Improved Bounds for the Chromatic Number of the Lexicographic Product of Graphs

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1 Improved Bounds for the Chromatic Number of the Lexicographic Product of Graphs Roland Kaschek University of Klagenfurt Department of Computer Science 9020 Klagenfurt, Austria Sandi Klavžar University of Maribor PF, Koroška cesta Maribor, Slovenia Abstract An upper bound for the chromatic number of the lexicographic product of graphs which unifies and generalizes several known results is proved. It is applied in particular to characterize graphs that have a complete core. An improved lower bound is also given. 1 Introduction We continue the investigation of the chromatic number of the strong and the lexicographic product of graphs. Geller, Stahl and Vesztergomby worked on the problem in the seventies [2, 3, 16, 17, 18]. Recently, several new motivations for studying these chromatic numbers appeared. The most important recent result is due to Feigenbaum and Schäffer [1]. They proved that the strong product admits a polynomial algorithm for decomposing a given connected graph into its factors. In [11] bounds for the chromatic number of the lexicographic product of graphs are used to study approximation algorithms. Results about the chromatic number of strong products turned out to be important in understanding retracts of strong products [7, 8, 10]. This work was supported in part by the Ministry of Science and Technology of Slovenia under the grant P

2 All graphs considered in this paper will be undirected, finite and simple graphs, i.e. graphs without loops or multiple edges. If G is a graph and X V (G), then the subgraph of G induced by X will be denoted by <X>. An n coloring of a graph G is a function f : V (G) {1, 2,..., n}, such that xy E(G) implies f(x) f(y). The smallest number n for which an n coloring exists is the chromatic number χ(g) of G. The size of a largest complete subgraph of a graph G will be denoted by ω(g) and the size of a largest independent set by α(g). Clearly, ω(g) χ(g). The strong product G H of graphs G and H is the graph with vertex set V (G) V (H) and (a, x)(b, y) E(G H) whenever ab E(G) and x = y, or a = b and xy E(H), or ab E(G) and xy E(H). The lexicographic product G[H] of graphs G and H is the graph with vertex set V (G) V (H) and (a, x)(b, y) E(G[H]) whenever ab E(G), or a = b and xy E(H). Whenever possible we shall denote the vertices of the first factor by a, b, c,... and the vertices of the other factor by x, y, z. The main contribution of this paper is an upper bound for the chromatic number of the lexicographic product of graphs (and hence of the strong product) which unifies and generalizes upper bounds from [3, 8, 9]. prove this upper bound in the next section. In Section 3 we give several consequences of the bound. It is shown in particular that if G is a χ critical graph, then for any graph H, χ(g[h]) χ(h) (χ(g) 1) + χ(h) α(g). We also characterize graphs that have a complete core. In the last section we extend a lower bound from [3, 10, 16]. We 2

3 2 An upper bound Note first that G[K n ] is isomorphic to G K n. Since furthermore, χ(g[h]) = χ(g[k n ]), if χ(h) = n (see [3, Theorem 3]), the results we obtain can also be applied to the strong product of graphs. This is our main result. Theorem 1 Let G and H be any graphs, χ(h) = m. Let {X i } i {1, 2,..., k} be a partition of a set X V (G). Let χ(g X i ) = n i, i {1, 2,..., k} and χ(<x>) = s. Then χ(g[h]) (n 1 + n n r ) m k + (n r+1 + n r n k + s) m k + χ(<x 1 X 2 X r >), where m = pk + r, 0 r < k. Proof. Let h be an m coloring of H and let f i be an n i coloring of G X i, i {1, 2,..., k}. Furthermore, for j {1, 2,..., m} write j = p j k + r j, where 1 r j k. Finally, let g be an s coloring of the graph <X>. For (a, x) V (G[H]) define: f(a, x) = { (g(a), ph(x), 0); a X, rh(x) (f rh(x) (a), p h(x), r h(x) ); otherwise. To prove that f is a proper coloring of G[H] let (a, x)(b, y) E(G[H]). We claim that f(a, x) f(b, y). There are four cases to consider: (i) a X rh(x), b X rh(y), (ii) a X rh(x), b X rh(y), (iii) a X rh(x), b X rh(y), (iv) a X rh(x), b X rh(y). 3

4 The claim is clearly true in the cases (ii) and (iii). Suppose ab E(G). Then g(a) g(b), hence the claim is trivially true in case (i). Consider now case (iv) and assume (p h(x), r h(x) ) = (p h(y), r h(y) ). It follows that h(x) = h(y) and thus a and b are in X X rh(x). Since ab E(G) we conclude that f rh(x) (a) f rh(x) (b). Suppose next a = b. Then xy E(H) and thus h(x) h(y), which in case (iv) immediately implies f(a, x) f(a, y). And in the case (i) we have r h(x) = r h(y) and thus p h(x) p h(y), which again implies f(a, x) f(b, y). From here we conclude that the claim is true. In the rest of the proof we show that f gives us the desired upper bound. Let {C 1, C 2,..., C m } be the color classes induced by h. It is easy to see that on the graph < V (G) (C 1 C 2... C pk ) > the image of the map f has p (n 1 + n n k + s) (1) elements, while on the graph G =< V (G) (C pk+1 C pk+2... C pk+r ) > it has n 1 + n n r + χ(< X 1 X 2 X r >) (2) elements. Summing (1) and (2) gives us the size of the image of f. If r = 0 then G is empty and therefore (1) gives us the size of f. But if r > 0 then p + 1 = m k and p = m k, which gives us together with (1) and (2) the desired result. One can extend the definition of the lexicographic product in the following way (see [15]). Let G be a graph and let H = {H a } a V (G) be a family of graphs. Then L = G[H] is the graph defined by V (L) = {(a, x) a V (G) and x V (H a )} E(L) = {(a, x)(b, y) ab V (G) or a = b and xy E(H a )}. 4

5 Clearly, if for every a V (G), H a is isomorphic to a graph H, then L is isomorphic to the lexicographic product G[H]. We can extend Theorem 1 a little more: Corollary 2 Let G be a graph and let H = {H a } a V (G) be a family of graphs. Let {X i } i {1, 2,..., k} be a partition of a set X V (G) and let χ(g X i ) = n i, i {1, 2,..., k}. If max{χ(h a ) a V (G)} = m and χ(<x>) = s, then χ(g[h]) (n 1 + n n r ) m k + (n r+1 + n r n k + s) m k + χ(<x 1 X 2 X r >), where m = pk + r, 0 r < k. Proof. Follows from the result that if χ(h) = m then χ(g[h]) = χ(g[k m ]) (see [3]) and an observation that χ(g[h]) χ(g[k m ]). 3 Applications of the upper bound A graph G is called χ critical if χ(g x) < χ(g) for every x V (G). We have: Corollary 3 [9, Theorem 4] If G is a χ critical graph, then for any graph H, χ(g[h]) χ(h) (χ(g) 1) + χ(h) α(g). Proof. Let χ(g) = n and let χ(h) = m. Let α(g) = k and let X = {x 1, x 2,..., x k } be an independent set of G. Let m = pk+r, 0 r < k. For 5

6 i {1, 2,..., k} set X i = {x i }. Clearly, χ(<x>) = 1 and χ(g X i ) = n 1, i {1, 2,..., k}. If r = 0 then if follows from Theorem 1 χ(g[h]) (k (n 1) + 1) p = m (n 1) + p. Suppose now r > 0. Then we have: χ(g[h]) r (n 1) m k + ((k r) (n 1) + 1) m k + 1 = (n 1) (r (p + 1) + (k r) p) + (p + 1) = m (n 1) + m k, and the proof is complete. Let G be a graph and let V (G) = {a 1, a 2,..., a p }. We construct the graph M(G) from G by adding p + 1 new vertices b 1, b 2,..., b p, b. The vertex b is joined to each vertex b i and the vertex b i is joined to every vertex to which a i is adjacent in G. The graph M(P 5 P 3 ) is shown on Fig Figure 1: The graph M(P 5 P 3 ) The construction is due to Mycielski [12] (and hence our notation). From now on we write MG instead of M(G). It is not hard to see that if G is 6

7 n chromatic, χ critical and triangle free, then M G is (n + 1) chromatic, χ critical and triangle free (see [4, p.277, Exercise 2]). If the Mycielski procedure is applied n times consecutively, the final graph will be denoted by M n G. Also, let M n G[H] denotes (M n G)[H]. It is shown in [8, Theorem 1] that for a bipartite graph H, χ(m n C 5 [H]) 2n + 4, n 1. However, this upper bound does not follow from Corollary 3, although the graphs M n C 5 are χ critical. In fact, Corollary 3 gives us χ(m n C 5 [H]) 2n + 5. We are going to show that 2n + 4 follows from Theorem 1 and moreover, the result can be generalized as follows. Theorem 4 Let H be a graph, χ(h) = m. Then for n 1 and l 2, χ(m n C 2l+1 [H]) { (n + 2) m; m is even, (n + 2) m + 1; m is odd. Proof. Throughout the proof let C denotes the cycle C 2l+1. We first prove the theorem for n = 1. Let a 0, a 1,..., a 2l be consecutive vertices of C and let V (MC) = {a 0, a 1,..., a 2l, b 0, b 1,..., b 2l, b}. Define: X 1 = {a 0, a 2,..., a 2l 2, a 2l 1 } and X 2 = {a 1, a 3,..., a 2l 3, b 1, b}. We claim that χ(mc X 1 ) = χ(mc X 2 ) = χ(<x 1 X 2 >) = 2. It is easy to check that {a 1, a 3,..., a 2l 3, a 2l, b} {b 0, b 1,..., b 2l }. is a bipartition of the graph MC X 1 and {a 0, a 2l 1, b 0, b 3, b 5,..., b 2l 1 } {a 2, a 4,..., a 2l, b 2, b 4,..., b 2l } 7

8 a bipartition of the graph MC X 2. Finally, {a 0, a 2,..., a 2l 2, b} {a 1, a 3,..., a 2l 1, b 1 } is a bipartition of the graph <X 1 X 2 >. The claim is proved. We now apply Theorem 1 letting X = X 1 X 2. For r = 0 we have χ(mc[h]) ( ) m 2 = 3m, and for r = 1 we obtain χ(mc[h]) 2 m m = 2(p + 1) + 4p + 2 = 6p + 4 = 3(2p + 1) + 1 = 3m + 1. This proves the theorem for n = 1. Let n 1 and suppose that m is even. Let f be an ((n + 2) m) coloring of M n C[H]. Let V (M n C) = {a 1, a 2,..., a p } and let b 1, b 2,..., b p, b be the corresponding vertices in the graph M n+1 C. We extend f to a coloring of M n+1 C[H] in the following way. Set f(b i, x) = f(a i, x), i {1, 2,..., p}, and x V (H). In addition, color vertices of the subgraph < {b} V (H) > with m new colors. It is straightforward to verify that we have a proper coloring with (n+3)m colors. If m is odd we proceed in the same way to get an ((n + 3)m + 1) coloring. Corollary 5 [8, Theorem 1] For n 1, χ(m n C 5 [H]) 2n + 4, for any bipartite graph H. A subgraph R of a graph G is a retract of G if there is a homomorphism (an edge preserving map) r : V (G) V (R) with r(x) = x, for all x V (R). 8

9 The map r is called a retraction. It is well known that if R is a retract of G then χ(r) = χ(g). A subgraph H of G is called a core of G if there is a homomorphism G H but no homomorphism G H for any proper subgraph H of H. Principal properties of graph cores are summarised in [5]. In particular, the core of a graph is unique up to isomorphism and the core of G is a retract of G (minimal with respect to inclusion). As an application of Theorem 4 to graph retracts one can give a new infinite sequence of pairs of graphs G and G such that G is not a retract of G while G[K n ] is a retract of G[K n ]. The construction is similar to those in [8, 10] hence we will not give it here. Here is another application. Theorem 6 Let X G be the core of a graph G. Then X G is complete if and only if X G is χ critical and χ(g[k 2 ]) = 2χ(G). Proof. Let X G be isomorphic to K n, for some n. Then it is χ critical. Since χ(g) = χ(x G ) = n, we also have χ(g[k 2 ]) χ(k n [K 2 ]) = 2n and hence χ(g[k 2 ]) = 2χ(G). Conversely, let X G be χ critical and let χ(g[k 2 ]) = 2χ(G). Since X G is a retract of G, it easily follows that X G [K 2 ] is a retract of G[K 2 ]. Hence χ(x G [K 2 ]) = χ(g[k 2 ]) = 2χ(G) = 2χ(X G ). Now Corollary 3 implies that χ(x G [K 2 ]) 2(χ(X G ) 1) + 2/α(X G ). It follows that α(x G ) = 1. Thus X G is complete. In connection with the last theorem we pose two problems. Let k 2. Call a graph G (χ, k) simple, if χ(g[k k ]) = kχ(g). 9

10 Problem 7 Characterize (χ, k) simple graphs. In particular, characterize (χ, 2) simple graphs. Clearly, a graph G with ω(g) = χ(g) is (χ, k) simple for every k. However, there are infinite series of 3 chromatic graphs without triangles that are (χ, 2) simple and (χ, 3) simple, respectively (see [9, Theorem 7]). By Corollary 3, χ critical, incomplete graphs are not (χ, k) simple for any k. Problem 8 Characterize the graphs G that have χ critical core. For example, perfect graphs have complete cores. On the other hand, every retract rigid graph (i.e. a graph without proper retracts) is a proper core of some graph. 4 A lower bound In this short section we slightly generalize a lower bound given in [3, 10, 16] by the following theorem. The proof idea is essentially the same as in [10]. For a, b V (G) we will write ab[h a, H b ] instead of < {a, b} > [H a, H b ] and a[h a ] instead of <{a}> [H a ]. Theorem 9 Let G be a graph with at least one edge and H = {H a } a V (G) be a family of graphs. Then χ(g[h]) χ(g) + min{χ(ab[h a, H b ]) ab E(G)} 2. Proof. Let L = G[H] and let c be a χ(l) coloring of L. Denote by m the number min{χ(h a ) a V (G)}. Let for a V (G), m a = min c(a[h a ]). Finally let m G = min{χ(ab[h a, H b ]) ab E(G)}. Define the mapping 10

11 γ : V (G) {1, 2..., χ(l) m G + 2}, in the following way: γ(a) = { ma ; m a < χ(l) m G + 2, χ(l) m G + 2; otherwise. For to show that γ is a coloring let ab E(G) and assume γ(a) = γ(b). We claim that γ(a) m a. Suppose not. Then γ(a) = m a = m b = γ(b). But because every vertex from a[h a ] is adjacent to every vertex from b[h b ], we have a contradiction. Now γ(a) m a implies m a χ(l) m G + 2. As γ(a) = γ(b) it follows also m b χ(l) m G + 2. Therefore, for to color ab[h a, H b ] only χ(l) (χ(l) m G + 1) = m G 1 colors are available, a contradiction. Thus γ is a coloring of G. It follows χ(g) χ(l) m G + 2 and hence the result. Corollary 10 [3, 10, 16] Let G and H be graphs and let E(G). Then χ(g[h]) χ(g) + 2χ(H) 2. 11

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