Width and Bounding Box of Imprecise Points

Transcription

1 Width nd Bounding Box of Impreise Points Vhideh Keikh Mrten Löffler Ali Mohdes Zhed Rhmti Astrt In this pper we study the following prolem: we re given set L = {l 1,..., l n } of prllel line segments, nd we wish to find set P = {p 1,..., p n }, where p i l i suh tht we mximize/minimize the width of P or the re of the ounding ox of P mong ll possile hoies for P. We design n O(n 2 ɛ 4.5 ) pproximtion lgorithm for omputing the lrgest width. We lso show tht the smllest width nd the smllest ounding ox n e omputed in O(n 2 ) time. We then proeed to present n O(n 6 ) time dynmi progrmming lgorithm for omputing the lrgest-re ounding ox. We lso present n FPTAS for this prolem whih runs in O(n 2 ɛ 5 ) time. 1 Introdution Shpe fitting is fundmentl prolem in omputtionl geometry, omputer vision, lustering, dt mining nd mny other res, whih sks the following question: suppose we re given set P of points in the plne, find shpe tht est fits P under some fitting riterion. In omputtionl geometry, mny prolems fit into the lss of shpe fitting, e.g., omputing the ounding ox, the width, the smllest enlosing irle, et. However, in the rel-world, the input is sujet to e impreise. Then the question is finding tight ounds on the size of the ojetive shpe. Impreise dt. Let P = {p 1,..., p n } e set of fixed points in the plne. In mny pplitions, eh element of P is sujet to e omputed with some errors, suh tht we do not know e.g., the ext oordintes of eh p i, or even the existene of p i. In this sitution we ll P set of impreise/unertin points. Mny studies hve foused on solving geometri prolems in the presene of impreise input. Depending on the informtion we hve out the input, different models of impreision re introdued. Here we riefly mention relted models: the Epsilon-geometry model, the Region-sed model, the Lotionl nd the Existentil model, where in these models, it is ssumed tht there exists set P i of points insted of eh p i, ut the ext lotion of p i in P i is unknown. See e.g., [3, 5]. Deprtment of Mthemti nd Computer Siene, Amirkir University, [v.keikh,mohdes,zrhmti]@ut..ir Deprtment of Informtion nd Computing Sienes, Utreht University, Utreht, The Netherlnds, m.loffler@uu.nl () () () (d) (e) Figure 1: An exmple. () A given set of prllel line segments. () The Lrgest possile width determined y 3 pirs simultneously. If we move ny of, or mong their line segments, we redue t lest one of the omputed width. () The smllest possile width. (d) The lrgest possile re ounding ox. (e) The smllest possile ounding ox. In this pper, we study our prolems in the Regionsed model. Let R e set of impreise points. An instne of R is set P of points seleted from distint regions of R. Then eh instne P of R will hve different onvex hull, width, ounding ox, et. Löffler nd vn Kreveld introdued frmework for omputing some tight lower nd upper ounds on the size of suh mesures, where they modeled the unertinty of the input y line segments, squres or disks [3]. Contriution. In this pper, we study the following prolems: given set L = {l 1,..., l n } of prllel line segments, hoose set P = {p 1,..., p n } of points, where p i l i, suh tht the size of width or the re of the ounding ox of P is s smll/lrge s possile mong ll possile hoies for P (see Figure 1(-e)). These prolems n e interpreted s finding the optiml filities in the form of ox or strip whih intersets eh line-segment-ustomer. Preliminries. Löffler nd vn Kreveld firstly studied the prolem of omputing the lrgest/smllest xisligned ounding ox of set of impreise points modeled s set of disks or squres in the plne, where their lgorithms vried from O(n log n) to O(n 2 ) [2]. In the sme pper, they proved tht omputing the lrgest possile width of set of impreise points modeled s set of ritrry line segments is NP-hrd. The sme prolem for prllel line segments, squres or disks ws posed s open question. The xis-ligned ounding ox of set P of fixed points in the plne is the minimum re ounding ox ontining P, sujet to the onstrint tht the edges

2 3 th Cndin Conferene on Computtionl Geometry, 218 y θ x d Figure 2: An exmple. The A(θ, d) digrms of the endpoints of three line segments nd two determined widths in diretion θ. Both the smllest nd lrgest possile width our in diretion θ. When we selet the lower endpoints of the lue nd green line segments, the lotion of the point on the red line segment determines the width: the lower endpoint of the red segment relizes the smllest possile width, while the upper endpoint relizes the mximum possile width. of the ounding ox re prllel to the x y oordinte xes. The smllest oriented ounding ox of P is the minimum re retngle ontining P. From now on, we simply ll it ounding ox. The width of set of points is the nrrowest strip ontining P. These prolems re extensively studied nd effiient lgorithms re known for them. One the onvex hull of P is known, ll these prolems n e solved in liner time sed on the rotting lipers method [6]. While the onvex hull of P is unknown there is n Ω(n log n) lower ound for oth prolems of omputing the ounding ox nd width of P in 2-D. Results. Let L = {l 1,..., l n } e set of prllel line segments. We otin the following results. We show tht the lrgest possile width of L n e pproximted within ftor (1 2ɛ) in O(n 2 ɛ 4.5 ) time (Setion 2.1). 1 We show tht the smllest ounding ox of L n e omputed in O(n 2 ) time (Setion 3.1). We present more involved O(n 6 ) time dynmi progrmming lgorithm for omputing the lrgest ounding ox of L. We lso present n FPTAS for this prolem whih runs in O(n 2 ɛ 5 ) time (Setion 3.2). We lso note tht ll missing proofs re in Appendix A.1. 2 Width We strt with the width prolem. Two prolems n e onsidered: finding n instne P on L, so tht P mximizes/minimizes the width of P. The minimum 1 Our method solves the smllest width prolem in O(n 2 ) time, however, there exists n O(n log n) time lgorithm for this prolem [4]. θ θ width of set of impreise points modeled s line segments (or ny other onvex regions), n e omputed in O(n log n) time [4], in whih the prolem is so-lled strip trnsversl, nd the uthors studied the prolem of omputing the thinnest strip tht intersets given set of onvex ojets. The mximum width prolem looks more diffiult, euse we should find n instne so tht our instne mximizes the width of the resulting point set. Sine width n e determined y multiple triples of points, nd eh point n tke prt in different triples, it looks diffiult to find the optiml position of the points (see Figure 1()). Let L = {l 1,..., l n } e set of prllel line segments in the plne. Let l i nd l + i, respetively, denote the lower nd upper endpoints of l i. Let E denote the set of ll the endpoints of segments in L. For point p, let l p denote the segment tht inludes p. For eh point p = (x, y), we define A(θ, d) digrm to e the plot of the funtion d θ = x sin θ + y os θ (1). It is the (signed) distne of p to line through the origin perpendiulr to the ry with ngle θ. [ ] [ ] [ ] xθ os θ sin θ x p θ = = = y θ sin θ os θ y [ x os θ y sin θ x sin θ + y os θ In Figure 2, the A(θ, d) digrm of the endpoints of set of segments is depited, the region etween sme olor digrms denotes the A(θ, d) digrms of the remining points of the segment. From now on, we use T (P ) to ddress the set of A(θ, d) digrms of set P of points. An exmple is depited in Figure 2, where eh T ({l + i, l i }) for i = 1,..., n is ssigned unique olor. Note tht the T ({l + i, l i }) of segment l i intersets ny other T ({l + j, l j }) for j i, j = 1,..., n in onstnt numer of intersetions. Thus there re qudrti numer of intersetion points. We ll I the set of intersetion points, where eh element of I is n intersetion point etween two digrms with distint olors. Notie the smllest possile width of L equls the vertil shortest distne etween point p I nd point f p on nother digrm with distint olor, so tht t lest one olor from eh digrm is interseted y vertil line segment pf p. Sine we wnt to minimize the length of segment pf p mong ll diretions θ, it will hve one endpoint on n intersetion point. Notie tht the instnes determined in this wy will introdue lrger width mong ll other diretions in A(θ, d) digrms of L. Thus this gives vlid width, nd further the smllest possile width. But omputing the smllest width y this method will ost O(n 2 ) time. 2.1 Lrgest width Let s e set of A(θ, d) digrms of distint olors. If s inludes extly one instne of eh olor, we ll s omplete set. For omplete set s of digrms, we define ]

3 optiml solution () A(θ, d) digrm D ɛ Figure 4: The rosses denote the points pproximting the line segments. the losest endpoint of its segment, or the width misses overing nother segment l, whih this hppens t n endpoint of l gin, then the new set relizes the sme width, ut one element of E is involved in the optiml solution () () Lemm 2 There exists n ɛ-kernel of size O( ɛ 1 ) for the mximum width prolem. Figure 3: () An exmple. The mximl width is determined y, nd. () The optiml solution. () The A(θ, d) digrms of the endpoints is denoted y solid urves, nd the optiml solution is shown in dsheddotted. The green dshed-dotted is not visile euse it is the sme s solid green urve. For etter visiility this figure is shown wider. w s s the shortest vertil distne etween the topmost nd ottommost digrms in s mong ll vlues of θ. Oviously w s determines the width of instnes in s. Oservtion 1 Let P e set of n points from distint line segments, then w T (P ) determines the width of P. Proof. The orretness omes from the ft tht w T (P ) is omputed mong ll vlues of θ. Let P denote the set of points whih mximizes the width. In the A(θ, d) digrm of L, the solution to the lrgest width prolem is equivlent to omplete set T (P ) of A(θ, d) digrms, so tht the vlue of w T (P ) is s lrge s possile mong ll possile hoies of s. In other words, for ny other omplete set s, w T (P ) w s. See Figure 3 s n exmple. As n e seen in Figure 3, the prolem proly hs lgeri issues. We design n pproximtion lgorithm to solve it. We strt stting our results with some oservtions. Lemm 1 Let L e given set of prllel line segments in the plne, nd let E denote the set of ll endpoints of L. There exists solution to the lrgest width prolem so tht one of the elements of E is involved in the optiml solution. Proof. It is esy to oserve tht we n trnslte the set of points relizing the lrgest width in up or down diretion until one of the points determining the strip of width (indeed, the point whih hs the smllest distne to one of the endpoints of its segment) rehes to Proof. Agrwl et l. [1] proved tht for ny point set in d-dimensionl spe, there is n ɛ-kernel of size O(1/ɛ d 1/2 ) nd it is lso worst se optiml. Let opt(l) denote the optiml solution to the mximum width prolem of L. Then there exists n ɛ-kernel Q opt for opt(l), tht is opt(l) < (1 + ɛ)widt(q opt ). Let S(Q opt ) denote the set of segments whih shre point on Q opt. Then oviously opt(s(q opt )) opt(l). Also we hve width(q opt ) opt(s(q opt )). Then opt(s(q opt )) opt(l) (1 + ɛ)opt(s(q opt )). Although we do not know wht is our ɛ-kernel, we still n use its size to design more effiient lgorithm. Let D denote the vertil distne etween the highest nd smllest y-oordintes of ny two endpoints of segments of L, s illustrted in Figure 4. Then we emnte set ρ of horizontl prllel rys, where the vertil distne etween ny two onseutive rys is ɛ. For ny l i L, the intersetion points ρ l i pproximte l i. We will ompute the A(θ, d) digrms of V = ρ L. Oviously V O(nɛ 1 ). We postpone the disussions of why this gives us the desired (1 ɛ) ftor, nd first disuss the solution on the pproximted points. 2 From Lemm 2 nd onsidering ny triple of points whih re potentilly involved in the mximum width, nive pproh solves the prolem in O(n ɛ 1/2 (ɛ 1/2 ɛ 1 ) 3 ) time. Corollry 3 There exists PTAS for the mximum width prolem whih runs in O(n ɛ 1/2 ɛ 4.5 ) time. Oservtion 2 If for eh l i L, T ({l + i, l i }) is lwys entirely loted etween two other A(θ, d) digrms in T (E) mong ll vlues of θ (shown in yellow in Figure 3()), then l i does not hve role in the onstitution of the optiml solution. error. 2 We hve supposed D = 1, sine we re onsidering the reltive

4 3 th Cndin Conferene on Computtionl Geometry, 218 The ove oservtion does not neessrily redue the omplexity of the lgorithm, ut still n redue the totl running time. 2.2 Dynmi progrmming lgorithm As onsequene of Oservtion 1 nd sine we look for the shortest vertil distne etween set of digrms, the shortest vertil distne will t lest use one intersetion point of two digrms with distint olors. Suppose we hve fixed one endpoint of segment, nd we hve omputed the A(θ, d) digrm of, T ({}) (for simpliity we denote it y T ()). Now the question is how to find omplete set s of digrms, so tht T () s nd other elements of s mximize their vertil distnes (w s ) from T (). First notie tht the numer of points in V is in O(nɛ 1 ). By onsidering ny triple of points in V whih re potentilly involved in the optiml solution, oviously the prolem n e solved in O(n 4 ɛ 3 ). In the extr O(n) we hek whether the strip of triple inludes one instne from eh segment or not. We will design DP lgorithm whih runs in O(n 2 ɛ 4.5 ) time. For fix endpoint, let w() denote the length of the shortest vertil segment whih is interseted y ll the trnsformtions of ɛ 1 1 other instnes (in the ɛ- kernel) mong ll diretions θ. Let w denote the mximum possile width of L. We should mximize the vlue w() for eh. Oviously w() in diretion θ n e defined y w() = Mx pi,d θ (T (p i)) d θ (T ()) d θ (T ()) d θ (T (p i )) + Mx pj,d θ (T (p j)) d θ (T ()) d θ (T ()) d θ (T (p j )), with i j, in whih T (p i ) nd T (p j ) hs the smllest nd lrgest vertil distnes from T () in diretion θ. W () = Mx [ w()], w = Mx E W (), where W () is the mximum over ll possile ɛ-kernels on. There only exist O( ɛ 1 ) ndidtes for eh of p i nd p j, sine they elong to n ɛ-kernel of this size. Also we only need to onsider the diretions θ whih is determined y the intersetion points of the A(θ, d) digrms of elements in the ɛ-kernel, sine the minimum vlue of w() tht needs to e mximized hppens there. Thus there exist O((ɛ 1/2 ɛ 1 ) 2 ) diretions θ in totl. Also there exist O(nɛ 1/2 ) different ɛ-kernels (of size O(ɛ 1/2 )) to e defined on. Consequently the dynmi progrm runs in (n 2 ɛ 4.5 ) time. Theorem 4 Let L e given set of prllel line segments in the plne. The lrgest possile width of L n e pproximted within ftor (1 2ɛ) in O(n 2 ɛ 4.5 ) time. 3 Bounding ox Similrly, for omputing the ounding ox of L two prolems n e onsidered, neither of these hs een studied yet: the smllest re ounding ox nd the lrgest re ounding ox. Let B denote the optiml solution to ny of these prolems. 3.1 Smllest ounding ox Now we extend our pproh for the minimum width prolem to design n lgorithm for the smllest-re ounding ox. Lemm 5 Let L e set of prllel line segments, nd let E e set of the endpoints of segments in L. There exists n optiml solution B to the smllest ounding ox of L, where eh edge of B psses through t lest one point of E, nd these points elong to distint segments. Proof. Suppose the lemm is flse. Then the miniml ounding ox B still hs n edge e whih is determined y point p i somewhere on the middle of l i. Consider line l through p i nd prllel to e. If we sweep l towrd the opposite side of e on B, it will interset l i, until it leves it t n endpoint p i (or B misses overing nother segment l j, whih hppens t n endpoint p j ). Then p i (or p j ) n e sustituted for p i to give us smller re ounding ox. Contrdition. Now we hve disretized the prolem on the endpoints. For ny set of fixed points, the smllest ounding ox n e determined y five points. Consequently, there exists nive O(n 6 ) time lgorithm for the smllest ounding ox prolem, where in the extr O(n) time we should hek whether n instne of ny segment is inluded in the solution. 3 Lemm 6 Let L e set of n prllel line segments. Only the diretions determined y the intersetion points of the elements of T (E) in the A(θ, d) digrm of L n e ndidtes to determine the diretion of two prllel edges of B. Proof. Only the intersetion points of T (E) in the A(θ, d) digrm of L denote the diretions in whih minimum width my exist. Suppose the lemm is flse. Then there exists solution B to the minimum ounding ox prolem, so tht none of the two diretions determined y edges of B, re determined y diretion in whih miniml width hppens (in n intersetion point). Then we find the losest diretion θ l (to the diretions of ny of two edges of B) whih is ndidte for the smllest width (whih hppens t n intersetion point) (see Figure 5). Let θ l denote the diretion of two prllel edges of B whih is loser to θ l. 3 Notie tht rotting liper tehnique does not look pplile here, sine we do not extly know the onvex hull.

5 θ l θ diff B B θ l Figure 5: The smllestre ounding ox will e onstruted in diretion in whih miniml width exists; if not we still n redue its re. Also first suppose the segments determined y diretion θ l re distint from the segments involved in the other edges of B. We define θ diff = θ l θ l. We sustitute the determined width in diretion θ l for the one in diretion θ l. We then rotte the two other edges of B with θ diff through the previously determined points. Oviously the hieved ox B hs smller re thn B. Contrdition. Now suppose the determined segments y the width in diretion θ l re not distint from the segments involved in the other edges of B. Then the shred point would e loted t orner of the new ox B, while the re of the B is smller thn B. Contrdition. In eh intersetion point p I in diretion θ, there my exist strip with miniml size whih inludes t lest one point from eh line segment. The other sides of the potentil optiml solution n e determined in diretion θ + π/2. Sine I O(n 2 ), the minimum re determined ox mong the intersetion points relizes B, nd the lgorithm works in O(n 2 ) time. Theorem 7 Let L e set of n prllel line segments. The optiml solution to the smllest ounding ox prolem of L n e omputed in O(n 2 ) time. 3.2 Lrgest ounding ox This prolem looks diffiult. Even rute-fore lgorithm is not strightforwrd, sine we nnot simply expnd the edges of possile ox (y using the endpoints of the segments), sine ollinerity of the points my redue the size of the optiml ox. See Figure 6(). Let θ denote the diretion of the lrgest ounding ox. Also notie tht t lest six points re involved in the optiml solution, sine the determined widths in oth diretion θ nd θ + π/2 hve the smllest size mong ll possile diretions in whih width n e determined, if not we still n redue its size, nd it is not vlid ounding ox. Also s n e seen in Figure 1(d), the lrgest re ounding ox does not neessrily use the orienttion of the mximum width. Lemm 8 Let L e set of n prllel line segments, nd let E e the set of the endpoints of segments in L. There exists solution to the lrgest-re ounding ox tht uses two points (from distint segments) of E on its two opposite sides, so tht eh edge inludes one of them. () l e t e l + t t 2 () Figure 6: () Expnding the edges of ox B to the endpoints of the segments determining the edges of B does not neessrily inrese the re of B. () The lrgest-re ounding ox t lest uses two elements of E on its two prllel sides. From the A(θ, d) digrm of L, the numer of different onfigurtions of hving two distint endpoints on two opposite sides of retngle is ounded y O(n 2 ), sine in the intersetion points the vertil order of two A(θ, d) digrms hnges, nd there only exists qudrti numer of intersetion points. Let t nd denote suh endpoints. Also n instne of ny other line segment is inluded in the determined strip y these two endpoints. Notie tht four suprolems need to e onsidered, sine we do not know whih of the upper or lower endpoints of l t nd l re the right ones. With the sme rgument we hd in Lemm 2, there exists n ɛ-kernel of size O(ɛ 1/2 ) for the lrgest ounding ox. By pproximting the set of line segments with set of prllel rys with ɛ differene etween onseutive rys, s disusses in Setion 2.1, for ny pir of endpoints we need to find triple of other points to onstrut ounding ox, where there re O(ɛ 1/2 ) ndidtes for eh point of triple nd O(ɛ 3 ) possile diretions for the optiml ox. Thus DP similr to the one presented in Setion 2.2 n solve the prolem in O(n 2 ɛ 5 ) time. In the following we try to solve it extly. Corollry 9 Let L e set of n prllel line segments. There exists n FPTAS for the lrgest ounding ox of L tht runs in O(n 2 ɛ 5 ) time. Algorithm. Let L e given set of n prllel line segments. Rell tht from A(θ, d) digrms of L we n ompute ll possile diretions whih there is strip S = d(l, θ), suh tht S inludes t lest one instne from eh element of L in diretion θ. From Lemm 8 we know two distint segments determine two opposite sides of B. As sid efore, from the A(θ, d) digrm of L we understnd t most O(n 2 ) ndidtes n determine two opposite sides of B, sine there re O(n 2 ) intersetion points nd thus the numer of onfigurtions in whih ll other digrms re resides etween two different digrms is ounded y O(n 2 ). Then we should find vlid width with these two points. Let θ denote diretion of suh vlid width. Notie tht there my exist O(n) possile diretions for θ. We will onsider omputing vlid solution from speifi θ, nd of ourse we will repet it for the remining possile diretions. Then ounding ox B in diretion θ n e defined y B = d(l, θ). d(l, θ + π 2 ) (2), where B is the smllest ox whih ounds S in θ + π 2 diretion, so tht B is retngle. With it using of the 1 r

6 3 th Cndin Conferene on Computtionl Geometry, 218 e t 4 3 e l 2 5 e r 6 1 e l i 3 v l () () () Figure 7: () A non-vlid ounding ox, where it is determined y 1 nd 5. () The mximum inner ngle t v l is determined t 3. () The orreted edge e l is denoted y e l denote the omputed vlid ox fter one step of the DP, whih is not ompletely vlid yet. Notie tht the hidden set of e r need to e updted for the next step. nottion, let S nd B lso denote size of the width nd re of the ounding ox, respetively. The ove definition of ounding ox B does not neessrily give us vlid ox, sine some segments my shre more thn one vertex on the oundry of B, or B my not e the smllest possile ox with these instnes, so tht we still n redue its re. But orreting this should e done in suh wy tht the removed re from B is minimized. This proedure is lled orreting B. Also we should do suh orretion for ll possile su-prolems in whih nonvlid ounding ox is determined y ny pir of endpoints. Finlly, the lrgest-re ounding ox mong ll determines the lrgest-re ounding ox of L. Oviously orreting ounding ox B (first omputed in diretion θ) my lso hnge the diretion of B. In the following we show tht we still n ompute the ext possile rottion of B. When looking for the lrgest smllest possile ox B, we onsider ll sets of six points whih my define ounding ox in Eqution(2), where two of them lredy determine vlid width nd lso the first diretion of ox B, ut not neessrily vlid ounding ox in tht diretion. And the other four points re omputed ordingly. In other words, for two fixed points, we first determine vlid width through them in diretion θ, nd then we ompute vlid width in diretion θ+π/2. Notie tht the omputed ox is superset for the optiml solution with these instnes. Finlly in the DP lgorithm we try to mke the iggest vlid ox whih is determined y these instnes. See Figure 7() for n exmple. Let A denote set onsisting of six suh points on the oundry of ox B. Let e, e l, e t nd e r denote the edges of B in lokwise diretion, nd let i for i = 1,..., 6 denote suh set A. Also let 1 e loted on edge e, 2, 3 e loted on edge e l, 4, 5 e loted on edge e t, et. Also let v l denote the ommon endpoint of edges e nd e l, s n e seen in Figure 7. W.l.o.g suppose B is defined on two endpoints 1 nd 5. For eh non-distint element of A, e.g., 3, distint line segment l i (whih is lredy interseted y B) will shre vertex 3 on e l, so tht 3 n e sustituted for 3 to give us smller (ut it more vlid) ounding ox B. We e l 1 ll 3 the hidden line segment y e l. Let H(A) denote ll the line segments hidden y the elements of A. In the worst se, orretion of B needs to find five hidden verties y the elements of A, ut suh sustitution should e done in suh wy tht removed re from B is s smll s possile, nd the resulting vlid B hs the lrgest possile re. (Notie tht sine H(A) hs onstnt omplexity, omputing the est onfigurtion n e done in onstnt time.) Also hidden line segment l i might simultneously e hidden y the points on two edges of B, e.g., in Figure 7, l i is hidden y oth e l nd e t. We will hek oth ses in different su-prolems, of whih there re onstntly mny. First suppose l i should shre vertex on e l. To do so, we should selet 3 suh tht the inner ngle 1v l 3 hs the mximum possile vlue mong ll possile hoies for 3. Further the hidden elements H(A) determine the ext vlue of the possile rottion of B. Let Θ denote ll possile diretions in whih vlid possile ox B is determined y two points 1 nd 5, s disussed efore, nd let α(a) denote the mximum ngle of rottion for orreting elements of A. Let A denote set A, where one distint element is sustituted for one non-distint element of A. Then we n write our DP s follows: d(l, θ) = width of L in diretion of θ (L, θ) = d(l, θ).d(l, θ + π 2 ), (L, Θ) = Min θ Θ (L, θ) Then we hve: V (A) = Mx (Min (A A, [α(a), α(a )]), V (A )) where V (A) denotes the lrgest ounding ox on possile set A in diretion θ. Then the mximum vlue mong ll possile V (A) denotes the optiml solution. The orretness of the lgorithm omes from the ft tht we onsider ll possile pirs tht n define ounding ox, nd then we find the lrgest possile ounding ox on this pir y our dynmi progrm. Finlly, the lrgest-re orreted ox determines the optiml solution. Notie tht we my need to rework on orreted set A, sine severl elements my need to e orreted. We will orret them lokwise. They only inrese onstnt numer of suprolems, whih t most equls 4 5. Notie tht there re onstnt possile diretions in Θ to mke vlid ox, nd it is needed to onsider O(n 2 ) pirs, nd for eh pir we need to onsider O(n) diretions, nd for eh diretion θ we should find the width in diretion θ + π/2. Then we ompute the hidden elements of the edges in O(n) time, nd we repet it for ny pirs etween ny two intersetion points in A(θ, d) of L. Thus the lgorithm runs in O(n 6 ) time nd spe. Theorem 1 Let L e given set of prllel line segments. The lrgest ounding ox of L n e omputed in O(n 6 ) time nd spe. 4 Conluding remrks nd open questions In this pper we present severl lgorithms for omputing n instne P on set of line segments, so tht P mximizes/minimizes the width or the re of the ounding ox of P. Solving mximum width prolem on set of squres remined open. We wish to extend our presented lgorithms to solve these prolems on set of squres.

7 Referenes [1] P. K. Agrwl, S. Hr-Peled, nd K. R. Vrdrjn. Approximting extent mesures of points. Journl of the ACM (JACM), 51(4):66 635, 24. [2] M. Löffler nd M. vn Kreveld. Lrgest ounding ox, smllest dimeter, nd relted prolems on impreise points. Computtionl Geometry, 43(4): , 21. [3] M. Löffler. Dt impreision in omputtionl geometry. PhD thesis, Utreht Univesity, 29. [4] J.-M. Roert nd G. Toussint. Computtionl geometry nd fility lotion. In Pro. Interntionl Conferene on Opertions Reserh nd Mngement Siene, pges B1 B19, 199. [5] D. Slesin, J. Stolfi, nd L. Guis. Epsilon geometry: Building roust lgorithms from impreise omputtions. In Pro. 5th Annul Symposium on Computtionl Geometry, pges , [6] G. T. Toussint. Solving geometri prolems with the rotting lipers. In IEEE Meleon, volume 83, pge A1, denote the points tht e t nd e re pssing through them, s illustrted in Figure 6(). Consider line l through t nd prllel to e t. If we sweep l wy from e t, it will interset the segment l t, so tht it leves it t n endpoint l + t. Then oviously the pentgon ll + t r 1 2 inludes the pentgon ltr 1 2. Sine with fixed length, the new ounding ox should now inlude some new point whih previously where loted outside the ounding ox, it must e expnded from the width. But then the hnges of the re of four oxes should e onsidered, if we sustitute l + t for t, the ox with n edge through 1, 2 will inrese its size, nd the sme rgument holds for the oxes with n edge through 1, r nd 2, l. Thus the ollinerity of l + t with existing verties nnot mke smller re ox. All together, we do not redue the size of ny other possile ounding ox of L, nd thus ny ounding ox whih is pssing through ll + t r 1 nd 2 will hve lrger re thn B. Thus B ould not e the lrgest re ounding ox. Contrdition. A Appendix A.1 Omitted proofs Theorem 4 Let L e given set of prllel line segments in the plne. The lrgest possile width of L n e pproximted within ftor (1 2ɛ) in O(n 2 ɛ 4.5 ) time. Proof. The only remining unproved prt is the (1 2ɛ) rtio of pproximtion. Let T (P pp) denote omplete set of digrms whih mximizes size of width. Let θ pp denote the diretion in whih T (P pp) gives the optiml solution. Then on the A(θ, d) digrms, using Eqution (1) we hve w w T (Ppp) + 2ɛ os θ pp, sine w T (Ppp) hs the lrgest vlue mong other omplete sets. Oviously w sin θ pp. But then if θ pp π/4, sin θ pp osθ pp nd then, w (1 2ɛ) w T (Ppp). In the se where θ pp < π/4 we oviously hve the sme width in diretion θ pp + π. The lemm follows. Lemm 8 Let L e set of n prllel line segments, nd let E e the set of the endpoints of segments in L. There exists solution to the lrgest-re ounding ox tht uses two points (from distint segments) of E on its two opposite sides, so tht eh edge inludes one of them. Proof. Like in the se of Lemm 1, there lwys exists n optiml solution B so tht t lest one element of E is involved on determining some edge of B. Let e denote suh edge. In the following, we disuss the existene of nother element of E on the opposite side of e. Let 1 denote the endpoint whih hs determined the edge e. W.l.o.g suppose 1 is loted on the ottom side of B. Suppose the lemm is flse. Then there exists mximl ounding ox B for L whih is pssing through some points l, t, r, 1 nd 2, so tht B hs mximl re nd B only uses one element of E. Let e t nd e denote the top nd ottom edges of B. And let t nd 1, 2, respetively,