Online Search for a Hyperplane in High-Dimensional Euclidean Space

We consider the online search problem in which a server starting at the origin of a $d$-dimensional Euclidean space has to find an arbitrary hyperplane. The best-possible competitive ratio and the length of the shortest curve from which each point on the $d$-dimensional unit sphere can be seen are within a constant factor of each other. We show that this length is in $\Omega(d)\cap O(d^{3/2})$.


Introduction
In d-dimensional Euclidean space, given a set of objects T ⊆ P(R d ), the online search problem (for an object in T ) asks for a curve (or search strategy) ζ : R 0 → R d with ζ(0) = 0 that minimizes the competitive ratio [14].We say that ζ is c-competitive if there exists an α ∈ R 0 such that length(ζ| [0,t ] ) c • dist(0, T ) + α for all T ∈ T , where t = inf{t : ζ(t) ∈ T }.The infimum over all such c is the competitive ratio of ζ.Problems of this type date back to Beck [4] and Bellman [6].
The arguably most basic and well-known such online search problem from this class is the version in which d = 1 and T = R, also known as the cow-path problem, where the best-possible competitive ratio is 9 [5], achieved by visiting the points (−2) 0 , (−2) 1 , . . .sequentially.In fact, the class of all 9-competitive strategies has been investigated more closely [1].Moreover, the competitive ratio can be improved to about 4.591 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origin [13], searching various types of objects in the plane or lattice [3], and many more scenarios.For a (slightly outdated) survey, see the book by Gal [11].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 d and T being equal to H d−1 , the set of all hyperplanes in R d .We call this version the d-Dimensional Hyperplane Search Problem.When T is a set of affine subspaces, this is arguably the most interesting case.Note that, if T contains all affine subspaces of dimension d ≤ d − 2, any constant competitive ratio is ruled out, even when d is fixed.Surprisingly, in addition to the case of d = 1, results only seem to be known for d = 2 and for offline versions [9,2], leaving a gap in the online-algorithms literature.For d = 2, it is conjectured that a logarithmic spiral that achieves a competitive ratio of about 13.811 is optimal [3,10].The present paper addresses the aforementioned gap by providing the first asymptotic results for d → ∞.
For the remainder of the paper, we will consider the essentially equivalent but arguably cleaner Sphere Inspection Problem: The goal is to find a d-dimensional minimum-length closed curve, γ, that inspects the unit sphere in R d , i.e., γ sees every point p on the unit sphere S d−1 .Here, we say that an object O ⊆ R d sees a point p on the surface of the unit sphere if there is a point p ∈ O such that the line segment pp intersects the unit ball exclusively at p .Note that the curve is not required to start in the origin any more.Such a minimum-length curve exists for any dimension [12].While this problem is trivial for d ∈ {1, 2}, it has only been shown recently that the best-possible length for d = 3 is 4π [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 I be an interval, and for β ∈ R, a set T ⊆ R d , and curve γ : Proof.We first show that (i) implies (ii).Fix d ∈ N, let γ be the length-curve for the Sphere Inspection Problem in R d , and let γ 0 be a point on γ.We show that there exists a (12 )-competitive strategy for the d-Dimensional Hyperplane Search Problem where the additive constant α is equal to 3 .Our strategy works in an infinite number of phases, starting with Phase 0. In each Phase i, the strategy ζ consists of a straight line from the origin to 2 i γ 0 , the curve 2 i • γ, and a straight line back from 2 i γ 0 to the origin.
In the following, we show that, in Phase i, all hyperplanes of distance at most 2 i from the origin are visited, and the total distance traversed in Phase i is at most 3 • 2 i • , which implies the claim.We show two parts separately: .Let H be the hyperplane γ 0 , x = 0. Note that γ must intersect H since otherwise γ does not see −γ 0 / γ 0 2 , thus dist(γ 0 , H) ≤ .Since this distance (of γ 0 to H) is precisely γ 0 2 , it follows that γ 0 2 .
We now show that (ii) implies (i).Again fix d ∈ N. Assume that there exists a curve γ starting in the origin and α ∈ R 0 such that, for all hyperplanes H ∈ H d−1 , γ visits H after traversing at most a length of c • dist(0, H) + α.We show that, for any ε > 0, there exists a, not necessarily closed, length-(c + ε) curve that inspects the sphere.This then implies the existence of a length-(2c + ε) closed curve that inspects the sphere.We define , where t := sup Note that the minimum exists because H is closed.Since for each H ∈ H d−1 with dist(0, H) α /ε, γ visits H after traversing at most a length of c In Section 2, we give an auxiliary lemma.In Sections 3 and 4 we will use that lemma and show the following two theorems.

An Auxiliary Lemma
The following lemma simplifies thinking about the Sphere Inspection Problem.Lemma 3. Let P ⊂ R d be the convex hull of some point set V .Then, the following two statements are equivalent.Proof.We start by showing that (i) implies (ii).Let H be the hyperplane that is tangent to the unit sphere in p.If H contains a point v ∈ V , then the lemma directly follows since the line segment vp lies completely within H and therefore does not intersect the unit sphere other than at p.So assume that there is no point v ∈ V that is contained in H.Then, and since P contains one point of the hyperplane (p ∈ S d−1 , which is contained in P by (i)), there must exist two points v 1 , v 2 ∈ V which are separated by H. Without loss of generality, let v 1 be the point in the halfspace defined by H whose interior is disjoint Now we show that (ii) implies (i).Towards a contradiction, assume that there exists p ∈ S d−1 with p / ∈ P .Then consider a hyperplane H that separates p from P , and let n be a unit normal vector of H pointing away from P .Consider the hyperplane H that is tangent to S d−1 in n.Clearly, both P and S d−1 \ {n} are contained in one open halfspace defined by H .Note that no point in this halfspace can see n.Therefore, no v ∈ V can see n; a contradiction.

Lower Bound
The goal of this section is to prove Theorem 1. Towards this, let γ be a curve in R d that inspects the unit sphere.We cut γ into a minimum number of contiguous portions of length at most δ for some fixed δ < 2. Let ξ 1 , . . ., ξ n be the resulting tour portions, where n = |γ|/δ .Choose a portion ξ, and let x be its midpoint.Clearly ξ is contained in the ball B that has center x and radius δ/2.Further define C to be the cone that is the intersection of all halfspaces that contain both B and S d−1 and whose defining hyperplanes are tangent to both B and S d−1 .Note that the set of points on the sphere that can be seen by the curve ξ can also be seen from the apex of C, as visualized by Figure 1.This holds since the radius of B is δ/2 < 1.Note that a single point p ∈ R d can see some subset of an open hemisphere H of the unit sphere.Let H 1 , . . ., H n denote a set of open hemispheres such that H i covers the portion of the sphere seen by ξ i .Since γ inspects the sphere, we have that (H i ) n i=1 covers the sphere.We need the following lemma.Let G( Cd ) be the graph1 of polytope Cd .Note that G( Cd ) is the so-called cocktail party graph which can be obtained by removing a perfect matching from a complete graph on 2d vertices (indeed, in Cd any vertex v has an edge to any other vertex except −v).We next prove that G( Cd ) is Hamiltonian.By Lemma 5 it directly follows that any closed curve that visits all the vertices of Cd inspects the unit sphere, and therefore so does the closed curve γ corresponding to the Hamiltonian cycle c d in the skeleton of Cd .To complete the proof of Theorem 2, note that each edge of Cd has a length of √ 2d and that γ traverses 2d such edges.

Conclusion
In this paper, we narrowed down the optimal competitive ratio for the d-Dimensional Hyperplane Search Problem to Ω(d) ∩ O(d 3/2 ).The obvious open problem is closing this gap.

Proposition 1 .
Let f : N → R 0 .The following statements are equivalent: (i) There exists a length-O(f (d)) closed curve in R d that inspects S d−1 .(ii) There exists an O(f (d))-competitive strategy for the d-Dimensional Hyperplane Search Problem.

Theorem 1 .Theorem 2 .
Any curve in R d that inspects S d−1 has length at least 2d.There exists a closed curve γ in R d of length (2d) 3/2 that inspects S d−1 .
(i) We have S d−1 ⊂ P .(ii) The set V sees every point p ∈ S d−1 .

B S d− 1 CξFigure 1 :
Figure 1: The apex of the cone C sees all points on the unit sphere that B does.

Lemma 4 .
The minimum number of open hemispheres that cover S d−1 is d + 1. Proof.To show that d + 1 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 p ∈ S d−1 .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 d.Clearly the circle S 1 needs at least three open half-circles to be covered.For S d−1 , we have that the boundary of the first hemisphere H 1 is S d−2 , and each hemisphere H i can cover at most an open hemisphere of ∂H 1 .So by induction at least (d − 2) + 2 = d hemispheres are needed to cover ∂H 1 , and thus we have that at least d + 1 hemispheres are needed to cover S d−1 .By Lemma 4 we have that n d + 1, implying length(γ)/δ d + 1 and therefore length(γ)/δ d.With δ = 2 − ε, we have that length(γ) (2 − ε) • d for all ε > 0, and thus length(γ) 2d.

Lemma 6 .
The graph G( Cd ) is Hamiltonian.Proof.The proof is by induction on d.The statement clearly holds for R 2 where the cross polytope is a square, and G( C2 ) itself is a Hamiltonian cycle that we denote by c 2 .Consider a Hamiltonian cycle c d−1 for G( Cd−1 ).Note that G( Cd ) can be constructed by G( Cd−1 ) and adding two nodes, v and −v that are connected to each of the nodes of G( Cd−1 ).To construct a cycle c d of G( Cd ), take any two distinct edges {v 1 , v 2 } and {v 3 , v 4 } contained in c d−1 and replace them with the edges {v 1 , v }, {v , v 2 } and {v 3 , −v }, {v 3 , −v }, respectively.The resulting tour is connected, visits all nodes of

Figure 2 :
Figure 2: The cross polytope that has vertices at distance √ d from the origin contains the unit sphere, and thus, a tour of its vertices (thick) inspects the unit sphere.

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We show that all claimed hyperplanes are visited by ζ.Consider Phase i and a hyperplane H ∈ H d−1 at distance at most 2 i from the origin.Let n be a unit normal vector of H.Note that 2 i • γ sees both the point 2 i n and the point −2 i n.Let p 1 be a point from which 2 i • γ sees 2 i n, and let p 2 be a point from which it sees −2 We next bound the distance traversed by ζ in Phase i.Note that it suffices to show that the start point of 2 i • γ, which is equal to its end point, has distance at most 2 i from the origin.By scaling, we can restrict to showing γ 0 2 i n.Note that, if dist(0, H) = 2 i , then p 1 or p 2 may be on H.In that case we are done.Otherwise p 1 and p 2 are on different sides of H, and, by the intermediate value theorem, 2 i • γ intersects H.•