Can the cannibals and regular people safely cross the river?

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The discussion revolves around a classic logic puzzle involving three cannibals and three regular people trying to cross a river using a boat that can carry only two at a time. The main challenge is to ensure that at no point are there more cannibals than regular people on either side of the river, as this would result in the cannibals eating the regular people. Participants explore various crossing strategies, emphasizing the need for careful planning to avoid violating the rules. The conversation also touches on a modified version of the puzzle where only one of the cannibals can row the boat, adding complexity to the scenario. Additionally, there are inquiries about starting new threads in the forum, with one user seeking assistance and another sharing a link to a related game for visual aid.
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There's a boat to cross a river, and it can carry 2 people at once. There are 3 cannibals and 3, well, regular people on one side of the river. They want to get across. If there are more cannibals on one side than regular people, you lose. The boat cannot travel empty.
Have fun! :biggrin:
 
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Two canibals cross.
One cannibal takes the boat back.
Two cannibals cross.
One takes the boat back.
Two non-cannibals cross.
A cannibal and a non-cannibal take the boat back.
Two non-cannibals cross.
A cannibal takes the boat back.
Two cannibals cross.
A cannibal takes the boat back
Two cannibals cross.
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sorry, should've mentioned that cannibals cannot pilot the boat
 
Answers

Get another boat.
Swim.

Can there be a canable on a side when no one else is there or do they start to eat them selves?
 
I must misunderstand the rules somehow..

C=Canibal c=Non-canibal

Initially, one C one c must cross. If two or one c cross the remaining will be eaten and zero c isn't an option (no one to use the boat). From there on, the c has to return (if he brought the C that'd be back to square one and nothing else is possible). One c from there would be back to previous state, two c would get the remaining c eaten and one c one C would get the traveling c eaten upon arival at the other side (which would then be two C one c). Unless we're talking some sort of wordplay like the one c staying in the boat with two C on the shore or something to that effect, it's impossible.
 
LarrrSDonald said:
I must misunderstand the rules somehow..

C=Canibal c=Non-canibal

Initially, one C one c must cross. If two or one c cross the remaining will be eaten and zero c isn't an option (no one to use the boat). From there on, the c has to return (if he brought the C that'd be back to square one and nothing else is possible). One c from there would be back to previous state, two c would get the remaining c eaten and one c one C would get the traveling c eaten upon arival at the other side (which would then be two C one c). Unless we're talking some sort of wordplay like the one c staying in the boat with two C on the shore or something to that effect, it's impossible.

Your answer is indeed correct. Congratulations. :approve:
 
hi friends, make the cannibal problem a little complicated. suppose, all the regular men can row the boat. and one out of three canibals can row the boat. the story remains same. now try!

kris
 
ckrishanu said:
hi friends, make the cannibal problem a little complicated. suppose, all the regular men can row the boat. and one out of three canibals can row the boat. the story remains same. now try!


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Here is a brain teaser: What do i need to do to start my onw threads? I can't work it out. PLEASE help.
 
  • #10
Tzar said:
Here is a brain teaser: What do i need to do to start my onw threads? I can't work it out. PLEASE help.

click on the green "new thread" button
 
  • #11
Cheers, I'm such an idiot for not seeing it...
 
  • #12
maybe i can help them solve thru visuals via this link: http://www.plastelina.net/games/game1.html

the logic game is called Cannibals and Missionaries :)
 
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