Straight-Line Grid Drawings of Planar Graphs

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1 Straight-Line Grid Drawings of Planar Graphs with Linear Area Md. Rezaul Karim and Md. Saidur Rahman August 28, 2006 Abstract A straight-line grid drawing of a planar graph G is a drawing of G on an integer grid such that each vertex is drawn as a grid point and each edge is drawn as a straight-line segment without edge crossings. It is well known that a planar graph of n vertices admits a straight-line grid drawing on a grid of area O(n 2 ). A lower bound of Ω(n 2 ) on the area-requirement for straight-line grid drawings of certain planar graphs are also known. In this paper, we introduce a fairly large class of planar graphs which admits a straightline grid drawing on a grid of area O(n). Our new class of planar graphs, which we call doughnut graphs, is a subclass of 5-connected planar graphs. We also show several interesting properties of doughnut graphs in this paper. Keywords: Planar Graph, Straight-Line Drawing, Grid Drawing, Linear Area Drawing. 1 Introduction Recently automatic aesthetic drawings of graphs have created intense interest due to their broad applications in computer networks, VLSI layout, information visualization etc., and as a consequence a number of drawing styles have come out [DETT99, JM03, KW01, NR04]. The most typical and widely studied drawing style is the straight-line drawing of a planar graph. A straight-line drawing of a planar graph G is a drawing of G such that each vertex is drawn as Dept. of Computer Science and Engineering, Bangladesh University of Engineering and Technology(BUET), Dhaka-1000, Dhaka, Bangladesh. Also Dept. of computer Science and Engineering, University of Dhaka, Dhaka- 1000, Bangladesh. rkarim@univdhaka.edu Dept. of Computer Science and Engineering, Bangladesh University of Engineering and Technology(BUET), Dhaka-1000, Dhaka, Bangladesh. saidurrahman@cse.buet.ac.bd 1

2 a point and each edge is drawn as a straight-line segment without edge crossings. A straightline grid drawing of a planar graph G is a straight-line drawing of G on an integer grid such that each vertex is drawn as a grid point as shown in Fig. 1(b). Wagner [Wag36], Fary [Far48] b h c a i b F1 c d j h p i b l k F2 o g (a) m e n f f g o n e j p m (b) d k l a F1 f g n h a e l m F2 p k d (c) o j i c b a h f g i n o m p e k j l (d) c d Figure 1: (a) A planar graph G, (b) a straight-line grid drawing of G with area O(n 2 ), (c) a doughnut embedding of G and (d) a straight-line grid drawing of G with area O(n). and Stein [Ste51] independently proved that every planar graph G has a straight-line drawing. Their proofs immediately yield polynomial-time algorithms to find a straight-line drawing of a given plane graph. However, the area of a rectangle enclosing a drawing on an integer grid obtained by these algorithms is not bounded by any polynomial in the number n of vertices in G. In fact, to obtain a drawing of area bounded by a polynomial remained as an open problem for long time. In 1990, de Fraysseix et al. [FPP90] and Schnyder [Sch90] showed by two different methods that every planar graph of n 3 vertices has a straight-line drawing on an integer grid of size (2n 4) (n 2) and (n 2) (n 2), respectively. Fig. 1(b) illustrates a straight-line grid drawing of the graph G in Fig. 1(a) with area O(n 2 ). A natural question arises: what is the minimum size of a grid required for a straight-line drawing? de Fraysseix et al. showed that, for each n 3, there exits a plane graph of n vertices, for example nested triangles, which needs a grid size of at least 2(n 1)/3 2(n 1)/3 for any grid drawing [CN98, FPP90]. It has been conjectured that every plane graph of n vertices has a grid drawing on a 2n/3 2n/3 grid, but it is still an open problem. For some restricted classes of graphs, more compact straight-line grid drawings are known. For example, a 4-connected plane graph G having at least four vertices on the outer face has a straight-line grid drawing 2

3 with area ( n/2 1) ( n/2 ) [MNN01]. Garg and Rusu showed that an n-node binary tree has a planar straight-line grid drawing with area O(n) [GR02, GR03b]. Although trees admit straight-line grid drawings with linear area, it is generally thought that triangulations may require a grid of quadratic size. Hence finding nontrivial classes of planar graphs of n vertices richer than trees that admit straight-line grid drawings with area o(n 2 ) is posted as an open problem in [BEGKLM03]. Garg and Rusu showed that an outerplanar graph with n vertices and the maximum degree d has a planar straight-line drawing with area O(dn 1.48 ) [GR03a]. Recently Di Battista and Frati showed that a balanced outerplanar graph of n vertices has a straight-line grid drawing with area O(n) and a general outerplanar graph of n vertices has a straight-line grid drawing with area O(n 1.48 ) [DF05]. In this paper, we introduce a new class of planar graphs which has a straight-line grid drawing on a grid of area O(n). We also give a linear-time algorithm to find such a drawing. Our new class of planar graphs is a subclass of 5-connected planar graphs, and we call the class doughnut graphs since a graph in this class has a doughnut-like embedding as illustrated in Fig. 1(c). In an embedding of a doughnut graph of n vertices there are two vertex-disjoint faces each having exactly n/4 vertices and each of all the other faces has exactly three vertices. Fig. 1(a) illustrates a doughnut graph of 16 vertices where each of the two faces F 1 and F 2 contains four vertices and each of all other faces contains exactly three vertices. Fig. 1(c) illustrates a doughnut-like embedding of G where F 1 is embedded as the outer face and F 2 is embedded as an inner face. A straight-line grid drawing of G with area O(n) is illustrated in Fig. 1(d). We also show several interesting properties of doughnut graphs in this paper. The remainder of the paper is organized as follows. In Section 2, we give some definitions. Section 3 provides some properties of the class of doughnut graphs. Section 4 deals with straight-line grid drawings of doughnut graphs. Finally Section 5 concludes the paper. 2 Preliminaries In this section we give some definitions. Let G = (V, E) be a connected simple graph with vertex set V and edge set E. Throughout the paper, we denote by n the number of vertices in G, that is, n = V, and denote by m the number of edges in G, that is, m= E. An edge joining vertices u and v is denoted by (u, v). The degree of a vertex v, denoted by d(v), is the number of edges incident to v in G. G is called r-regular if every vertex of G has degree r. We call a vertex v a neighbor of a vertex u in G if G has an edge (u, v). The connectivity κ(g) of a graph G is the minimum number of vertices whose removal results in a disconnected graph or a single-vertex graph K 1. G is called 3

4 k-connected if κ(g) k. We call a vertex of G a cut-vertex of G if its removal results in a disconnected or single-vertex graph. For W V, we denote by G W the graph obtained from G by deleting all vertices in W and all edges incident to them. A cut-set of G is a set S V (G) such that G S has more than one component or G S is a single vertex graph. A path in G is an ordered list of distinct vertices v 1, v 2,..., v q V such that (v i 1, v i ) E for all 2 i q. v 1 and v q are end-vertices of the path v 1, v 2,..., v q. Two paths are vertex-disjoint if they do not share any common vertex except their end vertices. A graph is planar if it can be embedded in the plane so that no two edges intersect geometrically except at a vertex to which they are both incident. A plane graph is a planar graph with a fixed embedding. A plane graph G divides the plane into connected regions called faces. The unbounded region is called the outer face. For a face F in G we denote by V (F ) the set of vertices of G on the boundary of face F. We call two faces F 1 and F 2 are vertex-disjoint if V (F 1 ) V (F 2 ) =. Let F be a face in a plane graph G with n 3. If the boundary of F has exactly three edges then we call F a triangulated face. One can divide a face F of p (p 3) vertices into p 2 triangulated faces by adding p 3 extra edges. The operation above is called triangulating a face. If every face of a graph is triangulated, then the graph is called a triangulated plane graph. We can obtain a triangulated plane graph G from a non-triangulated plane graph by triangulating all of its faces. Fig. 2(b) illustrates a triangulated plane graph obtained from the plane graph in Fig. 2(a). a a b c d j h i p l k o g (a) m e n f b c d h j i l k p o g (b) m e n f Figure 2: (a) Input graph G and (b) a triangulated plane graph G of graph G where extra edges are drawn by dotted lines A maximal planar graph is one to which no edge can be added without losing planarity. Thus in any embedding of a maximal planar graph G with n 3, the boundary of every face of G is a triangle, and hence an embedding of a maximal planar graph is often called a triangulated plane graph. It can be derived from the Euler s formula for planar graphs that if G is a maximal planar graph with n vertices and m edges then m = 3n - 6, for more details see [NR04]. A spanning subgraph of G is a subgraph with vertex set V (G). Let G be a 5-connected planar graph, let Γ be any planar embedding of G and let p be an integer such that p 3. We call G a p-doughnut graph if the following Conditions (d 1 ) and 4

5 (d 2 ) hold: (d 1 ) Γ has two vertex-disjoint faces each of which has exactly p vertices, and all the other faces of Γ has exactly three vertices; and (d 2 ) G has the minimum number of vertices satisfying Condition (d 1 ). In general, we call a p-doughnut graph for p 3 a doughnut graph. Since a doughnut graph is 5-connected, the following fact holds. Fact 2.1 Let G be a doughnut graph and let Γ and Γ be any two plane embeddings of G. Then any facial cycle of Γ is a facial cycle of Γ. Fact 2.1 implies that decomposition of a doughnut graph into its facial cycles is unique. Throughout the paper we often mention faces of a doughnut graph G without mentioning its plane embedding where description of the faces are valid for any plane embedding of G. 3 Properties of Doughnut Graphs In this section we will show some properties of a p-doughnut graph. We have the following lemma on the number of vertices of a graph satisfying Condition (d 1 ). Lemma 3.1 Let G be a 5-connected planar graph, let Γ be any planar embedding of G, and let p be an integer such that p 3. Assume that Γ has two vertex-disjoint faces each of which has exactly p vertices, and all the other faces of Γ has exactly three vertices. Then G has at least 4p vertices. Proof. Let F 1 and F 2 be the two faces of Γ each of which contains exactly p vertices. Let x be the number of vertices in G which are neither on F 1 nor on F 2. Then G has x + 2p vertices. We now calculate the number of edges in graph G. Faces F 1 and F 2 may not be triangulated, but all other faces of Γ are triangulated. If we triangulate both the faces F 1 and F 2 of Γ then the resulting graph G is a maximal planar graph. Using Euler formula for planar graphs, one can find that G has exactly 3(x + 2p) 6 = 3x + 6p 6 edges. To triangulate each of F 1 and F 2, we need to add p 3 edges and hence the number of edges in G is exactly (3x + 6p 6) 2(p 3) = 3x + 4p (1) Since G is 5-connected, degree of a vertex of G is at least 5. Hence, from the degree-sum formula we get 2(3x + 4p) 5(x + 2p). This relation implies x 2p (2) 5

6 Therefore G has at least 4p vertices. Q.E.D. Lemma 3.1 implies that a p-doughnut graph has 4p or more vertices. We now show that 4p vertices are sufficient to construct a p-doughnut graph as in the following lemma. Lemma 3.2 For an integer p, (p 3), one can construct a p-doughnut graph G with 4p vertices. Proof. We first construct a planar embedding Γ of G with 4p vertices and then show that G is a p-doughnut graph. Construction. Let C 1, C 2, C 3 be three cycles such that C 1 contains p vertices, C 2 contains 2p vertices and C 3 contains p vertices. Let, x 2,..., x p be the vertices on C 1, y 1, y 2,..., y p be the vertices on C 3, and z 1, z 2,.., z 2p be the vertices on C 2. Let R 1, R 2 and R 3 be three concentric circles on a plane with radius r 1, r 2 and r 3, respectively, such that r 1 > r 2 > r 3. We embed C 1, C 2 and C 3 on R 1, R 2 and R 3 respectively, as follows. We put the vertices, x 2...x p of C 1 on R 1 in clockwise order such that is put on the leftmost position among the vertices, x 2,..., x p. Similarly, we put vertices z 1, z 2,..., z 2p of C 2 on R 2 and y 1, y 2,..., y p of C 3 on R 3. We now add edges between the vertices on C 1 and C 2, and between the vertices on C 2 and C 3. We have two cases to consider. Case 1: k is even in z k. In this case, we add two edges (z k, x k/2 ), (z k, x i ) between C 2 and C 1, and one edge (z k, y i ) between C 2 and C 3 where i = 1 if k = 2p, i = k/2 + 1 otherwise. Case 2: k is odd in z k. In this case we add two edges (z k, y k/2 ), (z k, y i ) between C 2 and C 3, and one edge (z k, x k/2 ) between C 1 and C 2 where i = 1 if k = 2p 1, i = k/2 + 1 otherwise. We thus constructed a planar embedding Γ of G. Fig. 3 illustrates the construction above for the case of p = 4. We now prove that G is a p-doughnut graph. To prove this claim we need to prove that G satisfies the following properties (a) (c): (a) the graph G is a 5-connected planar graph; (b) any planar embedding Γ of G has exactly two vertex-disjoint faces each of which has exactly p vertices, and all the other faces are triangulated; and (c) G has the minimum number of vertices satisfying (a) and (b). 6

7 z 3 x 2 c 1 z 3 x 2 z 3 x 2 z 2 y 2 c c 2 3 z 4 z 2 y 2 z 4 z 2 y 2 z 4 x z z 1 1 y 5 1 y3 x 3 x z z 1 1 y 5 1 y 3 x 3 x z z 1 1 y 5 1 y 3 x 3 y 4 z 8 z 7 z 6 y 4 z 8 z 7 z 6 y 4 z 8 z 7 z 6 x 4 x 4 x 4 (a) (b) (c) Figure 3: Illustration for the construction of a planar embedding Γ of a p-doughnut graph G for p = 4; (a) embedding of the three cycles C 1, C 2 and C 3 on three concentric circles, (b) addition of edges for the case where k is even in z k and (c) Γ. (a) To prove that G is 5-connected, we show that there exist at least five vertex disjoint paths for any pair u, v of vertices in G. From the construction we can also prove that G is planar. The details of the proof are given in the appendix. (b) The construction of Γ implies that Cycle C 1 is the boundary of the outer face F 1 of Γ and Cycle C 3 is the boundary of an inner face F 2. Each of F 1 and F 2 has exactly p vertices. Clearly the two faces F 1 and F 2 of Γ are vertex-disjoint. Thus Γ has exactly two vertex-disjoint faces F 1 and F 2 each of which has exactly p vertices. We now show that the rest of the faces of Γ are triangulated. The rest of the faces can be divided into two groups; (i) faces having vertices on both the cycles C 1 and C 2 and (ii) faces having the vertices on both the cycles C 2 and C 3. We only prove that each face in Group (i) is triangulated, since similarly we can prove that each face in Group (ii) is triangulated. From our construction, each vertex z i with even i has exactly two neighbors on C 1 and the two neighbors of z i on C 1 are consecutive. Hence we get a triangulated face for each z i with even i which contains z i and the two neighbors of z i on C 1. We now show that the remaining faces in Group (i) are triangulated. Clearly each of the remaining faces in Group (i) must contain a vertex z i with odd i since a vertex on a face in Group (i) is either on C 1 or on C 2 and a vertex on C 2 has at most two neighbors on C 1. Let z i, z i+1 and z i+2 be three consecutive vertices on C 2 with even i. Then z i and z i+2 has a common neighbor x on C 1. One can observe from our construction that x is also the only neighbor of z i+1 on C 1. Then exactly two faces in Group (i) contains z i+1 and the two faces are triangulated. This implies that there are exactly two faces in Group (i) for each z i on C 2 with odd i which contains z i and are triangulated. Therefore each face in Group (i) are triangulated. 7

8 Thus Γ has exactly two vertex-disjoint faces F 1 and F 2 each of which has exactly p vertices, and the rest of the faces are triangulated. By Fact 2.1, any planar embedding Γ of G has exactly two vertex-disjoint faces each of which has exactly p vertices, and all other faces are triangulated. (c) We have constructed the graph G with 4p vertices and proofs for (a) and (b) imply that G satisfies properties (a) and (b). G is a 5-connected planar graph and hence satisfies (d 1 ) of the definition of a p-doughnut graph. By Lemma 3.1, 4p is the minimum number of vertices of such a graph. Since we have constructed G with 4p vertices, G has the minimum number of vertices satisfying properties (a) and (b). Q.E.D. Lemmas 3.1 and 3.2 imply that a p-doughnut graph G has exactly 4p vertices. Then the value of x in Eq. (2) is 2p in G. By Eq. (1), G has exactly 3x + 4p = 10p edges. Since G is 5-connected, every vertex has degree 5 or more. Then degree-sum formula implies that every vertex of G has degree exactly 5, since G has 4p vertices and 10p edges. Therefore the following theorem holds Theorem 3.3 Let G be a p-doughnut graph. Then G is 5-regular and has exactly 4p vertices. For a cycle C in a plane graph G, we denote by G(C) the plane subgraph of G inside C excluding C. Let C 1, C 2 and C 3 be three vertex-disjoint cycles in G such that V (C 1 ) V (C 2 ) V (C 3 )=V (G). Then we call an embedding Γ of G a doughnut embedding of G if C 1 is the outer face of Γ, G(C 1 ) contains C 2 and G(C 2 ) contains C 3. We call C 1 the outer cycle, C 2 the middle cycle and C 3 the inner cycle of Γ. We next show that a p-doughnut graph has a doughnut embedding. To prove the claim we need the following lemma. Lemma 3.4 Let G be a p-doughnut graph. Let F 1 and F 2 be the two faces of G each of which contains exactly p vertices. Then G {V (F 1 ) V (F 2 )} is connected and contains a cycle. Proof. Since G is 5-connected, G = G {V (F 1 ) V (F 2 )} is connected; otherwise, G would have a cut-set of 4 vertices - two of them are on F 1 and the other two are on F 2, a contradiction. G contains 2p vertices and at least 2p edges; if there is no edge between a vertex on F 1 and a vertex on F 2 in G then G contains exactly 2p edges, otherwise G contains more than 2p edges. Since G is connected, has 2p vertices and has at least 2p edges, G must have a cycle. Q.E.D. We now prove the following theorem. Theorem 3.5 A p-doughnut graph always has a doughnut embedding. 8

9 Proof. Let F 1 and F 2 be two faces of G each of which contains exactly p vertices. Let Γ be a plane embedding of G such that F 1 is embedded as the outer face. By Lemma 3.4 G {V (F 1 ) V (F 2 )} is connected and contains a cycle. Let C be a cycle in G {V (F 1 ) V (F 2 )}. One can observe that if G(C ) does not contain F 2, then there would be edge crossings among the edges from the vertices on C to the vertices on F 1 and F 2, and hence Γ would not be a plane embedding of G. Therefore G(C ) contains F 2. Furthermore, G {V (F 1 ) V (F 2 )} contains exactly one cycle C, and C contains all vertices of G {V (F 1 ) V (F 2 )}; otherwise, Γ would not be a planar embedding of G, a contradiction. Therefore in Γ, G(F 1 ) contains C and G(C ) contains F 2 and hence Γ is a doughnut embedding. Q.E.D. A 1-outerplanar graph is an embedded planar graph where all vertices are on the outer face. It is also called 1-outerplane graph. An embedded graph is a k-outerplanar (k > 1) if the embedded graph obtained by removing all vertices of the outer face is a (k 1)-outerplane graph. A graph is k-outerplanar if it admits a k-outerplanar embedding. A planar graph G has outerplanarity k ( k > 0) if it is k-outerplanar and it is not j-outerplanar for 0 < j < k. In the rest of this section, we show that the outerplanarity of a p-doughnut graph G is 3. Since none of the faces of G contains all vertices of G, G does not admit 1-outeplanar embedding. We now need to show that G does not admit a 2-outerplanar embedding. We have the following fact. Fact 3.6 A graph G having outerplanarity 2 has a cut-set of four or less vertices. Proof. Deleting vertices from the outer face of a 2-outerplanar graph leaves a 1-outerplanar graph. A 1-outerplanar graph has a cut-set of at most 2. From the definition of 2-outerplanar graph and the cut-set size of 1-outerplanar graph, one can observe that a graph G having outerplanarity 2 has a cut-set of four or less vertices. Q.E.D. Since G is 5-connected graph, G has no cut-set of four or less vertices. Hence by Fact 3.6 the graph G has outerplanarity greater than 2. Thus the following lemma holds. Lemma 3.7 Let G be a p-doughnut graph for p 3. Then G is neither an 1-outerplanar graph nor a 2-outerplanar graph. We now prove the following theorem. Theorem 3.8 The outerplanarity of a p-doughnut graph G is 3. Proof. A doughnut embedding of G immediately implies that G has a 3-outerplanar embedding. By Lemma 3.7 G is neither an 1-outerplanar graph nor a 2-outerplanar graph. Therefore the outerplanarity of a p-doughnut graph is 3. Q.E.D. 9

10 4 Straight-Line Grid Drawings of Doughnut Graphs In this section we give a linear-time algorithm for finding a straight-line grid drawing of a doughnut graph on a grid of linear area. Let G be a p-doughnut graph. By Theorem 3.5 G has a doughnut embedding. Let Γ be a doughnut embedding of G as illustrated in Fig. 4(a). Let C 1, C 2, and C 3 be the outer cycle, the middle cycle and the inner cycle of Γ, respectively. We now have the following facts. Lemma 4.1 Let G be a p-doughnut graph and let Γ be a doughnut embedding of G. Let C 1, C 2, and C 3 be the outer cycle, the middle cycle and the inner cycle of Γ, respectively. For any two consecutive vertices z i, z i+1 on C 2, one of z i, z i+1 has exactly one neighbor on C 1 and the other has exactly two neighbors on C 1. Proof. In Γ G(C 1 ) contains C 2 and G(C 2 ) contains C 3. By Theorem 3.3 G is 5-regular, and hence each vertex on C 1 has exactly three neighbors on C 2. Since each of the faces which contains vertices on both C 1 and C 2 is triangulated, one can observe that one of z i, z i+1 has exactly one neighbor on C 1 and the other has exactly two neighbors on C 1. Q.E.D. Lemma 4.2 Let G be a p-doughnut graph and let Γ be a doughnut embedding of G. Let C 1, C 2 and C 3 be the outer cycle, the middle cycle and the inner cycle of Γ, respectively. Let z i be a vertex on C 2, then either the following (a) or (b) holds. (a) z i has exactly one neighbor on C 1 and exactly two neighbors on C 3. (b) z i has exactly one neighbor on C 3 and exactly two neighbors on C 1. Proof. In Γ G(C 1 ) contains C 2 and G(C 2 ) contains C 3. By Theorem 3.3 G is 5-regular, and hence each vertex on C 2 has exactly three neighbors on C 1 and C 2. These neighbors are on C 1 or on C 3 or on both. Lemma 4.1 yields z i has either one or two neighbors on C 1. If z i has one neighbor on C 1 then (a) holds. Otherwise, (b) holds. Q.E.D. Before describing our algorithm we need some definitions. Let z i be a vertex on C 2 such that z i has two neighbors on C 1. Let x and x be the two neighbors of z i on C 1 such that x is the counter clockwise next vertex to x on C 1. We call x the left neighbor of z i on C 1 and x the right neighbor of z i on C 1. Similarly we define the left neighbor and the right neighbor of z i on C 3 if a vertex z i on C 2 has two neighbors on C 3. We are now ready to give our algorithm. We now embed C 1, C 2 and C 3 on three nested rectangles R 1, R 2 and R 3, respectively on a grid as illustrated in Fig. 4(b). We draw rectangle R 1 on grid with four corners on grid point 10

11 +1 C 1 C 2 x p yp C 3 z 2p x1 z1 y 1 zp z 2 y 2 y 3 z 3 x2 z z 5 4 x3 (0,5) xp z 2p (1,4) (2,3) (2,2) R 1 R 2 R 3 ( p+1,5) xi (a) (0,5) xp ( p,4) zp +1 z 2p (1,4) (p 1,3) (2,3) (p 1,2) (2,2) (p 1,3) (p 1,2) p ( p,4) z ( p+1,5) xi z1 (1,1) ( p,1) z p z1 (1,1) ( p,1) z p (0,0) (0,5) xp (b) ( p+1,0) xi ( p+1,5) xi +1 xp (0,0) (0,5) (c) ( p+1,0) xi ( p+1,5) xi +1 z 2p zp +1 z 2p zp +1 y p (2,3) (p 1,3) y i +1 y 1 (2,2) (p 1,2) y i y p (2,3) (p 1,3) y i +1 y 1 (2,2) (p 1,2) y i z1 z p z1 z p (0,0) (d) xi xi ( p +1,0) (0,0) ( p+1,0) (e) Figure 4: (a) A doughnut embedding of a p-doughnut graph of G, (b) edges between four corner vertices of R 1 and R 2 are drawn as straight-line segments, (c) edges between vertices on R 1 and R 2 are drawn, (d) edges between four corner vertices of R 2 and R 3 are drawn as straight-line segments, and (e) a straight-line grid drawing of G. 11

12 (0,0), (p + 1, 0), (p + 1, 5) and (0, 5). Similarly the four corners of R 2 are (1, 1), (p, 1), (p, 4), (1, 4) and the four corners of R 3 are (2,2), (p 1, 2), (p 1, 3), (2,3). We first embed C 2 on R 2 as follows. Let z 1, z 2,..., z 2p be the vertices on C 2 in counter clockwise order such that z 1 has exactly one neighbor in C 1. We put z 1 on (1, 1), z p on (p, 1), z p+1 on (p, 4) and z 2p on (1, 4). We put the other vertices of C 2 on grid points of R 2 preserving the relative positions of vertices of C 2. We now put vertices of C 1 on R 1 as follows. Let be the neighbor of z 1 on C 1 and let, x 2,..., x p be the vertices of C 1 in counter clockwise order. We put on (0, 0) and x p on (0, 5). Since z 1 has exactly one neighbor on C 1, by Lemma 4.1, z 2p has exactly two neighbors on C 1 since z 2p and z 1 are consecutive vertices on C 2. Since z 1 and z 2p is on a triangulated face of G having vertices on both C 1 and C 2, is a neighbor of z 2p. One can easily observe that x p is the other neighbor of z 2p on C 1. Clearly the edges (, z 1 ), (, z 2p ), (x p, z 2p ) can be drawn as straight-line segment without edge crossing as illustrated in Fig. 4(b). We next put neighbors of z p and z p+1. Let x i be the neighbor of z p on C 1 if z p has exactly one neighbor on C 1, otherwise let x i be the left neighbor of z p on C 1. We put x i on (p + 1, 0) and x i+1 on (p + 1, 5). We now show that the edges from z p and z p+1 to x i and x i+1 can be drawn as straight-line segments without edge crossings. We first consider the case where z p has exactly one neighbor on C 1. In this case x i is the only one neighbor of z p on C 1. Then, by Lemma 4.1, z p+1 has two neighbors on C 1. Since each face having vertices on both C 1 and C 2 is triangulated, x i and x i+1 are the two neighbors of z p+1 on C 1. Clearly the edges (z p, x i ), (z p+1, x i ) and (z p+1, x i+1 ) can be drawn as straight-line segments without edge crossings, as illustrated in Fig. 4(b). We next consider the case where z p has two neighbors on C 1. In this case let x i and x i+1 be the two neighbors of z p on C 1. By Lemma 4.1, z p+1 has exactly one neighbor on C 1. Since the face of G containing z p, z p+1 and the neighbor of z p+1 on C 1 is triangulated, x i+1 is the neighbor of z p+1. Clearly the edges (z p, x i ), (z p, x i+1 ) and (z p+1, x i+1 ) can be drawn as straight-line segments without edge crossings as illustrated in Fig. 5. We put the other vertices of C 1 on grid points of R 1 arbitrarily preserving their relative positions on C 1. x p (0,5) R 1 R 2 z p ( p+1,5) x +1 i +1 z 2 p (1,4) (2,3) (2,2) R 3 (p 1,3) (p 1,2) p,4) ( z 1 (1,1) ( p,1) zp (0,0) ( p+1,0) x i Figure 5: Illustration of case where z p has two neighbors on C 1. 12

13 Since vertices z 2, z 3,..., z p 1 are put in counter clockwise order on the horizontal segment between points (1, 1) and (p, 1) on R 2 preserving the relative positions of vertices on C 2 and x 2, x 3,.., x i 1 are put on the horizontal segment between points (0, 0) and (p + 1, 0) of R 1 preserving the relative positions of the vertices on C 1, all edges of G connecting vertices in {z 2, z 3,..., z p 1 } to vertices in {x 2, x 3,.., x i 1 } can be drawn as straight-line segments without edge crossings. Similarly, the edges of G connecting vertices in {z p+1, z p+2,..., z 2p 1 } to vertices in {x i+2, x i+3,.., x p 1 } can be drawn as straight-line segments without edge crossings. See Fig. 4(c). We now put the vertices of C 3 on R 3 as follows. Since z 1 has exactly one neighbor on C 1, by Lemma 4.2(a), z 1 has exactly two neighbors on C 3. Then, by Lemma 4.2(b), z 2p has exactly one neighbor on C 3 since z 2p has exactly two neighbors on C 1. Let y 1, y 2,..., y p be the vertices on C 3 in counter clockwise order such that y 1 is the right neighbor of z 1. Then y p is the left neighbor of z 1. We put y 1 on (2, 2) and y p on (2, 3). Clearly the edges (y 1, z 1 ), (y 1, z 2p ), (y p, z 1 ) can be drawn as straight-line segements without edge crossings, as illustrated in Fig. 4(d). We next put neighbors of z p and z p+1 on C 3. Let y i be the neighbor of z p on C 2 if z p has exactly one neighbor on C 3, otherwise y i be the left neighbor and y i+1 be the right neighbor of z p on C 3. We put y i on (p 1, 2) and y i+1 on (p 1, 3), and hence the edges of G from z p and z p+1 to y i and y i+1 can be drawn as straight-line segments without edge crossing as illustrated in Fig. 4(d). We put the other vertices of C 3 on grid points of R 3 arbitrarily preserving their relative positions on C 3. Since vertices z 2, z 3,..., z p 1 are put in counter clockwise order on a horizontal segment between points (1, 1) and (p, 1) on R 2 preserving the relative positions of vertices and y 2, y 3,.., y i 1 are put on a horizontal segment between points (2, 2) and (p 1, 2) of R 3, all edges of G connecting vertices in {z 2, z 3,..., z p 1 } to vertices in {y 2, y 3,.., y i 1 } can be drawn as straight-line segments without edge crossings. Similarly, the edges of G connecting vertices in {z p+1, z p+2,..., z 2p 1 } to vertices in {y i+2, y i+3,.., y p 1 } can be drawn as straight-line segments without edge crossings. Fig. 4(e) illustrates the complete straight-line grid drawing of a p-doughnut graph. The area requirement of the straight-line grid drawing of a p-doughnut graph G is equal to the area of rectangle R 1 and the area of R 1 is = (p + 1) 6 = (n/4 + 1) 6 = O(n), where n is the number of vertices in G. Thus we have a straight-line grid drawing of a p-doughnut graph on a grid of linear area. Clearly the algorithm takes linear time. Thus the following theorem holds. Theorem 4.3 A doughnut graph G of n vertices has a straight-line grid drawing on a grid of 13

14 area O(n). Furthermore, the drawing of G can be found in linear time. 5 Conclusion In this paper we introduced a new class of planar graphs, called doughnut graphs, which is a subclass of 5-connected planar graphs. A graph in this class has a straight-line grid drawing on a grid of linear area, and the drawing can be found in linear time. We also showed that a doughnut graph is a 5-regular graph and the outerplanarity of a doughnut graph is 3. Although the class doughnut graphs contains exactly one graph for each integer p 3, the class contains an infinite number of graphs. Since a doughnut graph G has a straight-line grid drawing on a grid of linear area, any spanning subgraph of G has a straight-line grid drawing on a grid of linear area. Thus doughnut graphs and their spanning subgraphs constitute a wide class of graphs which admits a straight-line grid drawing on a grid of linear area. Our linear algorithm finds a straight-line grid drawing of a doughnut graph. If we can augment a spanning subgraph G of a doughnut graph G to the doughnut graph G, then our algorithm can be used to find a straight-line grid drawing of G on a grid of linear area. However recognizing a spanning subgraph of a doughnut graph and its augmentation look a non-trivial problem and it is left as an open problem. Finding other nontrivial classes of planar graphs which admit straight-line grid drawings on grids of linear area is also left as an open problem. References [BEGKLM03] F. Brandenburg, D. Eppstein, M.T. Goodrich, S. Kobourov, G. Liotta and P. Mutzel, Selected open problems in graph drawing, Proc. of Graph Drawing 2003, Lect. Notes in Computer Science, Springer, 2912, pp , [CN98] M. Chrobak and S. Nakano, Minimum-width grid drawings of a plane graphs, Comp.Geom. Theory and Appl., 11, pp , [DF05] G. Di Battista and F. Frati, Small area drawings of outerplanar graphs, Proc. of Graph Drawing 2005, Lect. Notes in Computer Science, Springer, 3843, pp , [DETT99] G. Di Battista, P. Eades, R. Tamassia, I. G. Tollis, Graph Drawing: Algorithms for the Visualization of Graphs, Prentice-Hall Inc., Upper Saddle River, New Jersey, [Far48] I. Fary, On Straight line representation of planar graphs, Acta Sci.Math. Szeged, 11, pp ,

15 [FPP90] H. de Fraysseix, J. Pach and R. Pollack, How to draw a planar graph on a grid, Combinatorica, 10, pp , [GR02] A. Garg and A. Rusu, Straight-line drawings of binary trees with linear area and arbitrary aspect ration, Proc. of Graph Drawing 2002, Lect. Notes in Computer Science, Springer, 2528, pp , [GR03a] A. Garg and A. Rusu, Area-efficient drawings of outerplanar graphs, Proc. of Graph Drawing 2003, Lect. Notes in Computer Science, Springer, 2912, pp , [GR03b] A. Garg and A. Rusu, A more practical algorithm for drawing binary trees in linear area with arbitrary aspect ratio, Proc. of Graph Drawing 2003, Lect. Notes in Computer Science, Springer, 2912, pp , [JM03] M. Jünger and P. Mutzel (Editor), Graph Drawing Software (Mathematics and Visualization), Springer, Berlin, [KW01] M. Kaufmann and D. Wagner (Eds.), Drawing Graphs: Methods amd Models, Lect. Notes in Computer Science, Springer, 2025, Berlin, [MNN01] K. Miura, S. Nakano and T. Nishizeki, Grid drawings of four-connected planar graphs, Discrete and Computational Geometry, 26(1), pp , [NR04] T. Nishizeki and M. S. Rahman, Planar Graph Drawing, World Scientific, Singapore, [Sch90] W. Schnyder, Embedding planar graphs on the grid, Proc., First ACM-SIAM Symp. on Discrete Algorithms, San Francisco, pp , [Ste51] K. S. Stein, Convex maps, Proc. Amer Math. Soc., 2, pp , [Wag36] K. Wagner, Bemerkugen zum veierfarbenproblem, Jasresber. Deutsch. Math-Verien., 46, pp , A Appendix We give the details of the proof of Property (a) in the proof of Lemma 3.2 as in the following lemma. 15

16 Lemma A.1 Let G be a graph constructed using the construction in the proof of Lemma 3.2. Then G is a 5-connected planar graph. Proof. We first show that G is 5-connected. It is sufficient to prove that there are at least 5 vertex-disjoint paths between any pair u, v of vertices in G. We have the following four cases to consider. Case 1: Both vertices u, v are either on C 1 or on C 3. We assume that both of u and v are on C 1, since the case where both of u and v are on C 3 is similar. Let u = x i and v = x j. Without loss of generality, we assume that i < j. In this case we find five vertex-disjoint paths between u and v as follows. P 1 = x i, x i+1,..., x j 1, x j ; P 2 = x i, x i 1,...,, x p,..., x j+1, x j ; P 3 = x i, z 2i, z 2i+1,..., z 2j 3, z 2j 2, x j ; P 4 = x i, z 2i 2, z 2i 3,..., z 2j+1, z 2j, x j ; P 5 = x i, z 2i 1, y (2i 1)/2, y (2i 1)/2 1,..., y (2j 1)/2 +1, y (2j 1)/2, z 2j 1, x j. Fig. 6 illustrates the case where u = x 2 and v = x 4. In this example P 1 = x 2, x 3, x 4 ; P 2 = x 2,, x 5, x 4 ; P 3 = x 2, z 4, z 5, z 6, x 4 ; P 4 = x 2, z 2, z 1, z 10, z 9, z 8, x 4 ; P 5 = x 2, z 3, y 2, y 1, y 5, y 4, z 7, x 4. x2 z 3 z2 z1 z 4 x3 z 5 y y 3 2 y 41 y 1 y 5 z 6 z 7 P 1 P2 P3 P4 P5 x4 z 10 x5 z 9 z 8 Figure 6: Illustration for Case 1 in the proof of (a) Case 2: Both vertices u, v are on C 2. Let u = z i and v = z j. We have noticed that a vertex on C 2 has two neighbors on C 1 and one neighbor on C 3 if the index of the vertex is even, and a vertex on C 2 has two neighbors on 16

17 C 3 and one neighbor on C 1 if the index of the vertex is odd. Therefore we have two subcases to consider. Subcase 2(a): i and j are odd. In this case, we find five vertex-disjoint paths between z i and z j as follows. P 1 = z i, z i+1,..., z j 1, z j ; P 2 = z i, z i 1,..., z j+1, z j ; P 3 = z i, x i/2, x i/2 1,..., x j/2 +1, x j/2, z j ; P 4 = z i, y i/2, y i/2 1,..., y j/2 +2, y j/2 +1, z j ; P 5 = z i, y i/2 +1, y i/2 +2,..., y j/2 1, y j/2, z j. Fig. 7(a) illustrates the case where u = z 3 and v = z 7. In this example P 1 = z 3, z 4, z 5, z 6, z 7 ; P 2 = z 3, z 2, z 1, z 10, z 9, z 8, z 7 ; P 3 = z 3, x 2,, x 5, x 4, z 7 ; P 4 = z 3, y 2, y 1, y 5, z 7 ; P 5 = z 3, y 3, y 4, z 7. x2 z 3 z2 z 4 z 5 y y 3 2 x3 z 6 x2 z 3 z2 z 4 z 5 y y 3 2 x3 z 6 P 1 P2 P3 P4 P5 z1 y 1 y 5 y 41 z 7 x4 z1 y 1 y 5 y 41 z 7 x4 z 10 z 9 z 8 z 10 z 9 z 8 x5 (i) x5 (ii) Figure 7: Illustration for (i) Subcase 2(a) and (ii) Subcase 2(b) in the proof of (a) Subcase 2(b): i is even and j is odd. The vertex-disjoint five paths are as follows. P 1 = z i, z i+1,..., z j 1, z j ; P 2 = z i, x i/2+1, x i/2+2,..., x j/2 1, x j/2, z j ; P 3 = z i, x i/2, x i/2 1,..., x (j+1)/2 +1, x (j+1)/2, z j+1, z j ; P 4 = z i, z i 1, y (i 1)/2, y (i 1)/2 1,..., y j/2 +2, y j/2 +1, z j ; P 5 = z i, y i/2+1, y i/2+2,..., y j/2 1, y j/2, z j. Fig. 7(b) illustrates the case where u = z 2 and v = z 7. In this example P 1 = z 2, z 3, z 4, z 5,z 6, z 7 ; 17

18 P 2 = z 2, x 2, x 3, x 4, z 7 ; P 3 = z 2,, x 5, z 8, z 7 ; P 4 = z 2, z 1, y 1, y 5,z 7 ; P 5 = z 2, y 2, y 3, y 4, z 7. Case 3: The vertex u either on C 1 or C 3 and v on C 2. We assume that u is on C 1 and v is on C 2, since the case where u is on C 3 and v on C 2 is similar. Let u = x i and v = z j. In this case, we have two subcases to consider. Subcase 3(a): j is even. In this case, five vertex-disjoint paths are as follows. P 1 = x i, x i+1,..., x j/2 1, x j/2, z j ; P 2 = x i, x i 1,..., x j/2+2, x j/2+1, z j ; P 3 = x i, z 2i 2, z 2i 3,..., z j+1, z j ; P 4 = x i, z 2i, z 2i+1,..., z j 1, z j ; P 5 = x i, z 2i 1, y (2i 1)/2, y (2i 1)/2 +1,...y j/2, y j/2+1, z j. Fig. 8(a) illustrates the case where u = x 2 and v = z 8. In this example P 1 = x 2, x 3, x 4, z 8 ; P 2 = x 2,, x 5, z 8 ; P 3 = x 2, z 2, z 1, z 10, z 9, z 8 ; P 4 = x 2, z 4, z 5, z 6, z 7, z 8 ; P 5 = x 2, z 3, y 2, y 3, y 4, y 5, z 8. x2 z 3 z2 z1 z 4 x3 z 5 y y 3 2 y 41 y 1 y 5 z 6 z 7 x4 x2 z 3 z2 z1 z 4 x3 z 5 y y 3 2 y 41 y 1 y 5 z 6 z 7 x4 P 1 P2 P3 P4 P5 z 10 z 9 z 8 z 10 z 9 z 8 x5 (i) x5 (ii) Figure 8: Illustration for (i) Subcase 3(a) and (ii) Subcase 3(b) in the proof of (a) Subcase 3(b): j is odd. In this case, five vertex-disjoints paths are as follows. P 1 = x i, x i+1,..., x j/2 1, x j/2, z j ; P 2 = x i, x i 1,..., x (j+1)/2 +1, x (j+1)/2, z j+1, z j ; 18

19 P 3 = x i, z 2i, z 2i+1,..., z j 1, z j ; P 4 = x i, z 2i 2, z 2i 3, y (2i 3)/2, y (2i 3)/ , y j/2, y j/2 +1, z j ; P 5 = x i, z 2i 1, y (2i 1)/2 +1, y (2i 1)/2 +2,..., y j/2 1, y j/2, z j. Fig. 8(b) illustrates the case where u = x 2 and v = z 9. In this example P 1 = x 2, x 3, x 4, x 5, z 9 ; P 2 = x 2,, z 10, z 9 ; P 3 = x 2, z 4, z 5, z 6, z 7, z 8, z 9 ; P 4 = x 2, z 2, z 1, y 1, z 9 ; P 5 = x 2, z 3, y 3, y 4, y 5, z 9. Case 4: The vertex u on C 1 and v on C 3. Let us assume that u = x i and v = y j. In this case, vertex-disjoint five paths are as follows. P 1 = x i, x i+1,..., x j 2, x j 1, z 2j 2, y j ; P 2 = x i, x i 1,...x (2j 1)/2 +1, x (2j 1)/2, z 2j 1, y j ; P 3 = x i, z 2i, z 2i+1,..., z 2j 4, z 2j 3, y j ; P 4 = x i, z 2i 2, z 2i 3..., z 2j+2, z 2j+1, y j+1, y j ; P 5 = x i, z 2i 1, y (2i 1)/2, y (2i 1)/2 +1,..., y j 1, y j. Fig. 9 illustrates the case where u = x 2 and v = y 4. In this example P 1 = x 2, x 3, z 6, y 4 ; P 2 = x 2,, x 5, x 4, z 7, y 4 ; P 3 = x 2, z 4, z 5, y 4 ; P 4 = x 2, z 2, z 1, z 10, z 9, y 5, y 4 ; P 5 = x 2, z 3, y 2, y 3, y 4. x2 z 3 z2 z 4 z 5 y y 3 2 x3 z 6 P 1 P2 P3 P4 P5 y 41 z1 y 1 y 5 z 7 x4 z 10 x5 z 9 z 8 Figure 9: Illustration for Case 4 in the proof of (a) Thus we can find five vertex-disjoint paths for each pair of vertices in graph G. Therefore, the graph G is 5-connected. 19

20 It is remained to prove that G is planar. From the construction, we add edges between the vertices on C 1, C 2 and between the vertices on C 2, C 3. There is no edge between the vertices on C 1 and C 3. To prove that G is planar, we need to show that the edges between vertices on C 1 and C 2 can be drawn without edge crossings and edges between vertices on C 2 and C 3 can be drawn without edge crossings. Consider the edges between vertices on C 1 and C 2. For any two edges (z i, x i ) and (z j, x j ) between C 1 and C 2, if z i = z j then x j follows x i in clockwise order according to our construction and if z j appears later than z i in clockwise order of vertices on C 2 then either x i = x j or x j appears later than x i in clockwise order of vertices on C 1. Hence the two edges (z i, x i ) and (z j, x j ) can be drawn without edge crossings. Similarly we can prove that the edges between the vertices on C 2 and C 3 can be drawn without edge crossings. Therefore G is planar. Therefore the graph G is 5-connected planar graph. Q.E.D. 20