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8

If you mean exactly as likely, no, because the number of possible hashes is not a multiple of $100$. This is assuming all the hashes are exactly equally likely. You can come very close just by taking $SHA256 hash \pmod {100}$ This will be within one part in $\frac {2^{256}}{100}$, which is a very small number. If you want truly equal, check that the hash ...


5

I agree with the comments that SHA-256 should be fine here. However, if you already use HMAC-SHA-256 for PBKDF2, you could use HKDF Expand, which despite its name is defined even for output lengths shorter than input. In your case the output would be simply: $$\operatorname{HMAC-SHA-256}(\text{key}, \text{info} || \text{0x01}),$$ where 'info' is an ...


1

No, it is not possible to get exactly equal probabilities for a deterministic mapping from all 256-bit numbers to the range 0-99. However, you can ask whether it matters, since a bias on the order of $2^{-256}$ is undetectable. A mapping that took the 256-bit number modulo 100 and refused the inputs less than $2^{256} \bmod 100$ would be unbiased and would ...


1

Yes, according to NIST SP 800-56A revision 2, a KDF based on HMAC-SHA-256 is a suitable option. The basic idea behind using a Key Based Key Derivation Function KBKDF is that the output of the the primitive within the key agreement protocol (DH, ECDH) returns enough entropy for a key to be created. However that entropy may still be distinguishable from ...



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