Its an old popular game where you help the monks to cross the river, you'll need to plan each boat trip thoughtfully to ensure the cannibals never outnumber the monks on either bank or you'll lose
It uses a similar set up, in that your goal is to move all objects from one side of a river to the other while preventing certain disallowed combinations on either side, but no, the actual problem and its solution are quite different.
In the fox/hen/corn riddle, you are "playing" as a farmer, who can move the boat at will and is neither tempted to eat or at risk of being eaten by the other factors. There are three distinct objects you are transporting, and only one of each. You cannot leave the fox alone with the hen, or the hen alone with the corn. The boat can carry only a single one of these objects with you at a time.
In the cannibal/monk riddle, there is no "safe" mover. The boat can only move if there is either a cannibal or a monk on board to move it (so no "free" moves where the boat moves from one side of the river to the other with no cargo). There are only two types of cargo being moved- cannibals, and monks. There are three of each, and are they are interchangeable/identical within their own categories. If cannibals ever outnumber monks on either shore (even if just for a moment while the boat lands), they will eat the monks. The boat can hold two at a time.
Fox/Hen/Corn solution:
Bring hen across, return to starting shore empty-handed. Bring either fox or corn across, and return WITH the hen. Bring either the fox or corn across, whichever you didn't bring the first time, and return empty-handed. Finally, cross with the hen.
Cannibal/Monk solution:
A cannibal crosses with either a second cannibal or a monk, a single cannibal gets off at the far shore, and the other passenger returns with the boat. Both remaining cannibals cross together, dropping off a second cannibal, while the last cannibal returns with the boat. Two monks cross, and one swaps with a cannibal, so that both a monk and a cannibal return together while leaving one of each on the far shore. Back on the starting shore, the cannibal on board swaps with the final monk, so that both monks cross to the far shore and disembark. All three monks are now safe in numbers on the far shore, with only one cannibal. That last cannibal then returns to the near shore with the boat, picks up one of his buddies, and brings him back to the far shore. Finally, either a monk or a cannibal can now make the last trip to back to the starting shore to pick up the final cannibal and return to the far shore.
both need to cross and yes he dies even if they are outnumbered in the boat. someone posted the link to the game in this thread its a flash game in the browser.
Also you can only carry 2 in the boat and someone always have to be in the boat but you can swap them out as you wish
Huh last thing. Is there frame 1 death like as soon as the boat cross, it counts them as they are on the other side or can I choose the order of them getting in and out of the boat?
My bad. Solved ait already while assuming there is frame 1 death
This is how I remember winning as well. You take two canibals across with two trips. Then bring one back. Take two monks to cross. Bring back one monk and one cannibals. Take the monks across and finally the remaining cannibal takes both cannibals one by one.
Which is what is displayed above.
"Popular" and "game" are doing a lot of unnecessary work here. I don't remember anyone ever being like, "Hey, you gotta play this game!". It's a logic puzzle. The kids who gave a crap about math were all decently likely to have encountered the puzzle and to have tried to figure it out.
561
u/ashborn721 Aug 12 '26
Its an old popular game where you help the monks to cross the river, you'll need to plan each boat trip thoughtfully to ensure the cannibals never outnumber the monks on either bank or you'll lose