Math 575 Exam 3. (t). What is the chromatic number of G?

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1 Math 575 Exam 3 Name 1 (a) Draw the Grötsch graph See your notes (b) Suppose that G is a graph having 6 vertices and 9 edges and that the chromatic polynomial of G is given below Fill in the missing coefficients P G (t) = t 6! t t 4! 56t 3 + t 2! 16t P G (t) = t 6! 9t t 4! 56t t 2! 16t (c) If G is a graph on 50 vertices with!(g) = 7, then!(g) " (answer should be an integer and as large as possible)!(g) " 8 (d) How many ways are there to color the vertices of the 5-cycle below using red, blue and green if the vertex a must be colored green? P C 5 (3) 3 = 10 (e) Suppose that the chromatic polynomial of G is P G (t) = t(t! 1)(t! 2) 3 (t! 3) 4 P C4 (t) What is the chromatic number of G? 4 2 Let G be a planar graph with no induced K 1, 3 Show that deg(v)! 8 for every vertex v Suppose that v is a vertex of G such that deg(v)! 9 Let H be the graph induced by the neighbors of v Then since r(3, 4) = 9, there is either an independent set of 3 vertices in H or there is a clique of 4 vertices If there is an independent set of 3 vertices in H, then those vertices together with v would form an induced K 1, 3 in G So it must be that there is a clique of 4 vertices in H But then these 4 vertices together with v would form a K 5 in G which is impossible since G is planar

2 3 (a) Determine the chromatic polynomial of the graph below Hint: No need for any formulas here Name Just count the number of colorings directly, P G (t) = t(t! 1) 2 (t! 2) 5 (t! 3) (b) Find the chromatic polynomial of the graph below Consider the division of G into two graphs A and B as shown below with V(A)! V(B) = {a,b} Then P G (t) = P A(t)P B (t) t(t! 1) " P C4 (t) % = P A (t) $ # t(t! 1) ' & But doing a similar operation on A, gives P A (t) = P C 4 (t)p C5 (t) t(t! 1) P G (t) = P 2 C 4 (t)p C5 (t) t 2 (t! 1) 2 Thus, 4 Let G be the Grötsch graph Give an example to show that r(p 8,G)! 22 Justify your answer You may use the idea behind the Chromatic Lower Bound The graph 3K 7 = K 7!!K 7!!K 7 has 21 vertices and does not contain any P 8 Also since!(g) = 4, its complement can be colored with 3 colors and hence cannot contain a copy of G

3 5 Show that every red-blue-green-yellow coloring of the edges of K 66 must contain a monochromatic triangle Consider any red-blue-green-yellow coloring of the edges of K 66 and suppose that v is a vertex of K 66 Consider the 65 vertices adjacent to v and then consider any vertex v of the K 66 Since v is incident with 65 edges, some 17 of them must have the same color, say yellow Let H be the subgraph induced by these 17 vertices If any edge between two vertices of H is yellow, then we have a yellow triangle in our coloring Thus we may suppose that all the edges used in coloring the edges are H are red, blue and green But now since r(3, 3, 3) = 17, it follows that there is a monochromatic triangle in this coloring of the edges of H 6 Suppose that the numbers S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} are colored red, blue, and green as follows: Red:{2, 6, 10, 11}, Blue:{1, 4, 5, 9}, Green: {3, 7, 8, 12} Suppose that next that we color the edges of the complete graph on K = {0,!1,!2,!3,!4,!5,!6,!7,!8,!9,!10,!11,!12} as in the proof of Schur s Theorem (a) A portion of the graph K is given below Show how you would color the edges between these vertices (b) Is there a monochromatic equation in this coloring of S? Yes, = 5 and = 9 are blue equations

4 7 Fill in the missing details of the following proof that r(3, 4) 9 Proof Suppose that the edges of K 9 are colored red and blue We must show that this coloring contains either a red triangle or a blue K 4 Let R and B denote the red and blue subgraphs respectively of this coloring Suppose that first that there is some vertex v with deg B (v)! 6 Then in this case Since r(3, 3) = 6, there is a monochromatic triangle in this coloring If this triangle is red we are finished If it is blue, then together with v we get a blue K 4 So, we may suppose that every vertex in the blue graph has degree at most 5 But this means that each vertex in the red graph has degree at least 3 In fact there must be some vertex in the red graph that has degree at least 4 since otherwise there would be an odd number of vertices of odd degree in the blue graph So, let u be a vertex such that deg R (u)! 4 Then in this case Either one of the edges in the neighborhood of u is red and we have a red triangle or they are all blue and we have a blue K 4 8 Prove: For any graph G having n vertices,!(g)"(g) # n Consider any coloring of V(G) with k colors where k =!(G), and let V 1,!V 2,,!V k be the color classes of the coloring Then for each i, V i! "(G) k Thus, n =! V i "!#(G) = k#(g) = $(G)#(G) i=1 k i=1

5 9 (a) Label the vertices of the tree below with the a h so that the greedy coloring with respect to the alphabetical ordering requires 5 colors (b) Finish the following argument to show that r( P 4,!P 4 ) = 5 Consider any red-blue coloring of the edges of K 5 Let v be any vertex in this K 5 Then without loss of generality, two of the edges incident with v are colored red Say that the edges va and vb are red Let x any y be the remaining two vertices of G Then if any of xa, xb, ya or yb is red we have a red P 4 Otherwise all those edges are blue and we have a blue P 4 This shows that r( P 4,!P 4 )! 5 To see that r P 4,!P 4 ( ) > 4, consider the graph K 1,3 Neither it nor its complement contains a path on 4 vertices

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