46

If my Python programming is to be believed, the minimum number of moves required is 41:


38

I can get the knight trapped in 15 moves: $$\begin{array} {ccccccccc} \cdot & \cdot & \cdot & \cdot & 5 & \cdot & \cdot & \cdot & \cdot\\ \cdot & 3 & \cdot & \cdot & \cdot & \cdot & \cdot & 7 & \cdot\\ \cdot & \cdot & \cdot & 4 & \cdot & 6 & \cdot & \cdot & \cdot\...


26

Give these names to all the squares: 163 4 8 725 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.


26

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.


25

As shown below Alice gives the red moves and Bob gives blue moves... Moves 2 and 10 have symmetrical options which end up with same result. All other moves are forced.


22

The winner is: The winning strategy is to


21

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 ...


21

Here is my answer in text form. It's long, so I stored it on pastebin: 19 steps I don't have time to convert it to png (thanks @TheDarkTruth, see comments to the question), so maybe later. Gif animation :


19

Another solution, besides Rand al'Thor's accepted one, involves White can play After which Indeed With touchdown in each case.


18

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:


17

I think this works: The key realisation is that


16

It's possible to do it in


16

Here is my answer:


15

Edit: Now that @GOTO 0 got it in 16, I can at least prove that his solution is optimal. Proof: My best was:


15

It's relatively easy if you start at the exit and try to reach the starting point. (Apologies for my bad paint skills)


13

If you take the directions on a clock as the possible moves, 1,2,4,5,7,8,10,11: Will do it Which is 34 moves in total.


11

The trick is to place them Full list:


10

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 ...


10

Here is a solution


10

Why?


10

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 ...


10

This puzzle can be broken down relatively easily. Start with So now we can put it together to finish the whole thing:


9

Here is a solution to the maze:


9

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 ...


9

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: Mutlithreading:...


9

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 ...


8

Proof. Example.


8



7

There are some unreachable pellets, and there are safe squares for the ghosts.


6

I am 3 minutes late but I think I have a slightly different solution using a less obnoxious colour


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