Seriously, I don't even know where to begin to try to solve this. So much is going on in this that it makes my brain melt!

George Boolos, a United States logician, dreamt up this baffling exercise. There is a solution, and I will post a link to the EIGHT gd PAGE PDF if this really bothers you and you need to know the answer.

So here we go!

"Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter.

Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god.

The gods understand English, but will answer all questions in their own language in which the words for 'yes' and 'no' are 'da' and 'ja', in some order. You do not know which word means which."

Good luck. As I said, I had no idea how to even begin and even reading the PDF solution was confusing as heck!

Fuck this.

If you can ask all three questions to one of them it might be easier. Also, he cheated. This particular logic puzzle is varient of one thats been around for ages, the da and ja are red herrings. You can know which is which in the same manner of knowing which "god" is which. If there were two questions only though, it would be impossible. There's a similar logic puzzle in the labrynth b.t.w.

Yep, I was right. Read a simplified solution on wiki. Heres the link, have fun kiddos.

http://en.m.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

I think THIS is the hardest logic puzzle in the world.

Take a try at this.

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

Genetics, not logic... Blue is a recessive trait, so if guru married a blue eyed man her child would be 90% likely to have blue eyes. Or a child who had parents with both blue eyes... As for when most likely the next day at midnight.

why would you want to leave the island? Sounds magical. THAT IS THE REAL QUESTION

The real answer is on the 100th day, all 100 blue eyes people would leave.

If you consider the case of just one blue-eyed person on the island, you can show that he obviously leaves the first night, because he knows he's the only one the Guru could be talking about. He looks around and sees no one else, and knows he should leave. So: [THEOREM 1] If there is one blue-eyed person, he leaves the first night.

If there are two blue-eyed people, they will each look at the other. They will each realize that "if I don't have blue eyes [HYPOTHESIS 1], then that guy is the only blue-eyed person. And if he's the only person, by THEOREM 1 he will leave tonight." They each wait and see, and when neither of them leave the first night, each realizes "My HYPOTHESIS 1 was incorrect. I must have blue eyes." And each leaves the second night.

So: [THEOREM 2]: If there are two blue-eyed people on the island, they will each leave the 2nd night.

If there are three blue-eyed people, each one will look at the other two and go through a process similar to the one above. Each considers the two possibilities -- "I have blue eyes" or "I don't have blue eyes." He will know that if he doesn't have blue eyes, there are only two blue-eyed people on the island -- the two he sees. So he can wait two nights, and if no one leaves, he knows he must have blue eyes -- THEOREM 2 says that if he didn't, the other guys would have left. When he sees that they didn't, he knows his eyes are blue. All three of them are doing this same process, so they all figure it out on day 3 and leave.

This induction can continue all the way up to THEOREM 99, which each person on the island in the problem will of course know immediately. Then they'll each wait 99 days, see that the rest of the group hasn't gone anywhere, and on the 100th night, they all leave.

Here is the source where I got this.

http://xkcd.com/solution.html

I refute that as logic... But thats just the way I see it.

Does this island have beer pizza and video games....if so you all can go and ill stay

Milwaukee's best, goat cheese pizza, atari copies of e.t. and only movies by uwe boll..... Its why everyone wants out.

Thats not an island it Navada

LOL

Logic puzzles. Not sure when the last time I even attempted one. Probably high school. I'll pass.