CCCG 2020, Saskatoon, Canada, August 5{7, 2020 Efficient Folding Algorithms for Regular Polyhedra Tonan Kamata∗ Akira Kadoguchiy Takashi Horiyamaz Ryuhei Ueharax Abstract To understand unfolding, it is interesting to look at the inverse: what kind of polyhedra can be folded from a We investigate the folding problem that asks if a poly- given polygonal sheet of paper? For example, the Latin gon P can be folded to a polyhedron Q for given P and cross, which is one of the eleven nets of a cube, can fold Q. Recently, an efficient algorithm for this problem is to 23 different convex polyhedra by 85 distinct ways developed when Q is a box. We extend this idea to two of folding [6] and an infinite number of doubly covered different cases; (1) Q is a regular dodecahedron, and (2) concave polygons (Figure 1). Comprehensive surveys of Q is a convex polyhedron such that each face is formed folding and unfolding can be found in [6]. by regular triangles. Combining the known result, we We investigate the folding problem when a polygon can conclude that the folding problem can be solved ef- P and a polyhedron Q are given. That is, for a given ficiently when Q is a regular polyhedron, also known as polygon P and a polyhedron Q, the folding problem a Platonic solid. asks if P can fold to Q or not. This is a natural prob- lem, however, there are few results so far. When Q is a regular tetrahedron, we have a mathematical character- 1 Introduction ization of its net [4]; according to this result, P should be a kind of tiling, and hence the folding problem can In 1525, the German painter Albrecht D¨urerpublished be solved efficiently. Abel et al. investigated the fold- his masterwork on geometry [7], whose title translates ing problem of bumpy pyramids [1]: For a given petal as \On Teaching Measurement with a Compass and polygon P (convex n-gon B with n triangular petals), Straightedge for lines, planes, and whole bodies." In it asks if we can fold to a pyramid (with flat base B) the book, he presented each polyhedron by drawing a or a convex bumpy pyramid by folding along a certain net for it: an unfolding of the surface to a planar layout triangulation of B. In [1], they gave nontrivial linear by cutting along its edges. To this day, it remains an im- time algorithms for the problem. Recently, the fold- portant open problem whether every convex polyhedron ing problem was investigated for the case that Q is a has a (non-overlapping) net by cutting along its edges. box. Some special cases were investigated in [8] and On the other hand, when we allow to cut anywhere, any [11], and the problem for a box Q is solved in [10] in convex polyhedron can be unfolded to a planar layout general case. The running time of the algorithm in [10] without overlapping. There are two known algorithms; is O(D11n2(D5 + log n)), where D is the diameter of P . one is called source unfolding, and the other is called In these algorithms, Q is given as just a \box" without star unfolding (see [6]). size, and the algorithms try all possible sizes. If Q is explicitly given, the running time of the algorithm in 1/2 [10] is reduced to O(D7n2(D5 + log n)) time. x 2-x In this paper, we investigate two other cases. In the first case, we assume that Q is a regular dodecahedron. This is a very special case, however, it is one of the five regular polyhedra. The second case is that Q is a con- vex deltahedron whose faces consist of regular triangles. 1/2 A deltahedron is said to be strictly convex if it is con- vex and no two adjacent faces are coplanar. It is known Figure 1: A Latin cross made by six unit squares. For that there are eight strictly convex deltahedra1: a regu- any real number x with 0 < x < 1, folding along dotted lar tetrahedron, a regular octahedron, a regular icosahe- lines, we can obtain a doubly-covered fat cross. dron, a triangular bipyramid, a pentagonal bipyramid, a snub disphenoid, a triangulated triangular prism, and a ∗School of Information Science, JAIST, [email protected] gyroelongated square bipyramid. In this paper, we also y School of Information Science, JAIST, consider non-strictly-convex cases as a kind of deltahe- [email protected] zFaculty of Information Science and Technology, Hokkaido dron. That is, we allow each face to consist of coplanar University, [email protected] xSchool of Information Science, JAIST, [email protected] 1See, e.g., https://en.wikipedia.org/wiki/Deltahedron. 32nd Canadian Conference on Computational Geometry, 2020 regular triangles like a regular hexagon. Then there are See [6, Sec. 21.3] for details. infinite number of non-strictly-convex deltahedra. For Let Q be a convex polyhedron. A development of Q these two cases, we give pseudo-polynomial time algo- results when we cut Q along a set of polygonal lines, rithms: unfold on a plane, and obtain a polygon P . We assume that any cut ends at a point with curvature less than Theorem 1 Let P be a simple polygon with n vertices. 360◦. Otherwise, since Q is convex, it makes a redun- We denote by L the perimeter of P . Then the folding dant cut on P , which can be eliminated (the proof can problem of a regular dodecahedron from P can be solved be found in [9, Theorem 3]). The polygon P is called a 2 4 in O(n (n + L) ) time. net of Q if and only if P is a connected simple polygon, i.e., without self-overlap or hole. Let T be the set of cut Theorem 2 Let P be a simple polygon with n vertices lines on Q to obtain a net P . Then the following is well of perimeter L. Let Q be a non-concave deltahedron2 known (see [6, Sec. 22.1.3] for details): with m vertices. Then the folding problem of Q from P 2 2 can be solved in O(n m(L + n) ) time. Theorem 5 T is a spanning tree of the vertices of Q. Combining with the result in [10], we have the following: 2.1 Properties of Unfolding Corollary 3 The folding problem for the five regular polyhedra (also known as Platonic solids) can be solved A tetramonohedron is a tetrahedron that consists of four in pseudo-polynomial time. congruent triangles. This polyhedron is exceptional in the context of unfolding. To avoid this case, we first We here note that we use real RAM model, and the time show the following lemma. complexity is evaluated by the number of mathematical operations. Lemma 6 Let Q be a convex polyhedron. Then Q is a tetramonohedron if and only if the curvature of every ◦ 2 Preliminaries vertex is 180 . We first state the folding problem: the input is a poly- Proof. If Q is a tetramonohedron, by its symmetric gon P = (p0; p1; : : : ; pn−1; pn = p0) and a polyhedron property, each vertex q consists of three distinct angles Q, and the problem asks if P can fold to Q or not. of a congruent triangle. Thus the curvature at q is 180◦. Let xi and yi be the x-coordinate and y-coordinate of a In order to show the opposite, we assume that every ◦ point pi, respectively. We assume the real RAM model vertex of a polyhedron Q has curvature 180 . Then, for computation; each coordinate is an exact real num- by Theorem 4, Q has four vertices. Let q0; q1; q2; q3 be ber, and the running time is measured by the number these four vertices. We cut along three straight lines of mathematical operations. q0q1, q0q2, q0q3 on Q, respectively. Since the curvature ◦ When Q is a regular dodecahedron, we do not need at any point on Q except q0; q1; q2; q3 is 360 , we can to give it explicitly as a part of input. The length of take three non-crossing straight lines from q0 to q1, q2, the edges of Q can be computed from the area of P . and q3 on Q and they are the shortest lines from q0 to Without loss of generality, we assume that the length them. By developing Q from q0 along these three cut 0 00 of an edge of Q is 1. When Q is a non-concave deltahe- lines, we obtain a polygon P = (q0; q1; q0; q2; q0 ; q3; q0). dron, Q is represented in the standard form in compu- Then, by assumption, curvatures at q1, q2, q3 are all ◦ 0 tational geometry (see [5]). Precisely, Q consists of ver- 180 . That is, P is a triangle with three vertices q0, q0, 00 tices qi = (xi; yi; zi), edges fqi; qjg, and faces f1; : : : ; fk, and q0 . Moreover, each edge of the triangle consists of where each fi is represented by a cycle of vertices in two cut lines which form an edge on Q. Therefore, q1, 0 counterclockwise-order in relation to the normal vector q2, and q3 are all the middle points of three edges q0q0, 0 00 00 of the face. q0q0 , and q0 q0 of the triangle P , respectively. Thus all 0 00 Let Q be a convex polyhedron. Let q be a vertex of four triangles q0q1q3, q1q0q2, q3q2q0 , and q2q3q1 are con- Q. The curvature at q is the angle defined by the value gruent, which implies that Q is a tetramonohedron.