CAA: Christian Anime Alliance

Well I have another question I have yet to answer and this time it is a logic problem.This is how it goes.You have four missionaries and four cannibles on one side of a river.There is only one boat that can carry only two people at a time.At any time if the cannibles outnumber the missionaries on either side the missionaries are eaten and you fail the problem.At least one person has to come back across the river to drive the boat whether it be cannible or missionary.Two may come back although none can stay in the boat.The object is to get all missionaries and cannibles on the other side of the river with no harm to thim missionaries.There is a diagram below. Please help need help with this one.

MMMM R (other side of the river)
CCCC I
V
E
R

My teachers are ruthless when it comes to these type of questions. I am always feeling dumb when confronted with questions like these.>_<

i will figure it out for you.

I just have to do it tommorrow because it is 5.36 am.

I love logic problems!!!!


~Natsumi Lam~

I've figured it out, but since Natsumi got here first. I will wait to see if Natsumi can figure it out. Never mind I forgot about no one staying on the boat. I have to go back to the drawing board!

I was thinking may be a cannible could drive the boat and than drop off missionary than cannible and so on.

m c m c m c m at the destination
with c in the boat

Man this is hard. I think I will give up since I put my two cents in.

easy....

1. bring two Cannibals to the other side of the river
MMMMCC_______________CC

2. Have the Cannibal Go back
MMMMCCC___________________C

3. Have 2 cannibals go to the right
MMMMC____________________CCC

4. Have One cannibal go back to the left, and have 2 missionaries get on
MMCC ________________________MMCC

5. Have the 2 missionaries go back
MMMMCC_________________________CC

6. Have Missionary tie rope to boat
MMMMCC-----------|=>_______________CC

6. Have 2 cannibals go to the right
MMMM------------------------|=> ______________CCCC

7. Have missionaries pull the rope and the boat back...

MMMM-------|=>_____________________CCCC <ANGRY!>

8. Have missionaries pull out sniper rifles and kill cannibals
M--____________________________x_x
M--___________________________________x_x
M--____________________________________x_x
M--_______________________________________x_x
|=>

9. Have 2 missionaries go to the right, then bring one back, then pick up another, go to the other side, then go back, and pick up the last guy

you win

I tried ^^;; but I couldnt get it

Actually, from 4:
MMCC ________________________MMCC

5: Cannibal goes back and switches with both remaining missionaries, who go to the right (this should be safe because there are no longer any missionaries on the starting side)
CCC ________________________MMMMC

6: Missionary goes left and retrieves cannibal
CC ________________________MMMMCC

7: Missionary goes left and retrieves cannibal
C ________________________MMMMCCC

8: Missionary goes left and retrieves cannibal
________________________MMMMCCCC

9: ???

10: Profit!


...assuming that it's only the missionaries on that one side who would get eaten

I don't understand "none can stay in the boat," but I'm assuming that means that the people in the boat count in the ratio of cannibals to missionaries, so if you send one missionary over to a side with two or more cannibals, he'll get eaten. With that assumption, Kaligraphic's solution won't work.

If that's the case, the problem has no solution (although MSP's does have some promise) unless we can assume something like one of the following cases:
1) One of the Missionaries is Christ himself (and can thus walk on water).

2) If the missionaries out number the Cannibals two to one, the one cannibal is converted to a missionary.

3) Cannibals left alone on one bank or the other will wander off (or eat each other).

4) Three cannibals will not eat a single missionary because he either isn't enough to fill them up or they fear the three missionaries on the other side will do something to the single Cannibal.

Kaligraphic wrote:Actually, from 4:
MMCC ________________________MMCC

5: Cannibal goes back and switches with both remaining missionaries, who go to the right (this should be safe because there are no longer any missionaries on the starting side)
CCC ________________________MMMMC

Imposkible... so on the left there would be two missionaries and 2 cannibals, and 1 going... so the missionaries would be outnumbered

Only if the boat is not a separate container or the switch is not simultaneous.

If you count the return trips as separate steps with everyone exiting between steps, then you fail either the first time you move a missionary or the move immediately after.

Given this, MS Pants' solution breaks down at step 4 because the only possible move would be to send back both missionaries, and you cannot replace them with cannibals or the boat won't be large enough to balance them in a single trip. You can't have both sides balanced or emptied at the end of each move.

You sure it's four of each? Because there's a common puzzle that's basically the same situation but with only three of each.

That's because the puzzle can be solved for three each. With four of each you are right, you can never keep the missionaries from being out numbered.

Kaligraphic wrote:Only if the boat is not a separate container or the switch is not simultaneous.

If you count the return trips as separate steps with everyone exiting between steps, then you fail either the first time you move a missionary or the move immediately after.

Given this, MS Pants' solution breaks down at step 4 because the only possible move would be to send back both missionaries, and you cannot replace them with cannibals or the boat won't be large enough to balance them in a single trip. You can't have both sides balanced or emptied at the end of each move.

You sure it's four of each? Because there's a common puzzle that's basically the same situation but with only three of each.

I understand that if it were MMCC on both sides, the only choice would to bring back both Missionaries... i just put that because I couldnt solve it and gave up... so i let the missionaries turn bad and shoot the cannibals

But if it WERE MMCC... and a cannibal were to come to the left to switch with the two remaining M's... the cannibal could simply jump out and it would be MMCCC.... soon to be xx CCC <yummy>

I know the other puzzle, its like a farmer, a fox, and a rabbit, and carrots or something... if the farmer is not there the fox will eat the rabbit. and the rabbit will not eat the carrots. And the boat can only hold 2

or something like that, I dont remember. But the first variation ive heard I pretty much know how to solve (if i can remember it!)

I googled this..... basically, if it's 4 of each it can't be solved but it can be with 3 of each.

Anime Dad wrote:I googled this..... basically, if it's 4 of each it can't be solved but it can be with 3 of each.

So does that mean it's a trick question.

battletech wrote:So does that mean it's a trick question.

I guess so.

I TOLD YOU! SHOOT THE CANNIBALS! Gosh pay attention to me!

battletech wrote:I've figured it out, but since Natsumi got here first. I will wait to see if Natsumi can figure it out. Never mind I forgot about no one staying on the boat. I have to go back to the drawing board!

you can just say.. its all good.. i have been busy with family issues.

~NL~

Mr. SmartyPants wrote:I TOLD YOU! SHOOT THE CANNIBALS! Gosh pay attention to me!

or get them saved.... then you can all be missionaries!!!


Thats the God factor....he can save anyone

~NL~

Interesting, but it does become logically impossible if we assume that every individual moment must satisfy M >= C. However, considering that in reality a person can remain in a boat, we're operating from faulty assumptions.

Still, I'll keep this in mind the next time I'm among four missionaries with four cannibals, all needing to cross a river in one small boat.

uc pseudonym wrote:Interesting, but it does become logically impossible if we assume that every individual moment must satisfy M >= C. However, considering that in reality a person can remain in a boat, we're operating from faulty assumptions.

Still, I'll keep this in mind the next time I'm among four missionaries with four cannibals, all needing to cross a river in one small boat.

But can you depend on a cannibal to stay in the boat? The cannibal would likely look at out numbering the missionaries as a possitive thing. You can't count on them to stay in the boat to maintain a certain ratio.

But we're all working on one assumption...that the cannibals are hungry.

If the cannibals are currently well-fed, they'd have no reason to eat the missionaries.

So get some full cannibals and there's no problem.

Or what about vegetarian cannibals? Then your problem would be solved from the start.

termyt wrote:But can you depend on a cannibal to stay in the boat? The cannibal would likely look at out numbering the missionaries as a possitive thing. You can't count on them to stay in the boat to maintain a certain ratio.

Actually, I said that with the missionaries in mind. For example, if one missionary goes over to a side with two cannibals, he doesn't have to get out and be eaten, he could probably take just one (or stay away from the shore). That, of course, begs another question: why are the missionaries trying to get across the shore with these four cannibals?

They need to get the cannibals back to the missionary camp so that they can cook and eat them - because they are CANNIBAL MISSIONARIES!!!

Because that's where they left their rifles and they need to cleanse the Earth of these foul beings.

uc pseudonym wrote:Or what about vegetarian cannibals? Then your problem would be solved from the start.

um, my understanding is that cannibals eat human flesh, so this makes it a bit hard lol.

PS I know you're joking!