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Cannibals and Missionaries

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C= Cannibal

M= Missionary

/x/ = River

/x/CCC MMM

/CC/MMMC <

C/C/MMMC >

C/CM/MMC <

CM/C/MMC >

CM/CM/CM <

CMM/C/CM >

CMM/CM/C <

CMMM/C/C >

CMMM/CC/ <

CCCMMM/X/ FINISHED

THIS IS THE ANSWER. READ CAREFULLY AND YOU WILL UNDERSTAND

NOBODY ON ONE SIDE, NOBODY IN RIVER, ALL ON ONE SIDE

TWO CANNIBALS GO

ONE CANNIBAL DROPPED OFF, ONE GOES BACK

CANNIBAL ON ONE SIDE.CANNIBAL AND MISSIONARY IN RIVER, TWO MISSIONARY AND ONE CANNIBAL OTHER SIDE

C&M ONE SIDE, C IN R, TWO M AND ONE CANNIBAL OTHER SIDE

C&M ONE SIDE, C & M IN R, C & M OTHER SIDE

C & TWO M ONE SIDE, C IN R, C & M OTHER SIDE

C & TWO M ONE SIDE, C & M IN R, C OTHER SIDE

C& THREE M ONE SIDE, C IN R, C OTHER SIDE

C & 3 M ONE SIDE, TWO C IN R, NOBODY OTHER SIDE

3 C & 3 M ONE SIDE, NOBODY IN R, NOBODY OTHER SIDE

WAMMY! THAT'S IT!

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Posted · Report post

Cannible = ME

Missionary = YOU

You and i cross the river leaving two of me and two of you. I drop you off on the other side of the river. I go back and pick up a another me, leaving two of you and one of me. I drop myself off. Leaving only one of each of us. I go back and pick up you. We start crossing, leaving one me and two you. I drop you off and go back. i pick myself up and cross the river dropping myself off. there are two me and two you. I go back and pick you up the last time. no one is left. We get across and get off. leaving the boat... everyone there...!

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Posted · Report post

/x/CCC MMM

/CC/MMMC <

C/C/MMMC >

C/CM/MMC < once the CM arives <- you have 2 CC to 1 M = dead/eaten M

CM/C/MMC >

CM/CM/CM <

CMM/C/CM >

CMM/CM/C <

CMMM/C/C >

CMMM/CC/ <

CCCMMM/X/ FINISHED

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Posted · Report post

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

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Posted · Report post

Convert the Cannibals!!

Just kidding! I like the first solution. As for the question of wording, the riddle states that the point of not having the cannibals outnumber the missionaries is so the cannibals don't eat the missionaries. Therefore it's obvious that having 1 cannibal and 0 missionaries is safe!

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Posted · Report post

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)

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Posted · Report post

Most of your answers are terribly wrong...

The admin's is correct.

ablissfulgal said:

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

The bold part is where you would have a missionary eaten. What you’ve got is a cannibal on the left side as well as a cannibal and missionary in the boat. Once the boat gets to the left side to drop the missionary off you’ve got 2 cannibals and 1 missionary on the left side. (When the boat is on the left, you count all in it towards the left sides totals. Same if it’s on the right.)

Graphically, missionary about to be dropped off:

(can)_[(can)(mis)]………….._(can)(mis)(mis)

^

Missionary is outnumbered in the dropoff. Gonna get eaten.

The admin’s solution never has that scenario. This is the admin’s….

____.................................<-[(can)(mis)]_(can)(can)(mis)(mis)

(can)_.[(mis) ]->……….…...……..............……_ (can)(can)(mis)(mis)

(can)_................................<-[(can)(can)]_(mis)(mis)(mis)

(can)(can)_[(can)]->…….…………...............…_(mis)(mis)(mis)

(can)(can)_........................<-[(mis)(mis)]_(can)(mis)

(can)(mis)_[(can)(mis)]->…….…….........…..._(can)(mis)

(can)(mis)_........................<-[(mis)(mis)]_(can)(can)

(mis)(mis)(mis)_[(can)]->…….…….........…..._(can)(can)

Now all the missionaries are safe and across so the cannibal left with the boat can ferry the rest of the cannibals. At no point in time is a missionary with a greater number of cannibals (even including those sitting in the boat!)

artune, Thor7-10, stephcorbin, brownester, coolkid101, liz5000, seqee girl, theirish2121, and MangaMeggie have all made this mistake.

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Posted (edited) · Report post

They both start on the same side of the island right!! ok so ---

MM

----(MC)>->- MISSIONARY CARRIES CANIBAL ACROSS

CC

-------------------------

MM

-<-<-(M)----- AND RETURNS

CC.......................C

-------------------------

MM

----(MC)->->- MISSIONARY CARRIES SECOND CANIBAL ACROSS

C.........................C

------------------------

MM

-<-<-(M)----- AND RETURNS

C.........................CC

--------------------------

M

-----(MM)->->- MISSIONARY CARRIES MISSIONARY ACROSS

C........................CC

--------------------------

M....................... M

-<-<-(MC)----- AND RETURNS WITH A CANIBAL

C........................C

--------------------------

..........................M

-----(MM)->->- MISSIONARY PICKS UP THIRD MISSINARY AND CARRIES ACROSS

CC......................C

-------------------------

...........................MM

-<-<-(M)----- AND RETURNS

CC......................C

-------------------------

.........................MM

-----(MC)->->- MISSIONARY CARRIES CANNIBAL ACROSS

C.......................C

-------------------------

........................MM

-<-<-(M)----- AND RETURNS

C.......................CC

--------------------------

..........................MM

-----(MC)->->- MISSIONARY CARRIES THIRD AND FINAL CANIBAL ACROSS

.........................CC

--------------------------

..........................MMM

..........................CCC

YAY!!!!

Edited by Emapher
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Posted (edited) · Report post

C=Cannibal M=missionary. One of the cannibals drives everyone over...first a M, then C, M, C then M again

Edited by Dana
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Posted · Report post

This one is pretty simple too.

Firts, two cannibles go to the other side, and one returns.

Then the cannible and one missionarie go to the other side the missionarie stays on that side, and the cannible returns.

Next the cannible and the last missionarie go to the other side, the missionarie again stays on that side, and the cannible goes back and brings the third cannible to the other side. B))

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Posted · Report post

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.

Okay, now we're just getting picky... we can also read the part of the question that state "to avoid a potential tragedy" which inherently implies that 0 missionaries means 0% of a tragedy. Unless you want to consider that a missionary has to be present to keep the canibals from eating one another, but that eliminates all possible solutions.

Ovecomplicating things hardly ever makes them easier. :D

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In all actuality the solution is quite simple but is still outside of the box. They tie to each end of the boat a rope, then you send one cannibal and one missionary over to the other side of the river. When they get out of the boat the others pull the boat back to them and once again one cannibal and one missionary get on board. they continue doing this until they are all on the other side of the river. No breaks in numbers and no tragedies.

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Posted · Report post

two cannibals there

no one back

two missionaries there

no one back

one cannibal and one missionary there

the end!

At no point did the cannibals outnumber the missionaries.

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Okay, now we're just getting picky... we can also read the part of the question that state "to avoid a potential tragedy" which inherently implies that 0 missionaries means 0% of a tragedy. Unless you want to consider that a missionary has to be present to keep the canibals from eating one another, but that eliminates all possible solutions.

Ovecomplicating things hardly ever makes them easier. :D

1 missionary and 1 cannibal over

no one back

1 missionary and 1 cannibal over

no one back

1 missionary and 1 cannibal over

the end!

At no point did the cannibals outnumber the missionaries. Nor were the cannibals left to their own devices and allowed to eat each other.

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forget the two i previously mentioned, I was under the impression someone other than the missionaries and cannibals would be opperating the boat. the only way my other two would work would be to tie the boat to a rope and pull it back and forth between the sides of a river... so disregard my other posts on the issue... I just woke up :P

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Posted · Report post

one of each on each boat trip across

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another thing just to point out

if a CM goes over the first time, and M comes back

then another CM goes over, momentarily there would be 2 Cs and 1 M so would that not be game over too?

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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.

Thor7-10 by picking up 2 "M" you would leave 1 "M" with 3 "C" which fails the question

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this is easy. the question states that the should never be more cannibals than missionary but says nothing about having more missionarys!! it is also

assumed that there is always one in boat.

m = missionary

c = cannibal

one m and one c go across to left

one m comes back

(1 c on left and 2 c and 2 m on right, 1 m in boat)

one m and one c go across to left

one c comes back

(1 c and 1m on left and 2 m and 1 c on right, 1 c in boat)

one c and one m go across

one c comes back

(1 c and 2 m on the left and 1 c and 1 m on right, 1 c in boat)

one c and one m go across

one m comes back

(2 c and 2m on left and 1 c on right, 1 m in boat)

one c and one m go across

all c and m on left

one c and one m go across to left

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yeah :D

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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.

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.

I think got the idea of the question but is it worded wrong? It starts out with an even number of cannibals and missionaries, so why cant it be that way on the other side. It says that there cant be more cannibals so this is the easy thing I cane up with... what do you think?

CCC 1C and 1M go to side two, the boat returns to side one to grab another C and M, and so on. The numbers are always equal.

MMM

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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.

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.

I'm sorry but the question says there can never be more cannibals than missionaries on one side so if that means that these possible outcomes fit the question...then all you would have to do is take MC every time simple.

MM M

CC C

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let cannibals be A and the missinaries be M

so first one A comes with one M than A leaves M and comes back Where A gets down n two of M cross the river where one M stays and the other M goes back and than bring one A with him where the A goes back aned Brings another A with him than drops A and goes and picks the last A and they all cross the river

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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.

sorry the OP says clearly side indicating that in the boat there could be 1 cannibal and 0 missionaries.

otherwise I think it's unsolveable

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one of each on each boat trip across

which is this:

after the first batch of missionary/cannibal gets to the other side you have to find a way to get the boat back to the first side, remember: there are only 2 seats in boat, so there cannot be a third person manning the boat!!! It's important to read the OP carefully before you answer (me included)

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