On Area-Universal Quadrangulations

We study drawings of plane quadrangulations such that every inner face realizes a prescribed area. A plane graph is area-universal if for every assignment of non-negative weights to the inner faces, there exists a straight-line drawing such that the area of each inner face equals the weight of the face. It has been conjectured that all plane quadrangulations are area-universal. We develop methods to prove areauniversality via reduction to the area-universality of related graphs. This allows us to establish area-universality for large classes of plane quadrangulations. In particular, our methods are strong enough to prove area-universality of all plane quadrangulations with up to 13 vertices.


Introduction
A plane graph is a planar graph together with a crossing-free drawing. Let G be a plane graph and let F be its set of inner faces. An area assignment is an assignment of a non-negative real number to every face f ∈ F , i.e., a function A : F → R + 0 . A (potentially degenerate) planar straight-line drawing D of G realizes the area assignment A if for every f ∈ F the area of f in D is A(f ). A plane graph G is area-universal if it has a realizing drawing for every area assignment A.
Ringel [18] considered straight-line drawings of plane graphs such that all faces have the same area. He gave an example of a plane triangulation that has no equiareal drawing, hence, a triangulation which is not area-universal. Thomassen [19] proved that plane cubic graphs are area-universal. Biedl and Velázquez [3] showed area-universality for the class of plane 3-trees, also known as stacked triangulations and Apollonian networks. Concerning counter examples, Kleist [12,13]  a simple counting argument to show that no Eulerian triangulation is area-universal. Moreover, she showed that every plane graph is area-universal in the class of drawings where one bend per edge is allowed. For triangulations with special vertex orders, Kleist [14] presented a sucient criterion for their area-universality that only requires the investigation of one area assignment. Interestingly, if the sucient criterion applies to one plane triangulation, then all embeddings of the underlying planar graph are also area-universal. Dobbins et al. [6] investigated the complexity of deciding whether a given graph is area-universal and several related problems. They conjecture that the problem is complete for the complexity class ∀∃R.
In this paper, we focus on plane bipartite graphs. Because the property of area-universality is preserved under edge-deletions (see also Observation 1), we consider edge-maximal plane bipartite graphs known as quadrangulations. Regarding the area-universality of quadrangulations little is known. Evans et al. [9] showed that the m × n grid is area-universal for all m, n ≥ 2, even with the additional requirement that the outer face of the drawing is a rectangle. Kleist [12] showed that 2-degenerate quadrangulations are area-universal and that in the class of drawings where one bend per edge is allowed all quadrangulations have realizing drawings for all area assignments where only half of the edges have a bend.
The study of drawings in various drawing modes with prescribed face areas is summarized under the name cartograms. Cartograms date back to at least 1934 when Raisz [17] studied rectangular population cartograms, where the US population was visualized by representing the states with areas proportional to their population. This kind of visualization is particularly useful when showing geo-referenced statistical data in order to provide insight into patterns, trends and outliers in the world around us [22]. Cartograms have been intensely studied for duals of triangulations and rectilinear drawings with bends. The number of sides of the polygons representing a face has been improved in a series of papers from 40 sides [5], to 34 sides [11], to 12 sides [2]. Finally, Alam et al. [1] showed how to construct drawings with 8-sided faces, which is known to be optimal. Chang and Yen [4] studied contact representations of 2-connected outerplaner graphs and construct contact representations with 4-gons of prescribed area. Note that in the cartogram literature the problem is usually treated in the dual setting, i.e., weights are assigned to the vertices. We refer to Nusrat and Kobourov [16] for a survey of the cartogram literature. Area-universality has also been studied in the context of rectangular layouts, these are dissections of a rectangle into rectangles with prescribed contacts between the rectangles of the dissection. Eppstein et al. [8] showed that a rectangular layout is area-universal if and only if it is one-sided.
The key ingredient in their proof is that the weak equivalence class of any rectangular layout is area-universal. The weak equivalence class is obtained by prescribing the contacts between the segments. The area-universality of the weak equivalence class has been shown by dierent techniques [7,10,21]. This area-universality result is very special because, up to ane transformations, the rectangular layout realizing a given area assignment is actually unique.
The class of drawings: We study area-universality of plane quadrangulations. To realize nonnegative face areas, we extend the set of planar straight-line drawings of a plane quadrangulation by all drawings which can be obtained as the limit of a sequence of planar straight-line drawings (e.g. specied by the coordinates of the vertices). In particular, we allow degenerate drawings in which vertices and edges sharing a face may (partially) coincide; if two edges partially coincide their union forms a segment.
In various cases, considering this enriched set of drawings allows for simpler proofs [9,13,19]. For example, the counting argument by Kleist [13] greatly benets from allowing face area 0 and degenerate drawings. In the case of triangulations, degeneracies occur if and only if some face has area 0 and the set of realizable area assignments is closed [ Outline of this paper: In Section 2, we investigate operations that preserve area-universality. In Section 3, we use one of these operations, the edge contraction, to show area-universality of grids and large classes of angle graphs. In particular, we consider angle graphs of triangulations that are close to being area-universal. In Section 4, we study strong area-universality, i.e., area-universality within a prescribed outer face. Strong area-universal graphs may serve as building blocks for constructing area-universal quadrangulations. We show that not every plane bipartite graph is strongly area-universal and present families of strongly area-universal graphs. Shape restrictions are also the subject of Section 5 where we study convex drawings. We present both a large family of quadrangulations that are not convex area-universal and examples of strongly convex areauniversal graphs. In Section 6, we use our tools to show area-universality of all quadrangulations with at most 13 vertices. In some cases the argument relies on the known area-universality of the class of double stacking graphs.

Area-Universality Preserving Operations
We begin with an easy observation which can also be found in [3] and [13].
Observation 1 A subgraph of an area-universal plane graph is area-universal.
Therefore, a proof for the area-universality of plane quadrangulations, i.e., maximal plane bipartite graphs, would imply area-universality of all plane bipartite graphs. The following lemma extends Observation 1 with a new operation. A set of edge contractions in a plane graph G is face-maintaining if the contractions do not change the number of faces in G, i.e., for a face of degree d at most d − 3 edges are contracted. Lemma 1 Let G be a plane graph that can be transformed into an area-universal plane graph G by inserting vertices, inserting edges, and performing face-maintaining edge contractions. Then G is area-universal.
Proof: Let A denote an area assignment of G. A face f in G corresponds to a (non-empty) collection of faces C f in G . We dene A such that for each inner face f of G it holds that Simply deleting all vertices and edges of G which are not in G yields a (degenerate) drawing D of G. By denition of A , D is A-realizing.
There exists a further operation that preserves area-universality and is based on decomposition. For an illustration consider Figure 1. From a plane graph G with a simple cycle C, we obtain two plane graphs G i and G e by decomposing along C: G i is the subgraph of G consisting of C and its interior, while G e is the subgraph of G consisting of C and its exterior. Reversely, we obtain G from G i and G e by identifying the outer face of G i with the inner face of G e whose boundary is C.  Figure 1: Decomposing G along C yields two plane graphs G e and G i . If G e is area-universal and G i strongly area-universal, then G is area-universal.
i.e., ΣA := f ∈F A(f ). Since all triangles are ane equivalent, a plane graph with a triangular outer face has a realizing drawing (if it exists) within every triangle of correct area [13]. It follows that: Observation 2 Plane graphs with triangular outer faces (e.g. triangulations) are area-universal if and only if they are strongly area-universal.
A similar result for quadrangulations would be a pleasant surprise.
Lemma 2 Let G be a plane graph with a simple cycle C, and G i and G e obtained by decomposing G along C. If G e is area-universal and G i is strongly area-universal, then G is area-universal. Moreover, if G e is strongly area-universal, then G is also strongly area-universal.
Proof: Let A be an area assignment of G. For i ∈ {i,e}, A i denotes the induced area assignment of G i . Note that the interior of C is a face f of G e . Particularly, it holds that A e (f ) = ΣA i . Because G e is area-universal, there exists an A e -realizing drawing D e of G e . Since G i is strongly area-universal, we nd an A i -realizing drawing D i of G i whose outer face is the polygon representing C in D e . Thus, identifying D e and D i along C yields an A-realizing drawing of G.
The ideas of this lemma have been used in [13] to show the strong area-universality of 2degenerate quadrangulations. Recall that a graph is k-degenerate if and only if every subgraph contains a vertex of degree at most k.
It is easy to see that K 4 is area-universal, i.e., a vertex of degree 3 can be inserted into a triangle so that the three small triangles partition the big triangle in any prescribed ratio. This yields the following lemma. Lemma 4 Let T be a plane graph and T + the plane graph where a vertex of degree 3 is inserted (stacked) into a triangle of T . Then T is area-universal if and only if T + is area-universal.
Since a plane 3-tree is obtained from a triangle by iteratively stacking vertices into faces, Lemma 4 yields the result from [3]: Plane 3-trees are area-universal. 3 Area-Universality via Edge Contractions In this section, we discuss some implications of the edge contractions of Lemma 1.
Firstly, we show an alternative proof for the area-universality of grid graphs. The grid G(m, n) is the Cartesian product P m ×P n of paths P m and P n on m and n vertices, respectively. Figure 2(a) illustrates the grid G(9, 6). Area-universality of grid graphs was rst proved in the context of table cartograms where additionally the outer face is required to be a rectangle [9]. Our new proof does not yield a rectangular outer face; however, it is straight-forward and very simple. The reader is invited to check that Proposition 3 does not imply the area-universality of grid graphs, because G(m, n) is not a subgraph of a 2-degenerate quadrangulation if n, m ≥ 3. Then, we add vertices and edges to enhance the resulting graph to a triangulation G as depicted in Figure 2(b). The graph G is a stacked triangulation since the interior of each triangle (i, i + 1, C) with 1 ≤ i ≤ k −1, and the graph induced by A, B, C, 1, . . . , k is a stacked triangulation. Thus, G is area-universal. Therefore, every grid graph can be transformed into a subgraph of an area-universal graph using face-maintaining edge contractions. Hence, by Lemma 1, grids are area-universal.
The angle graph of a plane graph G is the graph Q G with vertex set consisting of the vertices and faces of G and edges corresponding to face-vertex incidences. If G is 2-connected, then Q G is a quadrangulation. Clearly, an angle graph is bipartite where the two bipartition classes are the vertices V and the faces F of G. In the following we consider angle graphs of triangulations.
For a plane graph G and its angle graph Q, their union (graph) G + Q, consists of the union of the vertex and edge sets of G and Q. Note that the union is again a plane graph: Indeed, the vertex set of G + Q coincides with the vertex set of Q. Hence, G + Q can be understood as the quadrangulation Q together with the edges between the vertices of one bipartition class, namely V . Proposition 6 The angle graph Q of an area-universal triangulation T is area-universal.
Proof: The graph T + Q can be seen as T where a vertex of degree 3 is inserted in every face. By Lemma 4, T + Q is area-universal. Thus, Observation 1 implies that Q is area-universal.
Note that the same approach shows that angle graphs of equiareal triangulations are equiareal. Moreover, a straightforward consequence of Proposition 6 is that angle graphs of stacked triangulations are area-universal.
In order for its angle graph to be area-universal, it suces for a triangulation to be close to areauniversal. As shown in [12], every plane graph has an area-universal subdivision. The subdivision number s(G) of a plane graph G is the minimum number of subdivision vertices to be inserted into G such that it becomes area-universal. If G is area-universal, then clearly s(G) = 0. To generalize Proposition 6, we introduce the notion of a rened area assignment. If G 1 is a subgraph of G 2 , then every face f of G 1 corresponds to a collection of faces C f in G 2 . An area assignment Theorem 7 The angle graph Q of a plane triangulation T with s(T ) ≤ 1 is area-universal. Proof: Figure 3(a) illustrates the proof. Let e be an edge of T such that subdividing e yields the area-universal graph T • . The strategy is as follows: For an area assignment A of Q, we dene a rening area assignment A of the union U :  For the denition of A , note that every face f of Q corresponds to two faces in U . Let q e denote the face of Q that is split by the subdivided edge e in U . We arbitrarily partition the area A(q) assigned to face q between the two corresponding faces in U , for all faces q in Q except the four faces q 1 , q 2 , q 3 , and q 4 sharing a boundary edge with q e . Let f 1 and f 2 denote the two faces adjacent to e in T . For q i , 1 ≤ i ≤ 4, assign area A(q i ) to the triangular face from U which is neither incident to f 1 nor to f 2 in U ; assign an area of 0 to the triangular faces incident to f 1 or f 2 . This denes the area assignment A and A • is the area assignment of We call the highlighted (thick, red) edges in Figure 3(b) incident to f 1 and f 2 , the red edges of f 1 and f 2 , respectively. By denition of A , the red edges must be contracted in every A -realizing drawing of U . Consequently, given a drawing T • , we easily insert f 1 and f 2 at the same location as the already placed vertex of the red edges, respectively. This yields an A -realizing drawing of U . Since A renes A, deleting the edges of T • yields an A-realizing drawing of Q.
This ties in with a result based on an operation called diamond addition. Let G be a plane graph and e an edge incident to two triangular faces consisting of e and the vertices u 1 and u 2 , respectively. Applying a diamond addition of order k on edge e of G results in a graph G in which the edge e is subdivided by vertices v 1 , . . . , v k which are also adjacent to u 1 and u 2 , as illustrated in Figure 4. Figure 4: A diamond addition on edge e.
Two diamond additions are disjoint if the partitioned triangles are dierent. For instance, in the left graph of Figure 5, diamond additions on the edges Au and vw are disjoint, while diamond additions on the edges Au and uv are not. Together with Theorem 7, the following theorem implies that the angle graphs of area-universal triangulations on which one diamond addition has been applied are area-universal. Theorem 8 (Kleist [15], Theorem 33) Let G be a graph obtained from an area-universal graph G by (multiple disjoint) diamond addition(s) adding k vertices in total. Then, s(G) ≤ k.
As a special case of graphs obtained by diamond additions, Kleist [14] studied double stacking graphs H ,k that can be obtained from the plane octahedron graph. Labeling the octahedron as in Figure 5, H ,k is obtained by applying one diamond addition of order − 1 on Au and one diamond addition of order k − 1 on vw. The octahedron graph is the smallest graph of this class and has parameters = k = 1.
In the following we use ideas similar to those used in the proof of Theorem 7, to show the areauniversality of angle graphs obtained from area-universal triangulations on which several diamond additions have been applied. We start by considering a special conguration that appears in the 178 Evans et al. On Area-Universal Quadrangulations neighborhood of an edge on which a diamond addition has been performed. A tent graph T k is a plane graph with the outer face v, x 0 , x 1 , x 2 , . . . , x k+1 and inner vertices y 0 , y 1 , . . . , y k where y i is incident to x i , x i+1 and v. Figure 6(a) depicts the tent graph Figure 6: Illustration of Lemma 9 and its proof. Lemma 9 Every area assignment A of a tent graph T k has an A-realizing drawing within each triangle that has area ΣA and corners v, x 0 , x k+1 . Moreover, the length of every segment x i x i+1 can be made proportional to the area of the incident triangle.
Proof: We denoted the assigned areas of T k by a i and b i as depicted in Figure 6(b). We position Then, in a realizing drawing the vertices y i lie on a line parallel to the segment x 0 x k+1 such that for each point x on the triangle x 0 x k+1 x has area i b i . Note that, by the placement of x i , each position of y i on realizes b i . Thus, we may use the freedom to realizing a i when placing y i on with the following procedure: Dening y −1 as the intersection of the segment vx 0 with the line , we suppose that y i−1 is placed already when we consider y i for i ≥ 0. Move y i rightwards on the line starting at y i−1 and observe the area of the face vy i−1 x i y i . Clearly, it starts at 0 and increases continuously. The intermediate value theorem guarantees a position, where the area equals a i . We place y i at the corresponding position and continue with y i+1 . Due to the correct total area, the area of a k+1 is realized if all other face areas are correct. Thus, we obtain an A-realizing drawing of T k .
In the following theorem, we not only consider several disjoint diamond additions, but also groups of non-disjoint diamond additions that are far apart. A set of edges of a plane triangulation T is called far apart if the subgraph of the dual graph induced by the duals of these edges and their vertices is a collection of stars; see Figure 7(a) for an example. Theorem 10 The angle graph Q of a plane triangulation T is area-universal if one of the following holds: (i) T is obtained from an area-universal triangulation T by several disjoint diamond additions of an arbitrary order. (ii) T has a set of edges that is far apart such that subdividing each of them at most once yields an area-universal subdivision T • .  Given an area assignment A of Q, we construct an area assignment A of T and an area assignment A U of U that renes both A and A . Observe that every face q of Q is either a face of U or corresponds to two faces of U .
In the latter case, we partition the prescribed area of q equally between its two faces in U and obtain the area assignment A U of U . The restriction of A U to H is denoted A H . We dene A of T as the area assignment rened by A U , where we identify the path P and the edge (u, w). From an A -realizing drawing D of T , we construct an A U -realizing drawing of U as follows: First, we add all face vertices not adjacent to P using Lemma 4; recall that they act as vertex of degree 3 in a triangle.
Observe that splitting H along P results in two tent graphs T k . Consequently, we may use Lemma 9 to reinsert each of the two tent graphs of H. By denition of A H , the subdivision 180 Evans et al. On Area-Universal Quadrangulations vertices on (u, w) are placed consistently when applying Lemma 9 to the two tent graphs when splitting H along P . Here we use the fact that the assigned areas of Q were split equally into two when dening A U . Since A U renes A, we obtain an A-realizing drawing of Q by deleting the edges of T • . This proves (i). Now, we show (ii). First, we consider the case when the set of subdivided edges of T forms only one dual star, i.e. T has a face f • such that subdividing each edge incident to f • (at most once) yields an area-universal graph T • . Theorem 7 shows the claim if exactly one edge of f • is subdivided. In the following, we show how to deal with the case of three subdivision vertices. The case of two subdivision vertices can be handled by a slight modication which is explained afterwards.
We denote the three faces incident to subdivision vertices in T • by f • , f 1 , f 2 , and f 3 as illustrated in Figure 9; thecorresponding vertices in Q are denoted by 3 be the vertices of f • and let w i be the third vertex of f i as depicted. Note that two of the w-vertices may coincide implying that a v-vertex has degree 3. However, at most one v-vertex has degree 3; otherwise T is the complete graph on four vertices which is area-universal. Therefore, it remains to consider two cases: One v vertex has degree 3 and none v-vertex has degree 3.  We rst consider the case that v 3 has degree 3. Figure 10(a) illustrates the neighborhood of f • in this case. Our strategy is as follows: For an area assignment A of Q, we dene an area assignment A U of the union U := Q + T • . This yields a unique area assignment A • of T • such that A U of U renes A • of T • . From an A • -realizing drawing of T • , we construct an A U -realizing drawing of U . Deleting the edges of T • results in an A-realizing drawing of Q.
For a given area assignment A of Q, we construct an area assignment A U of U such that the area of every face q of Q is partitioned between the two faces of T • . Let B denote the edges of T • bounding the region formed by the faces f 1 , f 2 , f 3 , f • in T • ; note that B has four edges since the common vertex of f • , f 2 , f 3 has degree 3. The edges of B divide four faces of Q in U . For all faces outside B (and not incident to B), the face areas of Q are partitioned arbitrarily between the two faces of U . The area of each of the four faces of Q divided by B is assigned to the subface outside B as indicated in Figure 10(b). We dene A • of T • as the unique area assignment rening A U .
Given an A • -realizing drawing of T • , note that every vertex of Q outside B is of degree 3 and can be inserted with Lemma 4. We will redraw all vertices inside B. By denition of A U , two incident triangular faces of f v 1 are supposed to be 0. Hence, we must place f v 1 at the same location • is correct, this is possible by Proposition 3. It remains to realize the graph G illustrated in Figure 10(c). We later show that every area assignment of G is realizable within any xed outer face of correct total area: This follows from Lemma 13 and the fact that G is the core of c(S 3 ). Now, we turn to the case that no v i has degree 3. This implies that all w i are distinct. The resulting neighborhood of f v • is illustrated in Figure 11(a). Let A be an area assignment of Q. In a rst step, we dene an area assignment A U of the union Q + T • =: U . Note that every face of Q corresponds to two faces in U . Except for the faces incident to f v we arbitrarily partition the area A(q) of a face q of Q between its two faces in U . For the faces incident to we assign their area as depicted in Figure 11(b). Since the set of subdivided edges is far apart, every subdivided edge belongs to a star. We handle each star separately as described above; in particular, the star consists of one, two or three edges since T is a triangulation. By the independence, for every two stars, the edges of T surrounding the regions of the stars are disjoint; these edges form a so-called boundary cycle of a star. For an example consider Figure 7(b). Note that in all cases, when dening A U from A, only the areas inside and adjacent to the boundary cycle are aected. Since these sets of faces in U are disjoint, the subdivision vertices can be handled independently. This nishes the proof.
The results of Proposition 6 and Theorems 7, 8 and 10 imply the area-universality of several classes of angle graphs. Corollary 11 The angle graph of a plane triangulation T is area-universal if • T is a stacked triangulation, • T is 4-connected and has at most ten vertices, or • any (possibly a dierent) embedding of T is a double stacking graph H ,k .
Proof: Stacked triangulations are area-universal, hence Proposition 6 implies the area-universality of its angle graphs. Theorem 8 can be used to show that triangulations with at most nine vertices and all embeddings of k-double stacking graphs have subdivision number at most 1 [15]. Consequently, Theorem 7 implies that their angle graphs are area-universal. Moreover, 4-connected plane triangulations on ten vertices can be obtained from area-universal triangulations by at most two disjoint diamond additions. Thus, their area-universality follows from Theorem 10(i).

Strongly Area-Universal Quadrangulations
In this section, we study strongly area-universal quadrangulations. Recall that a quadrangulation is strongly area-universal if it is area-universal within every xed outer face of the correct total area. A nice property of this class is that we can stack any strongly area-universal into a face of an (strongly) area-universal quadrangulation to obtain an (strongly) area-universal quadrangulation.
Therefore, few small strongly area-universal quadrangulations can serve as building blocks in order to construct innite and rich families of area-universal quadrangulations. Note that for n > k, the area of a convex n-gon strictly exceeds the area of any contained k-gon. Therefore, we immediately obtain that the plane bipartite graph depicted in Figure 12(a) is not strongly area-universal: Fixing the outer face as a regular hexagon, there exists no drawing in which the inner 4-face covers more than 2 /3 of the area.
Observation 3 Not all plane bipartite graphs are strongly area-universal.  In contrast, we do not know of a quadrangulation that is not strongly area-universal. Neither do we know of an area assignment that requires a convex outer face. In fact these two questions are closely related. Proposition 12 If there exists a plane quadrangulation G and an area assignment A such that every realizing drawing has a convex outer face, then there exists a plane quadrangulation H that is not area-universal. Moreover, if G is 3-connected, we can ensure that H is 3-connected.
Proof: Suppose we are given a quadrangulation G and an area assignment A with the described properties. Let H 0 denote the plane graph of K 2,4 , i.e., H 0 has three bounded faces each being a quadrangle. For each bounded face of H 0 , we take a copy of G and identify the outer 4-cycle of the copy of G with the boundary of the face. This yields the quadrangulation H as schematically illustrated in Figure 12(b). If G is 3-connected, then H is also 3-connected.
Assigning A to each copy of G, we claim that H has no realizing drawing. Suppose, by contradiction, that there exists a realizing drawing D. Due to the positive total area of the central copy of G, either the right or the left copy of G has a non-convex boundary cycle in D. Consequently, this copy induces an A-realizing drawing of G where the outer face is not convex and thus, we obtain a contradiction. Now, we present a family of plane graphs that have the property of being strongly area-universal.
The family consists of the angle graphs of wheels, which are also known as pseudo-double wheels. The pseudo-double wheel S k has 2k + 2 vertices and consists of a cycle with vertices v 1 , v 2 , . . . , v 2k and a vertex v adjacent to all vertices on the cycle with odd index and a vertex w adjacent to all vertices on the cycle with even index, see Figure 13(a). Up to the labeling, the plane embedding of S k is unique.
The smallest pseudo-double wheel S 3 is also known as the cube graph. In this section, we show that the cube graph and more generally, all pseudo-double wheels are strongly area-universal.
The core c(S k ) of S k with an area assignment.

Figure 13
We rst study a subgraph of S k , namely the plane graph c(S k ), called the core, which is obtained by deleting v 1 . Figure 13(b) illustrates the core of S 5 .
Lemma 13 Let c(S k ) be the core of a plane pseudo-double wheel with an area assignment A. Let q be a quadrangle of area ΣA containing the diagonal AB and whose corners are identied with the vertices A, w, B, v. Then, c(S k ) has a A-realizing drawing within q.
Proof: We distinguish two cases. We call the faces of c(S k ) incident to w the bottom faces and the faces incident to v the top faces of c(S k ). For simplicity, we denote the vertices v 2 and v 2k by A and B, respectively, and the face areas by a i for the top and by b i for the bottom faces. Consider also Figure 13(b).
Case (i): If i b i > area( AwB), we position the even vertices on the segment AB as illustrated in Figure 14(a). Note that adding the edges of consecutive even vertices and Av and Bv (and deleting w) results in a tent graph. We partition the face area b i of each bottom face into b 1 i and b 2 i such that the ratio b 1 i/b 2 i coincides for all i and i b 2 i = area( AwB). By Lemma 9, the tent graph has a realizing drawing within the triangle vAB. Due to the same ratio, the vertex   placement on AB also realizes the area for triangles incident to w. Figure 14(a) visualizes the realizing drawing of c(S k ). Case (ii): If i a i > area( vAB), we position the odd vertices on the segment AB as illustrated in Figure 14(b). Note that the graph in the bottom triangle is a tent graph. Therefore, we partition the area a i of a top faces into a 1 i and a 2 i such that the ratio a 1 i/a 2 i coincides for all i and i a 1 i = area( vAB). As in case (i), we use Lemma 9 to nd a realizing drawing of the tent graph. Figure 14(b) visualizes the realizing drawing of c(S k ).
This lemma helps us to settle three out of four cases of Theorem 14. Theorem 14 The pseudo-double wheel S k , k ≥ 3, is strongly area-universal.
Proof: For an area assignment A of S k , we consider an arbitrary but xed quadrangle q of area ΣA whose corners are identied with v 1 v 2 wv 2k . We distinguish two cases depending on the shape of q. Note that q can be triangulated by the segment v 2 v 2k or v 1 w (or both).
In case 1, the segment v 2 v 2k lies inside q. We distinguish two subcases based on the assigned areas a and b of the faces incident to v 1 relative to the area of the triangle v We can position v such that the triangles v 1 v 2 v and v 1 vv 2k realize a and b, respectively. The remaining graph corresponds to the core S k which we realize in the quadrangle vv 2 wv 2k containing the segment v 2 v 2k by applying Lemma 13. Figure 15 Figure 15: Illustration of case 1 in the proof of Theorem 14 in which the segment v 2 v 2k is contained in q. In case 1(i) the face areas incident to v 1 are small; in case 1(ii) the face areas incident to v 1 are big. Both cases reduce to Lemma 13.
. We position v on the segment v 2 v 2k such that v 3 and v 2k−1 are forced to be on a line parallel to the segment v 2 v 2k , i.e., v partitions v 2 v 2k according to the ratio of a and b. The positions of v 3 and v 2k−1 on the line are such that the areas of the triangles v 2 wv 3 and v 2k−1 wv 2k realize b 1 and b k . The graph induced by the vertices in the interior of vv 3 wv 2k−1 is the core of a smaller pseudo-double wheel and contains the diagonal v 3 , v 2k−1 . Consequently, Lemma 13 yields a realizing drawing, see Figure 15(b).
In case 2, the segment v 1 w lies inside q. We call the faces incident to v 2 the left faces and the faces incident to v 2k the right faces. We say the left (right) faces are small if their assigned area is at most the area of the triangle v 1 , v 2 , w (v 1 , w, v 2k ). Otherwise, we call the left (right) faces big. Note that the left or right faces must be small. By symmetry, we assume without loss of generality, that the left faces are small. Then we can realize the left faces by triangular faces, by positioning v 3 accordingly. Case 2(i): the right faces are small as well; this case is illustrated in Figure 16(a). We reduce this case to subgraphs of two stacked triangulations. To do so, we contract the edge vv 1 and realize the right faces by triangular faces with corner v 2k−1 . Denote the areas of the inner faces by x i as illustrated in Figure 16(a). There exists an i with 4 ≤ i ≤ 2k − 1 such that The exact layout depends on whether x i is the area of a top or bottom face of the core. If it is a top face, then its unique vertex with odd index is placed at w. If x i belongs to a bottom face, then the unique vertex with even index is placed at v. Afterwards we insert the remaining vertices. For j < i, we iteratively insert v j such that it realizes the face area x j by triangular face with a at angle at v j+1 from left to right. For j > i, we follow the same strategy but in decreasing order.
Case 2(ii): the right faces are big. Figure 16(b) depicts this case. Recall that v 3 has been xed already. Place v 2k−1 on v 1 w such that the area of the quadrangle v 1 v 3 v 2k−1 v 2k exceeds b but is not enough to also realize all top faces of the core, i.e., the striped faces incident to v in Figure 16(b). The remaining graph can be handled by Lemma 13: To do so, we choose vv 3 wv 2k−1 as the outer face and insert an articial vertex on wv 2k−1 .
We can combine Lemma 2 and Theorem 14 in order to construct further strongly area-universal graphs. A graph is a stacked pseudo-double wheel if there exists a set of cycles such that decomposition along these cycles yields several pseudo-double wheels. A generalized stacked pseudo-double wheel can be decomposed into pseudo-double wheels and copies of the unique plane quadrangulation Q 5 on ve vertices. Note that Q 5 is the plane graph that can be obtained by starting with a plane C 4 , inserting an edge between two non-incident vertices and then subdividing this new edge. It is easy to see that Q 5 is strongly area-universal [12]. Together with Lemma 2 and Theorem 14, it follows that Corollary 15 Generalized stacked pseudo-double wheels are strongly area-universal. Quadrangulations and Convexity A drawing of a planar graph is convex if each face is bounded by the boundary of a convex polygon.
Convexity is a visually appealing property of drawings of planar graphs which has therefore been studied extensively in graph drawing. For example, Tutte's spring embeddings [20] guarantee convex drawings for every 3-connected planar graph. In this section, we aim for convex realizing drawings, i.e., given an area assignment A of a quadrangulation Q, we want to nd an A-realizing drawing of Q which is also convex. A planar graph is convex area-universal if for every area assignment there exists a convex realizing drawing. Although convex area-universality seems to be a very strong property, there are examples of convex area-universal graphs, such as the cube graph.
Indeed, this result can be generalized in two directions. First, the cube graph is convex areauniversal for every convex outer face. Second, this holds not only for the cube graph but also for all pseudo-double wheels. We say a graph is strongly convex area-universal if for every area assignment A and every convex drawing of the outer face with total area ΣA, there exists a realizing drawing. Theorem 16 The pseudo-double wheel S k , k ≥ 3, is strongly convex area-universal.
In Theorem 14 we have shown that pseudo-double wheels are area-universal. The proof made ample use of Lemma 13. Since drawings obtained by using this lemma may contain non-convex faces we need an independent proof for Theorem 16; consider Figures 14(a) and 14(b) for an illustration of the possible appearance of non-convex faces.
Proof: Let A be a given area assignment of S k . We denote the areas assigned to the inner faces adjacent to the outer edges by a, b, c, d and the remaining areas by x 4 , x 5 , . . . , x 2k−1 where x 2k−1 = c as depicted in Figure 17. Figure 17: A pseudo-double wheel S k with a given area assignment.
Let q be a convex quadrangle of area ΣA with corners A, B, C, D which are identied with the outer vertices v 1 , v 2 , w, v 2k of S k , respectively. Considering the diagonal AC, shows that at least one of the two following inequalities hold: a + b ≤ area( ABC) or c + d ≤ area( ACD). By symmetry, we may assume that the rst inequality holds. Hence, we may place v 3 in the 188 Evans et al. On Area-Universal Quadrangulations triangle ABC such that the areas a and b are realized by the triangular faces ABv 3 and BCv 3 where v and v 4 are placed on Av 3 and Cv 3 , respectively. Now we distinguish two cases.
Case 1: d ≤ area( Av 3 D). By placing v on the segment Av 3 we realize area d by a triangular face. Split the quadrangle vv 3 CD into two parts by the diagonal vC and determine i ∈ [4, 2k − 1] such that For all j < i, we realize the area x j by a triangle, (namely uv j−1 v j for u = v if j is even and u = w if j is odd), by placing v j accordingly. Likewise for all j > i, we realize the area x j by a triangle, (namely uv j v j+1 for u = v if j is even and u = w if j is odd), by placing v j accordingly. Finally, by placing v i at v if i is odd and at w if i is even we the area x i with the convex quadrangle vv i−1 wv i+1 . Figure 18  Case 2: d > area( Av 3 D). In this case we place v at v 3 and v 4 at w. We place v 2k−1 such that the area c and d are realized. In decreasing order, we place v i such that x i is realized by the triangle uv i v i+1 with u = v if i is even and u = w if i is odd. This yields a realizing drawing within every convex outer face, see Figure 18(b).

5.1
Not all Quadrangulations are Convex Area-Universal Plane drawings of K 2,n have non-convex faces when n ≥ 4. Tutte's spring embedding theorem [20], however, warrants that 3-connected quadrangulations have convex drawings. In [12], it was asked whether all 3-connected quadrangulations are convex area-universal. Here we answer this question in the negative. Theorem 17 There is a 3-connected quadrangulation that is not convex area-universal.
Proof: We show that the 3-connected quadrangulation Q, depicted in Figure 19(b), has an area assignment which does not allow for a convex drawing in any (even non-convex) outer face. The construction is based on a non-realizable area assignment A of the octahedron graph G.
Theorem (Kleist [13], Theorem 1) For small enough ε > 0, the octahedron graph (with a white/gray-coloring of its faces as illustrated in Figure 19(a)) has no drawing where the white faces have area of at most ε and the gray faces have area of at least 1. (b) The quadrangulation Q and its black subgraph Q . There exists no convex drawing of Q in which each white face has area 0 and each gray face has area 1 /3. Figure 19: Illustration of Theorem 17 and its proof.
We show that a convex drawing of Q induces an A-realizing drawing of G, yielding a contradiction. Let Q be the subgraph of Q which is induced by the black vertices in Figure 19(b). Note that Q contains a 1-subdivision of the octahedron. We call the two bipartition classes of Q the squared and circled vertices.
For the purpose of a contradiction, suppose that for every ε > 0, Q has a convex drawing in which each white face has area ε /3 and each gray face area 1 /3. Then the induced drawing of Q has the following properties: the white faces have area ε, the gray faces have area 1, and each segment between two squared vertices is contained in some white face. In the remainder, we show that for small enough ε > 0, no such drawing of Q exists.
Suppose that Q has such a drawing D. Because each segment between two squared vertices is contained in some white face, these segments together with the squared vertices form a straight-line drawing D of G where the white faces have area at most ε and the gray faces have area of at least 1. The red dotted graph in Figure 19(b) illustrates D . The properties of D contradict the above stated theorem. Consequently, Q is not convex area-universal.
Remark. Because we did not use the shape of the outer face in the proof of Theorem 17, the quadrangulation Q does not even have an A-realizing drawing where we only require that each inner face is convex.
Remark. The construction of Q in the proof of Theorem 17 is based on a non-realizable area assignment of the octahedron graph. More generally, a white/gray-coloring of the faces of any Eulerian triangulation yields non-realizable area assignment [13,Theorem 1]. This fact allows the construction of a large family of quadrangulations that are not convex area-universal. 6 Small Quadrangulations In this section we show that our methods are strong enough to prove area-universality for quadrangulations with up to 13 vertices via reductions to know area-universal graphs. Theorem 18 Every quadrangulation on at most 13 vertices is area-universal.
Proof: First, it follows from Lemma 2 and Proposition 3 that a minimal non-area-universal quadrangulation has minimum degree 3. Thus, the smallest quadrangulation of interest is the cube graph on eight vertices. Figure 20 displays all quadrangulations on up to 13 vertices with minimum degree 3. We denote them by Q 1 , . . . , Q 9 as illustrated. Since all embeddings of a pseudo-double wheel S k are equivalent, Theorem 16 proves the (strong) area-universality of pseudo-double wheels. Specically, this shows the (strong) areauniversality of Q 1 = S 3 , Q 2 = S 4 and Q 4 = S 5 . Moreover, Q 5 is a stacked cube graph and Q 9 is obtained from the cube by rst adding a subdivided diagonal to a face and then stacking a cube graph in one of the two new faces. Since the cube graph is the double wheel S 3 , both quadrangulations are area-universal by Corollary 15. We reduce all remaining quadrangulations to area-universal triangulations. The quadrangulation Q 3 on eleven vertices is a subgraph of a stacked triangulation T , for which every embedding of T is area-universal. In Figure 20, the vertices and edges of the stacked triangulation are highlighted in red.
The three remaining quadrangulations are subgraphs of an area-universal graph family which was shown to be area-universal by Kleist [14].
Theorem (Kleist [14], Theorem 3) Any (embedding of a) double stacking graph H ,k is areauniversal if and only if · k is even.
More precisely, Q 6 , Q 7 , Q 8 are subgraphs of an area-universal double stacking graph with some additional vertices of degree 3 stacked into triangular faces. Thus, their area-universality follows from Observation 1 and [14,Theorem 3]. In Figure 20 the vertices which remain after iterative removal of degree-3 vertices are highlighted in red. The quadrangulation Q 6 on twelve vertices reduces to the double stacking graph H 2,2 ; the quadrangulations Q 7 and Q 8 on 13 vertices reduce to the double stacking graph H 2,1 . The vertices in the interior of red dotted curves in Figure 20 are added by diamond additions on the respective edge.