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

Revision as of 16:13, 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?

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]

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".