Vertex colorings (continued)
Recall...
Definition A graph G is k-critical if χ(G) = k and for all proper
subgraphs H of G, χ(H) < k.
Characterizations of k-critical graphs
• G is 2-critical ⇐⇒ G = K2.
• G is 3-critical ⇐⇒ G is an odd cycle.
• G is 4-critical ⇐⇒ ??
Remark No good characterization of 4-critical graphs or test for 3colorability is known.
Proposition. If G is k-critical, then δ(G) ≥ k − 1.
Proof. Suppose BWOC that G is k-critical and has a vertex v such that
d(v) < k − 1. Then G − v is (k − 1)-colorable.
Color G − v with k − 1 colors. Since v has fewer than k − 1 neighbors,
there is a leftover color that is not assigned to any neighbor and that can
be assigned to v so that we have a (k − 1)-coloring of G. ⇒⇐
Corollary. Let G be a k-chromatic graph (i.e. χ(G) = k). Then G
has at least k vertices of degree at least k − 1.
Proof 1. Throw away edges and/or vtcs of a k-chromatic graph G until
we reach a k-critical subgraph H. Then δ(H) ≥ k − 1 by previous prop.
Let v be a vertex of degree at least k − 1 in H. Then v and its k − 1
(or more) neighbors each have degree at least k − 1 in H and so the same
is true in G.
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Proof 2. Claim: In any k-coloring of a k-chromatic graph G, there is a
vertex of each color which is adjacent to vertices of every other color.
Pick such a vertex of each color. Then we have k vertices, each of
degree at least k − 1.
Proposition. For any graph G, χ(G) ≤ ∆(G) + 1.
Proof 1. If χ(G) ≥ ∆(G) + 2, then by previous corollary, G has vertices
of degree at least ∆(G) + 1. ⇒⇐
Proof 2. Greedy coloring algorithm gives a coloring that uses at most
∆(G) + 1 colors.
For an arbitrary ordering of vtcs, assign each vertex the lowest-indexed
color not already used on its already-colored neighbors. Each vertex has
at most ∆(G) neighbors that may have been colored before it, so we never
need more than ∆(G) + 1 colors.
Question When does the bound of this proposition hold with equality?
Complete graphs and odd cycles.
How can we take a slightly smarter approach?
=⇒ order vertices in a favorable manner
Problem 5.1.7 Construct a graph G that is neither
a complete graph nor an odd cycle but has a vertex
ordering relative to which greedy coloring uses ∆(G)+1
colors.
Theorem (Brooks’ Theorem). If G is a connected graph other than a
complete graph or an odd cycle, then χ(G) ≤ ∆(G).
Math 104 - Prof. Kindred - Lecture 12
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Proof. Let G be a connected graph on n vtcs, and let k = ∆(G). We
want to construct an ordering of the vtcs so that each vertex has at most
k − 1 lower-indexed neighbors. Then greedy algorithm yields a k-coloring
of G.
Handle small cases. If k ≤ 1, then G is a complete graph (K1 or K2). If
k = 2, then G is a cycle or a path. A path or an even cycle is bipartite,
so it is 2-colorable, meaning the given upper bound holds. Assume going
forward that k ≥ 3.
Case 1: G is not regular.
Color vtcs further from vertex v with d(v) < ∆(G) before closer
ones in greedy coloring algorithm.
Let vn be a vertex of degree < k. Grow a spanning tree T of G
from vn, assigning indices in decreasing order as we reach vtcs.
(CS terminology: consider a search tree of G rooted at vn using
BFS or DFS.)
When a vertex vi for i 6= n is about to be colored, it is
adjacent in T to at least one uncolored vertex, and so is adjacent
in G to at most d(vi) − 1 ≤ k − 1 colored vtcs. So it is assigned a
color from {1, 2, . . . , k}. Finally, when vn is about to be colored,
it has degree < k, so it can be colored with one of k colors.
Case 2: G is regular.
Case 2(a): G has a cut vertex v.
Color components of G − v separately and then combine.
Let H be induced subgraph on vtcs of component of G − v together with v. Then
dH (v) < dG(v) = k.
Use the same approach as in Case 1 to obtain a k-coloring of H.
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Repeat this process for each component of G − v. Permuting
names of colors as needed in component colorings, we can ensure
that the color of v is the same in all component colorings, thereby
yielding a k-coloring of G.
Case 2(b): G is 2-connected. Most difficult case.
Notice that no matter what order we choose for vtcs, the last
vertex vn has k lower-indexed neighbors, so we need to force two
neighbors of vn to be assigned the same color.
Idea: Find vertex vn w/neighbors v1, v2 such that v1 6∼ v2 and
G − {v1, v2} is connected. (?) As before, grow a spanning tree
from vn on G − {v1, v2} and index vtcs using indices {3, . . . , n}.
Then apply greedy coloring. (v1, v2 are assigned same color.)
Note that (?) is not possible for odd cycles or complete graphs.
Have to show that (?) is always possible. Trivial if G is 3connected (since G is not complete graph). Have to handle 2connected but not 3-connected more carefully.
v7
v3
v5
v4
Math 104 - Prof. Kindred - Lecture 12
v1
v6
v9
v10
v8
color 1
color 2
color 3
color 4
v2
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