Matching oints with geometric objects: Combinatorial results Bernardo Ábrego1, Esther Arkin, Silvia Fernández 1, Ferran Hurtado 3, Mikio Kano 4, Joseh S. B. Mitchell, and Jorge Urrutia 1 Deartment of Mathematics, California State University, Northridge Deartment of Alied Mathematics and Statistics, State University of New York at Stony Brook 3 Deartament de Matemàtica Alicada II, Universitat Politècnica de Catalunya 4 Ibaraki University Instituto de Matemáticas, Universidad Nacional Autónoma de México Abstract. Given a class C of geometric objects and a oint set P,aCmatching of P is a set M = {C 1,...,C k } of elements of C such that every C i contains exactly two elements of P. If all the elements of P belong to some C i, M is called a erfect matching; if in addition all the elements of M are airwise disjoint we say that this matching M is strong. In this aer we study the existence and characteristics of C-matchings for oint sets on the lane when C is the set of circles or the set of isothetic suares on the lane. 1 Introduction Let C be a class of geometric objects and let P be a oint set with n elements 1,..., n in general osition, n even. A C-matching of P is a set M = {C 1,...,C k } of elements of C, such that every C i contains exactly two elements of P. If all the elements of P belong to some C i, M is called a erfect matching. If in addition all the elements of M are airwise disjoint we say that the matching M is strong. If we define a grah G C (P ) in which the vertices are the elements of P, two of which are adjacent if there is an element of C containing them and no other element from P, a erfect matching in G C (P ) in the grah theory sense corresonds naturally with our definition of G C (P )-matchings. If C is the set of line segments or the set of all isothetic rectangles, then we get a segment-matching or a rectangle-matching, resectively. If C is the set of disks on the lane, M will be called a circle-matching. IfC is the set of all isothetic suares, M will be called a suare-matching. Notice that these four classes of objects have in common the shrinkability roerty: if there is an object C in the class that contains exactly two oints and in P, then there is an object C in the class such that C C, and lie on the boundary of C, and the relative interior of C is emty of oints from P. In the case of rectangles we can even assume the oints and to be oosite corners of C.
It is easy to see that P always admits a strong segment-matching and a strong rectangle-matching, which in fact are resectively non-crossing matchings in the comlete geometric grah induced by P (in the sense in which geometric grahs are defined in [4]) and in the rectangle of influence grah associated to P [3]. On the contrary, the situation is unclear for circles and suares, and gives rise to interesting roblems, That is the toic of this aer, in which we study the existence of erfect and non erfect, strong and non strong matchings for oint sets on the lane when C is the set of circles or the set of isothetic suares on the lane. Matching with disks In this section we study circle matchings. We show that a erfect circle matching is always ossible, but that there are collections of oints for which there is no erfect strong circle matching. We then give bounds on the size of the largest strong circle matching that any set P of n oints would admit. In the last art of the section we consider the secial case in which the oint set P is in convex osition..1 Existence of Circle Matchings First of all, notice that the fact that two oints from P can be covered by a disk that contains no other oint in P is euivalent to say that the two oints are neighbors in the Delaunay triangulation of P, DT(P ). In other words, when C is the set of all circles on the lane, the grah G C (P )isdt(p ). As a conseuence a oint set will admit a circle-matching if and only if the grah DT(P ) contains a erfect matching, which when P has an even number of oints is always the case, as roved by Dillencourt in 1990 []. Therefore we get the following result, which is a direct conseuence of Dillencourt s result: Theorem 1. Every oint set with an even number of elements admits a circlematching. Nevertheless a erfect strong circle-matching is not always ossible as we show next. Consider a circle C with unit radius and a oint set P with n elements 1,..., n, where 1 = a is the center of C and,..., n are oints evenly saced on the boundary of C. The oint a has to be matched with some oint b {,..., n }; this forces that the rest of oints are matched consecutively (see Figure 1), in articular the following and receding neighbors of b on the boundary have to be matched using large circles that are ushed outside of C and overla for n large enough. In fact, elementary trigonometric comutations show that this haen exactly for n 7. We don t describe the details of the receding construction, because the underlying basic ideas can we used for constructing an arbitrarily large set of oints such that at most a certain fraction of the oints can be strongly matched. More recisely, the following result holds:
3 a b C Fig. 1. A 7-oint set which does not admit a strong erfect circle-matching. Theorem. There is an n-element oint set in the lane, where n can be arbitrarily large, such that at most a fraction 7 73n of its oints can be strongly circle-matched. The roof of this result is omitted from this extended abstract, because it is very long and reuires several technical lemmas.. Subsets that can be matched strongly According to Theorem, not every oint set P admits a strong circle-matching. Here we rove that, at least, we can always find a linear number of disjoint disks each one covering exactly two oints from P : Theorem 3. For every P with n oints in general osition, there is a strong circle-matching using at least (n 1)/8 oints of P. Let M be the minimum suare-distance matching of P, that is M consists of m = n/ airs of oints 1 1,,..., m m where all i s and i s are different and the sum m i=1 ii is minimum among all ossible choices of the airs i i. Let C be the diametral disks determined by the airs i i in M. We denote by D i = DD( i i ) the closed disk with diameter i i and by o i the center of DD( i, i ).We first rove the following lemmas. Lemma 1. If DD(ab),DD(cd) C then {c, d} DD(ab). Proof. Suose that c, d DD(ab). Note that dcb + bdc < π, sowemay assume that dcb < π/. Thus bd <cd + bc, and since c DD(ab), then
4 bca π/ andbc +ac ab. Combining these ineualities we get bd +ac < ab + cd contradicting the minimality of M. Lemma. If DD(ab),DD(cd) Cand is in the intersection of the bounding circles of DD(ab) and DD(cd), then the triangles ab and dc do not overla. Proof. Suose that ab and dc overla. Assume d is between b and a as in Figure. Since ab and cd are diameters of their resective circles, then dc = ab = π/. So ad < π/ and bc<π/. Then ad <a + d, bc <b + c,and ad + bc <a + b + c + d = ab + cd, which contradicts the minimality of M. Fig.. ProofofLemma Lemma 3. No three disks in C have a common intersection. Proof. Suose I = DD( 1 1 ) DD( ) DD( 3 3 ). By Lemma 1, the boundary of I must contain sections of at least two of the bounding circles of I = DD( 1 1 ), DD( )anddd( 3 3 ). Thus we may assume there is a oint I such that is in the intersection of the bounding circles of DD( 1 1 )and DD( ). By Lemma the triangles 1 1 and do not overla. Now we consider three cases deending on the number of triangles that overla with 3 3. Case 1. 3 3 does not overla with 1 1 or. We may assume the relative order of the triangles i i is as in Figure 3. Then, since 1 1,, 3 3 π/, we have that 1 + 3 + 1 3 π/. So all these angles are at most π/ with at least two of them strictly acute (or zero). Thus 1 + 3 + 3 1 < 1 + + + 3 + 3 + 1.
Fig. 3. ProofofCase1:nooverla Also, since none of the angles 1 1,, 3 3 are acute then 1 + 1 + + + 3 + 3 1 1 + + 3 3. Thus 1 + 3 + 3 1 < 1 1 + + 3 3 which contradicts the minimality of M. Fig. 4. Proof of Case : 3 3 overlas Case. 3 3 overlas with but not with 1 1. Assume 3 is between and. We may also assume that 3 3 >π/, otherwise is in the bounding circle of DD( 3 3 ) and then by Lemma 3 3 and do not overla. Since 3 3 >π/ then 3 3 > 3 + 3. If 3 π/ (Figure 4a) then, same as in the roof of Lemma, 3 + 3, 3 + 3, and then 3 + 3 + + 3 + 3 < + 3 3, which contradicts the minimality of M.
6 If 3 >π/ (Figure 4b) then 1 3 + 1 <π/. Thus 1 3, 1 <π/ and then 1 3 < 1 + 3 and 1 < + 1. Also, since 3 is between and, then 3 <π/and 3 < + 3. Putting all these together we get 1 3 + 1 + 3 < 1 + 1 + + + 3 + 3. Moreover 1 + 1 = 1 1, + =,and 3 + 3 < 3 3.Thus 1 3 + 1 + 3 < 1 1 + + 3 3 which contradicts the minimality of M. Fig.. Proof of case 3: 3 3 overlas and 1 1. Case 3. 3 3 overlas and 1 1. We may assume 3 and 3 are between 1, 1 and, resectively (Figure ). Again by Lemma we may assume that 3 3 > π/and 3 3 > 3 + 3. If 1 π/ (see Figure a) then 1 1 +. From the locations of 3 and 3 we have that 1 3, 3 <π/and 3 1 <1 + 3, 3 < + 3. Then 1 + 3 1 + 3 < 1 + 1 + + + 3 + 3. In addition 1 + 1 = 1 1, + =,and 3 + 3 < 3 3.Thus 1 + 3 1 + 3 < 1 1 + + 3 3. If 1 >π/ then, (see Figure b) in a similar way, we get 1 + 1 3 + 3 < 1 1 + + 3 3. In both cases we contradict the minimality of M.
7 Lemma 4. If D 1,D,D 3,D 4 Cwith D 1 D and D 3 D 4 then the segments o 1 o and o 3 o 4 do not intersect. Proof. Suose o 1 o and o 3 o 4 intersect. Let x be the intersection of the two segments, D 1 D o 1 o,and D 3 D 4 o 3 o 4. Assume xo and xo 4. By the Triangle Ineuality o 1 o 1 x + x and o 3 o 3 x + x, then o 1 + o 3 o 1 x + x + o 3 x + x = o 1 + o 3. Thus either o 1 o 1 or o 3 o 3, which imlies that either D 1 or D 3. This is a contradiction to Lemma 3, since either D 1 D 3 D 4 or D 1 D D 3. Proof of Theorem 3. Let G be a grah with vertex set the centers of the disks in C, with two vertices connected by an edge if the corresonding disks intersect. By the last lemma, G is a lanar grah. Then by the Four Color Theorem the maximum indeendent set of G has at least m/4 = n/ /4 = (n 1)/8 vertices. Thus the corresonding diametral disks are airwise disjoint. Therefore P has a circle-matching using at least (n 1)/8 oints. It may haen that these diametral circles have oints of P in their interior. However it is always ossible to find a circle inside one of these diametral circles, containing only two oints of P..3 Convex osition When we have n oints on a line, with n even, it is obvious that a strong erfect matching with disks is always ossible, as we can simly take the diametral circles defined by consecutive airs. As a conseuence a strong erfect matching is also always ossible when we are given any set P of n oints lying on a circle C: using an inversion with center at any oint in C \ P the images of all oints from P become collinear and admit a matching, which alying again the same inversion gives the desired matching (because inversions are involutive and aly circles that don t ass through the center of inversion into circles). This may suggest that a similar result would hold for any set of oints in convex osition, but this is not the case as we show next using the same kind of arguments. Let Q be the oint set shown in Figure 1, consisting of the center a of a circle C, and 71 oints additional oints evenly distributed on C; as commented, Q does not admit a strong erfect circle-matching. Let P be the oint set obtained from Q by alying any inversion with center at some oint C which does not belong to Q; the oint set P does not admit a strong erfect circle matching. Notice that all the oints in P with the excetion of the image of a lie on a line. Alying an infinitesimal erturbation to the elements of P in such a way they remain in convex osition but no three are collinear roduces a oint set P in convex osition for which no strong erfect circle-matching exists, because the inverse set Q is an infinitesimal erturbation
8 of Q and therefore does not admit a strong erfect circle-matching. Therefore we have roved the following result: Proosition 1. There are oint sets in convex osition in the lane that cannot be strongly circle-matched. 3 Isothetic suares In this section we consider the following variation to our geometric matching roblem. Let P be a set of n oints in general osition on the lane. As in the revious section we define a grah G(P ) in which the oints are the vertices of G(P ) two of which are adjacent if there is an isothetic suare containing them that does not contain another element of P. 3.1 Existence of suare-matchings We show here that P always admits a suare-matching. We rove first: Lemma. G(P ) is lanar. Proof. Suose that i is adjacent to j,and k to l and that the segments joining i to j and i to j intersect. Let R i,j (res. R k,l ) be the smallest isothetic rectangle containing i an j (res. k and l ). It is straightforward to see that either R i,j contains k or l or R k,l contains i or j. In the first case k and l cannot be adjacent in G(P ). In the second i and j are not adjacent in G(P ) which in both cases roduce a contradiction. Let C be a suare that contains all the elements of P in its interior, and P be the oint set obtained by adding to P the vertices of C. LetG be the grah obtained from G(P ) by adding an extra oint adjacent to the vertices of C. We are going to see that G is 4-connected; before that we rove a technical lemma. Lemma 6. Let S be a oint set containing the origin O and a oint from the first uadrant, such that all the others oints in S lie in the interior of the rectangle R with corners at O and. Then there is ath in G(S) from O to such that every two consecutive vertices can be covered by an isothetic suare contained in R, emty of any other oint from S. Proof. The roof is by induction on S. If S = the result is obvious. If S > we grow homothetically from O a suare with bottom left corner at O until a first oint from S, different from O, is found. This suare is contained in R and gives an edge in G(S) between O and ; now we can aly induction to the oints from S covered by the rectangle with and as oosite corners. Obviously the receding result can be rehrased for any of the four uadrants to any oint taken as origin. We are now ready for roving the following result:
9 Lemma 7. G is 4-connected. Proof. Let us see that the grah G resulting from the removal of any three vertices from G is connected. Suose first that none of the suressed vertices is and let us see that can be reached from any vertex of G.Ifavertexv G is a corner of C, then it is adjacent to.ifv is not such a corner, consider the four uadrants it defines. In at least one of them no vertex from G has been suressed, then we can aly Lemma 6 to this uadrant and obtain a ath in G from v to a surviving corner of C; from there we arrive to. If we suress from G two oints from P and, then G contains the 4- cycle given by the corners of C. From any vertex v P in G we can reach one of these corners (and therefore any of them), because in at least two of the uadrants relative to v no vertex has been removed. The cases in which and one or two corners of C are suresses are handled similarly. 3 8 7 6 4 1 Fig. 6. Final ste for the existence of suare-matchings. As it is clear that G is lanar, it now follows using a classic result of Tutte [7] that G is Hamiltonian. This almost roves our result, since the removal of from G results in a grah that has a Hamiltonian ath. Using this ath, we
10 can now obtain a erfect matching in G(P ). A small roblem remains to find a matching in G(P ), namely the elements of P matched to the corners of C. To solve this difficulty we roceed in a way similar to that use in [1]. Consider the five shaded suares and eight oints 1,..., 8 (reresented by small circles) as shown in Figure 6. Within each of the shaded suares lace a coy of P, and let P be the oint set containing the oints of the five coies of P lus 1,..., 8. Consider the grah G(P ) and add to it a vertex adjacent to 1,, 3, 4. Once more G(P ) is lanar and four connected, and by Tutte s Theorem Hamiltonian. The removal of gives a Hamiltonian ath w in the resulting grah, with extremes in the set { 1,, 3, 4 }. At least one of the five coies of P does not contain any neighbor of 1 or 3 in the ath, because these two vertices have altogether at most four neighbors. Suose, for examle, that that is the case of the coy inside the box between and 6 ; as the ath has to arrive to the oints inside the box and leave the set, and this can only be done through and 6, the oints inside the box have to be comletely visited by the ath w before leaving the box. In this way we have obtained a Hamiltonian ath in G(P ), which gives a erfect matching in G(P ), and thus we have roved: Theorem 4. P has a erfect suare-matching. 3. Subsets that can be strongly suare-matched We show first a family of 1 oints that allows no erfect strong suare-matching. Consider the oint set with twelve oints shown in Figure 7: there are four extreme oints, labeled 1,..., 4 and eight more oints which are very close to the midoints of an auxiliary dashed suare as shown in the same figure; four of them, 1, 3,, 7,areinternal, while four of them,, 4, 6, 8,areexternal. The oints can all be drawn in general osition, with no two on a common vertical/horizontal line. Notice that no air can be matched inside { 1,..., 4 }; therefore two airs have to aear in any matching by icking oints from { 1,,..., 8 }.Notwo external oints can be matched, and matching a close air like 1, would leave an extreme oint ( 1 in the examle) without any artner for a matching. Distributing the four internal oints into two matched airs roduces always overlaing rectangles. All this leaves only two ossibilities for two airs taken from { 1,,..., 8 }: either using three internal oints and one external oint, or two internal oints and two external oints. In the first case the situation must be as in Figure 7, left, where 1 is matched to 7 and 4 to ; this forces ( 3, 6 ) and ( 4, 8 )to be matched airs and causes overla. In the second case the situation must be as in Figure 7, right, where is matched to 3 and 6 to 7 ; this forces ( 1, 1 ) and ( 4, 8 ) to be matched airs and causes overla. This concludes the roof. We now rove that the receding result can be used to construct arbitrarily large sets which do not admit erfect strong suare-matchings:
11 3 6 4 3 7 1 8 1 3 6 4 3 7 1 8 1 4 4 Fig. 7. Twelve oints that do not admit a erfect strong suare-matching Proosition. There are sets with 13m oints such that any strong suare matching of them contains at most 6m airs of matched oints. Proof. Take the line y = x, and consider the oints with coordinates (i, i), i =1,..., m +1, n even. For i =i +1, i =0,..., m 1 roceed as follows: Take an ɛ region around the oint (i +1, i + 1) and insert a coy of the 1 oint configuration P 1 (scaled down of to fit within this ɛ neighborhood). The remaining oints (i, i) stay as singletons. Let P be the oint set containing all these 1m+m oints, and let M be a strong suare-matching of P. See Figure 8 Observe that the 1 oint set close to the oint (1,1) cannot be matched within themselves. Then M matches at most 10 of these oints. This leaves two oints of ending. One of these oints can be matched with oint (.). The remaining oint cannot be suare matched with any oint in P. In a similar way one of the oints in the ɛ neighborhood of (i+1, i+1) cannot be matched to any element of P. This leaves at least n elements of P unmatched in M. Our result follows. We determine next a lower bound on the number of oints of a oint set that can always be strongly suare-matched. Theorem. For every P with n oints in general osition, there is a strong suare-matching using at least n oints of P. In fact, we rove a slightly stronger result, from which the reseding theorem is immediately derived: Lemma 8. Let S be a suare that contains a oint set P with at least two elements. Then it is always ossible to find a strong suare matching of P with n elements.
1 Fig. 8. Extending the twelve oints counterexamle for strong suare-matchings Proof. Our result is obviously true for n =. Suose then that it is true for n 1, and we now rove it for n, n 1. Observe first that if n =k + i, i =, 3, 4, then n = n 1, and by induction we are done. Suose then that n =k + 1 for some k. Partition S into for suares S 1,S,S 3,S 4 of eual size. Assume that each of them contains r 1,r,r 3,r 4 oints resectively. If all r i are greater than, or eual to zero, we are done since for any integers such that r 1 +... + r 4 = n we have: r 1 +... + r 4 n Suose then that some of the r i s are one. A case analysis follows. Case 1: Three elements of the set {r 1,r,r 3,r 4 },sayr = r 3 = r 4 =1are eual to 1; r 1 =(k 1) + 3. Let S 1 be the smallest suare, one of whose corners is, that contains all the elements of P in S 1 but one, say 1. Suose w.l.o.g that 1 lies below the horizontal line through the bottom edge of S 1. Then S 1 contains (k 1) + oints, and thus by induction we can find k disjoint suares in that suare containing exactly two elements of P n. It is easy to see that there is a suare contained in S S 1 that contains 1 and the element of P n in S 3. This suare contains a suare that contains exactly two elements of P. See Figure 9. Case : Two elements of {r 1,r,r 3,r 4 } are eual to 1. a) Suose that r i and r j are not 1. Observe that r i + r j =k 1 and that ri + rj n 1 ri.if + rj > n 1 = k we are done. Suose then that ri + rj = n 1 = k; this haens only if one of them, say r i =r and the other element r j =s 1 for some r and s greater than or eual to zero. U to symmetry two cases arise: a1) r 1 =r and r 3 =s 1. and a) r 1 =r and r 4 =s 1
13 S 1 S S 1 S 1 S 3 S 4 S 3 S 4 s r s r Fig. 9. In the first case let S 1 be the smallest suare contained in S 1 that contains all but three of the elements, say 1, and 3 of P in S 1, such that is a vertex of S 1. If two of these elements, say 1 and are below the horizontal line through the lower horizontal edge of S 1, then there is a suare S 3 contained in S S 1 that contains all the elements of P in S 3 and also contains 1 and, See Figure 10(a). Then by induction we can find in S 1 and S 3 r 3 = r and s+1 = s + 1disjoint suares, i.e. r + s +1=k + 1 disjoint suares contained in S each of which contains exactly two elements of P n. S 1 S S 1 S S 1 S 1 S 3 S 4 S 3 S 4 S 3 S 4 s (a) r s (b) r s (c) r Fig. 10. If no two elements of 1, and 3 lie below the horizontal through the lower horizontal edge of S 1, then there is a suare contained in S 1 S S 1 that contains two of these elements. Alying induction to the elements of P in S 1,
14 the elements of P in S 3 and the suare we just obtained rove our result. See Figure 10(b). If r =0,adthuss>0 choose S 3 such that it contains all but two oints of P n in S 3. If two oints in S 3 lie above line containing the to edge of S 3 or to the right of the line L containing the rightmost vertical edge of S 3, an analysis similar to the one above follows. Suose then that there is exactly one oint in S 3 to the right of L. Then S 3 contains s 3 oints, and there is a suare contained in S containing the oint of P n in S 4. See Figure refcase(c). By induction on the number of elements in S 3, and using the last suare we obtained our result follows. Case a) can be solved in a similar way. The remaining case, when only one of {r 1,r,r 3,r 4 } is 1 can be solved in a similar way to the revious cases. For examle the case when only r 4 = 1, (in which case r 1, r and r 3 are multiles of ) r 1 0,andr = 0 is solved almost the same way as case a1). We leave the details to the reader. 3.3 A erfect strong suare-matching for the convex case When several oints may have the same x-coordinate or the same y-coordinate, a erfect strong matching is not always ossible, as for examle haens when we try to match oint in Figure 11 to any oint on the line. Fig. 11. Point cannot be matched Nevertheless, we can rove that in convex osition, without reeated coordinates, a erfect strong matching always exists: Theorem 6. Any set of oints in the lane in convex osition with an even number of elements and such that no two oints are in the same vertical or horizontal line, admits a erfect strong suare-matching. The roof of this result is omitted from this extended abstract, because it is very long and reuires several technical lemmas.
1 4 Concluding remarks We have roved in this aer that (weak) matchings with circles and isothetic suares are always ossible. It is natural to ask which other classes of convex objects would enjoy the same roerty, and try to characterize them. On the comutational side, there are also decision and construction roblems that are very interesting. These issues are the main lines of our ongoing research on the toic. Acknowledgments This research was initiated at the International Worksho on Combinatorial Geometry at the Deartament de Matemàtica Alicada II, Universitat Politècnica de Catalunya, Jun. 30 Jul. 4, 003. The authors would like to thank the other worksho articiants, namely, Gabriela Araujo, Elsa Omaña-Pulido, Eduardo Rivera-Camo and Pilar Valencia for helful discussions. Ferran Hurtado is artially suorted by Projects MEC-MCYT-FEDER BFM003-00368 and Gen. Cat 001SGR004. Jorge Urrutia is artially suorted by CONACYT of Mexico grant 3740-A. References 1. J. Czyzowicz, E. Rivera-Camo, J. Urrutia and J. Zaks, Guarding rectangular art galleries, Discrete Mathematics, 0, 1994, 149-17.. M. Dillencourt, Toughness and Delaunay Triangulations, Discrete and Comutational Geometry, (6), 1990, 7-601. 3. G. Liotta, A. Lubiw, H. Meijer and S.H. Whitesides, The Rectangle of Influence Drawability Problem, Discrete and Comutational Geometry, (6), 1990, 7-601. 4. J. Pach, Editor, Towards a Theory of Geometric Grahs, Amer. Math. Soc., Contem. Math. Series/34, 004.. F. P. Prearata and M. I. Shamos, Comutational Geometry. An Introduction, Sringer-Verlag, 199. 6. M. Sharir, on k-sets in Arrangements of Curves and Surfaces, DiscreteandComutational Geometry, 6, 1991, 93-613. 7. W.T. Tutte. A theorem on lanar grahs. Trans. Amer. Math. Soc. 8,. 99-116, (196).