on the successive illumination parameters of convex ... - Springer Link
Periodica Mathematica Hungarica Vol. 53 (1–2), 2006, pp. 71–82
ON THE SUCCESSIVE ILLUMINATION PARAMETERS OF CONVEX BODIES ¨ ro ¨ czky2,∗∗ (Budapest) and K. Bezdek1,∗ (Budapest), K. Bo 3,∗∗∗ Gy. Kiss (Budapest) 1
Department of Mathematics and Statistics, University of Calgary, 2500 University Drive N. W., Calgary, AB Canada T2N 1N4, and Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any P. s. 1/c, H-1117 Budapest, Hungary, E-mail: [email protected] 2
Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any P. s. 1/c, H-1117 Budapest, Hungary, E-mail: [email protected] 3
Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any s. 1/c, H-1117 Budapest, Hungary, and Bolyai Institute, University of Szeged, Aradi v´ertan´ uk tere 1, H-6720 Szeged, Hungary, E-mail: [email protected] (Received: September 7, 2006; Accepted: September 30, 2006) Abstract The notion of successive illumination parameters of convex bodies is introduced. We prove some theorems in the plane and determine the exact values of the successive illumination parameters of spheres, cubes and crosspolytopes for some dimensions.
1. Introduction Let K be a convex body of Ed , d ≥ 1 (i.e. a compact convex set of the ddimensional Euclidean space Ed with non-empty interior). We say that a point i.e. Mathematics subject classification number: 52C17, 52B11. Key words and phrases: illumination by affine subspaces, convex body. ∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant No. T 043556, and by the Natural Sciences and Engineering Research Council of Canada. ∗∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant No. T 043556. ∗∗∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant Nos. T 043556 and T043758, and by the Slovenian–Hungarian Intergovernmental Scientific and Technological Cooperation Project, Grant No. SLO-9/05. 0031-5303/2006/$20.00 c Akad´ emiai Kiad´ o, Budapest
Akad´ emiai Kiad´ o, Budapest Springer, Dordrecht
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¨ ro ¨ czky and gy. kiss k. bezdek, k. bo
a light-source l ∈ Ed \ K illuminates the boundary point p of K if the half-line starting at l and passing through p intersects the interior of K not between l and p. Furthermore, we say that the light-sources {l1 , l2 , . . . , ln } ⊂ Ed \ K illuminate K if each boundary point of K is illuminated by at least one of the light-sources l1 , l2 , . . . , ln . The smallest number of light-sources that can illuminate K is called the illumination number I(K) of K. The well-known illumination conjecture phrased independently by Boltyanski (1960) and Hadwiger (1960) says that any d-dimensional convex body can be illuminated by 2d light-sources in Ed , that is the inequality I(K) ≤ 2d holds for any convex body K ∈ Ed . This is proved only for d ≤ 2, but there are quite a number of partial results supporting the conjecture for d > 2 (see [3] and [4]). This conjecture was generalized by K. Bezdek [4] in the following way: Let L ⊂ Ed \ K be an affine subspace of dimension , 0 ≤ ≤ d − 1. Then L illuminates the boundary point p of K if there exists a point q ∈ L that illuminates p. Furthermore, we say that the affine subspaces {L1 , L2 , . . . , Ln } ⊂ Ed \ K illuminate K if each boundary point of K is illuminated by at least one of the subspaces L1 , L2 , . . . , Ln . Now, the smallest number of affine subspaces of dimension that are disjoint from K and can illuminate K is called the -dimensional illumination number I (K) of the convex body K in Ed . K. Bezdek [4] conjectures that I (K) ≤ I (C) holds for any convex body K in Ed where C denotes a d-dimensional affine cube of Ed . The following quantitative version of the illumination numbers of convex bodies was introduced by K. Bezdek in [2]. If Ko is a convex body of Ed symmetric about the origin o of Ed , then Ko defines a norm xKo = inf λ : λ−1 x ∈ Ko , which turns Ed into a normed space. Then let the illumination parameter of Ko be defined as IP(Ko ) = inf pi Ko : {pi } illuminates Ko . i
This ensures that far-away light-sources are penalised. Let IP(d) = sup{IP(Ko ) : Ko is an o-symmetric convex body in Ed }. It has been outlined in [2] that IP(Ko ) ≤ 6 with equality for (affine) regular convex hexagons. Also, [2] has raised the fundamental problem of estimating (resp. computing) IP(d). Recently Swanepoel [7] has shown that IP(d) ≤ O(2d d2 log d). Perhaps IP(d) = O(2d ). Motivated by the above definitions we introduce the successive illumination parameters of any convex body Ko of Ed symmetric about the origin o of Ed . We say that the convex sets {S1 , S2 , . . . , Sn } ⊂ Ed \ Ko of dimension , 0 ≤ ≤ d − 1 illuminate Ko if each boundary point of Ko is illuminated by at least one point of S1 ∪ S2 ∪ · · · ∪ Sn . Let SKo = inf{λ : λ−1 S ⊂ Ko }.
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on the successive illumination parameters of convex bodies
Then the -dimensional illumination parameter of Ko is defined as IP (Ko ) = inf α · Si Ko : i
Si is convex and has dimension , Si illuminates Ko
where α is a parameter depending on only. The higher dimensional light-sources are penalised. If α0 = 1, then IP0 ( . ) = IP( . ). The natural choices for α are the following + 1 or α = 2 . In both cases α0 = 1 and α1 = 2. In this note we study the successive illumination parameters of convex bodies. We prove some theorems in the plane and determine the exact values of the successive illumination parameters of spheres, cubes and cross-polytopes for some dimensions.
2. On the successive illumination parameters in the plane A rather rough outline of a proof of the following statement was given in [2]. Here we give a detailed proof. Theorem 2.1. If Ko is a convex body of E2 symmetric about the origin o of E , then IP0 (Ko ) ≤ 6α0 with equality for any affine regular convex hexagon. 2
Proof. Without loss of generality we may assume that α0 = 1. Let K∗o = {x ∈ E2 | < x, y > ≤ 1 for all y ∈ Ko } be the polar set assigned to Ko where < ., . > denotes the standard inner product of E2 . Clearly K∗o is an o-symmetric convex body. Now, let P denote the parallelogram with consecutive vertices p1 , p2 , p3 and p4 having the largest area among the parallelograms lying in K∗o . Obviously, p1 , p2 , p3 and p4 are boundary points of K∗o with p1 and p3 (resp., p2 and p4 ) symmetric about o. Now, let Li be the line passing through pi and being parallel to the diagonal of P not containing pi , for i = 1, 2, 3, 4. The maximality of the area of P implies that Li is a supporting line of K∗o for all i = 1, 2, 3, 4. Let qi = Li ∩ Li+1 (i = 1, 2, 3, 4, 5 ≡ 1 ). Obviously, the points qi are the vertices of a parallelogram, say Q, symmetric about o and containing K∗o (see Figure 1). Finally, let hK∗o (resp. hP ) be the supporting function of K∗o (resp. P ), and let ni be the outer normal vector of the side pi pi+1 of P for i = 1, 2, 3, 4, with p5 ≡ p1 . (Here n1 = −n3 and n2 = −n4 .)
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Figure 1
The core part of our proof is the following inequality. Lemma 2.2. hK∗o (n1 ) hK∗o (n2 ) hK∗o (n3 ) hK∗o (n4 ) + + + ≤ 6. hP (n1 ) hP (n2 ) hP (n3 ) hP (n4 ) Proof. By applying a proper affinity, we can assume that P (resp. Q) is a squre of diagonal length 2 (resp. of side length 2) (Figure 1). Let Lni be the supporting line of K∗o with outer normal vector ni and let ri be a point of K∗o on Lni , i = 1, 2, 3, 4. By symmetry we can assume that r1 , r2 , r3 and r4 are consecutive vertices of a parallelogram R symmetric about o. By the choice of P we have that area(R) ≤ area(P ). Now let ri+ = Lni ∩ Li+1 and ri− = Lni ∩ Li for i = 1, 2, 3, 4 (see Figure 1). Starting with r1 (resp. r3 ) we note that either area(∆r1 r2 r4 ) ≥area(∆r1+ r2 r4 ) (resp. area(∆r3 r2 r4 ) ≥area(∆r3+ r2 r4 )) or area(∆r1 r2 r4 ) ≥area(∆r1− r2 r4 ) (resp. area(∆r3 r2 r4 ) ≥area(∆r3− r2 r4 ). In the first case we replace r1 (resp. r3 ) by r1+ (resp. r3+ ), and in the second case we replace r1 (resp. r3 ) by r1− (resp. r3− ). Then we perform the same procedure for the points r2 and r4 . As a result we get an o-symmetric parallelogram R lying in Q with area(R ) ≤ area(R) ≤ area(P ). In fact, it is not hard to see that the above area inequality implies that the vertices of R must lie on opposite sides of Q, say, the vertices of R are r1+ , r2− , r3+ and r4− (see Figure 1). As area(R ) ≤ area(P ), therefore clearly dist(r1+ , r2− ) = dist(r3+ , r4− ) ≤ 1. Let hK∗o (n1 ) hK∗o (n3 ) √ = = 2x hP (n1 ) hP (n3 )
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and hK∗o (n2 ) hK∗o (n4 ) √ = = 2y hP (n2 ) hP (n4 ) √ √ for √ some x ≥ 2/2, y ≥ 2/2, where hP (n1 ) = hP (n2 ) = hP (n3 ) = hP (n4 ) = 2/2. Then an easy computation shows that √ √ √ dist(r1+ , r2− ) = dist(r3+ , r4− ) = 2 − 2(2 2 − x − y) = 2(x + y) − 2. √ Thus the inequality dist(r1+ , r2− ) = dist(r3+ , r4− ) ≤ 1 implies that 2(x + y) − 2 ≤ 1, that is hK∗o (n1 ) hK∗o (n2 ) √ + = 2(x + y) ≤ 3, hP (n1 ) hP (n2 ) finishing the proof of Lemma 2.2.
In order to finish the proof Theorem 2.1 we make the following observation. On the one hand, P is not necessarily uniquelly determined by the condition of having the largest area among all parallelograms lying in K∗o , on the other hand, one can assume that none of the vertices of P is a relative interior point of some face of K∗o . (Here a face is the intersection of some supporting line of K∗o with K∗o .) In other words, we may assume that the points p1 , p2 , p3 and p4 divide the boundary of K∗o in four convex arcs and each face of K∗o belongs to (at least) one of them. Now let P ∗ = {x ∈ E2 | < x, y > ≤ 1 for all y ∈ P }. Clearly, P ∗ is an o-symmetric parallelogram containing Ko moreover, the above assumption easily implies that if we move the vertices of P ∗ slightly away from o along the diagonals of P ∗ , then the four points obtained illuminate Ko . (For a proof one can use the Separation Lemma of [1].) Thus it is sufficent to show that by labeling the vertices of P ∗ with p∗1 , p∗2 , p∗3 ∗ and p4 we have that p∗1 Ko + p∗2 Ko + p∗3 Ko + p∗4 Ko ≤ 6. As the definition of polarity easily implies that p∗1 Ko + p∗2 Ko + p∗3 Ko + p∗4 Ko =
hK∗o (n1 ) hK∗o (n2 ) hK∗o (n3 ) hK∗o (n4 ) + + + hP (n1 ) hP (n2 ) hP (n3 ) hP (n4 )
therefore Lemma 2.2 finishes the proof of the inequality in Theorem 2.1. Finally, let H be an affine regular convex hexagon symmetric about o with consecutive vertices x1 , x2 , x3 , x4 , x5 and x6 . First, make the observation that if l is a light-source that illuminates the vertices xi and xj , then xi and xj have to be consecutive vertices of H. (Thus a light-source can illuminate at most two vertices of H.) Second, take a system of light-sources {pi | i ∈ I} that illuminates H. We distinguish the following cases.
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Figure 2
(1) |I| = 3. Let {p1 , p2 , p3 } illuminate H. Then each pi must illuminate exactly two consecutive vertices of H, moreover pi H ≥ 2 for all i = 1, 2, 3 (see Figure 2). Thus p1 H + p2 H + p3 H ≥ 2 + 2 + 2 = 6. (2) |I| = 4. Let {p1 , p2 , p3 , p4 } illuminate H. Then there are at least two light-sources say, p1 and p2 illuminating two consecutive vertices of H, implying that p1 H ≥ 2, p2 H ≥ 2. Thus p1 H + p2 H + p3 H + p4 H ≥ 2 + 2 + 1 + 1 = 6. (3) |I| = 5. Let {p1 , p2 , p3 , p4 , p5 } illuminate H. Obviously, at least one light-source has to illuminate two consecutive vertices of H and therefore clearly p1 H + p2 H + p3 H + p4 H + p5 H ≥ 6. (4) |I| ≥ 6. Clearly i∈I pi H ≥ 6. All this mean that i∈I pi H ≥ 6, and therefore IP0 (H) ≥ 6. This together with the first part of Theorem 2.1 applied to H imply that IP0 (H) = 6 finishing the proof of Theorem 2.1. Theorem 2.3. If Ko is a convex body of E2 symmetric about the origin o of E , then IP1 (Ko ) ≤ 28/3α1 < 3.056α1 . 2
Proof. Without loss of generality we may assume that Ko is a convex polygon symmetric about o. Let E denote the the ellipse having the largest area among the ellipses contained in Ko . Obvoiusly E is symmetric about o. The maximality of the area of E implies that E ∩ bdKo contains at least four points. By applying an affinity we may assume that E is a circle with centre o and radius 1. Let {r1 , r2 , . . . , r2n } be the set of points in E ∩ bdKo , and let {L1 , L2 , . . . , L2n } be the set of the corresponding supporting lines of Ko , finally let qi = Li ∩ Li+1 with L2n+1 = L1 . By symmetry we can assume that ri and rn+i are opposite points of E and r1 r2 is the longest empty arc. It means that ∠r1 or2 is the largest among the angles ∠ri ori+1 . It follows from the maximality of the area of E that ∠r1 or2 ≤ 90◦ (otherwise the pencil of ellipses passing through
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Figure 3
r1 , r2 , rn+1 , rn+2 and having tangents L1 , L2 , Ln+1 , Ln+2 would contain an ellipse having greater area than E and lying in Ko ).
Let C be the circle with centre o and radius 4/3. Let p1 p2 p3 p4 p5 p6 be the regular convex hexagon inscribed in C in such a way that r1 is the midpoint of the line segment p6 p1 . First suppose that ∠r1 or2 ≤ 60◦ . Then C contains Ko because the polygon q1 q2 . . . q2n contains Ko and ∠ri oqi =
∠ri ori+1 ∠r1 or2 ≤ ≤ 30◦ , 2 2
so oqi ≤ 1/ cos 30◦ = 4/3. Let T1 and T2 denote the two tangents of C which are perpendicular to L1 , and let si,j = Ti ∩ Lj . Then s1,1 s1,n+1 s2,n+1 s2,1 is a rectangular box which contains Ko and the two line segments s1,1 s1,n+1 and s2,n+1 s2,1 eventually illuminate Ko . This means that line segments parallel and arbitrarily close to the line segments s1,1 s1,n+1 , s2,n+1 s2,1 (lying on the proper sides of them) illuminate Ko . From this the desired inequality follows. Now suppose that 60◦ < ∠r1 or2 < 90◦ . Let s1,1 s1,n+1 s2,n+1 s2,1 be the same rectangular box and suppose that the two line segments s1,1 s1,n+1 and s2,n+1 s2,1 do not illuminate Ko . This means that there exists a point u in Ko such that the line containing s1,1 s1,n+1 separates u and Ko . Then consider the rectangular box formed by L2 , Ln+2 and those two tangents of C, say T3 and T4 , which are perpendicular to L2 . Let s3,2 , s3,n+2 , s4,n+2 , s4,2 be the vertices of this box. If the line segments s3,2 s3,n+2 and s4,2 s4,n+2 do not illuminate Ko , then the same argument as above shows that there exists a point v such that the line containing s3,2 s3,n+2 separates v and Ko . The convexity of Ko implies that the line segment uv is in Ko . Let us rotate now r1 and r2 about o in such a way that the image of r1 , say t1 becomes
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¨ ro ¨ czky and gy. kiss k. bezdek, k. bo
Figure 4
a point on the half-line op2 . Let t2 be the image of r2 under this rotation. Let F denote the convex hull of the points u, v and the circle E. Then F ⊂ Ko , because u, v ∈ Ko and E ⊂ Ko and F contains t1 t2 , hence this is a longer empty arc than r1 r2 . This contradiction completes the proof. Theorem 2.4. If H is an affine-regular convex hexagon in E2 , then IP1 (H) = 3α1 . Proof. Without loss of generality we may assume that H is a regular convex hexagon of side length 1 with consecutive vertices p1 , p2 , . . . , p6 . Take three sides p1 p2 , p3 p4 and p5 p6 , translete them a bit away from H and enlarge them slightly. These three line segments clearly illuminate H implying that IP1 (H) ≤ 3α1 . Now we prove that if a line segment ab illuminates i consecutive vertices, then abH > i/2 for i = 3 and i = 4. We may assume that the illuminated vertices are p2 , p3 , p4 and also p5 for i = 4. Let L be the line passing through p3 and being parallel to p2 pi+1 , M be the line of p1 p2 and N be the line of pi+1 pi+2 . Finally let M ∩ ab = e, M ∩ L = c, N ∩ ab = f and N ∩ L = d. The points e and f are inner points of ab, because ab illuminates both p2 and pi+1 . The line L separates at least one of the points a and b from H, hence abH > max{eH , f H } ≥ cH = dH =
i . 2
In order to finish the proof make the observation that one line segment cannot illuminate H. If the number of the line segments in the set of the light-sources is greater than two, then the sum of their norms is at least 3. If we illuminate H by two line segments, say a1 b1 and a2 b2 , then there are two possibilities. First, if both of them illuminates three consecutive vertices of H, then ai bi H > 3/2 for i = 1, 2.
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Second, if one of the line segments illuminates at least four consecutive vertices, then its norm is at least 2, while the norm of the other line segment is at least 1. Hence the illumination parameter is at least 3α1 in both cases. The theorems just proved strongly support our final conjecture. Conjecture 2.5. If Ko is a convex body of E2 symmetric about the origin o of E , then IP1 (Ko ) ≤ 3α1 . 2
3. On the successive illumination parameters of spheres, cubes and cross-polytopes Throughout this section we assume that α0 = 1. First we make two observations about the successive illumination parameters. Proposition 3.1. If Ko is a smooth o-symmetric convex body in E2 , d ≥ 2 then α1 IP0 (Ko ) ≤ IP1 (Ko ) ≤ α1 IP0 (Ko ). 2 Proof. Let b be a boundary point of Ko . Then there exists a unique supporting hyperplane Hb of Ko at b. A light-source l illuminates b if and only if Hb separates l and Ko . Thus if a line-segment l1 l2 illuminates b, then at least one of its two endpoints is separated from Ko by Hb , hence illuminates b. This means that if a set of line segments illuminates Ko , then the set of endpoints of the line segments also illuminates Ko . As l1 Ko + l2 Ko l1 l2 Ko = max{l1 Ko , l2 Ko } ≥ , 2 therefore the above argument implies the first inequality. The second inequality is straightforward.
Proposition 3.2. For any o-symmetric convex body Ko in Ed , d ≥ 2, we have that 2αd−1 ≤ IPd−1 (Ko ) ≤ αd−1 IP0 (Ko ). Proof. The upper estimate on IPd−1 (Ko ) is obvious. For the lower estimate it is sufficient to prove that a hyperplane H cannot illuminate a convex body. Let H1 and H2 be the two supporting hyperplanes of Ko which are parallel to H. We may assume that H1 separates H and H2 . This means that if b ∈ H2 is a boundary point of Ko , then H cannot illuminate b.
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Let Cd , Bd and Xd be the d-dimensional cube, ball and cross-polytope, respectively. Theorem 3.3. For any with 0 ≤ ≤ d − 1, d ≥ 2 we have that IP (Cd ) ≤ 2d− α . Equality holds for = 0, 1 and d − 1. Proof. First, take 2d− pairwise disjoint and parallel -faces of Cd . Second, translate them a bit away from Cd and enlarge them slightly. The system of dimensional convex sets obtained this way, obviously illuminates Cd implying that IP (Cd ) ≤ 2d−α . This proves the inequality. Equality for = 0 follows from the fact that each vertex of Cd needs its own light-source, while equality for = d − 1 follows from Proposition 3.2. In the case = 1 first we prove that if the line segment ab illuminates k vertices of Cd then ABCd ≥ 2k − 3. It is obvious for k = 1. If k ≥ 2, then let the coordinate vectors of a and b be a = (ζ1 , ζ2 , . . . , ζd ) and b = (η1 , η2 , . . . , ηd ), respectively. Let am be the point with coordinate vector am = (k − 1 − m)a + mb /(k − 1) for m = 0, 1, . . . , k − 1. Then these points divide the line segment ab into k − 1 parts. There are k points e0 = a, e1 , e2 , . . . , ek−1 and ek = b on ab such that any two of them belong to different d-orthants. Hence there is at least one line segment ai ai+1 which contains at least two distinct points ej and el from the set {e0 , e1 , . . . , ek } We may assume without loss of generality that j < l, the first coordinate of ej is greater than 1 and the first coordinate of el is less than −1. The coordinate functions are monotonic ones along ab, thus this implies (k − 1 − i)ζ1 + iη1 > (k − 1), and (k − 1 − i − 1)ζ1 + (i + 1)η1 < −(k − 1). Hence ζ1 − η1 > 2k − 2, so ζ1 > 2k − 3. If k = 1 then abCd ≥ 1 > k/2, otherwise abCd ≥ 2k − 3 ≥ k/2, hence if the set of line segments {ai bi : i ∈ I} illuminates Cd and ai bi illuminates ki vertices of the cube, then ai bi Cd ≥ ki /2 ≥ 2d /2 = 2d−1 . i∈I
This implies the equality for = 1.
i∈I
If we illuminate Bd by the set of vertices of a slightly enlarged circumscribed √ cross-polytope, then we get IP0 (Bd ) ≤ 2d d. It was proved by K. Bezdek and A. E. Litvak [5] that in dimensions two and three this estimate is sharp. They proved the following Theorem:
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√ √ √ Theorem 3.4. IP0 (Bd ) ≤ 2d d for all d, and IP0 (B2 ) = 4 2, IP0 (B3 ) = 6 3. Applying their result and Propositions 3.1 and 3.2 we determine the exact values of all successive illumination parameters of the two and three dimensional spheres and give an estimate on IP1 (Bd ) for any dimension d. √ Theorem 3.5. IP1 (Bd ) ≤ d dα1 for all d, and √ √ √ IP1 (B2 ) = 2 2α1 , IP1 (B3 ) = 3 3α1 , IP2 (B3 ) = 2 2α2 . Proof. The balls are smooth bodies, hence the inequalities √ α1 IP0 (Bd ) ≤ IP1 (Bd ) ≤ d dα1 2 follow from Proposition 3.1 and from the following simple observation: any set of d pairwise disjoint edges of Xd (translate them a bit away and enlarge slightly) illuminates the inscribed ball. The two and three dimensional results for IP1 follow from the inequalities just proved and from Theorem 3.4. Finally, the three dimensional result for IP2 (B3 ): Consider two disks having the same radius as the ball. If first the disks touch the ball at the two endpoints of a diameter of it, then√translate them a bit away and enlarge slightly, we get the inequality IP2 (B3 ) ≤ 2 2α2 . If the ball is illuminated by more than two objects, then the sum of the norm of the light-sources is at least 3. If we use only two lightsources, then an easy argument (estimate on the spherical area of the illuminated surface) shows that the above construction is optimal. Theorem 3.6. IP0 (Xd ) = 2d for all d, and IP1 (X2 ) = 2α1 , IP1 (X3 ) = 3α1 , IP2 (X3 ) = 2α2 . Proof. Each vertex needs its own light-source, and if the light-sources are slightly away from the vertices along the axes, then the cross polytope is illuminated. Hence IP0 (Xd ) = 2d for all d. The remaining equalities follow from Proposition 3.1 and 3.2 in a straightforward way.
References [1] K. Bezdek, The problem of illumination of the boundary of a convex body by
affine subspaces, Mathematika 38 (1991), 362–375.
[2] K. Bezdek, Research problem 46, Periodica Math. Hungar. 24 (1992), 119–121. [3] K. Bezdek, Hadwiger–Levi’s covering problem revisited, New trends in discrete
and computational geometry, (ed. by J. Pach), Vol. 10, Springer (1993), 199–233.
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age zur Alg. und [4] K. Bezdek, On affine subspaces that illuminate a convex set, Beitr¨ Geom. 35 (1994), 131–139.
[5] K. Bezdek and A. E. Litvak, On the vertex index of convex bodies, 2006, 1–18,
(preprint). [6] H. Martini and V. Soltan, Combinatorial problems on the illumination of convex
bodies, Aequationes Math. 57 (1999), 121–152.
[7] K. J. Swanepoel, Quantitative illumination of convex bodies and vertex degrees
of geometric Steiner minimal trees, 2004, 1–5, (preprint).
ON THE SUCCESSIVE ILLUMINATION PARAMETERS OF CONVEX BODIES ¨ ro ¨ czky2,∗∗ (Budapest) and K. Bezdek1,∗ (Budapest), K. Bo 3,∗∗∗ Gy. Kiss (Budapest) 1
Department of Mathematics and Statistics, University of Calgary, 2500 University Drive N. W., Calgary, AB Canada T2N 1N4, and Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any P. s. 1/c, H-1117 Budapest, Hungary, E-mail: [email protected] 2
Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any P. s. 1/c, H-1117 Budapest, Hungary, E-mail: [email protected] 3
Department of Geometry, E¨ otv¨ os Lor´ and University, P´ azm´ any s. 1/c, H-1117 Budapest, Hungary, and Bolyai Institute, University of Szeged, Aradi v´ertan´ uk tere 1, H-6720 Szeged, Hungary, E-mail: [email protected] (Received: September 7, 2006; Accepted: September 30, 2006) Abstract The notion of successive illumination parameters of convex bodies is introduced. We prove some theorems in the plane and determine the exact values of the successive illumination parameters of spheres, cubes and crosspolytopes for some dimensions.
1. Introduction Let K be a convex body of Ed , d ≥ 1 (i.e. a compact convex set of the ddimensional Euclidean space Ed with non-empty interior). We say that a point i.e. Mathematics subject classification number: 52C17, 52B11. Key words and phrases: illumination by affine subspaces, convex body. ∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant No. T 043556, and by the Natural Sciences and Engineering Research Council of Canada. ∗∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant No. T 043556. ∗∗∗ The research was supported by the Hungarian National Foundation for Scientific Research, Grant Nos. T 043556 and T043758, and by the Slovenian–Hungarian Intergovernmental Scientific and Technological Cooperation Project, Grant No. SLO-9/05. 0031-5303/2006/$20.00 c Akad´ emiai Kiad´ o, Budapest
Akad´ emiai Kiad´ o, Budapest Springer, Dordrecht
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¨ ro ¨ czky and gy. kiss k. bezdek, k. bo
a light-source l ∈ Ed \ K illuminates the boundary point p of K if the half-line starting at l and passing through p intersects the interior of K not between l and p. Furthermore, we say that the light-sources {l1 , l2 , . . . , ln } ⊂ Ed \ K illuminate K if each boundary point of K is illuminated by at least one of the light-sources l1 , l2 , . . . , ln . The smallest number of light-sources that can illuminate K is called the illumination number I(K) of K. The well-known illumination conjecture phrased independently by Boltyanski (1960) and Hadwiger (1960) says that any d-dimensional convex body can be illuminated by 2d light-sources in Ed , that is the inequality I(K) ≤ 2d holds for any convex body K ∈ Ed . This is proved only for d ≤ 2, but there are quite a number of partial results supporting the conjecture for d > 2 (see [3] and [4]). This conjecture was generalized by K. Bezdek [4] in the following way: Let L ⊂ Ed \ K be an affine subspace of dimension , 0 ≤ ≤ d − 1. Then L illuminates the boundary point p of K if there exists a point q ∈ L that illuminates p. Furthermore, we say that the affine subspaces {L1 , L2 , . . . , Ln } ⊂ Ed \ K illuminate K if each boundary point of K is illuminated by at least one of the subspaces L1 , L2 , . . . , Ln . Now, the smallest number of affine subspaces of dimension that are disjoint from K and can illuminate K is called the -dimensional illumination number I (K) of the convex body K in Ed . K. Bezdek [4] conjectures that I (K) ≤ I (C) holds for any convex body K in Ed where C denotes a d-dimensional affine cube of Ed . The following quantitative version of the illumination numbers of convex bodies was introduced by K. Bezdek in [2]. If Ko is a convex body of Ed symmetric about the origin o of Ed , then Ko defines a norm xKo = inf λ : λ−1 x ∈ Ko , which turns Ed into a normed space. Then let the illumination parameter of Ko be defined as IP(Ko ) = inf pi Ko : {pi } illuminates Ko . i
This ensures that far-away light-sources are penalised. Let IP(d) = sup{IP(Ko ) : Ko is an o-symmetric convex body in Ed }. It has been outlined in [2] that IP(Ko ) ≤ 6 with equality for (affine) regular convex hexagons. Also, [2] has raised the fundamental problem of estimating (resp. computing) IP(d). Recently Swanepoel [7] has shown that IP(d) ≤ O(2d d2 log d). Perhaps IP(d) = O(2d ). Motivated by the above definitions we introduce the successive illumination parameters of any convex body Ko of Ed symmetric about the origin o of Ed . We say that the convex sets {S1 , S2 , . . . , Sn } ⊂ Ed \ Ko of dimension , 0 ≤ ≤ d − 1 illuminate Ko if each boundary point of Ko is illuminated by at least one point of S1 ∪ S2 ∪ · · · ∪ Sn . Let SKo = inf{λ : λ−1 S ⊂ Ko }.
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Then the -dimensional illumination parameter of Ko is defined as IP (Ko ) = inf α · Si Ko : i
Si is convex and has dimension , Si illuminates Ko
where α is a parameter depending on only. The higher dimensional light-sources are penalised. If α0 = 1, then IP0 ( . ) = IP( . ). The natural choices for α are the following + 1 or α = 2 . In both cases α0 = 1 and α1 = 2. In this note we study the successive illumination parameters of convex bodies. We prove some theorems in the plane and determine the exact values of the successive illumination parameters of spheres, cubes and cross-polytopes for some dimensions.
2. On the successive illumination parameters in the plane A rather rough outline of a proof of the following statement was given in [2]. Here we give a detailed proof. Theorem 2.1. If Ko is a convex body of E2 symmetric about the origin o of E , then IP0 (Ko ) ≤ 6α0 with equality for any affine regular convex hexagon. 2
Proof. Without loss of generality we may assume that α0 = 1. Let K∗o = {x ∈ E2 | < x, y > ≤ 1 for all y ∈ Ko } be the polar set assigned to Ko where < ., . > denotes the standard inner product of E2 . Clearly K∗o is an o-symmetric convex body. Now, let P denote the parallelogram with consecutive vertices p1 , p2 , p3 and p4 having the largest area among the parallelograms lying in K∗o . Obviously, p1 , p2 , p3 and p4 are boundary points of K∗o with p1 and p3 (resp., p2 and p4 ) symmetric about o. Now, let Li be the line passing through pi and being parallel to the diagonal of P not containing pi , for i = 1, 2, 3, 4. The maximality of the area of P implies that Li is a supporting line of K∗o for all i = 1, 2, 3, 4. Let qi = Li ∩ Li+1 (i = 1, 2, 3, 4, 5 ≡ 1 ). Obviously, the points qi are the vertices of a parallelogram, say Q, symmetric about o and containing K∗o (see Figure 1). Finally, let hK∗o (resp. hP ) be the supporting function of K∗o (resp. P ), and let ni be the outer normal vector of the side pi pi+1 of P for i = 1, 2, 3, 4, with p5 ≡ p1 . (Here n1 = −n3 and n2 = −n4 .)
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Figure 1
The core part of our proof is the following inequality. Lemma 2.2. hK∗o (n1 ) hK∗o (n2 ) hK∗o (n3 ) hK∗o (n4 ) + + + ≤ 6. hP (n1 ) hP (n2 ) hP (n3 ) hP (n4 ) Proof. By applying a proper affinity, we can assume that P (resp. Q) is a squre of diagonal length 2 (resp. of side length 2) (Figure 1). Let Lni be the supporting line of K∗o with outer normal vector ni and let ri be a point of K∗o on Lni , i = 1, 2, 3, 4. By symmetry we can assume that r1 , r2 , r3 and r4 are consecutive vertices of a parallelogram R symmetric about o. By the choice of P we have that area(R) ≤ area(P ). Now let ri+ = Lni ∩ Li+1 and ri− = Lni ∩ Li for i = 1, 2, 3, 4 (see Figure 1). Starting with r1 (resp. r3 ) we note that either area(∆r1 r2 r4 ) ≥area(∆r1+ r2 r4 ) (resp. area(∆r3 r2 r4 ) ≥area(∆r3+ r2 r4 )) or area(∆r1 r2 r4 ) ≥area(∆r1− r2 r4 ) (resp. area(∆r3 r2 r4 ) ≥area(∆r3− r2 r4 ). In the first case we replace r1 (resp. r3 ) by r1+ (resp. r3+ ), and in the second case we replace r1 (resp. r3 ) by r1− (resp. r3− ). Then we perform the same procedure for the points r2 and r4 . As a result we get an o-symmetric parallelogram R lying in Q with area(R ) ≤ area(R) ≤ area(P ). In fact, it is not hard to see that the above area inequality implies that the vertices of R must lie on opposite sides of Q, say, the vertices of R are r1+ , r2− , r3+ and r4− (see Figure 1). As area(R ) ≤ area(P ), therefore clearly dist(r1+ , r2− ) = dist(r3+ , r4− ) ≤ 1. Let hK∗o (n1 ) hK∗o (n3 ) √ = = 2x hP (n1 ) hP (n3 )
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and hK∗o (n2 ) hK∗o (n4 ) √ = = 2y hP (n2 ) hP (n4 ) √ √ for √ some x ≥ 2/2, y ≥ 2/2, where hP (n1 ) = hP (n2 ) = hP (n3 ) = hP (n4 ) = 2/2. Then an easy computation shows that √ √ √ dist(r1+ , r2− ) = dist(r3+ , r4− ) = 2 − 2(2 2 − x − y) = 2(x + y) − 2. √ Thus the inequality dist(r1+ , r2− ) = dist(r3+ , r4− ) ≤ 1 implies that 2(x + y) − 2 ≤ 1, that is hK∗o (n1 ) hK∗o (n2 ) √ + = 2(x + y) ≤ 3, hP (n1 ) hP (n2 ) finishing the proof of Lemma 2.2.
In order to finish the proof Theorem 2.1 we make the following observation. On the one hand, P is not necessarily uniquelly determined by the condition of having the largest area among all parallelograms lying in K∗o , on the other hand, one can assume that none of the vertices of P is a relative interior point of some face of K∗o . (Here a face is the intersection of some supporting line of K∗o with K∗o .) In other words, we may assume that the points p1 , p2 , p3 and p4 divide the boundary of K∗o in four convex arcs and each face of K∗o belongs to (at least) one of them. Now let P ∗ = {x ∈ E2 | < x, y > ≤ 1 for all y ∈ P }. Clearly, P ∗ is an o-symmetric parallelogram containing Ko moreover, the above assumption easily implies that if we move the vertices of P ∗ slightly away from o along the diagonals of P ∗ , then the four points obtained illuminate Ko . (For a proof one can use the Separation Lemma of [1].) Thus it is sufficent to show that by labeling the vertices of P ∗ with p∗1 , p∗2 , p∗3 ∗ and p4 we have that p∗1 Ko + p∗2 Ko + p∗3 Ko + p∗4 Ko ≤ 6. As the definition of polarity easily implies that p∗1 Ko + p∗2 Ko + p∗3 Ko + p∗4 Ko =
hK∗o (n1 ) hK∗o (n2 ) hK∗o (n3 ) hK∗o (n4 ) + + + hP (n1 ) hP (n2 ) hP (n3 ) hP (n4 )
therefore Lemma 2.2 finishes the proof of the inequality in Theorem 2.1. Finally, let H be an affine regular convex hexagon symmetric about o with consecutive vertices x1 , x2 , x3 , x4 , x5 and x6 . First, make the observation that if l is a light-source that illuminates the vertices xi and xj , then xi and xj have to be consecutive vertices of H. (Thus a light-source can illuminate at most two vertices of H.) Second, take a system of light-sources {pi | i ∈ I} that illuminates H. We distinguish the following cases.
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Figure 2
(1) |I| = 3. Let {p1 , p2 , p3 } illuminate H. Then each pi must illuminate exactly two consecutive vertices of H, moreover pi H ≥ 2 for all i = 1, 2, 3 (see Figure 2). Thus p1 H + p2 H + p3 H ≥ 2 + 2 + 2 = 6. (2) |I| = 4. Let {p1 , p2 , p3 , p4 } illuminate H. Then there are at least two light-sources say, p1 and p2 illuminating two consecutive vertices of H, implying that p1 H ≥ 2, p2 H ≥ 2. Thus p1 H + p2 H + p3 H + p4 H ≥ 2 + 2 + 1 + 1 = 6. (3) |I| = 5. Let {p1 , p2 , p3 , p4 , p5 } illuminate H. Obviously, at least one light-source has to illuminate two consecutive vertices of H and therefore clearly p1 H + p2 H + p3 H + p4 H + p5 H ≥ 6. (4) |I| ≥ 6. Clearly i∈I pi H ≥ 6. All this mean that i∈I pi H ≥ 6, and therefore IP0 (H) ≥ 6. This together with the first part of Theorem 2.1 applied to H imply that IP0 (H) = 6 finishing the proof of Theorem 2.1. Theorem 2.3. If Ko is a convex body of E2 symmetric about the origin o of E , then IP1 (Ko ) ≤ 28/3α1 < 3.056α1 . 2
Proof. Without loss of generality we may assume that Ko is a convex polygon symmetric about o. Let E denote the the ellipse having the largest area among the ellipses contained in Ko . Obvoiusly E is symmetric about o. The maximality of the area of E implies that E ∩ bdKo contains at least four points. By applying an affinity we may assume that E is a circle with centre o and radius 1. Let {r1 , r2 , . . . , r2n } be the set of points in E ∩ bdKo , and let {L1 , L2 , . . . , L2n } be the set of the corresponding supporting lines of Ko , finally let qi = Li ∩ Li+1 with L2n+1 = L1 . By symmetry we can assume that ri and rn+i are opposite points of E and r1 r2 is the longest empty arc. It means that ∠r1 or2 is the largest among the angles ∠ri ori+1 . It follows from the maximality of the area of E that ∠r1 or2 ≤ 90◦ (otherwise the pencil of ellipses passing through
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Figure 3
r1 , r2 , rn+1 , rn+2 and having tangents L1 , L2 , Ln+1 , Ln+2 would contain an ellipse having greater area than E and lying in Ko ).
Let C be the circle with centre o and radius 4/3. Let p1 p2 p3 p4 p5 p6 be the regular convex hexagon inscribed in C in such a way that r1 is the midpoint of the line segment p6 p1 . First suppose that ∠r1 or2 ≤ 60◦ . Then C contains Ko because the polygon q1 q2 . . . q2n contains Ko and ∠ri oqi =
∠ri ori+1 ∠r1 or2 ≤ ≤ 30◦ , 2 2
so oqi ≤ 1/ cos 30◦ = 4/3. Let T1 and T2 denote the two tangents of C which are perpendicular to L1 , and let si,j = Ti ∩ Lj . Then s1,1 s1,n+1 s2,n+1 s2,1 is a rectangular box which contains Ko and the two line segments s1,1 s1,n+1 and s2,n+1 s2,1 eventually illuminate Ko . This means that line segments parallel and arbitrarily close to the line segments s1,1 s1,n+1 , s2,n+1 s2,1 (lying on the proper sides of them) illuminate Ko . From this the desired inequality follows. Now suppose that 60◦ < ∠r1 or2 < 90◦ . Let s1,1 s1,n+1 s2,n+1 s2,1 be the same rectangular box and suppose that the two line segments s1,1 s1,n+1 and s2,n+1 s2,1 do not illuminate Ko . This means that there exists a point u in Ko such that the line containing s1,1 s1,n+1 separates u and Ko . Then consider the rectangular box formed by L2 , Ln+2 and those two tangents of C, say T3 and T4 , which are perpendicular to L2 . Let s3,2 , s3,n+2 , s4,n+2 , s4,2 be the vertices of this box. If the line segments s3,2 s3,n+2 and s4,2 s4,n+2 do not illuminate Ko , then the same argument as above shows that there exists a point v such that the line containing s3,2 s3,n+2 separates v and Ko . The convexity of Ko implies that the line segment uv is in Ko . Let us rotate now r1 and r2 about o in such a way that the image of r1 , say t1 becomes
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Figure 4
a point on the half-line op2 . Let t2 be the image of r2 under this rotation. Let F denote the convex hull of the points u, v and the circle E. Then F ⊂ Ko , because u, v ∈ Ko and E ⊂ Ko and F contains t1 t2 , hence this is a longer empty arc than r1 r2 . This contradiction completes the proof. Theorem 2.4. If H is an affine-regular convex hexagon in E2 , then IP1 (H) = 3α1 . Proof. Without loss of generality we may assume that H is a regular convex hexagon of side length 1 with consecutive vertices p1 , p2 , . . . , p6 . Take three sides p1 p2 , p3 p4 and p5 p6 , translete them a bit away from H and enlarge them slightly. These three line segments clearly illuminate H implying that IP1 (H) ≤ 3α1 . Now we prove that if a line segment ab illuminates i consecutive vertices, then abH > i/2 for i = 3 and i = 4. We may assume that the illuminated vertices are p2 , p3 , p4 and also p5 for i = 4. Let L be the line passing through p3 and being parallel to p2 pi+1 , M be the line of p1 p2 and N be the line of pi+1 pi+2 . Finally let M ∩ ab = e, M ∩ L = c, N ∩ ab = f and N ∩ L = d. The points e and f are inner points of ab, because ab illuminates both p2 and pi+1 . The line L separates at least one of the points a and b from H, hence abH > max{eH , f H } ≥ cH = dH =
i . 2
In order to finish the proof make the observation that one line segment cannot illuminate H. If the number of the line segments in the set of the light-sources is greater than two, then the sum of their norms is at least 3. If we illuminate H by two line segments, say a1 b1 and a2 b2 , then there are two possibilities. First, if both of them illuminates three consecutive vertices of H, then ai bi H > 3/2 for i = 1, 2.
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Second, if one of the line segments illuminates at least four consecutive vertices, then its norm is at least 2, while the norm of the other line segment is at least 1. Hence the illumination parameter is at least 3α1 in both cases. The theorems just proved strongly support our final conjecture. Conjecture 2.5. If Ko is a convex body of E2 symmetric about the origin o of E , then IP1 (Ko ) ≤ 3α1 . 2
3. On the successive illumination parameters of spheres, cubes and cross-polytopes Throughout this section we assume that α0 = 1. First we make two observations about the successive illumination parameters. Proposition 3.1. If Ko is a smooth o-symmetric convex body in E2 , d ≥ 2 then α1 IP0 (Ko ) ≤ IP1 (Ko ) ≤ α1 IP0 (Ko ). 2 Proof. Let b be a boundary point of Ko . Then there exists a unique supporting hyperplane Hb of Ko at b. A light-source l illuminates b if and only if Hb separates l and Ko . Thus if a line-segment l1 l2 illuminates b, then at least one of its two endpoints is separated from Ko by Hb , hence illuminates b. This means that if a set of line segments illuminates Ko , then the set of endpoints of the line segments also illuminates Ko . As l1 Ko + l2 Ko l1 l2 Ko = max{l1 Ko , l2 Ko } ≥ , 2 therefore the above argument implies the first inequality. The second inequality is straightforward.
Proposition 3.2. For any o-symmetric convex body Ko in Ed , d ≥ 2, we have that 2αd−1 ≤ IPd−1 (Ko ) ≤ αd−1 IP0 (Ko ). Proof. The upper estimate on IPd−1 (Ko ) is obvious. For the lower estimate it is sufficient to prove that a hyperplane H cannot illuminate a convex body. Let H1 and H2 be the two supporting hyperplanes of Ko which are parallel to H. We may assume that H1 separates H and H2 . This means that if b ∈ H2 is a boundary point of Ko , then H cannot illuminate b.
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Let Cd , Bd and Xd be the d-dimensional cube, ball and cross-polytope, respectively. Theorem 3.3. For any with 0 ≤ ≤ d − 1, d ≥ 2 we have that IP (Cd ) ≤ 2d− α . Equality holds for = 0, 1 and d − 1. Proof. First, take 2d− pairwise disjoint and parallel -faces of Cd . Second, translate them a bit away from Cd and enlarge them slightly. The system of dimensional convex sets obtained this way, obviously illuminates Cd implying that IP (Cd ) ≤ 2d−α . This proves the inequality. Equality for = 0 follows from the fact that each vertex of Cd needs its own light-source, while equality for = d − 1 follows from Proposition 3.2. In the case = 1 first we prove that if the line segment ab illuminates k vertices of Cd then ABCd ≥ 2k − 3. It is obvious for k = 1. If k ≥ 2, then let the coordinate vectors of a and b be a = (ζ1 , ζ2 , . . . , ζd ) and b = (η1 , η2 , . . . , ηd ), respectively. Let am be the point with coordinate vector am = (k − 1 − m)a + mb /(k − 1) for m = 0, 1, . . . , k − 1. Then these points divide the line segment ab into k − 1 parts. There are k points e0 = a, e1 , e2 , . . . , ek−1 and ek = b on ab such that any two of them belong to different d-orthants. Hence there is at least one line segment ai ai+1 which contains at least two distinct points ej and el from the set {e0 , e1 , . . . , ek } We may assume without loss of generality that j < l, the first coordinate of ej is greater than 1 and the first coordinate of el is less than −1. The coordinate functions are monotonic ones along ab, thus this implies (k − 1 − i)ζ1 + iη1 > (k − 1), and (k − 1 − i − 1)ζ1 + (i + 1)η1 < −(k − 1). Hence ζ1 − η1 > 2k − 2, so ζ1 > 2k − 3. If k = 1 then abCd ≥ 1 > k/2, otherwise abCd ≥ 2k − 3 ≥ k/2, hence if the set of line segments {ai bi : i ∈ I} illuminates Cd and ai bi illuminates ki vertices of the cube, then ai bi Cd ≥ ki /2 ≥ 2d /2 = 2d−1 . i∈I
This implies the equality for = 1.
i∈I
If we illuminate Bd by the set of vertices of a slightly enlarged circumscribed √ cross-polytope, then we get IP0 (Bd ) ≤ 2d d. It was proved by K. Bezdek and A. E. Litvak [5] that in dimensions two and three this estimate is sharp. They proved the following Theorem:
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√ √ √ Theorem 3.4. IP0 (Bd ) ≤ 2d d for all d, and IP0 (B2 ) = 4 2, IP0 (B3 ) = 6 3. Applying their result and Propositions 3.1 and 3.2 we determine the exact values of all successive illumination parameters of the two and three dimensional spheres and give an estimate on IP1 (Bd ) for any dimension d. √ Theorem 3.5. IP1 (Bd ) ≤ d dα1 for all d, and √ √ √ IP1 (B2 ) = 2 2α1 , IP1 (B3 ) = 3 3α1 , IP2 (B3 ) = 2 2α2 . Proof. The balls are smooth bodies, hence the inequalities √ α1 IP0 (Bd ) ≤ IP1 (Bd ) ≤ d dα1 2 follow from Proposition 3.1 and from the following simple observation: any set of d pairwise disjoint edges of Xd (translate them a bit away and enlarge slightly) illuminates the inscribed ball. The two and three dimensional results for IP1 follow from the inequalities just proved and from Theorem 3.4. Finally, the three dimensional result for IP2 (B3 ): Consider two disks having the same radius as the ball. If first the disks touch the ball at the two endpoints of a diameter of it, then√translate them a bit away and enlarge slightly, we get the inequality IP2 (B3 ) ≤ 2 2α2 . If the ball is illuminated by more than two objects, then the sum of the norm of the light-sources is at least 3. If we use only two lightsources, then an easy argument (estimate on the spherical area of the illuminated surface) shows that the above construction is optimal. Theorem 3.6. IP0 (Xd ) = 2d for all d, and IP1 (X2 ) = 2α1 , IP1 (X3 ) = 3α1 , IP2 (X3 ) = 2α2 . Proof. Each vertex needs its own light-source, and if the light-sources are slightly away from the vertices along the axes, then the cross polytope is illuminated. Hence IP0 (Xd ) = 2d for all d. The remaining equalities follow from Proposition 3.1 and 3.2 in a straightforward way.
References [1] K. Bezdek, The problem of illumination of the boundary of a convex body by
affine subspaces, Mathematika 38 (1991), 362–375.
[2] K. Bezdek, Research problem 46, Periodica Math. Hungar. 24 (1992), 119–121. [3] K. Bezdek, Hadwiger–Levi’s covering problem revisited, New trends in discrete
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age zur Alg. und [4] K. Bezdek, On affine subspaces that illuminate a convex set, Beitr¨ Geom. 35 (1994), 131–139.
[5] K. Bezdek and A. E. Litvak, On the vertex index of convex bodies, 2006, 1–18,
(preprint). [6] H. Martini and V. Soltan, Combinatorial problems on the illumination of convex
bodies, Aequationes Math. 57 (1999), 121–152.
[7] K. J. Swanepoel, Quantitative illumination of convex bodies and vertex degrees
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