[Bonus question, points 5]
Consider the protocol of the dining cryptographers with an arbitrary
number n (greater than 2) of cryptographers, and n coins. Does the
protocol that we have seen for the case of 3 cryptographers still work
for the general case, possibly with some adaptations? Or does it work
only for certain n? Or does it work only for the case n=3? Please
justify your answer.
Answer:It works for all n greater than 2,
in exactly the same way. In fact, if no cryptographers is paying, than all of them say "the truth"
about the agreement/disagreement of the coins, hence the number of "disagree" will be even.
If one of cryptographers is paying, than he says the opposite, hence the number of "disagree" will
be odd.