Journal of Graph Algorithms and Applications on Aligned Bar 1-visibility Graphs

A graph is called a bar 1-visibility graph if its vertices can be represented as horizontal segments, called bars, and each edge corresponds to a vertical line of sight which can traverse another bar. If all bars are aligned at one side, then the graph is an aligned bar 1-visibility graph, AB1V graph for short. We consider AB1V graphs from different angles. First, we study combinatorial properties and K5 subgraphs. Then, we establish a difference between maximal and optimal AB1V graphs, where optimal AB1V graphs have the maximum of 4n − 10 edges. We show that optimal AB1V graphs can be recognized in O(n 2) time and prove that an AB1V representation is determined by an ordering of the bars either from left to right or by length. Finally, we introduce a new operation, called path-addition, that admits the addition of vertex-disjoint paths to a given graph and show that AB1V graphs are closed under path-addition. This distinguishes AB1V graphs from other classes of graphs. In particular, we explore the relationship to other classes of beyond-planar graphs and show that every outer 1-planar graph is an AB1V graph, whereas AB1V graphs are incomparable, e.g., to planar, k-planar, outer fan-planar, outer fan-crossing free, fan-crossing free, bar (1, j)-visibility, and RAC graphs. Supported by the Deutsche Forschungsgemeinschaft (DFG), grant Br835/18-2. An extended abstract of this paper has been presented at WALCOM 2016 [11].

Visibility is a major topic in computational geometry [31] and graph drawing [19].A bar visibility representation displays each vertex as a horizontal bar and each edge as a vertical line of sight between the bars of the endvertices, which, in case of k-visibility, traverses up to k other bars.Bars are non-intersecting segments.There are several versions of visibility including distinct, strong and weak.In the distinct case, the endpoints of the bars must have different xcoordinates [29,25].In the distinct and strong versions there is an edge if and only if there is a visibility, whereas in the weak version there is a visibility if there is an edge.Thus edges can be omitted.Clearly, graphs in the weak version are exactly the subgraphs of graphs in the other versions.This assumption is relevant for a comparison with other classes of graphs which are generally closed under taking subgraphs.
Every weak visibility graph is planar and vice versa.Hence, weak visibility graphs can be recognized in linear time.Some planar graphs do not have a strong visibility representation, including K 2,3 [18] and some 3-connected graphs [3].The recognition problem for strong visibility graphs is N P-complete [3] and there is a characterization of strong visibility graphs [34,37,39].
In the non-planar case with k-visibility, a line of sight for an edge can traverse up to k other bars.Simply speaking, an edge can cross up to k vertices.The planar case corresponds to k = 0. Bar 1-visibility graphs were introduced by Dean et al. [18] and further investigated by Sultana et al. [35] and by Evans et al. [22], who also compared them with other classes of beyond-planar graphs and considered the strong and weak versions of 1-visibility.Bar 1-visibility graphs are specialized to bar (1, j)-visibility [13] and bar (1, 1)-visibility graphs [9] by restricting the number of edges that may traverse a bar to j > 0 and j = 1, respectively.In particular, bar 1-visibility graphs of size n have up to 6n − 20 edges and an N P-hard recognition problem.However, 1-planar graphs are a proper subclass of bar (1, 1)-visibility [9] and bar 1-visibility graphs [21].Hence, in contrast to the planar case, bar 1-visibility is stronger than 1-planarity.
An outer planar graph has a drawing with all vertices in the outer face.In particular, all vertices can be placed on a line with edges as circular arcs above this line.Outer planar graphs are an important subclass of the planar graphs.Each outer planar graph has at least one vertex of degree at most two.The graphs have book-thickness one and treewidth two, and K 4 and K 2,3 are the forbidden minors.The restriction from planar to outer planar transfers a general to an aligned bar visibility representation, where all bars start at a common line, e.g., at the bottom or at the left.In the planar case, outer planar and weak aligned bar visibility graphs coincide [17], whereas aligned bar 1-visibility is stronger than outer 1-planarity, as we shall show.
Aligned bar 1-visibility representations and graphs were introduced by Felsner and Massow [25] who used the distinct version of visibility and allow a line of sight to traverse up to k bars.A distinct AB1V representation, dAB1V for short, is characterized by two permutations, the t-and r-orders.The t-order is the left-to-right (or top-down) ordering of the bars of the vertices and the r-order is an ordering of the bars by length.Felsner and Massow established important properties of dAB1V graphs.For example, they showed that every dAB1V graph has a vertex of degree four and has at most 4n − 10 edges.This bound is tight for all n ≥ 4.Moreover, they established that AB1V graphs are 5-colorable, have clique number five and geometric thickness two, and they showed that an r-order can be computed from a t-order of a dAB1V graph.
In this work, we extend the research on AB1V graphs.There are two main features in AB1V graphs: clusters and paths.A cluster is a K 5 subgraph, i.e., the maximum complete AB1V graph.First, we show that every cluster has a special AB1V representation and can uniquely be associated with a vertex.Thereafter, we prove that maximal AB1V graphs have at least 3.5n − 9 edges and that there are sparse maximal AB1V graphs with 3.5n − 1 edges for every n ≥ 21.Then we establish a quadratic-time recognition algorithm for optimal AB1V graphs.Complementary to a result by Felsner and Massow, we show that the t-order can be computed from the r-order of a dAB1V graph in linear time.Given an AB1V representation, one can easily introduce a vertex-disjoint path between two vertices.This observation has led to the introduction of path-addition, which is a new operation on graphs and is further studied in [12].Finally, we explore the relationship of weak AB1V and other classes of beyond-planar graphs.First, there is a proper hierarchy for the versions of visibility and AB1V graphs.Then we show that every outer 1-planar graph is an AB1V graph but not conversely.An AB1V representation of an outer 1-planar graph can be constructed in linear time.However, AB1V graphs are incomparable, e.g., to planar, k-planar, outer fan-planar, outer fan-crossing free, fan-crossing free, bar (1, j)-visibility, and RAC graphs.
The paper is organized as follows.In Section 2 we introduce basic concepts.Structural properties and maximal graphs are studied in Section 3 and pathadditions in Section 4. In Section 5 we investigate recognition problems.The relationship to other graph classes is discussed in Section 6 and we conclude with some open problems.

Preliminaries
We consider simple, undirected graphs G = (V, E) with vertices listed in some arbitrary order.Let N (v) denote the set of neighbors of a vertex v including v and G[U ] the subgraph induced by U ⊆ V .
Graphs are defined by aligned bar 1-visibility representations, AB1V for short.As suggested by Felsner and Massow [25], we rotate the drawings, which, due to the alignment, are more intuitive and compact than ordinary visibility representations.Vertical bars were also used by Wismath [39].In an AB1V representation, each vertex is represented by a vertical bar, which is a closed interval with bottom at y = 0 and top at some y > 0. Each edge e = (u, v) corresponds to a horizontal line of sight between the bars of u and v, which can traverse another bar.Thus, there is a 1-visibility between the bars of u and v.
We distinguish between distinct, strong and weak visibility [19] and denote the respective classes of aligned bar 1-visibility graphs by dAB1V , sAB1V and wAB1V , respectively.In addition, we use mAB1V for maximal AB1V graphs, which cannot be extended by adding a further edge, and oAB1V for optimal AB1V graphs with the maximum of 4n − 10 edges for graphs of size n.Recall that there is an edge if and only if there is a visibility in the distinct and strong versions, where in the distinct case all bars have a different length.The weak version admits the omission of edges.
A partial AB1V representation is the AB1V representation of an induced subgraph G[U ].An extension of a partial AB1V representation is an AB1V representation of G[U ∪ W ] for sets of vertices U and W so that the restriction to U is the AB1V representation of G[U ].
From Felsner and Massow [25] we adopt the t-and r-orders for the description of AB1V representations, which are the orderings of the bars from left to right and by length, respectively.Then a partial AB1V representation is the restriction of the t-and r-orders to the vertices of U .For convenience, we say that vertex u is left of vertex v if the left to right ordering holds for their bars in an AB1V representation.Accordingly, we say that a vertex is between other vertices.Note that each AB1V representation has a reflection with an ordering from right to left representing the same graph.
We denote the t-and r-orders of two vertices by u < t v and u < r v, respectively, and we extend these relations to sets of vertices and bucket orders [24].For disjoint sets of vertices X and Y we write X < t Y if x < t y for every x ∈ X and y ∈ Y , however, the ordering of the vertices of X and Y is still unclear or may be exchanged.The sets X and Y are called buckets.A bucket order is extended to an order by ordering the elements in each bucket [24].The notation X < t Y is also used for induced subgraphs, and X < r Y is used accordingly.For convenience, we omit braces for singleton sets and inside brackets.
Aligned visibility representations were introduced by Cobos et al. [17].They considered the planar case with non-transparent bars and implicitly use the tand r-orders by assigning an n-tuple to an aligned bar visibility representation of a graph of size n, where the i th entry is the length of the i th bar from the left.
In general, we shall consider distinct AB1V representations with all bars of different length, that are represented by the t-and r-orders.For convenience, we omit the prefix d and explicitly refer to the other versions of visibility if this is relevant.For a comparison with other classes of graphs we use wAB1V graphs, since they are closed under taking subgraphs.

Combinatorial Properties
In their introductory paper on AB1V graphs, Felsner and Massow [25] observed that each AB1V graph has a vertex of degree at most four and that the clique number is five, i.e., K 5 is the largest complete AB1V (sub-) graph.We call a K 5 -subgraph a cluster.In fact, clusters and their AB1V representation play a prominent role in AB1V graphs.

Clusters
Forthcoming, we shall often consider K 5 -subgraphs and subgraphs that are induced by vertices with long bars in an AB1V representation, i.e., we consider the top-k vertices in r-order and a 1-visibility above a certain level.Two long bars prevent a 1-visibility between two short bars to their left and right.For AB1V graphs the following is immediate: has the second shortest bar, and there is no other vertex u of G with v 1 < t u < t v 5 and v 3 < r u.
An AB1V representation of K 5 is depicted in Fig. 1.The bars of the vertices of a cluster C are U-shaped with the shortest bar in the middle and the second shortest bar next to it.The top-3 vertices of C in r-order can be permuted and the longest bar may be next to the bar in the middle.This is excluded in a V -shape.Hence, a K 5 has six AB1V representations up to reflection, some of which may be excluded if more vertices are taken into account.The left to right order it tight in the sense that no other vertex u can be placed between the vertices of C if the bar of u is longer than the shortest bar of the vertices of C. In reverse, if u must be placed between the vertices of a cluster, then u has a short bar.In particular, if u is a neighbor of the vertex in the middle of C, then u is placed between the vertices of C and has a bar that is shorter than the bars of the vertices of C.
are two clusters of an AB1V graph G which differ in one vertex, namely x and y, then in any AB1V representation, either x is extreme in t-order and y is minimum in r-order, or conversely, see Fig. 2.
] is not a cluster by Lemma 1, since the longer bar of x and y is between the bars of the vertices of the other cluster.If x is to the left of v 2 , v 3 , v 4 , v 5 , then v 3 < r {x, v 2 , v 4 , v 5 } by Lemma 1 and v 3 < r y implies that there is no 1-visibility between v 3 and an extreme vertex.The case with x at the right is similar.
A cluster C may be associated with its vertex with the shortest bar, which may be taken as a representative of C. Lemma 3 For an AB1V graph G = (V, E), there is a one-to-one mapping κ : C(G) → V assigning each cluster to a vertex.
] is a cluster with vertices v 1 < t . . .< t v 5 in t-order.By Lemma 1, the bar of v 3 is the shortest of the bars of the vertices of X and v 3 is placed in the middle of the bars of the vertices of X.If two (or more) clusters X and Y were assigned to v 3 , then a long bar from a vertex of Y must be placed between the bars of X or vice versa, contradicting Lemma 1.
The assignment of clusters to vertices implies an upper bound on the number of clusters and all clusters can be computed in linear time from a given AB1V representation.Note that a subgraph induced by a vertex v and four neighbors of v with longer bars than v do not necessarily induce a cluster.The condition on the vertex with the second shortest bar from Lemma 1 must be satisfied.
Corollary 1 An AB1V graph of size n has at most n − 4 clusters, which, given an AB1V representation, can be listed in linear time.
Proof: A cluster can be assigned to all but the two vertices at the left and right ends in an AB1V representation, such that n − 4 vertices remain.The bound is achieved if the AB1V representation has a V -shape with a monotone t-order and a bitonic r-order.
For the computation, consider the vertices in decreasing r-order and check that κ −1 (v i ) is a cluster.
We can also consider clusters in decreasing r-order of their representatives.
Lemma 4 Let v 1 < r . . .< r v n be the r-order of the vertices of an AB1V graph.For i = 1, . . ., n − 4, there is a cluster κ −1 (v i ) if and only if v i has four neighbors with longer bars Proof: The "only-if" direction follows from Lemma 1 and for the "if-direction" observe that a vertex v j as specified hinders w i or v i to be 1-visible from x i , y i , and z i .
Once we know the clusters, we would like to know how they are interrelated.This is expressed by the clustered graphs CG w (G) that are parameterized by the number of common vertices.
A stripe is an outer planar graph with vertices of degree at most four that consists of two parallel paths with spokes as depicted in Fig. 3(b).
Lemma 5 The 4-clustered graph CG 4 (G) of an AB1V graph G consists of a set of paths and the 3-clustered graph CG 3 (G) is a subgraph of a stripe.
with four vertices in common with X and a further vertex x 6 ∈ Y − X with a longer bar than the bar of x 1 , then Lemma 2 implies that x 6 must be placed to the left or to the right of the vertices of X, , where x 2 is the neighbor of x 1 with the next longer bar.If x 6 < r x 1 , then κ(G[Y ]) = x 6 , and the roles of X and Y are exchanged.Thus, there may be an (incoming) edge to G[X] from a cluster whose representative has a shorter bar.Hence, a vertex of CG 4 (G) has degree at most two.Cycles are excluded by the increasing length of the bars of the assigned vertices.A path is interrupted at G[X] if there is no cluster that is assigned to vertex y, where y has the second longest bar of the vertices of X.
Accordingly, if clusters G[X] with X = {x 1 , . . ., x 5 } and x 1 < r {x 2 , . . ., x 5 } and G[Y ] have at least three common vertices, and the new vertices of Y − X have longer bars than those of X, then Y = {x 2 , . . ., x 6 } for some x 6 as stated in Lemma 2 or Y = {x 3 , x 4 , x 5 , y 1 , y 2 }, x 3 < r {y 1 , y 2 }, y 1 and y 2 are placed to the left or to the right of the vertices of X, and X ∩ Y = {x 3 , x 4 , x 5 }.Hence, there are at most two (outgoing) edges in CG 3 (G) from a cluster κ −1 (x 1 ) to the clusters of the two next neighbors of x 1 in r-order κ −1 (x 2 ) and κ −1 (x 3 ), provided these clusters exist.If they do, then there is an edge from κ −1 (x 1 ) to κ −1 (x 2 ) and to κ −1 (x 3 ) in CG 3 (G).Accordingly, there are edges in CG 3 (G) from clusters with representatives with shorter bars.
If x 1 < r . . .< r x n is the r-order of the vertices of G, then there are edges (κ −1 (x i ), κ −1 (x i+j )) for j = −2, −1, 1, 2 in CG 3 (G), provided there are clusters.Hence, CG 3 (G) is a subgraph of a stripe, which is an outer planar graph as depicted in Fig. 3(b).
If the AB1V representation of G has a monotone t-order and a bitonic rorder, then CG 4 (G) and CG 3 (G) have n − 4 vertices and are a path and a stripe, respectively, see Fig. 3. On the contrary, there are AB1V graphs such that the clustered graphs CG 1 (G) and CG 2 (G) are complete graphs on n − 4 vertices.If v 1 < t . . .< t v n and v 3 < r . . .< r v n < r v 2 < r v 1 are the t-and r-orders, then there are clusters κ −1 (v i ) for i = 3, . . ., n − 2 which each contain the two leftmost vertices with the longest bars v 1 and v 2 , and therefore are mutually connected in the clustered graphs.
Finally, we consider disjoint clusters.Two clusters G[X] and G[Y ] of an AB1V graph G are disjoint if in any AB1V representation of G the vertices of X are to the left of the vertices of Y , i.e., X < t Y , or vice versa.We say that G[Y ] nests in G[X] if there are two vertices x 1 and x 2 of X with x 1 < t Y < t x 2 .Thus (the bars of) the vertices of Y are placed between x 1 and x 2 and, by Lemma 1, x 1 and x 2 can be chosen so that no other vertex of X is placed between them.Proof: Otherwise, there are vertices x 1 < y 1 < x 2 < y 2 in t-order with x 1 , x 2 ∈ X and y 1 , y 2 ∈ Y , which contradicts Lemma 1.

Maximality
Next, we consider extremal graphs and Turán-type theorems on the maximum number of edges.An AB1V graph G is maximal if no further edge can be added without violating aligned bar 1-visibility, and G is optimal if it has 4n − 10 edges.Hence, a graph G is maximal if there is no AB1V supergraph with the same set of vertices and more edges, and G is optimal if there is no AB1V graph of the same size and more edges.These notions coincide for planar graphs and maximal and optimal planar graphs of size n have exactly 3n − 6 edges and are triangulated.However, for outer 1-planar [4], 1-planar [8], and (1, j)-visibility graphs [13], there are maximal graphs that are not optimal.Surprisingly, there are maximal 1-planar graphs with less than 2.65 n edges.
Optimal AB1V graphs were characterized by Felsner and Massow [25], who proved that a distinct AB1V representation results in an optimal graph if and only if the top-4 vertices in r-order are placed in pairs at the left and right ends and induce K 4 .In other words, the four longest bars are placed at the left and right ends and the shortest of them is not the extreme left or right.Thereafter, any other vertex has at least four neighbors and there are exactly four new neighbors if the vertices are taken in decreasing r-order.In consequence, AB1V graphs have a density of at most 4n − 10 [25], where the density is a function of the maximal number of edges of graphs of size n.However, there are sparser maximal AB1V graphs, and there are maximal AB1V representations with graphs with only 2n − 3 edges, e.g., if the bars are placed from left to right in increasing (or decreasing) r-order.
We aim at AB1V graphs with a unique AB1V representation.However, this is impossible, since the t-order can be reversed and the vertices with the three longest bars can be permuted in r-order.

Lemma 7
There is an AB1V graph B with a unique t-order of the vertices (up to reflection), and there are AB1V graphs G k for k ≥ 1 where the t-order is a unique bucket order with buckets of size two (up to reflection).
Proof: Graph B has 15 vertices from the sets V = {v 1 , . . ., v 10 } and U = {u 1 , . . ., u 5 }.For k ≥ 1, graph G k extends B by adding vertices from k + 2 pair sets Z = {z 1 , z 2 } and W i = {w i , w i } for i = 0, . . ., k.The sets V, U, Z and W 0 , . . ., W k are pairwise disjoint.For i = 0, . . ., k, the vertices of W i are added incrementally and are placed immediately to the right of v 1 if v 1 appears at the left.The vertices w i and w i of W i cannot be distinguished and can be swapped both with respect to their length and with respect to their position, except for i = k − 1.Hence, there is a bucket order W k < . . .< W 0 , both for the t-order (up to reflection) and the r-order.Both vertices of Z are placed one bar apart from the bar of v 1 .
An AB1V representation of G k is given in Fig. 4.
and {v 4 , v 5 , v 6 , v 7 , v 8 }, respectively.The first and the last clusters are disjoint, and two adjacent clusters C i and C i+1 with 1 ≤ i < 6 have four common vertices.Note that C 1 , . . ., C 6 forms a path-decomposition of width four of B[V ] [7].This graph shall be used later (Lemma 11) as an example of a non-fan-crossing free graph.
There is an edge between the vertices w 0 and w 0 of W 0 and both have v 1 , v 2 and u 1 as neighbors.For i = 1, . . ., k there is an edge between the vertices of W i and both have v 1 and the vertices of W i−1 as neighbors.Finally, each vertex z 1 , z 2 ∈ Z adds four edges, namely (z 1 , v 1 ), (z 1 , w k ), (z 1 , w k ), (z 1 , z 2 ) and (z 2 , v 1 ), (z 2 , w k ), (z 2 , w k ), (z 2 , w k−1 ).The edge (z 2 , w k−1 ) distinguishes z 1 from z 2 and w k−1 from w k−1 .
The r-order of the vertices of V in Fig. 4 is The r-order of B is not unique, since, e.g., the length of the top-4 bars can be exchanged and the length of the bars of {u 1 , u 2 , u 3 } and {u 4 , u 5 } are not related.Apart from this, B has the r-order from above.In addition, for G k we have W i < r W i−1 for i = 1, . . ., k and W 0 < r u 1 , as we shall show.However, the vertices of W i can be exchanged, except for i = k − 1.
Obviously, B and G k with k ≥ 0 are dAB1V graphs.The given visibility representation of G k is maximal, i.e., all edges are listed.The base B = G k [V ∪U ] is optimal and each set W i for i = 0, . . ., k adds two vertices and seven edges, since vertex w i has only vertex v 1 as a left neighbor with a bar that is longer than the one of w i for 0 = 1, . . ., k.Hence, one edge to a left neighbor is missing.Finally, Z adds two vertices and eight edges.
It remains to show that the t-order of B and of G k given in Fig. 4 is unique up to reflection and an exchange of w i and w i .Therefore, we use the properties of AB1V representations of clusters and prove four claims.
Claim 1.The vertices of V admit two t-orders with a bucket, namely, have four vertices in common.Then either v 1 or v 8 has the shortest bar among these vertices by Lemma 2. First, assume v 8 < r v 1 .Then v 8 is in the middle of v 2 , v 3 , v 9 , v 10 and v 8 < r {v 1 , v 2 , v 3 , v 9 , v 10 }, i.e., the bar of v 8 is short.Since v 4 is a neighbor of v 8 , it must be placed close to v 8 and it is placed between the extreme vertices of v 2 , v 3 , v 9 , v 10 .Then the bar of v 4 is shorter than the bar of v 8 by Lemma 1.By the same reasoning, the bars of v 5 , v 6 and v 7 are shorter than the bar of v 8 and are placed between the extreme bars of v 2 , v 3 , v 9 , v 10 .Next, consider cluster . Since v 4 has the shortest bar among these vertices, it is placed in the middle of them by Lemma 1.As before, the bar of v 7 (and also the bars of v 5 an v 6 ) must be shorter than the bar of v 4 .By the same reasoning, we obtain v 5 < r v 7 and v 6 < r v 5 from C 4 , C 5 and C 6 .Hence, assuming }.This is shown as follows: For the clusters C 1 and C 2 and v 1 < r v 8 , Lemma 2 implies that v 1 has the shortest bar and is placed in the middle of v 2 , v 3 , v 9 , v 10 .Now, the assumption v 4 < r v 10 leads to a contradiction.Then v 4 has the shortest bar of the vertices of cluster In consequence, we have v 1 < r v 10 < r v 4 .We can now proceed as before, and from the sequence of clusters C 1 , . . ., C 6 we obtain the r-order as claimed.
Next, we determine the t-order of B.
Proof: The clusters C 1 and C 6 are disjoint.Since v 3 ∈ C 1 has a shorter bar than its four neighbors in C 6 , v 3 must be placed between the vertices of C 6 and there are at least two vertices of C 6 on either side of v 3 .Then C 1 and C 6 nest by Lemmas 2 and 6 and the vertices of C 1 are placed between two vertices v x and v y of C 6 .Vertex v 10 has the second longest bar of the vertices of C 1 and therefore it is placed next to the middle bar v 1 in the U-shape of C 1 by Lemma 1.By the edge (v 8 , v 10 ) it must be placed close to v 8 , so that v x = v 8 and v 8 and v 10 are on the same side of v 1 .Fig. 5 depicts the AB1V representation with v 10 left of v 1 .Otherwise, the t-order is reversed.
Consider u 5 with neighbors v 7 , v 8 , v 9 , v 10 .Then u 5 must be placed immediately to the left of v 8 and it has a small bar that is shorter than the bar of v 10 .To see this, first u 5 must be placed between its four neighbors, since v 10 is not 1-visible from a bar to the left of the bars of C 6 that are long, or to the right of C 1 , since C 1 is U-shaped.Hence, u 5 must be placed immediately to the right of v 8 .Its bar must be shorter than the bar of v 10 , since the edge (v 8 , v 10 ) intersects the bar of vertex x 1 .Then x 1 = v 9 and y 1 = v 7 .Vertex u 4 cannot be placed left of y 1 = v 7 because of the edge (u 4 , u 5 ).Hence, u 4 must be placed left of v 8 and its bar is shorter than the bar of u 5 because of the edge (u 5 , v 7 ).Then u 6 must be placed furthest to the left.
By the same reasoning, u 1 must be placed between x 2 and x 3 and its bar is shorter than the one of v 1 .It determines y 2 = x 4 .Similarly, u 2 must be placed right of y 2 = v 4 and determines y 3 = v 5 , and u 3 must be placed right of y 3 = v 5 and determines y 4 = v 6 .However, there is a contradiction, since v 6 must be placed furthest to the left and to the right.
implies the t-order v 1 < t . . .< t v 10 or its reversal, and the vertices of U are placed between vertices of V as shown in Fig. 4.
Proof: By the same reasoning as in the proof of Claim 2, cluster C 6 must nest in C 1 , since v 8 has four neighbors in C 1 with longer bars, and therefore must be placed between the vertices of C 1 .Then all vertices of C 1 are placed between two vertices of C 1 by Lemmas 2 and 6.Vertex v 5 has the second longest bar of the vertices of C 6 and therefore must be placed next to the bar of v 6 and on the same side as v 3 .Assume that v 3 and v 5 are to the left of v 6 .Otherwise, the t-order is reversed.
As a neighbor of v 5 , vertex u 2 must be placed between v 3 and v 5 and its bar must be shorter than the one of v 5 because of the edge (v 3 , v 5 ), which intersects the leftmost bar y 1 of C 6 .Then y 1 = v 4 and u 3 must be placed between v 4 and v 5 .As a neighbor of v 4 , vertex u 1 cannot be placed to the far left.It is placed left of v 3 and enforces that v 1 and v 2 are to the left.The left to right order v 1 < t v 2 is determined by C 1 and C 2 , whose placement by Lemma 2 implies that v 1 is extreme for C 2 .Similarly, vertices u 4 and u 5 determine the t-order to the right.The vertices of U are inserted at fixed positions.Hence, the left to right order of the vertices of V ∪ U is as shown in Fig. 4, or its reversal.
Finally, we consider the extension of B to G k by the insertion of the pair sets of vertices W i = {w i , w i } for i = 1, . . ., k, which are buckets for the t-and r-orders.The ordering of the elements of a bucket is left open.Claim 4. If the left to right order of the vertices of V ∪ U is as depicted in Fig. 4 with v 1 < t v 2 < t u 1 < t u 3 at the left and u 1 < r {v 1 , v 2 , v 3 }, then the left to right order is v Proof: We proceed by induction and first consider W 0 = {w 0 , w 0 }.Since both vertices are neighbors of u 1 and the bar of u 1 is shorter than the bars of v 1 , v 2 , w 0 and w 0 must be placed between the extreme vertices of cluster C 1 .They are placed either between v 1 and v 2 or between v 2 and u 1 , and their bars must be shorter than the bar of u 1 because of the edge (v 1 , u 1 ).Placing one of them at either side of v 2 leads to a contradiction, since an edge incident to v 1 or to u 1 cannot be realized in an AB1V representation.
If the base B is extended only by W 0 , then both placements are possible, where an edge (w 0 , v 3 ) or (w 0 , v 3 ) must be added depending on the vertex with the longer bar.
The subgraphs G k [W i−1 ∪ W i ] are a K 4 for i = 1, . . ., k.First, consider W 0 ∪ W 1 .If w 0 and w 0 are placed between v 2 and u 1 , then w 1 and w 1 cannot be placed at all.As the bars of w 0 and w 0 are shorter than the ones of v 1 and v 2 , w 1 and w 1 must be placed to the right of v 1 .Each vertex between v 1 and The dotted lines represent visibility lines that traverse another bar.
v 2 must have a bar that is shorter than the shortest of the bars of w 0 and w 0 because of the edges incident to v 1 .If a vertex from W 1 were placed between v 1 and v 2 , then an edge from the vertex of W 1 with the shorter bar to v 1 or to the rightmost of the vertices of W 1 cannot be realized in an AB1V representation.If all of w 0 , w 1 , w 0 , w 1 were placed to the right of v 2 , then the edge from the one with the shortest bar to v 1 cannot be realized.Hence, both vertices of W 0 must be placed between v 1 and v 2 .Since u 2 is a neighbor of both, the leftmost of them must have the longer bar.Suppose that w i is left of w i for i = 1, . . ., k.The other case is symmetric and exchanges the roles.Then w 1 and w 1 cannot be placed to the left of v 1 , since the bar of w 0 is not 1-visible from there.Then {w 1 , w 1 } < r {w 0 , w 0 } by the edges from w 0 and w 0 to v 1 and u 1 .By the reasoning as before, w 1 and w 1 must be placed between v 1 and w 0 with bars that are shorter than the ones of W 0 and with the leftmost of the vertices of W 1 having the longer bar.
The same arguments are used in the inductive step with w i−1 and w i−1 taking the roles of v 2 and u 1 , where w i−1 is to the left of w i−1 and has the longer bar.
Finally, consider the vertices on the left and z 1 , z 2 .As before, z 1 and z 2 must be placed to the right of v 1 and their bars must be shorter than the ones of w k and w k .Then they must be placed between w k and w k , as depicted in Fig. 6.Now, the leftmost vertex v k−1 of W k−1 is 1-visible from z 2 and the edge (z 2 , w k−1 ) determines the t-and r-orders of the vertices of Z and W k−1 .
The proof of Lemma 7 now follows from Claims 1-4.
Next we show that the graphs from Fig. 4 are almost best possible if we aim at sparse maximal AB1V graphs.
Theorem 1 Every maximal AB1V graph of size n has at least 3.5n − 9 edges.
Proof: Consider an AB1V representation of a maximal AB1V graph G of size at least 6.For n < 6 the bound obviously holds.Let X = {v 1 , v 2 , v p , v q , v n−1 , v n } consist of the four extreme and the top-4 vertices such that (v 1 , v 2 , . . ., v p , . . ., v q , . . ., v n−1 , v n ) is the left to right order and v 1 , v p , v q , v n are the top-4 vertices in r-order.Here v 2 = v p and v q = v n−1 are allowed.Clearly, the vertices at the left and right are among the top-4 if G is maximal.
We count edges in decreasing r-order of the vertices and assign an edge (u, v) to the vertex with the shorter bar.
Consider vertices v ∈ V − X.If v is placed between v p and v q , then it has two neighbors with longer bars to the left and right, and therefore, v adds four edges to G. Every v ∈ V − X between v 1 and v p has at least v 1 to the left and two neighbors with longer bars to the right, and therefore adds at least three edges to G. The right side is similar.
Suppose that v ∈ V − X has only v 1 as a left neighbor with a longer bar.We call v a leftish vertex and denote the set of leftish vertices by L. The set R at the right is defined accordingly.By definition, all vertices u with u = v 1 to the left of a leftish vertex v in an AB1V representation of a maximal AB1V graph have a shorter bar than the bar of v.In addition, if we aim at a maximum number of leftish vertices, then the bar of u cannot be extended beyond the bar of v. Otherwise, u were a neighbor to the left of v with a longer bar and v were no longer a leftish vertex.Even more, if u is not a neighbor of v before the extension of the bar, then (u, v) is a new edge by the extension and G is not maximal.Hence, such extensions should be excluded.
A vertex u is called a left blocker of a leftish vertex v ∈ L in a given AB1V representation if v 1 < t v < t u < t v p in t-order, u < r v in r-order, and v 1 is a neighbor of both u and v. Thus, the blocker is to the right of v, it has a shorter bar, and the bar of v is traversed by the line of sight of the edge (v 1 , u).We denote the set of left blockers by B. Accordingly, consider right blockers and denote the set of right blockers by B .
For the lower bound on the number of edges, we show that we can assign a blocker to each vertex v ∈ L ∪ R. To see this, first, consider an AB1V representation with v 1 < t v < t v p for a leftish vertex v ∈ L. Let w ∈ V − X be the vertex between v 1 and v and with the longest bar, which is shorter than the bar of v. Then increase the length of the bar of w so that thereafter v < r w and there is no other vertex between v and w in r-order.The modified AB1V representation also represents G if v has no blocker, and v is no more a leftish vertex.Second, a vertex w cannot be a blocker for more than one vertex, since two or more vertices of L prevent the 1-visibility of the leftmost vertex v 1 from w.
Hence The graphs from Fig. 4 are maximal but not optimal, since edges are missing at the left, and they are close to sparsest maximal AB1V graphs.
Theorem 2 There are maximal AB1V graphs of size n ≥ 21 with 3.5 n − 1 edges.
Proof: Consider the graphs G k from Lemma 7. Since the left to right order given in Fig. 4 is a unique bucket order with buckets W i of size two (up to reflection) it suffices to show that any t-order does not allow the addition of further edge.The base B = G k [V ∪ U ] has 15 vertices and 50 edges and is optimal.Each bucket W i adds 2 vertices and 7 edges.One edge is missing towards optimality, since the leftmost vertex of W i in the given t-order has only the leftmost vertex v 1 as a neighbor with a longer bar to the left.Finally, Z adds two vertices and eight edges.Then, every G k with k ≥ 21 and k odd has 3.5 n − 1.5 edges.For every even k ≥ 22, we add a vertex with a short bar immediately to the left of v 8 in the AB1V representation of G k−1 in Fig. 4.This adds four edges.
Hence, there are tight bounds, both for densest and sparsest maximal AB1V graphs.

Corollary 2
The densest maximal AB1V graphs have 4n − 10 edges and the sparsest maximal AB1V graphs have 3.5n − c edges, and both bounds are tight up to a constant c with 1 ≤ c ≤ 9.
In consequence, the various versions of visibility induce a strict hierarchy of classes of AB1V graphs.Moreover, every wAB1V graph is an induced subgraph of an optimal (or m, d, s)AB1V graph.For the latter result, an AB1V representation of a graph G is turned into an optimal AB1V representation of a supergraph H by first varying the length of bars of the same length such that they are all different and then expanding the two bars at the left and right ends to the top-4 bars in r-order so that their vertices induce K 4 .Then, H is an optimal AB1V graph [25] and G is an induced subgraph.
Proof: The inclusions are obvious.The first proper inclusion is shown in Theorem 2. If v 1 < . . .< v n for the t-and r-orders, then the resulting graphs have only 2n − 3 edges and are distinct but not maximal, which proves the second inclusion.Third, the guarded double chains with vertices {1, . . ., n} and edges (i, i+1) and (i, i+2) for 1 ≤ i < n−1 and (1, n), (1, n−1), (2, n), (2, n−1), (n−1, n) and n ≥ 10 have an aligned bar 1-visibility representation, where the four longest bars are placed in pairs at the left and right and induce K 4 and all other bars have the same length.These graphs cannot be represented with all bars of distinct length, since the fifth longest bar can see the bars of the four extreme vertices 1, 2, n − 1, n.Finally, strong AB1V graphs of size at least five have degree at least four, which does not hold for weak AB1V graphs.
In this section we show that wAB1V graphs are closed under path-addition.Path-addition is a new operation on graphs which arises naturally from AB1V representations.The operation is very powerful and is the opposite of taking minors.Some classes of graphs are closed under path-addition, including bar 1-visibility, quasi-planar, fan-crossing free, RAC, and non-planar graphs, whereas others are not, including planar, treewidth-k, k-planar, bar (1, j)-visibility, and fan-planar graphs, see [12].
Definition 2 A path-addition takes a graph G = (V, E), two vertices u, v ∈ V , and an internally vertex-disjoint path P = (u, w 1 , . . ., w t , v) with w i ∈ V for 1 ≤ i ≤ t from u to v and results in G = (V ∪W, E ∪Q) where W = {w 1 , . . ., w t } and Q is the edges of P .We denote G by G ⊕ P .Definition 3 A class of graphs G is closed under path-addition if the graph G ⊕ P is in G for every graph G in G of size n and for every internally vertexdisjoint path P of length at least n − 1 between two vertices u and v of G.
For the closure of wAB1V (and other classes) under path-addition, we add a path P of sufficient length so that P can be added independent of the position of the endvertices in an AB1V representation.In consequence, the length of added paths increases at least exponentially.Path-additions allow the construction of any complete graph as a minor.First, add paths until there are k vertices, and then add a new path between any pair of vertices.Then we have a K k minor.
Theorem 3 The class of wAB1V graphs is closed under path-addition.
Proof: Consider an AB1V representation of G i after paths P 1 , . . ., P i have been added to a wAB1V graph G. Increase the length of the bars of the vertices of G i by k i+1 + 1, where k i+1 is the length of P i+1 , so that they are long, and let the bars of the internal vertices of P i+1 have length 1, . . ., k i+1 , respectively, so that they are short.Insert the bars of the internal vertices of P i+1 between the bars of G i such that there is a 1-visibility between two consecutive vertices of P i+1 .This can be done in many ways if P i+1 is longer than the distance of its endvertices in the t-order of G i .
The addition of a path P to a strong AB1V graph G does not preserve AB1V , since the inserted vertices of P induce 1-visibility to vertices of G. Then a set of supplementary edges F must be added to preserve aligned bar 1-visibility and F depends on the given AB1V representation.

Recognition of Optimal AB1V Graphs
The recognition problem of beyond-planar graphs is N P-hard, in general.However, it is known that outer 1-planar graphs can be recognized in linear time [4,26], map graphs in cubic time [15], maximal outer fan-planar graphs in polynomial time [5] and optimal 1-planar graphs in linear time [10].Here, we solve the recognition problem for optimal AB1V graphs.In addition, we show that the t-order can be computed in linear time from the r-order of a dAB1V graph.This complements a result by Felsner and Massow [25] who computed the r-order from the t-order.Their algorithm runs in quadratic time and can be implemented to run in linear time, although this is not stated.
The recognition algorithms use the fact that the vertex with the i-th longest bar either has four neighbors with longer bars and then must be placed between them, or it must be placed just outside the extreme vertices with the longest bars.This procedure is stepwise generalized from a given r-order via optimal graphs to distinct graphs and at the expense of the running time.
In particular, we use the following fact, which was mentioned above: Proposition 1 • Every AB1V graph has a vertex of degree at most four [23].
• The maximum complete subgraphs have size at most five [23].
• G is an optimal AB1V graph if and only if the vertices with the four longest bars are in pairs at the left and right ends and build K 4 [23].
• If G is an optimal AB1V graph, then all vertices have degree at least four.
• For i = 1, . . ., n − 4, the i-th vertex in r-order has four neighbors with a higher r-order, i.e., with longer bars, and the subgraph induced by the vertex and its neighbors is K 5 or K 5 − e.
Our recognition algorithm for optimal AB1V graphs proceeds in two phases.First, it computes all clusters.Then it incrementally adds a vertex at a time and chooses a vertex v so that v induces a cluster with the already inserted vertices.Thereby, it takes the vertices in decreasing r-order.Lemma 8 There is a linear-time algorithm that checks whether a partial AB1V representation of a cluster of an optimal AB1V graph G can be extended to an AB1V representation of G, and, if so, computes an AB1V representation, i.e., a t-and an r-order of G.
Proof: The algorithm implements Proposition 1.Let V = {v 1 , . . ., v n } be the set of vertices of G and let v 1 , . . ., v 5 be the vertices of the given cluster with according to Lemma 1.
For i = 5, . . ., n and V i = {v 1 , . . ., v i }, if there is a partial AB1V representation of G[V i ] and the vertices of V i are assigned the n + 1 − i longest bars, then the algorithm gets a vertex v i+1 ∈ V − V i which has four neighbors in V i , assigns the bar of length n − i and places the bar in the middle (at the median) of the bars of its four neighbors from V i .This is a partial AB1V representation of If such a vertex does not exist or the placement is invalid, then the algorithm stops and rejects.
Processing a vertex takes O(1) time.If the algorithm succeeds, it has computed a t-and an r-order of G and if it fails, then either G is not AB1V or the initial cluster did not represent the top-5 vertices in r-order.
Theorem 4 There is a quadratic-time algorithm that checks whether a graph G is an optimal AB1V graph, and, if so, computes an AB1V representation.
Proof: The top-5 vertices in r-order induce a cluster C.There are at most O(n) clusters by Corollary 1, which are computed in O(n) time.Since the bars of the vertices of C are longer than the bars of all other vertices, we take any r-order that defines a cluster.
The algorithm takes each cluster of G and successively tests all 120 t-orders of the vertices as a partial AB1V representation for an extension according to Lemma 8.The algorithm succeeds if it finds an extension and rejects otherwise.If G is an optimal AB1V graph, the algorithm must succeed and it cannot succeed otherwise.
Next, we show that the left-to-right order of the bars can be computed from an AB1V graph together with the knowledge of the length of the bars.The converse was shown in [25].
Theorem 5 There is a linear-time algorithm that computes a t-order given an r-order of a dAB1V graph G such that the t-and r-orders describe a dAB1V representation of G.For i = 5, . . ., n, if v i has four neighbors in W i−1 , then the partial AB1V representation is extended by placing v i in the middle of its four neighbors in W i−1 .If v i has three neighbors in W i−1 , then one of its neighbors is the first (resp.last) vertex in the actual partial AB1V representation, which is extended by placing v i in second (last but one) place.Finally, if v i has only two neighbors in W i−1 , then v i is an extreme vertex and is placed to the left (right) of the vertices of W i−1 if the former leftmost (rightmost) vertex is its neighbor.The algorithm stops if the placement was invalid.
The correctness is due to the fact that the bars of v i , . . ., v n are shorter than the bar of v i−1 .Hence, the bar of v i is 1-visible from the bars of its neighbors from W i−1 , and it is 1-visible from two, three, or four neighbors depending on its position in the partial AB1V representation.
Each run with an extension of a partial AB1V representation of G[W 4 ] takes linear time, which implies O(n) time in total.
Together with a result from [25] we can conclude: Corollary 4 If G is a dAB1V graph, then an AB1V representation can be computed in linear time if (i) a t-order or (ii) an r-order is given.

Relationship to other Classes of Graphs
For the comparison of AB1V graphs with other classes of graphs we consider weak AB1V graphs, since the other versions, e.g., sAB1V graphs, are too restrictive, since the graphs have vertices of degree four.Moreover, the other classes of graphs are generally closed under taking subgraphs, in particular, all classes of graphs listed in Fig. 13.

Inclusion Relations
Common drawings and visibility representations are equivalent in the planar case with non-transparent bars, both in the general case [37,39] and in the outer planar case with aligned visibility [17].However, visibility representations are more powerful than common drawings if crossings are allowed.The 1-planar graphs are a proper subset of bar (1, 1)-visibility graphs [9] and bar 1-visibility graphs [22].Hence, single edge-edge crossings are weaker than single edge-vertex crossings.A similar result holds for AB1V graphs.
Theorem 6 There is a linear time algorithm that constructs a wAB1V representation of an outer 1-planar graph.
Proof: Recall that a graph is outer 1-planar if it admits a drawing with all vertices in the outer face and at most one crossing per edge [4,26].
The linear time algorithm of Auer et al. [4] tests whether a graph G is outer 1-planar and, if so, augments G to a graph H by adding edges and constructs a (planar) maximal outer 1-planar embedding of H.The embedding of H consists of planar triangles and kites.A kite is a K 4 that is embedded with a pair of crossing edges.
Let X be the set of crossing edges in the embedding of H and let H − X be the graph after their removal.Then H − X is an outer planar graph with a Hamiltonian cycle where all inner faces are triangles or quadrangles.The dual of H − X is a tree T with t-and q-vertices corresponding to triangles and quadrangles, respectively.The outer face is ignored.We root T at a leaf r and thereby orient T from r.
First, we construct a (planar) bar visibility representation of H −X. Consider the root of T which corresponds to a triangle or a quadrangle (v 1 , . . ., v k ) with k = 3, 4 and with the edge (v 1 , v 2 ) in the outer face.Then the t-order is the ordering of the vertices of the Hamiltonian cycle of H with v 1 = 1 and v 2 = n.Now, two vertices u and v span an interval on the Hamiltonian cycle.
The length of each bar, i.e., the r-order of H − X, is computed by successively removing a leaf from T .This extends Mitchell's algorithm for the recognition of maximal outer planar graphs by successively removing a vertex of degree two [30].A leaf b of T is in one-to-one correspondence to a vertex v of degree two if b is a triangle, and to two vertices u and v of degree two with an edge (u, v) if b is a quadrangle.Suppose that i − 1 vertices have been removed so far, where i = 1, . . ., n − i − j for j = 0, 1 and j = 0 if and only if the root is a triangle.Assign v a bar of length i if b is removed and v is a degree-two vertex in the triangle b.The edges (u, v) and (v, w) incident to v are lines of sight at level i, i.e., at the top end of the bar of v. Increase i by one.Accordingly, assign bars of length i and i + 1 to u and v (in any order) if b is a quadrangle and u and v are the degree two vertices corresponding to b and draw the lines of sight at the top of the bars.Increase i by two.Finally, assign bars of length n and n − 1 to the vertices v 1 and v 2 corresponding to the root and bars of length n − 2 resp.n − 3 to the other vertices of the root.
The algorithm preserves the invariant that all bars have a distinct length and that the bars of all vertices in the interval between u and v have bars that are shorter than the bars of u and v if there is an edge (u, v).The latter is due to the fact that the vertices between u and v have been processed before u and v.It guarantees that an edge at level i is unobstructed.
Finally, we reinsert the crossing edges of X.For each pair of crossing edges (a, c) and (b, d) there is a quadrangle (a, b, c, d) with a < t b < t c < t d in t-order and {b, c} < r {a, d} in r-order.By the previous assignment, the bars of b and c have length i and i + 1 for some i, respectively.Then the bar of a vertex w in the interval from a to d has length at most i − 1 if w = b, c.Suppose the bar of b has length i, the case where the bar of c has length i is similar.Then draw the edge (a, c) as a planar line of sight at level i + 1 and draw the edge (b, d) as a line of sight at level (i − 0.5) which traverses the bar of c.No other bars are affected.Hereby, we adapt the technique of Brandenburg [9] for the bar (1, 1)-visibility representation of 1-planar graphs to AB1V representations.For integer coordinates, scale by two.
In total, we have obtained a weak distinct AB1V representation of G where lines of sight are ignored if there is no edge.All stages of the algorithm take linear time.
To establish proper inclusion relations we use the fact that outer 1-planar graphs are planar [4], that K 5 is an AB1V graph, and that K 6 is a bar 1-visibility graph and not an AB1V graph.Note that bar 1-visibility graphs are a proper subclass of quasi-planar graphs [22].
Corollary 5 Every outer 1-planar graph is a wAB1V graph, but not conversely, and every wAB1V graph is a bar 1-visibility graph, but not conversely.
Hence, AB1V graphs are quasi-planar, which is a large class of beyond-planar graphs containing graphs with up to 6.5n-20 edges [1].Quasi-planar graphs are defined via embeddings and exclude a mutual crossing of three edges.Therefore, they include all 1-planar, RAC, fan-planar, bar 1-visibility [22], and rectangle visibility graphs, and all graphs with geometric thickness two.Since AB1V graphs have geometric thickness two [25] and are bar 1-visibility graphs, we can conclude: Corollary 6 A graph G is quasi-planar if G is a wAB1V graph, but not conversely.

Incomparability Results
In this section we prove that wAB1V graphs are incomparable to some classes of beyond-planar graphs.Two graph classes are incomparable if there are mutual counterexamples, i.e., graphs that are in one class and not in the other.As aforesaid, we consider classes that are closed under taking subgraphs.The incomparability holds for any version of visibility, i.e., {o, m, d, s, w}AB1V graphs.
First, consider planar graphs.Since there are planar graphs that are not wAB1V graphs, a class of graphs G is not contained in wAB1V if G includes the planar graphs.

Theorem 7
The classes of planar and wAB1V graphs are incomparable.
Proof: Every wAB1V graph has a vertex of degree at most four, namely the one with the shortest bar in an AB1V representation.However, there are planar graphs of degree five, such as the dual of the dodecahedron graph or of the football graph C 60 .
Conversely, K 5 is an AB1V graph and not planar.Moreover, optimal AB1V graphs are denser than planar graphs.
For the incomparability of wAB1V and k-planar respectively bar (1, k)visibility graphs we use the fact that the closure of bipartite graphs K 2,m under path-addition results in graphs that are neither k-planar nor (1, k)-visibility graphs [12], but are wAB1V graphs.
Proof: There are planar graphs that are not wAB1V graphs, which proves one direction.Conversely, consider graphs that are obtained from K 2,n by at least one path-addition between each pair of vertices from the set of n vertices.Each such graph is a wAB1V graph by the closure of wAB1V under path-addition, as established in Theorem 3.However, for n ≥ 4k + 9 the graphs are not k-planar and not bar (1, k)-visibility graphs, as shown in [12].
Similarly, there are counterexamples for outer fan-planar and fan-planar graphs.
Theorem 9 The classes of wAB1V and outer fan-planar graphs are incomparable.
Proof: Outer fan-planar graphs of size n have at most 3n − 5 edges [6], but there are AB1V graphs with 4n − 10 edges.Figure 9: A fan-crossing-free embedding if vertex r 6 is placed in a bad face with only two neighbors in the face next vertex r 9 is placed in the triangle of r 6 , r 8 , x 3 , then there is no face for r 10 .Hence, r 9 must be placed in the triangle of r 6 , r 7 , r 8 and (r 9 , r 5 ) crosses (r 6 , r 8 ).Thereafter, r 10 can be placed in ∆(r 6 , r 7 , r 9 ) or in ∆(r 7 , r 8 , r 9 ).If the sequence is continued by some r 11 , r 12 , . .., then r 10 must be placed in ∆(r 7 , r 8 , r 9 ).In summary, the vertices r 1 , . . ., r 5 are determined and property (*) is preserved, which implies an almost fixed embedding.
Lemma 11 There are wAB1V graphs that are not fan-crossing free.
Proof: Our counterexample G consisting of the graph A = G k [V 0 ] from Fig. 4 together with a single vertex z which is connected to v 2 , . . ., v 5 or r 2 , r 4 , r 6 , r 8 and is placed between v 3 and v 4 or r 4 and r 6 with a short bar.Thus z = u 1 and u 2 , . . ., u 6 are removed.
As shown in Lemma 10, there is a unique fan-crossing free embedding of E(A) up to the choice of two triangles for the placement of r 10 .However, there is no face left for a placement of vertex z with the given neighbors without introducing a fan-crossing.A candidate triangle with vertices r 5 , r 7 , r 8 fails because of the edges (r 2 , z) and (r 6 , z) and similarly, the triangle with vertices r 5 , r 6 , r 8 fails because of the edges (r 4 , z) and (r 2 , z).
More counterexamples can be constructed by adding more vertices to G, as in Lemma 10.
In consequence, we can establish a further incomparability.Proof: There are planar graphs of minimum degree five which are not wAB1V graphs, and conversely the graph G from Lemma 11 is wAB1V and not fancrossing free and therefore not RAC, since every RAC graph is fan-crossing free.
JGAA, 21(3) 281-312 (2017) 307 x 1 x 2 x 3 r 6 x 4 r 1 Figure 10: A fan-crossing-free embedding after vertex r 6 is placed.Vertex r 7 can be placed in one of the shaded triangles and a placement in the darker shared triangle fails.The relationships of the class of wAB1V graphs to other classes of beyondplanar graphs are illustrated in Fig. 13.An "arrow" indicates a proper inclusion, whereas "no path" indicates incomparability.The incomparability has been proved for wAB1V and all shown classes except for fan-planar and outer fancrossing free graphs.Outer fan-crossing free graphs have not yet been studied in detail.They may be included in wAB1V , but not conversely, since K 5 is not outer fan-crossing free by Lemma 9. Accordingly, there are fan-planar graphs that are not wAB1V by Lemma 9 and fan-planar graphs are not closed under path-addition [12].We conjecture that all illustrated incomparabilities hold.
The bar (i, j)-visibility graphs [13] and the bar (1, 1)-visibility graphs [9] are shown to be incomparable with wAB1V .These classes range between the classes of 1-planar and bar 1-visibility graphs.
Clearly, the subclasses IC-planar [28] and N IC-planar [40] of the 1-planar graphs are also incomparable with wAB1V , since they include the planar graphs and do not include graphs with 4n − 10 edges.
Finally, wAB1V graphs have geometric thickness two [25], and graphs with geometric thickness two are quasi-planar.The relationship of wAB1V and RVG (rectangle visibility graphs) [27] is open and we conjecture incomparability.

Conclusion
In this work we extended the studies of Felsner and Massow [25] on AB1V graphs and added new properties and relationships.
Our studies have revealed new problems, such as the closure of classes of graphs under path-addition and the relation of AB1V graphs to other classes of beyond-planar graphs.Of interest are recognition problems, in particular for weak AB1V graphs, and the relationship of AB1V graphs to further classes of beyond-planar graphs.It might be worthwhile to study the relationship between classes of optimal or maximal graphs.

Figure 1 :
Figure 1: AB1V representation of a K 5 with planar and 1-visibility lines of sight drawn as bold and dotted horizontal lines in (b) has a vertex for each cluster of G and there is an edge between two clusters G[X] and G[Y ] if and only if |X ∩ Y | ≥ w, i.e., G[X] and G[Y ] have at least w vertices in common, where 1 ≤ w ≤ 4.

Figure 2 :
Figure 2: An AB1V representation of two clusters with four vertices in common

Lemma 6
If G[X] and G[Y ] are two vertex disjoint clusters of an AB1V graph G, then in any AB1V representation, G[X] and G[Y ] are disjoint or nest, and the bars of the nesting cluster are shorter than the bars of the other cluster.

Proof:
Consider the vertices of G in decreasing r-order and for i = 1, . . ., n let W i = {v 1 , . . ., v i } denote the set of i vertices with the longest bars.The algorithm successively extends a partial AB1V representation of G[W i ] to a partial AB1V representation of G[W i+1 ].It starts with an AB1V representation of G[W 4 ] and tries all 12 t-orders of the vertices up to reflection.

Theorem 10
The classes of wAB1V graphs and (a) RAC graphs and (b) fancrossing free graphs are incomparable.

Figure 13 :
Figure 13: Relationships between AB1V graphs and other classes of beyondplanar graphs.An arrow indicates proper inclusion and no path to/from AB1V indicates incomparability.The formula below the class name gives the maximum number of edges.Classes with a closed boundary are closed under path-addition, whereas classes with a dotted boundary are not closed under path-addition [12].