Discrete Mathematics with Graph Theory (3rd Edition) 273

Discrete Mathematics with Graph Theory (3rd Edition) 273 -...

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Section 10.2 271 (b) Such a tour is not possible because a 7 x 7 board has an odd number of squares. A knight's tour corresponds to an alternating set of distinct red and black squares. Since we require that the knight return to the square from which he left, the number of squares of the tour must be even. 7. No, it isn't. Label edges as shown. Two of the three middle edges 1, 2, 9 must be in any Hamiltonian cycle. By symmetry, we may assume that these are 1 and 2. (A solution for any other choice could be found with an argument similar to what follows.) Edge 9 is not in the cycle, hence, 5 and 6 both must be in. Now, one of 4 and 7 is in the cycle, and one of 3 and 8. Not both 7 and 8 can be in, however, or the cycle would contain the smaller cycle 1, 2, 7, 8. Also, as we saw in PAUSE 6, edges 3 and 4 cannot both be in the cycle. Hence, both 3 and 7 are in, or 4 and 8 are in. Assume that 3 and 7 are in. (The other case follows similarly.) 21 18 We now have 7, 1, 2, 3, 6 and 5 in the cycle, while 9, 4, 8 are not in the cycle. Since 3 and 6 are in,
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