Lemma. Let G be a graph and e an edge of G. Then e is a bridge of G if and only if e does not lie in any cycle of G.

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1 Lemma. Let G be a graph and e an edge of G. Then e is a bridge of G if and only if e does not lie in any cycle of G. Lemma. If e = xy is a bridge of the connected graph G, then G e consists of exactly two components one containing x and the other containing y. Lemma. If v is a cutvertex of the graph G. then δ( v) 2. So, an endvertex of a graph cannot be a cutvertex. Let P denote some property of graphs. Then A graph G is said to be minimal with respect to P if and only if G has P but for every edge e of G, G - e does not have P. If a property is worth studying at all, then it is worth studying those graphs that are minimal with respect to P. Graphs that are minimal with respect to connectedness are called trees. Thus a graph G is a tree if it is connected but for every edge e, G - e is not connected. That means that every edge of G must be a bridge. That in turn means that G cannot contain a cycle. So equivalently, we may define a graph G to be a tree if and only if G is connected and has no cycles. Examples of trees. Notice that there are two 'types' of trees on 4 vertices and three 'types' of trees on 5 vertices. We will make it clear soon just what we really mean by a 'type' of graph. (If we accidentally said there were 4 types of trees on 5 vertices - this is wrong- would we have made a 'type of graph'-ical error?) It is also interesting to notice that there is one tree on V = {a}, one tree on V = {a, b}, 3 distinct trees on V = {a, b, c}, 16 distinct trees on V = {a, b, c, d} and 125 distinct trees on V = {a, b, c, d, e}. How many distinct trees are there on V = {a, b, c, d, e, f}? How many 'types' of trees are there on six vertices?

2 A subgraph H of G that has the same vertex set as G (i.e., V(H) = V(G) ) is called a spanning subgraph of G. Theorem. If G is any connected graph, then G has a spanning subgraph that is a tree. (Such a subgraph is called, naturally enough, a spanning tree for G). Theorem. If G is a tree on n vertices, then G has n - 1 edges. Theorem. If G is a connected graph on n vertices and n - 1 edges, then G is a tree. Theorem. If G is a graph on n vertices having no cycles and n - 1 edges, then G is a tree. Theorem. Every tree on n 2 vertices contains at least two end-vertices. So we have used phrases like, 'G and H are the same type' to 'G and H are essentially the same,' and the time has come to make these ideas more precise. Two graphs G and H are identical if they have exactly the same set of vertices and edges. How the graph is drawn does not matter - the picture is just a representation of the graph. The two graphs below are identical. Example: Notice that the top two graphs in each collection yield a collection of 4 distinct graphs. However the bottom graphs in each group are identical to the one directly above. However, the top graph in each group of three is more than just different from its counterpoint in the other group. The graphs in each column have the same structure, which is quite different from the graphs in the other column. The graphs on the left differ from one-another only in names of vertices (similarly for the other group). We can turn any graph on the left into the other by simply renaming (we say relabeling) the vertices. But there is no way to rename the vertices in any graph on the on the left to produce one of the graphs on the right.

3 We say that two graphs are isomorphic if they differ only in their labels; i.e., we can turn one into the other by renaming the vertices. More precisely, A graph G is isomorphic to a graph H if there exists a bijection (i.e., 1-1 correspondence) φ: V ( G) V ( H) such that for every two vertices x and y of G, xy E( G) φ( x) φ( y) E( H) φ is called an isomorphism and it represents the relabeling that turns G into H. The idea is that if two graphs are isomorphic then they have exactly the same graphtheoretic properties. Two isomorphic graphs differ only in the names of the vertices and nothing else. In other words G and H are isomorphic if you can obtain H from G by renaming the vertices of G properly. The function φ describes how to rename the vertices. Note that the isomorphic relation is an equivalence realation lf the set of all graphs. For a given graph G, we say that the class of all graphs isomorphic to G (i.e., the equivalence class of G) forms an isomorphism class. Examples There are two isomorphism classes of trees of order 4 - we say up to isomorphism there are two trees of order 4. Up to isomorphism there are three trees of order 5. Up to isomorphism there are two graphs of order two. Up to isomorphism there are four graphs (not just trees) of order three. When we ask if two graphs are isomorphic we are essentially asking if we can reassign labels to the vertices so that the resulting graphs are identical. Of course two unlabeled graphs are isomorphic if we can assign labels to each of them so that the resulting graphs are identical. When we draw a graph without labels, we are referring to an isomorphism class of graphs. So, suppose that someone has promised you a great deal of money to determine if two graphs each with 30 vertices are isomorphic. If the graphs are isomorphic, then you can convince me of that by exhibiting a particular isomorphism. It might be quite difficult to determine an isomorphism - but it is likely doable. But at least you can potentially find something to convince me that you did your job. But what if the graphs are not isomorphic? What then? Well, consider this. Are the graphs below isomorphic? No - clearly not. Why not? Well, because of the triangle in the graph on the right.

4 Suppose that G and H are isomorphic graphs and G has a triangle abc. Then the images of a,b, and c under the isomorphism must be a triangle in H So if G has a triangle, then so does any graph isomorphic to it. In general, to show that two graphs are isomorphic we can just exhibit an isomorphism, How would you show that two graphs are not isomorphic? So we can point to the triangle as evidence that the isomorphism does not exist. In general any property of graphs that is preserved by isomorphisms could be used. Such properties are called graph-theoretic properties. Examples Here are some other such properties. If G and H are isomorphic then They have the same number of vertices They have the same number of edges. They have the same number of even-order cycles They have the same number of triangles They have the same degree sequence. If one of them contains 4 mutually adjacent vertices so does the other. If one has exactly two cut vertices, so does the other. If one contains a pair of adjacent vertices of degree 3, then so does the other. So to convince someone that there is no isomorphism you could provide a graph-theoretic property of one graph that is not shared by the other. Clearly, any two complete graphs on n vertices are isomorphic. Clearly any two paths on n vertices are isomorphic. We denote a complete graph on n vertices by K n and a path on n vertices by P n and a cycle on n vertices by C n. Here are a few more examples. If G has a (induced) subgraph isomorphic to F, then H has a (induced) subgraph isomorphic to F. If G has a vertex of degree 4 adjacent to a vertex of degree 7, then H also has a vertex of degree 4 adjacent to a vertex of degree 7. If G is isomorphic to H, then the subgraph of G induced by the vertices of degree 2 is isomorphic to the subgraph of H induced by the vertices of degree 2. If G is isomorphic to H and G and H both have exactly one vertex of degree 7, then subgraphs of G and H obtained by removing that vertex are also isomorphic. These are just a few of the infinite number of possible criteria that we might use to demonstrate that two graphs are not isomorphic. The Theorem below is frequently useful. Theorem. If G and H are isomorphic, then so are their complements.

5 If G and H have a large number of edges, it may be easier to deal with the their complements. Reconstruction Conjecture: Suppose that I take a graph and I delete a vertex one at a time and write down on an index card an unlabeled graph that is isomorphic to the vertex-deleted graph. Now I give you the index cards. Can you figure out, up to isomorphism, what the graph was? i.e., can you reconstruct the graph? The conjecture is that yes, that is always possible for graphs on n 2 vertices, But no one has proven it. And maybe it s not true. But it is known to be true for small and for graphs that are not connected. Questions: (i). Why do we require n 2 here? (ii). Can you prove that all regular graphs are reconstructible? It isn t too hard to figure out some of the properties of the graph from the set of vertexdeleted subgraphs. For example, you can find n, q the number of vertices and edges, and the degree sequence quite easily. You can also tell if the original graph was connected. Can you figure out how? Here is the official version of the Reconstruction Conjecture. Conjecture: Given the n 2 vertex deleted subgraphs G v1, G v2,, G v n (unlabelled), it is possible to reconstruct G. Effectively this asserts that any two non-isomorphic graphs will produce distinct sets of vertex deleted subgraphs. Theorem. Let G be a graph with n vertices and q edges. For each i = 1, 2,, n let q i denote the number of edges in G v i qi i= 1. Then q = n 2. n