(New page: ==How can Bob decrypt the message?== Bob can take the encrypted message vector, split it into sub-vectors of three elements each, multiply each sub-vector by the inverse of the encryption...) |
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Bob can take the encrypted message vector, split it into sub-vectors of three elements each, multiply each sub-vector by the inverse of the encryption matrix, then reassemble the resulting vectors into the original message. | Bob can take the encrypted message vector, split it into sub-vectors of three elements each, multiply each sub-vector by the inverse of the encryption matrix, then reassemble the resulting vectors into the original message. | ||
| − | == | + | ==Can Eve decrypt the message without finding the inverse matrix?== |
| + | |||
| + | |||
==What is the decrypted message corresponding to [ 2 23 3 ]?== | ==What is the decrypted message corresponding to [ 2 23 3 ]?== | ||
| + | |||
| + | By taking our three sets of inputs and outputs, we can form a system of three equations in three unknowns and solve for the encryption matrix. Taking the inverse gives | ||
| + | |||
| + | <math>A=\left[ \begin{array}{ccc} | ||
| + | -\frac{2}{3} & 0 & \frac{2}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 4 & 0 & -1 \end{array} \right] \rightarrow A^{-1}= | ||
| + | \left[ \begin{array}{ccc} | ||
| + | \ \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \end{array} \right] | ||
| + | </math> | ||
| + | |||
| + | Multiplying the inverse matrix by the three-element vector given in the homework statement gives | ||
| + | |||
| + | <math>\left[ \begin{array}{ccc} | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 2 & 0 & \frac{1}{3} \end{array} \right] | ||
| + | \left[ \begin{array}{c} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 3\end{array} \right] = | ||
| + | \left[ \begin{array}{c} | ||
| + | 2\\ | ||
| + | 23\\ | ||
| + | 5 \end{array} \right]</math>, | ||
| + | |||
| + | which is "BWE" using our letter-number mapping scheme. | ||
Revision as of 15:40, 16 September 2008
How can Bob decrypt the message?
Bob can take the encrypted message vector, split it into sub-vectors of three elements each, multiply each sub-vector by the inverse of the encryption matrix, then reassemble the resulting vectors into the original message.
Can Eve decrypt the message without finding the inverse matrix?
What is the decrypted message corresponding to [ 2 23 3 ]?
By taking our three sets of inputs and outputs, we can form a system of three equations in three unknowns and solve for the encryption matrix. Taking the inverse gives
$ A=\left[ \begin{array}{ccc} -\frac{2}{3} & 0 & \frac{2}{3} \\ 0 & 1 & 0 \\ 4 & 0 & -1 \end{array} \right] \rightarrow A^{-1}= \left[ \begin{array}{ccc} \ \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{array} \right] $
Multiplying the inverse matrix by the three-element vector given in the homework statement gives
$ \left[ \begin{array}{ccc} \frac{1}{2} & 0 & \frac{1}{3} \\ 0 & 1 & 0 \\ 2 & 0 & \frac{1}{3} \end{array} \right] \left[ \begin{array}{c} 2\\ 23\\ 3\end{array} \right] = \left[ \begin{array}{c} 2\\ 23\\ 5 \end{array} \right] $,
which is "BWE" using our letter-number mapping scheme.
