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Graph Theory through an example of Map Coloring
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Look at the map of an Island with 6 regions A, B, C, D, E, F Figure 1
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We want to give a unique color to each region. This can be put in mathematical terms as follows: V = {A, B, C, D, E, F}, C – a set of colors, f: V C We callfa coloring of the map. For obvious reasons, we decide to give distinct colors to two neighboring regions. Iff satisfies this condition, we call it aproper coloring.
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Now, two questions arise: What is the minimum number of colors required for a proper coloring? With a certain number of colors we have, in how many ways can a proper coloring be done? For the given map with only 6 regions it may be easy to answer these questions. We can make it easy for any map as follows.
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For the given map, represent each region by a point. If two regions are neighboring, draw a line segment joining them. In the resulting figure, we refer to the points asverticesand the line segments asedges. . . . . F D B A . . C E Figure 2
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Figure 2 represents what we call agraph in mathematics. Let us have its formal definition: V – a finite set, E – a set of pairs of elements in V G = (V, E) is called a graph. Elements of V are called vertices and those of E are called edges. If {a, b} ε E, we say the vertices a and b are adjacent. Now, the question of coloring the regions of the map becomes that of allotting a color to each vertex of the graph G. Map coloring becomes graph coloring.
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In Figure 2 the graph is G = (V, E), where V = {A, B, C, D, E, F} E = {AB, AC, BC, BD, BE, CE, DE, DF, EF}, Where AB stands for the edge {A, B}.
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Graph Coloring If G = (V, E) is a graph and C is a set of colors, then a function f: V C is called a coloring of the graph. It is a proper coloring if adjacent vertices get distinct colors. The least number of colors required for a proper coloring is called the chromatic number of the graph, denoted by (G). If is the number of colors available, the number of possible proper colorings will be a polynomial, denoted by p(G, ) – the chromatic polynomial.
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Two Special Cases: G = (V, E), E = - null graph G = (V, E), E is the set of all pairs of elements in V – complete graph. If |V| = n, the complete graph is denoted by Kn
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Examples Null Graph: G = (V, E), |V| = n, E = p(G, ) = n(G) = 1 Complete Graph: G = Kn p(G, ) = (n) = ( - 1)( - 2)…( - n + 1)(G) = n Linear Graph: G = (V, E), V = {v1, v2,…,vn}, E = {{vi, vi+1}, i = 1, 2, …, n-1}p(G, ) = ( -1)n-1(G) = 2
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Union and Intersection of graphs
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Example 1 Consider the graph G = (V, E), V = {A, B, C, D}, E = { AB, AC, AD, BC, CD} G1 = (V1, E1), V1 = {A, B, C}, E1 = {AB, AC, BC} G2 = (V2, E2), V2= {A, D, C}, E2 = {AD, AC, DC} G is the union of G1 and G2 Their intersection is H = ({A, C}, {AC}) . . D C . . A B Figure 4
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Theorem If G = (V, E) is a graph and G1, G2 are its subgraphs such that G = G1G2, G1G2 = Kn, then
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Example 2 In Example 1 G1 = K3, G2 = K3 G1G2 = K2 Therefore, Here p(G, 1) = p(G, 2) = 0. What does this mean? And p(G, 3) = 6 So, the chromatic number (G) = 3 There are 6 proper colorings possible with 3 colors.
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Example 3 Now, let us take up the initial problem of map coloring. In the corresponding graph, let V1 = {A, B, C, E}, E1 = {AB, AC, BC, BE, CE} V2 = {B,D,E,F}, E2 = {BD, BE, DE, DF, EF} G1 = (V1, E1), G2 = (V2, E2) G = G1G2, G1G2 = K2
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By Example 2, p(G1, ) = p(G2, ) = (-1)(-2)2 Also, p(K2, ) = (-1) Thus, (G) = 3, and there are 6 proper colorings possible with 3 colors. . . . . F D B A . . C E