  `On Fri, Aug 15, 2003 at 09:55:03AM +0200, Andries Brouwer wrote:> The lemma that you believe in, applied to x and z, says> >  entropy(x) >= entropy(x xor z)>  entropy(z) >= entropy(x xor z)[snip]> This "lemma", formulated in this generality, is just plain nonsense.Indeed.  In the context of x = first 80 bits of SHA-1 hash, y = second80 bits of SHA-1 hash, we are assuming that the entropy of x and y arebasically equal.  Then xoring them together gives us no benefit overtaking just one or the other, and may reduce the entropy if the outputof SHA-1 is correlated between the two halves in any way (which seemsmore likely than not).Including a little more context, then:If entropy(x) == entropy(y), then:entropy(x) >= entropy(x xor y)entropy(y) >= entropy(x xor y)Amusingly, someone already worked their way backwards from my postwithout the benefit of any context to find that it implied thatentropy(x) == entropy(y).  I should know better than to assume thatpeople have read the entire thread before posting.Apologies to Matt for sidetracking the discussion.  He has come upwith a nice solution despite our kibitzing.-VAL-To unsubscribe from this list: send the line "unsubscribe linux-kernel" inthe body of a message to majordomo@vger.kernel.orgMore majordomo info at  http://vger.kernel.org/majordomo-info.htmlPlease read the FAQ at  http://www.tux.org/lkml/`   