| 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 | ||
D=0 <br> | D=0 <br> | ||
Revision as of 14:24, 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
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(2,23,3) --> (1,23,1) --> AWE
