Zachary Monaco Georgia College Olympic Coloring: Go For The Gold


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1 Zachary Monaco Georgia College Olympic Coloring: Go For The Gold Coloring the vertices or edges of a graph leads to a variety of interesting applications in graph theory These applications include various scheduling questions and other optimization problems For this paper, we consider the idea of edgecoloring a graph, which is a coloring of the edges such that adjacent edges are distinctly colored The minimal number of colors required to edgecolor a graph is called the chromatic index This paper discusses the chromatic indices of graphs obtained from certain symmetric matrices 1 Introduction The goal of this paper is to determine the chromatic index of a graph given by a certain family of matrices The family that we are going to be working with are called Golden matrices The name comes from their connection with the golden fields described by Peter Steinbach which are further explained [FH11] In this paper, we discuss the adjacency matrices of these Golden matrices and go through different cases to find a connection between the dimension of its associated matrix and the chromatic index of the graph 2 Preliminaries In this section, we will identify terms and notation that will be used throughout this paper Any graph G consists of the nonempty set of elements called vertices and a set consisting of ordered pairs (u, v), where u and v are vertices, that are called edges Let the degree of a vertex v, denoted deg(v), be the number of edges incident with this vertex where a loop contributes two to the degree of v Adjacent edges are edges that share a common vertex and adjacent vertices are vertices that are connected by at least one edge Let a loop be an edge that connects a vertex to itself A graph has multiple edges if a vertex v has more than one edge connection to the another vertex We define a simple graph as a graph that contains no loops or multiple edges Throughout the paper we will be referring to the chromatic index of a graph, denoted χ (G), which is the number of distinct colors required to edgecolor a graph where the coloring of the edges requires that no two adjacent edges share the same color If we say our graph G is 1
2 2 disconnected that means it is possible to express the graph as a union of two subgraphs and G is connected if it cannot be so expressed See [Wil10] for more information about graphs Our graphs involve unions of certain families of basic graphs As such, we introduce some notation for these graphs Let C n represent a cycle graph on n vertices where the degree of each vertex is 2 Let P n represent a path graph on n vertices where P n is obtained by removing an edge from C n The graphs in our paper will be constructed using a family of matrices called Golden Matrices Golden Matrices are constructed using the following definition: let n 3 N, k = 1,, n 1, and let G[n, k] be the square matrix with dimension n 1 given by G[n, k] = (a u,v ) where a u,v = 1, if v = k u + 2i 1 for some positive integer i min{u, k, n u, n k}; a u,v = 0, otherwise These Golden Matrices serve as the adjacency matrices for the graphs we will consider 3 G[n, 1] and G[n, n 1] Proposition 31 For all n 3, χ (G[n, 1]) = 1 Proof Let G[n, 1] be our given matrix This implies that k = 1 which further implies that i = 1 because i =min {u, k, n u, n k} and k = 1 Using the defined values for k and i, we can solve for v Observe that v = 1 u + 2(1) 1 = 1 u = 1 u + 1 = 1 + u + 1 = u Then using the definition of the adjacency matrix for G[n, k], G[n, 1] = (a u,v ) where { a u,v = 1, if v = u; a u,v = 0, otherwise
3 Thus we have a value of 1 wherever a u,v when u = v Hence, the corresponding adjacency matrix is Note that in each row and column there is exactly one entry at a u,v where u = v Thus we know each vertex is connected only to itself with a loop creating n disjoint loops Since all n loops are disjoint, we can use one color to edgecolor the graph Hence χ (G[n, 1]) = 1 for all n 3 Now we will consider the k = n 1 case Proposition 32 For all n 3, χ (G[n, n 1]) = 1 Proof Let G[n, n 1] be our given matrix This implies that k = n 1 and it follows that i = 1 since i =min {u, k, n u, n k} and n k = n (n 1) = 1 Then using u n 1 and the definition of the adjacency matrices, we have Thus, v = (n 1) u + 2(1) 1 = n 1 u = n u G[n, n 1] = (a u,v ) where { a u,v = 1, if v = n u; a u,v = 0, otherwise, has the following adjacency matrix: From the adjacency matrix we will denote the following edge pairings
4 4 as u v, where u is the row and v are the position of the one entries The corresponding edge pairing from our matrix are 1 n 1, 2 n 2, 3 n 3, n n n n We know that we stop at because our matrix is square with dimension (n 1) and each entry will have one pair 2 We will show that the adjacency matrix of G[n, n 1] can give us two different cases when n is even or n is odd, but in both cases we will have χ (G[n, n 1]) = 1 Let n be even Then by definition of even, n = 2l for some l Z If we substitute this value for n into our pairing system observe that n 2l 2l n = = l l = 2l = Thus we have l l which creates a loop in our graph Thus our pairings and our adjacency matrix show that we will create multiple disjoint line segments and one disjoint loop Since each line segment and the loop are all disjoint, we say that χ (G[n, n 1]) = 1 if n is even Let n be odd Then by definition of even n = 2l + 1 for some l Z If we substitute this value for n into our pairing system this yields n 2l + 1 2l + 1 n = = l l + 1 = 2l + 1 = This yields the second graph we obtain which is multiple disjoint line segments Note all line segments are disjoint and we deduce that χ (G[n, n 1]) = 1, when n is odd Consequently, χ (G([n, n 1])) = 1 for all n 3 4 G[n, 2] and G[n, n 2] Now that we ahve dealt with the k = 1 and k = n 1 case, let us talk about the cases k = 2 nd k = n 2
5 5 Proposition 41 For all n 3, χ (G[n, 2]) = 2 Proof Note by the nature of the value of i in our equation to find v, we will end up having 4 different cases First we will break up the cases into when i = 1 and when i = 2 Then our subcases will be when u > 2 or when u 2 due to the absolute value of k u being either negative or nonnegative So let i = 1 Then when we solve for v we get: if u > 2, and if u 2, v = 2 u = 2 u + 1 = 2 + u + 1 = u 1, v = 2 u + 1 = 3 u Next we look at when i = 2 Observe the following values for v: if u > 2, and if u 2, v = 2 u = 2 u + 3 = 2 + u + 3 = u + 1, v = 2 u + 3 = 5 u Now we will use this information to construct our adjacency matrix for G[n, 2] Observe that now we must use the cases when i = 1 and i = 2 to complete our matrix Note that we must complete all the cases for when i = 1 to obtain the following edge pairings for u v: 1 2, 2 1, 3 3, 4 3, (n 1) (n 2) This creates a diagonal that runs from a 2,1 to a n 1,n 2 Now we must do the edge pairings for when i = 2 Using the apporpriate equations, we obtain (n 2) (n 1)
6 6 This creates a diagonal that runs from a 1,2 to a n 2,n 1 Putting this information together we obtain an adjacency matrix that looks like These two diagonals in our adjacency matrix create a path graph and we know that χ (P n ) = 2 Hence we have χ (G[n, 2]) = 2 Proposition 42 For all n 3, χ (G[n, n 2]) = 2 Proof To begin we find the possible values for i that will be used in calculating v We know that i min{u, k, n u, n k} Thus we plug in our given values for n and k and we see that i min{u, n 2, n u, 2} Hence, i 2 So since k = n 2, we say that when u = 1 or u = n 1 v = n = n 2, when u = 1, and v = (n 2) (n 1) + 1 = = 2, when u = n 1 Then when u = 2, 3, 4,, n 2, we must consider the cases when i = 1 and i = 2 First let i = 1 This yields v = n 2 u = n 2 u + 1 = n u 1 Next we check the values for v when i = 2 Given that i = 2, we solve v to get v = n 2 u = n 2 u + 3 = n u + 1 By calculating each v with each corresponding value of u, we deduce the following edge pairings of u v 1 n 2, 2 n 3, n 1, 3 n 4, n 2, n 2 1, 3, n 1 2
7 7 From these edge pairing we can produce the adjacency matrix below We know that the given graph for the matrix depends on when n is even or odd When n is even, we have a path graph which we have already shown has chromatic index of 2 Lastly when n is odd, we have two identical path graphs with loops on the endpoints Thus we can color the path graph with 2 colors and the loop can be colored with one of the remaining two colors Hence we have χ (G[n, n 2]) = 2 for all n N What we began to create was a pyramid of values that represent the chromatic indices based on the row and position in the row The row in the pyramid represents the n and the position represents k in G[n, k] We proved the boundary cases for k = 1, k = n 1, k = 2, and k = n 2 for all n N The pyramid is presented below along with our conjecture based on this pyramid of chromatic indices Conjecture 43 The formula for finding χ (G[n, k]) depends on when n is even or n is odd If n is odd, then {χ (G[n, 1]), χ (G[n, 2]), χ (G[n, 3])χ (G[n, n 2]), χ (G[n, n 1])} = {1, 2, 3,, ( n 2 1), ( n) (n) (n,, 2 1),, 3, 2, 1},
8 8 where the position of the index represents k in G[n, k] If n is even, then {χ (G[n, 1]), χ (G[n, 2]), χ (G[n, 3])χ (G[n, n 2]), χ (G[n, n 1])} = {1, 2, 3,, ( n ), ( n + 1) (n + 1, 1 ),, 3, 2, 1}, 2 where the position of the index represents k in G[n, k] 5 Conclusion In this paper we have found the chromatic indices for the first two boundary cases of G[n, 1], G[n, n 1], G[n, 2], and G[n, n 2] This method of proving the chromatic indices would not be sufficient for all n N since we know the set of natrual numbers is unbounded With this we would need to find a proof for the general case made in the conjecture, but this is a topic for further research We began to make connections between an algebraic family of matrices and graph theory and another topic that can be researched into more depth would be if the algebraic properties such as the productsum formula holds while using the chromatic indices that may be found from the conjecture found in this paper References [FH11] Anne Fontaine and Susan Hurley Golden matrix families The College Mathematics Journal, 42(2): , 2011 [Wil10] Robin James Wilson Introduction to Graph Theory, 5/e Pearson Education, 2010
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