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|seen||Nov 5 '13 at 14:51|
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 .) This violates the requirement that the hash function behave like a random mapping." -- The Skein Hash Function Family, Version 1.3.