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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 ?

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  • $\begingroup$ 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. $\endgroup$ – Dejan Govc Aug 28 '14 at 16:51
  • $\begingroup$ 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. $\endgroup$ – Peter Aug 28 '14 at 17:14
  • $\begingroup$ 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 ? $\endgroup$ – Peter Aug 30 '14 at 11:26
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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.

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    $\begingroup$ I accepted the answer because it is not likely that there is an easier method to proof the non-existence of a hamilton-cycle in this graph. Thanks for your effort. $\endgroup$ – Peter Aug 30 '14 at 11:30

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