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4

Your scheme is likely sound as the SHA-256 hash limits the input to the MD5 hashing function. If this was not so it could be possible (though very hard) to create a collision because of the break. Generally it is more secure to simply use the 128 leftmost bits of the output of SHA-256. SHA-3 - or rather SHAKE256 - would be even better if available. Keep in ...


4

We can easily find collisions for this system. Let $x=(x_1,...,x_n) $ be a sequence such that $h(x)=k$. For finding collision we can do this: Select $x'=(x_1',...,x_{n-1}')$ and then compute: $$x_n'=({a_n}^{-1}\cdot(k-\sum_{i=1}^{n-1}a_ix_i')) \bmod N$$ Now we have $h(x)=h(x')=k$.


3

It is not collision-resistant because the tag is only 64 bits, so on average only $2^{32}$ inputs are needed to find a collision. This is the classic birthday bound. This is completely insecure, so SipHash is not collision resistant. This is not a problem for hash tables because hash tables have collision-resolution mechanisms and because the odds of a ...


1

In general, you never want to use CRC/weak checksum for any computations on secret material (like keys). CRC is a linear function and by showing CRC of a key, you reveal a lot of equations that hold among the key bits. This is equivalent to showing the same number of bits of the key as the length of the checksum. The proper way of doing it has been ...



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