network profile
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| visits | member for | 7 months |
| seen | Mar 1 at 8:41 | |
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Jan 28 |
comment |
Why does Skein use an output transform, but other similar hashes don't? You do not need any output transformation at all. :-) In the case of SKEIN, the authors claim the following: "Originally we applied the output transformation only if the output size was larger than the state size. Unfortunately, without the output transform, you can construct two messages M and M' such that H(M) xor H(M) is the same as the XOR of the last blocks of M and M . (A similar property has recently been described for SHA-1 [101].) This violates the requirement that the hash function behave like a random mapping." -- The Skein Hash Function Family, Version 1.3. |
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Jan 28 |
answered | Same Size Crypto Algorithm? |
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Oct 24 |
answered | Can I use guids / uuids as counters for key derivation? |
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Oct 22 |
awarded | Supporter |
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Oct 22 |
revised |
Why does Skein use an output transform, but other similar hashes don't? added 1 characters in body |
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Oct 22 |
answered | Why does Skein use an output transform, but other similar hashes don't? |
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Oct 22 |
awarded | Editor |
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Oct 22 |
revised |
Is AES-256 weaker than 192 and 128 bit versions? added 65 characters in body |
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Oct 21 |
answered | Is AES-256 weaker than 192 and 128 bit versions? |
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Oct 5 |
awarded | Teacher |
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Oct 2 |
answered | Is 512-bit RSA still safe for signature generation? |
