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Rainbow Connection and Strong Rainbow Connection Numbers Of RAINBOW CONNECTION AND STRONG RAINBOW CONNECTION NUMBERS OF Srava Chrisdes Antoro Fakultas Ilmu Komputer, Universitas Gunadarma [email protected] Abstract A rainbow path in an edge coloring of graph is a path in which every two edges are assigned different colors. If a nontrivial connected graph contains a rainbow path for every two vertices in , then is rainbow-connected. The rainbow connection number of is the minimum integer such that is rainbow connnected in an edge- coloring with colors. If a nontrivial connected contains a rainbow path geodesic for every two vertices in , then is strongly rainbow-connected. The strong rainbow connection number of is the minimum integer such that is strongly rainbow-connected in an edge coloring with colors. The rainbow connection and strong rainbow connection number of graph , , and have been found. In this paper, the results will be generalized and the rainbow connection and strong rainbow connection number of graph will be determined. Keywords: rainbow coloring, rainbow connection number, strong rainbow coloring, strong rainbow connection number, graph INTRODUCTION distance between two distinct vertices in . All graphs considered in this paper In 2008, Chartand et. al. introduced the are finite, undirected, and simple. For following two new types of edge labeling. standard terminology and notation, we follow Bondy and Murty (2008). We use Definition 2 (Chartrand et. al., 2008). and to denote the set of Let be a graph and vertices and the set of edges, respectively. , then is called an edge coloring or -coloring on Definition 1 (Prabha and Rajasingh, 2012). where . A path is said to be a The distance between two distinct rainbow path if all edges in that path have vertices and of a connected simple different colors. A rainbow geodesic is a graph , denoted by , is the length geodesic which is also a rainbow path. We say that is a rainbow coloring (number of edges) of the shortest path in between and in . The radius of a if any two vertices in are connected by graph , denoted by , is the a rainbow path in . We say that is a minimum over all vertices of the strong rainbow coloring if any two vertices in are connected by a rainbow maximum distance from to another geodesic in . These two kinds of vertex in . The diameter of a graph , coloring give rise to the following two denoted by , is the maximum graph parameters. 120 Jurnal Teknologi Rekayasa Volume 21 No.2, Agustus 2016 Definition 3 (Chartrand et. al., 2008). Definition 4 (Bondy and Murty, 2008). The rainbow connection number of Let and be vertex-disjoint a graph , denoted by , is the graphs. The join is the graph smallest for which has a rainbow - obtained by adding a new edge from each coloring. The strong rainbow connection vertex of to each vertex of . number of , denoted by , is the smallest for which has a strong Definition 5 (Bondy and Murty, 2008). rainbow -coloring. Some basic proper- Let be a graph. The complement ties are collected in this following lemma. of is a new graph, denoted by , with Their proofs are omitted. the same set of vertices as , such that two vertices are adjacent in if and only Lemma 1 (Chartrand et. al., 2008). Let if they are not adjacent in . be a connected graph. Then: Next, some known results about the 1) . rainbow connection and strong rainbow 2) if and only if . connection numbers of , , and are given in the following 3) if and only if is a propositions. complete graph. 4) if and only if is a Proposition 1 (Chartrand et. al., 2008). tree. For every integer , 5) If is a connected spanning subgraph of , then . Proposition 2 (Chartrand et. al., 2008). If in addition , then For every integer , also . 1) 6) If , then 2) and . Proposition 3 (Antoro and Sugeng, In Chartrand et. al. (2008), the 2015). For every integer , rainbow connection and strong rainbow 1) connection of the wheel graph , which is the same as the join between the cycle 2) graph and a graph with only one vertex, i.e. , have been studied. Moreover, in Antoro and Sugeng MAIN RESULTS (2015) the rainbow connection and strong rainbow connection of graph In this section, the formula for the rainbow connection and strong rainbow also have been studied. In this paper, the connection of will be stated and more general graph is con- proved for all with and sidered. Let us recall the definition of join and complement. The reason for the condition is because a cycle is defined to have at least three vertices. Antoro, Rainbow Connection ... 121 The reason for requiring is because is Note that each color code is an - a complete graph (so we already know its tuple of numbers from . rainbow connection and strong rainbow So, there are at most different connection). It should be mentioned that codes. On the other hand, from the the proof that we present below is a definition of ceiling function , combination of several ideas in Chartrand et. al.’s original paper, especially the we have . So, . ideas that were used to find the rainbow Therefore, it is impossible that all vertices connection and strong rainbow in have different codes. So, we can find connection of complete bipartite graphs some with that and wheels. satisfies . Theorem. Let with and Let be a rainbow geodesic in . Then: from to . Clearly, . So, must have the form 1) for some . Since is rainbow, . This 2) means that and are different in the ’th entry. This contradicts to . Proof. So, we get . Or, Let , with , . Because and and . Note that since , then . , is not a complete graph. Claim 1 is proved. So, and also . For Claim 2: If , then convenience, we split the proof into . several claims. Each claim is immediately followed by its own proof. Claim 1 to 5 Let and . Then, prove the formula for , while . To prove Claim 2, it is Claim 6 to 8 prove the formula for . enough to produce a strong rainbow coloring on with colors. Let be the Claim 1: If , then . subgraph of induced by Let . Note that has a spanning . Then, subgraph isomorphic to . and by point 6 in Lemma 1. Since and , by Choose a strong rainbow coloring Lemma 1 ((5), (4), and (6)), we get . We will extend to . a strong rainbow coloring on . Define Now, suppose that . as follows This means there is a strong rainbow coloring on with colors. Let . For each We will show that is a strong rainbow , define the color code as coloring. Let be non-adjacent 122 Jurnal Teknologi Rekayasa Volume 21 No.2, Agustus 2016 vertices. We will find a rainbow geodesic From , we have . from to . Since and are non- So, . This adjacent, we only need to consider the means there are at least four vertices in following two cases. with the same color code (otherwise, Case 2.1 : , say and each code appears at most three times, so with . in total there would be If , then we can use codes, this is a contradiction because 1,2 path or 2,1 path in as a rainbow has n vertices). From these four vertices, geodesic. Now suppose . we can find at least two of them whose distance in is at least 3. So, we can Let and . Then, find with . Choose any rainbow geodesic and . Let be a for some , and let rainbow geodesic in from to . Then, is a geodesic in Since , (since ) and for some . Since is rainbow, . So, . But, it means is a rainbow geodesic from to . and are different in Case 2.2 : . the th entry. It contradicts In this case, . So, there . So, we get is a rainbow geodesic . Or, . where for some Claim 3 is proved. This is also geodesic in , and . Claim 4: If , then So, is a rainbow geodesic from to . Or, is a rainbow geodesic from to . Let . Note that has a With this, we conclude that spanning subgraph isomorphic to . Claim 2 is proved. Since and , by Lemma 1 ((5), (4), and Claim 3: If , then (6)), we get . From , we get . So, . Therefore, . If , Let and . Suppose then . So, would be a that . Choose a strong complete graph, which is not true because rainbow coloring . Hence, . For each , define the color Claim 4 is proved. code as Claim 5: There are no that Note that each color code is an r-tuple of satisfy . numbers from . So, there Suppose there are some that satisfy are at most different codes. the chain of inequalities. From From , we have . , we get , but also So, , i.e. So, . By induction, Antoro, Rainbow Connection ... 123 it can be shown that if , then The path has color . So, . If , then sequence 1, 3, 2. So, this is a rainbow becomes , which path. is impossible. If , then With this, we conclude that . becomes . So, . But, it Claim 6 is proved. implies or , a Claim 7: If or , contradiction. If , then then . becomes , which is impossible. Assume . Since , we have So, Claim 5 is proved. So, by Claim 1, we get Claim 6: For any , . From , we have . So, . Therefore, To prove this one, it is enough to find a , i.e. By Lemma rainbow coloring on that uses three 1 (2), . But, by colors. Let and Claim 6. Hence, . Now, . Define assume . Since , we as follows have . So, by Claims 2 and 3, we get . From , we have So, We will prove that is a rainbow . Therefore, , coloring. Take any non-adjacent vertices i.e. By Lemma 1 (2), we get in . We will find a rainbow path .
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