# 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,

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 G
_{1},…, G_{l}, with**2≤l**of G′ equal**G – v**. Each of**G – i’s**vertex vi has the value**VV**therefore_{i}∈ E(G),**dG**We can appropriately k-color each of_{i}(v_{i}) = k – 1.**G**using the previous way. In reality, we can tint G_{i}‘s_{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
**V**and that_{1}, V_{2}**V**and_{1}, V_{2}∉ E**G − {V**are connected. Find a spanning tree of_{1}, V_{2}}**G − {V**with a root in v and order it the same way you did before._{1}, V_{2}}**V**should be added at the beginning of the ordering. Greedy Coloring will then give both_{1}, V_{2}**V**the same color. Every vertex_{1}and V_{2}**w ≠ v**has a maximum of**k-1**neighbors in front of it. Vertex v has k but**V**_{1}, V_{2}_{ }have the same color and therefore we can give a color to**v.** - We shall now demonstrate that such a triple v, v
_{1}, v_{2 }exists at all times. Choose an arbitrary vertex x. If**κ(G − x) ≥ 2**is true, then**v**and v_{1}= x_{2}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, v_{1}and v_{2}have a common neighbour in v. Additionally,**G − {v**is related._{1}, v_{2}} = G − {x, v_{2}} - 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 v_{1}, v_{2}be the neighbors of x from two different leaf blocks of**G−x**. There is clearly no connection between v_{1}and v_{2}, and**G − {v**is connected as_{1}, v_{2}}**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.