eu STRONG RAINBOW EDGE-COLOURING OF VARIANTS OF CUBIC HALIN GRAPHS

A non trivial connected graph G = (V,E) is strongly rainbow connected if every two vertices u and v of G are connected by at least one shortest u− v path in which no two edges have same colours. The strong rainbow connection number of G, denoted by src(G) is the minimum number of colours that makes G strongly rainbow connected. In this paper we explore the strong rainbow connection number of variants of cubic Halin graphs. AMS Subject Classification: 05C15, 05C40


Introduction
The theory of rainbow colouring was put forward by Chartrand in 2008 [5].The rainbow connection number of an arbitrary graph is NP-Hard [4].An edge colouring c of a graph G is a function from its edge set to the set of natural numbers.A u − v path P in an edge coloured graph G is called a rainbow path if no two edges of it get the same colour.An edge coloured graph is strongly rainbow connected if every two vertices is connected by at least one shortest rainbow path.Such a colouring is called a strong rainbow edge colouring.The strong rainbow connection number src(G) is the minimum number of colours needed for strong rainbow edge colouring of a graph G.
This paper focuses on the strong rainbow connection number of variants of cubic Halin graphs.A Halin graph G = T ∪ C is a plane graph that consists of a plane embedding of a tree T and a cycle C connecting the leaves (vertices of degree 1) of the tree such that C is the boundary of the exterior face and the degree of each interior vertex of T is at least three [5].The tree T and the cycle C are in general referred to as characteristic tree and adjoint cycle of G respectively [5].A caterpillar is a tree where all vertices of degree ≥ 3 lie on a path, called the spine of the caterpillar.A cubic Halin graph is a Halin graph whose characteristic tree T is a caterpillar.In this paper, we study cubic Halin graphs whose characteristic tree T is a caterpillar and the degree of each vertex of T is 3.

Preliminaries
Graphs studied in this paper are simple, finite and undirected.The distance between two vertices u and V in G, denoted by d(u, v) is the length of a shortest path between them in G.The eccentricity e(v) of a vertex v in a connected graph G is ecc(v) = max u∈V (G) d(u, v).The maximum eccentricity of all vertices in a graph G is called the diameter of G and is denoted by diam(G).Two edges e = uv and f = xy are antipodal in G if d(v, x) = diam(G).If a graph has an edge colouring c, then c(G) denotes the set of colours appeared in G.The chromatic index χ(G), of a graph G is the fewest number of colours necessary to colour each edge of G such that no two edges incident on the same vertex have the same colour.

Main Results
Let G be a cubic Halin graph.The characteristic tree T of G is a caterpillar of order 2n+2, n ≥ 1.A pendent edge (leaf edge) is an edge of the caterpillar that is incident at a vertex of the outer cycle.An edge of G is called a cycle edge if it lies on the outer cycle.We denote the starting vertex(head) of the spine by u 0 , the intermediate vertices along the spine P n of the caterpillar by v 1 , v 2 , ..., v n and the end vertex(tail) of the spine by u n+1 .Other pendent vertices adjacent with v i is denoted by u i , 2 ≤ i ≤ n.Note that the vertices u 0 , u 1 , ..., u n , u n+1 lie on the adjoint cycle C n+2 .Moreover, if the vertices u 0 , u 1 , ..., u n , u n+1 in C n+2 are in order and the pendent edges of the caterpillar are in the same direction (either up or down), then G is called a necklace graph with n pendent edges and is denoted by N e n .The edges of N e n can be labeled as follows: The cycle edges are denoted by {e i |e i = u i u i+1 , 0 ≤ i ≤ n} ∪ {e n+1 = u n+1 u 0 }.The edges along the spine of the caterpillar are denoted by Since v 1 is adjacent to u 0 and v n is adjacent to u n+1 along the spine, denote the spine edges u 0 v 1 by s 0 and v n u n+1 by s n .The pendent edges are denoted by  Proof.It is easy to check that src(N e 1 ) = 1, src(N e 2 ) = 2. Case 1. n is even and n ≥ 3. The colouring algorithm is given as follows.c(s  Proof.We consider the following cases.Case 1: n is even Colouring Algorithm: We shall first colour the cycles above the spine as follows: 1.The three edges of the leftmost cycle C 3 can be assigned with a colour c 1 (same colour).This implies that c(e 0 ) = c(p 1 ) = c(s 0 ) = c 1 .find that the colours of the spine edges can be applied to the cycles having the spine edges as its edges.Antipodal edges of the cycles acquire the same colours.That is, c

The cycle adjacent with
. The uncoloured edge is the cycle edge e n 2 +1 in the rightmost cycle C 3 , which can be assigned with the colour c 1 .
2. We find that colours assigned for the cycles above the spine have been used for the cycles below the spine.
Thus, totally 2 + 2 1 + n − 4 2 colours are used to colour the edges of An illustration if given in Figure 3. 2. C 5 is the cycle adjacent to C 3 and hence both have a pendent edge in common.The remaining four edges of C 5 can be assigned with the colours c 2 and c 3 alternately since C 3 is already coloured with a colour c 1 .That is, the cycle C 5 constituted from the edges p 1 , e 1 , p 2 , s 2 , s 1 are assigned with the colours c 1 , c 2 , c 3 , c 2 , c 3 respectively.
3. There are totally n − 1 2 cycles of length 5 and any two consecutive cycles have a pendent edge in common.Each cycle requires exactly two colours as the common pendent edge is already assigned with a colour as given in step 2. Repeat this process until all the 5-cycles are coloured.
4. The two edges e ⌈ n 2 ⌉ , s n of the rightmost cycle C 3 are assigned with the colour of the common pendent edge p ⌈ n 2 ⌉ which is assigned in step 3.That is, c(e Thus all the cycles above the spine have been assigned with n colours.Now consider the cycles below the spine.We give an algorithm which uses the same set of n colours for the cycles below the spine.
1. We find that colours of the spine edges can be applied to the cycles having the spine edges as its edges.Antipodal edges of the cycles acquire the same colours.This implies that c( ).Now we study another class of cubic Halin graphs in which the characteristic tree is constructed by drawing Proof.We consider the following cases.Case 1: n is even.For n < 6, the colouring algorithms do not follow a general pattern.The strong rainbow connection numbers of G * 2 and G * 4 are found to be 2 and 3 respectively.Figure 5 illustrates the strong rainbow edge colouring of G * 4 .respectively.Below the spine, the leftmost cycle is constituted from the edges s n 2 , p n 2 +1 , e n+1 , s 0 , s 1 , ...s n 2 −1 .We note that the edge s n 2 is the common spine edge to both the cycles.The remaining edges p n 2 +1 , e n+1 , s 0 , s 1 , ...s n 2 −1 are assigned with respectively.The colour of the pendent edge p n 2 of the rightmost cycle above the spine can be assigned to the remaining pendent edges p 1 , p 2 , ..., p n After simplification we find that the minimum number of colours required for strong rainbow edge colouring is , respectively.
The colour of the pendent edge p n 2 of the rightmost cycle above the spine can be assigned to the remaining leaf edges p 1 , p 2 , ..., p n The proof is similar to case 1.In this case, the rightmost cycle above the spine is of length n 2 + 2 and leftmost cycle below the spine is of length n 2 + 3.For convenience, we denote the rightmost cycle above the spine and the leftmost cycle below the spine by respectively.First we colour the , which exhaust all the spine edges.Then the colour of the spine edges can be assigned to their corresponding antipodal cycle edges.

Conclusion
In this paper we characterized three non isomorphic cubic Halin graphs N e n , G 1 * n and G * n .We found that the strong rainbow connection numbers of N e n and G * n are closer to the diameter of the graph.However, the strong rainbow connection number of G 1 * n is n.In this process, we develop techniques that are useful in finding the strong rainbow connection numbers of cycle related graphs.

This gives a strong rainbow n 2 + 1 -Observation 2 .Observation 3 .
edge colouring for N e n , n is odd.Since diam(G) = n 2 + 1 when n is odd and n ≥ 3, This bound is sharp and src(G) = diam(G).Antipodal edges in each of C 4 and C n+2 acquire the same colour in strong rainbow edge colouring.If the vertices u 0 and u n+1 are removed from the necklace graph N e n , then the graph becomes a bipartite graph.Next, we study another class of cubic Halin graph in which the characteristic tree is constructed by drawing the pendent edges (leaf edges) up and down alternately starting from the leftmost to the rightmost on the spine.Here n represents the number of up and down pendent edges from the leftmost to the rightmost drawn vertically to the spine.Denote this class of graphs by G 1 * n .The diameter of G 1 * n is n + 3 2 .Label the cycle edges of G 1 * n from the leftmost, in clockwise direction by {e i |0 ≤ i ≤ n + 1}, that along the spine from the leftmost by {s i |0 ≤ i ≤ n}.Label the pendant edges above the spine from the leftmost to the rightmost by p i |1 ≤ i ≤ n 2 and that which is below the spine from rightmost to the leftmost by p i | n 2 + 1 ≤ i ≤ n .An illustration is given in Figure 2.

2 cycles C 5 . 4 . 5 . 1 .
C 3 is C 5 .The cycles C 3 and C 5 have a common pendent edge p 1 .Since C 3 is already coloured with a colour c 1 , the remaining four edges of C 5 can be assigned with the colours c 2 and c 3 alternately.That is, the cycle C 5 constituted from the edges p 1 , e 1 , p 2 , s 2 , s 1 are assigned with the colours c 1 , c 2 , c 3 , c 2 , c 3 respectively.3. Colour the cycle adjacent with C 5 .Since one of the edges of the cycle C 5 is already coloured as discussed in step 2, the remaining four edges of C 5 can be assigned with the colours c 4 and c 5 alternately.Continue this process until all the cycles of length 5 have been assigned with colours.There are totally 1 + n − 4 2 cycles of length 5 and any two consecutive cycles have a common pendent edge.We find that totally 2 1 + n − 4 2 colours are required for the 1 + n − 4 The rightmost cycle adjacent with C 5 is C 4 .Since one of the edges (pendent edge) of C 4 is already assigned with a colour, assign a new colour to the cycle edge adjacent to the coloured edge.The two colours can be assigned to the remaining spine edges s n , s n−1 alternately.That is, the edges p n 2 , e n 2 , s n , s n−1 receive the colours c n−1 , c n , c n−1 , c n respectively.All the cycles above the spine have been assigned with 1+2 1 + n − 4 2 + 1 colours totally which is exactly n.Now consider the cycles below the spine.The leftmost cycle is C 4 and the rightmost cycle is C 3 .The intermediate cycles are of length 5 and there are 1 + n − 4 2 intermediate cycles.We

2 − 1 2 
above the spine.Similarly, c(p n 2 +1 ) = c(p n 2 +2 ) = ... = c(p n ).The remaining uncoloured cycle edges acquire the colour of the spine edges as they are antipodal.Thus, we assign distinct colours to the edges of G * n .

2 − 1 2 + 3 .
above the spine.Similarly c(p n 2 +1 ) = c(p n 2 +2 ) = ... = c(p n ).The remaining uncoloured cycle edges acquire the colour of the spine edges as they are antipodal.Thus, we assign  to the edges of G * n .After simplification we find that which is equal to n Hence src(G * n ) = diam(G * n ) + 1 = n 2 + 3 for n ≥ 6. Case 2: n ≥ 3 is odd G * n .consists of n 2 pendant edges above the spine and n 2 pendant edges below the spine.
. , the leftmost cycle is constituted from the edges s n 2 , p n 2 +1 , e n+1 , s 0 , s 1 , ... s n 2 −1 .We note that the edge s n 2 is the common spine edge to both the cycles.The remaining p n 2 +1 , e n+1 , s 0 , s 1 , ...s n