(New page: == Part 1 == Since Bob does not know a specific output given a specific input, he must use the inverse of the secret matrix. == Part 2 == I believe that Eve should be able to decrypt th...) |
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== Part 2 == | == Part 2 == | ||
| − | I believe that Eve | + | I believe that Eve will not be able to decrypt the message because I think that the system is linear but not time invariant. The system is linear because it behaves like the picture in the link below: (the system is the secret matrix, a and b are scalars that are multiplied) |
| + | |||
| + | [[Media:301_schematic_ECE301Fall2008mboutin.pdf]] | ||
| + | |||
| + | The system is not time invariant because if the input is shifted before going through the system, the output is different than if the shift occurs after going through the system. The following is an example of how the system is not time invariant: | ||
| + | |||
| + | Shift after going through the system | ||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 & 3 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | 1/2 & 0 & 2 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1/3 & 0 & 1/3 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | 1 & 22 & 4 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
| + | |||
| + | |||
| + | Shift before going through the system | ||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 1 & 22 & 2 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | 1/2 & 0 & 2 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1/3 & 0 & 1/3 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | 1.167 & 22 & 2.667 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
| + | |||
| + | As you can see, a shift by one before going through the matrix yields a different value than a shift by one after going through the matrix. | ||
== Part 3 == | == Part 3 == | ||
| + | |||
| + | Using matlab, the secret matrix can be found in the following way: | ||
| + | |||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 0 & 0 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 0 & 0 & 3 | ||
| + | \end{bmatrix} | ||
| + | \div | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | S11 & S12 & S13 \\ | ||
| + | S21 & S22 & S23 \\ | ||
| + | S31 & S32 & S33 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
| + | |||
| + | The result is the secret matrix | ||
| + | <br> | ||
| + | :<math> | ||
| + | |||
| + | \begin{bmatrix} | ||
| + | 1/2 & 0 & 2 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1/3 & 0 & 1/3 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | |||
| + | <br> | ||
| + | |||
| + | Finally, applying the secret matrix to the decrypted message: | ||
| + | |||
| + | |||
| + | :<br> | ||
| + | |||
| + | :<math> | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 & 3 | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | 1/2 & 0 & 2 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1/3 & 0 & 1/3 | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | 2 & 23 & 5 | ||
| + | \end{bmatrix} | ||
| + | |||
| + | </math> | ||
| + | <br> | ||
| + | |||
| + | Which corresponds with the letters B W E | ||
Latest revision as of 10:08, 19 September 2008
Part 1
Since Bob does not know a specific output given a specific input, he must use the inverse of the secret matrix.
Part 2
I believe that Eve will not be able to decrypt the message because I think that the system is linear but not time invariant. The system is linear because it behaves like the picture in the link below: (the system is the secret matrix, a and b are scalars that are multiplied)
Media:301_schematic_ECE301Fall2008mboutin.pdf
The system is not time invariant because if the input is shifted before going through the system, the output is different than if the shift occurs after going through the system. The following is an example of how the system is not time invariant:
Shift after going through the system
- $ \begin{bmatrix} 2 & 23 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{bmatrix} = \begin{bmatrix} 1 & 22 & 4 \end{bmatrix} $
Shift before going through the system
- $ \begin{bmatrix} 1 & 22 & 2 \end{bmatrix} \cdot \begin{bmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{bmatrix} = \begin{bmatrix} 1.167 & 22 & 2.667 \end{bmatrix} $
As you can see, a shift by one before going through the matrix yields a different value than a shift by one after going through the matrix.
Part 3
Using matlab, the secret matrix can be found in the following way:
- $ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \div \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} S11 & S12 & S13 \\ S21 & S22 & S23 \\ S31 & S32 & S33 \end{bmatrix} $
The result is the secret matrix
- $ \begin{bmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{bmatrix} $
Finally, applying the secret matrix to the decrypted message:
- $ \begin{bmatrix} 2 & 23 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1/2 & 0 & 2 \\ 0 & 1 & 0 \\ 1/3 & 0 & 1/3 \end{bmatrix} = \begin{bmatrix} 2 & 23 & 5 \end{bmatrix} $
Which corresponds with the letters B W E
