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You sit down to play a “new” game of chess. In this game only one Knight is on the regular 8x8 chessboard.

Only regular legal moves of Knight are allowed. Game begins with computer going first to place the Knight at randomly selected place..say D6. It is your move to place it at any one of the 8 positions allowed in this instance.

As an eminent Game Theorist, you try to optimize and make the best move. You and computer take turns to move the Knight till all 64 squares are visited. Nobody is allowed to revisit a square.

Last one who cannot place the Knight in an empty square looses.

What are your chances? Can you win against the computer as long as he places the Knight first at the start of the game?

Give your reasoning.

enter image description here

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  • $\begingroup$ i think knight's notation is N, no? $\endgroup$ – Omega Krypton Jun 21 at 15:55
  • $\begingroup$ Regular notation is N..true..I just put K..the first letter..if it can be changed, it will be better $\endgroup$ – Uvc Jun 21 at 15:58
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According to this answer by @bof on Mathematics Stack Exchange:

you can win.

Proof:

The $m\times n$ knight's graph is the graph whose vertices are the squares of the $m\times n$ chessboard, two squares being adjacent if a knight can move from one to the other.

Player 2 has an obvious winning strategy on the $m\times n$ chessboard if the corresponding knight's graph has a perfect matching; namely, he moves the knight to the square which is matched with the knight's current square. In particular, assuming $mn$ is even, Player 2 has a winning strategy if the knight's graph has a Hamiltonian path; in chessic terms, if a knight's tour (open or closed) exists.

Hence Player 2 has a winning strategy on the normal $8\times8$ chessboard and many others. Namely, Player 2 has a winning strategy on boards of the following sizes:
(a) $m\times n$ where $m,n\ge5$ and $mn$ is even;
(b) $2\times n$ where $n$ is divisible by $4$;
(c) $3\times n$ where $n$ is even and $n\ge4$;
(d) $4\times n$ where $n\ge2$.

For case (a) we can use the fact that a knight's tour exists. For the remaining cases, it suffices to observe that the $2\times4$, $3\times4$, and $3\times6$ knight's graphs have perfect matchings, since the boards in cases (b), (c), and (d) can be tiled with $2\times4$, $3\times4$, and $3\times6$ boards.

In a similar way, it can be shown that Player 1 wins in all other cases. That is, Player 1 has a winning strategy on boards of the following sizes:
(e) $m\times n$ where $mn$ is odd;
(f) $1\times n$ where $n\ge1$;
(g) $2\times n$ where $n$ is not divisible by $4$.

Example. Here is a "perfect matching" for the ordinary $8\times8$ chessboard:

a1c2, b1d2, c1a2, d1b2, e1g2, f1h2, g1e2, h1f2,
a3c4, b3d4, c3a4, d3b4, e3g4, f3h4, g3e4, h3f4,
a5c6, b5d6, c5a6, d5b6, e5g6, f5h6, g5e6, h5f6,
a7c8, b7d8, c7a8, d7b8, e7g8, f7h8, g7e8, h7f8.

That is, square a1 is paired with square c2, b1 is paired with d2, and so on. Note that the 64 squares are partitioned into nonoverlapping pairs, and each set of paired squares are a knight's move apart. The winning strategy for Player 2 is: WHEREVER PLAYER 1 PUTS THE KNIGHT, MOVE IT TO THE OTHER SQUARE IN THE SAME PAIR. For instance, if Player 1 starts the game by dropping the knight on e8, Player 2 replies by playing Ng7. If Player 1 now plays Ne6, Player 2 plays Ng5, and so on. Here is a sample game, with Player 1 making random moves, and Player 2 following the winning strategy described above:

1. Ne8 Ng7 2. Ne6 Ng5 3. Nf3 Nh4 4. Nf5 Nh6 5. Nf7 Nh8 6. Ng6 Ne5 7. Nc6 Na5 8. Nb3 Nd4 9. Ne2 Ng1 10. Nh3 Nf4 11. Nh5 Nf6 12. Ng4 Ne3 13. Nc5 Na3 14. Nb1 Nd2 15. Nf1 Nh2 and Player 2 wins.

P.S. Of course the strategy still works if White is allowed to move the knight to any square not already visited, and only Black is limited to making knight moves.

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  • $\begingroup$ Very good strategy.. $\endgroup$ – Uvc Jun 21 at 16:17

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