[Guest]

There is three chicks and three wolves And they all need to get across. You can take only 2 at a time and there cant be more wolves then chick on the same side

[Guest]

Assuming that there can be the same number of chicks and wolves going at the same time:

C,W go across.

C,W go across.

C,W go across.

That's the most basic one.

Here's another one:

W,W go across.

C,C go across.

C,W go across.

In my opinion, there are many ways that can work like these.

Edit: Clarification - can the number of chicks be equal to the number of wolves, or does it have to have more chicks than wolves??

Edited July 6, 2008 by SillouhetteMind

[Guest]

I think this has already been posted already.

[Guest]

I think this has already been posted already.

I agree, I have heard this one numerous times. Try searching more, sagekid. :|

Edited July 6, 2008 by SillouhetteMind

[unreality]

SM: your solution doesn't work, as someone would have to come back with the boat. If it was the chick, there would be more wolves than chicks on the far side (1 > 0) and if was the wolf, there would be more wolves than chicks on the starting side (3 > 2)

[Guest]

SM: your solution doesn't work, as someone would have to come back with the boat. If it was the chick, there would be more wolves than chicks on the far side (1 > 0) and if was the wolf, there would be more wolves than chicks on the starting side (3 > 2)

The problem never stated anything about a boat. It just said that only two people went to the other side at a time. They could have chosen that two went at a time just for the sake of the problem. It didn't say one had to come back.

Edited July 6, 2008 by SillouhetteMind

[Guest]

Yeah, but I think that's the idea of the problem. That's what it usually is, anyway. Where's the challenge in getting your answer, anyway?

[unreality]

actually, ALL solutions are impossible

we've already shown how W,C is impossible

now if W,W go across, that leaves more wolves on the opposite shore, obviously that can't be allowed.

So the only remaining one is C, C

however if two Chicks leave the original side, that leaves more wolves there

everything is impossible!!! lol

[Guest]

actually, ALL solutions are impossible

we've already shown how W,C is impossible

now if W,W go across, that leaves more wolves on the opposite shore, obviously that can't be allowed.

So the only remaining one is C, C

however if two Chicks leave the original side, that leaves more wolves there

everything is impossible!!! lol

There can be more wolves on the oppisite shore, as long as there are no chicks right? So what about

*Means coming back on boat

w,w

w*

w,c

w*

One chick one wolf so far on the shore

w,c

w*

Chick hops on boat before wolf gets off...still 1-1 ratio right?

w,c

w*

w,w

[unreality]

There is three chicks and three wolves And they all need to get across. You can take only 2 at a time and there cant be more wolves then chick on the same side

[Guest]

SM: your solution doesn't work, as someone would have to come back with the boat.

You would come back with the boat.

You can take only 2 at a time

If that's not what the OP meant and the wolves and chicks travel on there own, then the solution can be found in this thread.

If it was the chick, there would be more wolves than chicks on the far side (1 > 0)

Wolves outnumbering chicks when the number of chicks is zero is generally okay in these type of riddles.

[unreality]

...? but they still outnumber them. 1 is bigger than 0

ohhhhh but I see! The OP never actually stated that the wolves would eat the chicks if there were more, he just said it isn't allowed. If he had clarified and said that, I think I would've understood thanks, Martini

[Guest]

Yes, if the number of chicks versus the number of wolves was equal and couldn't work, then unreality is right. There is no solution that can make sure that there is more chicks on one side than wolves. Therefore, you cannot solve this riddle in a way that will meet all the criteria required.

Edited July 6, 2008 by SillouhetteMind

[unreality]

no, you can now, since if there are 0 girls, it's okay to have wolves

thus, the solution must be:

2 wolves go across

1 wolf goes back

(3g2w, 1w)

2 girls go across

1 girl comes back

(2g2w, 1g1w)

but from that point, it's impossible. One more girl than wolf has to go across now, but if both go across, one has to go back or not everyone will get across- but if one girl goes across she will be meeting 2 wolves, not allowed. So one girl and one wolf have to go, and 1 girl has to go back- but this makes 2 wolves and 1 girl on the opposite shore. But if the wolf went back instead, there would be a wolf majority on the original side

hmmm, if it is solvable, then I messed up somewhere in the second step

but I don't think this riddle has a solution at all ;D

[Guest]

no, you can now, since if there are 0 girls, it's okay to have wolves

thus, the solution must be:

2 wolves go across

1 wolf goes back

(3g2w, 1w)

2 girls go across

1 girl comes back

(2g2w, 1g1w)

but from that point, it's impossible. One more girl than wolf has to go across now, but if both go across, one has to go back or not everyone will get across- but if one girl goes across she will be meeting 2 wolves, not allowed. So one girl and one wolf have to go, and 1 girl has to go back- but this makes 2 wolves and 1 girl on the opposite shore. But if the wolf went back instead, there would be a wolf majority on the original side

hmmm, if it is solvable, then I messed up somewhere in the second step

but I don't think this riddle has a solution at all ;D

In your second step, if two girls go across then that leaves one girl and two wolves on the starting shore.

[unreality]

oh yeah ;D

[Guest]

Yes It can have more chicks but u r rong it needs some thing on it to go back sorry forgot to clarrify

[Guest]

W=Wolf

C=Chick

A: [WW] [W] [WW] [W] [CC] [WC] [CC] [W] [WW] [W] [WW] = 11 moves

B: [WW] [W] [WW] [W] [CC] [WC] [CC] [W] [WW] [C] [WC] = 11 moves

[Guest]

The riddle has already been posted using missionaries and cannibals (linked to in my earlier post) and solved in nine moves.