Orcan Ogetbil wrote:
> I don't understand your point. The probability of a hash collision is
> many orders less than 10^{-4}. Yet this isn't acceptable for you.
> However you find the 10^{-4} probability of failure for simultaneous
> pushing to different branches acceptable. Aren't you contradicting
> yourself?
A procedure which otherwise has ZERO risk of breakage and to which you allow
any epsilon>0 risk has changed from an algorithm which is provably correct
to one which is not. A procedure which is already imperfect does not become
noticeably more imperfect if you add a much smaller second imperfection.
You can also see it in terms of the ratio between the risks of breakage. If
you go from p(breakage)=0 to p(breakage)=10^-100, you've increased your risk
by an infinite amount of times! If you go from p(breakage)=.01 to
p(breakage)=.0101, you've increased your risk by only a factor of 1.01. So
the 10^-4 risk in the second case is negligible compared to the already
existing risks, the 10^-100 risk in the first case is not!
Testing is already an imperfect procedure, you can only make it more or less
imperfect, and the difference (or really, the ratio, which is what matters
when computing relative risks) can be negligible or not.
Kevin Kofler
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