A new approach to the chromatic number of the square of Kneser graph K(2k+1,k)

Discrete Mathematics (

)

–

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A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k) Jeong-Hyun Kang Department of Mathematics, University of West Georgia, Carrollton, GA 30118, USA

article

info

Article history: Received 19 September 2016 Received in revised form 5 August 2017 Accepted 5 August 2017 Available online xxxx Keywords: Square of Kneser graph Chromatic number Kneser graph

a b s t r a c t The vertices of Kneser graph K (n, k) are the subsets of {1, 2, . . . , n} of cardinality k, two vertices are adjacent if and only if they are disjoint. The square G2 of a graph G is defined on the vertex set of G with two vertices adjacent if their distance in G is at most 2. Z. Füredi, in 2002, proposed the problem of determining the chromatic number of the square of the Kneser graph. The first non-trivial problem arises when n = 2k + 1. It is believed that χ (K 2 (2k + 1, k)) = 2k + c where c is a constant, and yet the problem remains open. The best known upper bounds are by Kim and Park: 8k/3 + 20/3 for 1k ≥ 3 (Kim and Park, 2014) and 32k/15 + 32 for k ≥ 7 (Kim and Park, 2016). In this paper, we develop a new approach to this coloring problem by employing graph homomorphisms, cartesian products of graphs, and linear congruences integrated with combinatorial arguments. These lead to χ (K 2 (2k + 1, k)) ≤ 5k/2 + c, where c is a constant in {5/2, 9/2, 5, 6}, depending on k ≥ 2. © 2017 Elsevier B.V. All rights reserved.

1. Introduction Let [n] = {1,(2, . ). . , n}. The Kneser graph K (n, k) is the graph with(vertex ) set consisting of all the k-subsets of an n-element [n ] [n ] set, denoted by k , and edge set consisting of all the pairs A, B ∈ k such that A ∩ B = ∅. The problem to determine the chromatic number of Kneser graph was proposed by Kneser in 1955 and settled by Lovasz [16] and Barany [2], independently in 1978. Since then Kneser graphs have been well-studied for their rich structural and extremal properties, especially with regard to coloring problems. The square G2 of a graph G is a graph with the vertex set V (G) such that two vertices are adjacent in G2 if and only if their distance in G is at most 2. The chromatic numbers of powers of graph (Gd defined analogously to G2 ) have been studied for many years. See for example general bounds on chromatic number of power of graphs in terms of their girth [1], and the rich literature on coloring square of a planar graph as discussed in [8] and [17]. In 2002, Z. Füredi proposed the problem to determine the chromatic number of the square of the Kneser graph. We will denote the square of the Kneser graph by K 2 (n, k). When n = 2k, the graph is a perfect matching, and when ( [2k+1]n) ≥ 3k − 1, the graph is a clique. Hence the problem is interesting when 2k + 1 ≤ n ≤ 3k − 2. Observe that A, B ∈ is an edge k in K 2 (2k + 1, k) if A ∩ B = ∅ or |A ∩ B| = k − 1. The problem has turned out to be surprisingly difficult even for the case n = 2k + 1. The exact values are known for k = 2, 3, 4 [11,12]. The trivial lower bound k + 2 on the chromatic number is easily obtained by observing the clique of the same size, and this lower bound has not been improved. The following upper bounds have been obtained: 4k + 2 by Kim and Nakprasit [12], 3k + 2 for k ≥ 3 by Chen, Lih, and Wu [4], 8k/3 + 20/3 for k ≥ 3 by Kim and Park [13], 32k/15 + 32 for k ≥ 7 by Kim and Park [14]. It is believed the true value is 2k + c for some constant c. In this paper, we prove the upper bound of 5k/2 + c where c ∈ {5/2, 9/2, 5, 6}, depending on k ≥ 2. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.disc.2017.08.008 0012-365X/© 2017 Elsevier B.V. All rights reserved.

Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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Theorem 1. For k ≥ 2,

χ (K 2 (2k + 1, k)) ≤

⎧ 5k ⎪ ⎪ + 6, if k ≡ 0 (mod 4), ⎪ ⎪2 ⎪ ⎪ ⎪ 5k 5 ⎪ ⎪ ⎨ + , if k ≡ 1 (mod 4), 2

2

5k ⎪ ⎪ ⎪ + 5, if k ≡ 2 (mod 4), ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎩ 5k + 9 , if k ≡ 3 (mod 4). 2

2

Remark. The additive constants 6 and

9 2

can be improved to 4 and 12 , respectively, with some additional argument.

Note that if k ≥ 82, then 5k/2 + 5/2 > 32k/15 + 32, hence the upper bounds in this paper are better only for a finite number of integers k ≤ 81. But we develop new tools that might be helpful to understand the properties of the square of Kneser graph as described below. This problem is closely related to L(2, 1)-labeling introduced by Girggs and Yeh [7] in 1992. An L(2, 1)-labeling of a graph is an integer labeling f on the vertices satisfying that |f (u) − f (v )| ≥ 2 if u, v ∈ V are adjacent in G, and |f (u) − f (v )| ≥ 1 if u, v ∈ V are at distance 2. The smallest value l that G has an L(2, 1)-labeling using labels {0, 1, 2, . . . , l} is denoted by λ(G). This problem has been actively studied, see [3,5,6,10,15]. Note that χ (G2 ) ≤ λ(G) + 1. The author [9] has shown that λ(K 2 (2k + 1, k)) ≤ 4k + 2. One of the main tools in the current paper is a refinement of the technique shown in [9], which is linear congruences integrated with combinatorial arguments so as to find the structure of Kneser graphs within and between congruence classes. In addition, we employ graph homomorphisms and cartesian products of graphs, and define intersection graphs that arise naturally as auxiliary graphs in our problem. We expect that the new tools appearing in this paper can be extended further to obtain an improved bound. In Section 2, we lay out the preliminaries—notation, known results, introduction to intersection graphs, and partition of the vertex set with direct sums. The proof of Theorem 1 is given in Sections 3 and 4. 2. Preliminaries

(X )

For a given set X and a positive integer s, |X | ≥ s, the notation s denotes the collection of the subsets of X of size s. Let X and Y be disjoint subsets of [n], and let X and Y be the collections of some of the subsets of X and Y , respectively. We define the direct sum X ⊕ Y := {X ′ ∪ Y ′ | X ′ ∈ X , Y ′ ∈ Y } as the set of the disjoint unions of X ′ ∈ X and Y ′ ∈ Y . The standard notation G[U ] is used for the induced subgraph of G on U ⊂ V (G). In this(paper,)K [U ] and K 2 [U ] denote the [2k+1] induced subgraphs of the Kneser graph K (2k + 1, k), and its square, respectively, on U ⊂ . Also, we write (K 2 \ K )[U ] k 2 2 for the subgraph of K [U ] induced by the edges only in K but not in K with both endpoints in U; for A, B ∈ U, AB ∈ E(K [U ]) means A ∩ B = ∅, and AB ∈ E((K 2 \ K )[U ]) means |A ∩ B| = k − 1. Let G1 := (V1 , E1 ), G2 := (V2 , E2 ) be two graphs. The union G1 ∪ G2 of the two graphs G1 and G2 is a graph having the vertex set of V1 ∪ V2 and the edge set of E1 ∪ E2 . The Cartesian product of G1 and G2 , denoted by G1 □G2 , has the vertex set V1 × V2 = {(v1 , v2 )| v1 ∈ V1 , v2 ∈ V2 }. The vertices (u1 , u2 ), (v1 , v2 ) ∈ V1 × V2 are adjacent in G1 □G2 if u1 = v1 and u2 v2 ∈ E2 , or u2 = v2 and u1 v1 ∈ E1 . It is known that

χ (G1 □G2 ) = max{χ (G1 ), χ (G2 )}.

(1)

A graph homomorphism f from G1 to G2 is a mapping f : V1 → V2 that preserves adjacency, i.e., if uv ∈ E1 , then f (u)f (v ) ∈ E2 . It is easy to see that, if there is a homomorphism from G1 to G2 ,

χ (G1 ) ≤ χ (G2 ).

(2)

If |V1 | = |V2 | and there is a graph homomorphism from G2 to G1 as well, G1 and G2 are isomorphic, denoted by G1 ∼ = G2 . 2.1. Intersection graphs Certain auxiliary graphs, namely intersection graphs defined below, arise naturally. The language of intersection graphs, together with a certain partition and the direct sums of the vertices stated in Section 2.2, will lead to Cartesian products of graphs that have been well-studied. For a given finite set X and positive integers s, l with |X | ≥ s > l ≥ 0, we define the intersection graph H(X , {s, s − 1, . . . , s − l)} with the vertex set of all the (s − i)-subsets of X , 0 ≤ i ≤ l, and the edge sets are defined as follows. The pair A, B ⊂ X is adjacent in H(X , {s, s − 1, . . . , s − l}) if one of the following two cases occur; (i) when A and B are of the same cardinality, they have all their elements in common except for one each, i.e., |A| = |B| = |A ∩ B| + 1; (ii) when A and B have different cardinalities, say |A| < |B|, it holds that A ⊂ B. When X = {1, 2, . . . , m}, we simply write H(m, {s, s − 1, . . . , s − l}). The intersection graph with l = 0 is known to be Johnson graph J(m, s). Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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It is not difficult to see that H(X , {s, ∑ s − 1}) ∼ x}, {s})(for) some ˜ x ̸ ∈ X . Also, H(X , {s}) has a proper coloring using = H(X ∪ {˜ X m := |X | colors via the coloring f (A) = a (mod m) for A ∈ ; a∈A s

χ (H(X , {s})) ≤ |X | for all s ≤ |X |.

(3)

2.2. Partition of the vertex set In this section, we partition the vertex set k , write the vertices as direct sums, and describe the edges of K 2 (2k + 1, k) accordingly. This point of view will enable us to interpret most of the graph in terms of Cartesian products of intersection graphs, whose chromatic numbers are known. Furthermore, the direct sums will be useful to analyze each part before we sum up the parts. ( [2k+1] ) Partition the point set [2k + 1] = X ∪ Y with |X | = k, |Y | = k + 1. For A ∈ V = , we write AX := A ∩ X , AY := A ∩ Y k and AX := X \ AX , AY = Y \ AY . Clearly, A = AX ⊕ AY . The vertex set V is the disjoint union of the direct sums

( [n ] )

V =

⋃ (( X ) 0≤s≤k

s

( ⊕

Y

))

k−s

.

For A, B ∈ V , it is trivial that A ∩ B = (AX ∩ BX ) ∪ (AY ∩ BY ). Let A ∈ s ⊕ k−s , B ∈ s′ ⊕ k−s′ . When s, s′ ≤ ⌊k/2⌋ − 1, both |AY | and |BY | are ≥ k − (⌊ 2k ⌋ − 1) = ⌈ 2k ⌉ + 1. It leads to the inequality |AY | + |BY | ≥ k + 2 > |Y |, therefore AY ∩ BY ̸ = ∅. Similarly, s, s′ ≥ ⌊k/2⌋ + 1 yields that |AX | + |BX | ≥ k + 1 > |X | therefore AX ∩ BX ̸ = ∅. The pair of A and B always intersect whenever both s, s′ ≤ ⌊k/2⌋ − 1 or both s, s′ ≥ ⌊k/2⌋ + 1.

(X )

Proposition 1. Let A ∈

(X ) s

⊕

(

Y k−s

)

,B ∈

(X ) s′

⊕

(

Y k−s′

)

(

Y

(X )

)

(

Y

)

for 0 ≤ s ≤ s′ ≤ k.

1. The pair of A and B is disjoint if and only if either (a) or (b) holds: (a) s + s′ = k, in which case AX = BX and AY ⊃ BY ;

(4)

(b) s + s′ = k − 1, in which case AX ⊂ BX and AY = BY .

(5)

2. The pair of A and B has k − 1 elements in common if and only if either (c) or (d) holds: (c) s = s′ , in which case either AX = BX and |AY ∩ BY | = k − s − 1,

(6)

|AX ∩ BX | = s − 1 and AY = BY ;

(7)

or

(d) s′ = s + 1, in which case AX ⊂ BX and AY ⊃ BY .

(8)

Proof. 1. It is trivial to see that A ∩ B = ∅ if and only if AX ∩ BX = ∅ in X , and AY ∩ BY = ∅ in Y . ′

(9) ′

′

It is necessary that s + s ≤ k and (k − s) + (k − s ) ≤ k + 1 . Therefore s + s = k or k − 1. When s + s′ = k, since |AX | = |BX | = s and |AY | = s + 1 ≥ s = |BY |, we should have (4) in order to satisfy (9). When s + s′ = k − 1, we have |AX | = s ≤ s + 1 = |BX | and |AY | = |BY | = s + 1, hence (5) follows. 2. When s′ ≥ s + 2, it is trivial |AX | < |BX | and |AY | > |BY |, hence |A ∩ B| ≤ |AX | + |BY | ≤ s + (k − s′ ) ≤ k − 2. Therefore s′ = s or s + 1 if |A ∩ B| = k − 1. When s′ = s, since |AX | = |BX | = s and |AY | = |BY | = k − s, hence (6) or (7) should be satisfied to achieve |A ∩ B| = k − 1. When s′ = s + 1, it means that |AX | = s ≤ s + 1 = |BX | and |AY | = k − s ≥ k − s − 1 = |BY |. The intersection size |A ∩ B| = k − 1 can be achieved only when (8) holds. □ Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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3. Proof: k is even ≥ 2 Let Vk/2 :=

(

X k/2

)

⊕

(

s + s′ = k, and set

( ) Vs :=

X

( ⊕

s

Y k/2

)

. For 0 ≤ s ̸ = s′ ≤ k, we pair up the subsets

)⋃(

Y k−s

)

X

( ) Y

⊕

k−s

s

(X ) s

⊕

(

Y k−s

)

and

(X ) s′

⊕

(

Y k−s′

)

of V satisfying

, for 0 ≤ s ≤ k/2 − 1.

Now then V = ∪0≤s≤k/2 Vs . Proposition 1 implies that, for s, s′ with |s − s′ | ≥ 2, there is no edge of K 2 (2k + 1, k) between the vertices Vs and Vs′ . Therefore, the subgraphs K 2 [Vk/2−(2i−1) ], i ≥ 1, can share the same set of colors, and the subgraphs K 2 [Vk/2−2i ], i ≥ 0, can another set of colors. We will show that each subgraph K 2 [Vs ] is properly colorable with k + 2 colors for 0 ≤ s ≤ k/2 − 1 ( Theorem 2), and K 2 [Vk/2 ] is with 3k/2 + c1 colors where c1 = 3 or 4 ( Theorem 3). We then have, for even k ≥ 2,

χ (K 2 (2k + 1, k)) ≤ (k + 2) + ≤

5 2

(

3 2

) k + c1

k + c,

where c = 5 or 6. 3.1. Coloring of K 2 [Vs ], for 0 ≤ s ≤ k/2 − 1 Theorem 2. Each subgraph K 2 [Vs ] has a proper coloring using k + 2 colors for 0 ≤ s ≤ k/2 − 1. Proof. For a given s, 0 ≤ s ≤ k/2 − 1, define a map φ : K 2 [Vs ] → H(X , {s})□H(Y , {s + 1, s}) as follows; for A := AX ⊕ AY ∈ Vs ,

( ) X

AX ⊕ AY ∈

(

Y

( )

)

X

(

Y

)

× , ⊕ ↦→ (AX , AY ) ∈ s s+1 k−s ) ( ) ( ) ( ) X Y X Y ⊕ ↦→ (AX , AY ) ∈ × . k−s s s s s

( AX ⊕ AY ∈

The mapping φ is well-defined. ( Y ) (Y ) A pair of A, B ∈ Vs is disjoint if and only if A, B satisfy (4). The inclusion AY ⊃ BY means that AY ∈ s+1 , BY ∈ s are

adjacent in the intersection graph H(Y , {s + 1, s}). Together with the equality AX = BX in s , their images φ (AX ⊕ AY ) = (AX , AY ) and φ (BX ⊕ BY ) = (BX , BY ) form an edge in the Cartesian product H(X , {s})□H(Y , {s + 1, s}). (X ) ( Y ) The vertices A, B of K 2 [Vs ] have k − 1 elements in common if and only if (6) or (7) are the cases. For A, B ∈ s ⊕ k−s ,

(X )

their images (AX , AY ), (BX , BY ) ∈

(X ) s

×

(

Y s+1

)

satisfy AX = BX , |AY ∩ BY | = s by (6) or |AX ∩ BX | = s − 1, AY = BY by (7).

When A, B ∈ k−s ⊕ s , (6) and (7) imply that the images (AX , AY ), (BX , BY ) ∈ s × s satisfy AX = BX , |AY ∩ BY | = s − 1 or |AX ∩ BX | = s − 1, AY = BY . Both cases, φ (A)φ (B) is an edge in H(X , {s})□H(Y , {s + 1, s}). We have shown that φ is a graph homomorphism. By (2), it is enough to show that

(

X

)

(Y )

(X )

(Y )

χ (H(X , {s}))□H(Y , {s + 1, s}) ≤ k + 2, which follows immediately from (1) and (3). □ Remark. The mapping φ is indeed an isomorphism. 3.2. Coloring of K 2 [Vk/2 ] We use the structure of Vk/2 =

(

X k/2

)

⊕

(

Y k/2

)

and employ the idea from [9] with a particular partition of [2k + 1] into X

and Y to obtain the improved upper bounds in Theorem 3. We set X to be the set of even numbers and Y to be the set of odd numbers of [2k + 1], respectively. For A ∈ Vk/2 , consider the coloring f (A) =

∑

a

(mod k + 2)

a∈A

=

∑ a∈AX

a+

∑

a (mod k + 2)

(10)

a∈AY

Observe that k + 2 is even. Set Rr := {A ∈ Vk/2 | f (A) ≡ r (mod k + 2)} where 0 ≤ r ≤ k + 1. Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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Lemma 1. The coloring f introduces k/2 + 1 color classes. Proof. Note that a∈A a and a∈A a in (10) are the sums of k/2 even numbers and sums of k/2 odd numbers, respectively. X Y Because k + 2 is even, reading the sum in modulo k + 2 results in the residue classes r (mod k + 2) such that r has the same parity as k/2. That is, Vk/2 is decomposed into R0 , R2 , R4 , . . . , Rk when k/2 is even; Vk/2 into R1 , R3 , R5 , . . . , Rk+1 when k/2 is odd. □

∑

∑

For a pair of disjoint A, B ∈ Vk/2 , the union A ∪ B leaves out exactly one element from [2k + 1], which in particular, should be one from Y , say y. Then f (A) + f (B) =

(2k+1 ) ∑ i

−y

i=1

= (2k + 1)(k + 1) − y = 3 − y (mod k + 2). If both A and B receive the same color r, i.e., A, B ∈ Rr , then r and y satisfy the linear congruence 2r = 3 − y (mod k + 2).

(11)

For a given r, the integers y, y ∈ Y are solutions of (11) if and only if y ≡ y (mod k + 2). Note that r = 2 if and only if y := k + 1 is the only solution in Y of (11). Put ′

′

Y1 := {1, k + 3}, Y3 := {3, k + 5}, . . . , Yk−1 := {k − 1, 2k + 1}, Yk+1 = {k + 1}.

(12)

For the rest of the proof, we will denote y,ˆ y ∈ Y with the property y ≡ ˆ y (mod k + 2) so that {y,ˆ y} = Yq for some odd q ̸ = k + 1. The disjoint sets A, B ∈ R2 uniquely determine each other; AX = BX , AY = BY ∪ {k + 1}. For r ̸ = 2 and for the disjoint sets A, B ∈ Rr with (A ∪ B)c = {y} assuming ˆ y ∈ A and y,ˆ y ̸ ∈ B without loss of generality, the set A′ := (A \ {ˆ y}) ∪ {y} is disjoint with B and should also receive the same color r. Consequently, A′ B is a monochromatic edge in K [Rr ] and ABA′ is a path in K [Rr ]. Furthermore, A and B uniquely determine each other, consequently A′ as well. We summarize these properties in K [Rr ] in Lemma 2. Lemma 2. 1. For each r ̸ = 2, the induced subgraph K [Rr ] of the Kneser graph consists of vertex–disjoint paths of length 2, if it contains an edge, and possibly isolated vertices. Furthermore, if ABA′ is a path in K [Rr ], where A, A′ , B ∈ Rr , then y} AX = A′X = BX , Y \ (BY ∪ AY ) = {y}, Y \ (B ∪ A′Y ) = {ˆ for y ∈ Y , y ̸ = k + 1, such that r , y satisfy the congruence (11). 2. When k/2 is even, the induced subgraph K [R2 ] of the Kneser graph, if not empty, consists of a matching and possibly isolated vertices. In order to characterize the monochromatic edges in (K 2 \ K )[Rr ], similar to (12), we pair up the elements of X : X2 := {2, k + 4}, X4 := {4, k + 6}, . . . , Xk−2 := {k − 2, 2k}, Xk := {k}, X0 := {k + 2}. We denote x,ˆ x ∈ X with the property x ≡ ˆ x (mod k + 2) so that {x,ˆ x} = Xp for some even p ̸ = 0, k. Lemma 3. For every r, let A, B ∈ Rr . The pair holds |A ∩ B| = k − 1 if and only if AX = BX , and AY \ BY = {y}, BY \ AY = {ˆ y} for some y ∈ Y , y ̸ = k + 1

(13)

AY = BY , and AX \ BX = {x}, BX \ AX = {ˆ x} for some x ∈ X , x ̸ = k, k + 2.

(14)

or

Proof. It is trivial that A, B ∈ Rr with |A ∩ B| = k − 1 if and only if AX = BX , AY \ BY = {y}, BY \ AY = {y′ } for some y, y′ ∈ Y , y ̸ = y′ or AY = BY , AX \ BX = {x}, BX \ AX = {x′ } for some x, x′ ∈ X , x ̸ = x′ . The coloring (10) of the sets A, B ∈ Rr with |A ∩ B| = k − 1 leads to y ≡ y′ (mod k + 2) or x ≡ x′ (mod k + 2), and the results (13) and (14) follow, respectively. □ Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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Lemma 4. Let r ̸ = 2. If ABA′ and CDC ′ are paths in the induced subgraph K [Rr ] of Kneser graph where A, A′ , B, C , C ′ , D ∈ Rr , then none of BC , BC ′ , DA, or DA′ is an edge in K 2 [Rr ]. Proof. Lemma 2 implies that there is no edge of K [Rr ] between ABA′ and CDC ′ . By symmetry, we will show that BC is not an edge in (K 2 \ K )[Rr ]. Suppose that BC is an edge in (K 2 \ K )[Rr ] so that |B ∩ C | = k − 1. Let y ∈ Y be the missing element on the disjoint pairs in Rr , and let (B ∪ A)c = (D ∪ C )c = {y}, (B ∪ A′ )c = (D ∪ C ′ )c = {ˆ y} without loss of generality. Since CY ∋ ˆ y, it follows that |CY ∩ BY | = k/2 − 1 and CX = BX , consequently C ∩ A′ = ∅, that is, A′ C ∈ E(K [Rr ]), a contradiction. □ Lemma 5. For every r, the cycles in (K 2 \ K )[Rr ] are even. Proof. Let A1 A2 . . . Aℓ , ℓ ≥ 3 be a cycle in (K 2 \ K )[Rr ] so that Ai ∈ Rr with |Ai ∩ Ai+1 | = k − 1 for 1 ≤ i ≤ l (Aℓ+1 = A1 ). We may assume that Ai ̸ = Aj for 1 ≤ i ̸ = j ≤ ℓ. Starting with A1 , Lemma 3 tells us that A2 is obtained by an iteration A2 = (A1 \ {a1 })∪{aˆ1 } for some a1 ∈ A1 and someˆ a ̸ ∈ A1 where the set of pairs {a1 , aˆ1 } is either Xp for some p = 2, 4, . . . , k−2 or Yq for some q = 1, 3, . . . , k − 1. We continue this process, and let Ai+1 = (Ai \ {ai }) ∪ {ˆ ai } for some {ai , ˆ ai } that is either Xp or Yq , for some p ̸ = k, k + 2, q ̸ = k + 1. We label the edge Ai Ai+1 by the unique element ai for each i. Observe that, for every i, exactly one of ai or ˆ ai belongs to indeed A1 . Let a ∈ A1 be one of ai , 1 ≤ i ≤ ℓ, and let i1 < i2 < · · · < im be the subsequence of the sequence 1, 2, . . . , ℓ such that a} and ai ̸ ∈ {a,ˆ a} if i1 , i2 , . . . , im are all the indices over which the edges are labeled by a or ˆ a, that is, ai1 , ai2 , . . . , aim ∈ {a,ˆ i ̸ = i1 , i2 , . . . , im . We may set ai1 = a without loss of generality. Then, after the i1 th iteration, none of Ai1 +1 , Ai1 +2 , . . . , Ai2 a. Similarly, ˆ a ̸ ∈ Ai2 +1 ∪ Ai2 +2 ∪ · · · ∪ Ai3 , contains a, and all of them contain ˆ a. Therefore the element ai2 should be ˆ a alternate a ∈ Ai2 +1 ∩ Ai2 +2 ∩ · · · ∩ Ai3 , and then ai3 = a. The process continues in the same fashion, therefore a and ˆ a} \ aim stays after the im th iteration and belongs to A1 while in the sequence ai1 , ai2 , . . . , aim . Observe that the element {a,ˆ aim ̸ ∈ A1 . It has to be that aim = ˆ a and m is even. The edges of the cycle are partitioned according to the labels a or ˆ a for some a ∈ A1 . Therefore, the length ℓ is a multiple of some even integer m. □ Lemma 6. 1. For each r ̸ = 2, the subgraph K 2 [Rr ] has a proper 3-coloring. 2. The subgraph K 2 [R2 ] has a proper 4-coloring. Proof. 1. Now we consider the case that r ̸ = 2. Let {Ai Bi A′i }i∈Λ be the collection of paths in K [Rr ], r ̸ = 2. Let G′ [Rr ] be the subgraph of K 2 [Rr ] removing the edges Ai Bi , Bi A′i , i ∈ Λ, of K [Rr ]. Then G′ [Rr ] is a subgraph of (K 2 \ K )[Rr ], and Lemma 5 implies that G′ [Rr ] is a bipartite graph, say with the colors α and β . Set Λα := {i ∈ Λ| Bi is colored with α} and Λβ := {i ∈ Λ| Bi is colored with β} so that Λ = Λα ∪ Λβ . Once we put back the edges Ai Bi , Bi A′i , i ∈ Λ, one of the two edges Ai Bi or Bi A′i is monochromatic in K 2 [Rr ]. We recolor some of the vertices Ai , Bi , A′i , i ∈ Λ, as follows.

• If i ∈ Λα , recolor Bi with a third color, say γ . • If i ∈ Λβ , turn the color of β assigned to one of Ai or A′i into γ . We claim this recoloring gives a proper coloring in K 2 [Rr ]. Since no vertex of Rr \∪i {Ai , Bi , A′i } changes its color, there is no monochromatic edge in the induced subgraph G[Rr \∪i {Ai , Bi , A′i }]. Furthermore, a vertex in Rr \∪i {Ai , Bi , A′i } would not be adjacent to a vertex in ∪i {Ai , Bi , A′i } of the same color because a vertex in ∪i {Ai , Bi , A′i } either keeps its original color determined under the proper coloring of G′ [Rr ] or has changed to the third color γ . We need to show that there is no monochromatic edge in the induced subgraph K 2 [∪i {Ai , Bi , A′i }]. For each i ∈ Λ, the vertices Ai , Bi , A′i are colored with α, β, γ , one each, hence no monochromatic edge within K 2 [{Ai , Bi , A′i }]. If Bi Bj , i ̸ = j, is an edge, it should be in (K 2 \ K )[Rr ] by Lemma 2, therefore one of Bi or Bj was colored with α and the other with β originally, and now with γ and β after recoloring. If Ai is adjacent to Aj or A′j where i ̸ = j, it occurs in G′ [Rr ]. Since Ai A′i is an edge in G′ [Rr ] and G′ [Rr ] is bipartite, at most one of Ai Aj or Ai A′j is an edge; otherwise Ai , Aj , A′j would form a triangle in G′ [Rr ]. Say Ai Aj is an edge. If neither of Ai nor Aj changes its colors under recoloring, they keep their different colors from the original proper coloring of G′ [Rr ]. If one of Ai or Aj has changed its color, one end is in the color γ and the other end in the colors {α, β}. If both Ai and Aj were recolored in γ , it would mean that i, j ∈ Γβ and both Ai and Aj had received the color β under the initial proper coloring of G′ [Rr ], which would be impossible because Ai Aj is an edge in G′ [Rr ]. For the pairs Bi , Aj or Bi , A′j for i ̸ = j, Lemma 4 gives us that they are not adjacent in G. 2. Both the subgraphs K [R2 ] and (K 2 \ K )[R2 ] are bipartite. For A ∈ R2 , the coloring by ordered pair (α (A), β (A)), where the α (A) and β (A) are proper 2-colorings for K [R2 ] and (K 2 \ K )[R2 ], respectively, yields a proper 4-coloring for K 2 [R2 ] = (K 2 \ K )[R2 ] ∪ K [R2 ]. □ Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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Remark. We can also show that the cycles in the subgraph K 2 [R2 ] have even lengths, which implies that K 2 [R2 ] is 2-colorable. The proof is a more delicate application of the argument in the proof of Lemma 5. However this increases the length of the paper by two pages (including the corresponding Lemma 11 for odd k) with only minor improvement to the additive constant of the bound, hence we settle for the 4-coloring of K 2 [R2 ]. Applying the results of Lemma 6 to the color classes in Lemma 1 yields a proper coloring of K 2 [Vk/2 ]: Theorem 3. The subgraph K 2 [Vk/2 ] has a proper coloring using

3 k 2

+ 3 colors when

k 2

is odd, and 32 k + 4 colors when

k 2

is even.

Proof. 3(k/2 + 1) = 3k/2 + 3 and 3(k/2 + 1 − 1) + 4 = 3k/2 + 4. □ 4. Proof: k is odd ≥ 3 The proof goes almost identically to the even case with slight modifications. We will point out the differences without details if the arguments are tedious modifications. ( ) The ( details ) will be given when necessary. Let V0 :=

(

X k−t ′

)

⊕

(Y ) t′

(X ) k

(Y ) 0

X (k−1)/2

and V(k+1)/2 :=

⊕

Y (k+1)/2

. For 1 ≤ t ̸ = t ′ ≤ k, we pair up the subsets

(

X k−t

)

⊕

(Y ) t

and

′

of V satisfying t + t = k + 1 so as to set

( Vt :=

⊕

X

)

k−t

⊕

( )⋃( ) ( ) Y X Y ⊕ , for 1 ≤ t ≤ (k − 1)/2. t t −1 k+1−t

Now then V = ∪0≤t ≤(k+1)/2 Vt , a disjoint union of Vt ’s. It is trivial to see that K 2 [V0 ] consists of an isolated vertex, namely X . There is no edge of K 2 (2k+1, k) between Vt and Vt ′ for |t − t ′ | ≥ 2, therefore the subgraphs K 2 [V(k+1)/2−(2i−1) ], i ≥ 1, share the same colors, and the subgraphs K 2 [V(k+1)/2−2i ], i ≥ 0, share the same colors. We will show that each subgraph K 2 [Vt ] is properly colorable with k + 1 colors for 1 ≤ t ≤ (k − 1)/2 ( Theorem 4), and K 2 [V(k+1)/2 ] is with 3k/2 + c2 colors where c2 = 1.5 or 3.5 ( Theorem 5). Therefore, for odd k ≥ 3,

χ (K 2 (2k + 1, k)) ≤ (k + 1) + ≤

5 2

(

3 2

) k + c2

k + c,

where c = 2.5 or 4.5. 4.1. Coloring of K 2 [Vt ], for 1 ≤ t ≤ (k − 1)/2 Theorem 4. Each subgraph K 2 [Vt ] has a proper coloring using k + 1 colors for 1 ≤ t ≤ (k − 1)/2. Proof. The mapping φ : K 2 [Vt ] → H(X , {t , t − 1})□H(Y , {t }) defined by, for A := AX ⊕ AY ∈ Vt ,

(

( ) Y × , k−t t t t ( ) ( ) ( ) ( ) X Y X Y × AX ⊕ AY ∈ ⊕ ↦→ (AX , AY ) ∈ t −1 t t −1 k+1−t AX ⊕ AY ∈

X

)

( ) Y

⊕

↦→ (AX , AY ) ∈

( ) X

is a graph homomorphism (indeed an isomorphism) with a proof similar to Theorem 2. Here, the pairs of the disjoint vertices in Vt correspond to s := k − t , s′ = k − t ′ with s + s′ = k − 1 in (5). The inequality of the upper bound

( ) χ H(X , {t , t − 1})□H(Y , {t }) ≤ k + 1 follows. □ 4.2. Coloring of K 2 [V(k+1)/2 ] In order to find a proper coloring for the subgraph K 2 [V(k+1)/2 ], we set X to be the even numbers and Y to be the odd numbers of [2k + 1] as before. For A ∈ V(k+1)/2 , using the even number k + 1, we consider the pre-coloring f (A) = ∑ Set Rr := {A ∈ V(k+1)/2 | f (A) ≡ r (mod k + 1)} where 0 ≤ r ≤ k. Since |AX | = k−2 1 and |AY | = k+2 1 , a∈A a (mod k + 1).∑ ∑ k+1 k+1 the coloring f (A) = a∈A a + a∈A a (mod k + 1) produces 2 colors classes Rr with r having the same parity as 2 X

(necessarily the opposite parity of

Y

k−1 ). 2

Lemma 7. The coloring f introduces

k+1 2

color classes.

Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.

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Analogous to (11), the pair of disjoint sets A, B ∈ Rr and the element x ∈ X such that (A ∪ B)c = {x} lead to the congruence 2r = −x

(mod k + 1).

(15)

Set X0 := {k + 1} and Xp := {p, p + (k + 1)} for even p, 2 ≤ p ≤ k − 1, and the pair x,ˆ x ∈ X denote the relation that {x,ˆ x} = X p for some even p ̸ = 0. For a given r, the linear congruence (15) has one of Xp as the solution set. Note that, when k+2 1 is even, the residues r = 0, k+2 1 if and only if x := k + 1 is the solution. Also set Yq := {q, q + (k + 1)} for odd q, 1 ≤ q ≤ k and {y,ˆ y} = Yq for some y,ˆ y ∈ Y. We have the following Lemmas 8–12 and Theorem 5 analogous to Lemmas 2–6 and Theorem 3. The proofs of Lemmas 8– 10 are routine adjustments between x,ˆ x and y,ˆ y in the arguments of Lemmas 2–4. The proofs of Lemmas 11–12 go along the lines of the proofs of Lemmas 5–6. Lemma 8. 1. For r ̸ = 0, k+2 1 , the induced subgraph K [Rr ] of Kneser graph consists of vertex–disjoint paths of length 2, if it contains an edge, and possibly isolated vertices. Furthermore, if ABA′ is a path in K [Rr ], where A, A′ , B ∈ Rr , then AY = A′Y = BY , X \ (BX ∪ AX ) = {x}, X \ (BX ∪ A′X ) = {ˆ x} for some x ∈ X , x ̸ = k + 1, such that r , x satisfy the congruence (15). 2. When k+2 1 is even, the induced subgraphs K [R0 ] and K [R(k+1)/2 ] of Kneser graphs, if not empty, consist of a matching and possibly isolated vertices. Lemma 9. For every r, let A, B ∈ Rr . The pair is of |A ∩ B| = k − 1. Then either AY = BY , and AX \ BX = {x}, BX \ AX = {ˆ x} for some x ∈ X , x ̸ = k + 1 or AX = BX , and AY \ BY = {y}, BY \ BX = {ˆ y} for some y ∈ Y . Lemma 10. Let r ̸ = 0, k+2 1 . If ABA′ and CDC ′ are paths in the induced subgraph K [Rr ] of Kneser graph where A, A′ , B, C , C ′ , D ∈ Rr , then none of BC , BC ′ , DA, or DA′ is an edge in K 2 [Rr ]. Lemma 11. For every r, the cycles in (K 2 \ K )[Rr ] are even. Lemma 12. 1. For each r ̸ = 0, k+2 1 , the subgraph K 2 [Rr ] has a proper 3-coloring. 2. The subgraphs K 2 [R0 ] and K 2 [R(k+1)/2 ] have proper 4-colorings. Theorem 5. The subgraph K 2 [V(k+1)/2 ] has a proper coloring using is even. Proof. 3( k+2 1 ) =

3 k 2

+

3 2

and 3( k+2 1 − 2) + 2 · 4 =

3 k 2

3 k 2

+

3 2

colors when

k+1 2

is odd, and 23 k +

7 2

colors when

k+1 2

+ 72 . □

Remarks. We can show that all the cycles in K 2 [R0 ] and K 2 [R(k+1)/2 ] are even by a more delicate argument, References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17]

N. Alon, B. Mohar, The chromatic number of graph powers, Combin. Probab. Comput. 11 (2002) 1–10. I. Bárány, A short proof of Kneser’s conjecture, J. Combin. Theory Ser. A 25 (1978) 325–326. G.J. Chang, D. Kuo, The L(2,1)-labelling problem on graphs, SIAM J. Discrete Math. 9 (1996) 309–316. J.-Y. Chen, K.-W. Lih, 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. R. Erman, S. Jurecic, D. Kral, K. Stopar, N. Stopar, Optimal real number graph labellings of a subfamily of Kneser graphs, SIAM J. Discrete Math. 23 (2009) 1372–1381. D. Goncalves, On the L(p, 1)-labeling of graphs, in: Proc. European Conference on Compbinatorics, Graph Theory, and Applicaitons, 2005, pp. 81–86. J.R. Griggs, R.K. Yeh, Labelling graphs with a condition at distance 2, SIAM J. Discrete Math. 5 (1992) 586–595. T.R. Jensen, B. Toft, Graph Coloring Problems, John Wiley & Sons, Inc., New York, 1995. J.-H. Kang, L(2, 1)-labeling of Kneser graphs. Ph.D. thesis, University of Illinois at Urbana-Champaign, 2004. J.-H. Kang, L(2,1)-labeling for Hamiltonian graphs of maximum degree 3, SIAM J. Discrete Math. 22 (2008) 213–230. A. Khodkar, D. Leach, The chromatic number of K 2 (9, 4) is 11, J. Combin. Math. Combin. Comput. 70 (2009) 217–220. S.-J. Kim, K. Nakprasit, On the chromatic number of the square of the Kneser graph K (2k + 1, k), Graphs Combin. 20 (2004) 79–90. S.-J. Kim, B. Park, Improved bounds on the chromatic numbers of the square of Kneser graphs, Discrete Math. 315 (2014) 69–74. S.-J. Kim, B. Park, Coloring of the square of Kneser graph K (2k + r , k), Graphs Combin. 32 (2016) 1461–1472. D. Kral, R. Skrekovski, A theorem about the channel assignment problem, SIAM J. Discrete Math. 16 (2003) 426–437. L. Lovász, Kneser’s conjecture, chromatic number, and homotopy, J. Combin. Theory Ser. A 25 (1978) 319–324. M. Molloy, M. Salavatipour, A bound on the chromatic number of the square of a planar graph, J. Combin. Theory Ser. B 94 (2005) 189–213.

Please cite this article in press as: J. Kang, A new approach to the chromatic number of the square of Kneser graph K (2k + 1, k), Discrete Mathematics (2017), http://dx.doi.org/10.1016/j.disc.2017.08.008.