>>>>>I may have already given this one, but what the heck, I'll give it again.
>>>>>
>>>>>There are 3 missionaries and 3 natives. They need to cross a river in a boat that only holds 2. The restraints are that the cannibals can't outnumber the Missionaries on either side because the missionaries feared that the natives would over-power and eat them.
>>>>
>>>>You always have a Cannibal [C] in the boat. Order of crossing:
>>>>
>>>>1. Missionary.(left 2M and 2C, crossed 1M and 0C, 1C in the boat)
>>>>2. Cannibal...(left 2M and 1C, crossed 1M and 1C, 1C in the boat)
>>>>3. Missionary.(left 1M and 1C, crossed 2M and 1C, 1C in the boat)
>>>>4. Missionary.(left 0M and 1C, crossed 3M and 1C, 1C in the boat)
>>>>5. Cannibal...(left 0M and 0C, crossed 3M and 2C, 1C in the boat)
>>>>
>>>>Now the Cannibal in the boat comes ashore.
>>>>
>>>>Either that, or the missionaries all stay together until they convert one of the cannibals to christianity before crossing. :^)
>>>
>>>That's close but the boat holds only 2. It looks like you've got 3 in the boat.
>>
>>Nope. I only have 2 in the boat. One is always a cannibal. If the cannibal takes a missionary over first, then goes back to pick up another passenger, does the cannibal in the boat count as being back on the other side? If so, I do not see a workable solution.
>
>The answer is yes. And there is a workable solution. If you want I can private message you the answer.

Does the boat always have to be piolted by a missionary or a cannibal? e.g., 1 of each can't just cross, both get out, and the ones still on the other side pull the empty boat back across with a rope.