MATH /2003. Assignment (1 pt) 7.3 #6 Represent the graph in Exercise 2 with an adjacency matrix. Solution:

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1 MATH /2003 Assignment 7 All answers must be justified. Show your work. 1. (1 pt) 7.3 #6 Represent the graph in Exercise 2 with an adjacency matrix. Solution: (1 pt) 7.3 #12 Draw a graph with the given adjacency matrix Solution: 3. (2 pts) 7.3 #38 Determine whether the given pair of graphs is isomorphic. Solution: They are isomorphic. An isomorphism is: f (u 1 )=v 1 f (u 2 )=v 3 f (u 3 )=v 2 f (u 4 )=v 5 f (u 5 )=v 4 4. (2 pts) 7.3 #42 Determine whether the given pair of graphs is isomorphic. Solution: They are not. One way to see that is that the two vertices of degree four in the first graph (u 2,u 3 ) are adjacent and the two vertices of degree four in the second graph (v 2,v 4 )arenotadjacent. 5. (3 pts) 7.3 #58 Determine whether the graphs without loops with the following incidence matrices are isomorphic. 1

2 (a) (b) , Solution: Both graphs are K 3. They are isomorphic , Solution: Both graphs are K 4 with one edge missing. They are isomorphic. 6. (3 pts) 7.4 #10 (Use Theorem 2) Find the number of paths between c and d in the graph in Figure 1 of length a) 2. b) 3. c) 4. d) 5. e) 6. f) 7. Solution: LetA be the adjacency matrix of G where the vertices are arranged in the alphabet order A = A 2 = A 3 = A 4 =

3 A 5 = A 6 = A 7 = The number of paths from c to d is the entry in the third row and fourth column. The answers are (a) 0; (b) 8; (c) 10; (d) 73; (e) 160; (f) (3 pts) 7.4 #18 Find all the cut edges in the graphs in Exercises Solution: 15. None. 16. cd. 17. ab,bc,cd,ce,ei,ih. 8. (2 pts) 7.4 #34 Use Theorem 2 to find the length of the shortest path from a to c in the directed 3

4 graph in Exercise 2. Solution: The adjacency matrix is A = A 2 = A 3 = Since the entries in row 1 and column 3 is zero in A and A 2 but 1 in A 3, there is no path of length 1 or 2 from a to c. Thereisonepathoflengththreefroma to c. Therefore, the shortest path from a to c has length (1 pt) 7.5 #4 Determine whether each graph has an Euler circuit. Construct such a circuit when one exists. Solution: Since both f and c have degree three, there is no Euler circuit. 10. (1 pt) 7.5 #12 Determine whether the graph in Exercise 5 has an Euler path. Construct such a path 4

5 if it exists. Solution: Every vertex has an even degree. There are Euler circuits. One of them is aeaebedcecdba. 11. (1 pt) 7.5 #20 Determine whether the picture shown can be drawn with a pencil in a continuous motion without lifting the pencil or retracing part of the picture. Solution: Yes. Since every vertex except two has even degree, the graph has an Euler path. 12. (3 pts) 7.5 #36 For which values of n do the following graphs have an Euler circuit a) K n b) C n c) W n d) Q n. Solution: (a) Every vertex in K n has degree n 1. K n has an Euler circuit if n is odd. (b) Every vertex in C n has degree two. C n has an Euler circuit for every n. (c) Every vertex except the center of W n has degree three. W n has an Euler circuit for no n. (d) Every vertex in Q n has degree n. Q n has an Euler circuit if n is even. 13. (1 pt) 7.5 #42 Determine whether the given graph has a Hamilton circuit. If it does, find such a circuit. If it does not, give an argument to show why no such circuit exists. Solution: A Hamilton circuit does not exist since f has degree 1. 5

6 14. (2 pts) 7.5 #44 Determine whether the given graph has a Hamilton circuit. If it does, find such a circuit. If it does not, give an argument to show why no such circuit exists. Solution: Suppose that there is a Hamilton circuit. Then the edges ab, da must be in the circuit for a to be in the circuit. Similarly, the edges bc, ch, hg, gf, fe, ed all have to be in the circuit. But these edges form a circuit already. Therefore, a Hamilton circuit cannot exist. 15. (1 pt) 7.5 #46 Determine whether the given graph has a Hamilton circuit. If it does, find such a circuit. If it does not, give an argument to show why no such circuit exists. Solution: There are Hamilton circuits. For example, a, b, c, e, f, i, h, g, d, a. 16. (2pts) 7.6#2Findthelengthoftheshortestpathbetweena and z in the given weighted graph. 6

7 Solution: a b c d e z a b c e d z The shortest distance from a to z is (4pts) 7.6#4Findthelengthoftheshortestpathbetweena and z in the given weighted graph. Solution: a b c d e f g h i j k l m n o p q r s t z a d b e c g f h k l i r j m n p q o t s z The shortest path from a to z is a, b, e, h, l, m, p, s, z and the length is (1 pt) 7.7 #10 Suppose that a connected planar graph has eight vertices, each of degree 3. Into how many regions is the plane divided by a planar representation of this graph? Solution: Since each vertex has degree 3 and 2e = thesumofthedegrees=24. We have e =12. Euler s formula, we have 8+r 12 = 2. 7

8 Therefore, r = (1 pt) 7.7 #12 Suppose that a connected planar graph has 30 edges. If a planar representation of this graph divides the plane into 20 regions, how many vertices does this graph have? Solution: Sincee =30and r =20, by Euler s formula, we have v = 2. Therefore, v = (2 pts) 7.7 #14 Suppose that a connected bipartite planar simple graph has e edges and v vertices. Show that e 2v 4 if v 3. Solution: In a bipartite planar simple graph, the smallest region would have at least 4 edges in its boundary. Since every edge can be used in the boundaries of two regions, we have Substituting this into Euler s formula, we have 4r 2e r e 2 v + e 2 e 2 v 2 e 2 Therefore e 2v (2pts) 7.8#6Findthechromaticnumberofthegivengraph. Solution: Since there are many triangles in G, the chromatic number of G is at least 3. G can be coloured with three colours: vertex a b c d e f colour The chromatic number of G is (3pts) 7.8#8Findthechromaticnumberofthegivengraph. 8

9 Solution: Since the vertices i, b, c, h form a K 4,thechromaticnumberofG is at least 4. G can be coloured with four colours: vertex a b c d e f g h i colour The chromatic number of G is four. 23. (3 pts) 7.8 #16 How many different channels are needed for six stations located at the distances shown in the table, if two stations cannot use the same channel when they are within 150 miles of each other? Solution: We construct a graph G. The vertices of G are the stations. Two vertices are adjacent if the stations are within 150 miles of each other. The adjacency matrix of G is The graph is It is easy to check that the chromatic number of G is 3. Therefore at least three channels are needed. 9