https://doi.org/10.1007/s00605-019-01355-w
From spherical to Euclidean illumination
Károly Bezdek1,2·Zsolt Lángi3
Received: 25 September 2019 / Accepted: 15 November 2019 / Published online: 20 November 2019
© Springer-Verlag GmbH Austria, part of Springer Nature 2019
Abstract
In this note we introduce the problem of illumination of convex bodies in spherical spaces and solve it for a large subfamily of convex bodies. We derive from it a combi- natorial version of the classical illumination problem for convex bodies in Euclidean spaces as well as a solution to that for a large subfamily of convex bodies, which in dimension three leads to special Koebe polyhedra.
Keywords Spherical space·Euclidean space·Convex body·Illumination number Mathematics Subject Classification 52A20·52A55
1 Introduction
LetEd denote thed-dimensional Euclidean space with the unit sphereSd−1= {x ∈ Ed| x,x =1}centered at the origino, where·stands for the standard inner product ofEd. We identifySd−1with the(d−1)-dimensional spherical space. A compact
Communicated by Adrian Constantin.
Károly Bezdek: Partially supported by a Natural Sciences and Engineering Research Council of Canada Discovery Grant. Zsolt Lángi: Partially supported by the National Research, Development and Innovation Office, NKFI, K-119670, the János Bolyai Research Scholarship of the Hungarian Academy of Sciences, and Grants BME FIKP-VÍZ of EMMI and ÚNKP-19-4 New National Excellence Program of the Ministry for Innovation and Technology.
B
Károly Bezdekbezdek@math.ucalgary.ca Zsolt Lángi
zlangi@math.bme.hu
1 Department of Mathematics and Statistics, University of Calgary, Calgary, Canada 2 Department of Mathematics, University of Pannonia, Veszprém, Hungary
3 MTA-BME Morphodynamics Research Group and Department of Geometry, Budapest University of Technology and Economics, Budapest, Hungary
convex set (resp., a compact spherically convex set) with nonempty interior is called aconvex bodyinEd(resp.,Sd−1). (Here, we call a subset ofSd−1spherically convex if it is contained in an open hemisphere ofSd−1moreover, for any two points of the set the spherical segment, i.e., the shorter unit circle arc connecting them belongs to the set.) Now, recall the following concept of illumination due to Boltyanski [4].
(For an equivalent notion of illumination using point sources instead of directions see Hadwiger [8].) LetKbe a convex body inEd, letp∈bdK, i.e., letpbe a boundary point ofK, and letv ∈ Sd−1be a direction. We say thatpisilluminated from the directionv, if the half-line with endpointpand directionv intersects the interior of the convex bodyK. We say that the directionsv1,v2, . . . ,vm ∈ Sd−1illuminateK, if every boundary point is illuminated from somevi for 1 ≤ i ≤ m. The smallest number of directions that illuminateKis called theillumination number ofK, and is denoted by I(K). It is easy to see that I(K) ≥ d +1 for any convex bodyK inEd. On the other hand, since no two distinct vertices of an affined-cube can be illuminated from the same direction, it follows that I(K)=2d holds for any affine d-cubeK. The following Illumination Conjecture [4,8] of Hadwiger and Boltyanski is a longstanding open problem in discrete geometry solved only in the plane. For a recent comprehensive survey on the numerous partial results on this conjecture see [3] (which surveys also the relevant results on the equivalent Covering Conjecture [6,7,10] as well as Separation Conjecture [1,2,12]).
Conjecture 1 (Illumination Conjecture) The illumination number I(K) of any d- dimensional convex bodyK, d ≥ 3, is at most2d and I(K) = 2d only ifKis an affine d-cube.
In this paper, we introduce the following notion of illumination (resp., illumination number) for convex bodies in spherical space.
Definition 1 LetK⊂Sdbe a convex body, and letp∈Sd\K. We say that a boundary pointq∈bdKisilluminated frompif it is not antipodal top, the spherical segment with endpointspandqdoes not intersect the interior intKofK, and the greatcircle throughpandqdoes. We say thatKisilluminatedfrom a set S ⊂Sd\K, if every boundary point ofKis illuminated from at least one point ofS. The smallest cardinality of a setSthat illuminatesKand lies on a(d−1)-dimensional greatsphere ofSdwhich is disjoint fromK, is called theillumination number ofKinSd, and is denoted by ISd(K).
We observe that dropping the seemingly artificial restriction that all light sources are contained in a(d −1)-dimensional greatsphere ofSd, disjoint from the convex bodyK, makes the problem of determiningISd(K)trivial. Indeed, choosing any point ofSd, antipodal to an arbitrary interior point ofK, illuminates every boundary point ofK.
We leave the easy proofs of the following three remarks to the reader.
Remark 1 If a setA⊂Sd\Killuminates the convex bodyKinSdandAis contained in a closed hemisphereH⊂Sdsuch thatK⊂intH, thenISd(K)≤card(A). (Here card(·)refers to the cardinality of the corresponding set.)
Remark 2 LetK∗be the polar body assigned to the convex bodyK ⊂Sd ⊂ Ed+1, i.e., letK∗:= {x∈Sd: x,y ≤0 for ally∈K}. It is easy to see thatK∗is a convex body inSd moreover,(K∗)∗ = K. Clearly, an open hemisphere containsKif and only if its center is in−intK∗. Note that if some setSofkpoints in the boundary of such an open hemisphere illuminatesK, and fxis the central projection to the tangent hyperplane ofSdat the centerxof this hemisphere, then fx(S)corresponds to a set ofkdirections which illuminate fx(K). In other words,
ISd(K)=min{I(fx(K)):x∈ −intK∗}.
Remark 3 Levi [10] proved thatI(K)=3 holds for any convex bodyKinE2which is not a parallelogram. Thus, Remark2implies thatIS2(K)=3 holds for any convex bodyKinS2.
In order to state the main illumination results of this paper we need
Definition 2 LetKbe a convex body inEd(resp.,Sd). Recall that a faceF ofKis a convex (resp., spherically convex) subset ofKsuch that for any segment (resp., spher- ical segment) ofKwhose relative interior intersectsF, the segment (resp., spherical segment) in question is contained inF. Then a sequence of facesFs ⊂ · · · ⊂Fd−1of Kwith dimFi =i fori =s, . . . ,d−1 is called apartial flagofKof lengthd−s.
Theorem 1 LetKbe a convex body inSd, d>2, whose boundary contains a partial flag of length d−2. Then ISd(K)=d +1. In particular, for any convex polytopeP inSd, d>2, we have ISd(P)=d+1.
Corollary 1 For any convex body Kin Sd, d > 2 and for any ε > 0, there is a convex bodyK ⊂Ksuch thatK\K is contained in a spherical cap of radiusε, and ISd(K)=d+1.
Although it is easy to see that for any smooth convex bodyKinSd,d >2 we have ISd(K)=d+1, the question of finding a proper extension of Theorem1to all convex bodies seems to raise an open problem.
Problem 1 Prove or disprove that ISd(K)=d +1holds for any convex bodyKin Sd,d >2.
Next, we apply Theorem1to illumination numbers of convex bodies inEd. Moti- vated by the notion of combinatorial equivalence for convex polytopes, we introduce the following notion. Note that, restricted to convex polytopes, this notion is equivalent to the usual concept of combinatorial equivalence.
Definition 3 LetK,K ⊂ Ed be convex bodies. If there is a homeomorphism h : bdK→bdK such that for anyX ⊂bdK,X is a face ofKif and only ifh(X)is a face ofK, then we say thatKandK are combinatorially equivalent.
Note that any two strictly convex bodies ofEd are combinatorially equivalent.
Furthermore, combinatorial equivalence is an equivalence relation on the family of convex bodies, the equivalences classes of which we callcombinatorial classes.
Example 1 Let 0< α < π, and for any positive integerk, letpk =
cos23αk,sin23αk
∈
E2, andqk=
cos3αk,sin3αk
∈E2. Furthermore, for any value ofk, defineHkas the
closed half plane inE2containingoin its interior andpkandqkon its boundary, and letH−kbe the reflected copy ofHkabouto. Finally, set
K=B2∩∞
k=1
Hk
and
K =B2∩
k∈Z\{0}
Hk,
where B2 is the closed Euclidean unit disk centered ato. Then, clearly, there is a bijection h:bdK→bdK such thatX ⊂bdKis a face ofKif and only ifh(X)is a face ofK, but there is nohomeomorphismwith the same property.
Definition 4 For any convex bodyKinEd, the smallest number ksuch that some element of the combinatorial class ofKcan be illuminated fromkdirections is called thecombinatorial illumination number of K, and is denoted byIc(K).
Theorem 2 For any convex bodyKinEd, d>2, whose boundary contains a partial flag of length d−2, we have Ic(K)=d+1. In particular, for any convex polytopeP inEd, d>2, we have Ic(P)=d+1.
InE3, we can prove more. In order to state our result, recall that the combinatorial class of every convex polyhedronP ⊂E3contains special convex polyhedra, called Koebe polyhedra, which are combinatorially equivalent toP and aremidscribed to S2, i.e., their edges are tangent toS2.
Theorem 3 If P is a convex polyhedron in E3, then the combinatorial class of P contains a Koebe polyhedronP with I(P)=4.
We close this section with the following polar description ofISd(K), which is the spherical analogue of the Separation Lemma in [1]. For the sake of completeness the Appendix of this paper contains a proof of Theorem4. Here we recall that an exposed face of the convex bodyKinSd(resp.,Ed) is the intersection of a supporting (d−1)-dimensional greatsphere (resp., supporting hyperplane) ofKwithK.
Theorem 4 ISd(K)is equal to the minimum number of open hemispheres ofSdwhose boundaries all pass through a common point in the interior of the polar convex body K∗and have the property that every exposed face ofK∗is contained in at least one of the open hemispheres.
Thus, Theorem4implies in a straightforward way that Problem1is equivalent to the following spherical question (resp., Euclidean question obtained from it via central projection), both of which one can regard as a natural counterpart of the Separation Conjecture [1,2,12].
Problem 2 Prove or disprove that ifK is an arbitrary convex body inSd(resp.,Ed), d >2, then there existx∈intK and d+1open hemispheres (resp., open halfspaces) ofSd (resp.,Ed) whose boundaries containxsuch that every exposed face ofK is contained in at least one of the open hemispheres (resp., open halfspaces).
On the one hand, ifK is a strictly convex body inSd(resp.,Ed),d >2, then there is an easy positive answer to Problem2. On the other hand, using Theorem4one can derive from Theorem1
Corollary 2 IfPis an arbitrary convex polytope inSd (resp.,Ed), d >2, then there existx∈intPand d+1open hemispheres (resp., open halfspaces) ofSd(resp.,Ed) whose boundaries contain xsuch that every (exposed) face ofP is contained in at least one of the open hemispheres (resp., open halfspaces).
In the rest of the paper we prove the theorems stated.
2 Proofs of Theorem1and Corollary1
First, observe that no convex body inSdcan be illuminated from fewer thand+1 points (lying on a(d−1)-dimensional greatsphere, which is disjoint from the convex body).
(This statement follows from the analogue Euclidean result via central projection betweenSdand its corresponding tangent hyperplane inEd+1.) Thus, we need to find a(d+1)-element set, contained in a(d−1)-dimensional greatsphere not intersecting K, that illuminatesKinSd.
We prove Theorem1by induction ond for alld >2. So, let us start by assuming that Theorem1holds for convex bodies with partial flags of lengthd −3 inSd−1, and letKbe a convex body inSd with a partial flag F2 ⊂ · · · ⊂ Fd−1 = F on its boundary. (Ifd=3, i.e., ifKis a convex body with partial flagF2inS3, thenF2is a convex body inS2and therefore Remark3impliesIS2(F2)=3, i.e., it guarantees the inductive assumption for this case.) For simplicity, we refer to the(d−1)-dimensional greatsphereHofSdcontainingFas theequator, and the open hemisphereHbounded by H and containing intKas thenorthern hemisphere. Furthermore, for any open neighborhoodUof a pointx ∈ H, we callU∩HandU∩(−H)thenorthernand southernhalves ofU.
Letpbe a relative interior point of F. Note that sincep∈relintF, and dimF = d −1, the northern half of a sufficently small neighborhood of p is contained in intK. We show that the point−p, antipodal top, illuminates every point of bdK\F.
Indeed,Hcan be decomposed into semicircles starting at−pand ending atp. As each such semicircle intersects the northern half of every neighborhood ofp, each of these semicircles contains interior points ofK. Thus, every point of bdK\H =bdK\Fis illuminated from−p.
Note that if somex∈bdKis illuminated from a pointy, thenyhas a neighborhood Vsuch thatxis illuminated from any point ofV. Thus, ifDis an arbitrary compact subset of bdK\F, then −p has a neighborhood V such that every point of D is illuminated from every point ofV.
SinceFis a convex body with a partial flag of lengthd−3 in the(d−1)-dimensional spherical spaceH, there is a setS ofd points, contained in a(d−2)-dimensional
greatsphereGofH, which illuminateF inH, i.e., for every relative boundary point q∈relbdF there is a pointx ∈ S such that the semicircle starting atxand passing throughqintersects relintF. SinceKis spherically convex and dimF =d−1, this also implies that ifx is in the southern half of a suitable neighborhood ofx, then x illuminatesKatqinSd, i.e., the semicircle starting atx and passing throughq intersects intK. Observe that (since dimF=d−1)Kis illuminated at every relative interior point of F from any point in the southern hemisphere−H. Thus, there is a family of setsU1,U2, . . . ,Ud, each being the southern half of a suitable neighborhood of a point ofS, such that anyd-tuplexi ∈Ui,i =1,2, . . . ,d, illuminatesKat every point ofF inSd. By compactness arguments, there is a setL ⊂bdKcontaining F and open in bdKthat has the same property asF, i.e., there are some suitable setsUi, i =1,2, . . . ,d, each being the southern half of a suitable neighborhood of a point of S such that anyd-tuplexi ∈Ui,i =1,2, . . . ,d, illuminatesKat every point ofLin Sd.
Note that bdK\Lis compact. Hence, we may choose a pointx0∈Hand sufficiently close to−psuch thatx0illuminatesKat every point of bdK\L. LetH be the(d−1)- dimensional greatsphere spanned byGandx0. Note that by our choice ofG, chosen as a(d−2)-dimensional greatsphereGof H withS ⊂G,Gdoes not intersectF, which implies thatGstrictly separatesFand−pinH. Sincex0is sufficiently close to−p, from this it follows that H does not intersectK. Since H is a rotated copy of H aroundG, it intersects the southern half of any neighborhood of any point of G. Thus, H intersectsUi for all values ofi. Pick some pointxi fromUi ∩H for i =1,2, . . . ,d, and setS= {x0, . . . ,xd} ⊂H . SinceS\{x0}illuminatesKat every point of L by the choice of theUis, we have constructed a set of (d +1)points, contained in the boundary of an open hemisphere containingK, that illuminatesK.
Thus,ISd(K)=d+1, finishing the proof of Theorem1.
Now, we prove Corollary1. Letpbe an exposed point of bdK, i.e., a boundary point ofKthat can be obtained as an intersection ofKwith a supporting(d−1)- dimensional greatsphere ofKinSd. (The existence of an exposed point is well known see for example, Theorem 1.4.7 in [11].) Then we can truncateKnearpby a(d− 1)-dimensional greatshere such that the closure of the part removed is contained in an open spherical cap of radiusε. Continuing the truncation process by subsequent greatspheres, we can construct a truncated convex bodyK whose boundary contains a partial flag of lengthd−2, and has the property thatK\K is covered by a spherical cap of radiusε. By Theorem1,ISd(K)=d+1.
3 Proof of Theorem2
Since for any convex bodyK, we haveI(K)≥d+1, therefore Ic(K)≥d+1. We show thatIc(K)≤d+1.
ImagineEd as a tangent hyperplane of the sphereSd, embedded inEd+1in the usual way. Leth :Ed→Sdbe the central projection ofEdtoSd. ThenK :=h(K)is a spherical convex body, having a partial flag of lengthd−2. By Theorem1, there is a greatsphereHofSddisjoint fromK, and a(d+1)-element point setS ⊂Hsuch thatS illuminatesK. LetHbe the open hemisphere bounded byHthat containsK,
and letcbe the spherical center ofH. Lethc : H→ TcSdbe the central projection ofHto the tangent hyperplane ofSd atc. SinceTcSd is ad-dimensional Euclidean space, K := hc(K)is a d-dimensional Euclidean convex body.hc ◦h|bdK is a homeomorphism, and hc◦h maps faces ofKto faces ofK . Thus,KandK are combinatorially equivalent. Furthermore, the images of the great circle arcs starting at a pointqofHare parallel lines inTcSd starting at the ‘ideal point’hc(q)ofTcSd. Hence, the set of ideal points ofTcSdcorresponding toS is a set ofd+1 directions that illuminatesK .
4 Proof of Theorem3
To prove the theorem, we adopt some ideas from the proof of Theorem 3 in [9]. Let Pbe a Koebe polyhedron, i.e., a convex polyhedron inE3whose edges are tangent to S2. Then there are two families of circles onS2associated toP[5,9]. The elements of the first family, calledface circlesare the incircles of the faces of P; each such circle touches the edges of a face ofPat the points where the edges touchS2. We denote these circles by fj, j =1,2, . . . ,m, wherem is the number of faces ofP.
The elements of the second family are calledvertex circles. These circles, denoted byvi,i =1,2, . . . ,n, wheren is the number of vertices ofP, are circles onS2that contain the tangency points on all the edges ofPstarting at a given vertex ofP. The tangency graphs of these two families are dual graphs. IfT :S2 → S2is a Möbius transformation thenT maps the two circle families ofPinto two circle families which are associated to another Koebe polyhedron, which, with a little abuse of notation, we denote byT(P). It is known [5] that ifPandP are two combinatorially equivalent Koebe polyhedra, then there is a Möbius transformation T : S2 → S2 satisfying T(P)=P.
In our proof, we regardS2as the set of the ‘points at infinity’ of the Poincaré ball model of the hyperbolic spaceH3, which is identified with the interior of the Euclidean unit ball bounded byS2. Then every face circle fj ofPis the set of the ideal points of a unique hyperbolic planeFj, and the same holds for every vertex circle ofP; we denote the hyperbolic plane associated to the vertex circlevi byVi. Furthermore, we denote byFj the closed hyperbolic halfspace bounded byFj which is disjoint from any hyperbolic plane associated to any other face circle ofP, and defineVi similarly for any vertex circlevi ofP. It is worth noting that every Möbius transformation of S2corresponds to a hyperbolic isometry in the Poincaré ball model, and vice versa.
LetD:=H3\n
i=1Vi∪m
j=1Fj
, and note that for every value ofj,Fj∩bdD is a closed hyperbolic polygonPjwith ideal vertices and nonempty relative interior in H3. Consider some pointpin relintP1. Lethp:H3→H3be a hyperbolic isometry that mapspintoo, and letTpbe the corresponding Möbius transformation. Then the first face FofTp(P)contains the origino. Letmbe the outer unit normal vector of the Euclidean plane throughF. Then the angle betweenmand the outer unit normal vector of any other face ofT(P)is obtuse, which implies that the projection ofT(P)\F onto the Euclidean plane throughF(or in other words, the projection in the direction ofm) is intF. In other words,milluminates every point of bd(T(P))\F. Note that if
there are three directionsm1,m2,m3which illuminate F in the plane containing it, then the vectorsmi −εm, wherei =1,2,3, illuminate every point of FinE3. On the other hand, by a result of Levi [10], apart from parallelograms, the illumination number of every plane convex body is 3.
Thus, to prove the assertion we need to show that, using a suitable pointpinF1, the first faceF ofTp(P)is not a parallelogram. Assume thatFis a parallelogram. Then clearly, the first face ofPis a quadrangle, and thus, there are four tangency points on the first face circle f1ofP. Let these points beq1,q2,q3,q4in cyclic order. Let the unique hyperbolic line with ideal pointsq1,q3be denoted byL1, and the line with ideal pointsq2,q4be denoted byL2. These lines are contained inF1.
SinceFis a parallelogram and is circumscribed about a circle,Fis a rhombus, and the tangent pointsT(q1),T(q2),T(q3)andT(q4)are the vertices of a rectangle. Thus, the midpoint of both Euclidean open segments(T(q1),T(q3))and(T(q2),T(q4)) is the origino. Note that these segments represent the hyperbolic lineshp(L1)and hp(L2). Thus, in this case ois the intersection point of hp(L1)andhp(L2). Since hyperbolic isometries preserve incidences, it follows thatpis the intersection point of L1andL2. On the other hand, asP1contains infinitely many relative interior points, we may choose some point in relintP1different from this point, implying that in this case the first faceFofTp(P)is not a parallelogram.
Acknowledgements We are indebted to the anonymous referee for careful reading and valuable comments.
5 Appendix: Proof of Theorem4
The following proof is a spherical analogue of the proof of the Separation Lemma in [1].
Definition 5 LetK⊂Sd ⊂Ed+1be a convex body andFbe an exposed face ofK.
We define the conjugate face of F as a subset of the polar convex bodyK∗ = {x ∈ Sd : x,y ≤0 for ally∈K}given by
Fˆ := {x∈K∗| x,y =0 for ally∈ F}. (1) One should keep in mind thatFˆdepends also onKand not only onF. So, if we write Fˆˆ, then it means(Fˆ)ˆ, where the right-hand circumflex refers to the spherical polar bodyK∗. Ifx∈Sd, then letHxdenote the open hemisphere ofSdwith centerx.
The following statement (which one can regard as a natural spherical analogue of Theorem 2.1.4 in [11]) shows that exposed faces behave well under polarity in spherical spaces.
Proposition 1 LetK ⊂Sd ⊂Ed+1be a convex body and F be an exposed face of K. ThenF is an exposed face ofˆ K∗withFˆ =
y∈F
bdHy∩K∗
, whereSd\Hyis a closed supporting hemisphere ofK∗ for everyy ∈ F . Moreover, Fˆˆ = F and so, F → ˆF is a bijection between the exposed faces ofKandK∗.
Proof Without loss of generality we may assume thatF is a proper exposed face of K, i.e., there exists an open hemisphereHx0 ofSdsuch that∅ =F =K∩bdHxo and K∩Hx0 = ∅. It follows thatx0∈ ˆFand so,Fˆ = ∅. Now, ify∈ F, thenK∗∩Hy= ∅ andx∈bdHyholds for allx∈ ˆF. Thus,Fˆ ⊆
y∈F(K∗∩bdHy), whereK∗∩Hy= ∅ holds for ally∈F. On the other hand, ifz∈
y∈F(K∗∩bdHy)withK∗∩Hy= ∅for ally∈ F, thenz∈K∗withz,y =0 for ally∈ F(and thereforez∈ ˆF) implying that
y∈F(K∗∩bdHy)⊆ ˆF. Thus,Fˆ =
y∈F(K∗∩bdHy)withK∗∩Hy= ∅for ally ∈ F. AsK∗∩bdHyis a proper exposed face ofK∗for ally∈ Ftherefore Fˆ is a proper exposed face ofK∗. Applying the above argument to the exposed faceFˆ ofK∗one obtains in a straightforward way thatFˆˆ =F. This completes the proof of Proposition1.
Proposition 2 LetK⊂Sd⊂Ed+1be a convex body and H be a(d−1)-dimensional greatsphere ofSdwith H∩K= ∅. Thenq∈ bdKis illuminated fromp∈ H with p= −qif and only ifFˆ ⊂Hp, where F denotes the exposed face ofKhaving smallest dimension and containingq∈bdK.
Proof Let Hh be the open hemisphere with center h and boundary H inSd such that K ⊂ Hh. Clearly, −h ∈ bdHp ∩intK∗. Proposition 1 implies that F =
x∈ ˆF(bdHx∩K∗), whereHx∩K= ∅for allx∈ ˆF. Let[p,q)denote the spher- ical segment ofSd with endpointspandqcontainingpand not containingq. Now, q ∈ bdK is illuminated fromp ∈ H if and only if [p,q) ⊂
x∈ ˆFHx, which is equivalent to p ∈
x∈ ˆFHx (because±q ∈
x∈ ˆFbdHx). Finally, p ∈
x∈ ˆFHx
holds if and only ifFˆ ⊂Hp. This finishes the proof of Proposition2.
Now, the following statement follows from Proposition2and its proof in a straight- forward way.
Corollary 3 LetK⊂Sd ⊂Ed+1be a convex body and H be a(d−1)-dimensional greatsphere ofSdwith H∩K= ∅. LetHhbe the open hemisphere with centerhand boundary H inSdsuch thatK⊂Hh. Then the point set{p1, . . . ,pn} ⊂ H illuminates Kif and only if the open hemispheresHp1, . . . ,Hpn with−h ∈bdHpi ∩intK∗for 1≤i ≤n have the property that every (proper) exposed face ofK∗is contained in at least one of the open hemispheres.
Finally, Corollary3yields Theorem4.
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