topological drawings of complete bipartite graphs · 1234512345 figure 1: two drawings of k 3;5...
TRANSCRIPT
Topological Drawingsof Complete Bipartite Graphs
Jean Cardinal and Stefan Felsner
April 10, 2019
Abstract
Topological drawings are natural representations of graphs in the plane, wherevertices are represented by points, and edges by curves connecting the points. Topo-logical drawings of complete graphs and of complete bipartite graphs have beenstudied extensively in the context of crossing number problems. We consider a nat-ural class of simple topological drawings of complete bipartite graphs, in which werequire that one side of the vertex set bipartition lies on the outer boundary of thedrawing.
We investigate the combinatorics of such drawings. For this purpose, we definecombinatorial encodings of the drawings by enumerating the distinct drawings ofsubgraphs isomorphic to K2,2 and K3,2, and investigate the constraints they mustsatisfy. We prove in particular that for complete bipartite graphs of the form K2,n
and K3,n, such an encoding corresponds to a drawing if and only if it obeys certainconsistency conditions on triples and quadruples.
For the case of K2,n, we make use of tools from the theory of pseudoline arrange-ments. For K3,n we show that drawings of the two induced K2,n can be manipulatedto yield a drawing of K3,n.
In the general case of Kk,n with k ≥ 2, we completely characterize and enumer-ate drawings in which the order of the edges around each vertex is the same forvertices on the same side of the bipartition. We also investigate straight-line draw-ings of K3,n and relate the realizability problem to that of stretchability of tropicalpseudoline arrangements.
1 Introduction
We consider topological graph drawings, which are drawings of simple undirected graphswhere vertices are represented by points in the plane, and edges are represented by sim-ple curves that connect the corresponding points. We typically restrict those drawingsto satisfy some natural nondegeneracy conditions. In particular, we consider simpledrawings, in which every pair of edges intersect at most once. A common vertex countsas an intersection.
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While being perhaps the most natural and the most used representations of graphs,simple drawings are far from being understood from the combinatorial point of view.For the smallest number of edge crossings in a simple topological drawing of Kn [5,2, 1] or of Kk,n [17, 3] there are long standing conjectures but the actual minimumremains unknown.
In order to cope with the inherent complexity of the drawings, it is useful to considercombinatorial abstractions. Those abstractions are discrete structures encoding somefeatures of a drawing. One such abstraction, introduced by Kratochvıl, Lubiw, andNesetril, is called abstract topological graphs (AT-graph) [6]. An AT-graph consists of agraph (V,E) together with a set X ⊆
(E2
). A topological drawing is said to realize an
AT-graph if the pairs of edges that cross are exactly those in X . Another abstractionof a topological drawing is called the rotation system. The rotation system associates acircular permutation with every vertex v, which in a realization must correspond to theorder in which the neighbors of v are connected to v. Natural realizability problemsare: 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 [7].
For simple topological drawings of complete graphs, the two abstractions are actu-ally equivalent [13]. It is possible to reconstruct the set of crossing pairs of edges bylooking at the rotation system, and vice-versa. Kyncl recently proved the remarkable re-sult that a complete AT-graph (an AT-graph for which the underlying graph is complete)can be realized as a simple topological drawing of Kn if and only if all the AT-subgraphson at most 6 vertices are realizable [8, 9]. This directly yields a polynomial-time al-gorithm for the realizability problem. While this provides a key insight on topologicaldrawings of complete graphs, similar realizability problems already appear much moredifficult when they involve complete bipartite graphs. In that case, knowing the rotationsystem is not sufficient for recontructing the intersecting pairs of edges.
We propose a fine-grained analysis of simple topological drawings of complete bi-partite graphs. In order to make the analysis more tractable, we introduce a naturalrestriction on the drawings, by requiring that one side of the vertex set bipartition lieson a circle at infinity. This gives rise to meaningful, yet complex enough, combinatorialstructures.
Definitions. We wish to draw the complete bipartite graph Kk,n in the plane in sucha way that:
1. vertices are represented by points,
2. edges are continuous curves that connect those points, and do not contain anyother vertices than their two endpoints
3. no more than two edges intersect in one point,
4. edges pairwise intersect at most once; in particular, edges incident to the samevertex intersect only at this vertex,
5. the k vertices of one side of the bipartition lie on the outer boundary of the draw-ing.
2
Properties 1–4 are the usual requirements for simple topological drawings also knownas good drawings. As we will see, property 5 leads to drawings with interesting combi-natorial structures. Throughout this paper, the term drawing always refers to drawingssatisfying the above properties.
The set of vertices of a bipartite graph Kk,n will be denoted by P ∪ V , where Pand V are the two sides of the bipartition, with |P | = k and |V | = n. When we considera given drawing, we will use the word “vertex” and “edge” to denote both the vertexor edge of the graph, and their representation as points and curves. Without loss ofgenerality, we can assume that the k outer vertices p1, . . . , pk lie in clockwise order onthe boundary of a disk that contains all the edges, or on the line at infinity. The verticesof V are labeled 1, . . . , n. An example of such a drawing is given in Figure 1.
13254
2143513254
21435
12345 12345
Figure 1: Two drawings of K3,5 satisfying the constraints. In both drawings the rotationsystem is (12345, 21435, 13254).
The rotation system of the drawing is a sequence of k permutations on n elements as-sociated with the vertices of P in clockwise order. For each vertex of P , its permutationencodes the (say) counterclockwise order in which the n vertices of V are connectedto it. Due to our last constraint on the drawings, the rotations of the k vertices of Paround each vertex of V are fixed and identical, they reflect the clockwise order ofp1, . . . , pk on the boundary.
Unlike for complete graphs, the rotation system of a drawing of a complete bipartitegraph does not completely determine which pairs of edges are intersecting. This isexemplified with the two drawings in Figure 1.
Results. The paper is organized as follows. In Section 2, we consider drawings with auniform rotation system, in which the k permutations of the vertices of P are all equalto the identity. In this case, we can state a general structure theorem that allows us tocompletely characterize and count drawings of arbitrary bipartite graphs Kk,n.
In Section 3, we consider drawings of K2,n with arbitrary rotation systems. Weconsider a natural combinatorial encoding of such drawings, and state two necessaryconsistency conditions involving triples and quadruples of points in V . We show thatthese conditions are also sufficient, yielding a polynomial-time algorithm for checkingconsistency of a drawing.
In Section 4, we consider drawings of K3,n and study a complete classification ofall drawings of K3,3. This directly gives a necessary consistency condition on triplesof vertices in V . We finally provide an additional necessary consistency condition onquadruples, and the proof that the consistency conditions on triples and quadruples aresufficient for drawings of K3,n.
3
2 Drawings with uniform rotation system
We first consider the case where k is arbitrary but the rotation system is uniform, that is,the permutation around each of the k vertices pi is the same. Without loss of generalitywe assume that this permutation is the identity permutation on [n].
In a given drawing, each of the n vertices of V splits the plane into k regionsQ1, Q2, . . . , Qk, where each Qi is bounded by the edges from v to pi and pi+1, withthe understanding that pk+1 = p1. We denote by Qi(v) the ith region defined by ver-tex v and further on call these regions quadrants. We let type(a, b) = i, for a, b ∈ V andi ∈ [k], whenever a ∈ Qi(b). This implies that b ∈ Qi(a), see Figure 2. Indeed if a < band j 6= i + 1, then edge pi+1b has to intersect all the edges pja, while edge pjb has toavoid pi+1b until they meet in b. It follows that none of the edges pjb can intersect pi+1a.This shows that a ∈ Qi(b).
a
b
Q4(a)
Q5(a) Q2(a)
Q3(a)
Q6(a)
Q1(a)
p1
p4
p2
p3p5
p6
Figure 2: Having placed b in Q4(a) the crossing pairs of edges and the order of crossingson each edge is prescribed. In particular a ∈ Q4(b). On the right a symmetric drawingof the pair.
Observation 1 (Symmetry).For all a, b in uniform rotation systems: type(a, b) = type(b, a).
For the case k = 2, we have exactly two types of pairs, that we will denote by A andB. The two types are illustrated on Figure 3.
The drawings of K2,n with uniform rotations can be viewed as colored pseudolinearrangements, where:
• each pseudoline is split into two segments of distinct colors,
• no crossing is monochromatic.
This is illustrated on Figure 4. The pseudoline of a vertex v ∈ V is denotedby `(v). The left (red) and right (blue) parts of this pseudoline are denote by `L(v)and `R(v). Now having type(a, b) = type(b, a) = A means that b lies above `(a) and alies above `(b). While having type(a, b) = type(b, a) = B means that b lies below `(a)and a lies below `(b).
4
a b
b aA B
Figure 3: The two types of pairs for drawings of K2,n with uniform rotation systems.
1
2
3
4
4
3
2
1
1 B B B
2 B A
3 A
4
Figure 4: Drawing K2,4 as a colored pseudoline arrangement. The type of each pair isgiven in the table on the right.
2.1 The triple rule
Lemma 2 (Triple rule1).For uniform 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 = 2. If type(a, b) 6= type(b, c) there is nothing to show since there areonly two types. Without loss of generality, suppose that type(a, b) = type(b, c) = B.This situation is illustrated in the left part of Figure 5. The pseudoline `(c) must cross`(b) on `R(b), otherwise we would have type(b, c) = A. Hence the point c is on theright of this intersection. Pseudoline `(a) must cross `(b) on `L(b), and a is left of thisintersection. It follows that `(a) and `(c) cross on `R(a) and `L(c), i.e., type(a, c) = B.
a
c
b
c a
b
ca
pj
Figure 5: Illustrations for the k = 2 case of Lemma 2 (left), and the k > 2 case ofLemma 2 (right).
Case k > 2. For the general case assume that type(a, b) = i and type(a, c) = j. Ifi = j there is nothing to show. Now suppose i 6= j. From c ∈ Qj(a) it follows that pj+1aand pjc are disjoint. Edges pjb and pjc only share the endpoint pj, hence c has to bein the region delimited by pjb and pj+1a, see the right part of Figure 5. This region iscontained in Qj(b), whence type(b, c) = j.
1Triples for uniform rotation systems are decomposable, see Lemma 6 for tables.
5
2.2 The quadruple rule
Lemma 3 (Quadruple rule).For four vertices a, b, c, d ∈ V with a < b < c < d and X ∈ {A,B}:
if type(a, c) = type(b, c) = type(b, d) = X then type(a, d) = X.
Proof. Case k = 2. Suppose, without loss of generality, that X = B. Consider thepseudolines representing b and c with their crossing at `R(b) ∪ `L(c). Coming fromthe left the edge `L(d) has to avoid `L(c) and therefore intersects `R(b). On `R(b) thecrossing with `L(c) is left of the crossing with `L(d), see Figure 6. Symmetrically fromthe right the edge `R(a) has to intersect `L(c) and this intersection is left of `R(b)∪`L(c).To reach the crossings with `L(c) and `R(b) edges `R(a) and `L(d) have to intersect,hence, type(a, d) = B.
a
b
c
d a
b
c
d
Figure 6: Illustration for the k > 2 case of Lemma 2.
Case k > 2. In the general case, we let X = i, and consider the pseudoline arrange-ment defined by the two successive vertices pi and pi+1 of P defining the quadrants Qi.Proving that type(a, d) = i, that is, that a ∈ Qi(d), can be done as above for k = 2 onthe drawing of K2,n induced by {pi, pi+1} and V .
2.3 Decomposability and Counting
We can now state a general structure theorem for all drawings of Kk,n with uniformrotation systems.
Theorem 4. Given a type for each pair of vertices in V , there exists a drawing realizingthose types if and only if:
1. there exists s ∈ {2, . . . , n} and X ∈ [k] such that type(a, b) = X for all pairs a, bwith a < s and b ≥ s, (in the table this corresponds to maximal rectangle whose cellshave all the same entry)
2. the same holds recursively when the interval [1, n] is replaced by any of the twointervals [1, s− 1] and [s, n].
Proof. (⇒) Let us first show that if there exists a drawing, then the types must satisfythe above structure. We proceed by induction on n. Pick the smallest s ∈ {2, . . . , n}such that type(1, b) = type(1, s) for all b ≥ s. Set X := type(1, s). We claim thattype(a, b) = X for all a, b such that 1 ≤ a < s ≤ b ≤ n. For a = 1 this is just thecondition on s. Now let 1 < a.
First suppose that type(1, a) 6= X. We can apply the triple rule on the indices 1, a, b.Since type(1, b) ∈ {type(1, a), type(a, b)}, we must have that type(a, b) = X.
6
Now suppose that type(1, a) = X. We have type(1, s − 1) = Y 6= X by definition.As in the previous case we obtain type(s − 1, b) = X from the triple rule for 1, s − 1, b.Applying the triple rule on 1, a, s− 1 yields that type(a, s− 1) = Y .
Now apply the quadruple rule on 1, a, s − 1, b. We know that type(1, s − 1) =type(a, s − 1) = Y , and by definition type(1, b) = X. Hence we must have thattype(a, b) 6= Y .
Finally, apply the triple rule 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 must have type(a, b) = X. This yields the claim.
(⇐) Now given the recursive structure, it is not difficult to construct a drawing.Consider the two subintervals as a single vertex, then recursively blow up these twovertices. (See Figure 7 for an illustration).
1 23 45 {1, 2, 3}
{4, 5}{1, 2, 3}
{1, 2, 3}
{4, 5}
{4, 5}
Figure 7: Illustration of the recursive structure of the drawings in the uniform case.
The recursive structure yields a corollary on the number of distinct drawings.
Corollary 5 (Counting drawings with uniform rotation systems). For every pair of in-tegers k, n > 0 denote by T (k, n) the number of topological drawings of the completebipartite graph isomorphic to Kk,n with uniform rotation systems. Then
T (n+ 1, k + 1) =n∑j=0
(n+j2j
)Cj k
j
where Cj is the jth Catalan number.
Proof. The recursive structure can be modeled in a labeled binary tree. The root cor-responds to [1, n], the subtrees correspond to the intervals [1, s − 1] and [s, n], and thelabel of the root is type(a, b) for a < s ≤ b. The definition implies that the label of theleft child of a node is different from the label of the node. Leaves have no label.
For the number T (k, n) of labeled binary trees we therfore get a Catalan like recur-sion T (k, n) = k
∑n−1i=1
(k−1kT (k, i)
)·T (k, n− i)+T (k, n−1) The factor k preceeding the
sum accounts for the choice of the label for the root. Using symmetry on the labels wefind that a k−1
kfraction of the candidates for the left subtree comply with the condition
on the labels. The case where the left subtree only consists of a single leaf node is
7
exceptional, in this case there is no label and we have one choice for this subtree, notjust (1− 1/k). This explains the additional summand. The recursion
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 numbers which is islisted as entry A103209 in the encyclopedia of integer sequences2 (OEIS). The statedexplicite expression for T (k, n) can be found there. It can be verified by induction.
3 Drawings with k = 2
In this section we deal with drawings with k = 2 and arbitrary rotation system. Wenow have three types of pairs, that we call N , A, and B, as illustrated on Figure 8.The type N (for noncrossing) is new, and is forced whenever the pair corresponds toan inversion in the two permutations.
a b
b a
a
bN A B
Figure 8: The three types of pairs for drawings of K2,n with arbitrary rotation systems.
Recall that a drawing of K2,n, in which no pair is of type N , can be seen as a coloredpseudoline arrangement as defined previously. Similarly, a drawing of K2,n in whichsome pairs are of type N can be seen as an arrangement of colored monotone curvescrossing pairwise at most once. We will refer to arrangement of monotone curves thatcross at most once as quasi-pseudoline arrangements. The pairs of type N correspond toparallel pseudolines. Without loss of generality, we can suppose that the first permuta-tion in the rotation system, that is, the order of the pseudolines on the left side, is theidentity. We denote by π the permutation on the right side.
The first question is whether every permutation π is feasible in the sense that thereis a drawing of K2,n such that the rotations are (id, π). The answer is yes, two easyconstructions are exemplified in Figure 9
3.1 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 a topological drawing of K2,n, such triples arecalled 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
.
2www.oeis.org
8
34152
34152
4321
54321
5
1 N N
2
3 N N
4 N
5
Figure 9: Two drawings with rotations (id5, [3, 4, 1, 5, 2]). On the left all non-N -typesare B on the right they are A.
Lemma 6 (decomposable triples3).A triple with Y ∈ {X,Z} is always legal. There are 15 triples of this kind.
Proof. If Y = X take a drawing of K2,2 of type X with vertices a, v and a < v. In thisdrawing double the pseudoline corresponding to v and cover vertex v by a small circle.Then plug a drawing of K2,2 of type Z with vertices b, c in this circle. This results in adrawing of K2,3 with the prescribed types. The construction is very much as in Figure 7.
There are 3 triplesa X X
b X
c
and in addition for each of the 6 pairs (X,Z) with
X 6= Z a triple of each the 2 typesa X X
b Z
c
anda X Z
b Z
c
.
Lemma 7.
There are exactly two non-decomposable legal triples:
a N A
b B
c
and
a A B
b N
c
.
Proof. From Lemma 2 we know that triples where all entries are A or B are decom-posable. If type(a, b) = N , then (a, b) is a non-inversion of π while pairs (a, b) withtype(a, b) ∈ {A,B} are inversions of π. Both, the set of inversion pairs and the set of
non-inversion pairs are transitive. Hence, triples of typeN
andN
Nwhere
empty cells represent inversion pairs are impossible. It remains to consider the caseswhere 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 classified all 17 legal triples, i.e., all topologicaldrawings of K2,3.
Observation 8 (Triple rule). Any three vertices of V in a drawing of K2,n must induceone of the 17 legal triples of types.
3These triples of this lemma are decomposable in the sense of Theorem 4
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3.2 Quadruples
We aim at a characterization of collections of types that correspond to drawings. Al-ready in the case of uniform rotations we had to add Lemma 3, a condition for quadru-ples. In the general case the situation is more complex than in the uniform case, seeFigure 10.
4
3
2
1
4
3
2
1
2
3
1
4
3
4
1
2
1 A N B
2 N N
3 B
4
1 N B A
2 B B
3 A
4
Figure 10: The quadruple rule from Lemma 3 does not hold in the presence of N types.
Reviewing the proof of Lemma 3 we see that in the case discussed there, wheregiven B types are intended to enforce type(a, d) = B, we need that in π element a isbefore b. This is equivalent to type(a, b) 6= N . Symmetrically, three A types enforcetype(a, d) = A when d is the last in π, i.e., if type(c, d) 6= N .
Lemma 9. Consider four vertices a, b, c, d ∈ V such 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 quadruples is enoughto grant the existence of a drawing.
Theorem 10 (Consistency of drawings for k = 2). Given a type for each pair of verticesin V , there exists a drawing realizing those types if and only if all triples are legal and thequadruple rule (Lemma 9) is satisfied.
The proof of this result uses the following known result on local sequences in pseudo-line arrangements. Given an arrangement of n pseudolines, the local sequences are thepermutations αi of [n] \ {i}, i ∈ [n], representing the order in which the ith pseudolineintersects the n− 1 others.
Lemma 11 (Thm. 6.17 in [4]). The set {αi}i∈[n] is the set of local sequences of an ar-rangement of n pseudolines if and only if
ij ∈ inv(αk)⇔ ik ∈ inv(αj)⇔ jk ∈ inv(αi),
for all triples i, j, k, where inv(α) is the set of inversions of the permutation α.
Proof of Theorem 10. The necessity of the condition was already stated in Observation 8We proceed by giving an algorithm for constructing an appropriate drawing. First
recall from the proof of Lemma 7 that having legal triples implies that the sets of in-version pairs and its complement, the set of non-inversion pairs, are both transitive.Hence, there is a well defined permutation π representing the rotation at p2.
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We aim at defining the local sequences αi that allow an application of Lemma 11.This will yield a pseudoline arrangement. A drawing of K2,n, however, will only cor-respond to a quasi-pseudoline arrangement. Therefore, we first construct a quasi-pseudoline arrangement T for the pair (π, id), i.e., only the quasi-pseudolines corre-sponding to i and j with type(i, j) = N cross in T . The idea is that appending T on theright side of the quasi-pseudoline arrangement of the drawing yields a full pseudolinearrangement.
Now fix i ∈ [n]. Depending on iwe partition the set [n]\i into five parts. For a typeXlet X<(i) = {j : j < i and type(j, i) = X} and X>(i) = {j : j > i and type(i, j) = X},the five relevant parts are A<(i), A>(i), B<(i), B>(i), and N(i) = N<(i) ∪ N>(i). Thepseudoline `i has three parts. The edge incident to p1 (the red edge) is crossed bypseudolines `j with j ∈ A>(i) ∪ B<(i). The edge incident to p2 (the blue edge) iscrossed by pseudolines `j with j ∈ A<(i) ∪ B>(i). The part of `i belonging to T iscrossed by pseudolines `j with j ∈ N(i). The order of the crossings in the third part,i.e., the order of crossings with pseudolines `j with j ∈ N(i), is prescribed by T .
Regarding the order of the crossings on the second part we know that the lines forj ∈ A<(i) have to cross `i from left to right in order of decreasing indices and thelines for j ∈ B>(i) have to cross `i from left to right in order of increasing indices,see Figure 11. If j ∈ A<(i) and j′ ∈ B>(i), then consistency of triples implies thattype(j, j′) ∈ {A,B}. If type(j, j′) = A, then on `i the crossing of j′ has to be left of thecrossing of j. If type(j, j′) = B, then on `i the crossing of j has to be left of the crossingof j′.
a
b
c
d e
i B B
BBA
A A
A A B
B
a
d
c
e
b
i
Figure 11: Crossings on the edge i p2.
The described conditions yield a “left–to–right” relation →i such 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 tournament it is enough to show that→i is transitive.
Suppose 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 gettype(x, z) = A, and type(y, z) = B. Since type(i, z) = B 6= N this is a violation of thesecond quadruple rule of Lemma 9.
If x < i and y, z > i, then we have type(i, y) = type(i, z) = B. From this togetherwith 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 first quadruple rule of Lemma 9.
Adding the corresponding arguments for the order of crossings on the first part ofline `i we conclude that the permutation αi is uniquely determined by the given typesand the choice of T .
11
The consistency condition on triples of local sequences needed for the applicationof Lemma 11 is trivially satisfied because legal triples of types correspond to draw-ings of K2,3 and each such drawing together with T consists of three pairwise crossingpseudolines.
Since the two rules we enforced only involve at most four vertices of V , we imme-diately get the following corollary.
Corollary 12. Consistency on all 4-tuples of V is sufficient and necessary for drawings ofK2,n, yielding an O(n4) time algorithm for checking consistency of an assignment of types.
A problem that remains is to find a good description of all drawings for a given pair(π1, π2) of rotations. We state the same problem differently:
Problem 1. For a given pair (π1, π2) of permutations characterize all consistent assign-ments of types with the property that if a < b in both permutations, then type(a, b) ∈{A,B}, and if a <1 b and b <2 a, then type(a, b) = N .
4 Drawings with k = 3
At the beginning of the previous section we have seen that any pair of rotations isfeasible for drawings of K2,n. This is not true in the case of k > 2. For k = 4 the systemof rotations ([1, 2], [2, 1], [1, 2], [2, 1]) is easily seen to be infeasible. In the case k = 3 it isless obvious that infeasible systems of rotations exist. We will show in Proposition 16that ([1, 2, 3, 4], [4, 2, 1, 3], [2, 4, 3, 1]) is infeasible.
We again start by looking at the types for pairs, i.e., at all possible drawings of K3,2.We already know that if the rotation system is uniform (id2, id2, id2), then there are threetypes of drawings. The other three options (id2, id2, id2), (id2, id2, id2), and (id2, id2, id2),each have a unique drawing. Figure 12 shows the six possible types and associatesthem to the symbols Bα, and Wα, for α = 1, 2, 3.
12
12
211
2 12
21
212121
21
12
12
1 21 2
12
12
12
1 2
B1
W1
B2 B3
W2 W3
Figure 12: The six types of drawings of K3,2.
12
The three edges emanating from a vertex i ∈ [n] partition the drawing area intothree regions. Let Rα(i) be the region bounded by the two edges i pα+1 and i pα−1 notcontaining pα (Rα(i) is the same as the quadrant Qα+1(i) from Section 2). When thetypes have been prescribed for all pairs of vertices we know which vertices are locatedin which region of i. The set of vertices in Rα(i) is denoted Sα(i). The conditions forj ∈ Sα(i) can be read from Figure 12. For example j ∈ S2(i) if either i, j is of type B2
or j < i and the type is W1 or W3.
4.1 An interesting special case: mixed rotations
A system of rotations is called mixed if all pairs are of type Wα for some α, i.e., if forall i, j ∈ [n] with i < j there is a rotation in the system where (i, j) is an inversionand one where (i, j) is a non-inversion. Note that in a mixed system each pair i, j ∈ Vcontributes a unique crossing pair of edges, moreover, if i <α j, i.e., π−1α (i) < π−1α (j),and type(i, j) = Wα, then the crossing edges are i pα+1 and j pα−1.
Since pairs of type Bα contribute 3 crossings we observe that mixed systems arethose with the minimum number of crossings, and this minimum is
(n2
).
Theorem 13. If (π1, π2, π3) is a mixed system of rotations, then there is a drawing realizingthe system of rotations.
Proof. Let (π1, π2, π3) be a mixed system of rotations. Use (π2, π3) to represent verticesas points in the first quadrant and edges to p2 = (0,−∞) and p3 = (−∞, 0) as rays,Figure 13(left) shows an example. The picture can be read as a 2-dimensional repre-sentation of the inclusion order of the sets S1(i).
We add the curves representing edges to p1 = (∞,∞) one by one in the order givenby π3. When it comes to determine the curve γ(i) for vertex i we have to make sure thatγ(i) is disjoint from the already drawn curves γ(j) and that for all points j that have nocurve yet if j <1 i the point representing j is on the left of γ(i) while if j >1 i the pointrepresenting j is on the right of γ(i).
Let j− = max{j : j <1 i and j <3 i} and j+ = min{j : j >1 i and j <3 i}. The curvesγ(j−) and γ(j+) are drawn already, they are the boundary of a tube that has to containγ(i). Vertices outside of the tube are on the prescribed side of γ(i) by transitivity.With a vertex j ∈ S2(i) with coordinates (x(j), y(j)) consider the north-east shadowF↖(j) = {x ∈ R2 : x1 ≤ x(j) and x2 ≥ y(j)} and with j ∈ S3(i) consider the south-westshadow F↘(j) = {x ∈ R2 : x1 ≥ x(j) and x2 ≤ y(j)}. We have to show that γ(i) can bedrawn in the tube such that it avoids all the shadows.
First we show that the shadows leave a canal for the curve γ(i). If not, then therewould be a pair a, b with a ∈ S2(i), b ∈ S3(i), and x(a) > x(b), and y(a) < y(b). Thetwo inequalities yield a <2 b and a <3 b, in addition we have a <1 i <3 b this is acontradiction because in a mixed system there must exist α ∈ 1, 2, 3 with a >α b. Itremains to show that the boundaries of the tube do not obstruct the canal between theshadows. From S2(i) ⊂ S2(j
+) we obtain that γ(j+) avoids all the shadows F↖(j) forj ∈ S2(i). From S3(i) ⊂ S3(j
−) we obtain that γ(j−) avoids all the shadows F↘(j) forj ∈ S3(i). Hence γ(i) can be drawn as required.
For a mixed system of rotations (π1, π2, π3) we define its companion system as (π1, π3, π2),see Figure 14.
13
65
4
3 7101 2118 9 45 6
1037
115986214
3 7101 2118 9 45 6
1037
115986214
1110
98
7
32
1
Figure 13: The initial placement of vertices depending on π2 and π3 (left). The con-straints for the edge γ(7) = γ(π−13 (3)) given by the already drawn edges and the shad-ows.
π1
π1
5
53
2
32
54
14
1 2 3 4 54
12
3
52
31
4
1
1 2 3 4 5
Figure 14: A straight line drawing of a mixed system and a companion drawing.
Proposition 14. The companion of a mixed system of rotations is realizable by a drawing.If the mixed system admits a straight line drawing, then the same holds for the companionsystem.
Proof. Given a drawing of the mixed system (π1, π2, π3) we extend the curves γ(i) inthe other direction, i.e., in the direction of (−∞,−∞). The condition is that the curvesremain pairwise disjoint, i.e., they become a system of pairwise parallel pseudolines.
If the mixed system (π1, π2, π3) admits a straight line drawing with p1 being a pointon the line at infinity, then a drawing of the companion system is obtained by revertingthe direction of the rays, see Figure 14.
A drawing of a graph is pseudorealizable if there is an arrangement of pseudolinessuch that the edges of the drawing are segments of the pseudolines. From Theorem 13and Proposition 14 it follows that drawings of K3,n with a mixed system of rotations are
14
pseudorealizable. Actually, the converse also holds. A drawing of K3,n is pseudorealiz-able exactly if the rotations form a mixed system or the companion of a mixed system.
We next give an example showing that there are mixed systems of rotations that donot admit a straight line drawing.
Theorem 15. The triple of rotations given by:
π1 = (b1, b2, b3, q1, a3, q6, q5, a2, q2, q3, a1, q4, c1, c2, c3)
π2 = (c1, c2, c3, q3, b3, q2, q1, b2, q4, q5, b1, q6, a1, a2, a3)
π3 = (a1, a2, a3, q5, c3, q4, q6, c2, q3, q1, c1, q2, b1, b2, b3)
is a mixed system and not realizable with straight lines.
q1
q4
q5
q6
q3q2
Figure 15: A drawing of the system from Theorem 15.
Proof. Figure 15 shows a drawing of the system. Suppose that the system is stretchable,then there is a drawing with p1 = (∞,∞), p2 = (0,−∞) and p3 = (−∞, 0), i.e., withhorizontal rays, vertical rays, and rays of slope 1. The coordinates of a vertex v aredenoted (x(v), y(v)).
From π1 and the fact that the edge a3 → p1 separates the edges q1 → p1 and q6 → p1we deduce the inequality y(q6)− x(q6) < y(q1)− x(q1). Similarly y(q2)− x(q2) < y(q5)−x(q5) and y(q4)− x(q4) < y(q3)− x(q3). We will refer to these inequalities as I1, I2, andI3 in this order.
From π2 and the separating edges bi → p2 we get x(q6) < x(q5), x(q4) < x(q1), andx(q2) < x(q3). Finally, from π3 we get y(q1) < y(q2), y(q3) < y(q6), and y(q5) < y(q4).These six inequalities are the helper inequalities.
Adding I1 and I2 and rearranging terms we obtain
y(q6) + y(q4) + x(q1) + x(q3) < y(q1) + y(q3) + x(q6) + x(q4)
15
Replacing y(q6), y(q4), and x(q3) on the left side and y(q3), x(q6), and x(q4) on the rightside using the six helper inequalities we obtain y(q5) + x(q2) < y(q2) + x(q5). Thiscontradicts I3.
A straight line drawing of a mixed system of rotations as in Figure 14 can be re-garded as a tropical arrangement of lines. We refer to [10] for an introduction to trop-ical geometry. Drawings of mixed systems of rotations can consequently be regardedas tropical arrangements of pseudolines. Restated as a result in tropical geometry Theo-rem 15 tells us that there are non-stretchable tropical arrangements of pseudolines.
The classical construction of a simple non-stretchable example of pseudolines dueto Ringel is based on the 9 lines of a Pappus configuration. In [15] tropical versionsof Pappus are discussed. They note that a configuration on 9 tropical lines which canbe obtained from the configuration of Theorem 15 by replacing points q` by triple in-cidences of lines ai, bj, ck with {i, j, k} = {1, 2, 3} is a valid tropical version of Pappus’Theorem.
Testing a given arrangement of pseudolines for stretchability is a hard problem.Shor [16] has shown that it is NP-complete and Mnev [12] showed that it is ∃R com-plete. Mnev’s result is also explained in [14] and [11]. In contrast, the stretchabilitytest for a tropical arrangements of pseudolines is polynomial. In this case stretchabilityis equivalent to the existence of a solution for a system of linear inequalities. The proofof Theorem 15 indicates how to set up such a system.
We close the section with two observations:• Every mixed system (id, id, π) is stretchable.• Mixed systems are in bijection with 2-element chains in the weak Bruhat order, i.e.,
pairs (π1, π2) such that every inversion of π1 is an inversion of π2. The correspon-dence is (π1, π2)←→ (id, π1, π2).
4.2 Classifying triples
We have seen that there are six drawings of K3,2, they correspond to the types Bα andWα for α = 1, 2, 3. In the following we classify the triples, i.e., the drawings of K3,3. Itturns out that there are 92 types.
4.2.1 Mixed systems
We choose a relabeling such that the system is (π1, id3, π3). Figure 16 shows the sixchoices for π3 together with the number of possible completions.
3 361 2 2
123
123
123
123
312
123
1232
2311
3
12 1 132 1 3 1 2
3 2 1 3 2 1
32
1
12
3
1232
13
Figure 16: Mixed systems on three vertices. The gray number is the count of compatiblepermutations π1.
16
In total we obtain 17 mixed systems. Among these are 15 decomposable types, i.e.,
of typesWαWα
Wβ
orWβ Wα
Wα
. The two non-decomposable types are shown in Figure 17.
1 2 3
231
231
W2 W1
W3
1 2 3
2
312
13
W3 W1
W2
Figure 17: The two non-decomposable tables with a mixed system of rotations.
4.2.2 Uniform systems
We have discussed uniform systems in Section 2. From there we know that they are
always decomposable. The 15 types areBα Bα
Bβor
Bβ Bα
Bα.
4.2.3 Two mixed and one uniform pair
In this class we have 18 decomposable types, they correspond toWαWα
Bβand
Bβ Wα
Wα
.
Figure 18 shows the six systems of rotations that yield a non-decomposable typetogether with the possible tables. Since α ∈ {1, 2, 3} can be chosen arbitrarily we have12 non-decomposable types.
1
3
1 2 3 321 1 3321
2
12
2
312
231
3 13
2
W3 B1
W2
W1 W2
Bα
Bα W3
W1
1 2 3
2312
13
1 2 3
213
312
W2 Bα
W3
W1 W3
B2
B3 W2
W1
2
21 3
213
13
Figure 18: Non-decomposable tables with two mixed and one uniform pair. Everychoice of a Bα drawing in the gray area is possible.
4.2.4 One mixed and two uniform pairs
Again we have 18 decomposable types, they correspond toBα Bα
Wβ
andWβ Bα
Bα.
Figure 19 shows the six systems of rotations that yield a non-decomposable typetogether with the possible tables. In each case α ∈ {1, 2, 3} can be chosen arbitrarily,however, one of the choices results in a decomposable table. Hence, we have 9 non-decomposable types.
17
1
Bα B2
W3
Bα B1
W2
W2 B3
Bα
W1 Bα
B2
2B3 Bα
W1
W3 B1
Bα
1 2 3
23
132
1 2 3
32
1
3
1 2 3
213 3
12
321321321
123
13
32
1
2132
13
213
12
Figure 19: Non-decomposable tables with one mixed and two uniform pairs. Everychoice of a Bα allows a drawing in the gray area, one of the choices yields a decompos-able table.
4.2.5 The classification and consequences
Summing up the numbers determined in each of the four cases we find that there are(15 + 2) + (15 + 0) + (18 + 12) + (18 + 12) = 92 types of drawings. The second numberin each pair of brackets is the number of non-decomposable types for that case. Hence,the total number of non-decomposable types is 26.
A drawing of K3,n induces three drawings of K2,n obtained by deleting pi and itsincident edges for i ∈ 1, 2, 3. Similarly a table T with entries from Bα,Wα with α =1, 2, 3 induces three tables T1,3, T2,1, and T3,2 with entries from A,B,N . The table T1,3is obtained by the substitution B1,W2, B3 → A, and B2 → A, and W1,W3 → N , seeFigure 12. The other two tables are obtained by similar substitutions but with theadditional condition that type(a, b) has to be written into the cell [π−1j (a), π−1j (b)] intable Tj,j−1. Clearly, if T corresponds to a drawing, then each of T1,3, T2,1, T3,2 has to beconsistent (Theorem 10). With the aid of a computer we have verified that there areexactly 92 tables of order 3 with the property that the three projections are consistent.In other words, consistency of the projections is is also sufficient for the existence of adrawing of K3,3. Later, with Theorem 18 we will see that this holds in general.
Now supposes that rotations (id, π2, π3) are prescribed. we want to decide whetherthere is a corresponding drawing. The first step would be to determine the type ofthe drawing for each pair of vertices. For all mixed pairs type(i, j) ∈ {W1,W2,W3} isuniquely given by the system.
The type of the remaining pairs is Bα for some α. Beforehand each α ∈ {1, 2, 3} ispossible but of course the types of every triple must also correspond to a drawing, i.e.,the types of each triple must be among the 92 drawable types of the classification. Thismay force the types of additional pairs.
Before giving a larger example we show that by looking at triples we can deducethat not all choices (id, π2, π3) of prescribed rotations are feasible, i.e., there are choicesthat have no corresponding drawing.
Proposition 16. The system (id4, [4, 2, 1, 3], [2, 4, 3, 1]) is an infeasible set of rotations.
Proof. The table of types for the given permutations is
18
1 W1W3W1
2 BαW2
3 W1
4
Looking at the subtableW1 W3
Bαcorresponding to {1, 2, 3} we obtain from the clas-
sification, in particular from Figure 17, that the only one choice for α, namely α = 2.
The same figure shows that the subtableBα W2
W1
of {2, 3, 4} again only allows a unique
choice of α, namely α = 3. This proves that there is no drawing for this set of rota-tions.
Let us look at the larger system (id6, [3, 2, 5, 1, 6, 4], [1, 3, 5, 4, 6, 2]). The left table inFigure 20 shows all the types of mixed pairs. To get to the right table we argue withsome forcings on triples. The triple {1, 4, 5} forces type(1, 4) = B3 (see Figure 18).For x = 4, 5, 6 the triple {2, 3, x} forces type(2, x) = B2 (see Figure 18). From thetriple {1, 3, 6} we obtain that either type(1, 6) = B3 or the triple is decomposable andtype(1, 6) = B2. From the triple {1, 4, 6} we obtain that either type(1, 6) = B1 or thetriple is decomposable and type(1, 6) = B3. Hence, the only choice that is consistentfor both triples is type(1, 6) = B3. For the remaining type(5, 6) there is no forcing,every choice of α for type(5, 6) = Bα is consistent with all triples. The middle part ofFigure 20 shows a drawing of the rotation system. The circular region around vertices 5and 6 shows that every choice of α for type(5, 6) = Bα has a corresponding drawing.
1 W2W2 W2
2 W1W3W3W3
3
4 W1W2
5
6
1 W2W2 B3 W2 B3
2 W1W3W3W3
3 B2 B2 B2
4 W1W2
5 Bα
6
1 2 3 4 5 6
46
15
232
64
53
1
Figure 20: The type of mixed pairs is given by the rotations. Consistency forces thetype of most uniform pairs. Every choice of a Bα corresponds to a drawing.
19
5 Consistency for k=3
In the k = 2 case we saw that all pairs of rotations are feasible for drawings of K2,n. Wealso had an efficient characterization of consistent assignments of types. As an openproblem we asked for a good description of all consistent assignments of types thatcorrespond to a given pair of rotations (Problem 1).
In the k = 3 case not all triples of rotations are feasible. For a triples of rotations wedo not know how to decide whether there is an assignments of types which is consistenton triples. Actually we want even more, we also want consistency on quadruples.
5.1 Quadruples
In the k = 2 case we encountered that in addition to the consistency on triples of typeswe also need some conditions for quadruples. This remains true for k = 3. Consider thetypes type(1, 2) = type(1, 4) = type(3, 4) = B1 and type(1, 3) = type(2, 3) = type(2, 4) =B2. Every triple is decomposable, i.e., we have consistency on triples, however, the fulltable is not decomposable. From Theorem 4 we know that there is no correspondingdrawing.
In Subsection 4.1 we saw that mixed systems of rotations are always feasible. Whenstarting from a table with only mixed types consistency of triples corresponds to transi-tivity conditions, they ensure that there are corresponding rotations.
The need for a condition on quadruples is not restricted to tables of uniform systems.The table below is consistent on all triples, still it is not realizable. This can be shownby looking at the table for green-blue which reveals a bad quadruple. Note that for thetable for green-blue the elements are sorted according to π3 = (2, 1, 3, 4).
1 W1 B1 B1
2 B1 B2
3 B2
4
2 A B A
1 B B
3 B
4
Let T be an assignment of types, e.g., in form of a table. From T we know thecorresponding system (π1, π2, π3) of rotations.
Definition 17. T is consistent on quadruples if for any four vertices a, b, c, d and i ∈{1, 2, 3} the assignment of types from A,B,N induced by the restriction of πi−1 and πi+1
to a, b, c, d satisfies the condition from Lemma 9.
Note that checking the condition requires sorting a, b, c, d according to πi−1.
Problem 2. For a given triple (π1, π2, π3) of rotations find a polynomial algorithm to decidewhether there is an assignment of types which is consistent on triples and quadruples.
The good news is that if an assignment is given which is consistent on triples andquadruples, then there is a corresponding drawing.
20
5.2 The consistency theorem
Theorem 18 (Consistency of drawings for k = 3). Given a type for each pair of verticesin V , there exists a drawing realizing those types if and only if all triples and quadruplesare consistent.
Proof. Let T be a consistent assignment for the pairs of V . And let (π1, π2, π3) be thecorresponding rotations. From T we “project” to two tables for the k = 2 case. Let T2be the table corresponding to (π1, π3) and T3 be the table corresponding to (π2, π1).In 4.2.5 we wrote T1,3 for the table T2 and T2,1 for T3, We strech notation and writeT3(i, j) for the entry in cell [π−12 (i), π−12 (j)]. Formally the projection is realized by thefollowing substitutions:
T2(i, j) =
A if T (i, j) ∈ W2, B1, B3
B if T (i, j) = B2
N if T (i, j) ∈ W1,W3
T3(i, j) =
A if T (i, j) ∈ W3, B1, B2
B if T (i, j) = B3
N if T (i, j) ∈ W1,W2
Consistency of T implies consistency of the tables T2 and T3 in the sense of Theo-rem 10. For triples this is because a K2,3 subdrawing of a drawing of K3,3 is drawable,for quadruples we have forced the property with Definition 17. Hence, there are draw-ings D2 and D3 of K2,n realizing the type assignments T2 and T3. Vertex p1, the outervertex with rotation π1, and its edges form a non-crossing star in both drawings.
Let drawing D2 live on plane Z2 and D3 live on plane Z3 and consider a fixedhomeomorphism φ between the planes. There is a homeomorphism ψ : Z2 → Z2 suchthat mapping D2 via φ ◦ ψ to Z3 yields a superposition of the two drawings with thefollowing properties.• Corresponding vertices are mapped onto each other.• The stars of p1 are mapped onto each other, i.e., the edges at p1 of the two drawings
are represented by the same curves.• At each vertex v ∈ V the rotation is correct, i.e., we see the edges to p1, p2, p3 in
clockwise order.• The drawing has no degeneracies like touching edges.
The drawing D obtained by superimposing D2 and D3 is a drawing of K3,n. We colorthe edges of D as in our figures, for example the edges incident to p2 are the greenedges. In D each color class of edges is a non-crossing star. For the blue and the greenthis is true because the edges come from only one of D2 and D3. For the red star itis true due to construction. Moreover, all the red-blue and red-green crossings are asprescribed by the original table T . The problem we face is that there is little control onblue-green crossings. Let ered(v), egreen(v), eblue(v) be the red, green, and blue edge of v.
Claim: For all v, w the parity of the number of crossings between egreen(v) and eblue(w)in D is as prescribed by T .
Consider the curves egreen(v)∪ ered(v) and eblue(w)∪ ered(w). The rotations prescribewhether the number of intersections of the two curves is odd or even. Hence, the parityis respected by D. The crossings of the pairs (egreen(v), ered(w)), (ered(v), eblue(w)) in D
21
are as prescribed by T (v, w). Hence, the parity of the number of crossings of egreen(v)and eblue(w) in D is also prescribed. 4
Because the rotation at v in D is correct we also note: If egreen(v) and eblue(v) cross,then the number of crossings is even. Hence, if D has no pair of a green and a blueedge crossing more than once, then D is a topological drawing that realizes the typesgiven by T .
Now consider a pair of edges egreen(v) and eblue(w) crossing more than once, v = w isallowed. Use a homeomorphism of the plane to make egreen(v) a horizontal straight linesegment, see Figure 21. (In the literature intersection patterns of two simple curves inthe plane are often called meanders. They are of interest in enumerative and algebraiccombinatorics).
The intersections with egreen(v) subdivide eblue(w) into a family of arcs and two ex-tremal pieces. A blue arc defines an interval on the green edge, this is the interval ofthe arc. An arc together with its interval enclose a bounded region, this is the region ofthe arc.
Fact. Apart from intersecting intervals any two regions of arcs over a fixed green edgeegreen(v) are either disjoint or nested. Because blue edges are pairwise disjoint this alsoholds if the arcs are defined by different blue edges.
v
w
Figure 21: A meander with 7 empty lenses.
An inclusionwise minimal region is a lens. Since D has a finite number of crossingsand hence a finite number 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 blue edge in D. Suppose that L
is an empty lens, i.e., there is no vertex u ∈ V inside of L. It follows that the boundaryof L is only intersected by red edges, moreover, if a red edge e intersects the boundaryof L on the green side, then e also intersects L on the blue side. Therefore, we canmake egreen(v) and eblue(w) switch sides at L and with small deformations at the twocrossings get rid of them. It is important to note that the switch at a lens does not changethe types. In particular after the swith the drawing still represents assignment T .
Apply switching operations until the drawingD′ obtained by switching has the prop-erty that every lens in D′ contains a vertex.
In the following we show that D′ has no lens. For the proof we use that the projec-tion T1, i.e., the table corresponding to (π3, π2), also has to be consistent. Since D′ has
22
no lens we conclude that it is a topological drawing that realizes the types given by T .The existence of such a drawing was the statement of the theorem.
B1 B2 B3 W1 W2 W3
T1 B A A A N NT2 A B A N A NT3 A A B N N A
Table 1: The projections of the six types of drawings of K3,2.
LetD′ be a drawing with the property that every lens contains a vertex. Before goinginto the proof that there is no lens in D′ we fix some additional notation. For a givengreen edge egreen(v) we classify the regions defined by blue arcs over egreen(v) as above,below, and wrapping depending on the side of egreen(v) where the arc touches the edge.For a region defined by an arc on eblue(w) we speak of a forward or backward regiondepending on the direction of the arc above egreen(v). Formally: label the t crossingsof egreen(v) and eblue(w) as 1, .., t according to the order on eblue(w). Arcs correspond toconsecutive crossings. An arc [i, i+1] is forward if crossing i is to the left of crossing i+1on egreen(v), otherwise it is backward. The permutation of 1, .., t obtained by readingthe crossings from v to p2 along egreen(v) is called the meander permutation and denotedσgreen(v).
We will occasionally make use of the following Fact. Above a point p on egreen(v) the
direction of the arcs alternates.
A region is a relative lense for egreen(v) and eblue(w) if it is above or below egreen(v) andminimal in the nesting order of regions defined by egreen(v) and eblue(w). In the sequelwe sometimes abuse notation by talking of lenses when we mean relative lenses.
Proposition 19. For all v ∈ V : the green and the blue edge of v do not cross in D′.
Proof. If egreen(v) and eblue(v) cross, then there is at least one blue arcs on eblue(w) overegreen(v) and, hence, there are regions. We consider a sequence of cases.
Suppose R is a wrapping region, then v ∈ R and the red edge of v has to intersectone of egreen(v) and eblue(v) to leave the region. This is not allowed. Hence, here is nowrapping region. A similar argument shows that at the last crossing the blue edge iscrossing downwards.
Since at the first crossing the blue edge is crossing upwards there is an arc andconsequently also a lense above egreen(v).
Suppose there is a forward lens above egreen(v). Let u be a vertex in the lens. Vertex uis below the line of v in the green-red arrangement. Hence, either T3(u, v) = B orT3(u, v) = N and v <1 u. At its last crossing the blue edge is crossing downwardsand there is a blue path from v to the blue arc of the lens. It follows that the highestarc above u is forward. From the alternation of directions of arcs above u it followsthat u is below the line of v in the red-blue arrangement. Hence, either T2(u, v) = Bor T2(u, v) = N and u <1 v. From table 1 we infer that the only legal assignment isT (u, v) = W1 and the projections are T2(u, v) = N and T3(u, v) = N . However, there is
23
no consistent choice for the order of u and v in <1. This shows that there is no forwardlens above egreen(v).
Now let [i, i+1] be a backward lens above egreen(v). We distinguish whether the lastcrossing on the blue edge is to the right/left of the lens on egreen(v), see Figure 22.
j i kv v iw
u uj
Figure 22: The two cases for a backward lens above egreen(v). Dashed curves indicatethe order of some crossings along the corresponding edges.
Suppose the last crossing k on the blue edge is to the right of the lens, Figure 22left. To get from the arc [i, i + 1] to k the edge eblue(v) has to cross egreen(v) upwardsat some crossing j left of i + 1. Let u be a vertex in the lens. The edge eblue(u) hasto stay disjoint from eblue(v), therefore, after crossing egreen(v) to leave the lense it hasa second crossing left of j. The edge egreen(u) has a crossing with eblue(v) in the arc[i, i + 1] and another one between j and k. Now consider the edge ered(u). If it leavesthe lens through egreen(v), then it has entered a region whose boundary consists of apiece of eblue(u) and a piece of egreen(v). Edge ered(u) is disjoint from eblue(u) and ithas already used its unique crossing with egreen(v). Hence, ered(u) has to leave the lensthrough the blue arc on eblue(v). In this case, however, ered(u) enters a region whoseboundary consists of a piece of eblue(v) and a piece of egreen(u). Again ered(u) is trapped.Hence, this configuration is impossible.
Suppose the last crossing k on the blue edge is to the left of the lens, Figure 22right. To get from v to the backward arc [i, i + 1] edge eblue(v) has to cross egreen(v)downwards at some crossing j right of i. Let u be a vertex in the lens and let w be avertex in the region of some blue arc [k, k + 1] below egreen(v) with the property thatj ≤ k < i. The edges ered(u) and ered(w) both need at least two crossings with theunion of egreen(v) and eblue(v). Hence, they both cross egreen(v) and eblue(v). The orderof the red edges at p1 implies u <1 v <1 w. From these data we find that T (u, v) = B3
and T (v, w) = B2. For a consistent assignment we need T (u,w) ∈ B2, B3. Since edgeeblue(w) has to follow the ‘tunnel’ prescribed by eblue(v) there is a crossing of edgesegreen(v) and eblue(w). There is also a crossing of egreen(v) and ered(w). However, neitherin B2 nor in B3 we see crossings of egreen(1) with eblue(2) and ered(2). Hence, again theconfiguration is impossible.
We now come to the discussion of the general case.
Proposition 20. For v, w ∈ V : there is no lens formed by egreen(v) and eblue(w) in D′.
Our proof of this proposition unfortunately depends on a lengthy case analysis. It isgiven in the following subsection.
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5.3 Proof of Proposition 20
Suppose that there is a lens L formed by egreen(v) and eblue(w) in D′. From Proposi-tion 19 we know that v 6= w. In fact the proposition implies that the star of every vertexis non-intersecting. For emphasis we collect this and the other restrictions on crossingsin D′ in a list.(1) Edges of the same color do not cross.(2) Edges of different 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 crucial throughout the argument. We make use of them forthe following two observations:• There is no wrapping region, i.e., there is no arc on eblue(w) touching egreen(v) from
above at one end and from below at the other end.This is because eblue(v) could not leave such a region. Dually egreen(w) could not leavea region formed by egreen(v) and eblue(w). Therefore:• Vertex w is not contained in the region defined by an arc of eblue(w) over egreen(v).
The proof of Proposition 20 has three parts.In the first part we discuss eight configurations that may appear in a meander. In
this discussion we impose additional conditions on vertices u that are contained inthe regions defined by arcs of eblue(w) over egreen(v). These conditions either ask foru ∈ S1(w) or for u 6∈ S1(w). The conditions are so that they are satisfied for freeif the respective regions are relative lenses of eblue(w) over egreen(v). In each of thecases we show that the T1 projections yield an illegal table. Hence the configurationsdo not appear in the drawing D′. Figure 23 depicts schematic views of these eightconfigurations.
Figure 23: The eight configurations for the first phase of the case analysis.
In the second part we use the results from the first part to show that meanders ofthe drawing D′ have no inflections, i.e., in the meander permutation we never see i− 1and i+ 1 on the same side of i.
The consequence is that the meanders in D′ are very simple, their meander permu-tations are either the identity permutation or the reverse of the identity. In particularevery arc defines a relative lense. It then follows that each meander in D′ containsone of the configurations that have been discussed in the first part and the additionalconditions are satisfied because the arcs define a relative lenses. Hence, there are nonontrivial meanders, i.e., every pair of a green and a blue edge crosses at most oncein D′.
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w
v
Figure 24:Illustration for Case I.
Case I. The first intersection along eblue(w) is downwards andthe last intersection is upwards.
Its last intersection with egreen(v) being upward forceseblue(w) to intersect ered(v) on its final piece. Hence, theinitial piece of eblue(w) has no intersection with ered(v) andw ∈ S3(v). Edge egreen(w) has to turn around v as shown inFigure 24. Now consider ered(w), its first crossing (amongthose relevant to the argument) has to be with egreen(v).With this crossing, however, ered(w) is trapped in a regionenclosed by parts of egreen(w), eblue(w), and egreen(v). There-fore, this case is impossible.
v
w
u
Figure 25:Illustration for Case II.
u B A
w N
v
.
Case II. The first intersection along eblue(w) is upwards andthe last intersection is downwards and to the right of the firstalong egreen(v) and there is a forward arc of eblue(w) aboveegreen(v) whose region contains a vertex u 6∈ S1(w).
Edge egreen(w) has to stay below egreen(v). Now considera vertex u in a backward lense of eblue(w) above egreen(v).Vertex u is not under an arc of eblue(u) and we exclude theconfiguration of Case I. Therfore, eblue(u) behaves as shownin Figure 25, i.e., u <3 w <3 v. It follows that egreen(u)has to stay above egreen(v) and u <2 v <2 w. From theintersections of green and blue edges we conclude that theprojections T1 are as given by the table on the right. Thistables is not legal. Hence, this case is impossible.
Case III. The first intersection along eblue(w) is upwards andthe last intersection is downwards and to the left of the first along egreen(v) and there is abackward arc of eblue(w) above egreen(v) whose region contains a vertex u ∈ S1(w).
v u
w
Figure 26:Illustration for Case III.
w A B
u A
v
.
Edge egreen(w) has to turn around v and also ered(w) hasto cross eblue(v), see Figure 26. Now consider a vertex u ina forward lense of eblue(w) above egreen(v). Vertex u is notunder an arc of eblue(u) and there is at most one crossingof eblue(u) and ered(w). Therfore, eblue(u) behaves as shownin the sketch, i.e., w <3 u <3 v. Also egreen(u) has to turnaround v as shown. In particular w <2 u <2 v. From theintersections of green and blue edges we conclude that theprojections T1 are as given by the table on the right. Thistable is not legal. Hence, this case is impossible.
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Case IV. The first intersection along eblue(w) is downwardsand the last intersection is upwards and to the left of the firstalong egreen(v) and there is a backward arc of eblue(w) belowegreen(v) whose region contains a vertex u 6∈ S1(w).
v
u
w
Figure 27:Illustration for Case IV.
v N B
w A
u
.
Edge eblue(w) has to cross ered(v) and the other edges ofw have no intersections with edges of v. Edge eblue(u) eitherturn around w or it crosses ered(v). In the first case egreen(u)has turn around v and after crossing eblue(w) edge ered(u)is ‘trapped’ in a region defined by parts of eblue(w), eblue(u),and egreen(u). Hence, there is no possible routing for edgeered(u) respecting the restrictions on crossings in D′. In thesecond case eblue(u) crosses ered(v) and v <3 w <3 u. Fromthe green edges we obtain w <2 v <2 u. The intersectionsof green and blue edges, see Figure 27 imply that the pro-jections T1 are as given by the table below the figure. Thistable is not legal. Hence, this case is impossible.
Case V. The first intersection along eblue(w) is downwardsand the last intersection is downward and to the right of the first. Moreover, along egreen(v)there is a forward arc of eblue(w) below egreen(v) whose region contains a vertex u ∈ S1(w)and further to the right a forward arc of eblue(w) above egreen(v) containing a vertex x 6∈S3(w).
u
w
v x
Figure 28:Illustration for Case V.
w A B
x A
u
.
If u <3 w, then edge eblue(w) makes sure that x ∈ S1(u).Therefore, we have Case II with v, u, and x.
If w <3 u <3 v, then w ∈ S1(u). Therefore, we have aCase III.
Now let w <3 v <3 u. If v <3 x <3 u, then we haveCase IV with x and u. Hence, w <3 x <3 v <3 u andthe intersections of green and blue edges are as shown inFigure 28. The projections T1 then are as given by the tablebelow the figure. This table is not legal. Hence, this case isimpossible.
Case VI. The first intersection along eblue(w) is downwardsand the last intersection is upward and to the left of the first.Moreover, along egreen(v) there is a backward arc of eblue(w)above egreen(v) whose region contains a vertex u ∈ S1(w) and further to the right a back-ward arc of eblue(w) below egreen(v) containing a vertex x 6∈ S1(w).
v u
x
w
Figure 29:Illustration for Case VI.
u A B
w A A
v B
x
.
If x <3 v, then edge eblue(x) either yields a Case II withw 6∈ S1(x) or a Case III with u ∈ S1(x). Therefore, v <3 x.
If w <3 u, then edge eblue(u) has an arc above u or itforms a Case I. Therefore, u <3 w and the total u <3 w <3
v <3 x.From green edges we get w <2 v <2 x and u <2 v, see
Figure 29. Independent of the entry T1(u,w) the resultingtable as shown below Figure 29 violates the quadruple rule,Lemma 9. Hence, this case is impossible. possible.
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Case VII. The first intersection along eblue(w) is upwards andthe last intersection is upward and to the left of the first.Along egreen(v) there is a backward arc of eblue(w) belowegreen(v) whose region contains a vertex u 6∈ S1(w) and further to the right a backwardarc of eblue(w) above egreen(v) containing a vertex x ∈ S1(w).
v
w
u
x
Figure 30:Illustration for Case VII.
x A B A
v B B
w u
.
If v <3 x, then edge eblue(x) either yields a Case I or aCase IV with u 6∈ S1(x). Therefore, x <3 v.
If u <3 v, then u <3 v <3 w and u <2 v <2 w. To-gether with u,w ∈ S1(v) and u 6∈ S3(w) this implies theillegal table with entries T1(u, v) = B, T1(u,w) = A, andT1(v, w) = B.
If v <3 u, then u <3 w <3 v <3 x and the situation is asshown in Figure 30. Irrespective of u <2 w or w <2 u the T1projections of the remaining 5 pairs yield the partial tableshown below the figure. This violates the quadruple rule,Lemma 9. Hence, this case is impossible.
Case VIII. The first intersection along eblue(w) is upwards andthe last intersection is upward and to the right of the first. Along egreen(v) there is a forwardarc of eblue(w) above egreen(v) whose region contains a vertex u 6∈ S1(w) and further to theright a forward arc of eblue(w) below egreen(v) containing a vertex x ∈ S1(w).
v
w
u
x
Figure 31:Illustration for Case VIII.
u A N∣∣B A
v B B
x w
.
If v <3 u, then edge eblue(u) either yields a Case I ora Case IV with w 6∈ S1(x). Therefore, u <3 v. If x <3 v,then edge eblue(x) yields a Case III with u ∈ S1(x). There-fore, v <3 x and in total u <3 v <3 x <3 w, see Figure 31.Also u <2 v <2 w and v <2 x only the order of x and win <2 is not determined. If w <2 x then T1(x,w) = N oth-erwise T1(x,w) = B. The resulting table is shown belowFigure 31. In the first case the triple u, v, x yields an ille-gal subtable. In the second case there is a violation of thequadruple rule, Lemma 9. Hence, this case is impossible.
Part 2: No inflections
In this part we use the results from the first part to showthat meanders of the drawing D′ have no inflections, i.e., in the meander permutationwe never see i−1 and i+1 on the same side of i. In particular none of the configurationsof Figure 32 appears in the intersection pattern of a green and a blue edge.
Suppose that a meander permutation σ = σgreen(v) has an inflection at i. We discussthe case where i+1 <σ i− 1 <σ i and the blue arc [i+1, i] of eblue(w) is above egreen(v).All the other cases can be treated with symmetrical arguments. First note that i−1 > 1,otherwise, w is in the region defined by the arc [i+ 1, i]. This is impossible.
Now suppose that i − 1 <σ i − 2 <σ i. Let u be a vertex in the region defined bythe arc [i− 1, i+ 1]. Confined by eblue(w) the edge eblue(x) has to cross egreen(v) at leastthree times and it contains an arc whose region contains x. This is impossible. Hence,i + 1 <σ i − 2 <σ i − 1 <σ i and the meander contains the second of the turn-back
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patterns shown in Figure 32. Other cases of inflections are related to the other cases ofturn-back patterns.
Figure 32: The four turn-back pattern.
Case TB 1. The turn back corresponds to a substring i, i− 1, i− 2, i + 1 or to a substringi, i+ 1, i+ 2, i− 1 and the turn-back arc is above egreen(v).
Let us assume that the substring is i, i− 1, i− 2, i+1 so that the turn-back is the arc[i, i− 1]. Let u be a vertex in the region of the turn-back arc and let x be a vertex in theregion of the arc [i− 1, i− 2].
The edge eblue(u) is confined by eblue(w) and has a forward arc below x. If eblue(u)has a crossing with ered(v) then we have a Case I. Hence, the last crossing of eblue(u)and egreen(v) is downward and to the right of i+ 1.
If x 6∈ S1(u), then eblue(u) has a backward arc a in the interval [i − 1, i − 2] suchthat x is in the region of arc a. The left part of Figure 33 shows a schematic viewof the situation. Let y be a vertex in the region of an arc of eblue(u) above egreen(v)and preceeding a along eblue(u). Note that y 6∈ S1(x). Edges eblue(u) and eblue(w) forceeblue(x) to have its last intersection with egreen(v) downwards and to the right of i + 1.Hence eblue(x) and y are in the configuration of Case II.
Now assume x ∈ S1(u). Let y be a vertex below the last forward arc of eblue(u)abowe egreen(v). If y 6∈ S1(u), then eblue(u) with x and y are in the configuration ofCase V. Otherwise, if y ∈ S1(u), then there is a backward arc a of eblue(v) above y.Choose a vertex z from backward arc below egreen(v) and to the right of a. The rightpart of Figure 33 shows a schematic view of the situation. Edges eblue(u) and egreen(z)force eblue(z) to have its last intersection with egreen(v) downwards and left of the lastintersection of eblue(u) and egreen(v). Hence eblue(y) and z are in the configuration ofCase II.
u
x
u
x
yy
z
Figure 33: Illustrations for case TB 1.
Case TB 2. The turn back corresponds to a substring i + 1, i− 2, i− 1, i or to a substringi− 1, i+ 2, i+ 1, i and the turn-back arc is below egreen(v).
Let us assume that the substring is i+1, i− 2, i− 1, i so that the turn-back is the arc[i, i− 1]. Let u be a vertex in the region of the turn-back arc and let x be a vertex in theregion of the arc [i− 2, i− 1].
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The edge eblue(u) is confined by eblue(w) and has a backward arc above x. We firstconsider the case x ∈ S1(u). If there is no crossing of eblue(u) and ered(v), then this isCase III. If there is a crossing with ered(v), then let y be a vertex from a backward arcbelow egreen(v) and left of i − 2. If y 6∈ S1(u), then eblue(u) with x and y is a Case VII.If y ∈ S1(u), then the situation is essentially as shown in the left part of Figure 33.There is a vertex z in a backward arc below egreen(v), such that eblue(u) with y and z isa Case VIII.
x
y
u
z
y
u
x
Figure 34: Illustrations for case TB 2.
Now let x 6∈ S1(u). There has to be a forward arc a on eblue(u) whose region containsx and we have u <3 x. On eblue(u) there is a forward arc below egreen(v) and before andto the left of a. let y be a vertex from the region of this arc. The right part of Figure 33shows a schematic view of the situation. Now consider eblue(y). this edge is confinedby eblue(u) to have a forward arc above x. If x 6∈ S1(y), then eblue(u) and x are a Case II.Otherwise, eblue(u) has to turn back so that u <3 x <3 y. The green edges of u, x, y have
to behave as shown in the figure. And the table of T1 projections isu N B
x A
y
. This is
not a legal table.
Case TB 3. The turn back corresponds to a substring i + 1, i− 2, i− 1, i or to a substringi− 1, i+ 2, i+ 1, i and the turn-back arc is above egreen(v).
Let us assume that the substring is i+1, i− 2, i− 1, i so that the turn-back is the arc[i, i− 1]. Let u be a vertex in the region of the turn-back arc and let x be a vertex in theregion of the arc [i− 2, i− 1].
If the last intersection of eblue(x) with egreen(v) is downward and to the right of i,then eblue(x) and u are in the configuration of Case II. If x <3 u, then eblue(u) containsa forward arc a below egreen(v) whose region contains x. Before and left of a thereis a forward arc on eblue(u). Let y be a vertex from the region of this arc. If y <3 x,then eblue(y) together with x and y are in the configuration of Case V. If x <3 y, thenx <3 u <3 y. The left part of Figure 35 shows a schematic view of the situation. The
table of T1 projections for this case isu N B
x A
y
. This is not a legal table.
If u <3 x, then eblue(u) contains a backward arc above egreen(v) and left of i− 2. Lety be a vertex from the region of this arc. If y ∈ S1(u), then eblue(x) with x and y arein the configuration of Case VI. If y 6∈ S1(u), then eblue(u) has a forward arc a whose
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x
yu y
x
u
z
Figure 35: Illustrations for case TB 3.
region contains. On eblue(u) here is an arc below egreen(v) preceeding and left of a. Letz be a vertex from the region of this arc. The right part of Figure 35 shows a schematicview of the situation. It follows that z <3 u <3 y whence eblue(z) with y are in theconfiguration of Case III.
Case TB 4. The turn back corresponds to a substring i, i− 1, i− 2, i + 1 or to a substringi, i+ 1, i+ 2, i− 1 and the turn-back arc is below egreen(v).
Let us assume that the substring is i, i− 1, i− 2, i+1 so that the turn-back is the arc[i, i− 1]. Let u be a vertex in the region of the turn-back arc and let x be a vertex in theregion of the arc [i− 2, i− 1].
If u <3 x then the two are in the configuration of Case II. If x <3 u and the lastintersection of eblue(u) with egreen(v) is left of i, then let y be a vertex from the region ofa forward arc of eblue(u) below egreen(v) and to the right of i− 2. We the have x 6∈ S1(u)and y ∈ S1(u). hence eblue(u) with x and y is in the configuration of Case VIII. This isshown in the left part of Figure 36.
yu
x
u
x
y
Figure 36: Illustrations for case TB 4.
Now let x <3 u and let the last intersection of eblue(u) with egreen(v) be to the right ofi− 1. In this case there is a backward arc a on eblue(u) whose region contains x. Beforeand to the right of a there is a backward arc on eblue(u) below egreen(u). let y be a vertexfrom the region of this arc. Edge eblue(y) is confined by eblue(u) to have a backward arcabove x. If u <3 y, then eblue(y) and x are in the configuration of Case II. Otherwise weget the situation shown in the right part of Figure 36. The table of T1 projections for
this case isy B A
x N
u
. This is not a legal table.
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6 Conclusion and open problems
• Extensions to drawings of Kk,n. Should be the same, a table is consistent if and onlyif all projections are consistent. It might be enough that projections to (πi, πi+1) areconsistent.
• A table with B and W positions prescribed (no indices). It is NP-complete to decidewhether there are corresponding permutations (id, π2, π3). There seems to be aconnection with order dimension 3.
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