0
$\begingroup$

I basically know whether the following statements are true, but I would like to know how they are proved.

  1. A knight kept anywhere on an empty chess board can not reach its adjacent square in exactly 8 moves.
  2. It is possible to fill the entire board with black and white knights such that no knight can kill another knight of the other colour.
  3. A knight that reaches a square in $n$ moves can not reach it in $n+1$ moves but can reach it in $n+2$ moves in atleast $n$ ways.

P.S. I'm not sure if my tags are ok.

$\endgroup$
1
$\begingroup$

Think about the coloring of a chess board. If a knight starts on a white square, what color of square is it on in one move? In two moves? In $2n$ moves? In $2n+1$ moves?

If you go through this carefully, this should answer all of the questions.

$\endgroup$
  • 1
    $\begingroup$ Thank you, but how does it answer part 2, or the no. of ways a knight reaches a position (as in part 3)? $\endgroup$ – ghosts_in_the_code Nov 28 '14 at 10:22
  • $\begingroup$ It doesn't admittedly answer the question about the number of ways that it can reach a square in n + 2 ways. However, I realize that what I Was thinking about question 2 makes no sense, so I retract that statement. $\endgroup$ – Simon Rose Nov 28 '14 at 13:09
0
$\begingroup$

For part three you need to think about the $n$ steps you take to get to the final square. How can you use each of these steps to make a new path containing two additional steps (that will give you $n$ ways). [sorry that doesn't quite do it, I'll think further, the paths I was thinking of need not be distinct, but you can use the fact that every square is adjacent by knight's move to at least two others to fix my idea]

Part $2$ should probably refer to attacking knights of the same colour. Some simple observations about a knights tour of the whole chessboard show that if there are knights of both colours, there must be a change of colour along the tour.

$\endgroup$
  • $\begingroup$ You don't really need to invoke the existence of a knights tour to show that all the knights would have to have the same color in Part 2, you just need the fact that you can get from any square to any other square in a finite number of knight's moves. $\endgroup$ – Barry Cipra Nov 30 '14 at 16:09
  • $\begingroup$ @BarryCipra Indeed so. $\endgroup$ – Mark Bennet Nov 30 '14 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.