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1

Request decryption for three blocks ( 2 is enough); \begin{align} P_1 =& Dec_k(C_1) \oplus IV\\ P_2 =& Dec_k(C_1) \oplus C_{0}\\ P_3 =& Dec_k(C_2) \oplus C_{1}\\ \end{align} Now remove the first ciphertext, and request decryption; \begin{align} P'_2 =& Dec_k(C_1) \oplus IV\\ P_3 =& Dec_k(C_2) \oplus C_{1}\\ \end{align} Now use the ...


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Given sufficient resources - maybe. If an attacker knows text that was encrypted and can associate that plaintext with an encrypted string then they can potentially work on figuring out what the other strings encrypted with the same IV are. Can they simply derive the plaintext using some formula though? No. If you're encrypting textual data it's easier ...


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There are two possibilities related to your "search" (partial or exact) and you do not specify which you're using. Let us imagine the plaintext is ATTACK AT DAWN. Let us suppose that the encryption translates this to QWWQTHXQWXBQNO: each letter is encoded as a different letter regardless of its position (so, A is always encrypted as Q and so on). ...


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Pick a key $K$. Pick a one-block message $m$ and the desired ciphertext $c$. Then compute the IV for CBC as follows: $$c=E_K(IV\oplus m)\iff D_K(c)\oplus m=IV$$ where $E_K$ is AES encryption under the key $K$ and $D_K$ is the corresponding decryption and $\oplus$ is bitwise XOR. Now you have the IV that maps $m$ to $c$ under the key $K$ with all three of ...


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They trick you with the IV. The IV must be the same for encryption and decryption so that the encryption has the functionality that is; $$ m = D_k(IV,E_k(IV,m))$$ To surprise you, they modify the IV according to their desire but you failed to see that. This is the more general case of the Bit-Flipping attack on the CBC mode. This attack works because the ...


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