Min-k -planar Drawings of Graphs ⋆

. The study of nonplanar drawings of graphs with restricted crossing configurations is a well-established topic in graph drawing, often referred to as beyond-planar graph drawing . One of the most studied types of drawings in this area are the k -planar drawings ( k ≥ 1), where each edge cannot cross more than k times. We generalize k -planar drawings, by introducing the new family of min-k -planar drawings . In a min-k -planar drawing edges can cross an arbitrary number of times, but for any two crossing edges, one of the two must have no more than k crossings. We prove a general upper bound on the number of edges of min-k -planar drawings, a finer upper bound for k = 3, and tight upper bounds for k = 1 , 2. Also, we study the inclusion relations between min-k -planar graphs (i.e., graphs admitting min-k -planar drawings) and k -planar graphs.


Introduction
Beyond planarity [20,26] is a recent area of focus in graph drawing and topological graph theory, having its foundations established in the 1970s and 1980s.It comprises works on graphs that go beyond planar graphs in the sense that several, mostly local, crossing configurations are forbidden.The simplest are 1-planar graphs, where at most one crossing per edge is allowed [28,32], and their generalization k-planar graphs, where at most k ≥ 1 crossings per edge are tolerated [13,20,24,30,31].Other prominent examples of graph classes are fan-planar graphs [11,12,15,16,27], where several edges might cross the same edge but they should be adjacent to the same vertex, and k-gap-planar graphs (k ≥ 1) [8,9,10], where for each pair of crossing edges one of the two edges contains a small gap through which the other edge can pass, and at most k gaps per edge are allowed.Another popular family is the one of k-quasiplanar graphs, which forbids k mutually crossing edges [2,3,4,5,23].Mostly, edge density and inclusion relations of different beyond-planar graph classes have been studied [5,20,26].
In this paper we introduce a new graph family that generalizes k-planar graphs by permitting certain edges to have more than k crossings.Namely, for each two crossing edges we require that at least one of them contains at most k crossings.Formally, this graph family is defined as follows: Definition 1 A graph G is min-k-planar (k ≥ 1) if it admits a drawing on the plane, called mink-planar drawing, such that for any two crossing edges e and e ′ it holds min{cr(e), cr(e ′ )} ≤ k , where cr(e) and cr(e ′ ) are the number of crossings of e and e ′ , respectively.
Clearly, every k-planar drawing Γ is also min-k-planar, but not vice versa.An edge of the graph that crosses in Γ is a heavy edge if it crosses more than k times, otherwise it is a light edge.There are two main motivations behind the study of min-k-planar graphs: (i) From a theoretical perspective, when a graph is not k-planar we may want to draw it by allowing some heavy edges, whose removal yields a k-planar drawing.To this regard, if m is the total number of edges in the graph, we will prove that the number of heavy edges in a min-k-planar drawing is at most k 2k+1 • m, whose value varies in the interval [ m 3 , m 2 ) for k ≥ 1. (ii) From a practical perspective, even if a graph is k-planar, allowing (few) pairwise-independent heavy edges may reduce the visual complexity of the layout, even when the total number of crossings grows.For example, Figure 1 shows two drawings of the same portion of a graph.Despite the drawing in Figure 1(a) being 2-planar and having fewer crossings in total, the one in Figure 1(b) appears more readable; it is not 2-planar, but it is min-2-planar.is min-2-planar, is not 2-planar, and has 12 crossings; it contains two "heavy" edges incident to vertex 6, each with several crossings.− Despite the maximum density of min-k-planar graphs for k = 1, 2 equals the one of k-planar graphs, we show that 1-planar and 2-planar graphs are proper sub-classes of min-1-planar and min-2-planar graphs (as for k = 3).However, the min-1-planar graphs that can reach the maximum density of 4n − 8 are also 1-planar (i.e., the two classes coincide), while this is not true for k = 2, 3.
Section 2 introduces notation and terminology; final remarks and open problems are in Section 5.

Basic Definitions
We only deal with connected graphs.A graph is simple if it does not contain multiple edges and self-loops.A graph with multiple edges but not self-loops is also called a multi-graph.Let G be any (not necessarily simple) graph.We denote by V (G) and E(G) the set of vertices and the set of edges of G, respectively.A drawing Γ of G maps each vertex v ∈ V (G) to a distinct point in the plane and each edge uv ∈ E(G) to a simple Jordan arc between the points corresponding to u and v.We always assume that Γ is a simple drawing, that is: (i) two adjacent edges (i.e., edges that share a vertex) do not intersect, except at their common endpoint (in particular, no edge is self-crossing); (ii) two independent (i.e.non-adjacent) edges intersect at most in one of their interior points, called a crossing point; (iii) no three edges intersect at a common crossing point.
Let Γ be a drawing of G.A vertex of Γ is either a point corresponding to a vertex of G, called a real-vertex, or a point corresponding to a crossing point, called a crossing-vertex or simply a crossing.We remark that in the literature a plane graph obtained by replacing crossing points with dummy vertices is often referred to as a planarization [19].We denote by V (Γ) the set of vertices of Γ.An edge of Γ is a curve connecting two vertices of Γ.We denote by E(Γ) the set of edges of Γ.An edge e ∈ E(Γ) is a portion of an edge in E(G), which we denote by e; if both the endpoints of e are real-vertices, then e and e coincide.
Drawing Γ subdivides the plane into topologically connected regions, called faces.The boundary of a face consists of a cyclical sequence of vertices (real-or crossing-vertices) and edges of Γ.We denote by F (Γ) the set of faces of Γ. Exactly one face in F (Γ) corresponds to an infinite region of the plane, called the external face of Γ; the other faces are the internal faces of Γ.If the boundary of a face f of Γ contains a vertex v (or an edge e), we say that f contains v (or e).
In the following, if not specified, we denote by n = |V (G)| and m = |E(G)| the number of vertices and the number of edges of G, respectively.Degree of vertices and faces.For a vertex v ∈ V (G), denote by deg G (v) the degree of v in G, i.e., the number of edges incident to v. Analogously, for a vertex v ∈ V (Γ), denote by deg Γ (v) the degree of v in Γ.Note that, if v ∈ V (G) then deg Γ (v) = deg G (v), while if v is a crossing-vertex then deg Γ (v) = 4.For a face f ∈ F (Γ), denote by deg Γ (f ) the degree of f , i.e., the number of times we traverse vertices (either real-or crossing-vertices) while walking on the boundary of f clockwise.Each vertex contributes to deg Γ (f ) the number of times we traverse it (possibly more than once if the boundary of f is not a simple cycle).Also, denote by deg r Γ (f ) the real-vertex degree of f , i.e., the number of times we traverse a real-vertex of Γ while walking on the boundary of f clockwise.Again, each real-vertex contributes to deg r Γ (f ) the number of times we traverse it.Finally, deg c Γ (f ) denotes the number of times we traverse a crossing-vertex of Γ while walking on the boundary of f clockwise.Clearly, deg Beyond-planar graphs.A family F of beyond-planar graphs is a set of (nonplanar) graphs that admit drawings with desired or forbidden edge-crossing configurations [20].The edge density of a graph G ∈ F is the ratio between its number m of edges and its number n of vertices.Graph G is maximally dense if it has the maximum edge density over all graphs of F with n vertices.Graph G is optimal if it has the maximum edge density over all graphs in F. Note that F might not contain optimal graphs for all values of n (see, e.g., [20]).

Edge Density of Min-k-planar Graphs
We start by proving some general bounds on the number of crossings in a min-k-planar drawing and on the number of edges of min-k-planar graphs.
Property 1 Any min-k-planar drawing Γ of a graph G (with k ≥ 1) has at most k • ℓ crossings, where ℓ is the number of light edges of G in Γ.
Proof: Two heavy edges cannot cross, thus each crossing in Γ belongs to at least one light edge.Since each light edge has at most k crossings, the bound follows.□ Property 2 Let Γ be a min-k-planar drawing of an m-edge graph G (with k ≥ 1).The number of heavy edges of G in Γ is at most k 2k+1 • m.Proof: Let h and ℓ be the number of heavy edges and the number of light edges of G in Γ, respectively.Observe that m ≥ h + ℓ.By definition, each heavy edge contains at least (k + 1) crossings, and two heavy edges do not cross.Hence, the number of crossings in Γ is at least h • (k + 1).By Property 1, we have h □ We now give a general bound on the edge density of min-k-planar simple graphs, for any k ≥ 2. Finer bounds for k = 1, 2, 3 are given in the next sections.
Theorem 1 For any min-k-planar simple graph G with n vertices and m edges it holds m ≤ min{5.39 Hence, we prove that for 2 ≤ k ≤ 3 we have m ≤ 5.39 Suppose now that k ≥ 4 and let Γ be any min-k-planar drawing of G with ℓ light edges.Since no two heavy edges cross, the subgraph of G consisting of all heavy and crossing-free edges in Γ has at most 3n − 6 edges, hence m ≤ ℓ + 3n − 6.Let G ′ be the subgraph of G consisting of the ℓ light edges of G only, and let Γ ′ be the restriction of G ′ in Γ.We show that ℓ ≤ 3.81 √ k • n.The relation trivially holds when ℓ < 6.95n, as k ≥ 4. If ℓ ≥ 6.95n, using Ackerman's version of the Crossing Lemma applied to G ′ , we have cr 2 crossings, because each light edge has at most k crossings and each crossing is shared by two edges of G ′ .It follows that

Density of Min-1-planar Graphs
Let Γ be a min-1-planar drawing of a graph G.We color each edge of E(G) either red or green with the following rule: (i) edges that are crossing-free in Γ are colored red; (ii) if {e 1 , e 2 } ∈ E(G) is a pair of edges that cross in Γ, with cr(e 1 ) ≥ cr(e 2 ), we color e 1 as green and e 2 as red (if cr(e 1 ) = cr(e 2 ) = 1, the red edge is chosen arbitrarily).Note that, since Γ is a min-1-planar drawing, each red edge is crossed at most once, hence the above coloring rule is well-defined.In particular, heavy edges are always colored green, while if two light edges cross, one is colored green and the other is colored red.Hence, the subgraph induced by the red edges is a plane graph, called the red subgraph of G defined by Γ, or simply the red subgraph of Γ.
Lemma 1 Let G be a simple graph and let Γ be a min-1-planar drawing of G.We can always augment Γ with edges in such a way that the new drawing is still min-1-planar and all faces of its red subgraph have degree three.
Proof: Let Γ r be the red subgraph of Γ.Since G is simple, every face of Γ has degree greater than two.Suppose that Γ r has at least one face f such that deg Γr (f ) ≥ 4. We augment Γ with new red edges in two steps, described below.The augmentation may introduce multiple edges, but it will guarantee that all faces of the new red subgraph have degree three.
Step 1. Suppose that there exists a face f of Γ r with deg Γr (f ) ≥ 4 and containing two vertices u and v that can be connected by an edge uv that splits f without crossing other edges of Γ.We add edge uv and color it as red (as it is crossing-free); we also say that this operation augments f .We repeat this procedure until no such a face f exists.The obtained drawing is still min-1-planar, since we added only crossing-free edges.
Step 2. Suppose that Γ r still contains a face f with deg Γr (f ) ≥ 4. Observe that: (i) Face f is traversed by a green edge in Γ, otherwise it would have been augmented in Step 1; see Figure 2(a).(ii) Every green edge e that traverses f is not incident to any vertex u of f .Namely, suppose for contradiction that e is incident to a vertex u of f , and let e r = vw be the red edge of f crossed by e.Since deg Γr (f ) ≥ 4, at least one among v and w, say for example v, is not adjacent to u. However this implies that either f can be augmented by adding a red edge uv, which contradicts that we completed Step 1, or there is another green edge that crosses e r , which contradicts that e r is crossed at most once; see Figure 2(b).
(iii) Face f cannot be traversed by two distinct green edges e 1 and e 2 (refer to Figure 2(c)).More precisely, if this is the case, these edges cannot cross each other and, by property (ii), each of e 1 and e 2 crosses two distinct red edges of f .Also, since each red edge is crossed at most once, e 1 and e 2 cross two disjoint pairs of red edges of f .Denote by c 1 and c ′ 1 (resp.c 2 and c ′ 2 ) the two crossing points of e 1 (resp.e 2 ) with the boundary of f .Assume that c 1 , c 2 , c ′ 2 , c ′ 1 occur in this clockwise order on the boundary of f .This implies that, while moving clockwise on the boundary of f , there is at least one vertex u of f between c 1 and c 2 , and at least one vertex v of f between c ′ 2 and c ′ 1 .Hence, we can augment f with a red edge uv, which contradicts that we completed Step 1.By properties (i), (ii), and (iii), f is traversed by exactly one green edge e; however this edge cannot leave on the same side two vertices of f that are not consecutive on its boundary, as otherwise they would have been connected in Step 1. Hence f is a quadrilateral and e splits f into two equal parts (see Figure 2(d)).We can then augment f by adding a diagonal red edge in the quadrilateral face.We repeat this procedure until Γ r contains no face f such that deg Γr (f ) ≥ 4. □ We now prove a tight bound on the edge density of min-1-planar graphs.
Theorem 2 Any n-vertex min-1-planar simple graph has at most 4n − 8 edges, and this bound is tight.
Proof: Let Γ be a min-1-planar drawing of a simple graph G with n vertices.By Lemma 1, we can augment Γ (and hence G) with new edges, in such a way that the new drawing Γ ′ (and the corresponding graph G ′ ) is min-1-planar and its red subgraph Γ ′ r is a triangulated planar graph.Hence, Γ ′ r has exactly 3n − 6 edges and 2n − 4 faces.Every green edge of G ′ (which is also a green edge of G) traverses at least two faces of Γ ′ r .Also, since Γ ′ is a min-1-planar drawing and the red subgraph has only triangular faces, each face of the red subgraph is crossed by at most one green edge.Hence the number of green edges is at most About the tightness of the bound, we recall that optimal 1-planar graphs with n vertices (which are also min-1-planar) have 4n − 8 edges [17,31,32].□ Plugging the bound of Theorem 2 into the bound of Property 2, we immediately get that a min-1-planar simple graph has at most 4  3 n − 8 3 heavy edges in any of its min-1-planar drawings.We considerably improve this bound in the next theorem.
Theorem 3 Let G be an n-vertex min-1-planar simple graph and let Γ be a min-1-planar drawing of G.There are at most 2  3 n − 1 heavy edges of G in Γ.Further, there exist min-1-planar drawings that contain 2  3 n − O(1) heavy edges.
Proof: Let Γ be a min-1-planar drawing of a simple graph G with n vertices.As in the proof of Theorem 2, by Lemma 1 we can augment Γ with new red edges, in such a way that the new drawing Γ ′ is min-1-planar and its red subgraph Γ ′ r has all faces of degree three.Hence, Γ ′ r has exactly 3n − 6 edges and 2n − 4 faces.Clearly, the number of heavy edges of G in Γ ′ is not smaller than the number of heavy edges of G in Γ.By definition, every heavy edge of G in Γ ′ is crossed at least twice, hence it traverses at least three faces of Γ ′ r .As before, each face of the red subgraph is crossed by at most one heavy edge.Hence the number of heavy edges is at most 2n−4 3 ≤ 2 3 n − 1.For the lower bound, consider a min-1-planar drawing constructed as follows.Start from a pentangulation P on n vertices, that is, an n-vertex planar drawing with all faces of degree five.Then, in each face of P , add two light edges incident to the same vertex and one heavy edge that crosses these two edges.Refer to Figure 3.By Euler's formula, P contains 2 3 n − 4 3 faces, and therefore 2  3 n − 4 3 heavy edges.□

Density of Min-2-planar Graphs
Proving a tight bound on the edge density of min-2-planar graphs is more challenging than for min-1-planar graphs.Observe that there are min-2-planar simple graphs with 5n − 10 edges, namely the optimal 2-planar graphs [13].Each optimal 2-planar drawing consists of a subset of planar edges forming faces of size five (i.e., pentagons), and each face is filled up with five more edges that cross each other twice.In the following we prove that 5n − 10 is also an upper bound to the number of edges of min-2-planar graphs.To this aim, for any k ≥ 1, we introduce a class of multi-graphs that generalize min-k-planar simple graphs.
Let G be a (multi-)graph (without self-loops) and let Γ be a (simple) drawing of G.A set of parallel edges of G between the same pair of vertices is called a bundle of G.We say that Γ is bundle-proper if for every bundle in G: (i) at most one of the edges of the bundle is involved in a crossing; and (ii) Γ has no face bounded only by two edges of the bundle (i.e., no face of Γ is a 2-real bigon).We remark that, in the literature, two parallel edges that form a face of degree two are called homotopic.Hence, property (ii) is equivalent to saying that a bundle-proper drawing does not contain homotopic parallel edges.
Graph G is bundle-proper min-k-planar if it admits a (simple) drawing Γ that is both min-kplanar and bundle-proper.If G has n vertices and has the maximum number of edges over all bundle-proper min-k-planar n-vertex graphs, then we say that G is a maximally-dense bundleproper min-k-planar graph.Consider a pair (G, Γ), where G is an n-vertex bundle-proper mink-planar graph and Γ is a bundle-proper min-k-planar drawing of G.We say that (G, Γ) is a maximally-dense crossing-minimal bundle-proper min-k-planar pair if G is maximally-dense and Γ has the minimum number of crossings over all bundle-proper min-k-planar drawings of maximallydense bundle-proper min-k-planar n-vertex graphs.
Lemma 2 Let (G, Γ) be a maximally-dense crossing-minimal bundle-proper min-k-planar pair.These properties hold: (a) If a face f of Γ contains two distinct real-vertices u and v, then f contains an edge uv.
Proof: We prove the three properties separately.(a) Suppose for contradiction that f does not contain an edge uv.Then we can add (another copy of) edge uv to Γ (and therefore to G) in the interior of f , without introducing any additional crossings or creating a 2-real bigon.Since the resulting drawing is bundle-proper min-k-planar, this contradicts the hypothesis that G is maximally-dense.(b) Let f be a face of Γ.Since G has no self-loops and Γ is a simple drawing, deg Also, since Γ is simple, f is neither a 0-real bigon nor a 1-real bigon.Finally, since Γ is also bundle-proper, f cannot be a 2-real bigon.It follows that deg then there would be two non-consecutive real vertices on the boundary of f that are not connected by an edge, which is impossible by (a).Then f is necessarily a 3-real triangle.□ To prove the upper bound we use discharging techniques.See [1,2,14,21] for previous works that use this tool.Define a charging function ch : The value ch(f ) is called the initial charge of f .Using Euler's formula, it is not difficult to see that the following equality holds (refer to [2] for details): The goal of a discharging technique is to derive from the initial charging function ch(•) a new function ch ′ (•) that satisfies two properties: If α > 0 is a number for which a function ch ′ (•) satisfies (C1) and (C2), by Eq. ( 2) we get: Thus, Eq. ( 3) can be exploited to prove upper bounds on the edge density of a graph for specific values of α, whenever we find a charging function ch ′ (•) that fulfills (C1) and (C2).We prove the following.
Theorem 4 Any n-vertex min-2-planar simple graph has at most 5n − 10 edges, and this bound is tight.

Proof:
We already observed at the beginning of this section that there exist min-2-planar simple graphs with 5n−10 edges (e.g., the optimal 2-planar).It remains to prove that min-2-planar simple graphs cannot have more than 5n−10 edges.Since any simple graph is also a bundle-proper graph, we can show that the upper bound holds more in general for multi-graphs that are bundle-proper min-2-planar.Also, since we want to find an upper bound on the number of edges, we can restrict our attention to maximally-dense bundle-proper min-2-planar graphs, and in particular to those having the minimum number of crossings.Let (G, Γ) be any maximally-dense crossing-minimal bundle-proper min-2-planar pair, with |V (G)| = n.We show the existence of a charging function ch ′ (•) that satisfies (C1) and (C2) for α = 2 5 , so the result will follow from Eq. ( 3).Consider the initial charging function ch(•) defined in Eq. (1).For each type of triangle t we analyze the value of ch(t) and the deficit/excess w.r.t. 2  5 deg r Therefore ch(•) only fails to satisfy (C1) at 0-real and 1-real triangles.We begin by setting ch ′ (f ) = ch(f ) for each face f of Γ and we explain how to modify ch ′ (•) in such a way that ch ′ (f ) ≥ 2 5 deg r Γ (f ) for each face f ∈ F (Γ), thus satisfying (C1), and the total charge remains the same, thus satisfying (C2).
Fixing 0-real triangles.Let t be a 0-real triangle in Γ with edges e 1 , e 2 , and e 3 .Refer to Figure 4.The edges e 1 , e 2 and e 3 are three pairwise crossing edges of G. Since Γ is a simple drawing, e 1 , e 2 and e 3 are independent edges of G (i.e., their six end-vertices are all distinct).Also, since Γ is min-2-planar, at least two of these three edges, say e 2 and e 3 , do not cross other edges of G in Γ.This implies that each of the two end-vertices of e 2 shares a face with an endvertex of e 3 .Hence, by Lemma 2(a), the four vertices of e 2 and e 3 form a 4-cycle e ′ e 2 e ′′ e 3 in G The charge of f 1 is ch ′ (f 1 ) = 2, with an excess of 6 5 w.r.t.
We reduce ch ′ (f 1 ) by 4  5 , reduce ch ′ (f 2 ) by 1 5 , and increase ch ′ (t) by 1.After that, the total charge is unchanged and all the three faces t, f 1 , and f 2 satisfy (C1).Namely, ch ′ (t) = 0 (it has no deficit/excess), ch ′ (f 1 ) = 6 5 (it has an excess of ), and ch ′ (f 2 ) = 4 5 (it has no deficit/excess).In the remainder of the proof, since we need a way to keep track of the 2-real triangles and 2-real quadrilaterals whose charge has been modified as described above, we call each of the faces f 1 and f 2 a 0-real triangle-neighboring face.Each 0-real triangle-neighboring face that is a 2-real triangle (as f 2 ) shares its unique crossing-vertex with a 0-real triangle; each 0-real triangle-neighboring face that is a 2-real quadrilateral (as f 1 ) shares its unique 0-real edge with a 0-real triangle.
Fixing 1-real triangles.Let t be a 1-real triangle, with real-vertex v 1 and crossing-vertices v 2 and v 3 .Refer to Figure 5 for an illustration.Let e 0 = v 2 v 3 be the 0-real edge of t, and let f 1 be the face of Γ that shares e 0 with t.If f 1 is a 0-real quadrilateral, denote by e 1 the 0-real edge of f 1 not adjacent to e 0 , and by f 2 the face of Γ that shares e 1 with f 1 .If f 2 is a 0-real quadrilateral, denote by e 2 the 0-real edge of f 2 not adjacent to e 1 , and by f 3 the face of Γ that shares e 2 with f 2 .We continue in this way until we encounter a face f p (p ≥ 1) that is not a 0-real quadrilateral.This procedure determines a sequence of faces f 0 , f 1 , f 2 , . . .f p , and a sequence of 0-real edges e 0 , e 1 , . . ., e p−1 such that f 0 = t, f i is a 0-real quadrilateral for each i ∈ {1, . . ., p − 1}, f p is not a 0-real quadrilateral, and the faces f i and f i−1 share edge e i−1 (i ∈ {1, . . ., p}).
Note that deg Γ (f p ) ≥ 4. Namely, let e = v 1 v 2 and e ′ = v 1 v 3 , and let e = v 1 u and e ′ = v 1 w be the edges of G that contain e and e ′ .Since f p has at least two crossing-vertices, if f p were a triangle then it would be either a 0-real triangle or a 1-real triangle.If f p were a 0-real triangle then e and e ′ would cross in Γ, which is impossible as e and e ′ are adjacent edges and Γ is a simple drawing.If f p were a 1-real triangle then u = w, i.e., e and e ′ would be parallel edges both involved in a crossing, which is impossible as Γ is bundle-proper.
Therefore, deg Γ (f p ) ≥ 4 and, as already observed at the beginning of this proof, ch ′ (f p ) ≥ 2 5 deg r Γ (f p ). Also, the charge excess of f p is larger than 2 5 .Namely, the charge excess of If f p has no real-vertices, then it must have at least five crossing-vertices (because f p is not a 0-real quadrilateral), which implies x ≥ 1 > 2 5 .If f p has at least one real-vertex then x ≥ 3 5 > 2 5 .Hence, since the charge excess of f p is larger than 2 5 , the idea is to fill the 2 5 charge deficit of t by moving an equivalent amount of charge from f p to t.We say that t demands from f p through edge e p−1 a charge amount of 2  5 .We call f 0 , . . ., f p (which is a path in the dual of Γ) the demand path for t.Therefore, for each 1-real triangle t of Γ whose demand path ends at a face f = f p , we decrease ch ′ (f ) by 2  5 and increase ch ′ (t) from 0 to 2 5 .Note that f cannot be a 0-real triangle-neighboring face.Indeed, f is not a triangle, and if f is a 2-real quadrilateral then its 0-real edge is shared either with a 0-real quadrilateral or directly with the 1-real triangle t.It follows that the set of faces whose charge is affected by fixing 1-real triangles does not intersect with the set of faces whose charge is affected by fixing 0-real triangles.
Due to the considerations above, after we have fixed all 1-real triangles, we may have problems only if multiple 1-real triangles demanded from the same face f .In this case, f might no longer satisfy (C1).In the remainder of the proof, we analyze which types of faces may be in this situation and, if so, we prove how to fix their charge.
Fixing faces that received multiple demands from 1-real triangles.Let f be a face of Γ of degree larger than three that received multiple demands from 1-real triangles.This is possible only if f has more than one 0-real edge, hence we can exclude that f is a 2-real quadrilateral.Note that, by Lemma 2(c), each face of Γ contains at most three real-vertices.If deg Γ (f ) ≥ 7 then f still satisfies (C1) even if it received a demand through each of its deg Γ (f ) edges when fixing 1-real triangles.Indeed, in the worst case, the new charge of f is ch , f has at least one real-vertex).Indeed, in this case, the number of 0-real edges of f is at most deg so f received at most this number of demands from 1-real triangles.Hence, in the worst case, the new charge of f is ch It follows that the only faces that may have received multiple demands from 1-real triangles and that (after we have fixed all 1-real triangles) no longer satisfy (C1) are the 1-real quadrilaterals, the 0-real pentagons, and the 0-real hexagons.Each face f of one of these types has at least two adjacent 0-real edges.If f no longer satisfies (C1), we show how to find extra charges that can be moved from some suitable faces with charge excess towards f , so to compensate the charge deficit of f .To this aim, we first prove the following claim; refer to Figure 6.
Claim 1 Let f be a face of Γ and let e 1 , e 2 , e 3 , e 4 be consecutive edges on the boundary of f for which a demand is made through both e 2 and e 3 .Let t 1 be the 1-real triangle that demanded from f through e 2 and let v 1 = e 1 ∩ e 3 be the real-vertex of t 1 .Analogously, let t 2 be the 1-real triangle that demanded from f through e 3 and let v 2 = e 2 ∩ e 4 be the real-vertex of t 2 .Then there is a curve C that begins in f , leaves f through the crossing-vertex common to e 2 and e 3 , passes through a sequence of zero or more 1-real triangles and 1-real edges, and ends in a face f * that is either a 2-real triangle containing v 1 and v 2 or a 2-real quadrilateral containing only one of v 1 and v 2 .
Proof: Observe that the closed region ∆ 123 bounded by e 1 , e 2 , e 3 does not contain any vertex of G other than v 1 since if it did, t 1 would be making a demand from some face other than f .Similarly, the closed region ∆ 234 bounded by e 2 , e 3 , e 4 contains no vertices of G.
We can construct a curve C that begins in f , passes through the crossing-vertex z of Γ common to e 2 and e 3 , and then enters the face f ′ opposite f at z. From the interior of f ′ the curve C then crosses a sequence of zero or more 1-real edges incident to v 2 and passes through zero or more 1-real triangles that contain v 2 until reaching some face f * that contains v 2 and is not a 1-real triangle.One of the following must occur: 1.The face f * contains v 1 (see Figure 6(c)).In this case Lemma 2(a) implies that f * also contains the 2-real edge v 1 v 2 .The crossing-minimality of Γ implies that f * is a 2-real triangle that contains v 1 and v 2 .(Otherwise e 3 has more than one crossing on the boundary of f * and could be rerouted to avoid all but one of these crossings.)2. The face f * has degree larger than three (see Figure 6(d)).Then f * contains an edge e of Γ that is not incident to v 1 and e crosses e 3 .No endpoint of e is in ∆ 123 , e does not cross e 2 , and since Γ is simple, e has only one crossing with e 3 , so e must cross e 1 .Since e 3 already crosses e 2 and e 4 , this implies that e has no additional crossings.Therefore one end-vertex v of e belongs to f * .By Lemma 2(a), the edge v 2 v is on f * .The crossing minimality of Γ then implies that f * is a 2-real quadrilateral that contains v 2 .This completes the proof of the claim.■ Let f be a 1-real quadrilateral, a 0-real pentagon, or 0-real hexagon with edges e 1 , e 2 , e 3 , e 4 that satisfy the conditions of the claim and let f * be the face whose existence is established by the claim.In each such case, we move a charge of 1  5 from f * to f .Based on the claim, there are two cases to consider: 1.The face f * is a 2-real triangle that contains v 1 and v 2 (see Figure 6(c)).Let x be the crossing-vertex of f * .Then the face g that shares x with f * but has no edge in common with f * is either a 0-real quadrilateral or it coincides with f (because g is one of the faces of the demand path for t 1 ending at f ).In particular, g is not a 0-real triangle, which implies that ch ′ (f * ) was not modified when fixing 0-real triangles.Therefore ch ′ (f * ) = 1 immediately after fixing 0-real triangles.The charge on 2-real triangles is never modified when fixing 1real triangles, hence ch ′ (f * ) = 1 even after fixing all 1-triangles.Since we reduce the charge of f * by 1 5 and increase the charge of f by 1  5 , we can think of this charge travelling along the suffix of the demand path t 1 ⇝ f that begins at g; we also say that the charge leaks out of f * through x.We show that charge leaks out of f * through x at most once.This is obviously the case if f * and f share the vertex x = z (i.e., g = f ).The only other possibility is that the charge leaks out of f * into the 0-real quadrilateral g that is part of another demand path t ⇝ f ′′ , with t ̸ = t 1 ; refer to Figure 7. Let e and e ′ be the two edges of g other than e 1 and e 3 .Then t is the 1-real triangle that contains v 2 and whose 1-real edges are portions of e and e ′ .Each of e and e ′ crosses e 1 and e 3 .Possibly e 2 ∈ {e, e ′ } but we can assume without loss of generality that e ̸ = e 2 .Therefore the edge e 3 crosses e 2 , e 4 , and e, for a total of at least 3 crossings.Hence, neither e nor e ′ is involved in any additional crossings, which implies that the face next to g on the demand path t ⇝ f ′′ contains end-vertices of e and e ′ , i.e., this face coincides with f ′′ .Hence, since Γ is simple (which excludes that e and e ′ cross), deg r Γ (f ′′ ) ≥ 2. It follows that f ′′ is neither a 1-real quadrilateral, nor a 0-real pentagon, nor a 0-real hexagon.Since the charge that leaks out of f * through x is always left at a 1-real quadrilateral, or at a 0-real pentagon, or at a 0-real hexagon, we conclude that charge leaks out of f * at x at most once.2. The face f * is a 2-real quadrilateral (see Figure 6(d)).In this case, f * has only one 0-real edge, shared with a 0-real quadrilateral.Again, this implies that ch ′ (f * ) was not modified when fixing 0-real triangles.Hence, immediately after fixing 0-real triangles we have ch ′ (f * ) = 2.
Since we reduce ch ′ (f * ) by 1  5 and increase ch ′ (f ) by 1  5 , we can again think of this as a charge of 1  5 leaking from f * through a crossing-vertex x of f * and then travelling down a suffix of the demand path t 1 ⇝ f .With the same reasoning as above, this can happen at most once for each of the two crossing-vertices of f * .Therefore, the total charge that leaves through these two vertices of f * is at most 2 5 .To summarize the discussion above, each face f * can give a charge of 1  5 for each of its crossingvertices, and after that it still satisfies (C1).In the following we call f * a supporting face.To complete the proof, we have to show that if f is either a 1-real quadrilateral, or a 0-real pentagon, or a 0-real hexagon, and if f received multiple demands from 1-real triangles, then f always finds a suitable number of supporting faces to satisfy (C1).We analyze separately the three different types of categories for f , and assume that f received more than one demand from some 1-real triangles.
If f is a 1-real quadrilateral, then it received exactly two demands of 2  5 , through its two consecutive 0-real edges.In this case, a charge of 1  5 leaks into f from one supporting face.Hence we have ch If f is a 0-real pentagon, the following cases are possible: f received exactly two demands.We have ch , thus f satisfies (C1) without needing supporting faces.
f received exactly three demands.Two of these demands necessarily occur through two consecutive edges of f , so a charge of at least 1 5 leaks into f from a supporting face.Therefore ch f received exactly four demands.There are three pairs of consecutive edges of f at which the demands occur, so a total charge of 3 5 leaks into f from three supporting faces.Therefore ch f received exactly five demands.There are five pairs of consecutive edges at which the demands occur, so a total charge of 5 5 = 1 leaks into f from five supporting faces.Therefore ch If f is a 0-real hexagon, we have two cases.If f received at most five demands, then ch Otherwise, a total charge of at least 6 5 leaks into f from six supporting faces, thus we have ch Hence, in both cases f satisfies (C1).
In conclusion, at the end of the discharging process, the new function ch ′ (•) satisfies (C1) for all faces of Γ, and the total charge is the same as the initial total charge, that is, ch(•) satisfies (C2).This completes the proof.
□ Proof: By definition, all heavy edges of G in Γ have at least three crossings with light edges.From Γ, we derive a graph G − and a corresponding drawing Γ − by removing all the heavy edges of G in Γ.Then we remove all light edges that have two crossings in Γ − and we call Γ ′ the resulting drawing.Note that the light edges that originally crossed the heavy edges are not removed in this phase.Drawing Γ ′ is 1-planar.All the faces of Γ ′ describe cyclic sequences of real-vertices and crossing-vertices, and in each sequence we do not have two subsequent crossing-vertices, as this would mean two crossings on a light edge.
Claim 2 Γ ′ can be augmented by adding light edges to get a 1-planar drawing Γ + whose faces are all triangles.Also, the drawing Γ * consisting of Γ + plus the heavy edges of G in Γ is still min-2-planar.
Proof: The augmentation of Γ ′ with the new light edges is done by considering the heavy edges of Γ.If u and v are vertices of two distinct 1-real edges with a crossing c, then we can add an edge between u and v that will have at most two crossings with heavy edges e and e ′ in Γ * ; see Figure 8(a).In this way, we can guarantee that the remaining faces that are not triangles have only real vertices.We now iteratively show how to split each non-triangular face f to complete the construction of Γ + .If f is not traversed by a heavy edge, we can triangulate it instantly.If f is traversed by a heavy edge e that is incident to a vertex u of f and crosses the boundary of f through the edge vv ′ , we can add an edge uv that has at most one crossing with a heavy edge e ′ in Γ * ; see Figure 8(b).Note that instead of uv we can choose uv ′ , if u and v are already adjacent.Assume now that no heavy edge e traversing f is incident to any vertex of f .Let u, u ′ , v, v ′ be vertices of f and uu ′ , vv ′ the edges that are crossed by e. Observe that uu ′ and vv ′ might have a common vertex (e.g., u ′ ≡ v ′ ).Then we can add the edge uv as it has at most two crossings with heavy edges e ′ , e ′′ in Γ * ; see Figure 8(c).Note again that instead of uv we may choose u ′ v ′ , if u and v are already adjacent.If u ′ and v ′ are also adjacent or they coincide, then deg In this case, since Γ is min-2-planar, there are at most four heavy edges traversing f and every possible configuration of these heavy edges allow adding a light edge between two vertices of f traversing at most two of these heavy edges in Γ * ; see Figures 8(d) to 8(f).Finally, observe that the edges added in Γ ′ to achieve Γ + can only cross heavy edges of G in Γ, hence Γ + is 1-planar.■ Since Γ + is 1-planar, it contains neither 0-real triangles nor 1-real triangles.On the other hand it may contain 2-real and 3-real triangles.Also, observe that a heavy edge in Γ * can cross several light edges in Γ + (see for example Figure 9(a)).The next two claims show some properties of 2-real and 3-real triangles of Γ + with respect to the heavy edges reinserted in Γ * .
Claim 3 Let f be a 2-real triangle in Γ + .The following properties hold: (i) If f is a start face for a heavy edge e (i.e., e is incident to a real-vertex of f ), then e is the only heavy edge traversing f in Γ * .(ii) At most two heavy edges traverse f in Γ * .(iii) If two heavy edges traverse f , then the face f ′ that shares only the crossing-vertex with f is not traversed by a heavy edge in Γ * .
Proof: (i) This is clear as one edge of f has already two crossings and a heavy edge through the other edges of f would imply two crossing heavy edges.(ii) Because of (i), we know that this is true if a heavy edge starts at f .Otherwise, three or more traversing heavy edges would imply at least six extra crossings with the border of f and this contradicts that Γ * is min-2-planar.(iii) Both 1-real edges of f belong to edges of the graph that also contain 1-real edges of f ′ .Because of (i), both heavy edges do not start at f and therefore the border of f is crossed four times.So the 1-real edges of f already have each two crossings and no heavy edge can traverse f ′ (see Figure 9 a light edge, there can be at most one heavy edge other than e that traverses f .(ii) The previous case implies that at most two heavy edges can traverse f if one of them has f as a starting face.
On the other hand, if a heavy edge traverses f and does not have f as a starting face, it causes two crossings along the boundary of f .Hence, since Γ * is min-2-planar, there can be at most three heavy edges of this type.■ Thanks to the claims above, each 2-real triangle of Γ + that is traversed by some heavy edges in Γ * is partitioned into either two or three faces of Γ * .Also each 3-real triangle of Γ + that is traversed by some heavy edges in Γ * is partitioned into two, three, or four faces of Γ * .Refer to Figure 10 for an illustration of all the cases.To count the maximum number of heavy edges in Γ * (and therefore in Γ), we apply a variant of the discharging technique given in Theorem 4. This variant is based on assigning some of the initial charge of the faces of Γ * , as defined in Eq. ( 1), to the edges of Γ * , without changing the overall charge, which is equal to 4n − 8 based on Eq. ( 2).Recall that a heavy edge of G in Γ * is partitioned into several portions (which correspond to edges of Γ * ).In particular, since a heavy edge e has at least three crossings, it is formed by at least four portions: two of these portions are 1-real edges of Γ * , which we call end-segments of e, and the others are 0-real edges of Γ * , which we call intermediate-segments of e.For example, the heavy edge in Figure 9(a) has two end-segments (those splitting the faces f 1 and f 5 ) and three intermediate-segments (those splitting the faces f 2 , f 3 , and f 4 ).For each heavy edge e, we assign charge 1 to each of its end-segments, by subtracting the same amount of charge from one of the faces of Γ * incident to the end-segment.Note that for each end-segment, there always exists a face of Γ * incident to it that has an initial charge equal to 1 (see Figures 10(a), 10(c) and 10(d)); hence, this subtraction of charge does not cause a negative charge in a face of Γ * .Also, for each intermediate-segment of e, we assign to this segment a charge of 2  3 , by subtracting the same amount of charge from one of the faces of Γ * incident to the intermediate-segment.Note that in all cases except for the case of 2-real triangles crossed by two heavy edges (see Figure 10(b)) each face of Γ * has always a sufficient amount of initial charge to support all the intermediate-segments of a heavy edge that are incident to it (see Figures 10(e) to 10(g)).In the last unsolved case of a 2-real triangle f that is crossed by two heavy edges, we apply Claim 3(iii) to the face f and therefore use the 1 charge of the face f ′ which uniquely corresponds to f (see Figure 9(b)).Hence, again, this subtraction of charge does not cause a negative charge in a face of Γ * .At the end of this assignment process, each heavy edge is assigned a charge of at least 1 + 1 + 2 3 + 2 3 = 10 3 and no faces of Γ * have a negative charge.Hence, denoted by h the number of heavy edges in Γ * , we have h 10  3 ≤ 4n − 8, which implies h ≤ 6 5 (n − 2).For the lower bound, consider a planar drawing consisting of n−2

Density of Min-3-planar Graphs
For the family of min-3-planar graphs we consider graphs that can contain non-homotopic parallel edges.Indeed, it is known that n-vertex 3-planar graphs that are simple have at most 5.5n − 15 edges [31], but this bound is not tight.On the other hand, a tight upper bound is known for 3-planar graphs that can contain non-homotopic multiple edges, namely 5.5n − 11 [29].Note that later we will present a class of min-3-planar graphs with 17 3 (n − 2) edges, cf.Theorem 10, which exceeds the upper bound for 3-planar graphs.Here, we give corresponding upper bounds on the edge density and on the density of the heavy edges in min-3-planar graphs.The proofs still exploit discharging techniques.
Theorem 6 Any min-3-planar graph with n vertices has at most 6n − 12 edges.
Proof: As in the proof of Theorem 4 we can assume a restriction to maximally-dense crossingsminimal bundle-proper min-3-planar pairs (G, Γ) with |V (G)| = n vertices, and we will use a discharging technique to prove the statement.Our discharging function ch ′ (•) is similar to that in [1], where the same bound for 4-planar graphs was proven, but the details differ, as here edges with more than four crossings can exist.We can assume that Γ is 2-connected and hence that the boundary of each face f of Γ is a simple cycle.Indeed, Ackerman proved that if Γ is not 2-connected it always has no more than 6n − 12 edges [1, Proposition 2.1].Although Ackerman concentrates on 4-planar graphs, he does not use this hypothesis to show this fact, thus his argument works also in our case.
We introduce the discharging steps of ch ′ (•) and show that all faces satisfy (C1) and (C2) for α = 1  3 , which implies with Eq. ( 3) the desired bound on the edge density.This is relatively difficult to see for 0-real pentagons, so the major part of the proof is reserved for these.For that we distinguish different cases depending on the structure of the graph near a 0-real pentagon.
Before we can write down the discharging function, we introduce some definitions.Let x be a crossing-vertex in a drawing Γ.We call the faces f and f ′ vertex-neighbors, if both their boundaries contain x but not a common edge e ∈ E(Γ).Recall the definition of demand path in the proof of Theorem 4 for 1-real triangles.We naturally extend the definition of demand path for arbitrary If f 1 is a 1-real quadrilateral, then t 4 is a 2-real quadrilateral and therefore ch ′ 4 (f ) ≥ 0, in case one of t 3 and t 5 is not a triangle.If both are triangles, one of them -say without loss of generality t 3 -is a 1-real triangle as otherwise e 1 and e 2 would have more than three crossings (see Figure 12(a)).This implies that t 3 receives 1 3 from the 2-real triangle f 3 in Step 2 and so ch ′ 4 (f ) ≥ 0.
If f 1 is a 2-real pentagon, then f 0 and f 2 can both not be 0-real pentagons, because then e 1 and e 2 would have more than three crossings (see Figure 12(b)).So f 1 contributes its excess of at least 4 3 in Step 4 only to f and therefore ch ′ 4 (f ) ≥ 0.
Assume now that without loss of generality e 0 has four or more crossings (see Figure 12(c)).Then e 1 and e 2 have exactly three crossings and f 0 and f 2 are 2-real quadrilaterals, which contribute each their excess of at least 2 3 to at most two faces in Step 4. So f receives 2 • 1 3 and therefore ch ′ 4 (f ) ≥ 0.
Then e 2 has four or more crossings and so e 0 and e 4 have exactly three crossings.So f 4 is a 2-real triangle and contributes in Step 2 and Step 4 in total 1 3 to f and its demand-path-neighbors.If f has four or less demand-path-neighbors that are 0-triangles or 1-triangles, then already ch ′ 4 (f ) ≥ 0. Assume that f has five such demand-path-neighbors and t 0 , t 2 , t 4 are 1-real triangles (see Figure 13) as otherwise we can refer to case 1.
If f 1 is a 2-real quadrilateral, then it contributes its excess of at least 2 3 to f in Step 4 and so ch ′ 4 (f ) ≥ 0. If f 1 is a 1-real quadrilateral, then it has an excess of 1  3 , because it contributes no charge in Step 1 to f 2 .This excess is contributed in Step 4 to only f , so ch ′ 4 (f ) ≥ 0. For deg Γ (f 1 ) ≥ 5 the excess of f 1 after Step 3 is at least because f 1 contributes through the 1-real edges in total at most 1  3 , it contributes no charge to f 2 and deg r Γ (f 1 ) ≥ 1.This is distributed over at most deg Γ (f 1 )−3 faces as f 1 has at most deg Γ (f 1 )−1 vertex-neighbors and does not contribute charge to f 0 and t 2 .For deg Γ (f 1 ) ≥ 5 the equation holds.This implies ch ′ 4 (f ) ≥ 0.
Case 3. The demand-path-neighbor t i is a 0-real triangle and the demand-pathneighbor t j , j ∈ {(i + 1) mod 5, (i − 1) mod 5} is a 1-real triangle with no 0-real quadrilaterals in its demand path.Fix i = 1 and j = 2.
Assume e 0 has more than three crossings.So e 2 has exactly three crossings.Then f 2 is a 2-real triangle contributing 1 3 to f and its demand-path-neighbors (see Figure 14(a)).So we can assume that all demand-path-neighbors of f are 0-real or 1-real triangles (otherwise ch ′ 4 (f ) ≥ 0) and t 0 , t 2 , t 3 , t 4 are 1-real triangles (otherwise we can refer to case 1-2).Note that if the demand path of t 4 contains no 0-real quadrilateral, then f 3 is a 2-real triangle and therefore ch ′ 4 (f ) ≥ 0. If it contains one or more 0-real quadrilaterals, then t 4 receives at least 1  6 by Proposition 1(e).Also f receives at least 1 6 from f 0 , as we can see with the following argument: If f 0 is a 1-real quadrilateral, then t 0 receives 1 3 charge from a 2-real triangle in Step 2. If f 0 is a 2-real quadrilateral, then f receives the excess of f 0 in Step 4, which is at least 2  3 .If f 0 has more than four vertices, then it distributes not less than deg vertex-neighbors and so f receives at least 1  6 charge in Step 4. So in total ch ′ 4 (f ) ≥ 0. Assume e 0 has exactly three crossings.Then f 1 is a 2-real quadrilateral, which contributes 1   6   in Step 2 to t 2 and has an excess of at least 2 3 after Step 3, which is distributed over at most two faces in Step 4 (see Figure 14(b)).If there is a demand-path-neighbor of f , which is not a 0-real or 1-real triangle, then ch ′ 4 (f ) ≥ 0. So we assume that t 0 , t 3 , t 4 are 1-real triangles (otherwise we can refer to case 1 or 2). Figure 14: Case 3: (a) If e 0 has more than three crossings, then f 2 contributes charge to t 2 and t 3 .Also charge is moved to t 4 and from f 0 to f .(b) If e 0 has exactly three crossings, then f 1 contributes charge to f and t 2 .If this is not 2 3 , then charge is moved to t 3 or from f 2 to f .
If now f 0 is not a 0-real pentagon, then f 1 contributes its excess in Step 4 only to f and ch ′ 4 (f ) ≥ 0 follows.If f 0 is a 0-real pentagon, then e 2 has more than three crossings.So if the demand path of t 3 contains one or more 0-real quadrilaterals, then t 3 receives at least 1  6 in by Proposition 1(e) and ch ′ 4 (f ) ≥ 0. If the demand path of t 3 contains no 0-real quadrilaterals, then f 2 is a 2-real triangle that contributes 1  6 both to t 2 and t 3 and so ch ′ 4 (f ) ≥ 0.
Case 4. The demand-path-neighbor t i is a 0-real triangle and the demand-pathneighbor t j , j ∈ {(i + 1) mod 5, (i − 1) mod 5} is a 1-real triangle with one or more 0-real quadrilaterals in its demand path.Fix i = 1 and j = 2.
Assume e 0 has more than three crossings.So e 1 and e 2 have exactly three crossings.It follows that f 0 is a 2-real quadrilateral with an excess of at least 2 3 after Step 3, so it contributes at least 1  3 to f in Step 4 (see Figure 15(a)).So if four or less demand-path-neighbors of f are 0-real or 1-real triangles, then ch ′ 4 (f ) ≥ 0. Otherwise we assume that all demand-path-neighbors except t 1 are 1-real triangles as with another 0-real triangle we can refer to case 1 or 2. But then t 0 is a 1-real triangle without a 0-real quadrilateral in its demand path and we can refer to case 3.
Assume e 0 has exactly three crossings and t 0 is a 1-real triangle.If the demand path of t 0 has no 0-real quadrilaterals, then we refer to case 3. Otherwise e 1 has more than three crossings and so by applying Proposition 1(e) both t 0 and t 2 receive each 1  6 charge in Step 2 (see Figure 15(b)).So if not all demand-path-neighbors of f are 0-real or 1-real triangles, then ch ′ 4 (f ) ≥ 0. So assume now the opposite and that t 3 and t 4 are 1-triangles (otherwise we can refer to case 1-2).Because e 0 and e 2 already have three crossings, t 3 and t 4 have no 0-real quadrilaterals in their demand paths.So f 3 is a 2-real triangle contributing 1  6 to each t 3 and t 4 .Therefore ch ′ 4 (f ) ≥ 0. Assume e 0 has exactly three crossings and t 0 is not a 1-real triangle.Then we can assume f 3 is a 0-real quadrilateral: Then f 2 and f 4 are 1-real triangles and therefore t 3 and t 4 receive both 1  6 from a 2-real triangle in Step 2. Further they receive also 1  6 from a 2-real quadrilateral in Step 2 and so ch ′ 4 (f ) ≥ 0. f 3 is a 1-real quadrilateral: Then t 3 and t 4 receive both 1  3 charge in Step 2 from a 2-real triangle and so ch ′ 4 (f ) ≥ 0. deg Γ (f 3 ) ≥ 5: We rename for this purpose the faces in the neighborhood of f := f 3 so that we denote for i ∈ {0, 1, . . ., deg Γ ( f )} by ẽi the edges of the boundary of f in clockwise order and we introduce further ẽ1 so that ẽ1 = e 3 , t1 := f 4 and so on.The demand-path-neighbors t1 and t2 are 1-real triangles, which receive both 1  6 from a 2-real triangle.If further t3 is a 1-real triangle, then it receives (together with t4 , if t4 is a 1-real triangle) 1  3 charge from a 2-real triangle.The same is true for t0 (together with tdeg Γ (f3)−1 ).So the excess of f 3 is at least and this is distributed over at at most deg Γ (f 3 ) − 4 faces.So for deg Γ (f 3 ) ≥ 5 the face f 3 contributes at least 1  6 to each of its vertex-neighbors and this implies with the other charges ch ′ 4 (f ) ≥ 0.
Summary Since we have shown for all 0-real pentagons f that we have ch ′ (f ) ≥ 0 in all cases, we know that (C1) holds for all faces w.r.t.α = 1 3 .Together with Eq. ( 3) and the observation that (C2) is true, Theorem 6 follows.□ Theorem 7 Let G be an n-vertex min-3-planar graph and let Γ be a min-3-planar drawing of G.
There are at most 2(n − 2) heavy edges of G in Γ.Further, there exist min-3-planar drawings that contain 6 5 n − 12 5 heavy edges.
Proof: Denote by h the number of heavy edges of G in Γ and by ℓ the number of light edges of G in Γ.To prove the upper bound, we first prove the following claim.
Proof: The proof uses a discharging technique.Namely, assume that each light edge is initially assigned charge 1, so that the total charge equals ℓ.We show how to move some charge from light edges to heavy edges, in such a way that: (i) each light edge keeps a non-negative charge; (ii) each heavy edge receives at least 2 charge.If we are able to do that, then the claim holds, as we have 2h ≤ ℓ, i.e., h ≤ ℓ 2 .In the first step, we observe that since light edges have at most three crossings, and heavy edges have at least four crossings (with light edges), each heavy edge may receive at least 4 × 1 3 charge, namely 1  3 from each crossing with the corresponding light edge.In a second step, we assign at least 1  3 charge to each of the two end segments of each heavy edge: Let e be a heavy edge (a, b) with end segments (a, x) and (y, b) where x and y denote the first and last crossings of e.We consider only the first segment, the other can be done analogously.Let e ′ = (c, d) be the light edge that crosses e at the first crossing x.Clearly, e ′ might be crossed by at most two other (potentially heavy) edges.We consider the edge (a, c) that starts at a, closely follows along the heavy edge e until the crossing x and then it follows the other edge e ′ to the JGAA, 28( 2 vertex c.The edge (a, d) has an analogous route.Note that these two edges (a, c) and (a, d) can be crossed by at most two edges in total (see Figure 18(a)), and by the assumption of maximality they do exist.
The edges (a, c) and (a, d) will transfer half of their remaining charge to the (at most three) heavy edges that start at a, that pass through the sector defined by the two edges and that cross e ′ .The other half of their charge might be used for a potential sector at the other side of the edges.We will distinguish three cases, depending on the number of heavy edges that start at a and will receive charges from (a, c) and (a, d) in this step.
If there is only one such heavy edge, i.e. e, we know that there is at least 4 3 charge left at the light edges (a, c) and (a, d) and half of that amount can be moved to the heavy edge e.This fulfills properties (i) and (ii).
In the case, that there are two heavy edges that start at a and have their first crossing with edge e ′ , we have at most one other heavy edge that might cross (a, c) or (a, d).Hence those two light edges still have 5  3 charge left and half of it can be assigned to the two heavy edges with the start segments between a and edge e ′ .In this case, both such heavy edges receive 5  12 > 1 3 charge, still fulfilling (i) and (ii).
In the last case, where we have three such heavy edges, we conclude that the edges (a, c) and (a, d) are not crossed at all (see Figure 18(b)).Hence we can assign half of their charge to those three heavy edges that start at a.They all receive 1  3 charge from the edges (a, c) and (a, d), as required by (i) and (ii).
For the lower bound, consider the graph H in Figure 19(a), which is a min-3-planar with 7 vertices and 6 heavy edges (in green).Construct a graph G obtained by composing in parallel a number p of subgraphs isomorphic to H, with poles u and v, as shown in Figure 19(b).The graph G consists of n = 5p + 2 vertices and h = 6p heavy edges.Hence h = 6 5 n − 12  5 .□

Relationships with k-planar Graphs
The next theorem shows that while the family of min-1-planar graphs properly contains the family of 1-planar graphs, the two classes coincide when we restrict to optimal graphs, i.e., those with  Theorem 8 1-planar graphs are a proper subset of min-1-planar graphs, while optimal min-1planar graphs are optimal 1-planar.
Proof: Any 1-planar graph is min-1-planar.By the NP-hardness of testing whether a given planar graph plus a single edge is 1-planar [18], we know that there are such graphs that are not 1-planar, while any planar graph that is extended by a single edge can be drawn min-1-planar.Hence, 1-planar graphs are a proper subset of min-1-planar graphs.Finally, as in the proof of Theorem 2, in every optimal min-1-planar drawing the red subgraph is maximal planar and each green edge traverses exactly two faces of the red subgraph.Hence, each green edge crosses exactly once, i.e., the drawing is also (optimal) 1-planar.□ Unlike min-1-planar graphs, we show that min-2-planar graphs are a proper superset of the 2-planar graphs even when we restrict to optimal graphs.Theorem 9 2-planar graphs are a proper subset of min-2-planar graphs, and there are optimal min-2-planar graphs that are not optimal 2-planar.
Proof: We first observe that there exist non-optimal min-2-planar graphs that are not 2-planar.For example, K 5,5 is not 2-planar [6], while Figure 20  where black edges have no crossings, orange edges have 1 crossing, blue edges have 2 crossings, green edges have 3 crossings, and the red edge has 4 crossings.In the following, we show how to construct optimal n-vertex min-2-planar graphs that are not 2-planar.Let G be the truncated icosahedral graph and let Γ be a planar drawing of G, as depicted in Figure 21(a).This drawing has 12 pentagonal faces, 20 hexagonal faces, 60 vertices and 90 edges.We enrich Γ by adding 5 edges inside each pentagonal face and 7 edges inside each hexagonal face.Denote the obtained graph and the obtained drawing as G ′ and Γ ′ , respectively.Γ ′ is depicted in Figure 21(b), where the edges inside pentagonal faces are colored orange and the edges inside hexagonal faces are colored blue.More precisely, for each pentagonal face we add an edge between each pair of vertices of the face that are not connected.For each hexagonal face f , we add 7 edges as follows; refer to Figure 21(b) for an illustration, where the vertices of f are denoted as u, v, w, x, y, and z .We add an edge between each pair of vertices having distance two on the boundary of f .Additionally, we arbitrarily choose two vertices having the maximum distance on the boundary of f (w and z in Figure 21(b)) and we add an edge between them, which we call the diagonal of f .Note that the end-vertices of the diagonal of f have degree 5 in f .All the diagonals are dashed in Figure 21 (iii) for each hexagonal face f of Γ, the two edges that cross the diagonal (bold in Figure 21(b)) have three crossings each, while the other edges added inside f have two crossings each; (iv) no two edges with three crossings cross each other.This implies that G ′ is optimal min-2-planar and Γ ′ is a min-2-planar drawing of G ′ .
To show that Γ ′ is not optimal 2-planar, we exploit a property on the degree distribution of optimal 2-planar graphs from [22], which states that the degree of each vertex of an optimal 2-u v planar graph is a multiple of three.In what follows, we show that G ′ contains vertices whose degree is not a multiple of three.
Each vertex of G belongs to the boundary of two hexagonal faces and one pentagonal face, and it has degree 3. We now show that each vertex of G ′ has degree 9, or 10, or 11, and that not all of them have degree 9. Let u be a vertex of G ′ whose degree is 9 (see, e.g., vertex u in Figure 21(b)).Since u belongs to the boundary of two hexagonal faces and one pentagonal face in G, and it has degree 9, it cannot be incident to any diagonals (otherwise it would have degree larger than 9).This implies that at least one of the vertices that are adjacent to u in G, call it v, is the end-vertex of at least one diagonal in G ′ .Two cases are possible: (a) v is the end-vertex of one diagonal; (b) v is the end-vertex of two diagonals.In case (a), v has degree 10 (see, e.g., vertex v in Figure 21 □ In contrast to 1-planar and 2-planar graphs, the maximum densities of 3-planar and min-3planar graphs differ. Theorem 10 There are min-3-planar (non-simple) graphs denser than optimal 3-planar (nonsimple) graphs.
Proof: First, consider a planar graph G, and a corresponding drawing Γ, consisting of h parallel chains (h ≥ 1), each with 8 vertices, sharing the two end-vertices u and v, and interleaved by h copies of edge uv; refer to Figure 22(a).Then, construct a new graph G ′ , and a corresponding drawing Γ ′ , obtained from G, and from Γ, by replacing each parallel chain with a copy of the graph G ′′ depicted in Figure 22(b).In the drawing Γ ′ , each copy of G ′′ has the same edge crossings as the drawing illustrated in Figure 22(b).Graph G ′′ has 8 vertices and 33 edges, and it is min-3planar.Indeed, only four edges in the drawing of G ′′ shown in the figure have more than three crossings and they do not cross each other (see the bold edges in Figure 22(b)).It follows that G ′ is min-3-planar; also it has n = 6h + 2 vertices and m = 33h + h = 34h edges.Since h = n−2 6 , we have m = 17  3 n − 34 3 = 5.6n − 11.3.Therefore, m > 5.5n − 11 for every n > 1 2 .Since a 3-planar graph has at most 5.5n − 11 edges [29], G ′ is not 3-planar.□ In this paper, we focused on simple drawings.A natural open research direction is to study mink-planar graphs of non-simple drawings.We remark that some considerations for this setting have recently been presented by P. Hliněnỳ [25].
About edge density, one can ask whether the bound of Theorem 6 for min-3-planar graphs is tight or if it can be further lowered.Providing finer bounds for k ≥ 4 is also interesting.Another classical research direction is to establish inclusion or incomparability relations between min-kplanar graphs and classes of beyond-planar graphs other than k-planar graphs.In the following, we state two lemmas providing initial results in this direction.
Lemma 3 leaves open to establish the relationship between min-2-planar graphs and 1-gapplanar graphs (which have the same maximum edge density).We recall that in a k-gap-planar drawing (k ≥ 1) it is possible to assign each crossing to one of the two edges that form it, in such a way that no more than k crossings are assigned to the same edge; see, e.g., [8,9,10].Also, in a k-quasiplanar drawing (k ≥ 3) there is no k mutually (pairwise) crossing edges; see, e.g., [2,3,4,5,23].
Lemma 3 Min-k-planar graphs are a subset of k-gap-planar graphs and of (k + 2)-quasiplanar graphs, for every k ≥ 1.
Proof: Let Γ be a min-k-planar drawing, for any given k ≥ 1.Each crossing in Γ involves at least one light edge, hence the set of light edges covers all crossings in Γ.Consider any light edge e and assign each crossing of e to e.Then, consider a second light edge e ′ and assign all unassigned crossings of e ′ to e ′ .Iterate this procedure until all crossings have been assigned to some light edge; refer to Figure 23(a) for an example.Since each light edge has at most k crossings, no more than k crossings are assigned to a single edge.Hence Γ is k-gap-planar.
We now prove that Γ is also (k + 2)-quasi planar.Suppose by contradiction that this is not the case.This means that Γ contains k + 2 mutually crossing edges.Since no two heavy edges cross, at least k + 1 ≥ 2 of these edges are light edges.But each of them cross k + 1 times, a contradiction.□Lemma 4 implies that min-2-planar graphs and fan-planar graphs are incomparable classes, even if they have the same maximum edge density.We recall that in a fan-planar drawing there cannot be two independent edges that cross a third one; see, e.g., [11,12,15,16,27].Lemma 4 For any given k ≥ 2, fan-planar and min-k-planar graphs are incomparable, i.e., each of the two classes contains graphs that are not in the other.
Proof: The existence of min-k-planar graphs that are not fan-planar is an immediate consequence of the fact that there exist 2-planar graphs that are not fan-planar [15].To show the existence of fan-planar graphs that are not min-k-planar, consider the graph K 1,3,h , for any h ≥ 1, and let n = 4 + h be its number of vertices.It is easy to see that this graph is fan-planar (see, e.g., [15] and Figure 23(b)).Also, it is known that any drawing of K 1,3,h has Ω(h 2 ) = Ω(n 2 ) crossings [7].On the other hand, by Theorem 1, any min-k-planar drawing with n vertices has at most c √ k • n edges (for a constant c) and therefore, by Property 1, it has at most ck 1.5 n crossings.Hence, K 1,3,h is not min-k-planar for sufficiently large values of n.

Figure 1 :
Figure 1: Two drawings of the same portion of a graph: (a) is 2-planar and has 10 crossings; (b) is min-2-planar, is not 2-planar, and has 12 crossings; it contains two "heavy" edges incident to vertex 6, each with several crossings.

3 n 2 .
95n, the relation m ≤ 5.39 √ k • n trivially holds, as 5.39 √ k • n ≥ 7.63n.If m ≥ 6.95n, let cr(G) be the minimum number of crossings required by any min-k-planar drawing Γ of G.The improved version by Ackerman of the popular Crossing Lemma (Theorem 6 in [1]) implies that cr(G) ≥ 1 29 m If ℓ is the number of light edges of G in Γ, by Property 1 we have cr

Figure 2 :
Figure 2: Augmentation of Γ as described in Step 2 of the proof of Lemma 1. Edges that can be added to Γ are represented as dashed red segments.

Figure 4 :
Figure 4: (a) A 0-real triangle t.(b) A 2-real quadrilateral f 1 and a 2-real triangle f 2 neighboring t.(c) The initial charges.(d) The charges after a redistribution.

Figure 5 :
Figure 5: The demand path for a 1-real triangle t, ending at a face f p .

Figure 6 :
Figure 6: A supporting face f * for a face f that receives charge demands through two consecutive 0-real edges of its boundary (e 2 and e 3 in the figure).

Figure 7 :
Figure 7: A supporting face f * supports at most one face f .

Figure 8 :Theorem 5
Figure 8: Illustration for the proof of Claim 2. Green edges are heavy edges, solid red edges are light edges of Γ ′ , and dashed edges are light edges that can be added to Γ ′ to get Γ + .

Figure 9 :
Figure 9: Heavy edges crossing 2-real and 3-real triangles.(a) f 1 and f 5 are start faces for the heavy edge e.(b) If f is traversed by two heavy edges, then f ′ is traversed by no heavy edges.

Claim 4 Figure 10 :
Figure 10: Illustration for the proof of Theorem 5: Different faces of Γ * .Light edges are red and heavy edges are green.Each face is labeled with its initial charge.

2Figure 11 :
Figure 11: (a) Construction for the lower bound of Theorem 5. (b) Addition of edges in the middle and the external face and (c) in all other faces of the drawing of Figure 11(a).

Figure 12 :
Figure 12: Case 1: If e 0 and e 3 have exactly three crossings, then f 1 is a (a) 1-real quadrilateral or (b) a 2-real pentagon.(c) If e 0 has more than three crossings, then f receives charge from f 0 and f 2 .(Here and in the following figures, edges that exist in some subcases are dotted and edges with more than three crossings are bold.)

Figure 13 :
Figure 13: Case 2: If t 1 and t 3 are 0-real triangles, then f 4 is a 2-real triangle.f receives also charge from f 1 .

Figure 18 :
Figure 18: The situation at the end segments of heavy edges in the proof of Theorem 7. (a) In total, at most two edges cross (a, c) and (a, d).(b) If three heavy edges start at a and cross e ′ , then (a, c) and (a, d) are planar.

Figure 19 :
Figure 19: Construction for the lower bound of Theorem 7.

Figure 21 :
Figure 21: (a) A planar drawing Γ of the truncated icosahedral graph G. (b) A min-2-planar drawing Γ ′ of the graph G ′ , obtained by adding 5 edges to each pentagonal face and 7 edges to each hexagonal face of Γ.

Figure 22 :
Figure 22: Illustration of the construction of Theorem 10.If in the graph of figure (a) we replace each shaded chain with a copy of the graph of figure (b), we get a min-3-planar graph that is not 3-planar.The bold edges are heavy edges.

□ 32 CFigure 23 :
Figure 23: (a) A 2-gap-planar drawing of the min-2-planar drawing of K 5,5 shown in Figure20; heavy edges are colored green; each crossing is assigned to a light edge (red edge), and it is represented as a small gap.(b) A fan-planar drawing of K 1,3,h , which is not min-k-planar as shown in the proof of Lemma 4.