On 5/18/06, Travis H. <[EMAIL PROTECTED]> wrote:
... There's 255 "other" permutations, so the chance that there is
at least one k' such that f_k'(x)=y is 255/256 = 99.6%.  The chance
that there is exactly one such k' is sampling with replacement and if
I am not mistaken P(|K|=1) = (255/256)^255 = 0.36.  Along those same
lines, P(|K|=2) = (255/256)^253 * 254 / 256^2 = 0.001, so it looks
like the expected number of equivocating keys is very small.

Oops, I left off a term in the recurrence.
P(|K|=2) = (255/256)^253 * ((254*255)/2)/(256^2) = 0.18

So the expected number of equivocating keys, given one byte of known
plaintext, is a bit under two.
"Curiousity killed the cat, but for a while I was a suspect" -- Steven Wright
Security Guru for Hire http://www.lightconsulting.com/~travis/ -><-
GPG fingerprint: 9D3F 395A DAC5 5CCC 9066  151D 0A6B 4098 0C55 1484

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