A system of disjoint representatives of line segments with given k directions
AA SYSTEM OF DISJOINT REPRESENTATIVES OFLINE SEGMENTS WITH GIVEN k DIRECTIONS
JINHA KIM, MINKI KIM, AND O-JOUNG KWON
Abstract.
We prove that for all positive integers n and k , thereexists an integer N = N ( n, k ) satisfying the following. If U is aset of k direction vectors in the plane and J U is the set of all linesegments in direction u for some u ∈ U , then for every N families F , . . . , F N , each consisting of n mutually disjoint segments in J U ,there is a set { A , . . . , A n } of n disjoint segments in (cid:83) ≤ i ≤ N F i anddistinct integers p , . . . , p n ∈ { , . . . , N } satisfying that A j ∈ F p j for all j ∈ { , . . . , n } . We generalize this property for underlyinglines on fixed k directions to k families of simple curves with certainconditions. Introduction
Given a positive integer m , let [ m ] := { j ∈ Z : 1 ≤ j ≤ m } .For positive integers n ≤ m and families F , . . . , F m of non-emptysets, a system of distinct representatives of size n is a set { A , . . . , A n } of mutually distinct sets in (cid:83) i ∈ [ m ] F i such that there exist n distinctnumbers p , . . . , p n ∈ [ m ] where A j ∈ F p j for all j ∈ [ n ]. A systemof disjoint representatives is a natural generalization of a system ofdistinct representatives, where we additionally require selected sets tobe mutually disjoint. Throughout this paper, we abbreviate a “systemof disjoint representatives” as an “SDR”. When the size n of an SDR R is the same as the number m of given families, we say that R is a complete SDR .Rainbow matching is a central concept in the study of systems ofdisjoint representatives. Given a set of matchings M , . . . , M x in agraph, a rainbow matching of size k is a matching with edges e , . . . , e k such that there exist distinct integers p , . . . , p k ∈ [ x ] where e j ∈ M p j for all j ∈ [ k ]. Considering the set of edges in a graph as objects, arainbow matching can be seen as a system of disjoint representatives. Date : January 11, 2021.This work was supported by the Institute for Basic Science (IBS-R029-C1). O-J.Kwon was supported by the National Research Foundation of Korea funded by theMinistry of Education (No. NRF-2018R1D1A1B07050294). a r X i v : . [ m a t h . C O ] J a n JINHA KIM, MINKI KIM, AND O-JOUNG KWON
Brualdi and Ryser [9], and Stein [18] conjectured that for every setof n mutually disjoint matchings of size n in K n,n , there is a rainbowmatching of size n −
1. When n is even, there is a set of n mutuallydisjoint matchings of size n in K n,n that has no rainbow matching of size n . In other words, we cannot always guarantee to find a complete SDR.Here is a more general conjecture, which was introduced by Aharoniand Berger [1]: for every set of n − n in a bipartite graph, there is a rainbow matching of size n − n so that we canalways find a rainbow matching of size n ? In this direction, Drisko [12]showed that every set of 2 n − n in K n,n has an SDRof size n . Bar´at, Gy´arf´as, and S´ark¨ozy [8] conjectured that withoutbipartiteness, every set of 2 n matchings of size n contains a rainbowmatching of size n . Aharoni et al. [2] proved that 3 n − n − n so that we can find a rainbowindependent set of size n ? In fact, systems of disjoint representativescan be written in terms of rainbow independent sets: given families F , . . . , F n , we create an intersection graph of objects in (cid:83) i ∈ [ m ] F i , andthen a system of disjoint representatives corresponds to a rainbow in-dependent set in the intersection graph.Generally, we may consider the following parameter. Given a family F of sets and a positive integer n , let f F ( n ) be the minimum integer k ≥ n satisfying the following: for every k subfamilies of F (not necessarilydistinct), each consisting of n disjoint members of F , there is an SDRof size n . For example, if F is the set of all edges in a graph G , then f F ( n ) is exactly the minimum number of matchings of size n in G thatguarantee to have a rainbow matching of size n .In this paper, we obtain lower and upper bounds of f F ( n ) when F isa set of simple curves in the plane under certain conditions. A simplecurve is an injective continuous function c from an interval to R . Fora simple curve c : I → R , the image of any subinterval of I is called a segment on c . For a direction vector u in the plane and v ∈ R , a lineof the form { ut + v : t ∈ R } is a line in direction u . A line segment isa simple curve on a line. A line segment is said to be in direction u ifit is a subset of a line in direction u . SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 3
We prove that for every n , f J U ( n ) is bounded from above by a func-tion of n and | U | , where J U is the set of all line segments in direction u for some u ∈ U . Theorem 1.1.
Given positive integers n and k , there exists a positiveinteger N = N ( n, k ) such that the following holds. Let U be a set of k direction vectors in the plane. If J U is the set of all line segments indirection u for some u ∈ U , then f J U ( n ) ≤ N . Aharoni, Briggs, Kim, and Kim [4] showed that f F ( n ) = n when F isa family of intervals in R . Thus, Theorem 1.1 generalizes the finitenessof f F ( n ) for the set F of all intervals in a line, to the set of all linesegments with finite possible direction vectors. Intersection graphs ofsegments with k directions are called k -DIR [16], and our result can betranslated to rainbow independent sets for k -DIR graphs. Corollary 1.2.
Given positive integers n and k , there exists a positiveinteger N = N ( n, k ) such that for every k -DIR graph G and a familyof N independent sets of size n in G , there is a rainbow independentset of size n . We essentially use the facts that every point in R may intersect onlyfinite number of lines with given direction vectors and that two distinctlines with same direction are disjoint. Based on this observation, wegeneralize Theorem 1.1 into a flexible form for simple curves as follows.Here, we relax the condition of Theorem 1.1 so that two simple curves indifferent groups meet bounded number of times. Clearly, Theorem 1.3implies Theorem 1.1. Theorem 1.3.
Given positive integers n, k and t , there exists a positiveinteger M = M ( n, k, t ) such that the following holds. Let β , . . . , β k besets of simple curves in R such that • for each i ∈ [ k ] , β i is a set of mutually disjoint simple curves(not necessarily finite), and • for every distinct i, j ∈ [ k ] , P ∈ β i , and Q ∈ β j , P and Q intersect on at most t points.If J is the set of all segments of simple curves in (cid:83) i ∈ [ k ] β i , then f J ( n ) ≤ M . In Remark 3.2, we discuss that if we remove the condition that twosimple curves from different groups meet bounded number of times,then f J ( n ) is not bounded.To prove Theorem 1.3, we need to argue that if there is a sufficientlylarge family A of disjoint unions of n simple curves where each simple JINHA KIM, MINKI KIM, AND O-JOUNG KWON curve is a segment of some curve in (cid:83) i ∈ [ k ] β i , then we can always find anSDR of size n . By perturbing endpoints of given segments if necessary,we can assume that all their endpoints are distinct. Therefore, we mayfurther assume that all the segments are closed.By the pigeonhole principle, we first collect a subset A of A whereeach set in A contains exactly same number of segments in each of β , . . . , β k . Then we focus on some β i and restrictions of A on β i .We show that if there are many simple curves of β i that contain somesegment of A , then we can find an SDR of size n . This part is sepa-rately discussed in Section 2. Thus, we can assume that there are onlyrestricted number of simple curves in β i that contain a segment from A . This means that we reduce the case where the size of each β i isbounded by some function of n and t .Now, we use the fact that every point in R is contained in at most t curves in (cid:83) i ∈ [ k ] β i . Let p be the total number of intersections ofthe simple curves in (cid:83) i ∈ [ k ] β i , and give the name v , v , . . . , v p to theintersection points. For each set X in A , we obtain a vector Q X ∈ Z p such that for each j ∈ [ p ], • if v j is contained in X and the segment of X containing v j isa segment of a simple curve in β z , then the j -th coordinate of Q X is z , and • otherwise the j -th coordinate of Q X is q for some curve β q containing v j .Applying the pigeonhole principle, we can obtain a large subset A of A whose corresponding vectors are all the same. Then it is notdifficult to see that all sets of A are contained in some disjoint unionof simple curves, and we derive the result by simple arguments.The details of the proof of Theorem 1.3 will be given in Section 3. InSection 4, we discuss some lower bounds and provide open problems.Throughout the paper, we use the following notation. For two realnumbers x < y , we write [ x, y ] := { z ∈ R : x ≤ z ≤ y } , ( x, y ) := { z ∈ R : x < z < y } , and [ x, y ] Z := [ x, y ] ∩ Z . Remark 1.4. f F ( n ) may not be bounded when F is a family of “fat”objects, that is, convex sets with non-empty interiors. For example, if B is the family of all axis-parallel boxes in the plane, then f B ( n ) = ∞ for any n ≥
4: for an arbitrary positive integer k , we can constructa family F of axis-parallel boxes whose intersection graph is the k -thpower of C k +1) . See Figure 1 for an example for k = 2.By [4, Theorem 5.5], we can find a collection of 3( k + 1) subfamiliesof F such that each consists of 4 disjoint boxes of F and there is noSDR of size 4. By adding n − SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 5
Figure 1.
A family of 12 axis-parallel boxes whose in-tersection graph is the C .that are disjoint from any of F to each of the subfamilies, we have3( k + 1) families without an SDR of size n , where each family consistsof n mutually disjoint axis-parallel boxes.2. Line segments in one direction
In this section, we prove Theorem 1.3 when t = 1. Also, as anintermediate step for the proof of Theorem 1.3, we prove a lemma onmutually disjoint segments of simple curves. Note that when all simplecurves are mutually disjoint, we may consider segments of the simplecurves as line segments in one direction by the following observation. Observation 2.1.
Let β = { c , . . . , c k } be a set of mutually disjointsimple curves in the plane, and let F be a finite family of segments ofcurves in β . For each A ∈ F , we define a horizontal line segment A (cid:48) as follows: if A is a segment of c i , then A (cid:48) = I A × { i } ⊂ R where I A is the preimage of A along the curve c i . Then two distinct segments A and B in F are disjoint if and only if A (cid:48) and B (cid:48) are disjoint. Given a family F of non-empty sets, a set F ∈ F is said to be simplicial if all members of F that meet F have a point in common.It is known that every family of n sets where each set consists of n disjoint line segments on a line has a complete SDR [4, Theorem 3.20].This implies f J U ( n ) = n when | U | = 1. The proof of this statementis based on the fact that every family of segments on a simple curvecontains a simplicial member. The case when t = 1 in Theorem 1.3immediately follows by Observation 2.1. JINHA KIM, MINKI KIM, AND O-JOUNG KWON
Proposition 2.2.
Let n be the positive integer and β be a set of mu-tually disjoint simple curves in R d . If J is the set of all segments ofcurves in β , then f J ( n ) = n . Next, we show a sufficient condition on a family of sets, each con-sisting of m disjoint horizontal line segments, to have an SDR of size n when m < n . This is one of the key ingredients for the proof ofTheorem 1.1 and Theorem 1.3. Theorem 2.3.
Let m and n be positive integers with n > m . Let A be a family of sets A , . . . , A n + m − , each consisting of m disjointhorizontal line segments in the plane. Let L be the set of all horizontallines that meet (cid:83) A ∈A A . If | L | ≥ m ( n − m ) + 1 , then A has an SDRof size n .Proof. Let R be an SDR of A such that(i) | R | is maximum, and(ii) subject to (i), the number of lines in L meeting a line segment of R is maximum.The statement is obviously true if | R | ≥ n , so we may assume | R | ≤ n − R is an SDR, there is an injection c : R → [ n + m −
1] such that I ∈ A c ( I ) for each I ∈ R . For each S ⊂ R , let c ( S ) = { c ( I ) : I ∈ S } .We take a partition of L into three parts L , L and L as follows: • L is the set of all lines of L that do not meet any of R . • L is the set of all lines l ∈ L \ L such that for any I ∈ R thatmeets l , every line segment in A c ( I ) lies on a line in L \ L . • L = L \ ( L ∪ L ).Note that L is the set of lines l ∈ L \ L such that for any J ∈ R thatmeets l , there is J (cid:48) ∈ A c ( J ) that lies on a line in L . Note also that L ∪ L is the set of all lines that contain at least one line segment of R . Claim 2.4.
Each line in L contains exactly one line segment of R .Proof. Assume that there is a line l ∈ L that contains two distinct linesegments of R , say I and J . By the definition of L , there is I (cid:48) ∈ A c ( I ) that meets some line l ∈ L . Then R (cid:48) = ( R ∪ { I (cid:48) } ) \ { I } is an SDR of A with | R (cid:48) | = | R | that meets all lines in { l } ∪ L ∪ L . This contradictsthe assumption (ii) for the choice of R . (cid:3) Claim 2.5.
Let A j be a set in A that is not represented by R . Thenevery element of A j lies on a line in L . SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 7
Proof.
Suppose there is I ∈ A j that does not lie on a line in L . If I lies on a line in L , then R ∪ { I } is a larger SDR of A , contradictingthe maximality assumption on R . So, we may assume that I lies on aline l ∈ L . Then by Claim 2.4, there is exactly one J ∈ R that lieson l . By the definition of L , there is J (cid:48) ∈ A c ( J ) that lies on l ∈ L .Now ( R ∪ { I, J (cid:48) } ) \ { J } is a larger SDR of A . It again contradicts themaximality assumption on R . (cid:3) Let R i be the set of line segments in R that lies on some line in L i for i = 1 ,
2. We claim that | R | ≥ m .Suppose | R | < m . Since | [ n + m − \ c ( R ) | ≥ m , by Proposition 2.2,there exists an SDR R (cid:48) of A \ { A i : i ∈ c ( R ) } with | R (cid:48) | = m . ByClaim 2.5, each line segment of R (cid:48) lies on a line of L . Then R (cid:48) =( R \ R ) ∪ R (cid:48) is an SDR of A with | R (cid:48) | > | R | , which contradicts themaximality of R . Therefore, we may assume | R | ≥ m .Now take l ∈ L . Then there exists i ∈ [ n + m −
1] such that A i contains some I l that lies on l . By Claim 2.5, A i must be representedby a line segment I i ∈ R . By the definition of L , I i should lie on aline of L . That is, every line segment that lies on a line of L belongsto some A i that is represented by R . This gives us an upper bound | L | ≤ ( m − | R | . Since R = R ∪ R and | R | ≥ m , we have | L | = | L | + | L | + | L |≤ ( m − | R | + | R | + | R | = m | R | − ( m − | R |≤ m ( n − − m ( m −
1) = m ( n − m ) , which is a contradiction to the assumption | L | ≥ m ( n − m ) + 1. There-fore, it must be | R | ≥ n , as required. (cid:3) The bound | L | ≥ m ( n − m ) + 1 is tight by the following example. Example 2.6.
Let X = { I , . . . , I m ( n − m ) } be a set of horizontal linesegments in the plane, where each lies on a distinct horizontal line.Consider a partition of X = X ∪ · · · ∪ X n − m into n − m parts, whereeach part consists of m line segments of X . For each positive integer i , let A i = (cid:40) X i if i ∈ [ n − m − X n − m if i ≥ n − m . Let A be the family of all A i ’s. Clearly, every SDR of A contains atmost one element from X j for each j ∈ [ n − m −
1] and at most m elements from X n − m . Therefore, A is an infinite family that does nothave an SDR of size n . JINHA KIM, MINKI KIM, AND O-JOUNG KWON Proof of Theorem 1.3
In this section, we discuss more general situation, giving the proofof the main theorems. In order to prove Theorem 1.3, we need thefollowing lemma which describes a sufficient condition for the existenceof an SDR for families of sets of segments in simple curves.For two distinct points x and y on a simple curve c : I → R where x = c ( t ) and y = c ( t ) with t < t , we denote by ( x, y ) c the image c (( t , t )). Lemma 3.1.
For each i ∈ [ m ] , let c i : [0 , → R be a simple curvesuch that the set V of all intersections of c i ’s is finite and every pointof V is not an endpoint of any of the curves. Let J be the set of allsegments of the form c i ([ x, y ]) , and let V = { v , . . . , v p } . If each v i iscontained in exactly q i curves, then f J ( n ) ≤ (cid:16)(cid:81) j ∈ [ p ] q j (cid:17) n .Proof. Let N = (cid:16)(cid:81) j ∈ [ p ] q j (cid:17) n and consider a family A = { A , . . . , A N } ,where each A j consists of n mutually disjoint sets from J . For each j ∈ [ N ], we assign x j := ( a , . . . , a p ) ∈ [ m ] p so that for each i ∈ [ p ], • if v i is contained in A j , then the segment of A j containing v i isa segment of c a i , and • otherwise a i = q for some curve c q containing v i .By the pigeonhole principle, there exist a subset A (cid:48) of A and ( n , . . . , n p ) ∈ [ m ] p such that • |A (cid:48) | ≥ n , and • for all A j ∈ A (cid:48) , x j = ( n , . . . , n p ).Let V (cid:48) = V ∪ { c i ( t ) : i ∈ [ m ] , t ∈ { , }} . For each k ∈ [ p ], let u k and w k be the points of V (cid:48) on the curve c n k such that u k , v k , w k appearconsecutively on the curve c n k . By definition, V (cid:48) ∩ ( u k , w k ) c nk = { v k } .Observe that if ( u k , w k ) c nk ∩ ( u k (cid:48) , w k (cid:48) ) c nk (cid:48) (cid:54) = ∅ then n k = n k (cid:48) . Thus weobtain that (cid:83) k ∈ [ p ] ( u k , w k ) c nk is a set of mutually disjoint simple curvesand every segment of a curve in A (cid:48) is contained in (cid:83) k ∈ [ p ] ( u k , w k ) c nk .By Proposition 2.2, there is an SDR of size n for A (cid:48) ⊂ A . (cid:3) Now we are ready to prove Theorem 1.3.
Proof of Theorem 1.3.
Let M = M ( n, k, t ) = (cid:18) n + k − k − (cid:19) nk t ( k − k n and A be a family of M sets, where each set in A consists of n disjointsegments in J . We will show that A has an SDR of size n . We may SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 9 assume that all simple curves are closed, i.e. they are of the form c : I → R for some closed interval I , because A is finite. We may alsoassume that all segments are closed.Consider a partition A = (cid:91) n + ··· + n k = n, ∀ j ∈ [ k ] , n j ∈ Z ≥ A n ,...,n k of A into (cid:0) n + k − k − (cid:1) parts, where A n ,...,n k is the family of all sets in A that contain exactly n i segments of curves in β i for each i ∈ [ k ]. Bythe pigeonhole principle, at least one of the parts, say A n ,...,n k , shouldhave cardinality at least M ( n + k − k − ) = nk t ( k − k n .Suppose that for some i ∈ [ k ], there are at least n i ( n − n i )+1 distinctcurves in β i such that each containing some segment in (cid:83) A ∈A n ,...,nk A .In this case, by Observation 2.1 and Theorem 2.3, we can find an SDR ofsize n . Therefore, we may assume that for all i ∈ [ k ], there are at most n i ( n − n i ) curves in β i containing any line segment in (cid:83) A ∈A n ,...,nk A .Let V be the set of all intersections of those curves, then we have | V | ≤ (cid:88) i In Theorem 1.3, the condition about the bound on thenumber of crossings between two curves is important, in the sense that,if the number of crossings is not bounded, one can construct an arbi-trarily large family of sets, each consisting of n disjoint segments ofsimple curves, with no SDR of size n . In the below, we show by anexplicit example that it can happen even when the number of crossingsis countably infinite. Namely, we will construct a family of size ( n − q without an SDR of size n when we allow the number of crossings to beat least 2 q for some positive integer q > X ⊂ R , y ∈ R , and a family F of sets in R , let X + y := ( x + y : x ∈ X ) and F + y := ( A + y : A ∈ F ). Take a positive integer n > and let (cid:15) = q +2 . Let I = [ (cid:15), − (cid:15) ] and for each i ∈ Z , A i = { I + ( x + (2 i + 1) (cid:15) ) : x ∈ { , . . . , n − }} × { } ⊂ R . Observe that the intersection graph of the family (cid:83) i ∈ [0 ,q − Z A i is the( q − n − q vertices.For each i ∈ [ q − u i and v i be point on the x -axis whose x -coordinates are 2 i(cid:15) and n − i − (cid:15) , respectively. For each positiveinteger i ∈ [2 q − W i as follows: • For i ∈ [ q − W i is the closed upper half circle having thesegment connecting v i and v i +1 as its diameter. • W q − is the closed upper half circle having the segment con-necting u and v q − as its diameter. • For i ∈ [ q, q − Z , W i is the the closed upper half circle havingthe segment connecting u i − q +1 and u i − q +2 as its diameter.Now for each i ∈ { , , . . . , q − } , let J i = (cid:83) j ∈ [ i +1 ,i + q − Z W j . Finallylet B i = A i ∪ { J i } . See Figure 2 for an illustration of the case n = 5and q = 4. : B : B : B : B u u u v v v Figure 2. An illustration when n = 5 and q = 4.Then we observe that the intersection graph of the family (cid:83) i ∈ [0 ,q − Z B i is the ( q − nq vertices, where the set of inde-pendent sets of size n equals to the set of B i ’s. Thus, as in [4, Theorem5.5], the family of size ( n − q consists of n − B i doesnot have an SDR of size n .The current upper bound of M ( n, k, t ) in Theorem 1.3 is exponentialin n , when k and t are fixed. We ask whether this bound can be reducedto a polynomial in n . SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 11 Question 3.3. For all fixed integers k and t , does there exist a poly-nomial function M ( n ) satisfying the statement of Theorem 1.3?4. Axis-parallel line segments The proof argument in the previous section shows that Theorem 1.1is true for N ( n, k ) = (cid:0) n + k − k − (cid:1) nk ( k − k n . However, our guess is that f J U ( n ) can be bounded by a polynomial about n when | U | is fixed.In this section, we give additional results about f J U ( n ) when | U | = 2.We suggest a possible direction to obtain a polynomial upper boundfor f J U ( n ), and we construct an example that gives a quadratic lowerbound on f J U ( n ). By rotating one direction if needed, we may assumeall line segments are axis-parallel , that is, each line segment is eitherhorizontal or vertical.Consider a family of sets of n disjoint axis-parallel line segmentssuch that each set has the same number of horizontal lines. If onecan prove that there exists a constant C and d such that every suchfamily with Cn d sets has an SDR of size n , then we can show that f J U ( n ) ≤ Cn d ( n + 1) when | U | = 2: given a family of size Cn d +1 , weapply the pigeonhole principle to find a subfamily of size Cn d whereeach set of the subfamily has the same number of horizontal lines. Here,we give the first step toward this direction. Theorem 4.1. Let A be a family of n − sets, each consisting of n − horizontal line segments and one vertical line segment in theplane. Then there is an SDR of size n for A .Proof. Let A = ( A , . . . , A n − ). For each A i ∈ A , let A (cid:48) i be the set ofall horizontal line segments of A i , and let A (cid:48) = ( A (cid:48) i : A i ∈ A ).We first construct a maximal SDR R for A (cid:48) by the following process.Let A = A (cid:48) , R , = ∅ , and A j, = A (cid:48) j for each A (cid:48) j ∈ A . In i -th step,we proceed the following:(1) Take the line segment in (cid:83) A i − such that the rightmost pointof it is leftmost, say I i ∈ A t i ,i − . Let R ,i = R ,i − ∪ { I i } .(2) For each A j,i − ∈ A i − , let A j,i be the set obtained from A j,i − by deleting all line segments in A j,i − that meet I i .(3) Let A i = ( A j,i : A j,i − ∈ A i − \ { A t i ,i − } ).Let R = R ,k where k is the minimum integer such that A j,k = ∅ for all A j,k ∈ A k . Note that k ≥ n − i , 0 ≤ | A j,i − | − | A j,i | ≤ t i = i for 1 ≤ i ≤ k . Let r i be therightmost point of I i . If k ≥ n , then R contains an SDR of size n for A . Otherwise, we have k = n − 1, and this implies that for each i ∈ [ n − 1] and j ∈ [ n, n − Z , A j lost exactly one element that contains r i , and does not contain r i (cid:48) for any i (cid:48) > i , in the i -th step.Now, we proceed the same process for B = ( A (cid:48) i : i ∈ [ n, n − Z ), R , = ∅ , and B j, = A (cid:48) j with the following modifications:(1’) Take the line segment in (cid:83) B i − such that the leftmost point ofit is rightmost, say J i ∈ B t i ,i − . Let R ,i = R ,i − ∪ { J i } .(2’) For each B j,i − ∈ B i − , let B j,i be the set obtained from B j,i − by deleting all line segments in B j,i − that meets J i .(3’) Let B i = ( B j,i : B j,i − ∈ B i − \ { B t i ,i − } ).Similarly as above, we may assume that the obtained SDR R has size n − 1. Without loss of generality, we may assume R = { J , . . . , J n − } where J j ∈ A (cid:48) n − j for each j ∈ [ n − J j ∈ R , let l j bethe left endpoint of J j . Then, by the choice of R and R , there is aninjection g : [ n − → [ n − 1] such that X j := I j ∩ J g ( j ) is the linesegment connecting the endpoints l g ( j ) and r j for each j ∈ [ n − I ∈ A n − that does not meet anyline segment of R , then R ∪ { I } is an SDR of size n for A . Then theunion of the horizontal line segments of A n − covers all of the X j ’s.Let I be the vertical line segment of A n − and B = R × I . Note that I does not meet any of X j ’s. We will construct an SDR R of size n − A (cid:48) such that R ⊂ R ∪ R and each line segment of R does not meet I . For each j ∈ [ n − I j or J g ( j ) to R as follows: • For every X j that is disjoint from B , we choose I j . • If X j ⊂ B and X j is on the left side of I , then we choose I j . • Otherwise, X j ⊂ B and X j is on the right side of I . In thiscase, we choose J g ( j ) .Now R ∪ { I } is an SDR of size n for A . (cid:3) In Theorem 4.1, 2 n − n with 2 n − Example 4.2. In Figure 3, X is a set of black line segments and Y is aset of gray line segments where each of them consists of n − A = A = · · · = A n − = X and A n = A n +1 = · · · = A n − = Y, and A be the family of A i ’s.Observe that if an SDR for A contains the black vertical line segment,then it cannot contain any of the gray horizontal line segments. Let R be a maximal SDR for A . If R contains both black and gray vertical linesegments, then it cannot contain any of the horizontal line segments. SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 13 ... : X : Y Figure 3. The family of n − X and n − Y does not have an SDR of size n .If R does not contain any vertical line segment, then it consists of atmost n − R contains the black verticalline segment and does not contain the gray vertical line segment, thenit cannot contain any of the gray line segments, thus R ⊂ X . Sincethere are only n − X , R can have at most n − | R | < n , i.e. A does not have an SDR of size n . (cid:3) Example 4.2 can be generalized to give a construction for a quadraticlower bound on f J U ( n ) when | U | = 2. Example 4.3. Let n, m be positive integers with 2 m − < n . For each i ∈ Z , let I i := { i }× [1 , n − m ] for i ∈ Z and J ij := [ − m + i +1 , i − ×{ j } for j ∈ Z . Now, for each i ∈ [ m − X i of disjointline segments, which consists of m vertical line segments and n − m horizontal line segments as follows: X i := { I j : j ∈ [ − m + 1 , − m + i ] Z ∪ [ i, m − Z } ∪ { J ik : k ∈ [ n − m ] } . See Figure 4 for an illustration of X i .For j ∈ [( m − n − m − A j = X q if ( q − n − m − < j ≤ q ( n − m − . We claim that the family A = ( A j : j ∈ [( m − n − m − n .Suppose that R is an SDR of size n for A . By the construction of A , (cid:83) j A j has at most n − m disjoint horizontal line segments and at most2 m − R contains t ≥ m verticalline segments and at least one horizontal line segment. Suppose the · · · · · · · · · · · · ... I − m +1 I − m + i J i , · · · , J i ( n − m ) I i I m − Figure 4. The set X i with n − m horizontal line seg-ments and m vertical line segments.vertical line segments of R are I j , . . . , I j t for some integers j , . . . , j t with − m + 1 ≤ j < · · · < j k < < j k +1 < · · · < j t ≤ m − . Consider a horizontal line segment J ab ∈ R . Observe that j k ≥ − m + k and j k +1 ≤ m − − ( t − k − 1) = m − t + k . Since J ab does not meetany of the vertical line segments of R , [ − m + k + 1 , m − t + k − 1] mustcontain [ − m + a + 1 , a − t = m and a = k , andin particular, this implies R ⊂ X k . Since there are only n − m − X k , R can contain at most n − m − R consists of m vertical line segments and at most n − m − | R | = n . Therefore, there isno SDR of size n for A . (cid:3) By setting m = (cid:4) n (cid:5) , Example 4.3 gives a family of ( (cid:4) n (cid:5) − (cid:6) n (cid:7) − m vertical line segments and n − m horizontalline segments, that does not have an SDR of size n . It is natural to askif the above lower bound is asymptotically best possible. Question 4.4. Does there exist a constant C such that for everyfamily of Cnk sets, each consisting of n − k horizontal line segmentsand k vertical line segments in the plane, there is an SDR of size n ?We conclude the discussion with the following stronger question. Question 4.5. Does there exist a constant C such that f J U ( n ) ≤ Cn when | U | = 2? References [1] R. Aharoni and E. Berger, Rainbow matchings in r -partite r -graphs, Elec-tron. J. Combin. , 16(1):R119, 2009.[2] R. Aharoni, E. Berger, M. Chudnovsky, D. Howard and P. Seymour, Largerainbow matchings in general graphs, European J. Combin. , 79:222-227,2019. SYSTEM OF DISJOINT REPRESENTATIVES OF LINE SEGMENTS 15 [3] R. Aharoni, J. Briggs, M. Cho and J. Kim, Cooperative conditions for theexistence of rainbow matchings, arXiv:2003.08247.[4] R. Aharoni, J. Briggs, J. Kim and M. 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