On Mar 22, 2005, at 11:51, Ben Laurie wrote:
This can be fixed quite easily:
H'(x)=H(H(x || H(x)) || H(x))
Doesn't this take us back to the original problem, by factoring in x only at the start of hash computations, so H'(x') will generate the same H(x') and the same internal state for H(x'||...) as for H(x||...) and thus the same H(x'||H(x')) etc, resulting in the same final value?
Doh. Yes. Slightly less elegantly, then:
H'(x)=H(H(x || H(0 || x) || H(0 || x))
Then you need two hashes running in parallel, but at least it is still online. Or, with three parallel streams:
H'(x)=H(H(x || H(0 || x) || H(1 || x))
I don't feel as comfortable with either as the original construction, though.
Cheers,
Ben.
-- http://www.apache-ssl.org/ben.html http://www.thebunker.net/
"There is no limit to what a man can do or how far he can go if he doesn't mind who gets the credit." - Robert Woodruff
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