(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
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.)
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,