A list of puns related to "The Hardest Logic Puzzle Ever"
The Hardest Logic Puzzle Ever, posed by Raymond Smullyan, goes like this:
Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for βyesβ and βnoβ are βdaβ and βja,β in some order. You do not know which word means which.
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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. If anyone has figured out the color of their own eyes, they [must] leave the island that midnight. 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.
"There are three gods A, B, and C. Their names are, in no particular order, True, False, and Random. As the names suggest, True always tells the truth; False always lies; Random lies sometimes and behaves randomly. You can ask three yes/no questions and in the end, you need to tell the names of A, B, and C. One challenging part is that these gods speak another language, so they can only answer da or ya. Unfortunately, you have no idea whether da means yes or no."
I came across this logic puzzle, and I can't wrap my head around it. I've read the explanation behind it, but I still don't fully understand it. Can someone please either explain/ELI5/break it down?
It would be most appreciated!
> Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
> Boolos provides the following clarifications: a single god may be asked more than one question, questions are permitted to depend on the answers to earlier questions, and the nature of Random's response should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
I found 2 things left unspecified.
1: What happens in the case of a paradoxical (unanswerable) question? Some have assumed the god's head with explode. I propose the God travels back in time and annihilates you before you pose the question so as to avoid any situation where they fail to answer as required.
2: Does Random "flip the coin" before or after determining the answer to the question posed? If before then you can include that in the question it self "if you're about to lie". If not, do such questions become paradoxes? Or can Random at one moment have an answer to whether they are about to lie, then change their mind before giving the answer without consequence? I'm not sure about this one.
Anyway, have fun with this crazy one.
Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are da and ja, in some order. You do not know which word means which.
This one really is brilliant, and is not a trick question. Credit to http://www.xkcd.com/blue_eyes.html
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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.
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