Question

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

Answer 1

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

Answer 2

Please, leave guidance here. will be back . Thanks in advance.

Answer 3

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

Answer 4

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?

Answer 5

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

Answer 6

But I don't know whether it is right or wrong since I don't have the answer sheet.

Answer 7

The net is on and off here. :(

Answer 8

It seems you get A back after every 3 steps

Answer 9

so it must be the case that
\(5xy5mn5\) must decrypt to \(2x'y'2m'n'2\)

look at the options!

Answer 10

Option C is indeed correct, notice that 4 is in the last row in matrix2, so it is fixed.

Answer 11

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.

Answer 12

Thanks for the work. :)