Difference between revisions of "HW3.C Jack Williams ECE301Fall2008mboutin" - Rhea

Latest revision as of 15:25, 19 September 2008

1. How can Bob decrypt the message?

Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.

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

Yes. Eve can solve the system of equations with the vectors she has in order to decrypt the message.

3. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)

By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where

a11 = -2/3 a12 = 0 a13 = 2/3

a21 = 0 a22 = 1 a23 = 0

a31 = 4 a32 = 0 a33 = -1

which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields

A =

  -0.6667         0    0.6667
        0    1.0000         0
   4.0000         0   -1.0000

B =

   0.5000         0    0.3333
        0    1.0000         0
   2.0000         0    0.3333

Message =

    2
   23
    5

which corresponds to the letters of the alphabet B W and E.

Answer = BWE

@@ Line 1: / Line 1: @@
-| 3 3 3 |
+. How can Bob decrypt the message?
-| 3 3 3 |
-| 3 3 3 |
-Alice is using a 3-by-3 secret matrix to encrypt a written text and send it to Bob. Her encryption method is as follows. First, Alice tells Bob what secret matrix she is going to use. To send a message, she replaces each letter by its corresponding order in the alphabet and puts a zero for a space. This yields a vector of integers which encodes the message. The 3-by-3 matrix is applied to the first three entries of the vector, then the next three entries, etc. This yields a new vector which carries the encrypted text. Alice then sends the encrypted vector in an email.
+Bob can use matrix algebra to find the inverse of the 3-by-3 matrix, which he can then multiply the encrypted vector by in order to obtain the 1-by-3 decrypted vector that corresponds to the numerical place of each letter of the alphabet.
-. How can Bob decrypt the message?
+. Can Eve decrypt the message without finding the inverse of the secret matrix?
+Yes. Eve can solve the system of equations with the vectors she has in order to decrypt the message.
-Eve is eavesdropping the conversation. Although she doesn’t know what the matrix is, she happens to know that the message (1,0,4,0,1,0,1,0,1) yields the encrypted vector (2,0,0,0,1,0,0,0,3).
+. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
+By solving the system equations given by the vector of decrypted numbers and vector of encrypted numbers one can derive the original matrix [A] where
-. Can Eve decrypt the message without finding the inverse of the secret matrix?
+a11 = -2/3
+a12 = 0
+a13 = 2/3
-. What is the decrypted message corresponding to (2,23,3)? (Write it as a text.)
+a21 = 0
+a22 = 1
+a23 = 0
+a31 = 4
+a32 = 0
+a33 = -1
+which can be used as matrix [A] in matlab to find the inverse matrix [B] which, when multiplied with the encrypted vector [2, 23, 3] yields
+A =
+   -0.6667         0    0.6667
+    1.0000         0
+.0000         0   -1.0000
+B =
+.5000         0    0.3333
+    1.0000         0
+.0000         0    0.3333
+Message =
+which corresponds to the letters of the alphabet B W and E.
-Post your answer on Rhea.
+Answer = BWE

Difference between revisions of "HW3.C Jack Williams ECE301Fall2008mboutin" - Rhea

Latest revision as of 15:25, 19 September 2008

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