Timeline for How to create the ciphertext that generates a desired plaintext in CBC (previous cipher- and plaintext are known)
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 15, 2021 at 21:39 | comment | added | kelalaka | @hello3123 to understand better the attack, looking the original source of the attack is better. Updated the answer. | |
| Mar 15, 2021 at 21:38 | history | edited | kelalaka | CC BY-SA 4.0 |
added the original attack. This is more clear.
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| Mar 14, 2021 at 19:50 | comment | added | kelalaka | Alas, it is correct, combine the eq (4) and (3) to see that. I know, not a good way to write it down. | |
| Mar 14, 2021 at 19:03 | comment | added | hello3123 | That you for your answer. i just have one question: Why is it: $P'_{i-1} = dec_{KAES}(C'_{i-1}) \oplus C_{i-2}$ instead of $P'_{i-1} = dec_{KAES}(C'_{i-1}) \oplus C'_{i-2}$? I though we want to repeat the whole process until we reach the first block | |
| Mar 14, 2021 at 16:24 | vote | accept | hello3123 | ||
| Mar 13, 2021 at 23:06 | history | edited | kelalaka | CC BY-SA 4.0 |
added 5 characters in body
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| Mar 13, 2021 at 21:05 | history | answered | kelalaka | CC BY-SA 4.0 |