d 2 -coloring of a Graph
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1 d -coloring of a Graph K. Selvakumar and S. Nithya Department of Mathematics Manonmaniam Sundaranar University Tirunelveli 67 01, Tamil Nadu, India selva 158@yahoo.co.in Abstract A subset S of V is called an i-set(i ) if no two vertices in S have the distance i. The -set number α (G) of a graph is the maximum cardinality among all -sets of G. A d -coloring of a graph is an assignment of colors to its vertices so that no two vertices have the distance two get the same color. The d -chromatic number χ d (G) of a graph G is the minimum number of d -colors need to G. In this paper, we initiate a study of these two new parameters. 1 Introduction By a graph G = (V, E) we mean a finite, undirected graph without loops and multiple edges. The order and size of G are denoted by p and q respectively. For graph theoretical terms we refer Chartrand and Lesniak [1]. A subset S of V is an independent set if no two vertices in S are adjacent. An assignment of colors to the vertices of a graph so that no two adjacent vertices get the same color is called a coloring of the graph and an assignment of colors to the vertices of a graph so that no two vertices have the distance 1 or get the same color is called a distant -coloring of G. This motivates to 000 Mathematics Subject Classification : 05C. Key words and phrases : d -coloring, d -chromatic number 1 10
2 K. Selvakumar and S. Nithya define two new parameters as follows: A subset S of V is called an i-set(i ) if no two vertices in S have the distance i. The -set number α (G) of a graph is the maximum cardinality among all -sets of G. A d -coloring of a graph is an assignment of colors to its vertices so that no two vertices have the distance two get the same color. The d -chromatic number χ d (G) of a graph G is the minimum number of d -colors need to G. The corona of two graphs G 1 and G is the graph G 1 G formed from one copy of G 1 and V (G 1 ) copies of G, where the i th vertex of G 1 is adjacent to every vertex in the i th copy of G. Any undefined term in this paper may be found in Chartrand and Lesniak [1]. In this paper, we initiate a study of these two new parameters. -set number of a Graph Definition.1. A subset S of V is called an i-set(i ) of G if no two vertices in S have the distance i. The -set number α (G) of a graph is the maximum cardinality among all -sets of G. Remark.. From the above definition, if i = 1, then S is an independent set of G. Proposition.3. α (K p ) = p. Proof. In K p, no two vertices have distance two and hence α (K p ) = p. p + 1 if p (mod 4) Proposition.4. α (P p ) =. p otherwise Proof. Consider P p : v 1 v v p. Case 1. p (mod 4) Let S = {v 1, v, v 5, v 6,..., v p 5, v p 4, v p 1, v p }. Then no two vertices in S have distance two and is a maximum -set of P p. Thus α (P p ) = p + 1. Case. If p 0(mod 4), then S = {v 1, v, v 5, v 6,..., v p 7, v p 6, v p 3, v p } is a maximum -set. If p 1(mod 4), then S = {v 1, v, v 5,..., v p 4, v p 3, v p } is a maximum -set. If p 3(mod 4), then S = {v 1, v, v 5, v 6,..., v p 6, v p 5, v p, v p 1 } is a maximum -set. In this case, we get α (P p ) = p. 103
3 d -coloring of a Graph 3 One can prove the following Proposition in analogous to the above. p 1 if p (mod 4) Proposition.5. α (C p ) = p otherwise, p > 3 Proposition.6. α (K m,n ) =. Proof. For any edge e = uv, whose end vertices form a maximum -set of K m,n and hence α (K m,n ) =. Lemma.7. Let G be a graph with p vertices. Then α (G) = p if and only if G is complete or a union of complete components. Proof. Suppose α (G) = p. Then no two vertices in G have distance two. If d(u, v) = k, < k <, for some u, v V (G), then d(w, u) = for some w V (G), a contradiction. If d(u, v) = 1, for all u, v V (G), then G is complete. If d(u, v) =, for some u, v V (G), then G = m i=1g i, where G i are connected components of G. Since α (G) = p, α (G i ) = V (G i ) for all 1 i m and so each G i is complete. Converse is obvious. The following Proposition is immediate from the definition. Proposition.8. For a connected graph G, α (G) ω(g). Theorem.9. Let G be a connected graph with p > vertices. Then α (G) = p 1 if and only if ω(g) = p 1. Proof. Suppose α (G) = p 1. Let S be a maximum -set of G with S = p 1. Then no two vertices in S have distance two. If d(u, v) = k > for some u, v S, then d(u, w) =, for some w S, a contradiction. Thus any two vertices in S have distance one, the subgraph induced by S is complete and hence ω(g) = p 1. Conversely, suppose ω(g) = p 1. Then G is not complete and by Lemma.7, α (G) p 1. If α (G) < p 1, then ω(g) < p 1, a contradiction. Hence α (G) = p 1. Proposition.10. For any connected graph G with p vertices, α (K m G) = mα (G). Proof. In K m G, for every i j, d(u, v) = 3, for every u in i th copy of G, v in j th copy of G and hence α (K m G) = mα (G). 104
4 4 K. Selvakumar and S. Nithya In view of Proposition.10, we have following corollary. Corollary.11. α (K m G) = m if and only if G = K 1 Lemma.1. Let G be a tree with p vertices. Then 1 α (G) p + 1. Proof. Obviously α (G) 1. Let S V (G) with d(u, v) = for all u, v S. Then S = and so maximum -set contains at most p ( 1) vertices. Thus α (G) p + 1. Theorem.13. Let G be a connected bipartite graph with (V 1, V )-partition of V (G), V 1 > 1 and V 1 V. Then, (i) α (G) = V 1 + V if and only if G is complete. (ii) V 1 is a maximum -set of G if and only if G is not complete. Proof. (i) Suppose α (G) = V 1 + V. Since G is connected and by Lemma.7, G is union of two complete components and hence G is complete. Converse is obvious. (ii) Suppose V 1 is a maximum -set of G. Then for every v V, d G (v, u) =, for some u V 1 and so at least one shortest path u w v is in G, u, w V 1, v V. Thus w v E(G) and so w v / E(G). Hence G is not complete. Conversely, suppose G is not complete. Then for each x V 1, y V, there exists x V, y V 1 such that d G (x, x ) = and d G (y, y ) = and so V 1 is a maximum -set of G. Hence α (G) = V 1. Lemma.14. For any connected graphs G and H, α (G+H) = ω(g)+ω(h). Proof. If G and H are complete, then G + H is complete and by Lemma.7, α (G + H) = ω(g) + ω(h). Suppose G is not complete. Then G + H is not complete, diam(g + H) = and by Proposition.8, α (G + H) ω(g + H). Suppose α (G + H) = k > ω(g + H). Then G + H contains K k, k > ω(g + H), a contradiction. Hence α (G + H) = ω(g + H) = ω(g) + ω(h). In view of Lemma.14, we have following corollary. Corollary.15. α (W p ) = 3, p > 4. Theorem.16. Let G and H be connected with V (G) > 1 and V (H) > 1. Then α (G + H) = 4 if and only if G and H are triangle free graphs. 105
5 d -coloring of a Graph 5 Proof. Suppose α (G + H) = 4. Then by Lemma.14, ω(g) + ω(h) = 4 and so ω(g) = ω(h) =. Thus G and H are triangle free graphs. Conversely, suppose G and H are triangle free graphs, then ω(g) = ω(h) = and by Lemma.14, α (G + H) = 4. Theorem.17. Let G be a connected graph with diam(g) =. Then α (G) = if and only if G is triangle free graph. Proof. Suppose α (G) =. If G contains K 3, then by Proposition.8, α (G) 3, a contradiction. Conversely, suppose G is triangle free graph. If α (G) = k >, then by hypothesis, G contains K k, a contradiction. In view of Theorems.16,.17, we have following corollary. Corollary.18. Let G and H be connected graphs with diameter is. If G and H are triangle free graphs then α (G + H) = α (G) + α (H). Theorem.19. Let G be a any tree. Then α (G) = if and only if G is a star or bistar. Proof. Suppose α (G) =. If G is not a star, then diam(g) = k 3. Suppose diam(g) = k > 3. Then G has a path of length k and so α (G) α (P k+1 ). By Proposition.4, this is not possible. Thus diam(g) = 3 and so G is a bistar. If G is not a bistar, then diam(g) 3 and so diam(g) = 1 or. Hence G is a star. Converse is obvious. 3 d -chromatic number of a Graph Definition 3.1. A d -coloring of a graph is an assignment of colors to its vertices so that no two vertices have the distance two get the same color. The d -chromatic number χ d (G) of a graph G is the minimum number of d -colors need to G. A graph G is k d -colorable(resp. d -chromatic) if χ d (G) k(resp. χ d (G) = k). Proposition 3.. χ d (K n ) = χ d (K n ) = 1. Proof. Since d(u, v) = 1 for all u, v V (K n ), every vertex in K n assigned the same color and hence χ d (K n ) = 1. Since d(u, v) = for all u, v V (K n ), χ d (K n ) =
6 6 K. Selvakumar and S. Nithya Proposition 3.3. χ d (P p ) = for p >. Proof. Consider P p : v 1 v v p. Case 1. p 0(mod 4) Let Ω 1 = {v 1, v, v 5, v 6,..., v p 3, v p } and Ω = {v 3, v 4, v 7, v 8,,..., v p 1, v p }. Then Ω 1 and Ω are -sets and every vertex in Ω 1 make a distance two to at least one vertex in Ω and vice versa. Hence we assign one color to Ω 1 and another one color to Ω and so χ d (P p ) =. Case. p 1(mod 4) Let Ω 1 = {v 1, v, v 5, v 6,..., v p 4, v p 3, v p } and Ω = {v 3, v 4, v 7, v 8,,..., v p, v p 1 }. Then Ω 1 and Ω are -sets and every vertex in Ω 1 make a distance two to at least one vertex in Ω and vice versa. Hence we assign one color to Ω 1 and another one color to Ω and so χ d (P p ) =. Case 3. p (mod 4) Let Ω 1 = {v 1, v, v 5, v 6,..., v p 1, v p } and Ω = {v 3, v 4, v 7, v 8,,..., v p 3, v p }. Then Ω 1 and Ω are -sets and every vertex in Ω 1 make a distance two to at least one vertex in Ω and vice versa. Hence we assign one color to Ω 1 and another one color to Ω and so χ d (P p ) =. Case 4. p 3(mod 4) Let Ω 1 = {v 1, v, v 5, v 6,..., v p, v p 1 } and Ω = {v 3, v 4, v 7, v 8,,..., v p 4, v p 3, v p }. Then Ω 1 and Ω are -sets and every vertex in Ω 1 make a distance two to at least one vertex in Ω and vice versa. Hence we assign one color to Ω 1 and another one color to Ω and so χ d (P p ) =. if p 0 (mod 4) Proposition 3.4. χ d (C p ) = 3 otherwise, p > 3 Proof. Consider C p : v 1 v v p v 1. Case 1. p 0(mod 4) Let Ω 1 = {v 1, v, v 5, v 6,..., v p 3, v p } and Ω = {v 3, v 4, v 7, v 8,,..., v p 1, v p }. Then Ω 1 and Ω are -sets and every vertex in Ω 1 make a distance two to at least one vertex in Ω and vice versa. Hence we assign one color to Ω 1 and another one color to Ω and so χ d (C p ) =. Case. p 1,, 3(mod 4) We consider first p 1 -vertices, where p 1 0(mod 4), as in case 1, assigning two colors to p 1 -vertices of C p. Remaining there are k(= 1 or or 3)vertices. If k = 1 then p = p and d(v p1 +1, v p1 1) = d(v p1 +1, v ) =, where v Ω 1 107
7 d -coloring of a Graph 7 and v p1 1 Ω. So assign a new color to v p1 +1. If k = then p = p 1 + and d(v p1 +, v ) = d(v p1 +, v p1 ) = d(v p1 +1, v 1 ) = d(v p1 +1, v p1 1) =,where v 1, v Ω 1 and v p1, v p1 1 Ω. So assign a new color to both v p1 +1 and v p1 +. If k = 3 then p = p 1 +3 and d(v p1 +, v 1 ) = d(v p1 +, v p1 ) =, where v 1 Ω 1 and v p1 Ω. So assign a new color to v p1 +. Also color of vertex set Ω 1 assign to v p1 +1 and color of vertex set Ω assign to v p1 +3 because Ω 1 {v p1 +1}, Ω {v p1 +3} are -sets and d(v p1 +1, v p1 +3) =. Hence in this cases, χ d (C p ) = 3. Proposition 3.5. χ d (K m,n ) = m for m n. Proof. Let V 1 = {u 1, u,..., u m } and V = {v 1, v,..., v n } be the partition of V (K m,n ). Clearly d(u i, u j ) = = d(v k, v l ) for every i j, k l. So we assign m-colors to vertices in V 1 with each vertex has different color. Since for every u V 1, d(u, v) = 1 for every v V, same set of colors of vertices in V 1 use to vertices in V.Hence χ d = m. Remark 3.6. In general, χ d (K n1,n,...,n t ) = max{n 1, n,..., n t }. p 1 if p is odd Proposition 3.7. For p 5, χ d (W p ) = p if p is even Proof. Note that W p = C p 1 + K 1. Let V (C p 1 ) = {v 1, v,..., v p 1 } and V (K 1 ) = v p. Consider C p 1 = v 1 v v p 1 v 1 Case 1. p is odd Consider the maximum edge independent set S = {e i : e i = (v i, v i+1 ) E(W p ) for i = 1, 3, 5,..., p } and S = β (W p ). Then V (C p ) = i {v i, v i+1 }. Since d(u, v) =,for some u {v i, v i+1 } and v {v j, v j+1 },for every i j,there are β colors assigning to V (C p ) with end vertices of each edge in S receive the same color. Also d(v p, v k ) = 1, k = 1 to p 1,center vertex v p receives any one color from β colors. Hence χ d (W p ) = β (W p ) = (p 1), if p is odd. Case. p is even Consider the maximum edge independent set S = {e i : e i = (v i, v i+1 ) E(W p ) for i = 1, 3, 5,..., p 1} and also V (W p ) = i {v i, v i+1 }. As in case 1, there are β colors assigned to V (W p ). Hence χ d (W p ) = β (W p ) = p,if p is even. 108
8 8 K. Selvakumar and S. Nithya Proposition 3.8. A connected graph G is one d -colorable if and only if G is complete. Proof. Suppose G is one d -colorable. Then no two vertices have distance two in G. Since G is connected,any two vertices must have distance one. Hence G is complete. Converse follows from Proposition 3.. Proposition 3.9. Let G be a connected graph with p > 1 vertices. χ d (G) = p 1 if and only if G is a star graph Then Proof. Suppose χ d (G) = p 1. Let S V (G) with d(u, v) =. By hypothesis, S = p 1 and S is an independent set of G. Since G is connected, there exists a vertex w V S such that d(w, u) = 1 for all u S. Thus G is a star graph. Converse is obvious. Theorem For any connected graph G with p > 1 vertices, 1 χ d (G) p 1. Proof. Obviously χ d (G) 1. Let S be a subset of V (G) such that any two vertices in S have distance two. Then S can have at most p 1 vertices and so χ d (G) p 1. Remark Let G be a connected graph with diameter two. Then χ d Theorem 3.1. If G has a perfect matching, then χ d (G) β. Proof. Let M = {e i : e i = (v i, v i+1 ) E(G), i = 1 to β } be a maximum matching of G. Then M = β (G) and V (G) = i {v i, v i+1 }. There are β colors assigning to the vertices of G with end vertices of each edge in M receive the same color because d(u, v) = for some u {v i, v i+1 } and v {v j, v j+1 }, for every i j. This is a maximum possible d -coloring of G. Hence χ d (G) β. Theorem Let G be a bipartite graph. Then χ d. Proof. Let (V 1, V ) be a partition of V (G) and let v V 1 with deg(v) =. Clearly the subgraph induces by N[v] in G is a star graph K 1, and so χ d (G) χ d (K 1, ) = in G. Hence χ d. Theorem If G is a connected bipartite graph, then G is two d colorable. But converse is not true. 109
9 d -coloring of a Graph 9 Proof. Let (V 1, V ) be a partition of V (G). Clearly V 1 and V induces a complete subgraphs of G. If G is complete, then χ d (G) = 1. If G is not complete, for every u V 1, d G (u, v) = for some v V and also for every v V, d G (v, u) = for some u V 1. Hence assigning one color to all the vertices in V 1 and another color to all the vertices in V. Hence χ d (G) =. Thus G is two d colorable. Conversely, consider the graph given in the Figure 1. Clearly χ d (G) = but G is not bipartite. Figure 1 Remark In general, if G is connected k-partite then G is k d -colorable. Theorem If G is a tree, then χ d = Proof. Since G is bipartite and by Theorem 3.13, χ. Suppose χ d = k >. Then V (G) contains S such that S = k and d(u, v) = for all u, v S. This is not possible because V (G) contains S such that d(u, v ) = for all u, v S and S =. Hence χ d =. Corollary If G is connected and χ d >, then G contains cycle. Proof. Suppose G contains no cycle. Then G is a tree and by Theorem 3.16, χ d =, a contradiction. Theorem Let G be a graph. Then χ d + χ d = p and χ d χ d = p 1 if and only if G = K 1,p 1 or K 1,p 1, p > 1. Proof. Assume that χ d +χ d = p and χ d χ d = p 1, then χ d p χ d +p 1 = 0 and χ d p χ d + p 1 = 0. Solving these equations, we get χ d = p 1, 1 and χ d = p 1, 1. By hypothesis, χ d = p 1 and χ d = 1 or χ d = 1 and χ d = p 1. By Proposition 3.9, G = K 1,p 1 or K 1,p 1. Conversely, suppose G = K 1,p 1 or K 1,p 1, p > 1. If G = K 1,p 1, then χ d = p 1 and χ d = 1 and hence χ d + χ d = p and χ d χ d = p 1. If G = K 1,p 1, then χ d = 1 and χ d = p 1 and hence χ d + χ d = p and χ d χ d = p
10 10 K. Selvakumar and S. Nithya Corollary Let G be a graph with p vertices. If p = p 1 + 1, where p 1 is prime and χ d χ d = p 1, then G = K 1,p1 or G = K 1,p1 Proof. Since χ d χ d = p 1, χ d = p 1 and χ d = 1 or χ d = 1 and χ d = p 1. If χ d = p 1 and χ d = 1, then χ d + χ d = p and χ d χ d = p 1. By Theorem 3.18, G = K 1,p1. Similarly, if χ d = 1 and χ d = p 1, then G = K 1,p1. Theorem 3.0. For any graph G with p vertices, α (G)+χ d (G) p+1. Proof. Obviously α (G) + χ d (G). Let S V (G) such that S contains exactly one vertex from each color class of G. Then S = χ d (G) and d(u, v) = for all u, v S. Thus any maximum -set contains at most p ( S 1) vertices and so α (G) p χ d (G) + 1. References [1] G. Chartrand and L. Lesniak, Graphs and Digraphs, Wadsworth and Brooks/Cole, Monterey, CA(1986). [] Fertin, Godard and Raspaud, Acyclic and k-distance coloring of the Grid, Information Processing Letters, 87, 1(003), [3] J. Van den Heuvel and S. McGuinness, Colouring the square of a planar graph, J. Graph Theory, 4(005),
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