List of Theorems. Mat 416, Introduction to Graph Theory. Theorem 1 The numbers R(p, q) exist and for p, q 2,

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1 List of Theorems Mat 416, Introduction to Graph Theory 1. Ramsey s Theorem for graphs Theorem 1 The numbers R(p, q) exist and for p, q 2, R(p, q) R(p 1, q) + R(p, q 1). If both summands on the right-hand side are even then the inequality is strict. Neighbors and nonneighbors of any vertex. To get strict: There exist a graph on R(p, q) 1 vertices with degree of each vertex R(p 1, q) Eulerian trails, Theorem Theorem 2 A multigraph G is Eulerian if and only if it has at most one nontrivial component and is even Nontrivial implication by induction on the number of edges. In a nontrivial component every vertex has degree at least 2. Find a cycle and obtain induction to each of the components left. Paste Euler circuits to obtain a circuit of the graph. 3. Dirac s Theorem, Theorem Theorem 3 If G is a simple graph on at least three vertices with the minimum degree δ(g) n(g) then G has a Hamiltonian cycle. Maximal counter-example G. G has a spanning path x 1,...,x n. There is no i such that x i x n, x i+1 x 1 E as otherwise a cycle is obtained. S = {i x 1 x i+1 E}, T = {i x n x i } {1,...,n 1} and so S T. 4. Basic results on trees, Lemma 2.1.3, Theorem Lemma 4 (2.1.3) Let G be a graph with at least two vertices. If G is a tree then it has at least two leaves. If G satisfies P C and P E then G has at least two leaves Follows from 2 E d(v). 1

2 Theorem 5 (Theorem 2.1.4) Let G be a multigraph on n vertices with n 1. Then the following conditions are equivalent. (a) G satisfies P C and P A. (b) G satisfies P C and P E. (c) G satisfies P E and P A. (d) G has no loops and for every u, v there is exactly one u, v-path. Show (a) iff (b) iff (c) by induction (all at the same time). Always want any two imply the third one: use a leaf, or (c) implying (a) by contradiction. Equivalence with (d) is a separate argument. Must show at most one path in one direction (by contradiction). Show (d) implies (a) to establish the second. 5. Bipartite graphs, Theorem Theorem 6 A loopless graph is bipartite if and only if it has no odd cycle. If bipartite then no odd cycles is obvious: x 1 y 1,..., y n x 1. Other direction (assume connected): Fix v. S is the set of vertices at even distance, T of odd. Show an edge inside S will give an odd walk and so an odd cycle. 6. Berge s Theorem Theorem 7 A matching M in G is maximum if and only if there is no M-augmenting path in G. ( ) Indirectly. If there is an M-augmenting path, exchange the edges on this path to get a bigger matching. ( ) Any matching M and maximum matching M with M < M. Analyze M M (symmetric difference) and show there is a component (a path) with more edges from M than M. This is your M-augmenting path. 7. Hall s Theorem and applications including Theorem 8 A bipartite multigraph with bipartition X, Y has a matching that saturates X if and only if for every set S X, N(S) S. In the nontrivial implication, consider the maximum matching M. Use Berge s to claim no M-augmenting paths and fix an M-unsaturated vertex v. 2

3 U-all vertices u such that there is an M-alternating v, u-path. S = U X, T = U Y. v S. Show S {v} = T and N(S) = T by matching M matching S {v} with T and using the fact all vertices from S {v} are M- saturated. Corollary 9 (3.1.13) For every positive integer k, every k-regular bipartite graph has a perfect matching. Show X = Y by counting edges. Take S X and count edges between S and N(S) in two ways. Theorem 10 (3.3.9) Every regular multigraph where each vertex has a positive even degree has a 2-factor. Use Euler circuit to get an orientation with d + (u) = d (u). Define the auxiliary bipartite graph (V +, V ). Use the previous corollary to get a perfect matching which translates back to a 2-factor. 8. König, Egerváry Theorem Theorem 11 If G is a bipartite multigraph, then β(g) = α (G). Always β(g) α (G). Take a minimum cover Q. Let R = Q X, T = Q Y. Let H 1 = G[R, Y T], H 2 = G[X R, T]. No edge from Y T to X R as Q is the cover. Check Hall s condition for H 1 with R and for H 2 with T. If it is voilated then you obtain (say S R with N(S) Y T < S ) a smaller cover (by exchanging S with N(S) Y T). 9. Tutte s Theorem 3.3.3, Theorem 12 (Tutte, 3.3.3) A simple graph G has a 1-factor if and only of o(g S) S for every S V (G). In the nontrivial implication assume G is maximal 1-factorless and the Tutte s condition holds. Your first goal is to find u, v, w, z such that uw, vw E and uv, wz / E. Look at vertices connected to every other vertex in graph- set U, and components in G U. If all are complete graphs then easy to find a matching. Otherwise you have uw, vw and find z by noting w / U (!). 3

4 u v w Figure 1: Proof of Tutte s Theorem (Case 2), u, v, w, z z Now look at G 1 = G+uv and G 2 = G+wz and find perfect matchings M 1, M 2 in each. Look at M 1 M 2 which are cycles and use the structure of above to obtain a new perfect matching that avoids uv and wz (two cases). Corollary 13 (Berge-Tutte Formula, 3.3.7) The largest number of vertices saturated by a matching in G is where d(s) = o(g S) S. min V (G) d(s) S V (G) The number is also equal to V max d(s). In any matching there will be at least max d(s) vertices unsaturated. To prove the second inequality. Take S that maximizes d( ) and note the parity of d := o(g S) S is the same as V Consider G K d. It has an even number of vertices. Check Tutte s condition for G K d : (1) When T is empty. (2) When V (K d ) is not a subset of T. When T contains K d. 10. Application Corollary 14 (Petersen, 3.3.8) Every 3-regular graph with no cut-edge has a 1-factor. Check Tutte s condition. Show that for set S and odd component C of G S there is an odd (and so at least 3) number of edges between C and S. Gives the total number of edges at least 3o(G S). The number of such edges is at most 3 S as 3-regular. 4

5 z u x y H H Figure 2: Theorem connected graphs and Theorem Theorem 15 (Whitney, 4.2.8) A graph is 2-connected if and only if it has an ear decomposition. Moreover the first cycle of the decomposition can be chosen to be an arbitrary cycle in G.. Induction on the number of ears. Take P 0,, P t+1 and H := P 0 P t is 2-connected by IH. Show H +P t+1 does not have a cut-vertex by a case by case analysis.. This is harder. Show by induction on m that the following statement is true: If m E(G) then there is a subgraph H of G which has an ear decomposition and with E(H) m. For m = 0 take the cycle. For inductive step : either E(H) m and done or there is an edge with both endpoints in H giving you a larger subgraph and the ear or H is induced. If H is induced then [V (H), V V (H)] is a cut (V V (H) is nonempty). Take an edge xy with x V (H), V V (H). As G is 2-connected find a path from y to H. The path plus xy gives you an ear and the subgraph connected graphs, Theorem Theorem 16 (Thomassen, 6.2.9) Every 3-connected graph G with V (G) 5 has an edge e = xy such that G e is 3-connected. By contradiction, no such edge. For each edge e = xy, G e is 2- connected. This leads to the notion of the mate. Select the edge e = xy such that G {x, y, z} has a component H of the largest order. Show H + xy is to 2-connected (otherwise the cut vertex plus z is a cut). 5

6 Look at zu and its mate w and show there is a component in G {x, u, w} which has H as a proper subset. 13. Menger s Theorem Theorem 17 (Menger, ) Let G be a graph. For two distinct vertices x, y with xy / E, κ(x, y) = λ(x, y). κ(x, y) λ(x, y) is clear. To prove the second inequality, proceed by induction on V (G). Let S be a minimum x, y-cut. S cannot contain N(x) (or N(y)) as a proper subset. Case 1: S is different than N(x) and N(y). Let V 1 be the vertices on all x, S-paths and V 2 vertices on all S, y paths. Show V 1 V 2 = S and V 1 (N(y) S) =. Consider auxiliary graphs H 1, H 2 and apply induction. Case 2: S is N(x) and N(y). First argue V (G) is N(x) N(y) {x, y} and N(x) N(y) = using induction. Now show that x, y-cuts are precisely coverings of (N(x), N(y)) and invoke König-Egervary. 14. Applications. Theorem 18 For a graph G = (V, E) let λ(g) = min x,y V λ(x, y). Then κ(g) = λ(g). κ(g) λ(g) is clear as a cut must destroy all internally disjoint vertices between a pair of two vertices. Case G = K n is trivial. To prove the second inequality. Take x, y and show that a cut separating x, y gives λ(g) paths. If xy / E(G) then this is Menger s. If xy E(G) then G xy has connectivity of at least κ(g) 1 and so there are at least λ(g) 1 paths. This, in connecting with xy, gives λ(g) paths. Theorem 19 (Fan Lemma, ) A graph is k-connected if and only if it has at least k+1 vertices and for any vertex x and set U with U k, there is an x, U-fan of size k. ( ). Let U V (G), x V (G). Add new vertex y to G and connect it to U. 6

7 The new graph is k-connected (easy observation) and by Menger s there are k pairwise internally disjoint paths between x and y in this graph. This gives an x, U-fan of size k. ( ). First observe δ(g) k. Let x, y be two vertices from V (G). There is a x, N(y)-fan of size k in G. This gives x, y-paths. 7