Graphs with Near Optimal Heights
Chieh-Yu Chen1, Ya-Fei Hung2, and Hsueh-I Lu1,2,
1Department of Computer Science and Information Engineering National Taiwan University
2Graduate Institute of Networking and Multimedia National Taiwan University
1 Roosevelt Road, Section 4, Taipei 106, Taiwan, ROC
f94922054@ntu.edu.tw, r94944014@ntu.edu.tw, hil@csie.ntu.edu.tw
Abstract. A visibility representation of a graph G is to represent the nodes of G with non-overlapping horizontal line segments such that the line segments repre- senting any two distinct adjacent nodes are vertically visible to each other. If G is a plane graph, i.e., a planar graph equipped with a planar embedding, a visibility representation of G has the additional requirement of reflecting the given planar embedding of G. For the case that G is an n-node four-connected plane graph, we give an O(n)-time algorithm to produce a visibility representation of G with height at mostn
2
+ 2
n−2 2
. To ensure that the first-order term of the up- per bound is optimal, we also show an n-node four-connected plane graph G, for infinite number of n, whose visibility representations require heights at leastn2.
1 Introduction
Unless clearly specified otherwise, all graphs in the present article are simple, i.e., hav- ing no self-loops and multiple edges. A visibility representation of a planar graph rep- resents the nodes of the graph by non-overlapping horizontal line segments such that, for any nodes u and v adjacent in the graph, the line segments representing u and v are vertically visible to each other. Observe that if G1is a subgraph of G2on the save node set, then any visibility representation of G2 is also a visibility representation of G1. Therefore, we may assume without loss of generality that the input graph is maximally planar. Let G be an n-node plane triangulation, i.e., a maximally planar graph equipped with a planar embedding. A visibility representation of G has an additional requirement of reflecting the given planar embedding of G. Figure 1(b), for instance, is a visibility representation of the four-connected plane graph shown in Fig. 1(a). Under the conven- tional restriction of placing the endpoints of horizontal line segments on the integral grid points, any visibility representation of G requires width no more than 3n−7 and height no more than n− 1. Otten and van Wijk [7] gave the first known algorithm for con- structing a visibility representation for any G. Rosenstiehl and Tarjan [8] and Tamassia
Corresponding author.http://www.csie.ntu.edu.tw/∼hil.This author also holds a joint appointment in the Graduate Institute of Biomedical Electronics and Bioinformatics, National Taiwan University. Research supported in part by NSC grant 96-2221-E-002-033.
I.G. Tollis and M. Patrignani (Eds.): GD 2008, LNCS 5417, pp. 67–77, 2009.
Springer-Verlag Berlin Heidelberg 2009c
1 (b)
2 6 7
8
5 4 3
(a) 5 4 2 3
7 8
6
1
Fig. 1. (a) A four-connected plane triangulation G. (b) A visibility representation of G.
and Tollis [9] independently gave algorithms to compute a visibility representation of G with height at most 2n− 5. Their work initiated a decade of competition on minimizing the width and height of the output visibility representation. All these algorithms run in linear time. In particular, the results of Fan, Lin, Lu, and Yen [2] and Zhang and He [16]
are optimal in that the upper bounds differ from the best known lower bounds by very small constants.
The present article focuses on four-connected plane G. The O(n)-time algorithm of Kant and He [5] provides the optimal upper bound n− 1 on the width. The best previously known upper bound on the height, ensured by the O(n)-time algorithm of Zhang and He [12], is3n
4
. In the present article, we obtain the following result with an improved upper bound on the required height.
Theorem 1. For any n-node four-connected plane graph G, it takes O(n) time to con- struct a visibility representation of G with height at mostn
2
+ 2
n−2 2
.
Table 1 compares our upper bound with previous results. All algorithms shown in Ta- ble 1 run in O(n) time. Our algorithm follows the approach of Zhang and He [10, 15–
17], originating from Rosenstiehl and Tarjan [8] and Tamassia and Tollis [9], that re- duces the problem of computing a visibility representation for G with small height to finding an appropriate st-ordering of G. To find such an st-ordering of G, we resort to three linear-time obtainable node orderings:
– four-canonical orderings of four-connected plane graphs (Kant and He [5]), – consistent orderings of ladder graphs (Zhang and He [15–17]), and
– post-orderings of canonical ordering spanning trees (He, Kao, and Lu [3]).
Our result is near optimal in that we can construct an n-node four-connected plane graph, for infinite number of n, whose visibility representations require heights at least
n
2
. That is, the first-order term of our upper bound is optimal.
The remainder of the paper is organized as follows. Section 2 gives the preliminaries.
Section 3 describes and analyzes our algorithm. Section 4 ensures that the first-order term of our upper bound on height is optimal. Section 5 concludes the paper.
Table 1. Previous upper bounds and our result for any n-node plane graph G
general G four-connected G
width height width height
Otten and van Wijk [7] 3n− 7 n− 1 Rosenstiehl and Tarjan [8],
Tamassia and Tollis [9] 2n− 5
Kant [4] 3n−6
2
Kant and He [5] n− 1
Lin, Lu, and Sun [6] 22n−24
15
Zhang and He [10] 15n
16
Zhang and He [14] 5n
6
Zhang and He [11, 13] 13n−24
9
Zhang and He [12] 3n
4
Zhang and He [15, 17] 4n3 + 2√n 2n3 + 2 n
2
Zhang and He [16] 2n3 + O(1)
Fan, Lin, Lu, and Yen [2] 4n
3
− 2
This paper n
2
+ 2
n−2 2
2 Preliminaries
2.1 Ordering and st-Ordering
Let G be an n-node plane graph. An ordering of G is a one-to-one mapping σ from the nodes of G to{1, 2, . . . , n}. A path of G is σ-increasing if σ(u) < σ(v) holds for any nodes u and v such that u precedes v in the path. Let length(G, σ) denote the maximum of the lengths of all σ-increasing paths in G. For instance, if G and σ are as shown in Fig. 1(a), then one can verify that (1, 2, 5, 6, 8) is a σ-increasing path with maximum length. Therefore, length(G, σ) = 4.
Let s and t be two distinct external nodes of G. An st-ordering [1] of G is an ordering σof G such that
– σ(s) = 1, σ(t) = n, and
– each node v of G other than s and t has neighbors u and w in G with σ(u) <
σ(v) < σ(w).
An example is shown in Fig. 1(a): the node labels form an st-ordering for the graph.
The following lemma reduces the problem of minimizing the height of visibility rep- resentation of G to that of finding an st-ordering σ of G with minimum length(G, σ).
Lemma 1 (See [2, 8–10, 15, 17]). If G admits an st-ordering σ for two distinct external nodes s and t of G, then it takes O(n) time to obtain a visibility representation of G with height exactly length(G, σ).
For instance, if G and σ are as shown in Fig. 1(a), then a visibility representation for Gwith height at most length(G, σ) = 4, as shown in Fig. 1(b), can be found in linear time.
2.2 Four-Canonical Ordering
Let G be an n-node four-connected plane triangulation. Let v1, v2, and vn be the ex- ternal nodes of G in counterclockwise order. Since G is a four-connected plane trian- gulation, G has exactly one internal node adjacent to both v2and vn. Let vn−1 be the internal node adjacent to v2 and vn in G. A four-canonical ordering [5] of G is an ordering φ in G such that
– φ(v1) = 1, φ(v2) = 2, φ(vn−1) = n− 1, φ(vn) = n, and
– each node v of G other than v1, v2, vn−1and vnhas neighbors u, u, w and w in Gwith φ(u) < φ(u) < φ(v) < φ(w) < φ(w).
An example is shown in Fig. 2(a): the node labels form a four-canonical ordering of the four-connected plane triangulation.
Lemma 2 (Kant and He [5]). It takes O(n) time to compute a four-canonical ordering for any n-node G.
2.3 Consistent Ordering of Ladder Graph Let L be ann
2
-node path. Let R be ann
2
-node path. Let X consist of edges with one endpoint in L and the other endpoint in R. Let (L, R, X) denote the n-node graph L∪R∪X. We say that (L, R, X) is a ladder graph [15, 17] if L∪R∪X is outerplanar.
A ladder graph is shown in Fig. 3(a).
An ordering σ of ladder graph (L, R, X) is consistent [15, 17] with respect to an outerplanar embeddingE of (L, R, X) if L (respectively, R) forms a σ-increasing path in clockwise (respectively, counterclockwise) order according toE. See Fig. 3(a) for an example: The node labels form a consistent ordering of the ladder graph with respect to the displayed outerplanar embedding.
Lemma 3 (He and Zhang [15, 17]). Let (L, R, X) be an n-node ladder graph. It takes O(n) time to compute a consistent ordering σ of (L, R, X) with respect to any given outerplanar embedding of (L, R, X) such that length((L, R, X), σ) ≤ n
2
+ 2 n
2
− 1.
For technical reason, we need a consistent ordering with additional properties, as stated in the next lemma, which is also illustrated by Fig. 3(a).
Lemma 4. Let (L, R, X) be an n-node ladder graph. It takes O(n) time to compute a consistent ordering σ of (L, R, X) with respect to any given outerplanar embeddingE of (L, R, X) such that
1
4 6
2
7 3
5
8 (a)
1
4 6 7 2
3 5
8
(b) (c)
1
2 3 4
2 1 3
4
G
RT
RG
LT
LFig. 2. (a) A four-canonical ordering φ of the four-connected plane triangulation G. (b) GLis the subgraph induced by the nodes v with 1≤ φ(v) ≤ 4 and GR is the subgraph induced by the nodes v with 5≤ φ(v) ≤ 8. (c) The counterclockwise post-ordering ψLof TLand the clockwise post-ordering ψRof TR.
– σ(1) = 1, σ(r1) = 2, and – length((L, R, X), σ)≤n
2
+ 2
n−2 2
,
where 1(respectively, r1) is the first (respectively, last) node of L (respectively, R) in clockwise order around the external boundary of (L, R, X) with respect toE.
Proof. Let L = L\ {1}. Let R = R\ {r1}. Let X = X \ {1, r1}. Clearly, (L, R, X) is a ladder graph of n− 2 nodes. Let σ be the consistent ordering of (L, R, X)with respect toE ensured by Lemma 3. We have
length((L, R, X), σ)≤n 2
+ 2
n− 2 2
− 2.
Let σ be the ordering of (L, R, X) such that
(a) L
X
R
(b) X∗
R L
1= v2
v1
r2= v5
r1= v7
4= v1
3= v4
2= v3
8 r4= v8
r3= v6 v4
v3 v5
v7
2 4 5 7
3 6
1
v8
v2
v6
Fig. 3. (a) A consistent ordering of a ladder graph (L, R, X) with respect to the displayed outer- planar embedding. (b) H∗= L∪ R ∪ X∗, where X∗= X∪ {(v2, v8)}.
– σ(1) = 1, σ(r1) = 2, and
– σ(u) = σ(u) + 2holds for each node u other than 1and r1. One can easily verify that the lemma holds.
3 Our Algorithm
Let G be the input n-node four-connected plane triangulation. According to Lemma 1, it suffices to describe our algorithm for computing an st-ordering σ for G in the following four steps.
3.1 Step 1
Let φ be a four-canonical ordering of G ensured by Lemma 2.
– Let GLbe the subgraph of G induced by the nodes v with 1≤ φ(v) ≤n
2
. – Let GRbe the subgraph of G induced by the nodes v withn
2
< φ(v)≤ n.
Figure 2(b) illustrates this step, which runs in O(n) time. Observe that each edge of G not in GL∪GRhas one endpoint on the external boundary of GLand the other endpoint on the external boundary of GR.
3.2 Step 2
For each i = 1, 2, . . . , n, let vi denote the node of G with φ(vi) = i. It follows from the definition of φ that v1, v2, and vnare the external nodes of G.
– For each i = 2, 3, . . . ,n
2
, let π(i) be the index j with j < i such that vjis the first neighbor of viin GLin counterclockwise order around vi. Let TLbe the spanning tree of GLrooted at v1such that each vπ(i)is the parent of viin TL. Let ψLbe the counterclockwise post-ordering of TL.
– For each i =n
2
+ 1,n
2
+ 2, . . . , n− 1, let π(i) be the index j with j > i such that vjis the first neighbor of viin GRin clockwise order around vi. Let TRbe the spanning tree of GRrooted at vnsuch that each vπ(i)is the parent of viin TR. Let ψRbe the clockwise post-ordering of TR.
Figure 2(c) illustrates this step, which runs in O(n) time. As a matter of fact, TLis the canonical ordering spanning tree of GLwith respect to φ, as defined by He, Kao, and Lu [3].
Lemma 5. ψL(v2) = 1, ψL(v1) =n
2
, ψR(vn−1) = 1, and ψR(vn) =n
2
.
Proof. Since φ is a four-canonical ordering of G, if (v2, vi)with i ≥ 3 is an edge of GL, then vihas to have a neighbor vkwith 2= k < i in GL. Observe that v2is the node immediately succeeding v1in counterclockwise order around the external boundary of GL. One can verify that v2cannot be the first neighbor of viin GLin counterclockwise order around vi. That is, we have π(i) = 2. Since v2cannot be the parent of viin TL, v2has to be a leaf of TL. By the relative position between v2and v1, it is clear that v2
is the first node in the counterclockwise post-ordering of TL, i.e., ψL(v2) = 1.
One can prove ψR(vn−1) = 1analogously, where vn (respectively, vn−1, ψR, TR, and GR) plays the role of v1(respectively, v2, ψL, TL, and GL). Since v1is the root of TLand ψLis a post-ordering of TL, we have ψL(v1) =n
2
. Since vnis the root of TR
and ψRis a post-ordering of TR, we have ψR(vn) =n
2
.
3.3 Step 3
Let L, R, and X be defined as follows.
– Let L be the path
1, 2, . . . , n/2
, where iis the node of GLwith ψL(i) = i.
– Let R be the path
r1, r2, . . . , rn/2
, where riis the node of GRwith ψR(ri) = i.
– Let X = X∗\ {(v2, vn)}, where X∗consists of the edges of G with one endpoint in L and the other endpoint in R.
Figure 3(a) illustrates Lemma 5 and this step, which runs in O(n) time. Figure 3(b) shows the corresponding L∪ R ∪ X∗.
Lemma 6. (L, R, X) is an n-node ladder graph.
Proof. Consider any edge (i, rj)of X. By definition of φ, ihas to be on the exter- nal boundary of GL and rj has to be on the external boundary of GR. By definition of TL, iis either a leaf of TL or on the rightmost path of TL. By definition of ψL, if
i1, i2, . . . , ipwith i1 = 1are the nodes on the external boundary of GL in counter- clockwise order, then i1 < i2 <· · · < ip. Similarly, by definition of TR, rj is either a leaf of TR or on the leftmost path of TR. By definition of ψR, if rj1, rj2, . . . , rjq
with j1 = 1are the nodes on the external boundary of GR in clockwise order, then j1< j2 <· · · < jq. Since G is a plane graph and the edges of X do not cross one an- other in G, the edges of X do not cross one another in (L, R, X). Therefore, (L, R, X) is outerplanar.
3.4 Step 4
Let H = (L, R, X). Lemma 6 ensures that H is an n-node ladder graph. Consider the outerplanar embeddingE of H such that
1, 2, . . . , n/2, rn/2, rn/2−1, . . . , r1
are the nodes in clockwise order around the external boundary of H. Let the output σ of our algorithm be the consistent ordering of H with respect toE ensured by Lemma 4.
Figure 3(a) illustrates this step, which also runs in O(n) time.
Lemma 7. The O(n)-time obtainable σ is an st-ordering of G with σ(v2) = 1 and max(σ(v1), σ(vn)) = n.
Proof. We first show that ψL is an st-ordering of GL. Let i be an index with 2 ≤ i <n
2
. Let k be the index such that k is the parent of iin TL. Since ψLis a post- ordering of TL, we know that kis a neighbor of iin GLwith i < k. Let j be the index such that jis the neighbor of iin GLimmediately succeeding kin counterclockwise order around i. Recall that k is the first neighbor of i in GL with φ(k) < φ(i) in counterclockwise order around i. Since φ is a four-canonical ordering of G, we also have φ(j) < φ(i). Since ψL is the counterclockwise post-ordering of TL, we have ψ(j) < ψ(i), i.e., j < i. Since j and k are two neighbors of i in GL with j < i < k, we know that ψLis an st-ordering of GL. It can be proved analogously that ψRis an st-ordering of GR.
Since σ is a consistent ordering of H with respect toE, we know that 1 ≤ i < j ≤
n
2
implies σ(i) < σ(j)and 1≤ i < j ≤n
2
implies σ(ri) < σ(rj). We have the following observations.
– Since ψLis an st-ordering of GL, for each i = 1, . . . ,n
2
− 1, ihas a neighbor
kin GLwith i < k. Since GLis a subgraph of G, kis a neighbor of iin G with σ(i) < σ(k).
– Since ψLis an st-ordering of GL, for each i = 2, . . . ,n
2
, ihas a neighbor j in GLwith j < i. Since GLis a subgraph of G, we know that jis a neighbor of iin Gwith σ(j) < σ(i).
– Since ψRis an st-ordering of GR, for each i = 1, . . . ,n
2
− 1, rihas a neighbor rkin GRwith i < k. Since GRis a subgraph of G, we know that rk is a neighbor of riin G with σ(ri) < σ(rk).
– Since ψRis an st-ordering of GR, for each i = 2, . . . ,n
2
, rihas a neighbor rj in GRwith j < i. Since GRis a subgraph of G, we know that rj is a neighbor of ri
in G with σ(rj) < σ(ri).
According to the above observations, it suffices to ensure that edges (1, r1) and (n/2, rn/2)belong to G. By Lemma 5, 1 = v2, r1 = vn−1, n/2 = v1, and rn/2 = vn. Since v1and vnare external nodes of the plane triangulation G, we know that (n/2, rn/2) = (v1, vn)is an edge of G. By definition of four-canonical order- ing φ, we know that vn−1is adjacent to v2. Therefore, (1, r1) = (v2, vn−1)is an edge of G.
Figure 1(a) shows the resulting st-ordering σ of G computed by our algorithm.
3.5 Proving Theorem 1
Proof. Note that v1, v2, and vn are the external nodes of G. By Lemmas 1 and 7, it suffices to ensure
length(G, σ)≤n 2
+ 2
n− 2 2
. (1)
By Step 4 and Lemmas 4 and 6, we have
length(H, σ)≤n 2
+ 2
n− 2 2
. (2)
Let H∗= L∪R∪X∗. That is, H∗= H∪{(v2, vn)}, as illustrated by Fig. 3(a) and 3(b).
By definition of σ and Lemma 5, we have σ(v2) = 1and σ(vn)≥ maxjσ(rj). There- fore, any σ-increasing path of H∗containing edge (v2, vn)contains exactly one node of R, i.e., vn, and thus has length at mostn
2
. It follows from Inequality (2) that
length(H∗, σ)≤n 2
+ 2
n− 2 2
. (3)
To prove Inequality (1), it remains to show that if P is a σ-increasing path of G, then there is a σ-increasing path Q of H∗such that the length of Q is no less than that of P . For each edge (u, v) of P with σ(u) < σ(v), let Q(u, v) be the σ-increasing path of H∗defined as follows.
– If u = i and v = rj, then let Q(u, v) = (u, v), which is a σ-increasing path of X∗.
– If u = ri and v = j, then let Q(u, v) = (u, v), which is a σ-increasing path of X∗.
– If u = iand v = j, then by σ(i) < σ(j)we know ψL(i) < ψL(j)and thus i < j. Let Q(u, v) = (i, i+1, . . . , j). Since σ is a consistent ordering of H with respect toE, Q(u, v) is a σ-increasing path of L.
– If u = riand v = rj, then by σ(ri) < σ(rj)we know ψR(ri) < ψR(rj)and thus i < j. Let Q(u, v) = (ri, ri+1, . . . , rj). Since σ is a consistent ordering of H with respect toE, Q(u, v) is a σ-increasing path of R.
Let Q be the union of Q(u, v) for all edges (u, v) of P . Since each Q(u, v) is a σ- increasing path of H∗, so is Q. The length of Q is no less than that of P . That is, we have
length(G, σ)≤ length(H∗, σ). (4)
Since Inequality (1) is immediate from Inequalities (3) and (4), the lemma is proved.
4 A Lower Bound
Let plane graph Nkbe defined recursively as follows.
– Let N1be the four-node internally triangulated plane graph with four external nodes.
(b) (a)
Dk
Nk+1
Nk
Dk+1
Fig. 4. (a) A four-connected plane graph Nk+1and its relation with Nk. (b) A visibility represen- tation Dk+1of Nk+1and its relation with Dk.
– Let Nk+1be obtained from Nkby adding four nodes and twelve edges in the way as shown in Fig. 4(a).
One can easily verify that each Nkwith k≥ 1 is indeed four-connected. The following lemma ensures that the the upper bound provided by Theorem 1 has an optimal first- order term.
Lemma 8. All visibility representations of Nkhave heights at least 2k.
Proof. We prove the lemma by induction on k. The lemma holds trivially for k = 1. As- sume for a contradiction that Nk+1admits a visibility representation Dk+1with height no more than 2k + 1. Let Dk be obtained from Dk+1by deleting all the horizontal segments representing those four external nodes of Nk+1. Since Dk+1has to reflect the planar embedding of Nk+1, Dk is a visibility representation of Nk. Since the external nodes of Nk are internal in Nk+1, the horizontal segments of Dk+1representing the external nodes of Nk+1have to wrap Dk completely. That is, Dk+1must have a hori- zontal segment above Dkand a horizontal segment below Dk. Therefore, the height of Dk+1is at least two more than that of Dk. It follows that the height of Dkis at most 2k− 1, contradicting the inductive hypothesis. Since Nk+1cannot admit a visibility representation with height less than 2k + 2, the lemma is proved.
5 Concluding Remarks
It would be of interest to close the Θ(√
n)gap between the upper and lower bounds on the required height for the visibility representation of any n-node four-connected plane graph. We conjecture that the Θ(√
n)term in our upper bound can be reduced to O(1).
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