Topological Drawings of Complete Bipartite Graphs

Topological drawings are natural representations of graphs in the plane, where vertices are represented by points, and edges by curves connecting the points. Topological drawings of complete graphs and of complete bipartite graphs have been studied extensively in the context of crossing number problems. We consider a natural class of simple topological drawings of complete bipartite graphs, in which we require that one side of the vertex set bipartition lies on the outer boundary of the drawing. We investigate the combinatorics of such drawings. For this purpose, we define combinatorial encodings of the drawings by enumerating the distinct drawings of subgraphs isomorphic to $K_{2,2}$ and $K_{3,2}$, and investigate the constraints they must satisfy. We prove that a drawing of $K_{k,n}$ exists if and only if some simple local conditions are satisfied by the encodings. This directly yields a polynomial-time algorithm for deciding the existence of such a drawing given the encoding. We show the encoding is equivalent 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 enumerate such drawings of $K_{k,n}$ in which the order of the edges around each vertex is the same for vertices on the same side of the bipartition. Finally, we investigate drawings of $K_{k,n}$ using straight lines and pseudolines, and consider the complexity of the corresponding realizability problems.


Introduction
We consider topological graph drawings, which are drawings of simple undirected graphs where vertices are represented by points in the plane, and edges are represented by simple curves that connect the corresponding points. We typically restrict those drawings to satisfy some natural nondegeneracy conditions. In particular, we consider simple drawings, in which every pair of edges intersect at most once. A common vertex counts as an intersection.
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. A prominent illustration is the problem of identifying the minimum number of edge crossings in a simple topological drawing of K n [9, 3, 1] or of K k,n [22,4], for which there are long standing conjectures.
In order to cope with the inherent complexity of the drawings, it is useful to consider combinatorial abstractions. Those abstractions are discrete structures encoding some features of a drawing. One such abstraction, introduced by Kratochv l, Lubiw, and Ne set ril, is called abstract topological graphs (AT-graph) [11]. An AT-graph consists of a graph (V, E) together with a set X E 2 . 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 circular permutation with every vertex v, which in a realization must correspond to the order in which the neighbors of v are connected to v. Natural 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 [12].
For simple topological drawings of complete graphs, the two abstractions are actually equivalent [18]. It is possible to reconstruct the set of crossing pairs of edges by looking at the rotation system, and vice-versa (up to reversal of all permutations). Kyn cl recently proved the remarkable result that a complete AT-graph (an AT-graph for which the underlying graph is complete) can be realized as a simple topological drawing of K n if and only if all the AT-subgraphs 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 much more dicult when they involve complete bipartite graphs. In that case, knowing the rotation system is not sucient for recontructing 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 introduce a natural restriction on the drawings, by requiring that one side of the vertex set bipartition lies on a circle at innity. This gives rise to meaningful, yet complex enough, combinatorial structures.
Definitions. We wish to draw the complete bipartite graph K k,n in the plane in such a way that: 1. vertices are represented by points, 2. edges are continuous curves 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 particular, edges incident to the same vertex intersect only at this vertex, 5. the k vertices of one side of the bipartition lie on the outer boundary of the drawing.
Properties 1{4 are the usual requirements for simple topological drawings also known as good drawings. As we will see, property 5 leads to drawings with interesting combinatorial structures. We will refer to drawings satisfying properties 1{5 as outer drawings. Since this is the only type of drawings we consider, we will use 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 use the word \vertex" and \edge" to denote both the vertex or edge of the graph, and their representation as points and curves. Without loss of generality, we can assume that the k outer vertices p 1 , . . . , p k lie in clockwise order on the boundary 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 such a drawing is given in Figure 1.
The rotation system of the drawing is a sequence of k permutations on n elements associated with the vertices of P in clockwise order. For each vertex of P, its permutation encodes the (say) counterclockwise order in which the n vertices of V are connected to it. Due to our last constraint on the drawings, the rotations of the k vertices of P 1  3  2  5  4  2  1  4  3  5  1  3  2  5  4  2  1  4  3  5 12345 12345 Figure 1: Two outer drawings of K 3,5 . In both drawings the rotation system is (12345,21435,13254).
around each vertex of V are xed and identical, they reect the clockwise order of p 1 , . . . , p k on the boundary. Unlike for complete graphs, the rotation system of an outer drawing of a complete bipartite graph does not completely determine which pairs of edges are intersecting. This is exemplied with the two drawings in Figure 1.
Results. The paper is organized as follows. In Section 2, we consider outer drawings with a uniform rotation system, in which the k permutations of the vertices of P are all equal to the identity. In this case, we can state a general structure theorem that allows us to completely characterize and count outer drawings of arbitrary bipartite graphs K k,n .
In Section 3, we consider outer drawings of K 2,n with arbitrary rotation systems. We consider a natural combinatorial encoding of such drawings, and state two necessary consistency conditions involving triples and quadruples of points in V. We show that these conditions are also sucient, yielding a polynomial-time algorithm for checking consistency of a drawing.
We also observe that our encoding is equivalent to specifying which pairs of edges must 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 underlying graph isomorphic to K 2,n in polynomial time.
In Section 4 and 5, we extend these results, rst to outer drawings of K 3,n , then to outer drawings of K k,n . We prove that simple consistency conditions on triples and quadruples are sucient for drawings of K k,n , yielding again a polynomial-time algorithm for consistency checking.
In Section 6, we consider outer drawings with the additional property that the edges can be extended into a pseudoline arrangement, which we refer to as extendable drawings. We give a necessary and sucient condition for the existence of an extendable outer drawing of a complete bipartite graph given the rotation system. We also touch upon the even more restricted problem of nding straight-line outer drawings with prescribed rotation systems.

Outer Drawings with uniform rotation system
We rst consider the case where k is arbitrary but the rotation system is uniform, that is, the permutation around each of the k vertices p i is the same. Without loss of generality we assume that this permutation is the identity permutation on [n].
In a given outer drawing, each of the n vertices of V splits the plane into k regions Q 1 , Q 2 , . . . , Q k , where each Q i is bounded by the edges from v to p i and p i+1 , with the understanding that p k+1 = p 1 . We denote by Q i (v) the ith region dened by vertex v and further on call these regions quadrants. We let type(a, b) = i, for a, b P V and i P [k], whenever a P Q i (b). This implies that b P Q i (a), see Figure 2. Indeed if a < b and j T = 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 until they meet in b. It follows that none of the edges p j b can intersect p i+1 a. This shows that a P Q i (b). For the case k = 2, we have exactly two types of pairs, that we will denote by A and B. The two types are illustrated on Figure 3. Note that the two types can be distinguished by specifying which are the pairs of intersecting edges. The outer drawings of K 2,n with uniform rotations can be viewed as colored pseudoline arrangements, where: each pseudoline is split into two segments of distinct colors, no crossing is monochromatic. Figure 3: The two types of pairs for outer drawings of K 2,n with uniform rotation systems.
This is illustrated on Figure 4. The pseudoline of a vertex v P V is denoted by (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 a lies above (b). While having type(a, b) = type(b, a) = B means that b lies below (a) and a lies below (b).
there is nothing to show since there are only 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 Hence the point c is on the right of this intersection. Pseudoline (a) must cross (b) on L (b), and a is left of this intersection. It follows that Case k > 2. For the general case assume that type(a, b) = i and type(a, c) = j. If i = j there is nothing to show. Now suppose i T = j. From c P 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 Figure 5. This region is contained in Q j (b), whence type(b, c) = j.

The quadruple rule
Lemma 3 (Quadruple rule). For four vertices a, b, c, d P V with a < b < c < d: Proof. Case k = 2. Suppose, without loss of generality, that X = B. Consider the pseudolines representing b and c with their crossing at R (b) L (c). Coming from the left the edge L (d) has to avoid L (c) and therefore intersects R (b). On R (b) the crossing with L (c) is left of the crossing with L (d), see Figure 6. Symmetrically from the 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 arrangement dened by the two successive vertices p i and p i+1 of P dening the quadrants Q i . Proving that type(a, d) = i, that is, that a P Q i (d), can be done as above for k = 2 on the drawing of K 2,n induced by {p i , p i+1 } and V.

Decomposability and Counting
We can now state a general structure theorem for all outer drawings of K k,n with uniform rotation systems. Theorem 4. Consider the complete bipartite graph G with vertex bipartition (P, V) such that |P| = k and |V| = n. Given a type in [k] for each pair of vertices in V, there exists an outer drawing of G realizing those types with a uniform rotation system if and only if: 1. there exists s P {2, . . . , n} and X P [k] such that type(a, b) = X for all pairs a, b with a < s and b ! s, (in the Proof. (⇒) Let us rst show that if there exists a drawing, then the types must satisfy the above structure. We proceed by induction on n. Pick the smallest s P {2, . . . , n} such 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 such that 1 a < s b n. For a = 1 this is just the condition on s. Now let 1 < a. First suppose that type(1, a) T = X. We can apply the triple rule on the indices 1, a, b. Since type(1, b) P {type(1, a), type(a, b)}, we must have that type(a, b) = X. Now suppose that type(1, a) = X. We have type(1, s − 1) = Y T = X by denition. 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 denition type(1, b) = X. Hence we must have that type(a, b) T = 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) T = Y, we must have type(a, b) = X. This yields the claim.
(⇐) Now given the recursive structure, it is not dicult to construct a drawing. Consider the two subintervals as a single vertex, then recursively blow up these two vertices. (See Figure 7 for an illustration). The recursive structure yields a corollary on the number of distinct drawings.
Corollary 5 (Counting outer drawings with uniform rotation systems). For every pair of integers k, n > 0 denote by T (k, n) the number of outer drawings of the complete bipartite graph isomorphic to K k,n with uniform rotation systems. Then T (n + 1, k + 1) = n j=0 n+j 2j C j k j where C j is the jth Catalan number.
Proof. The recursive structure can be modeled in a labeled binary tree. The root corresponds to [1, n], the subtrees 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 number T (k, n) of labeled binary trees we therefore get a Catalan-like recursion T (k, n) = k n−1 i=1 k−1 k T (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 we nd that a k−1 k fraction 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 exceptional, in this case there is no label and we have one choice for this subtree, not just (1 − 1/k). This explains the additional summand. The recursion together with the initial condition T (k, 1) = 1 yields an array of numbers which is is listed as entry A103209 in the encyclopedia of integer sequences 1 (OEIS). The stated explicite expression for T (k, n) can be found there. It can be veried by induction.
Note that in the case k = 2, Corollary 5 provides a bijection between outer drawings with uniform rotation systems and combinatorial structures counted by Schr oder numbers, such as separable permutations and guillotine partitions.

Outer Drawings with k = 2
In this section we deal with outer drawings with k = 2 and arbitrary rotation system. We now 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 to an inversion in the two permutations. Note again that the three types exactly encode which are the pairs of crossing edges. Recall that an outer drawing of K 2,n , in which no pair is of type N, can be seen as a colored pseudoline arrangement as dened previously. Similarly, an outer drawing of K 2,n in which some pairs are of type N can be seen as an arrangement of colored monotone curves crossing pairwise at most once. We will refer to arrangement of monotone curves that cross at most once as quasi-pseudoline arrangements. The pairs of type N correspond to parallel pseudolines. Without loss of generality, we can suppose that the rst permutation in the rotation system, that is, the order of the pseudolines on the left side, is the identity. We denote by π the permutation on the right side.
The rst question is whether every permutation π is feasible in the sense that there is a drawing of K 2,n such that the rotations are (id, π). The answer is yes, two easy constructions are exemplied in Figure 9 3.1 Triples For a, b, c P 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 outer drawing of K 2,n , such triples are 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 Lemma 6 (decomposable triples). A triple with Y P {X, Z} is always legal. There are 15 triples of this kind.
Proof. If Y = X take a drawing of K 2,2 of type X with vertices a, v and a < v. In this drawing double the pseudoline corresponding to v and cover vertex v by a small circle. Then plug a drawing of K 2,2 of type Z with vertices b, c in this circle. This results in a drawing of K 2,3 with the prescribed types. The construction is very much as in Note that the triples of the latter lemma are decomposable in the sense of Theorem 4.

Lemma 7.
There are exactly two non-decomposable legal triples: 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 outer drawings of K 2,3 .
Observation 8 (Triple rule). Any three vertices of V in an outer drawing of K 2,n must induce one of the 17 legal triples of types.

Quadruples
We aim at a characterization of collections of types that correspond to outer drawings. Already in the case of uniform rotations we had to add Lemma 3, a condition for quadruples. In the general case the situation is more complex than in the uniform case, see Figure 10. Reviewing the proof of Lemma 3 we see that in the case discussed 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 equivalent to type(a, b) T = N. Symmetrically, three A types enforce type(a, d) = A when d is the last in π, i.e., if type(c, d) T = N.

Consistency
With the next theorem we show that consistency on triples and quadruples is sucient to grant the existence of an outer drawing.
Theorem 10 (Consistency of outer drawings for k = 2). Consider the complete bipartite graph G with vertex bipartition (P, V) such that |P| = 2 and |V| = n. Given a type in {A, B, N} for each pair of vertices in V, there exists an outer drawing of G realizing those types if and only if all triples are legal and the quadruple rule (Lemma 9) is satised.
The proof of this result uses the following known result on local sequences in pseudoline arrangements. Given an arrangement of n pseudolines, the local sequences are the permutations α i of [n] \ {i}, i P [n], representing the order in which the ith pseudoline intersects the n − 1 others. Lemma 11 (Thm. 6.17 in [5]). The set {α i } i2[n] is the set of local sequences of an arrangement of n pseudolines if and only if 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 8 We 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 inversion pairs and its complement, the set of non-inversion pairs, are both transitive. Hence, there is a well dened permutation π representing the rotation at p 2 .
We aim at dening the local sequences α i that allow an application of Lemma 11. This will yield a pseudoline arrangement. A drawing of K 2,n , however, will only correspond to a quasi-pseudoline arrangement. Therefore, we rst construct a quasipseudoline arrangement T for the pair (π, id), i.e., only the quasi-pseudolines 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 quasi-pseudoline arrangement of the drawing yields a full pseudoline arrangement.
The pseudoline i has three parts. The edge incident to p 1 (the red edge) is crossed by pseudolines j with j P A > (i) B < (i). The edge incident to p 2 (the blue edge) is crossed by pseudolines j with j P A < (i) B > (i). The part of i belonging to T is crossed by pseudolines j with j P N(i). The order of the crossings in the third part, i.e., the order of crossings with pseudolines j with j P N(i), is prescribed by T .
Regarding the order of the crossings on the second part we know that the lines for j P A < (i) have to cross i from left to right in order of decreasing indices and the lines for j P B > (i) have to cross i from left to right in order of increasing indices, see Figure 11. If j P A < (i) and j 0 P B > (i), then consistency of triples implies that type(j, j 0 ) P {A, B}. If type(j, j 0 ) = A, then on i the crossing of j 0 has to be left of the crossing of j. If type(j, j 0 ) = B, then on i the crossing of j has to be left of the crossing of j 0 . The described conditions yield a \left{to{right" relation → i such that for all x, y P A < (i) B > (i) one of x → i y and y → i x holds. We have to show that → i is acyclic.
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 get type(x, z) = A, and type(y, z) = B. Since type(i, z) = B T = N this is a violation of the second quadruple rule 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 quadruple rule of Lemma 9.
Adding the corresponding arguments for the order of crossings on the rst part of line i we conclude that the permutation α i is uniquely determined by the given types and the choice of T .
The consistency condition on triples of local sequences needed for the application of Lemma 11 is trivially satised because legal triples of types correspond to drawings of K 2,3 and each such drawing together with T consists of three pairwise crossing pseudolines.
Since the condition only involves triples and quadruples of vertices in V, this directly yields a polynomial-time algorithm for consistency checking. By observing that the information given by the types is equivalent to specifying which pairs of edges cross, we directly get the analogue statement for AT-graphs.
Corollary 12 (AT-graph realizability). There exists an O(n 4 ) algorithm for deciding the existence of an outer drawing of an AT-graph whose underlying graph is of the form K 2,n .
Proof. We can check that the three types of pairs in Figure 8 exactly prescribe which pairs of edges cross. Furthermore, given the set of crossing pairs, we can reconstruct the type assignment. We can then check that every triple is legal and that the quadruple rule is satised in time proportional to the number of triples and quadruples, hence O(n 4 ).

Outer Drawings with k = 3
At the beginning of the previous section we have seen that any pair of rotations is feasible for outer drawings of K 2,n . This is not true in the case of k > 2. For k = 4 the system of rotations ( is easily seen to be infeasible. In the case k = 3 it is less obvious that infeasible systems of rotations exist. We also had an ecient characterization of consistent assignments of types for k = 2. We generalize this to k = 3. We again start by looking at the types for pairs, i.e., at all possible outer drawings of K 3,2 . We already know that if the rotation system is uniform (id 2 , id 2 , id 2 ), then there are three types of drawings. The other three options (id 2 , id 2 , id 2 ), (id 2 , id 2 , id 2 ), and (id 2 , id 2 , id 2 ), each have a unique drawing. Figure 12 shows the six possible types and associates them to the symbols B α , and W α , for α = 1, 2, 3.
This classication allows us to reason about the following simple example.  Proof. The table of types for the given permutations must be of the following form: Let T 1 be the assignment of types in {A, B, N} corresponding to the outer drawing of K 2,n with outer vertices (p 3 , p 2 ). Similarly, let T 2 be the table corresponding to (p 1 , p 3 ) and T 3 the table corresponding to (p 2 , p 1 ). Note that these assignments determine the rotation system (π 1 , π 2 , π 3 ). An assignment T of types in {B 1 , B 2 , B 3 , W 1 , W 2 , W 3 } to every pair of vertices in V translates to the following assignments T 1 , T 2 , T 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 instrumental in the upcoming consistency theorems. For an assignment T of types in {B 1 , B 2 , B 3 , W 1 , W 2 , W 3 } to pairs of vertices in V, we will refer to the induced assignments T 1 , T 2 , T 3 as the projections of T .
We are now ready to state our consistency theorem on outer drawings for k = 3.

The consistency theorem for k = 3
Theorem 14 (Consistency of outer drawings for k = 3). Consider the complete bipartite graph G with vertex bipartition (P, V) such that |P| = 3 and |V| = n. Given the assignments T 1 , T 2 , T 3 of types in {A, B, N} for the pairs (p 3 , p 2 ), (p 1 , p 3 ), and (p 2 , p 1 ) of vertices in P, respectively, there exists an outer drawing of G realizing those types if and only if all three assignments obey the triple and quadruple consistency rules in the sense of Theorem 10, for any pair of vertices in V, the assignments correspond to an outer drawing of K 3,2 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 outer drawing of an AT-graph whose underlying graph is of the form K 3,n .
Proof. Again, one can check that the three types of pairs in Figure 12 exactly prescribe which pairs of edges cross, and given the set of crossing pairs, we can reconstruct the type assignment. We can then check that every triple is legal and that the quadruple rule is satised in time proportional to the number of triples and quadruples, hence Proof of Theorem 14. Let us rst note that one direction of the Theorem is easy: if there exists an outer drawing, then the assignments must be consistent. We now show that consistency of the type assignments is sucient for the existence of an outer drawing. Let T be the assignment of types in {B 1 , B 2 , B 3 , W 1 , W 2 , W 3 } to the pairs of V given by Table 1, and let (π 1 , π 2 , π 3 ) be the corresponding rotations. The consistency of the tables T 2 and T 3 in the sense of Theorem 10 implies that there are drawings D 2 and D 3 of K 2,n realizing the type assignments T 2 and T 3 . The vertex p 1 (the outer vertex with rotation π 1 ) and its edges form a non-crossing star in both drawings.
Let the drawing D 2 live on plane Z 2 and D 3 live on plane Z 3 and consider a xed homeomorphism φ between the planes. There is a homeomorphism ψ : Z 2 → Z 2 such that mapping D 2 via φ ψ to Z 3 yields a superposition 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 curves.
At each vertex v P V the rotation is correct, i.e., we see the edges to p 1 , p 2 , p 3 in clockwise order.
The drawing has no touching edges, i.e., when two edges meet they properly cross in a single point. The drawing D obtained by superposing D 2 and D 3 is an outer drawing of K 3,n . We color the edges of D as in our gures, for example the edges incident to p 2 are the green edges. In D each color class of edges is a non-crossing star. For the blue and the green this is true because the edges come from only one of D 2 and D 3 . For the red star it is true due to construction. Moreover, all the red-blue and red-green crossings are as prescribed by the original table T . The problem we face is that there is little control on blue-green crossings. Let e red (v), e green (v), e blue (v) be the red, green, and blue edge of v.
Claim: For all v, w the parity of the number of crossings between e green (v) and e blue (w) in D is prescribed by T , i.e., if T requires a crossing between two edges, then they have an odd number of crossings and an even number of crossings otherwise.
Consider the curves e green (v)e red (v) and e blue (w)e red (w). The rotations prescribe whether the number of intersections of the two curves 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 blue (w)) in D are as prescribed by T (v, w). Hence, the parity of the number of crossings of e green (v) and e blue (w) in D is also prescribed.

R
Because the rotation at v in D is correct we also note: If e green (v) and e blue (v) cross, then the number of crossings is even. Hence, if D has no pair of a green and a blue edge crossing more than once, then D is an outer drawing realizing the types given by T .
Now consider a pair of edges e green (v) and e blue (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 Figure 13. (In the literature intersection patterns of two simple curves in the plane are often called meanders. They are of interest in enumerative and algebraic combinatorics.) The intersections with e green (v) subdivide e blue (w) into a family of arcs and two extremal pieces. A blue arc denes an interval on the green edge, this is the interval of the arc. An arc together with its interval enclose a bounded 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. Because blue edges are pairwise disjoint this also holds if the arcs are dened by dierent blue edges.
An inclusionwise minimal region of an arc is a lens. Since D has a nite number of crossings and hence a nite 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 P V inside of L. It follows that the boundary of L is only intersected by red edges, moreover, if a red edge e intersects the boundary of L on the green side, then e also intersects L on the blue side. Therefore, we can make e green (v) and e blue (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 particular after the switch the drawing still represents assignment T . Apply switching operations until 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 use that the table T 1 corresponding to (π 3 , π 2 ) also has to be consistent. Since D 0 has no lens we conclude that it is an outer drawing that realizes the types given by T . The existence of such 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 blue 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 blue (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 blue edge wrapping around e green (v) and note that v P R. If the wrapping blue edge is e blue (v), then e red (v) has to intersect one of e green (v) and e blue (v) to leave the region. This is not allowed. If the wrapping blue edge is e blue (u) with u T = v, then e blue (v) has to intersect one of e green (v) and e blue (u) 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 blue (w) over e green (v).
Proof. The green edge of w would be trapped in such a region.
For a region dened by an arc on e blue (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 blue (w) as 1, .., t according to the order on e blue (w). Arcs correspond to consecutive 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 permutation of 1, .., t obtained by reading the crossings from v to p 2 along e green (v) is called the meander permutation and denoted σ green (v).
A region is a relative lens for e green (v) and e blue (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 blue (w). In the sequel we sometimes abuse notation by talking of lenses when we mean relative lenses. Proof. If e green (v) and e blue (v) cross, then there is at least one blue arcs on e blue (w) over e green (v) and, hence, there are regions.
From the order of the three outer vertices and the fact that e red (v) has no crossing with the two other edges of v we conclude that at the last crossing of e blue (v) and e green (v) the blue edge is crossing e green (v) downwards.
Since at the rst crossing the blue edge is crossing upwards there is an arc and consequently also a lens above e green (v).
Suppose there is a forward lens above e green (v). Let u be a vertex in the lens. Vertex u is below the line of v in the green-red arrangement. Hence, either T 3 (u, v) = B or T 3 (u, v) = N and v < 1 u. Vertex u is below the forward arc of e blue (v) forming the lens. Hence, it is below the line of v in the red-blue arrangement, and either T 2 (u, v) = B or T 2 (u, v) = N and u < 1 v. From Table 1 we infer that the only legal assignment is T (u, v) = W 1 and the projections are T 2 (u, v) = N and T 3 (u, v) = N. However, there is no consistent choice for the order of u 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 u be a vertex in the lens. We distinguish whether the last crossing on the blue edge is to the right/left of the lens on e green (v), see Figure 14. Suppose the last crossing k on the blue edge is to the right of the lens, Figure 14 (left). To get from the arc [i, i + 1] to k the edge e blue (v) has to cross e green (v) upwards at some crossing j left of i + 1. The edge e blue (u) has to stay disjoint from e blue (v), therefore, after crossing e green (v) to leave the lens it has a second crossing left of j. The edge e green (u) has a crossing with e blue (v) in the arc [i, i + 1] and another one between j and k. Now consider the edge e red (u). If it leaves the lens through e green (v), then it has entered a region whose boundary consists of a piece of e blue (u) and a piece of e green (v). Edge e red (u) is disjoint from e blue (u) and it has already used its unique crossing with e green (v). Hence, e red (u) has to leave the lens through the blue arc on e blue (v). In this case, however, e red (u) enters a region whose boundary consists of a piece of e blue (v) and a piece of e green (u). Again e red (u) is trapped. Hence, this conguration is impossible.
Suppose the last crossing on the blue edge is to the left of the lens, Figure 14 (right). To get from v to the backward arc [i, i+1] edge e blue (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 blue arc [k, k + 1] below e green (v) with the property that j k < i. The edges e red (u) and e red (w) both need at least two crossings with the union of e green (v) and e blue (v). Hence, they both cross e green (v) and e blue (v). The order of the red edges at p 1 implies u < 1 v < 1 w. Matching these data with the drawings of Figure 12 we nd that T (u, v) = B 3 and T (v, w) = B 2 . These types together with the already known permutation π 1 imply that π 2 and π 3 also equal π 1 . Hence, the triple is a uniform system, and we must have T (u, w) P {B 2 , B 3 }. Since edge e blue (w) has to follow the`tunnel' prescribed by e blue (v) there is a crossing of edges e green (v) and e blue (w). There is also a crossing of e green (v) and e red (w). However, neither in B 2 nor in B 3 we see crossings of e green (1) with e blue (2) and e red (2). Hence, again the conguration is impossible.
We now come to the discussion of the general case.
Proposition 19. For all v, w P V, there is no lens formed by e green (v) and e blue (w) in D 0 .
Our proof of this proposition unfortunately depends on a lengthy case analysis, an outline of which is given in the following subsection.

Outline of the Proof of Proposition 19
Suppose that there is a lens L formed by e green (v) and e blue (w) in D 0 . From Proposition 18 we know that v T = 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.
(2) 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 crucial throughout the argument. 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 discuss eight congurations that may appear in a meander of e blue (w) over e green (v). The eight congurations shown in Figure 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 blue (w) is cutting e green (v) upward/downward. For example the meander labeled VII corresponds to below/left/up. In the case discussion we impose additional conditions on vertices u that are contained in the regions dened by arcs of e blue (w) over e green (v). These conditions either ask for u P S 1 (w) or for u T P S 1 (w). The conditions are so that they are satised for free if the respective regions are relative lenses of e blue (w) over e green (v). In each of the cases we show that the T 1 projections yield an illegal table. Hence the congurations do not appear in the drawing D 0 . In the second part we use the results from the rst part to show that meanders of the drawing D 0 have no inections, i.e., in the meander permutation 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 20. If a meander permutation σ green (v) has an inection, then it has one of the four turn-back patterns shown in Figure 16.
Proof. Suppose that a meander permutation σ = σ green (v) has an inection at i. We discuss the case where i + 1 < σ i − 1 < σ i and the blue arc [i + 1, i] of e blue (w) is above e green (v). All the other cases can be treated with symmetrical arguments. 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 suppose that i − 1 < σ i − 2 < σ i. Let x be a vertex in the region dened by the arc [i − 1, i − 2]. Conned by e blue (w) the edge e blue (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 − 2 < σ i − 1 < σ i and the meander contains the second of the turn-back patterns shown in Figure 16. Other cases of inections are related to the other cases of turn-back patterns.
The proof of the second lemma involves a delicate case analysis and is deferred to Subsection 4.3.2. Lemma 21. A meander permutation has no turn-back. TB 1 TB 2 TB 4 TB 3 Figure 16: The four turn-back pattern.
The consequence of Lemma 21 is that the meanders in D 0 are very simple, their meander permutations are either the identity permutation or the reverse of the identity. In particular 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 congurations that have been discussed in Subsection 4.3.1. The additional conditions are satised because the arcs all dene relative lenses. Hence, there are no nontrivial meanders, i.e., every pair of a green and a blue edge crosses at most once in D 0 . This concludes the proof of Proposition 19.

Case analysis for the proof of Proposition 19 4.3.1 The eight basic configurations
We now deal with special instances of the congurations from Figure 15. Case I. The rst intersection along e blue (w) is downwards and the last intersection is upwards and to the right of the rst intersection along e green (v).
Its last intersection with e green (v) being upward forces e blue (w) to intersect e red (v) on its nal piece. Therefore, edge e green (w) has to turn around v as shown in Figure 17. Now consider e red (w), its rst crossing (among those relevant to the argument) has to be with e green (v). With this crossing, however, e red (w) gets separated from its destination p 1 by the union of e blue (w) and e green (v). Therefore, this case is impossible. v w u Figure 18: Illustration for Case II.
Case II. The rst intersection along e blue (w) is upwards 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 blue (w) above e green (v) whose region contains a vertex u T P S 1 (w).
Edge e green (w) has to turn around v and also e red (w) has to cross e blue (v), see Figure 19. Now consider a vertex u in a forward region of e blue (w) above e green (v). Vertex u is not under an arc of e blue (u) (Lemma 17) and there is at most one crossing of e blue (u) and e red (w). Therefore, e blue (u) behaves as shown in the sketch, i.e., w < 3 u < 3 v. Also w < 2 u < 2 v. From the intersections of green and blue edges we conclude 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 blue (w) is upwards 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 blue (w) above e green (v) whose region contains a vertex Edge e green (w) has to stay below e green (v). Now consider a vertex u in a backward region of e blue (w) above e green (v). Vertex u is not under an arc of e blue (u) and we exclude the conguration of Case I. Therefore, e blue (u) behaves as shown in Figure 18, i.e., u < 3 w < 3 v. It follows that e green (u) has to stay above e green (v) and u < 2 v < 2 w. From the intersections of green and blue edges we conclude that the projections T 1 are as given by the table on the right. This tables is not legal. Hence, this case is impossible. Case IV. The rst intersection along e blue (w) is downwards and the last intersection is upwards and to the left of the rst along e green (v) and there is a backward arc of e blue (w) below e green (v) whose region contains a vertex u T P S 1 (w).
Edge e blue (w) has to cross e red (v) and the other edges of w have no intersections with edges of v. Edge e blue (u) either turn around w or it crosses e red (v). In the rst case e green (u) has turn around v. Now, e red (u) has to cross e blue (w) and gets separated from its destination p 1 by the union ofe blue (w) and e green (u). Hence, there is no possible routing for edge e red (u) respecting the restrictions on crossings in D 0 . In the second case e blue (u) crosses e red (v) and v < 3 w < 3 u. From the green edges we obtain w < 2 v < 2 u. The intersections of green and blue edges, see Figure 20 imply that the projections T 1 are as given by the table below the gure. This table is not legal. Hence, this case is impossible. Case V. The rst intersection along e blue (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 blue (w) below e green (v) whose region contains a vertex u P S 1 (w) and further to the right a forward arc of e blue (w) above e green (v) containing a vertex x T P S 1 (w).
If u < 3 w, then edge e blue (w) makes sure that x T P S 1 (u). Therefore, we have Case II with v, u, and x.
If w < 3 u < 3 v, then w P S 1 (u). Therefore, we have a Case III. Now let w < 3 v < 3 u. Since e blue (x) is crossing e green (v) downwards we have e green (x) above e green (v) and a crossing of e blue (w) with e green (x). Therefore w < 3 x.
w v u x Figure 22: Case V, 2nd illustration.
If w < 3 x < 3 v, then the intersections of green and blue edges areas shown in Figure 21. The projections T 1 are given by the table below the gure. This table is not legal. Hence, this case is impossible. If v < 3 x < 3 u, then we have Case IV with x and u. Finally, if u < 3 x, then the green and blue edges behave as shown in Figure 22. Now, e red (w) would have to cross one of e blue (x) and e green (v) twice to get to its destination p 1 .
Case VI. The rst intersection along e blue (w) is downwards and the last intersection is downward and to the left of the rst. Moreover, along e green (v) there is a backward arc of e blue (w) above e green (v) whose region contains a vertex u P S 1 (w) and further to the right a backward arc of e blue (w) below e green (v) containing a vertex x T P S 1 (w). If x < 3 v, then edge e blue (x) either yields a Case II with w T P S 1 (x) or a Case III with u P S 1 (x). Therefore, v < 3 x.
If w < 3 u, then edge e blue (u) has an arc above u or it forms a Case I. Therefore, u < 3 w and the complete order is u < 3 w < 3 v < 3 x.
From green edges we get w < 2 v < 2 x and u < 2 v, see Figure 23. Independent of the entry T 1 (u, w) the resulting table as shown below Figure 23 violates the quadruple rule, Lemma 9. Hence, this case is impossible.  Case VII. The rst intersection along e blue (w) is upwards and the last intersection is upward and to the left of the rst. Along e green (v) there is a backward arc of e blue (w) below e green (v) whose region contains a vertex u T P S 1 (w) and further to the right a backward arc of e blue (w) above e green (v) containing a vertex x P S 1 (w).
If v < 3 x, then edge e blue (x) either yields a Case I or a Case IV with u T P S 1 (x). Therefore, x < 3 v.
If u < 3 x, then vertices v, u and x are in the conguration of Case II. If x < 3 u < 3 w, then u < 3 v since otherwise u would be in a region of e blue (u), a violation of Lemma 17. Now u < 3 v < 3 w and T 1 (u, v) = B, T 1 (u, w) = A and T 1 (v, w) = B is an illegal table.
If v < 3 u, then x < 3 v < 3 w < 3 u and the situation is as sketched in Figure 24. Irrespective of u < 2 w or w < 2 u the T 1 projections of the remaining 5 pairs yield the partial table shown below the gure. This violates the quadruple rule, Lemma 9. Hence, this case is impossible.  Case VIII. The rst intersection along e blue (w) is upwards and the last intersection is upward and to the right of the rst. Along e green (v) there is a forward arc of e blue (w) above e green (v) whose region contains a vertex u T P S 1 (w) and further to the right a forward arc of e blue (w) below e green (v) containing a vertex x P S 1 (w).
If v < 3 u, then edge e blue (u) either yields a Case I or a Case IV with w T P S 1 (x). Therefore, u < 3 v.
If u < 3 x < 3 v, then edge e blue (x) yields a Case III with u P S 1 (x). If x < 3 u, then x < 3 u < 3 w and T 1 (x, u) = N, T 1 (x, w) = B and T 1 (u, w) = A is an illegal table.
Therefore, v < 3 x and in total u < 3 v < 3 x < 3 w, see Figure 25. Also u < 2 v < 2 w and v < 2 x only the order of x and w in < 2 is not determined. If w < 2 x then T 1 (x, w) = N otherwise T 1 (x, w) = B. The resulting table is shown below Figure 25. In the rst case the triple u, v, x yields an illegal subtable. In the second case there is a violation of the quadruple rule, Lemma 9. Hence, this case is impossible.

Proof of Lemma 21
Given v suppose that some e blue (w) has a meander over e green (v) with a turn-back. Among all the turn-backs over e green (v) choose the one with the turn-back point i as far as possible, where i is the crossing of the turn-back whose two adjacent crossings are on the same side.
It will be seen in the proof that this specic choice of turn-back substantially simplies the analysis of two of the cases.
Case TB 1. The turn back corresponds to a substring i, i − 1, i − 2, i + 1 or to a substring i, i + 1, i + 2, i − 1 in σ green (v) and the turn-back arc is above e green (v).
Let us assume that the substring is i, i − 1, i − 2, i + 1 so that the turn-back arc is The edge e blue (u) is conned by e blue (w) and has a forward arc below x. If e blue (u) has a crossing with e red (v) then we have a Case I. From Lemma 17 it follows that the last crossing of e blue (u) and e green (v) is downwards and to the right of i + 1.
If x T P S 1 (u), then e blue (u) has a backward arc a in the interval [i − 1, i − 2] such that x is in the region of arc a. The left part of Figure 26 shows a schematic view of the situation. Let y be a vertex in the region of an arc of e blue (u) above e green (v) and preceeding a along e blue (u) such that y T P S 1 (x). Edges e blue (u) and Lemma 17 force e blue (x) to have its last intersection with e green (v) downwards and to the right of i + 1. Hence e blue (x) and y are in the conguration of Case II.
We know that x P S 1 (u). Let y be a vertex below the last forward arc of e blue (u) above e green (v). If y T P S 1 (u), then e blue (u) with x and y are in the conguration of Case V. Otherwise, if y P S 1 (u), then there is a backward arc a of e blue (u) above y. Choose a vertex z from a backward arc below e green (v) and to the right of a. The right part of Figure 26 shows a schematic view of the situation. Edge e blue (u) is forcing e blue (z) to have an arc over y. If the last intersection of e blue (z) with e green (v) is downwards and left of the last intersection of e blue (u) and e green (v), then v with z and y are in the conguration of Case III.
If the last intersection of e blue (z) with e green (v) is downwards and right of the last intersection of e blue (u) with e green (v), then e red (z) has to cross e blue (u) twice to reach its destination p 1 without intersecting e blue (z) and e green (z). The left part of Figure 27 shows a schematic view of the situation, the light-blue arc is spanned by two arcs of e blue (z).
If the last intersection of e blue (z) with e green (v) is upwards and e blue (z) crosses e red (v), then there are two options. The rst is that x T P S i (z) and there is a conguration of Case VII with v, z, x and y. The second is x P S i (z), this requires that e blue (z) has a turn-back interior of the original turn-back of e blue (w) before crossing e red (v), in this situation e blue (x) shows a Case I or together with u a Case III. The right part of Figure 27 illustrates this case. Case TB 2. The turn back corresponds to a substring i + 1, i − 2, i − 1, i or to a substring i − 1, i + 2, i + 1, i and the turn-back arc is below e green (v).
Let us assume that the substring is i + 1, i − 2, i − 1, i so that the turn-back is the arc The rst intersection of e blue (u) with e green (v) is left of the turn back point i of the meander e blue (w). The choice of the turn-back as the leftmost implies that e blue (u) has no turn-back. Hence, σ green (u) is the reverse of the identity.
If the last intersection of e blue (u) with e green (v) is downwards, then this is a Case III with x. Otherwise the last intersection is upwards and e blue (u) intersects e red (v). Then there is a backward arc of e blue (u) below e green (v) and a vertex y from such an arc together with x shows that the situation of Case VII. See Figure 28. Case TB 3. The turn back corresponds to a substring i + 1, i − 2, i − 1, i or to a substring i − 1, i + 2, i + 1, i and the turn-back arc is above e green (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 the region of the arc [i − 2, i − 1].
The rst intersection of e blue (u) with e green (v) is left of the turn back point i of the meander e blue (w). The choice of the turn-back as the leftmost implies that e blue (u) has no turn-back. Hence, σ green (u) is the reverse of the identity.
If e blue (u) intersects e red (v), then with x this yields a Case IV. Otherwise there is a backward arc of e blue (u) above e green (v). Let y be a vertex from such an arc. Now e blue (u) together with x and y are a Case VI. Case TB 4. The turn back corresponds to a substring i, i − 1, i − 2, i + 1 or to a substring i, i + 1, i + 2, i − 1 and the turn-back arc is below e green (v).
Let us assume that the substring is i, i − 1, i − 2, i + 1 so that the turn-back is the arc If e blue (x) intersects e red (v) this is a Case I. It follows that that x < 2 v. If u < 3 x then u and x are in the conguration of Case II. If x < 3 u, then there is a forward arc of e blue (u) below e green (v) and to the right of i − 2. Let y be a vertex from the region of such an arc.
We rst consider the case y P S 1 (u). If e blue (u) intersects e red (v) then u with y yields a Case VIII. If x < 3 y, then it follows with Lemma 17 that x P S 1 (y) so that they form a Case III. For y < 3 x edge e blue (y) has to turn back as shown in the left part of Figure 29. In this conguration e red (u) has to cross one of e green (v) and e blue (y) twice to get to its destination p 1 . Now we are in the case x < 3 u and y P S 1 (u). This requires a backward arc of e blue (u) whose region contains y. To the right of this arc there has to be a backward arc above e green (v). Let z be a vertex from the region of this arc. The situation is shown in the right part of Figure 29. If y < 3 x, then z T P S 1 (y) so that y and z form a Case II. If x < 3 y, then either y is a Case I or it forms a Case III with x.

Outer Drawings with arbitrary k
We now generalize our result on consistency of outer drawings for k = 2, 3 to any k ! 2.
In the previous section, we dened types for each pair of vertices in V in an outer drawing of K 3,n . Fortunately, we do not need to go beyond the enumeration of the outer drawings of K 3,2 to be able to generalize our result to arbitrary values of k.
Theorem 22 (Consistency of outer drawings for k ! 2). Consider the complete bipartite graph G with vertex bipartition (P, V) such that |P| = k and |V| = n. Given assignments T ij of types in {A, B, N} for each pair of vertices (p i , p j ) in P and each pair of vertices in V, there exists an outer drawing of G realizing those types if and only if all assignments T ij for 1 i < j k obey the triple and quadruple consistency rules in the sense of Theorem 10, for any triple of vertices from P and and each pair of vertices in V the assignments correspond to an outer drawing of K 3,2 in the sense of Table 1.
Proof. We proceed by induction, suppose that the result holds for some k and prove that the result holds for k + 1. The base cases for k = 2, 3 were proved in the previous sections.
From the induction hypothesis, there exists a drawing D k−1 of K k−1,n realizing the types involving vertices p 1 , p 2 , . . . , p k−1 in P. Similarly, there exists a drawing D k of K 2,n realizing the types involving the two vertices p 1 and p k .
Similarly to what we did in the proof of Theorem 14, we can superpose the two drawings in such a way that the drawings of the two stars K 1,n corresponding to p 1 match. More precisely, let the drawing D k−1 live on plane Z k−1 and D k live on plane Z k . Consider a xed homeomorphism φ between the planes. There is a homeomorphism ψ : Z k−1 → Z k−1 such that mapping D k−1 via φ ψ to Z k yields a superposition 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 curves.
At each vertex v P V the rotation is correct, i.e., we see the edges to p 1 , p 2 , . . . , p k in clockwise order.
The drawing has no touching edges, i.e., when two edges meet they properly cross in a single point. The drawing D obtained by superposing D k−1 and D k is a drawing of K k,n , possibly with some multiple crossings. We color the edges of D according to the vertex p i P P it is incident to, using colors 1, 2, . . . , k. Each color class of edges is a non-crossing star, and the parity of the number of crossings between two edges of distinct colors is prescribed by the type assignments. Furthermore, since D k−1 and D k are proper outer drawings, all pairs of edges crossing more than once have respective colors i and k for some i P {2, 3, . . . , k − 1}.
Consider all lenses formed by an edge of color k and another edge of color i P {2, 3, . . . , k − 1}. Such a lens is said to be empty whenever it does not contain any vertex from V. It is inclusionwise minimal whenever it does not fully contain any other lens. Consider an empty inclusionwise minimal lens formed by an edge e k (v) of color k and another edge e i (w) of color i.
If an edge of color i 0 P {1, 2, 3, . . . , k − 1} \ {i} intersects the lens, it can only cross e i (w) once. Similarly, it cannot cross the boundary of the lens on e k (v) more than once, either because i 0 = 1, or because otherwise the lens would not be minimal. Another edge of color k cannot intersect the lens, because it could only do so by intersecting e i (w) at least twice, contradicting the minimality of the lens.
Hence all edges intersecting the lens intersect its boundary exactly once on each edge. Therefore we can safely get rid of the two crossings by making e k (v) and e i (w) switch sides. The switch at a lens does not change the type assignment. We can perform this iteratively until no empty lens is remaining. We call the nal drawing D 0 . This drawing has the property that every lens contains a vertex.
We now wish to show that D 0 has no lens. For the sake of contradiction, suppose that there is a lens formed by e i (v) and e k (w) in D 0 . We can apply our previous analysis for the case k = 3 on the restriction D 00 of D 0 formed by the edges of color 1, i, and k, letting red= 1, green= i, and blue= k. In particular Proposition 19 holds, and D 00 has no lens. Therefore, neither does D 0 , which is a suitable outer drawing of K k,n .
This yields the following corollary for AT-graphs.
Corollary 23 (AT-graph realizability). There exists an O(k 3 n 2 + k 2 n 4 ) algorithm for deciding the existence of an outer drawing of an AT-graph whose underlying graph is of the form K k,n .
Proof. We know that the three types of pairs A, B, N (see Figure 12) exactly prescribe which pairs of edges cross, hence given the set of crossing pairs, we can reconstruct the type assignments T ij . From these data, we can consider every 5-tuple (p a , p b , p c , u, v), with p a , p b , p c P P and u, v P V, and check that the corresponding drawing of the induced K 3,2 graph is one of the drawings given in Table 1. This can be done in time O(k 3 n 2 ). If this is correct, we can then check that every triple is legal and that the quadruple rule is satised for each T ij . This can be done in time O(k 2 n 4 ), hence the result.
In fact, this implies that checking consistency on all 6-tuples of vertices is sucient. This matches the result of Kyn cl for complete graphs [14].

Extendable and Straight-line Outer Drawings
A simple topological drawing of a graph is called extendable if its edges can be extended into a pseudoline arrangement. We consider the problem of the existence of an extendable outer drawing for a given rotation system. We also further restrict to straight-line drawings. We exploit a connection between topological drawings and generalized conguration of points which was also used by Kyn cl [14]. The main result uses the notion of suballowable sequence, as dened by Asinowski [2].

Extendable Outer Drawings
We will use the notion of allowable sequence. These sequences were introduced by Goodman and Pollack in a series of papers [6,7,8] as an abstraction of the sequences of successive permutations obtained by projecting a set of n points in R 2 on a rotating line. A good summary of this and related notions can be found in Goodman's article in the Handbook of Discrete and Computational Geometry [21]. We restrict to simple allowable sequences, corresponding to point sets with no three points on a line and such that no two pair of points determine the same slope.
A sequence π 1 , π 2 , . . . , π m+1 , . . . , π 2m of permutations of an n-element set, with m = n 2 , is an allowable sequence if for all i P [m], π i and π m+i are reverse of each other, each consecutive pair of permutations dier only by a single adjacent transposition, and every pair of element is reversed exactly twice in the sequence.
We will also need the following additional denition: A Generalized conguration is a pair (C, A) such that C is a collection of n points in the projective plane P 2 and A is an arrangement of pseudolines, one of which is the pseudoline at innity L ∞ , satisfying the following conditions: every pair of points of C lie on a pseudoline of A, each pseudoline in A contains exactly two points of C, except L ∞ , L ∞ does not contain any point of C. We say that an allowable sequence on an n-element set is realized by a generalized conguration on n points when the successive permutations of the sequence correspond to successive segments of the pseudoline L ∞ , and the transposition between a pair of permutations corresponds to the intersection of L ∞ with the pseudoline containing the two corresponding points. In particular, we observe that if a permutation π is realized at some point p P L ∞ , then p can be connected by additional pseudolines to the n other points in the radial order prescribed by π.
The following result is known.
This directly yields that allowable sequences exactly encode which are the achievable rotation systems for our drawings. We rst dene suballowable sequences, see Asinowski [2]: A sequence π 1 , π 2 , . . . , π k of permutations of an n-element set is said to be a suballowable sequence if there exists an allowable sequence admitting π 1 , π 2 , . . . , π k as a subsequence.
Theorem 25. Given a sequence π 1 , π 2 , . . . , π k of permutations of n elements, there exists an extendable outer drawing of K k,n with this rotation system if and only if π 1 , π 2 , . . . , π k is a suballowabe sequence.
Proof. First, suppose that the rotation system is a suballowable sequence. Then by denition there exists an allowable sequence admitting it as a subsequence. Then from Theorem 24 there exists a generalized conguration on n points realizing this sequence. The n points form the set V. For each permutation π i , there exists a point on L ∞ that can be connected by additional pseudolines to the n points of V in the order prescribed by π i . These k points form the set P. By keeping only the inner segments of the pseudolines connecting every point of P with all points of V, we obtain an extendable outer drawing satisfying our conditions. Now suppose there exists an extendable outer drawing. Consider the generalized conguration obtained as follows. First, we extend the edges between the points of P and V into pseudolines, which is possible from the extendability of the drawing. Then we connect the points of P by the pseudoline at innity. Finally, by iteratively applying Levi's enlargement Lemma, we can also connect every pair of points in V by a new pseudoline. The order in which each point of P is connected to all points of V realizes one permutation in the allowable sequence dened by the generalized conguration. The collection of such permutations therefore forms a suballowable sequence.
Note that since suballowable sequences can be recognized eciently (see [2]), this directly yields a polynomial-time algorithm for deciding whether there exists an extendable outer drawing realizing a given rotation system.

Straight-line Outer Drawings
We have a direct equivalent statement for straight-line outer drawings.
Theorem 26. Given a sequence π 1 , π 2 , . . . , π k of permutations of n elements, there exists an straight-line outer drawing of K k,n with this rotation system if and only if π 1 , π 2 , . . . , π k is a suballowable sequence that can be realized by an actual point arrangement.
This does not yield an ecient algorithm, since deciding whether a given allowable sequence can be realized by an arrangement of straight lines is a hard problem [20,17,10]. The problem, however, is tractable in the special case where k = 3. This is due to the fact that we can x the orientation of the three sets of edges. There are two cases to consider. If the suballowable sequence π 1 , π 2 , π 3 can be extended into an allowable sequence such that the three permutations π i appear in the rst m permutations of the whole sequence, then there exists a straight-line outer drawing with p 1 = (−∞, ∞), p 2 = (0, ∞), and p 3 = (∞, ∞). This corresponds to the case where the three projection directions span less than a half-circle. Otherwise there exists a drawing with p 1 = (∞, ∞), p 2 = (0, −∞) and p 3 = (−∞, 0). We rst give an example showing that there are rotation systems for k = 3 that are suballowable sequences but do not admit a straight line outer drawing. Up to renaming, this example is the same as Asinowski's example of nonrealizable suballowable sequence ( [2], Proposition 8). We give the proof for completeness. Furthermore, our proof directly indicates how to solve the straight-line realizability problem in the case k = 3.
A straight line outer drawing of such a rotation system can be regarded as a tropical arrangement of lines. We refer to [16] for an introduction to tropical geometry. Such outer drawings can consequently be regarded as tropical arrangements of pseudolines. Restated as a result in tropical geometry Theorem 27 tells us that there are nonstretchable tropical arrangements of pseudolines.
The classical construction of a simple non-stretchable example of pseudolines due to Ringel is based on the 9 lines of a Pappus conguration. In [19] tropical versions of Pappus are discussed. They note that a conguration on 9 tropical lines which can be obtained from the conguration of Theorem 27 by replacing points q by triple incidences of lines a i , b j , c k with {i, j, k} = {1, 2, 3} is a valid tropical version of Pappus' Theorem.
The stretchability problem for a tropical arrangements of pseudolines is polynomialtime solvable, as it boils down to the feasability of a linear program. The proof of Theorem 27 indicates how to set up such a linear program.

Open problems
A problem that remains for k = 2 is to nd a good description of all outer drawings for a given pair (π 1 , π 2 ) of rotations. We state the same problem dierently: Problem 1. For a given pair (π 1 , π 2 ) of permutations characterize all consistent assignments of types with the property that if a < b in both permutations, then type(a, b) P {A, B}, and if a < 1 b and b < 2 a, then type(a, b) = N.
The simpler case where π 1 = π 2 (hence all types are in {A, B}) is covered by Corollary 5, which provides a bijection with combinatorial structures counted by Schr oder numbers.
The following natural question also remains unsolved: Problem 2. For a given triple (π 1 , π 2 , π 3 ) of rotations, can we decide in polynomial time whether there exists a consistent assignment of types?
The problem could be generalized to arbitrary k: how hard is it to decide whether there exists an outer drawing of K k,n for a given rotation system? We answered this question partially, by showing that it is polynomial-time decidable when the drawing is further required to be extendable.
We also note that outer drawings are not really enlightening with respect to crossing numbers, as minimally crossing outer drawings of K k,n are easy to describe. It can be checked that by drawing the n vertices of V on a line so that k/2 vertices of P see them in some order, and the other dk/2e vertices see them in the reversed order, we achieve the minimum number of crossings. This minimum has value exactly n 2 k 2 − k/2dk/2e . Further insight into crossing numbers could be gained by considering the obvious remaining problem of generalizing our analysis to general simple topological drawings, dropping the outer property.