1. Mycielski’s Construction Mycielski’s Construction: From a simple graph G, Mycielski’s Construction produces a simple graph G’ containing G. Beginning with G having vertex set {v 1, v 2, …,v n }, add vertices U={u 1, u 2, …,u n } and one more vertex w. Add edges to make u i adjacent to all of N G (v i ), and finally let N G’ (w)=U. v2 v1 v2 v1 u2 u1 w v5 v2 v3v4 u1 u5 w u2 u3u4 v1

2. Theorem 5.2.3 From a k-chromatic triangle-free graph G, Mycielski’s construction produces a k+1-chromatic triangle-free graph G’. Proof. 1. Let V(G)={v 1, v 2, …,v n }, and let G’ be the graph produced from it by Mycielski’s construction. Let u 1, u 2, …,u n be the copies of v 1, v 2, …,v n, with the additional vertex w. Let U={u 1, u 2, …,u n }.

3. Theorem 5.2.3 2. Suppose G’ has a triangle. 3. The triangle contains at least one node in U, say u i, since G is triangle-free. 4. Since U is an independent set in G, the other vertices of the triangle belong to V(G), say v j, v k. 5. v j, v k are neighbors of v i. 6. There are a triangle v i, v j, v k in G, which is a contradiction. v2 v3v4 u1 u5 w u2 u3u4 v1 v5

4. Theorem 5.2.3 7. A proper k-coloring f of G extends to a proper k+1- coloring of G’ by setting f(u i )=f(v i ) and f(w)=k+1   (G’)<=  (G)+1. v2 v3v4 u1 u5 w u2 u3u4 v1 v5

5. Theorem 5.2.3 8. Suppose G’ has a proper k-coloring g. 9. By changing the names of colors, we may assume g(w)=k. This restricts g to {1, 2, …, k-1} on U. 10. For each v i on which g uses color k, we change the color of v i to g(u i ). 11. The modified coloring g’ of V(G) is a proper k-1- coloring of G, which is a contradiction. v2 v3v4 u1 u5 w u2 u3u4 v1 v5 v2 v3v4 u1 u3 v1 v5

6. Proposition 5.2.5 Every k-chromatic graph with n vertices has at least k*(k-1)/2 edges. Proof. 1.At least one edge with endpoints of colors i and j for each pair i, j of colors. Otherwise, colors i and j could be combined into a single color class and use fewer colors. 2. It implies at least k*(k-1)/2 edges in k-chromatic graph with n vertices.

7. Complete Multipartite Graph Complete Multipartite Graph: A complete multipartite graph is a simple graph G whose vertices can be partitioned into sets so that (u,v)  E(G) if and only if u and v belongs to different sets of the partition. Equivalently, every component of G is a complete graph. When k>=2, we write K n 1 n 2 n k for the complete k- partite graph with partite sets of size n 1, …, n k and complement K n 1 + …+K n k. K 1,3,4

8. Turan Graph Turan Graph: The Turan graph T n,r is the complete r- partite graph with n vertices whose partite sets differ in size by at most 1. That is, all partite sets have size  n/r  or  n/r . T 8,3

9. Lemma 5.2.8 Among simple r-partite graphs with n vertices, the Turan graph is the unique graph with the most edges. Proof. 1. We need only consider complete r-partite graphs. 2. Given a complete r-partite graph with partite sets differing by more than 1 in size, we move a vertex v from the largest size (size i) to the smallest class (size j). 3. It is easy to verify the number of edges increases. Gain i-1 neighbors Lose j neighbors

10. Theorem 5.2.9 Among the n-vertex simple graphs with no r+1-clique, T n,r has the maximum number of edges. Proof. 1. T n,r has no r+1-clique. 2. It suffices to show the maximum is achieved by an r- partite graph by Lemma 5.2.8. 3. It is proved that if G has no r+1-clique, then there is an r-partite graph H with the same vertex set as G and at least as many edges by induction on r. 4. When r=1, G and H have no edges.

11. Theorem 5.2.9 5. Consider r>1. Let G be an n-vertex graph with no r+1- clique, and let x  V(G) be a vertex of degree k=  (G). 6. Let G’ be the subgraph of G induced by the neighbors of x (N(x)) and let S=V(G)-N(x). 7. G’ has no r-clique because x is adjacent to every vertex in G’ and G has no r+1-clique. 8. By induction hypothesis, there is a r-1-partite graph H’ with vertex set V(G’) such that e(H’)>=e(G’). G’ G S-x x H’ H S x

12. Theorem 5.2.9 9. Let H be the graph formed from H’ by joining all of V(G’) to all of S. 10. H is r-partite because S is an independent set in H. 11. e(G)<=e(G’)+  v  S d G (v)<=e(G’)+k(n-k). 12. e(H)=e(H’)+k(n-k). 13. e(G)<=e(H) because e(G’)<=e(H’). G’ G S-x x H’ H S k(n-k) edges x

13. Lemma 5.2.15 Let G be a graph with  (G)>k, and let X,Y be a partition of V(G). If G[X] and G[Y] are k-colorable, then the edge cut [X,Y] has at least k edges. Proof. 1. Let X 1,…,X k and Y 1,…,Y k be the partitions of X and Y formed by the color class in proper k- colorings of G[X] and G[Y]. 2. If there is no edge between X i and Y j, then X i  Y j is an independent set in G. In this case, X i and Y j can have the same color.

14. Lemma 5.2.15 3. Form a bipartite graph H with vertices X 1,…,X k and Y 1,…,Y k, putting X i Y j  E(H) if in G there is no edge between the set X i and the set Y j. 4. It suffices to show if |[X,Y]|<k, then H has a perfect matching. (If H has a perfect matching, G has a proper k-coloring which constitutes a contradiction.)

15. Lemma 5.2.15 5. Suppose |[X,Y]|<k. 6. H has more than k(k-1) edges. 7. E(H) cannot be covered by k-1 vertices because m vertices can cover at most km edges in a subgraph of K k,k. 8. The minimum size of a vertex cover in H is at least k. 9. The maximum size of a matching in H is at least k by Theorem 3.1.16. 10. H has a perfect matching.

16. Theorem 5.2.16 Every k-critical graph is k-1-edge-connected. Proof. 1. Let G be a k-critical graph, and let [X,Y] be a minimum edge cut. 2. G is k-critical, G[X] and G[Y] are k-1-colorable. 3. |[X,Y]|>=k-1.