Zone Theorem for Arrangements in three dimensions

Zone Theorem fo r Arrangemen ts in three dimensions Sanjeev Saxena ∗ Dept. of Computer Science and Engineering, Indian Institute of T ec hnology , Kanpur, INDIA-20 8 016 F ebruary 17, 202 5 Abstract In this note, a simple description of zone theorem in three dimensions is given. Arrangements in three dimensions are useful for co nstructing higher-or der V o ronoi diagrams in plane. An elementary and very intuitiv e tre atment of this result is also g iven. Keywords: Computational Geometr y , Zone Theorem, Arra ngements, k -Nearest Neighbours 1. In tro duction Zone theor em is imp ortant in analysing incremental algo rithms for constructing arrang ements. Most popular text b o oks of Computationa l Geometry (see e.g. [8, 1]) describ e zone theorem fo r Arr angements in tw o dimensions. Sp ecialised bo oks like [6, 2] des c r ib e zone theore m for h yp erplanes in d -dimensio ns. Three dimensional a r rangements are useful for co nstructing higher or der V o r onoi dia - grams in plane [8, 7, 2, 6, 3]. An elemen tary a nd very intuit ive trea tment o f this is given in Section 4. Pro ofs o f Zone theorem in higher dimensions use Euler ’s relation: P d i =0 ( − 1) i F i ≥ 0 [6, 5, 4]. As most s tuden ts of Computationa l Ge- ometry are not familiar with these r esult, only zone theore m in tw o dimensions is taught in most Computational Geometr y courses . In this note, a pro of of zone theorem in three dimensions which can be ea sily taught in Computational Geometry course s is describ ed. The pro of uses zone theorem in tw o dimensions [8, 1, 6, 4, 9]. The pro o f is a essentially a simplified version of pro o f given by Edelsbrunner, Seidel and Sharir[4]. ∗ E-mail: ssax@iitk.ac.in 1 2. Definitions and basic pr op erties Arrangement in tw o dimensions is ba sically a set of n lines (infinite lines a nd not seg ments), and in three dimensio ns of n planes . W e will assume that lines in t wo dimensions and planes in three dimensions are in general pos ition. Thus, in tw o dimension no tw o line s are par allel and no three lines meet in a single po int [8, 1, 4]. Simila rly , in three dimensions w e will assume that • No tw o planes ar e para llel. Thus, each pair of pla nes meet in a line. And any three planes in a po int. • No three planes intersect in a common line and no four planes in a (com- mon) po int. Set of lines, in tw o dimensions, will partition the plane in to regions called faces. And set of planes in thr e e dimensions will partition the space int o regio ns , whic h we will call cells. Let C be a ny b ounded cell. As pla nes ar e in general pos ition, each p oint or vertex v in arra ngement is determined by three planes. Thus, there will b e three edges of C inciden t a t a ny vertex v of C . Mo r eov er, each edge of C is determined b y t wo vertices of C . If | V C | is the n umber of vertices of C and | E C | is the num be r of edges of C , then 2 | E C | = 3 | V C | . As C is o n one side of each plane, C will b e a conv ex p olytop e (3-dimensiona l analogue o f p olyg on). The set of edges, vertices a nd faces on b ounda ry o f C will form a planar g raph. In a planar graph, if | V | is the num ber of vertices, | E | the num b er of edg es and | F | the num b er o f faces , then by E uler’s formula | E | − | V | + 2 = | F | . If | F C | is the num ber of faces of C then | F C | = | E C | − | V C | + 2 = | E C | − 2 3 | E C | + 2 = 1 3 | E C | + 2. Thu s, | E C | < 3 | F C | and hence | V C | = 2 3 | E C | < 2 | F C | . O r, | V C | = O ( | F C | ) and | E C | = O ( | F C | ). In tw o dimensions, let S b e a line different from n giv en lines (also in genera l po sition). Then S will intersect (cut) s ome faces of the arrang e ment . Zone( S ) is defined as the set of faces through which line S passes. If C ∈ zone( S ), is a fa c e which is cut, then let | C | be the num ber of edges on the b oundary o f fac e C in the (original) arra ngement, then the size of zone( S ), | zone( S ) | = P C ∈ zone ( S ) | C | . It is known tha t | zone( S ) | = O ( n ) ([8, 1, 6, 4, 9]). Similarly , in thr ee dimensions, let S be a plane different fro m n giv en planes (also in general p ositio n). Then S will in tersect (cut) so me cells of the ar range- men t. Zone( S ) is defined as the set o f cells which the pla ne S in ters ects. If C ∈ zone( S ), is a cell which is cut, then let | F C | be the num b er of faces on the b oundary of cell C in the (orig inal) a rrang ement, then the size of zone( S ), | zone( S ) | = P C ∈ zone ( S ) | F C | . Remark: Normally the size of zone is defined a s th e sum o f n umber of e dg es, vertices a nd faces o f a ll cells in the zo ne, but as the num b er of vertices and edges in a c e ll are O ( | F i | ), the t wo de finitio ns a re equiv alent up to multiplicativ e constants. Mo reov er, as eac h face is on b oundary o f tw o cells, the total num b er of cells in a zone will also b e b o unded by O ( P C | F C | ). 2 3. Zone Th eorem in 3 -dimensions W e will assume that all planes o f the arrang ement (tog ether with S ) are in general po sition, as s ize of zone is not s maller in this case [6 , 4, 2]. F urther, let us enclose the arrangement in a “ b o unding b ox”[1, 6] b y having six planes x = ± A, y = ± A, z = ± A — basica lly we compute co or dinates of all  n 3  vertices of ar rangement (by taking every p ossible set of thr ee planes) a nd choosing A to b e large r than the (absolute v alue o f ) larg e st co o r dinate. Th us, all cells inside the b o unding b ox will b e b ounded. Let Q be any pla ne o f the arrange ment. Then a s all pla nes are in g eneral po sition, each of them will intersect Q in a line. All these lines will lie in the plane Q and for m a tw o-dimensional (planar) arrang ement of lines (say L Q ). Let us remove plane Q from the arrang ement A and let the res ulting ar- rangement be c a lled A − Q . Let C b e a cell in the ar r angement A − Q . If the plane Q do es no t cut (in tersect) cell C , then cell C and all its faces will b e pr esent (unchanged) in arrang ement A . If the plane Q cuts (intersects) cell C , then the cell C gets divided into tw o parts— s ay the par t o f C a b ove the plane Q and the par t of C b elow Q (or if Q is horizontal then left and rig ht of Q ). Let us call the tw o pa r ts as C 1 and C 2 . P art o f C in tersected by Q will lie in plane Q (definition of intersection) and hence will b e a face (say f Q ) in the t wo dimensional arra ngement L Q . If face f (of C in A − Q is not intersected by Q , then face f will be present (unc hanged) in either C 1 or C 2 . If face f is intersected by Q , then f will g et split in to tw o pa rts o ne ab ove Q and the o ther b elow Q (or one o n left and the other on right). Let the par t in C 1 be called f 1 and part in C 2 be ca lled f 2 . Le t the b oundary (part common to b oth) be called e Q . As e Q is (also ) in plane Q , e Q will b e an edge in tw o dimensional arra ng ement L Q . E dge e Q is in fac e f Q . T o prove the zone- theorem we need following in termediate result Theorem 1. Assume A is an arr angement of n planes, Q is a plane in A , and S is a plane not in A . L et C b e a c el l in zone ( S ) . L et f b e a fac e of c el l C not lying in plane Q . Then total numb er of such p airs ( f , C ) (of fac e f and c el l C ) is at most the sum of 1. size of zone ( S ) in arr angement A − Q and 2. size of zone ( S ) in two dimensional arr angement L Q Remark: T he first size is count of faces (alo ng with their mult iplicities) a nd second of edg es (along with their multiplicities). Pr o of. Assume that cell C is in zo ne( S ) (o f arrangement A − Q ) and f is a face of C , not ly ing in (part of ) plane Q . As C is in zo ne( S ), plane S passes through cell C . 3 Assume that ce ll C is in zone( S ) (o f ar rangement A − Q ) and f is a face of C , not lying in (part o f ) plane Q . As C is in zone( S ), plane S pa sses through cell C . If Q do es not cut cell C , then cell C (along with all its faces) will b e un- changed in ar rangement A (and as S pass es through C ), cell C will also b e in zone( S ) in ar rangement A . In this ca se f ′ = f and C ′ = C . O r the same pair is pr esent in bo th A and A − Q . Let us assume that Q cuts C . Then c e ll C gets div ided into t wo parts (say) C 1 and C 2 . If the plane Q do e s no t cut face f , then face f will remain in tact in one part (say) C i (for i = 1 or 2 ). If pa rt C i contains f , then there is o ne-to-one corres p o ndence b etw een the pair ( f , C ) and the pa ir ( f , C i ) (i.e., pair ( f , C ) corres p o nds to pair ( f , C i ) and conv ersely). No te that right hand side will b e larger if C i is not in the z o ne (s e e b elow). Since S pa sses through C , it will pass through either C 1 or C 2 or b oth. If S passes through only o ne part (say) C i (for i = 1 or 2 ), then only C i will b e in the zo ne( S ) in arra ngement A . In this case, fo r the pair ( f , C ) we will hav e the corresp o nding pa ir ( f , C i ) and conversely (or in case f is intersected by Q , then the pair ( f i , C i ) where f i is the part of f in C i ). W e ar e left with the ca se when face f is also cut by Q and bo th C 1 and C 2 are in the zo ne( S ). If S passes through b oth C 1 and C 2 , then both C 1 and C 2 will b e in the zone( S ) in arrangement A . As S pass es thro ug h b oth C 1 and C 2 , it will also int ersect the common bo unda ry o f C 1 and C 2 . But as Q passes through the common b ounda r y o f C 1 and C 2 , the commo n part will b e a face (say f Q ) in the tw o dimensio nal a rrang ement L Q . And as S intersects f Q , face f Q will be in the t wo dimensional z one( S T Q ). As face f is also cut by Q , in tersection of f and Q will be a line segment (say) e Q . As e lies in plane Q , e Q will b e an e dge in the t wo dimensional arrange ment L Q . Thus, for the tw o entries ( f 1 , C 1 ) and ( f 2 , C 2 ) on right hand side we have t wo entries: the pair ( f , C ) in three dimensiona l arr angement A − Q , and also hav e the pair ( e Q , f Q ) in the tw o dimensiona l a rrang ement L Q . Thus, for the t wo en tries ( f 1 , C 1 ) and ( f 2 , C 2 ) on the left hand side, we also hav e t wo entries ( f , C ) and ( e Q , f Q ) on the r ight hand side. Q.E.D. As eac h face of a cell in A is in exactly one plane of the ar rangement, it does not lie in remaining n − 1 planes. Thus, if w e take an y pa ir ( f , C ) (for face f in cell C lying in zone( S )), it will not lie in n − 1 planes. O r if we tak e each plane in turn a s pla ne Q and a dd we get ( n − 1) | zone( S ) | ≤ X Q ∈A  | zone A−Q ( S ) | + | zone L Q ( S \ Q ) |  T o get the bo unds, let z ( n ) b e the lar g est p oss ible v alue of | zone( S ) | for arrang ement of n -planes . Then, if we a re considering this arrangement and this set S (for which the v alue of | zone( S ) | is the largest), then we have 4 ( n − 1) z ( n ) ≤ X Q ∈A  | zone A−Q ( S ) | + | zone L Q ( S \ Q ) |  As A − Q is an ar rangement of n − 1 pla ne s ( Q is e x cluded), | zone A−Q ( S ) | ≤ z ( n − 1). F urther, as tw o dimensional ar rangement L Q is in plane Q and each line cor resp onds to one of the other plane, the num b er of lines in L Q is n − 1. By the tw o dimensional zo ne theorem ([8, 1, 6, 4, 9]), num b e r of edg es in zo ne will be linea r. Hence, for some c onstant c , | zone L Q ( S T Q ) | ≤ c ( n − 1). Thus, our equation be c o mes ( n − 1) z ( n ) ≤ X Q ∈A ( z ( n − 1) + c ( n − 1)) = n z ( n − 1) + cn ( n − 1) T o solve this, we put f ( n ) = z ( n ) /n or z ( n ) = nf ( n ), the equa tion be comes f ( n ) ≤ f ( n − 1) + c Or f ( n ) = cn , or z ( n ) = nf ( n ) = cn 2 . Hence we g et the Zone theorem: Theorem 2. Assume that we ar e given an arr angement A of n planes in thr e e dimension. L et S b e a plane differ ent fr om the planes of the arr angement. Then size of zone ( S ) , | zone ( S ) | = O ( n 2 ) 4. k -Nearest Neigh b ours and Arrangemen ts Assume S = { ( x i , y i } n i =1 is a set of n p oints in the pla ne. Assume p = ( h, k ) and q = ( r, s ) ar e tw o p oints of S . A query (or test) point ( x, y ) will be close r to p = ( h, k ) then to q = ( r, s ) iff ( x − h ) 2 + ( y − k ) 2 < ( x − r ) 2 + ( y − s ) 2 or x 2 − 2 hx + h 2 + y 2 − 2 k y + k 2 < x 2 − 2 rx + r 2 + y 2 − 2 sy + s 2 or 2 rx + 2 sy − r 2 − s 2 < 2 hx + 2 k y − h 2 − k 2 Define f ( u, v ) = 2 ux + 2 v y − u 2 − v 2 Then ab ov e condition b ecomes, f ( h, k ) > f ( r, s ). As equation z = f ( u, v ) is an equa tion of plane, the a bove co ndition is equiv alent to saying that plane z = hx + 2 k y − h 2 − k 2 is a b ove the pla ne z = 2 rx + 2 sy − r 2 − s 2 . Hence, if we dra w a n arra ng ement of n -pla nes with plane z = 2 x i x + 2 y i y − x 2 i − y 2 i for the i th p oint, then the nearest neighbour o f query p oint ( x, y ) will be the topmost plane ab ov e ( x, y ) (the one v isible fro m ( x, y , ∞ )). The second nearest neighbout will b e the second top mo st plane a nd so on. Hence, to find k th closest point, we need to only consider planes which hav e ( k − 1) planes ab ov e them. 5 Ac kno wledgemen ts I wish to thank the students who attended lectures of CS663 (20 19-2 0 and 2 024- 2025) for their comment s and reactions. Tha nks also to an a nonymous reviewer of [9 ] for po int ing an o ut error in the earlier version. References [1] M.de.B e r g, M.v an Kreveld, M.Overmars and O.Sch warzkops, Compu- tational Geo metry , Algorithms and Applications, second e ditio n, 2000, Springer. [2] H.E de ls brunner, Algorithms in Combinatorial Geometry , Spring er 19 87. [3] H.E de ls brunner, J.O’ Ro urke and R.Seidel, Constructing a rrang e ment s of lines and hyper planes with applica tions, SIAM J. Co mput., vol 1 5, no. 2, pp -, May 1986 . [4] H.E de ls brunner, R.Seidel a nd M.Sha rir, On the zone theorem for hyp e r- plane ar rangements, SIAM J. Comput., vol 22, no . 2, pp 4 18-4 2 9, April 1993. [5] L.Guiba s, Basic Algorithms and Combinatorics in Co mputational Geome- try , 2 011. [6] K .Mulmu ley , Computationa l Geo metry , An in tro duction thro ugh ra ndom- ized algor ithms, Prentice Hall, 199 4 [7] F.P .Prepar ata a nd M.I.Sha mo s, Computational Geometry , An in tro duc- tion, 1985 , Springer . [8] J .O’ Rour ke, Computationa l Geometry in C, 2 nd Ed, Cam bridge Univ. Press, 199 8. [9] S.Sa xena, Zone Theo rem for arrang e ment s in dimensio n three, Infor mation Pro cess ing Letters, 172 (2021 ), 10 6161 . 6