Chromatic Polynomial and Chromatic Uniqueness of Necklace Graph

For a graph G, let P (G,λ) be its chromatic polynomial. Two graphs G and H are said to be chromatically equivalent if P (G,λ) = P (H,λ). A graph is said to be chromatically unique if no other graph shares its chromatic polynomial. In this paper, chromatic polynomial of the necklace graph Nn, for n ≥ 2 has been determined. It is further shown that N3 is chromatically unique.


Introduction
A k −coloring of a graph G is a mapping φ : V (G) → {1, 2, 3, ..., k} such that φ(m) = φ(n) for any edge e = mn.A minimum number k such that G has a proper coloring is called chromatic number, and G is called k − colorable graph.Many useful tools had been developed for solving graph coloring problems.Birkhoff [1] was the first who proposed a method to determine kcolorable graphs by introducing a function P (G, λ) that represent the number of all proper coloring of a graph G with k colors.P (G, λ) is called chromatic polynomial of G.
Chao and Whithead Jr. [2] defined a graph to be chromatically unique if no other graphs share its chromatic polynomial and the two graphs are chromatically equivalent if they share a same chromatic polynomial.Let P n be a path of length n − 1 defined on the n vertices a 1 ...a n .The comb graph denoted by Cb n is the tree graph consisting of a path P n together with vertices b 0 , b 1 , ..., b n+1 and edges b i a i , with 0 ≤ i ≤ n.The necklace graph N n is a 3-regular Halin graph [6] and is obtained from comb graph Cb n by adding the edges a 1 b 0 , b 0 b n+1 , b n+1 b n and b i b i+1 , with 1 ≤ i ≤ n − 1.Alternatively, vertex and edge set of N n is defined as: The vertices b 0 and b n+1 are called end vertices of the Necklace graph.The N n has 2n + 2 vertices and 3n + 3 edges.It is important to note that the necklace graph N n is a 2-connected 3-regular graph (see Figure 1).The main objective of this paper is to study the chromatic polynomial and

Basic Results
In this section, some of the known results are stated that help in proving the main results of this paper.
Theorem 2.1 (Fundamental reduction theorem (FRT) [9]) Let G be a graph and e an edge of G. Then where G − e is the graph obtained from G by deleting e, and G. e is the graph obtained from G by contracting the end vertices of e and removing all but one of the multiple edges, if they arise.
Theorem 2.2 [4]) Let G be a graph with n vertices and m edges then is a polynomial in x of degree n such that 1.The coefficient a i are integers and alternate in sign; Theorem 2.3 [8] let G be a graph of order n and size m.Then where N (i, j) is the number of spanning subgraphs of G having i components and j edges.
• G is connected if and only if H is connected.
• G and H have the same number of shortest cycles; i.e. g(G) = g(H).
Theorem 2.5 [5] Let G and H be chromatically equivalent graphs.If G and H do not contain K 4 then they have the same number of pure C 4 .
Theorem 2.6 [3] Let G be a graph containing at least two triangles.If there is a vertex of a triangle having degree two in G, then (x−2) 2 divides P (G, x).

Main results
In this section, chromatic polynomial for necklace graph N n (n ≥ 2) is determined.It is further shown that the graph N 3 is chromatically unique.By using Fundamental reduction theorem, one can easily deduce that Proof: By using Fundamental reduction theorem for n ≥ 3 and 0 ≤ j ≤ n − 3, we get The graph L n−j is a polygone graph consisting of one C 3 and (n − j − 1) C 4 cycles.The graph G n−j is obtained by contracting the edge e of the necklace graph N n−j (see Figure 2).By using FRT recursively, one can deduce that: n b e a P(L n ) P(G n ) (3) Also from Figure 2, the chromatic polynomial of the graph H n can be computed and is: Substituting the value of P (G n−1 , x) from 1 in (4), we get (5) Similarly continuing in this way we have By replacing these values recursively in equations (3) for j = 0, we get Therefore by replacing the values of P (G n , x) and P (L n , x) in the equation ( 6) we conclude that In the next theorem, the chromatic uniqueness of the necklace graph N 3 is established which is also the main result of this paper.
Theorem 3.2 Necklace graph N 3 is chromatically unique.
Proof: Suppose N 3 is not chromatically unique and there is another graph H which has the same chromatic polynomial as N 3 .Then One can deduce that By comparing the coefficient of x 5 in (7) and a 4 , we get Since t 7 (H) < 2, otherwise number of triangles increased by 2. So we have two subcases Since H has exactly two triangles.Let k be the number of vertices covered by these two triangles then degree of each vertex of triangle must be ≥ 3, therefore remaining n − k vertices of H has at least degree 2. By Hand Shaking Lemma we have This implies that k ≤ 8 and clearly k = 3, 7, 8. Hence possible values of k are 4, 5 and 6.Therefore H must have one of the subgraph G 1 , or G 2 or G 3 .These subgraphs are shown in the Figure 3. From above information, there are three possible subgraphs of H in this case as shown in the Figure 4. We prove that H can not have any of these subgraph.Subcase 1.1 In subgraph (I), x and y are vertices of triangle so d(x) and d(y) ≥ 3.This implies that at most one of x and y can be adjacent to the vertex b otherwise there exist at least 3 triangles, a contradiction.
Also both x and y are not connected to the vertex b otherwise there exists at least three pure C 4 cycles.So without loss of generality, suppose that there is an edge between vertex x and b and there is a path of length 2 from y to x is a vertex of triangle, so d(x) ≥ 3.This implies that there must be an edge from x to any other vertex of H. Therefore, the subgraph (II) can be extended to two possible graphs shown in the Figure 6.If the second situation arises then there must exist at least one path of length 2 from vertex b to e as shown in the Figure 7. Otherwise t 2 (H) > 3. Let g be a new vertex in b − e path, then this new graph has two pure C 4 and a 5-vertex wheel graph with one spoke deleted.Since, one more path of length 2 has to be added.But by adding such path leads to a third pure C 4 , a contradiction.Hence H does not contain any of the subgraphs (I), (II) and (III).This implies that H does not contain the subgraph G 1 which is the edge gluing of 2-triangles.Case 2 In this case H contain the subgraph G 2 .Since G 2 has a cut vertex and two triangles, therefore there should be a path of length at least 2 between any two distinct vertices of 2 triangles.Also H has 8 vertices so we needed to add 3 more vertices in G 2 , therefore the length of this path is at most 4.So G 2 can extended to the three subgraphs G 21 , G 22 and G 33 shown in Figure 8.   x and y can not be adjacent to vertex 2 otherwise t 1 (G 21 ) > 2. So we have following three possible extension for the subgraph G 21 .
1.There is a path of length two between the vertices x and y.Let z be a vertex added to the subgraph G 21 .Since t 2 (H) = 2, therefore there must be a path of length two from vertex z to vertex 2. So the only possible graph in this case is G 211 shown in the Figure 9.
2. There are two paths of length two from the vertices x and y to the vertex 2. Let G 212 be the new graph (see Figure 9) formed by adding these two paths.Then, clearly t 2 (G 212 ) ≥ 3, a contradiction.
3. There is a path of length three from vertices x or y to any other vertex.Suppose there is a path of length 3 from vertex y to vertex 1.Let G 213 be the graph obtained by adding this path.Since d(x) ≥ 3 and x can be adjacent only to the vertex 5 or 6 but in either cases we have at least 3 pure C 4 , a contradiction. 1.There is a path of length two between the vertices x and y.Since one more edge is needed to be added.But by adding this edge, the resulting graph has at least 3 pure C 4 , a contradiction.
2. There is a path of length two from the vertex x or y to any vertex of graph G 22 .Suppose this path is from the vertex x to the vertex 3. Since d(y) ≥ 3 and y can only be adjacent to vertex 3 or z, otherwise number of triangles exceed from 2. Both these constructions gives at least 3 pure C 5 and no 5-vertex wheels with one spoke deleted, a contradiction.Since t 7 (H) = 0 → t 4 (H) = 2. contradiction.The graphs constructed in this case are shown in Figure 10. 1. Suppose x and y are adjacent to the vertices a and b respectively (See Figure 11).But in this construction t 4 (H) ≥ 3 and t 7 (h) = 0, a contradiction, since t 7 (H) = 0 → t 4 (H) = 2.

Figure 2 :
Figure 2: Reduction algorithm on N n graph

Figure 3 :Case 1
Figure 3: Possible subgraphs of H

Subcase 2 . 1 :
Let x and y are vertices of the triangles then d(x), d(y) ≥ 3.
2. Since there is no K 4 in H therefore by Theorem 2.5 t 2 (H) = 2. 3. H contain no separating(cut) vertex as H is 2-connected graph.4. H does not contain any vertex of degree 2 that lies in a triangle.Since otherwise (x − 2) 2 |P (H, x) by Theorem 2.6. 5.By Theorem 2.2, the coefficient a 4 of the P (H, x) is