- Email: [email protected]
When does the list-coloring function of a graph equal its chromatic polynomial
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.1 (1-7)
Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Combinatorial Theory, Series B www.elsevier.com/locate/jctb
Notes
When does the list-coloring function of a graph equal its chromatic polynomial ✩ Wei Wang a,b , Jianguo Qian a,∗ , Zhidan Yan b a b
School of Mathematical Sciences, Xiamen University, Xiamen 361005, PR China College of Information Engineering, Tarim University, Alar 843300, PR China
a r t i c l e
i n f o
Article history: Received 25 April 2016 Available online xxxx Keywords: List coloring Broken cycle Chromatic polynomial
a b s t r a c t Let G be a connected graph with n vertices and m edges. Using Whitney’s broken cycle theorem, we prove that if k > m−1 √ ≈ 1.135(m − 1) then for every k-list assignment L ln(1+ 2) of G, the number of L-colorings of G is at least that of ordinary k-colorings of G. This improves previous results of Donner (1992) and Thomassen (2009), who proved the result for k sufficiently large and k > n10 , respectively. © 2016 Published by Elsevier Inc.
1. Introduction For a positive integer k, a k-list assignment of a graph G = (V (G), E(G)) is a mapping L which assigns to each vertex v a set L(v) of k permissible colors. Given a k-list assignment L, an L-list-coloring, or L-coloring for short, is a mapping c : V (G) → ∪v∈V (G) L(v) such that c(v) ∈ L(v) for each vertex v, and c(u) = c(v) for any two adjacent vertices u and v. The notion of list coloring was introduced by Vizing [6] as well as by Erdős, Rubin and Taylor [3]. ✩ Supported by the National Natural Science Foundation of China under Grant Nos. 11471273 and 11561058. * Corresponding author. E-mail address: [email protected] (J. Qian).
http://dx.doi.org/10.1016/j.jctb.2016.08.002 0095-8956/© 2016 Published by Elsevier Inc.
JID:YJCTB 2
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.2 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
For a k-list assignment L, we use P (G, L) to denote the number of L-colorings of G and, moreover, we use Pl (G, k) to denote the minimum value of P (G, L) over all k-list assignments L of G. We note that, if L(v) = {1, 2, . . . , k} for all vertices v ∈ V (G), then an L-coloring is exactly an ordinary k-coloring [5] and therefore, P (G, L) agrees with the classic chromatic polynomial P (G, k) introduced by Birkhoff [1] in 1912. In this sense, Pl (G, k) is an analogue of the chromatic polynomial. However, it was shown that Pl (G, k) is in general not a polynomial [2], answering the problem of Kostochka and Sidorenko [4]. Following [5], we call Pl (G, k) the list-coloring function of G. This leads to an interesting question: ‘When does the list-coloring function Pl (G, x) equal the chromatic polynomial P (G, x) evaluated at k’. In [4] Kostochka and Sidorenko observed that if G is a chordal graph then Pl (G, k) = P (G, k) for any positive integer k. For a general graph G, Donner [2] and Thomassen [5] proved that Pl (G, k) = P (G, k) when k is sufficiently large. More specifically, Thomassen proved that Pl (G, k) = P (G, k) provided k > |V (G)|10 . In this note, we use Whitney’s broken cycle theorem to prove the following result. Theorem 1. For any connected graph G with m edges, if k>
m−1 √ ≈ 1.135(m − 1) ln(1 + 2)
(1)
then Pl (G, k) = P (G, k). 2. Proof of Theorem 1 Let G be a connected graph G with n vertices and m edges. Note that if m ≤ 1 then G is K1 or K2 and Theorem 1 trivially holds. In what follows we assume m ≥ 2 and, for the convenience of discussion, we label these m edges by 1, 2, . . . , m. A broken cycle of G is a set of edges obtained from the edge set of a cycle of G by removing its maximum edge. Define a set system B(G) = {S : S ⊆ E(G) and S contains no broken cycle}.
(2)
Such a system is also called a broken circuit complex; see [8] for details. We note that any cycle contains at least one broken cycle. So for each S ∈ B(G), the spanning subgraph (V (G), S) (the graph with vertex set V (G) and edge set S) contains no cycles and hence |S| ≤ n − 1. We write B(G) = B0 (G) ∪ B1 (G) ∪ · · · ∪ Bn−1 (G),
(3)
where Bi (G) = {S ∈ B(G) : |S| = i}. Note that for any S ∈ Bi (G), the subgraph (V (G), S) has exactly n − i components, all of which are trees. Now Whitney’s broken cycle theorem can be stated as follows.
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.3 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
3
Lemma 2. [7] n−1
i=0
S∈Bi (G)
(−1)i
P (G, k) =
kn−i .
(4)
Let L be a k-list assignment of G. For any tree T in G, we define β(T, L) =
L(v),
(5)
v∈V (T )
that is, β(T, L) is the number of colors common to all the lists that L assigns to the vertices of T . Whitney’s broken cycle theorem can extend easily to L-colorings as follows, which can be proved in the same way as for ordinary k-colorings. Lemma 3. P (G, L) =
n−1
i
(−1)
i=0
n−i
β(TjS , L),
(6)
S∈Bi (G) j=1
S where T1S , T2S , . . . , Tn−i are all components of the spanning subgraph (V (G), S).
For each edge e = uv ∈ E(G), let α(e, L) = k − |L(u) ∩ L(v)|.
(7)
Lemma 4. If T is a tree in G, then k ≥ β(T, L) ≥ k −
α(e, L),
(8)
e∈E(T )
where the right equality holds if |V (T )| = 1 or |V (T )| = 2. Proof. As |L(v)| = k for each v ∈ V (T ), the left inequality trivially holds. We prove the right inequality by induction on p = |V (T )|. It is easy to see that the right inequality becomes an equality when p = 1 or p = 2. Now consider the case that p > 2. Let u be a leaf vertex (a vertex with only one neighbor) and w be the only neighbor of u in T . Note that w ∈ V (T − u) and E(T ) = E(T − u) ∪ {uw}. So by the induction hypothesis, β(T, L) =
L(v)
v∈V (T )
=
v∈V (T −u)
L(v) ∩ L(u) ∩ L(w)
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.4 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
4
= ≥
L(v) + |L(u) ∩ L(w)| −
v∈V (T −u)
v∈V (T −u)
L(v) + |L(u) ∩ L(w)| − |L(w)|
v∈V (T −u)
α(e, L) + (k − α(uw, L)) − k
≥ k−
L(v) ∪ L(u) ∩ L(w)
e∈E(T −u)
=k−
α(e, L),
e∈E(T )
as desired. This completes the proof of the lemma. 2 We now estimate the difference between
S∈Bi (G)
n−i
kn−i in (4) and
S∈Bi (G)
S j=1 β(Tj , L)
in (6). To this end, we need an auxiliary inequality which can be proved by a simple induction on s. Lemma 5. Let t > 0 and a1 , a2 , . . . , as be s real numbers in [0, t]. Then s
(t − ai ) ≥ ts − ts−1
s
i=1
ai
(9)
i=1
with equality holding if at most one of these ai ’s is positive. Lemma 6. 0≤
k
n−i
−
n−i
β(TjS , L)
≤k
m−1 i−1
n−i−1
S∈Bi (G) j=1
S∈Bi (G)
with the right equality holding if i = 0, 1, where integer.
r s
α(e, L),
(10)
e∈E(G)
is treated as 0 if s is a negative
Proof. As k ≥ β(TjS , L), the left inequality in (10) is clear. As β(TjS , L) ≥ 0, the right inequality in (8) implies β(TjS , L) ≥ k − min k,
α(e, L) .
(11)
e∈E(TjS )
Thus, by Lemma 5, kn−i −
n−i j=1
β(TjS , L) ≤ kn−i −
n−i j=1
k − min k,
e∈E(TjS )
α(e, L)
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.5 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
≤ kn−i−1
n−i
min k,
j=1
≤ kn−i−1
n−i
5
α(e, L)
e∈E(TjS )
α(e, L)
j=1 e∈E(TjS )
= kn−i−1
α(e, L).
(12)
e∈S
Let Ei (G) = {S : S ⊆ E(G) and |S| = i} and for each e ∈ E(G), Eei (G) = {S : S ⊆ E(G) , |S| = i and e ∈ S}. Clearly Bi (G) ⊆ Ei (G) and |Eei (G)| =
(13)
m−1 i−1 . Therefore, by (12),
S∈Bi (G)
kn−i −
n−i
β(TjS , L) ≤
S∈Bi (G) j=1
kn−i−1
S∈Bi (G)
≤ kn−i−1
α(e, L)
e∈S
α(e, L)
S∈Ei (G) e∈S
= kn−i−1
α(e, L)
e∈E(G) S∈Eei (G)
= kn−i−1
m−1 i−1
α(e, L),
(14)
e∈E(G)
as desired. When i = 0, 1, the above argument becomes trivial. Indeed, all inequalities in (11)–(14) become equalities and hence the right equality in (10) holds. This completes the proof. 2 Proof of Theorem 1 By the definition of Pl (G, k), it trivially follows that Pl (G, k) ≤ P (G, k) for any graph G and any positive integer k. Let L be an arbitrary k-list assignment of G with k>
m−1 √ . ln(1 + 2)
(15)
JID:YJCTB 6
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.6 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
For simplicity, we write α =
e∈E(G)
fi =
α(e, L) and
kn−i −
n−i
By Lemma 6, f0 = 0, f1 = αkn−2 and 0 ≤ fi ≤ α Lemmas 2 and 3, n−1
(16)
S∈Bi (G) j=1
S∈Bi (G)
P (G, L) − P (G, k) =
β(TjS , L).
m−1 i−1
kn−i−1 for 2 ≤ i ≤ n − 1. By
(−1)i−1 fi
i=0
≥ f1 −
fi
2≤i≤n−1 i even
= f1 −
fi+1
1≤i≤n−2 i odd
≥ αkn−2 −
α
m − 1 n−i−2 k i
α
(m − 1)i n−i−2 k i!
1≤i≤n−2 i odd
≥ αkn−2 −
1≤i≤n−2 i odd
= αk
n−2
1−
1≤i≤n−2 i odd
i 1 m−1 i! k
1 m−1 m−1 exp − exp − . ≥ αkn−2 1 − 2 k k
(17)
Consider the function 1 φ(x) = 1 − (exp(x) − exp(−x)). 2 √ Let x0 = ln(1 + 2). It is easy to check that φ(x) is monotone decreasing and φ(x0 ) = 0. By (15), we have m−1 < x0 and hence P (G, L) − P (G, k) ≥ 0 by (17), which implies k Pl (G, k) ≥ P (G, k) from the arbitrariness of L. The proof of Theorem 1 is completed. Acknowledgments We thank the anonymous referee for her or his careful reading and valuable suggestions.
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.7 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
7
References [1] G.D. Birkhoff, A determinant formula for the number of ways of coloring a map, Ann. Math. 14 (1912) 42–46. [2] Q. Donner, On the number of list-colorings, J. Graph Theory 16 (3) (1992) 239–245. [3] P. Erdős, A.L. Rubin, H. Taylor, Choosability in graphs, Congr. Numer. 26 (1979) 125–157. [4] A.V. Kostochka, A.F. Sidorenko, Problems proposed at the problem session of the Prachatice conference on graph theory, Ann. Discrete Math. 51 (1992) 380. [5] C. Thomassen, The chromatic polynomial and list colorings, J. Combin. Theory Ser. B 99 (2) (2009) 474–479. [6] V.G. Vizing, Coloring the vertices of a graph in prescribed colors, Diskretn. Anal. 29 (1976) 3–10. [7] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc. 38 (1932) 572–579. [8] H.S. Wilf, Which polynomials are chromatic?, in: Proc. 1973 Rome International Colloq. Combinatorial Theory I, Accademia Nazionale dei Lincei, Rome, 1976, pp. 247–257.
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.1 (1-7)
Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
Contents lists available at ScienceDirect
Journal of Combinatorial Theory, Series B www.elsevier.com/locate/jctb
Notes
When does the list-coloring function of a graph equal its chromatic polynomial ✩ Wei Wang a,b , Jianguo Qian a,∗ , Zhidan Yan b a b
School of Mathematical Sciences, Xiamen University, Xiamen 361005, PR China College of Information Engineering, Tarim University, Alar 843300, PR China
a r t i c l e
i n f o
Article history: Received 25 April 2016 Available online xxxx Keywords: List coloring Broken cycle Chromatic polynomial
a b s t r a c t Let G be a connected graph with n vertices and m edges. Using Whitney’s broken cycle theorem, we prove that if k > m−1 √ ≈ 1.135(m − 1) then for every k-list assignment L ln(1+ 2) of G, the number of L-colorings of G is at least that of ordinary k-colorings of G. This improves previous results of Donner (1992) and Thomassen (2009), who proved the result for k sufficiently large and k > n10 , respectively. © 2016 Published by Elsevier Inc.
1. Introduction For a positive integer k, a k-list assignment of a graph G = (V (G), E(G)) is a mapping L which assigns to each vertex v a set L(v) of k permissible colors. Given a k-list assignment L, an L-list-coloring, or L-coloring for short, is a mapping c : V (G) → ∪v∈V (G) L(v) such that c(v) ∈ L(v) for each vertex v, and c(u) = c(v) for any two adjacent vertices u and v. The notion of list coloring was introduced by Vizing [6] as well as by Erdős, Rubin and Taylor [3]. ✩ Supported by the National Natural Science Foundation of China under Grant Nos. 11471273 and 11561058. * Corresponding author. E-mail address: [email protected] (J. Qian).
http://dx.doi.org/10.1016/j.jctb.2016.08.002 0095-8956/© 2016 Published by Elsevier Inc.
JID:YJCTB 2
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.2 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
For a k-list assignment L, we use P (G, L) to denote the number of L-colorings of G and, moreover, we use Pl (G, k) to denote the minimum value of P (G, L) over all k-list assignments L of G. We note that, if L(v) = {1, 2, . . . , k} for all vertices v ∈ V (G), then an L-coloring is exactly an ordinary k-coloring [5] and therefore, P (G, L) agrees with the classic chromatic polynomial P (G, k) introduced by Birkhoff [1] in 1912. In this sense, Pl (G, k) is an analogue of the chromatic polynomial. However, it was shown that Pl (G, k) is in general not a polynomial [2], answering the problem of Kostochka and Sidorenko [4]. Following [5], we call Pl (G, k) the list-coloring function of G. This leads to an interesting question: ‘When does the list-coloring function Pl (G, x) equal the chromatic polynomial P (G, x) evaluated at k’. In [4] Kostochka and Sidorenko observed that if G is a chordal graph then Pl (G, k) = P (G, k) for any positive integer k. For a general graph G, Donner [2] and Thomassen [5] proved that Pl (G, k) = P (G, k) when k is sufficiently large. More specifically, Thomassen proved that Pl (G, k) = P (G, k) provided k > |V (G)|10 . In this note, we use Whitney’s broken cycle theorem to prove the following result. Theorem 1. For any connected graph G with m edges, if k>
m−1 √ ≈ 1.135(m − 1) ln(1 + 2)
(1)
then Pl (G, k) = P (G, k). 2. Proof of Theorem 1 Let G be a connected graph G with n vertices and m edges. Note that if m ≤ 1 then G is K1 or K2 and Theorem 1 trivially holds. In what follows we assume m ≥ 2 and, for the convenience of discussion, we label these m edges by 1, 2, . . . , m. A broken cycle of G is a set of edges obtained from the edge set of a cycle of G by removing its maximum edge. Define a set system B(G) = {S : S ⊆ E(G) and S contains no broken cycle}.
(2)
Such a system is also called a broken circuit complex; see [8] for details. We note that any cycle contains at least one broken cycle. So for each S ∈ B(G), the spanning subgraph (V (G), S) (the graph with vertex set V (G) and edge set S) contains no cycles and hence |S| ≤ n − 1. We write B(G) = B0 (G) ∪ B1 (G) ∪ · · · ∪ Bn−1 (G),
(3)
where Bi (G) = {S ∈ B(G) : |S| = i}. Note that for any S ∈ Bi (G), the subgraph (V (G), S) has exactly n − i components, all of which are trees. Now Whitney’s broken cycle theorem can be stated as follows.
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.3 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
3
Lemma 2. [7] n−1
i=0
S∈Bi (G)
(−1)i
P (G, k) =
kn−i .
(4)
Let L be a k-list assignment of G. For any tree T in G, we define β(T, L) =
L(v),
(5)
v∈V (T )
that is, β(T, L) is the number of colors common to all the lists that L assigns to the vertices of T . Whitney’s broken cycle theorem can extend easily to L-colorings as follows, which can be proved in the same way as for ordinary k-colorings. Lemma 3. P (G, L) =
n−1
i
(−1)
i=0
n−i
β(TjS , L),
(6)
S∈Bi (G) j=1
S where T1S , T2S , . . . , Tn−i are all components of the spanning subgraph (V (G), S).
For each edge e = uv ∈ E(G), let α(e, L) = k − |L(u) ∩ L(v)|.
(7)
Lemma 4. If T is a tree in G, then k ≥ β(T, L) ≥ k −
α(e, L),
(8)
e∈E(T )
where the right equality holds if |V (T )| = 1 or |V (T )| = 2. Proof. As |L(v)| = k for each v ∈ V (T ), the left inequality trivially holds. We prove the right inequality by induction on p = |V (T )|. It is easy to see that the right inequality becomes an equality when p = 1 or p = 2. Now consider the case that p > 2. Let u be a leaf vertex (a vertex with only one neighbor) and w be the only neighbor of u in T . Note that w ∈ V (T − u) and E(T ) = E(T − u) ∪ {uw}. So by the induction hypothesis, β(T, L) =
L(v)
v∈V (T )
=
v∈V (T −u)
L(v) ∩ L(u) ∩ L(w)
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.4 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
4
= ≥
L(v) + |L(u) ∩ L(w)| −
v∈V (T −u)
v∈V (T −u)
L(v) + |L(u) ∩ L(w)| − |L(w)|
v∈V (T −u)
α(e, L) + (k − α(uw, L)) − k
≥ k−
L(v) ∪ L(u) ∩ L(w)
e∈E(T −u)
=k−
α(e, L),
e∈E(T )
as desired. This completes the proof of the lemma. 2 We now estimate the difference between
S∈Bi (G)
n−i
kn−i in (4) and
S∈Bi (G)
S j=1 β(Tj , L)
in (6). To this end, we need an auxiliary inequality which can be proved by a simple induction on s. Lemma 5. Let t > 0 and a1 , a2 , . . . , as be s real numbers in [0, t]. Then s
(t − ai ) ≥ ts − ts−1
s
i=1
ai
(9)
i=1
with equality holding if at most one of these ai ’s is positive. Lemma 6. 0≤
k
n−i
−
n−i
β(TjS , L)
≤k
m−1 i−1
n−i−1
S∈Bi (G) j=1
S∈Bi (G)
with the right equality holding if i = 0, 1, where integer.
r s
α(e, L),
(10)
e∈E(G)
is treated as 0 if s is a negative
Proof. As k ≥ β(TjS , L), the left inequality in (10) is clear. As β(TjS , L) ≥ 0, the right inequality in (8) implies β(TjS , L) ≥ k − min k,
α(e, L) .
(11)
e∈E(TjS )
Thus, by Lemma 5, kn−i −
n−i j=1
β(TjS , L) ≤ kn−i −
n−i j=1
k − min k,
e∈E(TjS )
α(e, L)
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.5 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
≤ kn−i−1
n−i
min k,
j=1
≤ kn−i−1
n−i
5
α(e, L)
e∈E(TjS )
α(e, L)
j=1 e∈E(TjS )
= kn−i−1
α(e, L).
(12)
e∈S
Let Ei (G) = {S : S ⊆ E(G) and |S| = i} and for each e ∈ E(G), Eei (G) = {S : S ⊆ E(G) , |S| = i and e ∈ S}. Clearly Bi (G) ⊆ Ei (G) and |Eei (G)| =
(13)
m−1 i−1 . Therefore, by (12),
S∈Bi (G)
kn−i −
n−i
β(TjS , L) ≤
S∈Bi (G) j=1
kn−i−1
S∈Bi (G)
≤ kn−i−1
α(e, L)
e∈S
α(e, L)
S∈Ei (G) e∈S
= kn−i−1
α(e, L)
e∈E(G) S∈Eei (G)
= kn−i−1
m−1 i−1
α(e, L),
(14)
e∈E(G)
as desired. When i = 0, 1, the above argument becomes trivial. Indeed, all inequalities in (11)–(14) become equalities and hence the right equality in (10) holds. This completes the proof. 2 Proof of Theorem 1 By the definition of Pl (G, k), it trivially follows that Pl (G, k) ≤ P (G, k) for any graph G and any positive integer k. Let L be an arbitrary k-list assignment of G with k>
m−1 √ . ln(1 + 2)
(15)
JID:YJCTB 6
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.6 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
For simplicity, we write α =
e∈E(G)
fi =
α(e, L) and
kn−i −
n−i
By Lemma 6, f0 = 0, f1 = αkn−2 and 0 ≤ fi ≤ α Lemmas 2 and 3, n−1
(16)
S∈Bi (G) j=1
S∈Bi (G)
P (G, L) − P (G, k) =
β(TjS , L).
m−1 i−1
kn−i−1 for 2 ≤ i ≤ n − 1. By
(−1)i−1 fi
i=0
≥ f1 −
fi
2≤i≤n−1 i even
= f1 −
fi+1
1≤i≤n−2 i odd
≥ αkn−2 −
α
m − 1 n−i−2 k i
α
(m − 1)i n−i−2 k i!
1≤i≤n−2 i odd
≥ αkn−2 −
1≤i≤n−2 i odd
= αk
n−2
1−
1≤i≤n−2 i odd
i 1 m−1 i! k
1 m−1 m−1 exp − exp − . ≥ αkn−2 1 − 2 k k
(17)
Consider the function 1 φ(x) = 1 − (exp(x) − exp(−x)). 2 √ Let x0 = ln(1 + 2). It is easy to check that φ(x) is monotone decreasing and φ(x0 ) = 0. By (15), we have m−1 < x0 and hence P (G, L) − P (G, k) ≥ 0 by (17), which implies k Pl (G, k) ≥ P (G, k) from the arbitrariness of L. The proof of Theorem 1 is completed. Acknowledgments We thank the anonymous referee for her or his careful reading and valuable suggestions.
JID:YJCTB
AID:3027 /SCO
[m1L; v1.185; Prn:29/08/2016; 13:41] P.7 (1-7)
W. Wang et al. / Journal of Combinatorial Theory, Series B ••• (••••) •••–•••
7
References [1] G.D. Birkhoff, A determinant formula for the number of ways of coloring a map, Ann. Math. 14 (1912) 42–46. [2] Q. Donner, On the number of list-colorings, J. Graph Theory 16 (3) (1992) 239–245. [3] P. Erdős, A.L. Rubin, H. Taylor, Choosability in graphs, Congr. Numer. 26 (1979) 125–157. [4] A.V. Kostochka, A.F. Sidorenko, Problems proposed at the problem session of the Prachatice conference on graph theory, Ann. Discrete Math. 51 (1992) 380. [5] C. Thomassen, The chromatic polynomial and list colorings, J. Combin. Theory Ser. B 99 (2) (2009) 474–479. [6] V.G. Vizing, Coloring the vertices of a graph in prescribed colors, Diskretn. Anal. 29 (1976) 3–10. [7] H. Whitney, A logical expansion in mathematics, Bull. Amer. Math. Soc. 38 (1932) 572–579. [8] H.S. Wilf, Which polynomials are chromatic?, in: Proc. 1973 Rome International Colloq. Combinatorial Theory I, Accademia Nazionale dei Lincei, Rome, 1976, pp. 247–257.