Straight Line Representations of Infinite Planar Graphs

STRAIGHT LINE REPRESENTATIONS OFINFINITE PLANAR GRAPHS

CARSTEN THOMASSEN

1. Introduction and summary

A plane graph is a graph whose vertex set is a point set in the Euclidean plane whilethe edges are Jordan curves such that two different edges have at most an end point incommon. A planar graph is an abstract graph isomorphic to a plane graph.

Finite, planar graphs were characterized by Kuratowski [8]. Dirac and Schuster[2] generalized Kuratowski's Theorem to countable graphs, and Wagner [15] charac-terized all planar graphs. Halin [4] characterized graphs isomorphic to locally finite,plane graphs with no vertex accumulation points.

Wagner [14], Fary [3], and Stein [10] independently proved that every finite,planar graph has a straight line representation, and Tutte [11, 12] described specialstraight line representations of planar graphs.

In §3 of this paper we prove that every infinite planar graph has a straight linerepresentation and solve thereby a problem of Adler [1]. We show in §4 that everyconnected, locally finite, plane graph with no vertex accumulation point isisomorphic to a subgraph of a straight line triangulation (to be defined in §2), and in§5 we apply this result to prove that every infinite, locally finite, plane graph, withno vertex accumulation point (or no edge accumulation point) has a straight linerepresentation with the same property.

2. Terminology and preliminaries

The graph terminology is that of Harary [6] with minor modifications.For a graph G, the set of vertices (respectively edges) is denoted by V(G)

(respectively E(G)). The edge joining the vertices x, y is denoted by xy. If A £ V(G),then G(A) denotes the subgraph of G induced by A, and if A £.£(G), then G[A]denotes the graph with vertex set V{G) and edge set A. If xy e£(G), then the graphobtained by contracting the edge xy is the graph G' obtained from G—{x, y} by addinga new vertex z and joining z to those vertices which in G are adjacent to x or y (orboth). We also say that G is obtained from G' by splitting z into x and y.

Abstract graphs are denoted by capital roman letters and plane graphs by capitalgreek letters. If the graph G is isomorphic to the plane graph F, then F is arepresentation of G. If the edges of F are polygonal arcs (respectively straight linesegments), then F is a polygonal arc representation (respectively straight line represen-tation).

A closed Jordan curve partitions the Euclidean plane into a bounded region calledthe interior and an unbounded region called the exterior of the curve. A finite, planegraph F partitions the plane into a finite number of regions one of which is

Received 4 November, 1976.

Research supported in part by the Danish Natural Science Research Council.

[J. LONDON MATH. SOC. (2), 16 (1977), 411-423]

412 CARSTEN THOMASSEN

unbounded. If the boundary of every region is a 3-cycle, then F is a triangulation. Twofinite, plane graphs F l s F2 are equivalent if there is a 1-1 incidence-preserving maptaking the set of regions of I \ onto the set of regions of F2, the set of vertices of I \onto the set of vertices of F2 , and the set of edges of I \ onto the set of edges of F2.If, furthermore, the unbounded region of I \ corresponds to the unbounded regionof F 2 , then I \ and F2 are strongly equivalent. Equivalence between plane graphsis generalised in the obvious way to equivalence between infinite plane graphs withfinite edge sets and vertex sets with no accumulation points.

For any point p in the plane and any positive real number e, we denote by D(p, e),C(p, e) and D(j>, e) the set of points of (Euclidean) distance less than e, equal to e, andless than or equal to e, respectively. If p and q are points, then [p, q] denotes thestraight line segment between p and q. The point set of a plane graph F is denoted byF*. If F is a plane graph and p is a point not in F*, then p is admissible with respect toF if and only if for every vertex q of F adjacent to the region containing p we have:[q, p] n F* = {q}. A plane graph F is star-shaped if every region contains an admis-sible point.

We need the following standard results about finite planar graphs:

(1). If G is a graph containing a vertex v such that G — v is a cycle, then any tworepresentations of G are equivalent.

(2). The boundary of any region of a 2-connected finite, plane graph is a cycle ofthe graph.

(3). If G is a finite, planar graph and x and y are vertices of G such that x and ybelong to distinct components of G or such that x and y are both adjacent to a vertex zand belong to distinct components of G—z, then G u {xy} is planar.

(4). If F is a representation of the finite, planar graph G and F' is apolygonal arc representation of G — e (where eeE(G)) such that F ' and thesubgraph of F representing G — e are (strongly) equivalent, then a polygonal arcrepresenting e can be added to F' such that the resulting plane graph is (strongly)equivalent to F.

A cycle S of a connected graph G is a separating cycle if and only if G— V(S) isdisconnected.

(5). A 3-cycle 9 of a triangulation A is a separating cycle if and only if it is not theboundary of any region.

(6). Let G be a finite, planar graph isomorphic to a triangulation, and let xybe an edge of G not contained in a separating 3-cycle of G. Let G' be the graphobtained by contracting xy into a vertex z. If F is a representation of G, then arepresentation of G' can be obtained from F such that only the edges incident with xare changed. Conversely, if F ' is a straight line representation of G', then a straightline representation of G can be obtained from F' such that only the edges incidentwith JC are changed and such that x is arbitrarily close to y.

Using the last part of (6) it is easy to modify the proof in [9; pp. 6-8] so as to obtainfollowing result.

(7). Let F be a finite triangulation such that the cycle xyzx is the boundary of theunbounded region. Let T be a triangle with corners p, q, r and let e be any positive

STRAIGHT LINE REPRESENTATIONS OF INFINITE PLANAR GRAPHS 413

real number. Then there exists a straight line triangulation A which is stronglyequivalent to F such that x, y, z are represented by p, q and r, respectively, and suchthat every vertex of A other thang and r has (Euclidean) distance less than e from p.

For further properties, of finite, planar graphs, see e.g. [9].

A vertex accumulation point (respectively edge accumulation point) of an infiniteplane graph F is a point p of the plane such that, for each e > 0, D(p,e) contains(respectively intersects) infinitely many vertices (respectively edges) of F. Vertexaccumulation point and edge accumulation point are abbreviated VAP and EAP,respectively. A VAP-free (respectively EAP-free) plane graph is a plane graph withno VAP (respectively EAP).

An infinite, connected, plane graph F is a triangulation if and only if F isVAP-free and, for every vertex x of F, F contains a cycle y¥x such that x is the onlyvertex of F in the interior of *FX and x is joined by an edge to every vertex of *F,<. If,in addition, every point not in F* is contained in the interior of some cycle of F,then F is a triangulation of the plane. We conclude this section with a characterizationof infinite triangulations.

(8). An infinite, plane, VAP-free graph F is a triangulation if and only if F containsa sequence of mutually disjoint cycles Qlt 0 2 , ... such that for each k ^ 2, the sub-graph of F induced by 0fc and the vertices in the interior of 0fc is finite andtriangulates the interior of 0 k and contains 0*-! , and furthermore every vertex of Fis in the interior of some 0m.

It is easy to see that a plane graph satisfying the assumption of (8) is a triangulation.To prove the converse, let F be any triangulation. We define the sequence of cycles0 1 } 0 2 , . . . recursively as follows: Let Qx be any cycle of F. Suppose that wehave already defined 0 l 5 ..., 0fc. Let Tk be the graph [) *FX, where the union is takenover all vertices x of 0fc. Then Ffc is 2-connected and, by (2), the boundary of theunbounded region of Ffc is a cycle 0 f c + v Since x is in the interior of ¥*, every vertexof 0fc is in the interior of 0fc+1. Since F has no VAP, there are only finitely manyvertices of F in the interior of 0*+ls and hence, by the definition of a triangulation, Ftriangulates the interior of 0 f c + v If y is a vertex of F in the exterior of 0f c + l s then the(graphic) distance from y to 0fc+ x is less than the distance from y to 0fc. Hence y is inthe interior of some 0m, and so F satisfies the conclusion of (8).

The Axiom of Choice is assumed throughout the paper.

3. Straight line representations of infinite planar graphs

00

PROPOSITION 1. Let G = (J Gi be an infinite graph such that the following hold:

(i) Gt is a cycle;

(ii) each Gt is an induced subgraph of G;(iii) for each i ^ 1, Gi+1 can be obtained from Gt by adding a vertex vi+1 and

joining vi+l to at least two vertices of Gt and then {possibly) subdividing someedges incident with vl+l;

and

(iv) each Gt is planar.

Then G has a straight line representation F such that the point set of F is bounded.


Proof. It follows from (i) and (iii) that each Gf is a finite, 2-connected subgraphof G. Since for every finite, planar graph there are only finitely many classes of repre-sentations of the graph in the plane, it follows by an application of Konig's Lemma[7; p. 81] (for details, see [2]) that there exists a sequence I \ , F 2 , . . . of plane graphssuch that for each i, Tt is isomorphic to Gh and the representation of Gt induced byF f + 1 is equivalent to Ff.

We define recursively a sequence of plane graphs x¥i,x¥2,... such that

VF1 £ ^ 2 — • • •> a n d such that for each i, ¥ ; is a star-shaped straight line representationof Gh equivalent to Ff. It is easy to describe a star-shaped straight line representation*?! of Gj. Suppose that we have already described M . A representation of GJ+1

equivalent to F i + 1 can be obtained from ^ by representing vi+l by a point in anappropriate region of % . This region is bounded by a cycle, say 0 , of ¥ ,(because Gf is 2-connected), and it has an admissible point, say p. Let su s2, ..-,sk

denote the vertices of 0 which are to be joined (by paths) to vi+i. We representvi+i by jp and the path from vi+l to Sj by the straight line segment [p, s j for each ;.This obviously defines a straight line representation *Fi+1 of Gi+l. Also, by (1),¥ , + i is equivalent to Ti+l. So it only remains to prove that ¥ l + 1 is star-shaped, andin order to do this it suffices to describe an admissible point in each region of ^Fi+1

adjacent to p. Since ¥,- is a straight line representation, there is a real number <5 > 0such that each point q within (Euclidean) distance 5 of p is admissible with respect toTj. Suppose that the notation is chosen such that [p, s j , [p, s2], •.., [p, sk] are the linesegments radiating from p in ¥ , + ! arranged in clockwise order (the indices beingexpressed modulo k), and let Rj denote the region whose boundary contains[P> Sj] u [p, sJ+1]. Let qs be a point of Rj of (Euclidean) distance less than 5 from psuch that any straight line starting at qs has at most one point in common with\P> sj\ v [p,Sj+1]. (If the angle between [p, Sj] and [p, sJ+l] does not exceed n, thelast condition is trivially satisfied.) We claim that #,- is admissible (with respect to¥,+ j). For consider any vertex q of *Fi+x adjacent to Rj. Ifq e [p, Sj] u [p, sJ+ J , thenclearly [#_/,#] n \P* + 1 = {q} (for convenience, we can assume that each vertex in¥ , + ! but not in ¥,• is within distance 5 from p), so assume q e F(0)\{5j , sj+l}. Sinceqj is admissible with respect to ¥ , , we conclude that [qj,q] n ^ j * = {q}, so we shallconsider the possibility that [qj,q] intersects some of the lines [p, sm]. Suppose there-fore that [qj, q] intersects the straight line starting at p and containing sJ+1 and supposefirst that the point of intersection is not p. Then let m be the largest integer such that[Qb # ] n tP> sj+m\ T* 0 ( 1 <>m ^ k). Since [qp q] does not intersect [p, Sj], we concludethat m < k. But then q is adjacent to the region of ¥ i + x containing qJ+m, that is, theregion RJ+m, and q is also adjacent to Rj. Since q is assumed to be in V(Q)\{sj, sJ+ J ,we have obtained a contradiction. If [qjt q] contains p, we argue similarly. This proves

' • 0 0

that %•+1 is a star-shaped representation of Gi+l. Now F = (J ¥,• is a straight linerepresentation of G. i = 1

The proof shows that if we select an admissible point in each of the two regions of*¥u then F can be constructed so that all other vertices are within Euclideandistance 1 from one of these points; so F can be chosen to be bounded.

PROPOSITION 2. Let <Sbea (possibly infinite) collection of finite graphs. Let Hbeagraph such that every finite subgraph ofH belongs to <S. Then there exists a graph Gsuch that

(i) V(G) = V(H) and E(G) 2 E(H);


(ii) every finite subgraph of G belongs to &;

and

(iii) G is maximal with respect to (i) and (ii).

Proof. Let M be the set of graphs H' such that V (#') = V (H), E(H') 2 E(H), andevery finite subgraph of if' belongs to ^ . The set M is partially ordered by inclusion.We shall show that M is inductively ordered. For this, let M' be any totally orderedsubset of M. Let H" be the graph which is the union of all graphs of M'. Since M'is totally ordered, every finite subgraph of H" is a subgraph of some graph of M';hence every finite subgraph of H" belongs to <$, that is, Ii" e M. So M' has a majorantin M. By Zorn's Lemma, M contains a maximal element, say G, and any suchgraph satisfies (i), (ii) and (iii) of the proposition.

As an immediate consequence of Proposition 2 we obtain the following result.

PROPOSITION 3. Let H be an infinite graph such that every finite subgraph is planar.Then there exists a graph G such that

(i) V(G) = V(H) and E{G) 2 E(H);

(ii) every finite subgraph of G is planar;

and

(iii) G is maximal with respect to (i) and (ii).

Furthermore, every graph G satisfying (i), (ii) and (iii) above is 2-connected.

Proof. The existence of G follows from Proposition 2 (by letting ^ be the set offinite, planar graphs). Suppose that G is not 2-connected. Then it contains two verticesx, y such that either x and y belong to distinct components of G, or belong to distinctcomponents of G —z where z is a common neighbour of x and y. But then by (3), everyfinite subgraph of H u {xy} is planar. This contradicts the maximality property of H,and the proof is complete.

It is well known that every finite, maximal planar graph is 3-connected. However,it is not true that every infinite graph G satisfying (i), (ii) and (iii) of Proposition 3 is3-connected. For example, take the union of two disjoint 2-way infinite paths, add twonew vertices and join each of these by edges to all other vertices. The resulting graph isclearly planar, it has connectivity 2, and the addition of any new edge results in agraph containing a K3, 3. Halin [5] has investigated the graphs G satisfying (i), (ii)and (iii) of Proposition 3. In particular, he has characterized those havingconnectivity 2 in terms of the 3-connected ones.

THEOREM 1. Let H be an enumerably infinite graph such that each finite subgraphis planar. Then H is planar and has a straight line representation T such that T*, thepoint set of T, is bounded.

Proof. Let G be a graph satisfying (i), (ii) and (iii) of Proposition 3. It is sufficientto prove that G is planar and has a straight line representation T such that T* is


bounded. We prove this by showing that G satisfies the assumption of Proposition 1.Let V(G) = {xu x2, x3,...} and let Gy be any cycle of G such that no edge of G joinstwo non-consecutive vertices of this cycle. Having already defined Gn, we define Gn+l

as follows: Let a(n) be the smallest integer k such that xk $ V(Gn). By Proposition 3, Gis 2-connected, and so G contains a path y0 yl y2...ym such that y0 e V(Gn), ym e V(Gn)and xa(n) = yj for some;. Suppose that among all paths with these properties we haveselected a shortest one, and put m = /?(«). Now let G'n+1 be the subgraph of G inducedby V(GJu{ylty2, •••,ym-i} and let Gn+1 be an induced subgraph of G'n+l (andhence also of G) such that Gn is a proper subgraph of GII+1, Gn+1 is 2-connected, andGn+! is minimal subject to these conditions. It is easy to see that Gn+l can be obtainedfrom Gn by adding a vertex yt and joining it to at least two vertices of Gn or by adding apath ZQZ! ...zh{h ^ 2) so that z o e V(Gn), zhe V(Gn), and zte{ylty2t . . . , ^ - J for1 ^ i < h—1. In either case Gn+1 is obtained from Gn as described in Proposition 1.Moreover, if a(« +1) = a(/i) (that is, *8(II) $ V(Gn+1)), then 0 < fi(n+1) < pin). From

oo

this it follows that a(n) -* oo as n -> oo, or, in other words, (J G,- = G, and now thetheorem follows from Proposition 1. / = 1

If we combine Theorem 1 with the theorems of Kuratowski [2,8] and Wagner [15],we get the following result.

THEOREM 2. For a graph G {finite or infinite) the following statements areequivalent:

(a) G is planar.

(b) G is planar and has a straight line representation.

(c) G has at most continuum many vertices and only countably many vertices ofdegree 3 or more; G contains no subdivision of Ks or K3i 3.

Proof. Wagner [15] demonstrated the equivalence between (a) and (c), so we shallprove that (c) implies (b).

Consider a graph satisfying (c). It is easy to dispose of those components of thegraph which contain no vertices of degree ^ 3. Each component with this propertyis a cycle or a path and there are at most continuum many such components, so wecan represent the union of all these in a bounded subset of the plane. There are at mostcountably many components containing a vertex of degree ^ 3, so it is now sufficientto show that a connected graph H' satisfying (c) has a straight line representation in abounded subset of the Euclidean plane. Let Vo denote the set of vertices of degree ^ 3in H'. By assumption, Vo is finite or enumerably infinite. We now define a graph Hwith vertex set Vo as follows: two vertices of Vo are adjacent in H if and only if theyare adjacent in H' or joined by a path in H' such that all vertices of this path (exceptfor the ends) have degree 2 in H'.

Obviously H contains no subdivision of K5 or K3t 3 and hence, by Kuratowski'stheorem, every finite subgraph off/ is planar. Let G be a graph satisfying (i), (ii) and(iii) of Proposition 3. By Proposition 3, G is 2-connected. Now we define a graph G'as follows: For every vertex x of G, add to G a 3-cycle xvx(x) vz(x) x such that v^x),v2(x) are two vertices not in G; and for every edge e = xy of G, add two new verticesux(e), u2{e) and five new edges u^x, u^y, u2{e)x, u2(e)y, «1(e)M2(e), that is, inG', the vertices x, y, y^ie), u2(e) induce a K4. Also, we assume that the new vertices

STRAIGHT LINE REPRESENTATIONS OF INFINTE PLANAR GRAPHS 417

corresponding to distinct vertices or edges of G are distinct. Clearly G' is finite orenumerably infinite, and every finite subgraph of G' is planar.

By Theorem 1, G' is planar and has a straight line representation F such that thepoint set of F is bounded. Let G" be the graph obtained from G in the following way:For each vertex x of G, add continuum many 1-way infinite paths starting at x andhaving only x in common with G, and for each k ^ 3, add continuum manycycles of length k having only x in common with G; for each edge xy of G and foreach k ^ 2, add continuum many paths of length k from x to y such that each ofthese paths has only x and y in common with G. Suppose, furthermore, that no twoof the new paths and cycles added intersect outside G. It is not difficult to show thatG" contains H' as a subgraph, so we shall show that a straight line representation ofG" can be obtained from F. Consider any vertex JC of G and consider the triangle of Fcorresponding to the 3-cycle xv^x) v2(x) x. Since G is 2-connected, there cannot bevertices of F both in the interior and in the exterior of this triangle, so either in theinterior or in the exterior of the triangle we can add continuum many cycles and1-way infinite paths. Now consider any edge e = xy of G and consider the K^ in Finduced by x, y, ux(e), u2(e). Denote this by x¥. For any vertexz of V(G)\{x, y}, thegraph G—x (respectively G—y) contains a path fromz to y (respectively fromz to x).This implies that vertices of G can only occur in the two regions of *F adjacent to theedge e. This makes it possible (after deleting the edge uy{e) u2(e)) to add continuummany paths of all lengths from x to y.

The proof is complete.

4. Embeddings of graphs into infinite straight line triangulations

In this section we show that a connected, locally finite, VAP-free plane graphis isomorphic to a subgraph of an infinite straight line triangulation of the plane.

LEMMA 1. Let T be a countably infinite, locally finite, VAP-free, plane graph.Then there exists a VAV-free, plane graph 0 isomorphic to T such that each edge of 0is a polygonal arc and such that no point of an edge of 0 is an EAP of 0.

Proof. Let G denote the abstract graph which is represented by F, and let= ieu e2t e3> •••}• F° r e a c n edge et of G we add two new edges e/, et" joining

the same vertices as eh and we denote the resulting multigraph by G'. To each vertex pof F we associate a positive real number e(p) such that, for any two distinct verticesp, q of F, we have that D(p, e(p)) n D(q, e(q)) = 0 . We define a sequence0O, 01} 02 , ... of plane multigraphs such that 0O £ 0X £ 0 2 S ... and such that,for each k,

(i) 0* is a polygonal arc representation of

G[{e1} et', ex", e2, e2', e2", ..., ek, ek', ek"}]

with the same vertex set as F;

(ii) for each j < k, the union of the arcs representing e/, e/' is a closed Jordancurve whose interior contains the arc representing ei and nothing else of thegraph;


(iii) for each vertex p of ®k, the intersection of D(p, e(p)) and the point set of ©fc

either consists of p only, or it consists of straight line segments radiatingfrom p;

(iv) the subgraphs of @fc (respectively F) representing G[{eu e2, ..., ek}] arestrongly equivalent.

The graph 0 O is defined as the graph with vertex set F(F) and with empty edge set.Having defined ©fc_! (k ^ 1), we define 0fc as follows: The subgraph © j ^ of 0fc_1

representing G[{el5 e2, •••5efc_1}] is equivalent to the corresponding subgraph of F.Hence, by (4), we can add to ©£_! a polygonal arc representing ek such that theresulting graph 0fc' is strongly equivalent to the subgraph of F representingG[{eu e2, ..., ek}]. If necessary, we can modify the new arc such that it does notintersect any of the arcs representing edges e/, e/' (J ^ k— 1), and we can furthermodify it so that condition (iii) is satisfied. Then we can add polygonal arcsrepresenting ek, ek" such that (ii) is satisfied. The resulting plane multigraph 0 k

00

satisfies conditions (i)—(iv). Now the multigraph ©' = U 0fc is a representation offc = O

G', and it contains a representation of G with the desired properties.

LEMMA 2. Let F be an infinite, locally finite, connected, \AV-free plane graphsuch that each edge is a polygonal arc and no point of an edge is an EAP of F. Then Tis a spanning subgraph of a polygonal arc triangulation.

Proof. Let G denote the abstract graph represented by F and let G' be themultigraph obtained from G by adding, for every edge e of G, two new edges e', e"with the same ends as e. Since F*, the point set of F, contains no EAP of F, we canextend F to a polygonal arc representation F ' of G'. If e is an edge of G', we denote bye the arc of F ' representing e. We can assume that for each edge e of G, the interiorof e' u e" (which we denote by ft(e)) contains e (except for the ends) and nothingelse of F', and to each vertex p of F ' we associate a positive real number E(J>) such thatD{p, e(p)) n D(q, e(q)) = 0 for any two distinct vertices p, q, and such thatD(p,e{p))n(Y')* consists of straight line segments radiating from p for eachvertex p of T'. Denote by Q the union of the sets D(p, £(p)), where p e V(T), and thesets Q(e), where eeE(G). The graph T has the property that if x and y are twonon-adjacent vertices with a common neighbour z and the edges joining z to x and j ;are consecutive in the clockwise ordering of the edges incident with z, then we canadd to F a polygonal arc representing the edge e = xy such that e £ Q. Any planegraph obtained from F by adding finitely many edges in Q satisfies the assumption ofLemma 2.

We now show that, for any vertex z of F, we can add finitely many edges in fiso as to obtain a plane graph 0 with the property that z is contained in the interiorof some cycle of 0 . Since F is connected, z has degree at least 1, and if z has degree 1,we can add an edge in ft joining z to a vertex of graphic distance two from z. So wecan assume that z has degree at least two. We can also assume that the ends(distinct from z) of any two edges which are incident with z and consecutive in theclockwise ordering of the edges incident with z are adjacent. So if z has degree atleast 3 in F, then F contains a cycle E containing the vertices adjacent to z and noother vertex. If z is in the interior of this cycle, we have finished, so assume the


opposite. Then F contains two adjacent vertices, x and y say, each of which isadjacent to z such that each vertex adjacent to z (other than y and x) is in theinterior of the 3-cycle zxyz. This is also true if z has degree 2. Since G is infinite,locally finite, and connected, it has, by Konig's Lemma, a 1-way infinite pathstarting at any vertex. In particular, F has a 1-way infinite path starting at z. Thispath contains either x or y (or both) since there are only finitely many vertices in theinterior of the 3-cycle xyzx. Therefore F has a 1-way infinite path starting at x or y(say y) and containing none of x, z. Suppose without loss of generality that the interiorof the cycle xyzx is on the left side when we traverse the cycle in the order indicated.Let yt be the vertex adjacent to y such that the edge yyt succeeds the edge yz in theclockwise ordering of the edges adjacent to y. Clearly yx # z. Also yx ^ x, for ify^ = x, then every 1-way infinite path starting at y must contain x. If x and yt arenon-adjacent in F, we can add an edge (in Q) joining y± and x such that z is in theinterior of the cycle xy^ yx. So assume that yt and x are adjacent and that z is in theexterior of the cycle xyyt x. Let y2 be the vertex adjacent to yt such that the edge

succeeds yt y in the clockwise ordering of the edges incident with yv As in theprevious case we conclude that y2 is distinct fromz, y and x, and also we may assumethat x is adjacent to y2 and that z is in the exterior of the cycle xyyx y2 x. Wecontinue like this defining the vertices y3, j>4, .... Since x has finite degree, there isa (smallest) k such that x is not adjacent to yk. Then we can add in Q an edge joiningyk and x such that z is in the interior of the cycle 0: xyy1y2 ... ykx.

We can now complete the proof of the lemma. Let {xlt x2,...} be the vertex setof F. By adding finitely many edges (in Q) to F we obtain a graph containing a cycle0 such that xt is in the interior of 0. The subgraph of the resulting graphconsisting of 0 and the vertices and edges in the interior of 0 is a finite, plane graph,so by adding finitely many edges in the interior of 0, we obtain a graph I \ containinga cycle x¥l whose vertices are precisely the neighbours of xx and whose interiorcontains xx and no other vertex of Fv The plane graph I \ has the same vertex set asF and no point of I \* is an EAP of Tv Hence we can repeat the argument with I \instead of F and x2 instead of xt. We then obtain a graph F2 and a cycle *F2 of F2

such that x2 and no other vertex of T2 is in the interior of x¥2. Since F2 has the samevertex set as Fx and F, none of the edges added to I \ in order to obtain F2 intersectsthe interior of Yj. We continue like this defining F3, F4, ..., and finally the union

oo

(J F,- is a polygonal arc triangulation containing F as a subgraph.t = l

In order to represent an infinite triangulation by a straight line triangulation, weneed the following result about finite triangulations:

PROPOSITION 4. Let T be a finite polygonal arc triangulation such that the boundaryof the unbounded region is the cycle 0 O : xoyozo x0, and suppose that T has a regionwhose boundary is a cycle Qt: x1y1z1x1 disjoint from 0O. Let To, 7\ be disjointtriangles of the plane with corners Pi,qh rjor i = 0, 1, such that Tt is contained in theinterior of To and such that each straight line segment connecting a corner of To with acorner of Tt has only its ends in common with To u Tv Let e be any positive real number.Then there is a straight line triangulation A which is strongly equivalent to F suchthat xx, yt andzv are represented by pu qt and rx respectively (in particular, 0X isrepresented by Tx) and 0O is represented by To, and such that each vertex of A other thanpu q1 and rx has Euclidean distance less than e from one of p0) q0, r0.


Proof. We prove the proposition by induction on n, the number of vertices of F.If n = 6, then F is strongly equivalent to the first or fourth triangulation of [11;Figure 3], and the statement is easily verified in this case. So assume that n > 6 andthat the proposition has been verified for triangulations with fewer than n vertices.

If F contains a separating 3-cycle 0 2 whose interior does not intersect 0 1 } then wedenote by F ' the triangulation obtained from F by deleting the interior of 0 2 . By theinduction hypothesis, F ' has a straight line representation A' with the propertiesdescribed in the proposition. In A' at least one of the vertices of 0 2 , say p2, is distinctfrom plt qu rx and has therefore Euclidean distance e' < e from one of p0) q0, r0.We can now extend A' to a straight line representation A with the desired propertiesby representing the interior of 0 2 in such a way that each vertex in A but not in A'has Euclidean distance less than e—s' from p2.

If F contains an edge e not in 0 O and not incident with any of xu yu zu such that eis not contained in any separating 3-cycle of F, then we define F ' as the triangulationobtained from F by contracting e. By the induction hypothesis, F ' has a straight linerepresentation A' with the properties described in the proposition, and by (6), we cantransform A' into a triangulation A with the desired properties.

So in order to conclude the proof it suffices to prove that at least one of the twocases described above occurs. Suppose therefore the opposite. Since Qi is theboundary of a region of F, it is not a separating cycle (by (5)), and hence F contains avertex x' which is distinct from x0, yo,zo, xt, yx,zy and which is adjacent to one of*o> o> ZOJ saY to x0. By assumption, the edge x0 x' is contained in a separating 3-cycleXQ X' X"X0 whose interior contains the interior of 0X. Let eu e2, ..., ek be the edgesincident with x0 in clockwise order, and suppose without loss of generality that

*o yo = eu *ozo = ek> *o x> = ei> *o x" = ej

where 1 <i <j ^k. Then e2 joins x0 to a vertex distinct from each of jq, yltzlt ande2 is by assumption an edge of a separating 3-cycle whose interior contains theinterior of Qx. So without loss of generality, we can assume i = 2. Then F containsthe 3-cycle y0 *o x'JV By assumption, the edge y0 x' is an edge of a separating 3-cycle0 ' whose interior contains the interior of 0 t . Then the interior of 0 ' must alsocontain the interior of the 3-cycle x0 x'x"x0, which implies that 0 ' contains the vertexx0. But there is only one 3-cycle of F containing the vertices x0, y0) x', namelyx0 y0 x'x0 and this 3-cycle is the boundary of a region of F and therefore not aseparating cycle. So we have reached a contradiction, and the proof is complete.

We now state and prove the main result of this section:

THEOREM 3. Let F be an infinite, locally finite, connected, YAV-free plane graph.Then there exists an infinite straight line triangulation A of the plane such that F isisomorphic to a subgraph of A.

Proof. By Lemmas 1 and 2, we can assume that F is an infinite polygonal arctriangulation, and we prove that there is a straight line triangulation A isomorphic toF. By (8), F contains a sequence of mutually disjoint cycles 0 l s 0 2 , 0 3 , . . . such that,for each k, the interior of 0fc and 0fc together form a finite, plane graph containing0fc_! and triangulating the interior of 0ft. For each k ^ \ , let Fk denote thesubgraph of F induced by the vertices of 03fc, 0 3 k -3 and the vertices in the interior of03fc and the exterior of 03*_3. We transform Tk into a plane graph Tk' as follows: Let


®3/t-i • viv2... vmvi. If the edge yx v2 is not contained in a separating 3-cycle of Tk

whose interior contains the interior of 03fc_2, then we delete the interior of everyseparating 3-cycle (if any) of Ffc containing the edge vt v2, and then we contract theedge v± v2 (in such a way that only the edges incident with v2 are affected). Werepeat this with the edge vx v3 instead of vt v2. In a finite number of steps wetransform Ffc into a graph Tk which has a separating 3-cycle 0 ^ - ! whose interiorcontains the interior of 03fc_2. All edges of Tk which are not incident with a vertexof 03fc_i are also present in Tk'. In particular, 0 3 k and 03fc_3 remain unchanged, so

00

the union \J Ffc' is an infinite triangulation F \ Using Proposition 4, we get a straightfc=i

line representation of F ' as follows: let Tk denote the triangle with corners (0, k),(k,-k) and (-k, -k). Let Tk" denote the subgraph of F ' induced by the vertices of0'3fe_i and 03ft_4 and all vertices in the interior of 0 ^ - ! and the exterior of 03fc_4.(Note that 03fc_i and 03fe_4 are disjoint). We represent r \ " such that 0 2 ' isrepresented by Tt and all other vertices of I Y ' are in the interior of 7\. UsingProposition 4 we extend this to a straight line representation of F / ' u F 2 " such that0 5 ' is represented by T2. Continuing like this, we represent F 3 " , F 4 " , . . . . The unionof the resulting straight line triangulations is a straight line triangulation A' of theplane, and A' is isomorphic to F' . For k = 0 (mod 3), let Ek denote the cycle of A'corresponding to 0fc in F', and let Ak denote the subgraph of A' induced by thevertices of 33fc, E3ft_ 3 and the vertices in the interior of S3fc and the exterior of S3fc_ 3.Then Afc' is isomorphic (and in fact strongly equivalent) to Ffc'. By splitting verticesof Ak and adding finite triangulations to Ak, we obtain in a finite number of steps astraight line representation Afc of Tk. Since the cycles H3fc and E3ft_3 remain

00

unchanged, the union (J Afc is a plane graph A. Moreover, A is a straight line

triangulation of the plane, and A and F are isomorphic.The proof is complete.

Since an infinite triangulation of the plane has no EAP, we get the following result:

COROLLARY 1. Let The an infinite, locally finite, connected, YAP-free plane graph.Then F has an EAP-free straight line representation.

The following example shows that a locally finite, disconnected, VAP-free planegraph need not have an EAP-free representation:

Let Gl5 G2 be abstract graphs such that Gt is isomorphic to an infinite triangulationand G2 is an infinite, connected graph having a VAP-free representation. By thetheorem of Halin [3], the disjoint union H = G1\JG2 has a VAP-free representation(and by the proof of Theorem 4 below, it has a VAP-free straight line representation).But if 0 is any representation of H such that each of Glt G2 is represented by VAP-free graphs F l 5 F2 , then for any vertexp of Tx and any vertexq of F2 , the line segment[p, q] intersects infinitely many edges of Tt because p is contained in the interior ofinfinitely many (mutually disjoint) cycles of F l 9 by (8), and F 2 * is not containedin the interior of any such cycle. Hence some point of [p,q] is an EAP of I \ .

5. Straight line representations of VAP- or EAP-free plane graphs

THEOREM 4. Let 0 be a locally finite WAV-free plane graph. Then 0 has a WAV-free straight line representation.


Proof. The graph 0 has at most countably many components. Let F be one ofthem. It is sufficient to show that F has a VAP-free straight line representation suchthat F* £ {(a} p)eR2\ - 1 < a < 1}. By Lemmas 1, 2, we may assume that F is atriangulation. We repeat the proof of Theorem 3. We define &k, Tk, 03fc_ls Tk, I",Tk" as in the proof of Theorem 3. But we define triangles Tk as follows: Tk has corners(0, /?ft), (<xki yk), (-af t, yk) where /?i = 1, ax =\, yx = 0 and such that, for eachk > 2

and each straight line segment joining a corner of Tk with a corner of Tk_t has onlythe ends in common with Tk^tuTk. It is easy to see that such numbersa*> Pk> fkik ^ 1) e x i s t - The set of corners of the triangles Tk has no accumulation point.We represent the graphs Tk" as in the proof of Theorem 3. However, care must beexercised to avoid VAPs with first coordinate ± 1. By Proposition 4, we can assumethat each vertex of the resulting straight line triangulation A' has Euclidean distanceless than 1 from one of the corners of some Tk. This ensures that A' is VAP-free.The graphs Ak are defined as in the proof of Theorem 3, and as in this proof wetransform Afc' into a straight line representation Afc of Fft. By (6) and (7), we canassume that each vertex if Ak has Euclidean distance less than 1 from some vertex

00

of Ak. This ensures that A = (J Afc is a VAP-free straight line triangulation

isomorphic to F. The proof is complete.

THEOREM 5. Let 0 be an EAP-free, plane graph. Then 0 has an EAP-freestraight line representation.

Proof. Since every bounded set of the plane intersects only finitely many edges of0 it is easy to see that 0 has an EAP-free polygonal arc representation 0 ' . Let 0 "denote the union of the infinite components of 0 ' . If 0 " = 0, there is no problem,so assume that 0 " # 0. We prove that 0 " is isomorphic to a subgraph of a straightline triangulation A of the plane. This will prove the theorem, since distinct finitecomponents of 0 ' then can be represented in distinct regions of A. So by Theorem 3,all we need to prove is that 0 " is a spanning subgraph of a connected, locallyfinite plane graph r . We define a sequence of real numbers als a2,... as follows:let ax be the distance from some point of (0")* to the origin 0. Having defined ak,we define ock+l such that afc+1 > afc+l and no edge of 0 " intersecting C(O, ak) hasan end in common with an edge intersecting C(O, afe+1). Since only finitely manyedges intersect C(O, ctk) and since 0 " is locally finite, ak+i exists. Since eachcomponent of 0 " is infinite, it is unbounded, and hence for each k, (0")*intersects C(O, ak) and partitions this circle into a finite number of segments. For eachsegment joining two edges eu e2 belonging to distinct components of 0 " , we selectan end of et and an end of e2 and draw a new edge between these two vertices (weleave the details to the reader). The resulting graph F is locally finite, by choice of thesequence a1} a2,..., and it is connected because each component of 0 " intersectssome C(O, <xk).

References

1. A. Adler, "/"-Planar Graphs ", / . Combinatorial Theory B,15 (1973), 207-210.2. G. A. Dirac and S. Schuster, " A theorem of Kuratowski", Indag. Math., 16 (1954), 343-348.3. I. Fary, " On straight line representation of planar graphs ", Ada Sci. Math. Szeged, 11 (1948),

229-233.


4. R. Halin, " Zur haufungspunktfreien Darstellung abzahlbarer Graphen in der Ebene ", Arch.Math., 17 (1966), 239-243.

5. R. Halin, " Zerlegungssatz fiir unendliche Graphen und seine Anwendung auf Homomorphie-basen ", Math. Nachr., 33 (1967), 91-105.

6. F. Harary, Graph Theory (Addison Wesley, Reading, Mass., 1969).7. D. Konig, Theorie der endlichen und unendlichen Graphen (Leipzig, 1963).8. K. Kuratowski, " Sur le probleme des courbes gauches en topologie ", Fund. Math., 15 (1930),

271-283.9. O. Ore, The Four Colour Problem (Academic Press, New York, 1967).

10. S. K. Stein, " Convex maps ", Proc. Amer. Math. Soc, 2 (1951), 464-466.11. W. T. Tutte," Convex representations of graphs ", Proc. London Math. Soc, 10 (1960), 304-320.12. W. T. Tutte, " How to draw a graph ", Proc. London Math. Soc, 13 (1963), 743-768.13. W. T. Tutte, " A census of planar triangulations ", Canad. J. Math., 14 (1962), 21-38.14. K. Wagner, " Bemerkungen zum Vierfarbenproblem ", Jber. Deutsch. Math. Ver., 46 (1936),

26-32.15. K. Wagner, " Fastplattbare Graphen ", / . Combinatorial Theory, 3 (1967), 326-365.

Matematisk Institut,Universitetsparken,

Ny Munkegade,8000 Aarhus C,

Denmark.

Straight Line Representations of Infinite Planar Graphs

Documents