Arie Girshson
306236399
Exercise 2: Brook’s Theorem
Question:
Proof that the cliques and the odd cycles are the only 𝑘 − 1 regular 𝑘 critical graphs.
Basic Notations:
𝑑(𝑉𝑛 ) – Degree of vertex 𝑛 (number of edges incident to the vertex).
𝑘 Regular – A graph where each vertex has the same number of neighbors (𝑘 neighbors).
𝑘 − 1 Regular – Every vertex is adjacent to 𝑘 − 1 others.
Critical element – A vertex is a critical element of graph 𝐺 if its deletion would decrease the chromatic
number (𝜒(𝐺)) of 𝐺.
𝑘 Critical graph – a graph in which every vertex or edge is a critical element (𝑘 elements), with chromatic
number (𝜒(𝐺)) being 𝑘.
The question presented above is implied directly from brook’s theorem. Therefore, I will start from
brook’s theorem.
Brook’s Theorem:
If 𝐺 is a connected graph which is neither complete (clique) nor an odd cycle, then 𝜒(𝐺) ≤ ∆(𝐺).
Proof:
Let 𝐺 be a connected graph with vertex set 𝑉 = {𝑣1 , 𝑣2 , … , 𝑣𝑛 } which is neither a complete graph
(clique), nor an odd cycle. Let ∆= 𝑘 ≥ 3.
For 𝑘 = 1 𝐺 is a complete graph (clique).
For 𝑘 = 2 𝐺 is an odd cycle or bipartite graph.
Assume 𝐺 is not 𝒌 Regular. Then there exists a vertex 𝑣 = 𝑣𝑛 such that 𝑑(𝑣𝑛 ) < 𝑘. Since 𝐺 is connected,
form a spanning tree of 𝐺 starting from 𝑣𝑛 , whose vertices are arranged in the order 𝑣𝑛 , 𝑣𝑛−1 , … , 𝑣1 .
𝑣1
𝑣2
𝑣 = 𝑣𝑛
Arie Girshson
306236399
Clearly, each vertex 𝑣𝑖 other than 𝑣𝑛 in the resulting order 𝑣𝑛 , 𝑣𝑛−1 , … , 𝑣1 has a higher indexed neighbor
along path to 𝑣𝑛 in the tree. Therefore each vertex 𝑣𝑖 has at most 𝑘 − 1 lower indexed neighbors.
Obviously, greedy coloring needs at most 𝑘 colors.
𝑣𝑛
𝑣𝑖
Assume 𝐺 is 𝒌 Regular.
Assume 𝐺 has a cut vertex 𝑥 (1-Connected) and 𝐺′ is a sub-graph containing a component of 𝐺 − 𝑥
together with the edges of 𝐺 − 𝑥 to 𝑥. Clearly 𝑑(𝑥|𝐺 ′ ) < 𝑘. By using the above argument, we have 𝑘coloring of 𝐺′. By making use of the permutations of the colors, it can be seen that this is true for all
such sub-graphs. Thus, 𝐺 is 𝑘-colorable.
Assume 𝐺 is 2-connected. 𝐺 Therefore has an induced 3-vertex path, with vertices say 𝑣1 , 𝑣𝑛 , 𝑣2 in
order, such that 𝐺 − {𝑣1 , 𝑣2 } is connected. To prove this claim, let 𝑥 be any vertex of 𝐺. If 𝑘(𝐺 − 𝑥) ≥ 2,
let 𝑣1 be 𝑥 and let 𝑣2 be a vertex with distance two from 𝑥, which clearly exists, as 𝐺 is regular and not a
complete graph (clique). If 𝑘(𝐺 − 𝑥) = 1, then 𝑥 has a neighbor in every end block of 𝐺 − 𝑥, since 𝐺 has
no cut-vertex. Let 𝑣1 and 𝑣2 be the neighbors of 𝑥 in two such blocks. Clearly 𝑣1 and 𝑣2 are nonadjacent. Also, since blocks have no cut vertices, 𝐺 − {𝑥, 𝑣1 , 𝑣2 } is connected. As 𝑘 ≥ 3, so 𝐺 − {𝑣1 , 𝑣2 }
is connected and let 𝑥 = 𝑣𝑛 , proving the claim.
Same as before, arrange the vertices of a spanning tree of 𝐺 − {𝑣1 , 𝑣2 } as 𝑣3 , 𝑣4 , … , 𝑣𝑛 . Again each
vertex before 𝑛 has at most 𝑘 − 1 lower indexed neighbors. The greedy coloring again uses at most
𝑘 − 1 colors on neighbors of 𝑣𝑛 , since 𝑣1 and 𝑣2 get the same color.
Implied directly from brook’s theorem, that cliques (complete graphs), which are 𝑘 − 1 regular (each
vertex has 𝑘 − 1 neighbors) 𝑘 critical (vertex removal decreases the chromatic number 𝜒(𝐺)) connected
graphs and odd cycles, which again are 𝑘 − 1 regular 𝑘 critical connected graphs, are the only graphs
which don’t apply 𝜒(𝐺) ≤ ∆(𝐺). These graphs definitely apply to 𝜒(𝐺) = ∆(𝐺) + 1.
The odd cycle and clique examples are shown below:
All the rest are covered by Brook’s Theorem.