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L(2,1) -Labeling of Kneser graphs and coloring squares of Kneser graphs
Discrete Applied Mathematics (
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L(2, 1)-Labeling of Kneser graphs and coloring squares of Kneser graphs Zhendong Shao a,∗ , Igor Averbakh a , Roberto Solis-Oba b a
Department of Management, University of Toronto Scarborough, Toronto, ON, Canada
b
Department of Computer Science, University of Western Ontario, London, ON, Canada
article
info
Article history: Received 2 November 2015 Received in revised form 27 November 2016 Accepted 4 January 2017 Available online xxxx Keywords: Frequency assignment Wireless network Graph coloring L(2, 1)-labeling Kneser graph Coloring square of graphs
abstract The frequency assignment problem is to assign a frequency to each radio transmitter so that transmitters are assigned frequencies with allowed separations. Motivated by a variation of the frequency assignment problem, the L(2, 1)-labeling problem was put forward. An L(2, 1)-labeling of a graph G is a function f from the vertex set V (G) to the set of all nonnegative integers such that |f (x) − f (y)| ≥ 2 if d(x, y) = 1 and |f (x) − f (y)| ≥ 1 if d(x, y) = 2, where d(x, y) denotes the distance between x and y in G. The L(2, 1)labeling number λ(G) of G is the smallest number k such that G has a L(2, 1)-labeling with max{f (v) : v ∈ V (G)} = k. Griggs and Yeh (1992) conjectured that λ(G) ≤ ∆2 for any graph with maximum degree ∆ ≥ 2. The Kneser graph K (a, b) is defined as the graph whose vertices correspond to all b-subsets of the a-set A = {1, 2, . . . , a}, with edges joining pairs of vertices that correspond to non-overlapping b-subsets. The chromatic number of the Kneser graph K (a, b) is a−2b+2. Füredi put forward an open problem: What is the value for the chromatic number of the square of any Kneser graph K (a, b) (the square of a graph is the graph obtained by adding edges joining vertices at distance 2)? In this article, the L(2, 1)labeling numbers of Kneser graphs K (a, b) are considered. The combined upper bounds for the L(2, 1)-labeling numbers of Kneser graphs are derived by using two approaches. The exact L(2, 1)-labeling number of Kneser graph K (a, b) for a ≥ 3b − 1 is obtained, and it is proved that Griggs and Yeh’s conjecture holds for Kneser graphs and the L(2, 1)-labeling numbers of Kneser graphs are much better than ∆2 in most cases. We also provide bounds for the chromatic number of the square of any Kneser graph K (a, b) using the proof for the upper bounds of the L(2, 1)-labeling numbers of Kneser graphs. © 2017 Published by Elsevier B.V.
1. Introduction In the frequency assignment problem, radio transmitters are assigned frequencies with some separation in order to reduce interference. This problem can be formulated as a graph coloring problem [11]. Roberts [24] proposed a new version of the frequency assignment problem with two restrictions: radio transmitters that are ‘‘close’’ must be assigned different frequencies; those that are ‘‘very close’’ must be assigned frequencies at least two apart. To formulate the problem in graph theoretic terms, radio transmitters are represented by vertices of a graph; adjacent vertices are considered ‘‘very close’’ and vertices at distance two are considered ‘‘close’’. Motivated by this problem, Griggs and Yeh [10] proposed the following labeling on a simple graph. Let d(x, y) be the distance between vertices x and y in a graph G. An L(2, 1)-labeling of a graph G
∗
Corresponding author. E-mail addresses: [email protected] (Z. Shao), [email protected] (I. Averbakh), [email protected] (R. Solis-Oba).
http://dx.doi.org/10.1016/j.dam.2017.01.003 0166-218X/© 2017 Published by Elsevier B.V.
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Fig. 1. Petersen graph.
is a function f from all vertices of G to nonnegative integers such that |f (x) − f (y)| ≥ 2 if d(x, y) = 1 and |f (x) − f (y)| ≥ 1 if d(x, y) = 2. For an L(2, 1)-labeling, if the maximum label is no greater than k, then it is called a k-L(2, 1)-labeling. The L(2, 1)-labeling number of G, denoted by λ(G), is the smallest number k such that G has a k-L(2, 1)-labeling. From then on, a large number of articles have been published devoted to the study of the frequency assignment problem and its connections to graph labelings, in particular, to the class of L(2, 1)-labelings and its generalizations. In addition to graph theory and combinatorial techniques (see references [1–4,6,9,10,12,14,17,20,23,25–31]), other interesting approaches in studying these labelings include: neural networks [8,18], genetic algorithms [19], and simulated annealing [7,22]. Since graph coloring problems are notoriously difficult, researchers have to consider the problem of obtaining λ(G) for particular classes of graphs. Griggs and Yeh [10] proved that it is NP-complete to decide whether a given graph G allows an L(2, 1)-labeling of span at most n. Thus, it is important to obtain good lower and upper bounds for λ. For a diameter two graph G, it is known that λ(G) ≤ ∆2 , where ∆ = ∆(G) is the maximum degree of G, and the upper bound can be attained by Moore graphs, that is, diameter 2 graphs of order ∆2 + 1 [10]. Based on the previous research, Griggs and Yeh [10] conjectured that λ(G) ≤ ∆2 holds for any graph G with ∆ ≥ 2. The conjecture is known as the ∆2 -conjecture and considered as the most important open problem in the area. Griggs and Yeh [10] proved originally that λ(G) ≤ ∆2 + 2∆, the best bound ∆2 + ∆ − 2 so far is due to Gonçalves [9]. Havet, Reed and Sereni [12] proved that the ∆2 -conjecture holds for sufficiently large ∆. Both the vertex coloring problem and the L(2, 1)-labeling problem are known to be strongly NP-hard. A number of studies have been devoted to the L(2, 1)-labeling problem (see references). Due to the inherent hardness of this problem, most papers have to consider only particular classes of graphs. Kneser graphs are an important graph class which has been extensively studied in the context of coloring problems. A set A of cardinality a is called an a-set. Similarly a subset B of A of cardinality b is called a b-subset of A. Given positive integers a, b with a ≥ b, the Kneser graph K (a, b) is defined as the graph whose vertices correspond to all b-subsets of the a-set A = {1, 2, . . . , a}. Two nodes x, y that correspond to b-subsets X and Y of A are joined by an edge if and only if X ∩ Y = ∅. It is easy to see that K (a, 1) = Ka , where Ka is the complete graph of a vertices. Note that K (5, 2) is the Petersen Graph (see Fig. 1). Since K (a, b) has no edges when a < 2b, we only consider here the case when a ≥ 2b. In 1978, Lovász ([13] and [21]) proved that χ (K (a, b)) = a − 2b + 2 where χ (G) denotes the chromatic number of a graph G. This was first conjectured by Kneser in 1955. A number of results have also been obtained about the related concept of circular chromatic numbers of Kneser graphs, see, e.g., [32]. Let G be a simple graph. The square of G, G2 is defined as follows: the vertex set of G2 is the same as the vertex set of G; any two distinct vertices of G2 are adjacent if and only if the distance between them in G is at most 2. For coloring squares of graphs, Füredi put forward an open problem: What is the value for the chromatic number of the square of any Kneser graph K (a, b)? In [16], Kim and Park obtained upper bounds for the chromatic number of the square of any Kneser graph K (2b + 1, b). In this article, we consider the L(2, 1)-labeling numbers of Kneser graphs K (a, b). We use two approaches to derive the combined upper bounds for the L(2, 1)-labeling numbers of Kneser graphs. We obtain the exact L(2, 1)-labeling number of Kneser graph K (a, b) for a ≥ 3b − 1, and prove that Griggs and Yeh’s conjecture holds for Kneser graphs and the L(2, 1)labeling numbers of Kneser graphs are much better than ∆2 in most cases. We also provide bounds for the chromatic number of the square of any Kneser graph K (a, b) using the proof for the upper bounds of the L(2, 1)-labeling numbers of Kneser graphs. 2. A labeling algorithm A subset X of V (G) is called an i-stable set (or i-independent set), if the distance between any two vertices in X is greater than i. A 1-stable set is a usual independent set. A maxima i-stable subset X of a set Y of vertices is an i-stable subset of Y such that X is not a proper subset of any other i-stable subset of Y .
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Chang and Kuo [4] proposed the following algorithm to compute an L(2, 1)-labeling on a given graph. The main idea is as follows. Initially, all vertices of G are unlabeled. In each step i of the algorithm, we find a maximal 2-stable subset of the set of all the unlabeled vertices which are at distance at least two from the vertices labeled in the previous step, and label all vertices in that 2-stable set with the index i of the current step. The index i starts from 0 and increases by 1 at each step. We stop when all vertices are labeled. Algorithm 2.1. Input: A graph G = (V , E ). Output: The value k of the maximum label. 1. Set X−1 = ∅; V ′ = V ; k = 0. 2. Let Yk = {x ∈ V ′ : x is unlabeled and d(x, y) ≥ 2 for all y ∈ Xk−1 }. If Yk = ∅ then set Xk ← ∅, otherwise compute a maximal 2-stable subset Xk of Yk . 3. Label the vertices in Xk (if there are any) with label k. 4. V ′ ← V ′ − Xk . 5. If V ′ ̸= ∅, then set k ← k + 1 and go to Step 2. 6. Return k. Note that the value k computed by the algorithm is greater than or equal to λ(G); thus, an upper bound for k would also be an upper bound for λ(G). Let x be a vertex with the largest label k computed by Algorithm 2.1.
• I1 = {i : 0 ≤ i ≤ k − 1 and d(x, y) = 1 for some y ∈ Xi }. This is the set of labels of the neighbors of x. • I2 = {i : 0 ≤ i ≤ k − 1 and d(x, y) ≤ 2 for some y ∈ Xi }. These are the labels of the vertices at distance at most 2 from x. • I3 = {i : 0 ≤ i ≤ k − 1 and d(x, y) ≥ 3 for all y ∈ Xi }. These are the labels not used by vertices at distance at most 2 from x. It is clear that |I2 | + |I3 | = k. For any i ∈ I3 , x ̸∈ Yi since otherwise Xi ∪ {x} would be a 2-stable subset of Yi , which contradicts the choice of Xi . That is, d(x, y) = 1 for some vertex y in Xi−1 ; i.e., i − 1 ∈ I1 . Since for every i ∈ I3 , i − 1 ∈ I1 , then |I3 | ≤ |I1 |. Hence k = |I2 | + |I3 | ≤ |I2 | + |I1 |. In order to find k, we need to bound B = |I1 | + |I2 |
(1)
in terms of ∆(G). 3. L (2, 1)-labeling of Kneser graphs In [10], Griggs and Yeh proved that for G = K (5, 2), λ(G) = 9. In [15], Kang considered the L(2, 1)-labeling problem on a particular Kneser graph, i.e., G = K (2b + 1, b). Kang’s main interest is to obtain upper bounds for the L(2, 1)-labeling numbers of the particular Kneser graph G = K (2b + 1, b) and the main contribution of that paper is Kang proved that λ(G) ≤ 4b + 2. In this section, we will consider any general Kneser graphs K (a, b) and prove the Griggs and Yeh’s conjecture is true for any general Kneser graphs K (a, b). We prove that L(2, 1)-labeling numbers of any Kneser graphs K (a, b) are much better than in the Griggs and Yeh’s conjecture in most a cases. Since K (a, 1) is the complete graph Ka , we assume b ≥ 2. We will first prove that the exact value for λ(K (a, b)) is b − 1 when a ≥ 3b − 1. We first give a property of K (a, b). Lemma 3.1. K (a, b) has diameter 2 if a ≥ 3b − 1. Proof. Consider any vertex x of K (a, b). Every other vertex v of K (a, b) represents a set V that is either disjoint or not with the set X represented by x. Every vertex v corresponding to a b-set V disjoint with X is a neighbor of x. Furthermore, for b-set V with X ∩ V ̸= ∅ there must be a b-set W that is disjoint with V and X since |V ∩ X | ≤ b − 1 and a ≥ 3b − 1. Therefore, the vertex w corresponding to W must be adjacent to the vertex v corresponding to V and to x and so v is at distance 2 from x. Theorem 3.2. The Kneser graph K (a, b) has L(2, 1)-labeling number λ(K (a, b)) ≤
a b
− 1. This bound is tight for a ≥ 3b − 1.
Proof. The collection C of b-subsets of the set A = {1, 2, . . . , a} can be represented as the leaves of a tree T of height b. The root r of T represents C and it has a children, each storing one value from A. Each one of these children is the root of a subtree Tj (1 ≤ j ≤ b) of height b − 1 in which an internal node storing value i has a − i children storing the values i + 1, . . . , a. In this tree we keep only those leaves at distance b from the root. Each leaf l of T represents a b-subset of A whose elements are the values stored in the nodes along the path from l to r. Consider the tree representation T of the family of b-subsets of A = {1, 2, . . . , a}. Let T1 , T2 , . . . , Ta−b+1 be the subtrees of T as described above. List the leaves of these subtrees in order in such a way that the leaves of subtrees Ti with i odd are listed from left to right and the leaves of subtrees Ti with i even are listed from right to left. For example for the tree in Fig. 2 the leaves would be listed in this order:
{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 4, 5}, {2, 3, 5}, {2, 3, 4}, {3, 4, 5}.
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Fig. 2. A tree representation for all 3-subsets of {1, 2, 3, 4, 5}.
We now show that any two adjacent subsets Sj , Sj+1 in the above order must have a non-empty intersection, i.e. Sj ∩ Sj+1 ̸= ∅. If Sj and Sj+1 are leaves of the same subtree Ti then both b-sets include i. On the other hand if Sj is a leaf of subtree Ti and Sj+1 is a leaf of subtree Ti+1 then (1) if i is odd then Sj and Sj+1 contain value a (2) if i is even then Sj and Sj+1 contain value i + 1. Hence, labeling the b-sets in the above order starting with label 0and using consecutive labels yields a feasible L(2, 1)a labeling for the vertices of K (a, b) and the maximum label used is b − 1, one less than the number of vertices in K (a, b). By Lemma 3.1, if a ≥ 3b − 1, then a the Kneser graph K (a, b) has diameter 2 which means any two vertices must have different labels. Hence λ(K (a, b)) = b − 1 in this case.
Note that by Theorem 3.2, for a = 3b − 1 and b = 2, we have λ(K (a, b)) = λ(K (5, 2)) =
5 2
− 1 = 9 which agrees
with Griggs and Yeh’s result in [10]. Let G be a graph with maximum degree ∆(G) and number of vertices ν(G). Note that if G has diameter 2, then ν(G) ≤ ∆(G)2 +1. Since the number of vertices of K (a, b) is ba , by Lemma 3.1 and Theorem 3.2, we have for a ≥ 3b−1, λ(K (a, b)) =
a b
− 1 ≤ ∆(K (a, b))2 . Since ∆(K (a, b)) =
a −b b
, in the following, we will prove for a ≥ 3b − 1, the bound is much better
than ∆(K (a, b)) in most cases. The following Lemmas 3.5, 3.8 and 3.10 provide results regarding this. 2
Lemma 3.3. For a ≥ 3b and b ≥ 3,
Proof. Since b 2 (( aa++11−−2b ) −
(
) >
a+1−b 2 a+1−2b
a+1−b b
)
=
Lemma 3.4. For b ≥ 3,
2b+2 b +1
=
a −b b
and
a+1 b
2b+2 b+1
a+1 b
b > 15.8( aa++11−−2b −
a −b a +1 a+1−b b a +1 . Thus, the a+1−b
Proof. Since
a+1−b a+1−2b
a+1−b 2 −15.8 a+1b−b b
conclusion follows. 2
−15.8
2b+2 b+1
3b+3 b+1
(2b+2)(2b+1) (b+1)2
2b b
>
2b b
to prove
2b b
>
b
,
> ··· >
3b−b 2 −15.8 3bb−b b
a+1−b 2 −15.8 a+1b−b b
a+1 b
>
3b b
−15.8
2b b
3b+3 b+1
2
=
6 3
> ··· >
−15.8 63 9 3
(3b+3)(3b+2)(3b+1) (b+1)(2b+2)(2b+1)
a b
−1<
=
2b 2 −15.8 2b b b
a−b 2 b −15.8 a− b b a b
3b b
.
is equivalent to
( )
a −b b
≥ 20 and
3b b
= 1.
,
2b+2 2 +2 −15.8 2b b+1 b+1
3b+3 b+1
(2b+2)(2b+1) 2 +3)(3b+2)(3b+1) +1) +3)(3b+2)(3b+1) ) − ((3b ) 2b > 15.8( (2b+(b2+)(12b − ((3b ). b+1)(2b+2)(2b+1) b b+1)(2b+2)(2b+1) (b+1)2 )2 +1) 2 +3)(3b+2)(3b+1) ≥ 20 and ( (2b+(b2+)(12b ) > ((3b . Thus, the conclusion follows. b+1)(2b+2)(2b+1) )2
Lemma 3.5. For a ≥ 3b and b ≥ 3,
3b b
and
a
). By a ≥ 3b and b ≥ 3, it is not difficult to prove
2
b a−b 2 −15.8 a− b b a b
( )
a +1 a+1−b
=
a+1 a+1−b
equivalent to ((
a −b b
2
− 15.8
a −b b
.
>
2b 2 −15.8 2b b b
3b b
is
By b ≥ 3, it is not difficult
Z. Shao et al. / Discrete Applied Mathematics (
Proof. By Lemmas 3.3 and 3.4, we have and the conclusion follows.
a−b b
2
−15.8
( )
a−b b
> 1. But
a b
a−b b
2
)
−15.8
( ) −1
–
a−b b
5
>
a b
a−b b
2
−15.8
( )
a−b b
for a ≥ 3b and b ≥ 3,
a b
Similarly to Lemmas 3.3–3.5, we can prove the following lemmas.
Lemma 3.6. For a ≥ 3b − 1 and b ≥ 3,
2b−1 2 −4.4 2bb−1 b
3b−1 b
a+1−b 2 −4.4 a+1b−b b
a+1 b
>
a−b 2 b −4.4 a− b b a b
> ··· >
( )
3b−1−b 2 −4.4 3b−b1−b b
3b−1 b
=
.
Lemma 3.7. For b ≥ 3,
2b+1 2 +1 −4.4 2b b+1 b+1
3b+2 b+1
>
Lemma 3.8. For a ≥ 3b − 1 and b ≥ 3,
Lemma 3.9. For a ≥ 3b and b = 2,
a b
−1<
3b−1 b
1 a−1 2 −3.5 a− 2 2
Lemma 3.10. For a ≥ 3b and b = 2,
2b−1 2 −4.4 2bb−1 b
a b
a+1 2
−1<
a−b b
>
a −b b
2
2
2
> ··· >
− 4.4
a −b b
2 a−2 2 −3.5 a− 2 2 a 2
( )
− 3.5
5 3
a −b b
−4.4
5 3
= 1.
8 3
. 2
> ··· >
4 2
−3.5
4 2
6 2
= 1.
.
In the following part, we will provide upper bounds for λ(K (a, b)) with 2b ≤ a ≤ 3b − 2. Theorem 3.11. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Then
a−2b b 2 − 1 + b < ∆2 if 2 ≤ a − 2b ≤ b − 2, < ∆ − ∆ b 1 − a−b λ(G) 2 ≤ ∆ − 2 if a − 2b = 1 and b ≥ 3, = 2 if a − 2b = 0. Proof. Let x be a vertex of K (a, b) which is labeled by k. Note that if a − 2b = 0, then the Kneser graph K (a, b) is a matching and hence, λ(G) = 2. Also note that in this case ∆ = 1. Thus, in the following, we can only consider the case when 0 < a − 2b ≤ b − 2. The particular case when a − 2b = 1 has been considered by Kang [15]. Kang proved that for G = K (2b + 1, b), λ(G) ≤ 4b + 2. Suppose a − 2b = 1, by 0 < a − 2b ≤ b − 2, we have b ≥ 3. If a − 2b = 1, then the maximum degree of G is ∆ = b + 1 ≥ 4, so λ(G) ≤ 4b + 2 = 4∆ − 2 ≤ ∆2 − 2 since b ≥ 3. So, let us assume 2 ≤ a − 2b ≤ b − 2. This implies b ≥ 4. The degree of any vertex in K (a, b) is
maximum degree is ∆ =
a−b b
a −b b
and hence the
. If xvw is a simple path of length 2 in K (a, b) then the corresponding b-sets X , V and W , for
vertices x, v , and w respectively, must satisfy X ∩ V = ∅, V ∩ W = ∅ and X ∩ W ̸= ∅. Considering X fixed, there are
a −b b
choices for V ; then, ifVis also fixed, since W must be disjoint with the b-set V but |X ∩ W | ≥ 1, the number of choices for W is
b−1
t =b−(a−2b)
a−b b
[
a−2b b −t
b t
b −1
t =b−(a−2b)
]. Thus, the number of vertices with distance 2 from x is not greater than
b
a − 2b
t
b−t
.
(2)
Note that the subgraph H induced in G by all vertices adjacent to x is a graph with no edges since a − 2b ≤ b − 2, and thus two neighbors of x cannot correspond to disjoint b-sets. However, Since some of the neighbors of x have common neighbors that do not belong to H, we can further tighten the above bound. Let xv z and xw z be two different paths of length 2 in K (a, b), where z is not a neighbor of x. The corresponding b-sets for vertices x, v, w , and z are X , V , W , and Z , respectively; these sets must satisfy the conditions X ∩ V = ∅,
X ∩ W = ∅,
V ∩ Z = ∅,
W ∩ Z = ∅,
and
X ∩ Z ̸= ∅.
(3)
Note that the sets X , V , W and Z are different (see Fig. 3). Consider X as fixed, and fix an arbitrary b-set V ⊂ (V (G) \ X ). Then, the bound (2) can be reduced by the number N ′ of distinct choices of b-sets W and Z that satisfy the conditions (3). Let us estimate N ′ .
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Fig. 3. The corresponding b-sets X , V , W , and Z for vertices x, v, w , and z.
Let |V ∩ W | = t, where 3 ≤ t ≤ b − 1. Then the number of choices for W is
b t
a−2b b−t
. When W is chosen, it is z’s turn
to choose its b-set Z . Since X ∩ Z ̸= ∅, z has to choose at least one element and at most b − 1elements in X. Summing up over these b − 1 possibilities, we get that given X , V , W , the number of choices for Z is
b−1 j=1
[
a−(3b−t ) b−j
b j
].
Adding up over all possible values of t, we get
b−1 b−1 b a − 2b b a − (3b − t ) N = . t b−t j b−j t =3 j =1 ′
In the summation we need to consider only values of t such that 1 ≤ a − (3b − t ), i.e., t ≥ 3b − a + 1. Since a − 2b ≤ b − 2, t ≥ 3b − a + 1 > 2. But 0 ≤ t ≤ b − 1, then 3 ≤ t ≤ b − 1. We have b −1
′
N =
b
a − 2b
t
b−t
t =3b−a+1
b
a − (3b − t )
j
b−j
b−1 j=1
b −1
≥b
b
a − 2b
t
b−t
t =3b−a+1
Therefore, the number of vertices with distance 2 from x is not greater than b
b−1
t =3b−a+1
c
Since
[
b t
a−b
b
a −b b
.
b−1 t =b−(a−2b)
[
b t
a−2b b −t
]. b c −b ] for all c ≥ b, we have t =0 [ t b−t
b
=
b
a−2b b −t
=
b
a − 2b
3b − a
+ ··· + b −1 t =b−(a−2b) b−1
3b − a + 1
b
a − 2b
t
b−t
b
a − 2b
b
0
b
a − 2b
b
0
+
+
b
a − 2b
t
b−t
b
a − 2b
t
b−t
+
a − 2b
+
a − 2b − 1
1
t =3b−a+1
b
a − 2b
b−1
+
b
+
a − 2b
=
b
a − 2b
(a − 2b − 2)
3b − a + 2
=
b
a − 2b
b
0
b
3b − a
a − 2b
a − 2b
.
Thus,
a−b b
b−1
t =b−(a−2b)
=
a−b
a−b
b
b
−b
a − 2b
t
b−t
b
t =3b−a+1
−1 −
b −1
a−b
b
−
b 3b − a
− 1 b = ∆(∆ − 1) − ∆ −
Also, the number of vertices with distance one from x is not greater than ∆ =
b 3b−a
a −b b
= <
b!
(3b − a)!(a − 2b)!
b a−b
a−2b
.
·
b!(a − 2b)!
(a − b)!
=
a −b b
b
3b − a
. But
(b!)2 b(b − 1) . . . (3b − a + 1) = (3b − a)!(a − b)! (a − b)(a − b − 1) . . . (b + 1)
− 1 b.
] −
Z. Shao et al. / Discrete Applied Mathematics ( b 3b−a a−2b
Hence
( a−b ) b
)
–
7
< ( a−b b )a−2b ∆, so the number of vertices with distance at most 2 from x is less than ∆ + ∆(∆ − 1) − (∆(1 −
) − 1)b = ∆2 − (∆(1 − ( a−b b )a−2b ) − 1)b.
Thus, by (1), we have
λ(G) ≤ k ≤ B = |I1 | + |I2 | < ∆ + ∆ − ∆ 1 − 2
= ∆ −∆ b 1− 2
a−2b
b a−b
a−2b
b
2
2
+ 1−
=
b +2
1 − ( a −b ) b
a−2b
b +2 b
>
1 b
1 b
+1+
+
3b−6 . (b+2)2
4 (b+2)2
−
4 b +2
= 1+
Thus, ∆(b(1 − ( a−b ) b
− 1 + b.
4 (b+2)2
a−2b
1 b
+
+ ( a−b b )a−2b ≤ < 1+
−4b+b+2 b(b+2)
) − 1) − b >
3b2 −6b (b+2)(b+1) 2 (b+2)2
3b2 −6b
. By b ≥ 4, we have, (b+2)2 ∆ − b ≥ b Thus, λ(G) < ∆2 − ∆(b(1 − ( a− )a−2b ) − 1) + b < ∆2 . b
then ∆ ≥
−1 b
a−b
Recall that we assumed 2 ≤ a − 2b ≤ b − 2 and b ≥ 4. Since 1 b
3b2 −6b ∆ (b+2)2
−b=
1 b
+
4 (b+2)2
b (2b+2)−b
+
−3b+2 (b+2 )2
− b. But ∆ =
b(3b2 −5b−10) 2(b+2)
(2b+2)−2b = 1+
a −b b
=
1 b
+
−3b+6 , (b+2)2
2 b b +2
=
we have
and a − b ≥ b + 2,
> 0.
By Theorem 3.2, Lemmas 3.5, 3.8, 3.10 and Theorem 3.11, we can see that the L(2, 1)-labeling numbers of Kneser graphs are much better than in the Griggs and Yeh’s conjecture in most cases. Combining Theorem 3.2, Lemmas 3.8, 3.10 and Theorem 3.11, we have the following result. Theorem 3.12. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Then
a = − 1 < ∆2 − 4.4∆ if a ≥ 3b − 1 and b ≥ 3, b a = − 1 < ∆2 − 3.5∆ if a > 3b − 1 and b = 2, b = 9 = ∆2 if a = 3b − 1 and b = 2, a−2b λ(G) b 2 − 1 + b < ∆2 if 2b + 2 ≤ a ≤ 3b − 2, <∆ −∆ b 1− a − b 2 ≤ ∆ − 2 if a = 2b + 1 and b ≥ 3, = 2 if a = 2b. 9(r +3) r 8
Remark. In [16], Kim and Park proved that χ (K 2 (2k + r , k)) ≤ 2( 94 k +
) for any integer r with 1 ≤ r ≤ k − 2.
Our upper bounds are stated in terms of the maximum degree and their upper bounds are stated in terms of different parameters from ours. Since coloring squares of a graph G requires that any two distinct vertices of G have different colors if 9(r +3) the distance between them in G is at most 2, we can deduce that λ(K (2k + r , k)) ≤ 2χ (K 2 (2k + r , k))− 2 ≤ 4( 94 k + 8 )r − 2 for any integer r with 1 ≤ r ≤ k − 2 (note that λ(G) starts from 0 and χ (G) starts from 1). a We have proved that the exact L(2, 1)-labeling number of Kneser graph K (a, b) with a ≥ 3b − 1 is b − 1. Note that when a ≥ 3b − 1, it implies r ≥ k − 1 when a = 2k + r and b = k. Thus, we have obtained the first known exact value for a ≥ 3b − 1. In the following we will make a conservative comparison between the above deduced bound and ours for 2 ≤ a − 2b < b − 1. A well-known inequality is
k+r k
≤(
(k+r )·e k k
n k
≤ ( nk·e )k for 1 ≤ k ≤ n. Since ∆(K (2k + r , k)) =
) . Thus, ∆ − ∆(b(1 − ( a−b ) b
2
is from Theorem 3.11). But 4( 49 k +
a−2b
) − 1) + b < ∆ ≤ ( 2
(k+r )·e 2k k
) ≤(
k+r k
(2k−2)·e 2k k
, we have ∆(K (2k + r , k)) = 2k
) < (2 · e) (the first inequality
9(r +3) r 8 r ln 94
) − 2 ≥ 4( 49 k + 29 )r − 2 > 4( 94 k)r . Note that ln((2 · e)2k ) = 2k ln(2 · e) and ln(4( ) ) = ln 4 + r ln( ) = ln 4 + ( ) + r ln k ≥ r ln k. Since ln(2 · e) < 1.694, when (r ln k)/(2k) > 1.7, our bound is better than the above deduced bound. For example, when r = k − 2 and k ≥ 37, (r ln k)/(2k) ≥ (35 ln 37)/74 > 1.7. 9 k r 4
9 k 4
4. Coloring square of Kneser graphs By Brook’s theorem, for any connected simple graph G with chromatic number χ (G) and maximum degree ∆(G), χ (G) ≤ ∆(G) unless G is a complete graph or an odd cycle where χ (G) ≤ ∆(G) + 1. Let the number of vertices with distance one and two of any vertex v in G be d1 (v) and d2 (v), respectively, then χ (G2 ) ≤ maxv∈V (G) {d1 (v) + d2 (v) + 1} = ∆(G2 ) + 1 ≤ ∆(G) + ∆(G)(∆(G) − 1) + 1 = (∆(G))2 + 1, and the upper bound ∆(G2 ) + 1 realizes if and only if G2 is a complete graph
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or an odd cycle. Note that (∆(G))2 + 1 is not always realizable even if G2 is a complete graph (the following Theorem 4.3 will provide a good example). In this section, a we consider the chromatic number of the square of any Kneser graphs K (a, b). By Lemma 3.1 and V (K (a, b)) = b , we can easily observe the following lemma. Lemma 4.1. The chromatic number of the square of any Kneser graphs K (a, b) are bounded by a ≥ 3b − 1.
a b
. This bound is tight for
By Lemma 3.1, we have for a ≥ 3b − 1, K (a, b) has diameter 2 and thus, ∆(G2 ) = V (K (a, b)) − 1 =
a b
− 1,
i.e., b = ∆(G2 ) + 1. In the following part, we will provide upper bounds for the chromatic number of the square of any Kneser graphs K (a, b) with 2b ≤ a ≤ 3b − 2.
a
Theorem 4.2. Let G be the Kneser graph K (a, b) with the maximum degree ∆. Then
a−2b b 2 − 1 b + 1 < ∆2 − ∆ + 1 if 2 ≤ a − 2b ≤ b − 2, < ∆ − ∆ 1 − a−b 2 χ (G ) ≤ ∆2 − ∆ − 1 if a − 2b = 1 and b ≥ 3, = 2 = ∆2 + 1 if a − 2b = 0. Proof. Note that if a − 2b = 0, then the Kneser graph K (a, b) is a matching and hence, λ(G) = 2. Also note that in this case ∆ = 1. Thus, in the following, we can only consider the case when 0 < a − 2b ≤ b − 2. The particular case when a − 2b = 1 has been considered by Chen, Lih and Wu [5]. They proved that for G = K (2b + 1, b), χ (G2 ) ≤ 3b + 2. Suppose a − 2b = 1, by 0 < a − 2b ≤ b − 2, we have b ≥ 3. If a − 2b = 1, then the maximum degree of G is ∆ = b + 1, so χ(G2 ) ≤ 3b + 2 ≤ ∆2 − ∆ − 1 since b ≥ 3. So, let us assume 2 ≤ a − 2b ≤ b − 2. This implies b ≥ 4. By Theorem 3.11, the maximum number of vertices at distance at most 2 from any vertex v in G can be at most
∆2 − ∆ 1 −
b a−b
a−2b
− 1 b.
Let the number of vertices with distance one and two of any vertex v in G be d1 (v) and d2 (v), respectively, then χ (G2 ) ≤ b )a−2b ) − 1)b + 1. Recall that we assumed 2 ≤ a − 2b ≤ b − 2 and b ≥ 4. maxv∈V (G) {d1 (v) + d2 (v) + 1} ≤ ∆2 − (∆(1 − ( a− b
b Thus, similarly to the proof in Case 1 in Theorem 3.11, we have χ (G2 ) < ∆2 −(∆(1 −( a− )a−2b )− 1)b + 1 < ∆2 − ∆ + 1. b
By Lemma 4.1 and Theorem 4.2, we can see that the chromatic number of the square of any Kneser graphs G with maximum degree ∆ are bounded by ∆2 − ∆ + 1 (may be much better in most cases) unless a = 5 and b = 2, or a − 2b = 0. Combining Lemma 3.8, Lemmas 3.10, 4.1 and Theorem 4.2, we have the following result. Theorem 4.3. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Let the maximum degree of G2 be ∆(G2 ). Then
a = = ∆(G2 ) + 1 < ∆2 − 4.4∆ + 1 if a ≥ 3b − 1 and b ≥ 3, b a = = ∆(G2 ) + 1 < ∆2 − 3.5∆ + 1 if a > 3b − 1 and b = 2, b = 10 = ∆(G2 ) + 1 = ∆2 + 1 if a = 3b − 1 and b = 2, a−2b χ (G2 ) b 2 <∆ − ∆ 1− − 1 b + 1 < ∆2 − ∆ + 1 if 2b + 2 ≤ a ≤ 3b − 2, a − b ≤ ∆2 − ∆ − 1 if a = 2b + 1 and b ≥ 3, = 2 = ∆2 + 1 if a = 2b. By Theorem 4.3, the upper bound ∆(G2 ) + 1 realizes if a ≥ 3b − 1. However, ∆2 + 1 is not always realizable even if a ≥ 3b − 1 which means G2 is a complete graph. 9(r +3)
Remark. In [16], Kim and Park proved that χ (K 2 (2k + r , k)) ≤ 2( 49 k + 8 )r for any integer r with 1 ≤ r ≤ k − 2. As in the previous remark, we can make a similar conservative comparison between the above deduced bound and our bound in this section.
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Acknowledgments Second author’s research was partially supported by Discovery Grant of the Natural Sciences and Engineering Research Council of Canada (NSERC). Third author’s research was partially supported by the Natural Sciences and Engineering Research Council of Canada (NSERC) (grant No.: 227829-2009). References [1] S.S. Adams, J. Cass, M. Tesch, D. Sakai Troxell, C. Wheeland, The minimum span of L(2, 1)-labelings of certain generalized Petersen graphs, Discrete Appl. Math. 155 (2007) 1314–1325. [2] T. Araki, Labeling bipartite permutation graphs with a condition at distance two, Discrete Appl. Math. 157 (2009) 1677–1686. [3] A.A. Bertossi, C.M. Pinotti, Mapping for conflict-free access of path in bidimensional arrays, circular lists, and complete trees, J. Parallel Distrib. Comput. 62 (2002) 1314–1333. [4] G.J. Chang, D. Kuo, The L(2, 1)-labeling on graphs, SIAM J. Discrete Math. 9 (1996) 309–316. [5] J.-Y. Chen, K.-W. Lih, J.-J. Wu, Coloring the square of the Kneser graph KG(2k + 1, k) and the Schrijver graph SG(2k + 2, k), Discrete Appl. Math. 157 (2009) 170–176. [6] S.-H. Chiang, J.-H. Yan, On L(d, 1)-labeling of Cartesian products of a cycle and a path, Discrete Appl. Math. 156 (15) (2008) 2867–2881. [7] M. Duque-Anton, N. Kunz, B. Ruber, Channel assignment for cellular radio using simulated annealing, IEEE Trans. Veh. Technol. 42 (1993) 14–21. [8] N. Funabiki, Y. Takefuji, A neural network parallel algorithm for channel assignment problem in cellular radio networks, IEEE Trans. Veh. Technol. 41 (1992) 430–437. [9] D. Gonçalves, On the L(p, 1)-labeling of graphs, Discrete Math. 308 (2008) 1405–1414. [10] J.R. Griggs, R.K. Yeh, Labeling graphs with a condition at distance two, SIAM J. Discrete Math. 5 (1992) 586–595. [11] W.K. Hale, Frequency assignment: Theory and application, Proc. IEEE 68 (1980) 1497–1514. [12] F. Havet, B. Reed, J.-S. Sereni, L(2, 1)-labeling of graphs, in: Proceedings of the ACM-SIAM Symposium on Discrete Algorithm (SODA 08), San Francisco, California, 20–22 January, 2008, pp. 621–630. [13] T.R. Jensen, B. Toft, Graph Coloring Problem, John Wiley & Sons, NY, NY, 1995. [14] J.S.-T. Juan, D.D.-F. Liu, L.-Y. Chen, L(j, k)-labelling and maximum ordering-degrees for trees, Discrete Appl. Math. 158 (2010) 692–698. [15] J.-H. Kang, L(2, 1)-labeling on Kneser graph, manuscript. [16] S.-J. Kim, B. Park, Coloring of the square of Kneser graph K (2k + r , k), manuscript. ˘ [17] S. Klav˘zar, S. Spacapan, The ∆2 -conjecture for L(2, 1)-labelings is true for direct and strong products of graphs, IEEE Trans. Circuits Syst. II 53 (2006) 274–277. [18] N. Kunz, Channel assignment for cellular radio using neural networks, IEEE Trans. Veh. Technol. 40 (1992) 188–193. [19] W.K. Lai, G.G. Coghill, Channel assignment through evolutionary optimization, IEEE Trans. Veh. Technol. 45 (1996) 91–96. [20] D.D.-F. Liu, R.K. Yeh, On distance-two labelings of graphs, Ars Combin. 47 (1997) 13–22. [21] L. Lovász, Knesers conjecture, chromatic number and homotopy, J. Combin. Theory Ser. A 25 (1978) 319–324. [22] R. Mathar, Channel assignment in cellular radio networks, IEEE Trans. Veh. Technol. 42 (1993) 647–656. [23] B.S. Panda, P. Goel, L(2, 1)-labeling of perfect elimination bipartite graphs, Discrete Appl. Math. 159 (16) (2011) 1878–1888. [24] F.S. Roberts, T-colorings of graphs: Recent results and open problems, Discrete Math. 93 (1991) 229–245. [25] D. Sakai, Labeling chordal graphs with a condition at distance two, SIAM J. Discrete Math. 7 (1994) 133–140. [26] C. Schwarz, D. Sakai Troxell, L(2, 1)-Labelings of products of two cycles, Discrete Appl. Math. 154 (2006) 1522–1540. [27] Z. Shao, S. Klavžar, W.C. Shiu, D. Zhang, Improved bounds on the L(2, 1)-number of direct and strong products of graphs, IEEE Trans. Circuits Syst. II 55 (2008) 685–689. [28] Z. Shao, R. Solis-Oba, L(2, 1)-labelings on the composition of n graphs, Theoret. Comput. Sci. 411 (2010) 3287–3292. [29] Z. Shao, R.K. Yeh, The L(2, 1)-labeling and operations of graphs, IEEE Trans. Circuits Syst. I 52 (2005) 668–671. [30] W.C. Shiu, Z. Shao, K.K. Poon, D. Zhang, A new approach to the L(2, 1)-labeling of some products of graphs, IEEE Trans. Circuits Syst. II 55 (8) (2008) 802–805. [31] R.K. Yeh, A survey on labeling graphs with a condition at distance two, Discrete Math. 306 (2006) 1217–1231. [32] X. Zhu, Circular chromatic number: a survey, Discrete Math. 229 (2001) 371–410.
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Contents lists available at ScienceDirect
Discrete Applied Mathematics journal homepage: www.elsevier.com/locate/dam
L(2, 1)-Labeling of Kneser graphs and coloring squares of Kneser graphs Zhendong Shao a,∗ , Igor Averbakh a , Roberto Solis-Oba b a
Department of Management, University of Toronto Scarborough, Toronto, ON, Canada
b
Department of Computer Science, University of Western Ontario, London, ON, Canada
article
info
Article history: Received 2 November 2015 Received in revised form 27 November 2016 Accepted 4 January 2017 Available online xxxx Keywords: Frequency assignment Wireless network Graph coloring L(2, 1)-labeling Kneser graph Coloring square of graphs
abstract The frequency assignment problem is to assign a frequency to each radio transmitter so that transmitters are assigned frequencies with allowed separations. Motivated by a variation of the frequency assignment problem, the L(2, 1)-labeling problem was put forward. An L(2, 1)-labeling of a graph G is a function f from the vertex set V (G) to the set of all nonnegative integers such that |f (x) − f (y)| ≥ 2 if d(x, y) = 1 and |f (x) − f (y)| ≥ 1 if d(x, y) = 2, where d(x, y) denotes the distance between x and y in G. The L(2, 1)labeling number λ(G) of G is the smallest number k such that G has a L(2, 1)-labeling with max{f (v) : v ∈ V (G)} = k. Griggs and Yeh (1992) conjectured that λ(G) ≤ ∆2 for any graph with maximum degree ∆ ≥ 2. The Kneser graph K (a, b) is defined as the graph whose vertices correspond to all b-subsets of the a-set A = {1, 2, . . . , a}, with edges joining pairs of vertices that correspond to non-overlapping b-subsets. The chromatic number of the Kneser graph K (a, b) is a−2b+2. Füredi put forward an open problem: What is the value for the chromatic number of the square of any Kneser graph K (a, b) (the square of a graph is the graph obtained by adding edges joining vertices at distance 2)? In this article, the L(2, 1)labeling numbers of Kneser graphs K (a, b) are considered. The combined upper bounds for the L(2, 1)-labeling numbers of Kneser graphs are derived by using two approaches. The exact L(2, 1)-labeling number of Kneser graph K (a, b) for a ≥ 3b − 1 is obtained, and it is proved that Griggs and Yeh’s conjecture holds for Kneser graphs and the L(2, 1)-labeling numbers of Kneser graphs are much better than ∆2 in most cases. We also provide bounds for the chromatic number of the square of any Kneser graph K (a, b) using the proof for the upper bounds of the L(2, 1)-labeling numbers of Kneser graphs. © 2017 Published by Elsevier B.V.
1. Introduction In the frequency assignment problem, radio transmitters are assigned frequencies with some separation in order to reduce interference. This problem can be formulated as a graph coloring problem [11]. Roberts [24] proposed a new version of the frequency assignment problem with two restrictions: radio transmitters that are ‘‘close’’ must be assigned different frequencies; those that are ‘‘very close’’ must be assigned frequencies at least two apart. To formulate the problem in graph theoretic terms, radio transmitters are represented by vertices of a graph; adjacent vertices are considered ‘‘very close’’ and vertices at distance two are considered ‘‘close’’. Motivated by this problem, Griggs and Yeh [10] proposed the following labeling on a simple graph. Let d(x, y) be the distance between vertices x and y in a graph G. An L(2, 1)-labeling of a graph G
∗
Corresponding author. E-mail addresses: [email protected] (Z. Shao), [email protected] (I. Averbakh), [email protected] (R. Solis-Oba).
http://dx.doi.org/10.1016/j.dam.2017.01.003 0166-218X/© 2017 Published by Elsevier B.V.
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Fig. 1. Petersen graph.
is a function f from all vertices of G to nonnegative integers such that |f (x) − f (y)| ≥ 2 if d(x, y) = 1 and |f (x) − f (y)| ≥ 1 if d(x, y) = 2. For an L(2, 1)-labeling, if the maximum label is no greater than k, then it is called a k-L(2, 1)-labeling. The L(2, 1)-labeling number of G, denoted by λ(G), is the smallest number k such that G has a k-L(2, 1)-labeling. From then on, a large number of articles have been published devoted to the study of the frequency assignment problem and its connections to graph labelings, in particular, to the class of L(2, 1)-labelings and its generalizations. In addition to graph theory and combinatorial techniques (see references [1–4,6,9,10,12,14,17,20,23,25–31]), other interesting approaches in studying these labelings include: neural networks [8,18], genetic algorithms [19], and simulated annealing [7,22]. Since graph coloring problems are notoriously difficult, researchers have to consider the problem of obtaining λ(G) for particular classes of graphs. Griggs and Yeh [10] proved that it is NP-complete to decide whether a given graph G allows an L(2, 1)-labeling of span at most n. Thus, it is important to obtain good lower and upper bounds for λ. For a diameter two graph G, it is known that λ(G) ≤ ∆2 , where ∆ = ∆(G) is the maximum degree of G, and the upper bound can be attained by Moore graphs, that is, diameter 2 graphs of order ∆2 + 1 [10]. Based on the previous research, Griggs and Yeh [10] conjectured that λ(G) ≤ ∆2 holds for any graph G with ∆ ≥ 2. The conjecture is known as the ∆2 -conjecture and considered as the most important open problem in the area. Griggs and Yeh [10] proved originally that λ(G) ≤ ∆2 + 2∆, the best bound ∆2 + ∆ − 2 so far is due to Gonçalves [9]. Havet, Reed and Sereni [12] proved that the ∆2 -conjecture holds for sufficiently large ∆. Both the vertex coloring problem and the L(2, 1)-labeling problem are known to be strongly NP-hard. A number of studies have been devoted to the L(2, 1)-labeling problem (see references). Due to the inherent hardness of this problem, most papers have to consider only particular classes of graphs. Kneser graphs are an important graph class which has been extensively studied in the context of coloring problems. A set A of cardinality a is called an a-set. Similarly a subset B of A of cardinality b is called a b-subset of A. Given positive integers a, b with a ≥ b, the Kneser graph K (a, b) is defined as the graph whose vertices correspond to all b-subsets of the a-set A = {1, 2, . . . , a}. Two nodes x, y that correspond to b-subsets X and Y of A are joined by an edge if and only if X ∩ Y = ∅. It is easy to see that K (a, 1) = Ka , where Ka is the complete graph of a vertices. Note that K (5, 2) is the Petersen Graph (see Fig. 1). Since K (a, b) has no edges when a < 2b, we only consider here the case when a ≥ 2b. In 1978, Lovász ([13] and [21]) proved that χ (K (a, b)) = a − 2b + 2 where χ (G) denotes the chromatic number of a graph G. This was first conjectured by Kneser in 1955. A number of results have also been obtained about the related concept of circular chromatic numbers of Kneser graphs, see, e.g., [32]. Let G be a simple graph. The square of G, G2 is defined as follows: the vertex set of G2 is the same as the vertex set of G; any two distinct vertices of G2 are adjacent if and only if the distance between them in G is at most 2. For coloring squares of graphs, Füredi put forward an open problem: What is the value for the chromatic number of the square of any Kneser graph K (a, b)? In [16], Kim and Park obtained upper bounds for the chromatic number of the square of any Kneser graph K (2b + 1, b). In this article, we consider the L(2, 1)-labeling numbers of Kneser graphs K (a, b). We use two approaches to derive the combined upper bounds for the L(2, 1)-labeling numbers of Kneser graphs. We obtain the exact L(2, 1)-labeling number of Kneser graph K (a, b) for a ≥ 3b − 1, and prove that Griggs and Yeh’s conjecture holds for Kneser graphs and the L(2, 1)labeling numbers of Kneser graphs are much better than ∆2 in most cases. We also provide bounds for the chromatic number of the square of any Kneser graph K (a, b) using the proof for the upper bounds of the L(2, 1)-labeling numbers of Kneser graphs. 2. A labeling algorithm A subset X of V (G) is called an i-stable set (or i-independent set), if the distance between any two vertices in X is greater than i. A 1-stable set is a usual independent set. A maxima i-stable subset X of a set Y of vertices is an i-stable subset of Y such that X is not a proper subset of any other i-stable subset of Y .
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Chang and Kuo [4] proposed the following algorithm to compute an L(2, 1)-labeling on a given graph. The main idea is as follows. Initially, all vertices of G are unlabeled. In each step i of the algorithm, we find a maximal 2-stable subset of the set of all the unlabeled vertices which are at distance at least two from the vertices labeled in the previous step, and label all vertices in that 2-stable set with the index i of the current step. The index i starts from 0 and increases by 1 at each step. We stop when all vertices are labeled. Algorithm 2.1. Input: A graph G = (V , E ). Output: The value k of the maximum label. 1. Set X−1 = ∅; V ′ = V ; k = 0. 2. Let Yk = {x ∈ V ′ : x is unlabeled and d(x, y) ≥ 2 for all y ∈ Xk−1 }. If Yk = ∅ then set Xk ← ∅, otherwise compute a maximal 2-stable subset Xk of Yk . 3. Label the vertices in Xk (if there are any) with label k. 4. V ′ ← V ′ − Xk . 5. If V ′ ̸= ∅, then set k ← k + 1 and go to Step 2. 6. Return k. Note that the value k computed by the algorithm is greater than or equal to λ(G); thus, an upper bound for k would also be an upper bound for λ(G). Let x be a vertex with the largest label k computed by Algorithm 2.1.
• I1 = {i : 0 ≤ i ≤ k − 1 and d(x, y) = 1 for some y ∈ Xi }. This is the set of labels of the neighbors of x. • I2 = {i : 0 ≤ i ≤ k − 1 and d(x, y) ≤ 2 for some y ∈ Xi }. These are the labels of the vertices at distance at most 2 from x. • I3 = {i : 0 ≤ i ≤ k − 1 and d(x, y) ≥ 3 for all y ∈ Xi }. These are the labels not used by vertices at distance at most 2 from x. It is clear that |I2 | + |I3 | = k. For any i ∈ I3 , x ̸∈ Yi since otherwise Xi ∪ {x} would be a 2-stable subset of Yi , which contradicts the choice of Xi . That is, d(x, y) = 1 for some vertex y in Xi−1 ; i.e., i − 1 ∈ I1 . Since for every i ∈ I3 , i − 1 ∈ I1 , then |I3 | ≤ |I1 |. Hence k = |I2 | + |I3 | ≤ |I2 | + |I1 |. In order to find k, we need to bound B = |I1 | + |I2 |
(1)
in terms of ∆(G). 3. L (2, 1)-labeling of Kneser graphs In [10], Griggs and Yeh proved that for G = K (5, 2), λ(G) = 9. In [15], Kang considered the L(2, 1)-labeling problem on a particular Kneser graph, i.e., G = K (2b + 1, b). Kang’s main interest is to obtain upper bounds for the L(2, 1)-labeling numbers of the particular Kneser graph G = K (2b + 1, b) and the main contribution of that paper is Kang proved that λ(G) ≤ 4b + 2. In this section, we will consider any general Kneser graphs K (a, b) and prove the Griggs and Yeh’s conjecture is true for any general Kneser graphs K (a, b). We prove that L(2, 1)-labeling numbers of any Kneser graphs K (a, b) are much better than in the Griggs and Yeh’s conjecture in most a cases. Since K (a, 1) is the complete graph Ka , we assume b ≥ 2. We will first prove that the exact value for λ(K (a, b)) is b − 1 when a ≥ 3b − 1. We first give a property of K (a, b). Lemma 3.1. K (a, b) has diameter 2 if a ≥ 3b − 1. Proof. Consider any vertex x of K (a, b). Every other vertex v of K (a, b) represents a set V that is either disjoint or not with the set X represented by x. Every vertex v corresponding to a b-set V disjoint with X is a neighbor of x. Furthermore, for b-set V with X ∩ V ̸= ∅ there must be a b-set W that is disjoint with V and X since |V ∩ X | ≤ b − 1 and a ≥ 3b − 1. Therefore, the vertex w corresponding to W must be adjacent to the vertex v corresponding to V and to x and so v is at distance 2 from x. Theorem 3.2. The Kneser graph K (a, b) has L(2, 1)-labeling number λ(K (a, b)) ≤
a b
− 1. This bound is tight for a ≥ 3b − 1.
Proof. The collection C of b-subsets of the set A = {1, 2, . . . , a} can be represented as the leaves of a tree T of height b. The root r of T represents C and it has a children, each storing one value from A. Each one of these children is the root of a subtree Tj (1 ≤ j ≤ b) of height b − 1 in which an internal node storing value i has a − i children storing the values i + 1, . . . , a. In this tree we keep only those leaves at distance b from the root. Each leaf l of T represents a b-subset of A whose elements are the values stored in the nodes along the path from l to r. Consider the tree representation T of the family of b-subsets of A = {1, 2, . . . , a}. Let T1 , T2 , . . . , Ta−b+1 be the subtrees of T as described above. List the leaves of these subtrees in order in such a way that the leaves of subtrees Ti with i odd are listed from left to right and the leaves of subtrees Ti with i even are listed from right to left. For example for the tree in Fig. 2 the leaves would be listed in this order:
{1, 2, 3}, {1, 2, 4}, {1, 2, 5}, {1, 3, 4}, {1, 3, 5}, {1, 4, 5}, {2, 4, 5}, {2, 3, 5}, {2, 3, 4}, {3, 4, 5}.
4
Z. Shao et al. / Discrete Applied Mathematics (
)
–
Fig. 2. A tree representation for all 3-subsets of {1, 2, 3, 4, 5}.
We now show that any two adjacent subsets Sj , Sj+1 in the above order must have a non-empty intersection, i.e. Sj ∩ Sj+1 ̸= ∅. If Sj and Sj+1 are leaves of the same subtree Ti then both b-sets include i. On the other hand if Sj is a leaf of subtree Ti and Sj+1 is a leaf of subtree Ti+1 then (1) if i is odd then Sj and Sj+1 contain value a (2) if i is even then Sj and Sj+1 contain value i + 1. Hence, labeling the b-sets in the above order starting with label 0and using consecutive labels yields a feasible L(2, 1)a labeling for the vertices of K (a, b) and the maximum label used is b − 1, one less than the number of vertices in K (a, b). By Lemma 3.1, if a ≥ 3b − 1, then a the Kneser graph K (a, b) has diameter 2 which means any two vertices must have different labels. Hence λ(K (a, b)) = b − 1 in this case.
Note that by Theorem 3.2, for a = 3b − 1 and b = 2, we have λ(K (a, b)) = λ(K (5, 2)) =
5 2
− 1 = 9 which agrees
with Griggs and Yeh’s result in [10]. Let G be a graph with maximum degree ∆(G) and number of vertices ν(G). Note that if G has diameter 2, then ν(G) ≤ ∆(G)2 +1. Since the number of vertices of K (a, b) is ba , by Lemma 3.1 and Theorem 3.2, we have for a ≥ 3b−1, λ(K (a, b)) =
a b
− 1 ≤ ∆(K (a, b))2 . Since ∆(K (a, b)) =
a −b b
, in the following, we will prove for a ≥ 3b − 1, the bound is much better
than ∆(K (a, b)) in most cases. The following Lemmas 3.5, 3.8 and 3.10 provide results regarding this. 2
Lemma 3.3. For a ≥ 3b and b ≥ 3,
Proof. Since b 2 (( aa++11−−2b ) −
(
) >
a+1−b 2 a+1−2b
a+1−b b
)
=
Lemma 3.4. For b ≥ 3,
2b+2 b +1
=
a −b b
and
a+1 b
2b+2 b+1
a+1 b
b > 15.8( aa++11−−2b −
a −b a +1 a+1−b b a +1 . Thus, the a+1−b
Proof. Since
a+1−b a+1−2b
a+1−b 2 −15.8 a+1b−b b
conclusion follows. 2
−15.8
2b+2 b+1
3b+3 b+1
(2b+2)(2b+1) (b+1)2
2b b
>
2b b
to prove
2b b
>
b
,
> ··· >
3b−b 2 −15.8 3bb−b b
a+1−b 2 −15.8 a+1b−b b
a+1 b
>
3b b
−15.8
2b b
3b+3 b+1
2
=
6 3
> ··· >
−15.8 63 9 3
(3b+3)(3b+2)(3b+1) (b+1)(2b+2)(2b+1)
a b
−1<
=
2b 2 −15.8 2b b b
a−b 2 b −15.8 a− b b a b
3b b
.
is equivalent to
( )
a −b b
≥ 20 and
3b b
= 1.
,
2b+2 2 +2 −15.8 2b b+1 b+1
3b+3 b+1
(2b+2)(2b+1) 2 +3)(3b+2)(3b+1) +1) +3)(3b+2)(3b+1) ) − ((3b ) 2b > 15.8( (2b+(b2+)(12b − ((3b ). b+1)(2b+2)(2b+1) b b+1)(2b+2)(2b+1) (b+1)2 )2 +1) 2 +3)(3b+2)(3b+1) ≥ 20 and ( (2b+(b2+)(12b ) > ((3b . Thus, the conclusion follows. b+1)(2b+2)(2b+1) )2
Lemma 3.5. For a ≥ 3b and b ≥ 3,
3b b
and
a
). By a ≥ 3b and b ≥ 3, it is not difficult to prove
2
b a−b 2 −15.8 a− b b a b
( )
a +1 a+1−b
=
a+1 a+1−b
equivalent to ((
a −b b
2
− 15.8
a −b b
.
>
2b 2 −15.8 2b b b
3b b
is
By b ≥ 3, it is not difficult
Z. Shao et al. / Discrete Applied Mathematics (
Proof. By Lemmas 3.3 and 3.4, we have and the conclusion follows.
a−b b
2
−15.8
( )
a−b b
> 1. But
a b
a−b b
2
)
−15.8
( ) −1
–
a−b b
5
>
a b
a−b b
2
−15.8
( )
a−b b
for a ≥ 3b and b ≥ 3,
a b
Similarly to Lemmas 3.3–3.5, we can prove the following lemmas.
Lemma 3.6. For a ≥ 3b − 1 and b ≥ 3,
2b−1 2 −4.4 2bb−1 b
3b−1 b
a+1−b 2 −4.4 a+1b−b b
a+1 b
>
a−b 2 b −4.4 a− b b a b
> ··· >
( )
3b−1−b 2 −4.4 3b−b1−b b
3b−1 b
=
.
Lemma 3.7. For b ≥ 3,
2b+1 2 +1 −4.4 2b b+1 b+1
3b+2 b+1
>
Lemma 3.8. For a ≥ 3b − 1 and b ≥ 3,
Lemma 3.9. For a ≥ 3b and b = 2,
a b
−1<
3b−1 b
1 a−1 2 −3.5 a− 2 2
Lemma 3.10. For a ≥ 3b and b = 2,
2b−1 2 −4.4 2bb−1 b
a b
a+1 2
−1<
a−b b
>
a −b b
2
2
2
> ··· >
− 4.4
a −b b
2 a−2 2 −3.5 a− 2 2 a 2
( )
− 3.5
5 3
a −b b
−4.4
5 3
= 1.
8 3
. 2
> ··· >
4 2
−3.5
4 2
6 2
= 1.
.
In the following part, we will provide upper bounds for λ(K (a, b)) with 2b ≤ a ≤ 3b − 2. Theorem 3.11. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Then
a−2b b 2 − 1 + b < ∆2 if 2 ≤ a − 2b ≤ b − 2, < ∆ − ∆ b 1 − a−b λ(G) 2 ≤ ∆ − 2 if a − 2b = 1 and b ≥ 3, = 2 if a − 2b = 0. Proof. Let x be a vertex of K (a, b) which is labeled by k. Note that if a − 2b = 0, then the Kneser graph K (a, b) is a matching and hence, λ(G) = 2. Also note that in this case ∆ = 1. Thus, in the following, we can only consider the case when 0 < a − 2b ≤ b − 2. The particular case when a − 2b = 1 has been considered by Kang [15]. Kang proved that for G = K (2b + 1, b), λ(G) ≤ 4b + 2. Suppose a − 2b = 1, by 0 < a − 2b ≤ b − 2, we have b ≥ 3. If a − 2b = 1, then the maximum degree of G is ∆ = b + 1 ≥ 4, so λ(G) ≤ 4b + 2 = 4∆ − 2 ≤ ∆2 − 2 since b ≥ 3. So, let us assume 2 ≤ a − 2b ≤ b − 2. This implies b ≥ 4. The degree of any vertex in K (a, b) is
maximum degree is ∆ =
a−b b
a −b b
and hence the
. If xvw is a simple path of length 2 in K (a, b) then the corresponding b-sets X , V and W , for
vertices x, v , and w respectively, must satisfy X ∩ V = ∅, V ∩ W = ∅ and X ∩ W ̸= ∅. Considering X fixed, there are
a −b b
choices for V ; then, ifVis also fixed, since W must be disjoint with the b-set V but |X ∩ W | ≥ 1, the number of choices for W is
b−1
t =b−(a−2b)
a−b b
[
a−2b b −t
b t
b −1
t =b−(a−2b)
]. Thus, the number of vertices with distance 2 from x is not greater than
b
a − 2b
t
b−t
.
(2)
Note that the subgraph H induced in G by all vertices adjacent to x is a graph with no edges since a − 2b ≤ b − 2, and thus two neighbors of x cannot correspond to disjoint b-sets. However, Since some of the neighbors of x have common neighbors that do not belong to H, we can further tighten the above bound. Let xv z and xw z be two different paths of length 2 in K (a, b), where z is not a neighbor of x. The corresponding b-sets for vertices x, v, w , and z are X , V , W , and Z , respectively; these sets must satisfy the conditions X ∩ V = ∅,
X ∩ W = ∅,
V ∩ Z = ∅,
W ∩ Z = ∅,
and
X ∩ Z ̸= ∅.
(3)
Note that the sets X , V , W and Z are different (see Fig. 3). Consider X as fixed, and fix an arbitrary b-set V ⊂ (V (G) \ X ). Then, the bound (2) can be reduced by the number N ′ of distinct choices of b-sets W and Z that satisfy the conditions (3). Let us estimate N ′ .
6
Z. Shao et al. / Discrete Applied Mathematics (
)
–
Fig. 3. The corresponding b-sets X , V , W , and Z for vertices x, v, w , and z.
Let |V ∩ W | = t, where 3 ≤ t ≤ b − 1. Then the number of choices for W is
b t
a−2b b−t
. When W is chosen, it is z’s turn
to choose its b-set Z . Since X ∩ Z ̸= ∅, z has to choose at least one element and at most b − 1elements in X. Summing up over these b − 1 possibilities, we get that given X , V , W , the number of choices for Z is
b−1 j=1
[
a−(3b−t ) b−j
b j
].
Adding up over all possible values of t, we get
b−1 b−1 b a − 2b b a − (3b − t ) N = . t b−t j b−j t =3 j =1 ′
In the summation we need to consider only values of t such that 1 ≤ a − (3b − t ), i.e., t ≥ 3b − a + 1. Since a − 2b ≤ b − 2, t ≥ 3b − a + 1 > 2. But 0 ≤ t ≤ b − 1, then 3 ≤ t ≤ b − 1. We have b −1
′
N =
b
a − 2b
t
b−t
t =3b−a+1
b
a − (3b − t )
j
b−j
b−1 j=1
b −1
≥b
b
a − 2b
t
b−t
t =3b−a+1
Therefore, the number of vertices with distance 2 from x is not greater than b
b−1
t =3b−a+1
c
Since
[
b t
a−b
b
a −b b
.
b−1 t =b−(a−2b)
[
b t
a−2b b −t
]. b c −b ] for all c ≥ b, we have t =0 [ t b−t
b
=
b
a−2b b −t
=
b
a − 2b
3b − a
+ ··· + b −1 t =b−(a−2b) b−1
3b − a + 1
b
a − 2b
t
b−t
b
a − 2b
b
0
b
a − 2b
b
0
+
+
b
a − 2b
t
b−t
b
a − 2b
t
b−t
+
a − 2b
+
a − 2b − 1
1
t =3b−a+1
b
a − 2b
b−1
+
b
+
a − 2b
=
b
a − 2b
(a − 2b − 2)
3b − a + 2
=
b
a − 2b
b
0
b
3b − a
a − 2b
a − 2b
.
Thus,
a−b b
b−1
t =b−(a−2b)
=
a−b
a−b
b
b
−b
a − 2b
t
b−t
b
t =3b−a+1
−1 −
b −1
a−b
b
−
b 3b − a
− 1 b = ∆(∆ − 1) − ∆ −
Also, the number of vertices with distance one from x is not greater than ∆ =
b 3b−a
a −b b
= <
b!
(3b − a)!(a − 2b)!
b a−b
a−2b
.
·
b!(a − 2b)!
(a − b)!
=
a −b b
b
3b − a
. But
(b!)2 b(b − 1) . . . (3b − a + 1) = (3b − a)!(a − b)! (a − b)(a − b − 1) . . . (b + 1)
− 1 b.
] −
Z. Shao et al. / Discrete Applied Mathematics ( b 3b−a a−2b
Hence
( a−b ) b
)
–
7
< ( a−b b )a−2b ∆, so the number of vertices with distance at most 2 from x is less than ∆ + ∆(∆ − 1) − (∆(1 −
) − 1)b = ∆2 − (∆(1 − ( a−b b )a−2b ) − 1)b.
Thus, by (1), we have
λ(G) ≤ k ≤ B = |I1 | + |I2 | < ∆ + ∆ − ∆ 1 − 2
= ∆ −∆ b 1− 2
a−2b
b a−b
a−2b
b
2
2
+ 1−
=
b +2
1 − ( a −b ) b
a−2b
b +2 b
>
1 b
1 b
+1+
+
3b−6 . (b+2)2
4 (b+2)2
−
4 b +2
= 1+
Thus, ∆(b(1 − ( a−b ) b
− 1 + b.
4 (b+2)2
a−2b
1 b
+
+ ( a−b b )a−2b ≤ < 1+
−4b+b+2 b(b+2)
) − 1) − b >
3b2 −6b (b+2)(b+1) 2 (b+2)2
3b2 −6b
. By b ≥ 4, we have, (b+2)2 ∆ − b ≥ b Thus, λ(G) < ∆2 − ∆(b(1 − ( a− )a−2b ) − 1) + b < ∆2 . b
then ∆ ≥
−1 b
a−b
Recall that we assumed 2 ≤ a − 2b ≤ b − 2 and b ≥ 4. Since 1 b
3b2 −6b ∆ (b+2)2
−b=
1 b
+
4 (b+2)2
b (2b+2)−b
+
−3b+2 (b+2 )2
− b. But ∆ =
b(3b2 −5b−10) 2(b+2)
(2b+2)−2b = 1+
a −b b
=
1 b
+
−3b+6 , (b+2)2
2 b b +2
=
we have
and a − b ≥ b + 2,
> 0.
By Theorem 3.2, Lemmas 3.5, 3.8, 3.10 and Theorem 3.11, we can see that the L(2, 1)-labeling numbers of Kneser graphs are much better than in the Griggs and Yeh’s conjecture in most cases. Combining Theorem 3.2, Lemmas 3.8, 3.10 and Theorem 3.11, we have the following result. Theorem 3.12. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Then
a = − 1 < ∆2 − 4.4∆ if a ≥ 3b − 1 and b ≥ 3, b a = − 1 < ∆2 − 3.5∆ if a > 3b − 1 and b = 2, b = 9 = ∆2 if a = 3b − 1 and b = 2, a−2b λ(G) b 2 − 1 + b < ∆2 if 2b + 2 ≤ a ≤ 3b − 2, <∆ −∆ b 1− a − b 2 ≤ ∆ − 2 if a = 2b + 1 and b ≥ 3, = 2 if a = 2b. 9(r +3) r 8
Remark. In [16], Kim and Park proved that χ (K 2 (2k + r , k)) ≤ 2( 94 k +
) for any integer r with 1 ≤ r ≤ k − 2.
Our upper bounds are stated in terms of the maximum degree and their upper bounds are stated in terms of different parameters from ours. Since coloring squares of a graph G requires that any two distinct vertices of G have different colors if 9(r +3) the distance between them in G is at most 2, we can deduce that λ(K (2k + r , k)) ≤ 2χ (K 2 (2k + r , k))− 2 ≤ 4( 94 k + 8 )r − 2 for any integer r with 1 ≤ r ≤ k − 2 (note that λ(G) starts from 0 and χ (G) starts from 1). a We have proved that the exact L(2, 1)-labeling number of Kneser graph K (a, b) with a ≥ 3b − 1 is b − 1. Note that when a ≥ 3b − 1, it implies r ≥ k − 1 when a = 2k + r and b = k. Thus, we have obtained the first known exact value for a ≥ 3b − 1. In the following we will make a conservative comparison between the above deduced bound and ours for 2 ≤ a − 2b < b − 1. A well-known inequality is
k+r k
≤(
(k+r )·e k k
n k
≤ ( nk·e )k for 1 ≤ k ≤ n. Since ∆(K (2k + r , k)) =
) . Thus, ∆ − ∆(b(1 − ( a−b ) b
2
is from Theorem 3.11). But 4( 49 k +
a−2b
) − 1) + b < ∆ ≤ ( 2
(k+r )·e 2k k
) ≤(
k+r k
(2k−2)·e 2k k
, we have ∆(K (2k + r , k)) = 2k
) < (2 · e) (the first inequality
9(r +3) r 8 r ln 94
) − 2 ≥ 4( 49 k + 29 )r − 2 > 4( 94 k)r . Note that ln((2 · e)2k ) = 2k ln(2 · e) and ln(4( ) ) = ln 4 + r ln( ) = ln 4 + ( ) + r ln k ≥ r ln k. Since ln(2 · e) < 1.694, when (r ln k)/(2k) > 1.7, our bound is better than the above deduced bound. For example, when r = k − 2 and k ≥ 37, (r ln k)/(2k) ≥ (35 ln 37)/74 > 1.7. 9 k r 4
9 k 4
4. Coloring square of Kneser graphs By Brook’s theorem, for any connected simple graph G with chromatic number χ (G) and maximum degree ∆(G), χ (G) ≤ ∆(G) unless G is a complete graph or an odd cycle where χ (G) ≤ ∆(G) + 1. Let the number of vertices with distance one and two of any vertex v in G be d1 (v) and d2 (v), respectively, then χ (G2 ) ≤ maxv∈V (G) {d1 (v) + d2 (v) + 1} = ∆(G2 ) + 1 ≤ ∆(G) + ∆(G)(∆(G) − 1) + 1 = (∆(G))2 + 1, and the upper bound ∆(G2 ) + 1 realizes if and only if G2 is a complete graph
8
Z. Shao et al. / Discrete Applied Mathematics (
)
–
or an odd cycle. Note that (∆(G))2 + 1 is not always realizable even if G2 is a complete graph (the following Theorem 4.3 will provide a good example). In this section, a we consider the chromatic number of the square of any Kneser graphs K (a, b). By Lemma 3.1 and V (K (a, b)) = b , we can easily observe the following lemma. Lemma 4.1. The chromatic number of the square of any Kneser graphs K (a, b) are bounded by a ≥ 3b − 1.
a b
. This bound is tight for
By Lemma 3.1, we have for a ≥ 3b − 1, K (a, b) has diameter 2 and thus, ∆(G2 ) = V (K (a, b)) − 1 =
a b
− 1,
i.e., b = ∆(G2 ) + 1. In the following part, we will provide upper bounds for the chromatic number of the square of any Kneser graphs K (a, b) with 2b ≤ a ≤ 3b − 2.
a
Theorem 4.2. Let G be the Kneser graph K (a, b) with the maximum degree ∆. Then
a−2b b 2 − 1 b + 1 < ∆2 − ∆ + 1 if 2 ≤ a − 2b ≤ b − 2, < ∆ − ∆ 1 − a−b 2 χ (G ) ≤ ∆2 − ∆ − 1 if a − 2b = 1 and b ≥ 3, = 2 = ∆2 + 1 if a − 2b = 0. Proof. Note that if a − 2b = 0, then the Kneser graph K (a, b) is a matching and hence, λ(G) = 2. Also note that in this case ∆ = 1. Thus, in the following, we can only consider the case when 0 < a − 2b ≤ b − 2. The particular case when a − 2b = 1 has been considered by Chen, Lih and Wu [5]. They proved that for G = K (2b + 1, b), χ (G2 ) ≤ 3b + 2. Suppose a − 2b = 1, by 0 < a − 2b ≤ b − 2, we have b ≥ 3. If a − 2b = 1, then the maximum degree of G is ∆ = b + 1, so χ(G2 ) ≤ 3b + 2 ≤ ∆2 − ∆ − 1 since b ≥ 3. So, let us assume 2 ≤ a − 2b ≤ b − 2. This implies b ≥ 4. By Theorem 3.11, the maximum number of vertices at distance at most 2 from any vertex v in G can be at most
∆2 − ∆ 1 −
b a−b
a−2b
− 1 b.
Let the number of vertices with distance one and two of any vertex v in G be d1 (v) and d2 (v), respectively, then χ (G2 ) ≤ b )a−2b ) − 1)b + 1. Recall that we assumed 2 ≤ a − 2b ≤ b − 2 and b ≥ 4. maxv∈V (G) {d1 (v) + d2 (v) + 1} ≤ ∆2 − (∆(1 − ( a− b
b Thus, similarly to the proof in Case 1 in Theorem 3.11, we have χ (G2 ) < ∆2 −(∆(1 −( a− )a−2b )− 1)b + 1 < ∆2 − ∆ + 1. b
By Lemma 4.1 and Theorem 4.2, we can see that the chromatic number of the square of any Kneser graphs G with maximum degree ∆ are bounded by ∆2 − ∆ + 1 (may be much better in most cases) unless a = 5 and b = 2, or a − 2b = 0. Combining Lemma 3.8, Lemmas 3.10, 4.1 and Theorem 4.2, we have the following result. Theorem 4.3. Let G be the Kneser graph K (a, b) with maximum degree ∆ ≥ 2. Let the maximum degree of G2 be ∆(G2 ). Then
a = = ∆(G2 ) + 1 < ∆2 − 4.4∆ + 1 if a ≥ 3b − 1 and b ≥ 3, b a = = ∆(G2 ) + 1 < ∆2 − 3.5∆ + 1 if a > 3b − 1 and b = 2, b = 10 = ∆(G2 ) + 1 = ∆2 + 1 if a = 3b − 1 and b = 2, a−2b χ (G2 ) b 2 <∆ − ∆ 1− − 1 b + 1 < ∆2 − ∆ + 1 if 2b + 2 ≤ a ≤ 3b − 2, a − b ≤ ∆2 − ∆ − 1 if a = 2b + 1 and b ≥ 3, = 2 = ∆2 + 1 if a = 2b. By Theorem 4.3, the upper bound ∆(G2 ) + 1 realizes if a ≥ 3b − 1. However, ∆2 + 1 is not always realizable even if a ≥ 3b − 1 which means G2 is a complete graph. 9(r +3)
Remark. In [16], Kim and Park proved that χ (K 2 (2k + r , k)) ≤ 2( 49 k + 8 )r for any integer r with 1 ≤ r ≤ k − 2. As in the previous remark, we can make a similar conservative comparison between the above deduced bound and our bound in this section.
Z. Shao et al. / Discrete Applied Mathematics (
)
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9
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