# 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 ?

• Actually, the hint I wrote before is not entirely sufficient. Hopefully this observation helps anyway: suppose the board is the A1-C8 rectangle on a standard chessboard. Consider the squares A1 and C1. A part of the cycle through them is basically forced. Next, it might be a good idea to consider the square B1. – Dejan Govc Aug 28 '14 at 16:51
• Maybe, there is another way to prove my claim, but the way I propose would be most elegant and easy to see, if it would work. A direct proof requires many cases here, and the danger is that some potential hamilton circle is overseen. – Peter Aug 28 '14 at 17:14
• It is well possible (even likely), that my method is not possible in this case. Are there any more useful NECESSARY conditions for the existence of a hamilton-cycle ? – Peter Aug 30 '14 at 11:26

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.