Hardest logic puzzle in the world

[Zeebu]

i still like my kill all people without blue eyes answer, and you can do that in just a little bit. not 100 days =)

but the long and short end of it was that pjot gave it to you too. thats exactly what the wiki solution told you to do. the key in the problem statement was "perfect logicians" or whatever. that means when you have figured something out, so did the guy right next to you... and the guy right next to him. they just happened to keep track of it and solve everything at one time on the 100th day


woo

[PjOtTeR]

Awesome puzzle, but the scale of it hinders the solving of the puzzle. The example with 2 "outsiders" would be understandable after some thought.

Reminds me of a similar puzzle someone told me a couple of years ago:


There are a couple a gnomes with either white or black hats arranged as on the picture. Each of the gnomes is a mute and totally paralyzed, so they can only look straight in the direction of the arrows. The wall is one of those walls that is totally impenetrable (does not reflect) and the gnomes cannot see through, over or below it, and neither do they have psychic powers.
Each of them in insanely gifted in logic, honest and want to win immediately. Each knows the arrangement and that there are two gnomes with a white hat and two with a black hat. If a gnome knows for sure which color of hat he wears he has to cry out loud and the contest is over. Which gnome knows first which color of hat he wears and why.
Answer:
The gnome in the middle of the three will be the first to know for sure and this is why; he knows the leftmost gnome can see both his hat and the hat of the person in front. Since that gnome can deduce the same he would know that if both had the same color he would have a different colored one, yet he did not scream out. This proves to the middle gnome has he a different colored hat from the gnome in front and thus he knows that since that is white, he wears a black hat.

Then I have another question for you. You ask which gnome knows first which color of hat he wears and why. I ask you what other gnome knows what color of hat he wears after that and why?

[Mantra-Slider]

browneyes > blueeyes

[Zeebu]

Then I have another question for you. You ask which gnome knows first which color of hat he wears and why. I ask you what other gnome knows what color of hat he wears after that and why?

dude next to the wall on the group side then knows hes the opposite color of what the guy in the middle said. same logic used: if the guy in back would have seen 2 hats of the same color, he woulda known right away. therefore, dude next to wall is opposite colored from the dude bangin him from behind

[Epinephrine]

By the "but they cannot otherwise communicate" part they mean: they cannot communicate any different way then looking what eyes the others have, and keep count of that... This implies they cannot communicate their count etc...

And later on in that same piece of text it says: "It doesn't depend on tricky wording...".
So it's a way it's not supposed to be done. So no it is not a solution.

They obviously don't teach you the meaning of the word 'communication' in your fancy euro school

[Hero*]

Then I have another question for you. You ask which gnome knows first which color of hat he wears and why. I ask you what other gnome knows what color of hat he wears after that and why?

no other gnome knows their color. if they do know their color they would have raised their hand in the first place. since they cannot speak or tell each other anything, the 2nd gnome raising their hand is irrelevant. The gnome closest to the wall has no way of knowing whats behind him and the gnome facing the two gnomes only knows hes wearing a black or white hat. the guy behind the wall is just fucked

[winipcfg]

the guy behind the wall is just fucked

lol

[PjOtTeR]

no other gnome knows their color. if they do know their color they would have raised their hand in the first place. since they cannot speak or tell each other anything, the 2nd gnome raising their hand is irrelevant. The gnome closest to the wall has no way of knowing whats behind him and the gnome facing the two gnomes only knows hes wearing a black or white hat. the guy behind the wall is just fucked

Well actually zeebu is right here...

dude next to the wall on the group side then knows hes the opposite color of what the guy in the middle said. same logic used: if the guy in back would have seen 2 hats of the same color, he woulda known right away. therefore, dude next to wall is opposite colored from the dude bangin him from behind...

that´s teh answer...

[Singularity]

Well actually zeebu is right here...

dude next to the wall on the group side then knows hes the opposite color of what the guy in the middle said. same logic used: if the guy in back would have seen 2 hats of the same color, he woulda known right away. therefore, dude next to wall is opposite colored from the dude bangin him from behind...

that´s teh answer...

Yes and no, indeed he knows he has the opposite color hat compared with the person behind him, but no, he cannot see or find out what color that person has.

[Hero*]

Well actually zeebu is right here...

dude next to the wall on the group side then knows hes the opposite color of what the guy in the middle said. same logic used: if the guy in back would have seen 2 hats of the same color, he woulda known right away. therefore, dude next to wall is opposite colored from the dude bangin him from behind...

that´s teh answer...

he has no way of knowing what colour he actually has because he cant see behind him nor communicate with anyone

[Xog]

The exact question was, "Who leaves, and what night?"


Well, here's your logical answer.. whoever was on the ferry coming to the island at midnight was leaving every single night.

[GuruMeditation]

You'd actually get partial credit for that

[Galleleo]

Unless it is a ghost ferry.

[Bioture]

guru has the right idea... there isn't really an easy one or two sentence answer here... it IS the "HARDEST" logic puzzle in the world, not the "trickiest." it doesn't depend on trying to trick you using words... it's not a riddle.

Also the killing idea is interesting, but hardly logical



for those of you who have smoke coming out of your ears, the answer is here: http://xkcd.com/solution.html

I'm still looking for anyone who's willing to go through at least several of the "theorems"

[genocidal]

That definitely isn't the hardest logic puzzle in the world and the answer is quite simple - as in it doesn't require a long explanation. I'd say it's tricky more than anything else, not hard. Here is my version that I think is more clear that some I've read, but others in this thread are good.

They all leave on day 100 because they can only see 99 blue-eyed people. They would all know on day one that there are at least 99 blue-eyed people so once nobody leaves on the 99th day they would all know that they make the 100th just like everyone else (meaning they all leave at once). That's the tricky part - remembering that each individual knows that everyone else is counting at least 98 blue-eyed people.