A Generalization of the Convex Kakeya Problem
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Abstract
Given a set of line segments in the plane, not necessarily finite, what is a convex region of smallest area that contains a translate of each input segment? This question can be seen as a generalization of Kakeya’s problem of finding a convex region of smallest area such that a needle can be rotated through 360 degrees within this region. We show that there is always an optimal region that is a triangle, and we give an optimal Θ(nlogn)time algorithm to compute such a triangle for a given set of n segments. We also show that, if the goal is to minimize the perimeter of the region instead of its area, then placing the segments with their midpoint at the origin and taking their convex hull results in an optimal solution. Finally, we show that for any compact convex figure G, the smallest enclosing disk of G is a smallestperimeter region containing a translate of every rotated copy of G.
Keywords
Computational geometry Discrete geometry Algorithms Kakeya1 Introduction
Let \(\mathfrak {F}\) be a family of objects in the plane. A translation cover for \(\mathfrak {F}\) is a set K such that any object in \(\mathfrak {F}\) is contained in a translate of K [28]. We are interested in determining a convex translation cover for \(\mathfrak {F}\) of smallest possible area or perimeter.
As a special case of translation covers, we can consider the situation where the family \(\mathfrak {F}\) consists of copies of a given compact convex figure G, rotated by all angles in [0,2π). In other words, we are asking for a smallest possible convex set K such that G can be placed in K in every possible orientation. We will call such a translation cover a keyhole for G (since a key can be turned fully in a keyhole, it can certainly be placed in every possible orientation).
For the general case, when the Kakeya set is not necessarily convex or even simply connected, the answer was thought to be a deltoid with area π/8. However, Besicovitch gave the surprising answer that one could rotate a needle using an arbitrary small area [3, 4].
Bezdek and Connelly [7] surveyed results on minimumperimeter and minimumarea translation covers. For the family of closed curves of length at most one, they proved that smallestperimeter translation covers are exactly the convex sets of constant width 1/2. The corresponding problem for minimizing the area, known as Wetzel’s problem, is still open [7, 28]. For the family of sets of diameter at most one, Bezdek and Connelly [8] proved that the unique minimumperimeter translation cover is the circle of radius \(1/\sqrt{3}\). More precisely, they proved that this circle is the unique smallestperimeter keyhole for the equilateral triangle of side length one. By Jung’s theorem [14], this circle contains any set of diameter one, and so the translation cover result follows.
Recently, Kakeyatype problems have received considerable attention due to their many applications. There are strong connections between Kakeyatype problems and problems in number theory [9], geometric combinatorics [29], arithmetic combinatorics [16], oscillatory integrals, and the analysis of dispersive and wave equations [25].
In this paper, we first generalize Pál’s result [18] in the following way: For any family \(\mathfrak {F}\) of line segments in the plane, there is a triangle that is a minimumarea translation cover for \(\mathfrak {F}\).
Theorem 1
Let \(\mathfrak {F}\) be a set of line segments in the plane, and let P be a convex translation cover for \(\mathfrak {F}\). Then there is a translation cover T for \(\mathfrak {F}\) which is a triangle, and such that the area of T is less than or equal to the area of P.
With this characterization in hand, we can efficiently compute a smallest area translation cover for a given family of n line segments. Our algorithm runs in time O(nlogn), which we prove to be optimal in the algebraic computation tree model. It is based on the problem of finding a smallestarea affineregular hexagon containing a given centrally symmetric polygon, a problem that is interesting in its own right. As far as we know, except for some trivial cases, previously known algorithms for finding smallestarea translation covers have a running time exponential in n, the number of input objects [1, 27].
As observed above, minimizing the perimeter of a translation cover is much easier. Let \(\mathfrak {F}\) be a family of centrally symmetric convex figures. We prove that if we translate each figure such that its center of symmetry is the origin, then the convex hull of their union is a smallestperimeter translation cover for \(\mathfrak {F}\).
This immediately implies that a circle with diameter 1 is a smallestperimeter keyhole for the unitlength segment. For figures G that are not centrally symmetric, this argument no longer works. We generalize the result by Bezdek and Connelly [8] mentioned above and prove the following theorem (Bezdek and Connelly’s result is the special case where G is an equilateral triangle):
Theorem 2
Let G be a compact convex set in the plane, and let \(\mathcal{G}\) be the family of all the rotated copies of G by angles in [0,2π). Then the smallest enclosing disk of G is a smallestperimeter translation cover for \(\mathcal{G}\).
2 Preliminaries
An oval is a compact convex figure in the plane. For an oval P, let \(w_{P}: [0,\pi] \rightarrow \mathbb {R}\) denote the width function of P. The value w _{ P }(θ) is the length of the projection of P on a line with slope θ (that is, a line that makes angle θ with the xaxis). Let P denote the area of P.
For two ovals P and Q, we write w _{ P }≥w _{ Q } or w _{ Q }≤w _{ P } to mean pointwise domination, that is for every θ∈[0,π) we have w _{ P }(θ)≥w _{ Q }(θ). We also write w _{ P }=w _{ Q } if and only if both w _{ P }≤w _{ Q } and w _{ Q }≤w _{ P } hold.
The Minkowski symmetrization of an oval P is the oval \(\bar {P}= \frac{1}{2} (P  P) = \{\frac{1}{2} (x  y) \mid x, y \in P\}\). It is well known and easy to show that \(\bar {P}\) is centrally symmetric around the origin, and that \(w_{\bar {P}} = w_{P}\).
Ohmann [17] and Chakerian [10] studied sets with a given fixed width function, and obtained the following result (see for instance Theorem 3′ in [10] for a proof):
Fact 1
Given an oval P, there is a trigonal disk D with D≤P such that w _{ D }=w _{ P }.
3 Minimum Area for a Family of Segments
In this section we will prove Theorem 1. The proof contains two parts. First we prove that for every oval P there exists a triangle T with T≤P and w _{ T }≥w _{ P } (Theorem 3). The second part is to prove that for an oval P and a closed segment s, if w _{ s }≤w _{ P } then P contains a translated copy of s (Lemma 1).
Theorem 3
Given an oval P, there exists a triangle T with T≤P and w _{ T }≥w _{ P }.
Proof
Let \(\mathfrak {D}\) be the set of trigonal disks D such that we have D≤P and w _{ D }≥w _{ P }. The set \(\mathfrak {D}\) is nonempty by Fact 1. Consider the three arcs connecting the main vertices of a trigonal disk in \(\mathfrak {D}\). Each arc can be straight, or not. We choose a trigonal disk \(D \in \mathfrak {D}\) with a maximum number of straight arcs. We show that D is a triangle.
Let AUBVCW be the hexagon from the definition of the trigonal disk D, and assume for a contradiction that D is not a triangle, that is, there is at least one nonstraight arc among the three arcs connecting A,B, and C. See Fig. 4(a). Without loss of generality, we assume that the arc connecting A and B is not straight.
Let the sides AW and BV be vertical, with C above the line AB. Let X be the point of D below AB with the largest vertical distance d from the line AB. Let C′ be the point vertically above C at distance d from C. Let D′ be the convex hull of the part of D above the line AB and the point C′. It is not difficult to see that D′ is also a trigonal disk: Let U′ be the point vertically below U at distance d from U. Then the hexagon AU′BVC′W is centrally symmetric and contains D′. Clearly D′ contains the triangle ABC′. See Fig. 4(b). We observe that if AC is a straight arc in D, then A′=A and A′C′ is a straight arc in D′. Similarly, if BC is a straight arc in D, then B′C′ is a straight arc in D′. Since AB is a straight arc in D′, the trigonal disk D′ has at least one straight arc more than D.
We show next that D′≤D. The area of D′∖D is bounded by the area of the two triangles A′C′C and B′C′C, where A′ and B′ are points on D such that A′C′ and B′C′ are tangent to D. This area is equal to d/2 times the horizontal distance between A′ and B′. But the horizontal distance between A′ and B′ is at most the horizontal distance between A and B, so the area of D′∖D is bounded by the area of the triangle AXB, and we have D′≤D.
We also need to argue that w _{ D′}≥w _{ D }. Consider a minimal strip \(\mathfrak {S}\) containing D, and a minimal strip \(\mathfrak {S}'\) with the same orientation for D′. If \(\mathfrak {S}\) does not touch D from below between A and B, then \(\mathfrak {S}'\) is at least as wide as \(\mathfrak {S}\). Otherwise, \(\mathfrak {S}\) touches D from below at a point Y between A and B, and touches from above at C, as C is the only antipodal point of D for Y. The strip \(\mathfrak {S}'\) is determined either by A and C′, or by B and C′. Since the top side of \(\mathfrak {S}'\) meets C′, it lies above the top side of \(\mathfrak {S}\) at distance d. The vertical distance between the bottom sides of \(\mathfrak {S}\) and \(\mathfrak {S}'\) is at most d. Hence, the width of \(\mathfrak {S}'\) is not less than the width \(\mathfrak {S}\).
Since w _{ D′}≥w _{ D }≥w _{ P } and D′≤D≤P the trigonal disk D′ is a member of \(\mathfrak {D}\), but has more straight arcs than D. This contradicts our choice of D, and so our assumption that D is not a triangle must be false. □
This finishes the first part. We need the following lemma, which shows that whether or not an oval P contains a translated copy of a given segment s can be determined by looking at the width functions of P and s alone:
Lemma 1
Let s be a segment in the plane, and let P be an oval. P contains a translated copy of s if and only if w _{ s }≤w _{ P }.
Proof
Without loss of generality, let s be a horizontal segment. The “only if” part follows immediately from the definition of the width function of an oval.
It remains to prove the “if” direction. Let pq be a horizontal segment of maximal length contained in P. Then P has a pair of parallel tangents ℓ _{1} and ℓ _{2} through p and q. By the assumption w _{ s }≤w _{ P }, the distance between ℓ _{1} and ℓ _{2} must be large enough to place s in between the two lines. But this implies that the segment pq is at least as long as s, and s can be placed on the segment pq in P. □
To prove Theorem 1, let P be an oval that contains a translated copy of every \(s \in \mathfrak {F}\). By Theorem 3 there is a triangle T such that T≤P and w _{ T }≥w _{ P }. Let \(s \in \mathfrak {F}\). Using the result in Lemma 1 we can argue as follows. Since there is a translated copy of s contained in P, we must have w _{ s }≤w _{ P }≤w _{ T }. Thus, there is then a translated copy of s contained in T.
4 From Triangles to Hexagons
We now turn to the computational problem: Given a family \(\mathfrak {F}\) of line segments, find a smallestarea convex set that contains a translated copy of every \(s \in \mathfrak {F}\).
By Theorem 1 we can choose the answer to be a triangle. In this section we show that this problem is equivalent to finding a smallestarea affineregular hexagon enclosing some centrally symmetric convex figure. An affineregular hexagon is the image of a regular hexagon under a nonsingular affine transformation. An affineregular hexagon is centrally symmetric, but not every centrally symmetric hexagon is affineregular: For instance, in an affineregular hexagon, the diagonals are parallel to a hexagon side, and have twice the length of that side. In this paper, we only consider affineregular hexagons that are centrally symmetric about the origin, so by abuse of terminology, we will write affineregular hexagon for an affineregular hexagon that is centrally symmetric about the origin.
The basic insight is that for centrally symmetric figures, comparing widthfunctions is equivalent to inclusion:
Lemma 2
Let P and Q be ovals centrally symmetric about the origin. Then w _{ P }≤w _{ Q } if and only if P⊂Q.
Proof
One direction is trivial, so consider for a contradiction the case where w _{ P }≤w _{ Q } and \(P \not\subset Q\). Then there is a point p∈P∖Q. Since Q is convex, there is a line ℓ that separates p from Q. Since P and Q are centrally symmetric, this means that Q is contained in the strip bounded by the lines ℓ and −ℓ, while P contains the points p and −p lying outside this strip. This implies that for the orientation θ orthogonal to ℓ we have w _{ P }(θ)>w _{ Q }(θ), a contradiction. □
Recall that \(\bar {P}\) denotes the Minkowski symmetrization of an oval P.
Lemma 3
Let T be a nondegenerate triangle. Then \(\bar {T}\) is an affineregular hexagon, and \(\bar {T} = \frac{3}{2} T\). Every affineregular hexagon H can be expressed in this form.
Proof
Since every nondegenerate triangle is the affine image of an equilateral triangle, it suffices to observe this relationship for the equilateral triangle and the regular hexagon. □
Since \(w_{P} = w_{\bar {P}}\), \(w_{T} = w_{\bar {T}}\), and by Lemmas 2 and 3, we immediately have
Lemma 4
Given an oval P, a triangle T is a smallestarea triangle with w _{ T }≥w _{ P } if and only if \(\bar {T}\) is a smallestarea affine regular hexagon with \(\bar {P}\subset \bar {T}\).
This leads us to the following algorithm. In Sect. 6, we will show that the time bound is tight.
Theorem 4
Let \(\mathfrak {F}\) be a set of n line segments in the plane. Then we can find in O(nlogn) time a triangle T that is a minimumarea convex translation cover for \(\mathfrak {F}\).
Proof
Given a family \(\mathfrak {F}\) of n line segments, place every \(s\in \mathfrak {F}\) with its center at the origin. Let P be the convex hull of these translated copies. P can be computed in O(nlogn) time, and is a centrally symmetric convex polygon with at most 2n vertices. We then compute a smallest area affineregular hexagon H containing P. In the next section we will show that this can be done in time O(n). Finally, we return a triangle T with \(\bar {T}= H\). The correctness of the algorithm follows from \(w_{P}(\theta ) = \max_{s \in \mathfrak {F}}w_{s}(\theta )\) and Lemma 4. □
5 Algorithm for Computing the Smallest Enclosing AffineRegular Hexagon
In this section we discuss the following problem: Given a convex polygon P, centrally symmetric about the origin, find a smallestarea affineregular hexagon H such that P⊂H.
Let us first sketch a simple quadratictime algorithm: The affineregular hexagons centered at the origin are exactly the images of a regular hexagon centered at the origin under a nonsingular linear transformation. Instead of minimizing the hexagon, we can fix a regular hexagon H with center at the origin, and find a linear transformation σ such that σP⊂H and such that the determinant of σ is maximized. The transformation σ can be expressed as a 2×2 matrix with coefficients a,b,c,d. The condition σP⊂H can then be written as a set of 6n linear inequalities in the four unknowns a,b,c,d. We want to find a feasible solution that maximizes the determinant ad−bc, a quadratic, nonconcave expression. This can be done by computing the 4dimensional polytope of feasible solutions, and considering every facet of this polytope in turn. We triangulate each facet, and solve the maximization problem on each simplex of the triangulation.
In the following, we show that the problem can in fact be solved in linear time.
For a set \(S \subset \mathbb {R}^{2}\), let S ^{∘}=−S denote the mirror image with respect to the origin. A strip is the area bounded by a line ℓ and its mirror image ℓ ^{∘}.
It is easy to see that if H is a minimumarea affineregular hexagon containing P, then two of the three strips must be touching P. Without loss of generality, we can assume these to be strips \(\mathfrak {S}_{1}\) and \(\mathfrak {S}_{2}\). Let us now observe what happens when we keep \(\mathfrak {S}_{1}\) fixed and let the orientation of \(\mathfrak {S}_{2}\) rotate through its possible range.
For convenience of presentation, let us choose a coordinate system where \(\mathfrak {S}_{1}\) is horizontal. Since \(\mathfrak {S}_{1}\) touches P, there is a vertex p of P on the side V ^{∘} B. For \(\mathfrak {S}_{2}\) to touch P, there must be a vertex q∈P on the side BU, and so \(\mathfrak {S}_{2}\)’s orientation can range from horizontal to the orientation of the edge of P that intersects the xaxis.
Furthermore, the point U moves horizontally along the xaxis to the right (or not at all, when q=U). The point A ^{∘} moves horizontally to the right with at least twice the speed of point U. As V is the midpoint of A ^{∘} and B ^{∘}, this implies that V moves horizontally to the right with at least the speed of U, and so the line UV is rotating counterclockwise. It follows that while strip \(\mathfrak {S}_{2}\) rotates counterclockwise, the part of H lying below the xaxis and to the left of the line pp ^{∘} is strictly shrinking.
We have shown:
Lemma 5
Keeping \(\mathfrak {S}_{1}\) horizontal and in contact with a pair of vertices p,p ^{∘} of P and rotating \(\mathfrak {S}_{2}\) counterclockwise through its possible range of orientations while in contact with P, the area H is nonincreasing. The quadrilateral oUVp ^{∘} is strictly shrinking by inclusion.
This implies immediately that there exists a minimumarea affineregular hexagon H such that every side of H contains a vertex of P. We can now show that we can in fact choose H such that one of its sides contains an edge of P:
Lemma 6
There exists a minimumarea affineregular hexagon H containing P such that a side of H contains an edge of P. In addition, if no minimumarea affineregular hexagon containing P shares a vertex with P, then every minimumarea affineregular hexagon has a side containing an edge P.
Proof
Consider now the case where at least one of the contact points a,b,c lies at a vertex of H. Since each side of H has a singlepoint intersection with P, two of a,b,c are identical. Using a suitable linear transformation, we can assume b=c in Fig. 8. We now use the shear transformation (x,y)↦(x,y+ρx), where ρ>0 is chosen such that the edge of P incident to b on the right becomes horizontal. Then σP⊂H, σP=P, and the horizontal sides of H contain an edge of P. □
We have now reduced the search for a minimumarea affineregular hexagon to a small set of candidates:
Lemma 7
There exists a minimumarea affineregular hexagon H defined by three strips \(\mathfrak {S}_{1}, \mathfrak {S}_{2}, \mathfrak {S}_{3}\), where \(\mathfrak {S}_{1}\) supports an edge of P, and \(\mathfrak {S}_{2} = \mathfrak {S}_{2}(\mathfrak {S}_{1})\) and \(\mathfrak {S}_{3} = \mathfrak {S}_{3}(\mathfrak {S}_{1})\).
It remains to find an efficient way to compute \(\mathfrak {S}_{2}(\mathfrak {S}_{1})\) for the n candidate strips \(\mathfrak {S}_{1}\). We will give a lineartime rotating calipers [26] type algorithm. Our algorithm enumerates the possible strips \(\mathfrak {S}_{1}\) by enumerating the edges of P in counterclockwise order, and maintains the strip \(\mathfrak {S}_{2}(\mathfrak {S}_{1})\) during the process.
The correctness of the algorithm hinges on the following description of the changes in the orientation of \(\mathfrak {S}_{2}\) and \(\mathfrak {S}_{3}\) as \(\mathfrak {S}_{1}\) rotates around P.
Lemma 8
As \(\mathfrak {S}_{1}\) rotates counterclockwise, the corresponding strips \(\mathfrak {S}_{2}(\mathfrak {S}_{1})\) and \(\mathfrak {S}_{3}(\mathfrak {S}_{1})\) rotate (nonstrictly) counterclockwise.
Proof
Furthermore, similar to the arguments above, we observe that A ^{∘} moves with speed at least twice the speed of V. Since U is the midpoint of BA ^{∘}, it moves with at least half the speed of A ^{∘}, so U moves with speed at least equal to the speed of V. Since U and V move on parallel lines, it follows that the line UV is rotating counterclockwise, and so \(\mathfrak {S}_{3}\) rotates counterclockwise during the rotation of \(\mathfrak {S}_{1}\). □
We are now ready to describe our algorithm.
Theorem 5
Given a centrallysymmetric convex 2ngon P, a smallestarea affineregular hexagon enclosing P can be found in time O(n).
Proof
As mentioned above, we use a rotating calipers [26] type algorithm. It maintains an edge e of P defining \(\mathfrak {S}_{1}\), a second strip \(\mathfrak {S}_{2}\) and the vertex q of P where \(\mathfrak {S}_{2}\) touches P, and a vertex r of P. Let H=BUVB ^{∘} U ^{∘} V ^{∘} be the hexagon defined by \(\mathfrak {S}_{1}\) and \(\mathfrak {S}_{2}\), as in Fig. 5. The algorithm maintains the invariant that P has a supporting line in r that is parallel to UV. The algorithm proceeds by rotating \(\mathfrak {S}_{2}\) counterclockwise (as in Fig. 6) until \(\mathfrak {S}_{2} = \mathfrak {S}_{2}(\mathfrak {S}_{1})\), and then proceeds to the next edge defining the next \(\mathfrak {S}_{1}\).
We initialize e to an arbitrary edge of P. Let e be horizontal for ease of presentation, with P below e. Let q be the left endpoint of e, and let r be the leftmost vertex of P (the lower one, if P has a vertical edge).
In the initial configuration, \(\mathfrak {S}_{2}\) is obtained from \(\mathfrak {S}_{1}\) by a counterclockwise rotation around q by an infinitely small amount. This implies that B=q, and that the side UV is nearly vertical, so the invariant for r holds.

If r no longer supports a tangent to P parallel to UV, replace r by the counterclockwise next vertex of P, and continue rotating \(\mathfrak {S}_{2}\).

If \(\mathfrak {S}_{2}\) supports an edge of P, then replace q by the counterclockwise next vertex of P, and continue rotating \(\mathfrak {S}_{2}\).

If UV touches r, or if q=U and UV supports the edge of P that starts counterclockwise from q, then we have found the strip \(\mathfrak {S}_{2}(\mathfrak {S}_{1})\). We compute its area and update a running minimum. Then we replace e with the counterclockwise next edge of P. As long as r does not support a tangent to P parallel to UV, we replace r by the counterclockwise next vertex of P. Then continue rotating \(\mathfrak {S}_{2}\).
6 Lower Bound for Computing a Translation Cover
Lemma 9
Let R denote a regular 6ngon centered at the origin, for some integer n≥1. Then any minimumarea affineregular hexagon enclosing R is a regular hexagon such that every side of this hexagon contains an edge of R.
Proof
The statement is trivial for n=1, so assume n≥2. Let H ^{∗} denote a regular hexagon enclosing R, and such that each side of H ^{∗} contains an edge of R. Let H denote another smallest affineregular hexagon enclosing R. We will argue that H is also a regular hexagon whose sides contain edges of R.
We first rule out the case where H shares a vertex with R. For sake of contradiction, assume that H shares two opposite vertices V and V ^{∘} of R. Without loss of generality, we assume that V,V ^{∘} are on the xaxis. The sides e,e ^{∘} of H that are not adjacent to V,V ^{∘} are parallel to VV ^{∘} and have half the length of \(\overline{VV^{\circ}}\). In addition, the sides of H that are adjacent to V and V ^{∘} make an angle at most π/12 with the yaxis. (See Fig. 10(b).) Then a direct calculation shows that H>H ^{∗}, a contradiction.
Thus, by Lemma 6, we know that a side e of H contains an edge of R. Without loss of generality we assume that this side is parallel to the xaxis. (See Fig. 10(c).) For sake of contradiction, assume that e is not symmetric with respect to the yaxis. Consider the hexagon H′ that is obtained from H by a horizontal shear transformation that moves e and the opposite side parallel to the xaxis, until they are centered at the yaxis. Then H′ (see Fig. 10(d)) is an affineregular hexagon containing R that is symmetric with respect to the yaxis and that only touches R along its top and bottom edges. This implies that H′ strictly contains a regular hexagon H ^{∗} enclosing R, and hence H=H′>H ^{∗}, a contradiction.
Therefore, e is symmetric with respect to the yaxis, and thus H is symmetric with respect to the yaxis. Only one such affineregular hexagon is circumscribed to R, so H=H ^{∗}. □
We are now able to prove our lower bound.
Theorem 6
In the algebraic computation tree model, and in the worst case, it takes Ω(nlogn) time to compute a minimumarea translation cover for a family \(\mathfrak {F}\) of n line segments in the plane.
Proof
For an interval \(I \subset \mathbb {R}\), we denote by C _{ I } the arc of the unit circle corresponding with polar angles in the interval I, that is C _{ I }={(cosθ,sinθ)∣θ∈I}. As C _{ I } is the intersection of a circle and a cone, a node of an algebraic computation tree can decide whether a point lies in C _{ I }.
We use a reduction from the following problem. The input is a set of points p _{1},…,p _{ n }∈C _{[0,π/3)}. The goal is to decide whether there exists an integer 0≤k<n such that C _{(kπ/3n,(k+1)π/3n)} is empty, that is, this arc does not contain any point p _{ i }. It follows from BenOr’s bound [2] that any algebraic computation tree that decides this problem has depth Ω(nlogn). (The set of negative instances has at least n! connected components: To each permutation σ of 1,…,n , we associate a negative instance where each p _{ i } lies in the σ _{ i }’s arc. In order to move continuously from one of these configuration to another, we must have a crossing p _{ i }=p _{ j }, which implies that one interval is empty by the pigeonhole principle, and thus the instance is positive.)
Our construction is as follows. Consider the (fixed) regular 6ngon R, whose vertices are r _{ k }=(cos(kπ/3n),sin(kπ/3n)) for k=1,…,6n. Let P denote the convex 12ngon whose vertices are the vertices of R and all the rotated copies of the points p _{1},…,p _{ n } by angles 0,π/3,…,5π/3 around the origin.
If there is an integer k=0,…,n−1 such that C _{(kπ/3n,(k+1)π/3n)} is empty, then by Lemma 6, the regular hexagon containing R whose sides contain the edge r _{ k } r _{ k+1} and its rotated copies by angles 0,π/3,…,5π/3 is a minimum area affineregular hexagon containing P.
If on the other hand, for every integer k∈{0,…,n−1} the arc C _{(kπ/3n,(k+1)π/3n)} is nonempty, then by Lemma 6, any minimumarea affine hexagon containing R is a regular hexagon whose sides contain edges of R, and thus it cannot contain P.
So we have proved that, when some arc C _{(kπ/3n,(k+1)π/3n)} is empty, then a minimumarea hexagon containing P has area H ^{∗}, where H ^{∗} is a minimumarea hexagon containing R. Otherwise, if all these arcs are nonempty, then the minimum area is larger than H ^{∗}.
Thus, if we could compute in o(nlogn) time a minimumarea convex translation cover for the diagonals of P, then by Lemma 4 we would also get in o(nlogn) time the area of a smallest enclosing affineregular hexagon containing P, and then we would be able to decide in o(nlogn) time whether there exists an empty arc C _{(kπ/3n,(k+1)π/3n)}, a contradiction. □
7 Minimizing the Perimeter
If we wish to minimize the perimeter instead of the area, the problem becomes much easier: it suffices to translate all segments so that their midpoints are at the origin, and take the convex hull of the translated segments. This follows from the following more general result.
Theorem 7
Let \(\mathcal{C}\) be a family of centrally symmetric convex figures. Under translations, the perimeter of the convex hull of their union is minimized when the centers coincide.
Proof
By the CauchyCrofton formula [11], the perimeter is the integral of the width of the projection over all directions. We argue that the width is minimized when the centers coincide, for all directions simultaneously, implying the claim. □
When the figures are not symmetric, our proof of Theorem 7 breaks down. However, we are able to solve the problem for a family consisting of all the rotated copies of a given oval. (Remember that an oval is a compact convex set.) The following theorem was already stated in the introduction.
Theorem 2
Let G be an oval, and let \(\mathcal{G}\) be the family of all the rotated copies of G by angles in [0,2π). Then the smallest enclosing disk of G is a smallestperimeter translation cover for \(\mathcal{G}\).
Proof
We observe first that, if G is a segment, then by Theorem 7, the smallest enclosing disk of G is a smallestperimeter translation cover for \(\mathcal{G}\).
Consider next the case where G is an acute triangle. Choose a coordinate system with origin at the center of the circumcircle of G, and such that the circumcircle has radius one. We wish to prove that any translation cover for \(\mathcal{G}\) must have perimeter at least 2π, implying that the circumcircle is optimal.
We borrow an idea of Bezdek and Connelly [8]. Let v _{1}, v _{2}, v _{3} be the three vertices of G. By our assumptions, the origin lies in the interior of their convex hull, and the three vectors have length one. The origin can be expressed as a convex combination \(0 =\sum_{i=1}^{3} \alpha_{i}v_{i}\) with α _{ i }≥0 and \(\sum_{i=1}^{3}\alpha_{i} = 1\). Let δ _{ i }, for i=1,2,3, be the angle formed by ov _{ i } and the positive xaxis.
Let K be a translation cover for \(\mathcal{G}\) and let h be the support function [21] of K. That is, h(u)=sup{〈x,u〉∣x∈K} for any unit vector u. We denote by u _{ θ }=(cosθ,sinθ) the unit vector making angle θ with the positive xaxis, so that \(v_{i} = u_{\delta_{i}}\).
Consider finally the general case where G is an arbitrary compact convex figure, and let D be the smallest enclosing disk of G. Either D touches G in two points that form a diameter of D, or D touches G in three points that form an acute triangle. In both cases, our previous results imply that D is a smallestperimeter translation cover for either the segment or the triangle, and therefore for G. □
The minimum enclosing circle is not always the unique minimumperimeter keyhole: For instance, when G is a unit line segment, then any set of constant width is a solution. In the theorem below, we show that when G is an acute triangle, then its circumcircle is the unique solution. This generalizes directly to any figure G that touches its circumcircle at 3 points.
Theorem 8
If G is an acute triangle, then its smallest enclosing disk is the unique smallestperimeter translation cover for the family of all rotated copies of G.
Proof
We use the same notations as in the proof of Theorem 2: K is a smallestperimeter translation cover for \(\mathcal{G}\). For any θ, it contains at least one copy of G rotated by angle θ. We call G(θ) one such rotated copy. The vertices of G(θ) are the points \(v_{i}(\theta)=c(\theta)+u_{\theta+\delta_{i}}\), for i=1,2,3.
We will prove that all the triangles G(θ) have the same circumcircle. Our strategy is to show that the function θ↦c(θ) is differentiable and its derivative is 0. Without loss of generality, we only prove that c′(0)=0, and we assume that c(0)=0.
For sake of contradiction, assume that c is not differentiable at 0, or it is differentiable at 0 and its derivative is nonzero. This means that we do not have lim_{ θ→0} c(θ)/θ=0. Hence, there exists an ε>0 such that for any integer n, there exists θ _{ n }∈(−1/n,0)∪(0,1/n) with ∥c(θ _{ n })/θ _{ n }∥>ε. This implies c(θ _{ n })≠0, and so c(θ _{ n })/∥c(θ _{ n })∥ is a sequence of unit vectors. Since the set of unit vectors is compact, there is a subsequence \((\theta_{n_{k}})\) such that \(c(\theta_{n_{k}})/\c(\theta_{n_{k}})\\) converges to a unit vector c _{0}. We denote this subsequence again as (θ _{ n }).
8 Conclusions
In practice, it is an important question to find the smallest convex container into which a family of ovals can be translated. For the perimeter, this is answered by Theorem 7 for centrally symmetric ovals. For general ovals, it is still not difficult, as the perimeter of the convex hull is a convex function under translations [1]. This means that the problem can be solved in practice by numerical methods.
For minimizing the area, the problem appears much harder, as there can be multiple local minima. The following lemma solves a very special case.
Lemma 10
Let \(\mathcal{R}\) be a family of axisparallel rectangles. The area of their convex hull is minimized if their bottom left corners coincide (or equivalently if their centers coincide).
Proof
Let C be the convex hull of some placement of the rectangles. For any x, let ℓ(x) be the length of the intersection of the vertical line at coordinate x with C. The function x↦ℓ(x) is concave (by the BrunnMinkowski theorem in two dimensions). For any z≥0, we define w(z) to be the length of the interval of all x where ℓ(x)≥z.
We observe that the area of C is equal to ∫ℓ(x)dx, which is again equal to \(\int_{0}^{\infty}w(z) d z\). We will now argue that w(z) is minimized for every z when the bottom left corners of the rectangles coincide, implying the claim.
To see this, consider the placement with coinciding bottom left corners at the origin, and the line y=z. It intersects the convex hull at x=0 and at some convex hull edge defined by two rectangles R _{1} and R _{2}. w(z) is equal to the length of this intersection. It remains to observe that for any placement of R _{1} and R _{2}, the convex hull of these two rectangle already enforces this value of w(z). □
Notes
Acknowledgements
We thank Helmut Alt, Tetsuo Asano, Jinhee Chun, Dong Hyun Kim, Mira Lee, Yoshio Okamoto, János Pach, Günter Rote, and Micha Sharir for helpful discussions.
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