Hardest logic puzzle in the world

[Zerzera]

I don't think they would need the guru if that was the case ^

[GuruMeditation]

I think the Guru's gone the first night :P

Did someone say my name?

[Zerzera]

P.S. I would certainly smack that guru if he could once speak one sentence and said such a stupid one that made all ppl stay longer then needed

But in this case it was a she, which makes it rather logical for her to put it this way instead. :wub:

[PjOtTeR]

I don't think they would need the guru if that was the case ^

what you mean? my solution? ofcourse they needed the guru for that. I'll tell you exactly why, in white text:


Every person sees atleast 99 blue-eyed man. If for example there were only 2 blue eyed ppl on the island, No one would leave the first night, and 2 ppl would leave the second night. One of the persons with blue-eyes would know that they had blue eyes because they can see all other eyes and the other person with blue eyes didn't leave because there was someone else with blue eyes. Knowing that is the only person with blue eyes you can see, would make it certain for yourself you have to have blue eyes aswell.

Bringing this to day 100... The other 99 guys haven't left yet, because they all see 99 guys with blue eyes... which means you yourself have to have blue eyes 2, cause else they would've left the island the night before. Since if you didn't have blue eyes, those people would've only seen 98 people with blue eyes as the opposing 99 they see... so everyone on day 100 with blue eyes knows they themselfs have blue eyes, cause there have to be 100 ppl with blue eyes cause else ppl would've left already. and since you only see 99 other ppl with blue eyes, you know you yourself have to have blue eyes.

[Zerzera]

what you mean? my solution?

No

[phata$$]

After reading that wiki page, it took me so long to figure it out even though I had the answer right in front of me. I finally understood it, I think. Here's what I think is correct:

Assuming that there are only 2 blue-eyed people on the island
-After the guru makes the announcement, let's say that one blue-eyed person(1) sees another person(2) with blue eyes.
-(1) will infer that if he himself doesn't have blue eyes, then (2) is the only one with blue eyes and (2) will realize that and leave. (2) infers the same thing.
-Upon realizing that (2) hasn't left the island, (1) will assume that he himself has blue eyes so he will leave. (2) assumes the same thing and leaves too.
-So that means that if there are 2 blue-eyed people, they will leave on the 2nd night.

Assuming that there are only 3 blue-eyed people on the island
-(1) will see (2) and (3). (1) will assume that if (2) and (3) don't leave then he himself has blue eyes. (2) and (3) assume the same thing.
-No one leaves the first night, so (1) will assume that the other one will leave the next night. (2) and (3) assume the same thing.
-No one leaves the second night, so (1) realizes that he has blue eyes. (2) and (3) realize as well. All three of them leave on the 3rd night.

Now for the answer:
-So from that, we can assume that since it will take two people two nights to realize they have blue eyes, it will take three people three nights, four people four nights, etc.
-The given is 100 blue-eyed people. So it will take 100 people 100 nights to realize that they have blue-eyes. So they will all leave on the 100th night.

EDIT: Pjotter beat me to it.
EDIT2:Oh and I kinda cheated.

[pascone]

how can one person deduce that because no one left (they can't see their own eyes either?)that they should leave?

this is gay. dont get stuck on an island with some bitch guru

[GuruMeditation]

The puzzle isn't comprehensible with 100 people. It's easier to understand when you understand the logic behind 1 blue eyed person, 2 blue eyed people and three blue eyed people.

For clarity we will dispose of the Guru being present on the island. Someone turned up, told them and left. We will also assume everyone with brown eyes has green eyes cause I typed it up after reading the wiki article and can't be bothered to correct myself.


First let's look at the case with 1 blue eyed person that we will call B. In this environment everyone around B has green eyes. B sees this. Therefore B knows two facts.

1) Everyone except B has green eyes.
2) There is someone in the group who has blue eyes.

In the logic the following phrases may be used for categorical knowledge, given S is the category of the subject and P is the category of what the subject may or may not be:

All S are P.
Some S are P.
Some S are not P.
No S are P.

The logical phrases for what B knows are:

Some people on the island are people with blue eyes.
All people who are not identical to B are people with green eyes.

Therefore he can deduce it must be him. So he leaves.

-------------------------------

Now look at it with two people. In this environment we have two people with blue eyes that we will call B and C. Everyone else we will call D. B and C have a different set of knowledge.

On day 1:

B knows:
1) Some people on the island are people with blue eyes.
2) All people identical to C are people with blue eyes.
3) All people not identical to B or C are people with green eyes.

C knows:
1) Some people on the island are people with blue eyes.
2) All people identical to B are people with blue eyes.
3) All people not identical to B or C are people with green eyes.

D1 (who is one of the D's) knows:
1) Some people on the island are people with blue eyes.
2) All people identical to B and C are people with blue eyes.
3) All people not identical to B, C or D1 are people with green eyes.

From this B can deduce the following:
If C can only see people with green eyes he will leave at the next dawn, because he knows he is the one with blue eyes.
(this next one is the tricky one)
If C does see someone with blue eyes he will not leave. He only knows that some people on the island have blue eyes. If he only saw people who had green eyes he would leave immediately, because he could conclude that he was the only person with blue eyes. However because he sees someone else with blue eyes he will not leave and wait until the next day.

C can deduce the same things (with B replacing C).

D1 can deduce the following:
B and C both have blue eyes.
B knows that C has blue eyes.
C knows B has blue eyes.
B and C also know what color eyes D1 has.

No one leaves the island.

B believes that C will leave the island because C has blue eyes, and everyone else that he knows the eye color of has green eyes. If C does not leave the island C must have seen someone else with blue eyes.

C believes that B will leave the island because B has blue eyes, and everyone else that he knows the eye color of has green eyes. If B does not leave the island B must have seen someone else with blue eyes.

D believes no one will leave the island. Being a logical person he is aware that neither B nor C can conclude what eye color they have, only what the other peoples eye color is.

On day 2:

B has seen that C has not left the island. C would have left the island if all he saw were people with green eyes. From there he can deduce the following.

All people not identical to C and not identical to B are people with green eyes.
All people identical to C are people with blue eyes.
Some people not-identical to C (the group that C can see) are people with blue eyes.

Therefore B can deduce that he must have blue eyes. C being the logical person he is would have left if and only if everyone else he saw had green eyes. This was not the case. Everyone who is not B and not C has green eyes. Therefore the only person who could have stopped C from leaving was B. B can conclude he has blue eyes.

I'll continue with 3 if I'm making any sense so far

Edit : Fixed the categorical statements.

[Singularity]

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.

[phata$$]

how can one person deduce that because no one left (they can't see their own eyes either?)that they should leave?

this is gay. dont get stuck on an island with some bitch guru

Because they have a record on how many people have blue eyes and brown eyes. If one person records only one person with blue eyes, then he will wait one night to see if that person leaves. The other person does the same. Since they are both logical people, they deduce that they both have blue eyes and leave the next day.

Yeah, it's really difficult to understand. Read the wiki that was posted a while ago, that's what I did.

[Epinephrine]

WTF Epi on drugs?
It's nowhere said you can see the count of other ppl... You can see every other person at all times and you can keep count of the number of people you see with each eye color(except for urself), but you cannot communicate any other way then looking and counting the eyes of others.. So you cannot exchange data...

P.S. I would certainly smack that guru if he could once speak one sentence and said such a stupid one that made all ppl stay longer then needed

It says 'but they cannot otherwise communicate', meaning that some part of communication is allowed. Since the only thing mentioned was that they could count, this implies by logic that the counting could be communicated. Which means you can tell other people about what you counted

But I agree that is a cheezy way to do it

[apt]

Pls this is stupid. If they were logical and intelligent they would just let other people know by body language that that person has blue eyes.

Like if Im there and I see my friend has blue eyes. I will grab that mother fucker and drag him to the boat and by doing that he will understand he has blue eyes. I will point at his fucking eyes, then he will point at mine and point at his and I will realize I have blue eyes too. And we both leave. And that shit will work for everyone.

You guys are going way to deep into this, but the answer is so simple. 100 night s my ass, no reason to get nerdy with this shit.

[GuruMeditation]

There's a huge difference between logical and practical.

[Noah]

Give more of these! :greedy:

[PjOtTeR]

It says 'but they cannot otherwise communicate', meaning that some part of communication is allowed. Since the only thing mentioned was that they could count, this implies by logic that the counting could be communicated. Which means you can tell other people about what you counted

But I agree that is a cheezy way to do it

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.