Mycielski

1
Mycielskis Construction

• Mycielskis Construction From a simple graph G,
  Mycielskis Construction produces a simple graph
  G containing G. Beginning with G having vertex
  set v1, v2, ,vn, add vertices Uu1, u2, ,un
  and one more vertex w. Add edges to make ui
  adjacent to all of NG(vi), and finally let
  NG(w)U.

2
Theorem 5.2.3

• From a k-chromatic triangle-free graph G,
  Mycielskis construction produces a k1-chromatic
  triangle-free graph G.
• Proof. 1. Let V(G)v1, v2, ,vn, and let G be
  the graph produced from it by Mycielskis
  construction. Let u1, u2, ,un be the copies of
  v1, v2, ,vn, with w the additional vertex. Let
  Uu1, u2, ,un.

3
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 ui, since G is triangle-free.
• ? Since U is an independent set in G, the
  other vertices of the triangle belong to V(G),
  say vj, vk.
• ? vj, vk are neighbors of vi.
• ? There are a triangle vi, vj, vk in G.
• ? It is a contradiction.

4
Theorem 5.2.3 (3/4)

• 3. A proper k-coloring f of G extends to a proper
  k1-coloring of G by setting f(ui)f(vi) and
  f(w)k1
• ? ?(G)lt ?(G)1.
• 4. The equality can be proved by showing ?(G)lt
  ?(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 vi?A, we change the color of vi to
  g(ui).
• 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 vi, vj be two adjacent vertices in G.
• 10. vi, vj have different colors under g. We need
  to prove vi, vj 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
  vi, vj is in A.
• 12. Case 1 vi, vj?A. The colors of vi, vj are
  not changed. ?vi, vj have different colors under
  g.
• 13. Case 2 vi?A and vj?A. By construction,
  (ui,vj)?E(G). ? ui, vj have different colors
  under g. ? vi, vj 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
  kgt2, we write Kn1n2nk for the complete k-partite
  graph with partite sets of size n1, , nk and
  complement Kn1 Knk.
• Turan Graph The Turan graph Tn,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-1gtj, the number of edges increases. ?
  We maximize the number of edges only by
  equalizing the size as in Tn,r.

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Theorem 5.2.9

• Among the n-vertex simple graphs with no
  r1-clique, Tn,r has the maximum number of edges.
• Proof. 1. Tn,r has no r1-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 Tn,r.
• 3. It suffices to prove that if G has no
  r1-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 r1, G and H have no edges.
• 6. Consider rgt1. Let G be an n-vertex graph with
  no r1-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 r1-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)lte(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 SV(G)-N(x).
• 11. S is an independent set. ? H is r-partite.
• 12. We need to prove e(H)gte(G).
• 13. By construction, e(H)e(H)k(n-k).
• 14. e(G)lte(G)?v?SdG(v)lte(H)k(n-k)lte(H).

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Lemma 5.2.15

• Let G be a graph with ?(G)gtk, and let X,Y be a
  partition of V(G). If GX and GY are
  k-colorable, then the edge cut X,Y has at least
  k edges.
• Proof. 1. Let X1,,Xk and Y1,,Yk be the
  partitions of X and Y formed by the color class
  in proper k-colorings of GX and GY.
• 2. If there is no edge between Xi and Yj, then
  Xi?Yj is an independent set in G. In this case,
  Xi and Yj can have the same color.
• 3. We show that if X,Yltk, then we can combine
  color classes from GX and GY 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
  X1,,Xk and Y1,,Yk, putting XiYj?E(H) if in G
  there is no edge between the set Xi and the set
  Yj.
• 5. Suppose that X,Yltk. 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 Kk,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 Xi and all of
  Yj to which it is matched by M.
• 8. There are no edges joining Xi and Yj, doing
  this for all i produces a proper k-coloring of G.
  ? It contradicts to the hypothesis that ?(G)gtk. ?
  X,Ygtk.

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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, GX and GY are
  k-1-colorable.
• 3. X,Ygtk-1.