A knight's nightmare
Imagine a chess board with $n\times n $ squares, $n$ on each side. Now imagine a knight moving around the board - only using the moves that are allowed to a knight of course - so that each square of the board is visited exactly once, and so that the knight ends up on the same square as it started. Such a tour is called a closed knight's tour (it's closed because the knight ends where it started). If you start experimenting on an ordinary chess board, you'll soon see that it's no easy feat to find a closed knight's tour. People have been entertaining themselves with this pursuit for centuries. The earliest recorded example of a knight's tour on the ordinary $ 8\times 8$ board came from al-Adli ar-Rumi, who lived in Baghdad around 840AD. There are also example of knight's tours of $10\times 10 $ and $12\times 12 $ boards.
But no-one has ever found a closed knight's tour on an $ n\times n $ board when $n$ is odd. Can you prove why this is, in fact, impossible?
If you're poetically minded, try this one: find a knight's tour on this $8\times 8$ board, so that the syllables on the squares, when read in the sequence of the tour, form a verse (note that this time you're not asked for a closed knight's tour - it does not have to end at the same place it started).
The solutionAssume that the knight starts out on a white square (the argument will be the same if it starts out on a black square). Because of the way a knight moves in Chess, the next square it lands on will be black. To complete a closed knight's tour, the knight has to make $n\times n $ moves. Since $n$ is an odd number, $ n \times n$ is also odd, so the knight has to make an odd number of moves. But this means that it will end on a black square, since the colour of the square changes with each move. This is a contradiction, because the knight has to start and end on a white square. The solution to the "cryptotour" is the verse With nerve of steel and heart of fire Back to main puzzle page |