(New page: 1. Bob needs to calculate the inverse of the secret matrix, and multiply it by the code given by Alice to get a vector. Then replaces each three entries by its corresponding letter in the ...) |
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:<math> | :<math> | ||
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\begin{bmatrix} | \begin{bmatrix} | ||
A & B & C \\ | A & B & C \\ | ||
D & E & F \\ | D & E & F \\ | ||
G & H & I | G & H & I | ||
| + | \end{bmatrix} | ||
| + | \cdot | ||
| + | \begin{bmatrix} | ||
| + | 1 & 0 & 4 \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | 1 & 0 & 1 | ||
\end{bmatrix} | \end{bmatrix} | ||
= | = | ||
| Line 27: | Line 27: | ||
<br> | <br> | ||
Thus we have | Thus we have | ||
| − | A+ | + | A+C=2 <br> |
| − | B | + | B=0 <br> |
| − | + | 4A+C=0 <br> | |
| − | D=0 <br> | + | D+F=0 <br> |
E=1 <br> | E=1 <br> | ||
| − | F=0 <br> | + | 4D+F=0 <br> |
| − | + | G+I=0 <br> | |
| − | + | H=0 <br> | |
| − | + | 4G+I=3 <br> | |
and so | and so | ||
| + | A=<math>-\frac{2}{3}</math> <br> | ||
| + | C=<math>\frac{8}{3}</math> <br> | ||
D=0 <br> | D=0 <br> | ||
| − | |||
F=0 <br> | F=0 <br> | ||
| − | + | G=1<br> | |
| − | G= | + | |
I=-1 <br> | I=-1 <br> | ||
| − | |||
i.e. | i.e. | ||
| Line 49: | Line 48: | ||
:<math> | :<math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | -\frac{2}{3} & 0 & \frac{ | + | -\frac{2}{3} & 0 & \frac{8}{3} \\ |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | + | 1 & 0 & -1 | |
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| − | + | 3. The inverse matrix is <br> | |
| − | + | :<math> | |
| + | \begin{bmatrix} | ||
| + | \frac{1}{2} & 0 & \frac{4}{3} \\ | ||
| + | 0 & 1 & 0 \\ | ||
| + | \frac{1}{2} & 0 & \frac{1}{3} | ||
| + | \end{bmatrix} | ||
| + | </math><br> | ||
| + | So(2,23,3) --> (5,23,2) --> EWB | ||
Latest revision as of 14:34, 18 September 2008
1. Bob needs to calculate the inverse of the secret matrix, and multiply it by the code given by Alice to get a vector. Then replaces each three entries by its corresponding letter in the alphabet.
2.Eve can get the secret matrix through calculation.
- $ \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \end{bmatrix} \cdot \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} $
Thus we have
A+C=2
B=0
4A+C=0
D+F=0
E=1
4D+F=0
G+I=0
H=0
4G+I=3
and so
A=$ -\frac{2}{3} $
C=$ \frac{8}{3} $
D=0
F=0
G=1
I=-1
i.e.
- $ \begin{bmatrix} -\frac{2}{3} & 0 & \frac{8}{3} \\ 0 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix} $
3. The inverse matrix is
- $ \begin{bmatrix} \frac{1}{2} & 0 & \frac{4}{3} \\ 0 & 1 & 0 \\ \frac{1}{2} & 0 & \frac{1}{3} \end{bmatrix} $
So(2,23,3) --> (5,23,2) --> EWB
