SmallArea Orthogonal Drawings of 3Connected Graphs
Abstract
It is wellknown that every graph with maximum degree 4 has an orthogonal drawing with area at most \(\frac{49}{64}n^2+O(n)\approx 0.76n^2\). In this paper, we show that if the graph is 3connected, then the area can be reduced even further to \(\frac{9}{16}n^2+O(n) \approx 0.56n^2\). The drawing uses the 3canonical order for (not necessarily planar) 3connected graphs, which is a special Mondshein sequence and can hence be computed in linear time. To our knowledge, this is the first application of a Mondshein sequence in graph drawing.
1 Introduction
An orthogonal drawing of a graph \(G=(V,E)\) is an assignment of vertices to points and edges to polygonal lines connecting their endpoints such that all edgesegments are horizontal or vertical. Edges are allowed to intersect, but only in single points that are not bends of the polygonal lines. Such an orthogonal drawing can exist only if every vertex has degree at most 4; we call such a graph a 4graph. It is easy to see that every 4graph has an orthogonal drawing with area \(O(n^2)\), and this is asymptotically optimal [17].
For planar 2connected graphs, several authors showed independently [10, 15] how to achieve area \(n\times n\), and this is optimal [16]. We measure the drawingsize as follows. Assume (as we do throughout the paper) that all vertices and bends are at points with integral coordinates. If H rows and W columns of the integer grid intersect the drawing, then we say that the drawing occupies a \(W\times H\) grid with width W, height H, halfperimeter \(H+W\) and area \(H\cdot W\). Some papers count as width/height the width/height of the smallest enclosing axisaligned box. This is one unit less than with our measure.
For arbitrary graphs (i.e., graphs that are not necessarily planar), improved bounds on the area of orthogonal drawings were developed much later, decreasing from \(4n^2\) [11] to \(n^2\) [1] to \(0.76n^2\) [9]. (In all these statements, we omit lowerorder terms for ease of notation.)
Our Results: In this paper, we decrease the areabound for orthogonal drawings further to \(0.56n^2+O(n)\) under the assumption that the graph is 3connected. The approach is similar to the one by Papakostas and Tollis [9]: add vertices to the drawing in a specific order, and pair some of these vertices so that in each pair one vertex reuses a row or column that was used by the other. The main difference in our paper is that 3connectivity allows the use of a different, stronger, vertex order.
It has been known for a long time that any planar 3connected graph has a socalled canonical order [7], which is useful for planar graph drawing algorithms. It was mentioned that such a canonical order also exists in nonplanar graphs (e.g. in [4, Remark on p.113]), but it was not clear how to find it efficiently, and it has to our knowledge not been used for graph drawing algorithms. Recently, the second author studied the socalled Mondshein sequence, which is an edge partition of a 3connected graph with special properties [8], and showed that it can be computed in linear time [13]. A Mondshein sequence is the appropriate generalization of the canonical order to (not necessarily planar) 3connected graphs [13] and is most naturally defined by ear decompositions. However, in order to highlight its relation to canonical orders, we define a Mondshein sequence here as a special vertex partition and call it a 3canonical order.
We use this 3canonical order to add vertices to the orthogonal drawing. This almost immediately lowers the resulting area, because vertices with one incoming edge can only occur in chains. We then mimic the pairingtechnique of Papakostas and Tollis, and pair groups of the 3canonical order in such a way that even more rows and columns can be saved, resulting in a halfperimeter of \(\frac{3}{2}n+O(1)\) and the areabound follows.
No previous algorithms were known that achieve smaller area for 3connected 4graphs than for 2connected 4graphs. For planar graphs, the orthogonal drawing algorithm by Kant [7] draws 3connected planar 4graphs with area \((\frac{2}{3}n)^2+O(n)\) [14], while the bestpossible area for planar 2connected graphs is \(n^2\) [16].
2 Preliminaries
Let \(G=(V,E)\) be a graph with \(n=V\) vertices and \(m=E\) edges. The degree of a vertex v is the number of incident edges. In this paper, all graphs are assumed to be 4graphs, i.e., all vertex degrees are at most 4. A graph is called 4regular if every vertex has degree exactly 4; such a graph has \(m=2n\) edges.
A graph G is called connected if, for any two vertices u, v, there is a path in G connecting u and v. It is called 3connected if \(n>3\) and, for any two vertices u, v, the graph \(G\{u,v\}\) is connected.
A loop is an edge (v, v) that connects an endpoint with itself. A multiedge is an edge (u, v) for which another copy of edge (u, v) exists. When not otherwise stated, the graph G that we want to draw is simple, i.e., it has neither loops nor multiedges. While modifying G, we will sometimes temporarily add a double edge, i.e., an edge for which exactly one other copy exists (we refer always to the added edge as double edge, the copy is not a double edge).
2.1 The 3Canonical Order
Definition 1

\(V_1=\{v_1,v_2\}\), where \((v_1,v_2)\) is an edge.

\(V_k=\{v_n\}\), where \((v_1,v_n)\) is an edge.
 For any \(1<i<k\), one of the following holds:

\(V_i=\{z\}\), where z has at least two predecessors and at least one successor.
 \(V_i=\{z_1,\dots ,z_\ell \}\) for some \(\ell \ge 2\), where

– \(z_1,\dots ,z_\ell \) is an induced path in G (i.e. edges \(z_1z_2\dots z_\ell \) exist, and there are no edges \((z_i,z_j)\) with \(i<j1\)),

– \(z_1\) and \(z_\ell \) have exactly one predecessor each, and these predecessors are different,

– \(z_j\) for \(1<j<\ell \) has no predecessor,

– \(z_j\in V_i\) for \(1\le j\le \ell \) has at least one successor.


Here, a predecessor [successor] of a vertex in \(V_i\) is a neighbor that occurs in a group \(V_h\) with \(h<i\) [\(h>i\)]. See Fig. 1 for a 3canonical order.
Numerous related methods of ordering vertices of 3connected graphs exist, e.g. (2,1)sequences [8], nonseparating ear decompositions [2, 13], and, limited to planar graphs, canonical orders for maximal planar graphs [6], canonical orders for 3connected planar graphs [7] and orderly spanning trees [3]. A Mondshein sequence (i.e. a 3canonical order) of a 3connected graph implies all these orders, up to minor subtleties.
The most efficient way known to compute a Mondshein sequence (proving in particular that one exists) uses nonseparating ear decompositions [2, 13]. This is a partition of the edges into ears \(P_1\cup \dots \cup P_k=E\) such that \(P_1\) is an induced cycle, \(P_i\) for \(i>1\) is a nonempty induced path that intersects \(P_1\cup \dots \cup P_{i1}\) in exactly its endpoints, and \(G\bigcup _{j=1}^{i} P_j\) is connected for every \(i < k\). Such a nonseparating ear decomposition exists for any 3connected graph [2], and we can even fix two edges \(v_1v_2\) and \(v_2v_n\) and require that \(v_1v_2\) is in the cycle \(P_1\) and that \(v_n\) is the only vertex in \(P_k\); hence, \(P_k\) will be a singleton.
Further, such a nonseparating ear decomposition can be computed in linear time [12, 13]. The sets of newly added vertices for each \(P_i\) will be the vertex groups of a 3canonical order (additionally, \(P_1\) is split into the groups \(V_1 := \{v_1,v_2\}\) and \(V_2 := V(P_1)\{v_1,v_2\}\)). Although vertices in \(V_i\) may have arbitrarily many incident edges in a nonseparating ear decomposition, we can easily get rid of these extra edges by a simple shortcutting routine in linear time (see Lemmas 8 and 12 in [12]). This gives a 3canonical order. Thus, a lineartime algorithm for computing a 3canonical order follows immediately from the one for nonseparating ear decompositions.
2.2 Making 3Connected 4Graphs 4Regular
It will greatly simplify the description of the algorithm if we only give it for 4regular graphs. Thus, we want to modify a 3connected 4graph G such that the resulting graph \(G'\) is 4regular, draw \(G'\), and then delete added edges to obtain a drawing of G. However, we must maintain a simple graph since the existence of 3canonical orders depends on simplicity. This turns out to be impossible (e.g. for the graph obtained from the octahedron by subdividing two distinct edges with a new vertex and joining the new vertices by an edge), but allowing one double edge is sufficient.
Lemma 2
Let G be a simple 3connected 4graph with \(n\ge 5\). Then we can add edges to \(G'\) such that the resulting graph \(G'\) is 3connected, 4regular, and has at most one double edge.
Proof
Since G is 3connected, any vertex has degree 3 or 4. If there are four or more vertices of degree 3, then they cannot be mutually adjacent (otherwise \(G=K_4\), which contradicts \(n\ge 5\)). Hence, we can add an edge between two nonadjacent vertices of degree 3; this maintains simplicity and 3connectivity.
We repeat until only two vertices of degree 3 are left (recall that the number of vertices of odd degree is even). Now an edge between these two vertices is added, even if one existed already; this edge is the only one that may become a double edge. The resulting graph is 4regular and satisfies all conditions. \(\square \)
3 Creating Orthogonal Drawings
From now on, let G be a 3connected 4regular graph that has no loops and at most one double edge. Compute a 3canonical order \(V=V_1\cup \dots \cup V_k\) of G with \(V_k=\{v_n\}\), choosing \(v_1v_n\) to be the double edge if there is one. Let \(x^{\text {short}}\) and \(x^{\text {long}}\) be the number of short and long chains. Let \(x^{j{\text{ }\ell }}\) be the number of vertices with indegree j and outdegree \(\ell \). Since G is 4regular, we must have \(j+\ell =4\). A j\(\ell \) singleton is a vertex z that constitutes a singleton group \(V_i\) for \(1 < i \le k\) and that has indegree j and outdegree \(\ell \).
Observation 3
 1.
\(x^{0{\text{ }4}}=x^{4{\text{ }0}}=1\)
 2.
\(x^{1{\text{ }3}}=x^{3{\text{ }1}}\)
 3.
Every chain \(V_i\) contributes one to \(x^{2{\text{ }2}}\) and \(V_i1\) to \(x^{1{\text{ }3}}\).
Proof
(1) holds, since every vertex that is different from \(v_n\) has a successor, and every vertex that is different from \(v_1\) has an incoming edge from either a predecessor or within its chain. For (2), observe that \(2n=m=\sum _{v} indeg (v)=x^{1{\text{ }3}}+2x^{2{\text{ }2}}+3x^{3{\text{ }1}}+4x^{4{\text{ }0}}\) and \(n=x^{0{\text{ }4}}+x^{1{\text{ }3}}+x^{2{\text{ }2}}+x^{3{\text{ }1}}+x^{4{\text{ }0}}\), and rearrange. For (3), say \(V_i=\{z_1,\dots ,z_\ell \}\) is directed from \(z_1\) to \(z_\ell \). Then \( indeg (z_\ell )=2\) and \( indeg (z_j)=1\) for all \(j<\ell \). \(\square \)
3.1 A Simple Algorithm
As in many previous orthogonal drawing papers [1, 7, 9], the idea is to draw the graph \(G_i\) induced by \(V_1\cup \dots \cup V_i\) in such a way that all unfinished edges (edges with one end in \(G_i\) and the other in \(GG_i\)) end in a column that is unused above the point where the drawing ends.
Embedding the First Two Vertices: If \((v_1,v_n)\) is a single edge, then \(v_1\) and \(v_2\) are embedded exactly as in [1]: refer to Fig. 2. If \((v_1,v_n)\) is a double edge, then it was added only for the purpose of making the graph 4regular and need not be drawn. In that case, we omit one of the outgoing edges of \(v_1\) that has a bend.
Observation 4

For the first and last vertexgroup: O(1)

For a 31singleton: \(+1\) (we add one row)

For a 22singleton: \(+2\) (we add one row and one column)

For a short chain: \(+3\) (we add one row and two columns)

For a long chain \(V_i\): \(+2V_i\) (we add two rows and \(2V_i2\) columns)
Corollary 5
The halfperimeter is at most \( \frac{3}{2}n + \frac{1}{2}x^{2{\text{ }2}}  x^\text {short} + O(1)\).
Proof
From Observation 4 and using Observation 3.3 the halfperimeter is at most \(x^{3{\text{ }1}}+2x^{2{\text{ }2}}+2x^{1{\text{ }3}}x^\text {short}+O(1)\). By Observation 3.2 this is at most \(\frac{3}{2}x^{3{\text{ }1}}+2x^{2{\text{ }2}}+\frac{3}{2}x^{1{\text{ }3}}x^\text {short}+O(1)\), which gives the result. \(\square \)
Theorem 6
Every simple 3connected 4graph has an orthogonal drawing of area at most \(\frac{25}{36}n^2+O(n)\approx 0.69n^2\).
Proof
First, make the graph 4regular, compute the 3canonical order, and consider the number \(x^{2{\text{ }2}}\) of 22vertices.
 1.
If \(x^{2{\text{ }2}}\le n/3\), apply the above algorithm. By Corollary 5, the halfperimeter is at most \(\frac{3}{2}n+\frac{1}{6}n+O(1) \le \frac{5}{3}n + O(1)\).
 2.
If \(x^{2{\text{ }2}}\ge n/3\), apply the algorithm from [9]. They state their area bound as \(0.76n^2+O(1)\), but one can observe (see [9, Theorem 3.1, ll.2–5]) that their halfperimeter is at most \(2n\frac{1}{2}(x^{1{\text{ }3}}+x^{2{\text{ }2}}) + O(1)\), since they pair at least \(x^{1{\text{ }3}}+x^{2{\text{ }2}}\) vertices. Using Observation 3.2 and ignoring O(1) terms, we have \(x^{1{\text{ }3}}+x^{2{\text{ }2}} = \frac{1}{2}x^{1{\text{ }3}} + x^{2{\text{ }2}} + \frac{1}{2}x^{3{\text{ }1}} = \frac{1}{2}n+\frac{1}{2}x^{2{\text{ }2}}\). Hence, the halfperimeter of their algorithm is at most \(\frac{7}{4}n\frac{1}{4}x^{2{\text{ }2}}+O(1)\le (\frac{7}{4}\frac{1}{12}) n +O(1) = \frac{5}{3}n+O(1)\).
In both cases, we get a drawing with halfperimeter \(\frac{5}{3}n + O(1)\). The area of it is maximal if the two sides are equally large and thus at most \((\frac{5}{6}n+O(1))^2\). \(\square \)
3.2 Improvement via Pairing
We already know a bound of \(\frac{3}{2}n+\frac{1}{2}x^{2{\text{ }2}}  x^\text {short}+O(1)\) on the halfperimeter. This section improves this further to halfperimeter \(\frac{3}{2}n+O(1)\). The idea is strongly inspired by the pairing technique of Papakostas and Tollis [9]. They created pairs of vertices with special properties such that at least \(\frac{1}{2}(x^{2{\text{ }2}}+x^{1{\text{ }3}})\) such pairs must exist. For each pair, they can save at least one gridline, compared to the \(2n+O(1)\) gridlines created with [1].
 1.
Initialize \(i:=k1\). (We ignore the last group, which is a 40singleton.)
 2.
While \(V_i\) is a 31singleton and \(i>2\), set \(i:=i1\).
 3.
If \(i=2\), break. Else, \(V_i\) is a chain or a 22singleton and we choose the partner of \(V_i\) as follows: Initialize \(j:=i1\). While \(V_j\) is a 31singleton whose successor is not in \(V_i\), set \(j:=j1\). Now, pair \(V_i\) with \(V_j\). Observe that such a \(V_j\) with \(j\ge 2\) always exists, since \(i>2\) and \(V_2\) is not a 31singleton.
 4.
Update \(i:=j1\) and repeat from Step (2) onwards.
In the small example in Fig. 1, the 22singleton \(V_5\) gets paired with the short chain \(V_4\), and all other groups are not paired.
Observe that, with the possible exception of \(V_2\), every chain is paired and every 22vertex is in a paired group (either as 22singleton or as part of a chain). Hence there are at least \(\frac{1}{2}(x^{2{\text{ }2}}1)\) pairs. The key observation is the following:
Lemma 7
Let \(V_i,V_j\) be two vertex groups that are paired. Then there exists a method of drawing \(V_i\) and \(V_j\) (without affecting the layout of any other vertices) such that the increase to rows and columns is at most \(2V_i\cup V_j1\).
We defer the (lengthy) proof of Lemma 7 to the next section, and study here first its consequences. We can draw \(V_1\) and \(V_k\) using O(1) gridlines. We can draw \(V_2\) using \(2V_2=2x^{2{\text{ }2}}_{V_2}+2x^{1{\text{ }3}}_{V_2}\) new gridlines, where \(x^{\ell {\text{ }k}}_W\) denotes the number of vertices of indegree \(\ell \) and outdegree k in vertex set W. We can draw any unpaired 31singleton using one new gridline. Finally, we can draw each pair using \( 2V_i\cup V_j1=2x^{2{\text{ }2}}_{V_i\cup V_j} + 2x^{1{\text{ }3}}_{V_i\cup V_j}  1\) new gridlines. This covers all vertices, since all 22singletons and all chains belong to pairs or are \(V_2\), and since there are no 13singletons.
Putting it all together and using Observation 3.2, the number of gridlines hence is \(2x^{1{\text{ }3}}+2x^{2{\text{ }2}}+x^{3{\text{ }1}}\#\text{ pairs }+O(1) \le 2x^{1{\text{ }3}}+\frac{3}{2}x^{2{\text{ }2}}+x^{3{\text{ }1}}+O(1) = \frac{3}{2}n + O(1)\) as desired. Since a drawing with halfperimeter \(\frac{3}{2}n\) has area at most \((\frac{3}{4}n)^2=\frac{9}{16}n^2\), we can conclude:
Theorem 8
Every simple 3connected 4graph has an orthogonal drawing of area at most \(\frac{9}{16}n^2+O(n)\approx 0.56n^2\).
We briefly discuss the runtime. The 3canonical order can be found in linear time. Most steps of the drawing algorithm work in constant time per vertex, hence O(n) time total. One difficulty is that to place a group we must know the relative order of the columns of the edges from the predecessors. As discussed extensively in [1], we can do this either by storing columns as a balanced binary search tree (which uses \(O(\log n)\) time per vertexaddition), or using the data structure by Dietz and Sleator [5] which allows to find the order in O(1) time per vertexaddition. Thus, the worstcase runtime to find the drawing is O(n).
4 Proof of Lemma 7
Recall that we must show that two paired vertex groups \(V_i\) and \(V_j\), with \(j<i\), can be embedded such that we use at most \(2V_i+2V_j1\) new gridlines. The proof of this is a massive case analysis, depending on which type of group \(V_i\) and \(V_j\) are, and whether there are edges between them or not.^{1} We first observe some properties of pairs.
Observation 9
 1.
For any pair \((V_i,V_j)\) such that \(j<i\), \(V_i\) is either a 22singleton or a chain.
 2.
If \(V_i\) is paired with \(V_j\) such that \(j<i\), then all predecessors of \(V_i\) are in \(V_j\) or occurred in a group before \(V_j\).
 1.
At least one of \(V_i\) and \(V_j\) is a short chain. Say \(V_i\) is the short chain, the other case is similar. Recall that the standard layout for a short chain uses 3 new gridlines, but \(x^{2{\text{ }2}}_{V_i}+x^{1{\text{ }3}}_{V_i}=2\). So the layout of a short chain automatically saves one gridline. We do not change the algorithm at all in this case; laying out \(V_i\) and \(V_j\) exactly as before results in at most \( 2x^{2{\text{ }2}}_{V_i\cup V_j} + 2x^{1{\text{ }3}}_{V_i\cup V_j}  1\) new gridlines. (This is what happens in the example of Fig. 1.)
 2.
One of \(V_i\) and \(V_j\) is a 31singleton. By Observation 9, the 31singleton must be \(V_j\). By the pairing algorithm, the unique outgoing edge of the 31singleton must lead to \(V_i\). Draw \(V_j\) as before. We can then draw \(V_i\) such that it reuses one of the columns that were freed by \(V_j\); see Fig. 4.
 3.
\(V_i\) and \(V_j\) are both long chains. In this case, both \(V_i\) and \(V_j\) can use the same extra row for the “detours” that their middle vertices (by which we mean vertices that are neither the first nor the last vertex of the chain) use. Since we can freely choose into which columns these middle vertices are placed, we can ensure that none of these “detours” overlap and, hence, one row suffices for both chains. This holds even if one or both of the predecessors of \(V_i\) are in \(V_j\), as these are distinct and the two corresponding incoming edges of \(V_i\) extend the edges that were already drawn for \(V_j\); see Fig. 5.
 4.None of the previous cases applies and \(V_j\) is a 22singleton. By Observation 9.1 and since Case (1) does not apply, \(V_i\) is either a 22singleton or a long chain. There are two columns reserved for edges from predecessors of \(V_j\). Since predecessors of \(V_i\) are distinct, at most one of them can be the 22singleton in \(V_j\). Thus, there also is at least one column reserved for an edge from a predecessor of \(V_i\) not in \(V_j\). We call these three or four columns the predecessorcolumns. We have three subcases depending on the relative location of these columns:
 (a)
The leftmost predecessorcolumn leads to \(V_j\). In this case, we save a column almost exactly as in [9]. Place \(V_j\) as before, in the right one of its predecessorcolumns. This leaves the leftmost predecessorcolumn free to be reused. Now no matter whether \(V_i\) is a 22singleton or a long chain, or whether \(V_i\) is adjacent to \(V_j\) or not, we can reuse this leftmost column for one outgoing edge of \(V_i\) with a suitable placement; see Fig. 6.
 (b)
The rightmost predecessorcolumn leads to \(V_j\). This case is symmetric to the previous one.
 (c)
The leftmost and rightmost predecessorcolumns lead to \(V_i\). This implies that \(V_i\) has two predecessors not in \(V_j\). Hence, \(V_i\) cannot be adjacent to \(V_j\). If \(V_i\) is a 22singleton, then (as discussed earlier) we can exchange the roles of \(V_i\) and \(V_j\), which brings us to Case 4(a). If \(V_i\) is a long chain, then place \(V_j\) in the standard fashion. We then place the long chain \(V_i\) such that the “detours” of its middle vertices reuse the row of \(V_j\). See Fig. 7.
 (a)
 5.
None of the previous cases applies. Then \(V_j\) is a chain, say \(V_j=\{z_1,\dots ,z_\ell \}\), and \(\ell \ge 3\) since Case (1) does not apply. We assume the naming is such that the predecessor column of \(z_1\) is left of the predecessor column of \(z_\ell \).
Since we are not in a previous case, \(V_i\) must be a 22singleton, say z. If \(V_i\) is not adjacent to \(V_j\), then we can again exchange the roles of \(V_i\) and \(V_j\), which brings us to Case (4). Hence, we may assume that there are edges between \(V_j\) and \(V_i\). We distinguish the following subcases depending on how many such edges there are and whether their ends are middle vertices.
 (a)
z has exactly one neighbor in \(V_j\), and it is either \(z_1\) or \(z_\ell \). We rearrange \(V_i\cup V_j\) into two different chains. Let z be adjacent to \(z_1\) (the other case is symmetric). Then \(\{z,z_1\}\) forms one chain and \(\{z_2,\dots ,z_\ell \}\) forms another. Embed these two chains as usual. Since \(\{z,z_1\}\) forms a short chain, this saves one gridline; see Fig. 8(left).
 (b)
z has exactly one neighbor in \(V_j\), and it is \(z_h\) for some \(1<h<\ell \). Embed the chain \(V_j\) as usual, but omit the new column next to \(z_h\). For embedding z, we place a new row below the rows for the chain. Using this new row, we can connect the bottom outgoing edge of \(z_h\) to the horizontal incoming edge of z; see Fig. 8(right).
 (c)
z has two neighbors in \(V_j\), and both of them are middle vertices \(z_{g},z_{h}\) for \(1<g<h<\ell \). Embed the chain \(V_j\) as usual, but omit the new columns next to \(z_g\) and \(z_h\). Place a new row between the two rows for the chain and use it to connect the two bottom outgoing edges of \(z_g\) and \(z_h\) to place z, reusing the row for the detours to place the bottom outgoing edge of z. This uses an extra column for z, but saved two columns at \(z_g\) and \(z_h\), so overall one gridline has been saved; see Fig. 9(top left).
 (d)
z is adjacent to \(z_1\) and \(z_2\) (the case of adjacency to \(z_{\ell 1}\) and \(z_\ell \) is symmetric). Embed \(z_2,\dots ,z_\ell \) as usual for a chain, then place \(z_1\) below \(z_2\). The horizontally outgoing edge of \(z_2\) intersects one outgoing edge of \(z_1\). Put z at this place to save a row and a column; see Fig. 9(top right).
 (e)
z is adjacent to \(z_1\) and \(z_h\) with \(h>2\) (the case of adjacency to \(z_\ell \) and \(z_h\) with \(h<\ell 1\) is symmetric). Draw the chain \(V_j\) with the modification that \(z_h\) is below \(z_{h1}\), but still all middle vertices use the same extra row for their downward outgoing edges. This uses 3 rows, but now z can be placed using the two left outgoing edges of \(z_1\) and \(z_h\), saving a row for z and a column for the left outgoing edge of \(z_h\); see Fig. 9(bottom), both for \(h<\ell \) and \(h=\ell \).
 (a)
5 Conclusion

Can we draw 2connected 4graphs with area less than \(0.76n^2\)? A natural approach would be to draw each 3connected component with area \(0.56n^2\) and to merge them suitably, but there are many cases depending on how the cutvertices and virtual edges are drawn, and so this is far from trivial.

Can we draw 3connected 4graphs with \((2\varepsilon )n\) bends, for some \(\varepsilon >0\)? With an entirely different algorithm (not given here), we have been able to prove a bound of \(2nx^{2{\text{ }2}}+O(1)\) bends, so an improved bound seems likely.

Our algorithm was strongly inspired by the one of Kant [7] for 3connected planar graphs. Are there other graph drawing algorithms for planar 3connected graphs that can be transferred to nonplanar 3connected graphs by using the 3canonical order?
Footnotes
 1.
The constructions we give have been designed as to keep the description simple; often even more gridlines could be saved by doing more complicated constructions.
Notes
Acknowledgments
We wish to thank the anonymous reviewers for their constructive comments.
References
 1.Biedl, T.C., Kant, G.: A better heuristic for orthogonal graph drawings. Comput. Geom. 9(3), 159–180 (1998)zbMATHMathSciNetCrossRefGoogle Scholar
 2.Cheriyan, J., Maheshwari, S.N.: Finding nonseparating induced cycles and independent spanning trees in 3connected graphs. J. Algorithms 9(4), 507–537 (1988)zbMATHMathSciNetCrossRefGoogle Scholar
 3.Chiang, Y.T., Lin, C.C., Lu, H.I.: Orderly spanning trees with applications. SIAM J. Comput. 34(4), 924–945 (2005)zbMATHMathSciNetCrossRefGoogle Scholar
 4.de Fraysseix, H., de Mendez, P.O.: Regular orientations, arboricity and augmentation. In: Tamassia, R., Tollis, I.G. (eds.) GD 1994. LNCS, vol. 894, pp. 111–118. Springer, Heidelberg (1995) CrossRefGoogle Scholar
 5.Dietz, P., Sleator, D.: Two algorithms for maintaining order in a list. In: 19th Annual ACM Symposium on Theory of Computing, pp. 365–372 (1987)Google Scholar
 6.de Fraysseix, H., Pollack, R., Pach, J.: How to draw a planar graph on a grid. Combinatorica 10, 41–51 (1990)zbMATHMathSciNetCrossRefGoogle Scholar
 7.Kant, G.: Drawing planar graphs using the canonical ordering. Algorithmica 16(1), 4–32 (1996)zbMATHMathSciNetCrossRefGoogle Scholar
 8.Mondshein, L.F.: Combinatorial ordering and the geometric embedding of graphs. Ph.D. thesis, M.I.T. Lincoln Laboratory / Harvard University (1971)Google Scholar
 9.Papakostas, A., Tollis, I.G.: Algorithms for areaefficient orthogonal drawings. Comput. Geom. 9(1–2), 83–110 (1998)zbMATHMathSciNetCrossRefGoogle Scholar
 10.Rosenstiehl, P., Tarjan, R.E.: Rectilinear planar layouts and bipolar orientation of planar graphs. Discrete Comput. Geom. 1, 343–353 (1986)zbMATHMathSciNetCrossRefGoogle Scholar
 11.Schäffter, M.: Drawing graphs on rectangular grids. Discrete Appl. Math. 63, 75–89 (1995)zbMATHMathSciNetCrossRefGoogle Scholar
 12.Schmidt, J.M.: The Mondshein sequence (2013). http://arxiv.org/pdf/1311.0750.pdf
 13.Schmidt, J.M.: The Mondshein sequence. In: Esparza, J., Fraigniaud, P., Husfeldt, T., Koutsoupias, E. (eds.) ICALP 2014. LNCS, vol. 8572, pp. 967–978. Springer, Heidelberg (2014) Google Scholar
 14.Biedl, T.: Optimal orthogonal drawings of triconnected plane graphs. In: McCune, W., Padmanabhan, R. (eds.) Automated Deduction in Equational Logic and Cubic Curves. LNCS, vol. 1095, pp. 333–344. Springer, Heidelberg (1996) Google Scholar
 15.Tamassia, R., Tollis, I.: A unified approach to visibility representations of planargraphs. Discrete Comput. Geom. 1, 321–341 (1986)zbMATHMathSciNetCrossRefGoogle Scholar
 16.Tamassia, R., Tollis, I.G., Vitter, J.S.: Lower bounds for planar orthogonal drawings of graphs. Inf. Process. Lett. 39, 35–40 (1991)zbMATHMathSciNetCrossRefGoogle Scholar
 17.Valiant, L.G.: Universality considerations in VLSI circuits. IEEE Trans. Comput. C–30(2), 135–140 (1981)MathSciNetCrossRefGoogle Scholar