Planar graphs and poset dimension

Planar Graphs and Poset Dimension

WALTER SCHNYDER
Department of Mathematics, Louisiana State L’nversity, Baton Rouge, LA 70803-4918, L’S A
Communicated by W. T. Trotter

(Received: 13 June 1987; accepted: 25 August 1988)

Abstract

We view the incudence relation of a graph $G=(V, E)$ as an order relation on its vertices and edges, i.e. $a<_{G} b$ if and only if $a$ is a vertex and $b$ is an edge incident on $a$. This leads to the definition of the order-dimension of $G$ as the minimum number of total orders on $1 \cup E$ whose intersection is $<_{G}$. Our main result is the characterization of planar graphs as the graphs whose order-dimension does not exceed three. Strong versions of several known properties of planar graphs are implied by this characterization These properties include: each planar graph has arboricity at most three and each planar graph has a plane embedding whose edges are straight line segments. A nice feature of this embedding is that the coordinates of the vertices have a purely combınatorial meaning.

AMS subject classifications (1980). Prımary 06A10: secondary 05C10, 05C75.
Key words. Planar graph. poset dimension, straight line embedding.

1. Introduction

For a poset $(X,<)$ consisting of a set $X$ and a (transitive, irreflexive) orderrelation $<$ on $X$, a realizer is a nonempty set of total orders on $X$ whose intersection is the relation $<$.
E. Szpilrajn [17] has proved that each poset has a realizer. B. Dushnik and E. W. Miller [3] have defined the dimension of a poset $P(\operatorname{dim} P)$ as the minimum cárdinality of its realizers.

Fig. 1.
For example, the diagram shown in Figure 1 defines an order $<$ (of height 1) on the set $X=\{a, b, c, d, e, f\}$ by $x<y$ if and only if ’ $x$ is in the lower part, $y$ is in the upper part and $x, y$ are adjacent’. The dimension of $(X,<)$ cannot be 1 ( $<$ is not total) and must therefore be 2 as $<$ is the intersection of the two

total orders

$$<_{1}: a b d c e f \text { and }<_{2}: c b f a e d$$

(Here as everywhere else in this paper, the order is from left to right.)
The importance of orders in mathematics has motivated numerous investigations of the dimension concept. For example, Komm [11] established very early that the dimension of the power-set of a set is the cardinality of this set and Hiraguchi [8] proved that, for $n \geqslant 4$, the dimension of a $n$-element poset is at most $n / 2$. Later, Gysin [7] and Trotter, Moore, and Sumner [20] applied results of Gallai [5] to show that the dimension of a poset $(X,<)$ depends only of the underlying comparability graph (the graph where vertices $x, y \in X$ are adjacent if and only if $x<y$ or $y<x$ ), i.e. two posets with the same comparability graph have the same dimension. These are but a few of many results in this area. For more information we refer to the overviews in [18, 12, 22].

Although two-dimensional posets are well understood [3, 10, 19] and can be recognized in polynomial time [6], no simple characterization of $d$-dimensional posets is known for $d \geqslant 3$. Indeed, Yannakakis [23] has shown that deciding whether a poset has dimension $d$ is a NP-complete problem, for all $d \geqslant 3$. Even the recognition of four-dimensional posets of height 1 is NPcomplete. The complexity of the recognition of three-dimensional posets of height 1 remains an open problem.

One of the simplest examples of posets of height 1 is the poset $P(n, k)$ induced by the inclusion relation on the class of all 1 - and $k$-element subsets of a $n$-element set. The study of this poset dates back to [3] proving $\operatorname{dim} P(n, n-1)=n$ and the lower bound $\operatorname{dim} P(n, 2) \geqslant \log _{2} \log _{2}(n)$. These investigations were continued in [2] with a formula for $\operatorname{dim} P(n, k)$ when $2\lfloor\chi \bar{n}\rfloor \leqslant k \leqslant n$. Eventually, Spencer [14] investigating the asymptotic behavior of $\operatorname{dim} P(n, k)$ has shown that $\operatorname{dim} P(n, 2)$ is asymptotically equal to $\log _{2} \log _{2}(n)$.

In this paper, posets of height 1 are viewed as hypergraphs whose incidence relation is interpreted as order relation. For example, $P(n, k)$ is the complete $k$-uniform hypergraph on $n$ vertices and $P(n, 2)$ is the complete graph $K_{n}$. We propose to extend the investigation of $P(n, 2)$ to general graphs, defining the (order-) dimension of a graph as the dimension of its incidence relation, and to try to relate the dimension of a graph to other graph-properties.

Our main result is that a graph has dimension at most three if and only if this graph is planar.

From this characterization we deduce that, given a maximal planar graph $G$ on at least four vertices the poset consisting of the vertices, the edges and the faces of $G$ ordered by the inclusion relation has dimension four. Removal of any face from this poset results in a three-dimensional poset. The hypergraph whose vertices are the vertices of $G$ and whose hyperedges are the edges and faces of $G$ has the same property.

Our result also implies strong versions of several known properties of planar graphs. For example, the property of having embeddings in the plane where edges are segments of straight lines $[4,16,21]$; here we can give a purely combinatorial meaning to the vertex coordinates of the embedding. Another property is the decomposition of maximal planar graphs in three edge disjoint trees $[13,9]$ that correspond to three-dimensional realizers.

Notice that the planarity of three-dimensional graphs (‘only if’ part of our characterization) has also been shown by Babai and Duffus [1] in a different context.

2. The Order-Dimension of a Graph

Throughout this paper, $V$ is a finite nonempty set and $|V|$ denotes the cardinality of $V$. Graphs are finite simple graphs, their edges are identified with 2 element sets. A triangle of a graph $G$ is a set of three pairwise adjacent vertices of $G$.

The symbols $G$ and $H$ are reserved for graphs, $V(G)$ is the vertex set of $G$ and $E(G)$ is the edge set of $G$; a formula $G=(V, E)$ means that $V=V(G)$ and $E=E(G) . G$ is a subgraph of $H$ if $V(G) \subseteq V(H)$ and $E(G) \subseteq E(H)$.

If $R$ is a binary relation on $V(G)$, an $R$-path in $G$ is a path $v_{0}, v_{1}, \ldots, v_{n}$ of $G$ such that $R\left(v_{i}, v_{i+1}\right)$ holds for all $i, 0 \leqslant i<n$. Tlte outdegree in $R$ of a vertex $x$ of $G$ is the number of neighbors $y$ of $x$ that satisfy $R(x, y)$.
DEFINITION. For a graph $G=(V, E)$ the partial order $<_{G}$ on $V \cup E$ is defined by $a<_{G} b \Leftrightarrow a \in V$ and $b \in E$ and $a \in b$.
The (order-) dimension $\operatorname{dim}(G)$ of $G$ is the dimension of the poset $\left(V \cup E,<_{G}\right)$.
Remark. If $H$ is a subgraph of $G$, the relation $<_{H}$ is the restriction of $<_{G}$ to $V(H) \cup E(H)$. Therefore $\operatorname{dim}(H) \leqslant \operatorname{dim}(G)$, i.e., the dimension is a monotone function of graphs.

The definition of $\operatorname{dim}(G)$ relates $<_{G}$ to its realizers, thus to total orders on $V(G) \cup E(G)$. It will be useful to consider the restrictions of the latter orders to $V(G)$. Their essential properties are stated in the following lemma whose straightforward proof is left to the reader. An analogous statement for $P(n . k)$ can be found in [2].
LEMMA 2.1. A graph $G$ with vertex set $V$ has $\operatorname{dim}(G) \leqslant d$ if and only if there exists a sequence $<_{1},<_{2}, \ldots,<_{d}$ of total orders on $V$ satisfying the following conditions:
(1) the intersection of $<_{1},<_{2}, \ldots,<_{d}$ is empty;
(2) for each edge $\{x, y\}$ and each vertex $z \notin\{x, y\}$ of $G$. there is at least one order $<_{i}$ in the sequence such that $x<_{i} z$ and $y<_{i} z$.

Condition (1) is a consequence of condition (2) when the minimum degree of

$G$ is at least 2. In this case, already the existence of $d$ partial orders satisfying condition (2) implies $\operatorname{dim}(G) \leqslant d$.

In the remainder of this paper, the names vertex property and edge property will refer to (1) and (2), respectively.

DEFINITION. A $d$-dimensional representation on a set $V$ is a sequence $<_{1}$, $<_{2}, \ldots,<_{d}$ of total orders on $V$ that has the vertex property. This sequence represents all graphs with vertex set $V$ for which the edge property holds. Among these graphs, the graph $G$ with maximal edge set is the graph induced by $<_{1},<_{2}, \ldots,<_{d}$.

The following examples and Proposition 2.4 illustrate the use of Lemma 2.1 and show some elementary properties of the dimension of graphs.

EXAMPLE 2.2. A graph has dimension less than 3 if and only if it is a subgraph of a path (notice that the only 1-dimensional graph is the isolated vertex): a two-dimensional representation on an $n$-element set $V$ has the form

$$<_{1}: v_{1} v_{2} \ldots v_{n-1} v_{n}, \quad<_{2}: v_{n} v_{n-1} \ldots v_{2} v_{1}$$

and represents a graph $G=(V, E)$ iff $E \subseteq\left\{\left\{v_{i}, v_{i+1}\right\} \mid 1 \leqslant i<n\right\}$.
EXAMPLE 2.3. A three-dimensional representation of the triangle with vertices $a, b, c$ is any sequence $<_{1},<_{2},<_{3}$ of total orders on $\{a, b, c\}$ such that each of $a, b$ and $c$ is the maximum of one of these orders.

PROPOSITION 2.4. Each 4-colorable graph has dimension at most 4.
Proof. Let $G=(V, E)$ be a four-colorable graph and consider a partition $X, Y, Z, W$ of $V$ in four color classes. Let $X^{+}, Y^{+}, Z^{+}, W^{+}$denote any total orderings of $X, Y, Z, W$ and $X^{-}, Y^{-}, Z^{-}, W^{-}$denote the inverse orderings. It is easy to see that the following four orderings of $V$ form a four-dimensional representation of $G$.

$$\begin{array}{lll} <_{1}: X^{+} & Y^{+} & Z^{+} W^{+} \\ <_{2}: Y^{-} & X^{-} & W^{-} Z^{-} \\ <_{3}: W^{+} & Z^{+} & X^{-} Y^{-} \\ <_{4}: Z^{-} & W^{-} & Y^{+} X^{+} \end{array}$$

(the superscripts chosen in column 1 are irrelevant).
The converse of Proposition 2.4 is false. For example, it can be shown that $\operatorname{dim}\left(K_{12}\right)=4$. ( $K_{13}$ is the first complete five-dimensional graph).

It will be proved in Section 4 that every three-dimensional graph is planar. Together with Proposition 2.4, this implies the existence of graphs of arbitrarily high genus in dimension 4.

The following observations (Lemma 2.5 and Corollary 2.6) show that each $d$-dimensional graph has standard representations. This fact has also been used in [2].

LEMMA 2.5. Let $<_{1},<_{2}, \ldots,<_{d}$ be a $d$-dimensional representation of $G=$ $(V, E)$ and $x$ be the maximum of $<_{k}$. For $i \neq k$, let $<_{i}^{\prime}$ be the total order on $V$ where $x$ precedes all elements of $V-\{x\}$ and the elements of $V-\{x\}$ are ordered by $<_{t}$. Then $<_{1}^{\prime}, \ldots,<_{k-1}^{\prime},<_{k},<_{k+1}^{\prime}, \ldots,<_{d}^{\prime}$ is also a $d$-dimensional representation of $G$.

DEFINITION. A $d$-dimensional representation $<_{1},<_{2}, \ldots,<_{d}$ on a set $V$ is standard if $|V| \geqslant d$ and, for all $i \neq j$, the maximum element of $<_{t}$ is one of the $d-1$ smallest elements of $<_{t}$. The maxima of the orders of a standard $d$ dimensional representation are the exterior elements of this representation. The other elements of $V$ are the interior elements.

Notice that, for $d \geqslant 3$, any two exterior elements of a standard $d$-dimensional representation are adjacent in the graph induced by this representation.

COROLLARY 2.6. Each Graph $G$ with at least $d$ vertices and $\operatorname{dim}(G) \leqslant d$ has a standard d-dimensional representation.

Proof. By $d$-fold application of Lemma 2.5 to an initial $d$-dimensional representation of $G$.

3. Three Dimensional Representations

For a sequence $R_{1}, R_{2}, R_{3}$ of binary relations on a set $V$, we define the dual sequence $R_{1}^{*}, R_{2}^{*}, R_{3}^{*}$ by

$$R_{k}^{*}(x, y) \Leftrightarrow R_{k}(x, y) \&\left[R_{i}(y ; x) \text { for } i \neq k\right]$$

In particular, if $<_{1},<_{2},<_{3}$ is a three-dimensional representation on $V$, the relations $<_{1}^{*},<_{2}^{*},<_{3}^{*}$ are partial orders on $V$ and any two distinct elements $x$ and $y$ of $V$ are comparable in exactly one of these orders. Each edge of the graph $G$ induced by $<_{1},<_{2},<_{3}$ receives therefore a unique label and direction. This decomposes $G$ in three arc disjoint digraphs whose arc-sets are denoted $A_{1}, A_{2}$, $A_{3}:$

$$A_{k}=\left\{(x, y) \mid\{x, y\} \in E(G) \& x<_{k}^{*} y\right\}$$

and $E(G)$ is the disjoint union of the three underlying edge sets $E_{1}, E_{2}, E_{3}$ defined by $E_{k}:=\left\{\{x, y\} \mid(x, y) \in A_{k}\right\}$. An immediate consequence of the edge and vertex properties is then Lemma 3.1.

LEMMA 3.1. For a 3-dimensional representation $<_{1},<_{2},<_{3}$ on the set $V$, there holds: $A_{k}=\left\{(x, y) \mid y\right.$ is the minimum in $\left.<_{k}\right.$ of $\left.\left\{v \in V \mid x<_{k}^{*} v\right\}\right\}$.

EXAMPLE 3.2. The three-dimensional representation $<_{1}: b$ c $x y \leq a,<_{2}$ : $c a z x y b,<_{3}: a b y z x c$ on $V=\{a, b, c, x, y, z\}$ induces the graph shown as Figure 2, where the thick lines correspond to $A_{1}$.

Fig. 2.

THEOREM 3.3. For $k=1,2,3$, the digraph $\left(V, A_{k}\right)$ determined by a threedimensional representation $<_{1},<_{2},<_{3}$ on $V$ is a rooted forest (whose arcs are directed toward the roots). If the representation is standard and $a_{1}, a_{2}, a_{3}$ denote the respective maxima of $<_{1},<_{2},<_{3}$, then the component $T_{k}$ of $a_{k}$ in $\left(V, A_{k}\right)$ has $a_{k}$ as root and includes at least all interior elements of the representation as further vertices. The other components, if any, are trivial.

Proof. By Lemma 3.1, the outdegree in $\left(V, A_{k}\right)$ of a vertex $x \in V$ is at most 1 (it is 1 iff there exists a vertex $y \in V$ such that $x<_{k}^{*} y$ ). Hence, to each undirected cycle of the underlying graph $\left(V, E_{k}\right)$ would correspond a directed cycle of $\left(V, A_{k}\right)$. Since $<_{k}^{*}$ is an (acyclic) order relation, $\left(V, E_{k}\right)$ is a forest. The same outdegree property of $\left(V, A_{k}\right)$ implies that each connected component of $\left(V, E_{k}\right)$ contains exactly one vertex whose outdegree in $\left(V, A_{k}\right)$ is 0 . This vertex is the root of the given component.

If $<_{1},<_{2},<_{3}$ is standard, the only vertices that may have outdegree 0 in $\left(V, A_{k}\right)$ are the respective maxima $a_{1}, a_{2}, a_{3}$ of $<_{1},<_{2},<_{3}$. Of these, only $a_{k}$ can have indegree greater than 0 in $\left(V, A_{k}\right)$. Hence, $\left(V, A_{k}\right)$ has at most one nontrivial component (rooted in $a_{k}$ ). This tree includes all interior elements of the representation.

Remark. For $i \neq k$ the element $a_{i}$ is a vertex of $T_{k}$ (the notation is the same as in Theorem 3.3) if and only if the edge $\left\{a_{i}, a_{k}\right\}$ is directed from $a_{i}$ to $a_{k}$. Therefore, if $|V|=n$, either the trees $T_{1}, T_{2}, T_{3}$ have each exactly $n-1$ vertices, or the cardinalities of their vertex sets form a permutation of the numbers $n, n-1$ and $n-2$.

We observe that Theorems 1 and 3 of [9] are a consequence of the last remark (combined with the characterization of planar graphs as graphs of dimension at most three).

COROLLARY 3.4. A three-dimensional representation $<_{1},<_{2},<_{3}$ on a set $V$ of cardinality $n \geqslant 3$ is standard if and only if the induced graph $G$ has exactly $3 n-6$ edges.

Proof. (1) The ‘only if’ statement is a direct consequence of the last remark. (2) Conversely, let $a_{1}, a_{2}, a_{3}$ denote the respective maxima of $<_{1},<_{2},<_{3}$. Suppose that $<_{1},<_{2},<_{3}$ is not standard. For example, there exist vertices $x, y \in V$ such that $x<_{2} y<_{2} a_{1}$. The vertex condition implies then that $a_{1} \neq a_{3}$. One of the vertices $x, y$ must be different from $a_{3}$. This vertex is in no order greater than both $a_{1}$ and $a_{3}$. Consequently $\left\{a_{1}, a_{3}\right\}$ is not an edge of $G$. However, applications of Lemma 2.5 to $<_{1}$ and $<_{3}$ result in a supergraph $H=$ ( $V, E^{\prime}$ ) of $G$ such that $\left\{a_{1}, a_{3}\right\} \in E^{\prime}$. Part (1) of this proof and Corollary 2.6 imply then that $|E(G)|<\left|E^{\prime}\right| \leqslant 3 n-6$.

4. Three Dimensional Graphs

The following notation and terminology concerning planar embeddings will be used in this and the next sections.

A triangular graph $G$ is a maximal planar graph on at least three vertices that is embedded in the plane (i.e. whose exterior face has been chosen). The triangle of $G$ whose edges form the boundary of the exterior face of $G$ is the exterior triangle of $G$; its vertices and edges are the exterior vertices and edges of $G$, the other vertices and edges are the interior vertices and edges of $G$. An elementary triangle of $G$ is a triangle whose edges form the boundary of an interior face of $G$. Given a cycle $Z$ of $G$, the region with boundary $Z$ is the subgraph of $G$ induced by the vertices of $Z$ and the vertices lying in the interior of $Z$. Therefore ‘face’ and ‘region’ have different meanings.

The notation $x y$ (for $x, y \in \mathbb{R}^{3}$ ) represents the straight line segment with endpoints $x$ and $y$. A straight line embedding of a graph $G$ in a plane is an injection $f$ of $V(G)$ in this plane such that for any two distinct edges $\{x, y\}$ and $\{u, v\}$ of $G: f(x) f(y) \cap f(u) f(v)=f(\{x, y\} \cap\{u, v\})$.
THEOREM 4.1. Each graph $G=(V, E)$ of dimension at most three is planar. Moreover, to each three-dimensional representation $<_{1},<_{2},<_{3}$ of $G$ corresponds a straight-line embedding $f: v \in V \rightarrow\left(v_{1}, v_{2}\right) \in \mathbb{R}^{2}$ of $G$ in the plane, such that for all $u, v \in V: u_{i}<v_{i} \Leftrightarrow u<_{i} v(i=1,2)$.

Proof. For $v \in V$ and $i \in\{1,2\}$, let $v_{i}$ be the power of 2 whose exponent is the ordinal of $v$ with respect to $<_{i}$.

It suffices to verify that the so defined mapping $f: V \rightarrow \mathbb{R}^{2}$ is a straight line embedding of $G$ in the plane. To simplify the notation, the same symbol will be used to denote a vertex of $V$ and its image under $f$.

Note that, by the definition of $f$, if $x, y \in V$ satisfy $y<_{1} x$ and $x<_{2} y$, no vertex $z \neq x, y$ will be mapped by $f$ in the (closed) triangle delimited by the points $x, y$ and $\left(x_{1}, y_{2}\right)$ shown in Figure 3, since each point $z$ of this triangle satisfies

$$\frac{x_{1}}{2}<z_{1} \leqslant x_{1} \quad \text { or } \quad \frac{y_{2}}{2}<z_{2} \leqslant y_{2}$$

Fig. 3.

Assume, for contradiction, that there exist two disjoint edges $\{x, y\}$ and $\{a, b\}$ such that $x y \cap a b \neq \emptyset$. (The case of nondisjoint edges has a similar proof.)

Let, for example, $x$ be the maximum of $\{x, y, a, b\}$ with respect to $<_{1}$. By the edge property, $y>_{1} a, b, a>_{j} x, y, b>_{k} x, y$ must hold for some choice of $i, j, k \in\{1,2,3\}$.
$x y \cap a b \neq \emptyset$ and the maximality of $x$ in $<_{1}$ imply $i, j, k \geqslant 2$. Thus $j=k$ and therefore $j=k=3$ and $i=2$. Then $y>_{2} a, b$ and $x y \cap a b \neq \emptyset$ imply $y>_{2} x$.

Consequently $x>_{1} y, y>_{2} x$ and $y_{2}>a_{2}, b_{2}, x_{1}>a_{1}, b_{1}$. Therefore, $a$ and $b$ are both on the same side (below) of the straight line through $x$ and $y$. Hence, $x y \cap a b=\emptyset$ in contradiction to the original assumption.

By Corollary 3.4 and Theorem 4.1, a three-dimensional representation $<_{1},<_{2},<_{3}$ on a set $V$ of cardinality $|V| \geqslant 3$ induces a maximal planar graph if and only if this representation is standard. In this case, we refer to the graph $G$ induced by $<_{1},<_{2},<_{3}$ together with its planar embedding defined in the proof of Theorem 4.1 as the triangular graph induced by $<_{1},<_{2},<_{3}$. Elementary geometric considerations show that the exterior vertices of $G$ are precisely the exterior elements of the representation. This justifies the terminology introduced in the definition of standard representations.

As Lemma 4.2, we now prove the converse of Theorem 3.3. For later use, it is convenient to formulate this lemma in terms of acyclic relations.

LEMMA 4.2. Let $G$ be a triangular graph whose vertex set $V$ has cardinality $|V| \geqslant 4$ and $<_{1},<_{2},<_{3}$ be three acyclic relations on $V$ such that every interior vertex of $G$ has outdegree exactly one in each of $<_{1}^{*},<_{2}^{*},<_{3}^{*}$. Then for each edge $\{x, y\}$ of $G$ and each vertex $z \notin\{x, y\}$ there exist, for some $k \in\{1,2,3\}, a<_{k}$-path from $x$ to $z$ and $a<_{k}$-path from $y$ to $z$. Therefore, any three total orders extending $<_{1},<_{2},<_{3}$ on $V$ form a three-dimensional representation of $G$.

Proof. Notice that for $i \neq j$, a $<_{i}^{*}$-path and a $<_{i}^{*}$-path starting in the same vertex $v$ have no common vertex except $v$ as the existence of such a common vertex $u$ would imply a $<_{i}$-cycle from $v$ to $v$ via $u$.

Let $v$ be any interior vertex of $G$. By the above, the (unique longest) $<_{1}^{*},<_{2}^{*}$ and $<_{3}^{*}$-paths starting at $v$ end in distinct exterior vertices that will be denoted

respectively by $a_{1}, a_{2}, a_{3}$. In particular, this implies the existence of some $<_{i}$-path from $a_{j}$ to $a_{i}$ for any choice of $j \neq i$.

If $u \neq v$ is any further interior vertex of $G$, the (unique longest) $<_{i}^{*}$-path starting at $u$ must also end at $a_{i}$, as an ending in $a_{j}$ for $j \neq i$ would imply the existence of some $<_{k}^{*}$-path $(k \neq i)$ from $u$ to $a_{i}$ and therefore the existence of a $<_{i}$-cycle from $a_{i}$ to $a_{i}$ via $u, a_{j}$ and $v$ (Figure 4).

Fig. 4.

Therefore there exists a $<_{i}$-path from each vertex $u \neq a_{i}$ of $G$ to $a_{i}$. This proves the lemma for the case where $z \in\left\{a_{1}, a_{2}, a_{3}\right\}$.

If $z \notin\left\{a_{1}, a_{2}, a_{3}\right\}$, the $<_{1}^{*},<_{2}^{*}$ and $<_{3}^{*}$-paths from $z$ to $a_{1}, a_{2}$ and $a_{3}$ divide $G$ in three regions $R_{1}, R_{2}$ and $R_{3}$ (where $R_{k}$ denotes the region opposite $a_{k}$, including its boundary) (Figure 5).

Fig. 5.

As the end-vertices $x$ and $y$ of each edge $\{x, y\}$ of $G$ are both in the same region, it suffices to show that there exists for each vertex $v$ of $G, v \in R_{k}$, a $<_{k}$-path from $v$ to $z$.

Let, for example, $v \in R_{3}$ and consider some $<_{3}$-path from $v$ to $a_{3}$. This path must intersect the $<_{1}^{*}$-path from $z$ to $a_{1}$ or the $<_{2}^{*}$-path from $z$ to $a_{2}$ in a vertex $u$. Suppose, for example, that $u$ belongs to the $<_{1}^{*}$-path from $z$ to $a_{1}$. Combining the two paths, there results a $<_{3}$-path from $v$ to $z$ via $u$.

5. Barycentric Representations

It has been shown in Section 4 that the graph $G$ induced by a three-dimensional representation $<_{1},<_{2},<_{3}$ is always planar. It is therefore tempting to look for an embedding of $G$ in a plane (or sphere) in $\mathbb{R}^{3}$ such that $<_{1},<_{2}$ and $<_{3}$ are precisely the orders defined on the vertices by their three coordinates. Such an embedding however, does not exist in general, as illustrated (for planes) by Example 5.1.

EXAMPLE 5.1. Let $V=\{a, b, c, x, y, z\}$ and $<_{1},<_{2},<_{3}$ be the three-dimensional representation defined on $V$ by
$<_{1}: a x b z c y$
$<_{2}: b$ y $c x a z$
$<_{3}: c z a y b x$.
There exists no mapping $f: v \in V \rightarrow\left(v_{1}, v_{2}, v_{3}\right) \in \mathbb{R}^{3}$ such that
(1) $f(V)$ is contained in a plane,
(2) for all $u, v \in V: u_{i}<v_{i} \Leftrightarrow u<_{i} v \quad(i=1,2,3)$.

Since for $i=1,2,3$ there exist elements $u, v \in V$ with $u<_{i}^{*} v$, the coefficients $\alpha_{1}, \alpha_{2}, \alpha_{3}$ of the equation $\alpha_{1} v_{1}+\alpha_{2} v_{2}+\alpha_{3} v_{3}=\delta$ of such a plane would have to be of the same sign. Replacing in each row $<_{i}$ of the above matrix every element $v$ by the product $\alpha_{i} v_{i}$ and comparing the sums of columns $1,3,5$ and columns $2,4,6$ would then lead to the contradiction $3 \delta \neq 3 \delta$.

Nevertheless, Lemma 4.2 implies that the edge and vertex properties of a standard three-dimensional representation are essentially determined by the restrictions of its orders to the edges of the induced graph. This suggests a relaxation of condition (2) of the last example and will result (Theorem 5.3 and Corollary 5.4) in the correct interpretation of three-dimensional representations by coordinates.

DEFINITION. A barycentric embedding of a graph $G$ is an injective function $v \in V(G) \rightarrow\left(v_{1}, v_{2}, v_{3}\right) \in \mathbb{R}^{3}$ that satisfies the conditions:
(1) $v_{1}+v_{2}+v_{3}=1$ for all vertices $v$,
(2) For each edge $\{x, y\}$ and each vertex $z \notin\{x, y\}$, there is some $k \in\{1,2,3\}$ such that $x_{k}<z_{k}$ and $y_{k}<z_{k}$.

LEMMA 5.2. Each barycentric embedding of a graph is a straight line embedding of this graph in the plane $v_{1}+v_{2}+v_{3}=1$.

Proof. By applying the proof of Theorem 4.1 to the ‘projection’ $f: v \in V(G) \rightarrow$ $\left(v_{1}, v_{2}\right) \in \mathbb{R}^{2}$ of an initial barycentric embedding of the graph $G$. Observe that if $\{x, y\}$ is an edge of $G$ and $y_{1} \leqslant x_{1}, x_{2} \leqslant y_{2}$, no vertex $z$ of $G$ will be mapped by $f$ in the convex hull of the points $\left(x_{1}, x_{2}\right),\left(y_{1}, y_{2}\right),\left(x_{1}, y_{2}\right)$.

THEOREM 5.3. Let $G$ be the triangular graph induced by a standard three-

dimensional representation $<_{1},<_{2},<_{3}$ on a set $V$ and let $a^{1}, a^{2}, a^{3}$ be the respective maxima of $<_{1},<_{2},<_{3}$. Then there exists a straight line embedding of $G$ such that $v_{1}, v_{2}, v_{3}$ denoting the barycentric coordinates of vertices $v \in V$ relative to $a^{1}, a^{2}, a^{3}$, there holds for each interior edge $\{u, v\}$ of $G: u<_{i} v \Leftrightarrow$ $u_{i}<v_{i}(i=1,2,3)$.

Proof. Let $n$ denote the cardinality of $V$. It suffices to define a mapping $f: v \in$ $V \rightarrow\left(v_{1}, v_{2}, v_{3}\right) \in \mathbb{R}^{3}$ with the properties:
(0) $f\left(a^{1}\right)=(2 n-5,0,0), f\left(a^{2}\right)=(0,2 n-5,0), f\left(a^{3}\right)=(0,0,2 n-5)$,
(1) $v_{1}, v_{2}, v_{3} \geqslant 0$ and $v_{1}+v_{2}+v_{3}=2 n-5$ for all vertices $v$,
(2) $u<_{i} v \Rightarrow u_{i}<v_{i}$ for all interior edges $\{u, v\}$ of $G$ and $i=1,2,3$.

By Lemmas 4.2 and 5.2 , the mapping resulting from $f$ upon division by $2 n-5$ satisfies the statement of the proposition.

Using the notation of the proof of Lemma 4.2, recall that each interior vertex $v$ divides $G$ in three regions $R_{1}(v), R_{2}(v), R_{3}(v)$ and that each vertex $u \neq v$ of region $R_{k}(v)$ satisfies $u<_{k} v(k=1,2,3)$.
$f\left(a^{1}\right), f\left(a^{2}\right), f\left(a^{3}\right)$ are defined by ( 0 ) above. For an interior vertex $v$ of $G$, we let $v_{i}(i=1,2,3)$ be the number of elementary triangles in region $R_{i}(v)$. This clearly satisfies condition (1).

To verify condition (2), consider an interior edge $\{u, v\}$ of $G$ and assume, for example, that $u<_{3}^{*} v$. To show: $u_{3}<v_{3}, v_{1}<u_{1}, v_{2}<u_{2}$. This is trivial if $v=a^{3}$. Suppose, therefore, that $v \neq a^{3}$. Then $v$ is an interior vertex of $G$ and $u$ must lie in the interior of region $R_{2}(v)$ (Figure 6).

Fig. 6.
For $i=1$ and $j=2$ (or $i=2$ and $j=1$ ), the $<_{i}^{*}$ path from $u$ to $a^{i}$ and the $<_{i}^{*}$ path from $v$ to $a^{j}$ have no common vertex, as the existence of such a vertex $x$ would imply $u \leqslant_{i} x \leqslant_{i} v$, contradicting $u<_{3}^{*} v$. The outdegree (at most 1) properties of $<_{1}^{*}$ and $<_{2}^{*}$ imply thus: $R_{3}(u) \subseteq R_{3}(v), R_{1}(v) \subseteq R_{1}(u)$, and $R_{2}(v) \subseteq$ $R_{2}(u)$. These inclusions being proper, there follows $u_{3}<v_{3}, v_{1}<u_{1}, v_{2}<u_{2}$.

As each three-dimensional representation of a graph $G$ can be extended, by adjunction of three new exterior elements, to a standard representation of a triangular supergraph of $G$, we obtain

COROLLARY 5.4. To each three-dimensional representation $<_{1},<_{2},<_{3}$ of a graph $G$ corresponds a barycentric embedding $v \in V(G) \rightarrow\left(v_{1}, v_{2}, v_{3}\right) \in \mathbb{R}^{3}$ of $G$ such that $u<_{i} v \Leftrightarrow u_{i}<v_{i}$ for all edges $\{u, v\}$ of $G$ and $i=1,2,3$.

6. Planar Graphs have Dimension at Most Three

We now show that each planar graph has dimension at most three. By Theorem 4.1 or Corollary 5.4 , this fact implies, in particular, that each planar graph has straight line embeddings, a result of Fáry [4], Stein [16] and Wagner [21]. Our proof of three-dimensionality is based on their methods, as analyzed by Kampen [9].

Two independent versions are given: construction of a straight line embedding in this section and a coordinate-free proof in Section 7.

We first review the method of edge contraction. For a vertex $x$ of a graph $G, N(x)$ denotes the set of neighbors of $x$ in $G$. If $\{x, y\}$ is an edge of $G$, the contracted graph $G /(x, y)$ is obtained from $G$ by removal of the vertex $y$ and the edges incident on $y$ and by introduction of an edge $\{x, z\}$ for each vertex $z \in$ $N(y)-N(x)$. The edge $\{x, y\}$ is contractible if $x$ and $y$ have exactly two common neighbors. If $G$ is a maximal planar graph on at least four vertices and $\{x, y\}$ is a contractible edge of $G$, then $G /(x, y)$ is a maximal planar graph.
LEMMA 6.1 [9]. Let $G$ be a triangular graph on $n \geqslant 4$ vertices. If $a, b$ and $c$ denote the exterior vertices of $G$, then there exists a neighbor $x \neq a, b$ of $c$ such that the edge $\{c, x\}$ is contractible.

Thus to each triangular graph corresponds a sequence of ‘allowed contractions’ transforming this graph into a triangle.

THEOREM 6.2. Each planar graph has dimension at most three.
Proof. By the monotonicity of the dimension, it suffices to prove the theorem for triangular graphs $G$. Let $V$ be the vertex set of $G$ and let $a, b, c$ denote the exterior vertices of $G$ in counterclockwise order. We show the existence of an embedding of $V$ in the plane such that (identifying the vertices and their images):
(a) $a, b, c$ are affinely independent and each vertex $v \in V$ lies in the triangle delimited by $a, b$ and $c$.
(b) The partial orders $<_{1},<_{2},<_{3}$ defined on $V$ by the barycentric coordinates $v_{1}, v_{2}, v_{3}$ of the vertices $v \in V$ with respect to $a, b, c\left(u<_{i} v \Leftrightarrow u_{i}<v_{i}\right)$ satisfy the hypothesis of Lemma 4.2 (if $|V| \geqslant 4$ ).
The proof proceeds by induction on the number $n=|V|$. The claim is trivial

if $n=3$. Let $n \geqslant 4$ and assume that the claim is true for all triangular graphs with less than $n$ vertices.

Consider some neighbor $x \neq a, b$ of $c$ such that the edge $\{c, x\}$ is contractible and let $p=v^{1}, v^{2}, \ldots, v^{r}=q$ be the neighbors of $x$, distinct from $c$, in counterclockwise order. The vertices $p$ and $q$ are the two common neighbors of $c$ and $x$ (Figure 7).

Fig. 7.

By induction hypothesis, there is an embedding $f$ of $V-\{x\}$ in the plane with the above properties (a) and (b). Lemma 4.2 implies that $f$ is a barycentric embedding. By Lemma 5.2, this embedding is therefore a straight line embedding of $G /(c, x)$ (Figure 8 ).

Fig. 8.

Notice that as $c_{3}>u_{3}$ for all $u \in V-\{c\}$ and $f$ is a straight line embedding of $G$, condition (2) of the definition of barycentric embeddings applied to the wheel of $c$ implies that

$$p_{2}=v_{2}^{1}<v_{2}^{2}<\cdots<v_{2}^{r}=q_{2} \quad \text { and } \quad q_{1}=v_{1}^{r}<v_{1}^{r-1}<\cdots<v_{1}^{1}=p_{1}$$

Let $d$ denote the point with barycentric coordinates $\left(q_{1}, p_{2}, 1-q_{1}-p_{2}\right)$. By the above inequalities, there is a choice of $x$ sufficiently close from $d$ in the quarterplane $u_{1}>q_{1}, u_{2}>p_{2}$ that will satisfy the relations: $x_{3}>v_{3}^{\prime}(i=1,2, \ldots, r)$, $x_{1}>q_{1}, x_{2}>p_{2}, x_{1}<v_{1}^{\prime}$ and $x_{2}<v_{2}^{\prime}(i=2, \ldots, r-1)$.