(→‎How can Bob Decrypt the Message?)
 
(7 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
== How can Bob Decrypt the Message? ==
 
== How can Bob Decrypt the Message? ==
 +
Bob can use the inverse of the secret matrix
 +
 +
Explaination:
  
 
Let A be the 3x3 secret matrix message.
 
Let A be the 3x3 secret matrix message.
Line 5: Line 8:
 
<math>\,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] \,</math>
 
<math>\,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] \,</math>
  
−
Let B be the 3x3 matrix for the unencrypted message.
+
Let B be the 3x3 matrix for the original message.
  
 
<math>\,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] \,</math>
 
<math>\,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] \,</math>
  
−
Correspondingly, let C be the decrypted message  
+
Correspondingly, let C be the crypted message  
  
 
From the poblem:  <math>\,C = B * A\,</math>
 
From the poblem:  <math>\,C = B * A\,</math>
Line 19: Line 22:
 
== 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? ==
  
−
Based on the multiplication of Matrix:
+
YES. Here is the explaination:
  
−
if
+
Let,
 +
<math>\,B=B_1+B_2+B_3\,</math> and <math>\,C=C_1+C_2+C_3\,</math>
  
−
<math>\,C_1 = B_1*A\,</math>, same for <math>C_2</math> and <math>C_3</math>
+
Thus,we can have:
  
 +
<math>\,B_1 = C_1*A^{-1}\,</math>
  
−
<math>\,e=mA\,</math>
+
<math>\,B_2 = C_2*A^{-1}\,</math>
  
−
which is how the message is being encrypted.  If we multiply both sides by the inverse of <math>\,A\,</math>, we get
+
<math>\,B_3 = C_3*A^{-1}\,</math>
  
−
<math>\,eA^{-1}=mAA^{-1}=mI=m\,</math>
+
if we can decompose <math>\,C_new=n*C_1+p*C_2+q*C_3\,</math>
 +
we have:
  
−
Therefore, we can get the original message back if we multiply the encrypted message by <math>\,A^{-1}\,</math>, given that the inverse of <math>\,A\,</math> exists.
+
<math>\,B_1new = n*C_1*A^{-1} = n*B_1\,</math>
  
−
== Can Eve Decrypt the Message Without Finding the Inverse of A? ==
+
<math>\,B_2new = p*C_2*A^{-1} = p*B_2\,</math>
−
Yes, because of the fact <math>\,e=mA\,</math> is linear and Eve was given three linearly independent vector responses to the system and their corresponding inputs.
+
  
 +
<math>\,B_3new = q*C_3*A^{-1} = q*B_3\,</math>
  
−
'''Proof of Linearity'''
+
So we can get:
  
−
Say we have two inputs <math>\,m_1\,</math> and <math>\,m_2\,</math> yielding outputs
+
<math>\,B_new = B_1new+B_2new+B_3new = n*B_1+p*B_2+q*B_3 \,</math>
  
−
<math>\,e_1=m_1A\,</math> and
+
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]
  
−
<math>\,e_2=m_2A\,</math>, respectively.
+
then the corresponding row of the original code is a*[1,0,4]+b*[0,1,0]+c*[1,0,1]
−
 
+
−
thus,
+
−
 
+
−
<math>\,ae_1+be_2=am_1A+bm_2A</math> for any <math>\,a,b\in \mathbb{R}\,</math>
+
−
 
+
−
 
+
−
Now, apply <math>\,am_1+bm_2\,</math> to the system
+
−
 
+
−
<math>\,(am_1+bm_2)A=am_1A+bm_2A\,</math>
+
−
 
+
−
 
+
−
Since the two results are equal
+
−
 
+
−
<math>\,am_1A+bm_2A=am_1A+bm_2A\,</math>
+
−
 
+
−
the system is linear.
+
−
 
+
−
 
+
−
''' Main Proof '''
+
−
 
+
−
Since Eve was given that for the system
+
−
 
+
−
<math>\,m_1=(1,0,4)\,</math> yields <math>\,e_1=(2,0,0)\,</math>
+
−
 
+
−
<math>\,m_2=(0,1,0)\,</math> yields <math>\,e_2=(0,1,0)\,</math>
+
−
 
+
−
<math>\,m_3=(1,0,1)\,</math> yields <math>\,e_3=(0,0,3)\,</math>
+
−
 
+
−
where <math>\,e_1, e_2, e_3\,</math> are clearly linearly independent vectors, Eve can take any encrypted message and write it as a linear combination of <math>\,e_1, e_2, e_3\,</math>
+
−
 
+
−
<math>\,\exists a,b,c\in \mathbb{R}\,</math> such that <math>\,e=ae_1+be_2+ce_3\,</math>, for any <math>\,e\in \mathbb{R}^{3}\,</math>
+
−
 
+
−
 
+
−
Because the system is linear, we can write the input as
+
−
 
+
−
<math>\,m=am_1+bm_2+cm_3\,</math>
+
−
 
+
−
thus, the message has been decrypted without knowing <math>\,A^{-1}\,</math>.
+
  
 
== What is the Decrypted Message? ==
 
== What is the Decrypted Message? ==
−
The given encrypted message is
+
The encrypted message is
−
 
+
−
<math>\,e=(2,23,3)\,</math>
+
−
 
+
−
 
+
−
This can be rewritten as a linear combination of the given system result vectors
+
−
 
+
−
<math>\,e=ae_1+be_2+ce_3\,</math>
+
  
−
<math>\,e=(2,23,3)=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\,</math>
+
<math>\,C=(2,23,3)\,</math>
  
 +
we can write it as
  
−
Because the system s linear, we can write the input as
+
<math>\,C=1\cdot (2,0,0)+23\cdot (0,1,0)+1\cdot (0,0,3)\,</math>
  
−
<math>\,m=am_1+bm_2+cm_3\,</math>
+
Given system is linear, the corresponding output is
  
−
<math>\,m=1\cdot (1,0,4)+23\cdot (0,1,0)+1\cdot (1,0,1)=(2,23,5)\,</math>
+
<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 17:17, 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".

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=HW3.C_Hetong_Li_ECE301Fall2008mboutin&oldid=12850"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

EISL lab graduate

Mu Qiao