# Tag Info

16

There's a good reason many democracies reserve mail voting to rare cases where that's the only option: it allows one's vote to be influenced by duress or bribery, because one can prove how one voted. No remote voting technology that I know has solved that issue. Ergo, physically going to a polling station with isoloirs should remain the normal voting ...

5

The Luby-Rackoff theorem says that a 3-4 round Feistel network is a pseudorandom permutation for some sufficiently large block size. As this paper by Patarin on Feistel networks with 5 or more rounds puts it: We will denote by $k$ the number of rounds and by $n$ the integer such that the Feistel cipher is a permutation of $2^n$ bits → $2^n$ bits. In [3] ...

4

As a society we have decided that "secret ballots" are better than "open ballots". When you use a blockchain you will issue public keys to individuals who can then verify their votes were counted. I'm aware that the voter's name is not on these keys (I could not look at the chain and determine how you voted), but regardless it is a way to prove how you voted ...

4

The question's code works for AES-128 and AES-256, but indeed does not for AES-192, for the reason pointed in the question. Yes changing if ( !( i % 36 ) ) into if((!(i%36)) && (key_words == 4)) as suggested in comment by the OP is enough to make it (and the rest of the code) pass the example of appendix A.2 and C.2 of FIPS 197 (as well as A.1, ...

3

Actually, that is depending on the padding applied. Maybe there is no padding at all. The implementation defines the padding applied. Check the code or documentation! The last 22, at first, indicates that might be PKCS#5 or PKCS#7 since PKCS#7 most common with CBC implementations. While PKCS#5 is limited to 8 bytes like DES, PKCS#7 works from 1 to 256 bytes....

1

The question likely really is: It is known $c$ with $c = E_{K_1}(m \oplus K_1 \oplus K_2)$, and a few distinct plaintext/ciphertext pairs $(m_i,c_i)$, that is with $c_i = E_{K_1}(m_i \oplus K_1 \oplus K_2)$. However, $c$ is not one of the $c_i$ (which would make finding $m$ trivial). Define a strategy to find $m$, despite the terms $\oplus K_1 \oplus K_2$ ...

1

The result (Luby-Rackoff) that using 3 rounds of a Feistel structure is enough depends on the $f$ function being a pseudorandom function. This is a theoretical idealized model and since you are using a specific single and concrete function, the result won't apply.

1

Is it really just a block cipher in ECB mode, so that each packet can be decoded without any prior packet⁰? No. ECB is bad and creating a secure real-time channel (that can even handle packet loss) is a solved problem. It has been for years. There's no good excuse for using ECB here. Is it a block cipher in some non-ECB mode, but it gets reset at the ...

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