(New page: ==Homework3 Part C== <br> <br> == How can Bob decode the message? == <br> Bob can decode the message by breaking the message into segments of three, <br> and then multiplying each 3 bit ...) |
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<br>No, Eve would have to find the inverse of the matrix in order to decode the messages. | <br>No, Eve would have to find the inverse of the matrix in order to decode the messages. | ||
<br> However, She does have enough information to determine the encryption vector. | <br> However, She does have enough information to determine the encryption vector. | ||
| − | + | <br> | |
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | 2 & 0 & 0 \\ | |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | + | 0 & 0 & 3 | |
| − | + | \end{bmatrix} | |
\cdot | \cdot | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | 1 & 0 & 1 \\ | |
| − | + | 0 & 1 & 0 \\ | |
| − | + | 4 & 0 & 1 | |
\end{bmatrix} | \end{bmatrix} | ||
| − | + | </math> <sup>-1</sup> | |
| + | |||
| + | |||
| + | Matlab will tell us that the encryption matrix, is:<br> | ||
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | 2 & 0 & | + | -\frac{2}{3} & 0 & \frac{2}{3} \\ |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | + | 4 & 0 & -1 \\ | |
\end{bmatrix} | \end{bmatrix} | ||
| − | |||
</math> | </math> | ||
| − | |||
| − | |||
| − | :<math> | + | |
| + | Now, find inverse of the encryption matrix and multiply it by the sample code in order to find the secret code:<br> | ||
| + | |||
| + | The inverse of the "secret" matrix:<br> | ||
| + | <math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | \frac{1}{2} & 0 & \frac{1}{3} \\ | |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | + | 2 & 0 & \frac{1}{3} \\ | |
\end{bmatrix} | \end{bmatrix} | ||
| + | </math> | ||
| + | <br> <br> <br> <br> | ||
| + | <math> | ||
| + | \begin{bmatrix} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \\ | ||
| + | \end{bmatrix} | ||
| + | * | ||
| + | \begin{bmatrix} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 3 \\ | ||
| + | \end{bmatrix} | ||
| + | = | ||
| + | \begin{bmatrix} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 5\\ | ||
| + | \end{bmatrix} | ||
| + | |||
</math> | </math> | ||
| − | + | <br> | |
| − | + | The super secret message of great importance is... | |
| + | <br>BWE | ||
| + | <br>How exciting... | ||
Latest revision as of 07:46, 18 September 2008
Homework3 Part C
How can Bob decode the message?
Bob can decode the message by breaking the message into segments of three,
and then multiplying each 3 bit segment by the inverse of the encryption vector.
Can Eve decrypt it without finding the inverse of the encryption matrix?
No, Eve would have to find the inverse of the matrix in order to decode the messages.
However, She does have enough information to determine the encryption vector.
$ \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 4 & 0 & 1 \end{bmatrix} $ -1
Matlab will tell us that the encryption matrix, is:
$ \begin{bmatrix} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \\ \end{bmatrix} $
Now, find inverse of the encryption matrix and multiply it by the sample code in order to find the secret code:
The inverse of the "secret" matrix:
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \\ \end{bmatrix} $
$ \begin{bmatrix} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \\ \end{bmatrix} * \begin{bmatrix} 2\\ 23\\ 3 \\ \end{bmatrix} = \begin{bmatrix} 2\\ 23\\ 5\\ \end{bmatrix} $
The super secret message of great importance is...
BWE
How exciting...
