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Loser66

  • one year ago

The following is an encryption algorithm. Denote (abc) at the permutation \(a\mapsto b\), \(b\mapsto c\) and \(c\mapsto a\) and fixing all the other elements. Denote A as the following matrix \[\left(\begin{matrix}1&4&7\\2&5&8\\3&6&9\end{matrix}\right)\] In each step the middle row of the matrix is used to define a permutation acting on a single digit. After each step, the first column is rotated downwards and the middle column upwards, resulting a new matrix to be used for the next step of encryption \[\left(\begin{matrix}3&5&7\\1&6&8\\2&4&9\end{matrix}\right)\] continue on comment

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  1. Loser66
    • one year ago
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    That is , given 2,5,8, the output is 5,5,3 Question: What was the input if the output is 5,4,4,5,7,8,5 Please, help

  2. Loser66
    • one year ago
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    Please, leave guidance here. will be back . Thanks in advance.

  3. Loser66
    • one year ago
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    Options A) 5,4,4,5,7,8,5 B) 8,4,8,8,7,3,8 C) 2,4,3,2,7,4,2 D) 8,4,4,5,7,1,8 E) Such input does not exists My choice E

  4. thomas5267
    • one year ago
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    If the matrix is \[ \left(\begin{matrix}1&4&7\\2&5&8\\3&6&9\end{matrix}\right) \] and the input is 8, the output is 2 and the new matrix is \[ \left(\begin{matrix}3&5&7\\1&6&8\\2&4&9\end{matrix}\right) \] Is that correct?

  5. Loser66
    • one year ago
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    Yes, My logic: From given information, if output is 5, input is 2 hence only C is a possible answer. However, after using the second matrix to find the input of 4, it must be 2, but the second digit of C is 4,that gives us C is not the correct answer --> E

  6. Loser66
    • one year ago
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    But I don't know whether it is right or wrong since I don't have the answer sheet.

  7. Loser66
    • one year ago
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    The net is on and off here. :(

  8. ganeshie8
    • one year ago
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    It seems you get A back after every 3 steps

  9. ganeshie8
    • one year ago
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    so it must be the case that \(5xy5mn5\) must decrypt to \(2x'y'2m'n'2\) look at the options!

  10. ganeshie8
    • one year ago
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    Option C is indeed correct, notice that 4 is in the last row in matrix2, so it is fixed.

  11. thomas5267
    • one year ago
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    I spent 2 hours to brute force a solution. Mathematica is very hard to use I must say. The permutations are: \[ \begin{align*} P_1&=(2,5,8)\\ P_2&=(1,6,8)\\ P_3&=(3,4,8)\\ P_4&=(2,5,8)\\ P_5&=(1,6,8)\\ P_6&=(3,4,8)\\ P_7&=(2,5,8)\\ \end{align*} \] The inverse permutations are: \[ \begin{align*} P_1^{-1}&=(8,5,2)\\ P_2^{-1}&=(8,6,1)\\ P_3^{-1}&=(8,4,3)\\ P_4^{-1}&=(8,5,2)\\ P_5^{-1}&=(8,6,1)\\ P_6^{-1}&=(8,4,3)\\ P_7^{-1}&=(8,5,2) \end{align*} \] Apply the inverse permutations to the output. The answer is C.

  12. Loser66
    • one year ago
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    Thanks for the work. :)

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