k-Coloring

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k-Coloring

• k-coloring A k-coloring of a graph G is a
  labeling f V(G)?S, where Sk. The labels are
  colors the vertices of one color form a color
  class.
• Proper k-coloring A k-coloring is proper if
  adjacent vertices have different labels.
• k-colorable graph A graph is k-colorable if it
  has a proper k-coloring.
• Chromatic number ?(G) The least k such that G is
  k-colorable.

• k-chromatic graph A graph G is k-chromatic if
  ?(G)k. A proper k-coloring of a k-chromatic
  graph is an optimal coloring.

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Greedy Coloring

• The greedy algorithm relative to a vertex
  ordering v1, v2, , vn of V(G) is obtained by
  coloring vertices in the order v1, v2, , vn,
  assigning to vi the smallest-indexed color not
  already used on its lower-indexed neighbors.

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Proposition 5.1.13

• ?(G)lt?(G)1.
• Proof. 1. In a vertex ordering, each vertex has
  at most ?(G) earlier neighbors.
• ? Greedy coloring cannot be forced to use more
  than ?(G)1 colors.

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Block

• Block A maximal connected subgraph of G that has
  no cut-vertex. If G itself is connected and has
  no cut-vertex, then G is a block. (Definition
  4.1.16)

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Block-cutpoint graph

• Block-cutpoint graph The block-cutpoint graph of
  a graph G is a bipartite graph H in which one
  partite set consists of the cut-vertices of G,
  and the other has a vertex bi for each block Bi
  of G. vbi is an edge of H if and only if v? Bi.

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Leaf Block

• Leaf Block A block that contains exactly one
  cut-vertex of G.
• When G is connected, its block-cutpoint graph is
  a tree (Exercise 34 of Sec. 4.1) whose leaves are
  blocks of G.
• ? A graph that is not a single block has at
  least two leaf blocks.

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Brooks Theorem

• If G is a connected graph other than a complete
  graph or an odd cycle, then ?(G)lt?(G).
• Proof. 1. Let k ?(G).
• 2. When klt1, G is a complete graph.
• 3. When k2, G is an odd cycle or is bipartite,
  in which case the bound holds.
• 4. We assume that kgt3.
• 5. The theorem holds if we can order the vertices
  such that each has at most k-1 lower-indexed
  neighbors.

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Brooks Theorem

• 6. Case 1 G is not k-regular. Let vn be the
  vertex of degree less than k.
• 5. Grow a spanning tree of G from vn, assigning
  indices in decreasing order as we reach vertices.

6. Each vertex other than vn in the resulting
ordering has v1, v2, , vn has a higher-indexed
neighbor along the path to vn in the tree. ? Each
vertex has at most k-1 lower-indexed neighbors.
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Brooks Theorem

• 4. Case 2 G is k-regular.
• 5. Case 2-1 G has a cut-vertex x.
• 6. Let G be a subgraph consisting of a component
  of G-x together with its edges to x.
• 7. The degree of x in G is less than k.
• 8. The method in case 1 provides a proper
  k-coloring of G.
• 9. By permuting the names of colors in the
  subgraphs resulting in this way from components
  of G-x, we can make the colorings agree on x to
  complete a proper k-coloring of G.

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Brooks Theorem

• 10. Case 2-2 G is 2-connected.
• 11. Suppose that some vertex vn has neighbors v1,
  v2 such that (v1,v2)?E(G) and G-v1,v2 is
  connected.
• 12. Index the vertices of a spanning tree of
  G-v1, v2 using 3, 4, , n such that labels
  increase along paths to the root vn.
• 13. Each of v1, v2, , vn-1 has at most k-1 lower
  indexed neighbors.
• 14. v1 and v2 receives the same color.
• ? At most k-1 colors are used on neighbors
  of vn.

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Brooks Theorem

• 15. It suffices to show that every 2-connected
  k-regular graph with kgt3 has such a triple
  v1,v2,vn in 10.
• 16. Choose a vertex x.
• 17. Case 2-2-1 ?(G-x)gt2.
• 18. Let v1 be x.
• 19. There exists a vertex v2 with distance 2 from
  x
• because G is not a complete graph and G is
  regular.
• 20. Let vn be a common neighbor of v1 and v2.
• 21. v1, v2, vn is the desired triple.
• 22. Case 2-2-2 ?(G-x)1.

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Brooks Theorem

• 23. Let vnx. Then, x has a neighbor in every
  leaf block of G-x.
• 24. G-x is not a single block
• 25. At least two leaf blocks in G-x
• 26. Clearly, neighbors v1 and v2 of x are not
  adjacent.
• 27. G-v1,v2,x is connected
• since blocks have no cut-vertices.
• 27. Vertex x has a neighbor other than v1 and v2
• because kgt3.
• 28. G-v1,v2 is connected.

Otherwise, G is not 2-connected.
because ?(G-x)1.
because G-x is connected.
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Definition 7.1.3

• k-edge-coloring of G A labeling f E(G) -gt S,
  where S k. The labels are colors the edges of
  one color form a color class.
• proper k-edge-coloring A k-edge-coloring such
  that incident edges have different labels.
• k-edge-colorable graph A graph having a proper
  k-edge-coloring.
• edge-chromatic number, x(G), of G The least k
  such that G is k-edge-colorable.

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Theorem 7.1.10

• If G is a simple graph, then x(G) ?(G)1.
• Proof 1. Let f be a proper ?(G)1-edge-coloring
  of a subgraph G of G.
• 2. If G? G, then some edge uv is uncolored by f.
• 3. It suffices to show we can extend the coloring
  to include uv after possibly recoloring some
  edges (an augmentation).

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Theorem 7.1.10
4. Every vertex has some color not appearing on
its incident edges because the number of
colors exceeds ?(G). 5. Let a0 be a color
missing at u. 6. Let a1 be a color missing at
v0(v). We may assume that a1 appears at u on
some edge uv1 otherwise, we would use a1 on uv0.
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Theorem 7.1.10
7. Let a2 be a color missing at v1. We may
assume that a2 appears at u on some edge uv2
otherwise, we would replace color a1 with a2 on
uv1 and then use a1 on uv0 to augment the
coloring.
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Theorem 7.1.10
8. Having selected uvi-1 with color ai-1, let ai
be a color missing at vi-1. If ai is missing
at u, then we use ai on uvi-1 and shift color aj
from uvj to uvj-1 for 1 j i-1 to complete the
augmentation. We call this downshifting from
vi-1. If ai appears at u (on some edges uvi),
then the process continues.
ai
ai-1
u
vi-1
ak
vk
v0 v
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Theorem 7.1.10
9. Since we have only ?(G)1 colors to choose
from, the list of selected colors eventually
repeats (or we complete the augmentation by
downshifting). 10. Let l be the smallest index
such that a color missing at vl is in the list
a1,,al let this color be ak. 11. The color ak
missing at vl is also missing at vk-1 and appears
on uvk. If a0 does not appear at vl, then we
downshift from vl and use color a0 on uvl to
complete the augmentation. Hence we may assume
that a0 appears at vl.
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Theorem 7.1.10
12. Let P be the maximal alternating path of
edges colored a0 and ak that begins at vl along
color a0. There is only one such path
because each vertex has at most one incident edge
in each color (we ignore edges not yet colored).
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Theorem 7.1.10
13. If P reaches vk, then it arrives at vk along
an edge with color a0, follows vku in color ak,
and stops at u, which lacks color a0. In this
case, we downshift from vk and switch colors on P.
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Theorem 7.1.10
14. If P reaches vk-1, then it reaches at vk-1 on
color a0, and stops there, because ak does not
appear at vk-1. In this case, we downshift
from vk-1, give color a0 to uvk-1, and switch
colors on P.
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Theorem 7.1.10
15. If P does not reach vk or vk-1, then it ends
at some vertex outside u, vl, vk, vk-1. In
this case, we downshift from vl, give color a0 to
uvl, and switch colors on P.