[rookie1ja]

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.

Spoiler for Solution:

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.

Spoiler for old wording:

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.

[ablissfulgal]

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

[artune]

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.

[Thor7-10]

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.

[cloud9ine]

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.

[almgtop]

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

[dak1530]

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

[aesculapius]

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

[dsu]

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

[stephcorbin]

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