Julia can you solve this IQ problem?
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| vigorous splenetic bawdyhouse feces | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Exhilarant Territorial Multi-billionaire Parlour | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | mint exciting pit | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | mint exciting pit | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | mint 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bawdyhouse feces | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | Thriller blood rage | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | Thriller blood rage | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | mint exciting pit | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | mint exciting pit | 10/06/14 | | Thriller blood rage | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | azure dopamine twinkling uncleanness | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Twisted useless brakes potus | 10/07/14 | | azure dopamine twinkling uncleanness | 10/06/14 | | poppy school cafeteria therapy | 10/06/14 | | Exhilarant Territorial Multi-billionaire Parlour | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | Thriller blood rage | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Thriller blood rage | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adulterous hominid | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Thriller blood rage | 10/06/14 | | adulterous hominid | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adulterous hominid | 10/06/14 | | mint exciting pit | 10/06/14 | | azure dopamine twinkling uncleanness | 10/06/14 | | Thriller blood rage | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Thriller blood rage | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Thriller blood rage | 10/06/14 | | Well-lubricated abode | 10/06/14 | | Swashbuckling brunch love of her life | 10/06/14 | | soggy sex offender | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Exhilarant Territorial Multi-billionaire Parlour | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adulterous hominid | 10/06/14 | | big sweet tailpipe | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | soggy sex offender | 10/06/14 | | Well-lubricated abode | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | Well-lubricated abode | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | Swashbuckling brunch love of her life | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Well-lubricated abode | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Well-lubricated abode | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Well-lubricated abode | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | Coiffed charcoal elastic band death wish | 10/07/14 | | Well-lubricated abode | 10/06/14 | | 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| | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | autistic bull headed filthpig giraffe | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | soggy sex offender | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | Charismatic impressive locale | 10/06/14 | | Irradiated Trust Fund Private Investor | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | Irradiated Trust Fund Private Investor | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | Thriller blood rage | 10/06/14 | | soggy sex offender | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | soggy sex offender | 10/06/14 | | Startling wonderful site boltzmann | 10/06/14 | | rebellious address | 10/06/14 | | Ruddy cuckoldry | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | frisky candlestick maker chad | 10/06/14 | | Aquamarine water buffalo university | 10/06/14 | | cyan effete queen of the night | 10/06/14 | | Vibrant spot laser beams | 10/06/14 | | Amethyst racy lodge pistol | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Maroon Incel Dragon | 10/07/14 | | Twisted useless brakes potus | 10/07/14 | | electric gas station police squad | 10/06/14 | | rebellious address | 10/06/14 | | electric gas station police squad | 10/06/14 | | rebellious address | 10/06/14 | | electric gas station police squad | 10/06/14 | | citrine bat-shit-crazy office mad-dog skullcap | 10/06/14 | | soggy sex offender | 10/06/14 | | Vibrant spot laser beams | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Heady Peach Trailer Park | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | coral public bath nibblets | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | soggy sex offender | 10/07/14 | | Ruddy cuckoldry | 10/06/14 | | vigorous splenetic bawdyhouse feces | 10/06/14 | | Startling wonderful site boltzmann | 10/06/14 | | pearl razzle-dazzle nowag pozpig | 12/14/16 | | multi-colored lettuce | 10/06/14 | | Crystalline Talented Principal's Office Digit Ratio | 10/06/14 | | adventurous area really tough guy | 10/06/14 | | Crystalline Talented Principal's Office Digit Ratio | 10/06/14 | | electric gas station police squad | 10/06/14 | | Crystalline Talented Principal's Office Digit Ratio | 10/06/14 | | spectacular sepia stage | 10/07/14 | | Crystalline Talented Principal's Office Digit Ratio | 10/07/14 | | Lavender infuriating coffee pot school | 10/06/14 | | electric gas station police squad | 10/06/14 | | sapphire cracking hospital | 10/06/14 | | Walnut Contagious Pervert Point | 10/06/14 | | electric gas station police squad | 10/06/14 | | sapphire cracking hospital | 10/06/14 | | anal toilet seat rigpig | 10/07/14 | | Maroon Incel Dragon | 10/07/14 | | Geriatric friendly grandma | 10/07/14 | | Maroon Incel Dragon | 10/07/14 | | jade self-centered market voyeur | 10/07/14 | | Maroon Incel Dragon | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Maroon Incel Dragon | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Well-lubricated abode | 10/07/14 | | jade self-centered market voyeur | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Maroon Incel Dragon | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Orange Mischievous Nursing Home | 10/07/14 | | Curious Immigrant | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Curious Immigrant | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Curious Immigrant | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | spectacular sepia stage | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | spectacular sepia stage | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | Thriller blood rage | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | Thriller blood rage | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Thriller blood rage | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | Thriller blood rage | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | vigorous splenetic bawdyhouse feces | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | soggy sex offender | 10/07/14 | | vigorous splenetic bawdyhouse feces | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | Well-lubricated abode | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Well-lubricated abode | 10/07/14 | | vigorous splenetic bawdyhouse feces | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | vigorous splenetic bawdyhouse feces | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | vigorous splenetic bawdyhouse feces | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | adventurous area really tough guy | 10/07/14 | | spectacular sepia stage | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Crystalline Talented Principal's Office Digit Ratio | 10/07/14 | | Well-lubricated abode | 10/07/14 | | mint exciting pit | 10/07/14 | | Blue sexy messiness | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Blue sexy messiness | 10/07/14 | | azure dopamine twinkling uncleanness | 10/07/14 | | Blue sexy messiness | 10/07/14 | | azure dopamine twinkling uncleanness | 10/07/14 | | spectacular sepia stage | 10/07/14 | | anal toilet seat rigpig | 10/07/14 | | Concupiscible Nudist Travel Guidebook | 10/07/14 | | Concupiscible Nudist Travel Guidebook | 10/07/14 | | mint exciting pit | 10/07/14 | | emerald aggressive locus | 10/09/14 | | vigorous splenetic bawdyhouse feces | 11/24/14 | | vigorous splenetic bawdyhouse feces | 11/26/14 | | cerebral rusted brethren tanning salon | 12/12/16 | | arousing very tactful organic girlfriend | 12/14/16 | | cerebral rusted brethren tanning salon | 12/14/16 | | arousing very tactful organic girlfriend | 12/14/16 | | cerebral rusted brethren tanning salon | 12/14/16 | | arousing very tactful organic girlfriend | 12/14/16 | | Well-lubricated abode | 08/09/19 | | sadistic vivacious persian turdskin | 08/09/19 | | abusive sanctuary | 08/09/19 |
Poast new message in this thread
Date: October 6th, 2014 11:30 AM Author: mint exciting pit
Link: http://io9.com/can-you-solve-the-hardest-logic-puzzle-in-the-world-1642492269?utm_source=recirculation&utm_medium=recirculation&utm_campaign=mondayAM
"
You visit a remote desert island inhabited by one hundred very friendly dragons, all of whom have green eyes. They haven't seen a human for many centuries and are very excited about your visit. They show you around their island and tell you all about their dragon way of life (dragons can talk, of course).
They seem to be quite normal, as far as dragons go, but then you find out something rather odd. They have a rule on the island which states that if a dragon ever finds out that he/she has green eyes, then at precisely midnight on the day of this discovery, he/she must relinquish all dragon powers and transform into a long-tailed sparrow. However, there are no mirrors on the island, and they never talk about eye color, so the dragons have been living in blissful ignorance throughout the ages.
Upon your departure, all the dragons get together to see you off, and in a tearful farewell you thank them for being such hospitable dragons. Then you decide to tell them something that they all already know (for each can see the colors of the eyes of the other dragons). You tell them all that at least one of them has green eyes. Then you leave, not thinking of the consequences (if any). Assuming that the dragons are (of course) infallibly logical, what happens?
If something interesting does happen, what exactly is the new information that you gave the dragons?"
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26465775) |
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Date: October 6th, 2014 12:43 PM Author: Heady Peach Trailer Park
fuck Julia. what did you write?
i was referring to the first question
the second question of the riddle is flawed, but again nothing happens
there is a third reason nothing happens, too
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466291) |
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Date: October 6th, 2014 12:46 PM Author: vigorous splenetic bawdyhouse feces
It's because you're so self-assured despite evidence that you're not very smart. Many XOers fall prey to this.
Okay. Put yourself in the mind of a green-eyed dragon. We know you're perfectly logical and will make all available deductions from all available evidence, which you will gather without any mistakes. We also know you and all other dragons now know that there is at least one green-eyed dragon.
1. First, say there are no other green-eyed dragons. Since you know there is at least one, it must be you. You will turn into a sparrow at midnight.
2. Next, say there is only one other green-eyed dragon. If that dragon didn't turn into a sparrow at midnight on the first day, despite being perfectly logical and rule-following, you know he saw another green-eyed dragon. You can see all the other dragons and none of them have green eyes. So you must be the second. The two of you turn into sparrows at midnight on the second day.
3. Well, say there are three. If you only see two and they didn't turn into sparrows at midnight on the second day, you know they saw a third and you're the only option left, similar to above.
.
.
.
x. If you see x-1 dragons with green eyes but nobody turned into a sparrow on the x-1th day, you know there are actually x dragons with green eyes and that you're one of them. On the xth day, at midnight, you will turn into a sparrow along with the x-1 perfectly logical dragons you see.
This would have been more deftly explained recursively, but that's harder in English (as opposed to in code).
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466321) |
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Date: October 6th, 2014 12:50 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 12:50 PM
Author: sealclubber
1 doesn't apply. every dragon knows that every other dragon has green eyes
2 you're still flawed, but i'll hold off on this one
you don't understand the riddle so you are making stupid assumptions, ass fuck
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466351) |
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Date: October 6th, 2014 12:52 PM Author: Heady Peach Trailer Park
you are just making yourself look foolish
read this and tell me what it means to you "if a dragon ever finds out that he/she has green eyes"
now, tell me how the person changed anything for the dragons?
we're still not done, btw
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466360) |
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Date: October 6th, 2014 12:53 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 12:52 PM
Author: sealclubber
you are just making yourself look foolish
read this and tell me what it means to you "if a dragon ever finds out that he/she has green eyes"
now, tell me how the person changed anything for the dragons?
we're still not done, btw
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466369) |
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Date: October 6th, 2014 1:02 PM Author: Exhilarant Territorial Multi-billionaire Parlour
u sound dumb but here goes
before the hiepo, no dragon knew for 100% certainty what eye color it had. the other dragons will never tell him and there are no mirrors and shit (i think tht's part of the hiepo). so prior to the dude coming to the island, a dragon will never turn into a sparrow
now, for simplicity's sake imagine there are only 2 dragons on the island. both dragons have green eyes but they don't know their own eye color, only the other dragon's. now, the dude says, one of you has green eyes. each dragon imagines it's the other dragon (b/c they don't know they have green eyes themselves, and they know for sure the other dragon has green eyes). so each dragon expects the other dragon to turn into a sparrow. the next day, when neither dragon has turned into the sparrow, each dragon realizes that the OTHER dragon expected it to turn because it has green eyes, ie they both have green eyes. so on day 2, both have to turn into sparrows
repeat for 3 dragons - dragon A will look at B and C, knows they have green eyes, and will expect them to both turn into sparrows (A thinks he does not have green eyes). B will look at A and C and expect the same, C will look at A and B. so on day one, none of them will change. So A will think - "why didn't B or C change? B doesn't know his eye color so he would have expected C to change. C doesn't know his eye color so he would have expected B to change. (remember A thinks he is not green)." so similar to the 2 dragon hypo, on day 2 both B and C should change (because if A was really not green eyed, B and C would realize it was just the two of them). when they don't, A realizes, oh shit, they didn't change because i am green eyed as well. so o nday 3, all 3 change
repeat for 100 dragons
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466430)
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Date: October 6th, 2014 1:10 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 1:09 PM
Author: sealclubber
thanks.
no dragon knows what color eyes it has now with 100% accuracy. because they are logically infallible, they will never inquire. they have no incentive to know. what they know is that no dragon can be 100% sure unless someone tells them.
"expects the other dragon to turn into a sparrow" wrong. it doesn't matter what color your eyes are, it only matters if you know what color your eyes are. if you are perfectly logical and assume your eyes are the same color as everyone else's, you will never want to know what color your eyes are and you will never tell anyone what color their eyes are in case they do the same to you.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466472) |
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Date: October 6th, 2014 1:20 PM Author: autistic bull headed filthpig giraffe
Yeah.
Yeah.
That was precisely my thought too.
TBF, I'm a math dood so I'm used to logic (though I thought/think it's stupid and therefore gave up on it), but still...the hubris is pretty bad combined with his complete lack of knowledge in this area.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466542) |
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Date: October 6th, 2014 1:36 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 1:36 PM
Author: sealclubber
yeah, and midnight starts a day, doesn't end it
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466659) |
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Date: October 6th, 2014 3:39 PM Author: Swashbuckling brunch love of her life
I think I understand the reasoning now, from reading other explanations (which I thought spelled this out better).
I still have a problem with the notion that there is any new knowledge.
Every dragon already knew that at least one dragon had green eyes. Also, every dragon already knew that every other dragon knew that there was at least one dragon with green eyes (barring some really unlikely even where dragons are isolated from others on the island).
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467622) |
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Date: October 6th, 2014 3:55 PM Author: Thriller blood rage
You need to go a step farther. For three dragons:
1. Every dragon knows at least one dragon has green eyes
2. Every dragon knows that every other dragon sees at least one pair of green eyes.
3. However, every dragon does NOT know that every other dragon knows that every dragon sees at least one pair of green eyes. That part is only confirmed after waiting two nights.
Still, having trouble for how this conclusion spirals with more dragons.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467764) |
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Date: October 6th, 2014 1:35 PM Author: poppy school cafeteria therapy
can you email me your post? 110 IQ bro here who can't think beyond the initial "Ok so yeah my dragon bros have green eyes." Curious to have someone else spell it out for me so that I can convince myself I would have figured it out on my own if I felt like it. tyia.
EDIT: nvm saw it elsewhere.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466643) |
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Date: October 6th, 2014 12:21 PM Author: autistic bull headed filthpig giraffe
I see two possibilities:
1. They realize all of them have the same eye color therefore, they all become sparrows. You've just identified to them that the color of their eyes is green.
2. Each individual dragon realizes that everyone but them (they cannot observe their own eyes) has the same color eyes. Therefore, they must be the weird one out and therefore they must have green eyes. Poof they become a sparrow.
Finally, the authors of this puzzle seem to have failed to include in your hypothetical that you are Julia. Who else would destroy a unique habitat by violating the one rule for no reason other than proving your intellectual superiority?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466150) |
Date: October 6th, 2014 12:39 PM Author: Well-lubricated abode
I know the answer to this because a similar problem came up in a linguistics class many moons ago when I was an undergrad.
But part of the commonly accepted answer has always bugged me.
Are we discussing this or still waiting for Julia?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466263) |
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Date: October 6th, 2014 12:48 PM Author: Well-lubricated abode
This is a common knowledge problem.
http://en.wikipedia.org/wiki/Common_knowledge_(logic)
http://plato.stanford.edu/entries/common-knowledge/
I haven't gone back and rethought about this in depth, but the new information is supposed to be the newly introduced common knowledge that the dragons (or the people, or whoever is in the particular version of the problem problem) now know that others know.
In other words, the public declaration makes every dragon realize, "I now know that those other dragons also know that there is at least one green eyed dragon."
But I have a hard time believing that they didn't have that knowledge before the announcement. Absent the announcement, it seems the dragons still would have said, "the other dragons are intelligent, speaking, reasoning beings and, therefore, they have the same knowledge that I have."
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466334)
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Date: October 6th, 2014 12:59 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 12:58 PM
Author: sealclubber
is it fair to assume that the dragons don't want to be sparrows?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466407)
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Date: October 6th, 2014 1:03 PM Author: Heady Peach Trailer Park
Date: October 6th, 2014 1:02 PM
Author: analysis per se
What they "want" has nothing to do with this logic problem, sealclubber. The fact that you even mentioned it shows that you have no idea what's going on. Among thinking people you are like some sort of carnival attraction, or a caged animal at a zoo mimicking human movements outside but failing utterly to reproduce them in a sensible or meaningful way.
wrong, bitch. they are infallibly logical. their own wants would be part of being logical. is it fair to assume they don't want to be sparrows?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466439) |
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Date: October 6th, 2014 12:59 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 12:59 PM
Author: sealclubber
yeah, I am. unless I am just missing one of those you are the driver of the bus parts, I am going to shred you
so wait
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466408) |
Date: October 6th, 2014 1:14 PM Author: Thriller blood rage
I'm still having a big problem with the "new information" aspect of this. I completely agree with how this would go with 1, 2, or 3 dragons, and I get that, rationally, this should be able to continue all the way to 100 dragons.
However, something seems off about this whole way of looking at the problem. In the case of only two or three dragons, the dragons DON'T automatically know that every dragon can see multiple green eyes. To look at the case of 3 dragons:
-Each dragon sees two pairs of green eyes.
-However, he know's it's POSSIBLE that the other two dragons only see a single pair of green eyes, because he doesn't know his own eyes are green.
-He knows a dragon that only sees one pair of green eyes would wait one night to see if the other leaves. Then, when the other pair doesn't, he realizes he has green eyes as well and both leave.
-Then this entire situation fails to come to pass after two nights, each dragon realizes that all of them were seeing two pairs of green eyes the entire time and all leave.
However, this seems to break down for me once we have 100 dragons. In this case, all of them know that every other dragon can see multiple sets of green eyes. Therefore, I'm not seeing how the cascade starts, because there is no dragon who could even potentially be seeing a single pair of green eyes.
Edit: To try to express this in a more straightforward manner- If there were only three dragons, the statement "at least one of you has green eyes" COULD, from the perspective of each dragon, have been new information to one of the dragons present. At a larger number of dragons, that possibility disappears.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466501) |
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Date: October 6th, 2014 1:32 PM Author: Thriller blood rage
In the three-dragon case, even if every dragon ACTUALLY sees two other green-eyes, the possibility exists to them that one of the dragons actually only sees one pair and could think one of them sees none.
Suppose I am Dragon A, and see that Dragons B and C have green eyes.
Now, the visitor tells us all that at least one of us has green eyes.
If I assume I don't have Green eyes, then I can put myself in the shoes of Dragon B. Dragon B would see C's green eyes, and non-green eyes for A. If that were the case this hypothetical Dragon B could imagine that C actually sees no green eyes, realizes he has them, and becomes a sparrow. When that doesn't happen for C, B realizes C can see Green eyes as well, they must be his, and he becomes a sparrow. When that doesn't happen after two nights, the entire hypo falls apart FOR DRAGON A, and he realizes that every dragon in the triad sees two pairs of green eyes, and he becomes a sparrow. Since this occurs for all three, they all change.
The problem I'm describing is that once we have FOUR different dragons, every dragon already knows with certainty that every other dragon can see multiple pairs of green eyes. Before, it was POSSIBLE that one of the other dragons could not, and now it is impossible. That's where I currently feel there may be a breaking point in the problem, pending a more in-depth explanation.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466615) |
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Date: October 6th, 2014 1:29 PM Author: azure dopamine twinkling uncleanness
You build out from the 2 dragon scenario:
2 Dragons, each with green eyes, each assumes he does not have green eyes. When neither turns sparrow after Day 1, both realize they have green eyes.
Add Dragon 3. He assumes he does not have green eyes. He sees 2 green eyed dragons. After Day 1 no one turns sparrow. Dragon 3 knows if he has non-green eyes, both Dragon 1 and Dragon 2 will realize they are green eyed and turn sparrow after Day 2 (the 2 dragon scenario). Each of Dragon 1 and Dragon 2 have the same perspective (each knows the two dragons other than himself must turn sparrow after day 2). Day 3 comes around and all are still dragons. Dragons 1, 2, and 3 realize they have green eyes.
Repeat as the numbers get bigger.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466603) |
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Date: October 6th, 2014 1:33 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 1:32 PM
Author: sealclubber
no, you are adding information that is illogical and more importantly, irrelevant
why would any dragon assume he doesn't have green eyes?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466626) |
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Date: October 6th, 2014 1:50 PM Author: vigorous splenetic bawdyhouse feces
Date: October 6th, 2014 1:49 PM
Author: sealclubber
his eye color doesn't matter
they all see at least one other dragon with green eyes, regardless of what color your own eyes are, regardless of which dragon you are. every dragon of the three has known this all along
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466784) |
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Date: October 6th, 2014 1:52 PM Author: Thriller blood rage
Yeah, but that in turn rebounds on the problem itself, and is making me suddenly doubt it.
Let's ignore people saying the dragons "assume" things, as I think that's a distraction. A better term is that they need to "make certain" of things.
When you see that the dragons need to "make certain" of their eye color, the small-number scenarios become divorced from the larger ones and so I'm not sure the whole N+1 chain works.
I'll try to explain this again, and I'm sorry that I'm kinda shitty at wording all this (it doesn't help that I haven't taken a college level class in logic or proofs):
Consider the case of 2 dragons, both green: Both know that at least one dragon has green eyes, but they do NOT know that every dragon knows that. The visitor's remark adds NEW information, making them aware that all dragons know there is a pair of green eyes. This prompts each dragon to consider the perspective of the other. If I'm Dragon A, until the visitor's remark I did not CONCLUSIVELY know that Dragon B knew there were any green eyes on the island. The first night then becomes a "test" of sorts. If Dragon B leaves, he had seen no other green eyes beforehand, deduced he was the only one, etc. When this doesn't happen, we both realize we have green eyes, blah blah you've heard all of this. The key here is that this is a problem of common knowledge. We all knew something, but we didn't know everybody else knew it.
Now, the same holds true for the 3-dragon case, as I described it above. Even if we all see two pairs of green eyes, it's possible for Dragon A to THINK that Dragon B sees only one pair, and that in his turn Dragon B would think Dragon C doesn't see any at all. Then the cascade can occur as that possibility is disproven.
But the thing is, it seems to me this can no longer happen if you have, say, 8 dragons. Even if A's eyes were brown, there's no way any other dragon can believe that another dragon ONLY sees one or zero sets of green eyes.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466804) |
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Date: October 6th, 2014 2:14 PM Author: adulterous hominid
"Even if A's eyes were brown, there's no way any other dragon can believe that another dragon ONLY sees one or zero sets of green eyes."
Did you read the problem closely? Right at the beginning, it says literally every dragon has green eyes.
"You visit a remote desert island inhabited by one hundred very friendly dragons, ALL OF WHOM HAVE GREEN EYES."
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466959) |
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Date: October 6th, 2014 2:36 PM Author: Well-lubricated abode
It seems to me that these two propositions are true:
(1) Before the announcement, every dragon already knew that at least one dragon had green eyes. All they had to do was look around.
(2) Before the announcement, every dragon already knew that every other dragon knew that at least one other dragon had green eyes. They had this knowledge about the other dragons' knowledge because they knew that the other dragons were seeing, thinking creatures.
Are those propositions true?
If so, then what information did the announcement add?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467152) |
Date: October 6th, 2014 1:47 PM Author: Exhilarant Territorial Multi-billionaire Parlour
here's a good explanation (similar hypo just without dragons)
http://www.explainxkcd.com/wiki/index.php/Blue_Eyes
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26466749) |
Date: October 6th, 2014 2:32 PM Author: frisky candlestick maker chad
Haven't read any comments.
Assuming that they know what the color 'green' means, here we go.
For one dragon, if you tell that dragon "at least one dragon has green eyes," the dragon is gonna go AW FUCK and turn into a sparrow. That's trivial.
For two dragons, if you tell that dragon "at least one dragon has green eyes," and they look at each other like FAGGOTS, D1 is gonna be like "lol this bitch has green eyes, and is going to turn into a sparrow tonight." D2 is gonna think the same thing. Night comes, nothing happens. They look at each other and they're like "aw shit, u faggot, I guess I have green eyes too if you were sure that someone else had green eyes" and then they turn into sparrows.
And then I'm pretty sure it works the same way up to 100, and in fact it's made more, not less, obv. If some faggot dragon looks around and sees a sea of 99 green eyes looking back at him, he's gonna think to himself "lol i'm in the clear, fuck these faggots." Every stupid ass dragon is going to think this. Then night is going to come and nothing is going to happen. Then every dragon is going to go "oh FML" and turn into a faggot ass sparrow.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467111)
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Date: October 6th, 2014 2:53 PM Author: frisky candlestick maker chad
i mean i get the confusion, i actually think this is going to take a fuckload of days. i am not a math guy but i assume it's going to take 99 or 100 days. twist?
like, they're all gonna sit around looking at each other's faggot asses for 99 days and be like "lol okay. it's like this. there are 100 of us faggots. we like to cum on dey face; we don't talk about eye color. that's not our thing. but we are all logical as FUARK. so what about this: what if we all wait around for as many days as there are dragons (or maybe 1 less i guess, since 2 dragons can figure this out in 1 day). if i literally sit here and wait for 99 days, i know that every other dragon is doing the same thing. waiting for 99 days. during those 99 days he's thinking, "i am aware that everyone else has green eyes. so each day, a counter ticks. each day every dragon is thinking to himself, 'YOLO i'm not a sparrow yet', because every dragon has concluded that there are 99 green eyes looking out there at him. but not me. it's not gonna be me. every day each fucking faggot dragon is ticking off another dragon from his list of, like, 'dragon 2, today, could have concluded that it wouldn't have been him because dragon 37 has green eyes, so THAT's why he's still around; not because *I* have green eyes.' so i know that by the 99th day, these guys are going to conclude, like me, that they have all of the information they need: either literally EVERYONE ELSE BUT ME has green eyes, or EVERYONE ELSE AND ME has green eyes."
this is actually kind of hard to explain verbally but do you kind of follow? it's like, idk, if i'm dragon A and I see dragon B has green eyes, figuring that shit out takes one night.
if i'm dragon A and i see dragon B and C have green eyes, i have to wait 2 nights. because on night 1, dragon B could have been like "well i know that dragon C has green eyes so i'm str8." but after 2 nights dragon B is still around, and still hasn't concluded that he has green eyes, so dragon A is like "oh FML"
does this make sense
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467287) |
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Date: October 6th, 2014 3:55 PM Author: adventurous area really tough guy
Let's take the three dragon hypo.
There are three green eyed dragons and no visitor. They each see two other green eyes and they each know that every dragon sees at least one set of green eyes.
Dragon A thinks he has brown eyes and sees Dragons B and C with green eyes.
The cascade only starts if Dragon B wonders why Dragon C hasn't disappeared since Dragon C see would otherwise see Dragon B's brown eyes (and Dragon A's). Thus Dragon B and C must disappear.
But without at AT LEAST prompting, there is no problem with a hypothetical Dragon C not disappearing because he sees brown eyes on Dragon A and B.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467768)
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Date: October 6th, 2014 3:44 PM Author: vigorous splenetic bawdyhouse feces
No, they didn't.
The base case is that if you don't see any green-eyed dragons, you have green eyes.
This wasn't true before the announcement.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467672) |
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Date: October 6th, 2014 3:57 PM Author: Well-lubricated abode
I don't mean to be dense, but I don't think I understand your argument.
Describe to me the actual, *new* thought that popped into each dragon's head upon the announcement.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467781)
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Date: October 6th, 2014 2:59 PM Author: adventurous area really tough guy
SO for each increasing number of dragons, we think of it as the previous number of dragons, plus a brown eyed observer dragon who fails to see the expected result.
SO to carry this to 4 dragons.
It's been explained well in this thread that if there were 3 dragons with all green eyes, they would all disappear on the third night because they each expect the other two to disappear after the second, which fails to happen.
The fourth dragon, knowing that this will happen to a group of three green eyed dragons, who see his brown eyes, expects all three to disappear the third night, when that doesn't happen he realizes he is pwned and disappears the next night.
This happens for each N+1 - they think they are observing the problem and expect all of them to disappear on night N.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467332)
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Date: October 6th, 2014 3:16 PM Author: adventurous area really tough guy Subject: Re. New Information
So if we have three or more dragons - why does the announcement set off the transformation? each dragon knows that he can see two other dragons with green eyes and he already knows that each dragon knows that there is at least one other dragon with green eyes.
Shouldn't the transformation happen by itself assuming all dragons know the rules about transforming and know that all other dragons know the rule?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467438)
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Date: October 6th, 2014 4:59 PM Author: 180 Generalized Bond Step-uncle's House
Isn't there a third possibility here? Everyone else has non-green eyes and I am the only one with green eyes? And doesn't this prevent the introduction of the new info being "what is green" from starting the cascade? And then aren't we back to the problem of why hasn't the cascade happened on its own?
When I don't turn sparrow the first night can I assume that the third possibility is not correct? There seems to be a lack of certainty with this scenario, though, where I'm assuming "if scenario three is true then I'll turn." (i.e. when I don't turn on the first night I can think that I'm not the only one with green eyes and that's why I didn't turn, or I can think that I didn't know, for sure whether I was the only one with green eyes and that's why I didn't turn)
The second part of this riddle is really bugging me.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26468098) |
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Date: October 6th, 2014 3:20 PM Author: frisky candlestick maker chad
Wait, why wouldn't it have already happened that? If they all already knew the rules, and each dragon knew that every other dragon had green eyes as soon as the rules were implemented, and the lady arrives 101 days later, why isn't she arriving to sparrow island?
Or if the lady arrives 51 days after they know this, and says 'at least one of you has green eyes', why aren't the dragons being like "look faggot lady we know that, we're working it out."
It's unclear to me why what she says can ever be what they already know.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467474) |
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Date: October 6th, 2014 3:32 PM Author: adventurous area really tough guy
if you look at the blue and green and brown eye hypo here
http://www.explainxkcd.com/wiki/index.php/Blue_Eyes
it assumes knowledge by the people of the different colors, otherwise the statement would not have any impact on the blue or brown eyed group.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467560) |
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Date: October 6th, 2014 3:38 PM Author: adventurous area really tough guy
I still dont see how knowing the color is relevant. Let's say the dragons already know what green is and that everyone else has green eyes.
The original question should be the same. Either the knowledge, which everyone knows, sets of the cascade, or it was already going to happen.
I can see why that statement is necessary in a two dragon scenario - which starts the cascade, but in a 100 dragon scenario, they all already know that statement to be true.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467613) |
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Date: October 6th, 2014 3:29 PM Author: adventurous area really tough guy
it relies on there being a two dragon scenario that triggers the cascade in the minds of the dragons.
Why I dont quite get is if there being numerous dragons does not affect the cascade, then why wouldn't it happen on its own.
I know its important for the base case to start, but how can adding no new information start it?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467534) |
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Date: October 6th, 2014 3:25 PM Author: adventurous area really tough guy
How does telling a group something they already know start an inductive chain?
I understand it goes back to there being two dragons both assuming they do not have green eyes.
But in a group of three, we rely on imagining that this green eyed dragon comes to the realization when he sees other dragons but they dont disappear. Why wont the inductive chain start on its own?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467503) |
Date: October 6th, 2014 3:21 PM Author: Charismatic impressive locale
hold on
*retrieves glove and turns it inside out*
ok, back (juli
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467481) |
Date: October 6th, 2014 3:30 PM Author: Irradiated Trust Fund Private Investor
Not going to read all the answers, but my thinking is that the dragons must not know what green is since they ALL have green eyes. Each one will see that every single other dragon has all the same eye color, and will assume that he himself is the different, and therefore green, one. All the dragons transform that night.
Yes?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467541) |
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Date: October 6th, 2014 3:43 PM Author: Irradiated Trust Fund Private Investor
I don't think it does. It is told from the perspective of you the visitor. YOU know they have green eyes.
That being said, as someone else noted below they are very friendly dragons. They may not want to declare anyone else to have green eyes. They may stay silent as they realize everyone else has green eyes, but assume perhaps they are an outlier who does not. So in this case none would change.
So if they DO know what green is, they are too nice to note it (and never speak of it), so no one changes.
If they DON'T know what green is, they do all self-report and all change.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467664) |
Date: October 6th, 2014 3:42 PM Author: Ruddy cuckoldry
tl;dr:
1. twist sets out a trap with a big ole hunk of julia bait in it
2. sealclubber, who lacks the capacity to distinguish between julia bait and a delicious well-done porterhouse slathered in heinz 57, jumps right on in and gets mangled beyond recognition
3. julia arrives late to the party, looks down to see sealclubber's grotesquely maimed body, and decides it would be a good idea to stick her fingers in the contraption
all in all a peak experience for everybody involved
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467652) |
Date: October 6th, 2014 4:22 PM Author: citrine bat-shit-crazy office mad-dog skullcap
I think you guys need to step back and consider this from the POV of 1 dragon (D1). We can assume from the question that D1:
1) believes he has non-green eyes, or he would've sparrowed out long ago
2) believes the other 99 dragons ALL have green eyes, because he knows what green eyes are
3) based on 1 and 2, he realizes that ALL of the other dragons with green eyes are EACH individually unaware that he himself actually has green eyes.
4) D1 is told that at least 1 dragon has green eyes. he already KNOWS this
therefore, the next morning, when nothing changes, none of D1's information has changed and none of his assumptions have changed either. unless D1 is a dick, he's not going to tell any other dragon they have green eyes. And why would he now? He could've done that ages ago. None of this will change as each day passes
TWIST: every single dragon is D1. therefore, nothing will ever change. HTFH
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26467931) |
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Date: October 6th, 2014 5:50 PM Author: adventurous area really tough guy
I think its because the statement leads to more than just the knowledge that there is one dragon with green eyes, which everyone knows.
It also means that after 99 days, there are at least 99 green dragons, which our dragon knows, but he also knows also that EVERY OTHER DRAGON also knows that there are 99 green dragons.
Before the statement, he was under the knowledge that only he knew that there were 99 green dragons and everyone else thought there were 98.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26468443) |
Date: October 6th, 2014 5:21 PM Author: adventurous area really tough guy
So it seems like "At least one person has blue eyes" is already COMMON KNOWLEDGE.
The problem is that that statement after 99 days turns into "at least 99 people have blue eyes" and when those people don't disappear on the 99th day, then everyone goes the 100th day.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26468225)
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Date: October 6th, 2014 6:18 PM Author: coral public bath nibblets
with these sorts of riddles, i can actually witness my iq topping out. i can proceed along an inquisitive path but then don't really know how to process the information that I receive in return.
for bros that are legit dumb, like sealclubber, this must occur every single day. for him this riddle is not extraordinarily difficult or easy--its just somewhere above his paygrade. akin to a riddle like "Three guys run into a bar, the fourth man ducks. Why does he duck?"
its insolvable. its premises must be tested. anyone who can answer it must be bluffing or bloviating.
oh and by the way, he is like one of the top 5 DUI attorneys in el paso, which gives him some grounding
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26468591) |
Date: October 6th, 2014 7:18 PM Author: Crystalline Talented Principal's Office Digit Ratio
How about doing it the other way around?
Pick one dragon, call him Dragon 1. Dragon 1 sees 99 green eyed dragons. Therefore he knows (at least from his POV) that there are either 99 or 100 green eyed dragons on the island. He also knows that no dragon has sparrowed to date.
Dragon 1 therefore has to assume there are 99 and not 100 green eyed dragons and that he is the sole non-green (lets call it red) eyed dragon otherwise he would have sparrowed long ago. He also has to assume that despite him being the only true red eyed dragon all 99 other green eyed dragons must (wrongly) not be convinced that they have green eyes otherwise they would have sparrowed.
What are the consequences of Dragon 1's necessary adoption of the beliefs that (1) he has red eyes and (2) that all the other dragons are mistakenly not convinced they have green eyes in terms of what that means he must believe about what other dragons believe?
Dragon 1 picks out another random dragon, call him Dragon 2, and tries to imagine what Dragon 2 must believe in order for Dragon 1's belief that he is red eyed to be true.
Dragon 1 must believe that when Dragon 2 looks at him he sees he has red eyes otherwise there would be 100 green eyed dragons. So Dragon 1 must believe that Dragon 2 has concluded that there are either 98 or 99 red eyed dragons on the island, and since Dragon 2 has not sparrowed Dragon 2 must also believe he himself has red eyes and that despite him and Dragon 1 being the only true red eyed dragons all 98 other green eyed dragons must (wrongly) not be convinced that they have green eyes otherwise they would have sparrowed.
Dragon 1 now picks out a second random dragon, call him Dragon 3 and tries to imagine what Dragon 2 must believe about what Dragon 3 believes in order for Dragon 2's belief about himself and therefore Dragon 1's belief about himself (ie that they each have red eyes) to both be true.
In order for Dragon 1's belief that he has red eyes to be true then Dragon 2 must also believe he has red eyes, so when Dragon 2 puts himself in the position of Dragon 3, Dragon 2 must believe that Dragon 3 sees two red eyed dragons (Dragon 1 and himself) and thus Dragon 3 must believe that there are either 98 or 97 green eyed dragons on the island. Since Dragon 3 has not sparrowed then Dragon 3 must also necessarily believe he has red eyes and that there are 3 red eyed dragons on the island and 97 green eyed dragons who mistakenly believe that they may have red eyes.
Continuing this on creates a chain of beliefs about what Dragon 1 must believe each succeeding dragon in the chain must believe about the next dragon's belief, such that Dragon 1 must believe Dragon N-1 in the chain must believe Dragon N believes there are (100-N) green eyed dragons on the island so ultimately the chain concludes with a Dragon 99 who must believe there is a Dragon 100 who believes there are no green eyed dragons on the island.
The visitor's statement that there is at least one green eyed dragon on the island means that Dragon 100's belief that there are no green eyed dragons cannot be true, therefore Dragon 99 cannot believe that Dragon 100 believes that there are no green eyed dragons, Dragon 98 cannot believe that Dragon 99 believes Dragon 100 believes there are no green eyed dragons etc etc and the whole logical chain comes crashing down, to Dragon 1's knowledge, and since Dragon 1's belief that there are 99 and not 100 green eyed dragons on the island depends on the validity of this chain of belief then Dragon 1 must conclude there are 100 green eyed dragons and sparrows.
Since the dragons are all equally clever they have all worked out that these chains of belief must exist starting from them and going to each possible Dragon 100 and thus they all realise at the same time there are 100 green eyed dragons on the island.
The new knowledge that the visitor imparted was the destruction of each dragon's belief that there must be dragon on the island who mistakenly believed that there were no green eyed dragons on the island.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26468930) |
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Date: October 6th, 2014 7:33 PM Author: Crystalline Talented Principal's Office Digit Ratio
I wondered about that. All we know is that each dragon sees 99 green eyed dragons. Since the dragons don't talk amongst themselves about eye colour do we know that each dragon knows that each other dragon actually sees 99 green eyed dragons?
The way I approached it doesn't assume each dragon knows each other dragon sees 99 green eyed dragons, it just proceeds on the basis of what each dragon must necessarily believe each other dragon believes in order for the first dragon's beliefs about itself to be true.
I agree that if we assume each dragon knows every other dragon sees 99 green eyed dragons, and that therefore every dragon knows no dragon can hold the belief that there are 98 or less green eyed dragons then this doesn't work.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26469049) |
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Date: October 6th, 2014 8:21 PM Author: Crystalline Talented Principal's Office Digit Ratio
I agree that dragons have to sparrow only if they know for certain they have green eyes so there may be some looseness in my approach if the difference between a dragon believing it must have red eyes and a dragon not being convinced it has green eyes is relevant to any of the steps.
For the extra bit of information the visitor imparts (i.e. at least one dragon has green eyes) to be meaningful and cause a mass sparrowing (which seems to be the end point the puzzle is headed towards) it seemed to me that new fact must somehow directly contradict a belief or assumption in the existing structure of knowledge and beliefs between the dragons so as to cause that structure to collapse.
That's why I think it must be an element of the existing structure that there is at least a necessary belief that one dragon believes (mistakenly) that there are no green eyed dragons on the island and this was the only way I could think of to get to that position.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26469317)
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Date: October 6th, 2014 9:28 PM Author: Walnut Contagious Pervert Point
The new information that you provide is not only that at least one dragon has green eyes, but that each dragon knows that the next one knows it as well, on down the line.
Consider 3 dragons before your statement. A knows that B has green eyes bc he can see them. A knows that B knows that C has green eyes bc he can see that B can see C. However, A does not know that B knows that C knows there is at least one dragon with green eyes. A assumes that it does not. Therefore if A thinks that B also thinks it does not then A would not know what B thought C thought. Since A thinks it does not have green eyes and that B thinks it does not have green eyes, A could think that B could think that C sees two non-green eyed dragons. When you say at least 1 dragon has green eyes, A now knows that B knows that C knows at least 1 dragon has green eyes.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26469703) |
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Date: October 7th, 2014 3:57 AM Author: Maroon Incel Dragon
Sealclubber was genuinely hilarious.
"yeah, and midnight starts a day, doesn't end it"
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26471591)
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Date: October 7th, 2014 4:13 AM Author: Orange Mischievous Nursing Home
Nothing happens at midnight. But when the dragons notice that no other dragons have turned into sparrows, each dragon individually realizes that he himself must have green eyes as well (because each dragon would have been able to avoid knowledge of their own eye color by seeing the other dragons, after you revealed at least one of them has green eyes)
So at midnight of the second day, they all turn to sparrows
Edit: my logic was on track but I didn't think it out enough to realize it would be the 100th midnight
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26471615) |
Date: October 7th, 2014 9:21 AM Author: Curious Immigrant
Lol this is easy. They still don't know what green is but they know everyone else has same eye color. So when no one else morphs a dragon will believe it is the odd one out w green eyes and will morph. When other dragons realize what happens they will all morph.
I have a 200 iq so this is the correct answer. Hth
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26471893) |
Date: October 7th, 2014 11:57 AM Author: spectacular sepia stage
This is the best way to explain it:
Before the announcement:
If there are 2 dragons, then Dragon A thinks that Dragon B thinks that there are no green eyes.
If there are 3 dragons, then Dragon A thinks that Dragon B thinks that Dragon C thinks that there are no green eyes.
If there are 4 dragons, then Dragon A thinks that Dragon B thinks that Dragon C thinks that Dragon D thinks that there are no green eyes.
...
The announcement negates those chains by adding new information.
I find it incredibly hard to visualize/conceptualize anything above the 3 dragon scenario, but it's logically sound.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26472408) |
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Date: October 7th, 2014 12:16 PM Author: adventurous area really tough guy
Once you get to 4 dragons, how can Dragon A think that any Dragon B thinks that Dragon C thinks that Dragon D thinks there are no green eyes?
Dragon A knows that Dragon B knows that Dragon C knows that Dragon D can see at least 2 green eyes. Each of those dragons can see 2 green eyes so how do we mentally attribute the incorrect though to any dragon even through another dragon's imagination?
It stops being clear after 3, though I get the process.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26472484) |
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Date: October 7th, 2014 12:36 PM Author: spectacular sepia stage
It's the same problem with the 3 dragon scenario:
Dragon A know that everyone sees at least 1 green eye. But Dragon A does NOT know that Dragon B/C knows that everyone sees at least 1 green eye. In fact, Dragon A thinks that Dragon B thinks that Dragon C sees no green eyes. The intermediary is the key.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26472592)
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Date: October 7th, 2014 2:38 PM Author: adventurous area really tough guy
No dragon can be construed to believe there are no green eyes by any dragon - but remember for our dragon A, at each step he is assuming the mistaken belief of the next dragon that such dragon does not have green eyes.
Our dragon A believes that there are 99 green eyes. (first tier)
Our dragon A believes that every other dragon thinks there are 98 eyes. (second tier)
If every other dragon at the second tier thinks there are 98 green eyes, it follows that those dragons (being imagined by A) imagine every other dragon to think that there are 97 green eyes. (third tier)
Even through IRL, the second tier dragons know that the third tier dragons know that dragon A in fact has green eyes, dragon A cannot possibly know or believe this. In dragon A's mind, the second tier dragons are idiots, believing that the third tier dragons are idiots.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473174) |
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Date: October 7th, 2014 2:43 PM Author: Thriller blood rage
Sure, but there is no rational way for them to extend this "presumed idiocy" all the way down to the necessary point of a super-idiot thinking he's the only dragon with green eyes. The chain falls apart.
Edit: I think I'm getting a better grasp on why I find this problematic. The supposed issue is that we can make all these nested beliefs descending all the way down, so that Dragon A can think Dragon YYY might think Dragon ZZZ only sees one pair of green eyes. But that doesn't work, because all dragons are thinking simultaneously. Dragon A is somebody else's Dragon B, who is somebody else's Dragon C, etc. We're basically counting on all of them to be having some collective delusion at the same time and I don't think that logically follows.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473192) |
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Date: October 7th, 2014 2:54 PM Author: adventurous area really tough guy
no one dragon is applying it to more than one other dragon. It's just necessary that each dragon, thinking about another dragon, apply another level of idiocy.
Using 4 is the best way to think about it I think, because with 4, Dragon A knows that each dragon can see at least 2 other green eyes.
Dragon A imagines, what B must think. B must think there are two dragons with green eyes, since B cant think he has green eyes, and A believes he himself does not have green eyes.
Dragon A, also thinks that Dragon C, must think there are two dragons with green eyes. However when Dragon A imagines Dragon B's thought process about Dragon C, he thinks that stupid Dragon B must think that Stupid Dragon C only thinks one dragon (Dragon D) has green eyes.
And so forth, A thinks, B thinks, C thinks that D, being the only dragon with green eyes (in C mind) (in B's mind) (in A's mind) thinks that there must be no dragons with green eyes. The information from the bitch makes D's belief (in C mind) (in B's mind) (in A's mind) impossible.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473235) |
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Date: October 7th, 2014 3:36 PM Author: Thriller blood rage
That much is obvious. I know that in the 4-dragon situation they all change at the same time, after three nights pass.
However, I don't see how this happens with eight dragons because I don't see how they ever conclude for a fact they have green eyes. When I talk about the chain of idiots, I'm talking about the assumptions each dragon has to make about the next succeeding dragon.
Dragon A presumes he has brown eyes. He then imagines he is Dragon B, believing that A and B both have brown eyes, who then imagines Dragon C believing A B and C all have brown eyes, who then imagines Dragon D assuming Dragons A B C and D all have brown eyes. But Dragon A already KNOWS this is impossible, because D would be able to look back and see B's brown eyes. The chain breaks down, without A having to realize he himself has green eyes. The same realization can play out with every dragon, with each being able to rely on the green-ness of other dragon's eyes without having to realize the greenness of their own. And this realization happens well before the cascade can begin.
Fuck, that description is hard to follow even for me.
Edit: To add on, I guess the biggest issue is that when we're descending down the cascade we're "cutting out" earlier dragons as if they don't exist. But they do, they continue to exist throughout and can constantly be mucking up the process. Every dragon knows that all other dragons are making the same calculations, and if I'm Dragon A I know that every other dragons calculations can be fucked up without including myself...and every dragon can reach the exact same conclusion.
People try to "simplify" things by asking us to work from the base case, but that's the exact problem I'm having. We're NOT working up from the easy case of two dragons. We're working DOWN from 100 when EVERY dragon automatically knows that every other dragon can see at least 1, 50, 95 pairs of green eyes.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473473) |
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Date: October 7th, 2014 3:03 PM Author: adventurous area really tough guy
That's the easiest way to explain how the cascade works and the timing. It's correct.
However my going backwards explanation is also correct (I think) and instead of dealing with the mechanics of the cascade, deals with the state of mind of the dragons who are each waiting to see what happens to the group per NS's explanation.
I think you need to think of it backwards towards idiots (he thinks, he thinks, he thinks) to understand why the visitor's information which is already universal knowledge, is not common knowledge. The status quo was the everyone thinking the chain of idiocy, notwithstanding having the knowledge themselves.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473286) |
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Date: October 7th, 2014 3:17 PM Author: Well-lubricated abode
With logic problems, I like to use counterexamples.
For instance:
(1) If it rains, the sidewalk will be wet.
(2) The sidewalk is wet, therefore it must have rained.
2 doesn't follow from 1. And an intuitive way to show that is to say that it's possible that the sidewalk in 2 could have gotten wet from a firehose, a pissing dog, or a bursting dam.
What's the intuitive logical possibility that existed before the dragon announcement?
I imagine it would take the form of "Dragon X was still permitted to believe Y" or "we could not be sure what Dragon X knew about Y."
Of course, those sorts of explanations tend to get harder as the problem gets more complex, and I can't come up with a good one in this case.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473386) |
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Date: October 7th, 2014 6:10 PM Author: Crystalline Talented Principal's Office Digit Ratio
I agree with this and I also found it hard to visualise the chains going up from 3 dragons so I tried doing it around the other way around from 100 down.
I think the keys to approaching this problem are:
1. the fact that no dragon has sparrowed to date is very important, because if we assume dragons don't want to be sparrows, it means that each dragon has, notwithstanding the fact that it can see every other dragon has green eyes, managed to construct an entirely logical scenario where it could nonetheless still have red eyes and each dragon is relying on that same scenario in order not to sparrow
2. the extra fact that the visitor introduces must somehow contradict this existing mass dragon state of belief because the puzzle would be boring if the fact didn't cause anything to happen, and since every dragon is identical there is no reason the new fact would cause only one or some dragons to sparrow, it must cause a mass sparrowing
3. the new fact only directly contradicts a belief that no dragon has green eyes but how could any dragon itself believe this when it can see 99 dragons with green eyes?
4. that leads to the conclusion that the relevant belief can only be something one dragon believes another dragon (mistakenly) believes, which leads you to the chains of belief idea that culminates in the "idiot dragon" who sees no green eyed dragons but in whose existence every other dragon must believe
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474397) |
Date: October 7th, 2014 4:44 PM Author: Blue sexy messiness
could someone explain to me like i'm an idiot (which I clearly am) why the transformation takes place with 3 dragons?
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26473852)
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Date: October 7th, 2014 5:07 PM Author: anal toilet seat rigpig
they know there only has to be one with green eyes to satisfy at least one.
they have to know they have green eyes to change.
dragon A looks to B and C and says ok no worries both green.
midnight 1 comes and nothing happens.
dragon A says ok they were just looking to each other and now both B and C will realize the other saw green in him and they'll both change.
midnight 2 comes and nothing happens.
it's on day 3 that dragon A says oh shit one of these guys had to think i was the at least one with green eyes.
each dragon is having this same realization on day 3
midnight 3 comes and all of them change
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474046) |
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Date: October 7th, 2014 5:13 PM Author: Blue sexy messiness
thakns. Can you explain why this can't be the case:
A sees that B and C both have green eyes --> A satisfied there's >=1 dragon with green eyes
A assumes B thinks that C has green eyes and A has brown eyes --> B satisfied there's >=1 dragon with green eyes
A assumes C thinks that B has green eyes and A has brown eyes --> C satisfied there's >=1 dragon with green eyes
So, since we know that all dragons actually have green eyes, then any dragon can be A, B, or C.
So none of them knows that they, themselves, have green eyes.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474084) |
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Date: October 7th, 2014 5:26 PM Author: Blue sexy messiness
I can't follow your reasoning on this for some reason.
Why is the below wrong:
A assumes B sees C's green eyes and A's brown eyes. Therefore B can't conclude that he (B) has green eyes because he sees someone else with green eyes.
A assumes C sees B's green eyes and A's brown eyes, therefore C can't conclude that he (C) has green eyes because he sees someone else with green eyes.
A KNOWS that both B and C are mistaken and that they actually do have green eyes, but this doesn't matter since they won't know this for certain unless someone tells them or they look in a mirror, and A can continue to assume that he has brown eyes for the same reason.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474159) |
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Date: October 7th, 2014 5:38 PM Author: azure dopamine twinkling uncleanness
This is A's perspective, which is based on the fact that he doesn't know his eyes are green:
B and C each see one brown eyed dragon (A) and one green eyed (the other of B/C) dragon. If A in fact has brown eyes, then B is expecting that C would have turned to a sparrow at 12am on night 1 (because B thinks that C would have been looking at 2 brown eyed dragons). So, B finds out on Day 2 that he must also have green eyes. Same thing with C. If A has brown eyes, then C is expecting B to turn into a sparrow at 12 am of the first night, or else it means C must also have green eyes.
A knows both B and C have green eyes. He knows neither will turn into a sparrow on night 1 because both see at least 1 green eyed dragon. But, if A himself had brown eyes, both B and C would have realized they were green eyed on Day 2, after the other failed to turn into a sparrow. So A expects both B and C to turn into sparrows on night 2. When they don't, A knows he must also have green eyes.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474210) |
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Date: October 7th, 2014 5:42 PM Author: spectacular sepia stage
It's not wrong, but it's incomplete.
A assumes B sees 1 green (C) and 1 brown (A),
but A also assumes that B assumes that C sees 2 brown eyes (A and B). This is because A knows that B thinks that B himself has brown eyes, and that A himself thinks he has brown eyes.
Therefore, A thinks that B thinks C will turn into sparrow on first night (after C realizes he must have the 1 green eye).
When this fails to happen, A thinks that B will know that B himself must also have green eyes along with C, and that on the second night, B and C will both turn into sparrows.
On this also fails to happen ton the second night, A will know that something is wrong, and that they all have green eyes. On the third night, they all turn into sparrows.
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26474224)
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Date: October 7th, 2014 8:24 PM Author: Concupiscible Nudist Travel Guidebook
Considering the 4-dragon case:
Dragon A knows that dragons B, C, and D each see at least 2 other green eyed dragons.
Dragon A also knows that dragon B knows that dragon C sees at least 1 green eyed dragon. Similarly, A knows that B knows that D sees at least 1 green eyed dragon. Yes, A knows that C sees at least two green eyed dragons, but A doesn't know that B knows that C sees at least two green eyed dragons because A knows that B doesn't know his own status. So, A only knows that B knows that C sees at least 1--even though B in fact knows that C sees at least 2. Stated slightly differently, A figures that he could be brown eyed and that B could assume he's brown eyed, so B could think that C could see only 1 green eyed dragon.
What if we go one more level deep? What does A think about B's knowledge of C's knowledge of D? A could actually think that B could think that C could think that D sees no green eyed dragons, even though A knows that everyone sees at least 2 other green eyed dragons. So, the announcement requires A to revise this assessment--within A's knowledge of B's knowledge, it is no longer possible for C to think that D sees no green eyed dragons.
I think. Does this level of meta knowledge even matter, considering that we already know that every dragon must know that every other dragon can see at least one green-eyed dragon?
Edit: yep, that's it, and the meta meta meta knowledge does matter--the change in this meta meta meta possibility means that under one set of knowledge assumptions dragon D would leave the first night, but since no dragon leaves then we have to move up the chain for the next night
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26475363) |
Date: October 9th, 2014 11:12 AM Author: emerald aggressive locus Subject: Finally got around to reading this
The faggotry and stupidity throughout this thread is an embarrassment to a self-proclaimed intelligent discussion board. Even after links to the proof were posted, people still seem unsure of the analysis/conclusion.
The following explanation goes into more detail than necessary for a neat proof, but it might help you fuckheads understand the reasoning.
The 1 dragon, 2 dragon and 3 dragon cases have been adequately explained in numerous posts above. People seem to have a problem going from 3 dragons to 4 dragons and beyond using elegant inductive reasoning, so let's consider the 4 dragon case (dragons D1, D2, D3 and D4) in greater detail. Please focus on the word POSSIBLE in the below description.
*PRIOR TO THE ANNOUNCEMENT*:
D1 does not know whether it has green eyes, so it therefore knows that there are either 3 or 4 dragons with green eyes on the island ("DWGEs"<- BOLD).
D1 then considers what D2 must think. *If* D1 has green eyes, then D2 would also think that there are either 3 or 4 DWGEs. However, *if* D1 does NOT have green eyes, then D2 would think that there are either 2 or 3 DWGEs (i.e. D3, D4 and maybe D2, but not D1).
Put another way: D1 thinks that it is possible that D2 thinks that there are 2 or 3 DWGEs.
D1 then considers what D2 thinks that D3 must think. From above, since D1 thinks that it is possible that D2 thinks that it is possible that there are only 2 DWGEs, it follows that D1 thinks that it is possible that D2 thinks that it is possible that D3 thinks that there are either 1 or 2 DWGEs (i.e. D4 and maybe D3, but not D1 or D2).
*IMPORTANT NOTE*: if you have lost the reasoning already, then your IQ is insufficient to go on. Please instead killself.
D1 then considers what D2 thinks that D3 thinks that D4 must think. From above, since D1 thinks that it is possible that D2 thinks that it is possible that D3 thinks that it is possible that there is only 1 DWGE, then it follows that D1 thinks that it is possible that D2 thinks that it is possible that D3 thinks that it is possible that D4 thinks that there are either 0 or 1 DWGEs (i.e. *maybe* D4, but not D1, D2 or D3). It follows that no dragon no cause to sparrow.
*AFTER THE ANNOUNCEMENT*
The new information (or perhaps better described as the "revelation") is that it is IMPOSSIBLE for a dragon to think that there are 0 DWGEs on the island. In other words, the final conclusion stated above ("... possible that D4 thinks that there are either 0 or 1 DWGEs ...") is rendered false.
Now, let's again consider what D1 thinks that D2 thinks that D3 must think. As concluded above: D1 thinks that it is possible that D2 thinks that it is possible that D3 thinks that it is possible that there is only 1 DWGE (being D4, which (in this hypothetical thought chain) will not see any other green eyes and will therefore (having now been disabused of the notion that there could be 0 or 1 DWGEs) conclude that it is the one with green eyes and, accordingly, will sparrow at midnight following the announcement). However, we know that in fact D4 will not automatically conclude that it is the only one with green eyes as a result of the announcement, because it can see 3 other sets of green eyes.
Accordingly, when D4 DOES NOT sparrow at midnight following the announcement, it becomes IMPOSSIBLE for D3 to think that there are any fewer than 2 DWGEs. (Please do remember that all dragons are undertaking the same thought process simultaneously and are identical with each other and arbitrarily numbered.)
Now, let's consider what D1 thinks that D2 thinks on the day after the announcement. As concluded above: D1 thinks that it is possible that D2 thinks that it is possible that there are only 2 DWGEs (being D3 and D4, which (in this hypothetical thought chain) will only see one other set of green eyes and will therefore conclude that they are the two with green eyes and, accordingly, will sparrow at the second midnight following the announcement). However, we know that in fact D3 and D4 will not automatically conclude that they are the only two with green eyes on the day following the announcement, because each of them can see 3 other sets of green eyes. Accordingly, when D3 and D4 do not sparrow at the second midnight following the announcement, it becomes IMPOSSIBLE for D2 to think that there are any fewer than 3 DWGEs.
Now, let's consider what D1 thinks on the second day after announcement. D1 still thinks that it is possible that there are only 3 DWGEs (being D2, D3 and D4, which (in this hypothetical thought chain) will only see two other sets of green eyes and will therefore conclude that they are the three with green eyes and, accordingly, will sparrow at the third midnight following the announcement). However, we know that in fact D2, D3 and D4 will not automatically conclude that they are the only three with green eyes on the second day following the announcement, because each of them can see 3 other sets of green eyes. Accordingly, when D2, D3 and D4 do not sparrow at the third midnight following the announcement, it becomes IMPOSSIBLE for D1 to think that there are any fewer than 4 DWGEs. All 4 dragons will therefore sparrow at the fourth midnight following the announcement.
It is trivial to go from "4" in the description above to "n".
(http://www.autoadmit.com/thread.php?thread_id=2693243&forum_id=2#26485354) |
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