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You do not have to divide the data into 16-byte blocks and individually encrypt each block. The encrypt_and_digest method will handle all of this division for you. Instead you should call the method on the full array of data. By individually enciphering each block you are creating a nonce and tag for each block rather than a single nonce and tag for all of ...

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$\text{AES-128}_d(C_1, K) \oplus \text{AES-128}_d(C_2, K) = P_1 \oplus P_2$ Doesn't that assume that the two ciphertexts share the same IV? Typically, they don't. In any case, even if you don't know any of the IVs, then a simpler method would be if you have a known two block message with plaintext $(P_a, P_b)$ and a corresponding ciphertext $(C_a, C_b)$, ...

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Grover's algorithm would work against CBC. Do you mean you have several thousand plaintext/ciphertext messages, each with their own IV? I'm not sure what you're doing in step 2. I think generally the IV is considered part of the ciphertext, so if you have the ciphertext, you have the IV, and then you end up with equations like: AES-128$_d$(C$_1$, K) = IV $\... 0 There are in fact two separate things to be discussed w.r.t. the paper: The details of how to apply a$\text{GF}(2)$-affine function to an SSS-shared variable. The correctness of the new SSS-multiplication scheme proposed in the paper. The paper On the Use of Shamir’s Secret Sharing Against Side-Channel Analysis mentioned by Kodlu suggests that there is a ... 1 The objection on the eprint.iacr forum is below and seems to be correct at a quick first look. Did you search literature that cites this paper? Have the authors published further on this? Edit: There is further work here which may validate the claim, in an MPC framework. Just putting this note up since I will not have time to look more deeply for a while. “... -1 Nearly reversible computation will eventually replace conventional computation since reversible computation will be more energy efficient than conventional computation. Furthermore, symmetric encryption and decryption are very well suited for reversible computing, so we should expect for symmetric encryption and decryption to eventually be computed on ... -3 Can you categorically prove that AES isn't broken today, or won't be in October 2021? See NOBUS. The one time pad with key size = message size. Facilitated via quantum key distribution. That's one of the the main areas of cryptographic research today in budgetary terms. 2 if we will change one bit in the plaintext, more than one bit will be changed in the ciphertext? In Output Feedback Mode, no, it does not. Specifically, if you change one bit of the plaintext and nothing else (e.g. you don't modify the IV), then only the corresponding bit of the ciphertext will change. Output Feedback Mode works by internally generating a ... 0 You want to work modulo$2^{32}$, except for shift counts where that should be modulo$32$. The following is generic and works in Python. code (result in z) operation z = (x+y)&0xffffffff 32-bit addition of x and y z = (x-y)&0xffffffff 32-bit subtraction x minus y z = (x*y)&0xffffffff 32-bit multiplication of x and y z = ((x<<(31&y)... 1 This will depend on the language that you implement in. Java and other C-like languages have a built-in data type to represent unsigned 32-bit integers (this is why RC6 chose to use this form of arithmetic, so that its implementation in these languages is relatively straightforward). In such cases +, -, and * all automatically work mod$2^{32}$. If you're ... 1 I'll assume that you have already performed PBKDF2 with the static 1000 iterations defined (which is, uh, on the low end of security), and that the two key check bytes have been validated. AES itself is no different, and it seems that Winzip doesn't use anything special. The only thing that is a bit weird is that the counter is in little endian. This is all ... 1 The only reason why you should be afraid of IV reuse is if the random number generator is off. Assuming a fully random IV you could encrypt$2^{64}$blocks within AES-CBC and still only have a one in$~2^{64}$chance that there is a collision (approximately). Note that the repeat input problem doesn't limit itself to the IV; each ciphertext is used as a &... 0 Can everything (that can be encrypted) alternatively be encrypted into ASCII? Yes. And it's common practice. The general technique is to encrypt, which for most modern cipher will output apparently random bits, which can be called "binary". Then re-encode that into ASCII, e.g. using Base64. That's built into some encryption software like PGP/GPG, ... 0 ASCII is an encoding, not an encryption method. ASCII turns certain characters (the English alphabet, some symbols, and some non-printable control characters like "tab" or "line feed") into 7-bit binary values. This is usually done encoding each ASCII character into one 8-bit byte, and setting the most significant bit of that byte to 0. ... 0 When a string like "Hello World" is encrypted, do the algorithms convert this string into binary or ASCII? Well, the answer depends on whether you're talking about historic or modern cryptography. Historically, encryption methods typically did take the string as a series of characters, and transformed them that way. Sometimes they did an internal ... 2 I've seen other people say Poly1305 has a 128-bit security level but haven't found much about the security level of either. Well, in terms of security level, there are two potential attacks: One in which you attempt as an evesdropper to recover the secret key; the security of both Poly1305 and GMAC is essentially the same as the underlying block cipher. ... 0 Of course$AES_{K_1} \oplus AES_{K_2}$will pass as that's a pseudorandom function. One$AES\$ isn't meant to be. But how can you prove that empirically? PractRand is written by one guy (sorry Chris). It's not used in the mainstream literature. TRNG's are principally validated using dieharder or NIST STS. Randomness has been well studied but PractRand hasn't. ...

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