New answers tagged xor
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Let us decrypt the CBC encrypted ciphertext.(IV and Key each of 16 bytes=128 bits)
key = 140b41b22a29beb4061bda66b6747e14
IV = 4ca00ff4c898d61e1edbf1800618fb28
Cipher Text = 28a226d160dad07883d04e008a7897ee2e4b7465d5290d0c0e6c6822236e1daafb94ffe0c5da05d9476be028ad7c1d81
Now if you put them in any CBC decryption function you will get plaintext as
Basic ...
1
For the CBC mode the encryption and decryption equations are as follows;
CBC encryption;
\begin{align}
C_1 &= E_k(P_1 \oplus IV)\\
C_i &= E_k(P_i \oplus C_{i-1}),\;\; 1 < i < nb,
\end{align}
CBC decryption;
\begin{align}
P_1 =& D_k(C_1) \oplus IV\\
P_i =& D_k(C_i) \oplus C_{i-1},\;\; 1 < i < nb,
\end{align}
But I'm not 100% ...
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