(→‎How can Bob Decrypt the Message?)
 
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== How can Bob Decrypt the Message? ==
 
== How can Bob Decrypt the Message? ==
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Bob can use the inverse of the secret matrix
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 +
Explaination:
  
 
Let A be the 3x3 secret matrix message.
 
Let A be the 3x3 secret matrix message.
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== Can Eve decrypt the message without finding the inverse of the secret matrix? ==
 
== Can Eve decrypt the message without finding the inverse of the secret matrix? ==
  
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if we write B as
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YES. Here is the explaination:
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<math>\,A=\left[ \begin{array}{ccc} A_{1} \\ A_{2} \\ A_{3} \end{array} \right] \,</math>
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(write in conbimation of column vector)
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Similarly, we write C as
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Let,
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<math>\,C=\left[ \begin{array}{ccc} C_{1} & C_{2} & C_{3} \end{array} \right] \,</math>
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<math>\,B=B_1+B_2+B_3\,</math> and <math>\,C=C_1+C_2+C_3\,</math>
  
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Thus,based on the multiplication of Matrix,we can have:
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Thus,we can have:
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<math>\,C_1 = B*A\,</math>, same for <math>\,C_2\,</math> and <math>\,C_3\,</math>
+
  
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which means
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<math>\,B_1 = C_1*A^{-1}\,</math>
  
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<math>\,C_1 = B*A_1 = B_1*A_1+B_2*A_2+B_3*A_3\,</math>
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<math>\,B_2 = C_2*A^{-1}\,</math>
  
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<math>\,C_2 = B*A = B_1*A_1+B_2*A_2+B_3*A_3\,</math>
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<math>\,B_3 = C_3*A^{-1}\,</math>
  
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<math>\,C_3 = B*A = B_1*A_1+B_2*A_2+B_3*A_3\,</math>
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if we can decompose <math>\,C_new=n*C_1+p*C_2+q*C_3\,</math>
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+
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if we multiply <math>B_1</math> by n,<math>B_2</math> by p,<math>B_3</math> by q,
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we have:
 
we have:
  
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<math>\,C_1new = B*A_1 = n*B_1*a_11+p*B_2*a_21+q*B_3*a_31\,</math>
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<math>\,B_1new = n*C_1*A^{-1} = n*B_1\,</math>
  
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<math>\,C_2new = B*A_2 = n*B_1*a_21+p*B_2*a_22+q*B_3*a_23\,</math>
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<math>\,B_2new = p*C_2*A^{-1} = p*B_2\,</math>
  
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<math>\,C_3new = B*A_3 = n*B_1*a_31+p*B_2*a_32+q*B_3*a_33\,</math>
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<math>\,B_3new = q*C_3*A^{-1} = q*B_3\,</math>
  
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and the reverse is also true
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So we can get:
  
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<math>\,B_new = B_1new+B_2new+B_3new = n*B_1+p*B_2+q*B_3 \,</math>
  
 +
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? ==
 
== What is the Decrypted Message? ==
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The given encrypted message is
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The encrypted message is
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<math>\,e=(2,23,3)\,</math>
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−
 
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This can be rewritten as a linear combination of the given system result vectors
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−
 
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<math>\,e=ae_1+be_2+ce_3\,</math>
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<math>\,e=(2,23,3)=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\,</math>
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<math>\,C=(2,23,3)\,</math>
  
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we can write it as
  
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Because the system s linear, we can write the input as
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<math>\,C=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\,</math>
  
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<math>\,m=am_1+bm_2+cm_3\,</math>
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Given system is linear, the corresponding output is
  
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<math>\,m=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\,</math>
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<math>\,B=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\,</math>
  
  
−
Therefore, the unencrypted message is "'''BWE'''".
+
So, the original message is "B,W,E".

Latest revision as of 16:17, 18 September 2008

How can Bob Decrypt the Message?

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.

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 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".

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