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Cannibals and Missionaries

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Cannibals and Missionaries - Back to the River Crossing Puzzles

Three missionaries and three cannibals want to get to the other side of a river. There is a small boat, which can fit only two. To prevent a tragedy, there can never be more cannibals than missionaries together.

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Cannibals and Missionaries - solution

1 cannibal and 1 missionary there, missionary back. 2 cannibals there, 1 cannibal back. 2 missionaries there, 1 missionary and 1 cannibal back. 2 missionaries there, 1 cannibal back. This one cannibal takes the remaining cannibals to the other side.

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.

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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

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c= canibal

m=missionary

c+m go across

m comes back gets another c

c drops m off and comes back to get another m

m drops of c and comes back for another c

repeat.

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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.

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all answers are wrong according to question.

The question says there should not be more cannibals than missionaries at one place at any time. In all answers, at some time, there is a cannibal (1) and no missionaries (0) failing the condition.

To fit the solution, the question should be rephrased as 'There should not be any missionaries present such that they are outnumbered.

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Yes, I've thought about the "outnumbering" issue myself. However, if there are no Missionaries with the Cannibals, then (perhaps, we can consider) there are no Missionaries to be outnumbered.

I've converted this puzzle to Algebraic Code, it's an Excel file (not allowed). I'll try again later. When I do, I'd appreciate feedback on this for pleasure or education

It's filled with coding and decoding opportunities.

Alan

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abliss has the best possible solution for speed although admin is correct also. at no time are there more cannibals than missionaries other than when there are no missionaries at which point it wouldn't matter because they can't be eaten by the cannibals if they are not there

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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

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Visualisation will help understand the solution:

At the beginning

CCC MMM [boat empty] [other coast empty]

"1 cannibal and 1 missionary there, missionary back"

CC MM......[CM]->......

CC MM ...... <-[M] ...... C

CC MMM ...... [] ...... C

"2 cannibals there, 1 cannibal back"

MMM ...... [CC]-> ...... C

MMM ...... <-[C] ...... CC

C MMM ...... [] ...... CC

"2 missionaries there, 1 missionary and 1 cannibal back"

C M ...... [MM]-> ...... CC

C M ...... <-[MC] ...... C M

CC MM ...... [] ...... C M

"2 missionaries there, 1 cannibal back"

CC ...... [MM]-> ...... C M

CC ...... <-[C] ...... MMM

CCC ...... [] ...... MMM

"This one cannibal takes the remaining cannibals to the other side"

C ...... [CC]-> ...... MMM

C ...... <-[C] ...... C MMM

...... [CC]-> ...... C MMM

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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

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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 ****.

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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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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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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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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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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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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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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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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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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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ok you give the cannibals vegetarian lessons then you wount have any cannibal prodlems

no fine

how says you cant put a rope on the boat so you can send a missionarie and a cannibal to cross. then the Cannibals and Missionaries on the other side pull the boat back and then the next missionarie and cannibal go. then the last missionarie and cannibal but the rope in the boat and cross the river there you go.

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duh!!!!

1m and 1 c

m go back

1m and 1 c

c go back

1m and 1c

c go back

wait!!what the

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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.

This one fails right off the bat. As soon as the two missionaries go over, one missionary is left w/ three cannibals.

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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.
Agreed. The only way this works is assuming that the cannibals are only cannibals while on shore, not in the boat. If they maintained their cannibal tendencies in the boat, here is how the given answer would break down...

Cannibals and Missionaries - solution

1 cannibal and 1 missionary there, missionary back. 2 cannibals there, all three cannibals stay and eat lone missionary.

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This is another one of those non-riddle riddles on this site. There was no question posed, just a few statements made.

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