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9

What choice did they have? F1 is a bitwise function with three inputs and one output. There are $2^8 = 256$ such functions. Only 70 of them are "unbiased" (i.e. have as many 0 and 1 outputs in their image). If you further require that each input, as well as the order of inputs, matters for the output, you are left with only 36. However, those 36 are all ...


6

Your doubts are absolutely valid. Disguising the algorithm is not a valid argument for security. It also contradicts to Kerckhoffs Law. It (the algorithm) should not require secrecy, and it should not be a problem if it falls into enemy hands; Designing cryptographic algorithms (ciphers, hashfunctions, ...) is a long and complicated process. In ...


5

Most standard-use iterative hash functions (including SHA-512) are build in a way that these types of operation are not possible (without breaking the hash function). They work generally in this way: The message is split in same-size blocks (usually with some padding at the end to fill the last block): $pad(M) = M_0 || M_1 || M_2 ... || M_n$. There is ...


3

Ignore the integer overflow issue I mentioned in a comment, for a moment. I don't see how this adds any security. For all $n>2$, the function you are calling Fibonacci is one-to-one, and since $n<256$, you could easily build a lookup table without much memory to invert the function. Therefore, to break this, all one has to do is invert the Fibonacci ...


2

No, that's not possible, as you calculate sha512(F2) without the state of sha512(F1). What you require is compress(mix(compress(mix(IH, F1)), F2)) while what you have is compress(mix(IH, F1)) and compress(mix(IH, F2)). So you would have to undo that last compression, which is obviously not possible. Here IH is the initial state (the values of $h_1$ etc.) ...


1

The multiplicative group $Z_m$ of integers modulo $m$ has exactly $\phi(m)$ elements by definition. Thus for any element $a$ of $Z_m$ the equation $$ a^{\phi(m)}=1~(mod~m) $$ holds, where $1$ is the multiplicative identity of $Z_m$ (the residue class $km+1$ of all integers congruent to 1 modulo $m$). In RSA, $e$ and $d$ are indeed inverses modulo ...



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