The adjacent vertex-distinguishing total chromatic number of 1-tree

Transcription

1 The adjacent ertex-distinguishing total chromatic number of 1-tree Haiying Wang The School of Information Engineering China Uniersity of Geosciences(Beijing) Beijing , P.R.China Abstract Let G = (V (G), E(G)) be a simple graph and T (G) be the set of ertices and edges of G. Let C be a k color set. A (proper) total k coloring f of G is a function f: T (G) C such that no adjacent or incident elements of T (G) receie the same color. For any u V (G), denote C(u) = {f(u)} {f(u) u E(G)}. The total k coloring f of G is called the adjacent ertex-distinguishing if C(u) C() for any edge u E(G). And the smallest number of colors is called the adjacent ertex-distinguishing total chromatic number χ at (G) of G. Let G be a connected graph. If there exists an ertex V (G) such that G is a tree then G is a 1 tree. In this paper, we will determine the adjacent ertex-distinguishing total chromatic number of 1-trees. MSC: 05C15 Keywords The adjacent ertex-distinguishing total coloring; The adjacent ertex-distinguishing total chromatic number; 1-tree. 1. Introduction Let G = (V (G), E(G)) be a simple graph and T (G) = V (G) E(G) be the set of ertices and edges of G. (G), δ(g) and V denote the maximum degree, the minimum degree and the set of the ertices with degree (G) respectiely. For V (G), we use N G () to denote the neighbor set of in V (G). For S V (G), G[S] denotes the subgraph of G induced by S. whycht@126.com. This research is supported by 2008 Foundation of China Uniersity of Geosciences (Beijing) 1

2 Let G be a connected graph. If there exists a ertex V (G) such that G is a tree, then G is a 1 tree. Let C be a k color set. A (proper) total k coloring f of G is a function f: T (G) C such that no adjacent or incident elements of T (G) receie the same color. For any u V (G), denote C(u) = {f(u)} {f(u) u E(G)}. The total k coloring f of G is called the adjacent ertex-distinguishing if C(u) C() for any edge u E(G). And the smallest integer k is called the adjacent ertex-distinguishing total chromatic number χ at (G) of G. It is obious that χ at (G) (G) + 1. Suppose that f is an adjacent ertex-distinguishing total coloring of G. For w V (G), f(n G (w)) denotes the color set of the edges incident to w in G and f G [w] = f(n G (w)) {f(w)}. If an element t is colored α, then we denote α t. After B.Bollóbas, A.C.Burris, R.H.Schelp, C.Bazgan and P.N.Balister discussed the ertex-distinguishing coloring in [1-3], Zhongfu Zhang, Linzhong Lin, Jianfang Wang and Xiangen Chen introduced the adjacent ertex-distinguishing edge coloring and the adjacent ertex-distinguishing total coloring in [6] and [7]. Some results on the subject hae been obtained in [4-7]. Lemma 1([6]) If G has two ertices of maximum degree which are adjacent, then χ at (G) (G) + 2. Lemma 2([6]) If G has m components G i (i = 1, 2,, m) and V (G i ) 2, i = 1, 2,, m, then χ at (G) = max{χ at (G 1 ), χ at (G 2 ),, χ at (G m )}. Lemma 3([6]) Let C n be a cycle with n 4. Then χ at (C n ) = 4. Lemma 4([6]) Let S n be a star with n 3. Then χ at (S n ) = n + 1. Lemma 5([6]) Let F n be a fan with n 4. Then χ at (F n ) = n + 1. Lemma 6([6]) Let T n be a tree with n 4. Then { (Tn ) + 1, if E(T χ at (T n ) = n [V ]) =, (T n ) + 2, if E(T n [V ]). Lemma 7([6]) Let S n,m be a double-star with n 4 and m 4. Then { (Sn,m ) + 1, if n m, χ at (S n,m ) = (S n,m ) + 2, if n = m. In this paper, we will determine the adjacent ertex-distinguishing total chromatic number of 1-trees. 2. Main results 2

3 The adjacent ertex-distinguishing total chromatic number of the graphs G with (G) 2 has been determined in [6] and [7]. So we only consider the graph G with (G) 3. Then V (G) 4. Theorem 2.1 If G is a 1-tree, then χ at (G) = { (G) + 1, if E(G[V ]) =, (G) + 2, if E(G[V ]). Proof: Suppose that G is a 1 tree. According to the definition of the adjacent ertex-distinguishing total chromatic number, χ at (G) (G) + 1. By Lemma 1, if E(G[V ]) then χ at (G) (G) + 2. Thus, { (G) + 1, if E(G[V ]) =, χ at (G) (G) + 2, if E(G[V ]). So we will only proe that χ at (G) { (G) + 1, if E(G[V ]) =, (G) + 2, if E(G[V ]). According to the definition of 1 tree, there exists V (G) such that G is a tree. Suppose that T = G. If T is a star, then it is easy to proe the conclusion. So we assume that T is not a star. Then there exists w V (T ) such that d T (w) 1 and it has only one adjacent ertex u with d T (u) 2. Let W (T ) denote the set of such w V (T ) aboe. We will proe the conclusion by induction on V (G). It is obious that Theorem 2.1 holds for V (G) = 4. Now assume that V (G) 5. Suppose that w is a ertex with the smallest degree in W (T ). Let N T (w) = {u, 1,, s } with d T (u) 2 and d T ( i ) = 1 for all i {1, 2, s} with s 1. Case 1 w E(G). Subcase 1.1 d G ( i ) = 2 and d G ( j ) = 1 for all i {1, 2,, k 1} and j {k,, s} with k 2 and s k. In this subcase, we consider the graph G 0 = G { k,, s }. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1-tree with (G 0 ) = (G) =. By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 { + 1, + 2}. In the following, we will extend f to an adjacent ertex-distinguishing total coloring f of G with the color set C with C 0 C and C { + 1, + 2}, respectiely. 3

4 s s 1 w u k k 1 1 There are four subcases in all, denoted (a)(b)(c)(d) below. (a) d G (w) d G () and d G (w) d G (u). (b) d G (w) d G () and d G (w) = d G (u). (c) d G (w) = d G () and d G (w) d G (u). (d) d G (w) = d G () = d G (u). Now we only deal with the case of d G (w) = d G () = d G (u) = = s + 2 with s 2 and k 2 (see Figure 1) (A similar arguments work for other cases). In this case, E(G 0 [V ]). Then C 0 = + 2. Figure 1 Firstly, let C 0 = C. Since f G 0 [u] = + 1, there exists at least one color c 1 C 0 such that c 1 f G 0 [u]. Similarly, there exists at least one color c 0 C such that c 0 f G 0 []. Secondly, we do it step by step below. Note: the elements which hae not yet colored, retain the colors gien by f when we extend f to f of G respectiely. In the proof below, we will not mention it again. Step 1 If c 1 f G 0 [w] and c 0 f G 0 [w] {c 1 } with d G () = 2 then d G (w) 3. Let c 1 w k. If c 1 f G 0 [w] and c 0 f G 0 [w] {c 1 } with d G () 2 then firstly select any color c {c 0, c 1 } f ( k 1 ) and let c w k 1 ; secondly, select any color α C {f (w), c} {f (), f ( k 1 )} and let α k 1. If c 1 f G 0 [w] and c 0 = c 1 then let c 0 (= c 1 ) w k. If c 1 f G 0 [w] and c 0 f G 0 [w] then let c 0 w k. If c 1 f G 0 [w] and c 0 f G 0 [w] then do Step 2 directly. Step 2 Let {f(w k ),, f(w s )} C {f (w), f (wu), f (w), f (w 1 ),, f (w k 2 ), f(w k 1 )}. 4

5 x w u s s Step 3 Select α i C {f(w i ), f (w)} and let α i i for all i {k,, s}. Subcase 1.2 d G ( i ) = 2 for all i {1, 2,, s} with s 1. (I) There exists one ertex x V (G) such that d G (x) = 1 (see Figure 2). Figure 2 We consider the graph G 0 = G 1 +x. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1-tree with (G 0 ) = (G) =. By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 { + 1, + 2}. In the following, we will extend f to an adjacent ertex-distinguishing total coloring f of G with the color set C with C 0 C and C { + 1, + 2}, respectiely. (A) If E(G 0 [V ]) then C 0 = (G) + 2 and let C = C 0. Since f G 0 [] = d G ()+1 (G)+1, there exists at least one color c 0 C such that c 0 f G 0 []. Similarly, there exists at least one color c 1 C such that c 1 f G 0 [u]. So c 0 f (x) and c 1 f (uw). Assume that f ( i ) = c 1 for i S {2,, s} below. (A.1) d G () 3. Step 1 Let f (x) 1. Step 2 If c 0 f G 0 [w] {c 1 } and c 1 f G 0 [w] {f (x)} then firstly let c 1 w and c 0 w 1 ; secondly, select any color a i C ({f (w i ), f ( i )} {f (w), f ()}) and let a i i for all i S. If c 0 = c 1 f G 0 [w] then c 0 = c 1 f (x). Let c 0 = c 1 w 1. If c 0 f G 0 [w] and c 1 = f (x) f G 0 [w] then c 0 c 1. Firstly, let c 0 w 1 and 5

6 c 1 w; secondly, select any color a i C ({f (w i ), f ( i )} {f (w), f ()}) and let a i i for all i S. If c 0 f G 0 [w] and c 1 f G 0 [w] then let c 0 w 1. If c 0 f G 0 [w] and c 1 f G 0 [w] {f (x)} then let c 1 w 1. If c 0 = f (w) f G 0 [w] and c 1 = f (x) f G 0 [w], then firstly let c 1 w and f (w)(= c 0 ) w 1 ; secondly, select any color a i C ({f (w i ), f ( i )} {f (w), f ()}) and let a i i for all i S. If c 0 f G 0 [w] and c 1 = f (x) f G 0 [w], but f (w) c 0, then firstly let c 1 w and f (w) w 1 ; secondly, select any color a i C ({f (w i ), f ( i )} {f (w), f ()}) and let a i i for all i S. If c 0 f G 0 [w] and c 1 f G 0 [w] then select any color α C (f G 0 [w] {f (x)}) and let α w 1. Step 3 Select any α 1 C ({f(w), f(w 1 )} {f (x), f ()}) and let α 1 1. (A.2) d G () = 2. In this subcase, s = 1. We reconsider the graph G 0 = G + wx. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1-tree with (G 0 ) = (G) =. By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 = +2. Let C = C 0. Firstly, select any color c C f G 0 [w] and let c 1. Secondly, f (wu) and f (w) 1. (B) If E(G 0 [V ]) = and E(G[V ]) then C 0 = (G) + 1. Let C 0 C and C = (G) + 2. Assume that c C C 0. Step 1 Let f (x) 1. Step 2 Let c w 1. Step 3 Select any α 1 C ({f (w), f(w 1 )} {f (x), f ()}) and let α 1 1. (C) If E(G 0 [V ]) = and E(G[V ]) = then C 0 = (G) + 1. Let C = C 0. (C.1) d G () 3. (C.1.1) If d G () = d G (u) = (G) then d G (w) (G) 4 and u E(G). Step 1 Let f (x) 1. Step 2 Select any α C (f G 0 [w] {f (x)}) and let α w 1. Step 3 Select any α 1 C ({f (w), f(w 1 )} {f (x), f ()}) and let α

7 w u s s (C.1.2) If d G () = (G) and d G (u) (G) then d G (w) (G) 4. So there exists at least one color c 1 C such that c 1 f G 0 [u]. Assume that f ( i ) = c 1 for i S {2,, s} below. Step 1 Let f (x) 1. Step 2 If c 1 f G 0 [w] and c 1 f (x) then let c 1 w 1. If c 1 f G 0 [w] and c 1 = f (x) then firstly let c 1 w and f (w) w 1 ; secondly, select any color a i C ({f (w i ), f ( i )} {f (w), f ()}) and let a i i for all i S. If c 1 f G 0 [w] then select any α C f G 0 [w] and let α w 1. Step 3 Select any α 1 C ({f (w), f(w 1 )} {f (x), f ()}) and let α 1 1. (C.1.3) If d G () (G) and d G (u) = (G) then d G (w) (G). So there exists one color c 0 C 0 such that c 0 f G 0 []. Step 1 Let f (x) 1. Step 2 If c 0 f G 0 [w] then let c 0 w 1. If c 0 f G 0 [w] then select any α C f G 0 [w] and let α w 1. Step 3 Select any α 1 C ({f (w), f(w 1 )} {f (x), f ()}) and let α 1 1. (C.1.4) If d G () (G) and d G (u) (G) then there exists at least one color c 0 C such that c 0 f G 0 []. Similarly, there exists at least one color c 1 C such that c 1 f G 0 [u] with c 1 f (uw). Its steps are similar to (A). (C.2) d G () = 2. In this subcase, s = 1. We reconsider the graph G 0 = G + wx. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1-tree with (G 0 ) = (G) =. By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 = + 1. Let C = C 0. Firstly, let f (wx) w and f (wx) 1. Secondly, f (wu) and f (w) 1. (II) d G (x) 2 for all x V (G) (see Figure 3). 7

8 s s 1 w 3 u 2 1 Figure 3 In this subcase, is the unique ertex with maximum degree in G. We consider the graph G 0 = G 1. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1 tree with (G) = (G 0 ) + 1. By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 = d G0 () + 1. In the following, we can extend f to an adjacent ertexdistinguishing total coloring f of G with the color set C with C 0 C step by step. Since d G0 (x) 2 for all x V (G 0 ), is also the unique ertex with maximum degree in G 0. So d G () = d G0 () + 1 and (G) = (G 0 ) + 1. Firstly, let C = C 0 + 1(= d G () + 1). Assume that α C C 0. Step 1 Let α 1. Step 2 Select any β C (f G 0 [w] {α}) and let β w 1. Step 3 Select any γ C ({α, β} {f (w), f ()}) and let γ 1. Subcase 1.3 d G ( i ) = 1 for all i = 1, 2,, s with s 1 (see Figure 4). Figure 4 In this subcase, we can consider the graph G 0 = G { 1,, s }. Then (G) = (G 0 ) or (G) = (G 0 ) + 1. It is easy to see that V (G 0 ) < V (G) and G 0 is also a 1-tree with (G 0 ) (G). By the induction assumption G 0 has an adjacent ertex-distinguishing total coloring f with the color set C 0 with C 0 { (G 0 ) + 1, (G 0 ) + 2}. In the following, we can extend f to an adjacent ertex-distinguishing total coloring f of G with the color set C with C { (G) + 1, (G) + 2} and C 0 C. Step 1 (A) If (G) = (G 0 ) + 1 then w is the unique ertex with maximum degree in G. Let C 0 C and C = (G) + 1(= (G 0 ) + 2). Firstly, let 8

9 {f(w 1 ),, f(w s )} C f G 0 [w]. Secondly, do Step 3 directly. (B) If (G) = (G 0 ) = then we may assume that w E(G) and u E(G). (B.1) If E(G 0 [V ]) then E(G[V ]) and C 0 = (G) + 2. Let C 0 = C. Assume that d G (w) = d G () = d G (u) = and {u, w} E(G). Assume that c 1 C 0 f G 0 [u] and c 0 C 0 f G 0 []. Then c 0 c 1. If c 1 {f (w), f ()} and f (wu) = c 0 then firstly let c 1 w; secondly, do Step 2(1). If c 1 {f (w), f ()} and f (wu) c 0 then firstly let c 0 w 1 ; secondly, do Step 2(2). If c 1 = f (w) then firstly let c 0 w 1 ; secondly, do Step 2(2). If c 1 = f () and c 0 {f (u), f (wu)} then firstly let c 0 w and c 1 w 1 ; secondly, do Step 2(2). If c 1 = f () and c 0 = f (wu) then firstly let c 1 w 1 ; secondly, do Step 2(2). If c 1 = f () and c 0 = f (u) with s 2, then firstly let c 0 w 1 and c 1 w 2 ; secondly, do Step 2(3). If c 1 = f () and c 0 = f (u) with s = 1, then = 3. Assume that N G () {u, w}. If f (u) f ( ) then firstly let f (u) and c 1 u; secondly, let c 1 w and c 0 w 1 ; finally, do Step 3. If f (u) = f ( ) then f (w) f ( ). Firstly, let f (w) and c 1 w; secondly, let c 0 w 1 ; thirdly, select any color α C ({c 0, c 1 } {f (wu)}) and let α w; finally, do Step 3. (B.2) If E(G 0 [V ]) = E(G[V ]) = then C 0 = + 1. Let C = C 0. If d G (w) = then d G (u) and d G (). Let {f(w 1 ),, f(w s )} C f G 0 [w]. Finally, do Step 3 directly. If d G (w) then we only deal with the case of d G (u) = d G () = d G (w) < with w E(G). Assume that c 0 C 0 f G 0 [] and c 1 C 0 f G 0 [u]. Its steps are similar to Subcase (B.1). (B.3) If E(G 0 [V ]) = and E(G[V ]) then C 0 = (G) + 1. Let C 0 C and C = C Assume that α C C 0. Firstly, let α w. Secondly, do Step 2(1). 9

10 Step 2 (1) Let {f(w 1 ),, f(w s )} C {f (uw), f (w), f(w)}. (2) Let {f(w 2 ),, f(w s )} C ({f (uw), f (w), f(w)} { f(w 1 )}). (3) Let {f(w 3 ),, f(w s )} C (f G 0 [w] { f(w 1 ), f(w 2 )}). Step 3 Select any α i C {f(w i ), f (w)} and let α i i for all i = 1,, s. It is easy to erify that the coloring f aboe is an adjacent ertex-distinguishing total coloring of G in the cases respectiely. Case 2 If w E(G) then its proof is ery similar to Case 1 by the definition of the adjacent ertex-distinguishing total coloring. Thus, Theorem 2.1 holds. Acknowledgements The author is ery grateful to Professor Liang Sun for his helpful comments and suggestions. References [1] P.N.Balister, O.M.Riordan & R.H.Schelp, Vertex distinguishing edge colorings of graphs, J.of Graph Theory, 42(2003): [2] P.N.Balister, B.Bollóbas & R.H.Schelp, Vertex-distinguishing colorings of graphs with (G) = 2, Discrete Mathematics, 252(2002): [3] A.C.Burns & R.H.Schelp, Vertex-distinguishing proper edge-coloring, J.of Graph Theory, 21,(1997): [4] Deshan Ma & Zhongfu Zhang, The adjacent ertex-distinguishing edge coloring of 1 tree, Journal of Mathematical Research and Exposition, 20(2000): [5] Zhongfu Zhang & Xiangen Chen, On the adjacent ertex-distinguishing total coloring of graphs, Science In China Ser.A Mathematics,34(2004): [6] Zhongfu Zhang, Linzhong Liu & Jianfang Wang, Adjacent strong edge coloring of graphs, Applied Mathematics Letters, 15(2002):