Difference between revisions of "HW3.C Hetong Li ECE301Fall2008mboutin" - Rhea

Latest revision as of 17:17, 18 September 2008

Bob can use the inverse of the secret matrix

Explaination:

Let A be the 3x3 secret matrix message.

$\,A=\left[ \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right] \,$

Let B be the 3x3 matrix for the original message.

$\,B=\left[ \begin{array}{ccc} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] \,$

Correspondingly, let C be the crypted message

From the poblem: $\,C = B * A\,$

So $\,C*A^{-1} = B * A * A^{-1} = B\,$ , i.e. $\,B = C*A^{-1}$

Thus Bob can decrypt the message by finding the inverse of the secret matrix.

YES. Here is the explaination:

Let, $\,B=B_1+B_2+B_3\,$ and $\,C=C_1+C_2+C_3\,$

Thus,we can have:

$\,B_1 = C_1*A^{-1}\,$

$\,B_2 = C_2*A^{-1}\,$

$\,B_3 = C_3*A^{-1}\,$

if we can decompose $\,C_new=n*C_1+p*C_2+q*C_3\,$ we have:

$\,B_1new = n*C_1*A^{-1} = n*B_1\,$

$\,B_2new = p*C_2*A^{-1} = p*B_2\,$

$\,B_3new = q*C_3*A^{-1} = q*B_3\,$

So we can get:

$\,B_new = B_1new+B_2new+B_3new = n*B_1+p*B_2+q*B_3 \,$

so all Eve need to do is to express each row of the encrypted message in terms of a*[2,0,0]+b*[0,1,0]+c*[0,0,3]

then the corresponding row of the original code is a*[1,0,4]+b*[0,1,0]+c*[1,0,1]

The encrypted message is

$\,C=(2,23,3)\,$

we can write it as

$\,C=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\,$

Given system is linear, the corresponding output is

$\,B=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\,$

So, the original message is "B,W,E".

@@ Line 1: / Line 1: @@
 == How can Bob Decrypt the Message? ==
+Bob can use the inverse of the secret matrix
+Explaination:
 Let A be the 3x3 secret matrix message.