Discrete Mathematics On b-coloring of the Kneser graphs

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4404 R. Javadi, B. Omoomi / Discrete Mathematics 309 (2009) 4399–4408 Case 2. n is odd. First we find an upper bound for ϕ(K(n, 2)). Let c be a b-coloring of K(n, 2) by ϕ = ϕ(K(n, 2)) colors and t starlike color classes with centers 1, . . . , t of sizes n 1 , . . . , n t , respectively. Then, ( n |V(K(n, 2))| = = 2) t∑ n i + 3(ϕ − t). (4) i=1 By Lemma 1, every 2-subset of the set {1, 2, . . . , t} is in the color classes 1 to t. Moreover, in the color class i we must have a b-dominating vertex, say {i, x}, where x ∈ {t + 1, t + 2, . . . , n}. Hence, t∑ t(t − 1) t(t + 1) n i ≥ + t = . 2 2 i=1 Therefore, ( n ) t(t + 1) ≥ 3ϕ + − 3t = 3ϕ + 2 2 So n(n − 1) t(t − 5) ϕ ≤ − . 6 6 t(t − 5) . 2 The minimum of the expression t(t − 5) occurs in t = 2 and t = 3, so ϕ ≤ n(n−1) + 1. 6 Now we prove that ϕ ≤ n(n−1) . Suppose ϕ = n(n−1) + 1, hence, t = 2 or t = 3. For every vertex i ∈ V(K 6 6 n ), the number of edges incident to i in the triangles of the decomposition is even. Since n is odd, the number of edges incident to i in the stars of the decomposition is also even. Equivalently, in the b-coloring of K(n, 2) the number of vertices containing i in the starlike color classes are even numbers. If t = 3 then by Lemma 1 (ii) and (iii), the vertices {1, 2}, {1, 3} and {2, 3} in K(n, 2) are in the starlike color classes with centers 1, 2, or 3 and for every i, 1 ≤ i ≤ 3, there is a vertex {i, x} in the starlike color classes which x ≠ 1, 2, 3. So by the discussion above, for every i, 1 ≤ i ≤ 3, at least two vertices {i, x} and {i, y}, where x, y ≠ 1, 2, 3, are in the starlike color classes. Therefore, ∑ 3 i=1 n i ≥ 3 + 2 × 3 = 9. So by Relation (4), ( n 2 ) ≥ 9 + 3(ϕ − 3) = 3ϕ. Hence, ϕ ≤ n(n−1) , which 6 contradicts our assumption. Now let t = 2. By Lemma 1 (ii) and (iii), the starlike color class with center 1 contains vertex {1, 2} and at least one more vertex, say {1, 3}. By the discussion above, if the vertex {1, i} in K(n, 2) is in the starlike color class with center 1, then the vertex {2, i} is in the starlike color class with center 2. If the vertices {1, 2}, {1, 3} and {2, 3} are the only vertices in the starlike color classes, then there is no b-dominating vertex in these classes. Therefore, the starlike color class with center 1 and consequently, the starlike color class with center 2 each one contains at least more two vertices. Hence, ∑ 2 i=1 n i = 1 + 2 × 3 = 7. Therefore, by Relation (4) ( n ) ≥ 7 + 3(ϕ − 2) = 3ϕ + 1. 2 So ϕ ≤ n(n−1) , which contradicts our assumption. 6 ⌊ ⌋ n(n−1) Therefore, ϕ ≤ . If n ≡ 1, 3(mod 6) then an STS(n) exists. Therefore, by Theorem 2, ϕ ≥ n(n−1) . If n ≡ 5(mod 6) 6 6 ⌊ ⌋ then by Proposition 3, ϕ ≥ n(n−1) − 1 . Hence, ϕ = n(n−1) . □ 6 3 6 Since the Petersen graph is Kneser graph K(5, 2), we get the following result. Corollary 1. If P is the Petersen graph, then ϕ(P) = 3. Kneser graph K(8, 2) is an exception. Proposition 4. ϕ(K(8, 2)) = 9. Proof. Consider the notations in the proof of Theorem 3 for Case 1. By Inequality (3), we have ϕ(K(8, 2)) ≤ 10 and the equality holds if and only if t = 4 or t = 5. Assume that a b-coloring of K(8, 2) exists with 10 colors and A 1 , A 2 , . . . , A t are starlike color classes with centers 1, 2, . . . , t, respectively. If t = ∑ 4 4 then by Equality (2), i=1 n i = 10. By Lemma 1 (ii) and (iii), every 2-subset of the set {1, 2, 3, 4} is in ∪ 4 i=1 A i and for each i, 1 ≤ i ≤ 4, there exists x i ∉ {1, 2, 3, 4}, where {i, x i } ∈ A i . On the other hand n − t and the number of vertices containing i in triangular color classes are even numbers. So there are at least two vertices {i, x i }, {i, y i } in the starlike color classes, where x i , y i ∉ {1, 2, 3, ∑ 4 4}. Hence, i=1 n i = 10 ≥ 6 + 4 × 2 = 14, which is contradiction.
R. Javadi, B. Omoomi / Discrete Mathematics 309 (2009) 4399–4408 4405 If t = ∑ 5 5 then by Equality (2), i=1 n i = 13. On the other hand, similar to the above by Lemma 1 (ii) and (iii), ∑ 5 i=1 n i = 13 ≥ 10 + 5, a contradiction. So ϕ(K(8, 2)) ≤ 9. Now we provide a b-coloring of K(8, 2) by 9 colors. First we consider an STS(7) and delete one point of it. What remains is a decomposition of K 6 into 4 triangles and a 1-factor called F = {{a 1 , b 1 }, {a 2 , b 2 }, {a 3 , b 3 }}. Now we add two new points a and b and define the color classes as all triangles in the decomposition above in addition to the triangular color classes {a, a 1 , b 1 }, {a, a 2 , b 2 } and {b, a 3 , b 3 } and the starlike color classes {{a, a 3 }, {a, b 3 }, {a, b}} and {{b, a 1 }, {b, b 1 }, {b, a 2 }, {b, b 2 }}. This is a proper coloring of K(8, 2) satisfying the conditions of Lemma 1, so is a b-coloring by 9 colors as desired. ( By Relation (1), ϕ(K(n, k)) ≤ ∆ + 1 = the following conjecture. Conjecture 1. For every integer k, we have ϕ(K(n, k)) = Θ(n k ). 4. b-continuity of the Kneser graph K(n, 2) In this section we prove that K(n, 2) is b-continuous when n ≥ 17. n−k k □ ) + 1. Hence ϕ(K(n, k)) = O(n k ). Theorems 2 and 3 motivate us to propose Lemma 2. (a) Let n = 6k + 1 or n = 6k + 3 and (S, B) be an STS(n). Also let T be a subset of S = {1, 2, . . . , n} and t be the number of blocks in B on the points of T , such that: (i) |T| = m ≥ 3, (ii) for each i ∈ T , there exists j ∈ T such that the third point of the block containing both i, j is not in T . Then there exists a b-coloring of K(n, 2) by ϕ − ( m(m−3) − 2t) colors, where ϕ = ϕ(K(n, 2)). 2 (b) Let n = 6k + 5 and (S, B) be a PBD(n) with one block of size 5, say {1, 2, n, n − 1, n − 2} and the others 3-blocks. Also let T be a subset of S = {1, 2, . . . , n} and t be the number of 3-blocks in B on the points of T , such that: (i) |T| = m ≥ 3, (ii) 1, 2 ∈ T and n − 2, n − 1, n ∉ T , (iii) for each i ∈ T , i ≠ 1, 2, there exists j ∈ T such that the third point of the 3-block containing both i, j is not in T . Then there exists a b-coloring of K(n, 2) by ϕ − ( m(m−3) − 2t + 1) colors, where ϕ = ϕ(K(n, 2)). 2 Proof. Let c be the b-coloring of K(n, 2) by ϕ colors corresponding to STS(n) or PBD(n) (see Theorem 2 and Proposition 3). In the case n = 6k + 5, we take the centers of starlike color classes as 1 and 2. Assume T = {1, 2, . . . , m}, consider the b-coloring c and delete all triangular color classes containing a vertex {i, j} ⊆ T . (a) Since each vertex {i, j} ⊆ T is contained in a triangular color class and there are exactly t triangles on the points of T , the number of deleted color classes (triangles) is m(m−1) − 3t + t. Now we define m new color classes as follows. New 2 color class i, 3 ≤ i ≤ m − 2, contains the set of vertices {{i, j} | i + 1 ≤ j ≤ m}. Also new color classes 1, 2, m − 1 and m contain respectively the sets {{1, j} | 2 ≤ j ≤ m − 2}, {{2, j} | 3 ≤ j ≤ m − 1}, {{m − 1, m}, {m − 1, 1}} and {{m, 1}, {m, 2}}. Moreover, if a vertex {i, x}, where i ∈ T and x ∉ T is in a deleted color class, then we add this vertex to the color class i. These m new color classes together with the old color classes give us a new proper coloring of K(n, 2) by ϕ − ( m(m−1) 2 − 2t) + m colors. (b) Since each vertex {i, j} ⊆ T except {1, 2} is contained in a triangular color class and there are exactly t triangular color classes on the points of T , the number of deleted triangles is m(m−1) 2 − 1 − 3t + t. Now we define m − 2 new color classes as follows. Color class i, 3 ≤ i ≤ m, contains the set of vertices {{i, j} | i + 1 ≤ j ≤ m} ∪ {{i, 1}, {i, 2}}. Moreover, if a vertex {i, x}, where i ∈ T and x ∉ T is in a deleted color class, then we add this vertex to the color class i. These m − 2 new color classes together with the old color classes give us a new proper coloring by ϕ − ( m(m−1) 2 − 1 − 2t) + m − 2 colors. The obtained colorings in (a) and (b) satisfy the conditions of Lemma 1, so they are b-colorings. Lemma 3. Let n ≥ 13 be an odd integer and let k = ⌊ n ⌋. For every odd integer m, 5 ≤ m ≤ k + 5 and for every integer t, 6 0 ≤ t ≤ 3m−11 , where (m, t) ≠ (5, 2), (7, 5), (k + 5, 0), there exists an STS(n) or PBD(n) and a set T satisfying the conditions 2 of Lemma 2. Proof. Let l = ⌊ n ⌋. Depending on n, using the Bose construction, the Skolem construction or the 6k + 5 construction given 3 in Section 2 and the quasigroups of Example 1, construct an STS(n) or a PBD(n). If t = 0, then it is easy to find a set T with parameters (m, t). Assume 5 ≤ m ≤ k + 5 and m is odd. (a) If 1 ≤ t ≤ m−5 , then define 2 T = {l 1 , i 1 , (l − i) 1 | 1 ≤ i ≤ t} ∪ {j 1 | t + 1 ≤ j ≤ m − 4 − t} ∪ {(σ (l)) 2 , 1 3 , (σ −1 (k + 2) − 1) 3 }. □
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Discrete Mathematics On b-coloring of the Kneser graphs

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