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If we only have one plaintext/ciphertext pair then we can pick a key at random, decrypt from the ciphertext, and then derive an IV which would make it correct. Not knowing the IV means that we don’t know what was XORed with the plaintext before encryption. If you do not know the IV, then you need more than one plaintext to have a chance of finding the correct key. The only difference between this and the ECB mode search is the XOR that must be performed with the \(i - 1\) ciphertext (or IV) before each check. You brute force until you find a key that matches your first chosen plaintext, then you check the next, and so on, until you find one that matches all your chosen plaintexts. If you know the IV, then it is basically no different from breaking an ECB mode cipher. In the worst case (that the last-checked of all possible keys is the right one), you will have to perform \(2^k\) checks.Ģ. To calculate the probability of finding a false-positive key (where \(t\) is the number of chosen plaintexts you have) you can use this formula: \ This is due to the fact that it’s unlikely that the false positive key would also map to the same ciphertext for another plaintext. For a given plaintext, there may be multiple keys that map to the same ciphertext.ĭepending on the actual values of \(k\) and \(n\), having a second (plaintext, ciphertext) pair can provide a high degree of certainty. Since the key length is smaller in bits than the block length, it’s possible (by the pigeonhole principle) that every keys maps to a unqique ciphertext. It depends on the level of confidence you’re happy with. If you spot any mistakes, please leave a comment in the Disqus box at the bottom of the page.ġ. I haven’t yet verified this solution independently. Is breaking a block cipher in CBC mode by means of an exhaustive key search considerably more difficult than breaking an ECB mode block cipher?.How many plaintexts and ciphertexts are necessary, if you do not know the IV?.How many plaintexts and ciphertexts are now needed to break the cipher by performing an exhaustive key search? How many steps need now maximally be done? Briefly describe the attack. Assume that the initialization vector IV for running the considered block cipher in CBC mode is known.How many plaintexts and ciphertexts are needed to successfully break a block cipher running in ECB mode? How many steps are done in the worst case?.
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The block length counts \(n\) bits with \(n > k\). We consider known-plaintext attacks on block ciphers by means of an exhaustive key search where the key is \(k\) bits long. Cryptography understanding-cryptography even-numbered-solutions