the other two edges of Q and Q0 to obtain 0 2 0 0 a0 0 0 0 0 0 (a ) = a b b0 > a b = ab, and 2a > a > b by a > a . the same doubly covered rectangle, resulting in a 2-step We cut two edges of Q of length a, resulting in a refolding sequence.  of height b and circumference 2a. Then we can cut the cylinder by a segment of length 2a0 because The technique in the proof of Theorem 3 works for 2a0 > b. The resulting polygon is a such any doubly covered triangle Q even if its faces are acute that two opposite sides have length 2a and the other or obtuse triangles. two opposite sides have length 2a0. Now we glue the 0 sides of length 2a and obtain a cylinder of height b and 4 Refoldability of a Regular Prismatoid to a Tetra- 0 0 circumference 2a . Then we can obtain Q by folding monohedron this cylinder suitably (the opposite of cutting two edges 0 0 of Q of length 2a ).  In this section, we give a 1-step refolding of any reg- ular or prismatoid to a tetramonohedron. We To complement the doubly covered rectangles han- extend the approach of Horiyama and Uehara [6], who dled by Theorem 2, we give a related result for doubly showed that the , the regular octa- covered triangles: hedron, and the regular hexahedron (cube) can be 1- step refolded into a tetramonohedron. As an example, Theorem 3 Any doubly covered triangle Q is 1-step re- Figure 3 shows their common unfolding for the regular foldable into a doubly covered rectangle. Thus, Q has a icosahedron. refolding sequence to any doubly covered triangle Q0 of A polygon P = (p , c , p , c , p , . . . , p , c , p , p ) length at most 3. If doubly covered triangles Q and Q0 0 1 1 2 2 2n 2n 2n+1 0 is called a spine polygon if it satisfies the following share at least one edge length, then the sequence has two conditions (refer to Figure 4): length at most 2.

1. Vertex pi is on the p0pn for each 0 < i < n; vertex p is on the line segment p p Proof. Let Q consist of a triangle T and its mirror i n+1 2n+1 for each n + 1 < i < 2n + 1; and the polygon image T 0. We first cut Q along any two edges, and B = (p , p , p , p , p ) is a parallelogram. We unfold along the remaining attached edge, resulting in 0 n n+1 2n+1 0 call B the base of P , and require it to have positive a quadrilateral unfolding as shown in Figure 2. Let b be area. the length of the uncut edge, which we call the base, and let h be the height of T with respect to the base. 2. The polygon Ti = (pi, ci+1, pi+1, pi) is an isosceles Let p and q be the midpoints of the two cut edges. Then triangle for each 0 ≤ i ≤ n − 1 and n + 1 ≤ i ≤ the line segment pq is parallel to the base and of length 2n. The triangles T0,T1,...,Tn−1 are congruent, 0 0 b/2. In the unfolding of Q, let p and q be the mirrors and Tn+1,Tn+2,...,T2n are also congruent. These of p and q, respectively. Then we can draw a grid based triangles are called spikes. on the rectangle pp0q0q as shown in Figure 2. By folding 33rd Canadian Conference on , 2021

Figure 5: A regular prismatoid and an augmented reg- Figure 3: A common unfolding of a regular icosahedron ular prismatoid and a tetramonohedron, from [6] covered regular polygons (prisms of height zero). c c c c 1 2 n−1 n A is the of a base convex poly- gon and an apex point. We call a pyramid regular if the base polygon is a , and the line passing through the apex and the center of the base is p p p pn pn pn 0 1 2 −2 −1 perpendicular to the base. (Note that the side faces of a regular pyramid do not need to be regular polygons.) p p p p p p 2n+1 2n 2n−1 n+3 n+2 n+1 A polyhedron is an augmented regular prismatoid if it can be obtained by attaching two regular pyramids to a regular prismatoid base-to-roof, where the bases of c c c c 2n 2n−1 n+2 n+1 the pyramids are congruent to the roofs of the prisma- toid and each roof is covered by the base of one of the Figure 4: A spine polygon with 2n spikes pyramids.

Theorem 5 Any regular prismatoid or augmented reg- Lemma 4 Any spine polygon P can be folded to a tetra- ular prismatoid of positive volume can be unfolded to a monohedron. spine polygon. Proof. Akiyama and Matsunaga [1] prove that a poly- Proof. Let Q be a regular prismatoid. Let c1 and c2 be gon P can be folded into a tetramonohedron if the the center points of two roofs P and P , respectively. boundary of P can be divided into six parts, two of 1 2 Cutting from ci to all vertices of Pi for each i = 1, 2 which are parallel and the other four of which are rota- and cutting along a line joining between any pair of tionally symmetric. We divide the boundary of a spine vertices of P1 and P2, we obtain a spine polygon. For an polygon P into l1 = (p0, c1, . . . , cn); l2 = (cn, pn), l3 = augmented regular prismatoid Q, we can similarly cut (pn, pn+1); l4 = (pn+1, cn+1, . . . , c2n), l5 = (c2n, p2n+1); from the apex ci of each pyramid to the other vertices and l6 = (p2n+1, p0). Then l3 and l6 are parallel be- of the pyramid, which are the vertices of the roof Pi of cause the base of P is a parallelogram. Each of l2 and the prismatoid.  l5 is trivially rotationally symmetric. Each of l1 and l4 is rotationally symmetric because each spike of P is an When the height of the regular prismatoid is zero (or isosceles triangle.  it is a doubly covered regular polygon), the proof of Theorem 5 does not work because the resulting polygon Now we introduce some classes of polyhedra; refer to is not connected. In this case, we need to add some Figure 5. twist. A prismatoid is the convex hull of parallel base and top convex polygons. We sometimes call the base and Theorem 6 Any doubly covered regular n-gon is 1-step the top roofs when they are not distinguished. We refoldable to a tetramonohedron for n > 2. call a prismatoid regular if (1) its base P1 and top P2 are congruent regular polygons and (2) the line passing Proof. First suppose that n is an even number 2k through the centers of P1 and P2 is perpendicular to P1 for some positive k > 1. We consider a spe- 2π and P2. (Note that the side faces of a regular prismatoid cial spine polygon where the top angles are k ; the do not need to be regular polygons.) The perpendicular vertices p0, p2n+1, p1 are on a circle centered at c1; distance between the planes containing P1 and P2 is the and the vertices p2n+1, p1, p2 are on a circle centered height of the prismatoid. The set of regular prismatoids at c2n; see Figure 6. Then we can obtain a dou- contains prisms and , as well as doubly bly covered n-gon by folding along the zig-zag path CCCG 2021, Halifax, Canada, August 10–12, 2021

5 Refoldability of a Regular Dodecahedron to a Tetramonohedron c1 cn

2π p In this section, we show that there is a refolding se- 1 p n n quence of the regular dodecahedron to a tetramonohe- p2n pn+1 p0 2π dron of length 4. Combining this result with Corol- p n 2 +1 n lary 7, we obtain refolding sequences between any two regular polyhedra of length at most 6. c2n cn +1 Demaine et al. [4] mention that the regular dodeca- hedron can be refolded to another convex polyhedron. Indeed, they show that any convex polyhedron can be Figure 6: The case of a doubly covered regular 8-gon refolded to at least one other convex polyhedron using an idea called “flipping a Z-shape”. We extend this idea.

c1 pn Definition 1 For a convex polyhedron Q and n, k ∈ N, let p = (s1, s2, . . . , s(2k+1)n) be a path that consists of p0 p1 pn+1 isometric and non-intersecting (2k + 1)n straight line p 2n+1 segments si on Q. We cut the surface of Q along p.

c2n Then each line segment is divided into two line segments cn+1 on the boundary of the cut. For each line segment si, l r Figure 7: The case of a doubly covered regular 5-gon let si and si correspond to the left and right sides on the boundary along the cut (Figure 8). Then p is a Z- flippable (n, k)-path on Q, and Q is Z-flippable by p, p2n+1, p1, p2n, p2, . . . , pn+2, pn shown in Figure 6. Thus if the following gluing satisfies Alexandrov’s conditions. when n = 2k for some positive integer k, we obtain the l l l l l l theorem. • Glue s1, s2, . . . , sn to s2n, s2n−1, . . . , sn+1. Now suppose that n is an odd number 2k + 1 for some positive integer k. We consider the spine polygon 4π r r whose top angles are 2k+1 ; the vertices p0, p2n+1, p1 are s s n+1 n l on a circle centered at c ; and the vertices p , p , p l s 1 2n+1 1 2 r l l s (2k+1)n r s s s 3n r s are on a circle centered at c2n. From this spine n+2 n n+1 s (2k+1)n l 3n l s l polygon, we cut off two triangles c1, p0, c2n+1 and s 3n−1 r s sl n+2 s (2k+1)n−1 sr c , p , p , as in Figure 7. Then we can obtain a n−1 sr 3n−1 (2k+1)n−1 n+1 n+1 n n−1 doubly covered n-gon by folding along the zig-zag path p2n+1, p1, p2n, p2, . . . , pn+2, pn shown in Figure 7. Al- l r l though the unfolding is no longer a spine polygon, it s r s r s r 2 s 2n−1 s n 2kn+2 s is easy to see that it can also fold into a tetramono- 2 2 +2 2kn+2 0 0 l r r r hedron by letting l1 = (c1, p1, . . . , pn), l2 = (pn, pn), s r l s s l s 1 s s 2n 2n+1 s kn 2kn+1 0 0 0 1 2n+2 2 +1 l3 = (pn, cn+1), l4 = (cn+1, pn+2 . . . , p2n+1), l5 = l l l 0 s s s (p2n+1, p2n+1), and l6 = (p2n+1, c1) in the proof of 2n+1 2n 2n−1 Lemma 4.  Z-fip The proof of Theorem 6 is effectively exploiting that sr sr sr n+1 n 2(k−1)n+1 a doubly covered regular 2k-gon (with k > 1) can be l r l r sr l sn s s s k n viewed as a degenerate regular prismatoid with two k- sn +1 n+2 3n 3n (2 +1) r r gon roofs, where each of the side triangles of this pris- l l l s r s s s s n r 3n−1 s (2k+1)n−1 matoid is on the plane of the roof sharing the base of n−1 n+2 3 −1 s 2(k−1)n+2 n−1 the triangle. Because the cube and the regular octahedron are reg- r ular prismatoids and the regular icosahedron is an aug- l r r r r s s s s n s n s 2kn+2 mented regular prismatoid, we obtain the following: 2 2 2 −1 2 +2 2kn−1 l r r r r s s s s kn s kn Corollary 7 Let Q and Q0 be regular polyhedra of the 1 1 2n 2 2 +1 sl sl sl sl sr same area, neither of which is a regular dodecahedron. 2n+2 2n+1 2n 2n−1 2n+1 Then there exists a refolding sequence of length at most 3 from Q to Q0. When one of Q or Q0 is a regular tetrahedron, the length of the sequence is at most 2. Figure 8: Z-flip 33rd Canadian Conference on Computational Geometry, 2021

r r r l l l • Glue s , s , . . . , s to s , s , . . . , s . s1 1 2 n 2n+1 2n+2 3n 6 r r r r r r 9 9 s1 9 9 3 s1 s1 • Glue sn+1, sn+2, . . . , s2n to s3n, s3n−1, . . . , s2n+1. 4 5 s1 2 9 9 9 9 9 s1 . s4 1 . 3 . s2 s2 s4 s4 9 9 1 2 2 1 s3 1 r r r s2 9 9 9 9 9 3 • Glue s2(k−1)n+1, s2(k−1)n+2, . . . , s2kn to s3 r r r s3 s3 2 s(2k+1)n, s(2k+1)n−1, . . . , s2kn+1. 5 4 s3 9 9 3 9 9 1 2 m s3 If there are Z-flippable paths p , p , . . . , p inducing 6 a forest on Q, we can flip them all at the same time. Then we say that Q is Z-flippable by p1, p2, . . . , pm.

Theorem 8 There exists a 4-step refolding sequence 10 10 9 10 8 between a regular dodecahedron and a tetramonohedron. 8 8 10 7 Proof. Let D be a regular dodecahedron. To simplify, 10 10 we assume that each edge of a regular is of 7 10 8 8 length 1. We show that there exists a refolding sequence 8 10 9 10 10 D,Q1,Q2,Q3,Q4 of length 4 for a tetramonohedron Q4. 9π All cocurvatures of the vertices of D are equal to 5 . For any vertices v, there are 3 vertices√ of distance 1 from 1+ 5 Figure 10: A refolding from D to Q1 v and 6 vertices of distance φ = 2 from v. Here- after, in figures, each circle describes a non-flat vertex s1 on a polyhedron and the number in the circle describes 3 π 10 9 10 8 s1 10 2 its cocurvature divided by 5 . Each pair of vertices of s1 distance 1 is connected by a solid line, and each pair 1 8 8 10 7 of vertices of distance φ is connected by a dotted line. 10 10 Figure 9 shows the initial state of D in this notation. 7 10 8 8 We note that solid and dotted lines do not necessarily s1 5 imply edges (or crease lines) on the polyhedron. s1 10 4 8 10 9 10

9 9 9 9 10 10 9 10 5 10 9 9 9 9 9 8 8 10 9 9 10 10 10 8 8 9 9 9 9 9 10 5 10 9 10 9 9 9 9 10

Figure 11: A refolding from Q1 to Q2 Figure 9: The initial regular dodecahedron

1 1 1 1 2 2 2 2 1 First, we choose p = (s1, s2, . . . , s6), p = (s1, s2, s3), (1, 3)-path. Thus, Q1 is Z-flippable by p to the next 3 3 3 3 4 4 4 4 p = (s1, s2, . . . , s6), and p = (s1, s2, s3) on the surface polyhedron Q2 on the right of Figure 11. 1 3 1 1 1 1 2 2 2 2 of D on the left of Figure 10. Then, p and p are Third, we choose p = (s1, s2, s3) and p = (s1, s2, s3) 2 4 Z-flippable (2, 1)-paths and p and p are Z-flippable on the surface of Q2 on the left of Figure 12. Then, 1 2 3 4 1 2 (1, 1)-paths. Thus, D is Z-flippable by p , p , p , p to p and p are Z-flippable (1, 1)-paths. Thus, Q2 is Z- 1 2 the polyhedron on the right of Figure 10. Let Q1 be the flippable by p and p to the polyhedron Q3 on the right resulting polyhedron. of Figure 12. 1 1 1 1 1 1 1 1 2 Second, we choose p = (s1, s2, . . . , s5) on the surface Fourth, we choose p = (s1, s2, . . . , s5), p = 1 2 2 2 3 3 3 3 of Q1 on the left of Figure 11. Then, p is a Z-flippable (s1, s2, . . . , s5), and p = (s1, s2, s3) on the surface of CCCG 2021, Halifax, Canada, August 10–12, 2021

π − θ s1 10 3 Π3 ∩ �5 ∈ s1 κ(vi) 2 p pm pj j × κ(vi) 10 9 10 5 10 λ 1 pm s1 1 1 c 1 vj 1 m v × c p i p 2 i 8 8 10 λ2 i p pm m2 2 c s σ v 10 10 π λ ×3 ( i) 3 θ

10 8 8

2 s ∈ 1 Smooth vertex 10 5 10 9 10 s2 p Π4 s2 2 10 m pj 3 p 1 m2 s pi Smooth vertex 8

10 7 10 7 10 Figure 14: A refolding of a polyhedron in Π3 ∩Q5 to Π4

10 8 10 of polyhedra in the refolding sequence. Appendix A 10 10 gives the (fully unfolded) common unfoldings.  10 8 10

10 7 10 7 10 6 Refoldability of a Tetrahedron to a Tetramonohe- dron 8 In this section, we prove that any tetrahedron can be refolded to a tetramonohedron. Let Q denote the class Figure 12: A refolding from Q2 to Q3 k of polyhedra with exactly k vertices.

s3 8 1 7 7 s1 6.1 Refoldability of Π3 to Π4 10 10 5 10 s2 1 s1 4 s3 First we show a technical lemma: any polyhedron Q in s2 2 10 2 8 10 Π3 can be refolded to a tetramonohedron by a refolding s2 s1 3 10 10 3 sequence of length linear in the number of vertices of Q. s1 10 8 2 s2 10 4 s1 1 Lemma 9 For any Q ∈ Π3 ∩ Qn with n ≥ 5, there is s2 7 7 10 5 10 10 a refolding sequence of length 2n − 9 from Q to some 8 0 s3 Q ∈ Π . 3 4 Proof. (Outline) We prove the claim by induction. As the base case, suppose n = 5; refer to Figure 14. Let 5 10 λ1, λ2, λ3 be the smooth vertices of Q and vi, vj be the 10 10 10 10 other vertices. Take a point m on the segment λ1λ2 5 with ∠vivjm = κ(vi) and cut the surface of Q along the 10 10 10 10 10 10 segments λ1λ2, vivj, and vjm. Then σ(vj) − κ(vi) = π 5 because κ(vi) + κ(vj) = π. The point vj is then di- vided into a point of degree κ(vi) and a point of de- 10 10 10 10 10 5 gree π on the boundary. the obtained boundary from vi counterclockwise and denote points correspond-

ing to vi, vj, m, m by pi, pj, pm1 , pm2 , respectively. Let c and c be the center points of the segments p p and Figure 13: A refolding from Q3 to Q4 1 2 i j pm1 pm2 , respectively. We take the point s which has the

same distance with pm1 from c1. Let c3 be the center of 1 2 Q3 on the left of Figure 13. Then, p and p are Z- the segment pm2 s. Now glue the segment sc1 to c1pm1 , 3 flippable (1, 3)-paths and p is a Z-flippable (1, 1)-path. the segment pm2 c2 to c2pm1 , and the segment pm2 c3 to 1 2 3 0 Thus, Q3 is Z-flippable by p , p , and p to the polyhe- c3, s. Let Q be the obtained polyhedron after the glu- 0 dron Q4 on the right of Figure 13. Finally, we obtain ing. Then Q is in Π4 because each curvature of every 0 a tetramonohedron Q4 from a regular dodecahedron D vertex of Q is π. by a 4-step refolding sequence. Now we turn to the inductive step. Let Q be any In this proof, we used partial unfolding between pairs polyhedron in Π3 ∩ Qk with k > 5. We prove that there 33rd Canadian Conference on Computational Geometry, 2021

σ(vj) − κ(vi) = 2π − (κ(vj) + κ(vi)) Π3 ∩ �k ∈ Π2 ∩ �5 ∈ π − θ λ × 1 π pm λ′ vi vj 1 pj c s pj m 1 λ 1 3 1 vi 1 v p vj j m1 λ  λ 2′ × p 2 m c i p 2 λ pm j2 p c1× λ λ 2 2 v m2 p 3 3 i j2 pi σ v κ v θ σ v κ v ( j) ( i) κ(vi) ( i) ( j) κ v

p × ( j) π

i1 c 1 ∈ pj p vi v m1 λ  j p p 1′ j j1 pm  Π3 ∩ �5 1′ p pi i2 1 pi p  2 p m1′ p i ×c pm  m 2 s λ  2′ 2 π − θ pm  c 2′ p 2′ 3× m  j σ v κ v π ′ λ 2 θ ( i) − ( j) − 3 Smooth vertex

= π − (κ(vi) + κ(vj)) ∈ Co-curvature p j 3π − (κ(vj) + κ(vi)) Π3 ∩ �k−1 p pi i pm  2 1 1′ Figure 16: Refolding from any tetrahedron to a tetra- pm  s 2′ monohedron

Smooth vertex Proof. (Outline) Let v, v0 be two vertices of Q with Figure 15: A refolding sequence from Π3 ∩ Qk to Π3 ∩ smallest cocurvature. (That is, σ(v), σ(v0) ≤ σ(v00) for Qk−1 the other two vertices v00 of Q.) We cut along the seg- ment vv0 and glue the point v to v0. Let Q0 be the resulting polyhedron. Then, because σ(v) + σ(v0) ≤ 2π 0 00 exists a refolding sequence (Q, Q ,Q ) for two polyhedra by the Gauss–Bonnet Theorem, Q0 satisfies the Alexan- 0 00 00 Q and Q with Q ∈ Π3 ∩ Qk−1; refer to Figure 15. drov’s conditions. That is, Q0 is a convex polyhedron in Let λ , λ , λ be the smooth vertices of Q and v , v be 1 2 3 i j Π2 ∪Q5. (We assume that the original Q has no smooth any other vertices on Q. Note that 0 < κ(vi)+κ(vj) < π vertex to simplify the arguments.) because k > 5. Take a point m on the segment λ λ 0 1 2 Let λ1 and λ2 be the two smooth vertices of Q (which with v v m = π, cut the surface of Q0 along the seg- 0 ∠ i j was generated by the gluing of v and v ), and vi, vj ment λ1λ2, and glue it again so that m is an endpoint. be two vertices of Q0 of larger cocurvature than others (This can be done because the cut produces a “rolling with κ(vi) < κ(vj). By the Gauss–Bonnet Theorem, belt” in terms of folding; see [5] for the details.) Let Q0 4π ≤ σ(vi) + σ(vj) ≤ 2π. We take the point m on the be the obtained polyhedron, and let λ0 (= m), λ0 , λ0 (= 3 1 2 3 segment λ1λ2 so that ∠(vi, vj, m) = κ(vi). Now we cut λ ) be the smooth vertices of Q0. Now take a point m0 3 along the segments λ1λ2, vivj, and vjm; see Figure 16. on the segment λ0 , λ0 such that λ0 v m0 = κ(v ) and 2 3 ∠ 1 i j Trace the obtained boundary from vi counterclockwise cut the surface of Q0 along the segments λ0 v , λ0 λ0 , 1 i 2 3 and denote points corresponding to vi, vj, m, m, vj by 0 vivj, and vim . Trace the obtained boundary from pi, pj1 , pm1 , pj2 , pm2 , respectively. Let c1 and c2 be the v counterclockwise and denote points corresponding j midpoints of the segments pipj and pm pm , respec- 0 0 2 1 2 to v , v , m , m , v by p , p , p 0 , p 0 , p , respectively. j i i j i1 m1 m2 i2 tively. Furthermore, we take the point s which has the 0 0 Let c1 and c2 be the midpoints of pi1 pj and pm pm , 1 2 same distance with pm2 from c1, and let c3 be the mid- respectively. Take the point s which has the same dis- point of spm1 . tance with pm0 from c1. Let c3 be the midpoint of the 1 Now glue segment pic1 to pj2 c1, segment pis to segment pm0 s. Now glue the segment pm0 c1 to c1s, 2 1 pj2 pm2 , segment pm1 c2 to pm2 c2, and segment sc3 to the segment p 0 c2 to c2p 0 , and the segment p 0 c3 00 m2 m1 m2 pm c3. Let Q be the resulting polyhedron. Then 00 00 1 to c3s. Let Q be the obtained polyhedron. Then Q the gluing to fold Q00 produces four vertices. Among is in Π ∩ Q because each curvature of points cor- 3 k−1 them, three vertices produced by the points c1, c2, c3 0 responding to c1, c2, c3 is π and 0 < σ(v4) < π be- are smooth vertices of curvature π. The cocurvature of 0 cause σ(v4) = σ(vi) − κ(vj) = σ(vi) + σ(vj) − 2π and 00 the vertex of Q generated by the gluing of pj1 to the 0 < κ(vi) + κ(vj) < π. 5π  boundary is 3π − (κ(vj) + κ(vi)), which is in [π, 3 ] by 4π 00 3 ≤ σ(vi)+σ(vj) ≤ 2π. Therefore, Q satisfies Alexan- 00 drov’s conditions, and hence we obtain Q ∈ Π3 ∪ Q. 000 Theorem 10 For any Q ∈ Q4, there is a 3-step refold- By Lemma 9, there exists Q ∈ Π4 such that there is a 000 000 ing sequence from Q to some Q ∈ Π4. 3-step refolding sequence from Q to Q .  CCCG 2021, Halifax, Canada, August 10–12, 2021

7 Conclusion Zalgaller solid. Journal of Graph and Ap- plications, 20(1):101–114, February 2016. In this paper, we give a partial answer to Open Problem [3] Amartya Shankha Biswas and Erik D. Demaine. Com- 25.6 in [5]. For every pair of regular polyhedra, we ob- mon development of prisms, anti-prisms, tetrahedra, tain a refolding sequence of length at most 6. Although and . In Proceedings of the 29th Canadian Confer- this is the first refolding result for the regular dodeca- ence on Computational Geometry (CCCG 2017), pages hedron, the number of refolding steps to other regular 202–207, 2017. polyhedra seems a bit large. Finding a shorter refolding [4] Erik D. Demaine, Martin L. Demaine, Jin-ichi Itoh, sequence than Theorem 8 is an open problem. Anna Lubiw, Chie Nara, and Joseph O’Rourke. Refold The notion of refolding sequence raises many open rigidity of convex polyhedra. Computational Geometry: problems. What pairs of convex polyhedra are con- Theory and Applications, 46(8):979–989, October 2013. nected by a refolding sequence of finite length? Is there [5] Erik D. Demaine and Joseph O’Rourke. Geomet- any pair of convex polyhedra that are not connected by ric Folding Algorithms: Linkages, Origami, Polyhedra. any refolding sequence? Cambridge University Press, 2007. At the center of our results is that the set of tetra- [6] Takashi Horiyama and Ryuhei Uehara. Nonexistence of monohedra induces a clique by the binary relation of common edge developments of regular tetrahedron and refoldability. Is the regular dodecahedron refoldable to other Platonic solids. In Proceedings of the China-Japan a tetramonohedron? Are all Archimedean and Johnson Joint Conference on Computational Geometry, Graphs solids refoldable to tetramonohedra? Is there any con- and Applications (CGGA 2010), pages 56–57, 2010. vex polyhedron not refoldable to a tetramonohedron? [7] Tonan Kamata, Akira Kadoguchi, Takashi Horiyama, (If not, we would obtain a 3-step refolding sequence be- and Ryuhei Uehara. Efficient folding algorithms for tween any pair of convex polyhedra.) regular polyhedra. In Proceedings of the 32nd Canadian Another open problem is the extent to which allowing Conference on Computational Geometry (CCCG 2020), or forbidding overlap in the common unfoldings affects pages 131–137, 2020. refoldability. While we have defined refoldability to al- [8] Yu. G. Reshetnyak and S. S. Kutateladze, editors. A. low overlap, in particular to follow [4] where it may be D. Alexandrov: Selected Works, chapter Existence of a necessary, most of the results in this paper would still convex polyhedron and a convex surface with a given apply if we forbade overlap. For example, Appendix A metric, pages 169–173. Part I. Gordon and Breach, confirms this for our refolding sequence from the regular 1996. dodecahedron to a tetramonohedron; while the general [9] Toshihiro Shirakawa, Takashi Horiyama, and Ryuhei approach of Lemma 9 is likely harder to generalize. Are Uehara. On common unfolding of a regular tetrahe- there two polyhedra that have a common unfolding but dron and a cube (in Japanese). Origami No Kagaku all such common unfoldings overlap? (If not, the two (Journal of Science Origami in Japanese), 4(1):45–54, notions of refolding are equivalent.) 2015. (See [10] for details.). [10] Ryuhei Uehara. Introduction to Computational Origami: The World of New Computational Geometry. Acknowledgments Springer, 2020. This work was initiated during MIT class 6.849: Geo- [11] Dawei Xu, Takashi Horiyama, Toshihiro Shirakawa, and metric Folding Algorithms, Fall 2020. Ryuhei Uehara. Common developments of three in- was partially supported by the Cornell congruent boxes of area 30. Computational Geometry: Center for Materials Research with funding from the Theory and Applications, 64:1–17, August 2017. NSF MRSEC program (DMR-1120296). Hanyu Zhang was primarily supported by the Cornell Center for Ma- A Common Unfoldings from Regular Dodecahedron terials Research with funding from the NSF MRSEC to Tetramonohedron in Theorem 8 program (DMR-1719875). Ryuhei Uehara was partially Figures 17, 18, 19, and 20 show the common unfoldings of supported by MEXT/JSPS Kakenhi Grant JP17H06287 each consecutive pair of polyhedra in the refolding sequence and JP18H04091. from the proof of Theorem 8.

References

[1] Jin Akiyama and Kiyoko Matsunaga. An for folding a Conway tile into an isotetrahedron or a rectangle . Journal of Information Processing, 28:750–758, December 2020. [2] Yoshiaki Araki, Takashi Horiyama, and Ryuhei Uehara. Common unfolding of regular tetrahedron and Johnson- 33rd Canadian Conference on Computational Geometry, 2021

Figure 17: A common unfolding of D and Q1

Figure 18: A common unfolding of Q1 and Q2 CCCG 2021, Halifax, Canada, August 10–12, 2021

Figure 19: A common unfolding of Q2 and Q3

Figure 20: A common unfolding of Q3 and Q4