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Well, the exact reason for an IV varies a bit between different modes that use IV. At a high level, what the IV does is act as a randomizer, so that each encrypted message appears to be encrypted to a random pattern, even if those messages are similar. In general, IVs disguise when you encrypt the same message twice (and more generally, when two messages ...

This is called ciphertext-only cryptanalysis*, and it's pretty difficult unless the cipher is quite weak. Therefore, the first priority for a cryptanalyst in such a situation is usually to try to find more information about the algorithm. Fortunately (for the cryptanalyst), as Kerckhoff's principle suggests, there are often ways to find out how the ...

Many cryptographic algorithms are expressed as iterative algorithms. E.g., when encrypting a message with a block cipher in CBC mode, each message "block" is first XORed with the previous encrypted block, and the result of the XOR is then encrypted. The first block has no "previous block" hence we must supply a conventional alternate "zero-th block" which we ...

Encryption using a block cypher such as AES by passing plaintext blocks directly to the encryption function is known as Electronic Code Book mode (ECB) and is not CPA secure as (as you say in your question) it is entirely deterministic and two identical plaintext blocks will result in two identical ciphertext blocks. To prevent this an initialisation ...

If you can't tell what function was applied to create the cipher text, your search space is as many bits as the message. It's the perfect secrecy achieved with a one time pad. (x + y = z, given z what are x and y?) During an exhaustive search the attacker could find as many messages as they were willing to compute, but they will never know which one was ...

To be secure against a chosen-plaintext attack, an encryption scheme must be non-deterministic — that is, its output must include a random element, so that e.g. encrypting the same plaintext twice will result in two different ciphertexts. Indeed, if that was not the case, an attacker could easily win the IND-CPA game just by using the encryption ...

In base 10 we write for example $133$ when we mean $$133 = 1 * 10^2 + 3*10^1 + 3*10^0. $$ If we want to write $49$ in base $2$ then note first that: $$49 = 1*2^5 + 1*2^4 + 0*2^3 + 0*2^2 + 0*2^1 + 1*2^0. $$ Because of this $49$ is $110001$. Noe obviously, you "don't know this", but I wanted to write it down so that you can see what happens as you divide by ...