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| Line 37: | Line 37: | ||
4G+I=3 <br> | 4G+I=3 <br> | ||
and so | and so | ||
| − | A=-\frac{2}{3} <br> | + | A=<math>-\frac{2}{3}</math> <br> |
| − | C=\frac{8}{3} <br> | + | C=<math>\frac{8}{3}</math> <br> |
D=0 <br> | D=0 <br> | ||
F=0 <br> | F=0 <br> | ||
| Line 48: | Line 48: | ||
:<math> | :<math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | -\frac{2}{3} & | + | -\frac{2}{3} & 0 & \frac{8}{3} \\ |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | + | 1 & 0 & -1 | |
\end{bmatrix} | \end{bmatrix} | ||
</math> | </math> | ||
| Line 57: | Line 57: | ||
:<math> | :<math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | \frac{1}{2} & | + | \frac{1}{2} & 0 & \frac{4}{3} \\ |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | \frac{1}{ | + | \frac{1}{2} & 0 & \frac{1}{3} |
\end{bmatrix} | \end{bmatrix} | ||
</math><br> | </math><br> | ||
| − | So(2,23,3) --> ( | + | 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
