Brook’s Theorem

Brook’s Theorem is one of the most well-known graph coloring theorems. Graph coloring is a subset of graph labeling, in graph theory. It involves the assignment of labels to elements of a graph, commonly referred to as “colors,” according to specific constraints. In its most basic form, vertex coloring is a method of coloring the vertices of a graph so that no two neighboring vertices are the same color. Now, let’s take a look at how Brook’s Theorem helps with graph coloring.

Statement of Brook’s theorem

In Graph Theory, Brook’s Theorem illustrates the relationship between a graph’s maximum degree and its chromatic number. Brook’s Theorem states that:

If G is a connected simple graph and is neither an odd cycle nor a complete graph i.e. χ(G)≥3 then 

χ(G) ≤ k, where k denotes the maximum degree of G and χ(G) denotes the chromatic number of G.

• An odd cycle means a cycle with an odd length, that is, with an odd number of edges.
• A complete graph is a graph in which a vertex should have edges with all other vertices.
• A graph’s chromatic number is the smallest number of colors required to color the vertices so that no two neighboring vertices have the same color.
• The biggest number of neighbors of a vertex in graph G is its maximum degree.

Brooks’ Theorem extends this assertion of the Greedy Algorithm which shows that χ(G) ≤ k +1, for any graph G. For example,

Fig. 1: Graph satisfying brook’s theorem

As you can see in fig. 1, the chromatic number of the graph i.e. χ(G) is 3 & the maximum degree i.e. k  is 3.  So the χ(G) = k, satisfying Brook’s theorem.

Proof of Brook’s theorem

Let’s say k = (G). G is either a cycle or a path if k = 2. We can suppose that k≥3. We’ll design an ordering in which each vertex has a maximum of k-1 vertices preceding it.

• Let us first assume that there is a vertex V with the value d(v)<k. Find a T spanning tree of G and root it at v. Put v as and the other vertices are ordered in T in such a way that all vertices at distance i are before all vertices at distance i+ 1 from v. In this arrangement, every vertex has at most k-1 vertices preceding it.
• Assume G has a cut vertex v and is k-regular. Consider the components G₁,…, G_l, with 2≤l of G′ equal G – v. Each of G – i’s vertex vi has the value VV_i ∈ E(G), therefore dG_i(v_i) = k – 1. We can appropriately k-color each of G_i‘s using the previous way. In reality, we can tint G_i so that the color of v_i is 1, if necessary, by swapping the colors.
• Assume G is 2-connected and k-regular.
• Assume that there is a vertex v with two neighbors V₁, V₂ and that V₁, V₂ ∉ E and  G − {V₁, V₂} are connected. Find a spanning tree of  G − {V₁, V₂} with a root in v and order it the same way you did before. V₁, V₂ should be added at the beginning of the ordering. Greedy Coloring will then give both V₁ and V₂ the same color. Every vertex w ≠ v has a maximum of k-1 neighbors in front of it. Vertex v has k but V₁, V₂ have the same color and therefore we can give a color to v.
• We shall now demonstrate that such a triple v, v₁, v₂ exists at all times. Choose an arbitrary vertex x. If κ(G − x) ≥ 2 is true, then v₁ = x and v₂ is a vertex at a distance of two times x. There must be such a vertex, because if there is an edge from x to every vertex from G−x, then G is complete because G is k-regular. Finally, v₁ and v₂ have a common neighbour in v. Additionally, G − {v₁, v₂} = G − {x, v₂} is related.
• Let κ(G − x) = 1 be the cut vertex of  G − x, and w be the cut vertex of G−x.
• Let H represent a block of G−x that is a leaf in the block graph, and a represent the cut vertex of G−x which is written in H. Then x has a different neighbor in H than a because otherwise, G-a is unconnected.
• Let v = x be the neighbors of x from two different leaf blocks of G-x, and v₁, v₂ be the neighbors of x from two different leaf blocks of G−x. There is clearly no connection between v₁ and v₂, and G − {v₁, v₂} is connected as d(x) ≥ 3.

Sample Problems

Question 1: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 3. 

The maximum degree of the graph i.e. k = 3.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.

Question 2: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 3.

The maximum degree of the graph i.e. k = 3.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.

Question 3: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 5.

The maximum degree of the graph i.e. k = 5.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.

Question 4: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 3.

The maximum degree of the graph i.e. k = 3.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.

Question 5: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 3.

The maximum degree of the graph i.e. k = 3.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.

Question 6: Prove Brook’s theorem for the given graph.

Solution:

The above graph,

The chromatic number of the graph, i.e. χ(G) = 4.

The maximum degree of the graph i.e. k = 4.

Therefore, χ(G) = k. Thus, the above graph proves Brook’s Theorem.