13. (a) G,G. A circuit of length 1 is a loop. 14. (a) E,E. (c) A,B,C,A. 16. (a) BF, FG

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1 13. (a) G,G. A circuit of length 1 is a loop. There are none. Such a circuit would consist of two vertices and two (different) edges connecting the vertices. 10. (a) 11. (a) C, B, A, H, F Other answers such as C, B, A, H, G, F are also possible. C, B, D, A, H, F Again, more than one answer is possible. C, B, A, H, F (d) C, D, B, A, H, G, G, F (e) 4 (It can be helpful to list them from shortest to longest as C, B, A; C, D, A; C, B, D, A; C, D, B, A) (f) 3 (H, F; H, G, F; H, G, G, F) (g) 12 (Any one of the 4 paths in (e) followed by edge AH followed by any one of the 3 paths in (f).) 12. (a) D, A, H, G, F, E D, B, A, H, G, G, F, E D, A, H, F, E (d) D, C, B, A, H, G, G, F, E (e) 5 (D, A; D, B, A; D, B, C, D, A; D, C, B, A; D, C, B, D, A) A,B,D,A; B,C,D,B; F,G,H,F (d) A,B,C,D,A; F,G,G,H,F (e) 6; The longest circuit has length 4. So using the results from (a)-(d) there are a total of 6 circuits in the graph. 14. (a) E,E F,H,F A,B,C,A (d) A,B,E,D,A (e) A,C,B,E,D,A; A,B,E,E,D,A (f) (a) AH, EF; If either of these edges were removed from the graph, the graph would be disconnected. There are none. This graph is just one circuit connecting five vertices. Circuits do not have bridges. AB, BC, BE, CD; That is, every edge is a bridge. 16. (a) BF, FG There are none. The graph consists of two cycles (A,B,C,E,A and A,D,E,A). AB, BC, CD, DE (i.e. every edge is a bridge) B. Graph Models 17. Such a graph will have five vertices (representing North Kingsburg, South Kingsburg, and islands A, B, and C) and seven edges (representing the seven bridges). (f) 3 (H, F, E; H, G, F, E; H, G, G, F, E) (g) 15 (Any one of the paths in (e) followed by AH followed by any one of the paths in (f).) 87

2 ISM: Excursions in Modern Mathematics C. Euler s Theorems 23. (a) (i); The graph has an Euler circuit because all vertices have even degree. (iii); The graph has neither an Euler circuit nor an Euler path because there are four vertices of odd degree. 20. (iv); See exercises 9(a) where an Euler circuit exists and 9 where an Euler circuit does not exist. 24. (a) (ii); exactly two vertices are odd (iii); 10 vertices are odd (v); the graph may be disconnected 21. Let the vertices represent the teams A, B, C, D, E, and F. The edges will correspond to the tournament pairings. Putting the vertices at the corners of a regular hexagon will make drawing and interpreting the graph a bit easier. 25. (a) (iii); The graph has neither an Euler circuit nor an Euler path because there are more than two (exactly 4 in fact) vertices of odd degree. (i); The graph has an Euler circuit because all vertices have even degree. (iii); The graph is disconnected. Disconnected graphs have neither Euler circuits nor Euler paths. 26. (a) (ii); exactly two vertices are odd 22. (a) (i); all vertices are even (iii); the graph is disconnected 27. (a) (ii); Exactly two vertices are odd. (i); All the vertices are even. (iii); More than two vertices (6 in this case) are odd. 28. (a) (ii); exactly two vertices are odd (iii); the graph is disconnected (iii); more than two vertices are odd 88

3 D. Finding Euler Circuits and Euler Paths Note that the starting and ending vertices of the Euler path are shown in black. 35. Again, there is more than one answer The starting and ending vertices of the Euler path (both having odd degree of course!) are shown in black. Naturally, other answers are possible. 89

4 F. Eulerizations and Semi-eulerizations 37. Adding as few duplicate edges as possible in eliminating the odd vertices gives an optimal eulerization. ISM: Excursions in Modern Mathematics 41. In an optimal semi-eulerization two vertices will remain odd. Again, since this graph has vertical symmetry, a mirror image of this answer is also an optimal semieulerization In an optimal semi-eulerization two vertices will remain odd. 39. Since this graph has vertical symmetry, a mirror image of this answer is also an optimal eulerization. G. Miscellaneous 43. (a) 40. (d) 44. (a) 90

5 None. If a vertex had degree 1, then the single edge incident to that vertex would be a bridge. 52. (a) If each vertex were of odd degree, then the graph has an odd number of odd vertices. This is not possible! So, it must be that each vertex is in fact of even degree. By Euler s Circuit Theorem, the graph would have an Euler circuit. The following graph is regular (all vertices are of degree 2) and it has an Euler circuit. 46. The graph below is also regular (all vertices have degree 3) but it does not have an Euler circuit. 47. You would need to lift your pencil 4 times. There are 10 vertices of odd degree. Two can be used as the starting and ending vertices. The remaining eight odd degree vertices can be paired so that each pair forces one lifting of the pencil ; There are 14 odd vertices. Two can be used as the starting and ending vertices. The remaining 12 can be paired so that each pair forces one lifting of the pencil. JOGGING 49. (a) An edge XY contributes 2 to the sum of the degrees of all vertices (1 to the degree of X, and 1 to the degree of Y). If there were an odd number of odd vertices, the sum of the degrees of all the vertices would be odd. That would contradict the result found in (a). 50. (k 2)/2. Two of the vertices of odd degree can be used as the starting and ending vertices. For the remaining vertices of odd degree the pencil will have to be lifted at least once for every two vertices of odd degree. 53. (a) Both m and n must be even. Why? If the graph has an Euler circuit, then the degree of each degree must be even. Suppose that m is odd (i.e. A has an odd number of vertices). Then the degree of each vertex in B would be an odd number (namely m). But this contradicts the existence of an Euler circuit. Similarly, if n is odd, the degree of each vertex in A would need to be odd (a contradiction). So, both m and n must be even. Either m = 1 and n = 1 or 2 or m = 2 and n is odd. Remember that a graph will have an Euler path if it has exactly two odd vertices. If m = 1 and n = 1 or 2, then it is easy to see that the graph has an Euler path. If m = 1 and n > 2, then the vertices in B are all odd (which can t happen since an Euler path has exactly 2 odd vertices). If m = 2, then n must be odd in order to have exactly two odd vertices (both will be in A). If m > 2, then so is n. If either is odd, then the graph has more than 2 odd vertices. If neither is odd, then the graph has no odd vertices (and hence no Euler path). So m > 2 is not possible. 54. (a) Keep adding edges as long as you can without creating any circuits. N 1 dollars 91

6 All graphs that have at least one vertex of degree 0 (i.e. isolated vertices). 62. The only possible value of k is 0. The graph G must have an Euler circuit since it is connected and every vertex is even. In a circuit it is possible to get from any vertex to any other vertex two different ways traveling the circuit forward and backward. Consequently, the removal of a single edge will not disconnect the graph. 63. The possible values of k are k = 0, 1, 2, 3,, N-3, and N-1. To have k = 0 bridges in G, place the N vertices at the corners of a polygon, connect adjacent corners with an edge and connect one pair of nonadjacent corners. To have k = 1 bridge, attach a single vertex to a graph formed by putting vertices at the corners of a regular N-1 polygon. To have k = 2 bridges, consider the following graph formed by attaching two vertices to a graph formed by putting vertices at the corners of a regular N-2 polygon. Adding more single vertices to the left also explains how to get k = 3, 4,, N-3 bridges. If the graph G is a complete graph, then it will clearly have an Euler circuit (since N is odd so that every vertex has even degree). However, the complement will not since it has no edges. The following example illustrates how such a graph G and its complement may both have an Euler circuit. 67. (a) The circuit kissing A, D, C, A is A, B, D, F, C, E, A. Two circuits that kiss A, B, D, A are C, D, F, C and A, C, E, A. Suppose C is a circuit in G. Consider the complement of C in G. If C is an Euler circuit then the complement is empty. So when C is not an Euler circuit, the complement must contain an edge leading to a vertex in C for otherwise G is not connected. Also, the complement of C has all even vertices. So, the complement has an Euler circuit and that circuit will be one that kisses C. 68. (a) There are many ways to implement this algorithm. The following figures illustrate one possibility. To get k = N-1 bridges, consider the following example: 64. Each component of the graph is a graph in its own right and so, according to Euler s Sum of Degrees Theorem, the number of vertices of odd degree (in each component) must be even. Therefore the 2 vertices of odd degree must be in the same component. 65. Suppose the graph has N vertices. Since there are no multiple edges or loops, the maximum degree a vertex can have is N 1. If the degrees of the N vertices are all different, they must be 0, 1, 2,..., N 1, but this is impossible because the vertex of degree N 1 would have to be adjacent to all the other vertices and then we couldn t have a vertex of degree (a) Since N is even, in the complement each vertex will have odd degree. So, the complement cannot have an Euler circuit. 93