# Chess King Tour 8x8 problem

Two squares on a chessboard are said to be neighbours if they have an edge or a corner on the board in common. This means that squares on the edge have 5 neighbours, on the corner have 3 neigbours, and central squares have 8 neighbours.

A king moving normally on a chessboard always moves from a square to one of its neighbours.

Is it possible for a king to make a tour of a chessboard, visiting each square exactly once, in such a way that (apart from the square visited on the first move) every square that is visited has an even number of neighbours that have already been visted?

• Could you, perhaps, give a small example to make the even condition clearer? – Michael Burr May 15 '16 at 17:00
• Starting from the end might be useful. The last move move has to be from an edge square to a central square. – joriki May 15 '16 at 17:07
• Do not vandalise your posts. – Daniel Fischer May 18 '16 at 9:58

• @LucyJ: The rules imply that each step must add an even number to the number of pairs of neighbours that have been visited. There is an even number of such pairs to be completed, we start out with $0$ and the first step adds $1$, an odd number. To get to the even number in the end, some step needs to add an odd number. – joriki May 15 '16 at 17:32