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25

I saw no implications that we aren't going by standard chess rules. Given that we only allow queens to attack the opposing color I propose:


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Here is the optimal answer with Shown as knights;


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If a rectangular piece of chessboard is $a\times b$ squares in size, then its diagonal squared is $a^2+b^2$ and its area squared is $a^2\cdot b^2$, and therefore the quantities $a^2,b^2$ are the roots of the quadratic equation $x^2-D^2x+A^2=0$ where $D,A$ are the diagonal and area. But this is enough to determine $\{a^2,b^2\}$ completely: these values are $\...


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HTM's answer is very nice and must be what's intended here. But I'm not quite seeing why the puzzle isn't cooked as follows:


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Note that without their king, White’s plan would be Thus, the aforementioned sequence of moves will The only square White’s king can be in that satisfies the constraint is


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Here is a solution with 15 knights: I could not find a solution with 14 knights.


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Following @Konstantin Pfennig's answer, using black and white queens, and squeezing in more, I have:


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There are


14

Here is my attempt at the solution. I don't have any mathematical proof that this is the minimum no. of knights, but the steps I followed suggest that. In the figures below, the yellow cells denote the knight's location, and the corresponding numbers denote their covering cells.


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Solution Methodology The first thing to realise is that I got as far as the following The next thing to realise is that The next step involves an assumption made in hope, namely that Checking through those, we see that Checking these with the board we've got so far, we find that Using this, the board becomes Uniqueness? Feedback section Great ...


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Here is a solution with pieces. The colouring of the empty squares ($2\times2$ in white, $4\times4$ in yellow, $6\times6$ in orange, $8\times8$ in red) shows how the pattern generalises to any $2k\times2k$ board. For $8\times 8$ this solution is optimal.


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Answer Mate in 4 moves: Answer for the cryptarithm: That is, using the FEN notation, The third part means:


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There are solutions for number of knights on each side equal to 4, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16. For example, here's one for 16: I used integer linear programming, with a binary decision variable $x_{i,j,k}$ to indicate if cell $(i,j)$ contains a knight of color $k$. For each cell $(i,j)$, let $N_{i,j}$ be the set of neighboring cells (one knight's ...


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Clues Putting it all together


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It could really be as simple as But it definitely seems very simple. I get the feeling we're missing a lot of context.


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Looks like after the black king has only bad options: and finally, after we have reached this position: All the escape squares of the black knight are blocked, so it will fall, which brings us to.. This assessment seems accurate, since the white bishop can sacrifice itself to stop black's e-pawn if necessary, after which black is left with a pawn on the ...


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Here's my go (click to enlarge): The pattern of reds is and the white and black patterns are so it's enough to count half the pieces only. :-)


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One solution to part A:


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Knights Bishops Pawns I cannot outdo JMP's answers for Kings and Rooks.


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Answer Attempt 1 At first I thought This can be achieved as follows, using the same technique as Daniel Mathias previously: Adjusting this a little, we can get After a lot more fiddling around with this, I realised it seems that Attempt 2 I then tried Attempt 3 Finally, the third option is The mistake was


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We can achieve the stated goals if Knight: Bishop (this one was harder): Doing a similar trick with a rook promotion is harder because


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Rook White to move: Explanation:


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Are you playing If so,


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Observation: Implication: Conclusion:


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Due to the 6, we can place the following few pieces: After a little bit of trial and error from here, I found the solution: 4 is attacked by: 6 is attacked by all but


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Solution for the $5$x$5$ grid Other solutions for the same grid are For the $6$x$6$ grid I noted that This leads to this conclusion for the $6$x$6$ grid


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