 # The Shadows of a Cycle Cannot All Be Paths

A "shadow" of a subset S of Euclidean space is an orthogonal projection of S into one of the coordinate hyperplanes. In this paper we show that it is not possible for all three shadows of a cycle (i.e., a simple closed curve) in R^3 to be paths (i.e., simple open curves). We also show two contrasting results: the three shadows of a path in R^3 can all be cycles (although not all convex) and, for every d≥ 1, there exists a d-sphere embedded in R^d+2 whose d+2 shadows have no holes (i.e., they deformation-retract onto a point).

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## 1 Introduction

Oskar’s maze, named after the Dutch puzzle designer Oskar van Deventer, who invented it in 1983, is a mechanical puzzle consisting of a hollow cube and three mutually orthogonal rods joined at their centers (see Figure 1). Each face of the cube has slits forming a maze, and the mazes on opposite faces are identical. Each rod is orthogonal to a pair of opposite faces, and it is able to slide in the slits, tracing out the maze. Hence, in order to move the rods around, one has to solve three mazes simultaneously.

In 1994, Hendrik W. Lenstra asked if the mazes could be chosen so that the common point of the three rods could trace a simple closed curve. Observe that none of the mazes may contain any cycles, or some pieces of the cube would fall out of the puzzle. So, what Lenstra was really asking for is a simple closed curve whose projections onto three pairwise orthogonal planes contain no cycles. In other words, he wanted the three shadows of a simple closed 3D curve to all be trees.

As Peter Winkler reported in his book Mathematical mind-benders , a solution had already been found some years before by John R. Rickard, who discovered the curve illustrated in Figure 2, also appearing on the front cover of Winkler’s book.

Several other, more complex solutions to Lenstra’s problem are known. Notably, in 2012 Adam P. Goucher constructed a simple closed curve having shadows that are all trees, which also happens to be a trefoil knot . The curve was therefore named Treefoil. Goucher also constructed a pair of linked cycles whose union has shadows that are all trees. Figure 2: Rickard’s curve, illustrated by Afra Zomorodian, and appearing on the front cover of Peter Winkler’s book Mathematical mind-benders.

Our research is motivated by the following two questions. Is it possible for the three shadows of a simple closed curve to be paths, i.e., have neither cycles nor branch points? Can the three shadows of a simple open curve be simple closed curves? Both these questions are related to Lenstra’s question, whose history is outlined in . These questions have also been posed independently (see [1, 2]).

### Our contribution.

In Section 2 we answer the first question in the negative: the three shadows of a simple closed curve in cannot all be paths.

In Section 3 we answer the second question in the affirmative: there exist simple open curves in whose three shadows are simple closed curves, although the shadows cannot all be convex. Furthermore, we exhibit a polygonal chain with this property having only six vertices, and we prove that six is the minimum.

In Section 4 we extend Rickard’s curve to higher dimensions, giving an inductive construction of a -sphere embedded in , for every , whose shadows are all contractible. (A contractible set is one that can be continuously shrunk to a point, and hence it has no holes.)

Section 5 concludes the paper with some remarks and suggestions for further work.

This research has obvious applications in computer vision and 3D object reconstruction, where the goal is to deduce properties of an unknown 3-dimensional object given its three projections. Specifically, we may want to study the topology of an object that projects to three given paths. It is easy to see that such an object may not be unique, and hence it makes sense to study the set of 3-dimensional objects that are

compatible with three given projections. Observe that any such set is closed under taking unions, and therefore it has a unique “largest” object, which is the union of all the objects in the set.

It is interesting to note that there are triplets of paths that are not compatible with any connected set, such as the one in Figure 3. This means that an Oskar’s-maze-like puzzle could be “unsolvable” even if it had no crossroads on any face. By “unsolvable” we mean that the set of locations that are reachable by the central point of the three rods depends on where the rods are located. Therefore, if we assign two points in the 3-dimensional maze determined by the three 2-dimensional mazes, it may be impossible to go from one to the other by moving the rods around.

## 2 The shadows of a cycle

In this section we prove that the shadows of a simple closed curve in cannot all be simple open curves. We start with some notation and definitions.

For a point and , we denote by the -th coordinate of . The -projection, or -shadow, of a set , denoted by , is the orthogonal projection of into the -th coordinate hyperplane, e.g., . If , we may simply write instead of .

A path is a (non-degenerate) simple open curve, and the interior of a path is a copy of the path with its endpoints removed. A cycle is a (non-degenerate) simple closed curve.

An -strand of a simple curve is a minimal path between the -extremes of the curve. That is, an -strand of a simple curve with -minimum and -maximum is a path whose endpoints and are such that and , and every internal point is such that .

The interiors of any two distinct -strands of a simple curve are disjoint. Hence any two distinct -strands of a path intersect at most at one common endpoint.

If is an -strand of a simple curve , then is an -strand of , for . If is a path, the converse of Observation 2 is also true, as stated in the next lemma.

If is an -strand of the -projection of a simple curve , with , and is a path, then there exists an -strand of whose -projection is . Let and be the endpoints of , let such that and , and let be a path with endpoints and . Since is a path, . Let and be the endpoints of , such that and belong to the same connected component of . Parameterizing from to , let be the last point of such that . Because separates and in , there are points of after whose -projection is . Letting be the first of such points, the sub-path of with endpoints and is an -strand of whose -projection is .

In the following lemma we show that, if two shadows of a non-degenerate cycle are paths, then each of the two shadows has at least two similarly-oriented strands.

If is a cycle in that is not contained in any -orthogonal plane, and and are paths, then has at least two distinct -strands. Since is not in an -orthogonal plane, and both have at least one -strand. Let be an -strand of , let , and assume for contradiction that has a unique -strand , as sketched in Figure 4.

Since is an -strand of by Observation 2, and the endpoints of a path cannot be in the interior of one of its strands, contains the endpoints of . Also, the -shadow of is a superset of , and hence it contains the endpoints of , as well. Moreover, is connected (because is a cycle), hence is a connected subset of the path containing its endpoints, and therefore it must be all of . By Lemma 2, has an -strand such that . Since has a unique -strand , we have , again by Observation 2.

Respectively parameterize , , , and each from the -minimum to the -maximum of . Choose some value strictly between these extremes, . Let , , , respectively be the first point of each strand where the coordinate attains the value . With this we have and , which implies . So the interiors of the strands and intersect, contradicting Observation 2. Thus our assumption must be wrong: must have at least two distinct -strands.

Next we prove that an -strand and an -strand of a planar path must intersect each other, and therefore their union must be a sub-path.

If and are respectively an -strand and an -strand of a path in , then is a path. Let be the bounding box of , and let and be the leftmost and rightmost points of , respectively. Consider the polygonal chain with vertices , , , , in this order. Then is a cycle which, by the Jordan Curve Theorem, disconnects the plane into two components: an interior and an exterior .

Let and be the lowest and highest points of , respectively. Note that lies on the bottom edge of , and hence it lies either in or on the curve . Similarly, lies on the top edge of , and hence it lies either in or on the curve . Thus, and . It follows that must intersect . Since and , must intersect .

Thus, is a connected subset of the path , and is therefore a path.

In our final lemma we show that a planar path cannot have two distinct -strands and two distinct -strands.

A path in has either a unique -strand or a unique -strand. Assume for a contradiction that is a path in with distinct -strands and distinct -strands . By Observation 2, and are either disjoint, or their intersection is precisely a common endpoint. Suppose for a contradiction that they are disjoint, and let be the minimal path connecting them. By Lemma 2, is a path, as well as , which implies that . Similarly, , and therefore , contradicting Observation 2. Thus is a single point , and by a symmetric argument , as well. Let be the bounding box of . Then must be a vertex of , and we may assume that . Also, by symmetry, we may assume that . It follows that , and either or .

Suppose that , as in Figure 5. Then is in the same connected component of as the edge . This implies that intersects , contradicting the fact that .

Suppose that , as in Figure 5. Then is in the same connected component of as the edge . This implies that intersects , contradicting the fact that .

Thus our assumption fails: has either a unique -strand or a unique -strand.

We are now able to prove the main result of this section.

There is no cycle in whose shadows are all paths. Assume for a contradiction that the three shadows of a cycle in are all paths. Note that cannot lie in any -orthogonal plane, or would not be a path. By Lemma 2, since the -shadow and the -shadow are both paths, the -shadow must have at least two distinct -strands. Likewise, since the -shadow and the -shadow are both paths, the -shadow must also have at least two distinct -strands. But by Lemma 2, a path in the -plane cannot have two distinct -strands and two distinct -strands, which is a contradiction.

## 3 The shadows of a path

Here we study the simple open curves in whose shadows are simple closed curves. In contrast with the similarly-defined curves of the previous section, in this case we can construct a wealth of such curves. An example is illustrated in Figure 6.

Note that the curve in Figure 6 is a polygonal path (i.e., a simple open polygonal chain) consisting of axis-parallel segments. If we allow arbitrarily oriented segments, we can find an example with only six vertices, which is the minimum possible.

There exists a polygonal path in with six vertices whose shadows are cycles. No such polygonal path exists with fewer than six vertices. An example of such a polygonal path is , which is shown in Figure 7.

Suppose for a contradiction that a polygonal path in with vertices exists such that its shadows are cycles. If , then clearly no shadow can be a cycle. Suppose that , and let the polygonal path be . Then each shadow must be a triangle, and hence for every . It follows that , which contradicts the fact that a polygonal path is an open curve.

Assume now that , and the polygonal path is . For every , the -projection of the polygonal path is either a triangle or a quadrilateral. In both cases, the -shadows of the segments and have a non-empty intersection. Since , the -shadows of and do not coincide for at least two ’s, say, and . Then the -shadow of the polygonal path must be a triangle, and are collinear, and hence the segments and lie on a plane that is orthogonal to the -plane. Similarly, the segments and lie on a plane that is orthogonal to the -plane, too. Hence and are either collinear or they lie on a common -orthogonal plane. If and are collinear (and disjoint), then their -shadows are disjoint for some , contradicting the fact that their intersection must be non-empty. If and lie on a common -orthogonal plane, then and are disjoint, which is again a contradiction.

Note that, in all the above examples, one of the shadows is a non-convex cycle. It is natural to ask whether a path exists whose shadows are all convex cycles. In the following theorem, we answer in the negative. (Due to space constraints, we only give a sketch of the proof.)

There is no path in whose shadows are convex cycles. Proof (sketch). Suppose for contradiction that there exists a path in whose shadows are convex cycles. For every , lies on the surface of a cylinder with section and -parallel axis.

The intersection of and is sketched in Figure 8. It consists of two horizontal axis-aligned rectangles and (assuming that the vertical direction is -parallel) whose vertices are joined by four paths , , , and . The rectangles and may be degenerate, i.e., they may be -parallel or -parallel segments, or points. Let a horizontal plane intersect the interior of the path in the point , for each . Then, for all , either or , where indices are taken modulo .

Further intersecting with , we reduce and to at most four horizontal curves each. Therefore, in total we have curves, whose union is an embedding in of a graph with edges. Also, we may assume without loss of generality that each endpoint of lies at a vertex of the embedding of , or at the midpoint of one of the edges. Hence there are only finitely many possible graphs to consider, and only finitely many choices of in each graph embedding. By exhaustively examining all the possible choices of , we conclude that none of them has shadows that are all convex cycles.

## 4 Shadows in higher dimensions

In this section we generalize Rickard’s curve to higher dimensions. We inductively construct an embedding of a -sphere in whose shadows are all contractible, i.e., they deformation-retract to a point.

An -slice of a set , with , is a non-empty intersection between and an -orthogonal hyperplane.

For every , there exists an embedding of a -sphere in whose shadows are all contractible. Let be Rickard’s curve, introduced in Section 1. Then, for all , we inductively define

 Sd+1=⋃λ∈[−1,1](1−|λ|)⋅Sd×{λ}.

It is easy to see that is an embedding of a -sphere in for every . We claim that all the shadows of deformation-retract to the point . This is true for , as suggested by Figure 2. Assume now the inductive hypothesis that the claim is true for , and therefore there exists a continuous map

 Fd,i:πi(Sd)×[0,1]→πi(Sd)

with and , for every . Now, for each , we can construct a continuous map

 Fd+1,i:πi(Sd+1)×[0,1]→πi(Sd+1)

with and .

If , we first define the auxiliary map

 F′:πi(Sd+1)×[0,1]→πi(Sd+1)

as follows. For every such that and , we let

 F′(x,λ)=(1−|xd+2|)⋅Fd,i(πd+2(x)1−|xd+2|,λ)×{xd+2}.

If with and , we let Observe that every -slice of is a scaled copy of . (Figure 9 shows for .) Informally, applies with parameter to a suitably scaled copy of each -slice, and then it rescales it back. Therefore, since is a deformation retraction of to the point , is a deformation retraction of to the segment . To obtain , one just has to compose with a deformation retraction of to the point . In formulas, for and ,

 Fd+1,i(x,λ)={F′(x,2λ)if% \ λ<1/2(2−2λ)⋅F′(x,1)if\ λ≥1/2.

If , we can simply set

 Fd+1,i(x,λ)=(1−λ)⋅x

for every and . This is easily seen to be a deformation retraction to . (Figure 10 shows for .)

Hence all the shadows of the -sphere deformation-retract to a point for every , meaning that they are contractible.

## 5 Concluding remarks

In this paper we studied the shadows of curves in (a shadow being an axis-parallel projection), also settling some long-standing open problems posed in [1, 2].

In Section 2 we proved that there is no cycle in whose shadows are all paths. Note that by applying a projective transformation, we may equivalently define shadows to be perspective projections, provided that the three viewpoints are not collinear, and the plane through them does not intersect the curve.

In Section 3 we proved that there exist paths in whose shadows are all cycles. We also showed that, if such a path is a polygonal chain, it must have at least six vertices, and we found an example with exactly six vertices. Then we proved that there is no path in whose shadows are all convex cycles.

Finally, in Section 4 we showed that there exists an embedding of a -sphere in whose shadows are all contractible, for every . This generalizes Rickard’s curve (see Figure 2), which is a cycle in whose shadows contain no cycles.

Our results can be expanded in several directions. A natural goal would be to minimize the total number of branch points of the shadows of a cycle in , assuming that all shadows are cycle-free. Because each shadow of Rickard’s curve has two branch points, such a minimum is at most six. On the other hand, by Theorem 2, the minimum is at least one. With the same proof technique employed in Section 2, we can prove the following generalized version of Theorem 2, which implies that the minimum number of branch points of the shadows must be at least three.

There is no cycle in with cycle-free shadows such that is a path, and and have at most one branch point each. We conjecture Rickard’s curve to be an optimal example in terms of branch points of its shadows.

If the shadows of a cycle in are all cycle-free, then each shadow has at least two branch points.

### Acknowledgments.

The authors are indebted to Giuseppe Antonio Di Luna, Pat Morin, and Joseph O’Rourke for stimulating discussions.

Prosenjit Bose and Jean-Lou De Carufel were supported in part by NSERC.

Michael G. Dobbins was supported by NRF grant 2011-0030044 (SRC-GAIA) funded by the government of South Korea.

Heuna Kim was supported by the Deutsche Forschungsgemeinschaft within the research training group “Methods for Discrete Structures” (GRK 1408).