I was having trouble figuring out how to get Ba and Da defined, so I figured maybe I should work with a higher level question (if I asked so and so, would you say "ba"/say "da"?). This would have solved the which is yes/no problem pretty easily, but wasn't getting me anywhere when I accounted for the random god, so I figured I had to tackle him. But I couldn't stop working with the meta-question idea. I came up with a couple questions, making simple tables of responses, and found one that I think works.
Question 1, directed to God #1: "If I asked you if God #3 was the random god, would you say 'ba'?"
I worked out some tables for this and found something interesting: if 1 answers "Ba", then you know one of two things for sure. Either 3 is in fact the random god, or 1 is the random god answering randomly. You only get the answer of "Ba" in rows of my table where 3 = random. So either way, an answer of "Ba" establishes that 2 is not the random god, it must be 1 or 3.
If 1 answers "Da", then you know that either 3 = not random, as that answer only comes up in rows where 3 = not random, or that 1 is the random god answering randomly. Either way, we know for a fact that God #3 is not the random god if God #1 answers "Da" to the first question.
Now that we can get an answer for sure without it being random, we can figure out who's the truth teller and who's the liar. If God 1 answered "Ba" to the first question, you ask #2. If he answered "Da", ask #3. The question is: "If I asked you if you were the truth telling god, would you say "Ba"?
If he's the truth teller, and Ba means yes, he'll say Ba. If he's the truth teller and Ba means no, he would answer "Are you the truth teller" with "Da", hence he would answer Ba to the question we actually asked him.
If he's the liar, Ba means yes, and we asked if he was the truth teller, he'd answer Ba. Hence, he'd answer "Da" to our question.
If he's the liar, Ba means no, and we asked if he was the truth teller, he'd answer Da. Hence, he'd answer "Da" to our question.
So from this question, we know that the 2nd God we ask is the truth teller if he says "Ba", the liar if he says "Da".
For the third question we just repeat question 1, since we know for a fact the one we're asking isn't random. So you ask the same god "If I asked you if God #1 was random, would you say 'ba'?"
Same kind of table as for question one. Ba = God 1 is random, Da = he's not, he's the opposite of the truth/lying God we asked the last two questions leaving the other God as the random one. Turns out the values of da/ba are red herrings, you don't even need to know them if you work with meta-questions.
Did I get it? It's kind of hard to explain my solution, since it branches after question one, but I'm pretty sure it'll solve any configuration of gods and da/ba values.
Also, that was the most fun puzzle I've ever worked on.
I was having trouble figuring out how to get Ba and Da defined, so I figured maybe I should work with a higher level question (if I asked so and so, would you say "ba"/say "da"?). This would have solved the which is yes/no problem pretty easily, but wasn't getting me anywhere when I accounted for the random god, so I figured I had to tackle him. But I couldn't stop working with the meta-question idea. I came up with a couple questions, making simple tables of responses, and found one that I think works.
Question 1, directed to God #1: "If I asked you if God #3 was the random god, would you say 'ba'?"
I worked out some tables for this and found something interesting: if 1 answers "Ba", then you know one of two things for sure. Either 3 is in fact the random god, or 1 is the random god answering randomly. You only get the answer of "Ba" in rows of my table where 3 = random. So either way, an answer of "Ba" establishes that 2 is not the random god, it must be 1 or 3.
If 1 answers "Da", then you know that either 3 = not random, as that answer only comes up in rows where 3 = not random, or that 1 is the random god answering randomly. Either way, we know for a fact that God #3 is not the random god if God #1 answers "Da" to the first question.
Now that we can get an answer for sure without it being random, we can figure out who's the truth teller and who's the liar. If God 1 answered "Ba" to the first question, you ask #2. If he answered "Da", ask #3. The question is: "If I asked you if you were the truth telling god, would you say "Ba"?
If he's the truth teller, and Ba means yes, he'll say Ba. If he's the truth teller and Ba means no, he would answer "Are you the truth teller" with "Da", hence he would answer Ba to the question we actually asked him.
If he's the liar, Ba means yes, and we asked if he was the truth teller, he'd answer Ba. Hence, he'd answer "Da" to our question.
If he's the liar, Ba means no, and we asked if he was the truth teller, he'd answer Da. Hence, he'd answer "Da" to our question.
So from this question, we know that the 2nd God we ask is the truth teller if he says "Ba", the liar if he says "Da".
For the third question we just repeat question 1, since we know for a fact the one we're asking isn't random. So you ask the same god "If I asked you if God #1 was random, would you say 'ba'?"
Same kind of table as for question one. Ba = God 1 is random, Da = he's not, he's the opposite of the truth/lying God we asked the last two questions leaving the other God as the random one. Turns out the values of da/ba are red herrings, you don't even need to know them if you work with meta-questions.
Did I get it? It's kind of hard to explain my solution, since it branches after question one, but I'm pretty sure it'll solve any configuration of gods and da/ba values.
Also, that was the most fun puzzle I've ever worked on.
I don't think I'm going to try to parse it out (and it's definitely more complicated than you need to be), but I'm pretty sure you got it. Or at least the key thing, which is to say
You have to ask them "If I asked if -whoever- was random would you say -whatever-?"
You are traveling to a place you've never been when you come to a fork in the road. There is a sign indicating that one path will be long and dangerous, the other short and safe. You obviously wish to take the short, safe path, but the sign does not indicate which is which. Next to the fork stand two men, one wearing a sign that says "I always lie" the other wearing a sign that says "I always tell the truth". You can only ask one question to one man. What do you ask and to which man (assuming the signs are both telling the truth)?
Unusual solution:
Go up to either of them and ask, "Can you repeat the sentence 'This statement is a lie' truthfully?"
Yes = liar, No = truth..person
Edit: I should read the whole thing next time before responding. Oh well, I'll leave my non-solution because I like it.
I hope you don't like it enough to ask that question if you were in that situation, haha. you can only ask one question, you'd still be stuck.
If I wanted to take the short path, which road would that guy tell me to go down?
Then take the opposite road. They both will answer with the long dangerous path.
edit: lol, I just watched the labyrinth scene to "spell check" since I've never seen it. Whew.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
You are standing in front of three light switches that control three light bulbs in an upstairs room. There is no way to see the three light bulbs without going upstairs. You can do whatever you want in your current room but may only go upstairs once, and you may not come back downstairs. You know that all three lights are off when you start. How can you determine which switch controls which light?
You can actually take this to 4 light bulbs without changing the setup or nature of the problem.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
There are way too many ways this could play out, but assuming that the knight is smart and the harpy is dumb:
Before the duel, the knight will drink from poisoned well #1. Whatever he drinks will cure him unless the harpy chooses well #1, which she wouldn't do. When the duel comes, he gives the harpy an unpoisoned glass of normal drinking water so that when she drinks from #8 to cure herself, she poisons herself.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
There are way too many ways this could play out, but assuming that the knight is smart and the harpy is dumb:
Before the duel, the knight will drink from poisoned well #1. Whatever he drinks will cure him unless the harpy chooses well #1, which she wouldn't do. When the duel comes, he gives the harpy an unpoisoned glass of normal drinking water so that when she drinks from #8 to cure herself, she poisons herself.
This feels wrong, though.
Probably because it is wrong. :P
Although I guess I could have written "both are extremely clever."
You are standing in front of three light switches that control three light bulbs in an upstairs room. There is no way to see the three light bulbs without going upstairs. You can do whatever you want in your current room but may only go upstairs once, and you may not come back downstairs. You know that all three lights are off when you start. How can you determine which switch controls which light?
Ah, I know this one:
You flip one switch on, wait a few minutes, then flip that one off and flip one other switch on. The light that's warm is your first choice, the light that's on is your second, and the remaining light is the remaining switch.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
The harpy can guarantee life by drinking from well 7 before the duel and well 8 after the duel, naturally. She will still be poisoned after the duel no matter what the knight gives her, so #8 will cure her.
The knight can guarantee life by drinking from well 1 before the duel, then well 1 followed by well 2 after the duel. If the harpy gives him pure water or #1, he will be poisoned after the duel, the #1 will have no effect, and the #2 will cure him. If the harpy gives him #2 or higher, he will be cured after the duel, the #1 will poison him, and the #2 will cure him.
Ten men are awaiting execution in the same cell and given the following situation.
The following day, they'll be lined up single file in a random order and have either a black or white hat placed on their heads. They'll easily be able to see the hats of everybody ahead of them, but not their own nor any behind them. They will then have to announce out loud, in any order they desire, what color hat they're wearing. If they get it correct, they can go free. If they get it incorrect, they die.
Each of their first priorities is to live, after that, assuming all else is equal, they will try to help the other prisoners as best they can. If they attempt to communicate in any way other than saying the color of their hat, then they'll all be immediately killed. They have all night to discuss it.
How many will go free the next day?
I liked this one, so two alternate versions.
Alternate puzzle A: Same as above, except they can also have a red hat.
Alternate puzzle B: Same as puzzle A, except they don't know what the three colors will be in advance.
I believe I've gotten my head around the first one; I'm not sure about the second one but I have a hunch what the answer will be.
The pitch of the voice corresponds to how high each colour is on the chromatic scale. Alternatively, if you want to get picky about the pitch-of-voice thing, what each colour means when the end guy announces it is still determines by the chromatic scale.
I suck at logic puzzles but I can lateral think like a motherfucker.
edit; if you think the colour system is sneaky bullshit and want me to use something else, then the code is based on whatever hat the first guy is wearing. The colour of his hat means odd and the other colour means even. In that case the first guy might die or he might be able to work something out, I don't know, and like I said, terrible at logic puzzles.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
There are way too many ways this could play out, but assuming that the knight is smart and the harpy is dumb:
Before the duel, the knight will drink from poisoned well #1. Whatever he drinks will cure him unless the harpy chooses well #1, which she wouldn't do. When the duel comes, he gives the harpy an unpoisoned glass of normal drinking water so that when she drinks from #8 to cure herself, she poisons herself.
This feels wrong, though.
Probably because it is wrong. :P
Although I guess I could have written "both are extremely clever."
Errr...since the harpy can bring water from well 8, he knight is fucked, right? That seems really obvious. There's no cure for well 8. Shit, even if she brings him 7 he's fucked since he can't get to 8.
If such lateral thinking is allowed that the knight can bring the harpy an unpoisoned glass, the harpy is fucked too, because she wouldn't know the number of times she needs to drink to cure herself. If it's not allowed, then the harpy can just drink from 8 and she's cool.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
There are way too many ways this could play out, but assuming that the knight is smart and the harpy is dumb:
Before the duel, the knight will drink from poisoned well #1. Whatever he drinks will cure him unless the harpy chooses well #1, which she wouldn't do. When the duel comes, he gives the harpy an unpoisoned glass of normal drinking water so that when she drinks from #8 to cure herself, she poisons herself.
This feels wrong, though.
Probably because it is wrong. :P
Although I guess I could have written "both are extremely clever."
Errr...since the harpy can bring water from well 8, he knight is fucked, right? That seems really obvious. There's no cure for well 8. Shit, even if she brings him 7 he's fucked since he can't get to 8.
If such lateral thinking is allowed that the knight can bring the harpy an unpoisoned glass, the harpy is fucked too, because she wouldn't know the number of times she needs to drink to cure herself. If it's not allowed, then the harpy can just drink from 8 and she's cool.
Hint: What happens to the harpy's poisoned state if she drinks from the same well (of any number) a dozen times after meeting the knight, regardless of the knight's choice?
a and b are two numbers (integers) in the interval [2,99]. Mr. P and Mr. S are told a*b and a+b respectively. They have the following conversation:
Mr. P: I don't know the two numbers.
Mr. S: I already knew you didn't.
Mr. P: Ah, now I know them!
Mr. S: And now, so do I.
Assuming they aren't lying or incorrect, what are a and b?
Note: Both Mr. P and Mr. S know that the numbers are in the interval [2,99]. That's important, if I recall correctly. Also, this can be done on pencil and paper but some very basic programming helps. If you think you have a method of getting the solution and don't want to program it or do it by hand, just post a description of the method and I'll tell you if it's correct or not.
Can we assume a & b are different from one another?
I'm going to have to think about this one, but I suspect that we can narrow the options a little bit:
it cannot be the case that both a and b are primes, because otherwise Mr P would know the two numbers immediately
we can also throw out any pairs where both numbers are so high or so low as to have the only other possible choices that match the product fall outside of the given range
a and b could be the same.
Your reasoning in the spoiler sounds good so far, but you (hint)
missed one other similar case.
Real spoiler:
If a and b are p and p^2 for a prime p, then Mr. P would know the two numbers off the bat.
Also, I didn't mention this in my first post but it's noteworthy: If a and b are distinct, you don't have to know which is which (and, indeed, it's pretty easy to show it's impossible for you or Mr. P or Mr. S to know which is which).
This is what I came up with:
My answer is the first line in here, I am only about 90% done with the proof though.
7 and 4.
Mr. P gets the number 28, and Mr. S gets 11.
Mr. S multiplies all the numbers that could add up to eleven.
9 x 2 = 18
8 x 3 = 24
7 x 4 = 28
6 x 5 = 30
And sees that all of these products have multiple factors. He thinks, there is no way Mr. P, with the number he received, could have enough information to guess the two numbers.
Mr. P says: I can't guess the two numbers.
Mr. S says, I already knew you wouldn't be able to guess the two numbers.
Mr. P looks at his twenty eight, and lists the possible factors for twenty eight
7 x 4 = 28
14 x 2 = 28
He then adds these factors, to see what Mr. S could have possible gotten, to see that Mr. S could have gotten:
(7 + 4 = ) 11
or
(14 + 2 16.
Mr. P looks at this, and says, well, if Mr. S got 16, then I could have gotten the number 55, as
11 x 5 is 55, and
11 + 5 = 16
HOWEVER, if I received the number 55, I would have known immediately that the two numbers were 11 and 5, as 55 has no other factors. Thus, because Mr. S said, I already knew you wouldn't know the two numbers, Mr. S must have gotten eleven, and not 16. Thus, the two numbers must be 7 and 4.
Mr. P says, "Well, now I know the numbers."
I just don't get how Mr. S figures out the numbers, but I have been looking at this for a LONG time, so I'll come back a little later.
Thats all I got, I'm in art school.
You're on the right track, although that's not the right answer.
Minor hint:
The fourth statement is very similar to the third, mathematically, but with an additional level of complexity.
Hint: What happens to the harpy's poisoned state if she drinks from the same well (of any number) a dozen times after meeting the knight, regardless of the knight's choice?
I don't like pruning the quote tree like that here but it was getting long.
A dozen times? It's an even number, so if the knight poisoned her, she'll be repoisoned. Right? If I'm being dense do tell.
EDIT:
Misread that. Drinking from the same well means you're poisoned. You have to drink from a higher well to get unpoisoned.
Apparently, reading about that 3 gods problem, there's like 7 different ways to solve it. I can't wait to see if someone else here comes up with one of the other solutions. If you do, please describe the thought process of how you figured it out, like I tried to.
There are three gods. One god speaks nothing but the truth. Another god speaks nothing but lies. The third god just gives a random answer (still either 'yes' or 'no'). While they understand English, they only speak Ancient God, with yes and no being "ba" or "da", but you're not certain which is which.
You can ask three questions, each addressed to a single god (but you do not need to ask all three gods), and by the answers to those questions, determine which god is which.
I'm still working on this, but I have a question for clarification:
What happens if I ask a question that's based on the random god's behavior? For example, if my question to one of them was something like, "If I asked the other 2 gods if 2 + 2 = 4, will they give me different answers?", the questioner cannot answer unless he can predict which answer the random god will give, or is the random god himself. Are these kind of questions not allowed?
There are three gods. One god speaks nothing but the truth. Another god speaks nothing but lies. The third god just gives a random answer (still either 'yes' or 'no'). While they understand English, they only speak Ancient God, with yes and no being "ba" or "da", but you're not certain which is which.
You can ask three questions, each addressed to a single god (but you do not need to ask all three gods), and by the answers to those questions, determine which god is which.
I'm still working on this, but I have a question for clarification:
What happens if I ask a question that's based on the random god's behavior? For example, if my question to one of them was something like, "If I asked the other 2 gods if 2 + 2 = 4, will they give me different answers?", the questioner cannot answer unless he can predict which answer the random god will give, or is the random god himself. Are these kind of questions not allowed?
From the wiki article regarding the problem:
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, Rabern and Rabern (2008) have pointed out that the puzzle does not actually state this.[2] And in fact, Boolos' third clarifying remark explicitly refutes this assumption.
* Whether Random speaks truly or not 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.
This says that Random randomly acts as a false-teller or a truth-teller, not that Random answers randomly.
And any question the god can answer with "Ba" or "Da" is allowed.
Very slight hint:
some solutions involve 'tricking' the random god, so you are on one right track.
You ask the first one if 1+1=2. Get his answer, and you ask the second God how the first God would answer. Then you ask the third God what the first God would answer.
There are three gods. One god speaks nothing but the truth. Another god speaks nothing but lies. The third god just gives a random answer (still either 'yes' or 'no'). While they understand English, they only speak Ancient God, with yes and no being "ba" or "da", but you're not certain which is which.
You can ask three questions, each addressed to a single god (but you do not need to ask all three gods), and by the answers to those questions, determine which god is which.
I'm still working on this, but I have a question for clarification:
What happens if I ask a question that's based on the random god's behavior? For example, if my question to one of them was something like, "If I asked the other 2 gods if 2 + 2 = 4, will they give me different answers?", the questioner cannot answer unless he can predict which answer the random god will give, or is the random god himself. Are these kind of questions not allowed?
From the wiki article regarding the problem:
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, Rabern and Rabern (2008) have pointed out that the puzzle does not actually state this.[2] And in fact, Boolos' third clarifying remark explicitly refutes this assumption.
* Whether Random speaks truly or not 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.
This says that Random randomly acts as a false-teller or a truth-teller, not that Random answers randomly.
And any question the god can answer with "Ba" or "Da" is allowed.
Very slight hint:
some solutions involve 'tricking' the random god, so you are on one right track.
So does that mean
the other gods can predict the result of the "coin flip", and tell me whether the random god will lie or tell the truth to the next question asked to him?
There are three gods. One god speaks nothing but the truth. Another god speaks nothing but lies. The third god just gives a random answer (still either 'yes' or 'no'). While they understand English, they only speak Ancient God, with yes and no being "ba" or "da", but you're not certain which is which.
You can ask three questions, each addressed to a single god (but you do not need to ask all three gods), and by the answers to those questions, determine which god is which.
I'm still working on this, but I have a question for clarification:
What happens if I ask a question that's based on the random god's behavior? For example, if my question to one of them was something like, "If I asked the other 2 gods if 2 + 2 = 4, will they give me different answers?", the questioner cannot answer unless he can predict which answer the random god will give, or is the random god himself. Are these kind of questions not allowed?
From the wiki article regarding the problem:
Most readers of the puzzle assume that Random will provide completely random answers to any question asked of him; however, Rabern and Rabern (2008) have pointed out that the puzzle does not actually state this.[2] And in fact, Boolos' third clarifying remark explicitly refutes this assumption.
* Whether Random speaks truly or not 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.
This says that Random randomly acts as a false-teller or a truth-teller, not that Random answers randomly.
And any question the god can answer with "Ba" or "Da" is allowed.
Very slight hint:
some solutions involve 'tricking' the random god, so you are on one right track.
So does that mean
the other gods can predict the result of the "coin flip", and tell me whether the random god will lie or tell the truth to the next question asked to him?
what if you pick the random god when you ask the question? or the lying one?
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Donkey KongPutting Nintendo out of business with AI nipsRegistered Userregular
edited November 2009
You ask one god:
"Are you the random god?" - Da
"Are you someone other than the truthful god?" - Ba
Consistent only with liar. And you learn Da means yes.
I didn't spoiler it because this strategy seems to be a dead end. The case where the responses are the same gives no useful info. But maybe it'll help someone form better questions.
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Thousands of hot, local singles are waiting to play at bubbulon.com.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
I thought about this one on my commute this morning
both live.
the harpy can avoid the poison by drinking twice from 7 and once from 8. if the knight brought him a cup from well 7, then the harpy is still poisoned after drinking twice and then cured at 8. if the knight brought him a cup from any other well, the harpy is cured then poisoned and then cured again. if the knight brought him plain water, the harpy is poisoned twice and then cured.
the Knight can avoid poison by drinking from well 1 before the harpy gives him the cup and then drinking from well 2 twice and finally 3 afterwards. if the harpy brings him any well water from wells 2-8, he will be cured of his initial poisoning then repoisoned at 2 twice and finally cured at 3. if the harpy brings him normal water or water from well 1, he'll be cured at well 2, poisoned at well 2 and finally cured at well 3.
Missing square in a triangle puzzle:
Both of these triangles are made up of the exact same shapes, just reconfigured. So where does that gap in the lower one come from?
3, 2, 1.. GO!
I never get why this is a puzzle. It's not a mystery, the gap is right there! Where else would it come from? It's not like you can squeeze the yellow piece into the green piece to get rid of it.
better way to phrase it: how is it possible for these four shapes, with a certain area, to be reconfigured within the exact same space to compose a different total area?
one shape is 32.5 square units, and the other is 31.5 square units - and yet they are constituted by the same shapes, within the same borders. how is that possible?
the answer being that neither shape is a triangle, and so neither shape has an area of 32.5 units; the "triangle" above is actually deformed, since the two small red and blue triangle shapes have different ratios and, if you were to fit both shapes within a larger, precisely triangular frame, you'd see missing space in both.
both live
No matter what poison either brings the other one can survive.
Knight drinks poison 1 before the fight. Harpy brings mystery water. This will either cure him or not, he won't know however. Go to well 2, drink once. This will cure you if the water the harpy gave you was just plain water. If the harpy gave you any poison water you are now infected with well 2. Drink from well 2 again. This will make you poisoned from well 2 no matter what now. Go drink from well 3, cured now.
Harpy can do the exact same thing. Hell the harpy can just drink from well 7 right before the fight then go to well 8 straight after. No matter what he will be poisoned after the duel because nothing the knight has can beat poison from well 7, only match it which does nothing. The harpy is invincible and the knight has the responsibility to drink from a fuckload of wells.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
I thought about this one on my commute this morning
both live.
the harpy can avoid the poison by drinking twice from 7 and once from 8. if the knight brought him a cup from well 7, then the harpy is still poisoned after drinking twice and then cured at 8. if the knight brought him a cup from any other well, the harpy is cured then poisoned and then cured again. if the knight brought him plain water, the harpy is poisoned twice and then cured.
the Knight can avoid poison by drinking from well 1 before the harpy gives him the cup and then drinking from well 2 twice and finally 3 afterwards. if the harpy brings him any well water from wells 2-8, he will be cured of his initial poisoning then repoisoned at 2 twice and finally cured at 3. if the harpy brings him normal water or water from well 1, he'll be cured at well 2, poisoned at well 2 and finally cured at well 3.
Man, I love puzzles and stuff, but I've spent too much time over the last week prepping for the lsat logic games section to want to do any more. Maybe in a few weeks I'll come back and work on some of these.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
He saw a black hat and a white hat on the other two guys' heads and thought "If I were wearing a black hat, then whitey over there would see two black hats, know that his is white and say something. As he hasn't that must mean my hat is white"
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
He saw a black hat and a white hat on the other two guys' heads and thought "If I were wearing a black hat, then whitey over there would see two black hats, know that his is white and say something. As he hasn't that must mean my hat is white"
how did he manage that before the lights were turned back on?
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
He saw a black hat and a white hat on the other two guys' heads and thought "If I were wearing a black hat, then whitey over there would see two black hats, know that his is white and say something. As he hasn't that must mean my hat is white"
how did he manage that before the lights were turned back on?
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
I thought about this one on my commute this morning
both live.
the harpy can avoid the poison by drinking twice from 7 and once from 8. if the knight brought him a cup from well 7, then the harpy is still poisoned after drinking twice and then cured at 8. if the knight brought him a cup from any other well, the harpy is cured then poisoned and then cured again. if the knight brought him plain water, the harpy is poisoned twice and then cured.
the Knight can avoid poison by drinking from well 1 before the harpy gives him the cup and then drinking from well 2 twice and finally 3 afterwards. if the harpy brings him any well water from wells 2-8, he will be cured of his initial poisoning then repoisoned at 2 twice and finally cured at 3. if the harpy brings him normal water or water from well 1, he'll be cured at well 2, poisoned at well 2 and finally cured at well 3.
I feel educated now. And a bit dense for basically being told the trick earlier and still not having gotten it.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
If you were a bad person, this would be an antecedent problem and
the prisoner called out the color of the warden's hat.
Of course, in that case it's possible that it was the warden that was set free, too.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
He saw a black hat and a white hat on the other two guys' heads and thought "If I were wearing a black hat, then whitey over there would see two black hats, know that his is white and say something. As he hasn't that must mean my hat is white"
how did he manage that before the lights were turned back on?
Or, alternately he thinks "We know that the guard is a dick. Therefore he won't go BBW, because then the guy in the white hat can see the two blacks and figure out his own colour. Similarly he won't go BWW, because then we can do what I did in the last spoiler. Therefore he must have gone WWW, the only combination where we would be unable to figure it out.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
He saw a black hat and a white hat on the other two guys' heads and thought "If I were wearing a black hat, then whitey over there would see two black hats, know that his is white and say something. As he hasn't that must mean my hat is white"
how did he manage that before the lights were turned back on?
He eats his carrots.
ok, my guess:
he figured the sheriff would try to make it as hard as possible, so obviously wouldn't use both black hats (because the one with the white hat would know for certain)
he assumed the sheriff would guess they would figure this out, and thus couldn't use any black hats
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
If you were a bad person, this would be an antecedent problem and
the prisoner called out the color of the warden's hat.
Of course, in that case it's possible that it was the warden that was set free, too.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
While I agree that the answers that came before mine are probably 'right' they depend on some assumptions
Namely that the warden isn't placing hats on heads truly randomly or to benefit one of the prisoners
. The best strategy given the stated rules is
For a prisoner to say "black, white!" As there is no stated penalty for being wrong one of those answers is right and therefore you'd go free.
Three prisoners were told they would each be given a fair chance to win their freedom. The warden showed them 5 hats, two black and three white. The warden explained that he would turn off the lights, place a hat on each of their heads and then turn the lights back on. The first prisoner to be able to name his hat color could go free. He then turned off the lights, and placed the hats on each of the prisoner's heads. Before he could turn the lights on, the cleverest prisoner announced his hat color. He was set free. How?
Assuming all 3 prisoners are equally smart, I'd have a 33% chance of getting my hat color right first. Just quickly randomly guessing white or black, I'd have a ~50% chance of getting my hat color right first. Seems cleverest just to guess.
While I agree that the answers that came before mine are probably 'right' they depend on some assumptions
Namely that the warden isn't placing hats on heads truly randomly or to benefit one of the prisoners
. The best strategy given the stated rules is
For a prisoner to say "black, white!" As there is no stated penalty for being wrong one of those answers is right and therefore you'd go free.
It seems like the key element here is saying "they would each be given a fair chance to win their freedom" which can only be done by having all three wear white hats since a prisoner in the only white hat would have an unfair advantage, while a prisoner in the only black hat would have an unfair disadvantage. However, even that is a little questionable since they can solve the fairness problem by simply randomizing the hats before they're distributed.
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It's when puzzles are discussed that I tend to mention my autistic diagnose. :P
Question 1, directed to God #1: "If I asked you if God #3 was the random god, would you say 'ba'?"
I worked out some tables for this and found something interesting: if 1 answers "Ba", then you know one of two things for sure. Either 3 is in fact the random god, or 1 is the random god answering randomly. You only get the answer of "Ba" in rows of my table where 3 = random. So either way, an answer of "Ba" establishes that 2 is not the random god, it must be 1 or 3.
If 1 answers "Da", then you know that either 3 = not random, as that answer only comes up in rows where 3 = not random, or that 1 is the random god answering randomly. Either way, we know for a fact that God #3 is not the random god if God #1 answers "Da" to the first question.
Now that we can get an answer for sure without it being random, we can figure out who's the truth teller and who's the liar. If God 1 answered "Ba" to the first question, you ask #2. If he answered "Da", ask #3. The question is: "If I asked you if you were the truth telling god, would you say "Ba"?
If he's the truth teller, and Ba means yes, he'll say Ba. If he's the truth teller and Ba means no, he would answer "Are you the truth teller" with "Da", hence he would answer Ba to the question we actually asked him.
If he's the liar, Ba means yes, and we asked if he was the truth teller, he'd answer Ba. Hence, he'd answer "Da" to our question.
If he's the liar, Ba means no, and we asked if he was the truth teller, he'd answer Da. Hence, he'd answer "Da" to our question.
So from this question, we know that the 2nd God we ask is the truth teller if he says "Ba", the liar if he says "Da".
For the third question we just repeat question 1, since we know for a fact the one we're asking isn't random. So you ask the same god "If I asked you if God #1 was random, would you say 'ba'?"
Same kind of table as for question one. Ba = God 1 is random, Da = he's not, he's the opposite of the truth/lying God we asked the last two questions leaving the other God as the random one. Turns out the values of da/ba are red herrings, you don't even need to know them if you work with meta-questions.
Did I get it? It's kind of hard to explain my solution, since it branches after question one, but I'm pretty sure it'll solve any configuration of gods and da/ba values.
Also, that was the most fun puzzle I've ever worked on.
I don't think I'm going to try to parse it out (and it's definitely more complicated than you need to be), but I'm pretty sure you got it. Or at least the key thing, which is to say
There are actually some framing things too, yada yada yada giant flowcharts etc etc
http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever
I hope you don't like it enough to ask that question if you were in that situation, haha. you can only ask one question, you'd still be stuck.
Then take the opposite road. They both will answer with the long dangerous path.
edit: lol, I just watched the labyrinth scene to "spell check" since I've never seen it. Whew.
A knight and a harpy live peacefully on a small island together, until one day that they decide that they cannot stand being around the other and want them dead. Also on this island are eight wells, each full of poisoned water. The wells are numbered 1-8. The only way to cure yourself from being poisoned is to drink from a higher numbered well. IE, if you drink from well #5, you must drink from well #6, 7, or 8 to be cured (you're not poisoned at that point). However, well #8 is at the top of the mountain where only the harpy can reach. The poison takes 2 days to act and there's no way to tell if you have been poisoned until you instantly keel over dead.
The two decided on their duel. Each will bring a glass of water. The other will drink it. After that, they are free to run off and do whatever they wish.
Who lives? Who dies? Why?
You can actually take this to 4 light bulbs without changing the setup or nature of the problem.
There are way too many ways this could play out, but assuming that the knight is smart and the harpy is dumb:
This feels wrong, though.
Probably because it is wrong. :P
Although I guess I could have written "both are extremely clever."
Ah, I know this one:
The knight can guarantee life by drinking from well 1 before the duel, then well 1 followed by well 2 after the duel. If the harpy gives him pure water or #1, he will be poisoned after the duel, the #1 will have no effect, and the #2 will cure him. If the harpy gives him #2 or higher, he will be cured after the duel, the #1 will poison him, and the #2 will cure him.
I suck at logic puzzles but I can lateral think like a motherfucker.
If such lateral thinking is allowed that the knight can bring the harpy an unpoisoned glass, the harpy is fucked too, because she wouldn't know the number of times she needs to drink to cure herself. If it's not allowed, then the harpy can just drink from 8 and she's cool.
You're on the right track, although that's not the right answer.
Minor hint:
I don't like pruning the quote tree like that here but it was getting long.
EDIT:
I'm still working on this, but I have a question for clarification:
From the wiki article regarding the problem:
* Whether Random speaks truly or not 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.
This says that Random randomly acts as a false-teller or a truth-teller, not that Random answers randomly.
And any question the god can answer with "Ba" or "Da" is allowed.
Very slight hint:
Getting closer?
So does that mean
"Are you the random god?" - Da
"Are you someone other than the truthful god?" - Ba
Consistent only with liar. And you learn Da means yes.
I didn't spoiler it because this strategy seems to be a dead end. The case where the responses are the same gives no useful info. But maybe it'll help someone form better questions.
I thought about this one on my commute this morning
the harpy can avoid the poison by drinking twice from 7 and once from 8. if the knight brought him a cup from well 7, then the harpy is still poisoned after drinking twice and then cured at 8. if the knight brought him a cup from any other well, the harpy is cured then poisoned and then cured again. if the knight brought him plain water, the harpy is poisoned twice and then cured.
the Knight can avoid poison by drinking from well 1 before the harpy gives him the cup and then drinking from well 2 twice and finally 3 afterwards. if the harpy brings him any well water from wells 2-8, he will be cured of his initial poisoning then repoisoned at 2 twice and finally cured at 3. if the harpy brings him normal water or water from well 1, he'll be cured at well 2, poisoned at well 2 and finally cured at well 3.
better way to phrase it: how is it possible for these four shapes, with a certain area, to be reconfigured within the exact same space to compose a different total area?
one shape is 32.5 square units, and the other is 31.5 square units - and yet they are constituted by the same shapes, within the same borders. how is that possible?
No matter what poison either brings the other one can survive.
Knight drinks poison 1 before the fight. Harpy brings mystery water. This will either cure him or not, he won't know however. Go to well 2, drink once. This will cure you if the water the harpy gave you was just plain water. If the harpy gave you any poison water you are now infected with well 2. Drink from well 2 again. This will make you poisoned from well 2 no matter what now. Go drink from well 3, cured now.
Harpy can do the exact same thing. Hell the harpy can just drink from well 7 right before the fight then go to well 8 straight after. No matter what he will be poisoned after the duel because nothing the knight has can beat poison from well 7, only match it which does nothing. The harpy is invincible and the knight has the responsibility to drink from a fuckload of wells.
this guy knows whats up
how did he manage that before the lights were turned back on?
He eats his carrots.
I feel educated now. And a bit dense for basically being told the trick earlier and still not having gotten it.
If you were a bad person, this would be an antecedent problem and
Of course, in that case it's possible that it was the warden that was set free, too.
ok, my guess:
he assumed the sheriff would guess they would figure this out, and thus couldn't use any black hats
so, they must all be wearing white hats
COLLECT HIS HAND
While I agree that the answers that came before mine are probably 'right' they depend on some assumptions
QEDMF xbl: PantsB G+