The degree chromatic polynomial Pm(G,k) of a graph G counts the number of k-colorings in which no vertex has m adjacent vertices of its same color.   We prove Humpert and Martin\u27s conjecture on the leading terms of the degree chromatic polynomial of a tree

Cifuentes, Diego

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Rose-Hulman Undergraduate Mathematics Journal Volume 12 Issue 2 Article 5 On the Degree-Chromatic Polynomial of a Tree Diego Cifuentes Universidad de los Andes, Bogota, Colombia, df.cifuentes30@uniandes.edu.co Follow this and additional works at: https://scholar.rose-hulman.edu/rhumj Recommended Citation Cifuentes, Diego (2011) "On the Degree-Chromatic Polynomial of a Tree," Rose-Hulman Undergraduate Mathematics Journal: Vol. 12 : Iss. 2 , Article 5. Available at: https://scholar.rose-hulman.edu/rhumj/vol12/iss2/5 Rose-HulmanUndergraduateMathematicsJournalSponsored byRose-Hulman Institute of TechnologyDepartment of MathematicsTerre Haute, IN 47803Email: mathjournal@rose-hulman.eduhttp://www.rose-hulman.edu/mathjournalOn the Degree-ChromaticPolynomial of a TreeDiego CifuentesaVolume 12, No. 2, Fall 2011aUniversidad de los Andes, Bogota, Colombia,df.cifuentes30@uniandes.edu.coRose-Hulman Undergraduate Mathematics JournalVolume 12, No. 2, Fall 2011On the Degree-Chromatic Polynomial of aTreeDiego CifuentesAbstract. The degree chromatic polynomial Pm(G, k) of a graph G counts thenumber of k-colorings in which no vertex has m adjacent vertices of its same color.We prove Humpert and Martin’s conjecture on the leading terms of the degreechromatic polynomial of a tree.Acknowledgements: I would like to thank Federico Ardila for bringing this problem tomy attention, and for helping me improve the presentation of this note. I would also liketo acknowledge the support of the SFSU-Colombia Combinatorics Initiative.Page 62 RHIT Undergrad. Math. J., Vol. 12, No. 21 IntroductionGeorge David Birkhoff defined the chromatic polynomial of a graph to attack the renownedfour color problem. The chromatic polynomial P (G, k) counts the k-colorings of a graph Gin which no two adjacent vertices have the same color [3].Given a graph G, Humpert and Martin defined its m-chromatic polynomial Pm(G, k) tobe the number of k-colorings of G such that no vertex has m adjacent vertices of its samecolor. They proved this is indeed a polynomial. When m = 1, we recover the usual chromaticpolynomial of the graph P (G, k).The chromatic polynomial is of the formP (G, k) = kn − ekn−1 + o(kn−1)where n is the number of vertices and e the number of edges of G. For m > 1 the formula isno longer true, but Humpert and Martin conjectured the following formula when the graphis a tree T :Pm(T, k) = kn −∑v∈V (T )(d(v)m)kn−m + o(kn−m) (1)where d(v) is the degree of v. Note that (1) is not true for m = 1 —we will see why in thecourse of proving Theorem 1.The goal of this paper is to prove this conjecture in Theorem 1. In section 2 we discussthe basic concepts required to understand the theorem, while in section 3 we provide theproof.2 BackgroundA finite graph G is an ordered pair (V,E), where V is a finite set of vertices and E is a setof edges, which are 2-element subsets of V .Figure 1 shows the graphic representation of graph.1 2 354Figure 1: Graphic representation of a graph with V = {1, 2, 3, 4, 5} and E ={{1, 2}, {2, 3}, {3, 4}, {3, 5}}.We now present some basic definitions of graph theory.RHIT Undergrad. Math. J., Vol. 12, No. 2 Page 63Definition 1. The degree of a vertex v is the number of edges which contain v, and isdenoted as d(v). Two vertices p, q ∈ V are said to be adjacent if the pair {p, q} ∈ E. A pathis a sequence of vertices v0, v1, . . . , vk where vi is adjacent to vi+1 for 0 ≤ i ≤ k − 1. A cycleis a path v0, . . . , vk with v0 = vk. A graph is connected if for any pair of vertices there existsa path containing both of them. A tree is a connected graph with no cycles.It is easy to see that the graph in Figure 1 is actually a tree.A coloring of a graph is an assignment of colors to each of its vertices. If σ is a coloring,we denote by σ(v) the color assigned to the vertex v. A k-coloring is one in which σ(v) ∈{1, 2, . . . , k} for all v, i.e. we may use at most k different colors. A graph with n verticesclearly has kn different k-colorings, as each of its n vertices has k possible choices for itscolor.A coloring is called proper if there is no edge connecting any two identically coloredvertices. Figure 2 shows all of these colorings with k = 3 for a 3-vertex tree.Figure 2: Proper 3-colorings of a tree with 3 vertices.The chromatic polynomial of a graph P (G, k) counts the proper k-colorings of G. It iswell-known to be a monic polynomial in k of degree n, the number of vertices.Example 1. The chromatic polynomial of a tree T with n vertices is P (T, k) = k(k−1)n−1.To prove this, fix an initial vertex v0. There are k possible choices for its color σ(v0). Then,consider a vertex v1 adjacent to v0. There are k − 1 ways to choose σ(v1), as it has to bedifferent from σ(v0). Now, consider a vertex v2 adjacent to v0 or to v1. Notice it cannot beadjacent to both of them, or there would be cycle. Thus, there are also k−1 possible choicesfor σ(v2). If we repeat this algorithm, we will always have a vertex adjacent to exactly oneof the previously colored vertices, so it can be colored in k− 1 ways. The result follows afterrepeating this procedure n− 1 times.3 ResultsNow, we prove the conjecture stated by Humpert and Martin.Theorem 1 ([1, 2], Conjecture). Let T be a tree with n vertices and let m be an integer with1 < m < n. Then the equation (1) holds, where Pm(G, k) counts the number of k-coloringsof T in which no vertex has m adjacent vertices of its same color.Page 64 RHIT Undergrad. Math. J., Vol. 12, No. 2Proof. For a given coloring of T , say vertices v1 and v2 are “friends” if they are adjacent andhave the same color. For each v, let Av be the set of colorings such that v has at least mfriends. We want to find the number of colorings which are not in any Av, and we will usethe inclusion-exclusion principle. As the total number of k-colorings is kn, we havePm(T, k) = kn −∑v∈V|Av|+∑v1,v2∈V|Av1 ∩ Av2 | − . . .We first show that |Av| =(d(v)m)kn−m + o(kn−m). Let A(l)v be the set of k-colorings suchthat v has exactly l friends. In order to obtain a coloring in A(l)v , we may choose the l friendsin(d(v)l)ways, the color of v and its friends in k ways, the color of the remaining adjacentvertices to v in (k− 1)d(v)−l ways, and the color of the rest of the vertices in kn−1−d(v) ways.Then|Av| =n−1∑l=m|A(l)v | =n−1∑l=m(d(v)l)kn−d(v)(k − 1)d(v)−l=(d(v)m)kn−m + o(kn−m).To complete the proof, it is sufficient to see that for any set S of at least 2 vertices|⋂v∈S Av| = o(kn−m); clearly we may assume S = {v1, v2}. Consider the following cases:Case 1 (v1 and v2 are not adjacent). Split Av1 into equivalence classes with the equivalencerelationσ1 ∼ σ2 ⇔ σ1(w) = σ2(w) for all w 6= v2.Note that each equivalence class C consists of k colorings, which only differ in the colorof v2. In addition, for each C at mostd(v2)mof its colorings are in Av2 , as if σ ∈ Av2 theremust be m vertices adjacent to v2 with the color σ(v2). Therefore|Av1 ∩ Av2 | =∑C|C ∩ Av2| ≤∑Cd(v2)m=|Av1 |k· d(v2)m.It follows that|Av1∩Av2 ||Av1 |goes to 0 as k goes to infinity, so |Av1 ∩ Av2 | = o(kn−m).Case 2 (v1 and v2 are adjacent). Let W be the set of adjacent vertices to v2 other than v1.They are not adjacent to v1 as T has no cycles. Split Av1 into equivalence classes with theequivalence relationσ1 ∼ σ2 ⇔ σ1(w) = σ2(w) for all w /∈ W.Each equivalence class C consists of k|W | colorings, which may only differ in the colors ofthe vertices in W . If v1 and v2 are friends in the colorings of C, then a coloring in |C ∩Av2|must contain at least m− 1 vertices in W of the same color as v2. ThereforeRHIT Undergrad. Math. J., Vol. 12, No. 2 Page 65|C ∩ Av2| =|W |∑l=m−1(|W |l)(k − 1)|W |−l <|W |∑l=0(|W |l)k|W |−1 = 2|W |k|W |−1.Notice that here we are using m ≥ 2 so that l ≥ 1. Otherwise, if v1 and v2 are not friendsin the colorings of C, then|C ∩ Av2| =|W |∑l=m(|W |l)(k − 1)|W |−l <|W |∑l=0(|W |l)k|W |−1 = 2|W |k|W |−1.Therefore|Av1 ∩ Av2| =∑C|C ∩ Av2| <∑C2|W |k|W |−1=|Av1|k|W |· 2|W |k|W |−1 = |Av1| · 2|W |kand |Av1 ∩ Av2| = o(kn−m) follows as in the first case.This completes the proof of the theorem.4 ConclusionsIn conclusion, the degree-chromatic polynomial is a natural generalization of the usual chro-matic polynomial, and it has a very particular structure when the graph is a tree. Theleading terms of the chromatic polynomial are determined by the number of edges. Like-wise, when m ≥ 2, the leading coefficients of the degree chromatic polynomial Pm(G) canbe described easily in terms of G, but now they depend on the degree of the vertices of G.References[1] B. Humpert and J. L Martin, The incidence Hopf algebra of graphs, PreprintarXiv:1012.4786 (2010).[2] , The incidence Hopf algebra of graphs, DMTCS Proceedings 0 (2011), no. 01.[3] R. C Read, An introduction to chromatic polynomials, Journal of Combinatorial Theory4 (1968), no. 1, 52–71.

On the Degree-Chromatic Polynomial of a Tree

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On the Degree-Chromatic Polynomial of a Tree

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