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

Revision as of 16:11, 18 September 2008

How can Bob Decrypt the Message?

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.

Can Eve decrypt the message without finding the inverse of the secret matrix?

$ \,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 the encrypted message in terms of a*[2,0,0]+b*[0,1,0]+c*[0,0,3]

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

What is the Decrypted Message?

The given encrypted message is

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

This can be rewritten as a linear combination of the given system result vectors

$ \,e=ae_1+be_2+ce_3\, $

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

Because the system s linear, we can write the input as

$ \,m=am_1+bm_2+cm_3\, $

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

Therefore, the unencrypted message is "BWE".