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Knights on the move

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This puzzle reportedly dates from the 13th century. So some of you older (ahem) Denizens may have encountered it already. Anticipating this possibility, I've added a competitive wrinkle to the mix.

On a 3x3 chessboard you are to place a knight on an unoccupied square and then move it to another unoccupied square using a legal knight's move. If you alternate turns with an opponent, the board eventually fills up to a point where there are not sufficient unoccupied squares for one of the players to complete a turn. The player who has that turn then loses the game. I.e., the last player to successfully complete a turn is the winner.

If first player is A and second player is B, which player wins the game if

1, Players cooperate to maximize the number of turns.

2. Players compete to win the game?

If you wish to describe a sequence of moves in your answer, it may be helpful to number the squares row-wise left to right starting with the top row

1 2 3

4 5 6

7 8 9

A legal move would then be 1 - 6, for example, provided that quares 1 and 6 are unoccupied prior to the move. Square 6 would remain occupied. Assume there are enough knights available to each player.

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Posted (edited) · Report post

The most I can get is 6 knights. Player B wins. 4-9 7-2 6-1 7-6 3-4 8-3

The least I can get is 4. Player B wins. 9-2 3-4 1-8 7-6 It seems like there should be a way with just 3 but I'm not seeing it right now.

Edited by Thalia
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Posted (edited) · Report post

7 turns, non competitive.

1-8, 8-3, 3-4, 4-9, 9-2, 2-7, 7-6 A wins.

Edited by Aaryan
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Posted · Report post

Oh, I misunderstood the OP. So it's NOT the same knight moving around the squares?

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Posted · Report post

Oh, I misunderstood the OP. So it's NOT the same knight moving around the squares?

Correct.

Each play leaves another knight on the board.

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Posted · Report post

The most I can get is 6 knights. Player B wins. 4-9 7-2 6-1 7-6 3-4 8-3

The least I can get is 4. Player B wins. 9-2 3-4 1-8 7-6 It seems like there should be a way with just 3 but I'm not seeing it right now.

Nice.

But you can do better on part 1.

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Posted (edited) · Report post

You can get 7 knights. I did it using linear logic.

Eliminate square 5. A knight can never occupy it. In one dimension, the board looks like 1 - 6 - 7 - 2 - 9 - 4 - 3 - 8 - 1.....

You need two adjacent spaces to place one knight. Start with 1 - 6. Eliminate 6 and move backward. 8 - 1. 3 - 8. 4 - 3. ...

You'll end with 7 - 2 and 7 will be the only open space (besides 5).

Edited by Molly Mae
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Posted · Report post

You can get 7 knights. I did it using linear logic.

Eliminate square 5. A knight can never occupy it. In one dimension, the board looks like 1 - 6 - 7 - 2 - 9 - 4 - 3 - 8 - 1.....

You need two adjacent spaces to place one knight. Start with 1 - 6. Eliminate 6 and move backward. 8 - 1. 3 - 8. 4 - 3. ...

You'll end with 7 - 2 and 7 will be the only open space (besides 5).

That's it, and very nice explanation.

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Posted · Report post

The most I can get is 6 knights. Player B wins. 4-9 7-2 6-1 7-6 3-4 8-3

The least I can get is 4. Player B wins. 9-2 3-4 1-8 7-6 It seems like there should be a way with just 3 but I'm not seeing it right now.

The second question was aimed at who would win if played competitively.

Is your "minimum" solution a statement that B wins if the players compete?

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Posted (edited) · Report post

The second question was aimed at who would win if played competitively.

Is your "minimum" solution a statement that B wins if the players compete?

The fewest that I can get on the board is 4 so yes but I'm guessing that there is a way to get 3. I'm not seeing it though.

Edited by Thalia
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Posted · Report post

Player 1 always wins with optimum play. I have a breakdown that I'll post in a bit.

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Posted (edited) · Report post

Player 1 always wins with optimum play. I have a breakdown that I'll post in a bit.

I actually miscalculated a case. I'll update shortly.

P2 wins in all cases.

1. (1, 9)

2. (x, y)

If x=6 or 2, y=4 or 8

if x=4 or 8, y=6 or 2

If x=7, y=3

If x=3, y=7.

P2 forces a win in all cases.

As established earlier, the positions (corner, side) are arbitrary as they all loop linearly.

If P1 forces the game to last as long as possible, there will be a total of 6 moves.

There are a few different ways this can happen, but here's an example game:

1. (1, 9)

2. (6, 4)

3. (7, 3)

4. P1 cannot play.

Edited by Molly Mae
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