47

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


42

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\...


29

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.


27

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.


24

I have a computer program for solving packing problems, and found a way to use it to solve this problem. One of the solutions it found is below: Note that this is very close to the attempted solution in the question, as it differs only in the top left quadrant. Modelling the problem in this way will not find all solutions (I'd need to allow other tile ...


22

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 :


22

The winner is: The winning strategy is to


22

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 ...


19

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:


19

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


18

I think this works: The key realisation is that


17

The easiest way to do this, I think, is The answer is and here are the calculations:


17

Here is my answer:


16

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


16

It's possible to do it in


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.


13

First, since we need to do four laps, then we need to make spaces for four routes since we cannot repeat squares. This means there should be four initial squares that would be the first square for each lap, and this would lead to four parallel routes. This would make it easier to find which square each route needs to go to next, while not blocking other ...


12

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 ...


12

First things first: let's check the divisibility. There are 64 squares, and the knight is standing in one of them. That leaves 63 squares to cover, and each move covers 3 squares, so that seems to work out. That means, however, that we won't be able to create a closed loop, so every solution we find only solves the puzzle for the starting point and, by ...


11

Here is a solution


11

The trick is to place them Full list:


11

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


11

Observation: Implication: Conclusion:


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 ...


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

Proof. Example.


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