Graph Theory
Spring 2012
Prof. Gábor Elek
Assist. Filip Morić
Exercise sheet 7: Solutions
Caveat emptor: These are merely extended hints, rather than complete solutions.
1. If T is a tree on n vertices, show that PT (t) = t(t − 1)n−1 , where PT (t)
is the chromatic polynomial of T .
Solution. By induction on n. Let v be a leaf of T and e the edge
incident to v. By Exercise 4 of Sheet 6, we have that PT (t) = PT −e (t)−
PT /e (t). By the inductive hypothesis, PT −e (t) = t × t(t − 1)n−2 and
PT /e (t) = t(t − 1)n−2 , so the conclusion follows.
2. Prove that for any connected graph G on n vertices PG (t) ≤ t(t−1)n−1
for t ∈ N (with equality for all t iff G is a tree) .
Solution. Since every connected graph has a spanning tree, the inequality follows from Ex.1. It remains to study the equality case. Suppose PG (t) = t(t − 1)n−1 . From Exercise 4 of Sheet 6 we can easily
obtain by induction that for any graph G we have
PG (t) = tn − |E(G)|tn−1 + . . . ,
i.e. the coefficient with the second highest term (of degree n − 1) is
equal −|E(G)|. Since
t(t − 1)n−1 = tn − (n − 1)tn−1 + . . . ,
it follows that |E(G)| = n − 1. Given that G is connected, we conclude
that it’s a tree.
3. Prove that the chromatic number of a connected graph is the maximum
of the chromatic numbers of its blocks.
Solution. Let k be the maximum chromatic number of a block in G.
Clearly, χ(G) ≥ k. Now we can color all the blocks of G by colors
1, 2, . . . , k, one by one. Since any two blocks share at most one vertex
and the blocks form a tree structure, we can do it in such a way as to
obtain a proper coloring of G.
Graph Theory
Spring 2012
Prof. Gábor Elek
Assist. Filip Morić
4. Suppose that G is a graph such that χ(G − {u, v}) = χ(G) − 2, for all
u, v ∈ V (G) with u 6= v. Prove that G is a complete graph.
Solution. Suppose there are two non-adjacent vertices u and v. Then
G − {u, v} can be colored properly by 1, 2, . . . , χ(G) − 2 and we can
color u and v by χ(G) − 1 to get a proper coloring of G by χ(G) − 1
colors. Contradiction.
5. Let (X, <) be a finite poset. Let G be a graph such that V (G) = X
and E(G) = {{u, v} : u, v ∈ X & u < v} . Prove that χ(G) = ω(G).
Solution. Certainly, χ(G) ≥ ω(G). We have to show that there is
a coloring with exactly ω(G) colors. Note that ω(G) is the size of a
longest chain in X. Then X can be decomposed into ω(G) antichains,
which corresponds to a decomposition of V (G) into ω(G) independent
subsets.
6. Let ` be the length of the longest path in a graph G. Prove that
χ(G) ≤ ` + 1.
Solution. First we give one short solution (update: scroll down to see
two even better solutions!) and then we prove two stronger statements
(from which the result follows trivially). Note that the length of a path
is by definition the number of edges in the path (which is not the same
as the number of vertices!).
Take a vertex v of G (which we assume to be connected) and consider
the spanning tree T of G produced by the Depth-first search algorithm
(click to see an explanation) rooted at v. By assumption the height of
the tree is at most `. Since all the vertices of depth i (i.e. at distance
i from the root in the tree) are independent (a property of depth-first
search), we can color them by color i for i = 1, 2, . . . , ` and color ` + 1
is saved for the root.
Stronger statement 1: If a graph G doesn’t contain a cycle of
length 1 modulo ` + 1, then χ(G) ≤ ` + 1 .
The proof is the same as before, except that in the depth-first search
tree we can have vertices of arbitrarily large depth. Nevertheless, for
each i = 0, 1, 2, . . . , ` we can color all the vertices of depth i modulo
` + 1 by color i, since they are independent by assumption (note that
Graph Theory
Spring 2012
Prof. Gábor Elek
Assist. Filip Morić
the root is colored 0).
Stronger statement 2: If ` is the length of the longest oriented
path in a digraph G, then χ(G) ≤ ` + 1 .
If G has no cycles, then for a fixed vertex v0 we can color all vertices
v by the maximum length of a path from v0 to v. In this coloring there
is no path that joins two vertices of the same color. If G has cycles
we remove a minimum set of edges to get an acyclic graph G0 . Then
a proper coloring of G0 as above is also good for G.
Alternative solution (Louis Bliss): By induction on `. Choose a maximal subset U ⊂ V (G) such that G[U ] has no path of length `. Then,
on one hand, V (G) \ U is an independent set (by the maximality of
U ), and, on the other hand, we can color G[U ] with at most ` colors
by induction. Thus, χ(G) ≤ ` + 1 .
Yet another solution (Corentin Perret): Observe that for any graph H
we have δ(H) ≤ `(H). Also, it’s known that χ(G) ≤ maxH⊆G δ(H)+1.
By choosing a subgraph H̄ of G such that δ(H̄) is maximum, we have
that
χ(G) ≤ δ(H̄) + 1 ≤ `(H̄) + 1 ≤ `(G) + 1 .
If you spot any mistakes on this sheet, please drop an email to [email protected].