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Cannibals and Missionaries


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77 replies to this topic

#31 jmllilshortman

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Posted 08 December 2007 - 01:43 PM

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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#32 Yoshia

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Posted 30 January 2008 - 08:27 PM

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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#33 Emapher

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Posted 05 February 2008 - 07:21 PM

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, 05 February 2008 - 07:25 PM.

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#34 Dana

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Posted 08 February 2008 - 05:52 PM

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

Edited by Dana, 08 February 2008 - 05:56 PM.

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#35 courtney ohlmann

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Posted 07 March 2008 - 02:19 AM

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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#36 txskydiver

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Posted 14 March 2008 - 10:06 PM

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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#37 Black_Silence

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Posted 20 March 2008 - 01:49 AM

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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#38 amandadeemcminn

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Posted 21 March 2008 - 10:16 PM

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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#39 amandadeemcminn

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Posted 21 March 2008 - 10:20 PM

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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#40 amandadeemcminn

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Posted 21 March 2008 - 10:52 PM

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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