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Cannibals and Missionaries


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#51 the viceroy

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Posted 11 June 2008 - 04:37 AM

A waaaaay easier way to do it would be to take 1 cannibal and one missionary to the other side, drop them both off and then go back, and then do the same thing 2 more times (send the boat back by pushing it)


Spoiler for really, it won't work and here's why

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#52 the viceroy

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Posted 11 June 2008 - 05:28 AM

either i misunderstood the question or the fact that all 6 of them must get across the river should be stated. the question just says all 6 of them "wants" to get across. didn't really specify that all 6 of them has to get across alive. I just took it the different way. though i did kind of eliminate the tragedy part of the question....
cannibal 1 eats cannibal 3 (3 missionary, 2 cannibal).
cannibal 2 eats missionary 3 (2 missionary 2 cannibal).
cannibal 1 eats cannibal 2 for eating missionary 3 (2 missionary, 1 cannibal)
missionary 2 tries to escape and gets eaten by cannibal 1 too. (1 missionary 1 cannibal)
cannibal 1 and missionary 1 get on the boat and cross the river happily..... the end


Spoiler for comeon read the OP!!!

Edited by the viceroy, 11 June 2008 - 05:29 AM.

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#53 rookie1ja

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Posted 11 June 2008 - 12:53 PM

Cannibals and Missionaries - Back to the River Crossing Puzzles
Three missionaries and three cannibals wanted to get on the other side of a river (Edited: all 6 of them have to get across alive). There was a little boat on which only two of them can fit. There can never be on one side more cannibals than missionaries because of a possible tragedy.


Spoiler for Solution

clarification: when 1 cannibal or missionary returns in the boat and does not have to get out (he stays in the boat), then he is not counted as a person on the side of the river.
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#54 Kla

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Posted 23 June 2008 - 10:52 PM

Who says cannibals are vicious and attack other people at will?
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#55 elixermixer

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Posted 03 July 2008 - 09:10 PM

either i misunderstood the question or the fact that all 6 of them must get across the river should be stated. the question just says all 6 of them "wants" to get across. didn't really specify that all 6 of them has to get across alive. I just took it the different way. though i did kind of eliminate the tragedy part of the question....
cannibal 1 eats cannibal 3 (3 missionary, 2 cannibal).
cannibal 2 eats missionary 3 (2 missionary 2 cannibal).
cannibal 1 eats cannibal 2 for eating missionary 3 (2 missionary, 1 cannibal)
missionary 2 tries to escape and gets eaten by cannibal 1 too. (1 missionary 1 cannibal)
cannibal 1 and missionary 1 get on the boat and cross the river happily..... the end



I must say, although I have not taken the time to solve this particular crisis. I love this guys insight to the problem, not to mention his sense of humor. It is obvious that cannibals have no respect for our laws or ways of doing things, not to mention the fact that they are apparently voracious eaters. After all, cannibal 1 has already eaten 2 fellow cannibals and a missionary. We can only hope this will tide him over long enough to get the last missionary across the river. If not than we are down to just one very full cannibal. And who cares what side of the river he's on. :D
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#56 Charlie87

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Posted 09 November 2008 - 10:40 PM

Ay ay ay!!! The reason y no one has gotten the solution is because the riddle was not stated clearly at all!!! and they forgot one very important rule ONLY ONE CANNIBAL KNOWS HOW TO ROW.
Ill give the riddle in a much better way, and I also know the solution, and some of you were close but not quite.

You have 3 cannibals and 3 humans and they all need to cross this river on a little boat that carries only 2 of them at a time, and of course someone has to bring the boat back. So here are the rules:

*only one of the cannibals knows how to row (which you have to keep track of, because some ppl sometimes want to forget what side they left him on)
*there can never EVER be more cannibals than humans on ANY side because the cannibals will eat the humans. There must be an EQUAL or HIGHER amount of humans than cannibals in order for the cannibals not to attack.
*Only 2 at a time and one has to bring the boat back.

And a little reminder... you can't have say for instance, one human takes one cannibal, he comes back, he takes another cannibal, then comes back. That can't happen because as soon as they get to the other side it becomes 2 cannibals on one human. It doesn't matter if he gets off the boat or not. There must ALWAYS be the same amount of humans per cannibals or more humans than cannibals.

I mention this because a lot of ppl think they can get away with that, and if that was possible this puzzle would be too easy.

small hint: it's a lot better to use some items to keep track of the ppl. (coins are good)
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#57 Charlie87

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Posted 09 November 2008 - 10:47 PM

clarification: when 1 cannibal or missionary returns in the boat and does not have to get out (he stays in the boat), then he is not counted as a person on the side of the river.

Actually, the person in the boat does count. I don't know where u read the rules, but all ppl count at all times whether their in the boat or not. And what the other guy was saying in his solution would make sense ONLY if all cannibals knew how to row, but only ONE knows how. And the one cannibal that was supposed to bring all the other cannibals back wasn't where he was supposed to be, so it wouldn't work out for his solution that way. If you disregard these circumstances this riddle would be way too easy. Which is why there are rules.
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#58 Charlie87

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Posted 09 November 2008 - 10:53 PM

I came up with a different solution, that seems to work. There are never more cannibals than missionaries at any one time and there are never 2 cannibals together either on one side of the river, with the assumption that a cannibal by itself can't eat anybody (but him/herself) and they don't eat each other crossing the river.

we'll go with m-missionary and c-cannibal:

MC
MC MC--> --

MC
MC <--- M C

MC
M MC-->

MC
M <--C MC

MC CM --> MC

MC <--C MMC

C MC --> MMC

C <--C MMMC

-- CC--> MMMC

-- MMMCCC

Ok if I'm understanding correctly ur saying that one man takes a cannibal, then the man goes back. Man takes another cannibal, then cannibal comes back. Well right there is a problem. 2 cannibals on one human as soon as they arrive. So I am sorry to say, you are incorrect.
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#59 Charlie87

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Posted 09 November 2008 - 10:56 PM

1> C+C and one C return
2> C+C and one C return
now situation is CMMM with boat and CC at destination end

3> M+M will go and C+M return
now CCMM with boat and CM is situation

4> now M+M will go and and C will return

5> C+C go and C return

6> C+C go ...end

Now let me see you do THAT with only one cannibal knowing how to row... you can't. the admin forgot to mention that.
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#60 Charlie87

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Posted 09 November 2008 - 11:07 PM

MMMCCC | |
MMMC   | |CC	 - CC cross the river
MMMCC  | |C	  - C goes back
MMM	| |CCC	- CC cross the river
MMMC   | |CC	 - C goes back
M  C   | |MMCC   - MM cross the river
MM CC  | |M  C   - MC goes back
   CC  | |MMMC   - MM cross the river
   CCC | |MMM	- C goes back
   C   | |MMMCC  - CC cross the river
   CC  | |MMMC   - C goes back
	   | |MMMCCC - CC cross the river

You almost had it, and i kno its not ur fault that u might not know that only ONE cannibal can row. and that cannibal that can row (following ur steps) is left on the left side of the river becuz he was the one to take himself and another cannibal back to the left side. Therefore the cannibal on the right can't take the boat back for the remaining 2.
You are super close though.
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