arXiv:1103.3524v1 [math.CO] 17 Mar 2011
A Fractional Analogue of Brooks’ Theorem Linyuan Lu
∗
Xing Peng
†
March 21, 2011
Abstract Let ∆(G) be the maximum degree of a graph G. Brooks’ theorem states that the only connected graphs with chromatic number χ(G) = ∆(G) + 1 are complete graphs and odd cycles. We prove a fractional analogue of Brooks’ theorem in this paper. Namely, we classify all connected graphs G such that the fractional chromatic number χf (G) is at least ∆(G). These graphs are complete graphs, odd cycles, C82 , C5 ⊠ K2 , and graphs whose clique number ω(G) equals the maximum degree ∆(G). Among the two sporadic graphs, the graph C82 is the square graph of cycle C8 while the other graph C5 ⊠ K2 is the strong product of C5 and K2 . In fact, we prove a stronger result; if a connected graph G with ∆(G) ≥ 3 is not one of the graphs listed above, then we have 1 χf (G) ≤ ∆(G) − 4(∆(G)−1) 5.
1
Introduction
The chromatic number of graphs with bounded degrees has been studied for many years. Brooks’ theorem perhaps is one of the most fundamental results; it is included by many textbooks on graph theory. Given a simple connected graph G, let ∆(G) be the maximum degree, ω(G) be the clique number, and χ(G) be the chromatic number. Brooks’ theorem states that χ(G) ≤ ∆(G) unless G is a complete graph or an odd cycle. Reed [8] proved that χ(G) ≤ ∆(G) − 1 if ω(G) ≤ ∆(G) − 1 and ∆(G) ≥ ∆0 for some large constant ∆0 . This excellent result was proved by probabilistic methods, and ∆0 is at least hundreds. Before this result, Borodin and Kostochka [1] made the following conjecture. Conjecture [1]: Suppose that G is a connected graph. If ω(G) ≤ ∆(G) − 1 and ∆(G) ≥ 9, then we have χ(G) ≤ ∆(G) − 1. If the conjecture is true, then it is best possible since there is a K8 -free graph G = C5 ⊠K3 (actually K7 -free, see Figure 1) with ∆(G) = 8 and χ(G) = 8. Here we use the following notation of the strong product. Given two graphs G and H, the strong product G ⊠ H is the graph with vertex set V (G) × V (H), and (a, x) is connected to (b, y) if one of the followings holds • a = b and xy ∈ E(H), • ab ∈ E(G) and x = y, • ab ∈ E(G) and xy ∈ E(H). ∗ University
of South Carolina, Columbia, SC 29208, (
[email protected]). This author was supported in part by NSF grant DMS 1000475. † University of South Carolina, Columbia, SC 29208, (
[email protected]). This author was supported in part by NSF grant DMS 1000475.
1
Figure 1: The graph C5 ⊠ K3 Reed’s result [8] settled Borodin and Kostochka’s conjecture for sufficiently large ∆(G), but the cases with small ∆(G) are hard to cover using the probabilistic method. In this paper we consider a fractional analogue of this problem. The fractional chromatic number χf (G) can be defined as follows. A b-fold coloring of G assigns a set of b colors to each vertex such that any two adjacent vertices receive disjoint sets of colors. We say a graph G is a:b-colorable if there is a b-fold coloring of G in which each color is drawn from a palette of a colors. We refer to such a coloring as an a:b-coloring. The b-fold coloring number, denoted by χb (G), is the smallest integer a such that G has an a:b-coloring. Note that χ1 (G) = χ(G). It was shown that χa+b (G) ≤ χa (G) + χb (G). The fractional chromatic number χf (G) is lim χb (G) b . b→∞
By the definition, we have χf (G) ≤ χ(G). The fractional chromatic number can be viewed as a relaxation of the chromatic number. Many problems involving the chromatic number can be asked again using the fractional chromatic number. The fractional analogue often has a simpler solution than the original problem. For example, the famous ω − ∆ − χ conjecture of Reed [7] states that for any simple graph G, we have ω(G) + ∆(G) + 1 . χ(G) ≤ 2 The fractional analogue of ω − ∆ − χ conjecture was proved by Molloy and Reed [6]; they actually proved a stronger result with ceiling removed. In this paper, we classify all connected graphs G with χf (G) ≥ ∆(G). Theorem 1 A connected graph G satisfies χf (G) ≥ ∆(G) if and only if G is one of the following 1. a complete graph, 2. an odd cycle, 3. a graph with ω(G) = ∆(G), 4. C82 , 5. C5 ⊠ K2 .
2
For the complete graph Kn , we have χf (Kn ) = n and ∆(Kn ) = n − 1. For the odd cycle C2k+1 , we have χf (C2k+1 ) = 2 + k1 and ∆(C2k+1 ) = 2. If G is neither a complete graph nor an odd cycle but contains a clique of size ∆(G), then we have ∆(G) ≤ ω(G) ≤ χf (G) ≤ χ(G) ≤ ∆(G).
(1)
The last inequality is from Brooks’ theorem. The sequence of inequalities above implies χf (G) = ∆(G). If G is a vertex-transitive graph, then we have [9] χf (G) =
|V (G)| , α(G)
where α(G) is the independence number of G. Note that both graphs C82 and C5 ⊠ K2 are vertex-transitive and have the independence number 2. Thus we have χf (C82 ) = 4 = ∆(C82 )
and
χf (C5 ⊠ K2 ) = 5 = ∆(C5 ⊠ K2 ).
C82
C5 ⊠ K2 Figure 2: The graph C82 and C5 ⊠ K2
Actually, Theorem 1 is a corollary of the following stronger result. Theorem 2 Assume that a connected graph G is neither C82 nor C5 ⊠ K2 . If ∆(G) ≥ 3 and ω(G) ≤ ∆(G) − 1, then we have χf (G) ≤ ∆(G) −
1 . 4(∆(G) − 1)5
Remark: The condition ∆(G) ≥ 3 is necessary because a path Pn and an even cycle Cn (for n ≥ 3) have maximum degree 2 and fractional chromatic number 2. Other conditions exclude those graphs listed in Theorem 1. For any connected graph G with sufficiently large ∆(G) and ω(G) ≤ ∆(G) − 1, Reed’s result [8] χ(G) ≤ ∆(G) − 1 implies χf (G) ≤ ∆(G) − 1. We make no effort to get the best bound on ∆(G) − χf (G). In the case ∆(G) = 3, Heckman and Thomas [4] conjectured that χf (G) ≤ 14/5 if G is triangle-free. Hatami and Zhu [3] 3 proved χf (G) ≤ 3 − 64 for any triangle-free graph G with ∆(G) = 3. The authors in a 3 previous paper [5] showed an improved result χf (G) ≤ 3 − 43 . Thus we need only consider the cases ∆(G) ≥ 4. The readers are encouraged to read these two papers [3, 5]. The method introduced in [3], and strengthened in [5], has a strong influence on this paper. Let f (k) = inf G {∆(G) − χf (G)}, where the infimum is taken over all connected graphs G with ∆(G) = k and not one of the graphs listed in theorem 1. Theorem 2 implies f (k) > 0 for all k ≥ 3. Reed’s result [8] implies f (k) ≥ 1 for sufficiently large k. Heckman and Thomas [4] conjectured f (3) = 1/5. It is an interesting problem to determine the value f (k) for small 3
k. Is f (4) = 13 ? We don’t know whether f (k) is an increasing function of k. If Borodin and Kostochka’s conjecture is true, then f (9) = 1. The rest of this paper is organized as follows. In section 2, we will introduce some notations and prove some Lemmas. The proof of Lemma 1 is quite long; we postpone the proof of cases ∆(G) = 4, 5 until Section 4. The proof of Theorem 2 will be given in section 3.
2
Notation and Lemmas
In this paper, we use the following notation. Let G be a simple graph with vertex set V (G) and edge set E(G). The neighborhood of a vertex v in G, denoted by ΓG (v), is the set {u : uv ∈ E(G)}. The degree dG (v) of v is the value of |ΓG (v)|. The independence number α(G) is the size of the largest independent set in G. When T ⊂ V (G), we use αG (T ) to denote the independence number of the induced subgraph of G on T . Let ∆(G) be the maximum degree of G. For any two vertex-sets S and T , we define EG (S, T ) as {uv ∈ E(G) : u ∈ S and v ∈ T }. Whenever G is clear under context, we will drop the subscript G for simplicity. If S is a subset of vertices in G, then contracting S means replacing vertices in S by a single fat vertex, denoted by S, whose incident edges are all edges that were incident to at least one vertex in S, except edges with both ends in S. The new graph obtained by contracting S is denoted by G/S. This operation is also known as identifying vertices of S in the literature. For completeness, we allow S to be a single vertex or even the empty set. If S only consists of a single vertex, then G/S = G; if S = ∅, then G/S is the union of G and an isolated vertex. When S consists of 2 or 3 vertices, for convenience, we write G/uv for G/{u, v} and G/uvw for G/{u, v, w}; the fat vertex will be denoted by uv and uvw, respectively. Given two disjoint subsets S1 and S2 , we can contract S1 and S2 sequentially. The order of contractions does not matter; let G/S1 /S2 be the resulted graph. To prove Theorem 2, we will use an approach which is similar to the one in [3, 5]. It suffices to prove that the minimum counter-example does not exist. For an integer k ≥ 4, let Gk be the set of connected graphs G satisfying 1. ∆(G) = k and ω(G) ≤ k − 1; 2. χf (G) > k −
1 4(k−1)5 ;
3. if k = 4, then G 6= C82 ; if k = 5, then G 6= C5 ⊠ K2 . For any G ∈ Gk and any subgraph H of G, if χf (H) ≥ χf (G), then H ∈ Gk . This observation is useful when we consider the minimal graphs in Gk . To prove Theorem 2, it is sufficient to show Gk is the empty set for all k ≥ 4. We can use proof by contradiction. Suppose that Gk 6= ∅ for some k ≥ 4. For this particular k, let G = G(k) be a graph with the smallest number of vertices which satisfies G ∈ Gk . By 1 minimality, each vertex in G has degree either k or k − 1. We will show χf (G) ≤ k − 4(k−1) 5, which gives us the desired contradiction. For a given vertex x in V (G), it is easy to color its neighborhood ΓG (x) using k − 2 colors. If dG (x) = k − 1, then we pick a non-edge S from ΓG (x) and color the two vertices in S using one color. If dG (x) = k and α(ΓG (x)) ≥ 3, then we pick an independent set S in ΓG (x) of size 3, and assign the same color to each vertex in S. If dG (x) = k and α(ΓG (x)) = 2, then we pick two disjoint non-edges S1 and S2 from ΓG (x) ; we assign one color to each vertex in S1 and another color to each vertex in S2 . Since G is Kk -free, this local coloring scheme on {x} ∪ ΓG (x) always works.
4
The following Lemma shows that G has a key property, which eventually implies that this local coloring scheme works simultaneously for x in a large subset of V (G). Lemma 1 Suppose that Gk 6= ∅ for some k ≥ 4. For this particular k, let G be a graph in Gk with the smallest number of vertices. For each x ∈ V (G) with dG (x) = k and α(ΓG (x)) = 2, there exist two vertex-disjoint non-edges S1 (x), S2 (x) ⊂ ΓG (x) satisfying the following property. If we contract S1 (x) and S2 (x), then the resulted graph G/S1 (x)/S2 (x) contains − − no subgraph Kk+1 . Here Kk+1 is the graph obtained by removing one edge from the complete graph Kk+1 . Whenever x is clear under the context, we write S1 for S1 (x) and S2 for S2 (x). Suppose that S1 and S2 are two disjoint non-edges in ΓG (x) and H is a clique of size k − 1 in V (G) \ ({x} ∪ Γ(x)). A triple (S1 , S2 , H) is bad if the induced subgraph of G/S1 /S2 on − H ∪ {S1 , S2 } contains a subgraph Kk+1 ; it is good otherwise. If (S1 , S2 , H) is a bad triple, then we have |E(S1 ∪ S2 , H)| ≥ 2k − 3. (2) − Since G is Kk -free, any Kk+1 in G/S1 /S2 must contain S1 and S2 . If Lemma 1 fails, then for any S1 and S2 , there is a bad triple (S1 , S2 , H). The subsequent lemmas will limit the possibilities of bad triples and eventually prove Lemma 1 for k ≥ 6.
Lemma 2 Suppose Lemma 1 fails for some vertex x. If (S11 , S21 , H1 ) and (S12 , S22 , H2 ) are two bad triples, then we have either H1 = H2 or H1 ∩ H2 = ∅. Proof: We prove the lemma by contradiction. Suppose H1 6= H2 and H1 ∩ H2 6= ∅. For any t ∈ H1 ∩ H2 , we have k ≥ dG (t) ≥ 2k − 3 − |H1 ∩ H2 |. We have |H1 ∩ H2 | = k − 2 or k − 3. If |H1 ∩ H2 | = k − 2, then each vertex t in H1 ∩ H2 has k − 1 neighbors in H1 ∪ H2 already. Thus t can have at most one neighbor in ΓG (x). In consequence, we have H1 ∪ {S11 , S21 } misses at least k − 2 edges. We have a contradiction as k − 2 ≥ 2 and the assumption of H. If |H1 ∩ H2 | = k − 3 ≥ 1, then each vertex t in H1 ∩ H2 has k neighbors in H1 ∪ H2 already. The vertex t has no neighbor in ΓG (x). Thus H1 ∪ {S11 , S21 } misses at least two edges tS11 and tS21 . Contradiction. Lemma 3 Suppose Lemma 1 fails for some vertex x. If (S11 , S21 , H1 ) and (S12 , S22 , H2 ) are two possible bad triples, then we have H1 = H2 . Proof: Suppose H1 6= H2 . In this case, H1 and H2 are disjoint by Lemma 2. Let S = S11 ∪ S21 ∪ S12 ∪ S22 . We assume G|S has s vertices and e edges. Now we count the edges from S to H1 ∪ H2 in two ways. On the one hand, from equation (2), every vertex in Hi (i = 1, 2) has at least 2k − 3 neighbor in S. We have |E(S, H1 ∪ H2 )| = |E(S, H1 )| + |E(S, H2 )| ≥ 2(2k − 3). On the other hand, every vertex in ΓG (x) \ S has at least 2 neighbors in S because of α(ΓG (x)) = 2. There are at least e edges with both ends in S. Together with s edges from x to S, we have |E(S, H1 ∪ H2 )| ≤ sk − 2(k − s) − 2 × e − s = (s − 2)k + s − 2e.
5
Thus 2(2k − 3) ≤ (s − 2)k + s − 2e.
(3)
Note that 4 ≤ s ≤ 8. We will consider the following cases according to the value of s. Case 1: s = 4. From inequality 3, we get k + e ≤ 5. Since k ≥ s = 4, we have e ≤ 1 and α(ΓG (x)) ≥ α(G|S ) ≥ 3. Contradiction. Case 2: s = 5. From inequality 3, we obtain k + 2e ≤ 11. Since k ≥ s = 5, we get e ≤ 3. Observe that every graph on 5 vertices with at most 3 edges contains an independent set of size 3. We have α(ΓG (x)) ≥ α(G|S ) ≥ 3. Contradiction. Case 3: s = 6. Inequality 3 indicates e ≤ 6. The only graph H on 6 vertices with α(H) ≤ 3 and |E(H)| ≤ 6 is the disjoint union of two triangles; the complement of G|S is K3,3 . By the symmetry of K3,3 , the pairs (S11 , S21 , S12 , S22 ) as a graph are isomorphic to P2 ∪ P2 ∪ P2 or P4 ∪ P2 . In either case, there exists a pair of two (non-)edges (S13 , S23 ) such that |S11 ∪ S21 ∪ S13 ∪ S23 | ≤ 5 and |S12 ∪ S22 ∪ S13 ∪ S23 | ≤ 5. By case 1 and 2, we have H1 = H(S11 , S21 ) = H(S13 , S23 ) = H(S12 , S22 ) = H2 . Contradiction. Case 4: s = 7. The union of four pairs (S11 , S21 , S12 , S22 ) as a graph is isomorphic to P3 ∪ P2 ∪ P2 . Without loss of generality, we can assume that |S11 ∩ S12 | = 1. Now (S11 , S22 ) forms a pair of disjoint non-edges. By Case 3, we have H1 = H(S11 , S21 ) = H(S11 , S22 ) = H(S12 , S22 ) = H2 . Contradiction. Case 5: s = 8. We observe that S11 , S21 , S12 , S22 are pairwise disjoint. By Case 3, we have H1 = H(S11 , S21 ) = H(S11 , S22 ) = H(S12 , S22 ) = H2 . Contradiction. If Lemma 1 fails for some vertex x, then Lemma 3 states that there is a unique H which can occur in all possible bad triples (S1 , S2 , H). Notice that every vertex in H can have at most 2 neighbors outside H. If there is more than one choice of pair (S1 , S2 ), then the arrangement of edges from H to ΓG (x) is quite restricted. The following lemma gives the structure of the graph H ∪ ΓG (x). Lemma 4 Suppose Lemma 1 fails for some vertex x and H is the unique Kk−1 in all bad triples. If ΓG (x) contains three non-edges uv, vw, and st, then k − 3 ≤ |ΓG (v) ∩ H| ≤ k − 2, |ΓG (u) ∩ H| ≤ 1, and |ΓG (w) ∩ H| ≤ 1. Moreover, if |ΓG (v) ∩ H| = k − 3, then we have |ΓG (u) ∩ H| = 1, |ΓG (w) ∩ H| = 1, and every vertex in H has exactly one neighbor in {s, t}. Proof: Since α(ΓG (x)) = 2, the vertex v has at least one neighbor in {s, t}. As x is also a neighbor of v, we have |ΓG (v) ∩ H| ≤ k − 2. Here, we notice that H ⊂ V (G) \ (x ∪ ΓG (x)). Case 1 : There is a vertex y in H such that y has no neighbors in {s, t}. Notice that we have two pairs of disjoint non-edges: (uv, st) and (vw, st). Every vertex z except y in H must have at least one neighbor in {s, t}, in {u, v}, and in {v, w}. Recall that z has at most 6
two neighbors in ΓG (x). This forces zv to be an edge but both zu and zw are non-edges. Hence, we get |ΓG (v) ∩ H| = k − 2, |ΓG (u) ∩ H| ≤ 1, and |ΓG (w) ∩ H| ≤ 1. Case 2 : Every vertex in H has at least one neighbors in {s, t}. In this case, every vertex in H can have at most one neighbor in {u, v, w}. If H has three vertices y1 , y2 , y3 not adjacent to v, then each yi (i ∈ {1, 2, 3}) has at most one neighbor in {u, w}. By the pigeonhole principle, one of {u, w}, say w, is not adjacent to two of the yi ’s, say y1 and y2 . When we contract {v, w} and {s, t}, we get two non-edges (y1 , vw) and (y2 , vw). Contradiction. Hence, we have |ΓG (v) ∩ H| ≥ k − 3. Moreover, if u has two neighbors y1 and y2 in H, then (y1 , vw) and (y2 , vw) are two non-edges when {v, w} and {s, t} are contracted, which is also impossible. Thus we have |ΓG (u) ∩ H| ≤ 1. By symmetry, we have |ΓG (w) ∩ H| ≤ 1. Moreover, the same argument shows that ΓG (u)∩H = ∅ is impossible. Thus |ΓG (u)∩H| = 1 and |ΓG (w) ∩ H| = 1. Furthermore, every vertex in H will have exactly one neighbor in {s, t}. ¯ G (x) be the complement If k ≥ 5, then we have|ΓG(v) ∩ H| ≥ k − 3 > 1 ≥ |ΓG (u)|. Let Γ ¯ G (x) contains more than one copy of P3 ∪ P2 , then the center of one P3 can of ΓG (x). If Γ not be a leaf of another P3 . Thus we have the following corollary. ¯ G (x) contains neither Corollary 1 If Lemma 1 fails for some vertex x, then the graph Γ P4 ∪ P2 nor Cm for m ≥ 5. ¯ G (x) is P2 ∪ P2 ∪ P2 -free. Lemma 5 If Lemma 1 fails for some vertex x, then the graph Γ Proof: Suppose that ΓG (x) contains three pairwise disjoint non-edges S1 , S2 , S3 . Let S = S1 ∪ S2 ∪ S3 . Assume that G|S has e edges. On the one hand, we have |E(S, H)| =
3 1 (|E(S1 ∪ S2 , H)| + |E(S2 ∪ S3 , H)| + |E(S1 ∪ S3 , H)|) ≥ (2k − 3). 2 2
On the other hand, every vertex in ΓG (x) \ S has at least three neighbors in S as α(ΓG (x)) = 2. Thus, we have |E(S, H)| ≤ 6k − 3(k − 6) − 6 − 2e = 3k + 12 − 2e. Combining two inequalities, we have 3 (2k − 3) ≤ |E(S, H)| ≤ 3k + 12 − 2e. 2 We get e ≤ 8.25. Since e is an integer, we have e ≤ 8. Now G|S has at least 62 −8 non-edges. ¯ G (x) is C3 -free and C5 -free. Thus the complement (denoted by G| ¯ S ) of G|S is Note that Γ ¯ ¯ S must be bipartite since it contains no odd cycle. Now G|S contains a perfect matching; G| a subgraph of K3,3 . Observe that every subgraph of K3,3 with at least 7 edges contains a ¯ G (x) contains a P4 ∪ P2 . This contradicts Corollary 1! P4 ∪ P2 . I.e., Γ ¯ G (x) has at most one nonLemma 6 If Lemma 1 fails for some vertex x, then the graph Γ trivial component. ¯ G (x) contains at most two non-trivial components. Suppose G has Proof: By Lemma 5, Γ two non-trivial components. Each non-trivial component has to be P4 -free by Corollary 1. ¯ G (x) is triangle-free, each non-trivial component must be a star. Let S be the set Since Γ ¯ of vertices in the two non-trivial components. Suppose that S has s vertices, then G|S has exactly s − 2 edges. Equivalently, G|S has exactly 2s − s + 2 edges. Moreover, uv ∈ E(G) 7
for any u ∈ S and v ∈ ΓG (x) \ S as αG (ΓG (x)) = 2. By considering the value of |E(S, H)|, we have s sk − (k − s)s − s − 2( − s + 2) ≥ 2k − 3. 2 We get s ≥ k + 21 , which contradicts s ≤ k.
¯ G (x) is P5 -free. Lemma 7 If Lemma 1 fails for some vertex x, then the graph Γ Proof: Suppose x1 x2 x3 x4 x5 is one of the P5 in ΓG (x) and H is the unique Kk−1 in all bad triples. Since α(ΓG (x)) = 2, the pairs x1 x3 , x3 x5 , and x2 x4 are all edges. Applying Lemma 4 with (u, v, w, s, t) = (x1 , x2 , x3 , x4 , x5 ), we have k−3 ≤ |ΓG (x2 )∩H| ≤ k−2, |ΓG (x1 )∩H| ≤ 1, and |ΓG (x3 ) ∩ H| ≤ 1. Similarly, applying Lemma 4 with (u, v, w, s, t) = (x3 , x4 , x5 , x1 , x2 ), we get k − 3 ≤ |ΓG (x4 ) ∩ H| ≤ k − 2, |ΓG (x3 ) ∩ H| ≤ 1, and |ΓG (x5 ) ∩ H| ≤ 1. Case 1: |ΓG (x2 ) ∩ H| = |ΓG (x4 ) ∩ H| = k − 3. We have |E({x1 , x2 , x3 , x4 }, H)| ≤
4 X
|ΓG (xi ) ∩ H|
i=1
≤ ≤
1 + (k − 3) + 1 + (k − 3) 2k − 4.
However, as ({x1 , x2 }, {x3 , x4 }, H) is a bad triple, we get |E({x1 , x2 , x3 , x4 }, H)| ≥ 2(k − 2) + 1 = 2k − 3. Contradiction. Case 2: |ΓG (x2 ) ∩ H| = k − 2 and |ΓG (x4 ) ∩ H| = k − 3. By counting the number of edges incident to x2 , the pair x2 x5 must be a non-edge. Note x3 and x5 are symmetric in this case. By Lemma 4, we also have |ΓG (x3 ) ∩ H| = |ΓG (x5 ) ∩ H| = 1; every vertex in H has exactly one neighbor in {x1 , x2 }. Let zi be the only neighbor of xi in H for i ∈ {1, 3, 5}. Note that every vertex in H has at most 2 neighbors outside H. If z1 = z3 , then z5 x2 is an edge. Both (z1 , x2 x5 ) and (z5 , x3 x4 ) are non-edges. This contradicts the assumption that ({x2 , x5 }, {x3 , x4 }, H) is a bad triple. If z1 = z5 , then we can get the contradiction by the symmetry of x3 and x5 . If z1 6= z3 and z1 6= z5 , then z3 is adjacent to x2 and x3 while z5 is adjacent to x2 and x5 . We must have z3 6= z5 . Both (z1 , x2 x3 ) and (z3 , x4 x5 ) are non-edges. This contradicts the assumption that ({x2 , x3 }, {x4 , x5 }, H) is a bad triple. Case 3: |ΓG (x2 ) ∩ H| = |ΓG (x4 ) ∩ H| = k − 2. By counting the number of edges incident to x2 and x4 , both x1 x4 and x2 x5 are non-edges. In this case, the induced subgraph of G on {x1 , x2 , x3 , x4 , x5 } is the complement of K2,3 . In particular, x1 , x3 , and x5 are symmetric; x2 and x4 are symmetric. Since x2 x4 is an edge and G is Kk -free, we have Γ(x2 ) ∩ H 6= Γ(x4 ) ∩ H. Let z2 be the only vertex in (Γ(x2 )∩H)\(Γ(x4 )∩H) and z4 be the only vertex in (Γ(x4 )∩H)\(Γ(x2 )∩H). Observe |E({x1 , x3 , x5 }, H)| ≤ 2. Without loss of generality, we can assume Γ(x1 ) ∩ H = ∅. Since |Γ(x3 ) ∩ H| ≤ 1, one of x3 z2 and x3 z4 must be a non-edge. By symmetry, we can assume z2 x3 is not an edge. Both (z4 , x1 x2 ) and (z2 , x3 x4 ) are non-edges after contracting {x1 , x2 } and {x3 , x4 }. This contradicts the assumption that ({x1 , x2 }, {x3 , x4 }, H) is a bad triple. ¯ G (x) is Lemma 8 If Lemma 1 fails for some vertex x, then the non-trivial component of Γ one of following.
8
1. P4 ; it occurs only when k = 4; 2. C4 , it occurs only when k = 4, 5; 3. a tree T5 as shown in Figure 3, it occurs only when k = 5. w
s
v
u
t
Figure 3: The tree T5 ¯ G (x) guaranteed by Lemma 6 and Proof: Let C be the unique non-trivial component of Γ H be the unique Kk−1 in all bad triples. Case 1: C contains a cycle. Since α(Γ(x)) = 2, C is triangle-free. By Lemma 7, C is P5 -free. Thus C must be isomorphic to C4 . Let x1 , x2 , x3 , x4 be the vertices of C4 in cyclic order. On the one hand, we have |Γ(xi ) ∩ H| ≤ 2, for each i in {1, 2, 3, 4}. Thus |E({x1 , x2 , x3 , x4 }, H)| ≤
4 X
|Γ(xi ) ∩ H| ≤ 8.
(4)
i=1
On the other hand, since ({x1 , x2 }, {x3 , x4 }, H) is a bad triple, equation (2) implies |E({x1 , x2 , x3 , x4 }, H)| ≥ 2k − 3.
(5)
Combining equation (4) and (5), we have 8 ≥ 2k − 3.
(6)
Thus k ≤ 5.5; it follows that k = 4 or 5. Case 2 : C is a tree. Since G is Kk -free and dG (x) = k, the component C is not a star. Let u and v be two leaves of C with the maximum distance possible; let u′ and v ′ be the unique neighbor of u and v in C, respectively. We have u′ 6= v ′ . By Lemma 7, C is P5 -free. It follows that ΓC (u′ ) ∩ ΓC (v ′ ) = ∅. Thus |Γ(u′ ) ∩ C| + |Γ(v ′ ) ∩ C| ≤ k.
(7)
Now we have |EG ({u, u′ , v ′ , v}, H)| ≤ dC (u) + dC (u′ ) + dC (v ′ ) + dC (v) ≤ k + 2.
(8)
From inequality (2), we get |EG ({u, u′ , v ′ , v}, H)| ≥ 2k − 3.
(9)
Combining inequalities (8) and (9), we have k + 2 ≥ 2k − 3.
(10)
Thus k ≤ 5; it follows that k = 4 or 5. If k = 4, then we have C = P4 , since C is not a star. If k = 5, equality holds in inequality ¯ G (x). Note T5 is the only tree on 5 vertices (7). It follows that C is a spanning tree of Γ which is neither a star nor a P5 . We have C = T5 . Remark: This lemma implies Lemma 1 for the case k ≥ 6. We will analyze the remaining cases k = 4 and k = 5 in section 4. 9
3
Proof of Theorem 2
In this section, we prove Theorem 2. Suppose that Gk 6= ∅ for some k ≥ 4. For this particular k, let G be a graph in Gk with the smallest number of vertices. Recall that each vertex in G either has degree k or has degree k − 1. For each vertex x in G, we associate a small set of vertices S(x) selected from ΓG (x) as follows. If dG (x) = k − 1, then let S(x) be the endpoints of a non-edge in ΓG (x); if dG (x) = k and α(ΓG (x)) ≥ 3, then let S(x) be any independent subset of size 3 in ΓG (x); if dG (x) = k and α(ΓG (x)) = 2, then let S(x) = S1 (x) ∪ S2 (x), where S1 (x) and S2 (x) are guaranteed by Lemma 1. If k = 4, we apply Lemma 13 to get S1 (x) and S2 (x) instead of Lemma 1. For any x ∈ V (G), we have |S(x)| = 2, 3, or 4. The following definitions depend on the choice of S(∗), which is assumed to be fixed through this section. For v ∈ G and each j ≥ 2, let j NG (v) = {u| there is a path vv0 . . . vj−2 u in G of length j such that v0 ∈ S(v) and vj−2 ∈ S(u)}. j Since |g(v)| ≤ 4, we have |NG (v)| ≤ 4(k − 1)j−1 for j ≥ 2. Consider an auxiliary graph G∗ on V (G), where two vertices u and v are adjacent in G∗ j 2 3 4 5 7 if v ∈ ΓG (u) ∪ NG (u) ∪ NG (u) ∪ NG (u) ∪ NG (u) ∪ NG (u). Note u ∈ NG (v) if and only if j v ∈ NG (u). The graph G∗ is well-defined.
Lemma 9 Let G∗ be the auxiliary graph defined above. If G∗ is m-colorable, then we have χf (G) ≤ k −
k . m+1
(11)
Proof of Theorem 2: For any vertex v, the degree of v in G∗ satisfies dG∗ (v)
2 3 4 5 7 ≤ dG (u) + |NG (u)| + |NG (u)| + |NG (u)| + |NG (u)| + |NG (u)|
≤ k + 4(k − 1) + 4(k − 1)2 + 4(k − 1)3 + 4(k − 1)4 + 4(k − 1)6 ≤ 4k(k − 1)5 − 2. The last step holds since k ≥ 4. Thus G∗ is (4k(k − 1)5 − 1)-colorable. Applying Lemma 9, we get 1 . χf (G) ≤ k − 4(k − 1)5 This concludes the proof of Theorem 2. Let X be an independent set of G∗ . By the definition of G∗ , X is also an independent 2 3 set of G. For any two vertices u and v in X, u 6∈ NG (v) implies S(u) ∩ S(v) = ∅; u 6∈ NG (v) implies that there is no edge between S(u) and S(v). Recall that every S(u) is either an independent set (of size 2 or 3) or the union of two independent set of size 2. Let W = ∪x∈X S(x). The set W can be partitioned into two independent sets of G, denoted by W1 and W2 , respectively. We can further require that the partition W1 ∪ W2 satisfies the following property. “For any vertex x ∈ X, if S(x) = 2 or 3, then either S(u) ⊆ W1 or S(u) ⊆ W2 .” Let G(X) = (G\X)/W1 /W2 . Namely, G(X) is a graph obtained by removing the vertices in X followed by contracting W1 and W2 . For simplicity, we write wi for Wi for i = 1, 2 in the remaining of this section. We have the following Lemma. Lemma 10 Let X be an independent set of G∗ and G(X) be the graph constructed above. We have χ(G(X)) ≤ k. 10
We postpone the proof of Lemma 10 until the end of this section and prove Lemma 9 first. Proof of Lemma 9. Since G∗ is m-colorable, we assume V (G) = ∪m i=1 Xi , where X1 , . . . , Xm are independent sets of G∗ . For integers i in {1, 2, . . . , m}, let W1i and W2i be the independent sets which are used to define G(Xi ). For each i ∈ {1, . . . , m}, Lemma 10 shows G(Xi ) is k-colorable; let ci : V (G(Xi )) → Ti be a proper k-coloring of the graph G(Xi ). Here T1 , T2 , . . . , Tm are pairwise disjoint; each of them consists of k colors. For i ∈ {1, . . . , m}, the k-coloring ci can be viewed as a k-coloring of G \ Xi such that each vertex in the independent set Wji a receives the color ci (Wji ) for j = 1, 2. Now we reuse the notation ci to denote this k-coloring of G \ Xi . For each v ∈ Xi , we have | ∪u∈ΓG (v) ci (u)| ≤ k − 2. We can assign two unused colors, denoted by the set Y (v), to v. We define fi : V (G) → P(Ti ) (the power set of Ti ) satisfying {ci (v)} if v ∈ V (G) \ Xi , fi (v) = Y (v) if v ∈ Xi . Observe that each vertex in Xi receives two colors from fi and every other vertex receives one color. Let σ : V (G) → P(∪ki=1 Ti ) be a mapping such that σ(v) = ∪m i=1 fi (v). It is easy to verify σ is a (m + 1)-fold coloring of G such that each color is drawn from a palette of mk colors; namely we have k mk =k− . χf (G) ≤ m+1 m+1 The proof of Lemma 9 is finished. A block of a graph is a maximal 2-connected induced subgraph. A Gallai tree is a connected graph in which all blocks are either complete graphs or odd cycles. A Gallai forest is a graph all of whose components are Gallai trees. A k-Gallai tree (forest) is a Gallai tree (forest) such that the degree of all vertices are at most k − 1. A k-critical graph is a graph G whose chromatic number is k and deleting any vertex can decrease the chromatic number. Gallai showed the following Lemma. Lemma 11 [2] If G is a k-critical graph, then the subgraph of G induced on the vertices of degree k − 1 is a k-Gallai forest. Now, we are ready to prove Lemma 10. Proof of Lemma 10: We use proof by contradiction. Suppose that G(X) is not k-colorable. The only possible vertices in G(X) with degree greater than k are the vertices w1 and w2 , which are obtained by contracting the sets W1 and W2 . The simple greedy algorithm shows that G(X) is always k + 1-colorable. Let G′ (X) be a k + 1-critical subgraph of G(X). Applying Lemma 11 to G′ (X), the subgraph of G′ (X) induced on the vertices of degree k is a k + 1-Gallai forest F . The vertex set of F may contain w1 or w2 . Delete w1 and w1 from F if F contains one of them. Let F ′ be the resulted Gallai forest. (Any induced subgraph of a Gallai forest is still a Gallai forest.) Since k ≥ 4, the Gallai forest F ′ is not empty. Let T be a connected component of F ′ and B be a leaf block of T . The block B is either a clique or an odd cycle from the definition of Gallai tree. Let v be a vertex in B. As v has at most two neighbors (w1 and w2 ) outside F ′ in G(X), we have dF ′ (v) ≥ k − 2. If v is not in other blocks of F ′ , then we have dB (v) ≥ k − 2. It follows that |B| ≥ k − 1. Since B is a subgraph of G and G is Kk -free, the block B is a Kk−1 if k ≥ 5 and is an odd cycle if k = 4. Let v1 v2 be an edge in B such that v1 and v2 are not in other blocks. The degree requirement implies vi wj are edges in G(X) for all i, j ∈ {1, 2}; i.e., ΓG (vi ) ∩ Wj 6= ∅. For 11
i = 1, 2, there are vertices xi , yi ∈ X satisfying S(xi ) ∩ ΓG (vi ) 6= ∅ and S(yi ) ∩ ΓG (vi ) 6= ∅. 4 If xi 6= yi , we get a 4-path from xi to yi in G; this contradicts yi 6∈ NG (xi ) as X is an ∗ independent set of G . For {i, j} = {1, 2}, if xi 6= yj , we get a 5-path from xi to yj in G. 5 Since yi 6∈ NG (xi ), this is a contradiction. Thus we have x1 = x2 = y1 = y2 . Let x denote this common vertex above. The requirement on W1 and W2 implies dG (x) = k and α(ΓG (x)) = 2. Let v0 be the only vertex in B shared by other blocks. Since B − v0 is connected, the argument above shows there is a common x for all edges in B − v0 . If ΓG (v0 ) ∩ W 6= ∅, then there exist x0 ∈ X such that ΓG (v0 ) ∩ S(x0 ) 6= ∅. By the similar argument, we must have x0 = x. Therefore, x depends only on B. In the sense that for any y ∈ X and any v ∈ B, if S(y) ∩ ΓG (v) 6= ∅, then y = x. Case 1: k ≥ 5. The block B is a clique of size k − 1. Both v0 w1 and v0 w2 are non-edges − in G(X); otherwise, we get a Kk+1 in G/S1 (x)/S2 (x) and it is a contradiction to Lemma 1. Thus v0 has no neighbor in W . Let u1 and u2 be the two neighbors of v0 in other blocks of F ′ . If u1 and u2 are in the same block, then this block is an odd cycle as v0 has only two neighbors in this block; otherwise, v0 u1 and v0 u2 are in two different blocks. The union of the non-leaf blocks of T is a Gallai-tree, denoted by T ′ . The argument above shows every leaf block of T ′ must be an odd cycle. Let C be such a leaf block of T ′ . Now C is an odd cycle, and C is connected to |C| − 1 leaf blocks of T . Let B and B ′ be two leaf blocks of T such that B ∩ C is adjacent to B ′ ∩ C. Without loss of generality, we may assume B is the one we considered before. By the same argument, B ′ is also a clique of size ′ be the vertices of B ′ and v0′ be the only vertex in B ′ ∩ C. For i k − 1. Let v0′ , v1′ , . . . , vk−2 ′ in {1, 2, . . . , k − 2}, vi has exactly two neighbors in W . There exists a vertex x′ ∈ X with dG (x′ ) = k and α(ΓG (x′ )) = 2 such that |E(vi′ , S1 (x′ ))| ≥ 1 and |E(vi′ , S2 (x′ ))| ≥ 1. We 7 must have x = x′ ; otherwise x′ ∈ NG (x) and it is a contradiction to the hypothesis that X ∗ is an independent set of G . Now we have |E(S1 (x), B)| ≥ k − 2 and |E(S1 (x), B ′ )| ≥ k − 2. Since α(ΓG (x)) = 2, there are at least k − 2 edges between S1 (x) and the rest of vertices in ΓG (x). For every vertex u in ΓG (x) \ (S1 (x) ∪ S2 (x)), the degree of u in G(X) is at most k − 1. Thus, u is not in the k + 1-critical subgraph G′ (X). The number of edges incident to S1 (x) is at least 3(k − 2) + 2 = 3k − 4 > 2k since k ≥ 5. Contradiction. Case 2: k = 4. The block B is an odd cycle. Suppose |B| = 2r + 1. Let v0 , v1 , . . . , v2r be the vertices of B in cyclic order and v0 be the only vertex which may be shared by other block. Let x ∈ X be the vertex determined by B. Recall dG (x) = 4 and α(ΓG (x)) = 2. Each vertex in Γ(x) can have at most 2 edges to B. We get 4r ≤ |E(B, Γ(x))| ≤ 8.
(12)
We have r ≤ 2. The block B is either a C5 or a K3 . We claim both v0 w1 and v0 w2 are non-edges of G(X). If B = C5 , then inequality (12) implies that v0 has no neighbor in Γ(x) and the claim holds. If B = K3 , then the claim also holds; otherwise B ∪ {S1 (x), S2 (x)} forms a K5− in G/S1 (x)/S2 (x), which is a contradiction to Lemma 1. Let u1 and u2 be the two neighbors of v0 in other blocks of F ′ . If u1 and u2 are in the same block, then this block is an odd cycle; otherwise, v0 u1 and v0 u2 are in two different blocks. The union of non-leaf blocks of T is a Gallai-tree, denoted by T ′ . The argument above shows every leaf block of T ′ must be an odd cycle. Let C be such a leaf block of T ′ . Now C is an odd cycle, and C is connected to |C| − 1 leaf blocks of T . Let B and B ′ be two 12
leaf blocks of T such that B ∩ C is adjacent to B ′ ∩ C. Without loss of generality, we may assume B is the one we considered before. By the same argument, B ′ is an odd cycle of size ′ ′ ′ 2r′ + 1 with r′ ∈ {1, 2}. Let v0′ , v1′ , . . . , v2r ′ be the vertices of B and v0 be the only vertex ′ ′ ′ ′ in B ∩ C. For i in {1, 2, . . . , 2r } and j in {1, 2}, vi wj are edges in G (X). Similarly, there exists a vertex x′ ∈ X with dG (x′ ) = k and α(ΓG (x′ )) = 2 such that |E(vi , S1 (x′ ))| ≥ 1 and 7 |E(vi , S2 (x′ ))| ≥ 1. We must have x = x′ ; otherwise x′ ∈ NG (x) and this contradicts the fact ∗ that X is an independent set in G . Now we have |E(Γ(x), B)| ≥ 4r and E(Γ(x), B ′ )| ≥ 4r′ . By counting the degrees of vertices in Γ(x) in G, we have 4r + 4r′ + 4 + 4 ≤ 16. We get r = r′ = 1. Both B and B ′ are K3 ’s. In this case, G/S1 (x)/S2 (x) contains the graph G0 , see figure 13. This contradicts Lemma 13. We can find the desired contradiction in each case; so the lemma follows.
4
Proof of Lemma 1 when k = 4 and k = 5
In this section, we will prove the remaining cases k = 4 and k = 5 for Lemma 1. We first review a Lemma in [5]. Lemma 12 Let G be a graph. Suppose that G1 and G2 are two subgraphs such that G1 ∪G2 = G and V (G1 ) ∩ V (G2 ) = {u, v}. 1. If uv is an edge of G, then we have χf (G) = max{χf (G1 ), χf (G2 )}. 2. If uv is not an edge of G, then we have χf (G) ≤ max{χf (G1 ), χf (G2 + uv), χf (G2 /uv)}, where G2 + uv is the graph obtained from G2 by adding edge uv and G2 /uv is the graph obtained from G2 by contracting {u, v}. Proof of Lemma 1 for k = 4: Recall that G is a connected K4 -free graph with minimum number of vertices such that G 6= C82 and χf (G) > 4 − 4(315 ) . Note that G is 2-connected. We will prove it by contradiction. ¯ G (x) is either a P4 or a Suppose Lemma 1 fails for some vertex x in G. By Lemma 8, Γ C4 (see Figure 4). Here we assume ΓG (x) = {a, b, c, d}. a
b
a
b
c
d
c
d
¯ G (x) = C4 Γ
¯ G (x) = P4 Γ
Figure 4: Two possible cases of ΓG (x). 13
¯ G (x) = P4 , then {a, d} and {b, c} is the only pair of disjoint non-edges. There is If Γ a triangle H with V (H) = {y, z, w} such that ({a, d}, {b, c}, H) is a bad triple. Note that |E({a, b, c, d}, {y, z, w})| = 5 or 6. By an exhaustive search, the induced subgraph of G on {x, a, b, c, d, y, z, w} is one of the following six graphs (see Figure 5). b
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¯ G (x) = P4 , then there are six possible induced subgraphs. Figure 5: If Γ ¯ G (x) = C4 , then ({a, c}, {b, d}) and ({a, d}, {b, c}) are two pairs of disjoint non-edges. If Γ There is a triangle H with V (H) = {y, z, w} such that ({a, c}, {b, d}, H) and ({a, d}, {b, c}, H) are two bad triples. By an exhaustive search, the induced subgraph of G on {x, a, b, c, d, y, z, w} is one of the following three graphs (see Figure 6). b
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¯ G (x) = C4 , then there are three possible induced subgraphs. Figure 6: If Γ It suffices to show that G cannot contain Hi for 1 ≤ i ≤ 9. Since all vertices in H1 (and H2 ) have degree 4, H1 (and H2 ) is the entire graph G. Observe that H1 is isomorphic to C82 and H2 is 11:3-colorable (see Figure 7). Contradiction! In H7 , the vertex d is the only vertex with degree less than 4. If H7 is not the entire graph G, then d is a cut vertex of G. This contradicts the fact that G is 2-connected. Thus G = H7 . The graph H7 is 11:3-colorable as shown by Figure 7. Contradiction. Now we consider the case H3 . Note H3 +bz is the graph H2 . We have χf (H3 ) ≤ χf (H2 ) ≤ 11/3. The graph H3 must be a proper induced subgraph of G, and the pair {b, z} is a vertex 14
a
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Figure 7: The graph H2 and H7 are 11:3-colorable. cut of G. Let G′ be the induced subgraph of G by deleting all vertices in H3 but b, z. We apply Lemma 12 to G + bz with G1 = H3 + bz = H2 and G2 = G′ + bz. We have χf (G + bz) ≤ max{χf (H2 ), χf (G′ + bz)}. Note χf (H2 ) ≤ 11/3 and 11/3 < χf (G) ≤ χf (G + bz). We have χf (G) ≤ χf (G′ + bz). Both b and z have at most 2 neighbors in G′ + bz. Thus G′ + bz is K4 -free; G′ + bz 6= C82 and has fewer vertices than G. This contradicts to the minimality of G. Note H5 + cy = H2 . The case H5 is similar to the case H3 . Note that H4 , H6 , and H8 are isomorphic to each other. It suffices to show G does not contain H4 . Suppose that H4 is a proper induced subgraph of G. Let G1 be the induced subgraph of G by deleting all vertices in H4 . Note C82 is not a proper subgraph of any graph in G4 . We have G1 6= C82 . Note that c and z have degree 3 while other vertices in H4 have degree 4. Since G is 2-connected, c has a unique neighbor, denoted by u, in V (G1 ). Similarly, z has a unique neighbor, denoted by v, in V (G1 ). Observe that the pair {u, v} forms a vertex cut of G. Let G2 be the induced graph of G on V (H4 ) ∪ {u, v}. Applying Lemma 12 to G with G1 and G2 , we have χf (G) ≤ max{χf (G1 ), χf (G2 + uv), χf (G2 /uv)}. Figure 8 shows χf (G2 + uv) and χf (G2 /uv) are at most 11/3. a
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Figure 8: Case H4 : both graph G2 + uv and G2 /uv are 11:3-colorable. Since χf (G) > 11/3, we have χf (G) ≤ χf (G1 ). Now G1 is K4 -free and has maximum degree at most 4; G1 has fewer vertices than G. This contradicts the minimality of G. If G = H4 , then χf (H4 ) ≤ 11/3, since H4 is a subgraph of G2 + uv in Figure 8. Now we consider the last case H9 . First, we contract b, c, z into a fat vertex denoted by bcz. We write G/bcz for the graph after this contraction. Observe that {bcz, d} is a vertex-cut of G/bcz. Let G4 and G′4 be two connected subgraphs of G/bcz such that G4 ∪ G′4 = G/bcz, G4 ∩ G′4 = {bcz, d}, and {u, v} ⊂ G′4 . Note that G4 is 11:3 colorable, see Figure 9. Now by Lemma 12, we have χf (G/bcz) ≤ max{χf (G4 ), χf (G′4 )}. 15
As {b, c, z} is an independent set, each a:b-coloring of G/bcz gives an a:b-coloring of G, that is χf (G/bcz) ≥ χf (G) > 11/3. The graph G4 is 11:3-colorable; see Figure 9. Thus we have χf (G′4 ) ≥ χf (G/bcz) ≥ χf (G). It is easy to check that G′4 has maximum degree 4, K4 -free, and it is not C82 . Hence G′4 must contain a K4 . Otherwise, it contradicts the minimality of G. Second, we contract {b, d, z} into a fat vertex bdz and denote the graph by G/bcz. Let G5 and G′5 be two connected subgraphs of G/bdz such that G5 ∪G′5 = G/bzd, G5 ∩G′5 = {bzd, c}, and {u, v} ⊂ G′5 . Note that G5 is 11:3-colorable; see Figure 9. By a similar argument, G′5 must contain a K4 . 456
456 u v
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Figure 9: Case H9 : the graphs G4 , G5 , and G6 are 11:3-colorable. The remaining case is that both G′4 and G′5 have a K4 when we contract b and z. Since the origin graph G is K4 -free, the K4 in G′4 (and in G′5 ) must contain the fat vertex bcz (or bdz), respectively. Note that each of the four vertices b, c, d, z has at most one edge leaving H9 . There must be a triangle uvp in G and these four outward edges are connected to some element of {u, v, p}. The graph G/bz must contain the subgraph G6 as drawn in Figure 9. Note that {u, v} is a vertex-cut in G/bz. Let G6 and G′6 be two connected subgraphs of G/bz, which satisfy G6 ∪ G′6 = G, G6 ∩ G′6 = {u, v}, and bz ∈ G6 . By Lemma 12, we have χf (G/bz) ≤ max{χf (G6 ), χf (G′6 )}. Note that G6 is 11:3-colorable; see Figure 9. We also have χf (G/bz) ≥ χf (G) > 11 3 . We obtain χf (G′6 ) ≥ χf (G/bz) ≥ χf (G). Observe that G′6 is a subgraph of G. We arrive at a contradiction of the minimality of G. In each case, we are able to find the desired contradiction, so the lemma follows. Proof of Lemma 1 for k = 5: Recall that G is a graph with minimum number of vertices in G5 . Suppose Lemma 1 fails for some vertex x. Applying Lemma 8, ΓG (x) is either T5 or the union of a C4 and an isolated vertex. Case 1: ΓG (x) is T5 ; see Figure 3. There is a 4-clique H such that ({u, v}, {s, t}, H) and ({w, v}, {s, t}, H) are two bad triples. Let V (H) = {a, b, c, d}. Notice that each of u, w, t has at most one neighbor in H, s has at most two neighbors in H, and v has at most three neighbors in H. Since ({u, v}, {s, t}, H) and ({w, v}, {s, t}, H) are two bad triples, each vertex in {x, u, v, s, t, w, a, b, c, d} has degree 5; the graph G is uniquely determined. We denote this graph by H10 ; see Figure 10. We have χf (G) ≤ 9/2; contradiction. Case 2: ΓG (x) is the union of a C4 and an isolated vertex. We assume that V (C4 ) = {u1 , u2 , u3 , u4 } in cyclic order and z is the isolated vertex. There is a 4-clique H such that ({u1 , u2 }, {u3 , u4 }, H) and ({u2 , u3 }, {u4 , u1 }, H) are two bad triples. Let V (H) = {a, b, c, d}. Up to isomorphism, the induced subgraphs of G on {x, u1 , u2 , u3 , u4 , a, b, c, d, z} is either C5 ⊠ K2 or H11 , which is one edge less than C5 ⊠ K2 ; see Figure 11. Since G is not equal to C5 ⊠ K2 , let us consider the case H11 . All vertices in H11 except u3 and a have degree 5. Suppose H11 is a proper induced subgraph of G. Let s (or t) be the 16
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Figure 10: Case T5 : the graph H10 and its 9:2-coloring. x
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Figure 11: Case C4 : two induced subgraphs of G. neighbor of u3 (or a) in G − H11 , respectively; see Figure 12. Note that {s, t} is a vertex-cut of G. Let G1 be the induced subgraph on G on V (G) \ V (H11 ) and G2 be the induced subgraph on G on V (H11 ∪ {s, t}. Applying Lemma 12 with G1 and G2 defined above, we have χf (G) ≤ max{χf (G1 ), χf (G2 + st), χf (G2 /st)}. Note χf (G2 + st) ≤ 9/2 and χf (G2 /st) ≤ 9/2 as shown by Figure 12. Since χf (G) > 9/2, we get χf (G1 ) ≥ χf (G). Notice that G1 is a proper subgraph of G. This contradicts the minimality of G. If G = H11 , then we have χf (H11 ) ≤ 9/2 as shown in Figure 12. Contradiction. x
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Figure 12: Case H11 : G2 + st and G2 /st are 9:2 colorable. The proof of Lemma 1 is completed. The following lemma strengthens Lemma 1 in the case k = 4.
17
Lemma 13 Suppose that G4 6= ∅. Let G be a graph in G4 with the smallest number of vertices. For each x ∈ V (G) with dG (x) = 4 and α(ΓG (x)) = 2, there exist two vertexdisjoint non-edges S1 (x), S2 (x) ⊂ ΓG (x) satisfying the following property. If we contract S1 (x) and S2 (x), then the resulted graph G/S1 (x)/S2 (x) contains neither K5− nor G0 . Here G0 is a graph defined in Figure 13.
x
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r
q
Figure 13: The graph G0 . Proof of Lemma 13: By Lemma 1, we can select two vertex disjoint non-edges S1 and S2 such that the graph G/S1 /S2 contains no K5− . For these particular S1 and S2 , if G/S1 /S2 contains no G0 , then Lemma 13 holds. Without loss of generality, we assume that G/S1 /S2 does contain G0 . Let si = Si for i = 1, 2. Observe that both s1 and s2 have four neighbors u, v, p, q other than x in G0 . It follows that |E(S1 ∪ S2 , {u, v, p, q})| ≥ 8. On the one hand, we have |E(G |S1 ∪S2 )| = ≤ =
1 2
X
d(v) − |E(S1 ∪ S2 , {u, v, p, q})| − 4
v∈S1 ∪S2
!
1 (16 − 8 − 4) 2 2.
On the other hand, α(Γ(x)) = 2 implies G |S1 ∪S2 contains at least two edges. Thus, we have ¯ G (x) = C4 . Label the vertices in ΓG (x) by a, b, c, d as in Figure 4. We assume ab and cd are Γ edges while ac, bd, ad, bc are non-edges. Observe that each vertex in {u, v, p, q} has exactly two neighbors in {a, b, c, d}. If one vertex, say u, has two neighbors forming a non-edge, say ac, then we can choose S1′ = {a, c} and S2′ = {b, d}. It is easy to check that G/S1′ /S2′ contains neither G0 nor K5− . We are done in this case. In the remaining case, we can assume that for each vertex y in {u, v, p, q}, the neighbors of y in {a, b, c, d} always form an edge. Up to relabeling vertices, there is only one arrangement for edges between {u, v, p, q} and {a, b, c, d}; see the graph H0 defined in Figure 14. The graph H10 is 11:3-colorable as shown in Figure 14. Since χf (G) > 11/3, H10 is a proper subgraph of G. Note in H10 , every vertices except w and r has degree 4; both w and r have degree 3. Thus, {w, r} is a vertex cut of G. Let G1 = H0 and G2 be the subgraph of G 18
x 159
a
u
b
d
c
v
237
p
w
q
456
r
8st
789 123
H0
234
6st
456
178
9st
an 11:3-coloring of H0
Figure 14: H0 and a 11:3-coloring of H0 . by deleting vertices in {x, a, b, c, d, p, q, u, v}. Applying Lemma 12 with G1 and G2 defined above, we have χf (G) ≤ max{χf (G1 ), χf (G2 )}. Since χf (G) > 11/3 and χf (G1 ) ≤ 11/3 (see Figure 12), we must have χf (G2 ) ≥ χf (G). Note that G2 has fewer number of vertices than G. This contradicts the minimality of G. Therefore, the lemma follows.
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