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Cannibals and Missionaries


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77 replies to this topic

#71 Lawjyk

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Posted 19 November 2008 - 02:44 AM

the missionaries starve the cannibals by putting them into a box, alone, then one eats the other two. The missionaries then let him out and a missionary and a cannibal go to one side, leaving two missionaries. The two missionaries then use god's divine power to make themselves another sturdy boat that carries them to the other side. Of course, the missionaries supposedly could just make a boat that could carry 6 people..
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#72 deadwood

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Posted 29 November 2008 - 08:58 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


This is harder if you presume that cannibals are unable to row the boat.
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#73 timi

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Posted 02 December 2008 - 11:08 AM

why cant 1 missionary and 1 cannibal cross at the same time,
there would always be the same amount of cannibals and missionaries on each side at any given time??
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#74 Petter90

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Posted 22 December 2008 - 02:27 AM

why cant 1 missionary and 1 cannibal cross at the same time,
there would always be the same amount of cannibals and missionaries on each side at any given time??

But Who would come back with the boat????? It does not have auto pilot on it. Or does it? :P
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#75 Petter90

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Posted 22 December 2008 - 02:39 AM

But with u do it like this. C= cannibal M= missonary

c+m -----> / <---m left c there/
c+c----->/ <----c left another c there/ ====== there is m m m c / c c
then m+m----->/ <----- m+c come back ====== there is m m c c / c m
then m+ m---->/ <---- c come back ====== there is c c c / m m m (and the boat is with the cannibals)
then c+c------>/ <------ c come back ====== there is c c / m m m c
then c+c------>/ nobody come back ====== there is _ / m m m c c c

Edited by Petter90, 22 December 2008 - 02:40 AM.

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#76 comander

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Posted 23 September 2009 - 02:10 AM

Once again, C=cannibal
M=missionary

M + M go over. M comes back, picks up another M brings it over, then comes back again to pick up C, brings him over. C comes back over and picks up another C, drops him off and comes back again with the last C.

no that is impossible there can NEVERbe more cannibals than missionarys or game over
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#77 savi

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Posted 21 January 2010 - 01:13 PM

Here is the correct answer
0-cannibles x-missionaries

east west
XX
000XXX --------------->
X
000X <--------------- X
XX
000 --------------> X
X
000 <--------------- XX
00
OX ----------------> XX
0X
0X <---------------- 0X
00
XX ----------------> 0X
X
XXX <--------------- 000
XX
X ----------------> 000
x
----------------> 000XX

---------------> 000XXX


Is that simple

Edited by savi, 21 January 2010 - 01:14 PM.

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#78 sikander

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Posted 17 December 2010 - 11:53 AM

Alternative Solution::

Cannibals are X's and Missionaries are O's

pick up two cannibals: in boat XX
leave one cannibal: left side of river X, right side of river X OOO
pick up one missionary: in boat XO
leave missionary: left side of river XO. right side X OO
pick up one missionary: in boat XO
leave missionary: left side of river XOO, right side XO
pick up one missionary: in boat XO
leave missionary: left side of river XOOO, right side of river X
pick up cannibal: in boat XX
leave both cannibals: left side of the river XXXOOO




misonary will die on second stage 2 canibles and one misonary
x was there and an xo went there
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