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EGRES Quick-Proof No. 2014-01
Yet another proof for Brooks’ theorem
Tamás Fleiner?
Abstract
We give a proof for Brooks’ theorem on the chromatic number of graphs
based on well-known properties of DFS trees.
Keywords: chromatic number, greedy coloring, DFS tree
The well-known theorem of Brooks [2] improves the greedy coloring based upper
bound of ∆(G) + 1 on the chromatic number χ(G) of graph G as follows.
Theorem 1. If G is a finite simple connected graph and G is neither complete nor
an odd cycle then χ(G) ≤ ∆(G) holds.
Lovász gave a short and elegant proof for Theorem 1 in [3] by greedy coloring the
vertices in an appropriate order. The proof is algorithmic and leans on the block
structure of non 2-connected graphs. Below we give another proof of Brooks’ theorem
that (beyond greedy coloring) uses the fact that DFS trees do not induce cross edges
and the following graph manipulation exercise which is interesting for its own sake.
Claim 2. Assume that G is a connected, finite simple graph and any DFS tree of G
rooted at v is a path starting at v for any vertex v of G. Then G is either a cycle or
a complete graph or G ∼
= Kn,n for some n.
We shall prove Claim 2 with the help of the following lemma.
Lemma 3. Assume that G is traceable and the terminals of any Hamiltonian path of
G are adjacent. Then G ∼
= Cn , G ∼
= Kn or G ∼
= Kn,n for some n.
Proof. Let C = (v1 , v2 , . . . , vn ) be a Hamiltonian cycle of G. If G is not a cycle then
G has some edge vi vj where j 6= i ± 1 (addition is modulo n). As
(vj+1 , vj+2 , . . . , vi−1 , vi , vj , vj−1 , . . . , vi+1 )
?
Department of Computer Science and Information Theory, Budapest University of Technology
and Economics, Magyar Tudósok körútja 2, Budapest, H-1117 and MTA-ELTE Egerváry Research
Group. Research was supported by the K109240 OTKA grant. E-mail:[email protected]
February 2014. Revised March 2014
Tamás Fleiner: Yet another proof for Brooks’ theorem
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is a Hamiltonian path of G, its terminals are adjacent, hence vi+1 vj+1 ∈ E(G). This
follows that vi+k vj+k ∈ E(G) holds for any k, that is, each “rotation” along C of any
edge of G also belongs to G. As
(vi−1 , vj−1 , vj , vi , vi+1 . . . , vj−2 , vi−2 , vi−3 , . . . , vj+1 )
is a Hamiltonian path of G, its terminals are adjacent, hence vi−1 vj+1 ∈ E(G). As
(vi−2 , vi−3 , . . . , vj+1 , vi−1 , vi , vj , vj−1 , . . . , vi+1 )
is a Hamiltonian path of G, its terminals are adjacent, hence vi−2 vi+1 ∈ E(G).
These observations imply that two vertices must be adjacent if their indices have
the same parity. If n is odd then the first observation yields that vertices with opposite
parity indices are also adjacent, so G is complete. For even n, we got that complete
bipartite graph with color classes {v1 , v3 , v5 , . . .} and {v2 , v4 , v6 , . . .} is a subgraph of
G. If G has further edges then G is complete by the first two observations.
Proof of Claim 2. As any DFS tree of G is a path, G is traceable. If P = (v1 , v2 , . . . , vn )
is a Hamiltonian path of G then there is a DFS tree T of G rooted at v2 that contains
path (v2 , v3 , . . . , vn ). As T is a path starting at v2 , vn v1 ∈ E(T ), hence the terminals
of any Hamiltonian path P are adjacent. Claim 2 follows from Lemma 3.
Proof of Theorem 1. We use induction on the number n of the vertices of G. Case
n = 1 is trivial. Assume that Theorem 1 holds for any graph with less than n = |V (G)|
vertices. If G ∼
= Kt,t for some t and G is not a complete graph then t > 1 hence
χ(G) = 2 ≤ t = ∆(G). If G is an even cycle then G ∼
= C2t and χ(G) = 2 = ∆(G).
Otherwise, by Claim 2, there is a DFS tree T of G such that that there are at least
two branches of T hanging on some vertex u of T . Let T1 and T2 be such branches
that are rooted at u1 and u2 , respectively. Let G1 and G2 be the subgraphs of G
spanned by V (T1 ) and V (T2 ), respectively. As uu1 , uu2 ∈ E(G), none of G1 and G2 is
∆(G)-regular, hence G1 and G2 are ∆(G)-colorable by induction. Pick ∆(G)-colorings
of G1 and of G2 such that u1 and u2 get the same color. This is a proper coloring of
V (T1 ) ∪ V (T2 ) as no cross edge connects two branches of a DFS tree.
Let v1 , v2 , . . . , u be an order of V (G) \ (V (T1 ) ∪ V (T2 )) such that vi , vi+1 , . . . , u
spans a connected sugraph of G for all i. As T − T1 − T2 is a spanning tree of
G − V (T1 ) − V (T2 ), such order exists. Now pick a color for vertices v1 , v2 , . . . , u in
this order in such a way that each vertex receives a color different from its previously
colored neighbors. As any vi has a neighbor vj for some j > i and u has two neighbors
u1 and u2 of the same color, there is at most ∆(G)−1 forbidden colors for any of these
vertices. Consequently, ∆(G) colors suffices to color G, hence χ(G) ≤ ∆(G).
Remark 4 (Added to the proof on March 14, 2014.). After the first version of this
quick proof have been published, the author learnt that the above proof is not new.
In the paper [1] by Böhme et al., the authors claim that a combination of their result
together with that of Stiebitz in [4] results in a new proof of Brooks’ theorem. Though
this proof is not given explicitly, most probably the authors refer to our argument
EGRES Quick-Proof No. 2014-01
Tamás Fleiner: Yet another proof for Brooks’ theorem
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above. In light of these, the goal of our present quick proof is that this perhaps not
so well-known proof is now available in a short and more or less self-contained form.
The author thanks Tibor Jordán for calling our attention to the work [1] by Böhme
et al..
References
[1] Böhme, T., Broersma, H. J., Göbel, F., Kostochka, A. V. and Stiebitz, M. it
Spanning trees with pairwise nonadjacent endvertices, Discrete Mathematics,
170(1-3), 219-222 (1997).
[2] Brooks, R. L. On colouring the nodes of a network, Proc. Cambridge Philosophical Society, Math. Phys. Sci. 37 194-197 (1941).
[3] Lovász, L. Three short proofs in graph theory, Journal of Combinatorial Theory,
Series B 19 269-271 (1975).
[4] Stiebitz, M The forest plus stars colouring problem Discrete Math. 126(1-3),
385-389 (1994).
EGRES Quick-Proof No. 2014-01