Maintaining Visibility Information of Planar Point Sets with a Moving Viewpoint

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Maintaining Visibility Information of Planar Point Sets with a Moving Viewpoint

Olivier Devillers, Vida Dujmovic, Hazel Everett, Samuel Hornus, Steve Wismath, Sue Whitesides

To cite this version:

Olivier Devillers, Vida Dujmovic, Hazel Everett, Samuel Hornus, Steve Wismath, et al.. Maintaining

Visibility Information of Planar Point Sets with a Moving Viewpoint. 17th Canadian Conference on

Computational Geometry - CCCG’2005, Aug 2005, Windsor, Canada. �inria-00000569�

Viewpoint

OlivierDevillers

VidaDujmovi y

HazelEverett z

SamuelHornus x

SueWhitesides {

SteveWismath k

Abstrat

Given a set of n points in the plane, we onsider the

problemofomputingtheirularorderingofthepoints

aboutaviewpointqandeÆientlymaintainingthisor-

deringinformationasqmoves.

1 Introdution

Let P = fp

0

;p

1

;:::;p

n 1

g be a set of points in the

plane. Apointpisvisiblefromaviewpointqiftheline

segmentpqontainsnootherpointofP. Ifpoinides

withqthenpisnotvisible. TheviewofP fromaview-

pointqisthelokwiseirularorderingofallpointsin

P that arevisiblefrom q. Given P, theviewfromq is

easilyomputedinO(nlogn)time.

In this paper we are interested in maintaining the

viewwhenqmoves. Morepreisely,wewouldliketoan-

swerthefollowingmoving viewpoint querieseÆiently:

Given a trajetory , report, as q movesalong , all

of the dierent views in the order that they are seen

by q. We assume that is given on-line. We don't

knowaheadoftimetheentiretrajetory;wewouldlike

to ompute the new views as the trajetory beomes

knowntous. Forsimpliity,hereweassumethetraje-

tory isasinglesegment, that doesnotpass through

any points of P, and we assumeno three points of P

areollinear.

This problem is motivated by the graphis problem

of maintaining the view during an interative walk-

throughofa3Dsene. HerewegiveaneÆientsolution

fora2Dseneonsistingofpoints.

Ourmainresultis:

Theorem1 LetPbeasetofnpointsintheplane. Af-

terO(nlogn)preproessingtimeandusingO(n)spae,

moving viewpoint queries along a speied line seg-

ment anbe answered inO(klogn+s)amortizedor

O(klog 2

n+s) worst-asetime where k is the number

INRIASophia-Antipolis,Olivier.Devillerssophia .in ria. fr

y

CarletonUniversity,vidass.arleton.a

z

LORIAUniversityNany2,everettloria.fr

x

INRIARh^one-Alpes,Samuel.Hornusinria.fr

{

MGillUniversity,sues.mgill.a

k

UniversityofLethbridge,wismaths.uleth.a

of dierent views seen from and s is the size of the

output.

Notiethat, ifeah timetheviewhangestheentire

view is output, then s = (kn); if the output is re-

strited to the twopointsthat must be swapped then

s=(k).

Similarproblemshavebeenonsideredinbothon-line

and o-line mode forobjetsother thanpoints. Ghali

and Stewart [3, 4℄ developed an on-line algorithm for

maintainingtheviewof aset oflinesegments. Pohi-

ola[7℄presentedasolutiontotheo-lineversionofthe

problem for a set of n objets. Roughly speaking, if

these tworesults aresaled down to npoints thenthe

spaerequirementsareO(n 2

)sinetheyonentrateon

ases with many olusions among the given objets.

ThealgorithmofGhaliandStewart[3℄ishoweversim-

ilarto oursandthey haveimplementedthe algorithm.

See[5℄foranoverviewofotherrelatedresults.

2 Preliminaries

Westartwithsomesimplepropertiespertainingtothe

viewofamovingpoint.

Lemma2 Letqandq 0

betwopointssuhthatsegment

qq 0

intersetslineab,a;b2P,andnootherlinethrough

two points of P. If segment qq 0

intersets segment ab

then the view from q is the same as the view from q 0

.

Otherwise the views are the same exept that a and b

areinverted. (Refer toFigure 1.)

The hipped line through two points a and b is the

linedened byaandb minusthe(losed)linesegment

ab. Thepointsaandbarenotonthehippedline. The

followingobservation states that if q movesarossthe

lineabthentheviewhangesifonlyifqdoesnotross

segmentab.

The CL-arrangement indued by a set of points is

onstrutedbydrawingahippedlinethrougheahpair

of points. The ells of this arrangement are verties,

openedgesandopenfaes. SeeFigure2. Thefollowing

lemmafollowsfromLemma 2.

Lemma3 Any two points in the same ell of a CL-

arrangement havethe sameview.

a

b c

d

e

f

q q’

w w’

Figure1: Theviewfromqandq 0

is(a;e;f;b;d;)from

witis(e;f;a;b;d;)andfrom w 0

itis(e;f;b;a;d;).

Figure2: ACL-arrangement

Lemma 3 leads to the following straightforward so-

lution to our problem. In a preproessing step, om-

pute the CL-arrangementof P. There are O(n 4

) ells

in this arrangementwhih an be omputed in O(n 4

)

time[2℄. Preproessthearrangementforpointloation

queries. Toansweramovingviewpointqueryforaseg-

ment =qr, rstomputetheviewfromq bysorting

the points aroundq. Find the ellC ontaining q us-

ing the point loation struture. Shoot a ray from q

along qr to nd the rst intersetion point of qr with

C; sayqr intersets C at q 0

on anedge dened bythe

linethroughd. Computetheviewfromthispointq 0

by

removingallthepointsonline dexeptfor thosevis-

ible to theviewpoint. Compute theviewfrom apoint

in the neighboring ell of C into whih the viewpoint

moves. Repeatthis proess until theviewpoint arrives

Clearlytheproblemwith thisapproahistheO(n 4

)

spae and preproessing time needed to ompute the

CL-arrangement. We show how to redue this signi-

antlyinsetion3.

Another, less naive approah to solve the moving

viewpoint query proeeds as follows: Reall that our

goalis to maintainthe irularly orderedset of points

around q. Theidea is to sort them rst (as a prepro-

ess), then, for eah of the n pairs of irularly on-

seutivepoints, insertaneventin an event-queuethat

tellswhentheswappingof2onseutivepointsshould

happen (if ever). The update is as follows: (A) Pop

an event. (B) Update the irular ordering by swap-

pingtwoonseutiveelements. (C)Insert upto3new

eventsinthequeue,whiledeleting2oldevents. Speial

areisrequiredwhentheviewpointqmovesoutofthe

onvexhull of P, but this does not inreasethe om-

plexity. Overall,foragivenlinesegmenttrajetory,the

omplexity is O(nlogn) preproessing time, and then

O(nlogn+klogn+s) time. The O(nlogn) term is

requiredbeause,asthe newtrajetoryis known, it is

neessarytoemptytheevent-queue,omputetheevent

times for n pairs of points, and insert O(n) events in

thequeue.

In the next setion, we show how to remove this

O(nlogn)term.

3 Proofof Theorem1

Ratherthanomputingtheentirearrangement,wewill

omputetheellsofthearrangementonlywhenweneed

them. SinethesizeofeahellisO(n),itistheability

to perform updates eÆiently without omputing the

entireCL-arrangementthatisritial.

It will be onvenient to rid ourselves of the non-

onvexells of the CL-arrangement whih, as the fol-

lowing lemmaindiates, anbe doneby addingto the

arrangementtheedgesoftheonvexhullofP. Forex-

ampleray shootingis moreeasilyperformedin onvex

regions.

Lemma4 A vertex of a CL-arrangement is reex in

someellifandonly ifitisapointonthe onvexhull.

Proof. Firstnotiethatareexvertexmustbeoneof

thenpoints;indeed, allotherarrangementvertiesare

atintersetionsoflines. Letpbeapointorresponding

toareexvertex. Now,sinepisreexinsomeell,all

ofthen 1raysemanatingfromplieononehalf-plane

dened by a line through p and thus all of the other

n 1other points lie in the other half-plane whih is

exatlytheonditionforptobeontheonvexhull.

In the following, we assume that the algorithm is

operating in theCL-arrangementwhih has been aug-

mentedtoinludetheedgesoftheonvexhullofp.

putingaellgivenaview.

Lemma5 The edges of a fully-dimensional ell C

of the CL-arrangement are determined by onseutive

pointsofthe view fromany pointqin C.

Proof. LetsbeapointonanedgeofCandletaandb

bethepointsgeneratingthelineontainings. Wewill

distinguish theaseswhere s ison ahipped line ora

onvexhulledge.

Case 1: Chipped line.

Without loss of generality, say s is on the ray em-

anating from b. It suÆes to show that a and b are

onseutivepointsoftheviewofq. Supposenot. Then

thereisapointxofP inthewedgeaqb. Werstshow

thatxisnotintriangleaqb. SeeFigure3. Ifitwasthen

therayemanatingfromx indiretionaxintersetsthe

segmentqs,whihis entirelyinside C byonvexity, at

apoint p. Butthen thepointsb, x and aare ordered

around any point on the segment qp in the order bxa

and about any point on segment ps in the order bax

whihontraditsLemma3. Similarly,thereisnopoint

xanywhereelseinthewedgeaqbsinethentherayem-

anating from x in diretion xb intersets the segment

qs. SeeFigure 4.

Case 2: Convexhull edge.

Thensisonthesegmentab. Firstifqisinsidetheon-

vexhulland if aand b arenot onseutivethen there

isapointx inthewedge,andsineabis aonvexhull

edge, x belongs to triangleaqb. A ray from x rosses

segment qs whih should be inside C givinga ontra-

dition;thisrayisissuedeitherfrombasinFigure5or

from a depending on theloation of x with respet to

segmentqs. Ifq isoutsidetheonvexhull,then either

allthepointsareinthewedgeaqb(outsidethetriangle)

and aand b areonseutive,orarayfrom b separates

q andagivingaontradition asin Figure 6(oraray

fromaseparatesqfrom b).

Lemma 5 implies that, in order to ompute a ell

it suÆes to know the view from a point in that ell;

eahpairof onseutivepoints intheirular ordering

ontributesahalf-plane,theintersetionofwhih gives

theell. NowfromLemma 2weknowthattoompute

a ellfrom a neighboring ell, it suÆes to delete two

half-planes andinsert twonew ones. These operations

anbeaomplishedinO(logn)amortizedorO(log 2

n)

worst-asetime[1,6℄.

Thealgorithm,whih impliesTheorem1,isgivenin

Figure 7. Theinput to thealgorithm is apointset P

andaquerysegmentqr. Thealgorithmissimilartothe

less eÆient version suggestedin theprevious setion,

howevernotethatthespaerequirementsareontained

toO(n)asthepointq 0

movesfromqto renountering

theellsoftheCL-arrangementwithoutrstomputing

thearrangement.

q

s

b x

a C p

Figure 3: If point s is on hipped line ab, there is no

pointintriangleaqb.

q

s

b

a

x

C p

Figure 4: If point s is on hipped line ab, there is no

pointinwedgeaqb.

q

s b x

a C

Figure 5: If point sis ononvexhulledge ab,there is

nopointintriangleaqb.

4 Extensions

Thesameapproahworkswhenthetrajetory isnot

alinesegment,theonlydierenebeingintheresulting

omplexity.

Theorem6 Let P be a set of n points in the plane.

After O(nlogn) preproessing time, moving viewpoint

queries along a speied urve an be answered in

O(f(n)+klogn+s)amortizedorO(f(n)+klog 2

n+s)

worst-asetime,wherekisthenumberofdierentviews

seen from,sisthesizeofthe output,andf(n)isthe

q

b x

a C

Figure 6: If point sis ononvexhulledge ab, there is

nopointoutsidewedgeaqb.

Algorithm: Moving Point Query (P, qr)

1. Setq 0

:=q.

2. Construttheviewaroundq. {O(nlogn).

3. ConstruttheellC ontainingq. {O(nlogn).

4. Whileq 0

<rdo

Shootarayfromq 0

inthediretionqrhitting

Cat pointq 00

. {O(logn).

q 0

:=q 00

Updateandoutputtheviewatq 0

. {O(logn).

UpdatetheellC. {O(logn)orO(log 2

n).

Figure7: Thealgorithmandanalysis

timerequiredtoperformray-shootinginaonvexn-gon

along urvesof the sametypeas.

Suppose as q moves from s to t along segment st,

the view at s is known, and that only the view at t

is required, andnot theintermediate views. The view

at t anbeomputed byrunning ouralgorithm and a

sortingalgorithminparallel.Thenthetimetoompute

theviewattisminfO(nlogn);O(mlogn)gwheremis

thenumberofellsthataretraversed.

Finally,ifP isthesetofendpointsofasetofdisjoint

linesegments,thentheatualviewofthesegmentsan

beobtainedeasilyfromtheviewofthepointsinlinear

time,thusmakingouralgorithmusefulintheasewhen

fewsegmentsareoluded.

Distane-sort maintenane. Interestingly, the same

algorithm an be used to maintain the set of points

in P sorted aording to their distane to a mov-

ing query point q. In this ordering, two onseutive

pointsareswappedwhenqrossestheirbisetor. Thus,

the CL-arrangementis replaed by an arrangementof

present only a ritial lemma to support our laim.

b a

c r

V C L

Lemma7 Eahedgeofa

ell in the B-arrangement

is supported by the bi-

setor of two onseutive

points in the distane-

ordering.

Proof. Let line L be the

bisetordened bypointsaandb supportingone edge

ofaellContainingthequerypointv. Assumethat,in

thedistane-ordering,there isapointrbetweenaand

b. Thentheirumirleof a,bandrhasenter ,the

pointof intersetionof the3bisetorsdened by(a,b)

(b,r)and(r,a). Wehavevb<vr<va. Thereforevlies

in the yellow shaded region. Hene the line L annot

supportanedgeofC,whih isaontradition.

Byredeningtheviewpointqueryoftheorem1toreport

hanges in the distane-ordering, we obtain the same

omplexityresult.

5 Conlusion

Wepresentheretherst solutionto the movingview-

pointqueryproblemamongpointsin2Dwithprovable

worst-aseandamortizedomplexity. Solvingtheprob-

lemin 3DremainsadiÆultopenproblem.

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VisibilityMapasSeenbyaMovingViewpointinTwo

Dimensions, Eurographis7thWorkshopon Computer

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StanfordUniversity,2002.

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204,1981.

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