Drawing Graphs on Few Lines and Few Planes

We investigate the problem of drawing graphs in 2D and 3D such that their edges (or only their vertices) can be covered by few lines or planes. We insist on straight-line edges and crossing-free drawings. This problem has many connections to other challenging graph-drawing problems such as small-area or small-volume drawings, layered or track drawings, and drawing graphs with low visual complexity. While some facts about our problem are implicit in previous work, this is the first treatment of the problem in its full generality. Our contribution is as follows. We show lower and upper bounds for the numbers of lines and planes needed for covering drawings of graphs in certain graph classes. In some cases our bounds are tight; in some cases we are able to determine exact values. We relate our parameters to standard combinatorial characteristics of graphs (such as the chromatic number, treewidth, maximum degree, or arboricity) and to parameters that have been studied in graph drawing (such as the track number or the number of segments appearing in a drawing). We pay special attention to planar graphs. For example, we show that there are planar graphs that can be drawn in 3-space on a lot fewer lines than in the plane.


Introduction
It is well known that any graph admits a straight-line drawing in 3-space.Suppose that we are allowed to draw edges only on a limited number of planes.How many planes do we need for a given graph G? For example, K 6 needs four planes; see Fig. 1.Note that this question is different from the well-known concept of a book embedding where all vertices lie on one line (the spine) and edges lie on a limited number of adjacent half-planes (the pages).In contrast, we put no restriction on the mutual position of planes, the vertices can be located in the planes arbitrarily, and the edges must be straight-line.
In a weaker setting, we require only the vertices to be located on a limited number of planes (or lines).For example, the graph in Fig. 2 can be drawn in 2D such that its  vertices are contained in three lines; we conjecture that it is the smallest planar graph that needs more than two lines even in 3D.This version of our problem is related to the well-studied problem of drawing a graph straight-line in a 3D grid of bounded volume [16,43]: If a graph can be drawn with all vertices on a grid of volume v, then v 1/3 planes and v 2/3 lines suffice.We now formalize the problem.Definition 1.Let 1 ≤ l < d, and let G be a graph.We define the l-dimensional affine complexity of G in R d , denoted by ρ l d (G), as the minimum number of l-dimensional planes in R d such that G has a drawing that is contained in the union of these planes.We define π l d (G), the weak l-dimensional affine complexity of G in R d , similarly to ρ l d (G), but under the weaker restriction that the vertices (and not necessarily the edges) of G are contained in the union of the planes.Finally, πl d (G) is a restricted version of π l d (G) in that we insist that the planes be parallel.We consider only straight-line and crossing-free drawings.
Clearly, for any combination of l and d, it holds that π l d (G) ≤ πl d (G) and π l d (G) ≤ ρ l d (G).Larger values of l and d give us more freedom for drawing graphs and, therefore, smaller π-and ρ-values.Formally, for any graph G, if l ≤ l and d ≤ d then π l d (G) ≤ π l d (G), ρ l d (G) ≤ ρ l d (G), and πl d (G) ≤ πl d (G).But in most cases this freedom is not essential.
First of all, it suffices to consider l ≤ 2 because otherwise ρ l d (G) = 1.More interestingly, we can actually focus on d ≤ 3 because every graph can be drawn in 3-space as efficiently as in higher dimensions.We prove this important fact in Section 2. Thus, our task is to investigate the cases 1 ≤ l < d ≤ 3.
Related work.We have already briefly mentioned 3D graph drawing on the grid, which has been surveyed by Wood [43] and by Dujmović and Whitesides [16].For example, Dujmović [12], improving on a result of Di Battista et al. [4], showed that any planar graph can be drawn into a 3D-grid of volume O(n log n).It is well-known that, in 2D, any planar graph admits a plane straight-line drawing on an O(n) × O(n) grid [38,10] and that the nested-triangles graph T k = K 3 × P k (see Fig. 6, p. 17) with 3k vertices needs Ω(k 2 ) area [10].ρ 1 2 (G) ≤ k are solvable in polynomial time for any fixed k.However, the versions of these problems with k being part of the input are complete for the complexity class ∃R that plays an important role in computational geometry [37].
Notation.For a graph G = (V, E), we use n and m to denote the numbers of vertices and edges of G, respectively.Let ∆(G) = max v∈V deg(v) denote the maximum degree of G. Furthermore, we will use the standard notation χ(G) for the chromatic number, tw(G) for the treewidth, and diam(G) for the diameter of G.The Cartesian product of graphs G and H is denoted by G × H.
Cubic graphs are graphs where all vertices have degree 3. Recall also that a graph is k-connected if it has more than k vertices and stays connected after removal of any set of up to (k − 1) vertices.A planar graph G is maximal if adding an edge between any two non-adjacent vertices of G violates planarity.Maximal planar graphs on more than three vertices are 3-connected.Clearly, all facial cycles in such graphs have length 3.By this reason maximal planar graphs are also called triangulations.Proof.The theorem follows from the following fact: For any finite family L of lines in R d , there exists a linear transformation A : R d → R 3 that is injective on L 0 = { : ∈ L}, the set of all points of the lines in L.

Collapse of the Multidimensional Affine Hierarchy
We prove this claim by induction on d.If d = 3, we let A be the identity map on R 3 .Suppose that d > 3.
Regarding two lines and in R d as 1-dimensional affine subspaces, we consider the Minkowski difference − = { l − l : l ∈ , l ∈ }.Note that this is a plane, i.e., a 2-dimensional affine subspace of R d .Denote L = L 0 − L 0 = { − : , ∈ L}.Let L = { tl : t ∈ R, l ∈ L} be the union of all lines going through the origin 0 of R d and intersecting the set L. Since the set L is contained in a union of finitely many planes in the space R d , the set L is contained in a union of finitely many 3-dimensional linear subspaces of R d (each of them contains the origin 0).Since d > 3, there exists a line 0 0 such that 0 ∩ L = {0}.Now, let A 0 : R d → R d−1 be an arbitrary linear transformation such that ker A 0 = 0 .Let x, y ∈ L 0 be arbitrary points such that By the inductive assumption applied to the family of lines { A 0 : ∈ L} in R d−1 , there exists a linear transformation A 1 : R d−1 → R 3 injective on the union of all ∈ L. It remains to take the composition A = A 1 A 0 .
3 The Affine Complexity in R 3   3.1 Placing Vertices on Few Lines or Planes (π 1  3 and π 2 3 ) A linear forest is a forest whose connected components are paths.The linear vertex arboricity lva(G) of a graph G equals the smallest size r of a partition such that every V i induces a linear forest.This notion, which is an induced version of the fruitful concept of linear arboricity (see Remark 10 below), appears very relevant to our topic.
Theorem 3.For any graph G, it holds that π 1 3 (G) = lva(G).Moreover, any graph G can be drawn with vertices on r lines in the 3D integer grid of size r × 4rn × 4r 2 n, where r = lva(G).
The proof is based on a construction of Pach et al. [32].A complete r-partite graph is called balanced if any two of its classes differ by at most one in size.Let K r (n) denote a balanced complete r-partite graph with n ≥ r vertices.In other words, the vertex set of Lemma 4 ([32, Lemma]).For any r ≥ 2 and for any n divisible by r, the balanced complete r-partite graph has a 3D-grid drawing that fits into a box of size r × 4n × 4rn.The drawing is such that any class V i is collinear.
We include the proof for the reader's convenience.
Proof.Let p be the smallest prime with p ≥ 2r − 1 and set N := pn/r.By Bertrand's postulate, p < 4r and, hence, N < 4n.For any 0 Note that V i is contained in the line i = {(i, 0, 0) + t(0, 1, i) : t ∈ R}.These sets are pairwise disjoint, and each of them has precisely N/p = n/r elements.Connect any two points belonging to different V i 's by a straight-line segment.The resulting drawing of K r (n) fits into a box of size r × 4n × 4rn, as desired.Pach et al. [32,Lemma] showed that no two edges of this drawing cross each other.Moreover, Case 2 of their proof implies that, for any i, no edge of this drawing crosses a segment between two consecutive vertices of V i placed along the line i .So we can join these vertices by edges without adding crossings to the drawing.
) such that, for each i, f maps adjacent vertices of the linear forest G[V i ] into consecutive vertices of the set V i placed along a line i .
Then the observation at the end of the proof of Lemma 4 implies that f (V (G)) induces a crossing-free straight-line drawing of G.
Proof.We have lva(G) ≤ χ(G) because any independent set is a linear forest.On the other hand, χ(G) ≤ 2 lva(G) because any linear forest is 2-colorable.
Corollary 5 readily implies that π 1 3 (G) ≤ ∆(G) + 1.This can be considerably improved using a relationship between the linear vertex arboricity and the maximum degree that is established by Matsumoto [29].Matsumoto's result implies that We now turn to the 2-dimensional affine complexity (weak version).The vertex thickness vt(G) of a graph G is the smallest size r of a partition are all planar.Theorem 6.For any graph G with m edges, it holds that π 2 3 (G) = π2 3 (G) = vt(G), and that it can be drawn with vertices on a 3D integer grid of size vt(G) Proof.The bounds vt(G) ≤ π 2 3 (G) ≤ π2 3 (G) are obvious.The bound π2 3 (G) ≤ vt(G) follows from the existence of a drawing with the specified volume, so we need to prove the last fact.
Let V 1 , . . ., V r be a partition of the vertex set of G such that each G[V i ] is a planar graph.As well known, every planar graph admits a plane straight-line drawing on an O(n) × O(n) grid [38,10].Let us fix such a drawing δ i for each G[V i ] and place it in the plane z = i.Call an edge uv horizontal if both u and v belong to the same V i for some i ≤ r.We now have to resolve two problems: • A non-horizontal edge can pass through a vertex of some δ i ; • Two edges that are not both horizontal can cross each other.
In order to remove all possible crossings, we replace each δ i with its random perturbation δ i (still in the same plane z = i) and prove that, with non-zero probability, no crossing occurs.Specifically, let s and t be parameters that will be chosen later.Let T a,b,p,q be an affine tranformation of the (x, y)-plane defined by where a, b, p, and q are integers such that 0 ≤ p, q < s, 1 ≤ b < a ≤ t, and a and b are coprime.Note that T a,b,p,q consists of a dilating rotation followed by a shift, and that it transforms integral points into integral points.The random drawings δ i are obtained by choosing a, b, p, and q at random and applying T a,b,p,q to δ i (this is done independently for different i ≤ r).Note that the resulting drawing occupies a 3D grid of size r × O(tn + s) × O(tn + s).
For each fixed edge uv and vertex w such that u ∈ V i , w ∈ V j , and v ∈ V k for some i < j < k, let us estimate the probability that uv passes through w in the drawing.
Conditioned on the positions of δ l for all l = j and on the choice of the parameters a and b in T a,b,p,q for δ j , this probability is clearly at most 1/s 2 .Therefore, this probability is at most 1/s 2 also if all δ l are chosen at random.It follows that there is a non-horizontal edge passing through some vertex with probability at most mn/s 2 .
Consider now two edges u 1 v 1 and u 2 v 2 .If there is an i ≤ r such that V i contains exactly one of the vertices u 1 , v 1 , u 2 , and v 2 , then an argument similar to the above shows that u 1 v 1 and u 2 v 2 cross with probability at most 1/s.It follows that some edges of this kind cross each other with probability at most m 2 /s.
Suppose now that u 1 , u 2 ∈ V i and v 1 , v 2 ∈ V j .Note that shifts cannot resolve the possible crossing of the edges u 1 v 1 and u 2 v 2 .Luckily, if we fix δ i and "rotate" δ j by means of T a,b,p,q with random a, b and fixed p, q, then u 1 v 1 and u 2 v 2 will cross in at most one case.The probability of this event is bounded by O(1/t 2 ) because the number of coprime a and b such that 1 ≤ b < a ≤ t is equal to the number of Farey fractions of order t, which is known to be asymptotically 3 π 2 t 2 + O(t log t) [23].It follows that some edges of this kind cross each other with probability at most O(m 2 /t 2 ).Summarizing, we see that the random drawing of G will have a crossing with probability bounded by This probability can be ensured to be strictly smaller than 1 by choosing parameters s = O(m 2 ) and t = O(m).We conclude that for such choice of s and t there is at least one crossing-free drawing.Since O(tn + s) = O(mn + m 2 ) = O(m 2 ) (the latter equality being true for G with no isolated vertex), such a drawing occupies volume r×O(m 2 )×O(m 2 ).
Proof.The lower bound is tight for 4-chromatic planar graphs and for complete graphs K n , see Example 8(c).The upper bound is tight for any r-partite complete graph K r (rs) where each of the r ≥ 2 vertex classes contains s > 2r − 1 vertices.Note that χ(K r (rs)) = r.
for every graph G. Let us supply this example with brief explanations.(a-b) The lower bound for π 1  3 (K n ) and the upper bound for π 1 3 (K p,q ) follow from Corollary 5.The upper bound π 1 3 (K n ) ≤ n/2 is given by any 3-dimensional drawing of K n ; we can split the vertices in pairs and draw a line through each pair.(c) By Theorem 6, such that every V i induces a planar subgraph of K n , that is iff every V i has size at most 4 (because K 4 is planar and K 5 is not).Such a partition exists iff r ≥ n/4 .

Placing Edges on Few Lines or Planes (ρ 1
3 and ρ 2 3 ) Proof.(a) In any drawing of a graph G, any essential vertex is shared by two edges not lying on the same line.Therefore, each such vertex is an intersection point of two lines, which implies that es(G) . Hence, The last inequality follows by the inequality between arithmetic and quadratic means.Hence, Part (a) of Lemma 9 implies that ρ 1 3 (G) > √ 2n if a graph G has no vertices of degree 1 and 2, while Part (b) yields ρ 1 3 (G) > √ m/2 for all such G.
Remark 10.The linear arboricity la(G) of a graph G is the minimum number of linear forests which partition the edge set of G; see [26].Clearly, we have ρ 1 3 (G) ≥ la(G).The gap between these two parameters is unbounded.Indeed, let G be an arbitrary cubic graph.Akiyama et al. [2] showed that la(G) = 2. On the other hand, any vertex of G is essential, so ρ 1 3 (G) > √ 2n by Lemma 9(a).Theorem 11 below shows an even larger gap.
We now prove a general lower bound for ρ 1 3 (G) in terms of the treewidth of G.Note for comparison that π 1 3 (G) ≤ χ(G) ≤ tw(G) + 1(and even π1 3 (G) is bounded from above by a function of tw(G); see Section 5.1).The relationship between ρ 1 3 (G) and tw(G) follows from the fact that graphs with low parameter ρ 1 3 (G) have small separators.This fact is interesting by itself and has yet another consequence: Graphs with bounded vertex degree can have linearly large value of ρ 1 3 (G) (hence, the factor of n in the trivial bound It is known [20,Theorem 11.17] The bisection width bw(G) of a graph G is the minimum possible number of edges between two sets of vertices W 1 and W 2 forming a partition of V (G) into two parts of sizes n/2 and n/2 .Note that sep * (G) ≤ bw(G) + 1.
for almost all cubic graphs with a given number of vertices.
Proof.(a) Fix a drawing of the graph G on r = ρ 1 3 (G) lines in R 3 .Choose a plane L that is not parallel to any line passing through two vertices of the drawing.Let us move L along the orthogonal direction until it separates the vertex set of G into two almost equal parts W 1 and W 2 .The plane L can intersect at most r edges of G, which implies that bw(G) ≤ r.
(b) follows from Part (a) and the fact that a random cubic graph on n vertices has bisection width at least n/4.95 with probability 1 − o(1) (Kostochka and Melnikov [27]). (c . Choose a plane L as in the proof of Part (a) and move it until it separates W into two equal parts W 1 and W 2 ; if |W | is odd, then L should contain one vertex w of W .If |W | is even, we can ensure that L does not contain any vertex of G.We now construct a set S as follows.If L contains a vertex w ∈ W , i.e. |W | is odd, we put w in S. Let E be the set of those edges which are intersected by L but are not incident to the vertex w (if it exists).Note that |E| < r if |W | is odd and |E| ≤ r if |W | is even.Each of the edges in E contributes one of its incident vertices into S.Note that |S| ≤ r.Set W 1 = W 1 \ S and W 2 = W 2 \ S and note that there is no edge between these sets of vertices.Thus, S is a strongly balanced W -separator.
(d) follows from (c) by the relationship between treewidth and balanced separators.
On the other hand, note that ρ 1 3 (G) cannot be bounded from above by any function of tw(G).Indeed, by Lemma 9(a) we have ρ 1 3 (T ) = Ω( √ n) for caterpillars with linearly many vertices of degree 3. The only possible relation in this direction is ρ 1 3 (G) ≤ m < n tw(G).The factor n cannot be improved here (take ) ≤ pq for any 1 ≤ p ≤ q.Brief comments on this example: (a) Any line contains at most one of the n 2 edges of K n , otherwise the line would contain a triangle.(b) In any drawing that realizes ρ 1 3 (K p,q ), each line contains at least one and at most two of the pq edges of K p,q .
We now turn to the 2-dimensional affine complexity in R 3 .
Example 13.For any integers 1 ≤ p ≤ q, it holds that ρ 2 3 (K p,q ) = p/2 .Indeed, let S be a set of planes underlying a drawing of K p,q .Every plane in S contains at most two points of either type p or of type q, otherwise it would contain the non-planar K 3,3 .Hence, every plane in S covers at most 2q edges.Given pq edges in total, we get |S| ≥ pq/2q = p/2 .This lower bound is tight.Place all points of type q on a line and introduce p/2 distinct planes containing .Line divides the planes into 2 p/2 ≥ p half-planes.Put every point of type p in one of these half-planes and connect it to the points on .
Since any planar graph has less than 3n edges, we easily conclude that ρ 1 3 (G) < 3nρ 2  3 (G).Examples 12 and 13 show that the factor of n is best possible here.
Determining the parameter ρ 2 3 (G) for complete graphs G = K n is a much more subtle issue.We are able to determine the asymptotics of ρ 2 3 (K n ) up to a factor of 2. By a combinatorial cover of a graph G we mean a set of subgraphs {G i } such that every edge of G belongs to G i for some i.A geometric cover of a drawing d : The asymptotics of the numbers c(K n , K s ) for s = 3, 4 can be determined based on the results about Steiner systems by Kirkman and Hanani [5,25].This yields the following bounds for ρ 2 3 (K n ).
Theorem 14.For all n ≥ 3, Proof.For the lower bound, take a drawing of K n with a geometric cover L using ρ 2 3 (K n ) planes.This geometric cover induces a combinatorial cover ).For the upper bound, let d : V (K n ) → R 3 be an arbitrary drawing of K n in 3-space (with non-crossing edges).Since, for any ).Now we show lower and upper bounds for c(K n , K 3 ) and c(K n , K 4 ) and determine their asymptotics.Since the graph K n has n 2 = n(n − 1)/2 edges and each copy of the graph , in particular we get c(K n , K 4 ) ≥ n 2 /12 − n/12 for k = 4.This lower bound is attained provided there exists a Steiner system S(2, k, n), so in this case c( . Hanani [25] (see also [36,Theorem 2.1]) showed that a Steiner system S(2, 4, n) exists iff n ≡ 1 (mod 12) or n ≡ 4 (mod 12).This implies that c(K n , K 4 ) = n 2 /12 + Θ(n) for any n.
Lemma 15.Let d : V (K n ) → R 3 be a drawing of K n and L a geometric cover of d.For each 4-vertex graph G ∈ K L , the set d(G ) not only belongs to a plane , but also is a triangle with an additional vertex in its interior.Also, all faces of the convex hull of d(K n ) are triangles.
Proof.In any planar drawing of K 4 , the four vertices cannot be drawn as vertices of a convex quadrilateral for else its diagonals would intersect.Proof.Indeed, assume the converse: both graphs K and K contain the same copy K 3 of K 3 .Since the triangle d(K 3 ) cannot be collinear, both d(K ) and d(K ) lie in the plane spanned by the set d(K 3 ).But then the plane contains five points of d(K n ), which is impossible.
where one of the covering planes contains exactly three vertices.
Proof.(a) Since each drawing of the graph K n can be extended to a drawing of the graph K n+1 by adding n segments with a common endpoint d(v n+1 ), which can be covered by n/2 planes, we see that ρ  Proof.n = 5.By Lemma 17(a), ρ 2 3 (K 5 ) ≤ ρ 2 3 (K 4 ) + 2 = 3.To obtain a lower bound, remark that ρ 2 3 (K 5 ) ≥ c(K 5 , K 4 ) = 3.To prove the last equality remark that although a graph K 5 has 10 < 2 • 6 edges, it cannot be covered by two copies K 4 and K 4 of a graph K 4 , because in this case they should have at least three common vertices, so their intersection should have at least three common edges, but 12 − 3 < 10.From the other side, each two different copies of K 4 cover all edges of K 5 but one, so c(K 5 , K 4 ) = 3. n = 6.Fig. 1 shows that ρ 2 3 (K 6 ) ≤ 4. Now we show that ρ 2 3 (K 6 ) ≥ 4. Assume that ρ 2 3 (K 6 ) < 4. Consider a combinatorial cover K L of K 6 by its complete planar subgraphs corresponding to a geometric cover L of its drawing by 3 planes.Graph K 6 has 15 edges, so to cover it by complete planar graphs we have to use at least two copies of K 4 and, additionally, a copy of K k for k ≥ 3. But, since each two copies of K 4 in K 6 have a common edge (and by Lemma 16 this edge is unique), the cover K L consists of three copies of K 4 .Denote these copies by K 1 4 , K 2 4 , and K 3 4 .By Lemma 15, for each i, d(K i 4 ) is a triangle with an additional vertex d(v i ) in its interior.Let V 0 = {v 1 , v 2 , v 3 }.By the Krein-Milman theorem [28,41], the convex hull conv d(K 6 ) of a set d(K 6 ) is the convex hull conv d(V ) \ d(V 0 ) of the set d(V ) \ d(V 0 ).If all the vertices v i are mutually distinct then the set d(V ) \ d(V 0 ) is a triangle, so the drawing d is planar, a contradiction.Hence, v i = v j for some i = j.Let k be the third index that is distinct from both i and j.Since graphs K i 4 and K j 4 have exactly one common edge, this is an edge (v i , v) for some vertex v of K 6 (see Fig. 1 with u 4 for v i and Therefore there exists a vertex v 0 ∈ V (K 7 ) covered by at most two members of the cover K L .Since each element K of cover K L covers at most three vertices incident to v 0 (and exactly three vertices only if K is a copy of graph K 4 ) and, in graph K 7 , there are 6 edges incident to vertex v 0 , we see that vertex v 0 belongs to exactly two members K 1  4 and K 2 4 of the cover K L , and each of these members is a copy of graph K 4 .Moreover, v 0 is the unique common vertex of the graphs K 1 4 and K 2 4 .For each i let ), there exists an edge (v, v ) of the graph K 7 \ {v 0 } which is not covered by the family K .Since a set {(u, u ) : of edges is covered by the family K , there exists an index i with V 0 = {v, v } ⊂ V i .Let j = i be the other index.We have there exists a vertex v 1 ∈ V 0 which belongs to at most one set K ∈ K .In fact, such a set K exists, because in in opposite case no edge (v 1 , w) for w ∈ V j is covered by K L .Since both K j 4 and K are members of the cover K L , by Lemma 16, there exists a vertex w ∈ V j \ V (K).Then an edge (v 1 , w) is not covered by K L , a contradiction.
We prove the last inequality.Since a graph K 9 has 36 = 6 • 6 edges, each cover of K 6 by six copies of K 4 generates a Steiner system S(2, 4, 9).The absence of such a system follows from the result of Hanani mentioned earlier, but we give a direct proof.Indeed, assume that c(K 9 , K 4 ) ≤ 6.Since degree of each vertex v of K 9 is 8, v belongs to at least 3 copies of K 4 from the cover K .Then But since each member K of the cover K contains 4 vertices of V (K 9 ), we have s ≤ 6 • 4 = 24, a contradiction.4 The Affine Complexity of Planar Graphs in R 2 and R 3   We now consider the affine complexity of planar graphs, both in R 2 and R 3 .
4.1 Placing Vertices on Few Lines (π 1 2 and π 1 3 ) Call a drawing outerplanar if all the vertices lie on the outer face.An outerplanar graph is a graph admitting an outerplanar drawing.Note that this definition does not depend on whether straight line or curved drawings are considered.Combining Corollary 5 with the 4-color theorem yields π 1 3 (G) ≤ 4 for planar graphs.Given that outerplanar graphs are 3-colorable (they are partial 2-trees), we obtain π 1 3 (G) ≤ 3 for these graphs.These bounds can be improved using the equality π 1 3 (G) = lva(G) of Theorem 3 and known results on the linear vertex arboricity: (a) For any planar graph G, it holds that π 1 3 (G) ≤ 3 [22,33].
According to Chen and He [9], the upper bound lva(G) ≤ 3 for planar graphs by Poh [33] is constructive and yields a polynomial-time algorithm for partitioning the vertex set of a given planar graph into three parts, each inducing a linear forest.By combining this with the construction given in Theorem 3, we obtain a randomized polynomial-time algorithm that draws a given planar graph such that the vertex set "sits" on three lines.
The example of Chartrand and Kronk [8] is a 21-vertex planar graph whose vertex arboricity is 3, which means that the vertex set of this graph cannot even be split into two parts both inducing (not necessarily linear) forests.Raspaud and Wang [34] showed that all 20-vertex planar graphs have vertex arboricity at most 2. We now observe that a smaller example of a planar graph attaining the extremal value π 1 3 (G) = 3 can be found by examining the linear vertex arboricity.
Example 19.The planar 9-vertex graph G in Fig. 2 has π 1 3 (G) = lva(G) = 3.Indeed, at the picture it is easy to see that 2 ≤ lva(G) ≤ 3.In order to show that lva(G) > 2, assume that the vertex set of G is colored black and white where each monochromatic component induces a linear forest.Without loss of generality, we may assume that the central vertex is white.Since the central vertex cannot have more than two white neighbors, at most two other vertices are white.Note that the neighbors of the central vertex form a square in Fig. 2 and that each side of the square contains a cycle.Hence, none of the sides of the square can be monochromatic; it must contain at least one white vertex.Therefore, the boundary of the square contains exactly two white vertices, which must be placed in opposite corners.If the white vertices are placed in the top left and the bottom right corners, then the two other corners, which are black, have three black neighbors.If the white vertices are placed in the top right and the bottom left corners, then the three white vertices induce a cycle.In both cases, we have a contradiction.Now we show lower bounds for the parameter π 1 2 (G).Recall that the dual of a 3connected planar graph G is a graph G * whose vertices are the faces of G (represented by their facial cycles).The definition does not depend on a particular embedding of G in the plane by the Whitney theorem, which says that all embeddings of a 3-connected planar graph in the sphere are equivalent up to a homeomorphism (therefore, the set of facial cycles of G does not depend on a particular plane embedding).Two faces are adjacent in G * iff they share a common edge.The dual graph G * is also a polyhedral graph, and (G * ) * , is isomorphic to G.In a cubic graph, every vertex has degree 3; the dual of any cubic 3-connected planar graph is a triangulation.Conversely, the dual of a triangulation is a cubic graph.
Recall that the circumference of a graph G, denoted by c(G), is the length of a longest cycle in G.If G is a 3-connected planar graph, let γ(G) denote the minimum number of cycles in the dual graph G * sharing a common vertex and covering every vertex of G * at least twice.Note that, if G is a triangulation, then γ(G) ≥ (2n − 4)/c(G * ), where 2n − 4 is the number of vertices in G * (as a consequence of Euler's formula).Proof.(a) Obviously, the number of vertices in G is bounded by π 1 2 (G)v(G).(b) Given a drawing realizing π 1 2 (G) with line set L, for every line ∈ L, draw two parallel lines , sufficiently close to such that they together intersect the interiors of all faces touched by and do not go through any vertex of the drawing.Note that and cross boundaries of faces only via inner points of edges.Each such crossing corresponds to a transition from one vertex to another along an edge in the dual graph G * .Since all faces are triangles, each of them is visited by each of and at most once.Therefore, the faces crossed along (resp.), among them the outer face of G, form a cycle in G * .It remains to note that every face f of the graph G is crossed at least twice, because f is intersected by at least two different lines from L and each of these two lines has a parallel copy that crosses f .(c) follows from (b).
An infinite family of triangulations G with v(G) ≤ n 0.99 is constructed in [35].By Part (a) of Lemma 20 this implies that there are infinitely many triangulations G with π 1 2 (G) ≥ n 0.01 .Part (c) of Lemma 20 along with an estimate of Grünbaum and Walther [24] (that was used also in [35]) yields a stronger result.
Proof.The shortness exponent σ G of a class G of graphs is the infimum of the set of the reals lim inf i→∞ log c(H i )/log v(H i ) for all sequences of where v(H) denotes the number of vertices in H. Thus, for each > 0, there are infinitely many graphs H ∈ G with c(H) < v(H) σ G + .The dual graphs of triangulations with maximum vertex degree at most 12 are exactly the cubic 3-connected planar graphs with each face incident to at most 12 edges (this parameter is well defined by the Whitney theorem).Let σ denote the shortness exponent for this class of graphs.It is known [24] that σ ≤ log 26 log 27 = 0.988 . ... The theorem follows from this bound by Lemma 20(c).
Problem 22.Does π 1 2 (G) = o(n) for all planar graphs G? A (planar) track drawing [19] of a graph is a drawing for which there are parallel lines, called tracks, such that every edge either lies on a track or its endpoints lie on two consecutive tracks.We call a graph track drawable if it has a track drawing.Let tn(G) be the minimum number of tracks of a track drawing of G.Note that π 1 2 (G) ≤ π1 2 (G) ≤ tn(G).The following proposition is similar to a lemma of Bannister et al. [3, Lemma 1] who say it is implicit in the earlier work of Felsner et al. [19].
Theorem 23 (cf.[19,3]).Let G be a track drawable graph.Then π 1 2 (G) ≤ 2. Proof.It suffices to put the tracks consecutively along a spiral with vertices placed on two intersecting lines.Tracks whose indices are equal modulo 4 are placed on the same half-line; see Fig. Observe that any tree is track drawable: two vertices are aligned on the same track iff they are at the same distance from an arbitrarily assigned root.Moreover, any outerplanar graph is track drawable [19].This yields an improvement over the bound π 1 3 (G) ≤ 2 for outerplanar graphs stated in the beginning of this section.Corollary 24.For any outerplanar graph G, it holds that π 1 2 (G) ≤ 2. Next, we consider a generalization of trees, the class of 2-trees, which is recursively defined as follows: • the graph consisting of two adjacent vertices is a 2-tree; • if G is a 2-tree and H is obtained from G by adding a new vertex and connecting it to two adjacent vertices of G, then H is a 2-tree.
A graph is a partial 2-tree if it is a subgraph of a 2-tree.It is well known that the class of partial 2-trees coincides with the class of graphs with treewidth at most 2. Any outerplanar graph is a partial 2-tree, and the same holds for series-parallel graphs (the latter class is sometimes defined so that it coincides with the class of partial 2-trees).Note that not all 2-trees are track drawable (for example, the graph consisting of three triangles that share one edge).Ravsky and Verbitsky [35,Theorem 4.5] have shown that any partial 2-tree admits a drawing with at least n/30 collinear vertices.This suggests the following question.
4.2 Placing Edges on Few Lines (ρ 1 2 and ρ 1 3 ) The parameter ρ 1 2 (G) is related to two parameters introduced by Dujmović et al. [13].They define a segment in a straight-line drawing of a graph G as an inclusion-maximal (connected) path of edges of G lying on a line.A slope is an inclusion-maximal set of parallel segments.The segment number (resp., slope number ) of a planar graph G is the minimum possible number of segments (resp., slopes) in a straight-line drawing of G.We denote these parameters by segm(G) (resp., slop(G)).Note that slop(G) ≤ ρ 1 2 (G) ≤ segm(G).
These parameters can be far away from each other.Figure 6 shows a graph with slop(G) = O(1) and ρ 1 2 (G) = Ω(n) (see the proof of Theorem 28).On the other hand, note that ρ 1 2 (mK 2 ) = 1 while segm(mK 2 ) = m where mK 2 denotes the graph consisting  of m isolated edges.The gap between ρ1 2 (G) and segm(G) can be large even for connected graphs.It is not hard to see that segm(G) is bounded from below by half the number of odd degree vertices (see [13] for details).Therefore, if we take a caterpillar G with k vertices of degree 3 and k + 2 leaves, then segm(G) ≥ n/2, while ρ 1 2 (G) = O( √ n) because G can easily be drawn in a square grid of area O(n).Note that, for the same G, the gap between slop(G) and It turns out that a large gap between ρ 1 2 (G) and segm(G) can be shown also for 3connected planar graphs and even for triangulations.Somewhat surprisingly, the parameter segm(G) can be bounded from above by a function of ρ 1 2 (G) for all connected graphs.
Proof.Call a vertex v of the graph G branching if its degree is at least three.A path between vertices of the graph G will be called straight if it has no branching vertices other than its endpoints.
Reduce the graph G to a graph G as follows.The set of vertices of G is the set of branching vertices in G, and two vertices are adjacent in G if they are connected by a straight path in G. Being a planar graph, G has a straight-line drawing.
If G is empty then G is a path or a cycle, and segm(G) ≤ 3. So, assume that G has a branching vertex.Since G is connected, every vertex in it is connected to a branching vertex by a straight path (possibly of zero length).Thus, we can construct a straight-line drawing of G from a straight-line drawing of G as follows.Note that an edge e of G corresponds to a bond of paths in G connecting the incident vertices of e.We draw one path in the bond on the segment e, and each other is drawn as a pair of two segments that are close to e.A branching vertex v in G can be connected by a straight path to a degree 1 vertex (which disappears in G ).We restore each such path by drawing it as a small segment.Moreover, v can belong to a cycle whose all vertices except v have degree 2. We draw each such cycle as a small triangle.
Note that the segments incident to the branching vertex v are split into three parts: b v segments which belong to the edge bonds, l v segments going to leaves, and t v segments that are sides of the small triangles.Unless l v = 0 and t v = 2, we can ensure that the last l v + t v segments are drawn in at most (l v + t v )/2 all crossing at the point v. Therefore, the vertex v is incident to at most b v + (l v + t v )/2 segments.This holds true even if l v = 0 and t v = 2; we need to use the fact that in this case b v = 0. We will say that these segments are related to v. We also relate to v the opposite sides of the corresponding small triangles; there are t v /2 of them.Thus, the vertex v has at most Since every segment of the constructed drawing of G is related to some vertex, the total number of the segments is bounded by The last estimate is the first inequality in the proof of Lemma 9(b).
Note that ∆(G)/2 ≤ ρ 1 3 (G) ≤ ρ 1 2 (G) ≤ segm(G) ≤ m for any planar graph G.For all inequalities here excepting the second one, we already know that the gap between the respective pair of parameters can be very large (by considering a caterpilar with linearly many degree 3 vertices and applying Lemma 9(a), by Example 26, and by considering the path graph P n , for which segm(P n ) = 1).Part (b) of the following theorem shows a large gap also between the parameters ρ 1 3 (G) and ρ 1 2 (G).Theorem 28.(a) There are infinitely many planar graphs with constant maximum degree, constant treewidth, and linear ρ 1 2 -value.
(b) Some planar graphs can be drawn much more efficiently, with respect to the 1-dimensional affine complexity, in 3-space than in the plane.Specifically, for infinitely many n there is a planar graph G on n vertices with ρ 1 2 (G) = Ω(n) and ρ 1 3 (G) = O(n 2/3 ).Proof.Consider the nested-triangles graph T k = K 3 × P k shown in Fig. 6.It suffices to prove the following bounds for statements (a) and (b), respectively: ).To see the linear lower bound (i), note that T k is 3-connected.Hence, Whitney's theorem implies that, in any plane drawing of T k , there is a sequence of nested triangles of length at least k/2.The sides of the triangles in this sequence must belong to pairwise different lines.Therefore, ρ 1 2 (T k ) ≥ 3k/2 = n/2.For the sublinear upper bound (ii), first consider the graph C 4 × P k .We can draw it in the 3D cubic grid of volume O(n) plus a number of additional lines to bend in each level and to move to the next, higher level.The first level is the grid of size consisting of two parallel slices of our cubic grid; see Fig. 7.The total number of lines involved is O(n 2/3 ).The same works for the graph C 3 × P k (= T k ).In addition to the standard 3D grid, here we need also its slanted, diagonal version (and, again, additional lines for bending in the cubic box of volume O(n)).The number of lines increases just by a constant factor.
Problem 29.Recall that, by Theorem 11, graphs of bounded degree can have linearly large parameter ρ 1 3 (G).If a planar graph G has bounded degree, does ρ 1 3 (G) = o(n)?We are able to determine the exact values of ρ 1 2 (G) for a complete bipartite graph G = K p,q whenever it is planar.
. The former equality is obvious.We have to prove for G = K 2,q that ρ 1 2 (G) = (3n − 7)/2 .Fig. 8a shows that ρ 1 2 (G) ≤ (n − 3)/2 + n − 2 = (3n − 7)/2 .It remains to show the lower bound ρ 1 2 (G) ≥ (3n − 7)/2 .Suppose that our bipartition is defined by 2 white vertices and q black vertices.Associate the graph G with its plane drawing.If there exist no line containing one white vertex and two black vertices of the graph G then we need m = 2n − 4 ≥ (3n − 7)/2 lines to cover all edges of the graph G. Assume from now that there exists a line containing a white vertex w and two black vertices b 1 and b 2 of the graph G. Then the vertex w lies on the line between the vertices b 1 and b 2 .Let w be the other white vertex.Since the point w sees all black points, no one of them can be placed inside the shaded area, see Fig. 8b.Then the point w cannot be an interior point of a segment between two black points.
Thus all lines which cover at least two edges of the graph G go through the point w, at most one of these lines go through the point w and the remaining lines can cover only the edges incident to the vertex w.Now let L be a family of lines such that |L| = ρ 1 2 (G) and each edge of the graph G belongs to some line ∈ L.
For any binary tree T , it holds ρ 1 2 (T ) = O( √ n log n).This follows from the known fact [7] that T has an orthogonal drawing on a grid of size O( otherwise.Indeed, since G has (n − 3)/2 vertices with degree 3, ρ 1 2 (G) > √ n − 3 by Lemma 9(a).
To obtain an upper bound, we prove by induction that we can draw the tree of height h on a m(h) × (m(h) + 1)-grid with the root placed at the top left corner (see also [21, Fig. 3(a)]).We can draw the tree of height 0 on a 0 × 0-grid and the tree of height 1 on a 1 × 1-grid, respectively, with the root placed at the top left corner.We can let m(2) = 2, m(3) = 4 and by induction we can let m(h + 2) = 2m(h) + 3 for each h ≥ 2 (we can easily show that m . Indeed, at the induction step, we place in m(h + 2) × (m(h + 2) + 1)-grid the root of the tree of height h + 2 on the top left grid point (0, 0), place its child nodes on points (1, 0) and (0, m(h) + 2), and place its grandchild nodes on points (m(h) + 3, 0), (1, 1), (1, m(h) + 2) and (0, m(h) + 3) (see Fig. 9).For each grandchild v, consider the intersection of m(h) + 1 gridlines to the left and the m(h) gridlines to the bottom of v (including the grid lines containing v).We reserve this part of the grid to the subtree of v.Note that every grid point is reserved to a most one grandchild due to the placement of the grandchild nodes.Since the subtree of a grandchild is a complete binary tree of height h, by induction, we can draw them inside each of reserved (m(h) + 1) × m(h)-parts of the grid with the roots placed at the top left corners.Thus all edges of the tree of height h belong to 2m(h) + 3 grid lines.Since n = 2 h+1 − 1, we have the claimed upper bounds.examples, note that π 1 2 (K 4 ) = 2 while π1 2 (K 4 ) = 3.In general, the gap is unbounded.In the following, we examine the gap for some interesting graph classes.
For any tree G, we have π 1 2 (G) ≤ 2 by Theorem 23.On the other hand, Felsner et al. [19], showed that π1 2 (G) ≥ log 3 (2n + 1) for every complete ternary tree G.We can show a much larger gap for graphs of vertex degree at most 3 with cycles.Beforehand, we need some preliminaries.
Let H be a plane graph, that is, a planar graph drawn without edge crossings in the plane.Removal of H splits the plane into connected components, which are called faces of H.We define H f to be the graph whose vertices are the faces of H; two faces are adjacent in H f if their boundaries have a common point.Note that this is not the same as the dual of G, which has an edge for each pair of faces that share an edge.For a planar graph G, let φ(G) = min H diam(H f ) where the minimum is taken over all plane representations H of G.
Proof.Let H be a drawing of G attaining the value r = π1 2 (G).The r underlying lines partition the plane into r + 1 parallel strips (including two half-planes).If a face of H intersects one of the bounded strips, then it is incident to a vertex lying in a line above this strip.This vertex is incident to another face intersecting a strip above this line.The same holds true also in the downward direction.It follows that from each face we can reach the outer face in H f along a path of length at most r/2 .Therefore, diam(H f ) ≤ 2 r/2 ≤ r.Proof.Consider the graph G = S k consisting of k = n/4 nested copies of C 4 connected as depicted in the drawing H k in Fig. 10.This drawing certifies that π 1 2 (S k ) = 2.Note that diam(H f k ) = k.In order to apply Lemma 36, we have to show that this equality holds true as well for any other drawing of S k .
Note that S k is 2-connected.We use general facts about plane embeddings of 2connected graphs; here we do not restrict ourselves to straight-line drawings only.A 2-connected planar graph G can have many plane representations, but all of them are obtainable from each other by a sequence of simple transformations.Specifically, let H be a plane version of G and C be a cycle in H containing only two vertices, u and v, that are incident to some edges outside C. We can obtain another plane embedding H of G by flipping G inside C, that is, by replacing the interior of C with its mirror version (up to homeomorphism).The rest of G is unchanged; in particular, u and v keep their location.It turns out [30,Theorem 2.6.8]that, for any other plane representation H 1 of G, H can be transformed into H 1 by a sequence of flippings that is followed, if needed, by re-assigning the outer face and applying a plane homeomorphism.
Let us apply this to G = S k and its plane representation H = H k .First we need to identify all cycles in H k for which the flipping operation is possible.Recall that such a cycle C is connected to its exterior only at two vertices u and v. Clearly, removal of these vertices disconnects the graph.This is possible only if u and v belong to two diagonal edges forming a centrally symmetric pair of edges.If the last condition is true for u and v, then an appropriate cycle C exists only if the pair {u, v} is centrally symmetric.It readily follows from here that flipping is possible only with respect to square cycles excepting the outer one.
Note now that, for each such cycle C, the interior of C is symmetric with respect to the axis passing through the corresponding vertices u and v.This implies that flipping of H k with respect to C does not change the graph H f k .Moreover, the flipped graph differs from H only by relabeling of vertices.Therefore, any further flipping also does not change H f k .Moreover, H f k stays the same up to isomorphism after re-assigning the outer face and applying a plane homeomorphism.We conclude that φ(S k ) = diam(H f k ) = k.
Lemma 38.For any graph G, it holds that π1 2 (G) ≤ area(G), where area(G) is the minimum number of vertices in a rectangular grid containing a stright-line drawing of G.
Di Battista and Frati [11] have shown that outerplanar graphs can be drawn straightline in area O(n 1.48 ), which yields the following corollary.

Figure 3 :
Figure 3: Combinatorial bounds for the numbers of K 3 's and K 4 's needed to cover K 7 and K 6 , respectively.

Lemma 16 .
Let d : V (K n ) → R 3 be a drawing of K n and L a geometric cover of d.No two different 4-vertex graphs G , G ∈ K L can have three common vertices.

2 3 (
K n+1 ) ≤ ρ 2 3 (K n ) + n/2 .(b)Let by the covering plane that contains exactly three vertices d(v), d(v ) and d(v ) of d(K n ).If we extend d to a drawing of the graph K n+1 by adding the endpoint d(v n+1 ) inside of the triangle with the vertices d(v), d(v ) and d(v ) then it suffices to cover by additional planes only n − 3 segments with a common endpoint d(v n+1 ), connecting it with vertices of d(V (K n )) \ .Theorem 18.For n ≤ 9, the value of ρ

6 .
Put a point d(u 7 ) inside of a triangle d(u 2 )d(u 5 )d(u 6 ) and a point d(u 8 ) inside of a triangle d(u 3 )d(u 5 )d(u 6 ) in the drawing of d(K 6 ) in Fig. 1 symmetrically with respect to the axis d(u 1 )d(u 4 ).Then, to cover all edges of the drawing d(K 8 ), it suffices to add the four planes of Fig. 1, an additional plane spanned by triangle d(u 1 )d(u 7 )d(u 8 ) and lines spanned by segments d(u 3 )d(u 7 ) and d

Example 26 .
There are triangulations with ρ 1 2 (G) = O( √ n) and segm(G) = Ω(n). 1 Note that this gap is best possible because any 3-connected graph G has minimum vertex degree 3 and, hence, ρ 1 2 (G) ≥ ρ 1 3 (G) > √ 2n by Lemma 9(a).Consider the pattern shown in Fig. 5.It is drawn in the standard orthogonal grid and two slanted grids, which implies that at most O( √ n) lines are involved.The pattern can be completed to a triangulation by adding three vertices around it and connecting them to the vertices on the pattern boundary.Since the pattern boundary contains O( √ n) vertices, O( √ n) new lines suffice for this.Thus, we have ρ 1 2 (G) = O( √ n) for the resulting triangulation G.Note that the vertices drawn fat in Fig. 5 have degree 5, and there are linearly many of them.This implies that segm(G) = Ω(n).

Figure 7 :
Figure 7: The graph C 4 × P k drawn into a 3D grid of linear volume on O(n 2/3 ) lines (the proof of Theorem 28(b)).

Figure 9 :
Figure 9: Drawing a complete binary tree of n nodes on an m 1 (n) × m 2 (n) grid.

Table 1 :
2 3 (K n ) is bounded by the numbers in Table 1.Lower and upper bounds for ρ 2 3 (K n ) for small values of n.