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Math Help - Distinct Square Questions

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    Distinct Square Questions

    Hello All, I have made a few threads already about Magic Squares and Regular Squares, but this thread is about Distinct Squares. As I had explained in my previous threads, I was reading through the chapter in my book about squares and had a few questions that I'm hoping you all might be able to help me with. Unlike previous threads, I have a discrete example from my book that I could use some help explaining.

    Here are two matrices, X and Y.

    X = \begin{pmatrix}<br />
15 & 2 & 1 & 12\\ <br />
4 & 9 & 10 & 7\\ <br />
8 & 5 & 6 & 11\\<br />
3 & 14 & 13 & 0<br />
\end{pmatrix} and Y = \begin{pmatrix}<br />
14 & 8 & 5 & 3\\ <br />
9 & 15 & 2 & 4\\ <br />
7 & 1 & 12 & 10\\<br />
0 & 6 & 11 & 13<br />
\end{pmatrix}

    How can I prove that X cannot be transformed into Y by way of exchanging rows and columns?

    Any help is appreciated!
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  2. #2
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    Sorry, my screenname was apparently illegal. Now that it is changed, please feel free to respond as normal.
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    ...
    Last edited by Samson; June 2nd 2010 at 05:00 AM. Reason: Double Posted by accident
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  4. #4
    Senior Member roninpro's Avatar
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    By "exchanging rows and columns", do you mean swapping rows with rows and columns with columns?
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    Quote Originally Posted by roninpro View Post
    By "exchanging rows and columns", do you mean swapping rows with rows and columns with columns?
    Yes, I mean swappring rows with rows, or columns with columns.
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Samson View Post
    Yes, I mean swappring rows with rows, or columns with columns.
    Just to clarify further, these would be considered legal moves, right?

    Step 0: Original

    X_0 = \begin{pmatrix}<br />
15 & 2 & 1 & 12\\ <br />
4 & 9 & 10 & 7\\ <br />
8 & 5 & 6 & 11\\<br />
3 & 14 & 13 & 0<br />
\end{pmatrix}

    Step 1: Switch rows 1 and 2

    X_1 = \begin{pmatrix}<br />
 4 & 9 & 10 & 7\\ <br />
 15 & 2 & 1 & 12\\ <br />
 8 & 5 & 6 & 11\\<br />
 3 & 14 & 13 & 0<br />
 \end{pmatrix}

    Step 2: Switch columns 1 and 2

    X_2 = \begin{pmatrix}<br />
9 & 4 & 10 & 7\\ <br />
  2 &15 & 1 & 12\\ <br />
  5 & 8 & 6 & 11\\<br />
  14 & 3 & 13 & 0<br />
  \end{pmatrix}

    If so, then this reminds me a lot of how on Rubik's cubes, not all configurations are possible (easily provable by just considering the center squares). Not sure about this one though.
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  7. #7
    Senior Member roninpro's Avatar
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    In that case, I would like to draw your attention to an important property: no matter how you swap rows and columns, you preserve the numbers in a particular (labeled) row or column. Let me give you an example.

    Consider the matrix

    <br />
A=\begin{pmatrix}<br />
15 & 2 & 1 & 12\\<br />
4 & 9 & 10 & 7\\<br />
8 & 5 & 6 & 11\\<br />
3 & 14 & 13 & 0<br />
\end{pmatrix}<br />

    The first column contains the numbers 15, 4, 8, 3. If I swap a few rows and columns, I might receive

    <br />
A'=\begin{pmatrix}<br />
9 & 4 & 10 & 7\\<br />
14 & 3 & 13 & 0\\<br />
5 & 8 & 6 & 11\\<br />
2 & 15 & 1 & 12<br />
\end{pmatrix}<br />

    Observe that I still have a column that contains 15, 4, 8, 3 - they are just permuted.

    Can you see how to use this property to show that your matrices cannot be transformed into one another?
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    Quote Originally Posted by roninpro View Post
    In that case, I would like to draw your attention to an important property: no matter how you swap rows and columns, you preserve the numbers in a particular (labeled) row or column. Let me give you an example.

    Consider the matrix

    <br />
A=\begin{pmatrix}<br />
15 & 2 & 1 & 12\\<br />
4 & 9 & 10 & 7\\<br />
8 & 5 & 6 & 11\\<br />
3 & 14 & 13 & 0<br />
\end{pmatrix}<br />

    The first column contains the numbers 15, 4, 8, 3. If I swap a few rows and columns, I might receive

    <br />
A'=\begin{pmatrix}<br />
9 & 4 & 10 & 7\\<br />
14 & 3 & 13 & 0\\<br />
5 & 8 & 6 & 11\\<br />
2 & 15 & 1 & 12<br />
\end{pmatrix}<br />

    Observe that I still have a column that contains 15, 4, 8, 3 - they are just permuted.

    Can you see how to use this property to show that your matrices cannot be transformed into one another?
    So do you think the following proof is substantial:

    By swapping out rows with rows and columns with columns, the final rows and columns will still contain the same elements within themselves, however they may be permutations of the original rows and columns and hence not appear in the same order. Because of this, it is impossible to produce the matrix shown here on the right because its rows and columns differe in content, not just in order.

    Is that correct/accurate?
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  9. #9
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Samson View Post
    So do you think the following proof is substantial:

    By swapping out rows with rows and columns with columns, the final rows and columns will still contain the same elements within themselves, however they may be permutations of the original rows and columns and hence not appear in the same order. Because of this, it is impossible to produce the matrix shown here on the right because its rows and columns differe in content, not just in order.

    Is that correct/accurate?
    Your wording is vague and the reasoning is not clear.

    Try this:

    For our purposes, the word "permutation" includes the identity permutation (which leaves the order of elements unchanged).

    Swapping a row with another row results in a matrix whose every row appears in the original matrix, and whose every column is a permutation of a column in the original matrix. Likewise with swapping columns. Therefore, any row of a transformed matrix will be a permutation of a row in the original matrix, and likewise with columns. (This is because a permutation of a permutation is another permutation.) The first row in Y is not a permutation of any row in X. Therefore, it is impossible to reach Y from X through the allowed transformations.
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  10. #10
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    Quote Originally Posted by undefined View Post
    Your wording is vague and the reasoning is not clear.

    Try this:

    For our purposes, the word "permutation" includes the identity permutation (which leaves the order of elements unchanged).

    Swapping a row with another row results in a matrix whose every row appears in the original matrix, and whose every column is a permutation of a column in the original matrix. Likewise with swapping columns. Therefore, any row of a transformed matrix will be a permutation of a row in the original matrix, and likewise with columns. (This is because a permutation of a permutation is another permutation.) The first row in Y is not a permutation of any row in X. Therefore, it is impossible to reach Y from X through the allowed transformations.
    That does make sense. Thank you! I was using the word wrong then, but I get what you're saying now. Thanks once again!
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