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| + | This was an interesting question Professor Boutin | ||
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==Part 1== | ==Part 1== | ||
How can Bob decrypt the message? <br><br> | How can Bob decrypt the message? <br><br> | ||
Bob can decrypt the message by multiplying the inverse of the 3-by-3 secret matrix with the coded message. | Bob can decrypt the message by multiplying the inverse of the 3-by-3 secret matrix with the coded message. | ||
<br> | <br> | ||
| + | |||
==Part 2== | ==Part 2== | ||
Can Eve decrypt the message without finding the inverse of the secret matrix?<br> | Can Eve decrypt the message without finding the inverse of the secret matrix?<br> | ||
| − | The asnwer is " | + | The asnwer is "yes." She can find the inverse of the secret matrix from the intercepted message. Or she can apply the intercepted information linearly to a wanted set of data. <br> |
The coded and decrypted message can be arranged in a 3-by-3 matrix form. <br><br> | The coded and decrypted message can be arranged in a 3-by-3 matrix form. <br><br> | ||
| + | ---------------------------------------------------------------------------------------------------------------------------- | ||
:<math> Coded = \begin{bmatrix} | :<math> Coded = \begin{bmatrix} | ||
1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ | 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ | ||
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| − | :<math> A = | + | :<math> A = \begin{bmatrix} |
1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ | 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ | ||
| − | \end{bmatrix}^{-1} | + | \end{bmatrix} * \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\\end{bmatrix}^{-1} = \begin{bmatrix} 1/2 & 0 & 4/3 \\ 0 & 1 & 0 \\ 1/2 & 0 & 1/3 \\\end{bmatrix} |
</math> | </math> | ||
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| + | <br> | ||
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| + | ==Part3== | ||
| + | Method 1.<br> | ||
| + | Previously, we observe that (1,0,4,0,1,0,1,0,1) yields (2,0,0,0,1,0,0,0,3).<br><br> | ||
| + | (1, 0, 4) -----> Matrix Encryption -----> (2, 0, 0) ......................... (1)<br> | ||
| + | (0, 1, 0) -----> Matrix Encryption -----> (0, 1, 0) ......................... (2)<br> | ||
| + | (1, 0, 1) -----> Matrix Encryption -----> (0, 0, 3) ......................... (3)<br><br> | ||
| + | |||
| + | (1) + (2) X 23 + (3) would yield the desired output(provided that the system is linear) <br> | ||
| + | <br> | ||
| + | (1) + (2) X 23 + (3) = (1, 0, 4) + (0, 23, 0) + (1, 0, 1) = (2, 23, 5) | ||
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| + | |||
| + | <br><br> | ||
| + | Method 2.<br> | ||
| + | Using the inverse matrix.<br> | ||
| + | |||
| + | Applying the secret matrix A to the message (2, 23, 3), we get,<br><br> | ||
| + | |||
| + | <math> Message\ = \begin{bmatrix} 2 & 23 & 3 \\\end{bmatrix} * \begin{bmatrix} 1/2 & 0 & 4/3 \\ 0 & 1 & 0 \\ 1/2 & 0 & 1/3 \\\end{bmatrix} = </math> | ||
Latest revision as of 17:00, 19 September 2008
This was an interesting question Professor Boutin
Part 1
How can Bob decrypt the message?
Bob can decrypt the message by multiplying the inverse of the 3-by-3 secret matrix with the coded message.
Part 2
Can Eve decrypt the message without finding the inverse of the secret matrix?
The asnwer is "yes." She can find the inverse of the secret matrix from the intercepted message. Or she can apply the intercepted information linearly to a wanted set of data.
The coded and decrypted message can be arranged in a 3-by-3 matrix form.
- $ Coded = \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} \ $
- $ Decrypted = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} \ $
Thus
$ Coded * A\ = Decrypted $
Or (A is the Secret Matrix)
$ \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix}\ * A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\ \end{bmatrix} $
- $ A = \begin{bmatrix} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ \end{bmatrix} * \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \\\end{bmatrix}^{-1} = \begin{bmatrix} 1/2 & 0 & 4/3 \\ 0 & 1 & 0 \\ 1/2 & 0 & 1/3 \\\end{bmatrix} $
Part3
Method 1.
Previously, we observe that (1,0,4,0,1,0,1,0,1) yields (2,0,0,0,1,0,0,0,3).
(1, 0, 4) -----> Matrix Encryption -----> (2, 0, 0) ......................... (1)
(0, 1, 0) -----> Matrix Encryption -----> (0, 1, 0) ......................... (2)
(1, 0, 1) -----> Matrix Encryption -----> (0, 0, 3) ......................... (3)
(1) + (2) X 23 + (3) would yield the desired output(provided that the system is linear)
(1) + (2) X 23 + (3) = (1, 0, 4) + (0, 23, 0) + (1, 0, 1) = (2, 23, 5)
Method 2.
Using the inverse matrix.
Applying the secret matrix A to the message (2, 23, 3), we get,
$ Message\ = \begin{bmatrix} 2 & 23 & 3 \\\end{bmatrix} * \begin{bmatrix} 1/2 & 0 & 4/3 \\ 0 & 1 & 0 \\ 1/2 & 0 & 1/3 \\\end{bmatrix} = $
