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# Cannibals and Missionaries

77 replies to this topic

### #11 Writersblock

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Posted 10 July 2007 - 04:57 AM

I misread the problem and therefore struggled a bit more to solve it. I was on the impression that neither the missionaries nor the cannibals can outnumber each other. Below the results:

Given that missionaries are 123 and cannibals abc.

```West				East
123abc

23bc	> 1a		a
< 1
123	 > bc		ab
< c
23	  > 1c		abc   ****
< 1
3	   > 12		12ab
< c
> 3c		123abc```

No, this is wrong. Like many of the others you are assuming that the missionary in the boat is safe, which is not the case. I have noted where the missionary is in trouble with ****.
• 0

### #12 brownester

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Posted 13 July 2007 - 09:56 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
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### #13 Sophiesmommy

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Posted 14 July 2007 - 07:37 AM

i think admin is wrong with taking 2 cannibals over at the same time and expecting on to return, if there is a m on the other side i think that both c's will stay and ,,,,well do what they do.
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### #14 coolkid101

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Posted 19 July 2007 - 05:24 PM

take 2 Cannibals to the right side of the river
then take 2 missionaries to the right and take 1 cannibal to the left,

the right side now has 1 C. and 2 M. and the left side has one cannibal and 1 misionary and there is 1 cannibal one the boat!

then take 1 missionary to the right
and then go back and bring the last 2 cannibals to the right

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### #15 liz5000

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Posted 25 July 2007 - 05:21 AM

c=cannibals
m=missionaries

one m or one c always stays in the boat.
c+m cross the river leaving c
m resturns in boat picking up c and cross the river
m returns in boat picking up m and cross the river but takes one c back with him
m drops off c picks up last m and cross the river
m returns picking c and cross the river
m returns picks up last c and return
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### #16 sourabhtheone

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Posted 27 July 2007 - 07:28 AM

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
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### #17 Jack

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Posted 01 August 2007 - 08:10 PM

Is it just me or are you guys making this far, far too complicated for yourselves?

All six need to get across? There are 3 of each? And just two can fit?

Surely the best way is to just put one cannibal and one missionary in the boat. (one of each on one side two of each on the other.
go back, and again just take one cannibal and one missionary.
Then again return and take the remaining missionary and cannibal.

Simple and easy.
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### #18 Em0_fRe@K

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Posted 02 August 2007 - 07:39 PM

asuming that one canibal has has row back and forth and ferry everyone across. he ids going to get petty mad though
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### #19 collisoo

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Posted 03 August 2007 - 05:15 PM

That makes no sense - no matter what way you start it, there's always got to be one person on the other side and 1>0...
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### #20 Jack

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Posted 04 August 2007 - 12:10 AM

ah, wait, I've made a fatal error ... I've had in my head that there is someone else rowing the boat. (other than the six)

... Now I feel like a right plum!
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