 # Some Polycubes Have No Edge-Unzipping

It is unknown whether or not every polycube has an edge-unfolding. A polycube is an object constructed by gluing cubes face-to-face. An edge-unfolding cuts edges on the surface and unfolds it to a net, a non-overlapping polygon in the plane. Here we explore the more restricted edge-unzippings where the cut edges form a path. We construct a polycube that has no edge-unzipping.

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## 1 Introduction

A polycube is an object constructed by gluing cubes whole-face to whole-face, such that its surface is a manifold. Thus the neighborhood of every surface point is a disk; so there are no edge-edge nor vertex-vertex nonmanifold surface touchings. Here we only consider polycubes of genus zero. The edges of a polycube are all the cube edges on the surface, even when those edges are shared between two coplanar faces. Similarly, the vertices of a polycube are all the cube vertices on the surface, even when those vertices are flat, incident to face angles. Such polycube flat vertices are degree-. It will be useful to distinguish these flat vertices from corner vertices, non-flat vertices with incident angles (degree-, -, or -). For a polycube , let its -skeleton graph include every vertex and edge of , with vertices marked as either corner or flat.

It is an open problem to determine whether every polycube has an edge-unfolding, a tree in the -skeleton that spans all corner vertices (but need not include flat vertices), which, when cut, unfolds the surface to a net, a planar, non-overlapping polygon [O’R19]. Here by non-overlapping is meant that no two points, each interior to a face, are mapped to same point in the plane. This allows two boundary edges to coincide in the net; so the polygon is “weakly simple.” The intent is that we want to be able to cut out the net and refold to . Henceforth “edge-unfolding” will mean: an edge-unfolding to a net.

It would be remarkable if it were true that every polycube could be edge-unfolded, but no counterexample is known. There has been considerable exploration of orthogonal polyhedra, a more general type of object, for which there are examples that cannot be edge-unfolded [BDD98]. (See [DF18] for citations to earlier work.) But polycubes have more edges in their -skeleton graphs for the cut tree to follow than do orthogonal polyhedra, so it is conceivably easier to edge-unfold polycubes.

A restriction of edge-unfolding has been studied in [DDL10], [O’R10], [DDU13]: edge-unzipping. This is an edge-unfolding whose cut tree is a path (so that the surface could be “unzipped”). It is apparently unknown if even this highly restricted edge-unzipping could unfold every polycube to a net. The result of this note is to settle this question in the negative: two different polycubes are constructed each of which has no edge-unzipping. They are shown in Figure 1 and will be described later.

## 2 Hamiltonian Paths

Shephard [She75] introduced Hamiltonian unfoldings of convex polyhedra, what we are now calling edge-unzippings, following the terminology of [DDL10].111 “Unzipping” is a slight variation on their “zipper unfoldings.” It is easy to see that not every convex polyhedron has an edge-unzipping, simply because the rhombic dodecahedron has no Hamiltonian path. This counterexample avoids confronting the difficult non-overlapping condition. We follow a similar strategy here, constructing a polycube with no Hamiltonian path. But there is a difference in that a polycube edge-unzipping need not include flat vertices, and so need not be a Hamiltonian path in . Thus identifying a polycube that has no Hamiltonian path does not immediately establish that has no edge-unzipping, if has flat vertices.

So one approach is to construct a polycube that has no flat vertices—every vertex is a corner vertex. Then if has no Hamiltonian path, then it has no edge-unzipping. A natural candidate is the polycube object shown in Fig. 2.

However, the -skeleton of does admit Hamiltonian paths, and indeed we found a path that unfolds to a net.

Let be the dual graph of : each cube is a node, and two nodes are connected if they are glued face-to-face. A polycube tree is a polycube whose dual graph is a tree. is a polycube tree. That it has a Hamiltonian path is an instance of a more general claim:

###### Lemma 1

The graph for any polycube tree has a Hamiltonian cycle.

Proof: It is easy to see by induction that every polycube tree can be built by gluing cubes each of which touches just one face at the time of gluing: never is there a need to glue a cube to more than one face of the previously built object.

A single cube has a Hamiltonian cycle. Now assume that every polycube tree of cubes has a Hamiltonian cycle. For a tree of cubes, remove a leaf-node cube , and apply the induction hypothesis. The exposed square face to which glues to make includes either or edges of the Hamiltonian cycle ( would close the cycle; or would imply the cycle misses some vertices of ). It is then easy to extend the Hamiltonian cycle to include , as shown in Figure 3. Figure 3: (a) f contains 3 edges of the cycle (blue); (b) f contains 2 edges of the cycle. The cycles are extended to C by replacing the blue with the the red paths.

So to prove that a polycube tree has no edge-unzipping would require an argument that confronted non-overlap. This leads to an open question:

###### Question 1

Does every polycube tree have an edge-unzipping?

## 3 Bipartite Gp

To guarantee the non-existence of Hamiltonian paths, we can exploit the bipartiteness of , using Lemma 2 below.

###### Lemma 2

A polycube graph is -colorable, and therefore bipartite.

Proof: Label each lattice point of with the -parity of the sum of the Cartesian coordinates of . A polycube ’s vertices are all lattice points of . This provides a -coloring of ; -colorable graphs are bipartite.

The parity imbalance in a -colored (bipartite) graph is the absolute value of the difference in the number of nodes of each color.

###### Lemma 3

A bipartite graph with a parity imbalance has no Hamiltonian path.2

Proof: The nodes in a Hamiltonian path alternate colors . Because by definition a Hamiltonian path includes every node, the parity imbalance in a bipartite graph with a Hamiltonian path is either (if of even length) or

(if of odd length).

So if we can construct a polycube that (a) has no flat vertices, and (b) has parity imbalance , then we will have established that has no Hamiltonian path, and therefore no edge-unzipping. We now show that the polycube , illustrated in Figure 4, meets these conditions Figure 4: The polycube P44, consisting of 44 cubes, has no Hamiltonian path.
###### Lemma 4

The polycube ’s graph has parity imbalance of .

Proof: Consider first the cube that is the core of ; call it . The front face has an extra ; see Fig. 5.

It is clear that the corners of are all colored . The midpoint vertices of the edges of are colored . Finally the face midpoints are colored . So vertices are colored and colored .

Next observe that attaching a cube to exactly one face of any polycube does not change the parity: the receiving face has colors , and the opposite face of has colors .

Now, can be constructed by attaching six copies of a -cube “cross,” call it , which in isolation is a polycube tree and so can be built by attaching cubes each to exactly one face. And each attaches to one corner cube of . Therefore retains ’s imbalance of .

The point of the attachments is to remove the flat vertices of . Note that when attached to , each has only corner vertices.

###### Theorem 1

There is no edge-unzipping of .

Proof: Although it takes some scrutiny of Figure 4 to verify, has no (degree-) flat vertices. Thus an edge-unzipping must pass through every vertex, and so be a Hamiltonian path. Lemma 4 says that has imbalance , and Lemma 2 says it therefore cannot have a Hamiltonian path.

## 4 Construction of P14

It turns out that the smaller polycube shown in Figure 6 also has no edge-unzipping, even though it has flat vertices.

To establish this, we still need an imbalance , which easily follows just as in Lemma 4:

###### Lemma 5

The polycube ’s graph has parity imbalance of .

But notice that has three flat vertices: , , and .

###### Theorem 2

There is no edge-unzipping of .

Proof: An edge-unzipping need not pass through the three flat vertices, , , and , but it could pass through one, two, or all three. We show that in all cases, an appropriately modified subgraph of has no Hamiltonian path. Let be a hypothetical edge-unzipping cut path. We consider four exhaustive possibilities, and show that each leads to a contradiction.

(0) includes .

So is a Hamiltonian path in . But Lemma 5 says that has imbalance , and Lemma 2 says that no such graph has a Hamiltonian path.

(1) excludes one flat vertex and includes .

(Because of the symmetry of , it is no loss of generality to assume that it is that is excluded.) If excludes , then it does not travel over any of the four edges incident to . Thus we can delete from ; say that . This graph is shown in Fig. 7. Figure 7: Schlegel diagram of G−a. We follow [DF18] in labeling the faces of a cube as F,K,R,L,T,B for Front, bacK, Right, Left, Top, Bottom respectively. The corners of P222 are labeled 0,1,2,3 around the bottom face B, and 4,5,6,7 around the top face T. m is the vertex in the middle of B. The edges deleted by removing a are shown dashed.

Following the coloring in Fig. 5, all corners of are colored , so each of the edge midpoints is colored . The parity imbalance of is extra ’s. Deleting maintains bipartiteness and increases the parity imbalance of to . Therefore by Lemma 2, has no Hamiltonian path, and such a cannot exist.

(2) includes just one flat vertex , and excludes .

(Again symmetry ensures there is no loss of generality in assuming the one included flat vertex is .) must include corner , which is only accessible in through the three flat vertices. If excludes , then it must include the edge . Let . In , has degree , so terminates there. It must be that is a Hamiltonian path in , but the deletion of increases the parity imbalance to , and so again such a Hamiltonian path cannot exist.

(3) excludes .

Because corner is only accessible through one of these flat vertices, never reaches and so cannot be an edge-unzipping

Thus the assumption that there is an edge-unzipping path for reaches a contradiction in all four cases. Therefore, there is no edge-unzipping path for .333 Just to verify this conclusion, we constructed these graphs in Mathematica and FindHamiltonianPath[] returned {} for each.

## 5 Edge-unfoldings of P14 and P44

Now that it is known that and each have no edge-unzipping, it is natural to wonder if either settles the edge-unfolding open problem: Can they be edge-unfolded? Indeed both can: see Figures 8

and 9. The colors in these layouts are those used by Origami Simulator [GDG18].