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The guru says that he sees someone with blue eyes. Ok, that means you either have Blue Eyes or you do not.
If there were two people on the island, then the person with blue eyes would know by the end of the second day and would leave.
If there were 10 people on the island with five blue eyes, it would take 5 days for everyone to leave.
Basically, if you see nobody else with blue eyes, then you know you have blue eyes.
If you see one person with blue eyes, and he doesn't disappear after the first day, then you know there are at least two people with blue eyes. If you see three people with blue eyes, and they do not disappear on the third day, then you know you must also have blue eyes.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
y-yes? this is true, which is why the puzzle specifies that everyone on the island is a perfect logician. i dunno, was that not said in the first version posted here? not sure what you're getting at
that's it, right? I can't connect the dots at all, even if I play the "pretend there are only 1, 2, 3, 4, 5 people on the island and so on" game... it feels like there's no way to connect them at all.
but I don't want to cheat because I know I'll feel like a baller if I figure it out
this was to me, right? could you explain in more detail? i must be dim because i still can't quite get how the guru's statement catalyzes the whole process.
This is the answer:
Say two people have blue eyes. Being perfectly logical, they each realise that if they don't have blue eyes, the other person is the only blue-eyed person on the island. So that person will leave the first night. If that person doesn't leave the first night, they know they have blue eyes too, so they both leave the second night.
The sequence continues, until you get 100 people leaving on the 100th night.
The way the fact pattern is set up, it's a throw-away at the setup. It should be the focal point of the entire discussion.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
If you see one person with blue eyes, and that person doesn't leave, you know that person sees someone else with blue eyes and it has to be you, so you both leave the next night.
If you see two people with blue eyes, and neither of them leaves the first night, it's because they at least see each other. If neither of them leaves the SECOND night, it's because they've each seen two people having blue eyes, so the third person has to be you, and you leave on the third night.
If there's a hundred people on the island with blue eyes, then all hundred of them leave after a hundred days.
WINNAR!
But the crux of the whole thing is that everyone else is thinking the same thing as you. It doesn't work if there is any question about that.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
you've read the xkcd solution, yeah?
the second theorem is only true because the first one is true, and so on, up to the 99th theorem. if these people don't all start looking around at the same time, there's no way for them to synchronize when they realize how many people have blue eyes and then leave.
augh, this is still fucking with my brain. i can't stop thinking "but they all already knew there was someone with blue eyes! they could just look around and see that!"
give me a second here, i'm sure i can make this make sense to myself
edit: thanks!
2 people on the island, 1 person with blue eyes
They see the other person does not have blue eyes. They leave that night.
3 people on the island, 2 people with blue eyes. Person 1 and 2 have blue eyes, person 3 does not.
Person 1 sees that person 2 has blue eyes. He knows that person 2 will leave after tonight if they are the only person with blue eyes. This is because they would see nobody else with blue eyes and therefore would be the only person with blue eyes and therefore would be able to leave. Person 2 does not leave tonight. Therefore the only answer is that Person 1 also has blue eyes. They leave the next night.
If there were 100 people with blue eyes, and Person 1 was one of them, then after the 99th night he would know that he also had blue eyes. Otherwise, everyone else would have figured out on the 99th night that they were the only ones with blue eyes and they would have left then.
The things that do not matter in this puzzle: how many people that are there who do not have blue eyes, the total number of people.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
yeahhh
i've solved this riddle before, forgotten it, been told it again and re-solved it, then told it to other people a bunch of times
and i never quite understood that til now
No, the brown eyed people don't get to ever leave, because they don't know if they are brown or some other color.
They rot in a brown eyed, muted world.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
The oracle's only purpose is to start the series of logical events that every other person is going through.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
I would call this the hardest logic problem in the world:
"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."
STEAM ID
I guess it depends on the rules. They way I usually encounter it is:
- People on the island have brown eyes or blue eyes
- People on the island are perfectly logical and rational
- People on the island know all of the above facts (including the fact that their eyes are either one or the other).
But yeah, if I go around pointing that out, I don't get to make a sad guru joke... spoilsport.
It's just a logic extrapolation.
It's not exactly mensa-level difficult once you get the premises straight.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
I don't know how you do it with three and not knowing yes or no.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
You have god 1, 2, and 3.
You ask god 1 whether god 3 would say god 3 is telling the truth.
You ask god 2 whether god 3 would say god 3 is telling the truth.
You ask god 3 whether god 1 is telling the truth.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
1) It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).
2) What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
3) Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
4) Random will answer 'da' or 'ja' when asked any yes-no question
Pay attention to number 3.
STEAM ID
The answer to the first question would be Yes for L god and R god, no for T god. The answer for question 2 would be Yes for L god and No for T god and R god.
So if the god changed his answer, then I would know he was the random god. If he did not change his answer, I would know he would either be the truthful god or the lying god.
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
There's also multiple ways to answer this problem, including one that is more efficient than the others.
STEAM ID
Question 2: Are you the random god who tells the truth? This would be yes for the random god and yes for the lying god.
Therefore: If the answer changed, then I know that the first response means no and the second response means yes. If he gives the same answer, then I know he is either the lying god or the truthful god
If he changed his answer, then I would ask the 2nd god if the 1st god was the random god. I would know the truth about the first god and I would know what Yes and No were in their language, so it would reveal all of their identities.
If he did not change his answer, I would ask the 2nd god if he was the random god who tells the truth. If his response matches the previous answe
"Humor can be dissected, as a frog can, but it dies in the process."
Imagine all of my posts being spoken by Alec Baldwin
GamerTag: MunkusBeaver ||||| Steam: munkus
Mensa level difficult isn't that difficult
Close, but you have to use all the numbers! That's what makes it a puzzle rather than a sum
Answer:
That is, divide three by seven, add three to the result, and then multiply that total by seven.
i was happy with that