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Example 3:

Suppose that you have received the following message from a friend,

L U P O Z M A E A E G I U A B J

using standard conversion table, it becomes

12 21 16 15 0 13 1 5 1 5 7 9 21 1 2 10,

which also the same as (due to mod 26),

12 47 16 67 52 65 27 83 79 135 33 113 47 53 80 140.

Unfortunatly, you do not recall the matrix $ A $ or $A^{-1}$. But you know that the plaintext of the fifth through eighth letters is GOOD, is it possible that you figure out what is the orginal message?

The answer is YES. Knowing the word GOOD, you can find the numerical equivalent of those letters. That is 7 15 for GO and 15 4 for OD. You also know the numerical equivalent of the fifth to eight letters in the encoded message are

52 65 for ZM and 27 83 for AE.

Therefore you can form the matrix $P$ and $Q$.

$p_1 = \left[ \begin{array}{r}
7\\
15\\
\end{array}\right] \leftrightarrow c_1 = \left[ \begin{array}{r}
52\\
65\\
\end{array}\right]$ $p_2 = \left[ \begin{array}{r}
15\\
4\\
\end{array}\right] \leftrightarrow c_1 = \left[ \begin{array}{r}
27\\
83\\
\end{array}\right]$



You can construct

\begin{displaymath}Q ^T =\left[ \begin{array}{r}
{c_1}^{T}\\
{c_2}^{T}\\
\end{...
...= \left[ \begin{array}{rr}
52&65\\
27&83\\
\end{array}\right]\end{displaymath}

and


\begin{displaymath}P^T = \left[ \begin{array}{r}
{p_1}^{T}\\
{p_2}^{T}\\
\end{...
...] = \left[ \begin{array}{rr}
7&15\\
15&4\\
\end{array}\right]\end{displaymath}

Form the matrix $ [Q^T\vert P^T] $ and use Gaussian Elimination to reduce $ Q^T $ to the identity matrix, the matrix $ [Q^T\vert P^T] $ will change to $[I\vert{(A^{-1})}^{T}]$.


\begin{displaymath}\left[ \begin{array}{rrrr}
52&65&7&15\\
27&83&15&4\\
\end{array}\right]\end{displaymath}

change to


\begin{displaymath}\left[ \begin{array}{rrrr}
1&0&-2/13&5/13\\
0&1&3/13&-1/13\\
\end{array}\right]\end{displaymath}

Since the matrix $\left[ \begin{array}{rr}
-2/13&5/13\\
3/13&-1/13\\
\end{array}\right]$ is the transpose of $(A^{-1})$, your need to find the transpose of tis matrix to get $A^{-1}$.

Therefore,


\begin{displaymath}A^{-1} = \left[ \begin{array}{rr}
-2/13&3/13\\
5/13&-1/13\\
\end{array}\right]\end{displaymath}

Multiply the secert message by $A^{-1}$, you will get

9 1 13 1 7 15 15 4 19 20 21 4 5 14 20 20

Use the coversion table to find the corresponding letters: IAMAG00DSTUDENTT, which will read "I am a good student".


next up previous
Next: Problems Up: How to break the Previous: How to break the
Ali A. Daddel 2000-09-18