Planar Drawings and Angular Resolution: Algorithms and ... - CiteSeerX
Planar Drawings and Angular Resolution: Algorithms and Bounds? (Extended Abstract)
Ashim Garg and Roberto Tamassia Department of Computer Science Brown University Providence, RI 02912{1910, USA
Abstract. We investigate the problem of constructing planar straightline drawings of graphs with large angles between the edges. Namely, we study the angular resolution of planar straight-line drawings, de ned as the smallest angle formed by two incident edges. We prove the rst nontrivial upper bound on the angular resolution of planar straight-line drawings, and show a continuous trade-o between the area and the angular resolution. We also give linear-time algorithms for constructing planar straight-line drawings with high angular resolution for various classes of graphs, such as series-parallel graphs, outerplanar graphs, and triangulations generated by nested triangles. Our results are obtained by new techniques that make extensive use of geometric constructions.
1 Introduction
Coping with the nite resolution of display devices and of the human eye is a fundamental problem in graph drawing. Namely, in visualization applications, it is important to construct drawings of graphs that avoid placing vertices and edges too close. In this paper we investigate the problem of constructing planar straight-line drawings with with large angles between the edges. Namely, we study the angular resolution of straight-line drawings, de ned as the smallest angle formed by two incident edges. Besides visualization applications (see, e.g., [5]), constructing drawing with high angular resolution is important in the design of optical communications networks (see, e.g., [8]). The study of the angular resolution of drawings has attracted considerable interest in the last years. Formann, Hagerup, Haralambides, Kaufmann, Leighton, Simvonis, Welzl, and Woeginger [8] were the rst to study the angular resolution of (generally nonplanar) straight-line drawings of various classes of graphs. They show that a degree-d graph has a straight-line drawing with angular resolution (1=d2), and every degree-d planar graph has a straight-line (nonplanar) ? Research supported in part by the National Science Foundation under grant CCR-
9007851, by the U.S. Army Research Oce under grants DAAL03-91-G-0035 and DAAH04{93{0134, and by the Oce of Naval Research and the Defense Advanced Research Projects Agency under contract N00014-91-J-4052, ARPA order 8225. The Email addresses of the authors are [email protected] and [email protected].
drawing with angular resolution (1=d). (Note that the above bounds are independent from the number of vertices of the graph.) They also pose the open problem of characterizing the angular resolution of planar straight-line drawings. Malitz and Papakostas [15] make further progress on the above open problem by proving a lower bound on the angular resolution dependent only on the degree of the graph. Namely, they show that a degree-d planar graph admits a planar straight line drawing with angular resolution (1=7d ), irrespectively of the number of vertices. They also analyze a linear program associated with two necessary conditions on the angles of a planar triangulation (i.e., the sum of the angles around a vertex is 2, and the sum of the angles inside each triangle is ), and nd that such a linear program admits a solution with (1=d) angles. This leads them to conjecture that the trivial (1=d) lower bound on the angular resolution may be achievable for planar straight-line drawings. Kant [10, 11] shows that testing whether a biconnected planar graph admits a planar straight-line drawing with angular resolution greater than or equal to a given constant is NP-hard, thus providing further motivation to the study of asymptotic bounds. He also considers polyline drawings (where edges are drawn as polygonal chains), and shows that a degree-d triconnected planar graph admits a planar polyline grid drawing with angular resolution (1=d). Di Battista and Vismara [7] provide a characterization of the angles in planar straight-line drawings of maximal planar graphs through a nonlinear system of inequalities. As a consequence, they show that testing whether a prescribed assignment of angles to a maximal planar graph is achievable in a planar straightline drawing can be done in linear time, and that constructing a planar straightline drawing with maximum angular resolution can be reduced to the solution of a nonlinear optimization problem. Our results can be summarized as follows: In Section 3, we prove the rst nontrivial upper bound on the angular resolution of planar straight-line drawings by p exhibiting a family of degree-d planar graphs that require angular resolution O( log d=d3) in any planar straight-line drawing. Previously, only the trivial O(1=d) bound was known, and determining whether it is achievable was posed as an open problem in [15, 11]. In Section 4, we show a continuous trade-o between the area and the angular resolution of planar straight-line drawings. Namely, there exists a constant c > 1 such that, for any n > d > 6, with n ? 3 a multiple of d ? 4, there exists a planar graph Gdn with n vertices and degree d such that any planar straight-line drawing of Gdn with angular resolution has area (cn ). In particular, this implies that there exist bounded-degree planar graphs for which asymptotically optimal resolution can be achieved only at the expense of an exponential increase in the area. Previously, area/trade-o results were proved for planar drawings (see, e.g., [1, 2, 3, 6, 4, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19]), depending on various restrictions on the representation (e.g., upward, straight-line, orthogonal, polyline). Our result is the rst \continuous" area/trade-o result for planar drawings. In Section 5, we give linear-time algorithms for constructing planar straight-
line drawings with high resolution of various classes of graphs, such as seriesparallel graphs, outerplanar graphs, and nested-star graphs (i.e., maximal planar graphs generated from a triangle by repeated insertions of stars). The angular resolution is shown to be (1=d2) for series-parallel and nested-star graphs, and =(d ? 1) for outerplanar graphs. Previously, Malitz and Papakostas [15] showed that angular resolution =2(d ? 1) can be achieved for outerplanar graphs. No polynomial bounds in 1=d were known for series-parallel graphs and nestedstar graphs. We leave it as an open problem either proving or disproving that
(1=d2) is a tight lower bound on the angular resolution of the series-parallel and nested-star graphs. Note that the family of graphs requiring angular resolution p O( log d=d3) in any planar straight-line drawing are actually nested-star graphs. The signi cance of our result on nested-star graphs is motivated by the following observation: Malitz and Papakostas [15] show that a disk packing of a planar degree-d graph yields a planar straight-line drawing with angular resolution (1=7d ), and they exhibit a family of nested-star graphs to prove that this exponential bound is the best that can be obtained with the disk packing method. Our results imply that the disk packing method yields suboptimal results and leave as an open problem whether the angular resolution of planar straight-line drawings can be bounded from below by a polynomial in 1=d. Our algorithms for drawing nested-star and series-parallel graphs exploit various geometric properties of the angles in certain triangulations. Since most existing algorithms for drawing planar graphs rely on graph-theoretic properties (e.g., orientation, coloring, ow), the techniques in this paper appear to be interesting in the way they blend geometric and graph-theoretic methods. We believe that our results will stimulate further research in this direction. Due to the space limitations, in this extended abstract we have either sketched the proofs or omitted them.
2 De nitions
First, we review basic graph drawing terminology [5]. A straight-line drawing maps each edge into a straight-line segment. The angular resolution of a straightline drawing is the smallest angle formed by two edges incident on the same vertex. A drawing is planar if no two edges intersect. The area of a drawing is the area of the smallest convex polygon covering the drawing. Whenever we give lower bounds on the area, we assume that the drawing is constrained by some rule that prevents it from being arbitrarily scaled down (e.g., integer coordinates for the vertices, or a minimum unit distance between any two vertices). The results of Section 4 are fully general, since they hold under any rule that implies a nite minimum area for the drawing of a graph. The degree of a graph is the maximumdegree of any vertex in the graph. The notation dG(u) indicates the degree of vertex u in graph (or subgraph) G. Given a line segment with endpoints A and B, AB denotes either the line segment itself or its length. Given two line segments AB and BC, 6 ABC denotes either the angle between AB and BC or its measure. Finally, ABC denotes the triangle with vertices A, B and C.
3 Upper Bound on the Angular Resolution
In this section we show that there p exists an in nite family of degree-d graphs that require angular resolution O( log d=d3 ) in any planar straight-line drawing. Thus, for these graphs the linear program of [15] gives an inconsistent drawing. First, we give the following geometric lemma (see Fig. 1(a)). uk
αk u0
A
B
β1 β2 β
α α2 α1 D
γ2 γ1
(a)
γ
v C 0
φk
θk
α0 G0 β0 γ0 w 0
x Gk-1
z
Gk
Gk-1 Gk-1
(b) vk
s
y
βk
γk
wk
(c)
Figure 1: (a) Dividing a Triangle; (b) Graph G0 ; (c) Graph Gk ; Lemma 1. Let D be a point inside a triangle ABC . Let 1, 2, 1 , 2 , 1 and 2 be the angles de ned in Fig. 1(a). If =2 and 2= 1 1, then 2 r 1
2 2 min( ; ) 4 (1) 1
1
2
We are now ready to present the main theorem of this section. Theorem 2. There exists an in nite family of planar p graphs with degree d and (3d=9 ) vertices that require angular resolution O( (log d)=d3 ) in any planar
straight-line drawing. Proof. Consider graph Gk , which is recursively de ned as follows: G0 is a triangle
with vertices u0, v0 and w0 (see Fig. 1(b)). Graph Gk is assembled from three copies of Gk?1, denoted H1, H2 and H3, as shown in Fig. 1(c)). Simple recurrence relations can be used to show that Gk has degree 9k and (3k ) vertices. Given a straight line drawing of Gk with uk vk wk on the external face, we de ne the following angles (see Fig. 1(c)): k = 6 wk uk vk , k = 6 uk vk wk , k = 6 vk wk uk , k = 6 wk uk s and k = 6 vk uk x. We now claim that the angular resolution of p any planar straight-line drawing of Gm with umvm wm as the external face is O( (log m)=m3 ). Assume without any loss of generality that m =2 because in any p triangle at least one internal angle has value not exceeding =2. If 0 is O( (log m)=m3 ) the claim holds. 3=2 So suppose p 0 = 3 (1=m ). We shall nd another angle in the drawing that is O( (logm)=m ). Namely, for any k Q m, let rk = k?1=( k + k ). Thus k?1 = k rk =(rk + 1). We have 0 = m k1 rk =(rk + 1). Since m < and 0 = (1=m3=2), we obtainm Y 3=2 k=1
rk =(rk + 1) = (1=m )
(2)
We now prove that there exists anQ i m=2 such that ri = (m= log m). For any k, from Eq. 2 we get that mj=k rj =(rj + 1) = (1=m3=2) because
rq =(rq + 1) < 1 for any q. If r(k) = maxmj=k (rj ), we get fr(k)=(r(k) + 1)gm?k =
(1=m3=2). Consequently m?k = O(m3=2 ) 1 + r(1k)
(3)
From Eq. 3, with simple manipulations we obtain r(m=2) = (m= logm). Thus there is anpindex i m=2 such that ri = (m= log m). By Lemma 1, either i?1 i 2=4( ri + 2=4) or i?1 i 2 =4(pri + 2=4). Since di?1(vi?1) = di?1(wi?1) = 3i ? 1, both i?1 and i?1 include 3i ? 2 3m=2 ? 2 angles in their interior. Hence p for suciently large m, there is an angle inside either i?1 or i?1 that is O( (log m)=m3 ). This concludes the proof of the claim. It can be readily seen that a planar straight-line drawing of Gm where um vm wm is not the external face contains as a subdrawing a drawing of Gm?1 with external face um?1vm?1 wpm?1. Hence, any planar straight-line drawing p of Gm has angular resolution O( (log(m ? 1))=(m ? 1)3 ) = O( (log m)=m3 ). We conclude by recalling that graph Gm has degree (m) and (3m ) vertices.
4 Tradeo between Area and Angular Resolution
In this section we show that there exists a continuous tradeo between the area requirement and the angular resolution of a planar straight-line drawing. Namely, for every n > d > 0, we show that there exists a planar graph Gdn with n vertices and degree d such that any planar straight-line drawing of Gdn with resolution requires area (cn ), where c is a constant greater than 1. For a xed d, we recursively de ne graph Gk as follows. Graph G0 is a triangle consisting of vertices A0 , B0 , and C0 . Graph Gk is obtained from Gk?1 as shown in Fig. 2(a). For k 2, Gk has k(d ? 4) + 3 vertices and degree d. Graph Gdn is de ned as G nd??43 , whenever n ? 3 is a multiple of d ? 4. First, we give a technical lemma, which uses the notation of Fig. 2. Ak
h0 a1 b2
f0 A0 Gk-1
b1
B0
f1
Gk
bi
C1 C2
d1
d2
e1
e2
Gk, i
Gk, i-1
c2
B2
fi
ai
c1
C0
B1 B
Ak
d-1 edges
a0
Bi
Bi-1
Ci-1
ei di
C
ci Ci
(b)
(a)
Figure 2: (a) Graph G(d; n); (b) Introducing vertices and edges Lemma 3. Given a planar straight line drawing of Gk , let ki be the area of Ak Bi Ci , with 1 i (d ? 4)=2, then inpat least one of the following pairs of
edges, both edges have length greater than 2ki : (ai ; ai+1), (fi ; fi+1), (di; ei+1 ), and (di ; bi+1).
We are now ready to state the main result of this section:
Theorem 4. There exists a constant c > 1 such that, for any n > d > 6, with n ? 3 a multiple of d ? 4, there exists a planar graph Gdn with n vertices and
degree d such that any planar straight-line drawing of Gdn with angular resolution has area (cn ). Proof. Consider a planar straight line drawing of Gk with Ak Bk Ck as the exter-
nal face for some k = (n ? 3)=(d ? 4). We denote with k the area of Gk , and with ki the area of triangle AkpBi Ci p (see Fig. 2). By Lemma 3, we have ki+1 > ki + 21 2ki 2ki sin > ki (1 + sin ):
(4)
and k1 = k?1, we obtain the recurrence d?4 2 +1 k > k?1 (1 + sin ) 2 : (5) Recalling the recursive dconstruction of G , we get k d?4 n?3 n?3 ?4 (6) k > 0 (11 + sin ) 2 k = 0 (1 + sin ) 2 d?4 = 0 (1 + sin ) 2 n Since sin 2 and =3, we have that k > r for some constant r > 1. Finally, we observe that in any planar drawing of Gk with Ak Bk Ck not being the external face, there is a subgraph of Gk isomorphic to Gk=2, that is drawn with Ak=2 Bk=2Ck=2 as the external face. We conclude that any planar straight line drawing of Gk has area (cn ) for some constant c > 1. Since k = kd?4
Note that Theorem 4 is fully general, since it holds under any rule that implies a nite minimum area for the drawing of a graph.
Corollary 5. There exists an in nite family of bounded-degree graphs that re-
quire exponential area in any planar straight-line drawing with bounded (from below) angular resolution.
5 Constructing Drawings with High Angular Resolution
In this section we give linear-time algorithms for constructing planar straightline drawings with good angular resolution for various classes of planar graphs. We show that every nested-star graph and series-parallel graph admits a planar straight-line drawing with angular resolution (1=d2), and that every outerplanar graph has a planar straight-line drawing with angular resolution =(d ? 1), where d denotes the degree of a graph.
5.1 A Geometric Lemma
In this section we give a geometric lemma that will be used in our algorithms for drawing nested-star and series-parallel graphs. C
γ/2
γ
∆ D1 α2
α0
D2
β2
β0
B
A α α1
p
β1 β
Figure 3: Triangle in Lemma 6
Lemma 6. Let = ABC be a triangle with internal angles , and where =2, as shown in Fig. 3. Let p be the bisector of angle , and D1 and D2
be any two points on p inside . Let 0 = 6 D1 AD2 and 0 = 6 D1 BD2 . Then
0 1 0 42
5.2 Nested-Star Graphs
A nested-star graph G is de ned recursively as follows: G has three distinguished vertices called its apex vertices (denoted by u, v and w in Fig. 4). G either consists of a single triangle (see Fig. 4(a)) or it is assembled from three nested-star graphs G1, G2 and G3, as shown in Fig. 4(b). Lemma 7, which can be easily proved by induction on the number of vertices in G, states a property of nested-star graphs that is used by our drawing algorithm. w
w
G
G1 z G2
v
u
G
G3
u
v
(a)
(b) γ/2
∆
α
∆ v (c)
γ
V x1 1 xm-1 Vm Um W xm p E
U1
w
u
w=x0
u
α1 αm
β1 βm
β v
(d)
Figure 4: A nested-star graph G: (a) as a single triangle; (b) formed by three nestedstar graphs; Drawing G when (c) it is a single triangle; (d) it consists of nested-star subgraphs
Lemma 7. Let G be a nested-star graph with apex vertices u, v and z. The vertices of G that are common neighbors of u and v form a path.
We now describe a recursive algorithm that constructs a drawing of a nestedstar graph G with angular resolution 1=(24d2), where d is the degree of G. Let u, v and w be the apex vertices of G. Given a triangle , which is said to be allocated to G, we show how to draw G inside (see Fig. 4(c){(d)). The vertices of are u, v and w. If G has three edges, then its drawing is (see Fig. 4(c)). Otherwise, we draw G as follows (see Fig. 4(d)): Let , and be the internal angles of at u, v and w respectively. We assume without loss of generality that angle . By Lemma 7, let x0 = w; x1; x2; : : :; xm be the path formed by the common neighbors of u and v. Let Ui be the nested-star
subgraph of G with xi?1, u and xi as its apex vertices. Let Vi be the nested-star subgraph of G with xi?1, v and xi as its apex vertices. Notice that uxm v is a face of G. Let W be the nested-star subgraph of G which consists only of the triangle uxm v. Let p be a line through w that bisects angle . For each i 1, place xi on line p in the interior of such that 6 xi?1uxi is (dUi (u) ? 1) =(dG(u) ? 1). (Recall that dG (u) denotes the degree of vertex u in graph G.) We allocate triangle xi?1uxi to Ui and triangle xi?1vxi to Vi , and then draw Ui and Vi recursively in their allocated triangles. Theorem 8. A nested-star graph with degree d admits a planar straight-line drawing with angular resolution 1=(24d2) that can be constructed in linear time. Proof. Allocate an equilateral triangle to the graph and then apply the above algorithm. We prove that the resulting drawing has angular resolution 1=(24d2). Intuitively, when allocating a triangle to a nested-star subgraph H of G, the algorithm allows the internal angle at an apex vertex of H to become \small",but no smaller than 1=(24d). The algorithm then divides the angle evenly between the edges incident on the vertex in H, so that the minimum angle is at least 1=(24d2). The proof is based over maintaining the following invariant: Invariant 1: Let uvw be a triangle allocated to a nested-star graph G with apex vertices u, v and w. If u is designated the critical vertex of G then { (dG(u) ? 1)=(24d2) 6 vuw =6, and { 6 uvw =6 and 6 uwv =6. Intuitively, the algorithm divides the angle at a critical vertex evenly between the incident edges by allocating an angle to each Ui proportional to the degree of the vertex in Ui . The key to proving Invariant 1 is showing that the angle allocated to each Vi is at least 1=(24d). We now prove that Invariant 1 holds for every nested-star subgraph of the input graph. Initially designate any apex vertex of the input nested-star graph to be the critical vertex. Invariant 1 holds trivially for the input graph. we now prove that the Invariant 1 is maintained at each recursion. Suppose Invariant 1 holds for a nested-star subgraph G with apex vertices u, v and w. Let uvw be the allocated triangle of G. Let u be the critical vertex of G. Suppose we denote the internal angles of uvw at u, v and w by , and respectively (see Fig. 4(d)). Let . Therefore =3 and =2. The algorithm allocates triangles xi?1uxi and xi?1vxi to the nested-star subgraphs Ui and Vi respectively. Let E be the intersection of the bisector p of with the line segment uv. We rst show that Invariant 1 holds for Ui . Let i = 6 xi?1uxi . 6 uxi?1xi 6 uwx1 = =2 =6. 6 uxixi?1 6 uEw = ? =2 ? =6. i = (dUi (u) ? 1) =(dG(u) ? 1). We designate u as the critical vertex of Ui . Because Invariant 1 holds for G, (dG (u) ? 1)=24d2. Hence i dUi (u)=24d2. Therefore Invariant 1 holds for Ui . Now we show that Invariant 1 holds for Vi . We designate v as the critical vertex of Vi . Let i = 6 xi?1vxi . 6 vxi?1 xi 6 vwx1 = =2 =6. 6 vxi xi?1
=6. From Lemma 6, i ( i =42 ) . Since =6 and i = (dUi (u) ? 1) =(dG(u) ? 1), (u) ? 1 i 41 2 ddUi(u) G ?1 6 since dG(u) d, dUi (u) 2 and dVi (v) d, We have i (1=(24d)
6 vEw = ? =2 ?
dVi (v)=(24d2 ) Therefore Invariant 1 holds for Vi . We can similarly show using Lemma 6 that Invariant 1 holds for W. Finally we show using induction over the number of vertices in G that the minimum angle in the drawing of G constructed by the algorithm is at least 1=24d2. The base case is when G consists of a single triangle . Each internal angle of is at least 1=24d2 because of Invariant 1. In a general inductive step let G consists of nested-star graphs W, U1; U2 ; : : :; Um and V1 ; V2 ; : : :; Vm , where Ui Vi , and W are as described in the algorithm. Since the drawing of each Ui and Vi , and W has minimum angle at least 1=24d2 by our inductive hypothesis, we conclude that the minimum angle in the drawing of G is at least 1=24d2.
5.3 Series-Parallel Graphs
A series-parallel graph G is de ned recursively as follows [1]: It contains a pair of distinguished vertices called its poles. It consists of either a single edge, or of a series composition, or of a parallel composition of two series-parallel graphs. For us the following equivalent de nition is more useful: It consists of either a single edge, or a series composition of two series-parallel graphs, or a parallel composition of series-parallel graphs G1; G2; : : :; Gm where each Gi is either a single edge or a series composition of two series-parallel graphs. A series-parallel subgraph of G is a subgraph of G that is also a series parallel graph. We now describe a recursive algorithm that constructs a drawing of a seriesparallel graph G with angular resolution 1=(48d2), where d is the degree of G. Let s and t be the poles of G. We allocate a triangle to G and draw G recursively inside . Two vertices of are s and t (see Fig 5(a)). Let the third vertex of be r. One internal angle of is =3 (6 rts in Fig 5(a)). We consider three cases: { G is a single edge (s; t): We draw it as a straight line between s and t (see Fig 5(b)). { G is a series composition of two series-parallel graphs G1 and G2, where s is a pole of G1 and t is a pole of G2: We draw G as shown in Fig. 5(c). We assume without any loss of generality that 6 rst 6 rts. Let r2 be the midpoint of the line segment rt. Let 2 be the line passing through r2 making an angle =3 with the line st. Let u be the intersection of 2 and st. Let 1 be the line parallel to the line rt passing through u. r1 is the intersection of 1 and the line sr. We allocate triangle sur1 to G1 and triangle tur2 to G2 and draw G1 and G2 recursively in their allocated triangles. { G is a parallel composition of the series-parallel graphs G1; G2; : : :; Gm : Since G does not have multiple edges, there can be only two subcases: (i) No Gi consists of a single edge, or (ii) one and only one Gi consists of a single edge
s
π/3
∆
G π/3
Hm t
(a)
Hi α
G
αi
t
(b)
pi
ri
H1 s
π/3
qi π/3
r1
r'1
π/3
π/3 q1 p
s (a) r
λ1
λ2
ηi
r1 ∆ G1 s
u β
G µi
qm r'm π/3 Km
λi
r'i
r ∆
γ/2
rm
r
∆
r
γ
ηi
r2
G
G2 π/3
Hm β
γ/2 π/3
∆
βi K1
t
G qm r'm π/3 pi
ri
Km
µi
λi
r'i
t
(c)
Figure 5: Drawing a
series-parallel graph G: (a) Triangle allocated to G; (b) Drawing G when G is a single edge; (c) Drawing G when G is a series composition;
β=π/3
r
γ rm
Ki
Hi α
s
αi
H2 α/(dG(s)-1) (b)
π/3
qi π/3
r2
r'2
π/3
π/3 q2 p
Ki βi
β=π/3
K2 t
Figure 6: Drawing a series-parallel graph G when it is a parallel composition of sp-graphs G1 ; G2 ; : : : ; Gm : (a) Case i; (b) Case ii.
and without any loss of generality let G1 be that subgraph. In both subcases i and ii, if Gi does not consist of a single edge, it is a series composition of two series parallel subgraphs. Let us denote the subgraph whose pole is s by Hi and the other subgraph whose pole is t by Ki . Let qi be the pole common to Hi and Ki . Fig. 6(a) shows the drawing of G in case i and Fig. 6(b) shows the drawing of G in case ii. In both the cases, if ui does not consist of a single edge (i.e., it is a series composition), we allocate a triangle sri qi to Hi and a triangle sri0 qi to Ki , and recursively draw Hi and Ki in their allocated triangles. The position of each ri, qi and ri0 is determined as follows: Assume without any loss of generality that 6 rst 6 rts. Let = 6 rst. Let p be the bisector of angle 6 srt. Draw lines i and i from s such that (a) the angle between i and i is (2dHi (s) ? 1) =(2dG(s) ? 1) and (b) angle between i and i+1 is =(2dG(s) ? 1). (Recall that we denote by dG(u), the degree of vertex u in G.) If u1 consists of a single edge then G1 is drawn as a straight line between s and t and 6 q2st is =(2dG(s) ? 1) (Fig. 6(b)). qi is the intersection of i with p. ri is a point on i such that 6 ri qis is =3. Let pi be the intersection of p and i . Let i be the line passing through t and
pi . ri0 is a point on i such that 6 ri0 qit is =3. Theorem9. A series-parallel graph with degree d admits a planar straight-line drawing with angular resolution 1=(48d2) that can be constructed in linear time. Proof. The proof is similar to the proof of Theorem 8 and uses Lemma 6.
5.4 Outerplanar Graphs
In this section we give a linear time algorithm that constructs a drawing of a maximal outerplanar graph G with angular resolution =(d ? 1), where d is the degree of G. This improves the lower bound given in [15] by a factor of two. Our algorithm constructs a breadth- rst search tree T(G) of G and places the vertices on concentric circles in the drawing so that the vertices equidistant from the root of T(G) are placed on the same circle. A sketch of the algorithm is as follows: π/(d-1)
w=v1 v2
vp
π/(d-1)
uj π/(d-1)
uj-1
π/(d-1)
uj+1
tangent of Ci-1 at uj+1
r Ci
C1
Ci-1
(a) (b)
Figure 7: Drawing an outerplanar graph G: (a) Level 1; (b) Level i 1. Construct a breadth- rst search tree and assign vertices to levels such that vertices equidistant from the root of the tree get placed at the same level. 2. Construct a drawing as follows (see Fig. 7): Let C1 be a circle. Place the root of the tree at the center of C1. Place the vertices of level 1 at equidistant positions on C1 such that the angle subtended on the center of the circle by consecutive vertices is =(d ? 1) (see Fig. 7(a)). Now, suppose we have already placed the vertices of levels 1 to i ? 1 on concentric circles C1 ; C1; : : :; Ci?1. Let u1 ; u2; : : :; um be the left-to-right order of the vertices on Ci?1 (see Fig. 7(b)). Draw lines from each uj making angles =(d ? 1) and ?=(d ? 1) respectively with the tangents at uj . If uj and uj ?1 have a common neighbor (e.g., vertex w in Fig. 7(b)) on level i, then place it at the point of intersection of the lines drawn from uj and uj ?1. Let Ci be the circle concentric to Ci?1 passing through these points. Let v1 ; v2; : : :; vp be the neighbors on level i of uj . Place each vk on Ci such that if there is an edge (vk ; vk+1) in G, then 6 vk uj vk+1 is =(d ? 1). Theorem10. A maximal outerplanar graph with degree d admits a planar straight-line drawing with angular resolution =(d ? 1) that can be constructed in linear time.
References
1. P. Bertolazzi, R. F. Cohen, G. Di Battista, R. Tamassia, and I. G. Tollis. How to draw a series-parallel digraph. Proc. 3rd Scand. Workshop Algorithm Theory, vol. 621 of Lecture Notes in Computer Science, pp. 272{283. Springer-Verlag, 1992. 2. R. P. Brent and H. T. Kung. On the area of binary tree layouts. Inform. Process. Lett., 11:521{534, 1980. 3. P. Crescenzi, G. Di Battista, and A. Piperno. A note on optimal area algorithms for upward drawings of binary trees. Comp. Geom. Theory Appl., 2:187{200, 1992. 4. H. de Fraysseix, J. Pach, and R. Pollack. Small sets supporting Fary embeddings of planar graphs. Proc. 20th ACM Sympos. Theory Comput., pp. 426{433, 1988. 5. G. Di Battista, P. Eades, R. Tamassia, and I. G. Tollis. Algorithms for drawing graphs: an annotated bibliography. Preprint, Dept. Comput. Sci., Brown Univ., Providence, RI, November 1993. To appear in Comput. Geom. Theory Appl. Preliminary version available via anonymous ftp from wilma.cs.brown.edu, gdbiblio.tex.Z and gdbiblio.ps.Z in /pub/papers/compgeo. 6. G. Di Battista, R. Tamassia, and I. G. Tollis. Area requirement and symmetry display of planar upward drawings. Discrete Comput. Geom., 7:381{401, 1992. 7. G. Di Battista and L. Vismara. Angles of planar triangular graphs. Proc. 25th ACM Sympos. Theory Comput. (STOC 93), pp. 431{437, 1993. 8. M. Formann, T. Hagerup, J. Haralambides, M. Kaufmann, F. T. Leighton, A. Simvonis, E. Welzl, and G. Woeginger. Drawing graphs in the plane with high resolution. Proc. 31th IEEE Sympos. Found. Comput. Sci., pp. 86{95, 1990. 9. A. Garg, M. T. Goodrich, and R. Tamassia. Area-ecient upward tree drawings. Proc. 9th ACM Sympos. Comput. Geom., pp. 359{368, 1993. 10. G. Kant. Drawing planar graphs using the lmc-ordering. Proc. 33th IEEE Sympos. Found. Comput. Sci., pp. 101{110, 1992. 11. G. Kant. Algorithms for Drawing Planar Graphs. PhD thesis, Dept. Comput. Sci., Univ. Utrecht, Utrecht, Netherlands, 1993. 12. G. Kant. A more compact visibility representation. Proc. 19th Internat. Workshop Graph-Theoret. Concepts Comput. Sci. (WG'93), 1993. 13. G. Kant, G. Liotta, R. Tamassia, and I. Tollis. Area requirement of visibility representations of trees. Proc. 5th Canad. Conf. Comp. Geom., pp. 192{197, 1993. 14. C. E. Leiserson. Area-ecient graph layouts (for VLSI). Proc. 21st IEEE Sympos. Found. Comput. Sci., pp. 270{281, 1980. 15. S. Malitz and A. Papakostas. On the angular resolution of planar graphs. Proc. 24th ACM Sympos. Theory Comput., pp. 527{538, 1992. 16. P. Rosenstiehl and R. E. Tarjan. Rectilinear planar layouts and bipolar orientations of planar graphs. Discrete Comput. Geom., 1(4):343{353, 1986. 17. W. Schnyder. Embedding planar graphs on the grid. Proc. 1st ACM-SIAM Sympos. Discrete Algorithms, pp. 138{148, 1990. 18. R. Tamassia and I. G. Tollis. A uni ed approach to visibility representations of planar graphs. Discrete Comput. Geom., 1(4):321{341, 1986. 19. L. Valiant. Universality considerations in VLSI circuits. IEEE Trans. Comput., C-30(2):135{140, 1981.
Ashim Garg and Roberto Tamassia Department of Computer Science Brown University Providence, RI 02912{1910, USA
Abstract. We investigate the problem of constructing planar straightline drawings of graphs with large angles between the edges. Namely, we study the angular resolution of planar straight-line drawings, de ned as the smallest angle formed by two incident edges. We prove the rst nontrivial upper bound on the angular resolution of planar straight-line drawings, and show a continuous trade-o between the area and the angular resolution. We also give linear-time algorithms for constructing planar straight-line drawings with high angular resolution for various classes of graphs, such as series-parallel graphs, outerplanar graphs, and triangulations generated by nested triangles. Our results are obtained by new techniques that make extensive use of geometric constructions.
1 Introduction
Coping with the nite resolution of display devices and of the human eye is a fundamental problem in graph drawing. Namely, in visualization applications, it is important to construct drawings of graphs that avoid placing vertices and edges too close. In this paper we investigate the problem of constructing planar straight-line drawings with with large angles between the edges. Namely, we study the angular resolution of straight-line drawings, de ned as the smallest angle formed by two incident edges. Besides visualization applications (see, e.g., [5]), constructing drawing with high angular resolution is important in the design of optical communications networks (see, e.g., [8]). The study of the angular resolution of drawings has attracted considerable interest in the last years. Formann, Hagerup, Haralambides, Kaufmann, Leighton, Simvonis, Welzl, and Woeginger [8] were the rst to study the angular resolution of (generally nonplanar) straight-line drawings of various classes of graphs. They show that a degree-d graph has a straight-line drawing with angular resolution (1=d2), and every degree-d planar graph has a straight-line (nonplanar) ? Research supported in part by the National Science Foundation under grant CCR-
9007851, by the U.S. Army Research Oce under grants DAAL03-91-G-0035 and DAAH04{93{0134, and by the Oce of Naval Research and the Defense Advanced Research Projects Agency under contract N00014-91-J-4052, ARPA order 8225. The Email addresses of the authors are [email protected] and [email protected].
drawing with angular resolution (1=d). (Note that the above bounds are independent from the number of vertices of the graph.) They also pose the open problem of characterizing the angular resolution of planar straight-line drawings. Malitz and Papakostas [15] make further progress on the above open problem by proving a lower bound on the angular resolution dependent only on the degree of the graph. Namely, they show that a degree-d planar graph admits a planar straight line drawing with angular resolution (1=7d ), irrespectively of the number of vertices. They also analyze a linear program associated with two necessary conditions on the angles of a planar triangulation (i.e., the sum of the angles around a vertex is 2, and the sum of the angles inside each triangle is ), and nd that such a linear program admits a solution with (1=d) angles. This leads them to conjecture that the trivial (1=d) lower bound on the angular resolution may be achievable for planar straight-line drawings. Kant [10, 11] shows that testing whether a biconnected planar graph admits a planar straight-line drawing with angular resolution greater than or equal to a given constant is NP-hard, thus providing further motivation to the study of asymptotic bounds. He also considers polyline drawings (where edges are drawn as polygonal chains), and shows that a degree-d triconnected planar graph admits a planar polyline grid drawing with angular resolution (1=d). Di Battista and Vismara [7] provide a characterization of the angles in planar straight-line drawings of maximal planar graphs through a nonlinear system of inequalities. As a consequence, they show that testing whether a prescribed assignment of angles to a maximal planar graph is achievable in a planar straightline drawing can be done in linear time, and that constructing a planar straightline drawing with maximum angular resolution can be reduced to the solution of a nonlinear optimization problem. Our results can be summarized as follows: In Section 3, we prove the rst nontrivial upper bound on the angular resolution of planar straight-line drawings by p exhibiting a family of degree-d planar graphs that require angular resolution O( log d=d3) in any planar straight-line drawing. Previously, only the trivial O(1=d) bound was known, and determining whether it is achievable was posed as an open problem in [15, 11]. In Section 4, we show a continuous trade-o between the area and the angular resolution of planar straight-line drawings. Namely, there exists a constant c > 1 such that, for any n > d > 6, with n ? 3 a multiple of d ? 4, there exists a planar graph Gdn with n vertices and degree d such that any planar straight-line drawing of Gdn with angular resolution has area (cn ). In particular, this implies that there exist bounded-degree planar graphs for which asymptotically optimal resolution can be achieved only at the expense of an exponential increase in the area. Previously, area/trade-o results were proved for planar drawings (see, e.g., [1, 2, 3, 6, 4, 9, 10, 11, 12, 13, 14, 16, 17, 18, 19]), depending on various restrictions on the representation (e.g., upward, straight-line, orthogonal, polyline). Our result is the rst \continuous" area/trade-o result for planar drawings. In Section 5, we give linear-time algorithms for constructing planar straight-
line drawings with high resolution of various classes of graphs, such as seriesparallel graphs, outerplanar graphs, and nested-star graphs (i.e., maximal planar graphs generated from a triangle by repeated insertions of stars). The angular resolution is shown to be (1=d2) for series-parallel and nested-star graphs, and =(d ? 1) for outerplanar graphs. Previously, Malitz and Papakostas [15] showed that angular resolution =2(d ? 1) can be achieved for outerplanar graphs. No polynomial bounds in 1=d were known for series-parallel graphs and nestedstar graphs. We leave it as an open problem either proving or disproving that
(1=d2) is a tight lower bound on the angular resolution of the series-parallel and nested-star graphs. Note that the family of graphs requiring angular resolution p O( log d=d3) in any planar straight-line drawing are actually nested-star graphs. The signi cance of our result on nested-star graphs is motivated by the following observation: Malitz and Papakostas [15] show that a disk packing of a planar degree-d graph yields a planar straight-line drawing with angular resolution (1=7d ), and they exhibit a family of nested-star graphs to prove that this exponential bound is the best that can be obtained with the disk packing method. Our results imply that the disk packing method yields suboptimal results and leave as an open problem whether the angular resolution of planar straight-line drawings can be bounded from below by a polynomial in 1=d. Our algorithms for drawing nested-star and series-parallel graphs exploit various geometric properties of the angles in certain triangulations. Since most existing algorithms for drawing planar graphs rely on graph-theoretic properties (e.g., orientation, coloring, ow), the techniques in this paper appear to be interesting in the way they blend geometric and graph-theoretic methods. We believe that our results will stimulate further research in this direction. Due to the space limitations, in this extended abstract we have either sketched the proofs or omitted them.
2 De nitions
First, we review basic graph drawing terminology [5]. A straight-line drawing maps each edge into a straight-line segment. The angular resolution of a straightline drawing is the smallest angle formed by two edges incident on the same vertex. A drawing is planar if no two edges intersect. The area of a drawing is the area of the smallest convex polygon covering the drawing. Whenever we give lower bounds on the area, we assume that the drawing is constrained by some rule that prevents it from being arbitrarily scaled down (e.g., integer coordinates for the vertices, or a minimum unit distance between any two vertices). The results of Section 4 are fully general, since they hold under any rule that implies a nite minimum area for the drawing of a graph. The degree of a graph is the maximumdegree of any vertex in the graph. The notation dG(u) indicates the degree of vertex u in graph (or subgraph) G. Given a line segment with endpoints A and B, AB denotes either the line segment itself or its length. Given two line segments AB and BC, 6 ABC denotes either the angle between AB and BC or its measure. Finally, ABC denotes the triangle with vertices A, B and C.
3 Upper Bound on the Angular Resolution
In this section we show that there p exists an in nite family of degree-d graphs that require angular resolution O( log d=d3 ) in any planar straight-line drawing. Thus, for these graphs the linear program of [15] gives an inconsistent drawing. First, we give the following geometric lemma (see Fig. 1(a)). uk
αk u0
A
B
β1 β2 β
α α2 α1 D
γ2 γ1
(a)
γ
v C 0
φk
θk
α0 G0 β0 γ0 w 0
x Gk-1
z
Gk
Gk-1 Gk-1
(b) vk
s
y
βk
γk
wk
(c)
Figure 1: (a) Dividing a Triangle; (b) Graph G0 ; (c) Graph Gk ; Lemma 1. Let D be a point inside a triangle ABC . Let 1, 2, 1 , 2 , 1 and 2 be the angles de ned in Fig. 1(a). If =2 and 2= 1 1, then 2 r 1
2 2 min( ; ) 4 (1) 1
1
2
We are now ready to present the main theorem of this section. Theorem 2. There exists an in nite family of planar p graphs with degree d and (3d=9 ) vertices that require angular resolution O( (log d)=d3 ) in any planar
straight-line drawing. Proof. Consider graph Gk , which is recursively de ned as follows: G0 is a triangle
with vertices u0, v0 and w0 (see Fig. 1(b)). Graph Gk is assembled from three copies of Gk?1, denoted H1, H2 and H3, as shown in Fig. 1(c)). Simple recurrence relations can be used to show that Gk has degree 9k and (3k ) vertices. Given a straight line drawing of Gk with uk vk wk on the external face, we de ne the following angles (see Fig. 1(c)): k = 6 wk uk vk , k = 6 uk vk wk , k = 6 vk wk uk , k = 6 wk uk s and k = 6 vk uk x. We now claim that the angular resolution of p any planar straight-line drawing of Gm with umvm wm as the external face is O( (log m)=m3 ). Assume without any loss of generality that m =2 because in any p triangle at least one internal angle has value not exceeding =2. If 0 is O( (log m)=m3 ) the claim holds. 3=2 So suppose p 0 = 3 (1=m ). We shall nd another angle in the drawing that is O( (logm)=m ). Namely, for any k Q m, let rk = k?1=( k + k ). Thus k?1 = k rk =(rk + 1). We have 0 = m k1 rk =(rk + 1). Since m < and 0 = (1=m3=2), we obtainm Y 3=2 k=1
rk =(rk + 1) = (1=m )
(2)
We now prove that there exists anQ i m=2 such that ri = (m= log m). For any k, from Eq. 2 we get that mj=k rj =(rj + 1) = (1=m3=2) because
rq =(rq + 1) < 1 for any q. If r(k) = maxmj=k (rj ), we get fr(k)=(r(k) + 1)gm?k =
(1=m3=2). Consequently m?k = O(m3=2 ) 1 + r(1k)
(3)
From Eq. 3, with simple manipulations we obtain r(m=2) = (m= logm). Thus there is anpindex i m=2 such that ri = (m= log m). By Lemma 1, either i?1 i 2=4( ri + 2=4) or i?1 i 2 =4(pri + 2=4). Since di?1(vi?1) = di?1(wi?1) = 3i ? 1, both i?1 and i?1 include 3i ? 2 3m=2 ? 2 angles in their interior. Hence p for suciently large m, there is an angle inside either i?1 or i?1 that is O( (log m)=m3 ). This concludes the proof of the claim. It can be readily seen that a planar straight-line drawing of Gm where um vm wm is not the external face contains as a subdrawing a drawing of Gm?1 with external face um?1vm?1 wpm?1. Hence, any planar straight-line drawing p of Gm has angular resolution O( (log(m ? 1))=(m ? 1)3 ) = O( (log m)=m3 ). We conclude by recalling that graph Gm has degree (m) and (3m ) vertices.
4 Tradeo between Area and Angular Resolution
In this section we show that there exists a continuous tradeo between the area requirement and the angular resolution of a planar straight-line drawing. Namely, for every n > d > 0, we show that there exists a planar graph Gdn with n vertices and degree d such that any planar straight-line drawing of Gdn with resolution requires area (cn ), where c is a constant greater than 1. For a xed d, we recursively de ne graph Gk as follows. Graph G0 is a triangle consisting of vertices A0 , B0 , and C0 . Graph Gk is obtained from Gk?1 as shown in Fig. 2(a). For k 2, Gk has k(d ? 4) + 3 vertices and degree d. Graph Gdn is de ned as G nd??43 , whenever n ? 3 is a multiple of d ? 4. First, we give a technical lemma, which uses the notation of Fig. 2. Ak
h0 a1 b2
f0 A0 Gk-1
b1
B0
f1
Gk
bi
C1 C2
d1
d2
e1
e2
Gk, i
Gk, i-1
c2
B2
fi
ai
c1
C0
B1 B
Ak
d-1 edges
a0
Bi
Bi-1
Ci-1
ei di
C
ci Ci
(b)
(a)
Figure 2: (a) Graph G(d; n); (b) Introducing vertices and edges Lemma 3. Given a planar straight line drawing of Gk , let ki be the area of Ak Bi Ci , with 1 i (d ? 4)=2, then inpat least one of the following pairs of
edges, both edges have length greater than 2ki : (ai ; ai+1), (fi ; fi+1), (di; ei+1 ), and (di ; bi+1).
We are now ready to state the main result of this section:
Theorem 4. There exists a constant c > 1 such that, for any n > d > 6, with n ? 3 a multiple of d ? 4, there exists a planar graph Gdn with n vertices and
degree d such that any planar straight-line drawing of Gdn with angular resolution has area (cn ). Proof. Consider a planar straight line drawing of Gk with Ak Bk Ck as the exter-
nal face for some k = (n ? 3)=(d ? 4). We denote with k the area of Gk , and with ki the area of triangle AkpBi Ci p (see Fig. 2). By Lemma 3, we have ki+1 > ki + 21 2ki 2ki sin > ki (1 + sin ):
(4)
and k1 = k?1, we obtain the recurrence d?4 2 +1 k > k?1 (1 + sin ) 2 : (5) Recalling the recursive dconstruction of G , we get k d?4 n?3 n?3 ?4 (6) k > 0 (11 + sin ) 2 k = 0 (1 + sin ) 2 d?4 = 0 (1 + sin ) 2 n Since sin 2 and =3, we have that k > r for some constant r > 1. Finally, we observe that in any planar drawing of Gk with Ak Bk Ck not being the external face, there is a subgraph of Gk isomorphic to Gk=2, that is drawn with Ak=2 Bk=2Ck=2 as the external face. We conclude that any planar straight line drawing of Gk has area (cn ) for some constant c > 1. Since k = kd?4
Note that Theorem 4 is fully general, since it holds under any rule that implies a nite minimum area for the drawing of a graph.
Corollary 5. There exists an in nite family of bounded-degree graphs that re-
quire exponential area in any planar straight-line drawing with bounded (from below) angular resolution.
5 Constructing Drawings with High Angular Resolution
In this section we give linear-time algorithms for constructing planar straightline drawings with good angular resolution for various classes of planar graphs. We show that every nested-star graph and series-parallel graph admits a planar straight-line drawing with angular resolution (1=d2), and that every outerplanar graph has a planar straight-line drawing with angular resolution =(d ? 1), where d denotes the degree of a graph.
5.1 A Geometric Lemma
In this section we give a geometric lemma that will be used in our algorithms for drawing nested-star and series-parallel graphs. C
γ/2
γ
∆ D1 α2
α0
D2
β2
β0
B
A α α1
p
β1 β
Figure 3: Triangle in Lemma 6
Lemma 6. Let = ABC be a triangle with internal angles , and where =2, as shown in Fig. 3. Let p be the bisector of angle , and D1 and D2
be any two points on p inside . Let 0 = 6 D1 AD2 and 0 = 6 D1 BD2 . Then
0 1 0 42
5.2 Nested-Star Graphs
A nested-star graph G is de ned recursively as follows: G has three distinguished vertices called its apex vertices (denoted by u, v and w in Fig. 4). G either consists of a single triangle (see Fig. 4(a)) or it is assembled from three nested-star graphs G1, G2 and G3, as shown in Fig. 4(b). Lemma 7, which can be easily proved by induction on the number of vertices in G, states a property of nested-star graphs that is used by our drawing algorithm. w
w
G
G1 z G2
v
u
G
G3
u
v
(a)
(b) γ/2
∆
α
∆ v (c)
γ
V x1 1 xm-1 Vm Um W xm p E
U1
w
u
w=x0
u
α1 αm
β1 βm
β v
(d)
Figure 4: A nested-star graph G: (a) as a single triangle; (b) formed by three nestedstar graphs; Drawing G when (c) it is a single triangle; (d) it consists of nested-star subgraphs
Lemma 7. Let G be a nested-star graph with apex vertices u, v and z. The vertices of G that are common neighbors of u and v form a path.
We now describe a recursive algorithm that constructs a drawing of a nestedstar graph G with angular resolution 1=(24d2), where d is the degree of G. Let u, v and w be the apex vertices of G. Given a triangle , which is said to be allocated to G, we show how to draw G inside (see Fig. 4(c){(d)). The vertices of are u, v and w. If G has three edges, then its drawing is (see Fig. 4(c)). Otherwise, we draw G as follows (see Fig. 4(d)): Let , and be the internal angles of at u, v and w respectively. We assume without loss of generality that angle . By Lemma 7, let x0 = w; x1; x2; : : :; xm be the path formed by the common neighbors of u and v. Let Ui be the nested-star
subgraph of G with xi?1, u and xi as its apex vertices. Let Vi be the nested-star subgraph of G with xi?1, v and xi as its apex vertices. Notice that uxm v is a face of G. Let W be the nested-star subgraph of G which consists only of the triangle uxm v. Let p be a line through w that bisects angle . For each i 1, place xi on line p in the interior of such that 6 xi?1uxi is (dUi (u) ? 1) =(dG(u) ? 1). (Recall that dG (u) denotes the degree of vertex u in graph G.) We allocate triangle xi?1uxi to Ui and triangle xi?1vxi to Vi , and then draw Ui and Vi recursively in their allocated triangles. Theorem 8. A nested-star graph with degree d admits a planar straight-line drawing with angular resolution 1=(24d2) that can be constructed in linear time. Proof. Allocate an equilateral triangle to the graph and then apply the above algorithm. We prove that the resulting drawing has angular resolution 1=(24d2). Intuitively, when allocating a triangle to a nested-star subgraph H of G, the algorithm allows the internal angle at an apex vertex of H to become \small",but no smaller than 1=(24d). The algorithm then divides the angle evenly between the edges incident on the vertex in H, so that the minimum angle is at least 1=(24d2). The proof is based over maintaining the following invariant: Invariant 1: Let uvw be a triangle allocated to a nested-star graph G with apex vertices u, v and w. If u is designated the critical vertex of G then { (dG(u) ? 1)=(24d2) 6 vuw =6, and { 6 uvw =6 and 6 uwv =6. Intuitively, the algorithm divides the angle at a critical vertex evenly between the incident edges by allocating an angle to each Ui proportional to the degree of the vertex in Ui . The key to proving Invariant 1 is showing that the angle allocated to each Vi is at least 1=(24d). We now prove that Invariant 1 holds for every nested-star subgraph of the input graph. Initially designate any apex vertex of the input nested-star graph to be the critical vertex. Invariant 1 holds trivially for the input graph. we now prove that the Invariant 1 is maintained at each recursion. Suppose Invariant 1 holds for a nested-star subgraph G with apex vertices u, v and w. Let uvw be the allocated triangle of G. Let u be the critical vertex of G. Suppose we denote the internal angles of uvw at u, v and w by , and respectively (see Fig. 4(d)). Let . Therefore =3 and =2. The algorithm allocates triangles xi?1uxi and xi?1vxi to the nested-star subgraphs Ui and Vi respectively. Let E be the intersection of the bisector p of with the line segment uv. We rst show that Invariant 1 holds for Ui . Let i = 6 xi?1uxi . 6 uxi?1xi 6 uwx1 = =2 =6. 6 uxixi?1 6 uEw = ? =2 ? =6. i = (dUi (u) ? 1) =(dG(u) ? 1). We designate u as the critical vertex of Ui . Because Invariant 1 holds for G, (dG (u) ? 1)=24d2. Hence i dUi (u)=24d2. Therefore Invariant 1 holds for Ui . Now we show that Invariant 1 holds for Vi . We designate v as the critical vertex of Vi . Let i = 6 xi?1vxi . 6 vxi?1 xi 6 vwx1 = =2 =6. 6 vxi xi?1
=6. From Lemma 6, i ( i =42 ) . Since =6 and i = (dUi (u) ? 1) =(dG(u) ? 1), (u) ? 1 i 41 2 ddUi(u) G ?1 6 since dG(u) d, dUi (u) 2 and dVi (v) d, We have i (1=(24d)
6 vEw = ? =2 ?
dVi (v)=(24d2 ) Therefore Invariant 1 holds for Vi . We can similarly show using Lemma 6 that Invariant 1 holds for W. Finally we show using induction over the number of vertices in G that the minimum angle in the drawing of G constructed by the algorithm is at least 1=24d2. The base case is when G consists of a single triangle . Each internal angle of is at least 1=24d2 because of Invariant 1. In a general inductive step let G consists of nested-star graphs W, U1; U2 ; : : :; Um and V1 ; V2 ; : : :; Vm , where Ui Vi , and W are as described in the algorithm. Since the drawing of each Ui and Vi , and W has minimum angle at least 1=24d2 by our inductive hypothesis, we conclude that the minimum angle in the drawing of G is at least 1=24d2.
5.3 Series-Parallel Graphs
A series-parallel graph G is de ned recursively as follows [1]: It contains a pair of distinguished vertices called its poles. It consists of either a single edge, or of a series composition, or of a parallel composition of two series-parallel graphs. For us the following equivalent de nition is more useful: It consists of either a single edge, or a series composition of two series-parallel graphs, or a parallel composition of series-parallel graphs G1; G2; : : :; Gm where each Gi is either a single edge or a series composition of two series-parallel graphs. A series-parallel subgraph of G is a subgraph of G that is also a series parallel graph. We now describe a recursive algorithm that constructs a drawing of a seriesparallel graph G with angular resolution 1=(48d2), where d is the degree of G. Let s and t be the poles of G. We allocate a triangle to G and draw G recursively inside . Two vertices of are s and t (see Fig 5(a)). Let the third vertex of be r. One internal angle of is =3 (6 rts in Fig 5(a)). We consider three cases: { G is a single edge (s; t): We draw it as a straight line between s and t (see Fig 5(b)). { G is a series composition of two series-parallel graphs G1 and G2, where s is a pole of G1 and t is a pole of G2: We draw G as shown in Fig. 5(c). We assume without any loss of generality that 6 rst 6 rts. Let r2 be the midpoint of the line segment rt. Let 2 be the line passing through r2 making an angle =3 with the line st. Let u be the intersection of 2 and st. Let 1 be the line parallel to the line rt passing through u. r1 is the intersection of 1 and the line sr. We allocate triangle sur1 to G1 and triangle tur2 to G2 and draw G1 and G2 recursively in their allocated triangles. { G is a parallel composition of the series-parallel graphs G1; G2; : : :; Gm : Since G does not have multiple edges, there can be only two subcases: (i) No Gi consists of a single edge, or (ii) one and only one Gi consists of a single edge
s
π/3
∆
G π/3
Hm t
(a)
Hi α
G
αi
t
(b)
pi
ri
H1 s
π/3
qi π/3
r1
r'1
π/3
π/3 q1 p
s (a) r
λ1
λ2
ηi
r1 ∆ G1 s
u β
G µi
qm r'm π/3 Km
λi
r'i
r ∆
γ/2
rm
r
∆
r
γ
ηi
r2
G
G2 π/3
Hm β
γ/2 π/3
∆
βi K1
t
G qm r'm π/3 pi
ri
Km
µi
λi
r'i
t
(c)
Figure 5: Drawing a
series-parallel graph G: (a) Triangle allocated to G; (b) Drawing G when G is a single edge; (c) Drawing G when G is a series composition;
β=π/3
r
γ rm
Ki
Hi α
s
αi
H2 α/(dG(s)-1) (b)
π/3
qi π/3
r2
r'2
π/3
π/3 q2 p
Ki βi
β=π/3
K2 t
Figure 6: Drawing a series-parallel graph G when it is a parallel composition of sp-graphs G1 ; G2 ; : : : ; Gm : (a) Case i; (b) Case ii.
and without any loss of generality let G1 be that subgraph. In both subcases i and ii, if Gi does not consist of a single edge, it is a series composition of two series parallel subgraphs. Let us denote the subgraph whose pole is s by Hi and the other subgraph whose pole is t by Ki . Let qi be the pole common to Hi and Ki . Fig. 6(a) shows the drawing of G in case i and Fig. 6(b) shows the drawing of G in case ii. In both the cases, if ui does not consist of a single edge (i.e., it is a series composition), we allocate a triangle sri qi to Hi and a triangle sri0 qi to Ki , and recursively draw Hi and Ki in their allocated triangles. The position of each ri, qi and ri0 is determined as follows: Assume without any loss of generality that 6 rst 6 rts. Let = 6 rst. Let p be the bisector of angle 6 srt. Draw lines i and i from s such that (a) the angle between i and i is (2dHi (s) ? 1) =(2dG(s) ? 1) and (b) angle between i and i+1 is =(2dG(s) ? 1). (Recall that we denote by dG(u), the degree of vertex u in G.) If u1 consists of a single edge then G1 is drawn as a straight line between s and t and 6 q2st is =(2dG(s) ? 1) (Fig. 6(b)). qi is the intersection of i with p. ri is a point on i such that 6 ri qis is =3. Let pi be the intersection of p and i . Let i be the line passing through t and
pi . ri0 is a point on i such that 6 ri0 qit is =3. Theorem9. A series-parallel graph with degree d admits a planar straight-line drawing with angular resolution 1=(48d2) that can be constructed in linear time. Proof. The proof is similar to the proof of Theorem 8 and uses Lemma 6.
5.4 Outerplanar Graphs
In this section we give a linear time algorithm that constructs a drawing of a maximal outerplanar graph G with angular resolution =(d ? 1), where d is the degree of G. This improves the lower bound given in [15] by a factor of two. Our algorithm constructs a breadth- rst search tree T(G) of G and places the vertices on concentric circles in the drawing so that the vertices equidistant from the root of T(G) are placed on the same circle. A sketch of the algorithm is as follows: π/(d-1)
w=v1 v2
vp
π/(d-1)
uj π/(d-1)
uj-1
π/(d-1)
uj+1
tangent of Ci-1 at uj+1
r Ci
C1
Ci-1
(a) (b)
Figure 7: Drawing an outerplanar graph G: (a) Level 1; (b) Level i 1. Construct a breadth- rst search tree and assign vertices to levels such that vertices equidistant from the root of the tree get placed at the same level. 2. Construct a drawing as follows (see Fig. 7): Let C1 be a circle. Place the root of the tree at the center of C1. Place the vertices of level 1 at equidistant positions on C1 such that the angle subtended on the center of the circle by consecutive vertices is =(d ? 1) (see Fig. 7(a)). Now, suppose we have already placed the vertices of levels 1 to i ? 1 on concentric circles C1 ; C1; : : :; Ci?1. Let u1 ; u2; : : :; um be the left-to-right order of the vertices on Ci?1 (see Fig. 7(b)). Draw lines from each uj making angles =(d ? 1) and ?=(d ? 1) respectively with the tangents at uj . If uj and uj ?1 have a common neighbor (e.g., vertex w in Fig. 7(b)) on level i, then place it at the point of intersection of the lines drawn from uj and uj ?1. Let Ci be the circle concentric to Ci?1 passing through these points. Let v1 ; v2; : : :; vp be the neighbors on level i of uj . Place each vk on Ci such that if there is an edge (vk ; vk+1) in G, then 6 vk uj vk+1 is =(d ? 1). Theorem10. A maximal outerplanar graph with degree d admits a planar straight-line drawing with angular resolution =(d ? 1) that can be constructed in linear time.
References
1. P. Bertolazzi, R. F. Cohen, G. Di Battista, R. Tamassia, and I. G. Tollis. How to draw a series-parallel digraph. Proc. 3rd Scand. Workshop Algorithm Theory, vol. 621 of Lecture Notes in Computer Science, pp. 272{283. Springer-Verlag, 1992. 2. R. P. Brent and H. T. Kung. On the area of binary tree layouts. Inform. Process. Lett., 11:521{534, 1980. 3. P. Crescenzi, G. Di Battista, and A. Piperno. A note on optimal area algorithms for upward drawings of binary trees. Comp. Geom. Theory Appl., 2:187{200, 1992. 4. H. de Fraysseix, J. Pach, and R. Pollack. Small sets supporting Fary embeddings of planar graphs. Proc. 20th ACM Sympos. Theory Comput., pp. 426{433, 1988. 5. G. Di Battista, P. Eades, R. Tamassia, and I. G. Tollis. Algorithms for drawing graphs: an annotated bibliography. Preprint, Dept. Comput. Sci., Brown Univ., Providence, RI, November 1993. To appear in Comput. Geom. Theory Appl. Preliminary version available via anonymous ftp from wilma.cs.brown.edu, gdbiblio.tex.Z and gdbiblio.ps.Z in /pub/papers/compgeo. 6. G. Di Battista, R. Tamassia, and I. G. Tollis. Area requirement and symmetry display of planar upward drawings. Discrete Comput. Geom., 7:381{401, 1992. 7. G. Di Battista and L. Vismara. Angles of planar triangular graphs. Proc. 25th ACM Sympos. Theory Comput. (STOC 93), pp. 431{437, 1993. 8. M. Formann, T. Hagerup, J. Haralambides, M. Kaufmann, F. T. Leighton, A. Simvonis, E. Welzl, and G. Woeginger. Drawing graphs in the plane with high resolution. Proc. 31th IEEE Sympos. Found. Comput. Sci., pp. 86{95, 1990. 9. A. Garg, M. T. Goodrich, and R. Tamassia. Area-ecient upward tree drawings. Proc. 9th ACM Sympos. Comput. Geom., pp. 359{368, 1993. 10. G. Kant. Drawing planar graphs using the lmc-ordering. Proc. 33th IEEE Sympos. Found. Comput. Sci., pp. 101{110, 1992. 11. G. Kant. Algorithms for Drawing Planar Graphs. PhD thesis, Dept. Comput. Sci., Univ. Utrecht, Utrecht, Netherlands, 1993. 12. G. Kant. A more compact visibility representation. Proc. 19th Internat. Workshop Graph-Theoret. Concepts Comput. Sci. (WG'93), 1993. 13. G. Kant, G. Liotta, R. Tamassia, and I. Tollis. Area requirement of visibility representations of trees. Proc. 5th Canad. Conf. Comp. Geom., pp. 192{197, 1993. 14. C. E. Leiserson. Area-ecient graph layouts (for VLSI). Proc. 21st IEEE Sympos. Found. Comput. Sci., pp. 270{281, 1980. 15. S. Malitz and A. Papakostas. On the angular resolution of planar graphs. Proc. 24th ACM Sympos. Theory Comput., pp. 527{538, 1992. 16. P. Rosenstiehl and R. E. Tarjan. Rectilinear planar layouts and bipolar orientations of planar graphs. Discrete Comput. Geom., 1(4):343{353, 1986. 17. W. Schnyder. Embedding planar graphs on the grid. Proc. 1st ACM-SIAM Sympos. Discrete Algorithms, pp. 138{148, 1990. 18. R. Tamassia and I. G. Tollis. A uni ed approach to visibility representations of planar graphs. Discrete Comput. Geom., 1(4):321{341, 1986. 19. L. Valiant. Universality considerations in VLSI circuits. IEEE Trans. Comput., C-30(2):135{140, 1981.
