Algorithmica (1996) 1 6 : 4 - 3 2
Algorithmica
c~. 1996Springer-VerlagNew YorkInc.
Drawing Planar Graphs Using the Canonical Ordering I
G. Kant 2
We introduce a new method to optimize the required area, minimum angle, and number of bends
of planar graph drawings on a grid. The main tool is a new type of ordering on the vertices and faces of
triconnected planar graphs. Using this method linear-time-and-space algorithms can be designed for many
graph-drawing problems. Our main results are as follows:
Abstract.
9 Every triconnected planar graph G admits a planar convex grid drawing with straight lines on a (2n - 4) •
(n - 2) grid, where n is the number of vertices.
9 Ever), triconnected planar graph with maximum degree 4 admits a planar orthogonal grid drawing on an
n • n grid with at most [3n/2] + 4 bends, and ifn > 6, then every edge has at most two bends.
9 Every planar graph with maximum degree 3 admits a planar orthogonal grid drawing with at most [ n / 2 J + I
bends on an [n/2J x [n/2J grid.
9 Every triconnected planar graph G admits a planar polyline grid drawing on a (2n - 6) x (3n - 9) grid
with minimum angle larger than 2/d radians and at most 5n - 15 bends, with d the maximum degree.
These results give in some cases considerable improvements over previous results, and give new bounds in
other cases. Several other results, e.g., concerning visibility representations, arc included.
Key Words.
Graph drawing, Canonical ordering, Linear time, Convex, Orthogonal, Visibility representation.
1. Introduction. The problem of"nicely" drawing a graph in the plane is an emerging
area of research that combines flavors of topological graph theory and computational
geometry. The large number of applications include VLSI layout, algorithm animation,
visual languages, and CASE tools. Several criteria to obtain a high aesthetic quality
have been established. Typically, vertices are represented by distinct points in a plane,
and are sometimes restricted to being grid points. Alternatively, vertices are sometimes
represented by line segments. Edges are often constrained to be drawn as straight lines
or as a contiguous set of line segments (e.g., when bends are allowed). The objective is to
find a layout for a graph that optimizes some cost function, such as area, minimum angle,
or the total number of bends. In the annotated bibliography an up-to-date overview of
the recent developments and optimization criteria in graph drawings is given, with more
than 300 references [9].
It is well known that every planar graph can be drawn planar with straight lines [ 16],
[40], [49], and by more recent algorithms this can be done in linear time and space on a
grid of size O(n) • O(n) (e.g., see [7], [181, [21 I, and [39]), where n denotes the number
I This work was supported by the ESPRIT Basic Research Actions program of the EC under Contract No. 7141
(project ALCOM II). An extended abstract of this paper was presented at the 33rd Annual IEEE Symposium
on the Foundations of Computer Science, Pittsburgh, 1992. This is a revised version of Technical Report
RUU-CS-92-33.
2 Department of Computer Science, Utrecht University, P.O. Box 80.089, 3508 TB Utrecht, The Netherlands.
[email protected].
Received January 29, 1993; revised April 12, 1994, September 8, 1994, and November 16, 1994. Commu nicated
by G. Di Battista and R. Tamassia.
Drawing PlanarGraphs Using the CanonicalOrdering
5
of vertices, f2 (n 2) is also a lower bound for the area of planar straight-line drawings [18].
However, a drawback of these drawing algorithms is that the minimum angle between
lines can be very small which makes the drawing unattractive.
A criterion to improve the aesthetic quality is convexity: every interior face must be
drawn convexly. Tutte showed that every triconneeted planar graph can be drawn with
convex interior faces [47]. Thomassen [46] characterized the class of planar graphs which
admit a convex drawing, and Chiba et al. [51 presented a linear-time drawing algorithm
for this class. However, the coordinates of the vertices can be reals and a huge number
of vertices can be clustered in a small area.
Another representation model is planar orthogonal drawing, in which the vertices
are represented by points and edges by alternatingly horizontal and vertical segments,
connecting the endpoints. Orthogonal drawings have numerous important applications in
the field of VLSI design and graphics. Storer 141] and Tamassia and Tollis [43] presented
heuristics to construct an orthogonal representation of a 4-planar (3-planar) graph with
at most 2n + 4 bends (resp. n + 2 bends). (d-planar means that the planar graph has
maximum degree d.) Every edge is bent at most four times. Tamassia presented an
O(n 2 log n)-time algorithm for drawing an embedded 4-planar graph on an n • n grid
with the minimum number of bends [42]. If the planar embedding is not given in advance,
then the problem is polynomial-time solvable for 3-planar graphs [10], and NP-complete
for 4-planar graphs [19]. In particular, Garg and Tamassia showed that it is even NP-hard
to approximate the minimum number of bends in a planar orthogonal drawing with an
O(n l-~) error, for any E > 0 [19].
In this paper we introduce a new ordering on the vertices and faces of a triconnected
planar graph, called the canonical ordering. The canonical ordering can be computed
in linear time. Refining the canonical ordering to a lefimost canonical ordering (or lmcordering) this leads to a general framework for drawing triconnected planar graphs on
a grid, and implies several drawing results (n denotes the number of vertices and d the
maximum degree):
1. Every triconnectcd planar graph with n vertices admits a planar straight-line convex
grid drawing of area at most (2n - 4) x (n - 2). This is the first algorithm combining
the aspects of grid size with convexity. It also implies a new and rather simple proof
that every triconnected planar graph admits a convex drawing.
2. Every triconnected 4-planar graph,admits a planar orthogonal grid drawing with at
most [3n] + 4 bends on an n • n grid. Ifn > 6, then every edge is bent at most twice.
3. Every 3-planar graph G admits a planar orthogonal grid drawing with at most [n/2J + 1
bends on an [n/2J • [n/2J grid. A nice characteristic is that G has a spanning tree
using n - I straight-line edges and all nontree edges have at most one bend.
4. Every triconnected planar graph admits a planar polyline grid drawing on a (2n 6) • (3n - 9) grid, with minimum angle > 2/d and at most 5n - I5 bends totally
(we call this the mLred model). Every edge has at most three bends. According to
our opinion, this is the first practical drawing algorithm, having good bounds on the
area, number of bends, and minimum angle. This result is extended to general planar
graphs as well.
The paper is organized as follows: in Section 2 we give some necessary definitions.
In Section 3 we present the canonical ordering and the general drawing framework.
In Sections 4, 5, and 6 we present the drawing results for convex drawing, orthogonal
6
G. Kant
drawing, and the mixed model, respectively. Section 7 contains some final remarks and
directions for further research.
2. Definitions. LetG = ( V , E)beaplanargraphwithn verticesandmedges. Agraph
is called planar if it can be drawn without any pair of crossing edges. An embedding of
a planar graph is the collection of circular permutations of the edges incident upon each
vertex in a planar drawing of the graph. An embedded graph is a planar graph equipped
with an embedding. An embedded graph divides the plane into a number of faces. The
unbounded face is the exterior face or outerface, all other faces are called interior faces.
Edges and vertices, belonging to the outerface, are called exterior edges and exterior
vertices, respectively; the other edges and vertices are called interior edges and interior
vertices. An interior edge, connecting two exterior vertices, is called a chord.
deg(v) denotes the degree of v, i.e., the number of neighbors of vertex v. A k-planar
graph is a planar graph with maximum degree k. The induced subgraph on the vertices
v~. . . . . vk consists of the vertices vl . . . . . vk and the edges (vi, vj) with 1 _< i, j _< k.
G - {v} denotes the graph after deleting vertex v and its incident edges from G. G is
called k-connected if deleting any k - 1 vertices and incident edges from G preserves the
connectivity. 2- and 3-connected are also called biconnected and triconnected, respectively. A vertex, whose deletion disconnects the graph, is called a cutvertex. A set of two
vertices, whose deletion disconnects a graph is called a separation pair. It is well known
that if the planar graph is triconnected, then its embedding is unique (up to chosing an
exterior face). A path between two vertices x and y is an alternating sequence of vertices
and edges such that x and y are at the end of this sequence and each edge in the sequence
is preceded and followed by its end vertices. If the vertices on the path have degree 2,
then the path is called a chain.
A drawing such that each edge is represented by a polygonal chain is a polyline
drawing. There are two common special cases of this standard. A straight-line drawing
maps each edge into a straight-line segment. An orthogonal drawing maps each edge
into a chain of horizontal and vertical segments. In a convex drawing every interior
face is mapped to a convex region, i.e., ever), angle of an interior face has size <_
n. Note that polyline drawings can be modified to give drawings with nicely curved
edges, also called splines. A polyline drawing is a grid drawing if the vertices and
the bends of the edges have integer coordinates. See also [9] for more definitions and
illustrating figures, not presented here. In our drawing "algorithms we denote the position
of vertex v by P(v) : (x(v), y(v)). All graphs considered in this paper are planar,
and all corresponding drawings are planar as well, and drawn on a grid. Several other
definitions and explaining figures, not presented here, are presented in the following
sections when we need them.
3. The Canonical Ordering and Drawing Framework.
3.1. The Canonical Ordering. In this section we introduce the canonical ordering
for triconnected planar graphs. In the next section we refine it to a leftmost canonical
ordering, to get a linear-time drawing framework.
Drawing Planar Graphs Using the Canonical Ordering
7
Canonical Ordering. Let G = (V, E) be a triconnected plane graph with a vertex
vl on the exterior face. Let rr = (V~ . . . . . VK) be an ordered partition of V, that is,
Vl U ... tO Vx = V and Vi A Vj = 0 for i =fi j. Define G~ to be the subgraph of G
induced by VI tO 99 9 t3 Vk, and denote by Ck the exterior face of Gk. We say that Jr is a
canonical ordering of G if:
9 V 1 consists of {vl, v~}, where v2 lies on the outefface and (UI, '/)2) E E.
9 Vx is a singleton {v.}, where v. lies on the outerface, (vl, v~) 6 E, and v. r v2.
9 Each Ck (k > 1) is a cycle containing (Vl, vz).
9 Each Gk is biconnected and internally triconnected, that is, removing two interior
vertices of Gk does not disconnect it.
9 For each k in 2 . . . . . K - 1, one of the two following conditions holds:
(a) Vk is a singleton, {z}, where z belongs to Ck and has at least one neighbor in
G -Gk.
(b) Vk is a chain, {zl . . . . . ze}, where each zi has at least one neighbor in G - G~,
and where Zl and ze each have one neighbor on Ck-l, and these are the only two
neighbors of Vk in Gk_ I.
THEOREM 3. !. Every triconnected planar graph G with predefined vertex Vl on the
exterior f a c e has a canonical ordering.
PROOF. Let G be a triconnected planar graph with vl given in advance. The decomposition of the vertices in VI . . . . . Vx will be defined by reverse induction. Let v2 and v~
be the neighbors of ol, also belonging to the exterior face. Notice that, by triconnectivity
of G, the graph G,,_I =- G - {v,,} is biconnected and the outcrface C~-I is a cycle,
containing (vl, v:).
Let 2 < k < K be fixed. Assume that Vi has already been determined for every
i > k such that the subgraph Gk satisfies the conditions of the canonical ordering. Let
Ck : v~ = Cl, ca . . . . . Cq-l, Cq = v2 be the corresponding outerface. Notice that if there
are vertices v c Gk of degree 2, then v E Ck. Notice also by triconnectivity of G that
there are at least three vertices c,~, c/~, c• ~ Ck having edges to vertices in G - G~.
Assume without loss of generality that c,~ ~ v~, v2. If Gk is triconnected, then we take
Vk = {c~ } because, by triconnectivity, c~ has at least three neighbors in Gk and Gk - c,~
is biconnccted.
Assume further that Gk is not triconnected, hence Gk contains separation pairs. Let
vx, Vy be a separation pair, and let Gl, G2 be two components of Gk -- {Vx, v~,}. Since
G is triconnected, there is a path P between G 1 and G2 in G, not visiting vx and vy. In
Gk, vx and Vy are forming a separation pair, hence the edges of path P are removed in
Gk. Since we defined the ordering by reverse induction, we removed only vertices and
edges from the outerface. Hence path P goes between two vertices c~,, cy,, belonging to
Ck with c~, ~ G l and cy, c G2. This yields that v~ and oy belong to Ck and one path
between v~ and Vy on Ck is part of G l ; the other path on Ck between v~ and v~, is part of
G2. This holds for every separation pair v~, v,., hence all vertices of the separation pairs
belong to Ck.
Let ca, c~, be a separation pair such that b - a is minimal. If deg(ca+l) > 2, then
there is a vertex c,~, a < ot < b, with at least one edge to a vertex deleted in step j > k,
8
G. Kant
otherwise the graph G - {c~, cb} is disconnected,which contradicts the triconnectivity
of G. By minimality of b - a, c,, is not part of a separation pair in Gk, hence Gk - c~ does
not have a cutvertex and the outerface of Gk - c~ is biconnected. We take Vk = {c,,}.
Assume now that there is no separation pair c~, Cb with deg(c~+t) > 2. Then b = a + 2 ,
because c, and c,+2 are the only neighbors of c,,+j in Gk. Let now ! _< a ' _< a and
b < b' < q besuchthat all verticesc~,+l, c,,+2 . . . . . Cb'-l have degree 2, anddeg(c~,) > 2
i f a ' > 1 and deg(c~,,) > 2 i f b ' < r. Notice that v l , w r {c,,+l . . . . . Cb'--l} and
every v e r t e x c i , a ' < i < b', has an edge to G - Gk. If edge (c,,,ch,) ~ G, then
Gk - {c~,+l . . . . . Ch'-l} is biconnected and we set Vk = {c~,+l . . . . . cl,,-1 }.
Assume finally that (ca', Cb,) r G. Let F be the face in Gk, containing the vertices
ca,, ca,+1 . . . . . cu. Wc claim that the path P in F between ca' and oh, not containing
ca'+l . . . . . Cb'+l, does not contain another vertex on the outerface of Gk. Suppose not,
i.e, the path P contains a vertex Cd on the outerface of Gk. Suppose without loss of
generality 1 < d < a'. However, it now follows that Cd+l . . . . . c,,_ I is a chain of vertices
of degree 2. Since (Cd, Ca') q{ G it follows that deg(c~,) = 2, which is a contradiction.
Hence Gk - {c~,f 1.....
Cb; } is biconnected and we set Vk = {c~,+l . . . . . co,- 1 }'
[]
If Vk = {zl . . . . . ze} with e > 1, then precisely one face is added, otherwise one
vertex is added to G k - l . The algorithm for computing the canonical ordering is based on
the proof of Theorem 3.1: we start with the entire graph G, and in each step we delete
a face or vertex. For this we introduce the variables outv(F) and oute(F) for each face
F , denoting the number of vertices and edges of F belonging to the current outerface.
We also introduce a variable sepf(v) for every vertex v, denoting the number of incident
faces of v, containing a separation pair. We call a face with a separation pair a separation
face. Using these variables we can prove the following theorem.
THEOREM 3.2. For ever)' triconnected planar graph a canonical ordering can be computed in linear time and space.
PROOF. Let an embedding of the triconnected planar graph G be given. Let every
vertex v and edge e have pointers to the faces they belong to. Initially all variables
outv(F), oute(F), and sepf(v) are set to 0. We take an arbitrary face F,,,,, as the outerface
during the algorithm. Assign vl with neighbors v2 and v~ on Four. In every step we
remove vertices from G and update F,,,,. For every vertex v c Fo,t and every edge
e ~ Fo,,, e ~ (vl, v2), we increase outv(F~.) and oute(Fe) for every F,: ~ F,,,t to which
v belongs, and Fe ~ Fo~t, to which e belongs.
For every vertex v which becomes part of Fo,t we have to compute sepf(v). Consider
for this problem a face F to which v belongs. We claim that F is a separation face if
and only if outv(F) > 3 or if outv(F) = 2 and oute(F) = 0. This follows because,
precisely in these cases, F has at least two nonadjacent vertices on the outerface which
makes this face a separating face. To compute sepf(v) we simply inspect outwand oute
of the incident laces.
A face becomes at most once a separation face, because when outv(F) or oute(F)
decreases, then a vertex or edge from F is deleted and F is added to F,,,,t. Every face
F with outv(F) = oute(F) + 1 and oute(F) > 2 can be the next face in our ordering,
becausc in this case the vertices of F , belonging to the outerface, form a consecutive
Drawing Planar Graphs Using the Canonical Ordering
///
/
/ /
/
/
I
/IB
9
"'-\
' ~<13
~
9 ~
~ II
//
6.,/"/ I/
~
12
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Fig. 1. A graph with canonical ordering and corresponding variable-values at some step.
sequence. If no such face exists, then a vertex v, with v 5~ Vl, v2 and sepf(v) = 0 and
visited(v) > 1 (with visited(v) the number of deleted neighbors of v), can be the next
vertex vk in our ordering. By Theorem 3.1, such a face or vertex exists.
The time complexity of the algorithm is the following: every vertex v has deg(v)
neighbors and belongs to deg(v) faces. When v becomes part of Four,then updating outv
for all incident faces of v requires O(deg(v)) time in total. Computing sepf(v) requires
O(deg(v)) time if v becomes part of Four, and O(1) time otherwise. Updating oute of
the incident face of an edge e, which becomes part of Fo~t, requires O(1) time. When a
face becomes a separation face, then sepf of the other vertices of this face, part of the
outerface, must be increased by one. There are at most two vertices and this happens
only once, hence requires constant time. Deleting a vertex or face can be done in time,
constant in the number of deleted edges. Since Y~.deg(v) = 2m, and m = O(n), this
yields a linear time-and-space algorithm.
[]
See Figure 1 for a graph with corresponding values of the variables visited(v), sepf(v),
outv( F), and oute( F).
3.2. The Drawing Framework. The general idea for drawing the graph is to start with
edge (Vl, v2), and add in step k the vertices of Vk. In step K vertex Vn is added. Assume
without loss of generality that in step 1 vl is drawn furthest left and v2 furthest right.
Let Ck-l: c~ = Vl, c2 . . . . . Cq = Vz be the vertices from left to fight on the outefface
of Gk_j. When adding the vertices of Vk let ct and Cr be two neighbors of Vk on Gk-l,
with l and r as small and as big as possible, respectively. We call ct the leftvertex and
Cr the rightvertex. If v ~ Vk, then edges between w, w r Vk,, k' <_ k, and v are called
incoming edges of v. The remaining incident edges of v are called outgoing edges, in(v)
and out(v) denote the corresponding number.
We place the vertices in such a way on the grid that when adding Vk, the corresponding
incoming edges have a downward direction. Moreover, we want to maintain the invariant
that the vertices cl . . . . . Cq of Ck remain "visible from the top" during each step. This
implies that after adding Vk, vertices c. . . . . . Cq must be "shifted to the right," as well as
several interior vertices of Gk-l. However, updating all x-coordinates of the vertices in
Gk in each step implies a quadratic running time. To avoid this, we use lazy evaluation:
10
G. Kant
The exact coordinates of vertices in the current drawing are only computed when they
are necessary to compute the insertcoordinates of vertices added in this step. This means
that only the exact coordinates of the vertices on the outerface are essential during the
insertions. As a first step toward this process, we refine the canonical ordering to the
lefimost canonical ordering, which we call the hnc-ordering from now on.
DEFINITION 3.1. A canonical ordering is a leftmost canonical (lmc-)ordering if we can
add in any step a vertex set Vk with leftvertex ct or a vertex set Vk, with leftvertex ct,,
and l < l' holds, then k < k'.
In other words, we take this vertex set Vk, for which the corresponding leftvertex cl
is minimal with respect to l. By planarity it follows that the corresponding rightvertex,
say c~, is also minimal with respect to r.
To compute the lmc-ordering, we maintain a list Outerface-Stack for the vertices on
the outerface from left to right, implemented as a stack, and initialized as {v2}. Also,
the vertex sets Vk of the canonical ordering, with pointers to its left- and rightvertex are
stored. Notice that Vk can be added in step k', if all incoming edges of Vk are part of Ck,- i.
We now delete vertices from the top from Outerface-Stack until we find a vertex Cr on
top, which is the rightvertex of a vertex set V~, not added yet. Let Vk = {zl . . . . . ze} from
left to right, then we add ze . . . . . zt in this order to Outerface-Stack. We repeat this step
with the updated Outerface-Stack until all sets Vl . . . . . Vr are added. Notice that when
rightvertex Cr of some set V~ of the canonical ordering is on top of Outerface-Stack, then
all other incoming edges of Vk are left from cr on the current outerface, i.e., Vk can be
added to the ordering. Hence every vertex set Vk will be added once to the ordering. This
implies that every vertex will be added once and deleted once from Outerface-Stack.
Since the vertices are added from right to left to Outerface-Stack the vertex sets Vk are
added in a leftmost order.
THEOREM 3.3.
Given a canonical ordering, an lmc-ordering can be computed in 1inear
time.
PROOF. The correctness is shown above. Regarding the time complexity, every vertex
v is added to once and deleted once from Outerface-Stack. Testing whether vertex cr on
top of Outerface-Stack is the rightmost vertex of some set V~ (not added yet) requires
constant time, which completes the proof.
[]
In Figure 2 an example of the lmc-ordering is given which will serve as an example
for almost all drawing algorithms presented in this paper.
In the drawing algorithms we distinguish the insertcoordinates of vk (when we insert
vk by the lmc-ordering) and the endcoordinates of vk (in the complete drawing). We
introduce a boolean variable correct(v), denoting whether x(v), with v c Ck, has been
recalculated. We also introduce a counter for each vertex v, called shift(v), shift(cr)
denotes the value which must be added to all x(ck), with r < k < q, where cl . . . . . Cq
is the current outerface Ck. When a vertex v is added by the lmc-ordering, we set
correct(v) = .false and shift(v) = 0. Inspect step k. Let cr be the rightvertex of Vk.
Drawing Planar Graphs Using the Canonical Ordering
11
15
15
2
i
i
52
Fig. 2. From a canonical ordering of the graph in Figure I to an lmc-ordering.
We walk along the outerface from cr toward Cl until we find the first true marked
correct(ca). Then we walk back from c,~ to Cr. When visiting ct~ (or < /3 < r) we add
Y'~<i<~ shift(ci) to x(c~) and set correct(c~) to true, because x(c~) is recalculated.
We add ~-~'~,~<i<~shift(ci) to shift(c~). This approach is correct since the following two
lemmas hold in step k.
LEMMA 3.4.
All vertices ct~, ot < [3 < r, have correct(ctj) =false.
PROOF. Suppose not. Inspect the first time that a vertex cy on the outerface is encountered for which correct(cy) = false and correct(cy+l) = true. correct(cy+l) = true
means that in a step k' < k, x(Cr+l ) and correct(cy+l ) are recomputed, due to the insertion of a vertex set Vk' with leftvertex cr, and l' >_ y + 1. However, correct(cy) =false
means that in step k we add V, with rightvertex c v. This contradicts the definition of the
lmc-ordering.
LEMMA 3.5.
[]
r > ct holds for rightvertex c,..
PROOF. Suppose not. correct(cc~) = true means that in a step k' < k we updated x(c,,),
due to the insertion of Vk, with leftvertex ct,, l' > or. Adding Vk with rightvertex Cr in
step k implies l' > r. Since k' < k, this contradicts the definition of the lmc-ordering. []
LEMMA 3.6. The total time for visiting the vertices marked false and updating shift(v),
x(v), and correct(v).for all vertices v is O(n).
PROOF. When we insert Vk in step k with leftvertex ct and rightvertex Cr, extra timc
is required for walking toward ct to find the first true marked correct(ca). All correctvalues of the vertices c . . . . . . c r - i are marked true after visiting them. If, in a step
k' > k, correct(c~) becomes false again (with ~ < /3 < r - i), then a vertex set Vk,
with rightvertex c~, = c~ is added, with k' < I. This contradicts the lmc-ordering, hence
every correct(cl~) becomes false once and becomes true after visiting ct~ again. Updating
requires constant time, hence the total time for visiting the vertices marked .false and
updating shift(v), x(v), and correct(v) for all vertices is O(n).
[]
12
G. Kant
These three lemmas show that in any step k we can compute the up-to-date xcoordinates of the vertices Cl . . . . . cr of Ck_l when adding Vk, where cr is the rightvertex
of Vk. Let P ( v ) = (xin~.e,(v), yinse,~(V)) be the coordinates of v at the time of inserting v.
However, how can we compute the final x-coordinates of the vertices? To this end we
have to traverse the vertices of Vk in decreasing order, i.e., from Vr to Vz, and set initially
shift(v) = 0 for all v ~ V. When considering the vertices of Vk = {Zl . . . . . z~}, we set
shift(el) = shift(z l), with l < i < r, and ct and c~ the left- and rightvertex, respectively,
of Vk, and Cl . . . . . Cq the outerface Ck-l of Gk-~. shift(ct) is not updated (because ct
is not shifted when adding V~ initially). Since Cr is also part of some outerface C~,-I,
k' > k, shift(c~) could already be greater than zero at the moment of visiting Vk. The
question arises whether this value was also added to the vertices of Vk or not. If this was
the case, then this shift-value should not be added to shift(c~) again. How can we solve
this problem?
The solution is as follows: to compute the right shift of c~ we distinguish the shifts
added to ct+l . . . . . Cr-l and to Cr, by introducing a new variable, rshift(v). When considering Vk for computing the final x-coordinates, c~ must be shifted a value x' more to
the right than c~_ l, then we add rshift(zl) + x ' to rshift(Cr). The final coordinates of the
vertices zi . . . . . zt of Vk is now given by x i ~ , ~ ( z i ) + shift(zi) + rshift(zi), 1 < i < ~.
The precise values for shift and rshift are given in the next sections, when we present
the applications for several graph-drawing representations. We call this method the shiftmethod.
4. Convex Drawings. The lmc-ordering is a generalization of the canonical ordering
of de Fraysseix et al. [ 18]. We can apply the lmc-ordering and the shift-method to get a
linear implementation of the straight-line grid-drawing algorithm of triangulated planar
graphs [18]. (In [7] another linear implementation of [18] is described, assuming that
the input graph is triangulated.) Moreover, we show that this algorithm can be modified such that we can draw every triconnected planar graph with convex faces on a
grid.
The algorithm of [18] is as follows: it maintains a straight-line embedding during
every step k of the line-ordering such that:
1. v~ is at (0, 0), v2 is at (2k - 4, 0).
2. If ol = cl, c2 . . . . . Cq = 02 is the outerface of Gk in step k, then X(CI) "< X(C2) <
9 .. < X(Cq).
3. The edges (ci, ci+l) have slopes + 1 or - 1.
Assume first that G is triangulated, in which case we can add a vertex vk in every step
k of the lmc-ordering [18]. Let L (v) be a set of vertices. The idea of the algorithm is the
following: when we add vertex vk with leftvertex ct and rightvertex cr then all vertices
Ct+l . . . . . Cr-l are shifted one to the right, and the vertices cr, . . . , Cq are shifted two to
the right (and, of course, several interior vertices of Gk-i have to be shifted to the right
as well). The crossing point of the line with slope +1 from c/ and the line with slope
- 1 from Cr denotes the place for vertex vk. We denote this by/z(ct, c~). All vertices
ct . . . . . cr are visible from this point. See Figure 3 for the corresponding picture. Let
Drawing Planar Graphs Using the Canonical Ordering
13
Ok
/~(cl, r
CI+I)~
"'.
ct
c<
I) 1 :
. . . . . .
-:~:-, I C.
,,,.
C1
U2 :
C,1
I,ll :
C1
7) 2 ----" Cq
Fig. 3. Idea of the straight-line drawing algorithm.
CI . . . . .
Cq
be the outerface in each step k. The algorithm becomes now as follows:
Let a canonical ordering Vl . . . . . vn be given
P(vl) : = (0, 0); L(vt) : = {Vl};
P(v2) : = (2, 0); L(V2) : = {v2};
P(V3) : = (1, 1); L(V3) : = {v3};
fork : = 4 t o n do
r-I
for v ~ [-Ji=t+l L(ci)do x(v) : = x ( v ) + 1 rof;
for v E [,-Jq=r L(ci) do x ( v ) : : x(v) + 2 rof;
P(vk) := l~((x(cl), y(c~)), (x(cr), y(c,)));
r-i
L(Vk) : : {'Ok}I._J[--Ji:,'+l L(ci)
rof
The correctness of this algorithm is proved in [18]. Using the variables shift(v) and
rshift(v), the complete algorithm can be implemented as follows:
LINEARSTRAIGHT-LINEDRAW(G);
Let an lmc-ordering vl . . . . . v,, be given
P(Vl) : = P(v2) : = (0, 0);
fork : = 3 t o n do
update x(cl) and shift(or);
shift(Cr) : = shift(c~) "4- 2;
P(vk) : = U((x(ct), y(ct)), (x(cr) + shifi(c~), y(Cr)))
rof;
shift(v) : = rshift(v) := 0 for all v c V;
for k := n downto 2 do
for every interior neighbor vi of vk do shift(vi) : = shift(vk) + rshift(vk) + lrof;
rshift(Cr) : : rshift(Cr) + rshift(vk) + 2;
X(Vk) : = Xi,,ser,(Vg) + shift(vk) + rshift(v~)
rof;
END LINEARSTRAIGHT-LINEDRAW
Moreover, using the line-ordering it is already sufficient that the planar graph is triconnected, because when adding Vk --- {zl . . . . . ze}, we can draw z~ . . . . . ze on a horizontal
line with distance 2 in between. This yields that edge (el, zl) has a slope + 1 , edges
(Zl, z2) . . . . . (ze-l, ze) have slope 0 and length 2, and edge (z~, Cr) has a slope - I. It is
easy to see that this still gives a correct straight-line drawing on a (2n - 4) x (n - 2) grid.
14
G. Kant
In the remaining part of this section we modify this new algorithm a little such that
all interior faces are convex.
Let Vk be {z l}, and let ci . . . . . . ci, be the vertices on Ck_ i, adjacent to Zl. (il =- l and
i~ = r.) Let Fs. (1 _< j < s) be the face formed by the edges (Zl, cis), (zl, ci,_,) and the
path B/between ci~ and cir,,.
LEMMA 4.1.
Each path Bj has the following pattern:
9 From c,/to some vertex cc~, a sequence D j, IDj I > l, o f vertices with strictly decreasing
y-coordinate.
9 Two vertices c~,, c/jj with the same y-coordinate.
9 From c~j to c#+,, a sequence Uj o f vertices with strictly increasing y-coordinate.
PROOF. By definition of the lmc-ordering, every vertex v ~ Vk has a neighbor w c Vk,,
with k' > k. By definition of the algorithm LINEARSTRAIGHT-LINEDRAW it follows that
y ( w ) > y(v). In step k the vertices c6+l . . . . . %+,-1 have higher-placed neighbors. Let
c% be the lowest-placed vertex, with i9 _< or/ < i/+~ and % minimal. If there is a vertex
c,~; 4: c~ with y ( c ~ ) = y(c,,,), then fl; = c~j + 1, because otherwise there would be
a vertex c 7, eej < F < fij, which does not have a higher-placed neighbor. From % to
c% the vertices have strictly decreasing y-coordinate and from c~,,+l to c~., the vertices
have strictly increasing y-coordinate.
[]
Notice that all edges on C~_l have slope +1, 0, and - 1 before adding V~. When
adding z l, we shift ca, . . . . . c/~_, to the right by one, and shift c~,_, +l . . . . . ci~ to the right
by two. As explained above, we draw Zl at point I2(ci,, c~).
When Vk = {Zl . . . . . z~ }, we add only one face/"1. Let B1 be the path of Ck-1 between
Q and cr. B~ also has the pattern of L e m m a 4.1. We shift c,,~ and c~ to the right by one,
and c~+1 . . . . . cr to the right by 2 . g. zl . . . . . ze are placed as explained above. This
yields the following slopes after adding Vk:
9
9
9
9
The slope of edge (ca,-~, ca, ) is in the range [ - 1,0).
The slope of edge (ct~,_ , , c~_,+1) is in the range (0, +1).
All other slopes on Ck-l are not changed.
The slopes of the incident edges of Vk are in the range ( - ~ ,
- 1 ] U [+1, ~ ) .
This implies that the faces Fl . . . . . F.~-l are convex when inserting Vk at step k. To
preserve convexity during the other steps k' > k, we add edges from cij to c~j . . . . . % - 2
(1 < j < s). This does not destroy planarity and implies that ifci, is shifted to the right
in some step k' > k, then c~,+l . . . . . c#_ I is also shifted to the right with the same value.
The modified graph is still called G. Now we can prove the following lemma.
LEMMA 4.2.
1he faces remain convex during the algorithm.
PROOF. AssumeVk = { z l l , and let ci, (l < j < s ) , c , , ; , c ~ j , a n d F i j ( l < j < s ) b e a s
defined above. (The proof is analogous when [Vkl > 1.) Consider a step k' > k. If ci~ is
shifted to the right, then, by the added dummy edges, c~,_ ,+j, ci~-j are also shifted to the
right with the same value, thereby preserving planarity in the only relevant face, F~- 1- If
Drawing Planar Graphs Using the Canonical Ordering
15
zl is shifted to the right, then the vertices ci~ . . . . . ci, are shifted to the right as well, and
if ci, is shifted, then ct%,+~ . . . . . c,_~ and %+~, c~j~_~ are shifted to the fight with the
same value. This yields planarity in the faces F1 . . . . . F.,.-l. It also has the consequence
that if c~ is shifted to the right (il < i < is), then cv, i < i' < i~ is shifted to the right
with at least the same value.
We use this observation for the case that c~. is shifted to the right in some later step
k. Then z~ is also shifted to the right by at least the same value. Hence the vertices
Cfl, ~ 1 . . . . . Ci2 are shifted to the right by at least the value of the shift of ci, . . . . . ct~, . This
preserves the planarity in Fl and completes the proof.
[]
Finally we remove the added dummy edges from cij to ct~j , c~j + l . . . . . cij-2 ( 1 <__j _<s).
THEOREM 4.3. There is a linear-time-and-space algorithm to construct a planar
straight-line convex drawing o f a triconnected planar graph with n vertices on a
(2n - 4) • 01 - 2) grid.
Our algorithm not only outperforms the algorithms of [48] and [5], but is also much
easier to implement than the algorithm of [5]. However, a drawback of the algorithm is
that the drawings are not strictly convex, compared with those of [48] and [5]. On the
positive side, this algorithm gives a new proof that every triconnected planar graph admits
a planar drawing, in which every interior face is convex. The outerface is a triangle. With
respect to the tightness of the grid size we note that every strictly convex drawing of a
cycle with n vertices requires a (-)(n 3) grid 135]. On the positive side, Chrobak and Kant
[6] improved the algorithm above such that the required grid is (n - 2) • (n - 2). In
Figure 4 the straight-line convex drawing of the graph in Figure 2 is given.
13
3
4
5
i_-~_
Step 3
2
7
I
I2
2
7
s
I
Step 9
2
Step 5
6
4
~
I
9
13
I0
_2
Step 7
I
2
Step 11
Fig. 4. Convexdrawing of the graph of Figure 2.
16
G. Kant
1
L
I
I
(a)
(b)
Fig. 5. l.ower bound of ~ (n - 1) + 2 bends. (a) The graph G5 and (b) the graph G~.
5. Orthogonal Drawings.
5.1. 4-Planar Graphs. In this section we consider the problem of drawing a planar
graph G on a rectilinear grid with orthogonal edges, i.e., the edges are polygonal chains
of horizontal and vertical segments. The vertices are represented by points.
There are embedded triconnected 4-planar graphs Gk with 3k + 1
vertices and 6k + 1 edges, for which an), layout requires at least 4k + 2 bends, for any
THEOREM 5. I.
k>l.
PROOF. Consider the triconnected plane graph Gk with 3k + 1 vertices, and its layout in
Figure 5(a), which has 4k + 3 bends. Notice that there are no bends in the edges between
two white vertices. We delete these edges from the graph. The vertices, which had degree
4 initially, have degree 2 now, and are deleted, while connecting the two incident edges.
This leads to a biconnected planar graph G~ with 2k + 2 vertices (see Figure 5(b)). It is
shown in Corollary 4 in [45] that the layout in Figure 5(b) of G~ is best possible with
respect to the minimum number of bends, which is 4k + 2. If there was a layout for Gk
with fewer than 4k + 2 bends, then there was a better layout of G~ with fewer than 4k + 2
bends, which contradicts Corollary 4 of [451,
[]
Let G be a triconnected 4-planar graph. Let a canonical ordering of G be given. It is
not necessary to compute the lmc-ordering of G, since we do not use the shift-method.
We introduce a variable mark(v) for each vertex v, which is important when adding
Vk = {v} to Gk-i. v has at most two outgoing edges, say to u l and u2 (from left to right).
If v is the rightvenex of u l, and v is not the leftvertex of u2, then we set mark(v) = left,
otherwise we set mark(v) = right.
There are four directions to connect an edge at v, namely at left(v), right(v), up(v),
and down(v). A direction is called free if there is no edge connected in that direction of v
yet. The idea for the algorithm is as follows: we add v to Gk-I such that down(v) is not
free in Gk. Let c/and cr be the left- and rightvertex of v. We connect (ct, v) at right(ct),
if it is free, otherwise at up(or), if it is free, otherwise at left(or). The opposite direction
is followed for cr. We want to add v such that when mark(v) = left then left(v) is free
Drawing Planar Graphs Using the Canonical Ordering
17
after addition. Since v is the rightvertex of Ul, we can use left(v) for the edge (v, Ul)
to the left. When right(u1) is used for (v, ul), then no bends occur in (v, u]), otherwise
up(u i) is used, yielding one bend. Similarly for mark(v) = right.
Let v 6 Vk. Recall from Section 3 that in(v) denotes the number of incident edges
of v, belonging to sets Vk,, k' < k. out(v) is defined similarly. The algorithm can be
described in a more elaborate way as follows:
4-ORTHOGONAL(G);
Let a canonical ordering VI . . . . . Vk of G be given;
edge (vl, v2) via down(vl) and down(v2);
for k : = 2 to K - 1 do
Let Vk = {z] . . . . . ze};
9 if/? = 1 and in(zl) = {ct, ci, cr} then
(ct, zl) via left(zl);
(ci, z]) via down(zl);
(cr, za) via right(zl );
9 i f e = 1 and in(zl) = {cl, Cr} then
if mark(z l) = left or (lefl(Cr) free and right(Q) not free) then
(zl, ct) via down(zl) and (z], c~) via right(Zl)
else
(zl, cD via left(z]) and (zl, Cr) via down(zl);
9 otherwise (g > 1)
if right(Q) is free then (zl, ct) via left(z]) else via down(zl);
f o r / : = 2 t o e do
if down (zi- l) is free then (zi- j, zi) via down (zi_ 1) else via right(zi_ ~);
if mark(zi) = left then (zi-l, zi) via down(zi) else via left(zi);
rof;
If left(c~) is free then (zt-l, z~) via down(z~); (zt, cr) via right(zt) else
(Ze-l, ze) via left(zD; (zt, Cr) via down(zD
rof;
let cl, ca, ctj, and cr be the neighbors of v.;
edges from ct, ca, cp, Cr to V. via left(v.), down(v.), right(v.), and up(v.), resp.;
END 4-ORTHOGONAL
See Figure 6 for an illustration of the different cases.
There are several ways of computing the coordinates. Here we briefly describe the
method given by Biedl and Kant [2]: Note that the y-coordinate of a vertex is never
changed later, so we only have to worry about the x-coordinates. The crucial observation
is that we need not know the values of the x-coordinates of the incoming edges of vi
when adding vi. We can use the following strategy: throughout the algorithm maintain
a list Columns. Every embedded vertex v contains a pointer x(v) to one element of
Columns. Whenever we want to add a column, we add a new element in Columns. By
storing a list as a sequence of pointers we can do so without changing any of the x-values
of vertices already visited. The final x-coordinates are computed by traversing Columns
and assigning ascending values to each element. Every vertex and bend then checks the
value of the element it points to and stores it as its x-coordinate. This yields a planar
orthogonal drawing.
18
G, Kant
cl
T vl
Z1
T v2
I
Ci
Starting with (vl, v2).
I---
Cr
Adding zl with in(za)= {ct,ci,c,.}.
-'I iz'
ct
leJ~ o - -
-
Cr
7
c~
right
c,;l--r
c
Adding zl with i n ( z t ) = {cl,cr}.
~?1 f.... Q
" ,-
. . . .
I
9
t~ __teyt
left te'~ 4g~t 3 te~ -*~z5 *' z6
O----
I zj
Ii
2
. . . .
4
i
i
.
Cr
cl3 ~ c r - -
ct =v!
Adding z l , . . . , ze.
Adding vn.
Fig. 6. Adding vertices and faces to obtain an orthogonal drawing.
LEMMA 5.2. The number of bends is at most r3n] + 3. One edge has at most three
bends, all other edges have at truest two bends.
PROOE Let Vk = {zl . . . . . Ze}.Assumefirste > 1. For every vertex zi, down(zi) and
either left(zi) o r right(zi) are used. The edge to down(zi) always requires one bend, the
others do not. If mark(zl) = right, and right(zl) is used by (zl, z2), then later a bend
is required for the outgoing edge via left(z1). A similar condition holds for z~ when
mark(zD = left. Hence this implies at most s bends.
l f g = 1, then we have at most two bends if in(z1) = 3, so assume in(zl) = 2. The
incoming edge, using down(zl), gets no bends, the other edge gets one more bend. If
right(ct) and up(cr) are both free, then both edges (ct, z l) and (cr, z i) are straight lines
(using left(z l) and down(zl ), resp). However, if mark(zj ) = left, then a bend is required
for the outgoing edge of z l, using right(zj ). We assign this bend to step k. Similarly when
mark(Zl) = right, hence in all cases at most one bend is introduced when in(zl ) = 2.
A similar assignment also follows for edge (vl, v2): we assign the bends, where the
Drawing Planar Graphs Using the Canonical Ordering
19
edges use the connections left(l) 1) or right(v2), to step 1. S u m m a r i z i n g this leads to the
following table:
Step
Number of
vertices
Number of
edges
Number of
bends
2
1
1
e
1
1
2
3
g+ 1
4
4
1
2
e
4
1. adding (vl. v2)
k,s
I,in(zl)=2
k,s = 1, in(zj) = 3
k, zl . . . . . zt
K, adding vn
Consider the steps 2 . . . . . K - 1 in which n - 3 vertices and at most 2n - 5 edges arc
added (since m _< 2n). A d d i n g Vt = {z I } with in(z l) = 3 occurs at most I n / 2 ] - 2 times,
because then at most 13/2n] - 6 edges are added, and at most [n/2J + 1 edges are added
by at most Ln/21 - 1 vertices. This yields at most 2 ( [ n / 2 1 - 2) + Ln/2J - 1 + 8 = [ 3 n ] + 3
bends. The edge using up(v,,) has at most three bends, all other edges have at most two
bends.
[]
LEMMA 5.3.
The gridsize is at m o s t n x n.
PROOF. The increase in height in step k, 1 < k < K, is at most f., where Vk =
{zl . . . . . ze}. In step i the increase in height is at most one, and in step K two, which
proves the total height of n.
For the total width, we consider the different cases for step k, 1 < k < K. Let
Vk : {Zl . . . . . ze}.Ifg. > I , t h e n x ( z i ) - x ( z i - I )
= 1(1 < i < s
> 1,
then there is no increase in width at all in this step, since then X(Cr) - x ( c i ) > s - 1. If
(ct, Zl) is horizontal, then this means an increase of one in width. If (ct, zl) is vertical
and m a r k ( z l ) = right, then one outgoing edge of zl has to go via left(zl), hence this
also means an increase of one in width later. We assign this increase to step k. A similar
condition holds for (z~, cr). Since X(Cr) > X(CD + i in step k - 1, it follows that the
increase in width is at most s in step k.
l f e = 1 and i n ( z l ) = 2, then the width increases by one, due to the fact that an extra
c o l u m n might be necessary for the outgoing edge ofzt via left(zl ) when m a r k ( z l) = right
(similarly when m a r k ( z l ) = left). If e = 1 and i n ( z l ) > 3, then the width does not increase.
For (v~, v~) we assign the extra columns, required by the outgoing edges via left(vl)
and right(v2) to step 1, yielding a starting width of three units. This gives the following
table:
Step
1, adding (vl. v2)
k, g -- 1 and in(zl) = 2
k , e = landin(7..l)=3
k,e>l
K, adding v,~
This leads to a total width of at most n
Number of vertices
Increase in width
2
1
1
~
1
3
1
0
g
0
[]
20
G. Kant
r
~g
I
14
12
13
II
12
I1
ql c
d
9
II
kI
m
P
I
(a)
(b)
Fig. 7. Orthogonal drawing of the 4-planar graph of Figure 2. (a) Orthogonal drawing and (b) at most 2 bends
in every edge.
Indeed, in our solution, the edge using up(vn) has at most three bends, all other edges
have at most two bends. How can we avoid the edge with three bends? Even and Granot
proved that any orthogonaI drawing of the 4-planar triangulated planar graph on six
vertices (octahedron) requires at least one edge with at least three bends [14]. In our
case, if there is a vertex v with deg(v) = 3, then we can set v,, = v. Otherwise let n > 6.
Then there is a face with at least four vertices, which we choose to be the outerface.
Let ct, c,~, c~, cr be the neighbors of v,, from left to right. If edge (v,,, c,.) uses right(or),
then we change the four directions of Cr such that up(c~) is used for (c~, v~). Since
mark(c~) = right (by definition), it follows that at most one extra bend is introduced.
Moreover, since c~ ~ v2 it follows that all other edges still have at most two bends. This
completes the following theorem.
THEOREM 5.4. There is a linear-time-and-space algorithm to construct a planar orthogonal grid drawing o f every triconnected 4-planar graph G with n vertices on an
area o f size at most n x n and at most [3n] + 4 bends, such that every edge has at most
two bends and length O(n) if n > 6.
In Figure 7 thc orthogonal drawing of the graph of Figure 2 is given. In particular, in
Figure 7(b) it is shown how to change cr such that all edges have at most two bends. The
previous bound on the number of bends was 2n + 4, given by Tamassia and Tollis [43].
Hence our algorithm improves this result considerably for triconnected planar graphs. In
the algorithm of Tamassia and Tollis, cvery edge gets at most four bends, hence we also
improve this bound. Very recently, Biedl and Kant [21 presented a linear-time algorithm
for constructing an orthogonal repesentation of a connected planar graph on an n • n
grid, having at most 2n + 2 bends, and every edge is bent at most twice. Notice that
at most two bends in every edge is best possible, because if the planar graph contains
a separating triangle on the vertices vi, vj, vk, then at least one edge of the separating
triangle has at least two bends in any orthogonal drawing.
5.2. Triconnected 3-Planar Graphs. In this section we present a linear-time-and-space
algorithm to draw every 3-planar graph with at most Ln/2J + 1 bends on an Ln/2J • Ln/2J
Drawing PlanarGraphs Usingthe CanonicalOrdering
21
grid. This improves all previous bounds (from [41] and [43]) and matches the worst-case
lower bounds and, hence, is best possible. An interesting side-effect is that there is a
spanning tree using n - 1 straight-line edges. All m - n + 1 < Ln/2J + 1 nontree edges
have at most one bend.
3
Assume first that G is triconnected. By Euler's formula, n is even, m = i n , and
f = n/2 + 2. Let an line-ordering of G be given. Similarly as in Section 5.1 there are
four directions to connect an edge at v, namely at left(v), up(v), right(v), and down(v).
Every vertex v (except v~, v2, and v,,) has one outgoing edge, and we connect this edge
via up(v) at v. We start with placing Vl and v2 at (0, 1) and (1, 1). Edge (vl, v2) goes
via down(vl) and down(v2), hence via (0, 0) and (1,0). In step 2 the vertices zl . . . . . z~
of V2 are placed on the horizontal linc between vl and ve, i.e., via right(vl) and left(v2).
In every step k, 3 < k < K, we place zj . . . . . zt: also on a horizontal line of height 1 +
max{y(ct), y(cr)}, with ct and c~ the left- and rightvertex of Vk. If ~ > 1, then we shift
the drawing such thatX(Zl) = x(ct) andx(ze) = X(Cr). Since in(Vk) = 2 for2 < k < K
and in(v,,) = 3, it follows that K = f , with f the number of faces in G. Notice that
f = n/2 + 2 (n is even). The complete algorithm can now be described as follows (recall
the definitions of P(vi) and Xi,,.~e~t(Vi)from Section 3.2):
3-ORTHOGONAL
Let an lmc-ordering Vj . . . . . VK of G be given.
P(vl) : = (0, 1); P(v2) : = (I, 1);
fork:=2toK-ldo
assume we add z~ . . . . . ze (e > 1), from c/to Cr;
y(zl) :. . . . : = y(ze) : = 1 + max{y(c/), y(cr)};
update x(ct) and shift(or);
x(zi) : = x(ct);
forl:=2togldox(z/):=x(zl)+l-lrof;
i f g > I then x(zD : = max{x(zl) + e - 1, x(cr)+ shift(Cr)};
shift(Cr) : = max{shift(Or), x(ze) - x(cr)}
rof;
Let in(v,) = {ct, c~, c~}, P(v,,) : = (x(c~), I + max{y(ct), y(c~), y(cD});
for k := K downto 2 do
assume we added z~ . . . . . z~ from ct to c,.;
f o r / : = 1 to g do x(zi) : = xi,~.ert(zi) + Y~l<j<i shift(z)) rof;
i f e = 1 thenshift(c~) := shift(zl) elseshift(cr) :-- x(ze) - xi,,.,.e,~(cr)
to f;
END 3-ORTHOGONAL
LEMMA 5.5.
The number of bends is at most n /2 + 2.
3
PROOF. Since m = 7n,
we add at most n/2 - 2 times a vertex v with in(v) = 2,
each one introduces one bend. The edge (vl, v2) introduces two bends, as well as
adding vn.
[]
LEMMA 5.6.
The gridsize is at most n/2 x n/2.
22
G. Kant
20
19
20
19
18
't
I S .
5
8
16
~-'
I3
17
I
]
15 - - 1 1 4 113
6
10
•
_
')f
l
2
(a)
(b)
(c)
Fig. 8. Orthogonal drawing of a triconnected 3-planar graph.
PROOF. Edge (1)], v2) gives 1 unit in the X- and Y-direction. Then we add n / 2 - 1
times a face with ~ _> I vertices, increasing the X-direction with at most e - t units and
the Y-direction (except the first time) by 1 unit. Adding v~ increases the Y-direction by
1 unit. Counting this together leads to at most n/2 units in X-direction and n / 2 units in
Y-direction.
[]
In Figure 8(b) an example is given of a triconnected 3-planar graph.
We can change the drawing as follows, such that there is one bend less, and there is a
spanning tree, using only straight-line edges (ifn > 4). Let the vertices of the first drawn
face be numbered v~, vi, v i - j . . . . . v3, Vz. We place v~, vi, ui .1
1)3 on a horizontal
line, and place v2 on (x(v3), y(v3) - 1). Let F ' be the other face, to which (v2, v3)
belongs. Let vj . . . . . Vk b e the other vertices of F'. We draw vj . . . . . vk on a horizontal
line on height y ( v 3 ) , as shown in Figure 8(c). The remaining faces are drawn similarly as
before. Notice that using this strategy, every triconnected planar graph G with n vertices
can be drawn orthogonally on a grid of size at most n / 2 x (n/2 - 1), with at most n / 2 + 1
bends (n is even), in which there is a spanning tree, using only straight-line horizontal
and vertical edges. All nontree edges have at most one bend (ifn > 4).
We notice that better bounds can be obtained if the dual graph G* of G is a 4-connected
planar graph in which all interior faces are triangles. Bhasker and Sahni [ i ] showed that
in this case G admits a planar orthogonal drawing with at most four bends on an n x n
grid. Their algorithm works in linear time.
. . . . .
5.3. C o n n e c t e d 3 - P l a n a r G r a p h s . To apply the algorithm toconnected 3-planargraphs,
we have to split up the graph into its biconnected and triconnected components.
The triconnected components of a biconnected graph G are defined as follows: If G
is triconnected itself it is the unique triconnected component. Otherwise let {u, v} be a
separation pair of G. Wc partition G into two subgraphs G1 and G2 which have only
vertices u and v in common. We continue the decomposition process recursively on
G'l = GI + (u, 1)) and G~ = G2 Jr (u, v) until no decomposition is possible. The added
Drawing Planar Graphs Using the Canonical Ordering
23
:
N
!
13
f
16
18 i
~
~
~
4
$
~
6--%_
/15,,,~ \
" /"
\\
',
',
'
1
~5
_
/99~...8\
! "~7)
--
11
\.2fJ
(a)
(b)
Fig. 9. Exampleof a biconnectcd3-planar graph (a) and its SPQR-tree (b).
edges are called virtual edges. The resulting graphs are each either a triconnected simple
graph, or a set of three multiple edges (triple bond), or a cycle of length 3 (triangle).
The triconnected components of G are obtained from such graphs by merging the triple
bonds into maximal sets of multiple edges (bonds), and the triangles into maximal simple
cycles (polygons). The triconnected components are unique.
Next we need the SPQR-tree, a versatile data structure that represents the decomposition of a biconnected graph into its triconnected components [ 11 ]. The SPQR-tree T
is defined as follows: for every triconnected component we create an R-node, for every
polygon an S-node, for every bond a P-node, and for every edge a Q-node. The edges in
T are defined as follows: Let u, v be nodes in T. If u is a Q-node, then there is an edge
between u and v if the edge represented by u belongs to the triconnected component
represented by v. Otherwise there is an edge between u and v if and only if they contain
the same virtual edgc added in the same step of the decomposition process. In Figure 9
an example is given of a biconnected graph and the corresponding SPQR-tree.
Let the triconnected component of node bi in T be defined by Bi, and let si and t i be
the two vertices of Bi, called the poles, also shared by the triconnected component of
parent(bi). Since the maximum degree of G is 3, it follows that the neighbors of a P-node
are not R-nodes. Root T at an arbitrary S-node b,., and let bi, and bi2 be the two children
of a P-node bi in T. We merge B~, and Bi2 as described above, yielding a cycle. We
represent the resulting cycle by an S'-node. In this way we remove all P-nodes from T.
We also remove all leaves (which are Q-nodes) from T. Notice that since the maximum
degree is 3, the neighbors of an R-node and the parent of an S'-node are S-nodes. The
neighbors of an S-node and the children of an S'-node are R- or S'-nodes.
The algorithm for drawing a biconnected 3-planar graph G draws the triconnected
components of G in order of visiting the corresponding nodes in the SPQR-tree bottomup. Let si and ti be the poles of a leaf bi in T:
24
G. Kant
bi is an R-node.
B~ is drawn by the algorithm 3-ORTHOGONAI. with ~q = s~ and
02 ~ ti.
bi is an S'-node.
We draw Bi as a rectangle such that s~ and ti are the lower-left and
lower-right corners. If one path between si and ti in B~ contains at least two other
vertices, then vertices are placed in the other two comer points.
bi is an S-node. Let (si, s;) and (ti, t~) ~ B~, s~ r tl and t{ # s~. Let B; be B~ - {s~, t~}.
Draw B{ as a rectangle with s~ as the lower-left corner, t; as a right comer, and
with the other vertices of B; (if any) placed in the other comers. We draw si one
below s~ and ti one fight of t{.
Let IBil = n'. It follows that the used area of the described drawings is at most
[n'/2J x (Ln'/2J - 1). Let x(bi) = x(ti) - x(si) and y(bi) = y(ti) - y(si).
The idea now is to draw a triconnected component Bj, after all descendants of bj in
T are already drawn. Let bi be a child of by. We want to stretch the drawing of Bj such
that the edge (si, ti) has width x(bi) and height y(bi), and that Bi can be placed inside
without crossing edges.
If hi is an S'- or R-node, then bj is an S-node. We can draw Bj such that (si, ti) is a
straight line, and since Bi c a n be drawn such that si and ti have the same y-coordinate,
the problem is easily solved. Hence we now assume that bj is an R-node, thus Bj is a
triconnected graph and b i is an S-node. We now place B i inside Bj as shown by the
templates in Figure 10. By a tedious but rather easy case analysis, the following lemma
can be proved:
LEMMA 5.7. After replacing a virtual edge (si, ti) by the corresponding orthogonal
drawing of Bi in an orthogonal drawing of Bj, the total required grid size is at most
L(IBil + IBjl - 2)/2] x (L(IBi[ + I B j l - 2)/2J - 1).
This means that after replacing all virtual edges of Bj by the triconnected components, we obtain an orthogonal drawing of size Ln'/2J x ([n'/2J - !), where n' is the
number of nodes of the subgraph of G, corresponding to the subtree of T, rooted at bj.
We continue this approach until we are at root br of T. If G is not triconnccted, then T
contains an S-node, and we assumed that br is an S-node. It is easily observed that no
bends are introduced, when we consider an S- or S'-node. Hence the following theorem is
obtained:
THEOREM 5.8. There is a linear-time-and-space algorithm to construct a planar orthogonal drawing of a biconnected 3-planar graph with n vertices on a grid of size at
most Ln/2J • Ln/2J, with at most {_n/2J + 1 bends, and the property that there is a
spanning tree of n - 1 straight-line edges, while all nontree edges have at most one bend
(ifn > 4).
In Figure 11 the orthogonal drawing of the graph in Figure 9 is given.
We extend the algorithm 3-ORTHOGONAL to draw arbitrary 3-planar graphs orthogonally. Assume all biconnected components B~ of G are drawn orthogonally, with the
cutvertcx, say vi, in the upper-left comer. (Since vi has degree 2 in Bi and the root
of the corresponding SPQR-tree was assumed to be an S-node, this can be reached in
Drawing Planar Graphs Using the Canonical Ordering
25
Si
(a)
(b)
l+
,
t~
ti
s; !si
(c)
~
~si
N
= ~
ti
(d)
t~ ]
sit-~,i
ti
s~
, ~
(g)
(h)
t
- 4J,
ti
Fig. 10. Replacing virtual edges by triconnected components.
14
15
17_.... 16
2
1
4
3
18
5
6
9 8
i7
11
13
10
12
(a)
(b)
Fig. 11. Orthogonal drawing of a biconnected 3-planar graph. (a) Drawing the triconneCted components and
(b) the complete drawing.
26
G. Kant
r
T!
9
"-rI
-
-I
-
-
FT
L
I
Fig. 12. Orthogonal drawing of the blocks.
every step.) We draw the blocks recursively, and merge it into one drawing, as shown in
Figure 12.
No extra bends are introduced in the drawing. Also the required area for drawing
blocks B1 and Bj is at most [(IBil + IBjl)/2J x [(Inil + IBjl)/2], which completes the
following theorem:
THEOREM 5.9. There is a linear-time-and-space algorithm to construct a planar orthogonal drawing of a connected 3-planar graph on an Ln/2] x [n/2] grid with at most
[;7/21 + 1 bends, and the property that there is a spanning tree of n - 1 straight-line
edges, while all nontree edges have at most one bend (/fn > 4).
In [331 Liu et al. independently proved the same result for biconnected 3-planar
graphs, and presented an O(n)-time algorithm for this. In [34] theoretical results are
given for planar graphs, which can be embedded with at most one bend per edge.
6. The Mixed Model, In this section we use the drawing framework, introduced in
Section 3. l, to draw any triconnected d-planar graph G on a (2n - 6) x (3n - 9) grid
such that there arc at most 5n - 15 bends and the minimum angle is at least 2 / d radians.
All. vertices and bend coordinates will be placed on grid points only. Every edge will
have at most three bends and length (D(n).
To this end we define, for each vertex, inpoints and outpoints. The idea is that an edge,
say (u, v) goes from u to an outpoint of u, say uo~t, from Uo,t in the vertical direction to
a point, say w. From w it goes in the horizontal direction to an inpoint of v, say vi~, and
from vin to v. Important question for this are the coordinates of the in- and outpoints.
Recall the definition of in(v) and out(v) of Section 3.2. Let outl (v) = [(out(v) - I ) / 2 j
and OUtr(V) = [(out(v) - 1)/2]. Similarly for in(v) we define int(v) = [(in(v) - 3)/2.1
and inr(v) = [(in(v) - 3)/2]. The idea is to place the outpoints of v on the following
places: the diagonal lines from (x(v) - outz(v), l) to (x(v) - l, y(v) + outl(v)), and
from (x(v) Jr-OUtr(V), 1) to (X(V) Jr 1, y(v) + outr(v)), and using (x(v), y(v) +outt(v)).
For the inpoints it is defined similarly: the diagonal lines from (x(v) - inl(v), - I ) to
( x ( v ) - l, y ( v ) - o u t l ( v ) ) , and from (x(v)+out~(v), - I) to ( x ( v ) + 1, y(v)-OUtr(V)), and
the three points ( x ( v ) - o u t t ( v ) , y(v)), (x (v), y(v) - o u t z ( v ) ) and ( x ( v ) + o u t r ( v ) , y(v)).
See Figure 13 for some examples and the corresponding bounding boxes.
Drawing Planar Graphs Using the Canonical Ordering
I
27
IIII
I
~ 1
I I
I I I I I
I
Fig. 13. Examples of bounding boxes.
The width of the bounding box is max{out(v) - 1, in(v) - 3}, the height is int (v) +
outt (v). The idea is to insert the vertices of the canonical ordering such that the bounding
boxes do not intersect or touch. Also, outgoing edges may not cross or overlap, i.e., this
means that, for every pair of consecutive vertices ci, Ci+l on the outerface, x(ci+O >
x (ci) + OUtr(ci) + outt (ci+1) must hold. We can now explain the different adding steps
as follows:
Adding a Vertex v. For the y-direction we simply set y(v) = 1 + max{y(ct) +
outl(Cl), y(Cr) + OUtl(Cr)}. In the x-direction the problem is a little more difficult: let
u~ . . . . . uin(~) be neighbors of v, corresponding to the left-to-right order of incoming edges of v. Let u = uint(v), then we want to have x(v) = x(u), or, in other
words, we want to place x(v) right above the most central incoming edge. For the
outcoming edges, we want to have x(v) > x(ct) + outr(cl) + ouh(v). Hence we set
x(v) = max{x(u), x(c~) + OUtr(Cl) + OUtlO))}, and shift Uinz(v). . . . . Uin(v)_1 to the
right (if x(cL) + out~(cl) + outl(v) > x(u)). For the rightvertex Cr we set X(Cr) =
max{x(cr), x(v) + out~(v) + OUtl(C~) + 1}.
Adding Zl . . . . . ze, s > 1.
Now every vertex zl . . . . . ze has precisely two incoming
edges, hence they are placed on a horizontal line at height max{y(ct) + out1(ct), y(cr) +
outt (Cr)} + 1. In the x-direction we do not have to deal with the incoming edges, hence
we set x(zl) = x(cl) + OUtr(Ct) + outl(Zl) + 1, we set x(zi) = x(zi-1) + outr(Zi_l) +
OUtl(Zi) -t- 1 (1 < i < p), and we set x(c~) = max{x(Cr), x(zs +OUtr (ze) -t-OUtl(Cr)q- 1}.
Figure 14 makes this more precise.
Analysis of the Algorithm.
Using the lmc-ordering and the shift-technique, explained in
Section 3.2, it is not difficult to obtain a linear-time-and-space algorithm, satisfying the
constraints with respect to width and height, as given in the two relevant steps. Therefore
we now consider in detail the number of bends, the size of the minimum angle, and the
total grid size.
LEMMA 6.1.
of G.
The size of the minimum angle is 2/d, where d is the maximum degree
G. Kant
28
r~
/'-\
/
1:1 \
(a)
(b)
Fig. 14. Adding vertices in the mixed model. (a) Adding one vertex zl and (b) adding {zl . . . . . ze}.
PROOF. Let v have maximum degree d. The minimum angle, say a , is reached at an
outpoint which is neighbored to a horizontal line. If d < 6, then it is easily proved,
hence assume d > 6. The size of the angle is arctan(1/out~ (v)) > a r c t a n ( / 2 / ( d - 2)J).
Using the potence series of the arctan we know that, for Ixl < 1, arctan(x) = x I 3 + -~x
I 5 - ~x
i 7 + . . . > x - ~x
i 3. Since L2/(d - 2)J >_ 2 / ( d - 1), we o b t a i n t h a t
~x
a > 2 / ( d - 1) - .~(2/(d - 1)) 3 > 2 / d , which completes the proof.
[]
LEMMA 6.2.
The gridsize is at most (2n - 6) • (3n - 6).
PROOF. For the width notice t h a t x ( c i + l ) = x ( c i ) +OUtr(Ci) +outl(Ci+l) holds in every
step on the outerface if y(c,+l) ~ y(ci) and x(ci+l) = x ( c i ) + out~(ci) + outt(ci+l)
otherwise. If y(ci+l) = y(ct), then (ci, ci+l) is not an outgoing edge of any vertex.
We call (ct, ci+l) in this case unmarked. Counting leads to a horizontal distance of at
most ~ 1_<i<n outt (vi) + out~ (vi) + n u m b e r of unmarked edges = ~ l_<i<n (out(vi) - 1) =
2n - 6.
Adding one vertex vk requires a larger increase in height per vertex than adding
a face, hence assume we add a vertex Ok in every step. Let the incoming edges of
vertex vk come from vertices ul . . . . . Up, then y ( v k ) > maxlsi<_p{y(ui)} + max{1 +
inr(V),out~(ul),OUtr(Ur)}. The increase for every vertex ok during the insertions is at
most 1 + inr(Vl<) + OUtr(Vk). Summarizing this for all vertices leads to a total distance
in the Y-direction of at most 3n - 6 units.
[]
LEMMA 6.3.
length (,9(n).
There are at most 5n - 15 bends. Every edge has at most three bends and
PROOF. All outgoing edges of vertex v, except the one going straight upward, requires
one bend in the worst case to go in a vertical direction. We assign these bends to the
insertion step of v. Adding a face requires less bends per vertex than adding one vertex,
so assume we only add vertices Ok. If in(vk) = 2 and y(ct) >_ y(cr), then there will be
at most one bend in (cl, vk) and two bends in (Cr, Ok). (A similar situation holds when
y(ct) < y(cr).) In each edge one bend was already assigned to the insertion step o f c t and
Cr, hence adding ok with in(ok) = 2 requires at most one bend for the incoming edges. If
in(v~) >_ 3, then at most 2 - in(vk) -- 4 extra bends are required for the incoming edges.
Drawing Planar Graphs Usingthe Canonical Ordering
29
15
\
14
12
6
/- 13
7
S
~
4
5/
8
11
13
/
9
-,, X
Fig. 15. Drawingthe graph of Figure 2 with bends.
Edge (vl, v2) requires no bends. Counting this leads to at most a total of 5n - 15 bends.
Every edge goes at most once vertically and at most once horizontally, hence requiring
three bends in the worst case and, by Lemma 6.2, has length O(n).
[]
THEOREM 6.4. There is a linear-time-and-space algorithm to construct a planar polyline grid drawing of a triconnected planar graph with n vertices and maximum degree d
ona (2n - 6 ) x (3n - 9 ) grid with at most 5n - 15 bends and minimum angle > 1 / ( d - 2 ) ,
in which ever3' edge has at most three bends and length O(n).
In Figure 15 the drawing of the graph of Figure 2 is given.
Notice that Kant [291 proved that every graph G can be augmented by adding edges
to a triconnected planar graph G' such that d' < [23-d] § 3, with d and d' the maximum
degree of G and G', respectively.
THEOREM 6.5. There is a linear-time-and-space algorithm to construct a planar polyline grid drawing of a planar graph with n vertices and maximum degree d on a
(2n - 6) x (3n - 9) grid with at most 5n - 15 bends and minimum angle > 4/(3d + 1),
in which every" edge has at most three bends and length O(n).
7. Final R e m a r k s and Open Problems. In this paper a new ordering of the vertices
and faces of a triconnected planar graph is introduced. This ordering leads to various
algorithms for drawing a planar graph on a grid. In some cases considerable improvements on existing results are obtained, in other cases new bounds are achieved. All the
algorithms can be implemented by straightforward techniques to run in linear time and
space. Instead of using the shift-method, explained in Section 2, we could also use the
sophisticated technique of Chrobak and Paync. They build a tree on the vertices, where
the edges contain the amount of shift for every edge. When applying this techique for
our canonical ordering, there is no reason to compute a lefimost canonical ordering.
This is described in more detail in [6]. In this paper the line-ordering is required in all
algorithms, except the algorithm 4-ORTHOGONAL,where a canonical ordering is already
sufficient.
30
G. Kant
: 4
21
,
:Z
Fig. 16. The visibility representation of the graph of Figure 2.
Since every canonical ordering is also an st-ordering, wc can use the canonical ordering in various drawing applications, where the st-ordering is used. As an example, consider the visibility representations. In a visibility representation every vertex is mapped
to a horizontal segment, and every edge is mapped to a vertical line, only touching the
two vertex segments of its endpoints. See [38] and [44] for linear-time algorithms of this
and Figure 16 for an example. By using the lmc-ordering with shift-values on the edges
we are able to obtain a linear-time framework as well. The description of the complete
algorithm is not significantly different from the other algorithms, described in this paper,
and is left to the reader. (Independently, Nummenmaa [37] also presented a visibility
presentation algorithm, based on the canonical ordering of triangular planar graphs.) As
in the other algorithms for visibility representations (e.g., see [26], [37], [38], and [44])
the grid size is at most (2n - 5) x (n - 1).
If, for every vertex vi, OUt(•'i) >_ 2 holds, then our algorithm leads to a visibility
representation on a grid of size at most (n - 1) • (n - 1). Kant and He [31 ] proved that
such a canonical ordering is possible for 4-connected triangular planar graphs. Moreover,
using this result, Kant [30] proved that a visibility representation of a general planar graph
can be constructed on a grid of size at most (L-~nJ - 3) • (n - 1).
We conclude the paper by mentioning some open problems and directions for further
research.
9 Decrease some of the bounds with respect to the grid size, number of bends, or
minimum angle given in this paper. As an example, very recently Chrobak and Kant
[6] improved the grid bound for convex drawing of triconnected planar graphs to
(n - 2) • (n - 2). New bounds on the grid size and the number of bends for orthogonal
drawings are given by Biedl and Kant [2].
9 Is it possible to use the canonical ordering in other drawing representations as well,
to obtain better results in planar graph drawings on a grid?
9 Inspect the more combinatorial aspects of the canonical ordering. It is not very hard
to prove that a canonical ordering for G also directly defines a canonical ordering of
the dual graph, but maybe other combinatorial observations can be made.
9 Can an arbitrary planar graph be drawn planar with straight lines such that the minimum
angle is > ~ ( 1 / d ) ? (See also [36], [13], and [20] for research in this question.)
9 Devise dynamic algorithms for the sequential algorithms of this paper. In [8] a dynamic
framework for graph-drawing problems is described, but this approach seems not to
work here. Very recently, He and Kao [23] presented a parallel implementation for
Drawing Planar Graphs Using the Canonical Ordering
3I
finding a canonical ordering, and c o m p u t i n g the convex drawing in (Q(log 4 n) time,
requiring O ( n 2) processors.
Acknowledgments.
The author wishes to thank Hans Bodlaender, Jan van L e e u w e n ,
Therese Biedl, and the a n o n y m o u s referees for giving a lot o f helpful c o m m e n t s .
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