| Line 58: | Line 58: | ||
:<math> | :<math> | ||
\begin{bmatrix} | \begin{bmatrix} | ||
| − | + | \frac{1}{2} & -1 & 2 \\ | |
0 & 1 & 0 \\ | 0 & 1 & 0 \\ | ||
| − | \frac{ | + | \frac{1}{3} & 0 & \frac{1}{3} |
\end{bmatrix} | \end{bmatrix} | ||
</math><br> | </math><br> | ||
So(2,23,5) --> BWE | So(2,23,5) --> BWE | ||
Revision as of 19:18, 17 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} 1 & 0 & 4 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} A & B & C \\ D & E & F \\ G & H & I \end{bmatrix} = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{bmatrix} $
Thus we have
A+4G=2
B+4H=0
C+4I=0
D=0
E=1
F=0
A+G=0
B+H=0
C+I=3
and so
D=0
E=1
F=0
A=B=-2/3
G=H=2/3
I=-1
C=4
i.e.
- $ \begin{bmatrix} -\frac{2}{3} & -\frac{2}{3} & 4 \\ 0 & 1 & 0 \\ \frac{2}{3} & \frac{2}{3} & -1 \end{bmatrix} $
3. The inverse matrix is
- $ \begin{bmatrix} \frac{1}{2} & -1 & 2 \\ 0 & 1 & 0 \\ \frac{1}{3} & 0 & \frac{1}{3} \end{bmatrix} $
So(2,23,5) --> BWE
