You are a pussy
There is no answer and you are lying.
You are a damn coward.
You are so transparent!
Well if I am lying then when I don’t post the answer you will know I am lying. But since I will be posting the answer at some point, you might not want to throw around the ad homs so much. I am not offended but calling someone a liar can get you a ban, not from me, but report monkeys do like to report. They have long arms smaller brains than most humans and a real talent for polishing apples. I would like to ad I in no way am saying you shouldn’t report people, you should, but not because you are a kiss ass. 
If you do get banned for calling me a liar, understand that I will contest the ban, because I could not care less. 
Ironically I think it is you that needs to behave himself. And I could pm you the answer, you could look it up on google too. But I am not going to because you didn’t ask nicely. 
Somebody seems a little upset! 
I know. I SAID IT! I said that no monk knows.
The fuck? No I’m not. I’m taking the fact that the wording is what it is to prove that no monk knows.
You need to read the post again very slowly. I demonstrate very carefully why no monk knows he has the disease, given the wording, and how changing the wording could make the monks know they have it. I start from the example case of if 1 monk had the disease, extrapolate from that what the case would be like if 2monks had the disease, and we can extrapolate from that the case of 4 monks, 5, etc. With the current wording, in the case of 1 monk, he wouldn’t know that he had the disease, and we can extrapolate from that that neither would 2 monks, neither would 3, etc. If the wording is changed to my suggestion, they would know.
You don’t understand my reasoning. You make that clear with the ridiculous things you said above that I quoted. I never said that ‘it says no monks know.’ I said no monks know. I said it. Me. Not ‘it’. Me.
And my logic isn’t based on the fact that some monks die. My logic is only based on the wording. The fact that you don’t understand that yet tells me that you don’t understand my argument, so you’re hardly in a position to dismiss it.
This is a similar problem with correct wording, which I thought about when writing my response to this puzzle.
The logic should work the same in both puzzles, but because of the problem in the wording of the monk puzzle, the logic fails.
Listen buddy. It’s not a ad hominem - its a straight insult. And I was only giving out what I originally get from you calling me an ORC.
So save your breath to chill your porridge.
As fo the problem - either deal with the poorly worded and internally incoherence of it or put a solution that managed to get over the problems. Problems, I might add that others have pointed out to you.
As for the solution: put up or shut up.
I’m done with you until you give the solution.
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Hellandhighwater is on a hiding to nothing. I’ve told him, you told him, but he still claims there is a coherent solution.
Until he has the guts to put up, then he won’t get it.
If he has a legitimate explanation of a solution, I’ll certainly read it and if it’s correct I’ll change my mind.
I’m just bothered that (a) he clearly hasn’t even tried to understand my argument, and yet still feels qualified to dismiss it, and (b) he’s taking it all so personally. As if some part of his identity is invested in the wording of the problem being sufficient. Just chill out bro, just chill out.
I’m not sure how the green eyed guru telling the rest that there is a person with blues eyes helps anyone in particular.
Care to share the secret?
Sure. Spoilers following:
Prior to the guru saying that, nobody knows if they have blue eyes, right? We know, from the outside, that there are 100 people with blue eyes, so presumably each blue-eyed person sees 99 other people with blue eyes and doesn’t know if they do or not, and anybody without blue eyes sees 100 people with blue eyes and also doesn’t know what their own color is. Much in the same way I talked about before, the solution is to start not with 100 blue-eyed people, but with 1.
Imagine the island had 1 blue eyed person. Everyone on the island except him (group A) sees 1 blue-eyed person, and he himself sees no blue-eyed people. So when the guru says ‘I see a blue-eyed person’, the guy with blue eyes knows it has to be him. It couldn’t possibly be anybody else – he’s the only person who has the possibility of having blue eyes. So he leaves that night.
Imagine the island had 2 blue-eyed people. You’ll notice that I labeled ‘group A’ in the above scenario – each of the 2 blue-eyed people, after hearing the guru say ‘I see a blue-eyed person’, has 2 options: either they are in group A, and seeing the 1 blue-eyed person, or they are blue-eyed. Everyone else on the island is in Group B – non-blue eyed, seeing 2 blue-eyed people. So, these are perfect logicians, and as such, they know that if they are in fact in group A, then after the first night, they would expect the 1 blue-eyed person they see to be gone. So, each of them wakes up the next morning and sees that the blue-eyed guy hasn’t left, and there’s only 1 explanation for the 1 blue-eyed got not leaving – he must himself have seen someone with blue eyes. So there are 2 people on the island with blue eyes, but I only see 1, so the other must be me, so I leave on the second night. Everyone in Group B sees the 2 blue-eyed gone by the 3rd morning.
Imagine the island has 3 blue-eyed people. Each blue-eyed person, like above, has two options: either they are in group B, and there are 2 blue-eyed people on the island who will leave the second night and be gone the third morning, or they are themselves blue-eyed. So they wait the first night, nobody leaves, the second night, nobody leaves, so they can deduce that they aren’t in group B, so they must have blue eyes, so they leave on the third night.
This pattern extends indefinitely. There are 100 blue-eyed people, they all leave on the 100th night after the guru speaks, the following morning everyone else sees that all the blue-eyed people are gone.
You read it and read it, and pretty soon you realise that it’s not you , it’s the solution that is at fault.
It’s not the wording it the conceptualisation of reality. It is a fault for the same reason that people think that Xeno’s paradox is a paradox.
The fact is that a green eyed man says someone has blue eyes., this is not NEW information. Everyone knew that already.
It could only work if there were only one with blue eyes. Case closed.
If there were 2 blue eyed persons then each one would not know if the person with the blue eyes was himself or the other one. Case closed. He could not infer that it was not him as he had not left nor that it was himself and leave. They do not form a class or group, as much as you want them to as neither knows that they are part of a group. And that fact blows the solution.
With 100 the problem is much more difficult.
So - back to you. the words of the guru make no difference. Thus the problem is also bogus.
Over to you
There’s your problem I don’t think Xeno’s paradox is a paradox. I also think this has a solution because I have seen it, look it up if you don’t believe me. Then stop badgering people because your logic is flawed. If you dispute the answer go talk to the person who made the damn question up, what is the point telling me that you think the answer is illogical, it wont change anything. Thinking Xeno’s paradox is a paradox is the result of poor maths skills. Thinking there is no solution to this is the result of you stubbornly refusing to admit that thousands of people have answered the question to the satisfaction of the person setting the problem on thousands of forums, without 9 pages of it’s all wank!!! You can’t solve it I forbid you!!!
If you got a grip on yourself and took some time to read what people actually write you would know that I know Xeno’s paradox is not a paradox. You would also have realised that I was not directing that post at you, but at Flannel Jesus. You seem to do this quite a lot.
You need to focus.
Xeno’s paradoxes are solved by shooting an arrow at a target and by winning a race. The problem is with the conception of the problem.
I imagine that the Monk bollocks is due to you misconceiving the problem too.
Stop stalking me. You have not shown that my logic is faulty because you have not had the courage to print your solution.
Saying that my logic is faulty is not an argument; it is a empty accusation.
Put up or shut up!
Lol let’s get personal shall we that will solve the problem.
ok lets present the exact same puzzle but with a variation so I don’t have to here any more of this inane crap:
It’s basically the same problem but the answer is now a different multiple.
I prefer the monk puzzle just because it results in more interest than the blue eye version.
Just to let you know, my solution isn’t just something I came up with. This is the correct solution. The solution agreed upon by people…you know…with IQs greater than yours and mine put together. Just putting that out there, so maybe you’ll rethink this response.
Now, I find it interesting that you agree that it would work with only 1 blue-eyed person, but not for the rest. Because…well…if it works with 1 blue-eyed person, then it necessarily works with 2.
[tab]You see, you don’t know your eye color, right? You’re some dude on the island, you don’t know. But, BUT, you see someone with blue eyes. You see 1 person with Blue Eyes. Right. Now you, Hobbes, have already agreed that it works with 1 blue-eyed person, right? You agreed to that. So, that being the case, you see 1 blue eyed person, you know that IF he’s the only blue-eyed person, he must leave the first night, right? It’s logically necessary. You agreed that it works with 1 blue-eyed person, so if you see 1 blue-eyed person AND it’s true that he’s the only blue-eyed person, he must leave the first night.
So you wake up after the first night, and still see him there. What can you conclude? You can conclude that you have blue eyes also. You can leave the second night. So, if you agree that it works for 1 blue-eyed person, we’ve logically deduced that it must also work for 2. You following? I don’t mean this condescendingly, but as an open invitation: if there’s any step in reasoning there that you didn’t think was good enough, point it out and I’ll work through it.
This is the actual solution, mind you. Not something I’m making up. I need to stress that. This is a real problem with a real solution, and this is the answer. It’s only a matter of whether or not you follow the logic. I think it’s a fun problem, so I’m enjoying sharing the logic with you.[/tab]
And, idk why we’re talking about Xeno’s paradox. This doesn’t have anything to do with this for me. I don’t see the relationship. This is an independent problem.
[tab]http://xkcd.com/solution.html
The answer is that on the 100th day, all 100 blue-eyed people will leave. It’s pretty convoluted logic and it took me a while to believe the solution, but here’s a rough guide to how to get there. Note – while the text of the puzzle is very carefully worded to be as clear and unambiguous as possible (thanks to countless discussions with confused readers), this solution is pretty thrown-together. It’s correct, but the explanation/wording might not be the best. If you’re really confused by something, let me know.
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.
Before you email me to argue or question: This solution is correct. My explanation may not be the clearest, and it’s very difficult to wrap your head around (at least, it was for me), but the facts of it are accurate. I’ve talked the problem over with many logic/math professors, worked through it with students, and analyzed from a number of different angles. The answer is correct and proven, even if my explanations aren’t as clear as they could be.
User lolbifrons on reddit posted an inductive proof.
If you’re satisfied with this answer, here are a couple questions that may force you to further explore the structure of the puzzle:
What is the quantified piece of information that the Guru provides that each person did not already have?
Each person knows, from the beginning, that there are no less than 99 blue-eyed people on the island. How, then, is considering the 1 and 2-person cases relevant, if they can all rule them out immediately as possibilities?
Why do they have to wait 99 nights if, on the first 98 or so of these nights, they're simply verifying something that they already know?
These are just to give you something to think about if you enjoyed the main solution. They have answers, but please don't email me asking for them. They're meant to prompt thought on the solution, and each can be answered by considering the solution from the right angle, in the right terms. There's a different way to think of the solution involving hypotheticals inside hypotheticals, and it is much more concrete, if a little harder to discuss. But in it lies the key to answering the four questions above.
==BASE CASE==
TO PROVE: F(1) = G(1). If exactly one person had blue eyes, he would leave on day one
1 GIVEN at least one person on the island has blue eyes (from the oracle, who speaks only truth)
2 BASE HYPOTHESIS exactly one person has blue eyes
3 BY EXCLUSION AND 2 The person with blue eyes sees no one else with blue eyes
4 BY ELIMINATION AND 3 AND 1 The person with blue eyes concludes that he must have blue eyes
5 QED: If exactly one person had blue eyes, he would leave on day one. F(1) = G(1)
==INDUCTIVE CASE==
TO PROVE: If F(n) = G(n) then F(n+1) = G(n+1). (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1)
6 GIVEN No one leaves until he knows he has blue eyes
7 ASSUMING n+1 people have blue eyes
8 INDUCTIVE HYPOTHESIS n people with blue eyes will leave on day n
9 BY EXCLUSION AND 7 Each person with blue eyes will see n people with blue eyes
10 BY 6 AND 7 AND 8 No one leaves on day n
11 _BY 8 AND 10 _The number of people with blue eyes cannot be n
12 BY ELIMINATION AND 9 AND 11 Each person with blue eyes reasons that he must have blue eyes (to make the total not equal to n)
13 QED: (If n people have blue eyes, they will leave on day n) implies (If n+1 people have blue eyes, they will leave on day n+1). If F(n) = G(n) then F(n+1) = G(n+1)
==INDUCTIVE CONCLUSION==
14 BY INDUCTION AND 5 AND 13 x people with blue eyes will leave on day x
reddit.com/r/AskReddit/comme … le/c2kdlr6[/tab]
The answer.
Thanks. The reasoning is quite similar to my own (probably because I read that a week ago)
I have to say that I am disappointed in your as well as hellandhighwater. For both of you failing to understand the that the solutions are bogus because the problems are at fault.
And then doubly disappointed by you resorting to an argumentum vericundiam.
Yes this is a independent problem from Xeno’s but what Xeno problematises with integers or “units” your problem is the inverse but for “sets.”
2 blue eyed logicians do not form a set. Neither do two diseased monks. And it is upon recognition of those sets that both problem’s solutions lie.
Try to use your own brain and think it through.
Put yourself in place of a person of blue or brown eyes, and ask yourself what the guru’s words add to your knowledge. The answer is fuck all. Everyone on the Island knows that there is a person with blues, they only have to look around. The fact that the guru has green eyes makes no difference whatever.
Taking the sinlge point argument of one person might work. A person who is the only one with blues eyes would have to know it is himself. That cannot apply to the next paragraph in which there are two persons.
This test, is not “solving the problem”. The test here is how gullible are you to accept the solution that does not work!
Both you and HNHWater have fallen for it. Just like many fall for the Xeno problems.
You’ve both been had.
If there were only one with blues eyes, then he could look around to see no others with blues eyes; conclusion he has to be the one.
With 2 people, each can see that the other has blue eyes and would have to assume that either; 1) the other OR 2) 3) himself OR both had blues eyes. That does not amount to anyone knowing what colour their eyes were. Sorry that is case closed.
QED
It can’t work with two, or more - you have 100
Tell me how in your own words!
There are 100b, and 100B remember Everyone already knows there is a person with blue eyes. All the b see 99b, and see 100B. The B sees 99B,a and 100b. Unless they can communicate, the information from the guru can’t be put to any use.
Over to you!
The guru is the external reference who identifies which group can leave and starts the clock - that’s the new information. Once the statement is made, they start counting days which is critical to the solution.
A blue-eye person (or any other color) could have stood up and said the same thing but then he would not be able to leave because he can’t identify his own eye color.