(→Part C.3) |
|||
| (One intermediate revision by the same user not shown) | |||
| Line 17: | Line 17: | ||
Homework #3 Part C.3 | Homework #3 Part C.3 | ||
%} | %} | ||
| + | clear all | ||
| + | clc | ||
message = [1 0 4; 0 1 0; 1 0 1]; | message = [1 0 4; 0 1 0; 1 0 1]; | ||
| Line 22: | Line 24: | ||
encryptedVector = [2 0 0; 0 1 0; 0 0 3]; | encryptedVector = [2 0 0; 0 1 0; 0 0 3]; | ||
| − | secretMatrix = ((inv(message)) * encryptedVector) | + | secretMatrix = ((inv(message)) * encryptedVector) |
%{ | %{ | ||
| Line 34: | Line 36: | ||
%} | %} | ||
| + | |||
| + | newMessage = [2 23 3]; | ||
| + | |||
| + | decryptedMessage = newMessage * inv(secretMatrix) | ||
| + | |||
| + | %{ | ||
| + | This code produces the matrix: | ||
| + | |||
| + | decryptedMessage = | ||
| + | |||
| + | 2 23 5 | ||
| + | |||
| + | %} | ||
| + | |||
</pre> | </pre> | ||
| + | |||
| + | So the decrypted message is BWE! | ||
Latest revision as of 14:47, 19 September 2008
Application of Linearity
Part C.1
If Bob knows the secret matrix used to encrypt the message, He can simply take the inverse of that matrix and multiply the encrypted vector by the inverted matrix.
Part C.2
Yes she can just solve the system of equations represented by the matrix, but in general it is easier to just use matrices to solve such equations.
Part C.3
%{
Jeremiah Wise
Homework #3 Part C.3
%}
clear all
clc
message = [1 0 4; 0 1 0; 1 0 1];
encryptedVector = [2 0 0; 0 1 0; 0 0 3];
secretMatrix = ((inv(message)) * encryptedVector)
%{
This code produces the matrix:
secretMatrix =
-0.6667 0 0.6667
0 1.0000 0
4.0000 0 -1.0000
%}
newMessage = [2 23 3];
decryptedMessage = newMessage * inv(secretMatrix)
%{
This code produces the matrix:
decryptedMessage =
2 23 5
%}
So the decrypted message is BWE!
