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| − | == | + | == Can Bob decrypt the message? == |
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| + | Yes, Bob can in fact decode the hidden message if he is given an identity matrix. | ||
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| + | The way he can do this is that he first needs to take the inverse of the matrix. From this he can take the first row and its three column entries as the first matrix to be multiplied by the inverse of the matrix, this will give him the first entry in the Alphabet for the secret message. This process is to be repeated using the next row of the same amount of columns until the message has been decoded. | ||
| + | |||
| + | It has also been told that a zero counts as a space. | ||
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| + | |||
| + | == Can Eve decode the secret message without finding the inverse of the matrix? == | ||
| + | |||
| + | Yes, Eve can find out what the secret message is saying, but she will have to use a different way to find it. | ||
| + | |||
| + | She knows the original message that is the output when it is passed through the encoder. | ||
| + | |||
| + | The answer ends up being, referencing Max Paganini's source to input a matrix, | ||
| + | |||
| + | <math>\begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} </math> | ||
Revision as of 14:04, 18 September 2008
Can Bob decrypt the message?
Yes, Bob can in fact decode the hidden message if he is given an identity matrix.
The way he can do this is that he first needs to take the inverse of the matrix. From this he can take the first row and its three column entries as the first matrix to be multiplied by the inverse of the matrix, this will give him the first entry in the Alphabet for the secret message. This process is to be repeated using the next row of the same amount of columns until the message has been decoded.
It has also been told that a zero counts as a space.
Can Eve decode the secret message without finding the inverse of the matrix?
Yes, Eve can find out what the secret message is saying, but she will have to use a different way to find it.
She knows the original message that is the output when it is passed through the encoder.
The answer ends up being, referencing Max Paganini's source to input a matrix,
$ \begin{matrix} - \frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 &0 & -1 \end{matrix} $
