# Knight on a $3\times 4$ board: Hamiltonian graphs

A chess knight sits on a $$3\times 4$$ board. Is it possible for the knight to jump into the $$12$$ squares without jumping twice in any of them and ending and starting in the same box? What if it starts and ends in the different boxes?

I have drawn the graph that represents this problem and by looking at it I know that the answer to the first question is that it is impossible, but the second one is possible. However I can't find a mathematical reasoning to prove this. I know that my problem is equivalent to finding a hamiltonian cycle in the first case and a hamiltonian path in the second, but I don't know how to do this in any other way that trying to draw different paths. Could someone please help me with the mathematical reasoning?

Let's number the squares on the board in the conventional way:

a4 b4 c4
a3 b3 c3
a2 b2 c2
a1 b1 c1


If we look at squares a1, b4, and c1, there are only two possible moves from each of them. If these squares are in the middle of a tour - and that's always the case for a closed tour - then both of those moves have to be part of the tour: one entering the square and one leaving.

However, putting those two moves together gives a cycle of length $$6$$: a1 - b3 - c1 - a2 - b4 - c2 - a1. Instead of a tour that visits all $$12$$ squares, we end up with two subtours: this cycle, and its mirror image a4 - b2 - c4 - a3 - b1 - c3 - a4. So a closed tour is impossible.

For an open tour, we can just pick a move, such as a3 - c2, that goes between the two cycles of length $$6$$, and use it to stitch them together into a path. Just finding a path that works should be enough for a complete solution to this part.

It is well-known that a closed knight's tour on $$3\times 4$$-board is impossible. I'll try to recall the proof.

An open tour exists, see, for instance, the picture below.