1 Introduction
Given two polygons and of equal area, a dissection is a decomposition of into pieces that can be reassembled (by translation and rotation) to form . In a (chain) hinged dissection, the pieces are hinged together at their corners to form a chain, which can fold into both and , while maintaining connectivity between pieces at the hinge points. Many known hinged dissections are reversible (originally called Dudeney dissection [3]), meaning that the outside boundary of goes inside of after the reconfiguration, while the portion of the boundaries of the dissection inside of become the exterior boundary of . In particular, the hinges must all be on the boundary of both and . Other papers [4, 2] call the pair of polygons reversible.
Without the reversibility restriction, Abbott et al. [1] showed that any two polygons of same area have a hinged dissection. Properties of reversible pairs of polygons were studied by Akiyama et al. [3, 4]. In a recent paper [2], it was shown that reversible pairs of polygons can be generated by unfolding a polyhedron using two noncrossing nets. The purpose of this paper is to show that this characterization is in some sense complete.
An unfolding of a polyhedron cuts the surface of using a cut tree ,^{1}^{1}1For simplicity we assume that the edges of are drawn using segments along the surface of , and that vertices of degree 2 can be used in to draw any polygonal path. spanning all vertices of , such that the cut surface can be unfolded into the plane without overlap by opening all dihedral angles between the (possibly cut) faces. The planar polygon that results from this unfolding is called a net of . Two trees and drawn on a surface are noncrossing if pairs of edges of and intersect only at common endpoints and, for any vertex of both and , the edges of (respectively, ) incident to are contiguous in clockwise order around . Two nets are noncrossing if their cut trees are noncrossing.
Lemma 1.
Let be noncrossing trees drawn on a polyhedron , each of which spans all vertices of . Then there is a cycle passing through all vertices of such that separates the edges of from edges of , i.e., the (closed) interior (yellow region) of includes all edges of and the (closed) exterior of includes all edges of . Furthermore, all such cycles visit the vertices of in the same order.
Proof.
Let be the union of and . Because and are noncrossing, is a planar graph. Let be a third of the smallest angle between any two incident edges in , or , whichever is smaller. Let be a third of the smallest distance between any edge of and a vertex not incident to that edge. View each edge of as the union of two directed halfedges. For every halfedge in , the sidewalk of is a polygonal path composed of three segments such that

the clockwise angles and are both (placing and on the left of the directed line ); and

both and are at distance from the segment .
By construction, is the unique closest edge of from any point on its sidewalk. Thus, no two distinct sidewalks intersect and sidewalks do not intersect edges of .
Construct an Euler tour of (Figure 0(a)) that is noncrossing and traverses clockwise around , and replace each step from to in the tour by the sidewalk of . The concatenation of all these sidewalks forms a clockwise cycle that visits each vertex as many times as the degree of (Figure 0(b)). For any two consecutive sidewalks where the wedge does not contain an edge of incident to , shorten the walk by using the crosswalk to obtain , thereby avoiding a duplicate visit of . Because and are noncrossing, all but one of the visits of each vertex will be removed by using crosswalks (Figure 0(c)).
The resulting walk is a simple closed Jordan curve that visits each vertex of exactly once. Because does not intersect and , and locally separates the edges of and at each vertex, and because and are connected, the curve separates into two regions, one containing and the other containing .
Finally, we show that the order of vertices of visited by any such cycle is determined by , , and . Consider the planar graph drawn on . We claim that every face of this graph consists of at most one path of edges from and at most one path of edges from . Otherwise, we would have at least two components of and at least two components of , neither of which could be connected interior to the face (because the face is empty), and at most one of which could be connected exterior to the face (by planarity and the noncrossing property), contradicting that and are both trees. Therefore, every face with at least one edge from and at least one edge from locally forces where must go, connecting the two vertices with incident face edges from both and . Every vertex of has at least one incident edge from each of the spanning trees and , so has two incident such faces. In this way, we obtain the forced vertex ordering of . ∎
We can now state our first characterization.
Theorem 2.
Two polygons and have a reversible hinged dissection if and only if and are two noncrossing nets of a common polyhedron.
Proof.
To prove one direction (“only if”), it suffices to glue both sides of the pieces of the dissection as they are glued in both and to obtain a polyhedral metric homeomorphic to a sphere, and note that this metric corresponds to the surface of some polyhedron [2]. In the other direction (“if”), we use Lemma 1 to define the sequence of hinges. Now the cut tree of net is completely contained in the net and determines the hinged dissection. ∎
Often times, reversible hinged dissections are also monotone, meaning that the turn angles at all hinges in increase to produce . Figure 4 shows a hinged dissection that is reversible but not monotone. Monotone reversible hinged dissections also have a nice characterization:
Theorem 3.
Two polygons and have a monotone reversible hinged dissection if and only if and are two noncrossing nets of a common convex polyhedron.
Proof.
Let be a hinge of the monotone reversible hinged dissection. Pick two reference points and in the neighborhood of and in the pieces before and after hinge , respectively^{2}^{2}2in counterclockwise order. Let be the angle when the dissection forms polygon , and let be the same angle when the dissection forms polygon . Since the dissection is monotone, for all .
Just as in Theorem 2, glue both sides of the dissection as they are glued in both and to obtain a polyhedral metric homeomorphic to a sphere. Observe now that the total angle glued at vertex is exactly . Therefore by Alexandrov’s theorem, there exists a unique convex polyhedron (up to rigid transformations) whose surface has this intrinsic metric.
In the other direction, suppose we have two noncrossing nets of a convex polyhedron . Use Lemma 1 to find a cycle separating and on the surface of , and to define the sequence of hinges and cut both trees to obtain the dissection. Pick points and before and after on and in the neighborhood of . Let be the angle in net and on the surface of , and be the angle in net and on the surface of . Since is convex, . The angle of when the dissection forms polygon is exactly for every hinge , and so the dissection is monotone. ∎
An interesting special case of a monotone reversible hinged dissection is when every hinge touches only its two adjacent pieces in both its and configurations, and thus and are only possible such configurations. We call these simple reversible hinged dissections. (For example, Figure 4 is not simple.)
Lemma 4.
Every simple reversible hinged dissection is monotone.
Proof.
Pick the reference points and and define angle as in Theorem 3. Since the dissection is simple, the two pieces attached to hinge touch on the inside of . Therefore for any hinge angle less than , those two pieces would intersect. Since no two piece intersect when the dissection forms polygon , for all and the dissection is monotone. ∎
Corollary 5.
If two polygons and have a simple reversible hinged dissection, then and are two noncrossing nets of a common convex polyhedron.
Figure 4 shows two examples of hinged dissections resulting from these techniques. Historically, many hinged dissections (e.g., in [5]) have been designed by overlaying tessellations of the plane by shapes and . This connection to tiling is formalized by the results of this paper, combined with the characterization of shapes that tile the plane isohedrally as unfoldings of certain convex polyhedra [6].
References
 [1] T. G. Abbott, Z. Abel, D. Charlton, E. D. Demaine, M. L. Demaine, and S. D. Kominers. Hinged dissections exist. Discrete & Computational Geometry, 47(1):150–186, 2012.
 [2] J. Akiyama, S. Langerman, and K. Matsunaga. Reversible nets of polyhedra. In Revised Papers from the 18th Japan Conference on Discrete and Computational Geometry and Graphs, pages 13–23, Kyoto, Japan, September 2015.
 [3] J. Akiyama and G. Nakamura. Dudeney dissection of polygons. In Revised Papers from the Japan Conference on Discrete and Computational Geometry, volume 1763 of Lecture Notes in Computer Science, pages 14–29, Tokyo, Japan, December 1998.
 [4] J. Akiyama and H. Seong. An algorithm for determining whether a pair of polygons is reversible. In Proceedings of the 3rd Joint International Conference on Frontiers in Algorithmics and Algorithmic Aspects in Information and Management, pages 2–3, Dalian, China, June 2013.
 [5] G. N. Frederickson. Hinged Dissections: Swinging & Twisting. Cambridge University Press, August 2002.
 [6] S. Langerman and A. Winslow. A complete classification of tilemakers. In Abstracts from the 18th Japan Conference on Discrete and Computational Geometry and Graphs, Kyoto, Japan, September 2015.
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