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The equitable vertex arboricity of complete tripartite graphs
Information Processing Letters 115 (2015) 977–982
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Information Processing Letters www.elsevier.com/locate/ipl
The equitable vertex arboricity of complete tripartite graphs ✩ Zhiwei Guo a,∗ , Haixing Zhao b , Yaping Mao a a b
Department of Mathematics, Qinghai Normal University, Xining, Qinghai 810008, China School of Computer, Qinghai Normal University, Xining, Qinghai 810008, China
a r t i c l e
i n f o
Article history: Received 11 January 2015 Received in revised form 13 June 2015 Accepted 26 June 2015 Available online 2 July 2015 Communicated by Ł. Kowalik Keywords: Combinatorial problems Equitable coloring Vertex k-arboricity k-Tree-coloring Complete tripartite graph
a b s t r a c t The equitable coloring problem, introduced by Meyer in 1973, has received considerable attention and research. Recently, Wu et al. introduced the concept of equitable (t , k)-treecoloring, which can be viewed as a generalization of proper equitable t-coloring. Wu et al. investigated the strong equitable vertex k-arboricity of complete bipartite graphs. In this paper, we mainly investigate the strong equitable vertex 2-arboricity of complete tripartite graphs. © 2015 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are finite and simple. For a real number x, x is the least integer not less than x and x is the largest integer not larger than x. We use V (G ), E (G ), δ(G ) and (G ) to denote the vertex set, edge set, minimum degree and maximum degree of G, respectively. For a vertex v ∈ V (G ), let N G ( v ) denote the set of neighbors of v in G and d G ( v ) = | N G ( v )| denote the degree of v. We use d( v ) instead of d G ( v ) for brevity. For other undefined concepts, we refer the reader to [3]. We associate positive integers 1, 2, . . . , t with colors and call f a t-coloring of a graph G if f is a mapping
✩ Research supported by the National Natural Science Foundation of China (Nos. 61164005, 61460005, 11161037 and 11461054), Key Laboratory of Tibetan intelligent information processing and machine translation of Qinghai P.R.C (No.2013-Z-Y17) and the Program for Changjiang Scholars and Innovative Research Team in Universities (No. IRT1068), the Research Fund for the Chunhui Program of Ministry of Education of China (No. Z2014022) and the Nature Science Foundation from Qinghai Province (No. 2014-ZJ-721 and No. 2014-ZJ-907). Corresponding author. E-mail addresses: [email protected] (Z. Guo), [email protected] (H. Zhao), [email protected] (Y. Mao).
*
http://dx.doi.org/10.1016/j.ipl.2015.06.016 0020-0190/© 2015 Elsevier B.V. All rights reserved.
from V (G ) to {1, 2, . . . , t }. A t-coloring of G is proper if any two adjacent vertices have different colors. For 1 ≤ i ≤ t, let V i = { v | f ( v ) = i }. A t-coloring of a graph G is said to be equitable if || V i | − | V j || ≤ 1 for all i and j, that is, every color class has size | V (G )|/t or | V (G )|/t . A graph G is said to be properly equitably t-colorable if G has a proper equitable t-coloring. The smallest number t for which G is properly equitably t-colorable is called the equitable chromatic number of G, denoted by χ= (G ). The equitable coloring problem, introduced by Meyer [8], was motivated by a practical application to municipal garbage collection [10]. In this context, the vertices of a graph represent garbage collection routes. A pair of vertices share an edge if the corresponding routes should not be run on the same day. It is desirable that the number of routes ran on each day be approximately the same. Therefore, the problem of assigning one of the six weekly working days to each route reduces to finding a proper equitable 6-coloring. For more applications such as scheduling, constructing timetables and load balance in parallel memory systems, we refer to [1,2,4,6,7,9]. Note that a properly equitably t-colorable graph may admit no proper equitable t -colorings for some t > t. For example, the complete bipartite graph H := K 2m+1,2m+1
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has no proper equitable (2m + 1)-colorings, although it satisfies χ= ( H ) = 2. This fact motivates another interesting parameter for proper equitable coloring. The equitable ∗ (G ), is the smallchromatic threshold of G, denoted by χ= est integer t such that G has a proper equitable t -coloring for all t ≥ t. For a graph G, a d-relaxed k-coloring, also known as a d-defective coloring, of G is a function f from V (G ) to {1, 2, . . . , k} such that each subgraph G [ V i ] is a graph of maximum degree at most d. A d-relaxed equitable k-coloring is a d-relaxed k-coloring that is equitable. In Fan et al. [5], relaxed equitable coloring of graphs was considered. They proved that every graph has a proper equitable (G )-coloring such that each color class induces a forest with maximum degree at most one. On the basis of this research, Wu et al. [11] introduced the notion of equitable (t , k)-tree-coloring, which can be viewed as a generalization of proper equitable t-coloring. A (t , k)-tree-coloring is a t-coloring f of a graph G such that each component of G [ V i ] is a tree of maximum degree at most k. A (t , ∞)-tree-coloring is called a t-tree-coloring for short. An equitable (t , k)-tree-coloring is a (t , k)-tree-coloring that is equitable. The equitable vertex k-arboricity of a graph G, denoted by vak = (G ), is the smallest integer t such that G has an equitable (t , k)-treecoloring. The strong equitable vertex k-arboricity of G, denoted by vak ≡ (G ), is the smallest integer t such that G has an equitable (t , k)-tree-coloring for every t ≥ t. It is ∗ (G ) for evclear that va0 = (G ) = χ= (G ) and va0 ≡ (G ) = χ= ery graph G, and vak = (G ) and vak ≡ (G ) may vary a lot. Wu et al. [11] investigated the strong equitable vertex k-arboricity of complete bipartite graphs. In this paper, we mainly investigate the strong equitable vertex 2-arboricity of complete tripartite graph. Theorem 1. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 3 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ . 4 Theorem 2. Let k and m be positive integer.
(i ) (ii) (iii) (iv) (v )
For k ≡ 1 (mod 5), va2 ≡ ( K 4k,4k,4k ) = For k ≡ For k ≡
12k+3 . 5
2 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . 3 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . 4 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+12 .
For k ≡ For k ≡ 0 (mod 5), if m ≥ 2q + 1 where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3;
if m ≥ 2q where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Theorem 3. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 1 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ + 2. 4 Theorem 4. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 2 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ + 2. 4
2. Preliminary In this section, we give some basic results, which will be used later. Proposition 1. The complete -partite graph K n,...,n has an equitable (t , k)-tree-coloring for every integer t satisfying the following condition:
t = h (h ≥ 1). Proof. One can easily construct an equitable (t , k)-treecoloring of K n,...,n by dividing each partite set into h classes equitably and coloring each class with different colors. Note that the coloring is a proper equitable (t , k)-treecoloring. 2 Theorem 5. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. Then
va2 ≡ ( K n,n,n ) ≤ 3
n+1
4
+ 2.
Proof. By Proposition 1, in order to show va2 ≡ ( K n,n,n ) ≤ 1 3 n+ + 2, we only need to prove that K n,n,n has an eq4 uitable (q, 2)-tree-coloring for every q satisfying the fol1 + 2 ≤ q ≤ 3n and 3 q. Let X , Y lowing condition: 3 n+ 4 and Z be the partite sets of K n,n,n . 1 Let q = 3a + 1 and 3 n+ + 4 ≤ q ≤ 3(n − 1) + 1. Note 4 1 that 4q − 3n ≥ 12 n+ + 16 − 3n ≥ 12( n−4 2 ) + 16 − 3n = 10. 4
10 Suppose 3n+ ≤ q ≤ n. Observe that n > 4. Let {x1 , x2 , 4 x3 , x4 } ⊂ X , { y 1 , y 2 , y 3 , y 4 } ⊂ Y and { z1 , z2 , z3 , z4 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 , z2 and y 2 with 2 and color x3 , x4 and y 3 with 3 and color y 4 , z3 and z4 with 4. We divide each of X \ {x1 , x2 , x3 , x4 }, q−4 Y \ { y 1 , y 2 , y 3 , y 4 } and Z \ { z1 , z2 , z3 , z4 } into 3 classes equitably and color the of each class vertices with a color
in {5, . . . , q}. Since
n−4
= 3 and
q−4 3
n−4 q −4 3
= 4, it fol-
lows that the resulting coloring is an equitable (q, 2)-treecoloring of K n,n,n with the size of each color class being 3 or 4. Suppose n + 1 ≤ q ≤ 3n2−1 . Note that n > 1. Let e = xy be edge of K n,n,n with x ∈ X and y ∈ Y . We color x, y with 1 q−1 and divide each of X \ {x}, Y \ { y } and Z into 3 classes equitably and color the vertices of each class with a color in {2, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 2 or 3. Claim 1. q cannot be in ( 3n2−1 ,
3n+2 ). 2
Proof of Claim 1. Assume, to the contrary, that there exists an integer q such that q is in ( 3n2−1 , 3n2+2 ). Then
3n − 1 2
< 3a + 1 <
3n + 2 2
.
We find that n − 1 < 2a < n, a contradiction.
2
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Suppose 3n2+2 ≤ q ≤ 3(n − 1) + 1. Observe that n ≥ 1. Let e = xy be an edge of K n,n,n with x ∈ X and y ∈ Y . We color x and y with 1 and divide each of X \ {x}, Y \ { y } and q−1 Z into 3 classes equitably and color the vertices of each class with a color in {2, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 1 or 2. 1 Let q = 3a + 2 and 3 n+ + 2 ≤ q ≤ 3(n − 1) + 2. Note 4 1 that 4q − 3n ≥ 12 n+ + 8 − 3n ≥ 12( n−4 2 ) + 8 − 3n = 2. 4
Suppose 3n4+2 ≤ q ≤ n. Observe that n ≥ 2. Let {x1 , x2 } ⊂ X , { y 1 , y 2 } ⊂ Y and { z1 , z2 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color y 2 , z1 and z2 with 2 and divide each of q−2 X \ {x1 , x2 }, Y \ { y 1 , y 2 } and Z \ { z1 , z2 } into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 3 or 4. Suppose n ≤ q ≤ 3n2−2 . Note that n ≥ 2. Let {x1 , x2 } ⊂ X , { y 1 , y 2 } ⊂ Y and { z1 , z2 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color y 2 , z1 and z2 with 2 and divide each q−2 of X \ {x1 , x2 }, Y \ { y 1 , y 2 } and Z \ { z1 , Z 2 } into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 2 or 3. Claim 2. q cannot be in ( 3n2−2 ,
2
< 3a + 2 <
3n + 1 2
We now in a position to give our main results. 3.1. The strong equitable vertex 2-arboricity of K 4k+3,4k+3,4k+3 Proof of Theorem 1. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 3 for some non4 negative integer k. Firstly, we prove that va2 ≡ ( K n,n,n ) ≤ 1 3 n+ . By Proposition 1, we need to show that K n,n,n has 4 an equitable (3k + 4, 2)-tree-coloring. If k = 0, then K n,n,n = K 3,3,3 . We need to show that K 3,3,3 has an equitable (4, 2)-tree-coloring. Let {x1 , x2 , x3 } ⊂ X , { y 1 , y 2 , y 3 } ⊂ Y and { z1 , z2 , z3 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 and y 2 with 2 and color x3 and y 3 with 3 and color z2 and z3 with 4. It is easy to check that the resulting coloring of K 3,3,3 is an equitable (4, 2)-tree-coloring with the size of each color class being 2 or 3. If k ≥ 1, we need to prove that K 4k+3,4k+3,4k+3 has an equitable (3k + 4, 2)-tree-coloring. Then the size of every +9 color class is at least 3 because 12k = 3. Let c i de3k+4 note the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
c 3 + c 4 = 3k + 4.
.
We find that n − 2 < 2a < n − 1, a contradiction.
3. Main results
3c 3 + 4c 4 = 12k + 9
3n+1 ). 2
Proof of Claim 2. Assume, to the contrary, that there exists an integer q such that q is in ( 3n2−2 , 3n2+1 ). Then
3n − 2
979
2
Suppose 3n2+1 ≤ q ≤ 3(n − 1) + 2. Note that n ≥ 1. Let e = xy be an edge of K n,n,n with x ∈ X and y ∈ Y . Let z be a vertex of Z . We color x and y with 1 and color z with 2 q−2 and divide each of X \ {x}, Y \ { y } and Z \ { z} into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 1 or 2. 2 Lemma 1. If the cardinality of color classes is at least 4 in any equitable (t , 2)-tree-coloring of K n,n,n , then all vertices of every color class must belong to the same partite set. Proof. Assume, to the contrary, that all the vertices of some color class belong to at least two partite sets of K n,n,n . Then the graph induced by all the vertices of one color class has a cycle or the maximum degree of the graph induced by all the vertices of one color class is at least 3. For the former case, it contradicts the fact that the graph induced by all the vertices of one color class is a tree. For the latter case, it contradicts to the definition of the equitable (t , 2)-tree-coloring of K n,n,n, . So, the conclusion holds. 2
We have the following unique solution c 3 = 7, c 4 = 3k − 3. Let {x1 , x2 , x3 , x4 , x5 , x6 , x7 } ⊂ X , { y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 } ⊂ Y and { z1 , z2 , z3 , z4 , z5 , z6 , z7 } ⊂ Z . We color x1 , x2 and x3 with 1 and color y 1 , y 2 and y 3 with 2 and color z1 , z2 and z3 with 3 and color x4 , x5 and y 4 with 4 and color y 5 , z4 and z5 with 5 and color x6 , x7 and y 6 with 6 and color y 7 , z6 and z7 with 7. We divide each of X \ {x1 , x2 , x3 , x4 , x5 , x6 , x7 }, Y \ { y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 } and Z \ { z1 , z2 , z3 , z4 , z5 , z6 , z7 } into k − 1 classes equitably and color the vertices of each class with a color in {8, . . . , 3k + 4}. It is easy to check that the resulting coloring of K 4k+3,4k+3,4k+3 is an equitable (3k + 4, 2)-treecoloring with the size of each color class being 3 or 4. 1 Next, we prove that va2 ≡ ( K n,n,n ) ≥ 3 n+ . By Propo4 sition 1, we need to show that K n,n,n has no equitable (3k + 2, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 2, 2)-tree-coloring, then the size of every color class is at least 4 because 3k3n +2 = +9 12k = 4. By Lemma 1, we have that all the vertices 3k+2
of every color class belong to some partite set of K n,n,n . In fact, there exists a partite set in which there are at least k + 1 colors appearing. Without loss of generality, assume that there are at least k + 1 colors appearing in X . We then have | X | ≥ 4(k + 1) = (4k + 3) + 1 = | X | + 1, a con1 tradiction. This implies va2 ≡ ( K n,n,n ) ≥ 3k + 3 = 3 n+ . So, 4 1 va2 ≡ ( K n,n,n ) = 3 n+ . 4
2
3.2. The strong equitable vertex 2-arboricity of K 4k,4k,4k We investigate the strong equitable vertex 2-arboricity of the complete tripartite graph K 4k,4k,4k . To prove the first
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four equalities of Theorem 2, we first give an upper bound of va2 ≡ ( K 4k,4k,4k ), and then give a lower bound of it. By such a method, we can derive the exact value of K 4k,4k,4k for k ≡ i (mod 5) (i = 1, 2, 3, 4). The following upper bound of K 4k,4k,4k can be proved. Proposition 2. Let k be a positive integer. If k ≥ 5m (m ≥ 0), then va2 ≡ ( K 4k,4k,4k ) ≤ 3k − 3m. Proof. We prove the proposition by induction on m. If m = 0, then by Theorem 5, we have va2 ≡ ( K 4k,4k,4k ) ≤ 3k + 2. In order to prove that va2 ≡ ( K 4k,4k,4k ) ≤ 3k, we need to prove that K 4k,4k,4k has an equitable (3k + 1, 2)-treecoloring, by Proposition 1. Then the size of every color class is at least 3 because 3k12k +1 = 3. Let c i denote the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
3c 3 + 4c 4 = 12k c 3 + c 4 = 3k + 1. We have the following unique solution c 3 = 4, c 4 = 3k − 3. Let {x1 , x2 , x3 , x4 } ⊂ X , { y 1 , y 2 , y 3 , y 4 } ⊂ Y and { z1 , z2 , z3 , z4 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 , z2 and y 2 with 2 and color x3 , x4 and y 3 with 3 and color y 4 , z3 and z4 with 4. We divide each of X \ {x1 , x2 , x3 , x4 }, Y \ { y 1 , y 2 , y 3 , y 4 } and Z \ { z1 , z2 , z3 , z4 } into k − 1 classes equitably and color the vertices of each class with a color in {5, . . . , 3k + 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k + 1, 2)-tree-coloring with the size of each color class being 3 or 4. Suppose m ≥ 1. Since k ≥ 5m > 5(m − 1), by the induction hypothesis and Proposition 1, we need to prove that K 4k,4k,4k has an equitable (3k − 3m + 1, 2)-treecoloring and an equitable (3k − 3m + 2, 2)-tree-coloring. Divide X into k − m + 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , k − m + 1}. Divide Y into k − m classes equitably and color the vertices of each class with a color in {k − m + 2, . . . , 2k − 2m + 1}. Divide Z into k − m classes equitably and color the vertices of each class with a color in {2k − 2m + 2, . . . , 3k − 3m + 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k − 3m + 1, 2)-tree-coloring with the size of each color class being 4 or 5. Divide X into k − m + 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , k − m + 1}. Divide Y into k − m + 1 classes equitably and color the vertices of each class with a color in {k − m + 2, . . . , 2k − 2m + 2}. Divide Z into k − m classes equitably and color the vertices of each class with a color in {2k − 2m + 3, . . . , 3k − 3m + 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k − 3m + 2, 2)-tree-coloring with the size of each color class being 4 or 5. 2 The following corollaries are immediate. Corollary 1. Let k be a positive integer.
(i ) If k ≡ 1 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤
12k+3 . 5
(ii) If k ≡ 2 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+6 . (iii) If k ≡ 3 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+9 . (iv) If k ≡ 4 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+12 . We now distinguish the following four cases to give a lower bound of va2 ≡ ( K 4k,4k,4k ). Lemma 2. If k ≡ 1 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+3 . 5
Proof. Let k = 5m + 1. We only need to show that K 4k,4k,4k has no equitable (12m + 2, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 2, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at 12k 60m+12 least 5 because 12m +2 = 12m+2 = 5 for m ≥ 1 and
12k 60m+12 12m +2 = 12m+2 = 6 for m = 0. By Lemma 1, we have
that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at least 4m + 1 colors appearing. Without loss of generality, assume that there are at least 4m + 1 colors appearing in X . We then have | X | ≥ 5(4m + 1) > (20m + 4) = 4(5m + 1) = | X |, a contradiction. 2 Lemma 3. If k ≡ 2 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+6 . 5
Proof. Let k = 5m + 2. We only need to show that K 4k,4k,4k has no equitable (12m + 5, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 5, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at least 12k 60m+24 4 because 12m +5 = 12m+5 = 4. By Lemma 1, we have that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 1 colors appearing. Without loss of generality, assume that there are at most 4m + 1 colors appearing in X . We then have | X | ≤ 5(4m + 1) < (20m + 8) = 4(5m + 2) = | X |, a contradiction. 2 Lemma 4. If k ≡ 3 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+9 . 5
Proof. Let k = 5m + 3. We only need to show that K 4k,4k,4k has no equitable (12m + 8, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 8, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at least 12k 60m+36 4 because 12m +8 = 12m+8 = 4. By Lemma 1, we have that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 2 colors appearing. Without loss of generality, assume that there are at most 4m + 2 colors appearing in X . We then have | X | ≤ 5(4m + 2) < (20m + 12) = 4(5m + 3) = | X |, a contradiction. 2 Lemma 5. If k ≡ 4 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+12 . 5
Proof. Let k = 5m + 4. We only need to show that K 4k,4k,4k has no equitable (12m + 11, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 11, 2)-treecoloring of K 4k,4k,4k . Then the size of every color class 12k 60m+48 in c is at least 4 because 12m +11 = 12m+11 = 4. By Lemma 1, we have that all the vertices of every color class
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belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 3 colors appearing. Without loss of generality, assume that there are at most 4m + 3 colors appearing in X . We then have | X | ≤ 5(4m + 3) < (20m + 16) = 4(5m + 4) = | X |, a contradiction. 2 In the following, we consider the remaining case k ≡ 0 (mod 5) to give the proof of ( v ) of Theorem 2. Lemma 6. If k = 5m and m ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3. Proof. By Proposition 2, va2 ≡ ( K 4k,4k,4k ) ≤ 12m. So we need to prove that K 4k,4k,4k has an equitable (12m − 2, 2)tree-coloring and an equitable (12m − 1, 2)-tree-coloring by Proposition 1. Divide X into 4m − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 1}. Divide Y into 4m − 1 classes equitably and color the vertices of each class with a color in {4m, . . . , 8m − 2}. Divide Z into 4m classes equitably and color the vertices of each class with a color in {8m − 1, . . . , 12m − 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 2, 2)-tree-coloring with the size of each color class being 5 or 6. Divide X into 4m − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 1}. Divide Y into 4m classes equitably and color the vertices of each class with a color in {4m, . . . , 8m − 1}. Divide Z into 4m classes equitably and color the vertices of each class with a color in {8m, . . . , 12m − 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 1, 2)-tree-coloring with the size of each color class being 5 or 6. 2 Lemma 7. If k = 5m and m ≥ 12m − 6.
4, then va2 ≡ ( K 4k,4k,4k )
≤
Proof. We can prove it by using an argument similar to that of Lemma 6. 2 Lemma 8. If k = 5m and m ≥ 2q + 1, where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3. Proof. We prove the theorem by induction on q. If q = 2, it is equivalent to prove that if m ≥ 5, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 9. Since m ≥ 5 > 4, va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 6 by Lemma 7. We need prove that K 4k,4k,4k has an equitable (12m − 8, 2)-tree-coloring and an equitable (12m − 7, 2)tree-coloring, by Proposition 1. Divide X into 4m − 3 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 3}. Divide Y into 4m − 3 classes equitably and color the vertices of each class with a color in {4m − 2, . . . , 8m − 6}. Divide Z into 4m − 2 classes equitably and color the vertices of each class with a color in {8m − 5, . . . , 12m − 8}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 8, 2)-tree-coloring with the size of each color class being 5 or 6.
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Divide X into 4m − 3 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 3}. Divide Y into 4m − 2 classes equitably and color the vertices of each class with a color in {4m − 2, . . . , 8m − 5}. Divide Z into 4m − 2 classes equitably and color the vertices of each class with a color in {8m − 4, . . . , 12m − 7}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 7, 2)-tree-coloring with the size of each color class being 5 or 6. Assume q ≥ 3. Since m ≥ 2q + 1 > 2(q − 1) + 1, by the induction hypothesis and Proposition 1, we need to prove that K 4k,4k,4k has an equitable (12m − 3q − 1, 2)-treecoloring and an equitable (12m − 3q − 2, 2)-tree-coloring. Divide X into 4m − q − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − q − 1}. Divide Y into 4m − q − 1 classes equitably and color the vertices of each class with a color in {4m − q, . . . , 8m − 2q − 2}. Divide Z into 4m − q classes equitably and color the vertices of each class with a color in {8m − 2q − 1, . . . , 12m − 3q − 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 3q − 2, 2)-treecoloring with the size of each color class being 5 or 6. Divide X into 4m − q − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − q − 1}. Divide Y into 4m − q classes equitably and color the vertices of each class with a color in {4m − q, . . . , 8m − 2q − 1}. Divide Z into 4m − q classes equitably and color the vertices of each class with a color in {8m − 2q, . . . , 12m − 3q − 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 3q − 1, 2)-tree-coloring with the size of each color class being 5 or 6. 2 Lemma 9. If k = 5m and m ≥ 2q, where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Proof. When q = 0, 1, 2, the result holds from Proposition 2, Lemma 6, Lemma 7, respectively. If q ≥ 3, then q − 1 ≥ 2 and since m ≥ 2q > 2(q − 1) + 1, va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3(q − 1) − 3 = 12m − 3q. 2 Proof of Theorem 2. Suppose k ≡ 1 (mod 5). From Lemma 2 and (i ) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+3 . Suppose k ≡ 2 (mod 5). From Lemma 3 and (ii) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . Suppose k ≡ 3 (mod 5). From Lemma 4 and (iii) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . Suppose k ≡ 4 (mod 5). From Lemma 5 and (iv) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+12 . Suppose k ≡ 0 (mod 5). From Lemma 8, if k = 5m and m ≥ 2q + 1, where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3. From Lemma 9, if k = 5m and m ≥ 2q, where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. 2 3.3. The strong equitable vertex 2-arboricity of K 4k+1,4k+1,4k+1 We investigate the strong equitable vertex 2-arboricity of complete tripartite graph K 4k+1,4k+1,4k+1 . To prove Theorem 3, we give a lower bound of va2 ≡ ( K 4k+1,4k+1,4k+1 ). Combining with the upper bound from Theorem 5, we can obtain the exact value of va2 ≡ ( K 4k+1,4k+1,4k+1 ).
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Proof of Theorem 3. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 1. In order to prove 4
1 that va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2, we only need to show that 4 K n,n,n has no equitable (3k + 1, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 1, 2)-treecoloring. Then the size of every color class is at least 3 12k+3 because 3k3n +1 = 3k+1 = 3. Let c i denote the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
3c 3 + 4c 4 = 12k + 3 c 3 + c 4 = 3k + 1. We have the unique solution c 3 = 1, c 4 = 3k. Since each color class that contains exactly 4 vertices must appear in some partite set of K n,n,n , it follows that K n,n,n has no equitable (3k + 1, 2)-tree-coloring satisfying the above 1 conditions. Then va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2 and hence 4 1 va2 ≡ ( K n,n,n ) = 3 n+ + 2. 4
2
3.4. The strong equitable vertex 2-arboricity of K 4k+2,4k+2,4k+2 We investigate the strong equitable vertex 2-arboricity of the complete tripartite graph K 4k+2,4k+2,4k+2 . We prove Theorem 4 by using an argument similar to that of Theorem 3.
4. Conclusions In this paper, we discuss the strong equitable vertex 2-arboricity of complete tripartite graph by considering four cases. For the case of n ≡ 3 (mod 4), we obtain 1 that the exact value of va2 ≡ ( K n,n,n ) is 3 n+ . For the 4 case of n ≡ 1 (mod 4), the exact value of va2 ≡ ( K n,n,n ) 1 is 3 n+ + 2. For the case of n ≡ 2 (mod 4), the ex4
1 act value of va2 ≡ ( K n,n,n ) is 3 n+ + 2. For the case of 4 n ≡ 0 (mod 4), we investigate the va2 ≡ ( K 4k,4k,4k ) by considering the following five subcases. Let k and m be positive integer.
(i ) (ii) (iii) (iv) (v )
For k ≡ 1 (mod 5), va2 ≡ ( K 4k,4k,4k ) = For k ≡ For k ≡
12k+3 . 5
2 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . 3 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . 4 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+12 .
For k ≡ For k ≡ 0 (mod 5), if m ≥ 2q + 1 where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3;
if m ≥ 2q where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Acknowledgements The authors are very grateful to the editor and two referees’ valuable comments and suggestions, which helped greatly to improve the presentation of this paper. References
Proof of Theorem 4. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 2. In order to 4
1 prove that va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2, we only need to 4 show that K n,n,n has no equitable (3k + 1, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 1, 2)-tree-coloring. Then the size of every color class 12k+6 is at least 4 because 3k3n +1 = 3k+1 = 4 for k ≥ 1 and 12k+6 3k3n +1 = 3k+1 = 6 for k = 0. Let c i denote the number
of those color classes such that each color class contains exactly i vertices, where i = 4, 5. Then we have the following two equations:
4c 4 + 5c 5 = 12k + 6 c 4 + c 5 = 3k + 1. We have the unique solution c 4 = 3k − 1, c 5 = 2. Since each color class containing exactly 4 or 5 vertices must appear in some partite set of K n,n,n , it follows that K n,n,n has no equitable (3k + 1, 2)-tree-coloring satisfying the above 1 conditions. Then va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2 and hence 4 1 va2 ≡ ( K n,n,n ) = 3 n+ + 2. 4
2
[1] B. Baker, E. Coffman, Mutual exclusion scheduling, Theor. Comput. Sci. 162 (1996) 225–243. [2] J. Blazewicz, K. Ecker, E. Pesch, G. Schmidt, J. Weglarz, Scheduling Computer and Manufacturing Processes, Springer, Berlin, 2001. [3] Béla Bollobás, Modern Graph Theory, Springer-Verlag, New York, 1998. [4] S.K. Das, I. Finocchi, R. Petreschi, Conflict-free star-access in parallel memory systems, J. Parallel Distrib. Comput. 66 (2006) 1431–1441. [5] H. Fan, H.A. Kierstead, G.Z. Liu, T. Molla, J.L. Wu, X. Zhang, A note on relaxed equitable coloring of graphs, Inf. Process. Lett. 111 (2011) 1062–1066. [6] S. Irani, V. Leung, Scheduling with conflicts and applications to traffic signal control, in: Proceedings of the Seventh Annual ACM–SIAM Symposium on Discrete Algorithms, Atlanta, 1996, pp. 85–94. [7] F. Kitagawa, H. Ikeda, An existential problem of a weight-controlled subset and its application to schedule timetable construction, Discrete Math. 72 (1988) 195–211. [8] W. Meyer, Equitable coloring, Am. Math. Mon. 80 (1973) 143–149. [9] B.F. Smith, P.E. Bjorstad, W.D. Gropp, Domain decomposition, in: Parallel Multilevel Methods for Elliptic Partial Differential Equations, Cambridge University Press, Cambridge, 1996, p. 224. [10] A. Tucker, Perfect graphs and an application to optimizing municipal services, SIAM Rev. 15 (1973) 585–590. [11] J.L. Wu, X. Zhang, H.L. Li, Equitable vertex arboricity of graphs, Discrete Math. 313 (2013) 2696–2701.
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
The equitable vertex arboricity of complete tripartite graphs ✩ Zhiwei Guo a,∗ , Haixing Zhao b , Yaping Mao a a b
Department of Mathematics, Qinghai Normal University, Xining, Qinghai 810008, China School of Computer, Qinghai Normal University, Xining, Qinghai 810008, China
a r t i c l e
i n f o
Article history: Received 11 January 2015 Received in revised form 13 June 2015 Accepted 26 June 2015 Available online 2 July 2015 Communicated by Ł. Kowalik Keywords: Combinatorial problems Equitable coloring Vertex k-arboricity k-Tree-coloring Complete tripartite graph
a b s t r a c t The equitable coloring problem, introduced by Meyer in 1973, has received considerable attention and research. Recently, Wu et al. introduced the concept of equitable (t , k)-treecoloring, which can be viewed as a generalization of proper equitable t-coloring. Wu et al. investigated the strong equitable vertex k-arboricity of complete bipartite graphs. In this paper, we mainly investigate the strong equitable vertex 2-arboricity of complete tripartite graphs. © 2015 Elsevier B.V. All rights reserved.
1. Introduction All graphs considered in this paper are finite and simple. For a real number x, x is the least integer not less than x and x is the largest integer not larger than x. We use V (G ), E (G ), δ(G ) and (G ) to denote the vertex set, edge set, minimum degree and maximum degree of G, respectively. For a vertex v ∈ V (G ), let N G ( v ) denote the set of neighbors of v in G and d G ( v ) = | N G ( v )| denote the degree of v. We use d( v ) instead of d G ( v ) for brevity. For other undefined concepts, we refer the reader to [3]. We associate positive integers 1, 2, . . . , t with colors and call f a t-coloring of a graph G if f is a mapping
✩ Research supported by the National Natural Science Foundation of China (Nos. 61164005, 61460005, 11161037 and 11461054), Key Laboratory of Tibetan intelligent information processing and machine translation of Qinghai P.R.C (No.2013-Z-Y17) and the Program for Changjiang Scholars and Innovative Research Team in Universities (No. IRT1068), the Research Fund for the Chunhui Program of Ministry of Education of China (No. Z2014022) and the Nature Science Foundation from Qinghai Province (No. 2014-ZJ-721 and No. 2014-ZJ-907). Corresponding author. E-mail addresses: [email protected] (Z. Guo), [email protected] (H. Zhao), [email protected] (Y. Mao).
*
http://dx.doi.org/10.1016/j.ipl.2015.06.016 0020-0190/© 2015 Elsevier B.V. All rights reserved.
from V (G ) to {1, 2, . . . , t }. A t-coloring of G is proper if any two adjacent vertices have different colors. For 1 ≤ i ≤ t, let V i = { v | f ( v ) = i }. A t-coloring of a graph G is said to be equitable if || V i | − | V j || ≤ 1 for all i and j, that is, every color class has size | V (G )|/t or | V (G )|/t . A graph G is said to be properly equitably t-colorable if G has a proper equitable t-coloring. The smallest number t for which G is properly equitably t-colorable is called the equitable chromatic number of G, denoted by χ= (G ). The equitable coloring problem, introduced by Meyer [8], was motivated by a practical application to municipal garbage collection [10]. In this context, the vertices of a graph represent garbage collection routes. A pair of vertices share an edge if the corresponding routes should not be run on the same day. It is desirable that the number of routes ran on each day be approximately the same. Therefore, the problem of assigning one of the six weekly working days to each route reduces to finding a proper equitable 6-coloring. For more applications such as scheduling, constructing timetables and load balance in parallel memory systems, we refer to [1,2,4,6,7,9]. Note that a properly equitably t-colorable graph may admit no proper equitable t -colorings for some t > t. For example, the complete bipartite graph H := K 2m+1,2m+1
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has no proper equitable (2m + 1)-colorings, although it satisfies χ= ( H ) = 2. This fact motivates another interesting parameter for proper equitable coloring. The equitable ∗ (G ), is the smallchromatic threshold of G, denoted by χ= est integer t such that G has a proper equitable t -coloring for all t ≥ t. For a graph G, a d-relaxed k-coloring, also known as a d-defective coloring, of G is a function f from V (G ) to {1, 2, . . . , k} such that each subgraph G [ V i ] is a graph of maximum degree at most d. A d-relaxed equitable k-coloring is a d-relaxed k-coloring that is equitable. In Fan et al. [5], relaxed equitable coloring of graphs was considered. They proved that every graph has a proper equitable (G )-coloring such that each color class induces a forest with maximum degree at most one. On the basis of this research, Wu et al. [11] introduced the notion of equitable (t , k)-tree-coloring, which can be viewed as a generalization of proper equitable t-coloring. A (t , k)-tree-coloring is a t-coloring f of a graph G such that each component of G [ V i ] is a tree of maximum degree at most k. A (t , ∞)-tree-coloring is called a t-tree-coloring for short. An equitable (t , k)-tree-coloring is a (t , k)-tree-coloring that is equitable. The equitable vertex k-arboricity of a graph G, denoted by vak = (G ), is the smallest integer t such that G has an equitable (t , k)-treecoloring. The strong equitable vertex k-arboricity of G, denoted by vak ≡ (G ), is the smallest integer t such that G has an equitable (t , k)-tree-coloring for every t ≥ t. It is ∗ (G ) for evclear that va0 = (G ) = χ= (G ) and va0 ≡ (G ) = χ= ery graph G, and vak = (G ) and vak ≡ (G ) may vary a lot. Wu et al. [11] investigated the strong equitable vertex k-arboricity of complete bipartite graphs. In this paper, we mainly investigate the strong equitable vertex 2-arboricity of complete tripartite graph. Theorem 1. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 3 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ . 4 Theorem 2. Let k and m be positive integer.
(i ) (ii) (iii) (iv) (v )
For k ≡ 1 (mod 5), va2 ≡ ( K 4k,4k,4k ) = For k ≡ For k ≡
12k+3 . 5
2 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . 3 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . 4 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+12 .
For k ≡ For k ≡ 0 (mod 5), if m ≥ 2q + 1 where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3;
if m ≥ 2q where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Theorem 3. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 1 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ + 2. 4 Theorem 4. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. If n ≡ 2 (mod 4), then va2 ≡ ( K n,n,n ) = 1 3 n+ + 2. 4
2. Preliminary In this section, we give some basic results, which will be used later. Proposition 1. The complete -partite graph K n,...,n has an equitable (t , k)-tree-coloring for every integer t satisfying the following condition:
t = h (h ≥ 1). Proof. One can easily construct an equitable (t , k)-treecoloring of K n,...,n by dividing each partite set into h classes equitably and coloring each class with different colors. Note that the coloring is a proper equitable (t , k)-treecoloring. 2 Theorem 5. Let n be a positive integer, and let K n,n,n be a complete tripartite graph. Then
va2 ≡ ( K n,n,n ) ≤ 3
n+1
4
+ 2.
Proof. By Proposition 1, in order to show va2 ≡ ( K n,n,n ) ≤ 1 3 n+ + 2, we only need to prove that K n,n,n has an eq4 uitable (q, 2)-tree-coloring for every q satisfying the fol1 + 2 ≤ q ≤ 3n and 3 q. Let X , Y lowing condition: 3 n+ 4 and Z be the partite sets of K n,n,n . 1 Let q = 3a + 1 and 3 n+ + 4 ≤ q ≤ 3(n − 1) + 1. Note 4 1 that 4q − 3n ≥ 12 n+ + 16 − 3n ≥ 12( n−4 2 ) + 16 − 3n = 10. 4
10 Suppose 3n+ ≤ q ≤ n. Observe that n > 4. Let {x1 , x2 , 4 x3 , x4 } ⊂ X , { y 1 , y 2 , y 3 , y 4 } ⊂ Y and { z1 , z2 , z3 , z4 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 , z2 and y 2 with 2 and color x3 , x4 and y 3 with 3 and color y 4 , z3 and z4 with 4. We divide each of X \ {x1 , x2 , x3 , x4 }, q−4 Y \ { y 1 , y 2 , y 3 , y 4 } and Z \ { z1 , z2 , z3 , z4 } into 3 classes equitably and color the of each class vertices with a color
in {5, . . . , q}. Since
n−4
= 3 and
q−4 3
n−4 q −4 3
= 4, it fol-
lows that the resulting coloring is an equitable (q, 2)-treecoloring of K n,n,n with the size of each color class being 3 or 4. Suppose n + 1 ≤ q ≤ 3n2−1 . Note that n > 1. Let e = xy be edge of K n,n,n with x ∈ X and y ∈ Y . We color x, y with 1 q−1 and divide each of X \ {x}, Y \ { y } and Z into 3 classes equitably and color the vertices of each class with a color in {2, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 2 or 3. Claim 1. q cannot be in ( 3n2−1 ,
3n+2 ). 2
Proof of Claim 1. Assume, to the contrary, that there exists an integer q such that q is in ( 3n2−1 , 3n2+2 ). Then
3n − 1 2
< 3a + 1 <
3n + 2 2
.
We find that n − 1 < 2a < n, a contradiction.
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Suppose 3n2+2 ≤ q ≤ 3(n − 1) + 1. Observe that n ≥ 1. Let e = xy be an edge of K n,n,n with x ∈ X and y ∈ Y . We color x and y with 1 and divide each of X \ {x}, Y \ { y } and q−1 Z into 3 classes equitably and color the vertices of each class with a color in {2, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 1 or 2. 1 Let q = 3a + 2 and 3 n+ + 2 ≤ q ≤ 3(n − 1) + 2. Note 4 1 that 4q − 3n ≥ 12 n+ + 8 − 3n ≥ 12( n−4 2 ) + 8 − 3n = 2. 4
Suppose 3n4+2 ≤ q ≤ n. Observe that n ≥ 2. Let {x1 , x2 } ⊂ X , { y 1 , y 2 } ⊂ Y and { z1 , z2 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color y 2 , z1 and z2 with 2 and divide each of q−2 X \ {x1 , x2 }, Y \ { y 1 , y 2 } and Z \ { z1 , z2 } into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 3 or 4. Suppose n ≤ q ≤ 3n2−2 . Note that n ≥ 2. Let {x1 , x2 } ⊂ X , { y 1 , y 2 } ⊂ Y and { z1 , z2 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color y 2 , z1 and z2 with 2 and divide each q−2 of X \ {x1 , x2 }, Y \ { y 1 , y 2 } and Z \ { z1 , Z 2 } into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 2 or 3. Claim 2. q cannot be in ( 3n2−2 ,
2
< 3a + 2 <
3n + 1 2
We now in a position to give our main results. 3.1. The strong equitable vertex 2-arboricity of K 4k+3,4k+3,4k+3 Proof of Theorem 1. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 3 for some non4 negative integer k. Firstly, we prove that va2 ≡ ( K n,n,n ) ≤ 1 3 n+ . By Proposition 1, we need to show that K n,n,n has 4 an equitable (3k + 4, 2)-tree-coloring. If k = 0, then K n,n,n = K 3,3,3 . We need to show that K 3,3,3 has an equitable (4, 2)-tree-coloring. Let {x1 , x2 , x3 } ⊂ X , { y 1 , y 2 , y 3 } ⊂ Y and { z1 , z2 , z3 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 and y 2 with 2 and color x3 and y 3 with 3 and color z2 and z3 with 4. It is easy to check that the resulting coloring of K 3,3,3 is an equitable (4, 2)-tree-coloring with the size of each color class being 2 or 3. If k ≥ 1, we need to prove that K 4k+3,4k+3,4k+3 has an equitable (3k + 4, 2)-tree-coloring. Then the size of every +9 color class is at least 3 because 12k = 3. Let c i de3k+4 note the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
c 3 + c 4 = 3k + 4.
.
We find that n − 2 < 2a < n − 1, a contradiction.
3. Main results
3c 3 + 4c 4 = 12k + 9
3n+1 ). 2
Proof of Claim 2. Assume, to the contrary, that there exists an integer q such that q is in ( 3n2−2 , 3n2+1 ). Then
3n − 2
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2
Suppose 3n2+1 ≤ q ≤ 3(n − 1) + 2. Note that n ≥ 1. Let e = xy be an edge of K n,n,n with x ∈ X and y ∈ Y . Let z be a vertex of Z . We color x and y with 1 and color z with 2 q−2 and divide each of X \ {x}, Y \ { y } and Z \ { z} into 3 classes equitably and color the vertices of each class with a color in {3, . . . , q}. One can easily check that the resulting coloring is an equitable (q, 2)-tree-coloring of K n,n,n with the size of each color class being 1 or 2. 2 Lemma 1. If the cardinality of color classes is at least 4 in any equitable (t , 2)-tree-coloring of K n,n,n , then all vertices of every color class must belong to the same partite set. Proof. Assume, to the contrary, that all the vertices of some color class belong to at least two partite sets of K n,n,n . Then the graph induced by all the vertices of one color class has a cycle or the maximum degree of the graph induced by all the vertices of one color class is at least 3. For the former case, it contradicts the fact that the graph induced by all the vertices of one color class is a tree. For the latter case, it contradicts to the definition of the equitable (t , 2)-tree-coloring of K n,n,n, . So, the conclusion holds. 2
We have the following unique solution c 3 = 7, c 4 = 3k − 3. Let {x1 , x2 , x3 , x4 , x5 , x6 , x7 } ⊂ X , { y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 } ⊂ Y and { z1 , z2 , z3 , z4 , z5 , z6 , z7 } ⊂ Z . We color x1 , x2 and x3 with 1 and color y 1 , y 2 and y 3 with 2 and color z1 , z2 and z3 with 3 and color x4 , x5 and y 4 with 4 and color y 5 , z4 and z5 with 5 and color x6 , x7 and y 6 with 6 and color y 7 , z6 and z7 with 7. We divide each of X \ {x1 , x2 , x3 , x4 , x5 , x6 , x7 }, Y \ { y 1 , y 2 , y 3 , y 4 , y 5 , y 6 , y 7 } and Z \ { z1 , z2 , z3 , z4 , z5 , z6 , z7 } into k − 1 classes equitably and color the vertices of each class with a color in {8, . . . , 3k + 4}. It is easy to check that the resulting coloring of K 4k+3,4k+3,4k+3 is an equitable (3k + 4, 2)-treecoloring with the size of each color class being 3 or 4. 1 Next, we prove that va2 ≡ ( K n,n,n ) ≥ 3 n+ . By Propo4 sition 1, we need to show that K n,n,n has no equitable (3k + 2, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 2, 2)-tree-coloring, then the size of every color class is at least 4 because 3k3n +2 = +9 12k = 4. By Lemma 1, we have that all the vertices 3k+2
of every color class belong to some partite set of K n,n,n . In fact, there exists a partite set in which there are at least k + 1 colors appearing. Without loss of generality, assume that there are at least k + 1 colors appearing in X . We then have | X | ≥ 4(k + 1) = (4k + 3) + 1 = | X | + 1, a con1 tradiction. This implies va2 ≡ ( K n,n,n ) ≥ 3k + 3 = 3 n+ . So, 4 1 va2 ≡ ( K n,n,n ) = 3 n+ . 4
2
3.2. The strong equitable vertex 2-arboricity of K 4k,4k,4k We investigate the strong equitable vertex 2-arboricity of the complete tripartite graph K 4k,4k,4k . To prove the first
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four equalities of Theorem 2, we first give an upper bound of va2 ≡ ( K 4k,4k,4k ), and then give a lower bound of it. By such a method, we can derive the exact value of K 4k,4k,4k for k ≡ i (mod 5) (i = 1, 2, 3, 4). The following upper bound of K 4k,4k,4k can be proved. Proposition 2. Let k be a positive integer. If k ≥ 5m (m ≥ 0), then va2 ≡ ( K 4k,4k,4k ) ≤ 3k − 3m. Proof. We prove the proposition by induction on m. If m = 0, then by Theorem 5, we have va2 ≡ ( K 4k,4k,4k ) ≤ 3k + 2. In order to prove that va2 ≡ ( K 4k,4k,4k ) ≤ 3k, we need to prove that K 4k,4k,4k has an equitable (3k + 1, 2)-treecoloring, by Proposition 1. Then the size of every color class is at least 3 because 3k12k +1 = 3. Let c i denote the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
3c 3 + 4c 4 = 12k c 3 + c 4 = 3k + 1. We have the following unique solution c 3 = 4, c 4 = 3k − 3. Let {x1 , x2 , x3 , x4 } ⊂ X , { y 1 , y 2 , y 3 , y 4 } ⊂ Y and { z1 , z2 , z3 , z4 } ⊂ Z . We color x1 , x2 and y 1 with 1 and color z1 , z2 and y 2 with 2 and color x3 , x4 and y 3 with 3 and color y 4 , z3 and z4 with 4. We divide each of X \ {x1 , x2 , x3 , x4 }, Y \ { y 1 , y 2 , y 3 , y 4 } and Z \ { z1 , z2 , z3 , z4 } into k − 1 classes equitably and color the vertices of each class with a color in {5, . . . , 3k + 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k + 1, 2)-tree-coloring with the size of each color class being 3 or 4. Suppose m ≥ 1. Since k ≥ 5m > 5(m − 1), by the induction hypothesis and Proposition 1, we need to prove that K 4k,4k,4k has an equitable (3k − 3m + 1, 2)-treecoloring and an equitable (3k − 3m + 2, 2)-tree-coloring. Divide X into k − m + 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , k − m + 1}. Divide Y into k − m classes equitably and color the vertices of each class with a color in {k − m + 2, . . . , 2k − 2m + 1}. Divide Z into k − m classes equitably and color the vertices of each class with a color in {2k − 2m + 2, . . . , 3k − 3m + 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k − 3m + 1, 2)-tree-coloring with the size of each color class being 4 or 5. Divide X into k − m + 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , k − m + 1}. Divide Y into k − m + 1 classes equitably and color the vertices of each class with a color in {k − m + 2, . . . , 2k − 2m + 2}. Divide Z into k − m classes equitably and color the vertices of each class with a color in {2k − 2m + 3, . . . , 3k − 3m + 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (3k − 3m + 2, 2)-tree-coloring with the size of each color class being 4 or 5. 2 The following corollaries are immediate. Corollary 1. Let k be a positive integer.
(i ) If k ≡ 1 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤
12k+3 . 5
(ii) If k ≡ 2 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+6 . (iii) If k ≡ 3 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+9 . (iv) If k ≡ 4 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≤ 12k5+12 . We now distinguish the following four cases to give a lower bound of va2 ≡ ( K 4k,4k,4k ). Lemma 2. If k ≡ 1 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+3 . 5
Proof. Let k = 5m + 1. We only need to show that K 4k,4k,4k has no equitable (12m + 2, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 2, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at 12k 60m+12 least 5 because 12m +2 = 12m+2 = 5 for m ≥ 1 and
12k 60m+12 12m +2 = 12m+2 = 6 for m = 0. By Lemma 1, we have
that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at least 4m + 1 colors appearing. Without loss of generality, assume that there are at least 4m + 1 colors appearing in X . We then have | X | ≥ 5(4m + 1) > (20m + 4) = 4(5m + 1) = | X |, a contradiction. 2 Lemma 3. If k ≡ 2 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+6 . 5
Proof. Let k = 5m + 2. We only need to show that K 4k,4k,4k has no equitable (12m + 5, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 5, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at least 12k 60m+24 4 because 12m +5 = 12m+5 = 4. By Lemma 1, we have that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 1 colors appearing. Without loss of generality, assume that there are at most 4m + 1 colors appearing in X . We then have | X | ≤ 5(4m + 1) < (20m + 8) = 4(5m + 2) = | X |, a contradiction. 2 Lemma 4. If k ≡ 3 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+9 . 5
Proof. Let k = 5m + 3. We only need to show that K 4k,4k,4k has no equitable (12m + 8, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 8, 2)-tree-coloring of K 4k,4k,4k . Then the size of every color class in c is at least 12k 60m+36 4 because 12m +8 = 12m+8 = 4. By Lemma 1, we have that all the vertices of every color class belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 2 colors appearing. Without loss of generality, assume that there are at most 4m + 2 colors appearing in X . We then have | X | ≤ 5(4m + 2) < (20m + 12) = 4(5m + 3) = | X |, a contradiction. 2 Lemma 5. If k ≡ 4 (mod 5), then va2 ≡ ( K 4k,4k,4k ) ≥
12k+12 . 5
Proof. Let k = 5m + 4. We only need to show that K 4k,4k,4k has no equitable (12m + 11, 2)-tree-coloring. Assume, to the contrary, that c is an equitable (12m + 11, 2)-treecoloring of K 4k,4k,4k . Then the size of every color class 12k 60m+48 in c is at least 4 because 12m +11 = 12m+11 = 4. By Lemma 1, we have that all the vertices of every color class
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belong to some partite set of K 4k,4k,4k . In fact, there exists a partite set in which there are at most 4m + 3 colors appearing. Without loss of generality, assume that there are at most 4m + 3 colors appearing in X . We then have | X | ≤ 5(4m + 3) < (20m + 16) = 4(5m + 4) = | X |, a contradiction. 2 In the following, we consider the remaining case k ≡ 0 (mod 5) to give the proof of ( v ) of Theorem 2. Lemma 6. If k = 5m and m ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3. Proof. By Proposition 2, va2 ≡ ( K 4k,4k,4k ) ≤ 12m. So we need to prove that K 4k,4k,4k has an equitable (12m − 2, 2)tree-coloring and an equitable (12m − 1, 2)-tree-coloring by Proposition 1. Divide X into 4m − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 1}. Divide Y into 4m − 1 classes equitably and color the vertices of each class with a color in {4m, . . . , 8m − 2}. Divide Z into 4m classes equitably and color the vertices of each class with a color in {8m − 1, . . . , 12m − 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 2, 2)-tree-coloring with the size of each color class being 5 or 6. Divide X into 4m − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 1}. Divide Y into 4m classes equitably and color the vertices of each class with a color in {4m, . . . , 8m − 1}. Divide Z into 4m classes equitably and color the vertices of each class with a color in {8m, . . . , 12m − 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 1, 2)-tree-coloring with the size of each color class being 5 or 6. 2 Lemma 7. If k = 5m and m ≥ 12m − 6.
4, then va2 ≡ ( K 4k,4k,4k )
≤
Proof. We can prove it by using an argument similar to that of Lemma 6. 2 Lemma 8. If k = 5m and m ≥ 2q + 1, where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3. Proof. We prove the theorem by induction on q. If q = 2, it is equivalent to prove that if m ≥ 5, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 9. Since m ≥ 5 > 4, va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 6 by Lemma 7. We need prove that K 4k,4k,4k has an equitable (12m − 8, 2)-tree-coloring and an equitable (12m − 7, 2)tree-coloring, by Proposition 1. Divide X into 4m − 3 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 3}. Divide Y into 4m − 3 classes equitably and color the vertices of each class with a color in {4m − 2, . . . , 8m − 6}. Divide Z into 4m − 2 classes equitably and color the vertices of each class with a color in {8m − 5, . . . , 12m − 8}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 8, 2)-tree-coloring with the size of each color class being 5 or 6.
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Divide X into 4m − 3 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − 3}. Divide Y into 4m − 2 classes equitably and color the vertices of each class with a color in {4m − 2, . . . , 8m − 5}. Divide Z into 4m − 2 classes equitably and color the vertices of each class with a color in {8m − 4, . . . , 12m − 7}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 7, 2)-tree-coloring with the size of each color class being 5 or 6. Assume q ≥ 3. Since m ≥ 2q + 1 > 2(q − 1) + 1, by the induction hypothesis and Proposition 1, we need to prove that K 4k,4k,4k has an equitable (12m − 3q − 1, 2)-treecoloring and an equitable (12m − 3q − 2, 2)-tree-coloring. Divide X into 4m − q − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − q − 1}. Divide Y into 4m − q − 1 classes equitably and color the vertices of each class with a color in {4m − q, . . . , 8m − 2q − 2}. Divide Z into 4m − q classes equitably and color the vertices of each class with a color in {8m − 2q − 1, . . . , 12m − 3q − 2}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 3q − 2, 2)-treecoloring with the size of each color class being 5 or 6. Divide X into 4m − q − 1 classes equitably and color the vertices of each class with a color in {1, 2, . . . , 4m − q − 1}. Divide Y into 4m − q classes equitably and color the vertices of each class with a color in {4m − q, . . . , 8m − 2q − 1}. Divide Z into 4m − q classes equitably and color the vertices of each class with a color in {8m − 2q, . . . , 12m − 3q − 1}. It is easy to check that the resulting coloring of K 4k,4k,4k is an equitable (12m − 3q − 1, 2)-tree-coloring with the size of each color class being 5 or 6. 2 Lemma 9. If k = 5m and m ≥ 2q, where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Proof. When q = 0, 1, 2, the result holds from Proposition 2, Lemma 6, Lemma 7, respectively. If q ≥ 3, then q − 1 ≥ 2 and since m ≥ 2q > 2(q − 1) + 1, va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3(q − 1) − 3 = 12m − 3q. 2 Proof of Theorem 2. Suppose k ≡ 1 (mod 5). From Lemma 2 and (i ) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+3 . Suppose k ≡ 2 (mod 5). From Lemma 3 and (ii) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . Suppose k ≡ 3 (mod 5). From Lemma 4 and (iii) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . Suppose k ≡ 4 (mod 5). From Lemma 5 and (iv) of Corollary 1, va2 ≡ ( K 4k,4k,4k ) = 12k5+12 . Suppose k ≡ 0 (mod 5). From Lemma 8, if k = 5m and m ≥ 2q + 1, where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3. From Lemma 9, if k = 5m and m ≥ 2q, where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. 2 3.3. The strong equitable vertex 2-arboricity of K 4k+1,4k+1,4k+1 We investigate the strong equitable vertex 2-arboricity of complete tripartite graph K 4k+1,4k+1,4k+1 . To prove Theorem 3, we give a lower bound of va2 ≡ ( K 4k+1,4k+1,4k+1 ). Combining with the upper bound from Theorem 5, we can obtain the exact value of va2 ≡ ( K 4k+1,4k+1,4k+1 ).
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Proof of Theorem 3. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 1. In order to prove 4
1 that va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2, we only need to show that 4 K n,n,n has no equitable (3k + 1, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 1, 2)-treecoloring. Then the size of every color class is at least 3 12k+3 because 3k3n +1 = 3k+1 = 3. Let c i denote the number of those color classes such that each color class contains exactly i vertices, where i = 3, 4. Then we have the following two equations:
3c 3 + 4c 4 = 12k + 3 c 3 + c 4 = 3k + 1. We have the unique solution c 3 = 1, c 4 = 3k. Since each color class that contains exactly 4 vertices must appear in some partite set of K n,n,n , it follows that K n,n,n has no equitable (3k + 1, 2)-tree-coloring satisfying the above 1 conditions. Then va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2 and hence 4 1 va2 ≡ ( K n,n,n ) = 3 n+ + 2. 4
2
3.4. The strong equitable vertex 2-arboricity of K 4k+2,4k+2,4k+2 We investigate the strong equitable vertex 2-arboricity of the complete tripartite graph K 4k+2,4k+2,4k+2 . We prove Theorem 4 by using an argument similar to that of Theorem 3.
4. Conclusions In this paper, we discuss the strong equitable vertex 2-arboricity of complete tripartite graph by considering four cases. For the case of n ≡ 3 (mod 4), we obtain 1 that the exact value of va2 ≡ ( K n,n,n ) is 3 n+ . For the 4 case of n ≡ 1 (mod 4), the exact value of va2 ≡ ( K n,n,n ) 1 is 3 n+ + 2. For the case of n ≡ 2 (mod 4), the ex4
1 act value of va2 ≡ ( K n,n,n ) is 3 n+ + 2. For the case of 4 n ≡ 0 (mod 4), we investigate the va2 ≡ ( K 4k,4k,4k ) by considering the following five subcases. Let k and m be positive integer.
(i ) (ii) (iii) (iv) (v )
For k ≡ 1 (mod 5), va2 ≡ ( K 4k,4k,4k ) = For k ≡ For k ≡
12k+3 . 5
2 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+6 . 3 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+9 . 4 (mod 5), va2 ≡ ( K 4k,4k,4k ) = 12k5+12 .
For k ≡ For k ≡ 0 (mod 5), if m ≥ 2q + 1 where q ≥ 2, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q − 3;
if m ≥ 2q where q ≥ 0, then va2 ≡ ( K 4k,4k,4k ) ≤ 12m − 3q. Acknowledgements The authors are very grateful to the editor and two referees’ valuable comments and suggestions, which helped greatly to improve the presentation of this paper. References
Proof of Theorem 4. From Theorem 5, we have 1 va2 ≡ ( K n,n,n ) ≤ 3 n+ + 2. Let n = 4k + 2. In order to 4
1 prove that va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2, we only need to 4 show that K n,n,n has no equitable (3k + 1, 2)-tree-coloring. Assume, to the contrary, that K n,n,n has an equitable (3k + 1, 2)-tree-coloring. Then the size of every color class 12k+6 is at least 4 because 3k3n +1 = 3k+1 = 4 for k ≥ 1 and 12k+6 3k3n +1 = 3k+1 = 6 for k = 0. Let c i denote the number
of those color classes such that each color class contains exactly i vertices, where i = 4, 5. Then we have the following two equations:
4c 4 + 5c 5 = 12k + 6 c 4 + c 5 = 3k + 1. We have the unique solution c 4 = 3k − 1, c 5 = 2. Since each color class containing exactly 4 or 5 vertices must appear in some partite set of K n,n,n , it follows that K n,n,n has no equitable (3k + 1, 2)-tree-coloring satisfying the above 1 conditions. Then va2 ≡ ( K n,n,n ) ≥ 3 n+ + 2 and hence 4 1 va2 ≡ ( K n,n,n ) = 3 n+ + 2. 4
2
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