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KingGeorge

  • one year ago

[UNSOLVED] Combinatorics challenge problem! Suppose you have a chessboard of size \(n\times n\), and you want to place some pawns on it. How many ways are there to place these pawns such that each row and column has an even number of pawns?

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  1. vf321
    • one year ago
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    Does chessboard orientation matter?

  2. KingGeorge
    • one year ago
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    No.

  3. vf321
    • one year ago
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    OK, well I don't want to make you think that I'm still working on it. I tried for a couple days, but then I gave up.

  4. andriod09
    • one year ago
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    wouldn't you have to fill the entire board so that way every column and every row have the same number of pawns in it?

  5. KingGeorge
    • one year ago
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    Note necessarily. For example, in a 4x4 chessboard, you can have a 4x2 rectangle filled with pawns.

  6. andriod09
    • one year ago
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    Ah, that's true.

  7. jasmine2334
    • one year ago
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    7

  8. Numb3r1
    • one year ago
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    I'm going to try a few n's and see what patterns I notice. n=1, x=0 (1 if the zero-board counts) n=2, x=1 (2 if the zero-board counts) n=3, x=3 (4 if the zero-board counts) assuming that orientation does not matter, therefore rotations and reflections count as the same thing. n=4, x=7 (8 if the zero-board counts) Based on this, x=\[2^{n-1}-1, 2^{n-1} if the zero-board counts.\]

  9. KingGeorge
    • one year ago
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    @Numb3r1 the zero board is does count, and \(2^{n-1}\) is correct. Can you prove it?

  10. Numb3r1
    • one year ago
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    I have to prove that adding one more row and one more column doubles the number of possibilities, or find a recursive relation between previous boards. The recursion option seems simpler to me. One moment...

  11. Numb3r1
    • one year ago
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    A zero-by-zero board does not exist, so it will obviously have 0 options. The one-by-one board has one option. The two-by-two board has two options (and even-numbered boards will always have a new option in terms of how many pieces are in a row). Adding one row and one column means adding one new option for each option that was available on the previous board (one expanding sideways, one expanding both sideways and upwards). From the 3x3 board, the 4x4 has the same double-expansion factor on the square sub-boards (the 2x2 board and 3x3 with only the corners) but the two rectangular sub-boards can only be lengthened (expanding upwards will lead to the same result as another rectangle or square being lengthened). The 2x2 board in the middle and the 4x4 board are the new options corresponding to the rectangles... This approach isn't leading to an elegant proof, so I'll have to reevaluate.

  12. KingGeorge
    • one year ago
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    I think your method could lead to a relatively good proof, but it's definitely not what I ended up with. Let me know if you want a hint about my method.

  13. Numb3r1
    • one year ago
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    \[\sum_{1}^{n}n!+(n-1)!+....+1!\] seems to work.

  14. KingGeorge
    • one year ago
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    I'm not sure I understand what that's supposed to be doing.

  15. KingGeorge
    • one year ago
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    @Numb3r1 I apologize, I misread my previous work. \(2^{n-1}\) is not the correct answer. It should be \[\large 2^{(n-1)^2}.\] This is because we consider a rotation of the board to potentially be a different solution.

  16. Numb3r1
    • one year ago
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    I had already ignored rotations, and that means that I was correct as well! Actually, are you sure it wasn't \[2^{2(n-1)}\]? I don't see how a 7x7 board would have six times as many options if you can rotate it.

  17. KingGeorge
    • one year ago
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    I am absolutely positive this is the correct solution. Consider the smaller \((n-1)\times(n-1)\) chessboard that is a subboard of the larger one, and start putting pawns on that. What can you say about the placement of pawns on the rest of the board?

  18. Numb3r1
    • one year ago
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    My point is that it could be \[(2^{n-1})^{2}=2^{2(n-1)} but 2^{(n-1)^{2}}wont work.\]

  19. Numb3r1
    • one year ago
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    To clarify, the first one will, the second will not.

  20. KingGeorge
    • one year ago
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    A 7x7 board will have \(2^{36}\) possible solutions, not \(2^{12}\) or \(2^6\). Again, I apologize for leading you down the wrong path for a while :(

  21. Numb3r1
    • one year ago
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    For a 3x3 board, \[\left[\begin{matrix} 1& 0& 1 \\ 0&0 &0\\ 1 &0 &1 \end{matrix}\right],\left[\begin{matrix} 1& 1& 0 \\ 1& 1&0\\ 0 &0 & 0\end{matrix}\right]\times4,\left[\begin{matrix} 1&0 & 1\\ 1& 0&1\\ 0&0 &0 \end{matrix}\right]\times4,\left[\begin{matrix} 0& 0&0 \\ 0& 0&0\\ 0 &0 & 0\end{matrix}\right], and what else?\]

  22. KingGeorge
    • one year ago
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    \[\begin{bmatrix}0&1&1\\1&1&0\\ 1&0&1\end{bmatrix}\]for example.

  23. Numb3r1
    • one year ago
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    Ah, that times four, and then there are two more? I'll have to try another approach to the problem.

  24. KingGeorge
    • one year ago
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    the last two I believe are given by \[\begin{bmatrix}1&1&0\\1&0&1\\0&1&1\end{bmatrix}\]and a 90 degree rotation.

  25. Numb3r1
    • one year ago
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    That'll do it.

  26. KingGeorge
    • one year ago
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    Well, I've got to sleep now, but I would strongly consider looking at the smaller board of dimensions \((n-1)\times(n-1)\) (which has \((n-1)^2\) boxes), placing some pawns on that, and then looking at where you can place pawns on the one row/column not included in this smaller board.

  27. mathslover
    • one year ago
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    Wait for 5 minutes more kg, I am trying.

  28. KingGeorge
    • one year ago
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    I should correct what I said near the top. Orientation does actually matter. I just had a slight misunderstanding when I read vf321's comment.

  29. mathslover
    • one year ago
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    What should be the correct answer @KingGeorge ?

  30. KingGeorge
    • one year ago
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    \[\large 2^{(n-1)^2}\](The \(n-1\) is squared)

  31. mathslover
    • one year ago
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    But if n = 1, then it is 1, but it can not exist ?

  32. KingGeorge
    • one year ago
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    If \(n=1\), then \(2^{(n-1)^2}=2^0=1\). The only way would be to have the empty board.

  33. mathslover
    • one year ago
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    lol, empty board also allowed :

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