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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.

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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 w the additional vertex. Let U={u 1, u 2, …,u n }.

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Theorem 5.2.3 (2/4) 2. G’ is triangle-free. Suppose G’ has a triangle. The triangle contains at least one node in U, say u i, since G is triangle-free. Since U is an independent set in G, the other vertices of the triangle belong to V(G), say v j, v k. v j, v k are neighbors of v i. There are a triangle v i, v j, v k in G. It is a contradiction.

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Theorem 5.2.3 (3/4) 3. 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. 4. The equality can be proved by showing (G)< (G’). To prove this we consider any proper coloring of G’ and obtain from it a proper coloring of G using fewer colors.

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Theorem 5.2.3 (3/4) 5. Let g be a proper k-coloring of G’. By changing the names of colors, we may assume g(w)=k. This restricts g to {1, 2, …, k-1} on U. 6. On V(G), it may use all k colors. Let A be the set of vertices in G on which g uses color k. It suffices to change the colors used on A to obtain a proper k-1-coloring of G. 7. For each v i A, we change the color of v i to g(u i ). 8. We need to prove the modified coloring g’ of V(G) is a proper k-1-coloring of G.

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Theorem 5.2.3 (4/4) 9. Let v i, v j be two adjacent vertices in G. 10. v i, v j have different colors under g. We need to prove v i, v j have different colors under g’. 11. All vertices of A have color k under g. No two vertices of A are adjacent. At most one of v i, v j is in A. 12. Case 1: v i, v j A. The colors of v i, v j are not changed. v i, v j have different colors under g’. 13. Case 2: v i A and v j A. By construction, (u i,v j ) E(G). u i, v j have different colors under g. v i, v j have different colors under g’.

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Proposition 5.2.5 Every k-chromatic graph with n vertices has at least k*(k-1)/2 edges. Proof. 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. At least k*(k-1)/2 edges in k- chromatic graph with n vertices.

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Turan 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. 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 .

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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. The edges not involving v are the same as before, but v gains i-1 neighbors in its old class and loses j neighbors in its new class. 4. Since i-1>j, the number of edges increases. We maximize the number of edges only by equalizing the size as in T n,r.

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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. If we can prove that the maximum is achieved by an r-partite graph, then Lemma 5.2.8 implies that the maximum is achieved by T n,r. 3. It suffices to prove 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.

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Theorem 5.2.9 4. This is proved by induction on r. 5. When r=1, G and H have no edges. 6. 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). 7. Let G’ be the subgraph of G induced by the neighbors of x. 8. x is adjacent to every vertex in G’ and G has no r+1- clique. The graph G’ has no r-clique. By induction hypothesis, there is a r-1-partite graph H’ with vertex set N(x) such that e(H’)<=e(G’).

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Theorem 5.2.9 10. Let H be the graph formed from H’ by joining all of N(x) to all of S=V(G)-N(x). 11. S is an independent set. H is r-partite. 12. We need to prove e(H)>=e(G). 13. By construction, e(H)=e(H’)+k(n-k). 14. e(G)<=e(G’)+ v S d G (v)<=e(H’)+k(n-k)<=e(H).

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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. 3. We show that if |[X,Y]|<k, then we can combine color classes from G[X] and G[Y] in pairs to form a proper k-coloring of G.

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Lemma 5.2.15 (2/3) 4. 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. 5. Suppose that |[X,Y]|<k. Then, H has more than k(k-1) edges.

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Lemma 5.2.15 (3/3) 6. m vertices can cover at most km edges in a subgraph of K k,k E(H) cannot be covered by k-1 vertices. The minimum size of a vertex cover in H is at least k. The maximum size of a matching in H is at least k by Theorem 3.1.16. H has a perfect matching M. 7. In G, we give color i to all of X i and all of Y j to which it is matched by M. 8. There are no edges joining X i and Y j, doing this for all i produces a proper k-coloring of G. It contradicts to the hypothesis that (G)>k. |[X,Y]|>=k.

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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.

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