Proof that there is no closed knight tour on a $3\ \times\ 8$ - board I want to prove that there is no closed knight tour on a $3\ \times\ 8$ - board by deleting $s$ vertices
 of the corresponding knight graph to get a graph with more than $s$ connected
 components (which would prove that it is not hamiltonian), but I did not succeed so far.
Who can help ?
 A: Do you have to prove it that way? It is not difficult to prove directly that there is no closed knights tour of a 3 x 8 board. You start with squares that have only two moves available, so you know both those moves must be part of any tour.
Label the squares $$\begin{array}{cccccccc}1 & 4 & 7 & 10 & 13 & 16 & 19 & 22\\ 2 & 5 & 8 & 11 & 14 & 17 & 20 & 23\\ 3 & 6 & 9 & 12 & 15 & 18 & 21 & 24\end{array}$$ There are only two moves from 1, so the path must include 8 1 6. Similarly, there are only two from 3, so it must include 8 3 4, and hence 4 3 8 1 6. Similarly, there are only two from 2, so it must include 7 2 9. If the path includes 4 9 and 6 7, then we have the closed path 4 3 8 1 6 7 2 9 4. Contradiction, so it can include at most one of 4 9 and 6 7. Similarly, it can include at most one of 4 11 and 6 11. But it must include one of 4 9 and 4 11 and one of 6 7 and 6 11. So wlog it includes 6 11 and 4 9. So we have 11 6 1 8 3 4 9 2 7.
Exactly the same argument in the right hand half of the board shows that 14 cannot be connected to both 19 and 21. It cannot be connected to 9 because that is already connected to 2 and 4, so it must be connected to 7.
Again by the same argument in the right hand half, 17 is connected to 22 and 24. Hence the only squares that can be connected to 10 are 5 and 15. Obviously 5 must also be connected to 12. 12 must also be connected to 13 (7 and 17 are unavailable). But 20 must be connected to 13 and 15, so we have the closed loop 20, 13, 12, 5, 10, 15, 20. Contradiction.
