g=good vote
b=bad vote
ggb    system ok
gggbb    system ok
ggggbbb  system ok

I "fail" to see the Kevin's logic below,
2 x (design maximum failures) +1 works above, 5 unit votes is the required numbe
r
to achieve a correct consensus with 0 to 2 vote failures.   Perhaps Kevin refers
to safety in the event of incorrect implementation of Murphy's Law ?   I think t
he
real issue is specification of the maximum concievable failures, costs and risks
,
also susceptability of units to voting empathy (ie common power supply ! fails)
Kevin Maciunas wrote:

> As a CS academic I have to raise a caution here: 5 computers is NOT ENOUGH to
> decide/vote.  If you wish to engineer a fail-safe/fail-operational system, it
> takes 3k+1 entities to achieve concensus in the presence of k failures.  It
> seems to be pretty common practice to engineer in "a" redundant system -
> unfortunately "a" redundant system doesn't buy you much.  There was an
> excellent survey of the concensus problem in the early 1990's in IEEE Computer
> (even made the front page, I believe).
>
> /Kevin