This question was posed to me, and I was surprised that I could not find a solution (as I thought that all rook tours [open or closed] were possible). Starting from a8, could a rook visit every square on the board once, ending on f3?
I tried a few times, with a few different strategies, but I always ended up missing one square.
It's really easy to burn pairs of rows or columns, so the problem space could be reduced...
...but at some point (4x4), I was able to convince myself that it is impossible (at least at this size and state):
...but it might be possible that shaving off column or row pairs is also discarding a solution?
I am afraid that I didnโt apply any rigor here and thought I solved it. Been playing with it for a bit and see there is likely not a solution. Good fun for me.
Seems this is related to Hamiltonian path problems. The issue is there is always one square that canโt be picked up. Why could this be?
So the rook has to move from a white square to a black square or a black square to a white one. This would mean the sequence would go white, black, white, black and so on for all squares. Since there is an even number of squares if the rook starts on white it must end on black but the problem states the start and end squares are both white, thus impossible to solve. Doesn't really have anything to do with hamiltonian paths because they are loops that will fill a space. It does relate more broadly to space filling curves in general but I think a graphical approach to this problem can be a bit misleading.
Interestingly you can pick any two white squares on the chess board and you couldn't make a path between them in the way op is trying to.
If you take the rooks movement, and reduce each longer move to a series of single steps a1a2a3a4 for example, this becomes a Hamiltonian problem. The corners have 2 nodes, the edges three, and the core has four. Black and white indicates even or uneven distance in nodes.
Ahh yes sorry my bad I mixed up hamiltonian and hilbert in my head