Solutions to the Missionaries and Cannibals problem ================================================== Note that at the start we really only have two choices: either send to cannibals in the boat or send one missionary and one cannibal (sending one person is not useful, the only thing that can happen next is that that person comes right back again.) A solution can be found with either start. Notice too that there is usually very little choice once you get started. Remember, returning to a situation you have already had is wasteful. Some comments about choosing are indicated below. M denotes a missionary C denotes a cannibal <> denotes the boat near side of river far side of river comments about choices ================== ================= ====================== MMM CCC <> other option listed below MMM C <> CC someone has to bring the boat back MMM CC <> C send M across leaves trouble next trip MMM <> CCC someone has to bring the boat back MMM C <> CC must send 2M's if any M C <> MM CC must send equal # of M's and C's ----------------------------- now we can just reverse everything MM CC <> M C CC <> MMM C CCC <> MMM C <> MMM CC CC <> MMM C <> MMM CC Here is a second solution that starts differently: MMM CCC <> here's the other start MM CC <> C M can't send C back, so M goes MMM CC <> C sending M's leads to trouble MMM <> CCC hmm. looks just like above now MMM C <> CC M C <> CC MM ----------------------------- now we can either reverse this MM CC <> C M CC <> C MMM CCC <> MMM C <> CC MMM M C <> CC MM <> CCC MMM note that we could mix and match tops and bottoms, so we have actually found 4 solutions!