In this section we prove the second part of the main theorem, namely:
Theorem 5
LetGbe a graph with girth at least 9. IfGhas a good set, thenχ
b
(G)=m(G).
Let W={v1,…,vm(G)} be a good set of G. Our aim is to construct a b-coloring of G with m(G) colors such that, for each i∈{1,…,m(G)}, vertex v
i
is a b-vertex of color i. We start by assigning color i to v
i
, for each i∈{1,…,m(G)}. Next, we extend this partial coloring to the rest of the graph in several steps. Before explaining each step, we need to introduce some other terminology and notation.
A link is any path of length two or three whose extremities are in W and whose internal vertices are not in W. Any interior vertex of a link is called a link vertex. Let L be the set of all link vertices.
We first color G[W∪L] in a way not to repeat too many colors in N(w), for all w∈W, and at the end we extend the obtained partial coloring to a b-coloring of G with m(G) colors. Let G′=G[W∪L], L1 be the set of vertices of L that have at least one neighbor in L and L2 be the set of vertices in L that have at least two neighbors in W. The steps below are followed in order in such a way that we only move on to the next step when all the possible vertices are iterated.
-
1.
For each x∈L1, let x′∈NL(x). Since x′∈L, there must exist v
i
∈NW(x′); color x with i;
-
2.
For each v
i
∈W, let
. Also, let
. If q>1, then use colors i1,…,i
q
to color the uncolored vertices in
in a way that x
j
is not colored with i
j
(it suffices to make a derangement of those colors on the vertices);
-
3.
Let x∈L2 still uncolored be such that there exists v
i
∈NW(x) that has some neighbor y∈L1. Let c be the color of y; color x with c and recolor y with j, for any v
j
∈NW(x)∖{v
i
};
-
4.
Finally, if x∈L2 is still uncolored, we know that NL(v
i
)={x}, for all v
i
∈NW(x). Since NL(x)=∅, we can color x with i, for any v
i
that is not adjacent to x and has no common neighbor with x in W of degree m(G)−1, which exists as x is not encircled by W.
Suppose that the algorithm above produces a partial coloring that colors every vertex in L in such a way that, at the end, each v
i
∈W has at least as many uncolored neighbors as missing colors in its neighborhood. Since L is colored, we know that the uncolored neighbors of W form a stable set. Thus, we can independently color N(v
i
) in such a way that v
i
sees every other color, for all v
i
∈W. By the definition of a good set, we know that if d(v)≥m(G), then v is already colored; hence, the partial coloring can be greedily transformed into a b-coloring with m(G) colors. Now, to prove that the algorithm works, we show that after these steps the obtained partial coloring ψ satisfies:
P1 ψ is proper; and
P2 the number of uncolored neighbors of v
i
is at least the number of missing colors in N(v
i
), for each v
i
∈W.
Proof of Theorem 5
First, we make some observations concerning the coloring procedure. Note that L1∩L2 is not necessarily empty, but all vertices in this subset are colored in Step 1. However, a vertex x∈L1∩L2 may play a role in Step 2 in the following way: if x∈N(v
i
) and there exists
, then x′ may be colored with color j for some v
j
∈NW(x)∖{v
i
}, while the color of x remains unchanged. Also, note that, in Step 3, since
, we have y∈L1∖L2. Hence, NW(y)={v
i
} and, consequently, the color of y cannot be changed again. Thus (*) the color of y is changed at most once, for every y∈L1. Finally, if x receives color i in Step 1, 2 or 3, then one of the following holds (fact (iii) holds because of (*)):
-
(i)
x receives color i in Step 1 and there exists a path 〈x,x′,v
i
〉, for some x′∈L1; or
-
(ii)
x receives color i in Step 2 and there exists a path 〈x,v
j
,x′,v
i
〉, for some v
j
∈W and x′∈L2; or
-
(iii)
x receives color i in Step 3 and there exists a path 〈x,v
j
,y,y′,v
i
〉, for some v
j
∈W, y∈L1∖L2 and y′∈L1; or
-
(iv)
x is recolored with color i in Step 3 and there exists a path 〈x,v
j
,x′,v
i
〉, for some v
j
∈W and x′∈L2∖L1.
We first prove that P1 holds after Step 3. Suppose that there exists an edge wz such that ψ(w)=ψ(z)=i. Since G has no cycle of length at most 7, the paths defined in (i)–(iv) are shortest paths. Therefore, vertex v
i
has no neighbor colored i and hence, w,z∈L. Also, as wz∈E(G), we have w,z∈L1 and they are colored in Step 1 and maybe recolored in Step 3. By (i) and (iv), there exist a w,v
i
-path P
w
and a z,v
i
-path P
z
, both of length at most 3. Note that either P
w
+P
z
+wz contains a cycle of length at most 7 or one of these paths consists of the edge wz followed by the other path. Because G has girth at least 9, the latter case occurs. We get as contradiction as this implies that at least one path is defined by (i) and, thus, vertex v
i
has a neighbor colored i.
Now, we prove that P2 also holds after Step 3. We actually prove that, after Step 3, no color is repeated in N(v
i
), for each v
i
∈W. Suppose there exist a vertex v
j
∈W and w,z∈N(v
j
) such that ψ(w)=ψ(z)=i. First, consider the case v
i
∈{w,z}. Since the paths defined by (i)–(iv) are shortest paths, we see that (i) occurs for the vertex in {w,z}∖{v
i
}. We get a contradiction as this implies G has a cycle of length 4. Therefore we may assume v
i
∉{w,z}.
Now, by (i)–(iv), there exist a w,v
i
-path P
w
and a z,v
i
-path P
z
. Let ℓ
w
and ℓ
z
be the length of P
w
and P
z
, respectively. Clearly ℓ
w
,ℓ
z
≤4. Note that either P
w
+P
z
+〈w,v
j
,z〉 contains a cycle of length at most ℓ
w
+ℓ
z
+2 or either P
w
or P
z
consists of the path 〈w,v
j
,z〉 followed by the other path. Since both w and z are at distance at least 2 from v
i
and ℓ
w
,ℓ
z
≤4, the latter can only occur if one of the paths is defined by (i), say P
w
, and the other is defined by (iii), say P
z
. We get a contradiction as P
z
=〈z,v
j
,w,y,v
i
〉 implies w is recolored in Step 3 and therefore, P
w
must be defined by (iv). Now, suppose that the former occurs, i.e., P
w
+P
z
+〈w,v
j
,z〉 contains a cycle of length at most ℓ
w
+ℓ
z
+2. Because G has girth at least 9, we have ℓ
w
+ℓ
z
≥7. This implies that at least one of P
w
and P
z
, say P
z
, is defined by (iii), and the other is not defined by (i). Therefore z is colored in Step 3 and
. Furthermore, w∈L1∖L2 and NW(w)={v
j
}. Therefore, since (i) does not occur for w, we find that P
w
must be defined by (iv). Thus the only choice for P
w
is 〈w,v
j
,z,v
i
〉, a contradiction as P1 holds.
Finally, consider x to be colored during Step 4 with color i. By the choice of i we know that v
i
∉N(x). Thus, since NL(x)=∅, Property P1 holds. Now, suppose that some v
j
∈N(x) is such that color i already appears in N(v
j
). Since NL(v
j
)={x} we must have v
i
∈N(v
j
) and, by the choice of i, d(v
j
)>m(G)−1. Property P2 thus follows as i is the only repeated color in the neighborhood of v
j
. □