On Fri, 17 Oct 2003 10:16:29 EDT, Jeff Garzik said:

> > This assumes that the probability of there being a bug in the code which
> > checks for identical blocks is less than the probability of a hash collision.
> > I am not sure that is a good assumption.
> 
> The complexity of a memcmp() is pretty low...

What's the probability of a non-ECC-corrected error on a memory read during the
computation of the hash or the memcmp()? (Remember that Linux *does* boot on
machines that don't support ECC memory by default, such as many Macs).