Báo cáo toán học: " Long heterochromatic paths in edge-colored graphs" potx

Long heterochromatic paths in edge-colored graphs He Chen and Xueliang Li Center for Combinatorics and LPMC Nankai University Tianjin 300071, P.R. China lxl@nankai.edu.cn Submitted: Jan 12, 2005; Accepted: Jun 10, 2005; Published: Jul 29, 2005 AMS Subject Classification (2000): 05C38, 05C15 Abstract Let G be an edge-colored graph. A heterochromatic path of G is such a path in which no two edges have the same color. d c (v) denotes the color degree of a vertex v of G. In a previous paper, we showed that if d c (v) ≥ k for every vertex v of G,thenG has a heterochromatic path of length at least  k+1 2 .Itiseasy to see that if k =1, 2, G has a heterochromatic path of length at least k.Saito conjectured that under the color degree condition G has a heterochromatic path of length at least  2k+1 3 . Even if this is true, no one knows if it is a best possible lower bound. Although we cannot prove Saito’s conjecture, we can show in this paper that if 3 ≤ k ≤ 7, G has a heterochromatic path of length at least k − 1, and if k ≥ 8, G has a heterochromatic path of length at least  3k 5  + 1. Actually, we can show that for 1 ≤ k ≤ 5 any graph G under the color degree condition has a heterochromatic path of length at least k, with only one exceptional graph K 4 for k = 3, one exceptional graph for k = 4 and three exceptional graphs for k =5,for which G has a heterochromatic path of length at least k−1. Our experience suggests us to conjecture that under the color degree condition G has a heterochromatic path of length at least k − 1. 1. Introduction We use Bondy and Murty [3] for terminology and notations not defined here and consider simple graphs only. Let G =(V, E) be a graph. By an edge-coloring of G we will mean a function C : E → N, the set of nonnegative integers. If G is assigned such a coloring, then we say that G is an edge-colored graph. Denote the colored graph by (G, C), and call C(e)thecolor of the the electronic journal of combinatorics 12 (2005), #R33 1 edge e ∈ E and C(uv)=∅ if uv /∈ E(G) for any u, v ∈ V (G). All edges with the same color form a color class of the graph. For a subgraph H of G,weletC(H)={C(e) | e ∈ E(H)} and c(H)=|C(H)|. For a vertex v of G,thecolor neighborhood CN(v)ofv is defined as the set {C(e) | e is incident with v} and the color degree is d c (v)=|CN(v)|.Apath is called heterochromatic if any two edges of it have different colors. If u and v are two vertices on a path P , uPv denotes the segment of P from u to v. There are many existing literature dealing with the existence of paths and cycles with special properties in edge-colored graphs. In [5], the authors showed that for a 2-edge- colored graph G and three specified vertices x, y and z, to decide whether there exists a color-alternating path from x to y passing through z is NP-complete. The heterochromatic Hamiltonian cycle or path problem was studied by Hahn and Thomassen [9], R¨odl and Winkler (see [8]), Frieze and Reed [8], and Albert, Frieze, Reed [1]. For more references, see [2, 6, 7, 10, 11]. Many results in these papers are proved by using probabilistic methods. In [4], the authors showed that if G is an edge-colored graph with d c (v) ≥ k for every v of G,thenG has a heterochromatic path with length at least  k+1 2 . It is easy to see that if k =1, 2, G has a heterochromatic path of length at least k. Saito conjectured that under the color degree condition G has a heterochromatic path of length at least  2k+1 3 .Even if this is true, no one knows if it is a best possible lower bound in general. Although we cannot prove Saito’s conjecture, we can show in this paper that if 3 ≤ k ≤ 7, then G has a heterochromatic path of length at least k −1, and if k ≥ 8, then G has a heterochromatic path of length at least  3k 5  + 1. Actually, we can show that for 1 ≤ k ≤ 5 any graph G under the color degree condition has a heterochromatic path of length at least k,with only one exceptional graph K 4 for k = 3, one exceptional graphs for k =4andthree exceptional graphs for k = 5, for which G has a heterochromatic path of length at least k − 1. Our experience suggests us to conjecture that under the color degree condition G has a heterochromatic path of length at least k − 1. 2. Long heterochromatic paths for k ≤ 7 We consider the case when 1 ≤ k ≤ 7, first. For the case when k = 1 or 2, it is obvious that there is a heterochromatic path of length k in G. In fact, for k =1anyedgeofG is a required heterochromatic path of length k; for k = 2, at each vertex there exist two adjacent edges with different colors, and they form a required heterochromatic path of length k. Next, we consider the case when 3 ≤ k ≤ 7 and get the following result. Theorem 2.1 Let G be an edge-colored graph and 3 ≤ k ≤ 7 an integer. Suppose that d c (v) ≥ k for every vertex v of G. Then G has a heterochromatic path of length at least k − 1. Proof. (1) k =3. Sincek =3> 2, there is a heterochromatic path of length 2 in G. the electronic journal of combinatorics 12 (2005), #R33 2 (2) k =4. Sincek =4> 3, there is a heterochromatic path of length 2 in G.Let P = u 1 u 2 u 3 be such a path that u 1 u 2 has color i 1 and u 2 u 3 has color i 2 with i 1 = i 2 . Since d c (u 3 ) ≥ k = 4, there are two vertices v, w ∈ G such that u 3 v, u 3 w ∈ E(G) have two different colors i 3 ,i 4 /∈{i 1 ,i 2 }.Letu 4 be a vertex in {v, w}\{u 1 }.Then {u 4 }∩{u 1 ,u 2 ,u 3 } = ∅ and P  = u 1 u 2 u 3 u 4 is a heterochromatic path of length 3. (3) k =5. Sincek =5> 4, there is a heterochromatic path of length 3 in G.LetP = u 1 u 2 u 3 u 4 be such a path that u x u x+1 has color i x for x =1, 2, 3. If there exists a v/∈{u 1 ,u 2 ,u 3 ,u 4 } such that C(u 4 v) /∈{i 1 ,i 2 ,i 3 },thenP  = u 1 u 2 u 3 u 4 v is a heterochromatic path of length 4. Otherwise, since d c (u 4 ) ≥ 5, we have that |C({u 1 u 4 ,u 2 u 4 }) −{i 1 ,i 2 ,i 3 }| = 2 and there exists a v 1 ∈ V (G) such that C(u 4 v 1 )=i 1 . Since d c (v 1 ) ≥ 5, we have that |CN(v 1 ) −{i 1 ,i 2 ,i 3 }| ≥ 2. If there exists a v 2 ∈ V (G) such that C(v 1 v 2 ) /∈{i 1 ,i 2 ,i 3 },thenP  = u 2 u 3 u 4 v 1 v 2 is a heterochromatic path of length 4; if C(u 1 v 1 ) /∈{i 1 ,i 2 ,i 3 },thenP  = v 1 u 1 u 2 u 3 u 4 is a heterochromatic path of length 4. So, the only remaining case is when |C({u 2 v 1 ,u 3 v 1 }) −{i 1 ,i 2 ,i 3 }| = 2, and in this case we have that P  = u 1 u 2 v 1 u 3 u 4 is a heterochromatic path of length 4. (4) k =6. Sincek =6> 5, there is a heterochromatic path of length 4 in G.Let P = u 1 u 2 u 3 u 4 u 5 be such a path that u x u x+1 has color i x for x =1, 2, 3, 4. If there exists a v/∈{u 1 ,u 2 ,u 3 ,u 4 ,u 5 } such that C(u 5 v) /∈{i 1 , ,i 4 },thenu 1 u 2 u 3 u 4 u 5 v is a heterochro- matic path of length 5. Next we consider the case when there is no such a vertex v, in other words, |C({u 1 u 5 ,u 2 u 5 ,u 3 u 5 }) −{i 1 ,i 2 ,i 3 ,i 4 }| ≥ 2. Since d c (u 5 ) ≥ 6, there is a vertex v 1 such that C(u 5 v 1 )=i 1 or i 2 . (4.1) C(u 5 v 1 )=i 1 .Sinced c (u 5 ) ≥ 6and|C({u 1 u 5 ,u 2 u 5 ,u 3 u 5 }) −{i 1 ,i 2 ,i 3 ,i 4 }| ≥ 2, there exists a u 6 ∈ V (G) such that C(u 5 u 6 )=i 2 or i 3 . If there is a vertex v/∈ {u 2 ,u 3 ,u 4 ,u 5 } such that C(v 1 v) /∈{i 1 , ,i 4 },thenu 2 u 3 u 4 u 5 v 1 v is a heterochromatic path of length 5. And, since d c (v 1 ) ≥ k =6,wehavethat|C({u 2 v 1 ,u 3 v 1 ,u 4 v 1 }) − {i 1 , ,i 4 }| ≥ 2. If there exists an 2 ≤ i ≤ 3 such that |C({u i v 1 ,u i+1 v 1 }) −{i 1 ,i 2 ,i 3 ,i 4 }| =2,thenP  = u 1 Pu i v 1 u i+1 Pu 5 is a heterochromatic path of length 5. Otherwise, |C({u 2 v 1 ,u 4 v 1 }) −{i 1 ,i 2 ,i 3 ,i 4 }| =2andu 1 u 2 v 1 u 4 u 5 u 6 is a heterochromatic path of length 5. (4.2) C(u 5 v 1 )=i 2 .Sinced c (u 5 ) ≥ 6, there exists a u 6 ∈ V (G) such that C(u 5 u 6 )=i 3 . Then we have the following three cases: (i) There is no vertex v/∈{u 1 , ,u 5 } such that C(v 1 v) /∈{i 1 , ,i 4 }.Sinced c (v 1 ) ≥ 6, we have that |C({u 1 v 1 ,u 2 v 1 ,u 3 v 1 ,u 4 v 1 }) − {i 1 ,i 2 ,i 3 ,i 4 }| ≥ 2. If C(u 1 v 1 ) /∈{i 1 ,i 2 ,i 3 ,i 4 },thenv 1 u 1 u 2 u 3 u 4 u 5 is a heterochromatic path of length 5. If there exists an i = 2 or 3 such that |C({u i v 1 ,u i+1 v 1 }) −{i 1 ,i 2 ,i 3 ,i 4 }| =2, then u 1 Pu i v 1 u i+1 Pu 5 is a heterochromatic path of length 5. Otherwise, |C({u 2 v 1 ,u 4 v 1 })− {i 1 ,i 2 ,i 3 ,i 4 }| =2,andthenu 1 u 2 v 1 u 4 u 5 u 6 is a heterochromatic path of length 5. (ii) There is a v 2 /∈{u 1 , ,u 5 } such that C(v 1 v 2 ) −{i 1 , ,i 4 }= ∅ and there is no ver- tex v/∈{u 1 , ,u 5 } such that |C({v 1 v 2 ,v 2 v}) −{i 1 , ,i 4 }| =2. Leti 5 = C(v 1 v 2 ). Since d c (v 2 ) ≥ 6, we have that C({u 1 v 2 ,u 2 v 2 ,u 3 v 2 ,u 4 v 2 }) −{i 1 , ,i 5 }= ∅.IfC(u i v 2 ) − {i 1 , ,i 5 }= ∅ for i = 1 or 2, then u 3 u 4 u 5 v 1 v 2 u i is a heterochromatic path of length 5. If C(u 4 v 2 ) −{i 1 , ,i 5 }= ∅,thenu 1 u 2 u 3 u 4 v 2 v 1 is a heterochromatic path of length 5. Otherwise, C(u 3 v 2 ) −{i 1 , ,i 5 }= ∅.Sinceu 1 u 2 u 3 u 4 u 5 is a heterochromatic path of length 4, we have that |C({u 1 u 3 ,u 1 u 4 ,u 1 u 5 }) −{i 1 ,i 2 ,i 3 ,i 4 }| ≥ 2 which implies that the electronic journal of combinatorics 12 (2005), #R33 3 C({u 1 u 3 ,u 1 u 4 }) −{i 1 ,i 2 ,i 3 ,i 4 }= ∅.IfC(u 1 u 3 ) −{i 1 ,i 2 ,i 3 ,i 4 }= ∅,thenu 2 u 1 u 3 u 4 u 5 v 1 is a heterochromatic path of length 5. If C(u 1 u 4 ) /∈{i 1 ,i 2 ,i 3 ,i 4 },thenu 5 u 4 u 1 u 2 u 3 v 2 or u 2 u 1 u 4 u 5 v 1 v 2 is a heterochromatic path of length 5. (iii) There exist v 2 ,v 3 /∈{u 1 , ,u 5 } such that |C({v 1 v 2 ,v 2 v 3 }) −{i 1 , ,i 4 }| =2. Thenu 3 u 4 u 5 v 1 v 2 v 3 is a heterochromatic path of length 5. (5) k =7. Sincek =7> 6, there is a heterochromatic path of length 5 in G.Let P = u 1 u 2 u 3 u 4 u 5 u 6 be such a path that u x u x+1 has color i x for x =1, ,5. If there is a vertex v/∈ V (P ) such that C(u 6 v) /∈ C(P ), then u 1 Pu 6 v is a heterochromatic path of length 6. Next we consider the case when there is no such v.Sinced c (u 6 ) ≥ 7, we have that |C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 1 , ,i 5 }| ≥ 2. Let k 0 =min{k|there is a vertex v/∈ V (P ) such that C(u 6 v)=i k },andv 1 be a vertex /∈ V (P ) such that C(u 6 v)=i k 0 . Then k 0 =1or2or3. (5.1) k 0 = 1. If there exists a v 2 /∈ V (P ) such that C(v 1 v 2 ) /∈ C(P ), then u 2 Pu 6 v 1 v 2 is a heterochromatic path of length 6. Or, |C({u 1 v 1 , ,u 5 v 1 }) − C(P )|≥2. If there is a vertex u/∈ V (P) such that C(uu 1 ) /∈ C(P ), then uu 1 Pu 6 is a heterochromatic path of length 6. Next we consider the case when |C({u 2 v 1 ,u 3 v 1 ,u 4 v 1 ,u 5 v 1 }) − C(P )|≥2 and |C({u 1 u 3 , ,u 1 u 6 }) − C(P )|≥2. If C(u 2 v 1 ) /∈ C(P ), then there is an 3 ≤ i ≤ 6 such that C(u 1 u i ) /∈ C(P ) ∪ C(u 2 v 1 ), and so u i−1 P −1 u 2 v 1 u 6 P −1 u i u 1 is a heterochro- matic path of length 6. If there exists an i = 3 or 4 such that |C({u i v 1 ,u i+1 v 1 }) − C(P )| =2,thenu 1 Pu i v 1 u i+1 Pu 6 is a heterochromatic path of length 6. In the rest we shall only consider the case when |C({u 3 v 1 ,u 5 v 1 }) − C(P )| =2,andleti 6 = C(u 3 v 1 ) and i 7 = C(u 5 v 1 ). If there exists a v/∈ V (P) such that C(u 6 v) /∈{i 1 ,i 2 ,i 5 }, i.e., C(u 6 v) ∈{i 3 ,i 4 },thenu 1 u 2 u 3 v 1 u 5 u 6 v is a heterochromatic path of length 6. Otherwise, |C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 1 ,i 2 ,i 5 }| = 4. On the other hand, if C(u 1 u 6 ) −{i 1 ,i 2 ,i 5 } = ∅,thenv 1 u 3 u 2 u 1 u 6 u 5 u 4 or v 1 u 5 u 6 u 1 u 2 u 3 u 4 is a heterochromatic path of length 6, a contradiction. (5.2) k 0 =2. So,|C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 2 ,i 3 ,i 4 ,i 5 }| ≥ 3. We have the fol- lowing three cases: (i) There is no vertex v/∈ V (P ) such that C(v 1 v) /∈ C(P ). Since d c (v 1 ) ≥ 7, we have that |C({u 1 v 1 ,u 2 v 1 ,u 3 v 1 ,u 4 v 1 ,u 5 v 1 }) − C(P )|≥2. If there exists a u/∈ V (P ) such that C(uu 1 ) /∈ C(P ), then uu 1 Pu 6 is a heterochromatic path of length 6. Next we consider the case when |C({u 2 v 1 ,u 3 v 1 ,u 4 v 1 ,u 5 v 1 })−C(P )|≥2and|C({u 1 u 3 , ,u 1 u 6 })−C(P )|≥2. If C(u 2 v 1 ) /∈ C(P ), then there is an 3 ≤ i ≤ 6 such that C(u 1 u i ) /∈ C(P )∪C(u 2 v 1 ), and so u 3 Pu i u 1 u 2 v 1 u 6 P −1 u i+1 is a heterochromatic path of length 6. If there exists an i =3or4 such that |C({u i v 1 ,u i+1 v 1 })−C(P )| =2,thenu 1 Pu i v 1 u i+1 Pu 6 is a heterochromatic path of length 6. In the rest we shall only consider the case when |C({u 3 v 1 ,u 5 v 1 })−C(P )| =2. Since |C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 2 ,i 3 ,i 4 ,i 5 }| ≥ 3, there exists a v/∈ V (P ) such that C(u 6 v) ∈{i 3 ,i 4 },andsou 1 u 2 u 3 v 1 u 5 u 6 v is a heterochromatic path of length 6. (ii) There exists a v 2 /∈ V (P ) such that C(v 1 v 2 ) /∈ C(P ), and there is no v/∈ V (P ) ∪ {v 1 } such that C(v 2 v) /∈ C(P ) ∪ C(v 1 v 2 ). Let i 6 = C(v 1 v 2 ). Then C({u 1 v 2 , ,u 5 v 2 }) − {i 1 , ,i 6 }= ∅. If there exists a u/∈{u 1 , ,u 6 } such that C(uu 1 ) /∈ C(P ), then uu 1 Pu 6 is a heterochromatic path of length 6. If C(u 2 v 2 ) /∈{i 1 , ,i 6 },thenv 1 v 2 u 2 Pu 6 is a het- erochromatic path of length 6. If C(u 5 v 2 ) /∈{i 1 , ,i 6 },thenu 1 Pu 5 v 2 v 1 is a heterochro- the electronic journal of combinatorics 12 (2005), #R33 4 matic path of length 6. So we shall only consider the case when |C({u 1 u 3 , ,u 1 u 6 }) − C(P )|≥2andC(u 3 v 2 ) /∈{i 1 , ,i 6 } or C(u 4 v 2 ) /∈{i 1 , ,i 6 }. (ii.1) C(u 3 v 2 ) /∈{i 1 , ,i 6 },andleti 7 = C(u 3 v 2 ). Since P is a heterochromatic path of length 5, we have that C({u 1 u 3 , ,u 1 u 6 }) −{i 1 , ,i 5 ,i 7 }= ∅.IfC(u 1 u 3 ) /∈ {i 1 , ,i 5 ,i 7 },letP  = u 2 u 1 u 3 Pu 6 v 1 ;ifC(u 1 u 4 ) /∈{i 1 , ,i 5 ,i 7 },letP  = v 2 u 3 u 2 u 1 u 4 u 5 u 6 ;ifC(u 1 u 5 ) /∈{i 1 , ,i 5 ,i 6 ,i 7 },letP  = v 1 v 2 u 3 u 2 u 1 u 5 u 6 ;ifC(u 1 u 6 ) /∈{i 1 , ,i 5 , i 7 },letP  = v 2 u 3 u 2 u 1 u 6 u 5 u 4 . Then, P  is a heterochromatic path of length 6 in all these cases. It remains to show that when C(u 1 u 5 )=i 6 , there is a heterochromatic path of length 6. Since d c (u 6 ) ≥ 7and|C({u 1 u 6 , ,u 4 u 6 }) −{i 2 ,i 3 ,i 4 ,i 5 }| ≥ 3, we have that C(u 6 v 2 ) ∈{i 3 ,i 4 } or there exists a v/∈{u 1 , ,u 6 ,v 1 ,v 2 } such that C(u 6 v) ∈{i 3 ,i 4 }.IfC(u 6 v 2 )=i 3 ,andsou 1 u 2 u 3 v 2 u 6 u 5 u 4 is a heterochromatic path of length 6; if C(u 6 v 2 )=i 4 ,thenu 4 u 3 u 2 u 1 u 5 u 6 v 2 is a heterochromatic path of length 6. If there is a vertex v/∈{u 1 , ,u 6 ,v 1 ,v 2 } such that C(u 6 v) ∈{i 3 ,i 4 },thenv 2 u 3 u 2 u 1 u 5 u 6 v is a heterochromatic path of length 6. (ii.2) C(u 4 v 2 ) /∈{i 1 , ,i 6 },andleti 7 = C(u 4 v 2 ). Since P is a heterochromatic path of length 5, we have that C({u 1 u 3 , ,u 1 u 6 }) −{i 1 , ,i 5 ,i 7 }= ∅.IfC(u 1 u 3 ) /∈ {i 1 , ,i 5 ,i 7 },letP  = u 2 u 1 u 3 u 4 u 5 u 6 v 1 ;ifC(u 1 u 4 ) /∈{i 1 , ,i 5 ,i 6 ,i 7 },letP  = u 2 u 1 u 4 u 5 u 6 v 1 v 2 ;ifC(u 1 u 5 ) /∈{i 1 , ,i 5 ,i 7 },letP  = v 2 u 4 u 3 u 2 u 1 u 5 u 6 ;ifC(u 1 u 6 ) /∈{i 1 , ,i 5 ,i 7 }, let P  = v 2 u 4 u 3 u 2 u 1 u 6 u 5 . Then, P  is a heterochromatic path of length 6 in all these cases. It remains to show that when C(u 1 u 4 )=i 6 , there is a heterochromatic path of length 6. Since u 1 u 2 u 3 u 4 v 2 v 1 is a heterochromatic path of length 5, we have that C({u 1 u 3 ,u 1 v 1 ,u 1 v 2 }) −{i 1 ,i 2 ,i 3 ,i 4 ,i 6 ,i 7 }= ∅.IfC(u 1 u 3 ) /∈{i 1 ,i 2 ,i 3 ,i 4 ,i 6 ,i 7 },andso u 2 u 1 u 3 u 4 v 2 v 1 u 6 is a heterochromatic path of length 6; if C(u 1 v 1 ) /∈{i 1 ,i 2 ,i 3 ,i 4 ,i 6 ,i 7 }, then v 2 v 1 u 1 u 2 u 3 u 4 u 5 is a heterochromatic path of length 6; if C(u 1 v 2 ) /∈{i 1 ,i 2 ,i 3 ,i 4 ,i 6 , i 7 },thenv 1 v 2 u 1 u 2 u 3 u 4 u 5 is a heterochromatic path of length 6. (iii) There are vertices v 2 ,v 3 /∈{u 1 , ,u 6 ,v 1 } such that |C({v 1 v 2 ,v 2 v 3 }) −C(P )| =2, and u 3 u 4 u 5 u 6 v 1 v 2 v 3 is a heterochromatic path of length 6. (5.3) k 0 =3. So,|C({u 1 u 6 , ,u 4 u 6 }) −{i 3 ,i 4 ,i 5 }| = 4. We have the following three cases: (i) There is no vertex v/∈ V (P ) such that C(v 1 v) /∈ C(P ), and so |C({u 1 v 1 , , u 5 v 1 }) − C(P )|≥2. Since |C({u 1 u 6 , ,u 4 u 6 }) −{i 3 ,i 4 ,i 5 }| =4,thereisau 7 /∈ V (P ) such that C(u 6 u 7 )=i 4 . If there exists a u/∈ V (P ) such that C(uu 1 ) /∈ C(P ), then uu 1 Pu 6 is a heterochromatic path of length 6. If C(u 3 v 1 ) /∈ C(P ), then u 1 u 2 u 3 v 1 u 6 u 5 u 4 is a hete- rochromatic path of length 6. If there exists an 2 ≤ i ≤ 4 such that |C({u i v 1 ,u i+1 v 1 }) − C(P )| =2,thenu 1 Pu i v 1 u i+1 Pu 6 is a heterochromatic path of length 6. So we shall only show that there is a heterochromatic path of length 6 when |C({u 1 u 3 , ,u 1 u 6 })−C(P )|≥ 2andC(u 2 v 1 ) /∈ C(P ). Let i 6 = C(u 2 v 1 ). Then C({u 1 u 3 , ,u 1 u 6 }) −{i 1 , ,i 5 ,i 6 }= ∅.IfC(u 1 u 3 ) /∈{i 1 , ,i 6 },letP  = v 1 u 2 u 1 u 3 u 4 u 5 u 6 ;ifC(u 1 u 4 ) /∈{i 1 , ,i 6 },let P  = u 3 u 2 u 1 u 4 u 5 u 6 v 1 ;ifC(u 1 u 5 ) /∈{i 1 , ,i 6 },letP  = u 4 u 3 u 2 u 1 u 5 u 6 u 7 ;ifC(u 1 u 6 ) /∈ {i 1 , ,i 6 },letP  = v 1 u 2 u 1 u 6 u 5 u 4 u 3 . Then, P  is a heterochromatic path of length 6 in all these cases. (ii) There is a vertex v 2 such that C(v 1 v 2 ) /∈ C(P ), and there is no vertex v/∈ {u 1 , ,u 6 ,v 1 ,v 2 } such that C(v 1 v 2 ) /∈{i 1 , ,i 5 }∪C(v 1 v 2 ). Let i 6 = C(v 1 v 2 ). Then the electronic journal of combinatorics 12 (2005), #R33 5 C({u 1 v 2 , ,u 5 v 2 }) −{i 1 , ,i 6 }= ∅.IfC(u 1 v 2 ) /∈{i 1 , ,i 6 },thenv 2 u 1 Pu 6 is a hete- rochromatic path of length 6. If C(u 2 v 2 ) /∈{i 1 , ,i 6 },thenv 1 v 2 u 2 Pu 6 is a heterochro- matic path of length 6. If C(u 3 v 2 ) /∈{i 1 , ,i 6 },thenu 1 u 2 u 3 v 2 v 1 u 6 u 5 is a heterochromatic path of length 6. If C(u 5 v 2 ) /∈{i 1 , ,i 6 },thenu 1 Pu 5 v 2 v 1 is a heterochromatic path of length 6. Next we shall consider the case when C(u 4 v 2 ) /∈{i 1 , ,i 6 }.Leti 7 = C(u 4 v 2 ). Since |C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 3 ,i 4 ,i 5 }| = 4, there is a vertex u/∈ V (P ) such that C(u 6 u)=i 4 .Ifu = v 2 , i.e., C(u 6 v 2 )=i 4 ,thenu 1 u 2 u 3 u 4 v 2 u 6 u 5 is a heterochromatic path of length 6. It remains to show that there is a heterochromatic path of length 6 if there is a vertex u/∈{u 1 , ,u 6 ,v 1 ,v 2 } such that C(u 6 u)=i 4 . In this case, since |C({u 1 u 3 , ,u 1 u 6 }) − C(P )|≥2, we have that C({ u 1 u 4 ,u 1 u 5 ,u 1 u 6 }) − C(P ) = ∅.If C(u 1 u 4 ) /∈ C(P ), let P  = u 3 u 2 u 1 u 4 u 5 u 6 v 1 ;ifC(u 1 u 5 ) /∈ C(P ), let P  = u 4 u 3 u 2 u 1 u 5 u 6 u;if C(u 1 u 6 ) /∈{i 1 , ,i 5 ,i 7 },letP  = v 2 u 4 u 3 u 2 u 1 u 6 u 5 . Then, P  is a heterochromatic path of length 6. Last, we consider the case when C(u 1 u 6 )=i 7 .Sinceu 3 u 2 u 1 u 6 v 1 v 2 is a hete- rochromatic path of length 5, we have |C({u 1 v 2 ,u 2 v 2 ,u 3 v 2 ,u 6 v 2 }) −{i 1 ,i 2 ,i 3 ,i 6 ,i 7 }| ≥ 2, and so C({u 1 v 2 ,u 3 v 2 ,u 6 v 2 })−{i 1 ,i 2 ,i 3 ,i 6 ,i 7 }= ∅.IfC(u 1 v 2 )−{i 1 ,i 2 ,i 3 ,i 4 ,i 6 ,i 7 }= ∅,let P  = v 1 v 2 u 1 Pu 5 ;ifC(u 1 v 2 )=i 4 ,letP  = u 3 u 2 u 1 v 2 v 1 u 6 u 5 ;ifC(u 3 v 2 ) −{i 1 ,i 2 ,i 3 ,i 4 ,i 6 ,i 7 } = ∅,letP  = u 1 u 2 u 3 v 2 v 1 u 6 u;ifC(u 3 v 2 )=i 4 ,letP  = u 1 u 2 u 3 v 2 v 1 u 6 u 5 ;ifC(u 6 v 2 ) − {i 1 ,i 2 ,i 3 ,i 6 ,i 7 }= ∅,letP  = u 4 u 3 u 2 u 1 u 6 v 2 v 1 . Then, P  is a heterochromatic path of length 6 in all these cases. (iii) There are vertices v 2 ,v 3 /∈{u 1 , ,u 6 ,v 1 } such that |C({v 1 v 2 ,v 2 v 3 }) −C(P )| =2. Let i 6 = C(v 1 v 2 )andi 7 = C(v 2 v 3 ). If there exists a v/∈{u 4 ,u 5 ,v 1 } such that C(v 3 v) /∈ {i 3 , ,i 7 }, i.e., there exists a v/∈{u 4 ,u 5 ,u 6 ,v 1 ,v 2 } such that C(v 3 v) /∈{i 3 , ,i 7 }, then u 4 u 5 u 6 v 1 v 2 v 3 v is a heterochromatic path of length 6. Next we shall only consider the case when |C({u 4 v 3 ,u 5 v 3 ,v 1 v 3 }) −{i 3 , ,i 7 }| ≥ 2. If C(u 5 v 3 ) /∈{i 2 , ,i 7 },then u 2 u 3 u 4 u 5 v 3 v 2 v 1 is a heterochromatic path of length 6. If C(u 5 v 3 )=i 2 ,thenu 3 u 4 u 5 v 3 v 2 v 1 is a heterochromatic path of length 5 and C(v 1 u 6 )=C(u 3 u 4 ), and so there is a heterochro- matic path of length 6 from the cases discussed above. If C(u 4 v 3 ) /∈{i 1 , ,i 7 },then v 1 v 2 v 3 u 4 u 3 u 2 u 1 is a heterochromatic path of length 6. If C(u 4 v 3 )=i 1 ,thenu 2 u 3 u 4 v 3 v 2 v 1 is a heterochromatic path of length 5 and C(v 1 u 6 )=C(u 3 u 4 )=i 3 ,andsothereisa heterochromatic path of length 6 because of (5.1) and (5.2). Now it remains to show that there is a heterochromatic path of length 6 when C(u 4 v 3 )=i 2 and C(v 1 v 3 ) /∈{i 2 , ,i 7 }. Since |C({u 1 u 6 ,u 2 u 6 ,u 3 u 6 ,u 4 u 6 }) −{i 3 , ,i 5 }| =4,thereisan1≤ x ≤ 3 such that C(u x u 6 ) /∈{i 2 , ,i 6 }.IfC(u x u 6 ) /∈{i 2 , ,i 7 },thenv 1 v 2 v 3 u 4 u 5 u 6 u x is a heterochro- matic path of length 6; if C(u x u 6 )=i 7 ,thenv 2 v 1 v 3 u 4 u 5 u 6 u x is a heterochromatic path of length 6. The proof is now complete. Actually, we can show that for 1 ≤ k ≤ 5 any graph G under the color degree condition has a heterochromatic path of length at least k, with only one exceptional graph K 4 for k = 3, one exceptional graph for k = 4 and three exceptional graphs for k = 5, for which G has a heterochromatic path of length at least k − 1. the electronic journal of combinatorics 12 (2005), #R33 6 3. Long heterochromatic paths for k ≥ 8 From the above section we know that when 1 ≤ k ≤ 4, under the color degree condition G always has a heterochromatic path of length  3k 5 ,andwhen5≤ k ≤ 7, G always has a heterochromatic path of length  3k 5  + 1. In this section we give our main result and do some preparations for its proof. The detailed proof is left in the next section. Theorem 3.1 Let G be an edge-colored graph and k ≥ 8 an integer. Suppose that d c (v) ≥ k for every vertex v of G. Then G has a heterochromatic path of length at least  3k 5  +1. Before proving the result, we will do some preparations, first. Let G be an edge-colored graph and k ≥ 8 an integer. Suppose that d c (v) ≥ k for every vertex v of G.LetP = u 1 u 2 u 3 u l−1 u l u l+1 v 1 v 2 v s be a path in G such that (a) u 1 Pu l+1 is a longest heterochromatic path in G; (b) C(u l+1 v 1 )=C(u k 0 u k 0 +1 )and1≤ k 0 ≤ l is as small as possible, subject to (a); (c) v 1 Pv s is a heterochromatic path in G with C(u 1 Pu l+1 ) ∩ C(v 1 Pv s )=∅ and v 1 Pv s is as long as possible, subject to (a) and (b). Let i j = C(u j u j+1 ) for 1 ≤ j ≤ l and i l+j = C(v j v j+1 ) for 1 ≤ j ≤ s − 1, then C(u l+1 v 1 )=i k 0 .Thereexistt 1 ≥ 0and1≤ x 1 <x 2 < < x t 1 ≤ l such that c({u x 1 v s ,u x 2 v s , ,u x t 1 v s })=t 1 and C({u x 1 v s ,u x 2 v s , ,u x t 1 v s })=C({u 1 v s , ,u l v s }) −{i 1 , ,i l+s−1 }.Leti l+s+j−1 = C(u x j v s ) for all 1 ≤ j ≤ t 1 .Therealsoexist0≤ t 2 ≤ s − 2and1≤ y 1 <y 2 < < y t 2 ≤ s − 2 such that C({v 1 v s ,v 2 v s , ,v s−2 v s }) − {i 1 ,i 2 , ,i l+s+t 1 −1 } = C({v y 1 v s ,v y 2 v s , ,v y t 2 v s })andc({v y 1 v s ,v y 2 v s , ,v y t 2 v s })=t 2 . Let i l+s+t 1 +j−1 = C(v y j v s ) for all 1 ≤ j ≤ t 2 . Then it is easy to get the following Lemmas. In these lemmas we assume that l =  3k 5 . Lemma 3.2 s ≤ k 0 ≤ 2l − k. Proof. Since d c (u l+1 ) ≥ k and u 1 Pu l+1 is a longest heterochromatic path in G,thereareat least k − l different edges in {u 1 u l+1 ,u 2 u l+1 , u l−1 u l+1 } that have different colors which are not in {i 1 ,i 2 , ,i l }.Thenwehavethatk 0 ∈{1, 2, ,(l − 1) − (k − l)+1=2l − k}. On the other hand, if s>k 0 ,thenP  = u k 0 +1 Pv k 0 +1 is a heterochromatic path of length l + 1, a contradiction to the choice of P .So,s ≤ k 0 ≤ 2l − k. Lemma 3.3 There are at least k − l + k 0 − 1 different colors not in {i k 0 , ,i l } that belong to C({u 1 u l+1 , ,u l−1 u l+1 }), and C({u 1 v s , ,u s v s }∪{u k 0 −s+1 v s , ,u k 0 v s }∪ {u l−s+2 v s , ,u l v s }) ⊆{i 1 , ,i l+s−1 }. Proof. By the choice of P ,wehaveCN(u l+1 ) − C({u 1 u l+1 , ,u l−1 u l+1 }) ⊆{i k 0 , ,i l }. Since d c (u l+1 ) ≥ k,thereareatleastk − (l − k 0 +1)=k − l + k 0 − 1 different colors not in {i k 0 , ,i l } that belong to C({u 1 u l+1 , ,u l−1 u l+1 }). If there exists an x ∈{1, 2, ,s}∪{k 0 − s +1, ,k 0 }∪{l − s +2,l− s +3, ,l} the electronic journal of combinatorics 12 (2005), #R33 7 such that u x v s has a color not in {i 1 , ,i l+s−1 },then P  =    v 1 Pv s u x Pu l+x−s+1 if x ∈{1, 2, ,s}; u 1 Pu x v s P −1 u x+s if x ∈{k 0 − s +1, ,k 0 }; u x−(l−s+1) Pu x v s P −1 v 1 if x ∈{l − s +2,l− s +3, ,l}. is a heterochromatic path of length l + 1, a contradiction to the choice of P .So, C({u 1 v s , ,u s v s ,u k 0 −s+1 v s , ,u k 0 v s ,u l v s , ,u l−s+2 v s }) ⊆{i 1 , ,i l+s−1 }. Lemma 3.4 s<x 1 <x 1 +1 <x 2 <x 2 +1 < <x t 1 ≤ l −s+1, t 1 +t 2 ≥ k −(l +s−1) and max{k − l − 2s +3, 0}≤t 1 ≤ l−2s+1 2 , 0 ≤ t 2 ≤ s − 2. Proof. It is obvious that t 1 + t 2 ≥ k − (l + s − 1) and 0 ≤ t 2 ≤ s − 2, and so t 1 ≥ k − (l + s − 1) − (s − 2) = k − l − 2s + 3. From Lemma 3.3, we have that s<x 1 <x 2 < <x t 1 ≤ l − s + 1. If there exists a j with 1 ≤ j ≤ t 1 − 1 such that u x j +1 = u x j+1 ,let P  = u 1 Pu x j v s u x j+1 Pu l+1 ,thenP  a heterochromatic path of length l + 1, a contradiction to the choice of P.So,s ≤ k 0 <x 1 <x 1 +1<x 2 <x 2 +1< < x t 1 ≤ l − s +1, t 1 + t 2 ≥ k − (l + s − 1) and max{k − l − 2s +3, 0}≤t 1 ≤ l−2s+1 2 ,0≤ t 2 ≤ s − 2. Lemma 3.5 Let t 1 =0. Then k ≡ 2, 4(mod 5), k 0 = s =2l − k and t 2 = s − 2 if k ≡ 4(mod 5); t 2 ≥ s − 3 if k ≡ 2(mod 5). There are exactly l − 1 differ- ent colors not in {i k 0 ,i k 0 +1 , ,i l } that belong to C(u 1 u l+1 , ,u l−1 u l+1 ), and CN(v s ) − {u s+1 v s , ,u l+1 v s ,v 1 v s , ,v s−2 v s }⊆{i k 0 , ,i l+s−1 }. Proof. Since 0 = t 1 ≥ k − l − 2s +3,we have k − l +3≤ 2s ≤ 2(2l − k)=4l − 2k.On the other hand, from (4l − 2k) − (k − l +3)=5l − 3k − 3, we have that 5l − 3k − 3 < 0if k ≡ 0, 1, 3(mod 5), 5l − 3k − 3=0ifk ≡ 4(mod 5) and 5l − 3k −3=1ifk ≡ 2(mod 5), which implies that k ≡ 2, 4(mod 5) and k 0 = s =2l − k.Sincet 2 ≥ k − l − s +1, we have that t 2 = s − 2ifk ≡ 4(mod 5) and t 2 ≥ s − 3ifk ≡ 2(mod 5). From Lemma 3.3, there are at least k − l + k 0 − 1=k − l + s − 1=k − l +2l − k − 1=l − 1 different colors not in {i k 0 ,i k 0 +1 , ,i l } that belong to C(u 1 u l+1 , ,u l−1 u l+1 ). If there exists a v/∈{u s+1 ,u s+2 , ,u l+1 ,v 1 , ,v s } such that v s v has a color not in {i k 0 , ,i l+s−1 },then P  = u s+1 Pv s v is a heterochromatic path of length l + 1, a contradiction to the choice of P . Lemma 3.6 If there exists an 1 ≤ x ≤ x 1 − 1 such that u x u l+1 has a color in {i l+1 , , i l+s− x 1 −1 2  }∪{i l+ s+x 1 −t 2 2 −2 ,i l+ s+x 1 −t 2 2 −1 , ,i l+s−1 }, then there is a heterochromatic path P  of length l +1in G. Proof. First, note that (l +  s+x 1 −t 2 2 −2) −(l + s − x 1 −1 2 )= s+x 1 −t 2 2 −2 −s+ x 1 −1 2 ≥  x 1 +2 2  +  x 1 −1 2 −s − 2=x 1 +1− s − 2 ≥ 0. If there exists an 1 ≤ x ≤ x 1 − 1 such that u x u l+1 has a color in {i l+1 , ,i l+s− x 1 −1 2  }, then let P  =  v s− x 1 −1 2 +1 Pv s u x 1 Pu l+1 u x P −1 u x−( x 1 −1 2 +1)+1 if  x 1 −1 2  +1≤ x ≤ x 1 − 1; v s− x 1 −1 2 +1 Pv s u x 1 Pu l+1 u x Pu x+( x 1 −1 2 +1)−1 if 1 ≤ x ≤ x 1 −1 2 . the electronic journal of combinatorics 12 (2005), #R33 8 Since  x 1 −1 2  +( x 1 −1 2  +1)− 1=2 x 1 −1 2 ≤x 1 −1, P  is a heterochromatic path of length l +1. If there exists an 1 ≤ x ≤ x 1 − 1 such that u x u l+1 has a color i l+y ∈{i l+ s+x 1 −t 2 2 −2 , i l+ s+x 1 −t 2 2 −1 , ,i l+s−1 }, then since t 2 − [(s − 2) − ( s+x 1 −t 2 2 −2)] = t 2 − s +  s+x 1 −t 2 2 ≥t 2 − s + s+x 1 −t 2 2 = t 2 −s+x 1 2 ≥ k−l−s+1−t 1 −s+s+1 2 ≥ k−l−s+2 2 − 1 2  l−2s+1 2  ≥ k−l−s+2 2 − l−2s+2 4 = 2k− 3l+2 4 > 0 and y t 2 −[(s−2)−( s+x 1 −t 2 2 −2)] ≤ s+x 1 −t 2 2 −2, there are some y  ∈{1, 2, , s−t 2 −1 2 } ∪ { s+x 1 −t 2 2 −2,  s+x 1 −t 2 2 −3, , s+x 1 −t 2 2 − s−t 2 −1 2 −1} such that y  ∈{y 1 ,y 2 , ,y t 2 }. Let P 1 =      v y  + s+x 1 −t 2 2 − s−t 2 −1 2 −2 P −1 v y  v s if y  ∈{1, 2, , s−t 2 −1 2 }; v y  − s+x 1 −t 2 2 + s−t 2 −1 2 +2 Pv y  v s if y  ∈{ s+x 1 −t 2 2 −2, ,  s+x 1 −t 2 2 − s−t 2 −1 2 −1}. Note that if y  ∈{1, 2, , s−t 2 −1 2 },wehavethaty  + s+x 1 −t 2 2 − s−t 2 −1 2 −2 ≤ s−t 2 −1 2 +  s+x 1 −t 2 2 − s−t 2 −1 2 −2= s+x 1 −t 2 2 −2, and if y  ∈{ s+x 1 −t 2 2 −2, , s+i 1 −t 2 2 − s−t 2 −1 2 − 1},wehavethaty  − s+x 1 −t 2 2  +  s−t 2 −1 2  +2≥ s+x 1 −t 2 2 − s−t 2 −1 2 −1 − s+x 1 −t 2 2  +  s−t 2 −1 2  +2≥ 1. Then, P 1 is a heterochromatic path of length  s+x 1 −t 2 2 − s−t 2 −1 2 −1 with colors not in {i 1 ,i 2 , ,i l ,i l+y ,i l+s }.Let P  =          P 1 u x 1 Pu l+1 u x P −1 u x−(x 1 − s+x 1 −t 2 2 + s−t 2 −1 2 )+1 if x 1 − s+x 1 −t 2 2 +  s−t 2 −1 2 ≤x ≤ x 1 − 1; P 1 u x 1 Pu l+1 u x Pu x+(x 1 − s+x 1 −t 2 2 + s−t 2 −1 2 )−1 if 1 ≤ x ≤ x 1 −  s+x 1 −t 2 2  +  s−t 2 −1 2 −1. Since 2(x 1 − s+x 1 −t 2 2  +  s−t 2 −1 2 −1) = 2x 1 − 2 s+x 1 −t 2 2  +2 s−t 2 −1 2 −2 ≤ 2x 1 − (s + x 1 − t 2 )+(s − t 2 ) − 2=x 1 − 2, P  is a heterochromatic path of length l +1. Lemma 3.7 Let t 1 =0, k ≥ 8, s +1≤ x ≤ l − s +2and C(u x v s )=i 1 . If there exists an 2 ≤ x  ≤ x − 1 such that u x  u l+1 has a color in {i l+1 , ,i l+s− x 2  ,i l+ x 2 −1 , ,i l+s−1 } then there is a heterochromatic path P  of length l +1in G. Proof. Since t 1 =0,wehavethatk ≡ 2, 4(mod 5) and s =2l − k,andthatt 2 = s − 2if k ≡ 4(mod 5); t 2 ≥ s − 3ifk ≡ 2(mod 5) from Lemma 3.5. If there exists an 2 ≤ x  ≤ x − 1 such that u x  u l+1 has a color in {i l+1 , ,i l+s− x 2  }, then let P  =  v s− x 2 +1 Pv s u x Pu l+1 u  x P −1 u x  − x 2 +1 if  x 2  +1≤ x  ≤ x − 1; v s− x 2 +1 Pv s u x Pu l+1 u  x Pu x  + x 2 −1 if 2 ≤ x  ≤ x 2 . the electronic journal of combinatorics 12 (2005), #R33 9 Note that if 2 ≤ x  ≤ x 2 ,thenx  +  x 2 −1 ≤ 2 x 2 −1 ≤ x − 1, and so P  is a heterochromatic path of length l +1. If there exists an 2 ≤ x  ≤ x − 1 such that u x  u l+1 has a color in {i l+ x 2 −1 , ,i l+s−1 }, since t 2 = s − 2ifk ≡ 4(mod 5) and t 2 ≥ s − 3ifk ≡ 2(mod 5), and k ≥ 8, we have t 2 ≥ 1. On the other hand, if k ≡ 2(mod 5), then 2 ≤ s+1 2 −1 ≤ x 2 −1 ≤  l−s+2 2 −1= k−l+2 2 −1= 2s+(k−l−2(2l−k)+2) 2 −1=s − 2, and so C({ v 1 v s ,v  x 2 −1 v s })  {i 1 ,i 2 , ,i l+s−1 }.Let P 1 =  v 1 Pv  x 2 −1 v s if v  x 2 −1 v s has a color not in {i 1 ,i 2 , ,i l+s−1 }; v  x 2 −1 P −1 v 1 v s if v 1 v s has a color not in {i 1 ,i 2 , ,i l+s−1 }. P  =  P 1 u x Pu l+1 u x  P −1 u x  − x 2 +1 if  x 2  +1≤ x  ≤ x − 1; P 1 u x Pu l+1 u x  Pu x  + x 2 −1 if 2 ≤ x  ≤ x 2 . Note that if 2 ≤ x  ≤ x 2 ,thenx  +  x 2 −1 ≤ 2 x 2 −1 ≤ x − 1, and so P  is a heterochromatic path of length l +1. Lemma 3.8 Let t 1 =0, |C(v 1 v s , ,v s−2 v s ) −{i 2 , ,i l+s−1 }| = s − 2, k ≥ 8 and C(u x v s )=i 2 . If there exists an 3 ≤ x  ≤ x − 1 such that u x  u l+1 has a color in {i l+1 , ,i l+s− x+1 2  ,i l+ x+1 2 −1 , ,i l+s−1 }, then there is a heterochromatic path of length l +1. Proof. If there exists an 3 ≤ x  ≤ x − 1 such that u x  u l+1 has a color in {i l+1 , , i l+s− x+1 2  },thenlet P  =  v s− x+1 2 +1 Pv s u x Pu l+1 u x  P −1 u x  − x+1 2 +2 if  x+1 2  +1≤ x  ≤ x − 1; v s− x+1 2 +1 Pv s u x Pu l+1 u x  Pu x  + x+1 2 −2 if 3 ≤ x  ≤ x+1 2 . Note that if 3 ≤ x  ≤ x+1 2 ,thenx  +  x+1 2 −2 ≤ 2 x+1 2 −2 ≤ x − 1, and so P  is a heterochromatic path of length l +1. If there exists an x  with 3 ≤ x  ≤ x−1 such that u x u l+1 has a color in {i l+ x+1 2 −1 , , i l+s−1 },thenlet P  =  v  x+1 2 −1 P −1 v 1 v s u x Pu l+1 u x  P −1 u x  − x+1 2 +2 if  x+1 2  +1≤ x  ≤ x − 1; v  x+1 2 −1 P −1 v 1 v s u x Pu l+1 u x  Pu x  + x+1 2 −2 if 3 ≤ x  ≤ x+1 2 , and so P  is a heterochromatic path of length l +1. Lemma 3.9 Let t 1 ≥ 1, k ≥ 8 and k ≡ 1, 2, 4(mod 5). Then there is no x j (1 ≤ j ≤ t 1 ) such that C({u 1 u l+1 ,u 2 u l+1 , ,u x j −1 u l+1 }) −{i 1 ,i 2 , ,i l ,i l+1 , ,i l+s−1 ,i l+s+j−1 }= ∅. Proof. Suppose there is some x j (1 ≤ j ≤ t 1 ) such that C({u 1 u l+1 ,u 2 u l+1 , ,u x j −1 u l+1 }) −{i 1 ,i 2 , ,i l ,i l+1 , ,i l+s−1 ,i l+s+j−1 }= ∅.Letx j 0 be the one of such x j with the smallest subscript, and u x  u l+1 (1 ≤ x  ≤ x j 0 − 1) has a color not in {i 1 ,i 2 , ,i l ,i l+1 , , the electronic journal of combinatorics 12 (2005), #R33 10 [...]... that any longest heterochromatic path of G is of length k − 1 In fact, let Gk be an edge-colored graph whose vertices are the ordered (k − 1)-tuples of 0’s and 1’s; two vertices are joined by an edge if and only if they differ in exactly one coordinate or they differ in all coordinates An edge is in color j (1 ≤ j ≤ k − 1) if and only if its two ends differ in exactly the jth coordinate, or in color k... M Spyratos and Zs Tuza, Paths through fixed vertices in edge-colored graphs, Math Inf Sci Hun 32 (1994), 49–58 the electronic journal of combinatorics 12 (2005), #R33 31 [6] P Erd˝s and Zs Tuza, Rainbow Hamiltonian paths and canonically colored subo graphs in infinite complete graphs, Mathematica Pannonica 1 (1990), 5–13 [7] P Erd˝s and Zs Tuza, Rainbow subgraphs in edge-colorings of complete graphs,... vertex v in G with 8 ≤ k ≤ k − 1 In the following, all the notations are the same as in Section 3 Since dc (v) ≥ k > k − 1 for every vertex v of G, G has a heterochromatic path of length 3(k−1) + 1 It remains to show that l ≥ 3k + 1, which implies that 5 5 P = u1 P u 3k +2 is a heterochromatic path of length 3k + 1 We will proceed by con5 5 tradictions Suppose that l ≤ 3k On the other hand, by induction... the longest heterochromatic path in H is of length 4 Then, by exhausting all the possible adjacency and colorings of edges for 1 ≤ k ≤ 5, we can get that any graph G under the color degree condition has a heterochromatic path of length at least k, with only one exceptional graph K4 for k = 3, one exceptional graph G4 for k = 4 and three exceptional graphs G5 , G5 and H for k = 5, for which G has a heterochromatic. .. in color k if and only if its two ends differ in all the coordinates Then it is not difficult to check that Gk is a graph we want Another class of graphs is given as follows Since Kk is (k − 1)-edge-colorable when k is even, we can get a proper k-edge coloring for Kk+1 when k is odd Denote it by Gk Then, it is obvious that the longest heterochromatic path in Gk is of length k − 1 when k is odd Remark... P is a heterochromatic path of length l + 1, a contradiction to the choice of P The proof is now complete 5 Concluding remarks Finally, we consider whether our lower bound is best possible It is obvious that when k = 1, 2, the bound is best possible Next we consider the case when k ≥ 3 Remark 5.1 For any integer k ≥ 3, there is an edge-colored graph Gk with dc (v) ≥ k for all the vertices v in G such... then x + x ≤ 2 x = x − 1, and so P is a heterochromatic 2 2 2 path of length l + 1, a contradiction to the choice of P So, we get that i1 ∈ C({u1 vs , u2vs , , ul vs }) / Now we turn to proving our main theorem 4 Proof of Theorem 3.1 Proof We will use induction on k to prove the theorem For k = 8, by Theorem 2.1 there is a heterochromatic path of length 6 = in G 3k 5 +1 Suppose now k ≥ 9, and the... (z − s) + 1 = 4, and so P is a heterochromatic path of length l + 1, a contradiction to the choice of P So, there are at least (z − 4) − (k0 − 3) = z − s − 1 = s − 2 colors in {il+1 , , il+s−1} that belong to C({u4 ul+1, , uz−1ul+1 }), and there are at least 2 colors in {il+1 , il+s−2 , il+s−1} that the electronic journal of combinatorics 12 (2005), #R33 27 belong to C({u4 ul+1 , , us ul+1,... color il+s−2 If s ≥ 5, P is a heterochromatic path of length l + 1, a contradiction to the choice of P So, s = 4 Since there are exactly l − 1 different colors not in {ik0 , ik0 +1 , , il } that belong to C({u1 ul+1, , ul−1ul+1 }), v1 P vs us+2P ul+1u1 u2 is a heterochromatic path of length l + 1 if C(u1 ul+1 ) ∈ {i1 , , il+s−1 }; v1 P vs us+2 P ul+1u2 u1 is a heterochromatic / path of length... cycles, Electronic J Combin 2 (1995), Research Paper R10 [2] M Axenovich, T Jiang and Zs Tuza, Local anti-Ramsey numbers of graphs, Combin Probab Comput 12 (2003), 495–511 [3] J.A Bondy and U.S.R Murty, Graph Theory with Applications, Macmillan London and Elsevier, New York (1976) [4] H.J Broersma, X.L Li, G Woeginger and S.G Zhang, Paths and cycles in colored graphs, Australasian J Combin 31 (2005), 297–309 . Long heterochromatic paths in edge-colored graphs He Chen and Xueliang Li Center for Combinatorics and LPMC Nankai University Tianjin 300071, P.R. China lxl@nankai.edu.cn Submitted:. segment of P from u to v. There are many existing literature dealing with the existence of paths and cycles with special properties in edge-colored graphs. In [5], the authors showed that for a 2-edge- colored. 1. Proof. (1) k =3. Sincek =3> 2, there is a heterochromatic path of length 2 in G. the electronic journal of combinatorics 12 (2005), #R33 2 (2) k =4. Sincek =4> 3, there is a heterochromatic