Give these names to all the squares:
Each number can only be accessed by way of the numbers before and after it (where 8 wraps around to 1). That means they form a loop. Since they can never pass each other up on the loop, their relative ordering cannot change. Therefore it is impossible.
I found a solution that uses 16 moves.
After exhaustively checking that there is no solution in 14 moves, I conclude that 16 moves is optimal, because after any odd number of moves the number of white and black squares occupied by knights cannot be equal.
You need at least 16 Moves.
Let's make the task visually more simple. The initial board is:
a4 b4 c4
a3 b3 c3
a2 b2 c2
a1 b1 c1
We cut it into 12 cells and connect only those, which are separated exactly by one move of a knight. Easy to check that the result is the following:
c4 - a3 - c2 - a1
| | | |
b2 b1 b4 b3
Here is my answer in text form. It's long, so I stored it on pastebin:
I don't have time to convert it to png (thanks @TheDarkTruth, see comments to the question), so maybe later.
Gif animation :
Here are all 48 solutions to the maze with no repeated squares (shortest first):
The maze actually has some isolated or unreachable components, and one component that is isolated unless you pass through the goal cell:
All 64 knights are needed, in which case any setup is stuck. We prove that with 63 knights, any position is reachable from any other position.
With 63 knights, the puzzle is much like a sliding puzzle, except pieces slide as knight moves rather than into orthogonality adjacent cells. We can think of a move as swapping the position of the hole and the knight ...
Full solution with explanation
Of the numbers from one to nine:
ONE, TWO, and SIX have three letters; we already know which region is ONE, and TWO and SIX are then also given by the W and X already in place.
FOUR, FIVE, and NINE have four letters; FOUR is given by the U already in place, NINE must be the bottom right region, so FIVE must be the top middle ...
Besides the starting square,
The images below show one possible tour:
This tour uses the "domino pattern" from this paper: K. McGown and A. Leininger. "Knight's Tour." REU at Oregon State
University. August, 2002. In that paper, they also show that
Though the above technique only works for $4 \times m$ boards where $m$ is a multiple of $3$, it should be ...
The player who will win this game on an $8 \times 8$ board is:
Computer based proof
I managed to significantly increase the speed of my previously posted program, so I was able to calculate the result for an $8 \times 8$ board. It's much bigger now, but I have added some comments this time. The speed up was achieved through following steps:
Coded b1 as initial position, bn as position before n'th jump
As for methodology, I started by deducing the following:
From there it seemed as if there were three points of high tension,
From this conclusion I started working on solving the 6-area at the top right, because I had to work around a fair amount of filled squares, knowing I needed to head ...