1 Introduction
In dimensional Euclidean space, given a set of objects , the online search problem (for an object in ) asks for a curve (or search strategy) with that minimizes the competitive ratio [14]. We say that is competitive if there exists an such that
for all , where . The infimum over all such is the competitive ratio of . Problems of this type date back to Beck [4] and Bellman [6].
The arguably most basic and wellknown such online search problem from this class is the version in which and , also known as the cowpath problem, where the bestpossible competitive ratio is [5], achieved by visiting the points sequentially. In fact, the class of all competitive strategies has been investigated more closely [1]. Moreover, the competitive ratio can be improved to about using randomization [5]. The problem has been generalized to more than two paths starting at the origin [13], searching various types of objects in the plane or lattice [3], and many more scenarios. For a (slightly outdated) survey, see the book by Gal [11].
Another very natural generalization is the case of general and being equal to , the set of all hyperplanes in . We call this version the Dimensional Hyperplane Search Problem. When is a set of affine subspaces, this is arguably the most interesting case. Note that, if contains all affine subspaces of dimension , any constant competitive ratio is ruled out, even when is fixed. Surprisingly, in addition to the case of , results only seem to be known for and for offline versions [9, 2], leaving a gap in the onlinealgorithms literature. For , it is conjectured that a logarithmic spiral that achieves a competitive ratio of about is optimal [3, 10]. The present paper addresses the aforementioned gap by providing the first asymptotic results for .
For the remainder of the paper, we will consider the essentially equivalent but arguably cleaner Sphere Inspection Problem: The goal is to find a dimensional minimumlength closed curve, , that inspects the unit sphere in , i.e., sees every point on the unit sphere . Here, we say that an object sees a point on the surface of the unit sphere if there is a point such that the line segment intersects the unit ball exclusively at . Note that the curve is not required to start in the origin any more. Such a minimumlength curve exists for any dimension [12]. While this problem is trivial for , it has only been shown recently that the bestpossible length for is [12]. No results for higher dimensions are known. Such visibility problems have also been considered from an algorithmic point of view, e.g., [7, 8].
While the connection between hyperplane search and sphere inspection seems to be folklore (e.g., [12]), we state it formally and provide a short proof for completeness. Let be an interval, and for , a set , and curve , we denote by the set and by the curve .
Proposition 1.
Let . The following statements are equivalent:

There exists a length closed curve in that inspects .

There exists an competitive strategy for the Dimensional Hyperplane Search Problem.
Proof.
We first show that (i) implies (ii). Fix , let be the length curve for the Sphere Inspection Problem in , and let be a point on . We show that there exists a competitive strategy for the Dimensional Hyperplane Search Problem where the additive constant is equal to . Our strategy works in an infinite number of phases, starting with Phase . In each Phase , the strategy consists of a straight line from the origin to , the curve , and a straight line back from to the origin.
In the following, we show that, in Phase , all hyperplanes of distance at most from the origin are visited, and the total distance traversed in Phase is at most , which implies the claim. We show two parts separately:

We show that all claimed hyperplanes are visited by . Consider Phase and a hyperplane at distance at most from the origin. Let
be a unit normal vector of
. Note that sees both the point and the point . Let be a point from which sees , and let be a point from which it sees . Note that, if , then or may be on . In that case we are done. Otherwise and are on different sides of , and, by the intermediate value theorem, intersects . 
We next bound the distance traversed by in Phase . Note that it suffices to show that the start point of , which is equal to its end point, has distance at most from the origin. By scaling, we can restrict to showing . Let be the hyperplane . Note that must intersect since otherwise does not see , thus . Since this distance (of to ) is precisely , it follows that .
We now show that (ii) implies (i). Again fix . Assume that there exists a curve starting in the origin and such that, for all hyperplanes , visits after traversing at most a length of . We show that, for any , there exists a, not necessarily closed, length curve that inspects the sphere. This then implies the existence of a length closed curve that inspects the sphere. We define
Note that the minimum exists because is closed. Since for each with , visits after traversing at most a length of , the length of is at most , so the length of is at most .
It remains to be shown that indeed sees every point . Let be the hyperplane tangent to the unit sphere in . Further let . By definition of , we have . Further, by definition of , we have , implying that sees . ∎
In Section 2, we give an auxiliary lemma. In Sections 3 and 4 we will use that lemma and show the following two theorems.
Theorem 1.
Any curve in that inspects has length at least .
Theorem 2.
There exists a closed curve in of length that inspects .
2 An Auxiliary Lemma
The following lemma simplifies thinking about the Sphere Inspection Problem.
Lemma 3.
Let be the convex hull of some point set . Then, the following two statements are equivalent.

We have .

The set sees every point .
Proof.
We start by showing that (i) implies (ii). Let be the hyperplane that is tangent to the unit sphere in . If contains a point , then the lemma directly follows since the line segment lies completely within and therefore does not intersect the unit sphere other than at . So assume that there is no point that is contained in . Then, and since contains one point of the hyperplane (, which is contained in by a), there must exist two points which are separated by . Without loss of generality, let be the point in the halfspace defined by whose interior is disjoint from . Then the line segment is completely contained in that halfspace, and since the interior of the halfspace is disjoint from , can see .
Now we show that (ii) implies (i). Towards a contradiction, assume that there exists with . Then consider a hyperplane that separates from , and let be a unit normal vector of pointing away from . Consider the hyperplane that is tangent to in . Clearly, both and are contained in one open halfspace defined by . Note that no point in this halfspace can see . Therefore, no can see ; a contradiction. ∎
3 Lower Bound
The goal of this section is to prove Theorem 1. Towards this, let be a curve in that inspects the unit sphere. We cut into a minimum number of contiguous portions of length at most for some fixed . Let be the resulting tour portions, where . Choose a portion , and let be its midpoint. Clearly is contained in the ball that has center and radius . Further define to be the cone that is the intersection of all halfspaces that contain both and and whose defining hyperplanes are tangent to both and . Note that the set of points on the sphere that can be seen by the curve can also be seen from the apex of , as visualized by Figure 1. This holds since the radius of is . Note that a single point can see some subset of an open hemisphere of the unit sphere. Let denote a set of open hemispheres such that covers the portion of the sphere seen by . Since inspects the sphere, we have that covers the sphere.
We need the following lemma.
Lemma 4.
The minimum number of open hemispheres that cover is .
Proof.
To show that points are sufficient, choose a simplex containing the sphere. Now consider the set of open hemispheres whose poles are colinear with the origin and a vertex of that simplex. Indeed, since the simplex contains the sphere, by Lemma 3, the set of vertices of the simplex sees every point . Since each point sees only a subset of the corresponding open hemisphere, the upper bound follows.
For the lower bound, we can use induction on . Clearly the circle needs at least three open halfcircles to be covered. For , we have that the boundary of the first hemisphere is , and each hemisphere can cover at most an open hemisphere of . So by induction at least hemispheres are needed to cover , and thus we have that at least hemispheres are needed to cover . ∎
By Lemma 4 we have that , implying and therefore . With , we have that for all , and thus .
4 Upper Bound
In this section we prove Theorem 2. Let be the dimensional crosspolytope, i.e., the polytope . Define , as shown in Figure 2. We claim that the vertices of inspect the unit sphere.
Lemma 5.
For any point , there exists a vertex of that sees .
Proof.
We prove that contains the unit sphere; then Lemma 3 shows the claim. In other words we prove that for any point with (so, ), it also holds that . Indeed, by the CauchySchwartz inequality we have that for any which in turn can be upper bounded by for any with . ∎
Let be the graph^{1}^{1}1The graph of a polytope is a graph with a node for each vertex of and an edge connecting those nodes if has an edge between the corresponding vertices. The graph of is also referred to as its skeleton. of polytope . Note that is the socalled cocktail party graph which can be obtained by removing a perfect matching from a complete graph on vertices (indeed, in any vertex has an edge to any other vertex except ). We next prove that is Hamiltonian.
Lemma 6.
The graph is Hamiltonian.
Proof.
The proof is by induction on . The statement clearly holds for where the cross polytope is a square, and itself is a Hamiltonian cycle that we denote by . Consider a Hamiltonian cycle for . Note that can be constructed by and adding two nodes, and that are connected to each of the nodes of . To construct a cycle of , take any two distinct edges and contained in and replace them with the edges and , respectively. The resulting tour is connected, visits all nodes of exactly once and the used edges are contained in the edge set of , thus ensuring the feasibility of . ∎
By Lemma 5 it directly follows that any closed curve that visits all the vertices of inspects the unit sphere, and therefore so does the closed curve corresponding to the Hamiltonian cycle in the skeleton of . To complete the proof of Theorem 2, note that each edge of has a length of and that traverses such edges.
5 Conclusion
In this paper, we narrowed down the optimal competitive ratio for the Dimensional Hyperplane Search Problem to . The obvious open problem is closing this gap.
Acknowledgements.
We thank Paula Roth for helpful discussions.
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