Representing 3-trees as Unit Rectangle-Visibility Graphs

Transcription

1 Representing 3-trees as Unit Rectangle-Visibility Graphs Alice M. Dean Department of Mathematics and Computer Science Skidmore College Saratoga Springs, NY 12866, USA Joan P. Hutchinson Department of Mathematics, Statistics, and Computer Science Macalester College St. Paul, MN 55105, USA Abstract In [2] it is shown that, for 1 k 4, every k-tree is a rectanglevisibility graph (RVG). In contrast, Dean, et al. [3] show that not every such k-tree is a unit rectangle visibility graph (URVG). The results in that paper give necessary and sufficient conditions for a tree to be a URVG, and they also imply that no 4-tree is a URVG, but no results are given for 2-trees or 3-trees. In this paper we consider 3- trees, giving necessary and partially sufficient conditions for a 3-tree to be a URVG. 1 Introduction In this paper we combine the study of two classes of graphs, both important for their theoretical and their applied interest. The class of k-trees, k > 0, consists of graphs constructed in a tree-like fashion from the complete graph K k, by repeatedly forming the join of a new vertex with an existing k-clique. The k-trees are a subclass of perfect elimination graphs with clique cutsets and have been enumerated, characterized, and analyzed algorithmically. See, for example, [1, 11, 16]. Applications include questions about reliability of communication networks [14] and sparse linear systems [11]. The complexity of recognizing k-trees for arbitrary k is 1

2 NP-hard [1]. For each k-tree there is an associated decomposition tree that gives the underlying tree structure of the (k + 1)-cliques formed by each appended vertex. Such graph decompositions are the subject of [7], and the related concept of tree-width is a key ingredient of the Robertson-Seymour graph-minors project; see [9, 10]. The study of rectangle-visibility graphs (RVGs) is an active one in the area of graph-visualization and graph-drawing [6]. RVGs are graphs that can be realized with vertices represented by axis-aligned, nonintersecting rectangles in the plane and with edges represented by unobstructed horizontal and vertical visibility between rectangles; see [2, 3, 12]. A variant of RVGs in which the rectangles must all be unit squares was studied in [3]. Also representation by unit squares in three dimensions, parallel to the x-y plane with z-visibility, was studied in [8]. RVGs, URVGs, and their layouts are of particular applicability in the field of VLSI design and two-layer routing [13]. The question of which k-trees are RVGs was considered in [2], in which it was shown that all k-trees with 1 k 4 are RVGS, while not all 5-trees need be RVGs. Although the recognition of RVGs is in general known to be NP-hard [12], efficient algorithms to lay out k-trees with 1 k 4 are given in [2]. In [3] a characterization is given of the 1-trees that are URVGs, and it follows from other results in that paper that no 4-trees are URVGs. Thus a natural question to ask is which 2-trees and 3-trees are URVGs, since all such graphs are known to be RVGs. In this paper we focus on the question of which 3-trees are URVGs, although we also say a little about the case of 2-trees. We show that if G is a 3-tree that is a URVG, then the decomposition tree T(G) of G has maximum degree 4 and is a union of three caterpillars with additional properties. These conditions are similar to, but more restrictive than, the necessary and sufficient conditions given in [3] for a tree to be a URVG. However the conditions given here for a 3-tree to be a URVG are not sufficient, since we show that there are trees T that satisfy these conditions such that the set of 3-trees G for which T(G) = T contains both URVGs and graphs that are not URVGs. 2 Background and Definitions We mainly follow the definitions of [15, 9]. Definition 2.1. A k-tree, k > 0, is a graph G with n k + 1 vertices that can be numbered 1, 2,..., n so that each vertex i, i = k +1,..., n, has exactly k neighbors in {1, 2,..., i 1}, and these neighbors induce a K k. We call such a numbering of V (G) a k-tree numbering. Let G be a k-tree with k-tree numbering {1, 2, 3,..., n}, where n k+1. 2

3 For k + 1 i n, K k+1 (i) denotes the unique K k+1 containing i and its k lower-numbered neighbors. Thus K k+1 (k+1) is the initial K k+1 on vertices {1, 2,..., k + 1}. For 1 i k, we set K k+1 (i) = K k+1 (k + 1). We define a tree, T = T(G), with vertex set {t k+1, t k+2,..., t n }, that encodes the k-tree structure of G. For 1 i k+1, vertex t k+1 represents K k+1 (i) = K k+1 (k + 1), and for k + 1 < i n, t i represents K k+1 (i). The edges of T(G) are defined as follows: For k + 2 i n, suppose that {a 1, a 2,..., a k }, with a 1 < a 2 <... < a k < i, are the k lower-numbered neighbors of i in G. If a k = k or k + 1, then the vertex t i is adjacent to t k+1 in T; if k + 1 < a k n, then t i is adjacent to t ak in T. We say that the successive K k+1 (i) s, i = k + 1,..., n, form a tree-decomposition of G, denoted (T(G), (K k+1 (i)) i=k+1,...,n ). The tree T(G) is called the decomposition tree. Definition 2.2. A graph G is a unit rectangle-visibility graph or URVG if its vertices can be represented in the plane by axis-aligned, closed unit squares with pairwise disjoint interiors, in such a way that two vertices are adjacent if and only if there is an unobstructed nondegenerate (positive width) horizontal or vertical band of visibility joining the two squares. Such a layout of squares is called a URV layout and is denoted L(G). G is a rectangle-visibility graph (RVG) if it can be so represented with arbitrarily sized rectangles. If the URVG G has vertices 1, 2,..., n and layout L(G), then the squares that correspond to the vertices of G are also labeled 1, 2,..., n, respectively. If square i is in the layout L(G), we let x 1 (i) equal the x-coordinate of the left edge of i and x 2 (i) the x-coordinate of the right edge of i, which equals x 1 (i) + 1. Similarly y 1 (i) and y 2 (i) = y 1 (i) + 1 equal the y-coordinates of the bottom and the top edge of i. Fig. 1a shows a URV layout of the 3-tree shown in Fig. 1b, whose decomposition tree, shown in Fig. 1c, is a path of 7 vertices, t 4, t 5,..., t t 4 t 5 t 6 t 7 t 8 t 9 t 10 Figure 1: (a) The URV-layout of (b) a 3-tree whose decomposition tree is (c) a path of seven vertices t 4,..., t 10 A caterpillar is a tree in which a single path P is incident with (or contains) every edge, and a maximal such path is called a spine of the 3

4 caterpillar. The vertices and edges off the path are called legs. A caterpillar forest is a disjoint union of caterpillars and, for k > 0, a k-caterpillar is a caterpillar with maximum degree at most k. A subdivided caterpillar is one in which each edge may be replaced by a path of arbitrary length. Theorem 2.3. [3] A tree is a URVG if and only if it is the union of two subdivided 3-caterpillar forests. It follows from Thm. 2.3 that the maximum degree in a URV tree is 6; the bipartite graph K 1,7 is the smallest tree that is not a URVG. Our result for 3-trees also involves caterpillars, and it depends on the fundamental observation that there are essentially three ways to lay out a K 4 in a URVG and that there are restrictions on the layouts of pairs of K 4 s that represent adjacent vertices in the decomposition tree. Definition 2.4. A URV layout of K 4 is called an h-v layout, where 0 h, v and h + v = 6, if h edges correspond to horizontal visibilities and v edges correspond to vertical visibilities. In [5] it is shown that no unit bar-visibility layout contains K 4 ; thus there can be no 6-0 or 0-6 URV layout of K 4. Fig. 2 shows 3-3, 4-2, and 5-1 layouts of K 4, and these can be rotated through π to obtain 2-4 and 1-5 layouts. In Fig. 1a the successive triples of squares form alternately 3-3 and 5-1 layouts. a b c d a b c d a b c d Figure 2: 3-3, 4-2, and 5-1 URV layouts of K 4 Let r and s be two vertices in a 3-tree G with decomposition tree T, 4 r < s. If the vertices t r and t s of T are adjacent, we also say that the 4-cliques K 4 (r) and K 4 (s) are adjacent. Thus, K 4 (r) and K 4 (s) are adjacent if either r = 4 and they have three vertices in common, or if r > 4 and they have three vertices in common, one of which is r. For example, in the 3-trees of Fig. 1a, K 4 (5) and K 4 (6) are adjacent. Prop. 2.5 gives additional restrictions, which follow easily from consideration of the layouts shown in Fig. 2, on the layouts of adjacent K 4 s. Proposition 2.5. In a URV layout of a 3-tree, 1. A K 4 with a 3-3 layout is adjacent to at most four other K 4 s, none of which has a 3-3 layout. 4

5 2. A K 4 with a 2-4 or 4-2 layout is adjacent to at most three other K 4 s, at most one of which can have a 3-3 layout. 3. A K 4 with a 1-5 or 5-1 layout is adjacent to at most two other K 4 s. It follows from Prop. 2.5 that if G is a URV 3-tree, then its decomposition tree T(G) has maximum degree 4. Definition 2.6. Suppose in a URV layout that K 4 (r) contains a square s. If s is positioned far enough to the west so that its east edge is at least one unit from the furthest west edge of every other square in K 4 (r), then we say that s is a far west square of K 4 (r). Similarly we define a far east square, a far north square, and a far south square of K 4 (r). We may also call each a far square, and a square that is not far in its K 4 is called a near square. Fig. 3 shows two 4-2 layouts of a K 4, one with all squares near, and the other with one far square. At most one square in a 4-2 or 2-4 layout can be far; a 5-1 or 1-5 layout can have two far squares, and a 3-3 layout cannot have any far squares. a b c d a b c d Figure 3: Near and far 4-2 URV layouts of K 4 3 Necessity The focus of this section is Thm. 3.4, which gives necessary conditions for a tree T to be the decomposition tree T(G) of a URV 3-tree G. More related results on 2-tree and 3-tree URV representations appear in [4]. We begin with some definitions concerning paths and their layouts that will be used in what follows. Definition 3.1. Given a URV 3-tree G with decomposition tree T, consider the layouts of two adjacent K 4 s, K 4 (r) and K 4 (s), and let s K 4 (s) \ K 4 (r). Then in the URV layout of G, the square s may or may not intersect the bounding box B of K 4 (r) (that is, the smallest rectangle containing K 4 (r)), but it is easy to see that s cannot intersect two sides of B. If s intersects or lies fully to the north (resp., east, south, west) of B, we say that K 4 (s) is a north neighbor (resp., east, south, west neighbor) of K 4 (r); similarly t s is a north neighbor (resp., east, south, west neighbor) of t r. 5

6 Note that this relationship is not, in general, symmetric. For example, in Fig. 1a K 4 (7) on {4, 5, 6, 7} is an east neighbor of K 4 (6), but K 4 (6) is a north neighbor of K 4 (7). Definition 3.2. Suppose s 1, s 2, s 3,..., s k, k 5, is a sequence of squares in a URVG G, in which {s 1, s 2, s 3, s 4 } is the vertex set of K 4 (v) with v {s 1, s 2, s 3, s 4 }. Suppose further that, for each i = 1,..., k 4, {s i+1, s i+2, s i+3, s i+4 } is an east neighbor of {s i, s i+1, s i+2, s i+3 }. By relabeling the initial four squares if necessary, we may assume that x 1 (s 1 ) < x 1 (s 2 ) < x 1 (s 3 ) <... < x 1 (s k ), as illustrated in Fig. 4. Such a sequence of squares is called an east corridor from s 1 ; south, west, and north corridors are defined analogously. More generally, a corridor is a south, west, north, or east corridor. s 1 s 3 s5 s 4 s 2 s 6 s 7 s 8 s 9 s 10 Figure 4: An east corridor at s 1 An east corridor from s 1, as described above, also corresponds to a path of k 3 vertices in the decomposition tree T(G). The initial K 4 in Fig. 4 has a 3-3 layout, but in general the initial K 4 of an east corridor can have a 3-3, 4-2, 2-4, or 1-5 layout; since a K 4 with a 5-1 layout cannot have an east neighbor, only the last K 4 in an east corridor can have a 5-1 layout. Definition 3.3. A path P 2 in a graph is said to abut another path P 1 at vertex v of P 1 if they are disjoint except that an endpoint of P 2 coincides with v; v is called the abutment vertex. Theorem 3.4. Let G be a URV 3-tree with decomposition tree T = T(G). Then T has maximum degree 4 and can be written as a union of three caterpillars, T 1, T 2, and T 3 (with T 2 or T 3 possibly empty) having the following properties. (In each of the conditions 1-5, the degree of a vertex means its degree in T.) 1. If S 1 is a spine of T 1, then all the degree-4 vertices lie on S If v and v are two degree-4 vertices, then there is at least one degree-2 vertex between v and v on S The three caterpillars T 1, T 2, and T 3 are mutually vertex-disjoint except that T 2 and T 3 each have spines S 2 and S 3 that abut S 1 at vertices 6

7 v 2 and v 3, resp., with each of v 2 and v 3 at distance at least two from both endpoints of S 1, and possibly with v 2 = v If T 2 and T 3 are both nonempty, then all the degree-4 vertices lie between (and possibly include) v 2 and v If T 3 is empty, then all the degree-4 vertices lie on one side of (and possibly include) v 2. Note that condition 1 implies that T 1 is a 4-caterpillar and T 2 and T 3 are 3-caterpillars. Also, condition 3 implies that the endpoints of S 1, S 2, and S 3, with the exception of v 2 and v 3, are leaves of T; see Fig. 15a for an example. Lemma 3.5, a key technical result used in the proof of Thm. 3.4, says that under certain conditions some neighbors of degree-3 or degree-4 vertices must be leaves of T. By the length of a path, we mean the number of edges. Lemma 3.5. Let T be the decomposition tree of a 3-tree G with a URV layout L(G). Suppose T contains a vertex t u that has degree 3 or 4 and suppose, without loss of generality, that K 4 (u) has a 3-3 or 4-2 layout in L(G). Suppose that there is a path P of length 2 or more from t u to another vertex t v with degree 3 or 4. Suppose further that if t v has degree 3, then its two other neighbors are not leaves of T. If the neighbor t e of t u on P is an east (resp., south, west, north) neighbor of t u, and t u also has a south (resp., west, north, east) neighbor t s, then t s is a leaf of T. Proof. Suppose the squares of K 4 (u) are {s 1, s 2, s 3, s 4 }, with u {s 1, s 2, s 3, s 4 }, by Prop. 2.5 and without loss of generality in a 3-3 or 4-2 layout with squares labeled as shown in Fig. 5. We assume that the neighbor t e of t u on the path P is an east neighbor and that t u also has a south neighbor t s, and that K 4 (s) is represented in the layout by {s 1, s 2, s 3, s 5 }, with s {s 1, s 2, s 3, s 5 }. Let F top denote the region, indicated in dark gray in each layout in Fig. 5, that is bounded below by y 2 (s 4 ), above by y 2 (s 2 ), on the left by x 2 (s 2 ), and extending infinitely to the east. Since t e is an east neighbor of t u, its K 4 does not contain square s 2. Each K 4 corresponding to a subsequent vertex of P contains a new square, together with three squares from the preceding K 4. Thus t u is the only vertex on P whose corresponding K 4 contains square s 2, and hence F top is a forbidden region, in the sense that no square in the layout of P may intersect it. We prove first that the fourth square s 5 in K 4 (s) lies fully to the south of square s 3, and then we show that t s is a leaf of T. If the square s 5 does not lie fully to the south of s 3, then there is a second forbidden region F bot, analogous to F top, bounded on the left by x 2 (s 5 ), above by y 1 (s 3 ), 7

8 s 2 s 4 s 1 s 3 s 2 s 4 s1 s 2 s4 s 3 Figure 5: Layouts of K(u) in Lemma 3.5 and below by y 1 (s 5 ), as shown in Fig. 6, that cannot be intersected by the squares comprising the path P. s 2 s 4 s 1 s 3 s 5 s 2 s3 s 1 s 4 s 5 Figure 6: Lower forbidden region if y 2 (s 5 ) > y 1 (s 3 ) Since the neighbor t e of t u on the path P is an east neighbor of t u, its fourth square lies between the upper and lower forbidden regions, forming a K 4 with squares s 1, s 3, s 4, and these squares begin an east corridor from s 1. If this east corridor comprises the complete layout of the path P, then all of the squares lie between the upper and lower forbidden regions, and K 4 (v), corresponding to the final vertex t v on P, has either a 4-2 or a 5-1 layout. Since t v has degree 3 or 4, K 4 (v) cannot have a 5-1 layout, and if it has a 4-2 layout, at least one of its other two neighbors has a layout containing a square that is either fully above the upper forbidden region or fully below the lower forbidden region. This neighbor must be a leaf, which contradicts the assumptions about t v. Thus not all the squares K 4 (v) are in the east corridor from s 1. Hence a square of K 4 (v) must lie fully above the upper forbidden region or fully below the lower forbidden region, and t v has maximum degree 2, again contradicting the assumptions about t v. Therefore square s 5 must lie fully to the south of square s 3. Now suppose that s 5 lies fully to the south of s 3, but that it is not a leaf, so that it has an additional neighbor s 6. Fig. 7 shows all possible configurations of s 5 near and far from {s 1, s 2, s 3 } and s 6 near and far from {s 1, s 3, s 5 }, for both the 3-3 and 4-2 layouts of {s 1, s 2, s 3, s 4 }. In all cases there is a lower forbidden region, so we can argue as we did when we assumed that s 5 was not fully to the south of s 3. Thus, if s 5 does lie fully to the south of s 3, it cannot have a neighbor s 6, which completes the proof of the lemma. 8

9 3 3 layout with s 5 near, s 6 near s 2 s 4 s 1 s 3 s 6 s layout with s 5 near, s 6 near s 2 s4 s 1 s 3 s 6 s layout with s 5 near, s 6 far s 2 s 4 s 1 s 6 s 3 s layout with s 5 near, s 6 far s 2 s4 s 1 s s 3 6 s layout with s 5 far, s 6 near s 2 s 4 s 1 s 3 s 6 s layout with s 5 far, s 6 near s s 2 s 4 1 s 3 s 6 s 5 Figure 7: Lower forbidden region if y 2 (s 5 ) < y 1 (s 3 ) Lemmas are fairly straightforward from Lemma 3.5. Lemma 3.6. Let T be the decomposition tree of a 3-tree G with a URV layout L(G). Then all degree-4 vertices of T lie on a path. Furthermore, if u is a vertex lying on a path P between two degree-4 vertices, then every neighbor of u not on P must be a leaf. Lemma 3.7. Let T be the decomposition tree of a 3-tree G with a URV layout L(G), and let P be a path in T. Then there are at most two paths of length two or more that each abut P at distance two or more from both endpoints of P and that are mutually disjoint except possibly for a common abutment vertex on P. Lemma 3.8. Let T be the decomposition tree of a 3-tree G with a URV layout L(G), and let P be a path containing all degree-4 vertices of T, and let P 1 and P 2 be two paths, each of length at least two, that abut P at vertices p 1 and p 2 of P, resp., with p 1 and p 2 each at distance at least two from the endpoints of P. Suppose P, P 1, and P 2 are pairwise disjoint except for meeting at p 1 and p If P 1 and P 2 are both nonempty, then all degree-4 vertices lie on the subpath of P between p 1 and p 2, possibly including p 1 or p 2, and possibly with p 1 = p If P 2 is empty, then all degree-4 vertices lie on the same side of p 1 on P, possibly including p 1. The last lemma we need is that every path in T joining two degree-4 vertices must contain a degree-2 vertex. We do this by showing that if we have a path P = t n1, t n2,..., t nk, in which K 4 (n 1 ) has a 3-3 layout and no K 4 (n i ) has a 5-1 or 1-5 layout, then no K 4 (n i ) with i 2 has a 3-3 layout. 9

10 Lemma 3.9. Let T be the decomposition tree of a 3-tree G with a URV layout L(G). Then every path in T joining two degree-4 vertices must contain at least one degree-2 vertex. Proof. We prove by contradiction that the layout of such a path must contain a K 4 with a 5-1 or 1-5 layout, which by Prop. 2.5 must correspond to a vertex with degree 2. So suppose that P = t n1, t n2,..., t nk is a path in T whose endpoints have degree 4 and, for 1 i k, no K 4 (n i ) has a 5-1 or 1-5 layout. By Prop. 2.5, K 4 (n 1 ) and K 4 (n k ) have 3-3 layouts. If we assume without loss of generality that t n1 and t nk are the first and second vertices on P whose K 4 s have 3-3 layouts, then for 1 < i < k, K 4 (n i ) has a 4-2 or 2-4 layout. Also by Prop. 2.5, for 1 < i < k, at most one neighboring K 4 of K 4 (n i ) can have a 3-3 layout. Furthermore, at most one triple of squares from the four squares in K 4 (n i ) can be part of a neighboring K 4 with a 3-3 layout; we call this the 3-3 triple of K 4 (n i ). For example, in the 4-2 layout shown in Fig. 2b, the 3-3 triple is {a, b, c}. In general, if K 4 (n i ) on {a, b, c, d} has 3-3 triple {a, b, c}, then a square can be appended to {a, b, c} to give a neighboring K 4 with either a 3-3 or a 4-2/2-4 layout. If that neighbor has a 4-2/2-4 layout then K 4 (n i ) has no 3-3 neighbor. We prove by induction that for 2 i < k, the 3-3 triple of K 4 (n i ) is contained in K 4 (n i 1 ). For the base case, observe that since K 4 (n 1 ) has a 3-3 layout, the 3-3 triple of K 4 (n 2 ) must be contained in K 4 (n 1 ). For the inductive case, assume i 2 and the 3-3 triple of K 4 (n i ) is contained in K 4 (n i 1 ). Without loss of generality, we assume the vertices of K 4 (n i ) are {a, b, c, d} in a 4-2 layout as shown in Fig. 2b, with 3-3 triple {a, b, c}. By the inductive assumption the 3-3 triple {a, b, c} is contained in K 4 (n i 1 ), together with a fourth square s i 1. If s i+1 is the fourth square of K 4 (n i+1 ), then the four squares of K 4 (n i+1 ) are either {a, c, d, s i+1 } or {b, c, d, s i+1 } in a 4-2/2-4 layout, as shown in Fig 8. But in either case the 3-3 triple of K 4 (n i+1 ) is contained in K 4 (n i ), proving the inductive case. Thus the 3-3 triple of K 4 (n k 1 ) is contained in K 4 (n k 2 ). Since K 4 (n k ) is also a neighbor of K 4 (n k 1 ), this contradicts the assumption that K 4 (n k ) has a 3-3 layout. We conclude that unless some K 4 (n i ), 2 i k 1, has a 5-1 or 1-5 layout, and thus t ni has degree 2 in P, K 4 (n k ) cannot have a 3-3 layout and t nk cannot have degree four. We re now ready to prove the main result, Theorem 3.4. Proof of Thm T has maximum degree 4 by Prop. 2.5, and all degree- 4 vertices of T lie on a path by Lemma 3.6. Let S 1 = p 1, p 2,..., p k be a path in T containing all degree-4 vertices of T (if there are any) and with maximum length among all such paths. Let p α,..., p β, 1 α... β k, be the shortest subpath of S 1 that contains all degree-4 vertices. Suppose P is a path that abuts S 1 at vertex p j of S 1 and that is otherwise 10

11 a b c d s i 1 a b c s i 1 d Figure 8: Squares of K 4 (n i ) (on {a, b, c, d}) and neighbor K 4 (n i+1 ) disjoint from S 1. Then Lemma 3.8 and the assumptions about S 1 imply that either j α and the length of P is at most j, or j β and the length of P is at most k j. In particular, all additional neighbors of p 1 and p k not on S 1 must be leaves. So suppose the abutment vertex p j is at distance at least two from p 1 and p k. Let w be a vertex on P that is not on S 1. Then w has degree at most 3, since S 1 contains all degree-4 vertices. If w does have degree 3, then since there is a path from w to p j, Lemma 3.5 implies that w has a neighbor that is a leaf. This implies that the union of the path P and all subtrees incident with vertices of P, but not on S 1, is a caterpillar. By Lemma 3.7, there are at most two such caterpillars, T 2 and T 3, abutting S 1 at one or two vertices. If t is another vertex of S 1, then all neighbors of t not on S 1 must be leaves. Thus the union of S 1 together with all leaves of T incident with S 1 forms a 4-caterpillar T 1 with spine S 1, and T 1 T 2 T 3 = T. Furthermore S 1 satisfies condition 1, and T 1, T 2, and T 3 satisfy condition 3-5. By Lemma 3.9, S 1 also satisfies condition 2, completing the proof of Thm Partial Sufficiency In this section we show that for every tree T that satisfies the conditions of Thm. 3.4 there is some URV 3-tree G with decomposition tree T(G) = T. We do this by describing and proving the correctness of an algorithm to construct a URV layout of such a 3-tree. Theorem 4.1. Suppose that T is a tree that satisfies the hypotheses of Thm 3.4. Then there is at least one URV 3-tree G whose decomposition tree is T. For T a caterpillar satisfying the conditions given in Thm. 3.4, Lemma 4.2 gives technical details of a layout from one degree-4 vertex to the next one on a path. It s proved by giving an algorithm for creating the layout. The algorithm lays out the spine S 1 in a southeast direction by first placing an initial K 4 in a 3-3 layout and then appending squares alternately to the south or the east of the existing layout. The 3-3 layout of the initial K 4 11

12 determines whether the next square is placed to its east or south. With that in mind, we call a 3-3 layout either an east-first or south-first layout, resp., depending on whether its furthest east square is placed to the east or south, resp., of the furthest north square. For example, the layout of {1, 2, 3, 4} in Fig. 9a is a south-first layout, because square 4 is placed to the south of square Figure 9: A south-first and an east-first layout of the squares representing the start of a path t 4, t 5, t 6. The next lemma gives an algorithm to lay out a path joining two consecutive degree-4 vertices in a caterpillar satisfying the conditions of Thm The algorithm is then used multiple times in the proof of the theorem, each time for a path P = t n1, t n2,..., t nk that joins two consecutive degree-4 vertices, t n1 and t nk. In the statement and proof of Lemma 4.2 we use the path P = t 4,..., t k on squares 1, 2,..., k without loss of generality so as to simplify notation. The layout of the Lemma also prepares for the subsequent layout of abutting paths as corridors (see Def. 3.2). Lemma 4.2. Let T be a caterpillar satisfying the conditions given in Thm. 3.4, and let P = t 4, t 5,..., t k be a path in S 1 joining two consecutive degree-4 vertices, t 4 and t k. Let T = P plus every leaf of T that is adjacent to a vertex on P. Then there is a 3-tree G with decomposition tree T, such that G has a URV layout with the following properties, where V (G) = {1, 2,..., k} L(G), with L(G) the set of additional vertices needed to represent the leaves of T : 1. For i = 4,..., k, K 4 (i) is the K 4 on {i 3, i 2, i 1, i} and corresponds to t i in T. 2. K 4 (4) has a south-first 3-3 layout, and there are unobstructed visibility bands to its west, north, and east, permitting the placement of squares so that t 4 can have three neighbors in addition to its neighbor t 5. To the north and west there can be near neighbors that begin corridors; to the east there can be a far neighbor that is a leaf. 3. For 5 i k with i odd, K 4 (i) is a south neighbor of K 4 (i 1). 12

13 4. For 6 i k with i even, K 4 (i) is an east neighbor of K 4 (i 1). 5. There is a unique K 4 (α), 4 < α < k, such that K 4 (α) has either a 1-5 layout (if α is odd) or a 5-1 layout (if α is even). 6. For 5 i < α, i odd (resp., i even) K 4 (i) has a 2-4 layout (resp., 4-2 layout), and there is an unobstructed visibility band to its south (resp., east), permitting the placement of a square, sufficiently far to avoid unwanted visibilities, so that t i can have one leaf neighbor in addition to its two neighbors on P. 7. For α < i < k, i odd (resp., i even) K 4 (i) has a 2-4 layout (resp., 4-2 layout), and there is an unobstructed visibility band to its north (resp., west), permitting the placement of a square, sufficiently far to avoid unwanted visibilities, so that t i can have one leaf neighbor in addition to its two neighbors on P. 8. K 4 (k) has a 3-3 layout, and there are three unobstructed visibility bands permitting the placement of squares so that t k can have three neighbors in addition to its neighbor t k 1. Two of these can be near neighbors that can begin corridors to the east and south, and the other can be a far neighbor that is a leaf. If k is odd, (resp., even), the visibility corridors are to the east, south, and north (resp., east, south, and west) of K 4 (k). Proof. Since T and hence T satisfy Thm. 3.4, condition 2 of that theorem implies the existence of a degree-2 vertex t α in T, with 4 < α < k. We choose such a vertex t α, which we call the pivot vertex for the two consecutive degree-4 vertices t 4 and t k. We begin by placing the squares 1, 2, 3, and 4 in a south-first 3-3 layout as shown in Fig. 9a. The lower left-hand corners of the squares are (x 1 (1), y 1 (1)) = (0, 0), (x 1 (2), y 1 (2)) = (1.5, 0.5), (x 1 (3), y 1 (3)) = (0.75, 1.25), and (x 1 (4), y 1 (4)) = (2.25, 0.75). The layout of subsequent squares creates each K 4 (i) alternately as a south or east neighbor of K 4 (i 1), depending on the parity of i, as specified in conditions 2 and 3. The squares from 5 to α 2 are placed increasingly close together, so we must show that, in addition to the correct visibilities holding, the squares do not overlap one another. After that we begin again with a separate layout of squares k, k 1,..., α, and α 1 using the same algorithm. The visibilities are seen to be correct with the same calculations as for the first layout of squares 1,..., α 2. Then in the third and final step we rotate the second layout by π, and place the rotated layout at the end of the first, after checking that we can align it to achieve the visibilities we need. To this end we define two constants, one for each of the two layouts, λ 1 = 0.5/α and λ 2 = 0.5/(k α), where α is the index of the pivot vertex and k is the length of the path to be laid out. 13

14 Step 1. The layout of P from t 5 to t α 2 : For 5 i α 1, we use squares i 2 and i 1 to define a small rectangle r i (see below), and we tuck the upper left-hand corner of the square i partially into that rectangle r i. The rectangles r 5 and r 6 are indicated using dashed lines in Fig. 9, assuming α > 6; note that the upper left-hand corners of squares 5 and 6 protrude into r 5 and r 6, respectively. The constant λ 1 = 0.5/α is used to specify the amount by which square i protrudes into the small rectangle r i. Since α > 4, 0 < λ 1 < 0.5. Recall that K 4 (4) has a south-first layout. For i is odd, we define r i to be the rectangle whose lower-left corner equals the lower-right corner of i 2 and whose upper-right corner equals the lower-left corner of i 1. Square i is placed with its upper-left corner within r i, λ 1 to the right of r i s left side and λ 1 above the bottom of r i. For i even, we define r i to be the rectangle whose lower-left corner equals the upper-right corner of i 1 and whose upper-right corner equals the lower-right corner of i 2. Square i is placed with its upper-left corner within r i, λ 1 to the left of r i s right side and λ 1 below the top of r i. It is routine to check that for i = 5, 6,..., α 1, r i has width w i = 0.5 (i 5)λ 1 and height h i = 0.5 (i 5)λ 1. In order that we can place square i α 2 as described above to form K 4 (i) with {i 3, i 2, i 1, i}, we must have 1 > w i > λ 1. But this is clear, since the values of w i are decreasing from w 5 = 0.5 to w α 1 = 6λ 1. Thus the protrusion of square i into r i does not block any previous visibilities through r i. Equally important, notice that for i odd, we can place an additional square far south that is visible to squares i, i 2, and i 3, and for i even, we can place an additional square far east that is visible to squares i, i 2, and i 3 so that in all cases t i in T can have one leaf neighbor in addition to its two neighbors on P. Though square α 1 is not placed now, we define r α 1, which has width w α 1 = h α 1 = 6λ 1. See Fig. 10. Α 5 Α 4 Α 4 Α 3 Α 5 Α 3 Α 2 Α 2 Figure 10: The end of the first layout when α is odd or even, including r α 1. Step 2. The placement of squares k, k 1,..., α, α 1: We proceed with 14

15 the same algorithm as in Step 1, only using the parameter λ 2 = 0.5/(k α); however, first we check to see whether the parity of the number of squares to be laid out now, k α +2, is the same as that of α 2, the number laid out in Step 1. For example, if α 2 is odd, then K 4 (α 2) has a 2-4 layout, as shown in Fig. 10a, and similarly since α 1 is the k α 2 square to be laid out in Step 2, then if k α 2 = α 1 is odd, it is the final square of a 2-4 layout in Step 2, as shown in Figs. 11a, 12a, and 13a. Thus when k α has the same parity as α, we follow the algorithm exactly, beginning with a south-first 3-3 layout for {k, k 1, k 2, k 3}. Then we will attempt to merge the two layouts so that K 4 (α) on {α 3, α 2, α 1, α} will form the needed 1-5 or 5-1 layout to represent t α. On the other hand, if the parities of α and k α differ, then we begin our second layout with an east-first 3-3 layout of {k, k 1, k 2, k 3}. In this case the lower left-hand corners of the squares are (x 1 (k), y 1 (k)) = (0, 0), (x 1 (k 1), y 1 (k 1)) = ( 0.5, 1.5), (x 1 (k 2), y 1 (k 2)) = (1.25, 0.75), and (x 1 (k 3), y 1 (k 3)) = (0.75, 2.25), which correspond to the reflection of a south-first 3-3 layout in the line y = x. The roles of south and east are interchanged as are the roles of north and west; in particular, an x-y layout becomes a y-x layout. In addition, for i even, the first definition of r i holds, and for i odd, the second definition holds, as illustrated in Fig. 9b. k... Α 2 Α 1 Α Α 1 k... Α 2 Α 1 Α Α 1 Figure 11: The second layout when k α is odd or even. In both cases we define the same rectangles, calling them r i with width w i = h i = 0.5 (k 4 i)λ 2 for i = k 4, k 5,..., α, α 1, with λ 2 = 0.5/(k α). Again 1 > w i > λ 2, and the needed visibilities are achieved. For future use we also define r α 2 with side length 2λ 2. Step 3. Finishing the layout of P: We combine the two layouts as follows. We rotate the layout of Step 2 through π; notice that this rotation will change the visibility directions of east and west, also north and south, but will not otherwise affect any visibility corridor. Furthermore, if the 3-3 layout of {k, k 1, k 2, k 3} was south-first (resp., east-first), then the rotated layout is still south-first (resp. east-first) with the same coordinates 15

16 Α 1 Α Α 1 Α 2... Α 1 Α Α 1 Α 2... k k Figure 12: The rotated second layout when k α is odd or even. up to translation. See Fig. 12 for the rotations of the layouts shown in Fig. 11. There are cases now to consider, depending on the parity of α and of k α. When α is odd, we have the end of the first layout as in Fig. 10a. If k α is odd, we have the layouts of Figs. 11a and 12a, that merge as shown in Fig. 13a. When k α is even, we begin Step 2 with an east-first layout of {k, k 1, k 2, k 3}, which creates a layout also as in Figs. 11a and 12a, so that we merge as in Fig. 13a. Similarly when α is even, regardless of the parity of k α, we use layouts as in Figs. 10b, 11b, 12b, to merge as shown in Fig. 13b. Α 4 Α 3 Α 4 Α 2 Α 1 Α Α 1 Α 3 Α 2 Α 1 Α Α 1 Figure 13: Merging two layouts Suppose α is odd. To achieve the merge illustrated in Fig. 13a, we slide the rotated layout so that square α 1 protrudes partially into r α 1 and square α 2 protrudes partially into r α 2, though the amount of protrusion doesn t matter. This alignment is possible since the left side of α lies to the west of the left side of α 1 and the top of α lies south of the bottom of α 1. Then K 4 (α) has the desired 1-5 layout, α 4 continues to have visibility to the east, and α+1 continues to have visibility to the west. The 16

17 case when α is even is completed similarly. Recall that in the statement and proof of Lemma 4.2 we let P = t 4,..., t k, so as to simplify notation, and we began the layout of P with a south-first 3-3 layout of K 4 (4). Cor. 4.3 states that we may use the more general notation, P = t n1,..., t nk, and that we may begin the layout of P with either a south-first or an east-first 3-3 layout of K 4 (n 1 ). Corollary 4.3. Let T be a caterpillar satisfying the conditions of Thm. 3.4, let P = t n1,..., t nk be a path joining two consecutive vertices in the spine of T, and T = P plus every leaf of T that is adjacent to a vertex on P. If G is the 3-tree with URV layout L(G) that results from applying the algorithm in the proof of Lemma 4.2, in which K 4 (n 1 ) has a south-first 3-3 layout, then there is another layout for G in which K 4 (n 1 ) has an east-first 3-3 layout and such that all other properties of Lemma 4.2 hold with south and east, north and west interchanged. Proof. Given the layout L(G) produced by the algorithm in the proof of Lemma 4.2, we obtain a new layout that satisfies the conditions of the corollary simply by taking the reflection of L(G) in the line y = x. In light of Cor. 4.3 we assume henceforth that Lemma 4.2 can be applied to produce a layout beginning with either a south-first or an east-first 3-3 layout, as needed. Fig. 14 illustrates two successive applications of Lemma 4.2, first from t 4 to t 8, then from t 8 to t 10, and also their concatenation. K 4 (4) has an east-first 3-3 layout, α = 6, and both K 4 (8) and K 4 (10) also have east-first 3-3 layouts. The degree-4 vertices have squares placed far to the north and to the south, and the degree-3 vertices, t 5 and t 7, have a far square to the east and to the west, respectively. Now we present the proof of Thm Proof of Theorem 4.1. Let S 1 be a spine of T 1 with its two end vertices deleted. The proof is done in cases depending on whether the abutment vertices v 2 and v 3 exist and where they are with respect to degree-4 vertices on S 1. Case 1: T is a caterpillar T 1 with spine S 1. In this case there are no abutment vertices. We begin by making several assumptions that do not affect the generality of our argument. First we assume that S 1 has more than one vertex, since otherwise T 1 is comprised of a single vertex v with at most four degree-1 neighbors. In that case we give K 4 (v) a 3-3 layout, and we place squares as needed, to the north, east, south, and west, to lay out the neighbors of v. Without loss of generality we label the vertices of S 1 successively, t 4, t 5,..., t k, where 4 < k. Finally we assume without loss of generality that deg(t 4 ) = deg(t k ) = 4. We can do this because if, 17

18 Figure 14: A layout using two applications of the algorithm of Lemma 4.2 say, deg(t k ) < 4, we could insert two vertices that follow t k on the spine of T 1 : t k+1 with degree 2 and t k+2 with degree 4. We could then lay out the extended caterpillar, and then delete any squares that correspond only to t k+1 or t k+2. We would proceed in an analogous way if deg(t 4 ) < 4. So suppose the vertices of degree 4 on S 1 are t 4 = t i0, t i1,..., t ij = t k, where possibly j = 1 and t i1 = t k. We lay out the caterpillar T 1 by applying Lemma 4.2 to the subpaths of S 1 from t il 1 to t il, for l = 1, 2,..., j. By Cor. 4.3 we may assume that after laying out the subpath from t il 1 to t il, then we can lay out the subpath from t il to t il+1 so that the two layouts of K 4 (i l ) have the same sort of 3-3 layout, either both south-first or both eastfirst. Thus we can concatenate the layouts of the two successive subpaths by superimposing the two 3-3 layouts of K 4 (i l ). Note that when two layouts are concatenated at the 3-3 layout of K 4 (i l ), 0 < l < j, K 4 (i l ) retains visibility bands in two directions, in each of which a far neighbor can be placed to represent a leaf. By conditions 2 and 6 8 of Lemma 4.2, the spine representation can be extended to include squares representing all leaves of T. Case 2: Each abutment vertex has degree 4. By conditions 4 and 5 of Thm. 3.4, v 2 is the first degree-4 vertex on S 1, and v 3, if it exists, is the last. Label the subpath of S 1 from v 2 to v 3, in order, v 2 = t 4,..., t k = v 3 ; if v 3 doesn t exist, let t k be the final endpoint of S 1. Using the same technique as in Case 1, we create a layout of the subpath of S 1 from t 4 to t k. The layout algorithm gives K 4 (4) and K 4 (k) 3-3 layouts with unobstructed visibility bands in three directions. We label the subpath of S 1 preceding v 2 = t 4 sequentially beginning with t k+1 for the preceding 18

19 neighbor of t 4, and we lay out this section as a west-facing corridor based at square 3 or 4, depending on the layout of K 4 (4). (It can be seen in Fig. 9a that we could lay out a west-facing corridor at 4, and in Fig. 9b that we could lay out a west-facing corridor at 3.) We label the section of S 1 following t k sequentially from the neighbor of t k, continuing the numbering already in use, and we lay out this section as an east-facing corridor based at a square of K 4 (k). Next we label the vertices of S 2 in order, beginning with the neighbor of v 2, and continuing the labeling already in use. We lay out S 2 as a northfacing corridor based at a square of K 4 (4) (again see Fig. 9). If S 3 exists, we label its vertices in order, beginning with the neighbor of v 3, and continuing the labeling in use already. We lay out S 3 as a south-facing corridor based at a square of K 4 (k). The completed layout of S 1 S 2 S 3 permits the placement of additional far squares to create a K 4 for each leaf of T. Specifically, for degree-4 vertices of S 1, we can place far squares north and south, and for degree-3 vertices of S 1, S 2 and S 3 we can place far squares in a suitable direction. Case 3: At least one abutment vertex, say v 3, has degree less than 4. Similarly to Case 2, label the subpath of S 1 from v 2 to v 3, in order, v 2 = t 4,..., t k = v 3, and lay out this subpath of S 1 as described in Case 1. Label the rest of S 1, S 2, and S 3 as in Case 2. The vertices preceding t 4 on S 1 and following t k can be laid out as a west-facing and an east-facing corridor, respectively. By Lemma 4.2 it is possible to lay out squares for S 2 as a north-facing corridor and for S 3 as a south-facing corridor, much as in Case 2. The remaining leaves can be labeled in arbitrary order and their squares placed far to create the needed representation. In Fig. 15a the example of Fig. 14 is extended to include abutting caterpillars at t 4 and at t 13, shown in darker gray, and the layout created using the algorithm of the proof is shown in Fig. 15b. 5 Counterexamples to Sufficiency Here we show that the path on three vertices is the decomposition tree of a 3-tree (on six vertices) that is not a URVG; by the example of Fig. 1b it also is the decomposition tree of a different 3-tree that is a URVG. We also show that every 3-tree G with decomposition tree T(G) that is a path on five or fewer vertices with t 4 as a leaf is a URVG, but that G may or may not be a URVG when such a path has six or more vertices. It is routine to check the following (independent of Thm. 3.4). Proposition 5.1. five K 4 s. 1. Two squares forming a K 2 can be a part of at most 19

20 Figure 15: A layout using the algorithm described in the proof of Thm Three squares forming a K 3 can be a part of at most two K 4 s. In Fig. 15b consider squares {1, 2, 3, 4, 5,11,12, 28} for an example that attains these two bounds. Corollary For each n 3, there is a 3-tree G on n+3 vertices, with decomposition tree T(G) a path of n vertices, such that G is not a URVG. 2. For each n 6, there is a 3-tree G on n + 3 vertices, with decomposition tree T(G) a path of n vertices that begins with t 4, such that G is not a URVG. Proof. Consider the 3-tree with K 4 s with vertex sets {1, 2, 3, 4}, {1, 2, 3, 5}, and {1, 2, 3, 6}, which is not a URVG by Prop Its tree is the path P = t 5, t 4, t 6. Then extend P at either t 5 or t 6. Similarly consider the 3-tree with K 4 s with vertex sets {1, 2, 3, 4}, {1, 2, 4, 5}, {1, 2, 5, 6}, {1, 2, 6, 7}, {1, 2, 7, 8}, {1, 2, 8, 9},..., so that each K 4 contains {1, 2}. It follows from Prop. 5.1 that G cannot be a URVG when it has nine or more vertices. It is routine to check every 3-tree with five or fewer vertices is a URVG and also that the bound of Cor is best possible. 20

21 6 Conclusion A characterization of the 3-trees that have a URV layout seems to be more difficult than characterizing their decomposition trees. We have an example of a 3-tree on 16 vertices that satisfies Prop. 5.1, and its decomposition tree satisfies Thm. 3.4, yet is not a URVG. In the case of URV 2-trees, even characterizing the decomposition trees has thus far proved intractable. In [4] we determine the intersections of the sets of decomposition trees of URV 1-trees, 2-trees, and 3-trees, but there still remains much left open for further study. References [1] S. Arnborg, D. Corneil, and A. Proskurowski. Complexity of finding embeddings in a k-tree. SIAM J. Alg. and Discr. Methods, 8: , [2] P. Bose, A. Dean, J. Hutchinson, and T. Shermer. On rectangle visibility graphs. In Lecture Notes in Computer Science 1190: Graph Drawing, pages Springer-Verlag, [3] A. M. Dean, J. A. Ellis-Monaghan, S. Hamilton, and G. Pangborn. Unit rectangle visibility graphs. Electronic Journal of Combinatorics, 15(#R79):1 24, [4] A. M. Dean and J. P. Hutchinson. On k-trees that are unit rectanglevisibility graphs. In preparation, [5] A. M. Dean and N. Veytsel. Unit bar-visibility graphs. Congr. Numer., 160: , [6] G. Di Battista, P. Eades, R. Tamassia, and I. G. Tollis. Graph Drawing. Prentice Hall, Upper Saddle River, NJ, [7] R. Diestel. Graph Decompositions, a study in infinite graph theory. Clarendon Press, Oxford, [8] S. Fekete, M. Houle, and S. Whitesides. New results on a visibility representation of graphs in 3d. In Lecture Notes in Computer Science 1027: Graph Drawing, pages Springer, [9] B. Mohar and C. Thomassen. Graphs on Surfaces, chapter 7. Johns Hopkins University Press, Baltimore, [10] N. Robertson and P. Seymour. Graph minors II, algorithmic aspects of tree-width. J. Algorithms, 7: ,

22 [11] D. J. Rose. On simple characterizations of k-trees. Discrete Math., 7: , [12] T. Shermer. On rectangle visibility graphs III. External visibility and complexity. In Proc. 8th Canad. Conf. on Comp. Geom., pages , [13] J. Ullman. Computational Aspects of VLSI. Computer Science Press, Rockville, MD, [14] J. Wald and C. Colbourn. Steiner trees, partial 2-trees, and minimum IFI networks. Networks, 13: , [15] D. West. Introduction to Graph Theory, 2nd ed. Prentice-Hall, Upper Saddle River, NJ, [16] S. H. Whitesides. An algorithm for finding clique cut-sets. Inf. Proc. Letters, 12(1):31 32,