Section 1.1: What is a Graph? Graph, directed graph, vertex degree, complete graph, complement graph, bipartite graph, proper coloring, subgraph, path, cycle, connected graph. Cliques and bicliques. Independent set of vertices. Given a subset of subgraphs, one can define maximal subgraphs and max- imum subgraphs. Maximal: not a subgraph of any other. Maximum: has largest possible size of some sort. Connected component: maximal connected subgraph. Adjacency matrix. Graph isomorphism: bijection of vertex sets inducing bijection of edge sets. Graphs can be superimposed. Proving non-isomorphic: find structure in one that does not exist in the other. For example: subgraph with certain number of vertices or edges or vertex degrees etc. The number of graphs. The number of connected graphs? Trees. The number of trees? The Petersen graph. Properties: bottom of page 13. Homework: 4, 13, 14, 18, 24, 25, 26, 27, 28, 34. Section 1.2: Paths, Cycles, and Trails. Walk: sequence of successive edges. Length is number of edges. Path: a walk in which no vertices are encountered more than once. Endpoints and internal vertices. Trail: a walk in which no edges are encountered more than once. Cycle: a path plus a distinct returning edge. cut-edge, cut-vertex, induced subgraph. Property of cut-edge: bottom of page 23. Property of closed walk: page 24. Characterization of bipartite graph: top of page 25. See also Exercise 25. Union of subgraphs: bottom of page 25. 1 Characterization of Kn as union of bipartite graphs: top of page 26. Appli- cation at very top of page. Eulerian graph: top of page 27. Characterization of Eulerian graph: bottom of page 27. Proof: we can prove that each component of a graph with even degrees has an Euler Cycle by induction on edges. No edges: done. At least one edge: find cycle, cut out, leaving fewer edges. There's an Euler Cycle for each component. Add pack the cycle edges, gluing some of these together. The ones that are glued together constitute one component which has an Euler Cycle. Graphs with large minimal degrees have large paths and cycles: bottom of page 28. Just read the rest yourselves. Homework: 6, 17, 20, 25, 29, 31, 38, 39, 40, 41. Section 1.3: Vertex Degrees and Counting Degree, k-regular, NG(v) neighborhood of v, n(G) order of G (number of vertices), e(G) size of G (number of edges). Degree-Sum Formula. Corollaries: average vertex degree, even number of odd degree vertices, number of edges in a k-regular graph. Structure of the hypercube Qk. k-regular bipartite graph has same number vertices on each side of vertex partition. The Petersen graph has ten 6-cycles. Computing the number of edges and the degree sequence of a graph from its vertex-deleted subgraphs, page 37. Reconstruction Conjecture, page 38. The minimal number of edges in a connected graph with n vertices is n − 1. Minimal degree in an n vertex graph ≥ (n − 1)=2 implies graph is connected. Every loopless graph G has a bipartite subgraph with at least e(G)=2 edges. An algorithm is provided. Example 1.3.20 on page 41 shows that this algo- rithm is not optimal for all graphs. 2 n2 The maximum number of edges in an n-vertex triangle-free graph is b 4 c. The bound is attainable. Proof employs vertex-degree sum formula. Skip the material on graphic sequences. Homework: 1, 10, 20, 31, 32, 37, 38, 40, 45, 51. Section 1.4: Directed Graphs Directed graph. Isomorphism. Strongly connected components. Kernel of digraph: independent set of vertices, every vertex outside this set has edge to this set. Theorem 1.4.16: Every digraph having no odd cycle has a kernel. Proof: First assume strongly connected. Let y be an arbitrary vertex. Let S be the set of vertices with an even length directed walk to y. Then every vertex outside S has an edge into S. To show that S is independent, suppose that u ! v in S. Since closed odd walks contain closed odd cycles, there are no closed odd walks in S. Hence there is an even walk from y to u, an odd walk from u to v, and an even walk from v to y, which creates a closed odd walk. Contradiction. Therefore u ! v cannot exist. Now drop the strongly connected hypothesis and use induction on number of vertices. Base case is 1 vertex: true. Now assume true for all n vertex digraphs. Let D be digraph with n + 1 vertices, no odd directed cycles. If strongly connected, use construction above. If not, find component C with no exiting edges. Form kernel S. Now form kernel T on what remains (D0) after discarding S and its predecessors in D. Then there are no edges among vertices in T or among vertices in S or from S to T or from T to S, and every outside S and T has an edge into S or T , so the kernel for D is the union of S and T . In-degree, out-degree, in-neighborhood, out-neighborhood. Sum of in-degrees = sum of out-degrees = number of vertices. If all out-degrees in a digraph are at least one then the digraph has a directed cycle. Maximal path argument. There are Eulerian circuits through the components in a directed graph if and only if in-degree = out-degree at all vertices. De Bruijn Cycle: a circular arrangement of 2n 1s and 0s in such a way that every consecutive n-string is distinct. Construction: create digraph with 3 vertex set the collection of binary strings of length n−1. Edge: (b(x); c; (x)c). Out-degrees are 2. In-degrees are 2. Graph is (strongly) connected: To get from b1 ··· bn−1 to c1 ··· cn−1, follow the edges (b1(b2 ··· bn−1); c1; (b2 ··· bn−1)c1) (b2(b3 ··· bn−2c1); c2; (b3 ··· bn−2c1)c2) and so forth. There are 2n edges, they are all distinct, and the order in which they are encountered is the same as the order they would be encountered along a cycle of 0s and 1s, properly laid out according to the Eulerian Circuit. Orientation of a graph: form digraph. Tournament: orientation of complete graph. Number of tournaments: pick orientation for each graph. Number of di- graphs: for each pair of edges, one of three possibilities: don't form edge, form edge in one direction, form edge on other direction. King: vertex in digraph which has path of length ≤ 2 to every vertex. Ever tournament has a king. Proof: find vertex x of maximal out-degree. Suppose x does not have an edge to y. Then y has edge to x. None of the successors of x point to y, so y points to all of these, so y has at least as many successors (in fact one more, namely x { contradiction). Therefore x points to y. Homework: 4, 6, 11, 14, 17, 19, 21, 25, 28, 39 Section 2.1: Basic Properties of Trees A tree with at least two vertices has at least 2 leaves. Proof: maximal path. Deleting leaf vertex leaves a tree. Proof: connected, no cycles. Equivalent to being a tree on n vertices: connected and no edges, connected and n − 1 vertices, n − 1 edges and no cycles, unique paths. Every edge of tree is cut edge. Adding edge to tree creates 1 cycle only. Every graph contains a spanning tree. Let T and T 0 be trees. Whenever you delete an edge from T that doesn't belong to T 0, you can join the two components using an edge in T 0 that doesn't belong to T . Proof: Pick a vertex in each component. There is a 4 path joining them in T 0. Hence there is an edge joining them in T 0. This is not the edge you took out, so it is some other edge. Hence it cannot belong to T . Results require that both trees have same number of edges. Let T and T 0 be trees. Add an edge to T 0, not in T 0 but in T . This creates one cycle. The cycle cannot live entirely in T , so we can break it by removing an edge from T 0, leaving a tree in place (by a count of edges and by connectivity). Proposition 2.1.8: If T is any tree with k edges and every vertex in G has degree ≥ k, then there is an isomorphic copy of T in G. Proof: By induction on k. Base case k = 1 is trivial. Now consider G with minimum degree k. Let u − v be any leaf edge in T . T − v has an isomorphic copy in G. We just need to find one so that the degree of the image of u in G is greater than the degree of u. Since T has k + 1 vertices, the worst-case scenario is that the degree of u is k. But in the worst-case scenario, T is star-shaped, and the image of u can be anywhere in G! Distance between vertices. Largest distance is diameter. Eccentricity of a vertex is distance to farthest vertex. Radius is minimum eccentricity. Theorem: For a simple graph G, diameter ≥ 3 implies the conjugate diameter is ≤ 3. Proof: Diameter > 1 implies the existence of two non-adjacent vertices u and v. For every other vertex x, x cannot be connected to both, so there is at least one edge from x to fu; vg in the complement.