5. GRAPHS ON SURFACES


 Verity Francis
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1 . GRPHS ON SURCS.. Graphs graph, by itself, is a combinatorial object rather than a topological one. But when we relate a graph to a surface through the process of embedding we move into the realm of topology. graph is a set, X, of elements together with a relation on X. We often draw pictures of graphs, where the elements of X are represented by dots and the relation by a set of arrows connecting certain dots to others. The elements of X are called vertices and the arrows joining vertices related by the relation are called edges. xample : The following is a graph on the set {,,, }: {(, ), (, ), (, ), (, ), (, ), (, ), (, )}. These are the (directed) edges and they can be represented by the following diagram. There are vertices and edges. However we ll only consider undirected graphs with no loops. loop is an edge (x, x) from a vertex to itself. n undirected graph is one where the relation is symmetric, that is, if x is related to y then y is related to x. So if there s an arrow in one direction there s always one in the opposite direction. In an undirected graph there s no need to use arrows since we know that the relation goes in both directions, so we simply use edges without arrows. xample : The following is an undirected graph with no loops. rom now on, when we use the word graph, we ll mean that it s undirected and has no loops. graph is a combinatorial structure where the only consideration is which vertices are adjacent to which. When we draw a graph the positions of the points representing the vertices are arbitrary. So are the routes of the edges. The edges needn t be straight, they re allowed to cross over other edges, and they could even wind around in more complicated ways. However we usually draw a graph in such a way that it gives as simple a picture as possible. xample : The graph in example could be redrawn as in the diagram on the left, but would look much better when drawn as the one on the right.
2 Notice that in the above example it s possible to draw the graph without any of the edges crossing. This isn t always possible, and we ll be very much concerned with the problem of when it is possible and when it isn t. In a graph we say that two vertices are adjacent if there s an edge between them. xample : In example, vertices and are adjacent but and are not. Two graphs X and Y are equivalent if there s a  and onto map f: X Y such that x and x are adjacent in X if and only if f(x ) and f(x ) are adjacent in Y. xample : The two graphs in example are equivalent. nd both are equivalent to the following graph. (It s essentially the same graph but with the vertices labelled differently.) B C D xample : The following list contains all the graphs with vertices or less. very graph with up to vertices is equivalent to exactly one of these. They ve been systematically classified according to the number of vertices, V and edges,. V = = 0 V = = 0 V = = V = = 0 V = = V = = V = = V = = 0 V = = V = = V = =
3 . V = = V = = V = = The complete graph on n vertices, denoted by K n, is the graph where every vertex is n adjacent to every other vertex. The number of edges in K n is clearly the binomial coefficient xample : The following are the complete graphs on vertices or less. K K K K K K nother important family of graphs consists of the graphs K m,n for various values of m and n (they don t have a name, just a symbol). The graph K m,n has m + n vertices divided into two subsets, one of size m and the other of size n. very vertex in one subset is adjacent to every vertex in the other, but there are no edges connecting two vertices within the same subset. xample : The following is K, : This graph was once featured in an ir New Zealand advertisement, where the vertices consisted of the cities Brisbane, Sydney, Melbourne, uckland, Wellington and Christchurch. The edges represented the transtasman routes... mbedding a Graph in a Surface graph has vertices and edges. So has a map, so what s the difference? Maps have faces, while graphs don t. It s the existence of faces on a map that give it its topological significance. Consider the following famous puzzle, called the Utilities Puzzle. You have three houses and three utilities. The utilities are a gasworks, a power station and a water reservoir. They have to pump gas, electricity and water to each of the three houses. But they have to do this so that the pipes and cables don t intersect. You see this is a dimensional problem. In real life (dimensional) there s no problem at all. Gas pipes can be routed over electrical cables or under water pipes. But we have to solve the puzzle in dimensions. The problem is to draw the three houses and three utilities on a sheet of paper, and draw lines to represent the pipes and cables, in such a way that they only meet one another at endpoints. You might like to have a go at this problem. 9
4 xample 9: The following is a near solution. Clearly we can t put in the remaining water pipe without crossing the lines we ve drawn already. But that doesn t prove that the problem has no solution. Perhaps you can think of better places to put the six vertices or perhaps you can route the first eight edges in a really clever way so that the last one can be drawn without crossing the others. Don t spend too long on this puzzle, because it s impossible! If you ve tried to solve it for a few minutes you ll come to this conclusion, though you won t have a proof. But beware! Haven t you ever attempted puzzles where, after trying in vain for many minutes, you become quite convinced that it s impossible, only to have someone come along and show you a really clever solution? Not in this case, though. We re going to prove that this puzzle is impossible! In the language of graph theory the network of pipes and cables is the graph K,. We want to draw this graph in the plane so that edges meet only at vertices. Or, to use a new concept that we re just about to define, the problem is to embed K, on a disk. graph can be embedded (is embeddable) on a surface if it can be drawn on that surface so that edges meet only at vertices. So we ll be proving that K, is not embeddable on a disk... Planarity graph is defined to be planar if it can be embedded in a disk. So K, isn t planar. But K, is: So is K : Since a disk can be cut out of a sphere, any planar graph can be embedded in a sphere. On the other hand, if we can embed a graph in a sphere, we can cut out a small hole in the middle of one of the faces and we have an embedding of the graph in a disk. (Remember that a disk is homeomorphic to a sphere with one hole.) 0
5 and = So a graph is planar if and only if it can be embedded in a sphere. This is useful because often a sphere is more convenient to work with. To prove that a graph is planar we can simply draw it, with edges not crossing. But how do we show that a graph, such as, K, or K is not planar? The technique discussed here is to work out the average number of edges per face and compare this to the smallest number of edges per face. But wait a minute. Graphs don t have faces! That s true, but a graph embedded in a surface becomes a map, and maps have faces. So we suppose that the graph is planar, that is, it can be embedded in a sphere. But how can we count the number of faces if we re only supposing that the graph can be embedded? The answer is to use uler s formula: V + = χ or planarity we use χ =, the uler characteristic of the sphere. Why not χ= for the disk? The answer is that we ll be assuming that there s a face on both sides of each edge of the map. If we have boundaries this will not be so. xample 0: K is not planar. Proof: or K we have V = and = 0. Suppose that K is planar. Then embedding it in a sphere we can deduce that the number of faces must be: = + V =. 0 The average number of edges per face must therefore be =. Why not just? The reason is that every edge is associated with two faces one on each side. So if you were to split each edge lengthwise, so that each half edge was associated with only one face, you d have half edges to share among the faces. Now we wanted to prove that K can t be embedded in a sphere and we started out by supposing that it can be. We re clearly looking for a contradiction. So what s contradictory about the average number of edges per face being? What s wrong is that it s less than. very face must be surrounded by at least edges ( face bounded by edges would require that the two edges connect the same two vertices, and a face bounded by just edge would mean that the graph has a loop.) Now the average of a collection of numbers can t be less than the smallest of them. So here we have our contradiction! xample : K, is not planar.
6 < = Proof: Here V = and = 9. Suppose that K, can be embedded in a sphere. The resulting map would have to have faces where + 9 =, that is, it must have faces. The average number of edges per face would therefore be. This isn t less than, so where s the contradiction? The contradiction is that it s less than. You see, in this graph there are no cycles of length. ach edge takes you from one set of vertices to the other. Going along another edge must take you back to a different vertex in the first set. The smallest cycles in this graph therefore have length. The boundary of a face must be a cycle in the graph. So the smallest number of edges for each face is. But the average of these numbers is less than. This can t be, and so we have our contradiction. The girth of a graph is the length of the shortest cycle. The girth of K is but the girth of K, is. We get a contradiction if the average number of edges per face is less than the girth... Proving that a Graph is not mbeddable in a Surface. Suppose a graph has edges and V vertices and the girth is g. If the graph can be embedded in a surface with uler characteristic χ then it will produce a map with faces where = + V. If < g then we have a contradiction. That graph can t possibly be embedded in that surface. But beware. The g is a oneway test. If, on the contrary, we find that g we d be wrong to conclude that the graph is embeddable. It may be embeddable, or on the other hand it may not be embeddable but this test isn t powerful enough to show it. This is one of the most common errors in graph embedding. Remember that essentially the only way to prove that a graph IS embeddable is to actually draw an embedding. 0BTest for Nonmbeddablility only applies to surfaces with no boundaries. B = χ + V < g graph with edges, V vertices and girth g is NOT g The graph may be embeddable. On the other hand it might not be. In this case the TST ILS. embeddable in any surface with uler characteristic χ If the surface has boundaries, remove them and use the corresponding surface with no boundaries. xample : The graph K is embeddable in the torus. b a b b K a torus
7 = =. NonProof: or K, V = and = 0. Suppose that K can be embedded in a torus. The resulting map will have faces where: + 0 = 0 (remember that χ = 0 for the torus). So =. Now. The girth of K is. So > g. What does this prove? bsolutely nothing! This whole calculation achieves nothing beyond confirming that the claimed result might be true. Proof: To prove that K is embeddable in a torus we simply take a torus and draw K on it (in such a way that edges don t cross). Of course it s easier to use a polygon with identified edges rather than the surface of a real dimensional torus. a b b a Note that there are alternative routes we could have followed for the  connection. We should only include one of them. (The apparent doubleup of the  connection is an illusion as the  edge at the top and the bottom of the square are, in reality, the same edge.) The duplication of the labels of the vertices comes about because of the identification of the edges. Check that every vertex, from to, is adjacent to every other. Theorem : The largest value of n for which K n is embeddable in the torus is. Solution: or K n, V = n and = ½ n(n ). Suppose that K n can be embedded in a torus. The resulting map will have faces where: n n n + ½ n(n ) = 0. So = n(n ) n(n ) n. The girth of K n n is. So n, in which case n 9n n n. n Hence n n 0. Since n > 0 we may divide this inequality by n to obtain n. We now show that K can be embedded in a torus.
8 < = = < = < < Suppose you have a graph G and a surface S an you want to determine whether or not G is embeddable in S. The first thing you might try is the g test. But before you do, there are a couple of things you should do first to both the graph and the surface: () Clearly holes are irrelevant to the embedding problem. graph is embeddable in a surface with holes if and only if it is embeddable in the corresponding surface with no holes. But the < g test assumes that the surface has no holes. So the very first thing you must do is to: replace the surface by the corresponding surface with no holes. If you re asked whether a graph is embeddable in a disk, or a cylinder, you replace that surface by a sphere. In testing embeddability in a Möbius Band you replace it by a projective plane. () If the graph has a vertex of degree you can clearly remove that vertex, and the edge joined to it, without affecting embedability. urthermore, you can remove a vertex of degree and join the edges on either side into a single edge, without affecting embedability. However removing vertices of degree may reduce the girth, which will weaken the test. So before you apply the g test you should: remove vertices of degrees and remove any vertices of degree if so doing does not change the girth. Having modified the graph and surface appropriately you can now apply the < g test. If it holds, the graph is not embeddable in the surface. If g then the test fails, that is, it is inconclusive. The next thing to do would be to attempt to draw an embedding. xample : The following graph is known as the Petersen Graph: V = 0 = If embedded in a sphere = + 0 = 0 f = g = So the Petersen graph can t be embedded in a sphere. In other words it isn t planar. or a projective plane, χ = and so if this graph can be embedded in a PP, = + 0 = 0 and f = = g. The g test fails. So it might be possible to embed the Petersen graph in a projective plane. On the other hand it might be impossible, which isn t much help! But in fact we can embed the Petersen graph in a projective plane by actually displaying such an embedding:
9 is and.. b 9 9 a 0 a 0 9 b Whenever you display such an embedding it s important to do what we ve done here label the vertices of the original graph and the embedded graph so that it s easy to check that it s indeed the same graph. (Vertex is adjacent to in both graphs, it s not adjacent to vertex in either graph, and so on.) + m + Theorem : If K n is embeddable in a sum of m toruses then n Proof: Suppose n. Then for K n, V = n, = ½ n(n ) and the girth is g =. The uler characteristic of the sum of m toruses is m. So = + χ V = ½ [n n m + ]. n(n ) Since K n can be embedded, = n g =, that is, n n (m ) 0. n m + The zeros of n ± m + ± m + n (m ) are so n + m + The smallest value of, occurring when m = 0. The sum of zero toruses is a sphere and since K and K can be embedded in a sphere the cases n = and n = automatically hold... Printed Circuit Boards. printed circuit board has electronic components laid out on both sides of a board with the connecting tracks printed on the board. They can be considered as graphs where the vertices occur on both sides of the surface and the edges on each side form a planar graph. We insist that each edge lies entirely on one side or the other. graph is embeddable in a sided (orientable) surface S if it is the union of two subgraphs, each of which is embeddable in S.
10 and m Theorem : If K n is embeddable in a sum of m toruses then n Proof: Suppose that K n is embeddable in mt. Let the numbers of vertices in the subgraphs, the graphs representing the two sides, be V and V repectively and let the number of edges be and. Clearly each V i n and the uler characteristic of the sum of m toruses is m. Suppose that the numbers of faces of the corresponding maps are and. Since each i subgraph must be planar we must have, for i =,. Since every vertex must be connected to every other, on one side or the other, we have + = ½ n(n ). Hence, for each i, i i = ( i + m V i ) ( i + m n) So i n + m. We therefore have ½ n(n ) = + n + m. It follows that n n (m ) 0. The zeros of n ± 9 + 9(m ) n (m ) are i + + 9m so n xample : The largest value of n for which K n can be embedded in a plane is, 9 or 0. If K n is embeddable in a plane then from Theorem, n 0 (taking m = 0). The following is a embedding of K in a plane. Is it possible to embed K 9 or even K 0?
11 XRCISS OR CHPTR xercise : Show that the following graph may be embedded in a projective plane, but not in a disk. xercise : Determine (with reasons) which of the following graphs are planar, which can be embedded in a Möbius Band and which can be embedded in a torus. (a) (b) xercise : (a) What is the girth of the following graph? (b) Prove that this graph is not planar. (c) Can this graph be embedded in a projective plane? (d) Can this graph be embedded in a torus? (e) Can this graph be embedded in a Möbius Band? Give reasons for your answers to (c) to (e).
12 . xercise : or each of the following graphs determine whether or not it can be embedded in a cylinder and whether or not it can be embedded in a torus with holes. Graph Graph B xercise : (i) Draw a diagram for K,, as a graph on vertices. Draw a square with identified edges that represents the Klein bottle. (ii) mbed K, in the Klein bottle (so that edges don t cross). xercise : or each of the following graphs determine whether or not it can be embedded in (i) a cylinder; (ii) a torus. graph graph B xercise : (a) Suppose the following graph was embedded in a sphere. Calculate the average number of edges per face in the resulting map. xplain how this leads to a contradiction. (b) Can the graph in (b) be embedded in a cylinder (Give reasons.) xercise : (a) Show that if K n can be embedded in a surface with no boundaries and uler characteristic χ then + 9 χ n (b) Hence find the largest value of n for which K n can be embedded in the connected sum of a torus, a projective plane and a disk.
13 = xercise 9: The graph K mn (where m n ) consists of two sets S, T of sizes m, n respectively where every vertex in S is joined to every vertex in T but where vertices within the same set are not joined to each other. (a) Draw pictures of K mn for all m, n with m + n =. (b) Prove that if K mn is embeddable in a surface of weight W then, unless m = n =, W ½(m )(n ). (c) ind all values of m, n for which K mn is planar. (Illustrate these cases with suitable picture.) (d) ind all values of m, n for which K mn is not planar but can be embedded in a torus. (Illustrate each of these with a suitable picture, using a polygon with identified edges to represent the torus.) SOLUTIONS OR CHPTR xercise : Deleting the vertices of degree and we obtain the following graph: But this decreases the girth from to. By adding an extra vertex of degree on the edge from to 0 we increase the girth back to. This graph has V = vertices and = 9 edges. Moreover this graph can be embedded in a sphere if and only 0 9 if the original graph can be. But if it this new graph could be embedded in a sphere it would result in a map with = + V = 0 faces. The average number of edges per face would therefore be 0 =. which is less than the girth, a contradiction. Hence the new graph is not embeddable in a sphere and so the original graph is not embeddable in a sphere. But since a disk is a sphere with one hole, and holes do not affect embedability, the original graph can not be embedded in a disk. The following gives an embedding of the graph in a projective plane. 9 0 xercise : (a) By inspection the girth =. The number of edges,, is and the number of vertices, V, is 0. graph is planar if and only if it can be embedded in a sphere. Suppose it is planar. By uler s formula V + = and so =. 9
14 = < The average number of edges per face is = > so it might be planar. Let s try, 0 9 is equivalent to 9 0 and hence the graph is indeed planar. Hence it is embeddable in a Möbius Band and a torus. (b) If it could be embedded in a Möbius Band plane it could be embedded in a Projective Plane, with χ =. g =, V =, = 9, so if embedded in a Projective Plane, = + V =.. The average number of edges per face would be less than the minimum number of edges per face, a contradiction. Hence the second graph is not planar. (c) If embeddable in the Projective Plane = + = 9. Since f = /9 < we get a contradiction, so the graph is not embeddable in a projective plane. (d) If embeddable in the torus = + 0 =. Since f = / = the graph may be embeddable in the torus. To prove that it is requires a specific embedding, which we show below. (e) Since the Möbius Band is D + P, a graph is embeddable in the Möbius Band if and only if it is embeddable in the Projective Plane. So by (c) it is not embeddable in the Möbius Band. 90
15 = xercise : (a) girth = (b) V =, = (rather than count them all, simply notice that each vertex has degree ( edges to each vertex). Multiplying by the number of vertices, and dividing by (since each edge connects two vertices) we get. If the graph is planar it can be embedded in a sphere, and the number of faces that would result would be = + V = 0. 0 <, so we get a contradiction. This graph is not planar. (c) If it was embedded in a projective plane, = 9 and = 9 <. So it cannot be embedded in a projective plane. (d) or the torus, = and =. The test fails. Perhaps it can be embedded in a torus. If it can t it will need another argument to prove this. So we attempt to draw it on a torus. It s a good idea to locate one point at the corners of the diagram (all identified) and some others on the edges. We don t have to use the edges on the diagram made up of edges in the graph but it makes life simpler if we do. So let s put at the four corners and look for a couple of cycles of edges that intersect only in that vertex. There s a cycle  and So we plot these points along the edges in appropriate places. There s still vertex to locate. It will go somewhere in the middle of the square. We ll ignore it for the moment. Now we put in the remaining edges, except for those having as an endpoint. Because vertices have two locations we have choices as to which locations we use. We make choices, with an eye to allowing to be able to access the four vertices it will need to be joined to. 9
16 = Clearly, if we locate in the largest face in this diagram it will be able to reach,, and. xercise : Graph : girth = V =, =. Suppose graph can be embedded in a sphere. Let the number of faces be. Then = + =. Hence < girth. So graph cannot be embedded in a sphere. B C D G H can be embedded in a projective plane as follows: B C a G b D H D b G C B a 9
17 Graph B: The graph can be redrawn as: so it is planar. Hence it can be embedded in any surface, in particular the sphere and the Möbius Band. xercise : Graph can be drawn as: so it is embeddable in the disk and hence in all surfaces, including the cylinder and the torus. or graph B, V =, = and if embedded in the sphere the number of faces would be = + =. So f = = =. < = girth. Hence graph B cannot be embedded in the sphere and hence not in the cylinder. But we can embed B in a torus: can be redrawn as There s just one crossing left. But we can join to by going up and to the right to the middle of the vertical and reappear at the corresponding point on the left (since the rectangle represents a torus). 9
18 = <. xercise : (a) V =, =. If embedded in a sphere the resulting map would have = + = faces and so the average number of edges per face would be f = / = ½. Since this is less than the girth (which is ), we get a contradiction. Hence this graph cannot be embedded in a sphere. (b) Since the cylinder is a sphere with two holes and since holes do not affect embeddability, a surface is embeddable in a cylinder if and only if it is embeddable in a sphere. Hence the above graph is not embeddable in a cylinder. However it is embeddable in a torus. xercise : (a) The number of vertices of K n is V = n and the number of edges is = ½ n(n ). Its girth is g =. Suppose K n can be embedded in S. The number of faces would be = V χ = ½ (n n + χ). n n Then n. n + χ n n Hence n and so n n + χ 0. n + χ + 9 χ Hence n (b) χ 0 (T + P) = + = so χ(t + P) = = so by part (a), if K n can be embedded in T + P + then n. So K n cannot be embedded in T + P if n. But K can be embedded in a torus and hence can be embedded in T + P so the largest value of n for which K n can be embedded in T + P is n =. xercise 9: (a) K K K 9
19 (b) Suppose K mn is embeddable in a surface (with no holes) of weight W. The uler characteristic of this surface is χ = W. Case I: n. Here g =, V = m + n, = mn. Thus = ( W) + mn m n and so mn f =. Thus mn m n + W mn and so mn m n + W W mn m n + = (m )(n ) and so W ½(m )(n ). Case II: n =, m >. Then ½(m )(n ) 0 W. (c) If n =, K mn is embeddable in a plane (and hence a torus). Suppose n and that K mn is embeddable in a torus (with weight W = ). rom (b) (m )(n ). The only solutions of this inequality for positive integers m, n with m n are: n =, m = any value (These are planar so can be ignored); n =, m = ; n =, m = ; n =, m = ; n =, m = ; n =, m =. The last five all include K, which is not planar so none of these is planar. It remains to prove that they are indeed embeddable in the torus. K can be embedded in a torus and hence so can K and K and K. B C That just leaves K. D B C 9
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