Researchchnical Report

The Furthest-Site Geodesic Voronoi Diagram

Technical Report No. 420

Robotics Report No. 186

December, 1988

New"ork UniversityInstituteofMathematicalSciences

The Furthest-Site Geodesic Voronoi Diagram

Technical Report No. 420

Robotics Report No. 186

December, 1988

New "ork UniversityDept. of Computer Science

Courant Institute of Mathematical Sciences

251 Mercer Street

New "ork , New "ork 10012

W ork on this paper by the first author has been supported in part by O ffice of Naval

Research Grant N00014-87-K -0129 , by National Science Foundation G rant No. NSF-DCR

83 -20085 , and by grants from the Digital Equipment Corporation, the IBM Corporation,the

U .S.-Israeli Binational Science Foundation, the NCRD the Israeli National Council for

Research and Development, and the Fund for Research in E lectronics , Computers and Com

munication of the Israeli Academy of Sciences and Humanities . The work of the first authorwas also supported by an AT"T Bell L aboratories Ph.D . Scholarship .

The Furthos t-SiteGeodesic Voronoi Diagram

Boris Aronov

Courant Institute of Mathematical Sciences , New"ork University

Steven Fortune

AT"T Bell Laboratories

Gordon W ilfong

AT"T Bell Laboratories

ABSTRACT

We present an +k)) time , 0 (n+k) space algorithm for computing the furthest-site geodesic Voronoi diagram of I:point sites within a simple n- sided polygon .

November 9 , 1988

The Furthest-SiteGeodes ic Voronoi Diagram

Boris Aronov

Courant Institute of Mathematical Sciences , New"ork University

Steven Fortune

AT"T Bell Laboratories

Gordon W ilfong

AT"T Bell Laboratories

1. Introduction

A common goal of much recent research in computational geometry is to extendalgorithm s that have been developed for the Euclidean metric to the more complicated geodesic metric inside a simple polygon . The geodesic distance between twopoints in a simple polygon is the length of the shortest path connecting the points thatremains inside the polygon . For example , Toussaint [T86] gives an algorithm for the"relative convex hull" of a set of points inside a simple polygon ; Aronov [AS7] givesan algorithm for the nearest-neighbor geodesic Voronoi diagram ; and Pollack , Sharirand Rote [PSR87] give an algorithm for the geodesic center

" of a simple polygon .

A classic structure in the Euclidean metric is the "furthest-site Voronoi

diagram . Given a finite co llection of point sites in the plane , the furthest- site Voronoi diagram partitions the plane into Voronoi cells , one cell per site . The site thatowns a cell is the site that is furthest from every point in the cell . Using well-knownalgorithms , the Euclidean furthest- site Voronoi diagram of k sites can be computed intime 0 (k logk) and space 0 (k) [P585] .

The content of this paper is an efficient algorithm for computing the furthest- siteVoronoi diagram , defined by the geodesic metric inside a simple polygon . The algorithm uses 0 ((n+k) log (n+k)) time and 0 (n+k) space , where n is the number ofbounding edges of the polygon and k is the number of sites . The best previous algorithm for this problem had running time o(u 3 log logn) [ATS6] , and just computed (asuperset of) the vertices of the furthest- site Voronoi diagram of the n corners of thepolygon . We remark that our furthest-site geodesic Voronoi diagram algorithm is afactor of 0 (logn) faster than the best known nearest-site geodesic Voronoi diagramalgorithm [AS7] .

A preliminary version of this paper appeared in Proceedings of the FourthACH Symposium

on Computational Geometry , 1988 .

The work of the first author was supported by an AT"T Bell Laboratories Ph.D . Scholar

ship . Part of the work was performed while the first author was at AT"T Bell Laboratories ,Murray Hill , NJ .

The problem of computing the furthest-site Voronoi diagram is an extension ofthe "furthest neighbor problem which is "

Given a finite co llection of points , foreach point identify the element in the collection that is maximally distant from it .Suri [887] shows how to solve a special case of the furthest neighbor problem in thegeodesic metric inside a simple polygon . Specifically , he gives an algorithm that foreach corner of the polygon computes the corner that is max imally distant from it inthe geodesic metric. H is algorithm runs in time 0 (n logu) and space o(u) , where nis the number of bounding edges of the polygon .

The geodesic furthest-site Voronoi diagram generalizes the geodesic furthestneighbor mapping of Suri [887] in two ways . First , the Voronoi diagram provides aplanar partition of the polygon together with its interior into furthest-site Voronoicells . Consequently , arbitrary furthest-site queries can be answered using a planarpoint-location algorithm . Second , the set of sites is not restricted to the corners ofthe polygon . Rather , the sites can be arbitrarily situated in the polygon . Both ofthese generalizations have substantial technical impact on the algorithm for computing furthest-site Voronoi diagrams .

There are many analogies between the Euclidean furthest-site Voronoi diagramand the geodesic furthest- site Voronoi diagram . In the Euclidean case , if a site hasnonempty Voronoi cell , then it lies on the convex hull of the set of sites and it doesnot appear on the line segment between two sites . The counterclockwise sequence ofVoronoi cells (at infinity) is the same as the counterclockwise sequence of sites on theconvex hull . In the geodesic case , we show that a site with nonempty Voronoi celllies on the relative convex hull of the set of sites and does not appear on a geodesicbetween two sites . The counterclockwise order of Voronoi cells along the boundaryof the polygon is a suborder of the counterclockwise order of sites on the relativeconvex hull . In the Euclidean case this characterization is exact: any site on the convex hull not between two other sites has nonempty Voronoi cell . In the geodesic casethe characterization is not exact , roughly because the polygon may not be not largeenough for the cell to appear .

A further analogy between the two cases is the structure of the Voronoi diagramitself . In the Euclidean case the Voronoi diagram forms a tree with root at theEuclidean center of the set of sites . (The center of a set of point sites is the pointthat minimizes the maximum distance to any site .) If edges are directed towards theroot , then this orientation is consistent with geometric direction towards the centerIn the geodesic case exactly the same properties hold , substituting

"

geodesic centerfor "Euclidean center and

"geodesic direction" for "direction .

"

The algorithm for computing the furthest-site Voronoi diagram consists of twosteps . First, we compute the restriction of the Voronoi diagram to the boundary ofthe polygon . Intuitively , the boundary of the polygon in the geodesic casecorresponds to points "

at infinity in the Euclidean case . Second ,we extend the

diagram to the interior of the polygon . Because the Voronoi diagram forms a treewith root at the geodesic center , the second step is easy . It can be performed by a"reverse geodesic sweep towards the geodesic center .

- 3

The first step , the computation of the Voronoi diagram on the boundary of thepolygon , is much more involved . We use a technique developed by Suri [587] fordetermining furthest neighbors . We reduce the problem to three instances of the"two-fragment problem

”; an instance of the two-fragment problem consists of a frag

ment of the boundary of the polygon and a fragment of the relative convex hull ofthe set of sites . The relative convex hull fragment contains the furthest sites of allpoints on the polygon boundary fragment . We solve an instance of the two-fragmentproblem using divide and conquer in the following manner . The polygon boundaryfragment is split at its midpoint ; this implies a corresponding split of the convex hullfragment . Thus one instance of the two-fragment problem results in two simplerinstances . Eventually instances become small enough to be solved directly .

We refine Suri ’s two-fragment technique in two ways . First , "1ri ’ s algorithmalways splits the polygon boundary fragment at a com er of the polygon . This is sufficient for the furthest-neighbor problem , because furthest-neighbor information forpoints along a wall is not of interest . For the furthest-site Voronoi diagram , simplysplitting at corners is insufficient , since potentially many Voronoi cells meet a singlewall of the polygon . If necessary , we further split each wall into subsegments so thatthe shortest path tree from a point in a subsegment to the sites is combinatoriallyinvariant over the entire subsegment . The combinatorial invariance of the shortestpath tree implies that the Voronoi partition of the subsegment can be easily com

puted.

The second refinement of Suri ’s technique concerns the complexity analysis ofthe recursion . Suri ’s original algorithm required a step called "trimming

"; trimming a

two-fragment instance introduces a different subproblem that could be solveddirectly . This O peration is necessary in Suri ’s analysis in order to maintain the linearity of the total size of all subproblems at a particular level of recursion . We showthat even without trimming , the total size of all subproblems at a particular level islinear . This observation simplifies the recursive structure of the two-fragmentalgorithm so that it actually matches the description given above . The analysis hasbeen incorporated into Suri ’ s furthest-neighbor algorithm .

The best lower bound that we know for computing the furthest-site geodesicVoronoi diagram is 0 (n+k log It) . It follows from known lower bounds for diametercomputation in the Euclidean case . Conceivably , the current algorithm could beimproved to match this lower bound .

2. The Furthest-Point Geodesic Voronoi Diagram

This Section contains the definition of the furthest-point geodesic Voronoidiagram and some of its basic properties . We begin by discussing geodesics in Section Section contains a fairly extensive treatment of relative convex bulls.

The notion of a far side" of a relative convex hull is developed in Section this isa technical idea used to prove the Ordering Lemmain Section Sites are assumedto be in "

general position"; this assumption and its consequences are discussed in Sec

tion In Section we actually give the definition of the Voronoi diagram .

- 4

Section contains the Ordering Lemma , which states that the order of Voronoicells around the boundary of the containing polygon is the same as the order of sitesaround the relative convex hull of the set of sites . In Section we define a refinedform of the Voronoi diagram and use it to show a linear bound on the descriptivecomplexity of the (unrefined) Voronoi diagram . Finally , in Section we show thatthe Voronoi diagram forms a tree directed towards the geodesic center of the set ofsites . The algorithm for computing the diagram appears in Section 3 .

Preliminaries

The universe U is a compact region in the plane whose boundary 60 is a simplen- sided polygon . S , the set of sites , is a collection of it points of U. A vertex of aUis called a corner and a segment of aU is a wall . A corner is reflex if the measure ofits interior angle is more than 11 and convex if it is less than 17 . If x and y are distinctpoints of OU, then boundary fragment aU[x,y ] is the portion of OU counterclockwisefrom x to y inclusive . The symbol a denotes the boundary of a set relative to thewhole p lane , rather than to any proper subset of the plane . The terms "relativeboundary" and "relative interior" used without any other qualification mean "relativeto U.

A polygonal path is a simple path comprised of a sequence of line segments . Ifp is a polygonal path , then the size of p , Ip I , is the number of maximal segmentscontained in p and not containing a corner of OU in their interiors , and the length of pis the sum of their (Euclidean) lengths . A polygonal reg ion is a compact set whoseboundary is the union of a finite number of line segments . We allow points as(degenerate) line segments , so for example a finite set of points is also a polygonalregion .

Geodesics

For points x,y 6U, the geodesic path g (r ,y) is the shortest path in U connecting xand y . Such a shortest path is called simply a geodesic . This path is unique . In fact,it is a polygonal path with interior vertices only at reflex corners of aU [LPS4] . Weoften consider g (x ,y) directed from x to y . A link of g (x ,y) is a maximal segment ofg (x ,y) not containing any corners in its interior ; clearly an endpoint of a link is eitherx , y , or a corner . The first endpoint of the last link of g (r ,y) is the anchor of y withrespect to x; it is either a reflex com er of aU or x itself.

We make heavy use of the fact that the intersection of any two geodesics is connected (and is itself a geodesic) . This fact follows immediately from uniqueness ofgeodesics . Two geodesics overlap if they intersect in more than a single point.

The geodesic distance d (r ,y) between points x and y is the length ofThe geodesic distance is a metric; in particular , it is continuous as a function of both xand y and satisfies the triangle inequality Furthermore , byuniqueness of geodesics , if and only if y lies on Weoften write d"for the function defined by

- 5

Lemma [PSR87 , Lemma 1] For any d" is a convex function ong (v,w) wi th unique local minimum (possibly at v or w) . In particular , for any

zr vnv.

The geodesic direction 0(x ,y) from x to y¢x is the direction ( i .e unit vector)from x towards the anchor of x with respect to y . For fixed y , 6(x,y)

—Vd, ,where

d is the gradient of d,with respect to x, evaluated at x [AS7 At any point x

not an anchor wi th respect to y , d, is differentiable as a function of x and 0(x,y) iscontinuous as a function of both x and y . For H a closed subset of U, let 0(x,H) be

habx} .

The geodesic anglen a is the angle counterclockwise from 9(y ,x) to Themeasure of n z wi ll be written mn z. The angle between 9(y ,x) and 0(y ,z) is thesmaller of the two anglesn z and Lzyx.

Lemma [PSR87 , Corollary 2] If y¢x,z , and the angle between 9(y ,x)

and 0(y ,z) is at least tr/2, then d (x , z)

The shortest path tree from s , T(s) , is the union of the sets of links of g (s ,y)taken over corners y of av. It has n— 1 or n links , depending on whether or not sitself is a corner of U

Let Pa(s) be the set of points in U that have anchor a,with respect to s . The

shortest pathpartition of U from s is the co llection It is a planarpolygonal subdivision of U; it can be computed and in fact triangulated in linear timegiven a triangulation of U We can describe the bounding edges of theshortest path partition as follows . Suppose Pa(s) is not empty , where a sé s . Let abbe the first link of the geodesic g (a ,s) (clearly ab is the second link of all geodesicsg (x,s) for Since Pa(s) is not empty and a is a reflex corner of aU, we canextend link ab past a into U. First suppose that neither wall of aU incident to a overlaps this extension . Let y be the first point past a of the extension so that y éaU.

Then segment ay is the shortest path partition edge (from s with anchor a) , denotedpa(s) . Now suppose some wall of aU overlaps the extension of segment ab; then wesimply define pa(s) to be this wall . It can be checked that the boundary of a cell ofthe shortest path partition consists of an alternating sequence of shortest path partition edges and sections of aU.

A set AQU is relatively convex (with respect to U) if g (x,y)gA whenever x,y EA;

the relative convex hull of set F , denoted R (F ) , is the smallest relatively convex set

contain ing F (that is , the intersection of all relatively convex sets containing F) [T86] .Relatively convex sets are discussed in detail in the next section . A set is degenerateif it is contained in a single geodesic.

If x and y are distinct points of U,then a shadow of g (x ,y) is a point year] so

that segment yy extends the last link of g (x,y) while staying in U. Clearly y is a sha

dow of g (x ,y) only if yGaU. Shadows are not unique , indeed it is possible that everypoint on a subsegment of a wall is a shadow of Similar ly , a foreshadow ofg (x ,y) is a point x

'

GaU lying on a segment contained in U extending the first link ofg (x ,y) backwards .

- 6

A boundary geodesic is a geodesic connecting two distinct points of aU. Let x '

and y be the foreshadow closest to x and shadow‘closest to y of respectively .

Then g(x,y) denotes the boundary geodesic Geodesic g(x,y) splits U intotwo simply connected polygonal regions U[x ,y ] and U[y ,x l with disjoint interiors ;

is and is (Note thatis distinct from Intuitively , U [x,y ] contains points lying on or to

the right of while points on the geodesic or to the left of it constituteU Notice that g(x,y) is exactly the common boundary of U [x,y ] and Uhence any geodesic from a point in U [x,y ] to a point in U [y ,x ] must intersectBoth U[r ,y l and U [y ,x ] are relatively convex , since a geodesic connecting two pointsof, say , U [x,y ] must have connected intersection wi th

Lemma Suppose u, v€aU, u-iév, and w¢x.

( 1) then any shadow f of g (w ,x) lies in(2) then some shadow f of g (w ,x) lies inProof: ( 1) Suppose Then g (w ,B cannot intersect g(u,v) again after x , so

(2) Suppose without loss of generality assume w, x , v are in that orderalong We can choose 3c

'

= v unless g (u,v) bends at or after x . If g (u, v) bendsright at some point e at or after x , then since U [a,vl lies locally to the right of

c must be a reflex corner of and we can choose 3t'

=c. If g(u, v)bends left at some point c then the straight- line continuation of g (w ,c) at c enters theinterior of U [u, v ] and thus will not intersect g (u,v) again . Hence we can choose x tobe any shadow of g (w,e) distinct from c . 0

Lemma Suppose u,v € 6U, uafiv , w¢x , and W the closest

shadow of( 1) If w i aU and g (x ,w) does not overlap then(2) If v is the closest shadow of g (a,w) and g (a ,w) is not an initial portion of g (u ,x)

(i .e. thenProof: ( 1) If g (x,w) does not overlap then 0(w,x) cannot be 9(w ,u) or

Since w éaU, Waéw and must enter the interior of U[v ,u ]

at w . As g (x ,w) cannot intersect g (u,v) again ,

(2) The statement is trivial if weaU. If not , then W ¢W. We might havein which case w=v. We cannot have else

Otherwise 9(w ,W) must enter the interior of U [v,u ] and as beforeD

Boundary geodesic g (x ,y) separates points a and b if and b éULy ,x ] ,

or vice versa. Similarly , boundary geodesic g (x ,y) separates sets A and B ifAQU[x ,y ] and BQUIy ,x ] , or vice versa. Note that separation does not imply disjointness , indeed if (degenerate) set A is contained in boundary geodesic theng (x ,y) separates A from itself.

For a compact set FCU and zGU, let rad The center of F isI

the point 2 that minimizes Pollack , Shat ir , and RotefPSR87] show that the

- 7

center of the set of vertices of U is unique . In fact with minor modifications theirproof shows that the center of any compact set FQU is unique .

For points x,y ,z €U we define the geodesic triangle Axyz as follows [PSR] . If

is degenerate , then Arya is the smallest geodesic containing x , y , and z . Otherwise we can choose points x', y

'

, and z' so that x' is the point at which g (x,y) andg (x,z) diverge , and similarly for y

' and z'. Then the circuit

is a simple polygon We define Axyz to be the union ofand Ax'

y'z' together with its interior . All of the interior angles of

Ax’

y'

z’are reflex except at x ’ , y ’ , z ’ [PSR ] . The geodesic triangle is a special case of

the relative convex hull of a finite set of points (discussed in Section The proofof the following Lemma is in the same spirit as the proofs of Lemmas and

Lemma (Triangle Lemma): Suppose is not degenerate . Let y and'

z

'

beshadows of g (x ,y) and g (x ,z) respectively . Assume If uéAxyz and

then there is a shadow E of g (x ,u) so that 3 6 80 523 and g (x ,zDintersects If u is in the interior of Axyz, then for any shadow u ofEGGUEE} and g (x,D intersects If then g (x,u)

intersects g (y ,z) and there is a shadow u of g (x ,u) in BUWEJ .

Relatively Convex Sets

This section develops properties of relatively convex sets . The main result isLemma It states that the "extreme elements of a set F can be ordered so thatthe relative convex hull of F is the intersection of all "cones defined by consecutivetriples of extreme elements . An immediate consequence of Lemma is a decomposition of a relatively convex set into a collection of simple polygons and connectinggeodesics . Also , the order of extreme elements extends to a natural notion of atraversal of the boundary of a relatively convex set . This ordering has a number ofuseful consequences , given in Lemmas through

Lemma Any relatively convex set R is simply connected .

Proof: We show that the region enclosed by a simple cycle 1 in R lies completely inR. Suppose x 67 , y is in the interior of y , and y is a shadow of Since yy con

nects a point inside 7 to a point on or outside 1 , there is a point w in the intersectionof y"and 7 . Since R is relatively convex , so y éR. G

Let F be a nonempty polygonal region contained in U. For x 6U, we definespan(x,F ) to be the smallest set of directions containing 0(x ,F ) so that whenever

a t: —a'

, and the angle between a. and a' is contained in U near x ,

then every direction in this angle is in If x EU is not a reflex com er ofaU, then it is easy to see that either span(x ,F ) has a single component or span (x ,F )

consists of two O pposite directions . If x is a reflex corner of amthen span (x,F ) may

have two components that are not opposite directions ; in fact one or both componentsmay have positive measure . Since F is a polygonal region , it is easy to check that theendpoint of any component of span (x ,F ) is 0(x,y) for some y €F . We show below(Lemma that span

- 3

If span(x,F ) consists of a single component of measure less than 21r, then x isan exterior point of F. Notice span (x,F ) can have measure exceeding 17 but less than21: only if x is a reflex corner of GM If span (x,F) consists of a single component ofmeasure less than tr and x €F then x is an extreme point of F. If span(x,F ) consistsof two connected components , then x is a thin point of F. In Figure for

point 5 and all points of aU except a are exterior (but none areextreme) , points 1, 2 , and 3 are extreme points , and every point on g ( l ,a) except 1is a thin point. Set F is extreme if every element of it is extreme . In Figure{1 , 2 , 3} is extreme .

Figure Relative convex hull

For the remainder of this section , F g U is a finite set of points containing at

least two elements .

Let x be an exterior point of F or a point of aU. If xéaU define oppp (x) to bex,otherwise define oppp (x) to be the first point of aU intersected by the ray with

endpoint x directed opposite the bisector of span The clockwise extreme pointof F from x , denoted r (x) , is the point fEF so that there is no geodesic extendingg (x,f ) to a point f

'GF , and among all such points the closest shadow of g (x ,f ) is as

clockwise as possible in aU {opp Sim ilarly define the counterclockwise extremepoint of F from x , denoted l (x) . For example , in Figure l (2) 1, r (2) 3 ,

l (a) = 1, There are two subtleties to these definitions . First, r and ldepend upon the set F , but except in the proof of Lemma we leave this dependence implicit . Second , there are two distinct ‘

(though overlapping) cases in thedefinition: either x is exterior (and span (x ,F ) consists of a single component) orxEBU (and even if span (x ,F ) consists of two components , there is stil l a natural wayto define r (x) and I Notice there is no natural definition of r (x) and l (x) ifspan (x,F ) has measure 21r or if span (x ,F ) has two components and xGaU. The following Lemma describes properties of functions r and 1; its proof is quite technicaland detailed .

Lemma Let x be an exterior point of F or a point of aU,Fthe

closest shadow of and l the closest shadow of1. If x is exterior then If x éaU or e xl afi 'n' then

if e xl= 1r then

- 9

2 . If r ifi l , then r"and ll are disjoint; if also x éaU then x , r , l are in that counterclockwi se order around GU.

3 . Both r and l are extreme points of F.

4 . If x is an extreme point of F , then5 . Let x €aU or e xl< 1t . Then f=r (x) if and only if fGF , and g (x,f )cannot be extended to for any other f

'

€F

Proof: ( 1) We must always have span (x,F )ge l by definition of span(x,F ) and ofpoints r and I; certainly If x is exterior then span(x ,F ) isconnected , so we must have e l=span For the second statement , first sup

pose x €aU or e xl $ 17 . Let x‘ be the closest foreshadow of We claimthis follows immediately from the definition of r and I if e xl< 1r,

x 66U, or e xl> 1t (since in the last case necessarily x éaU) . For any f 6F , eitheror the closest shadow of g (x ,f ) lies in so

fGU and FC; U Similarly FQU [x , l The argument that e xl= 1r impliesFQU[r, l ] is similar .(2) Suppose l¢r. If ll and r"met , either I would lie on g (x , r) or vice versa, contraryto the definition of r , l . The ordering of x , r , and 1 follows immediately by definition .

(3) We assume FQU otherwise a similar argument works using U Toshow r extreme , we need to show span (r ,F ) is connected and has measure less than17 . Whether or not réaU, is connected . As FCUspan Hence it suffices to show that for fGF , 6(r ,f ) lies in the

angle from 0(r ,x) clockwi se to but not including Now 0(r ,f ) ca nnot beelse g (x,f ) would extend Also 9(r ,f ) can be clockwise of

only in the case that r is a reflex corner of aU lying to the right of again thisis impossib le because g (x,f would extend(4) By FQU For any f €F not appearing on f is in the relative interior of U[r ,x l . Thus , by Lemma the closest shadow of g (r,f ) lies on

where x ‘ is the closest foreshadow of This impliesby definition of counterclockwise extreme point.(5) Clearly there can be at most one point f in F satisfy ing FQU[f ,x ] and g (x,fcannot be extended to for any f

'

€F Since f=r (x) is one such point, theclaim follows . 0

Lemma For x 6U, span

Proof: Suppose y ,z €F are such that 0<mLyxz< 1r and Lyxz is ( locally around x) contained in U. By examining Axyz and using g (y , z)g R (F) we see that

Hence span

If span (x ,F ) has measure 21r, then it must be that span (F Sup

pose x is an exterior point of F; let r=r (r ) and l = l (x) . If e xl$ 17 thensince by Lemma and U[r ,x l and

U [x , l ] are relatively convex . Hence 0(x,R (F g e l

span where the first equality follows by definition and the second by LemmaIf e xl= tr then so g,

e l

- lo

span

Finally ,if x is a thin point of F , we can find a segment through x splitting U into

two simple polygons , each containing one of the components of span Let F 1and F2 be the intersections of these two polygons with F , and let H 1 , H 2 be theirrespective convex hulls (relative to U) . It is easy to verify that x is an exterior pointof both F 1 and F2 and thus , by the first part of the proof,

span(x ,F 1) U span (r ,Fz) 0(x ,H 1) U 0(x,H 2) c_; 0(x ,R (F

The inclusion is proper only if there are two directions , 01 span (x,F 1) and

92 6 span (x ,F g) the angle between which has measure less than 17 and is contained inU locally near x . But then this angle would be included in contradictingthe assumption that x is a thin point , i .e. , span(x,F ) is disconnected . D

It is immediate that any two sets with the same relative convex hull have thesame extreme , exterior , and thin points .

The next Lemma is the main result of this section . To state it we need the notionof a geodesic cone . Suppose mn z< 1r and Let s be the segment from y

to the closest foreshadow y' of open at y and closed at y

'

. The geodesic cone

is —s . See Figure Notice s intersects

only if s is also a foreshadow segment of It is easily checked that U isrelatively convex and that the boundary of U consists of g (ypfl , andaUEfl , where x and E are the closest shadows of g (y ,x) and respectively .

Furthermore , if f is an extreme point of F , then

Figure U ( solid) and U (dashed)

Lemma H=R (F is a simply connected polygonal region . The extreme elements O f F can be labelled f l . fo=fm= f (fm SO

rn

that H = fl ULfi+ t.fl.ft- i ] and aH =i =1 i =1

Proof: F must contain some extreme element , since r (u) is extreme for any aéaU.

Let f l be an extreme element of F and consider the sequenceSince F is finite and In is the identity (Lemma it must be that

fm + 1=f1 for some m with distinct. If F is degenerate , then m = 2 and the

Lemma follows easily . We assume F is nondegenerate , hence m>2 and points in

https://www.forgottenbooks.com/join

- 12

point x of R (F) is an interior point of a bridge (if span (x,F ) consists of exactly twodirections) , an endpoint

'

of a bridge of positive length ( if span (x,F consists of twocomponents only one of which has positive measure) , or a bridge by itself ( ifspan(x ,F ) consists of two components both of positive measure) .

For F labelled as in Lemma and H we define the counterclockwisetraversal of GH to be the circuit g (f b fz) g (fm ,f 1) . The counterclockwise traversal visits every thin point twice and other points once . Hence for x ,y not thin pointswe can unambiguously define aH [x,y] to be the section of the counterclockwisetraversal of OH from x to y . Similarly if the points in H '

g 8H are not thin points ,they have an unambiguous counterclockwise ordering on aH . If F is extreme , thenthe counterclockwise ordering covers every point of F. Even if the sequence ofpoints x l , xi on aH includes thin points , it is still meaningful to say that theyappear in counterclockwise order , if they are visited in that order in a single counterclockwise traversal of aH . Of course , another order differing in the position of thinpoints may be consistent wi th a counterclockwi se traversal as well .

Corollary The center of F is the center of R (F) .Proof: It is sufficient to show that for any x €U, rad (x ,R A point ofH=R (F max imally distant from x must lie on GR By the previous Lemma anypoint on OH lies on a geodesic connecting two points of F hence by

.

Lemmathe maximally distant point must be an endpoint of the geodesic, that is , a point of F .

Thus a point of H furthest from x is necessarily a point of F and rad (x ,H )= rad

[3

Lemma If f ,f are extreme points of F , then every extreme point counterclockwi se from f to f is in UIf ,f and all but f and f lie in the relative interior ofU [f sf

’

l

Proof: The proof is by induction on the counterclockwise order around F using r. If

then since f' must lie in the relative interior of r (f ) must lie in

the relative interior of U [f ,f Now suppose f is counterclockwise of f before f’

and r (f"hef By inductive assumption , f is in the relative interior of U [f ,f

Again we have f' in the relative interior of U [r (f

' so r(f") must be in the

relative interior of Now is contained in U except possiblyfor a region bounded by segments f

's' and f

’

s" and where s ' and s" are

the closest shadows of and respectively . But cannotintersect segment f

'

s'

, since f' extreme . Hence Since r (f cannot

lie on as f, f r (f") are extreme and distinct , in fact r(f

' is in the relativeinterior of U [f,f B

An immediate consequence of Lemma is that counterclockwise order ofextreme points is an absolute order , not depending on other extreme points: if a ,b , c

are extreme points of both sets A and B , then their order around A is the same as it isaround B .

Lemma If points x , y , z , w occur in that order (not necessarily all distinct) in acounterclockwise traversal of the boundary of a relatively convex polygonal region R,

then g (x , z) intersects

- 13

Proof: By creating dummy plateaus if necessary , we can assume that x , y ,z,and w

each lie in a plateau. We proceed by induction on the number of plateaus in thedecomposition of R. If there is only a single plateau , then R is a simple polygon andthe claim is immediate . Otherwise there is some bridge g (a,b) whose removal SplitsR into relatively convex polygonal regions R 1 and R2 . If all of x,y ,z ,w are containedin one of R 1 or R2 , the claim follows by induction . Otherwi se , since x,y ,z ,w appearin that order in a counterclockwise traversal of the boundary of R, there are up tosymmetry two cases: x,y ,a in R 1 and z,w ,b in R2 , or x ,y ,z ,a in R 1 and b ,w in R2 . In

the first case g (x,z) and g (y ,w) both contain In the second case x ,y ,z ,a

appear in that order in a counterclockwise traversal of R1 , hence by inductionhypothesis , g (x,z) intersects Since g (x ,z) must intersectsow ) . D

We say segment f n connects a relatively convex set R to GU if f is an extremepoint of R, xGGU, and fx intersects R only at f (possibly f =x or some point of fitbesides x lies on GU) .

Lemma (Connecti on Lemma) Suppose are extreme points of relatively convex set R and segments fix; are pairwi se disjoint and connect R to GU. Thenthe order of the fi

’

s around GR is the same as the order of x,-’s around GU.

Proof: Suppose fi ,_fj ,fk are in counterclockwise order around GR; we showi .e. x; ,xj ,xk are in counterclockwise order around GU. By Lemma

f, is in the relative interior of UU} ,ft] . Since g (fbfk)QR and fix; meets R onlyat fj , fjxj cannot intersect Clearly f, is in the region bounded bysegment k k , g (fb fg) , and segment fixg. Since i GU and fix, does not intersect k k ,

g (fb fg) or fixi , and xj#x¢and xi -‘

text . 0

Lemma Mappings r and l restricted to GU preserve order .Proof: Suppose u 1 ,u2 ,u3 €GU are in that counterclockwi se order and

and r3=r(u3) are distinct . Let F, be the closest shadow of

We show that 71 , F2 , F3 lie in that counterclockwise order on GU; this followsfrom the claim that for all i iej. To establish this claim , supposefor the sake of contradiction that GU for some i atej . Thenrj €U[u,

-

,r, since but -

,u. by so In

fact rn (a, since rj cannot lie past r; on If e GU thenso either rg=rj or g (uj ,rj) can be extended to a contradiction . If y

'

,-( GU,

then neither g (u‘f i) nor g (ujf j) can bend at rj . Since -

, r, the two geodesics must overlap in some link containing ri . Furthermore the link must betraversed in the same direction in both geodesics , since u] , ui , r, r; appear in thatcounterclockwise order on GU. But then g (q j) could be extended to g (uf , r, a con

tradiction.

We claim that rm and rjrj are disjoint whenever i j; the Lemma then followsimmediately by the Connection Lemma. Suppose r,-r, intersects ri j . The intersectionmust be a single point distinct from both r; and ri . Hence , either rj €U[r, or

contradicting D

- 14

Far Sides

We consider the notion of the far side of a set F from a point . Informally , anextreme point f€F is on the far side of F from x if the geodesic g(x,f leaves R (F )after f. There are two main reasons for studying far sides . First , we use this notionto relate the order of extreme points of F with the order of points on GU (Corollary

Second , we will be able to compute relative convex hulls efficiently using farsides (Lemma

Let F be a finite set of points in U and xGGU. The far side of F from x is the setof all extreme points of F counterclockwise from r (x) to l (x) , inclusive . Here r (x)and l (x) are the clockwise and counterclockwise extreme points of F from x , as

before .

Lemma Suppose f is an extreme point of F , x €GU, and flex. Then the following are equivalent .( 1) f is on the far side of F from x .

(2) f is an extreme point of FU{x} .(3) ff connects R (F ) to G U, for any shadow f of and there is no geodesicextending g (x ,f to an extreme point f of F distinct from f.

Proof: If x €R (F ) then all three conditions are satisfied , so assume Then xis an exterior point of F. Set r=r(x) and By LemmafE

( l 2) Iff is on the far side of F from x , then féU[r, l l by Lemma If fser andfeel , then by the Triangle Lemma , g (x,f ) intersects g (l , r) at some point x

'

, wherex’aef since f is extreme . Hence 0(f ,x) g span and

span(f ,F ,F so f is extreme in FU{x} . If f=r, then by the definitionof clockwise extreme point , r (x) is unchanged if F is replaced wi th FU{x} , so r isextreme by Lemma The case f =I is similar .(2 3) Trivial .(3 1) We argue the contrapositive . If f is not on the far side of F , then by Lemma

f € Since f cannot be on fGArxl . If or theng (x,f can be extended to an extreme point of F distinct from f. Otherwise by theTriangle Lemma , for some shadow f of g (x ,f g (x ,f) intersects g (r, l) at a point distinct from f, thus ff cannot connectR (F) to GU. 0

Corollary Suppose {fl} is a set of distinct points on the far side of F from xand , for each i , fl is a shadow of Then the order of f,~ on GU is the same asthe order offl on GR (F) .Proof: If i aej then g (x,fi) and g (x ,fj) must diverge before reaching f} or f} . Henceff , does not intersect fif} . The result then follows from the Connection Lemma.

0

Lemma Suppose u ,v €GU, uaev, and Then

constitute a counterclockwise traversalof the boundary ofProof: In this proof, we write for example rH(x) and lH (x) for the clockwise andcounterclockwise extreme points of H from x . This indicates the dependence upon Hexplicitly . Thus in the statement of the Lemma, r(x) really refers to rp (x) . We also

- 15

assume GU[u,v ]g R (F) , hence some point of GU[u,v ] is extreme inIf GU (F) , a similar and easier argument suffices .

Let G be {u, v} together with the set of convex corners of G U ClearlyWe first show that we will use Lemma

Some point of F is in the relative interior of U [v,u ] by assumption ; it iseasy to check that in fact rp(v) must be in the relative interior of U HenceFUGU[u, v]g U Any geodesic extending must stay in the relativeinterior of U hence must avoid and must also avoid any f

'GF distinct

from rp (v) since rp (v) is extreme in F. Hence by LemmaBy Lemma rp (v) is extreme in FUG . By a similar argument,

is extreme in FUG .

We claim is the most clockwise point of G (ordered along GU)that is extreme in FUG: h is either v, the convex corner immediately clockwise of v,or possib ly u if GU[u,v ] contains no convex corners . We have and

This is only possib le if v, h and rp (v) lie on a common geodesic.Since h and rp (v) are extreme in FUG , they must be the endpoints of the geodesic.Hence either or extends to Certainly

so v €GU and force and not

extendib le to another point of FUG forces h to be u or the convex corner of Gimmediately clockwise of v.

Set - 1) Let G'

{S u n -net} be thesubset of G—F extreme in FUG , where the index is given by the ordering along

We want to show that e G maps f l ~ f2 -o f;, ~ g 1 g, ~ f1 .

We have already shown that lpug maps f l to g ) , hence By a sim ilarargument rpugcfh)=g 1 . It is easy to see that rpug (g ;) can only be g. for 1< i<l .

It remains to show that rpugm) for 1< i<h. This follows if we establishspan (b (which also establishes f, extreme in F Sincefh= lp (u) is extreme in FUG , ft. is extreme in FU{v} , hence fh is on the far side of Ffrom v

,and hence are on the far side of F from v. Similarly are on

the far side of F from u. Since r and l preserve order on GU, the far side of F fromany includes In particular , for any and any i ,

1< i<h, we have (since f, extreme in F and and (as

in the proof of Lemma g 9(f,,R (F ) )Hence

Now if g 1=u and g,=v, we are done . If, say , g l aeu, then u must appear on

To obtain the Lemma Split into andthen merge g (u,g 1) with GU[g h gk] . A similar split applies if gfiev. D

The General Posi tionAssumption

For the remainder of the paper we mak e the following general position assumption: no reflex corner of GU is equidistant from two sites . This condition can alwaysbe satisfied by applying a slight perturbation to the positions of the sites or corners .If this assumption is not made , the set of points equidistant from two sites may

- 16

include a two-dimensional region [A87] . This section contains some consequences ofthe general position assumption that are critical to the rest of the paper . Weemphasize that none of Lemmas and hold if the assumption isremoved .

Lemma If s and t are distinct sites , x 6U, and thenProof: Suppose Then the geodesics g (s ,x) and g (t,x) share their finallink yx. Point y must be the anchor of x wi th respect to both of s and t. Now y¢ s , tsince otherwi se would imply s =t. Hence y must be a reflex corner ofGU equidistant from s and t, contradicting the general position assumption . D

Lemma Si ppose u, v €U, utev , and each of u and v is equidistant from distinct

sites s and t. Then g (s ,a) does not intersectProof: Suppose x lies on both g (s , a) and Without loss of generality , assumed (x,s) 5 d (x, t) Observe that

dv(t)=dv(S) 5 4 0 ) +4 (L S)d (x) +4 (x , t)=dy (t)

Hence path is in fact geodesic and similarlyis Since x lies on both geodesics g (s ,a) and x must be

a reflex corner of GU or an interior point of a comm on link of the two geodesics connecting two reflex corners of GU. In the second case the link must be common tog (t,a) and g (t,v) as well , hence in either case we can find a reflex corner of GUequidistant from s and t, violating the general position assumption . 0

Lemma There is at most one point equidistant from three distinct sites .Proof: Suppose to the contrary that points it and v are both equidistant from sitesr ,s , t. First note that r,s , t cannot lie on a common geodesic, for if say thenby Lemma Hence r , s , t are extreme points of

We claim that r,s , t are extreme points of (and also of In

order to demonstrate this , we show Now r does not lie on g (a , s ) orelse s or t would be further from u than r . As argued before , r does not lie

on If r is in the interior of then by the Triangle Lemma , g (uflintersects where is a shadow of But then using Lemma again ,

we would have

Now suppose one of u and v, say u, is extreme in Hence r,s , t,u

are extreme in assume that they appear in that counterclockwise order .By Lemma and It must he

‘

thatintersects either g (v, r) or assume it is To obtain a

contradiction of Lemma we show in fact g (u,s ) intersects Now uu'

intersects only at u since u is extreme in Hence uu' is

disjoint from A lso s"intersects only at s ,since s is extreme

in and some portion of g (a, s ) must lie in hence inSince g (v , t)QR g (v, t) must intersects

If neither it nor v is extremal in then both and thePr°°f is sim ilar . Geodesics and g (u, t) split into three geodesic triangles . Hence v lies in one of the triangles which again implies a

- 17

contradiction of Lemma D

Voronoi Cells

The bisector b (s , t) of distinct sites s and t is and the halfspace closer to s , H is Clearly H and H (t,s)form a partition of U. A breakpoint of b (s ,t) is the intersection of b (s , t) with a shortest path partition edge from s or t. Figure indicates the bisectors of threepoints , wi th breakpoints marked .

Figure Bisectors ; dots indicate breakpoints

Lemma [AS7 , Bisector b (s , t) is a smooth curve connecting two pointson GU and having no other points in common with GU. It is the concatenation of0 (n) straight and hyperbolic arcs ; the points where such arcs meet are precisely thebreakpoints of The tangent to b (s , t) at x bisects the angle between 0(x,s) and

In particular , together wi th Lemma this implies that, given9(x ,s) and enter H (s , t) at x , while and 0(x, t) enter H (t,s) (if theystay within U, that is) .

Corollary The relative boundary of H (s , t) and H (t,s) is

Lemma [AS7 , H (s , t) is path-connected .

Recall that S is the set of sites . The (geodesic furthest-site) Voronoi cell of site sis The (geodesic fitrthest-site) Voronoi diagram V is

and Figure indicates the Voronoi diagram and Voronoi cellsI’

of the three points depicted in Figure A Voronoi edge e(s , t) is ifthe intersection consists of more than one point (else we wi ll say that e (s , t) does notexist) . A (Voronoi) vertex is a point x €V which has three or more sites furthest fromit. By Lemma above , there is at most one such point x for each triple of sites .A hitpoint is the intersection of a Voronoi edge with GU. Intuitively , a hitpoint

correspond to the"point at infinity" of an infinite Voronoi edge in a Euclidean

furthest-site Voronoi diagram .

Lemma Voronoi edge e (s , t) is connected and has vertices or hitpoints as endpoints .

- 13

Figure Voronoi cellsDashed lines appear in refined Voronoi diagram

Suppose r ,s , t are distinct sites . Since d, is continuous , a connected componentof must have for each of its endpoints either an endpoint of

i .e. , a hitpoint, or a point equidistant from s , t,r . But by Lemma .3 , thereis at most one point equidistant from s , t, r. so consists of asingle connected component . Hence also D‘

(b (r ,s)UH is conr s,

nected and has hitpoints or vertices for endpoints . 0

Lemma Suppose are the sites furthest (and thus equidistant)from vertex v, and directions are in counterclockwi se order . Then ,

for each i , edge is incident to v and extends away from v in directionbisecting Lsi x l vsi , as long as that direction ( locally) stays inside U.

Proof: E lementary analysis , using D

If vertex v appears on GU, then there is only one edge of V incident to v: as vcannot be a com er of GU by the general position assumption , it must be an interiorpoint of a wall . Hence only the edge bisecting s t remains within U, where dif eotions 0(v, s ) and 0(v, t) are the most clockwise and most counterclockwi se directionstowards sites furthest from v, respectively .

Lemma (Extension Lemma): If x lies on g (s ,y) and x €V(s) or forsome other site t, then all of g (s ,y) past x lies in V(s) .Proof: Suppose xEV(s) . For any site rat s and for any d, (z)

so Suppose and y aex . Bythe general position assumption x is not a corner so Since y¢x ,

direction 9(x,y) stays in U at x . We show 0(x ,y) enters V(s)= QH (r ,s ) at x; ther 3

result follows as before . Since for any rat s , If

enters H if d, (x) any direction ( locally) stays in H

An immediate consequence of this Lemma is that every point in a Voronoi cellis connected to GU: if then segment where 55 is a shadow of

Lemma Both V(s)fl GU and V(s ) are path-connected .

Proof: Since every point of V(s) is connected to a point of V(s)flGU, it suffices to

https://www.forgottenbooks.com/join

- 20

respectively . We show that x , s , t are distinct and appear in that order counterclockwise around GU. Note that by Lemma and are

geodesics emanating from x with distinct initial directions ; hence fs'

iet. Near x ,and also O(x ,s) enters H (s , t) and

0(x , t) enters H Since g(x, s) does not intersect g(x , t) again , the ordering of x ,s , tmust be counterclockwise around GU.

We claim 5, E, and t are in that counterclockwise order on GU, where E is theclosest shadow of Let r and I be the clockwi se and counterclockwise extremepoints of S from x , with 7 and I closest shadows of g (x, r) and respectively .

By Lemma GU[FJ ] does not contain x. By Lemma s , t are on the farside of S from x . Hence by Corollary and in fact the counterclockwise order must be 7 r, t, 1, since GUE t] does not contain x . By Corollary

r , s , t, I appear in that counterclockwise order on C ; since it appears between sand t, u is on the far side of S from x . Again by Corollary 37, u, t appear inthat counterclockwi se order on GU.

Now suppose contrary to the Lemma V(u) is not empty ; we will obtain a contradiction. Since Voronoi cells are connected to GU, there is Now yaexsince Voronoi cells are relatively O pen , and x lies on the boundary of V(s) (and

As s , u, and t are distinct extreme elements of S, theset is extreme . As

x €GU, it is impossible for x to lie in the interior of Asut. Moreover , s , u, and t areon the far side of from x ( in that counterclockwise order) . Thus either

or is extreme (with s , u, t, x , occurring in that counterclockwiseorder) . In either case g (x,u) intersects g (s , t) and enters U [s , t] at some point u'

(as

u is an extreme point of C lying counterclockwise between s and t) . Assumethe case is similar . Lemma implies that g (y , t) does not

intersect As t lies in the relative interior of U [u,x ] and y lies in Ug (y , t) must intersect an at some point u”$ u. This implies that the portion of g(x ,u)

from u' to u

" inclusive is contained in the triangle Ayst; in particular uEAyst. ByLemma this contradicts the choice of uses , t as a site furthest from y . D

Corollary (Ordering Lemma): The ordering of sites with nonempty Voronoicells around GC is the same as the ordering of Voronoi cells around GU.

Lemma Suppose and are separated by boundary geodesic Then there is a total of at most o(m+n) distinct links in pathsi 1, ,m .

Proof: By a proof essentially identical to Lemma 4 of Suri for each i there areat most three links of not in T (Three links are needed rather thanone because u, v need not be corners of GU.) D

Lemma There are three boundary geodesics g 1 ,g 2 ,g 3 so that for any pointaGGU and any site b furthest from a one of the geodesics separates a from b .

Proof: Pick xGS arbitrarily . Let y be a site furthest from x and z a site furthest fromy . We argue the case that x ,y ,z are distinct and appear in that counterclockwise order

- 21

around GC; the case that x =z is easier and the case that the order is x ,z ,y is similar .

We claim that we canchoose a site w furthest from 2 in ac We first showthat w can be chosen to lie in GC if we can only choose a w furthest from 2 inGC [z ,x ] then in particular We also have and

so adding all three we get Thiscontradicts the triangle inequality as can be seen by considering a point in the intersection of g (y ,w) and Now w cannot be in GC [y , z l else g (x,w) intersects a contradiction of Lemma taking u=x , t

=w , v=z, and s =y .

Hence we can choose w in

Let and y , z ,w be the closest foreshadows and closest shadows ofand respectively . We claim y

‘ this is immediate ify €GU or g (x,y) and g (z ,y) share a common final link ; otherwi se it follows fromLemma since y

' is the closest shadow of Similarly z ‘EGU[y°

,fl and

Now by the Extension Lemma , x'

, y‘

, 2. have y , z ,w as their respective furthest

sites . By the Ordering Lemma , all points in have furthest sites in GCLy, z] ;clearly g(x,y) separates GU from Sim ilarly the sites furthest from

lie in GC [z ,w ] and these two sets are separated by Finally , thesites furthest from lie in GC [w ,y] and these two sets are separated by

Note that in fact x ‘ can have furthest sites both in ac [y , z ] and ac andsimilarly for y ‘ and z ‘, while each of the remaining points of GU has its furthest sitesin exactly one of GC D

TheRefined Voronoi Diagram

The Voronoi diagram V clearly has 0 (k) edges , since together with GU it formsa planar graph with at most k bounded regi ons and all Voronoi vertices have degreethree or more . However , this is not an accurate description of the size complexity ofthe Voronoi diagram , since each Voronoi edge may consist of sections of several different hyperbolic arcs . This section discusses a refinement of the Voronoi partition ,

obtained by further subdividing each Voronoi cell V(s) by the shortest path partitionfrom s . Each bounding edge of a refined Voronoi cell is a line segment or a sectionof a single hyperbolic arc. The main theorem is a linear bound on the size complexity of the refined Voronoi diagram . This implies an 0 (n+k) bound on the size com

plexity of the Voronoi diagram itself.

The refined Voronoi cell of site s with anchor a , is V(s )flPa(s) . Therefined bisector edge is flPb(t) . The refined partition edge (from

s with anchor a) , is V(s ) O pa(s) . A refined Voronoi edge is a refined bisectoredge or a refined partition edge . Observe that distinct refined Voronoi edges are disjoint (except possibly at their endpoints) . Empty refined Voronoi cells and edges andrefined Voronoi edges consisting of a single point are disregarded .

Suppose is not empty . It is easy to see that each endpoint of iseither a vertex of V, a hitpoint, or a breakpoint. Moreover , does not contain

- 22

breakpoints (except possibly as endpoints) . Consequently is a hyperbo lic areor a line segment .

Lemma E ither p;(s) is empty , or it is all of pa(s) , or it has an O pen endpointat a breakpoint of e(s , t) for some site t and closed endpoint on GU. (The latter twocases are illustrated in FigureProof: Suppose p ; (s) is not empty and is not all of pa(s)=ay (where yGGU) . Thenpa(s) must intersect some edge e(s , t) before first entering V(s) . By the ExtensionLemma, the intersection is a single point x , and —{x} is contained in V(s) .

El

Suppose p ;(s) is a refined partition edge of V(s ) . Let be the first linkof the geodesic directed towards s . Then by definition of the refined partition edge pa(s) and are collinear and meet at a. We claim that if

p;(s) and are distinct refined partition edges , then and aredistinct

,at least as directed links . This claim certainly holds if aaea'

. If a=a'

, thenwe must have s¢s ’ by the previous Lemma . Since andwe must have Since and have these two directions , they must be distinct .

Lemma Link is a link of where v can be chosen to be a cornerlying in GUflV(s) or a hitpoint of V(s) (i .e. an endpoint ofProof: Let Segment ay partitions U into two polygonalregions U1 and U2 so that U1 contains s and every geodesic to s from a point in U;contains If we traverse GU0 U2 starting at y , we must encounter either theendpoint of GUflV(s ) or a corner of GU lying in V(s) . D

The refined Voronoi diagram , V‘

, is the union of all refined Voronoi edges . Avertex of V' is a vertex of V or a breakpoint . A hitpoint of V

‘ is a hitpoint of V orthe point of intersection of a refined partition edge with GU ( in the case when therefined partition edge coincides with a shortest path partition edge , only its nonanchor endpoint is considered a hitpoint) .

Lemma There are at most 0 (n+k) refined Voronoi edges and vertices .Proof: Clearly V has k cells , hence k hitpoints and 0 (k) edges . We show that thereare only 0 (n+k) refined partition edges . Since each refined partition edge contributes a single breakpoint , it follows that there are only O (n+k) refined bisectoredges . By planarity , there are only o(u +k) vertices in V

‘ as well .

By Lemma there are three boundary geodesics g 1 , 3 2 , g 3 and a partitionof GU into three fragments UI , U2 , U3 so that for i g ; separates every pointin U.from its furthest sites . (This is not strictly true for the endpoints of Up, but theargument is similar .) Let W , be the union of the sets of links of geodesicswhere v is a corner of GU lying in U; and w is the unique site furthest from v orwhere v is a hitpoint of V lying in U, and w is one of the two sites whose Voronoi cellboundary contains v. Since there are k hitpoints and n corners in all of GU, there arecertainly at most as many in each U, By Lemma there are 0 (n+k) links inW , Now modify W , so that it contains for each link two oppositely directed links ;this doubles its size . By Lemma if p; is a refined partition edge , then is

- 23

in W ; , for some i Since each t(p;) corresponds to a unique refined partitionedge , there are at most o (h +k) such edges . D

Directing Edges of theRefined Voronoi Diagram

We let c be the center of C . Necessarily c lies on some Voronoi edge , sincethere must be (at least) two sites attaining the maximum distance from c .

Lemma Suppose s is a site furthest from xEU. Then the angle between 0(x,s )and 9(x , c ) is less thanProof: Suppose the angle between 0(x ,s) and 9(x ,c) is at least By Lemma

But since s is a furthest site from x , Thiscontradicts the choice of c as the center of C . D

We use this Lemma to direct each refined Voronoi edge towards c ( if there is aVoronoi edge containing c in its interior , we split the edge at e) . We show how todirect an edge e(s , t) of the (unrefined) Voronoi diagram ; this direction extends toeach refined bisector edge . A similar argument directs each refined partition edge .

Suppose xaec . Since e(s , t) bisects Lsxt and both the angle between 9(x,c)

and 0(x , s ) and the angle between 0(x ,c) and 0(x, t) are less than it is not possible that 9(x ,c) is perpendicular to e(s ,t) at x . Hence we can direct e(s ,t) towards clocally at x . If this direction extends globally to e(s , t) (since directiontowards c is a continuous function away from the corners and can never be perpendicular to If then e(s , t) is split at c into two pieces , each of whichis consistently directed towards c .

Lemma Suppose vEGU is a vertex of V‘

. Then there are at least two edges ofV

' entering v and exactly one edge of V‘ leaving v.

Proof: We consider the case that v is in fact a vertex of V; the case that v is a breakpoint is similar . For simplicity assume there are exactly three sites equidistant fromv. Label the sites r ,s , t so that and 8(v, t) are in counterclockwi seorder leaving v so that e vt< 17 . (This is possible by the previous Lemma as each ofthe angles formed between 0(v,e) and e(v,s) or e(v, t) has measure less than

By Lemma Voronoi edges and c (t, r) are incident to vand extend in directions bisecting angles s r, Atvs , and Lrvt, respectively . Henceedges e(r ,s) and e(s ,t) enter v and e(t, r) leaves v. The general case of a vertex ofarbitrary high degree is handled analogously . Vertices vGGU of degree two or lessdo not occur by definition of V‘

. 0

Suppose Voronoi edge e(s , t) intersects GU at hitpoint x ; recall that by the general position assumption x is not a corner of GU. Then e(s , t) is directed into the interior of U, because 0(x, c) makes an angle strictly less than with both 9(x ,s ) and

and none of 0(x, t) and 0(x ,c) can leave U at x . A similar argumentshows that refined partition edges are directed away from GU at hitpoints.

Coroll ary The unrefined Voronoi diagram V forms a (directed) tree with rootc and edges directed towards c .

Proof: No cycles are possible because otherwise some Voronoi cell would beseparated from GU. O nly c has out-degree zero ; every other vertex or hitpoint of V

- 24

has out-degree 1, so V is a root-directed tree .0

The refined Voronoi diagram V' consists of V together with refined partition

edges . Each such edge must lie entirely within a single Voronoi cell , having one endpoint on GU and the other endpoint at a reflex corner or on V (by Lemma

3 . TheAlgorithm

Procedure gfvInput: A polygon U with n sides and a set SCU of k sites .O utput: The refined furthest-site geodesic Voronoi diagram V

'

.

1. Triangulate U.

2 . Compute C , the relative convex hull of S, and discard all non-extreme sites of S .

3 . Determine two or three two-fragment instances so that the union of the source fragme4 . Compute V

‘

flGU by calling for each two-fragment instance5 . Call sweep to extend V

'

to the interior of U.

Figure

This section contains the algorithm for computing V'

. An outline of the algorithm is given in Figure Step 1 can be performed in time 0 (nlogn) [GJ PT78] .The relative convex hull computation of the second step can be accomplished in time0 ((n+k) log (n+k)) [T86] . The third step also takes time o((h +k) log (n+k)) and isdescribed in Lemma below . The fourth step , the most difficult of the algorithm , is the computation of V

' restricted to G U. It is discussed in Sectionsthrough The last step is the extension of V‘ to the interior of U. This is doneusing a "reverse geodesic sweeping" algorithm , discussed in Section below . Boththe fourth and fifth steps take time 0 ((n+k) log (n

The computation of V‘ restricted to GU is quite similar in outline to Suri ’ s algorithm for furthest geodesic neighbors inside a simple polygon[387] . We first reducethe computation of V

' on GU to at most three instances of the "two-fragment problem .

"Roughly , an instance of the two-fragment problem consists of a fragm ent of

GU and a fragment of GC so that all furthest sites of points in the fragment of GU arecontained in the fragment of GC . Such a pair must also sati sfy a technical conditiongiven below ; this reduction appears in Section The algorithm to solve the twofragm ent problem is based on a divide-and-conquer schema that splits an instanceinto two smaller instances . The basic properties of the divide-and-conquer schemaappear in Section Section contains the exact splitting method and the procedures for handling the base cases of the recursion . The complexity analysis appearsin Section We show that the sum of all instance sizes at each level of recursionis linear in n+k. This implies over all 0 ( (n +k) log(h + k) ) running time .

- 25

We work with polygonal relatively convex sets in addition to simple polygons .Any such relatively convex set Q can be decomposed into a collection of plateaus andbridges . Clearly a triangulation of Q can be obtained just by triangulating each plateau in the decomposition . This is easily done in time 0 (m logm) [GJ PT78] , if m isthe number of segments in the boundary of Q. Similarly a shortest path partition ofQ from an arbitrary point in it can be obtained by using a shortest path partitionalgorithm in each plateau of the decomposition . If Q has been triangulated , thistakes time 0 (m)

The Two-Fragment Problem

A two-fragment instance is a quadruple where u,véGU, uaev , s is a site

furthest from u, t is a site furthest from v (possibly s = t) , and g (u, v) separatesGU[u, v] from The two-fragment problem is Given two-fragment instance

compute Observe that by the Ordering Lemma , only theVoronoi cells of sites in GC [s , t] can intersect GU The sourcefragment of twofragment instance is GU the targetfragment is GC

Lemma There exists a set of at most three instances of the two-fragment problem so that the union of the source fragments is GU. The instances each have sizeo (h +k) and can be computed in time o ((h +k) log (n+k)) given a triangulation ofU.

Proof: Choose x , y , z, w , x‘

, y‘

, z' as in the proof of Lemma It is clear that

and are two-fragment instances each of size atmost 0 (n+k) ; their source fragments cover GU. As for computing them , the choiceof x was arbitrary . Site y can be determined in time o((u +k) log(h +k)) by computing the shortest path tree from x , then determining the distance from every site to xusing a planar point location algorithm in the resulting shortest path partition . Sites 2and w can be determined similarly . Surely can be computed in additionaltime o(u) . The case when x =z is handled similarly . G

Let D be the relative convex hull of the sites on GC Clearly the orderingof sites on D is the same as on C , with the addition that s immediately follows t incounterclockwi se order . Let be the relative convex hull of GU [u,v ] and D .

SeeFigure

We say is degenerate if D is contained in If is degencrate , there can be no sites in D besides s and t and , by Lemma the orderalong g (u, v) must be u, t, s , v. In this case we define l =t and r =s .

Suppose is not degenerate . Let I be the counterclockwise extreme pointof D from u and r the clockwise extreme point of D from v. By LemmaGU GD [r , l and g (l ,a) constitute a counterclockwise traversal of theboundary of It is an imm ediate consequence of the following Lemmathat GD [r , l ] is a subpath of

Lemma Sites s , r , l , t are in that counterclockwise order on D , not necessarilyall distinct .Proof: The Lemma is trivial if r = l , s= t, or if is degenerate , so assume s aet

- 26

and is not degenerate . We show that if s appears on GD [r, l ] then r=s andif t appears on GD [r, l then t= l . Since s is the extreme point of D immediatelycounterclockwi se from t, this implies the Lemma .

1

Suppose t appears on GD [r, l ] and t iel . Since s is immediately counterclockwisefrom t, s is also an extreme point of D lying on GD [r, l ] and thus an extreme point of

Then t, s , u, and v appear in that order on a counterclockwise traversalof GR which by Lemma implies that g (t,a) meets contrary toLemma D

We wi sh to give a definition of "left side connector and "right side connectorto capture the bounding edges of not in GC [r, l ] and GU The obvious definitions are g (u, l) and respectively . Unfortunately , these definitionsare not adequate . In Section we analyze the size of side connectors ; one crucialproperty used in our argument is that if s iet, then the left and right side connectorsare disjoint except possibly at their endpoints (Lemma below) . Unfortunately ,this is not true for side connectors defined as g (u, l ) and See Figure

(a) (b)Figure So lid outlines D ; dashes outline R

It is clear that geodesic g (v, r) has connected intersection with GD ; furthermoreif the intersection is more than a point it must be some final portion of geodesic

r’ the site of D immediately clockwise of r. Let f be r if r=s , otherwise let

be the first point of intersected by The right side connector ofis Sim ilarly , we define the left side connector of to be

g (u ,t) where l is I if l = t, otherwise I is the first point of intersected bywhere l' is the site of D immediately counterclockwise from I. See FigureA connector edge is a link in either the left or right connector .The size of denoted I , is plus the sizes

of the side connectors . Since GD [r , l ] is a subpath of it is clear thatIGR IS I .

Lemma If the side connectors of a two-fragment instance meet at apoint other than one of their endpoints , then s= r.

https://www.forgottenbooks.com/join

- 28

partition of from w containing r is known . The cell containing r for allsites r in GC [s ,t ] can be determined in total time as follows . Weassume the shortest path partition of from w has been computed andrefined to a triangulation (this takes only linear addi tional time) ; hence each cell is atriangle . First loca te the triangle containing s ; this clearly can be done in the allowedtime bound . Then traverse GC in one step moving to the next vertex of GC [s , t]

or to the next intersection of the current edge of GC [s , t] wi th the boundary GA of thecurrent triangle A of the shortest path partition from w. Notice that the intersectionof GC with A has at most three connected components , since C is relatively convex .

Hence the traversal of GC [s , t ] takes total time 0 ( I ) , since the charge for astep to a vertex of GC [s , t] can be allotted to the vertex and the step to an intersectionwith A can be allotted to one of the three connected components of AflGC

We compute the boundary and triangulation of f (w)) as follows (handling R (w , v,f is similar) . If is degenerate , both the boundaryand triangulation of f (w)) can be easily obtained from the shortest path partition of from w. Otherwi se compute r’ , the clockwise extreme point ofGC [s ,f (w)] from w using the shortest path partition from w. Similarly l ' , the counterclockwise extreme point of GC [s ,f (w) ] from u, can be determined by computingthe shortest path partition of from it . Now since I' is extreme in

sp lits into two pieces , one piece lying to theleft and one to the right . (Possibly one or the other is just A triangulationof the piece lying to the right can be obtained by refining the shortest path partitionof from w . Similarly a triangulation of the piece lying to the left can beobtained by refining the shortest path partition of from l

'

. Notice thelinks in and are used as triangle edges in this triangulation .

The left and right side connectors of are easily determined fromand This computation can clearly be done in time 0 ( I ) .

Choosing splitting points and the base cases

If u and v do not lie on the same wall of GU, then spli tting point w is chosensimply as the corner of GU[u,v ] so that IGU[u,w] I is within one of

If GC [s , t] is a single site , i .e. , s= t, then all of GU[u, v ] lies in V(s) . We need to

find the refined parti tion edges of V' intersecting GU it is sufficient to compute

the shortest path partition of from s , which can be done in time0 ( I ) given the triangulation of

It is possible that s iet but u and v lie on the same wall of GU. In this case thereis no obvious splitting point w . We perform a ”partition step: segment uv is splitinto subsegments so that within each subsegment the shortest path tree from any

point on the subsegment to the sites on GC [s , t ] is combinatorially invariant . Thispartitioning is described below ; it results in subsegments and takestime 0 ( I I ) . We introduce the partition points as dummy vertices , and use them as splitting points in the divide and conquer . Notice that the partitioning is performed at most once on a path from the topmost instance to a leaf

- 29

instance in the recursion tree . Hence we introduce only o(u +k) such points , sincethe sum of instance sizes at each level of the recursion tree is 0 (n+k) (see theanalysis in Section Similarly , the total time to compute the partitioning points iso ( (h +k) log(n

The remaining problem is to handle a base case instance where s aetand the shortest path tree is combinatorially invariant on segmentBecause of this invariance , no refined partition edge of V

' intersects segment uv,in

other words uvflV‘=uvflV is the set of bisector hitpoints on uv. Notice that, the

partition induced on segment uv by V' is exactly the partition induced by the upper

envelope of the functions d, , where r is a site in Again because of the combinatorial invar iance of the shortest path tree , each function d, is

"simple" on segment uv; specifical ly d, (x) is of the form c l + Vc 2(x) where C 1 is constant and c 2(x)depends quadratically upon the position of x on segment uv. (Observe that the purpose of partitioning the original wall was to ensure that c l and c ; are fixed over thelength of uv. Their values for each site r €GC [s , t] are defined by the identity of theanchor of x €uv wi th respect to r and the distance from r to this anchor ; all of thisinformation can be determined in linear time from the shortest-path tree of

from , say , u.) Thus in constant time it is possible to determine , for a pairof sites r and r

' the (unique) point x 6uv, if any , for which Now by theOrdering Lemma , Voronoi cells appear along segment uv in the same order as thecorresponding sites appear along This implies that the upper envelope of thefunctions d, on segment uv can be computed in time proportional to the number ofsites (which is certainly For example , an incremental algorithm issufficient . Suppose that the partition of segment uv induced by an initial subsequenceof the sites on GC [s , t] has been computed . Then the partition induced by adding thenext site in order can be determined in constant time plus time proportional to thenumber of cells deleted from the partition of segment uv computed so far .

We now describe the partition step , which uses a technique similar to that ofWe actually partition segment uv so that within each subsegment, the

shortest path tree to every site of GC [s , t] and every vertex of GR is combinatorially invariant. To do this , compute T(u) , the shortest path tree offrom u. Include in T(u) the link s of geodesics from u to sites of GC [s , t ] not appearing on (there can be at most one such link per site not already inThe augmented T(u) can be computed in total time using the technique described for computation of f (w) in Section Similarly compute T(v) , theaugmented shortest path tree from v.

For each 2 , 2 either a site of GC [s , t] or a vertex of compare 9(z ,u)with This is possible using T(u) and T(v) . If do nothing . If 2

lies on segment uv do nothing . O therwise geodesic triangle Azuv must actually be asimple polygon (since uv is a wall) . Pass a line through the first link of itmust hit segment uv since the interior angles of Azuv are reflex except at z ,a , v. Theintersection point is a partition point. Similarly obtain a partition point from the firstlink of It is easy to see that the partition points generated in this fashion

- 30

form the required partition of segment uv. Generating the partition points takes0 ( I ) time and sorting them requires 0 ( I log I ) time .

Complexity analysis

Lemma Let be a topmost two-fragment instance . There are atmost 0 (n+k) distinct connector edges among all subinstances ofProof: By the discussion of the preceding section , there are , over the course of algorithm execution , only o (h +k) subinstances of the two-fragment problem , hence asmany connectors . Since g (ub vj) separates GU[aim] from it separates theendpoints of each connector as well . Hence by Lemma there are at most0 (n+k) distinct connector edges . 0

Lemma Let be a topmost two-fragment instance . At each level ofrecursion , each connector edge appears in only a constant number of subinstances of(ui svi sSi J i) .

Proof: We count the number of times edge ab can appear as a left connector edgedirected from a to b . Notice a and b may be assumed to be reflex corners of GU, forotherwi se ab is necessarily a first or last link on a side connector , and hence can

appear in only one left connector (three if ab is the last link and the target fragmentconsists of the single site b in two of the instances) . Let Q, be the set of points x ofU for which g (x ,b) passes through a. Similarly , define Q , as the set of points thatcan reach a only through b. Clearly Q, and Q .are non-empty and disjoint . Let A be

and B be GC It can be checked that A is a single fragmentof GU and that B is a single fragment of GC . Then for and t€GC [s ,-, t,link ab appears in g (u, t) exactly if uEA and t€B .

We first claim that at each level of recursion there are at most two nonleafinstances of the two-fragment problem for which both the source fragment intersectsA and the target fragment intersects B . To see this supposeare instances at the same level , GU[uj ,vj]flA4 and u 1 ,u2 ,u3 (andhence s 1 ,s 2 ,s 3) are in that counterclockwi se order . Since these instances are all atthe same level of recursion , v2 appears between u2 and u3 (possibly v2

= u3) and s ;appears between t1 and t; (possib ly s 2

= t1) . But then v2 €A, széB , so the geodesicg (v2 , s 2) contains link ab . Recalling that the relative interior of ab does not lie in Q1,and hence is disjoint from we deduce that the right connector of

contains the relative interior of ab . Since g (u2 , t2) also contains ab , the

relative interior of ab lies in the left connector of this instance as well . Hence s2=t2by Lemma and is a leaf instance .

We now claim that at each level of recursion there are at most four instances ofthe two-fragment problem ( leaf and nonleaf) with ab appearing as a left connectoredge . Such an instance must have 14 6A and tEB . This is only possib le if ,for its parent instance , the source fragm ent intersects A and the target fragment intersects B . As just argued there are only two such parent instances . 0

Theorem V'

flGU can be computed in time 0 ( (n +k) logProof: Clearly U can be triangulated in time 0 (n log h ) [GJ PT78] . By Lemma

- 3 1

there are three two-fragment instances with union of source fragments equal to GUthat can be computed in time 0 (n+k) . We show that rgfs so lves the two-fragmentproblem in total time 0 ((n+k) log(u +k)) for each (top- level) instance , proving theTheorem .

Consider the work performed by rgfs for all instances at a particular level ofrecursion , ignoring recursive calls and the time required to partition walls as discussed in Section It is linear in instance size which is the sum of the sizes ofsource and target fragments and the sizes of the side connectors . The total size ofsource and target fragments at a particular level of recursion is 0 (n+k) , becausesource and target fragments are partitioned disjointly except for endpoints , and thereare only o (h +k) possible endpoints . By Lemmas and the total size ofall connectors at a particular level of recursion is also o (h +k) . Hence the total workat a particular level of recursion , summed over all instances at the level , is o (h +k) .

The total depth of recursion is 0 ( log (n at each step except for partitionsteps , the size of a source fragment is split in half. At a partition step , the size of thesource fragment increases to at most o(u +k) , and a partition step happens at mostonce on a path in the recursion tree from topmost instance to leaf instance .

The total work required for partitioning is o( (u +k) log (n Hence the totalwork to solve a single top- level two-fragment instance is o ((h +k) log (n CI

Computing V'

.

V‘ is computed by the procedure sweep (Figure which is a reverse geo

desic sweeping algorithm”; it progresses from GU towards c , the center of C .

Theorem Procedure sweep computes V'

. It can be implemented to run in time0 ((n+k) log (n+k)) and SpaceProof: For positive real 2 , let Dc(z) be the geodesic disc of radius 2 centered at c , i .e

the set of all points of U at geodesic distance at most 2 from e. We claim that thewhile loop maintains the invariant that L is exactly the refined Voronoi edges intersected by GDc(r) , in order around GDc(r) . This follows from Lemma usingstandard sweepline arguments Hence procedure sweep computes V

'

.

List L can be implemented simply as a circular doubly-linked list , so each listoperation takes constant time . Q can be implemented as a heap , so that each operation takes time 0 (log(n The geodesic center c of C can be computed in time0 ( (n+k) log (n+k)) as follows . Pollack , Sharit , and Rote [PSR87] show how to

compute the center of (the set of vertices of) a simple polygon ; their algorithmextends easily to a polygonal relatively convex set as well . Since geodesics restrictedto be inside a relatively convex set (with respect to U) are identical to geodesicsinside U, it suffices to compute the center of C

The shortest path partition of U from c can be computed in time 0 (n) since U istriangulated [GHLST] . Given the shortest path partition from c , the geodesic distance from a point x €U to c can be computed in time 0 ( logn) , using a planar subdivision search algorithm to locate the shortest path partition cell containing x ( such as

- 32

Input: Triangulated polygon U, refined Voronoi diagram V‘

restricted to GU.

O utput: v‘

.

Data structures

L: a doubly- linked circular list of refined Voronoi edges .Q: a priority queue of points of U, ordered by decreasing geodesic distance from c .

procedure sweep

Compute c , the center of C .

Compute the shortest path partition of U from c .

Initialize L to be dunv‘

Initialize Q to contain all pairwi se intersections of refined Voronoi edges immediately adjacent in L and all anchors of the refined partition edges appearingin L .

whileQaeQ

Extract from Q the point v of maximum geodesic distance from c .

Delete from L all refined Voronoi edges with head v.Delete from Q any intersections and/or anchors involving just-deletededges .if v is not an anchor thenInsert into L the refined bisector edge e with tail v.

Insert into Q any new intersections of e wi th adjacent refined Voronoiedges .end if

end whileend sweep

Figure Procedure sweep

that of By Theorem GU(1V'

can be computed in time0 ((n+k) log (n Hence L and Q can be initialized in total time0 ((n+k) log (n

Each iteration of the while loop uses a constant number of operations on Q andL and one geodesic-distance computation for each item inserted in Q; hence eachiteration takes time 0 ( log (n By Lemma there are only 0 (n+k) iterations of the while loop . Thus the running time of the entire algorithm is0 ((n+k) log (n Clearly the space usage is 0 (n+k) . D

Acknowledgement

We wish to thank M icha Sharit for suggesting the idea of the proof for Lemma

- 33

References

IAS7]

[AT86]

IBO 79]

[CG85]

B . Aronov , "On the geodesic Voronoi diagram of point sites in a simplepolygon to appear , Algorithmica. A prelim inary version appeared inProceedings of the Third Annual Symposium on Computational Geometry,

1987 , pp . 39-49 .

T . Asano , G.T. Toussaint , Computing the Geodesic Center of a SimplePolygon , in Perspectives in Computing: D iscreteAlgorithms and Complexity,

Proceedings of Japan-US Joint Seminar , D .S. Johnson , A . Nozaki , T . Nishizeki , H . Willis , eds , June 1986 , pp. 65-79 .

J .L . Bentley , T.A . O ttmann,

"Algorithms for Reporting and Counting

Geometric Intersections", IEEE Transactions on Computers , C-28 , pp . 643

647 ,

B . Chazelle , L . Guibas , "Visibility and intersection problems in plane

geometry ,"Proceedings of the Symposium on Computational Geometry , 1985 ,

pp . 135-146 .

[GHLSl‘

] L . Guibas , J . Herschberger , D . Leven , M . Shatir , R .E . Tarjan ,

"Lineartime algorithms for visibility and shortest path problems inside triangulatedsimple polygons ,"Algorithmica 2 , 1987 , pp . 209-233 .

Garey , D .S. Johnson , F .P . Preparata , R .E . Tarjan ,

"Triangulating a

[LP84 ]

[PSR87]

[P885]

[S W]

FT86]

simple polygon ," Information Processing Letters , Vol 1978 , pp . 8-21.

D .T. Lee , F.P . Preparata,

"Euclidean shortest paths in the presence of rec

tilinear barriers ,"Networks pp . 393-4 10 .

R . Pollack , M . Sharir , G . Rote , "Computing the Geodesic Center of a Simple Polygon ,” to appear , D iscrete and Computational Geometry.

F.P . Preparata , M .I . Shamos , Computational Geometry: an Introduction,

Springer-Verlag , 1985 .

S. Suri , "Computing geodesic furthest neighbors in simple polygons , to

appear , Journal Computer and Systems Sciences . A preliminary versionappeared in Proceedings of the Third Annual Symposium on Computational

Geometry , 1987 , pp . 64-75 .

N . Sarnak , R .E . Tarjan ,

"Planar Point Location using Persistent Search

Trees ,"CACM Vol pp . 669-679 , July 1986 .

G . Toussaint , "An optimal algorithm for computing the relative convex hullof a set of points in a polygon ,

"Signal Processing III: Theories and Applica

tions , Proc . of EUSIPCO -86 , Part 2 , North-Holland , 1986 , pp . 853-856 .

https://www.forgottenbooks.com/join