Topological Drawings of Complete Bipartite Graphs

Transcription

1 Topological Drawings of Complete Bipartite Graphs Jean Cardinal Stefan Felsner y Febrary 017 Abstract Topological drawings are natral representations of graphs in the plane, where vertices are represented by points, and edges by crves connecting the points. Topological drawings of complete graphs and of complete bipartite graphs have been stdied extensively in the context of crossing nmber problems. We consider a natral class of simple topological drawings of complete bipartite graphs, in which we reqire that one side of the vertex set bipartition lies on the oter bondary of the drawing. We investigate the combinatorics of sch drawings. For this prpose, we dene combinatorial encodings of the drawings by enmerating the distinct drawings of sbgraphs isomorphic to K, and K 3,, and investigate the constraints they mst satisfy. We prove that a drawing of K k,n exists if and only if some simple local conditions are satised by the encodings. This directly yields a polynomial-time algorithm for deciding the existence of sch a drawing given the encoding. We show the encoding is eqivalent to specifying which pairs of edges cross, yielding a similar polynomial-time algorithm for the realizability of abstract topological graphs. We also completely characterize and enmerate sch drawings of K k,n in which the order of the edges arond each vertex is the same for vertices on the same side of the bipartition. Finally, we investigate drawings of K k,n sing straight lines and psedolines, and consider the complexity of the corresponding realizability problems. Contents 1 Introdction Oter Drawings with niform rotation system 4.1 The triple rle The qadrple rle Decomposability and Conting Universite libre de Brxelles (ULB), Brssels, Belgim. jcardin@lb.ac.be. y Technische Universitat Berlin, Germany. felsner@math.t-berlin.de. 1

2 3 Oter Drawings with k = Triples Qadrples Consistency Oter Drawings with k = The consistency theorem for k = Otline of the Proof of Proposition Case analysis for the proof of Proposition The eight basic congrations Proof of Lemma Oter Drawings with arbitrary k 6 6 Extendable and Straight-line Oter Drawings Extendable Oter Drawings Straight-line Oter Drawings Open problems 3 1 Introdction We consider topological graph drawings, which are drawings of simple ndirected graphs where vertices are represented by points in the plane, and edges are represented by simple crves that connect the corresponding points. We typically restrict those drawings to satisfy some natral nondegeneracy conditions. In particlar, we consider simple drawings, in which every pair of edges intersect at most once. A common vertex conts as an intersection. While being perhaps the most natral and the most sed representations of graphs, simple drawings are far from being nderstood from the combinatorial point of view. A prominent illstration is the problem of identifying the minimm nmber of edge crossings in a simple topological drawing of K n [9, 3, 1] or of K k,n [, 4], for which there are long standing conjectres. In order to cope with the inherent complexity of the drawings, it is sefl to consider combinatorial abstractions. Those abstractions are discrete strctres encoding some featres of a drawing. One sch abstraction, introdced by Kratochvl, Lbiw, and Nesetril, is called abstract topological graphs (AT-graph) [11]. An AT-graph consists of a graph (V, E) together with a set X E. A topological drawing is said to realize an AT-graph if the pairs of edges that cross are exactly those in X. Another abstraction of a topological drawing is called the rotation system. The rotation system associates a circlar permtation with every vertex v, which in a realization mst correspond to the order in which the neighbors of v are connected to v. Natral realizability problems are: given an AT-graph or a rotation system, is it realizable as a topological drawing? The realizability problem for AT-graphs is known to be NP-complete [1]. For simple topological drawings of complete graphs, the two abstractions are actally eqivalent [18]. It is possible to reconstrct the set of crossing pairs of edges

3 by looking at the rotation system, and vice-versa (p to reversal of all permtations). Kyncl recently proved the remarkable reslt that a complete AT-graph (an AT-graph for which the nderlying graph is complete) can be realized as a simple topological drawing of K n if and only if all the AT-sbgraphs on at most 6 vertices are realizable [13, 15]. This directly yields a polynomial-time algorithm for the realizability problem. While this provides a key insight on topological drawings of complete graphs, similar realizability problems already appear mch more diclt when they involve complete bipartite graphs. In that case, knowing the rotation system is not scient for recontrcting the intersecting pairs of edges. We propose a ne-grained analysis of simple topological drawings of complete bipartite graphs. In order to make the analysis more tractable, we introdce a natral restriction on the drawings, by reqiring that one side of the vertex set bipartition lies on a circle at innity. This gives rise to meaningfl, yet complex enogh, combinatorial strctres. Definitions. way that: We wish to draw the complete bipartite graph K k,n in the plane in sch a 1. vertices are represented by points,. edges are continos crves that connect those points, and do not contain any other vertices than their two endpoints 3. no more than two edges intersect in one point, 4. edges pairwise intersect at most once; in particlar, edges incident to the same vertex intersect only at this vertex, 5. the k vertices of one side of the bipartition lie on the oter bondary of the drawing. Properties 1{4 are the sal reqirements for simple topological drawings also known as good drawings. As we will see, property 5 leads to drawings with interesting combinatorial strctres. We will refer to drawings satisfying properties 1{5 as oter drawings. Since this is the only type of drawings we consider, we will se the single term drawing instead when the context is clear. The set of vertices of a bipartite graph K k,n will be denoted by P [V, where P and V are the two sides of the bipartition, with P = k and V = n. When we consider a given drawing, we will se the word \vertex" and \edge" to denote both the vertex or edge of the graph, and their representation as points and crves. Withot loss of generality, we can assme that the k oter vertices p 1,..., p k lie in clockwise order on the bondary of a disk that contains all the edges, or on the line at innity. The vertices of V are labeled 1,..., n. An example of sch a drawing is given in Figre 1. The rotation system of the drawing is a seqence of k permtations on n elements associated with the vertices of P in clockwise order. For each vertex of P, its permtation encodes the (say) conterclockwise order in which the n vertices of V are connected to it. De to or last constraint on the drawings, the rotations of the k vertices of P 3

4 Figre 1: Two oter drawings of K 3,5. In both drawings the rotation system is (1345, 1435, 1354). arond each vertex of V are xed and identical, they reect the clockwise order of p 1,..., p k on the bondary. Unlike for complete graphs, the rotation system of an oter drawing of a complete bipartite graph does not completely determine which pairs of edges are intersecting. This is exemplied with the two drawings in Figre 1. Reslts. The paper is organized as follows. In Section, we consider oter drawings with a niform rotation system, in which the k permtations of the vertices of P are all eqal to the identity. In this case, we can state a general strctre theorem that allows s to completely characterize and cont oter drawings of arbitrary bipartite graphs K k,n. In Section 3, we consider oter drawings of K,n with arbitrary rotation systems. We consider a natral combinatorial encoding of sch drawings, and state two necessary consistency conditions involving triples and qadrples of points in V. We show that these conditions are also scient, yielding a polynomial-time algorithm for checking consistency of a drawing. We also observe that or encoding is eqivalent to specifying which pairs of edges mst intersect in the drawing, hence exactly encodes the corresponding AT-graph. Therefore we show as a corollary that we can decide the realizability of a given ATgraph with nderlying graph isomorphic to K,n in polynomial time. In Section 4 and 5, we extend these reslts, rst to oter drawings of K 3,n, then to oter drawings of K k,n. We prove that simple consistency conditions on triples and qadrples are scient for drawings of K k,n, yielding again a polynomial-time algorithm for consistency checking. In Section 6, we consider oter drawings with the additional property that the edges can be extended into a psedoline arrangement, which we refer to as extendable drawings. We give a necessary and scient condition for the existence of an extendable oter drawing of a complete bipartite graph given the rotation system. We also toch pon the even more restricted problem of nding straight-line oter drawings with prescribed rotation systems. Oter Drawings with niform rotation system We rst consider the case where k is arbitrary bt the rotation system is niform, that is, the permtation arond each of the k vertices p i is the same. Withot loss of generality we assme that this permtation is the identity permtation on [n]. 4

5 In a given oter drawing, each of the n vertices of V splits the plane into k regions Q 1, Q,..., Q k, where each Q i is bonded by the edges from v to p i and p i+1, with the nderstanding that p k+1 = p 1. We denote by Q i (v) the ith region dened by vertex v and frther on call these regions qadrants. We let type(a, b) = i, for a, b V and i [k], whenever a Q i (b). This implies that b Q i (a), see Figre. Indeed if a < b and j 6= i + 1, then edge p i+1 b has to intersect all the edges p j a, while edge p j b has to avoid p i+1 b ntil they meet in b. It follows that none of the edges p j b can intersect p i+1 a. This shows that a Q i (b). p 1 p 6 Q 1 (a) Q 6 (a) a Q 5 (a) Q (a) p p 5 b Q 4 (a) Q 3 (a) p 3 p 4 Figre : Having placed b in Q 4 (a) the crossing pairs of edges and the order of crossings on each edge is prescribed. In particlar a Q 4 (b). On the right a symmetric oter drawing of the pair. Observation 1 (Symmetry). For all a, b in niform rotation systems: type(a, b) = type(b, a). For the case k =, we have exactly two types of pairs, that we will denote by A and B. The two types are illstrated on Figre 3. Note that the two types can be distingished by specifying which are the pairs of intersecting edges. The oter drawings of K,n with niform rotations can be viewed as colored psedoline arrangements, where: each psedoline is split into two segments of distinct colors, no crossing is monochromatic. A b a a b B Figre 3: The two types of pairs for oter drawings of K,n systems. with niform rotation This is illstrated on Figre 4. The psedoline of a vertex v V is denoted by l(v). The left (red) and right (ble) parts of this psedoline are denote by l L (v) and l R (v). 5

6 Now having type(a, b) = type(b, a) = A means that b lies above l(a) and a lies above l(b). While having type(a, b) = type(b, a) = B means that b lies below l(a) and a lies below l(b) B B B B A 3 A 4 Figre 4: Drawing K,4 as a colored psedoline arrangement. The type of each pair is given in the table on the right..1 The triple rle Lemma (Triple rle). For niform rotation systems and three vertices a, b, c V with a < b < c type(a, c) {type(a, b), type(b, c)}. Proof. Case k =. If type(a, b) 6= type(b, c) there is nothing to show since there are only two types. Withot loss of generality, sppose that type(a, b) = type(b, c) = B. This sitation is illstrated in the left part of Figre 5. The psedoline l(c) mst cross l(b) on l R (b), otherwise we wold have type(b, c) = A. Hence the point c is on the right of this intersection. Psedoline l(a) mst cross l(b) on l L (b), and a is left of this intersection. It follows that l(a) and l(c) cross on l R (a) and l L (c), i.e., type(a, c) = B. p j c b a a b c a c Figre 5: Illstrations for the k = case of Lemma (left), and the k > case of Lemma (right). Case k >. For the general case assme that type(a, b) = i and type(a, c) = j. If i = j there is nothing to show. Now sppose i 6= j. From c Q j (a) it follows that p j+1 a and p j c are disjoint. Edges p j b and p j c only share the endpoint p j, hence c has to be in the region delimited by p j b and p j+1 a, see the right part of Figre 5. This region is contained in Q j (b), whence type(b, c) = j.. The qadrple rle Lemma 3 (Qadrple rle). For for vertices a, b, c, d V with a < b < c < d: 6

7 if type(a, c) = type(b, c) = type(b, d) = X then type(a, d) = X. Proof. Case k =. Sppose, withot loss of generality, that X = B. Consider the psedolines representing b and c with their crossing at l R (b) [ l L (c). Coming from the left the edge l L (d) has to avoid l L (c) and therefore intersects l R (b). On l R (b) the crossing with l L (c) is left of the crossing with l L (d), see Figre 6. Symmetrically from the right the edge l R (a) has to intersect l L (c) and this intersection is left of l R (b) [ l L (c). To reach the crossings with l L (c) and l R (b) edges l R (a) and l L (d) have to intersect, hence, type(a, d) = B. d c b a a b c d Figre 6: Illstration for the k > case of Lemma. Case k >. In the general case, we let X = i, and consider the psedoline arrangement dened by the two sccessive vertices p i and p i+1 of P dening the qadrants Q i. Proving that type(a, d) = i, that is, that a Q i (d), can be done as above for k = on the drawing of K,n indced by {p i, p i+1 } and V..3 Decomposability and Conting We can now state a general strctre theorem for all oter drawings of K k,n with niform rotation systems. Theorem 4. Consider the complete bipartite graph G with vertex bipartition (P, V) sch that P = k and V = n. Given a type in [k] for each pair of vertices in V, there exists an oter drawing of G realizing those types with a niform rotation system if and only if: 1. there exists s {,..., n} and X [k] sch that type(a, b) = X for all pairs a, b with a < s and b s, (in the table this corresponds to maximal rectangle whose cells have all the same entry). the same holds recrsively when the interval [1, n] is replaced by any of the two intervals [1, s 1] and [s, n]. Proof. ( ) Let s rst show that if there exists a drawing, then the types mst satisfy the above strctre. We proceed by indction on n. Pick the smallest s {,..., n} sch that type(1, b) = type(1, s) for all b s. Set X := type(1, s). We claim that type(a, b) = X for all a, b sch that 1 a < s b n. For a = 1 this is jst the condition on s. Now let 1 < a. First sppose that type(1, a) 6= X. We can apply the triple rle on the indices 1, a, b. Since type(1, b) {type(1, a), type(a, b)}, we mst have that type(a, b) = X. 7

8 Now sppose that type(1, a) = X. We have type(1, s 1) = Y 6= X by denition. As in the previos case we obtain type(s 1, b) = X from the triple rle for 1, s 1, b. Applying the triple rle on 1, a, s 1 yields that type(a, s 1) = Y. Now apply the qadrple rle on 1, a, s 1, b. We know that type(1, s 1) = type(a, s 1) = Y, and by denition type(1, b) = X. Hence we mst have that type(a, b) 6= Y. Finally, apply the triple rle on a, s 1, b. We know that type(a, s 1) = Y, type(s 1, b) = X. Since type(a, b) 6= Y, we mst have type(a, b) = X. This yields the claim. ( ) Now given the recrsive strctre, it is not diclt to constrct a drawing. Consider the two sbintervals as a single vertex, then recrsively blow p these two vertices. (See Figre 7 for an illstration). {1,, 3} {4, 5} {4, 5} {1,, 3} {1,, 3} 4 {4, 5} Figre 7: Illstration of the recrsive strctre of the oter drawings in the niform case. The recrsive strctre yields a corollary on the nmber of distinct drawings. Corollary 5 (Conting oter drawings with niform rotation systems). For every pair of integers k, n > 0 denote by T(k, n) the nmber of oter drawings of the complete bipartite graph isomorphic to K k,n with niform rotation systems. Then T(n + 1, k + 1) = where C j is the jth Catalan nmber. n j=0 n+j Cj k j j Proof. The recrsive strctre can be modeled in a labeled binary tree. The root corresponds to [1, n], the sbtrees correspond to the intervals [1, s 1] and [s, n], and the label of the root is type(a, b) for a < s b. The denition implies that the label of the left child of a node is dierent from the label of the node. Leaves have no label. For the nmber T(k, n) of labeled binary trees we therefore get a Catalan-like recrsion T(k, n) = k n 1 k 1 i=1 T(k, i) T(k, n i)+t(k, n 1). The factor k preceeding the k sm acconts for the choice of the label for the root. Using symmetry on the labels we nd that a k 1 fraction of the candidates for the left sbtree comply with the condition k on the labels. The case where the left sbtree only consists of a single leaf node is 8

9 exceptional, in this case there is no label and we have one choice for this sbtree, not jst (1 1/k). This explains the additional smmand. The recrsion T(k, n) = T(k, n 1) + (k 1) n 1 i=1 T(k, i) T(k, n i) together with the initial condition T(k, 1) = 1 yields an array of nmbers which is is listed as entry A10309 in the encyclopedia of integer seqences 1 (OEIS). The stated explicite expression for T(k, n) can be fond there. It can be veried by indction. Note that in the case k =, Corollary 5 provides a bijection between oter drawings with niform rotation systems and combinatorial strctres conted by Schroder nmbers, sch as separable permtations and gillotine partitions. 3 Oter Drawings with k = In this section we deal with oter drawings with k = and arbitrary rotation system. We now have three types of pairs, that we call N, A, and B, as illstrated on Figre 8. The type N (for noncrossing) is new, and is forced whenever the pair corresponds to an inversion in the two permtations. Note again that the three types exactly encode which are the pairs of crossing edges. N b A b a B a a b Figre 8: The three types of pairs for oter drawings of K,n with arbitrary rotation systems. Recall that an oter drawing of K,n, in which no pair is of type N, can be seen as a colored psedoline arrangement as dened previosly. Similarly, an oter drawing of K,n in which some pairs are of type N can be seen as an arrangement of colored monotone crves crossing pairwise at most once. We will refer to arrangement of monotone crves that cross at most once as qasi-psedoline arrangements. The pairs of type N correspond to parallel psedolines. Withot loss of generality, we can sppose that the rst permtation in the rotation system, that is, the order of the psedolines on the left side, is the identity. We denote by π the permtation on the right side. The rst qestion is whether every permtation π is feasible in the sense that there is a drawing of K,n sch that the rotations are (id, π). The answer is yes, two easy constrctions are exemplied in Figre Triples For a, b, c V, with a < b < c, we are interested in the triples of types (type(a, b), type(a, c), type(b, c)) that are possible in an oter drawing of K,n, sch triples are 1 9

10 N N 3 N N 4 N 5 Figre 9: Two oter drawings with rotations (id 5, [3, 4, 1, 5, ]). On the left all non-ntypes are B on the right they are A. called legal. We like to display triples in little tables, e.g., the triple type(a, b) = X, type(a, c) = Y, and type(b, c) = Z is represented as a X Y b Z c. Lemma 6 (decomposable triples). A triple with Y {X, Z} is always legal. There are 15 triples of this kind. Proof. If Y = X take a drawing of K, of type X with vertices a, v and a < v. In this drawing doble the psedoline corresponding to v and cover vertex v by a small circle. Then plg a drawing of K, of type Z with vertices b, c in this circle. This reslts in a drawing of K,3 with the prescribed types. The constrction is very mch as in Figre 7. a X X There are 3 triples b X and in addition for each of the 6 pairs (X, Z) with X 6= Z a triple of each the types c a X X b Z c and a X Z Note that the triples of the latter lemma are decomposable in the sense of Theorem 4. b Z c. Lemma 7. a N A a A B There are exactly two non-decomposable legal triples: b B c and b N c. Proof. From Lemma we know that triples where all entries are A or B are decomposable. If type(a, b) = N, then (a, b) is a non-inversion of π while pairs (a, b) with type(a, b) {A, B} are inversions of π. Both, the set of inversion pairs and the set N N of non-inversion pairs are transitive. Hence, triples of type and N where empty cells represent inversion pairs are impossible. It remains to consider the cases 10

11 where exactly one of X and Z is N and the other two symbols in the triple are A and B. Only the two triples shown in the statement of the lemma remain. With the two lemmas we have classied all 17 legal triples, i.e., all oter drawings of K,3. Observation 8 (Triple rle). Any three vertices of V in an oter drawing of K,n mst indce one of the 17 legal triples of types. 3. Qadrples We aim at a characterization of collections of types that correspond to oter drawings. Already in the case of niform rotations we had to add Lemma 3, a condition for qadrples. In the general case the sitation is more complex than in the niform case, see Figre A N B N N 3 B N B A B B 3 A Figre 10: The qadrple rle from Lemma 3 does not hold in the presence of N types. Reviewing the proof of Lemma 3 we see that in the case discssed there, where given B types are intended to enforce type(a, d) = B, we need that in π element a is before b. This is eqivalent to type(a, b) 6= N. Symmetrically, three A types enforce type(a, d) = A when d is the last in π, i.e., if type(c, d) 6= N. Lemma 9. Consider for vertices a, b, c, d V sch that a < b < c < d. If type(a, b) 6= N and type(a, c) = type(b, c) = type(b, d) = B then type(a, d) = B. If type(c, d) 6= N and type(a, c) = type(b, c) = type(b, d) = A then type(a, d) = A. 3.3 Consistency With the next theorem we show that consistency on triples and qadrples is scient to grant the existence of an oter drawing. Theorem 10 (Consistency of oter drawings for k = ). Consider the complete bipartite graph G with vertex bipartition (P, V) sch that P = and V = n. Given a type in {A, B, N} for each pair of vertices in V, there exists an oter drawing of G realizing those types if and only if all triples are legal and the qadrple rle (Lemma 9) is satised. The proof of this reslt ses the following known reslt on local seqences in psedoline arrangements. Given an arrangement of n psedolines, the local seqences are the permtations α i of [n] \ {i}, i [n], representing the order in which the ith psedoline intersects the n 1 others. 11

12 Lemma 11 (Thm in [5]). The set {α i } i[n] is the set of local seqences of an arrangement of n psedolines if and only if ij inv(α k ) ik inv(α j ) jk inv(α i ), for all triples i, j, k, where inv(α) is the set of inversions of the permtation α. Proof of Theorem 10. The necessity of the condition was already stated in Observation 8 We proceed by giving an algorithm for constrcting an appropriate drawing. First recall from the proof of Lemma 7 that having legal triples implies that the sets of inversion pairs and its complement, the set of non-inversion pairs, are both transitive. Hence, there is a well dened permtation π representing the rotation at p. We aim at dening the local seqences α i that allow an application of Lemma 11. This will yield a psedoline arrangement. A drawing of K,n, however, will only correspond to a qasi-psedoline arrangement. Therefore, we rst constrct a qasipsedoline arrangement T for the pair (π, id), i.e., only the qasi-psedolines corresponding to i and j with type(i, j) = N cross in T. The idea is that appending T on the right side of the qasi-psedoline arrangement of the drawing yields a fll psedoline arrangement. Now x i [n]. Depending on i we partition the set [n] \ i into ve parts. For a type X let X < (i) = {j : j < i and type(j, i) = X} and X > (i) = {j : j > i and type(i, j) = X}, the ve relevant parts are A < (i), A > (i), B < (i), B > (i), and N(i) = N < (i) [ N > (i). The psedoline l i has three parts. The edge incident to p 1 (the red edge) is crossed by psedolines l j with j A > (i) [B < (i). The edge incident to p (the ble edge) is crossed by psedolines l j with j A < (i) [ B > (i). The part of l i belonging to T is crossed by psedolines l j with j N(i). The order of the crossings in the third part, i.e., the order of crossings with psedolines l j with j N(i), is prescribed by T. Regarding the order of the crossings on the second part we know that the lines for j A < (i) have to cross l i from left to right in order of decreasing indices and the lines for j B > (i) have to cross l i from left to right in order of increasing indices, see Figre 11. If j A < (i) and j 0 B > (i), then consistency of triples implies that type(j, j 0 ) {A, B}. If type(j, j 0 ) = A, then on l i the crossing of j 0 has to be left of the crossing of j. If type(j, j 0 ) = B, then on l i the crossing of j has to be left of the crossing of j 0. e d c b a a A A B i b A A B c A B B i B B d e Figre 11: Crossings on the edge i p. The described conditions yield a \left{to{right" relation i sch that for all x, y A < (i) [ B > (i) one of x i y and y i x holds. We have to show that i is acyclic. Since i is a tornament it is enogh to show that i is transitive. 1

13 Sppose there is a cycle x i y i z i x. If x, y < i and z > i, then type(x, i) = type(y, i) = A, moreover, from x i y we get y < x and from y i z i x we get type(x, z) = A, and type(y, z) = B. Since type(i, z) = B 6= N this is a violation of the second qadrple rle of Lemma 9. If x < i and y, z > i, then we have type(i, y) = type(i, z) = B. From this together with y i z we obtain y < z, and z i x i y yields type(x, y) = B, and type(x, z) = A. This is a violation of the rst qadrple rle of Lemma 9. Adding the corresponding argments for the order of crossings on the rst part of line l i we conclde that the permtation α i is niqely determined by the given types and the choice of T. The consistency condition on triples of local seqences needed for the application of Lemma 11 is trivially satised becase legal triples of types correspond to drawings of K,3 and each sch drawing together with T consists of three pairwise crossing psedolines. Since the condition only involves triples and qadrples of vertices in V, this directly yields a polynomial-time algorithm for consistency checking. By observing that the information given by the types is eqivalent to specifying which pairs of edges cross, we directly get the analoge statement for AT-graphs. Corollary 1 (AT-graph realizability). There exists an O(n 4 ) algorithm for deciding the existence of an oter drawing of an AT-graph whose nderlying graph is of the form K,n. Proof. We can check that the three types of pairs in Figre 8 exactly prescribe which pairs of edges cross. Frthermore, given the set of crossing pairs, we can reconstrct the type assignment. We can then check that every triple is legal and that the qadrple rle is satised in time proportional to the nmber of triples and qadrples, hence O(n 4 ). 4 Oter Drawings with k = 3 At the beginning of the previos section we have seen that any pair of rotations is feasible for oter drawings of K,n. This is not tre in the case of k >. For k = 4 the system of rotations ([1, ], [, 1], [1, ], [, 1]) is easily seen to be infeasible. In the case k = 3 it is less obvios that infeasible systems of rotations exist. We also had an ecient characterization of consistent assignments of types for k =. We generalize this to k = 3. We again start by looking at the types for pairs, i.e., at all possible oter drawings of K 3,. We already know that if the rotation system is niform (id, id, id ), then there are three types of drawings. The other three options (id, id, id ), (id, id, id ), and (id, id, id ), each have a niqe drawing. Figre 1 shows the six possible types and associates them to the symbols B α, and W α, for α = 1,, 3. This classication allows s to reason abot the following simple example. Proposition 13. The system (id 4, [4,, 1, 3], [, 4, 3, 1]) is an infeasible set of rotations. 13

14 1 1 1 B B 3 1 B W W 3 1 W Figre 1: The six types of oter drawings of K 3,. Proof. The table of types for the given permtations mst be of the following form: 1 W 1 W 3 W 1 B α W 3 W 1 4 Looking at the sbtable W 1 W 3 B α corresponding to {1,, 3} one can realize that there is only one choice for α, namely α =. The same gre shows that the sbtable B α W W 1 of {, 3, 4} again only allows a niqe choice of α, namely α = 3. This proves that there is no drawing for this set of rotations. Let T 1 be the assignment of types in {A, B, N} corresponding to the oter drawing of K,n with oter vertices (p 3, p ). Similarly, let T be the table corresponding to (p 1, p 3 ) and T 3 the table corresponding to (p, p 1 ). Note that these assignments determine the rotation system (π 1, π, π 3 ). An assignment T of types in {B 1, B, B 3, W 1, W, W 3 } to every pair of vertices in V translates to the following assignments T 1, T, T 3 : B 1 B B 3 W 1 W W 3 T 1 B A A A N N T A B A N A N T 3 A A B N N A Table 1: The projections of the six types of oter drawings of K 3,. This table reveals a dependency between the types in {A, B, N} of every pair of vertices in V for dierent pairs of vertices in P. This dependency is instrmental in the pcoming consistency theorems. For an assignment T of types in {B 1, B, B 3, W 1, W, W 3 } 14

15 to pairs of vertices in V, we will refer to the indced assignments T 1, T, T 3 as the projections of T. We are now ready to state or consistency theorem on oter drawings for k = The consistency theorem for k = 3 Theorem 14 (Consistency of oter drawings for k = 3). Consider the complete bipartite graph G with vertex bipartition (P, V) sch that P = 3 and V = n. Given the assignments T 1, T, T 3 of types in {A, B, N} for the pairs (p 3, p ), (p 1, p 3 ), and (p, p 1 ) of vertices in P, respectively, there exists an oter drawing of G realizing those types if and only if all three assignments obey the triple and qadrple consistency rles in the sense of Theorem 10, for any pair of vertices in V, the assignments correspond to an oter drawing of K 3, in the sense of Table 1. This directly yields the following corollary. Corollary 15 (AT-graph realizability). There exists an O(n 4 ) algorithm for deciding the existence of an oter drawing of an AT-graph whose nderlying graph is of the form K 3,n. Proof. Again, one can check that the three types of pairs in Figre 1 exactly prescribe which pairs of edges cross, and given the set of crossing pairs, we can reconstrct the type assignment. We can then check that every triple is legal and that the qadrple rle is satised in time proportional to the nmber of triples and qadrples, hence O(n 4 ). Proof of Theorem 14. Let s rst note that one direction of the Theorem is easy: if there exists an oter drawing, then the assignments mst be consistent. We now show that consistency of the type assignments is scient for the existence of an oter drawing. Let T be the assignment of types in {B 1, B, B 3, W 1, W, W 3 } to the pairs of V given by Table 1, and let (π 1, π, π 3 ) be the corresponding rotations. The consistency of the tables T and T 3 in the sense of Theorem 10 implies that there are drawings D and D 3 of K,n realizing the type assignments T and T 3. The vertex p 1 (the oter vertex with rotation π 1 ) and its edges form a non-crossing star in both drawings. Let the drawing D live on plane Z and D 3 live on plane Z 3 and consider a xed homeomorphism φ between the planes. There is a homeomorphism ψ : Z Z sch that mapping D via φ ψ to Z 3 yields a sperposition of the two drawings with the following properties. Corresponding vertices are mapped onto each other. The stars of p 1 are mapped onto each other, i.e., the edges at p 1 of the two drawings are represented by the same crves. At each vertex v V the rotation is correct, i.e., we see the edges to p 1, p, p 3 in clockwise order. 15

16 The drawing has no toching edges, i.e., when two edges meet they properly cross in a single point. The drawing D obtained by sperposing D and D 3 is an oter drawing of K 3,n. We color the edges of D as in or gres, for example the edges incident to p are the green edges. In D each color class of edges is a non-crossing star. For the ble and the green this is tre becase the edges come from only one of D and D 3. For the red star it is tre de to constrction. Moreover, all the red-ble and red-green crossings are as prescribed by the original table T. The problem we face is that there is little control on ble-green crossings. Let e red (v), e green (v), e ble (v) be the red, green, and ble edge of v. Claim: For all v, w the parity of the nmber of crossings between e green (v) and e ble (w) in D is prescribed by T, i.e., if T reqires a crossing between two edges, then they have an odd nmber of crossings and an even nmber of crossings otherwise. Consider the crves e green (v)[e red (v) and e ble (w)[e red (w). The rotations prescribe whether the nmber of intersections of the two crves is odd or even. Hence, the parity is respected by D. The crossings of the pairs (e green (v), e red (w)), (e red (v), e ble (w)) in D are as prescribed by T(v, w). Hence, the parity of the nmber of crossings of e green (v) and e ble (w) in D is also prescribed. 4 Becase the rotation at v in D is correct we also note: If e green (v) and e ble (v) cross, then the nmber of crossings is even. Hence, if D has no pair of a green and a ble edge crossing more than once, then D is an oter drawing realizing the types given by T. Now consider a pair of edges e green (v) and e ble (w) crossing more than once, v = w is allowed. Use a homeomorphism of the plane to make e green (v) a horizontal straight line segment, see Figre 13. (In the literatre intersection patterns of two simple crves in the plane are often called meanders. They are of interest in enmerative and algebraic combinatorics.) The intersections with e green (v) sbdivide e ble (w) into a family of arcs and two extremal pieces. A ble arc denes an interval on the green edge, this is the interval of the arc. An arc together with its interval enclose a bonded region, this is the region of the arc. Fact. Apart from intersecting intervals any two regions of arcs over a xed green edge e green (v) are either disjoint or nested. Becase ble edges are pairwise disjoint this also holds if the arcs are dened by dierent ble edges. An inclsionwise minimal region of an arc is a lens. Since D has a nite nmber of crossings and hence a nite nmber of regions we have: Fact. Every region of an arc contains a lens. Consider a lens L formed by pair of a green and a ble edge in D. Sppose that L is an empty lens, i.e., there is no vertex V inside of L. It follows that the bondary of L is only intersected by red edges, moreover, if a red edge e intersects the bondary of L on the green side, then e also intersects L on the ble side. Therefore, we can make e green (v) and e ble (w) switch sides at L and with small deformations at the two crossings get rid of them. It is important to note that the switch at a lens does not change the types. In particlar after the switch the drawing still represents assignment T. 16

17 w v Figre 13: A meander with 7 empty lenses. Apply switching operations ntil the drawing D 0 obtained by switching has the property that every lens in D 0 contains a vertex. In the following we show that D 0 has no lens. For the proof we se that the table T 1 corresponding to (π 3, π ) also has to be consistent. Since D 0 has no lens we conclde that it is an oter drawing that realizes the types given by T. The existence of sch a drawing was the statement of the theorem. Let D 0 be a drawing with the property that every lens contains a vertex. Before going into the proof that there is no lens in D 0 we x some additional notation. For a given green edge e green (v) the regions dened by ble arcs over e green (v) can be classied as above, below, and wrapping. A wrapping region is a region with one contact between e green (v) and e ble (w) from above and one contact from below. Lemma 16. There is no wrapping region. Proof. Let R be the wrapping region formed by a ble edge wrapping arond e green (v) and note that v R. If the wrapping ble edge is e ble (v), then e red (v) has to intersect one of e green (v) and e ble (v) to leave the region. This is not allowed. If the wrapping ble edge is e ble () with 6= v, then e ble (v) has to intersect one of e green (v) and e ble () to leave the region. Again, this is not allowed. With a similar proof we get: Lemma 17. Vertex w is not contained in the region dened by an arc of e ble (w) over e green (v). Proof. The green edge of w wold be trapped in sch a region. For a region dened by an arc on e ble (w) we speak of a forward or backward region depending on the direction of the arc above e green (v). Formally: label the t crossings of e green (v) and e ble (w) as 1,.., t according to the order on e ble (w). Arcs correspond to consective crossings. An arc [i, i + 1] is forward if crossing i is to the left of crossing i + 1 on e green (v), otherwise it is backward. The permtation of 1,.., t obtained by reading the crossings from v to p along e green (v) is called the meander permtation and denoted σ green (v). A region is a relative lens for e green (v) and e ble (w) if it is above or below e green (v) and minimal in the nesting order of regions dened by e green (v) and e ble (w). In the seqel we sometimes abse notation by talking of lenses when we mean relative lenses. 17

18 Proposition 18. For all v V, the green and ble edges of v do not cross in D 0. Proof. If e green (v) and e ble (v) cross, then there is at least one ble arcs on e ble (w) over e green (v) and, hence, there are regions. From the order of the three oter vertices and the fact that e red (v) has no crossing with the two other edges of v we conclde that at the last crossing of e ble (v) and e green (v) the ble edge is crossing e green (v) downwards. Since at the rst crossing the ble edge is crossing pwards there is an arc and conseqently also a lens above e green (v). Sppose there is a forward lens above e green (v). Let be a vertex in the lens. Vertex is below the line of v in the green-red arrangement. Hence, either T 3 (, v) = B or T 3 (, v) = N and v < 1. Vertex is below the forward arc of e ble (v) forming the lens. Hence, it is below the line of v in the red-ble arrangement, and either T (, v) = B or T (, v) = N and < 1 v. From Table 1 we infer that the only legal assignment is T(, v) = W 1 and the projections are T (, v) = N and T 3 (, v) = N. However, there is no consistent choice for the order of and v in < 1. This shows that there is no forward lens above e green (v). Now let [i, i + 1] be a backward lens above e green (v). Let be a vertex in the lens. We distingish whether the last crossing on the ble edge is to the right/left of the lens on e green (v), see Figre 14. v j i k v i w j Figre 14: The two cases for a backward lens above e green (v). Dashed crves indicate the order of some crossings along the corresponding edges. Sppose the last crossing k on the ble edge is to the right of the lens, Figre 14 (left). To get from the arc [i, i + 1] to k the edge e ble (v) has to cross e green (v) pwards at some crossing j left of i + 1. The edge e ble () has to stay disjoint from e ble (v), therefore, after crossing e green (v) to leave the lens it has a second crossing left of j. The edge e green () has a crossing with e ble (v) in the arc [i, i + 1] and another one between j and k. Now consider the edge e red (). If it leaves the lens throgh e green (v), then it has entered a region whose bondary consists of a piece of e ble () and a piece of e green (v). Edge e red () is disjoint from e ble () and it has already sed its niqe crossing with e green (v). Hence, e red () has to leave the lens throgh the ble arc on e ble (v). In this case, however, e red () enters a region whose bondary consists of a piece of e ble (v) and a piece of e green (). Again e red () is trapped. Hence, this congration is impossible. Sppose the last crossing on the ble edge is to the left of the lens, Figre 14 (right). To get from v to the backward arc [i, i+1] edge e ble (v) has to cross e green (v) downwards at some crossing j right of i. Let w be a vertex in the region of some ble arc [k, k + 1] 18

19 below e green (v) with the property that j k < i. The edges e red () and e red (w) both need at least two crossings with the nion of e green (v) and e ble (v). Hence, they both cross e green (v) and e ble (v). The order of the red edges at p 1 implies < 1 v < 1 w. Matching these data with the drawings of Figre 1 we nd that T(, v) = B 3 and T(v, w) = B. These types together with the already known permtation π 1 imply that π and π 3 also eqal π 1. Hence, the triple is a niform system, and we mst have T(, w) {B, B 3 }. Since edge e ble (w) has to follow the `tnnel' prescribed by e ble (v) there is a crossing of edges e green (v) and e ble (w). There is also a crossing of e green (v) and e red (w). However, neither in B nor in B 3 we see crossings of e green (1) with e ble () and e red (). Hence, again the congration is impossible. We now come to the discssion of the general case. Proposition 19. For all v, w V, there is no lens formed by e green (v) and e ble (w) in D 0. Or proof of this proposition nfortnately depends on a lengthy case analysis, an otline of which is given in the following sbsection. 4. Otline of the Proof of Proposition 19 Sppose that there is a lens L formed by e green (v) and e ble (w) in D 0. From Proposition 18 we know that v 6= w. In fact the proposition implies that the star of every vertex is non-intersecting. For emphasis we collect this and the other restrictions on crossings in D 0 in a list. (1) Edges of the same color do not cross. () Edges of dierent color that belong to the same vertex do not cross. (3) Red edges have at most one intersection with any other edge. These properties will be crcial throghot the argment. We also know that there are no wrapping regions (Lemma 16) and the vertex in a region or lens is always dierent from the vertices of the edges dening the region (Lemma 17). In section 4.3.1, we discss eight congrations that may appear in a meander of e ble (w) over e green (v). The eight congrations shown in Figre 15 correspond to the simplest meanders for the following three binary decisions: Vertex w is above/below edge e green (v). The last crossing is to the left/right of the rst crossing. At the last crossing e ble (w) is ctting e green (v) pward/downward. For example the meander labeled VII corresponds to below/left/p. In the case discssion we impose additional conditions on vertices that are contained in the regions dened by arcs of e ble (w) over e green (v). These conditions either ask for S 1 (w) or for 6 S 1 (w). The conditions are so that they are satised for free if the respective regions are relative lenses of e ble (w) over e green (v). In each of the cases we show that the T 1 projections yield an illegal table. Hence the congrations do not appear in the drawing D 0. 19

20 I II III IV V VI VII VIII Figre 15: The eight congrations for the rst phase of the case analysis. In the second part we se the reslts from the rst part to show that meanders of the drawing D 0 have no inections, i.e., in the meander permtation we never see i 1 and i + 1 on the same side of i. This is split in the following two lemmas. The rst, easy one, is the following. Lemma 0. If a meander permtation σ green (v) has an inection, then it has one of the for trn-back patterns shown in Figre 16. Proof. Sppose that a meander permtation σ = σ green (v) has an inection at i. We discss the case where i + 1 < σ i 1 < σ i and the ble arc [i + 1, i] of e ble (w) is above e green (v). All the other cases can be treated with symmetrical argments. First note that i 1 > 1, otherwise, w is in the region dened by the arc [i + 1, i]. This is impossible (Lemma 17). Now sppose that i 1 < σ i < σ i. Let x be a vertex in the region dened by the arc [i 1, i ]. Conned by e ble (w) the edge e ble (x) has to cross e green (v) at least three times and it contains an arc whose region contains x. This is impossible (Lemma 17). Hence, i + 1 < σ i < σ i 1 < σ i and the meander contains the second of the trn-back patterns shown in Figre 16. Other cases of inections are related to the other cases of trn-back patterns. The proof of the second lemma involves a delicate case analysis and is deferred to Sbsection Lemma 1. A meander permtation has no trn-back. TB 1 TB TB 4 TB 3 Figre 16: The for trn-back pattern. The conseqence of Lemma 1 is that the meanders in D 0 are very simple, their meander permtations are either the identity permtation or the reverse of the identity. In particlar every arc denes a relative lens. It then follows that each meander in D 0 can be classied according to the three binary decisions mentioned above. Hence, it falls into one of the congrations that have been discssed in Sbsection The additional conditions are satised becase the arcs all dene relative lenses. Hence, there are no nontrivial meanders, i.e., every pair of a green and a ble edge crosses at most once in D 0. This concldes the proof of Proposition 19. 0

21 4.3 Case analysis for the proof of Proposition The eight basic configrations We now deal with special instances of the congrations from Figre 15. Case I. The rst intersection along e ble (w) is downwards and the last intersection is pwards and to the right of the rst intersection along e green (v). Its last intersection with e green (v) being pward forces e ble (w) to intersect e red (v) on its nal piece. Therefore, edge e green (w) has to trn arond v as shown in Figre 17. Now consider e red (w), its rst crossing (among those relevant to the argment) has to be with e green (v). With this crossing, however, e red (w) gets separated from its destination p 1 by the nion of e ble (w) and e green (v). Therefore, this case is impossible. Case II. The rst intersection along e ble (w) is pwards and the last intersection is downwards and to the right of the rst along e green (v) and there is a forward arc of e ble (w) above e green (v) whose region contains a vertex 6 S 1 (w). Edge e green (w) has to trn arond v and also e red (w) has to cross e ble (v), see Figre 19. Now consider a vertex in a forward region of e ble (w) above e green (v). Vertex is not nder an arc of e ble () (Lemma 17) and there is at most one crossing of e ble () and e red (w). Therefore, e ble () behaves as shown in the sketch, i.e., w < 3 < 3 v. Also w < < v. From the intersections of green and ble edges we conclde that the projections T 1 are as given by the table on the right. This table is not legal. Hence, this case is impossible. Case III. The rst intersection along e ble (w) is pwards and the last intersection is downwards and to the left of the rst along e green (v) and there is a backward arc of e ble (w) above e green (v) whose region contains a vertex S 1 (w). Edge e green (w) has to stay below e green (v). Now consider a vertex in a backward region of e ble (w) above e green (v). Vertex is not nder an arc of e ble () and we exclde the congration of Case I. Therefore, e ble () behaves as shown in Figre 18, i.e., < 3 w < 3 v. It follows that e green () has to stay above e green (v) and < v < w. From the intersections of green and ble edges we conclde that the projections T 1 are as given by the table on the right. This tables is not legal. Hence, this case is impossible. 1 v w Figre 17: Illstration for Case I. v w Figre 18: Illstration for Case II. w A v B A v w Figre 19: Illstration for Case III. B A w N v..

22 Case IV. The rst intersection along e ble (w) is downwards and the last intersection is pwards and to the left of the rst along e green (v) and there is a backward arc of e ble (w) below e green (v) whose region contains a vertex 6 S 1 (w). Edge e ble (w) has to cross e red (v) and the other edges of w have no intersections with edges of v. Edge e ble () either trn arond w or it crosses e red (v). In the rst case e green () has trn arond v. Now, e red () has to cross e ble (w) and gets separated from its destination p 1 by the nion ofe ble (w) and e green (). Hence, there is no possible roting for edge e red () respecting the restrictions on crossings in D 0. In the second case e ble () crosses e red (v) v Figre 0: Illstration for Case IV. v N B w A and v < 3 w < 3. From the green edges we obtain w < v <. The intersections of green and ble edges, see Figre 0 imply that the projections T 1 are as given by the table below the gre. This table is not legal. Hence, this case is impossible. Case V. The rst intersection along e ble (w) is downwards and the last intersection is downward and to the right of the rst. Moreover, along e green (v) there is a forward arc of e ble (w) below e green (v) whose region contains a vertex S 1 (w) and frther to the right a forward arc of e ble (w) above e green (v) containing a vertex x 6 S 1 (w). If < 3 w, then edge e ble (w) makes sre that x 6 S 1 (). Therefore, we have Case II with v,, and x. If w < 3 < 3 v, then w S 1 (). Therefore, we have a Case III. Now let w < 3 v < 3. Since e ble (x) is crossing e green (v) downwards we have e green (x) above e green (v) and a crossing of e ble (w) with e green (x). Therefore w < 3 x. If w < 3 x < 3 v, then the intersections of green and ble edges areas shown in Figre 1. The projections T 1 are given by the table below the gre. This table is not legal. Hence, this case is impossible. If v < 3 x < 3, then we have Case IV with x and. Finally, if < 3 x, then the green and ble edges behave as shown in Figre. Now, e red (w) wold have to cross one of e ble (x) and e green (v) twice to get to its destination p 1. w v Figre 1: Illstration for Case V. w v w A x B A Figre : Case V, nd illstration. w.. x x