1) changes to W at position 0x0c do not affect 11 of the remaining 64 words in W, and
2) if you only change positions 0x0c and 0x0d of a sha-1 block, then
you can calculate A,B,C,D, and E after the first 12 iterations of
the sha-1 loop once, use those values instead of H0,H1,H2,H3, and
H4 as initializers and skip the first 12 (out of 80) iterations of
the sha-1 loop.
I'm already using both of these ideas. Sorry. :)
-------------------------------------------------------------- from: Jonathan "Chromatix" Morton mail: chromi@xxxxxxxxxxxxxxxxxxxxx website: http://www.chromatix.uklinux.net/ tagline: The key to knowledge is not to rely on people to teach you it.