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Cannibals and Missionaries

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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)

ye, I'd like to see you push a boat to the other shore of a fast running river. In all probability, unless you're very lucky, the boat will be pulled along downsteam by the fast current. Bet you've never seen a river or one with a fast current, at any rate.

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Posted (edited) · Report post

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

it says in parenthesis, ok maybe you didn't read them,but

you have to read the whole OP to get the riddle or puzzle right, it says clearly that all 6 need to get there alive.

Edited by the viceroy
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Posted · Report post

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.

clarification: when 1 cannibal or missionary returns in the boat and does not have to get out (he stays in the boat), then he is not counted as a person on the side of the river.

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Posted · Report post

Who says cannibals are vicious and attack other people at will?

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Posted · Report post

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

I must say, although I have not taken the time to solve this particular crisis. I love this guys insight to the problem, not to mention his sense of humor. It is obvious that cannibals have no respect for our laws or ways of doing things, not to mention the fact that they are apparently voracious eaters. After all, cannibal 1 has already eaten 2 fellow cannibals and a missionary. We can only hope this will tide him over long enough to get the last missionary across the river. If not than we are down to just one very full cannibal. And who cares what side of the river he's on. :D

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Posted · Report post

Ay ay ay!!! The reason y no one has gotten the solution is because the riddle was not stated clearly at all!!! and they forgot one very important rule ONLY ONE CANNIBAL KNOWS HOW TO ROW.

Ill give the riddle in a much better way, and I also know the solution, and some of you were close but not quite.

You have 3 cannibals and 3 humans and they all need to cross this river on a little boat that carries only 2 of them at a time, and of course someone has to bring the boat back. So here are the rules:

*only one of the cannibals knows how to row (which you have to keep track of, because some ppl sometimes want to forget what side they left him on)

*there can never EVER be more cannibals than humans on ANY side because the cannibals will eat the humans. There must be an EQUAL or HIGHER amount of humans than cannibals in order for the cannibals not to attack.

*Only 2 at a time and one has to bring the boat back.

And a little reminder... you can't have say for instance, one human takes one cannibal, he comes back, he takes another cannibal, then comes back. That can't happen because as soon as they get to the other side it becomes 2 cannibals on one human. It doesn't matter if he gets off the boat or not. There must ALWAYS be the same amount of humans per cannibals or more humans than cannibals.

I mention this because a lot of ppl think they can get away with that, and if that was possible this puzzle would be too easy.

small hint: it's a lot better to use some items to keep track of the ppl. (coins are good)

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Posted · Report post

clarification: when 1 cannibal or missionary returns in the boat and does not have to get out (he stays in the boat), then he is not counted as a person on the side of the river.

Actually, the person in the boat does count. I don't know where u read the rules, but all ppl count at all times whether their in the boat or not. And what the other guy was saying in his solution would make sense ONLY if all cannibals knew how to row, but only ONE knows how. And the one cannibal that was supposed to bring all the other cannibals back wasn't where he was supposed to be, so it wouldn't work out for his solution that way. If you disregard these circumstances this riddle would be way too easy. Which is why there are rules.

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Posted · Report post

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

Ok if I'm understanding correctly ur saying that one man takes a cannibal, then the man goes back. Man takes another cannibal, then cannibal comes back. Well right there is a problem. 2 cannibals on one human as soon as they arrive. So I am sorry to say, you are incorrect.

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Posted · Report post

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

Now let me see you do THAT with only one cannibal knowing how to row... you can't. the admin forgot to mention that.

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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

You almost had it, and i kno its not ur fault that u might not know that only ONE cannibal can row. and that cannibal that can row (following ur steps) is left on the left side of the river becuz he was the one to take himself and another cannibal back to the left side. Therefore the cannibal on the right can't take the boat back for the remaining 2.

You are super close though.

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Posted · Report post

Now let me see you do THAT with only one cannibal knowing how to row... you can't. the admin forgot to mention that.

the admin did not forget to mention anything ... in the spirit of other river crossing puzzles the wording of this puzzle is sufficient

btw, if you want to quote several posts, you can do that in one single reply and not increasing your post count by replying to more posts in several posts

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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.

Nooooo, I can't believe even the ADMIN got it wrong!!!! thats not the answer. Only ONE cannibal knows how to row and again, he's on the wrong side!!! Thats gotta be embarrassing.

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Posted · Report post

the admin did not forget to mention anything ... in the spirit of other river crossing puzzles the wording of this puzzle is sufficient

btw, if you want to quote several posts, you can do that in one single reply and not increasing your post count by replying to more posts in several posts

Oh ok, i didnt know that. I wasn't trying to make my post count higher though. But yo, only one cannibal knows how to row man, and I do have the correct solution for that.

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Posted · Report post

Oh ok, i didnt know that. I wasn't trying to make my post count higher though. But yo, only one cannibal knows how to row man, and I do have the correct solution for that.

I read your topic before it was deleted and I found the solution to the situation you mentioned (3 human - 3 cannibals - only 1 cannibal can row)

There are 3 cannibals: C1 C2 C3

3 humans H1 H2 H3

Able to row: H1 H2 H3 C3

Solution:

Boat H1 C1

Leaves C1 on the other side of the river

H1 returns

Boat C3 C2

Leaves C2 on other side

C3 returns

Boat H1 H2

Leaves H1

H2 returns

Boat H2H3

Leaves H2

H3 returns

Boat C3H3

All alive, on the other side of the river

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Posted · Report post

I read your topic before it was deleted and I found the solution to the situation you mentioned (3 human - 3 cannibals - only 1 cannibal can row)

There are 3 cannibals: C1 C2 C3

3 humans H1 H2 H3

Able to row: H1 H2 H3 C3

Solution:

Boat H1 C1

Leaves C1 on the other side of the river

H1 returns

Boat C3 C2

Leaves C2 on other side

C3 returns

Boat H1 H2

Leaves H1

H2 returns

Boat H2H3

Leaves H2

H3 returns

Boat C3H3

All alive, on the other side of the river

Nope, sorry. Your second step leaves one human with 2 cannibals on the other side. All that would be too easy.

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Nope, sorry. Your second step leaves one human with 2 cannibals on the other side. All that would be too easy.

Oh yeah I see >.<

The fault was made due solving it too quickly and not paying enough attention.. Spent about 3 mins to solve it, explains a lot I guess..

Anyway, I can't seem to find the answer, I'd say it's impossible if only 1 cannibal knows how to row, unless you can give me the solution.

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Oh yeah I see >.<

The fault was made due solving it too quickly and not paying enough attention.. Spent about 3 mins to solve it, explains a lot I guess..

Anyway, I can't seem to find the answer, I'd say it's impossible if only 1 cannibal knows how to row, unless you can give me the solution.

I could provide the solution, and i can assure there IS a solution, but if i tell you... what fun would that be? It took me for EVER to figure it out but i finally did. I knew of this a couple of years ago, i had given up on it quite a few times, but always took a stab at it every now and then.

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Posted · Report post

Nooooo, I can't believe even the ADMIN got it wrong!!!! thats not the answer. Only ONE cannibal knows how to row and again, he's on the wrong side!!! Thats gotta be embarrassing.

would you kindly show me in which step I made the fatal error?

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.

the implied information:

1. each cannibal and each missionary can row

2. whenever there are more cannibals than missionaries on one bank, the missionaries on that bank will be killed

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.

the same in numbered steps:


1. CCMM ... CM > ... 0
2. CCMM ... < M ... C
3. MMM ... CC > ... C
4. MMM ... < C ... CC
5. MC ... MM > ... CC
6. MC ... < MC ... CM
7. CC ... MM > ... CM
8. CC ... < C ... MMM
9. C ... CC > ... MMM
10. C ... < C ... MMMC
11. 0 ... CC > ... MMMC
end ... MMMCCC
0. CCCMMM 

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Posted · Report post

I could provide the solution, and i can assure there IS a solution, but if i tell you... what fun would that be? It took me for EVER to figure it out but i finally did. I knew of this a couple of years ago, i had given up on it quite a few times, but always took a stab at it every now and then.

Hmm yeah true that, I'll take another look at it today.. Actually this could be the same puzzle a friend of my moms gave me a few years ago, since it was also a river cross puzzle which at first sight seemed impossible, and after searching for a loooong time, I finally found it..

would you kindly show me in which step I made the fatal error?

There is no error in your solution, but I think what charlie was pointing out, is that ALL the cannibals can row in your situation, in charlies situation, only 1 cannibal can row, which makes it a lot harder to find the solution >.<

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would you kindly show me in which step I made the fatal error?

Well, what maggi said is correct. There actually is no error in your solution given the fact that they don't state that only one cannibal knows how to row in the site where you got this puzzle from. Maybe there's different versions of it and i was aware of only the version i knew. I guess it just kinda made me mad that this one was simpler, and I was stuck on the idea that only one knows how to row, which does make the problem a lot harder.

I suppose I can apologize for saying that you were wrong for your particular version of this puzzle. But it would be nice if you solved the harder version :). Also, if you try to do it with the only-one-rowing-cannibal version you would then see where the "fatal error" would be at that I was talking about. I'm sure it wont take you long to figure this one out since you already did the other one :).

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Posted · Report post

the missionaries starve the cannibals by putting them into a box, alone, then one eats the other two. The missionaries then let him out and a missionary and a cannibal go to one side, leaving two missionaries. The two missionaries then use god's divine power to make themselves another sturdy boat that carries them to the other side. Of course, the missionaries supposedly could just make a boat that could carry 6 people..

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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.

This is harder if you presume that cannibals are unable to row the boat.

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Posted · Report post

why cant 1 missionary and 1 cannibal cross at the same time,

there would always be the same amount of cannibals and missionaries on each side at any given time??

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Posted · Report post

why cant 1 missionary and 1 cannibal cross at the same time,

there would always be the same amount of cannibals and missionaries on each side at any given time??

But Who would come back with the boat????? It does not have auto pilot on it. Or does it? :P

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Posted (edited) · Report post

But with u do it like this. C= cannibal M= missonary

c+m -----> / <---m left c there/

c+c----->/ <----c left another c there/ ====== there is m m m c / c c

then m+m----->/ <----- m+c come back ====== there is m m c c / c m

then m+ m---->/ <---- c come back ====== there is c c c / m m m (and the boat is with the cannibals)

then c+c------>/ <------ c come back ====== there is c c / m m m c

then c+c------>/ nobody come back ====== there is _ / m m m c c c

Edited by Petter90
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