Das Paradox

[ShakyFiend]

Spoiler: Hardest Puzzle [b

Ever[/b]]Blue Eyes:
The Hardest Logic Puzzle in the World

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."

I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort.

Nicked this from my avatar, not strictly on topic I know but its on the general theme, if anyone can answer it without looking up the answer I will be incredibly impressed.

Also I've always loved the Euripides paradox, mainly cos I was thinking one day, came to the same conclusions myself, and only discovered it was actually a 'thing' many years later.

 

[Shadow Geo]

This... statement... is... false!!!!
*don't think about it, don't think about it, don't think about it...*

 

[ManOwaRrior]

Can omni-potent god (an all powerful god) create a stone which he cannot lift.

If he can't create it then he isn't all powerful.

If he creates it but then is unable to lift it he still isn't all powerful.

Not a paradox, you just prooved the lack of existence of an Omni-potent God.

No, an omni-potent God could still exist, he would be able to defy logic.
So he could in fact create this stone. And he could still lift it. While this would be logically impossible, if he is omni-potent, he can do it regardless.

 

[ipop@you]

Well I like the old going back in time with a purpose thing. If you go back in time to let's say kill Hitler then you go back and shoot him but then in your time he is already dead so there is no reason to come back and kill him so you don't go, but if you don't go then he is still alive so you have a reason to go and kill him and the cycle starts again.

Only if you go back with the intention of killing him. If you were travelling past Hitler in your car and your car spontaneously combusted killing Hitler then you would be caught in an endless loop of going back in time and accidentally killing Hitler, it would become History and nobody would know any different. Unless I've made a mistake in my thought over this.

Good point, hence why I said the travelling with a purpose thing.

 

[SextusMaximus]

Can omni-potent god (an all powerful god) create a stone which he cannot lift.

If he can't create it then he isn't all powerful.

If he creates it but then is unable to lift it he still isn't all powerful.

Not a paradox, you just prooved the lack of existence of an Omni-potent God.

No, an omni-potent God could still exist, he would be able to defy logic.
So he could in fact create this stone. And he could still lift it. While this would be logically impossible, if he is omni-potent, he can do it regardless.

No, it just means an Omni Potent God wouldn't exist because it would defy logic.

 

[Hosker]

I think this is called Curry's paradox:

If this statement is true, then Santa Claus exists.

It can be used to prove anything.

And what if it isn't true? Not much of a proof, is it?

It's irrelevant what the second phrase refers to. Wikipedia can explain it much better than I can: http://en.wikipedia.org/wiki/Curry's_paradox

I just exploded with laughter cause you linked to the Wiki article on curry.

OT: I always lie.

Damn links, not doing as they're told!

 

[Judgement101]

Zeno's Paradox is a thing?! I always thought it was just a cheat code for Age of Mythology.....

OT:The sword that breaks all against the shield that cannot be peirced.

 

[SextusMaximus]

I think this is called Curry's paradox:

If this statement is true, then Santa Claus exists.

It can be used to prove anything.

And what if it isn't true? Not much of a proof, is it?

It's irrelevant what the second phrase refers to. Wikipedia can explain it much better than I can: http://en.wikipedia.org/wiki/Curry's_paradox

I just exploded with laughter cause you linked to the Wiki article on curry.

OT: I always lie.

Damn links, not doing as they're told!

Gave me a good giggle anyhow!

 

[The Unworthy Gentleman]

Well I like the old going back in time with a purpose thing. If you go back in time to let's say kill Hitler then you go back and shoot him but then in your time he is already dead so there is no reason to come back and kill him so you don't go, but if you don't go then he is still alive so you have a reason to go and kill him and the cycle starts again.

Only if you go back with the intention of killing him. If you were travelling past Hitler in your car and your car spontaneously combusted killing Hitler then you would be caught in an endless loop of going back in time and accidentally killing Hitler, it would become History and nobody would know any different. Unless I've made a mistake in my thought over this.

Good point, hence why I said the travelling with a purpose thing.

Shit, I missed that bit. And there I was, thinking I was all smart.

 

[Death God]

The following statement is true
The previous statement is false
PARADOX!

i like paradoxes that involve time.

WHAT HAVE YOU DONE SNAKE?! YOU'VE CAUSED A TIME PARADOX!

ps. i've got a time paradox. Say you travel back in time to before you were born and killed your grandfather. Logically he'd be dead and you would never have been born. But now, if you were never born, then how was the grandfather killed? He was killed by you, but since you were never born he should be perfectly fine...

TIME PARADOX

<youtube=rXU-ZdmzNmo>

You, sir, have raped my brain with that song!

OT: Since we are talking about Portal, my favorite paradox is this:

 

[TonyVonTonyus]

This statement is false.

 

[Nieroshai]

The first paradox is only true in a digital world. In fact, the object can NEVER come to rest and will never BE at a state of rest because there is always a force acting upon it. "rest" to us is relative, the APPARENT lack of motion as observed to our eyes. Rest is the paradox, everything is in constant motion.

 

[Not G. Ivingname]

1/0

 

[Lyx]

Solution: There is no such thing as "a point in time". It is simply a theoretical invention, that 1. is used to deny interaction, and 2. is a necessary evil of how our thought structures work: To define a range/area, we need to define where that range/area begins and ends... so, boundaries.... the problem is that those boundaries do not really exist, and are simply an arbitrary aspect of the process of "defining ranges/areas".

Yup, i have just proven the premises of the concept of the atom/particle false and untrue. Then again, did anyone really believe that a mathematical point is anything more than an idea? Yes? Oh dear.....

As for the "at rest"-thingie.... that concept is only true if it is treated as a comparison of interactions - so, an *active* relationship. When you think of something being at rest, you should compare movements... so, interactions - ranges, not "points in time".

 

[Condor219]

Spoiler

Blue Eyes:
The Hardest Logic Puzzle in the World

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."

I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort.

Nicked this from my avatar, not strictly on topic I know but its on the general theme, if anyone can answer it without looking up the answer I will be incredibly impressed.

Also I've always loved the Euripides paradox, mainly cos I was thinking one day, came to the same conclusions myself, and only discovered it was actually a 'thing' many years later.

I got real dang close. I didn't consider the logic of the browns would be applied in the same way, which would lead the browns to false information. But I figure it's a pretty good effort, considering basically the whole process was in line.
Spoilered so other people can give it a try.

Spoiler: My EVER SO CLOSE but ultimately incorrect reasoning

All of the blues leave on that night. No other person leaves, they just know that their eyes aren't blue.
Alright, so nobody on the island can deduce their own eye color, but each blue (let's call this specific guy Bob) knows that there are 99 blues, and 100 browns, with his eye color being an unknown factor. Then, he knows that another blue can determine that there are 98 blues on the island, and 100 browns, with the unknown factors being Bob's eye color and this other blue's eye color (call him Terry). Bob figures that Terry will also make this deductive statement, coming to the conclusion that Terry's eyes are an unknown factor, some other blue knows his respective eyes are an unknown factor. Through that, Bob can determine that this blue knows for sure that there are at least 97 blues on the island, and 100 browns. From this, Bob can go through that same process all the way down to 0 (considering his knowledge started at 99 + unknown factor, then decreased to 98 + 2 unknown factors at the 99th blue, etc). Bob then knows, that the final person who's making this logical statement (knowing that everyone's sense of logic will go down this road) can in fact be Bob himself, considering his logical process is identical to all the other blues. So, Bob can deduce that there are at least 0 blues, and 100 browns. Considering there are 101 people left, 100 of them definitavely brown, Bob can figure out that his eyes are blue in order to satisfy the Guru's statement.

Bob also knows that every blue will also go through this process, and determine their true eye color, and they will all leave that night.

 

[deserteagleeye]

Taping a slice of buttered toast on the back of a cat and dropping it.

 

[ShakyFiend]

Spoiler: My EVER SO CLOSE but ultimately incorrect reasoning

All of the blues leave on that night. No other person leaves, they just know that their eyes aren't blue.
Alright, so nobody on the island can deduce their own eye color, but each blue (let's call this specific guy Bob) knows that there are 99 blues, and 100 browns, with his eye color being an unknown factor. Then, he knows that another blue can determine that there are 98 blues on the island, and 100 browns, with the unknown factors being Bob's eye color and this other blue's eye color (call him Terry). Bob figures that Terry will also make this deductive statement, coming to the conclusion that Terry's eyes are an unknown factor, some other blue knows his respective eyes are an unknown factor. Through that, Bob can determine that this blue knows for sure that there are at least 97 blues on the island, and 100 browns. From this, Bob can go through that same process all the way down to 0 (considering his knowledge started at 99 + unknown factor, then decreased to 98 + 2 unknown factors at the 99th blue, etc). Bob then knows, that the final person who's making this logical statement (knowing that everyone's sense of logic will go down this road) can in fact be Bob himself, considering his logical process is identical to all the other blues. So, Bob can deduce that there are at least 0 blues, and 100 browns. Considering there are 101 people left, 100 of them definitavely brown, Bob can figure out that his eyes are blue in order to satisfy the Guru's statement.

Bob also knows that every blue will also go through this process, and determine their true eye color, and they will all leave that night.

[/quote]

This is the 'official' solution, you got a lot closer than I did

Spoiler: Anyone else dont read this until youve given it a go

Solution to the Blue Eyes puzzle

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. If you're not convinced, you can contact me (info available on my website) for further discussion. 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.

 

[ShakyFiend]

Spoiler: My EVER SO CLOSE but ultimately incorrect reasoning

All of the blues leave on that night. No other person leaves, they just know that their eyes aren't blue.
Alright, so nobody on the island can deduce their own eye color, but each blue (let's call this specific guy Bob) knows that there are 99 blues, and 100 browns, with his eye color being an unknown factor. Then, he knows that another blue can determine that there are 98 blues on the island, and 100 browns, with the unknown factors being Bob's eye color and this other blue's eye color (call him Terry). Bob figures that Terry will also make this deductive statement, coming to the conclusion that Terry's eyes are an unknown factor, some other blue knows his respective eyes are an unknown factor. Through that, Bob can determine that this blue knows for sure that there are at least 97 blues on the island, and 100 browns. From this, Bob can go through that same process all the way down to 0 (considering his knowledge started at 99 + unknown factor, then decreased to 98 + 2 unknown factors at the 99th blue, etc). Bob then knows, that the final person who's making this logical statement (knowing that everyone's sense of logic will go down this road) can in fact be Bob himself, considering his logical process is identical to all the other blues. So, Bob can deduce that there are at least 0 blues, and 100 browns. Considering there are 101 people left, 100 of them definitavely brown, Bob can figure out that his eyes are blue in order to satisfy the Guru's statement.

Bob also knows that every blue will also go through this process, and determine their true eye color, and they will all leave that night.

This is the 'official' solution, you got a lot closer than I did

Spoiler: Anyone else dont read this until youve given it a go

Solution to the Blue Eyes puzzle

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. If you're not convinced, you can contact me (info available on my website) for further discussion. 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.

[/quote]

 

[DrummerM]

Not sure if it's been said yet, but the one where you strap buttered toast to a cat's back with the butter facing up.

Anti-gravity!

 

[legion431]

You've got a sword that can penetrate any shield and a shield invincible to any sword or sharp object. What happens when your special sword hits your special shield?

What happens when an unstoppable force meets an immovable object. They contradict each other so therefore they either cannot exist or something much worse would happen.