Stumbled across this fiendish puzzle — I don’t have the patience to work it out, anyone else do better than that? (Without writing a program to solve it, wimble… oh OK then, you can if you like…)
Rules of the game:
You have to get all of the characters on the other side of the river: a woman with 2 daughters, a man with 2 sons, a prisoner and a police officer.
To reach your goal, you have to use the raft. The raft can carry one or two persons and can only be steered by the man, the woman or the police officer.
There are some restrictions:
- The little girls can not be left alone with the man
- The little boys can’t be left alone with the woman
- The prisoner can not be left alone with anyone that’s not the police officer
You may be wondering why the little girls cannot be left alone with the man, etc. It is not for the reason you might think.
Edit: For the benefit of those who can’t see the Flash, “X left alone with Y” seems to mean “X is on the same side of the river as Y, without X’s ‘supervisor’ also being present”.
31 replies on “River puzzle”
It all depends on what “left alone” means.
I assume that the little girls must not be left in any location such that the only other person present is the man. Same with the boys and the woman.
The prisoner meanwhile should not be anywhere that the police officer is not present.
If this is correct, the solution is:
Woman steers daughters across one at a time, then returns.
Man steers sons across one at a time, then returns.
Man and woman go across together. Man returns.
Police officer and prisoner go across together. Woman returns.
Man and woman go across together. The end.
I did try to test whether this was right by clicking on the link you provide, but annoyingly there is no blue button. Is this perhaps the IQ element of the test?
The blue button is to the lower right of the Flash animation of the puzzle, is circular and has Japanese characters on it. If you’re not seeing it, maybe you have Flash blocked or something? It makes attempting the puzzle a lot easier, as you can see the consequences of the forbidden combinations (and the official resolution of the ambiguity of ‘left alone’, which unfortunately is far more restrictive than you hypothesize).
I must not have flash then :o(
As you say it depends what “left alone with” means. You could interpret it as the prisoner can be completely alone 🙂 but not with (one or more people none of whom are the copper).
Equally one assumes the kids count as people.
In that event you could shovel the prisoner over first, the cop second, and then either the man and sons or woman and daughters…. I think 😉
See edit above for what “left alone with” seems to mean…
Do you have flash type things installed? I get a nice big flash thing with a blue button as described…
It appears that when I tried your solution by sending the woman and a daughter across that the man then punched the other daughter and I failed. Go figure, eh?
So I believe that the little girls can’t be somewhere with the man but not the woman and vice verca. Don’t quote me on that though.
But you are allowed to leave the prisoner alone!
Yeah, for some reason she doesn’t just run off…
I preferred this trip when there was only a tiger, a goat and a cabbage to ferry across.
I have a life you know, I can’t spend hours schlepping back and forth across a river all the time. Tell them to take the train. And tell the police officer next time he should bring some handcuffs. And why can neither of the adults be trusted with someone else’s children for five minutes? They all sound like a bunch of crims to me.
Hm. Having deduced the rules from the above discussion, I am confused. Surely the only way for either the woman or the man to cross the river without the one beating up the other’s kids is if they go across together. But now the only way to continue is to send one of them back over with the boat, at which point the other’s kids are toast.
I shall have to wait until I get home to see how to do this.
Maybe you can leave the man or woman with the other’s kids if there is another adult there?
I wonder whether you can teach the children how to drive the raft? Or even just wait until they are big enough to fend off the nasty parents?
I got as far as send policeman and prisoner across. Leave the prisoner. policeman takes one kind across and comes back with the prisoner (leaving one kid on the other side). The same gender kid and adult then go across to join him and the adult comes back. this leaves two kids over there.
To simplify things I’m going to assume two boys over there. Man and woman go across and woman comes back. Woman goes over with girl and man and woman come back. We now have two boys and a girl on the far side and two adults and a prisoner and a girl on the near side. I’m a bit stuck again though.
Dunno if that helps at all though…
That’s pretty much the point where I gave up 😉
I can give clues if you want. 🙂
Ha! Got it! Now just to see if I can recreate it… 🙂
Do you want the answer posted here if I can? I’ll try to do it white on white or something if you do…
Go on, white on white sounds good. And clues please!
OK. Here is the solution in white on white spoiler spaceness. I hope. Mo: if you get e-mail notification in plain text stop reading here or you will get the whole solution. 🙂
One line per transfer if you want to just selectively highlight to check how you are doing so far. And this is from memory partly rather than as I go through it in the game so there are possible errors.
-> Copper + boy
<- Copper + Prisoner
-> Man + boy
-> Man + woman
-> Copper + prisoner
-> Man + Woman
-> Woman + girl
<- Copper + Prisoner
-> Copper + girl
-> Copper + Prisoner
for a clue take “The cop and robber cross over in the middle (but not just then)“.
So the prisoner can *cross* with someone who is not the copper, provided they are either then left completely alone on the bank or with the copper?
well if the prisoner crossed with not the copper then either a) he ends up on the opposite side of the river to the copper. You lose b) he started off on the opposite side to the copper. You already lost.
a) should have read “he ends up on the opposite side of the river to the copper and with somebody else” and a similar ammendment to b.
Oh, I thought the prisoner could be left *completely* alone. So the man couldn’t take the prisoner over to the (empty) other side and leave them there then.
the prisoner can be left completely alone but if he crossed with somebody that wasn’t the copper then you would lose before you got a chance to abandon him there. That make sense? I can tell you that I have a solution that doesn’t involve that anyway. 🙂
Got it in one. <smugness>
Two (big) hints in white text follow:
1) You have to start by ferrying the Prisoner to the other side with the cop, then leaving the prisoner alone in a very unrealistic way.
2) The entire solution is symmetric about its own halfway point (not surprising, since the puzzle is symmetric). Therefore your real goal is just to get the man and the two boys onto the other bank, at which point you’ve got the solution.
yes. I think 2 gives it away pretty much entirely. I only really noticed that after I’d solved it though. 🙂
Aha! – yes indeed. Thanks for that!
Next time, bigger ferry.
What has Little My done to get herself in trouble with the law though, that’s what I want to know?
From what I remember (and this is quite a long time ago. Mental note: must reread the books), isn’t the answer, “pretty much everything?”
And have now read the comments. So pace them, you quickly get your head around the fact that people who should not be left alone very much are useful when left by themselves, it’s just that they mustn’t be with the person they’re not supposed to be with (and for the prisoner that’s anyone, but they can be by themself, oddly). Nice evil grin on the prisoner. Thanks; very addictive.
Mm, I suspect translation from the Japanese may have shifted the emphasis of “alone with” slightly.