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Yes, she can set an equation that <math>decrypted*secret=encrypted</math> and put nine variables in the secret matrix. | Yes, she can set an equation that <math>decrypted*secret=encrypted</math> and put nine variables in the secret matrix. | ||
Then peform matrix multiplication between decrypted and secret matrix. This will yield nine equations with nine variables. However, some of variables will just be zeros, so it would not take too long. | Then peform matrix multiplication between decrypted and secret matrix. This will yield nine equations with nine variables. However, some of variables will just be zeros, so it would not take too long. | ||
| + | |||
| + | |||
| + | == What is the decrypted message corresponding to (2,23,3)? == | ||
| + | <pre> | ||
| + | C=[2 23 3]'; | ||
| + | Cde=inv(secret)*C | ||
| + | % | ||
| + | % Cde = | ||
| + | % | ||
| + | % 5 | ||
| + | % 23 | ||
| + | % 2 | ||
| + | </pre> | ||
| + | The decrypted text is 'ewb' | ||
Latest revision as of 16:14, 19 September 2008
How can Bob decrypt the message?
Since Alice specified that she was going to use 3-by-3 secret matrix and we have a decrypted matrix and its pre-decrypted matrix, we can solve for secret matirx by multiplying encrypted matrix by inverse of decrypted matrix
format rat A=[1 0 4;0 1 0;1 0 1]; B=[2 0 0;0 1 0;0 0 3]; secret= B*inv(A) % secret = % % -2/3 0 8/3 % 0 1 0 % 1 0 -1 %
Can Eve decrypt the message without finding the inverse of the secret matrix?
Yes, she can set an equation that $ decrypted*secret=encrypted $ and put nine variables in the secret matrix. Then peform matrix multiplication between decrypted and secret matrix. This will yield nine equations with nine variables. However, some of variables will just be zeros, so it would not take too long.
What is the decrypted message corresponding to (2,23,3)?
C=[2 23 3]'; Cde=inv(secret)*C % % Cde = % % 5 % 23 % 2
The decrypted text is 'ewb'
