6 Planar Graphs. 6.1 Euler's Formula

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1 6 Planar Graphs Roughly speaking, a graph is planar if and only if it can be drawn in the plane without any two of its edges crossing. More formally, an embedding of a graph G = (V; E) is a function f : V! R 2 together with a set of continuous curves C e : e 2 E such that f is one-to-one and the curve C e for e = fu; vg has endpoints f(u) and f(v). The graph G is planar if the curves C e : e 2 E intersect only at their endpoints { that is any point in the intersection of two of the curves is an endpoint of both of the curves. A drawing of G without crossings is called a plane embedding of G, or a plane graph. Thus a graph is planar if and only if it has a plane embedding. These notions can be made more rigorous using the terminology of topology and, more generally, we can talk about embeddings of a graph in a topological space. For example, a geometric realization of a graph is a topological space whose points are the vertices and such that the edges are homeomorphic to the closed interval [0; 1]. For details, the reader is referred to the survey of topological graph theory by Archdeacon. The main theorem of this section, which we will prove, is a necessary and sucient condition for a graph to be planar { and a characterization of planar graphs. Recall that a subdivision of a graph G is any graph obtained from G by replacing each edge of G with a path with the same endpoints as the edge, such that paths may meet only at their endpoints. Theorem (Kuratowski's Theorem) A graph is planar if and only if it contains no subdivision of K 5 K 3;3. and no subdivision of The proof of this theorem is in Section Euler's Formula Throughout this section, we deal only with connected graphs. If G is a plane graph, then R 2 ng consists of a union of disjoint connected plane regions, which are called faces of G. The of a face F of G is the set of points in the closure of F which are not in the interior of F. Each plane graph has a face which is innite, which we refer to as the innite face. The boundary walk of F, which we also denote is the closed walk consisting of edges and vertices in the boundary of F. We denote by F (G) the set of faces of a plane graph G. The degree of a face F 2 F (G) is the length of the and denoted deg(f ).

2 Figure 16 The graph on the left in Figure 16 has six faces, all boundary walks of which are cycles of length four { so every face has degree four. The tree in the centre has only one face { the innite face { and since a tree on n vertices has n 1 edges and the boundary walk goes through each edge twice, the degree of the innite face is 2(n 1). In the graph on the right, there are two faces, one of degree six and one of degree ten. The degrees of the faces in a plane graph depend very much on the way the graph is drawn in the plane: for example, the graph on the right in Figure 16 can be redrawn as a new plane graph by ipping one of the bridges into the innite face, thereby producing two new faces, both of degree eight. In Section 1.2, we will come back to the issue of uniqueness of the face degrees. There is a very useful analog of the handshaking lemma for face degrees in a plane graph. If we add up the degree of every face F 2 F (G), we observe that every edge of the graph is counted exactly twice. This is true since an edge in a cycle is counted once for each of the faces on either side of it, and an edge which is not in a cycle is a bridge (Lemma 1.7.1), and therefore counted twice in one boundary walk. These observations give the following useful fact: Theorem Let G be a plane graph. Then X F 2F (G) deg(f ) = 2jE(G)j: For the examples in Figure 16, one checks that the theorem holds. In general, note the a bridge on the boundary of a face is counted twice in the boundary walk of that face, whereas all other edges in the boundary are counted once. Theorem is very useful in conjunction with Euler's Formula and the handshaking lemma for proving non-existence of planar graphs with given face and vertex degrees. Euler's Formula relates jf (G)j, je(g)j and jv (G)j as follows:

3 Theorem (Euler's Formula) Let G be a connected plane graph. Then jv (G)j je(g)j + jf (G)j = 2: The proof of this theorem is quite straightforward: if one proceeds by induction on je(g)j, starting with trees, where je(g)j = jv (G)j 1, then the induction step goes as follows: for je(g)j > jv (G)j 1, if e 2 E(G), then by induction jv (G e)j je(g e)j + jf (G e)j = 2: Now since G is not a tree, we can choose e to be part of a cycle in G, and so e is not a bridge. Therefore G e is connected, and je(g e)j = je(g)j 1 and jf (G e)j = jf (G)j 1. It follows that jv (G)j (je(g)j 1) + (jf (G)j 1) = 2 and this gives Euler's Formula As an example of an application, it has been known for millenia that there are only platonic solids { these are the unique r-regular plane graphs in which all faces have degree s, where (r; s) 2 f(3; 3); (3; 4); (4; 3); (5; 3); (3; 5)g. For example, lets see why (r; s) = (4; 4) is not possible. By Theorem 6.1.1, we have 4jF (G)j = 2jE(G)j. By the handshaking lemma, 4jV (G)j = 2jE(G)j. By Euler's Formula 6.1.2, jv (G)j je(g)j + jf (G)j = 2. But the left hand side is zero, so this is a contradiction { so there is no 4-regular plane graph with all faces of degree four. The platonic solids are shown below. Another useful application is to give a sucient condition for non-planarity, rather than trying to draw the graph in every possible way: Theorem Let G be a planar graph containing a cycle. Then je(g)j g (jv (G)j 2), where g is the g 2 length of a shortest cycle in G. In particular, for any planar graph G, je(g)j 3jV (G)j 6, and therefore G is 5-degenerate. Proof. Since every face has degree at least g, Theore gives gjf (G)j 2jE(G)j. Putting this in Euler's Formula 6.1.2, we get jv (G)j je(g)j + 2 g je(g)j 2 which, rearranged, gives the required bound on je(g)j. The right side of the formula is maximized when g = 3, in which case we get je(g)j 3jV (G)j 6 for all planar graphs G. By the handshaking lemma, if all vertices of G had degree at least six, then je(g)j 3jV (G)j, a contradiction to what we just proved. So every planar graph has a

4 vertex of degree at most ve. Since every subgraph of G is also planar, this means that every subgraph of G has a vertex of degree at most ve, so G is 5-degenerate. By Theorem 6.1.3, any graph satisfying je(g)j > (jv (G)j 2) can't be planar. In particular, K 5 is not planar since je(k 5 )j = 10 and g = 3, and K 3;3 is not planar since je(k 3;3 )j = 9 and g = 4. If a graph is not planar, then neither is any subdivision of it. So in particular any subdivision of K 5 or K 3;3 is not planar, and Kuratowski's Theorem states that these are the two reasons for a graph not being planar: either it contains a subdivision of K 5, or a subdivision of K 3;3. We will prove this in Section 6.3. g g Colouring Planar Graphs Euler's Formula also can be applied to vertex-colouring of planar graphs. Recall that a graph is d-degenerate if every subgraph of G (including G itself) has minimum degree at most d. Also, any d-degenerate graph is (d + 1)-colourable, by Proposition By Theorem 6.1.3, every planar graph is 5-degenerate, so this means that every planar graph is 6-colourable. Here is another example: suppose we have a planar graph G of girth at least six. Then je(g)j 3 (jv (G)j 2) by Theorem 6.1.3, so every subgraph of G 2 must have a vertex of degree at most two, by the handshaking lemma. Therefore G is 2-degenerate, which means that G is 3-colourable. In fact, every planar graph is 5-colourable. We prove this theorem, since the ideas can be used later in Kuratowski's Theorem Theorem Every planar graph is 5-colourable. Proof.* Proceed by induction on jv (G)j. If jv (G)j 5, then the theorem is true: just assign all vertices dierent colours. Now suppose jv (G)j > 5. If G has a vertex v of degree at most four, then G v is 5-colourable by induction, and we can extend this colouring to v by assigning to v a colour which does not appear on any of its neighbours, since there were ve colours but at most four neighbours of v. So now we assume G has no vertex of degree at most four. Since G is 5-degenerate, by Theorem 6.1.3, G has a vertex v of degree ve. If all neighbours of v are adjacent to each other, then they form a K 5, but K 5 is not planar, so that is a contradiction. Therefore we can pick neighbours a and b of v which are not adjacent. Now consider the graph H = G=fa; b; vg { that is, we contract a; b and v to a single vertex w. By induction, H is 5-colourable. For each u 2 V (G)nfa; b; vg, assign to u the colour it received in H. Assign a and b the colour that w received in H { we can do that since a and b are not adjacent. Finally, we've used four colours on the neighbours of v, since a and b got the same colour, so there is a fth colour which can be assigned to G. This proves that G is 5-colourable.

5 Perhaps the most famous theorem in all of graph theory is the 4-colour theorem, proved by Appel and Haken (1976): every planar graph is 4-colourable. Unfortunately, there is no proof known which is not computer assisted. The shortest proof is currently the one in Robertson and Seymour (1997). Theorem (4-Colour Theorem) Every planar graph is 4-colourable. 6.3 Drawing Planar Graphs* We remarked earlier that there are many plane embeddings for a given planar graph G; even the degrees of the faces can change with dierent embeddings. In fact, we can go from any plane embedding of G to any other plane embedding of G using the notion of Stereographic Projection. In particular, we can make any face of a plane embedding the innite face: Proposition Every face of a plane embedding G 0 of a graph G is the innite face of some plane embedding of G. Furthermore, if every edge of G 0 is a straight line, then we can ensure that every edge of the new embedding is also a straight line. Proof.* Let S denote a sphere of diameter one placed so that the xy-plane is tangent to S at the origin. Then wrap the plane embedding G 0 of G around the sphere. Formally, consider the function f which maps a point (x; y) to the point (x; y; z) 2 S which is at height 1 1=(1 + x 2 + y 2 ) in the plane dened by the line through the origin and (x; y) and the z-axis. Note that f is a bijection between R 2 and SnN, where N = (0; 0; 1) denotes the north pole of S. Let H be the image of G 0 under f. Keeping H xed, rotate the sphere until some face F of H contains the north pole of S. Now apply f 1 to get an embedding of G, namely f 1 (H), with the property that the face F of f 1 (H) is the innite face. The second statement of the theorem is left as an excercise. Figure 17 : Stereographic projection

6 Perhaps the most natural embedding would be to try to draw the edges as striaght lines. This can be done, by the following theorem: Theorem (Fary's Theorem) Every simple planar graph has a plane embedding in which all edges are straight line segments. Proof. By Theorem 6.1.3, every planar graph is 5-degenerate. Now we proceed by induction on jv (G)j, the number of vertices in a planar graph G. If jv (G)j 3, then the result is obvious. Suppose G is a planar graph with jv (G)j > 3. We may assume that G is maximal planar { so G + e is not planar anymore for any edge e. Then if G 0 is a plane embedding of G, all faces of G 0 have degree three, otherwise we could add a diagonal edge in some face. Now let v be a vertex of degree at most ve in G 0. Then G 0 v has a plane embedding, call it H, such that all the edges are straight lines. Let v 1 ; v 2 ; : : : ; v k be the neighbours of v, where k 5. Then there is a cycle C H such that V (C) = fv 1 ; v 2 ; : : : ; v k g { since every face of G 0 is of degree three, every face of H containing non-neighbours of v on its boundary is a triangle. This means that C is the boundary of a face of H. By Proposition 6.3.1, we can make C the boundary of a nite face of H. Now place v in the interior of C so that v sees all vertices of C { that is, we can draw a straight line segment from v to each vertex v 1 ; v 2 ; : : : ; v k. This is an embedding of G in which all edges are straight line segments. Concerning properties of the drawing of a planar graph, we have seen that there are, in general, plane embeddings with dierent face degrees (Figure 16). Furthermore, we can't ensure that the faces are convex, even for 2-connected planar graphs, for example K 2;3 has no embedding in which all faces are convex. However, Tutte showed that every 3-connected planar graph can be drawn with convex faces and straight line edges, and Whitney's Theorem states that the embedding is unique. Theorem (Tutte-Whitney Theorem) Every 3-connected planar graph has a unique embedding in the plane in which every face is a convex polygon and every edge is a straight line segment. A natural physical interpretation is to nail down the edges of a cycle which is a face in a plane embedding of G, and replace the edges with rubber bands. Then, allowing this dynamical system to reach equilibrium in terms of the laws of physics, Tutte proved that the plane embedding at equilibrium is a convex straight line embedding. We do not prove this or Theorem here.

7 6.4 Duality* Let G be a plane graph, and let G denote the pseudograph obtained by placing a vertex v f in the interior of each face f 2 F (G) and whose edges are dened as follows: (1) for each bridge on the boundary of a face f 2 F (G), join v f to v f with a loop in G passing through the bridge. (2) for each edge e 2 E(G) on the boundary of distinct faces f; g 2 F (G), join v f and v g by an edge in G which crosses e. Then G is referred to as the plane (or combinatorial) dual of G. Examples of duals are shown in Figure 18: Figure 19 : Duality There are many uses of duality in planar graphs, but for brevity we mention one example in colouring. The map colouring problem is to colour the faces of a plane graph in such a way that whenever two faces share an edge, they have dierent colours. Now by drawing the dual of a planar graph, we see that the map colouring problem on a plane graph is equivalent to the vertex colouring problem in the dual, except that we have to remove all loops in the dual. By the 4-colour theorem, this means that the regions of any map can be coloured in four colours in such a way that adjacent regions have dierent colours. In fact, even more is true: if we want to prove the 4-colour theorem for plane graphs, it is sucient to consider maximal plane graphs on at least three vertices(i.e. if we add any edge we get a non-planar graph). In a maximal plane graph, all faces are bounded by triangles (exercise), and therefore the dual of a maximal plane graph is a cubic graph. It can also be checked that every maximal plane graph, except a triangle, is 3-connected, and that the dual is therefore also 3-connected. The oldest approach to the 4-colour theorem is to try to prove that every cubic graph is 3-edge-colourable: in fact this is equivalent to the 4-colour theorem. Theorem Every planar graph is 4-colourable if and only if every cubic planar 3-connected graph has edge-chromatic number three. Proof. Let G be a planar graph and let G 0 be a plane embedding of G. Then G 0 is contained in a maximal plane graph G 1. If every planar graph is 4-colourable, then G 1 is

8 4-colourable which means that the map G 1 is 4-face-colourable and cubic. Since G 1 is 3- connected, no edge of G 1 is a bridge so every edge of G 1 is on the boundary of exactly two faces. Now assign edge-colour 1 to those edges of G 1 on the boundary of faces of colour 1 and 2, or colour 3 and 4, assign edge-colour 2 to those edges of G 1 on the boundary of faces of colours 1 and 3, or colours 2 and 4, and assign edge-colour 3 to all remaining edges of G. One checks that this is a proper 3-edge-colouring of G, as required. Dene G; G 0 ; G 1 ; G 1 as in the rst part of the proof. If every cubic planar graph is 3-edgecolourable, then G 1 has a proper 3-edge-colouring, with colours 1, 2 and 3. That is to say that G 1 = M 1 [ M 2 [ M 3 where M i is the perfect matching consisting of edges of colour i. Then H 1 = M 1 [ M 2 is a plane graph and H 2 = M 1 [ M 3 is a plane graph. Colour the faces of H 1 with colours 1 and 2, and colour the faces of H 2 with colours 1' and 2'. To get a colouring of the faces of G 1, and hence a colour of G, colour a face F with colour (i; j 0 ) if it is contained in a region of colour i in H 1 and a region of colour j 0 in H 2. Then the number of colours we used is four, and one checks that this a proper colouring of the faces of G 1. Tait conjectured that all 3-connected cubic planar graphs are hamiltonian { i.e. contain a spanning cycle { but this is false, as a counterexample of Tutte on 46 vertices showed. If Tait's conjecture had been true, then we could colour the hamiltonian cycle red and blue, and the remaining matching with green to get a proper 3-colouring of every cubic 3-connected graph. 6.5 Kuratowski's Theorem In this section, we'll give a proof of Kuratowski's Theorem Before doing so, we dene -graphs: a -graph is any graph consisting of the union of three internally disjoint paths between two points. Recall that a block of a graph is a maximal 2-connected subgraph of G, or a bridge of G. To prove Kuratowski's Theorem 6.0.1, we need the following decomoposition theorem for graphs with no -subgraphs: Proposition Let G be a connected graph containing no -graph. Then every block of G is a cycle or K 2 and G is a tree of cycles and K 2 s, as shown in Figure 19 below. Proof.* Let B be a block of G. If B 6= K 2, then B has an ear-decomposition, by Theorem If B is not a cycle, then we can nd a cycle C B and an ear P B internally disjoint from C such that P [ C is a -graph, a contradiction. So every block is a cycle. We prove the second statement by induction on the number of blocks in G. It is clear if G has only one block, since then G is a cycle or K 2. Suppose G has at least two blocks, and let v be a cutvertex of G. Then G = G 1 [ G 2 where G 1 and G 2 share only the vertex v. By induction, G 1 and G 2 are trees of cycles and K 2 s, and joining G 1 and G 2 at v, we get that G itself is a tree of cycles and K 2 s.

9 Figure 19 : Tree of cycles and K 2 s In the chapter on connectivity, we saw that it is true that every connected graph has a \tree of blocks" structure { instead of cycles and K 2 s, we allow any pattern of blocks intersecting in a tree like manner. An endblock of a graph is a block of G containing at most one cutvertex of G. In Proposition 6.5.1, these correspond to the leaves of the tree in the tree structure of blocks. In Figure 19, there are exactly fourteen endblocks. A nal bit of notation is the following: if C is a cycle in a plane graph G, then int(c) denotes the set of vertices of G inside C, and ext(c) denotes the set of vertices of G outside C. Proof.* We already saw that if a graph contains a subdivision of K 5 or K 3;3, then it is not planar. Now we prove that if G is not planar, then G contains a subdivision of K 5 or K 3;3. Let G be a minimal non-planar graph which contains no subdivision of K 5 or K 3;3. This means that every proper subgraph of G is planar. We may assume G is connected. Also we assume jv (G)j > 6 { this is a matter of checking some cases. Part 1 (G) 3. Suppose, for a contradiction, that v 2 V (G) has degree less than three. Contract an edge fu; vg 2 E(G). Then G=fu; vg is planar, by minimality of G, and since v has degree at most two, it follows easily that G is planar. Therefore (G) 3. Part 2 For e 2 E(G), G(e) = G=e fz e g contains no -graph, where e is contracted to z e. Suppose T G=e fz e g is a -graph. Now G=e has a plane embedding, H, by minimality of G. Now T H fz e g splits the plane into three connected regions. So by Proposition 6.3.1, we can assume z e 2 int(c) and ext(c) is non-empty, for some cycle C H fz e g. By minimality of G, G ext(c) has a plane embedding, G 1 (see Figure 20). By Proposition 6.3.1, we can ensure C is on the boundary of the innite face in G 1. Now combine the plane graphs H int(c) (dotted in Figure 20) and G 1 to get a plane embedding of G.

10 Figure 20 : Plane embedding of G 1 Part 3 For e 2 E(G), G(e) has at most one vertex of degree one. Since (G) 3, by Part 1, G(e) has no isolated vertices and if G(e) has two leaves, u and v, then u and v are both adjacent to both endpoints of e. So u and v have degree exactly three. If e = fx; yg, then the edges fx; yg; fu; xg; fu; yg; fv; xg; fv; yg form a -graph in G. By Part 2, this implies G fu; v; x; yg has no edges, and so the neighbourhood of every vertex of G fu; v; x; yg is a subset of fu; v; x; yg. But u and v have degree exactly three, so there can be at most two vertices in G fu; v; x; yg, and G has at most six vertices, a contradiction to jv (G)j > 6 (see Figure 21). Figure 21 : Vertices of degree one in G(e) Now we complete the proof. By Part 2 and Proposition 6.5.1, every block of G(e) fz e g is K 2 or a cycle, for any e 2 E(G). By Part 3, G(e) has at most one vertex of degree one, so some endblock of G(e) is a cycle, C. Let fu; vg 2 E(C), where neither u nor v is a cutvertex of G(e). Then u and v have degree two in G(e), so they are each adjacent to an endpoint of e. Therefore the subgraph of G induced by V (C) [ e contains a -graph. By Part 2, this means no edge of G(e) is vertex-disjoint from C. By Part 3, this means G(e) = C [ fw; ag for some a 62 V (C), or G(e) = C. Now since jv (G)j > 6, there is

11 a path P C fwg with jv (P )j 3 (see Figure 22). The subgraph of G induced by V (P ) [ e contains a -graph, by the same argument as above. But if G(e) = C [ fw; ag, then fw; ag is disjoint from this -graph since jv (C)j 4, and if G(e) = C, some edge of C is disjoint from this -graph since jv (C)j 5, contradicting Part 2. This completes the proof. 6.6 Graphs on Surfaces Figure 22 : The path P C In this section, we study drawing graphs on general surfaces without crossings. Rather than assume background in general topology, we will dene everything in elementary terms, keeping in mind that everything can be made rigorous through topology. The surfaces we look at will all be orientable: these are closed surfaces which consist of a sphere with a nite number of handles (or tube) attached. For example, the torus consists in attaching one handle to the sphere, and the double torus consists in attaching two handles to the sphere. The number of handles attached to S is called the genus of S, and denoted (S). The Euler characteristic of S is (S), dened by 2 2(S). For example, the Euler characteristic of the sphere is two, whereas the euler characteristic of the torus is zero. Let G be a graph. An embedding of G in S is a map f : V (G)! S together with a continuous curve on S joining f(u) to f(v) whenever fu; vg 2 E(G). For example, an embedding of K 5 on the torus is shown below. It is convenient to call a graph toroidal if it can be embedded on the torus without crossings. Figure 23 : Toroidal embedding of K 5.

12 The denition of faces in a drawing without crossings, and their boundaries and degrees, are the same as for plane graphs. For example, in the toroidal embedding of K 5 in Figure 23, there are ve faces, four having degree three and one having degree four. It is a good exercise to try to embed other graphs, for example K 4;4, in the torus. The genus of a graph G, denoted (G), is the minimum possible value of such that G embeds without crossings in a surface of genus. So (K 5 ) = 1. The Euler characteristic of G is (G) = 2 2(G) )not to be confused with the chromatic number). The Euler- Poincare formula is an analog of Euler's Formula for surfaces, and it says that the number of faces in any embedding of a graph on a surface of characteristic does not depend on the embedding: Theorem (Euler-Poincare Formula) Let G be a graph of Euler characteristic embedded without crossings in a surface S of characteristic, then jv (G)j je(g)j + jf (G)j = : The proof of this theorem can be achieved by induction on e. Actually it is convenient rather to prove that jv (G)j je(g)j jf (G)j = (G) + c(g) 1 where c(g) is the number of components of G. For instance, if G has a bridge e, then (G) = (G e) and jf (G)j = jf (G e)j and c(g e) = c(g) + 1 so the induction step works. Now suppose G has no bridges. Then for an edge e 2 E(G) such that (G e) = (G), we have jf (G e)j = jf (G)j 1 and again the induction step works. Finally if (G e) = (G) 1, then e must have been the only edge on some handle of S. Then one veries that jf (G e)j = jf (G)j + 1, so the induction step works. The details are left to the reader. The formula is called the Euler-Poncare Formula 6.6.1, since Poincare gave a topological generalization of it, where edges are replaced by simplices. The rst proof of the Euler-Poincare Formula for general was given by Lhuilier. A natural consequence of the Euler-Poincare Formula is that je(g)j 3jV (G)j 3 whenever G is a graph which can be embedding in a surface of characteristic without crossings. This shows, for example, that K 8 cannot be embedding on the torus without crossings, since je(k 8 )j = 8 2 = 28 whereas 3jV (K8 )j 3 = 24. Another consequence is that the degeneracy of a graph G embedded in S without crossings satises d(g) b6 6 jv (G)j c: So we can extend our result about planar graphs being 6-colourable to higher genus surfaces.

13 Theorem (Heawood's Map-Colouring Theorem) The chromatic number of a graph embedding without crossings in a surface of characteristic 0 is at most h() = b 1 2 (7 + p 49 24)c Furthermore, if G is a minimal h()-chromatic graph drawn on a surface of characteristic 6= 2, then G = K h(). Proof. First we prove the upper bound h() b 1(7+p 49 24)c. Let G be embedded 2 on a surface S of characteristic without crossings. We may assume that G is a minimal k-chromatic graph on S, where k = h(). By Proposition (or by just deleting a vertex of small degree) it follows that (G) k 1. Therefore je(g)j k 1 jv (G)j, by 2 the handshaking lemma. On the other hand je(g)j 3jV (G)j 3, by the Euler-Poincare Formula Therefore which is the same as k 1 jv (G)j 3jV (G)j 2 (k 7)jV (G)j + 6 0: Since we assumed 0, k 7 follows from the fact that K 7 can be embedded on the torus. Now jv (G)j k since (G) k 1, so k 2 7k + 6 0: This gives 2k 7 p 49 24, and we clearly must take the positive square root. This proves the upper bound on h(). * The second part of the proof is to show G = K k. If G 6= K k then jv (G)j k + 2, and it is an excercise to prove that if jv (G)j = k + 2, then G consists of a pentagon disjoint from K k 3 together with all edges between the pentagon on the K k 3. In particular, je(g)j = k+2 5. This is greater than 3(jV (G)j ), contradicting the Euler-Poincare 2 Formula Therefore jv (G)j k + 3. Now (G) k 1 6, and by Brook's Theorem, G is not k 1 regular, otherwise it would be k 1 colourable. Therefore je(g)j > jv (G)j(k 1) : 2 This gives k 2 4k and so k 2 + p This is not possible unless = 2. Note that we do not include the case of plane graphs, where h() = 4. Surprisingly, this case also agrees with the above formula for h(). Also, we avoided = 2. The full classication, for all and even for all surfaces (not only orientable ones) was given by 3

14 Ringel and Youngs. It should be notes that better results can be obtained for a graph of girth g > 3: in that case, the Euler-Poincare Formula can be used to give je(g)j g (jv (G)j ) g 2 and then one can colour these graphs with fewer colours, by repeating the pattern of the proof of Theorem Characterization of Graph on Surfaces* We give the following quote from the book of Mohar and Thomassen: \Graphs on surfaces form a natural link between discrete and continuous mathematics. They enable us to understand both graphs and surfaces better. It would be dicult to prove the celebrated classication theorem for (compact) surfaces without the use of graphs. Map color problems are usually formulated and solved as problems concerning graphs." Which graphs are embed in a surface of genus without crossings? A consequence of one of the deepest theorems in mathematics, called the Graph Minors Theorem, due to Robertson and Seymour, is that for any, there exists a nite list of graphs H such that any graph which does not embed on a surface of genus contains a graph in H as a minor. Kuratowski's Theorem is such a result for planar graphs, where H 0 = fk 5 ; K 3;3 g. Unfortunately, for surfaces of higher genus, the known list is prohibitively long, and the smallest possible list is not known. The shortest proof that there is a nite list is due to Thomassen. A very natural question on embedding graphs is the following: in the last section, we gave embeddings of graphs on surfaces where some of the faces were a bit unorthodox: they are not homeomorphic to plane disks. While it is true that every 2-connected planar graph has an embedding where every face is homeomorphic to a disk, it is not known if every 2-connected graph can be embedded in some surface in such a way that all faces are homeomorphic to disks. This is known as the strong embeddability conjecture, and it remains open It has many other consequences in graph theory; for example, does every 2-connected graph contain a set of cycles such that every edge is in either one or two of the cycles? For planar graphs, this is clear: the boundaries of the faces will do, and then every edge is in exactly two cycles. The conjecture also has implications for the existence of nowhere zero 5-ows. For more information on graphs on surfaces, see Archdeacon or Mohar and Thomassen.

15 6.8 Exercises 1: A maximal plane graph is a plane graph G = (V; E) with n 3 vertices such that if we join any two non-adjacent vertices in G, we obtain a non-plane graph. (a) Draw a maximal plane graphs on six vertices. (b) Show that a maximal plane graph on n points has 3n 6 edges and 2n 4 faces. 2: Which of the graphs in Figure 1 is planar? Justify your answer. Figure 1 3: Show that every planar graph with no triangles is 3-degenerate, and 4-colourable. 4. (a) Prove that if G is a graph of girth g which can be embedded in a surface of characteristic without crossings, then je(g)j g(jv (G)j )=(g 2). (b) Use (a) to prove that the graph in Figure 2 is not toroidal. [2] Figure 2

16 5: Prove that K 5;5 does not embed on the torus, and give a drawing of K 4;4 on the torus. 6: (a) The crossing number of a graph G is the smallest number of pairs of edges which cross, taken over all drawings of G in the plane, where no three edges cross at a single point. Find the crossing number of K 5 and the crossing number of K 3;3. (b) Prove that the crossing number of K n;n is at most k 2: j n 2k 2 j n 1 2 [Hint: draw the vertices in one part of K n;n along the y-axis and the vertices in the other part of K n;n along the x-axis.]