|
This is a logic problem, with a light bit of 1st grade algebra in it (yes that is a hint ;D).
The Sultan is displeased with his Viziers and summons them to test if all of them are worthy of keeping. He tells them they will be tested the next morning, and until then they have to devise a strategy for the following conditions:
1. The Sultan will place a hat on the head of each Vizier. The hat will either be red or blue.
2. The Sultan will then ask the Viziers, one by one, to name the color of the hat on their head correctly; only one can be wrong about the color of the hat on his head.
3. They are not allowed to communicate with each other in any way, or use any tricks or mirrors to see the hat on their head.
4. They have the night to come up with a strategy to this problem.
How do they do it? I will answer any questions with yes/no or irrelevant/relevant.
 
|
It's when I'm presented with a problem like this that I realize how stupid I am.
|
how many viziers are there?
|
+ Show Spoiler +If there are an equal amount of hats
Each person sees x white hats and y black hats, but they don't know what their hat is.
A person sees lets say 5 white hats 4 black hats. He thinks he has a black hat and guesses. If correct, then the next people know that they must have a hat that they can't see, i.e. a white/black hat, based on how many hats they see..
|
On January 31 2008 09:26 Caller wrote:+ Show Spoiler +If there are an equal amount of hats
Each person sees x white hats and y black hats, but they don't know what their hat is.
A person sees lets say 5 white hats 4 black hats. He thinks he has a black hat and guesses. If correct, then the next people know that they must have a hat that they can't see, i.e. a white/black hat, based on how many hats they see..
Question for Caller (spoiler for others  ):
+ Show Spoiler +But what about not equal amount of hat? And the "only one can be wrong about the color of the hat on his head" - part - because in your case - everyone guessed right
|
are the vizjens isolated from each other and can they see each other's hats?
|
On January 31 2008 09:56 Rev0lution wrote:
are the vizjens isolated from each other and can they see each other's hats?
If they are isolated from each other, there is exactly 50% chance to guess their hat right - that means - a lot of them will be wrong - I don't think that's the case here, they should be able to see each other - something else is the tricky part  .
|
On January 31 2008 09:26 Caller wrote:+ Show Spoiler +If there are an equal amount of hats
Each person sees x white hats and y black hats, but they don't know what their hat is.
A person sees lets say 5 white hats 4 black hats. He thinks he has a black hat and guesses. If correct, then the next people know that they must have a hat that they can't see, i.e. a white/black hat, based on how many hats they see..
The amount of hats is unknown.
|
On January 31 2008 08:51 .MistiK wrote:
how many viziers are there?
Irrelevant.
|
On January 31 2008 10:01 Emptyness wrote:Show nested quote +On January 31 2008 09:56 Rev0lution wrote:
are the vizjens isolated from each other and can they see each other's hats? If they are isolated from each other, there is exactly 50% chance to guess their hat right - that means - a lot of them will be wrong - I don't think that's the case here, they should be able to see each other - something else is the tricky part .
Hint warning:
+ Show Spoiler +Seriously, free minds and problem solvers should not read this: + Show Spoiler +That is correct. You have to include that in your strategy somehow.
|
On January 31 2008 08:46 Pwntrucci[sR] wrote:
It's when I'm presented with a problem like this that I realize how stupid I am.
Don't be disheartened, even a little effort put to it makes you smarter than most [: It's not the answer that is important, it's using your brain to get to the solution - even a small attempt is worthy of praise.
|
Is there an equal number of red and blue hats or almost equal like if there's an odd number of viziers for example if there were 19 would there be 10 red hats and 9 blue hats?
|
On January 31 2008 10:56 Oceanic wrote:
Is there an equal number of red and blue hats or almost equal like if there's an odd number of viziers for example if there were 19 would there be 10 red hats and 9 blue hats?
The amount of hats is unknown. So is the ratio of the hat colors.
If you were referring to the other post someone made, then yea, if you have a red hat, you'd have a 50/50 shot at it.
|
I'm working on a solution but can you tell me if I'm on the right track?
+ Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong.
|
On January 31 2008 11:07 Oceanic wrote:I'm working on a solution but can you tell me if I'm on the right track? + Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong.
+ Show Spoiler [For Oceanic] +That breaks down at the third person, as the second person must repeat the answer of the first, giving the third no information. It also isn't stated that the viziers know which order they'll be asked in.
I somehow read that there were 4 viziers the first time I read the problem and solved that.
+ Show Spoiler [Special case for 4 hats] +The first vizier looks at the three hats visible to him. If he sees a 2-1 split, he announces the color he sees 2 of; if the split is 3-0 he announces the color he sees 0 of. From this, the other 3 can each infer their own hat color - If I'm not first and see the same color on the other two who aren't first, if the first guy calls that color I'm the opposite, whereas if he calls the opposite I'm the same. If I see each color on the other 2, I know I'm whatever color he calls. If the first person's call is wrong, the other 3 all announce their colors correctly, if he happened to be correct with his signal color the last simply "guesses" wrong.
I'm trying to work out a way to generalize that to an unspecified number of hats now.
+ Show Spoiler +Not knowing whether there's an even number of viziers - and thus whether I can count on the first guy seeing some level of imbalance in the number of hats - is kinda screwing with me at the moment.
Edit: Ok, I got it.
+ Show Spoiler [Solution] +The first vizier announces red if he sees an odd number of red hats, and blue if he sees an even number of red hats. Since all other viziers can count whether the number of red hats they see not counting the first vizier is odd or even, they can all determine the color of hat they wear from this information. Again, if the first vizier announced his color incorrectly, all the following announce their own correctly; otherwise the last vizier intentionally announces his color incorrectly to meet that condition.
|
On January 31 2008 11:07 Oceanic wrote:I'm working on a solution but can you tell me if I'm on the right track? + Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong.
+ Show Spoiler +Wrong. What would the third person say? He would not be able to say both his hat color and the color of the next person in line.
|
On January 31 2008 11:47 Macavenger wrote:Show nested quote +On January 31 2008 11:07 Oceanic wrote:I'm working on a solution but can you tell me if I'm on the right track? + Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong. + Show Spoiler [For Oceanic] +That breaks down at the third person, as the second person must repeat the answer of the first, giving the third no information. It also isn't stated that the viziers know which order they'll be asked in. I somehow read that there were 4 viziers the first time I read the problem and solved that. + Show Spoiler [Special case for 4 hats] +The first vizier looks at the three hats visible to him. If he sees a 2-1 split, he announces the color he sees 2 of; if the split is 3-0 he announces the color he sees 0 of. From this, the other 3 can each infer their own hat color - If I'm not first and see the same color on the other two who aren't first, if the first guy calls that color I'm the opposite, whereas if he calls the opposite I'm the same. If I see each color on the other 2, I know I'm whatever color he calls. If the first person's call is wrong, the other 3 all announce their colors correctly, if he happened to be correct with his signal color the last simply "guesses" wrong. I'm trying to work out a way to generalize that to an unspecified number of hats now. + Show Spoiler +Not knowing whether there's an even number of viziers - and thus whether I can count on the first guy seeing some level of imbalance in the number of hats - is kinda screwing with me at the moment. Edit: Ok, I got it. + Show Spoiler [Solution] +The first vizier announces red if he sees an odd number of red hats, and blue if he sees an even number of red hats. Since all other viziers can count whether the number of red hats they see not counting the first vizier is odd or even, they can all determine the color of hat they wear from this information. Again, if the first vizier announced his color incorrectly, all the following announce their own correctly; otherwise the last vizier intentionally announces his color incorrectly to meet that condition.
+ Show Spoiler +That last sentence is unnecessary - only one CAN be wrong - but yes you got it correct, nicely done [:
|
The first person's answer is the color of the hats of everyone around him in order (assuming they're standing in some sort of formation)?
|
On January 31 2008 12:41 Motiva wrote:
The first person's answer is the color of the hats of everyone around him in order (assuming they're standing in some sort of formation)?
... What? They are only allowed to say the color of the hat on their own head.
|
the first vizier looks at someone who has a red hat and says: "red". one can be wrong, so it doesn't matter which colour the hat of the first person has. then the one the first vizier looked at looks at another vizier with a red hat and says "red". this goes on until the perons with the last red hat cannot see any remaining red hats. this person closes his eyes and says "red". this is the signal for the other viziers, that the remaining hats are blue.
Is this wrong?
|
On February 01 2008 06:07 DaasEuGen wrote:
the first vizier looks at someone who has a red hat and says: "red". one can be wrong, so it doesn't matter which colour the hat of the first person has. then the one the first vizier looked at looks at another vizier with a red hat and says "red". this goes on until the perons with the last red hat cannot see any remaining red hats. this person closes his eyes and says "red". this is the signal for the other viziers, that the remaining hats are blue.
Is this wrong?
Yes, because then you get 2 possible mistakes. And making eye contact is under communication n_n.
|
On January 31 2008 12:18 fanatacist wrote:Show nested quote +On January 31 2008 11:47 Macavenger wrote:On January 31 2008 11:07 Oceanic wrote:I'm working on a solution but can you tell me if I'm on the right track? + Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong. + Show Spoiler [For Oceanic] +That breaks down at the third person, as the second person must repeat the answer of the first, giving the third no information. It also isn't stated that the viziers know which order they'll be asked in. I somehow read that there were 4 viziers the first time I read the problem and solved that. + Show Spoiler [Special case for 4 hats] +The first vizier looks at the three hats visible to him. If he sees a 2-1 split, he announces the color he sees 2 of; if the split is 3-0 he announces the color he sees 0 of. From this, the other 3 can each infer their own hat color - If I'm not first and see the same color on the other two who aren't first, if the first guy calls that color I'm the opposite, whereas if he calls the opposite I'm the same. If I see each color on the other 2, I know I'm whatever color he calls. If the first person's call is wrong, the other 3 all announce their colors correctly, if he happened to be correct with his signal color the last simply "guesses" wrong. I'm trying to work out a way to generalize that to an unspecified number of hats now. + Show Spoiler +Not knowing whether there's an even number of viziers - and thus whether I can count on the first guy seeing some level of imbalance in the number of hats - is kinda screwing with me at the moment. Edit: Ok, I got it. + Show Spoiler [Solution] +The first vizier announces red if he sees an odd number of red hats, and blue if he sees an even number of red hats. Since all other viziers can count whether the number of red hats they see not counting the first vizier is odd or even, they can all determine the color of hat they wear from this information. Again, if the first vizier announced his color incorrectly, all the following announce their own correctly; otherwise the last vizier intentionally announces his color incorrectly to meet that condition. + Show Spoiler +That last sentence is unnecessary - only one CAN be wrong - but yes you got it correct, nicely done [:
I really have a problem with these riddles. In this case it's ok cause you offered help on understanding the question correctly. But without further explanation it's really a matter of luck (how u interpret the question at first glance) and not so much an indication of intelligence.
+ Show Spoiler + I thought for example they'd not be allowed to communicate with each other concerning every aspect, therefore even the part of making a plan. Also I didn't realize what the actual question was. Wether to name the plan they come up with or to determine in which way the sultan places the hats.
So my solution was: They all wear the same colour. Hence when one of them answers incorrectly,
the others know (because they know there can only be one incorrect answer) that they all wear the same colour.
In other words in my interpretation of the problem each visier is on his own and you have to solve it from the point of view of one of them
|
On February 01 2008 09:15 Desade wrote:Show nested quote +On January 31 2008 12:18 fanatacist wrote:On January 31 2008 11:47 Macavenger wrote:On January 31 2008 11:07 Oceanic wrote:I'm working on a solution but can you tell me if I'm on the right track? + Show Spoiler +The 1st person says the color of the person after him and he is the person who gets it wrong. + Show Spoiler [For Oceanic] +That breaks down at the third person, as the second person must repeat the answer of the first, giving the third no information. It also isn't stated that the viziers know which order they'll be asked in. I somehow read that there were 4 viziers the first time I read the problem and solved that. + Show Spoiler [Special case for 4 hats] +The first vizier looks at the three hats visible to him. If he sees a 2-1 split, he announces the color he sees 2 of; if the split is 3-0 he announces the color he sees 0 of. From this, the other 3 can each infer their own hat color - If I'm not first and see the same color on the other two who aren't first, if the first guy calls that color I'm the opposite, whereas if he calls the opposite I'm the same. If I see each color on the other 2, I know I'm whatever color he calls. If the first person's call is wrong, the other 3 all announce their colors correctly, if he happened to be correct with his signal color the last simply "guesses" wrong. I'm trying to work out a way to generalize that to an unspecified number of hats now. + Show Spoiler +Not knowing whether there's an even number of viziers - and thus whether I can count on the first guy seeing some level of imbalance in the number of hats - is kinda screwing with me at the moment. Edit: Ok, I got it. + Show Spoiler [Solution] +The first vizier announces red if he sees an odd number of red hats, and blue if he sees an even number of red hats. Since all other viziers can count whether the number of red hats they see not counting the first vizier is odd or even, they can all determine the color of hat they wear from this information. Again, if the first vizier announced his color incorrectly, all the following announce their own correctly; otherwise the last vizier intentionally announces his color incorrectly to meet that condition. + Show Spoiler +That last sentence is unnecessary - only one CAN be wrong - but yes you got it correct, nicely done [: I really have a problem with these riddles. In this case it's ok cause you offered help on understanding the question correctly. But without further explanation it's really a matter of luck (how u interpret the question at first glance) and not so much an indication of intelligence. + Show Spoiler + I thought for example they'd not be allowed to communicate with each other concerning every aspect, therefore even the part of making a plan. Also I didn't realize what the actual question was. Wether to name the plan they come up with or to determine in which way the sultan places the hats.
So my solution was: They all wear the same colour. Hence when one of them answers incorrectly,
the others know (because they know there can only be one incorrect answer) that they all wear the same colour.
In other words in my interpretation of the problem each visier is on his own and you have to solve it from the point of view of one of them
+ Show Spoiler +That's a pretty twisted interpretation o;
|
On February 01 2008 06:43 fanatacist wrote:Show nested quote +On February 01 2008 06:07 DaasEuGen wrote:
the first vizier looks at someone who has a red hat and says: "red". one can be wrong, so it doesn't matter which colour the hat of the first person has. then the one the first vizier looked at looks at another vizier with a red hat and says "red". this goes on until the perons with the last red hat cannot see any remaining red hats. this person closes his eyes and says "red". this is the signal for the other viziers, that the remaining hats are blue.
Is this wrong? Yes, because then you get 2 possible mistakes. And making eye contact is under communication n_n.
1. Where do you see 2 mistakes? i see only one, the first vizier
2. Then you should have made clear what you mean when you define as communication. if i one vizier tells the others + Show Spoiler + if there is a odd or even number of red hats
|
On February 01 2008 20:58 DaasEuGen wrote:Show nested quote +On February 01 2008 06:43 fanatacist wrote:On February 01 2008 06:07 DaasEuGen wrote:
the first vizier looks at someone who has a red hat and says: "red". one can be wrong, so it doesn't matter which colour the hat of the first person has. then the one the first vizier looked at looks at another vizier with a red hat and says "red". this goes on until the perons with the last red hat cannot see any remaining red hats. this person closes his eyes and says "red". this is the signal for the other viziers, that the remaining hats are blue.
Is this wrong? Yes, because then you get 2 possible mistakes. And making eye contact is under communication n_n. 1. Where do you see 2 mistakes? i see only one, the first vizier 2. Then you should have made clear what you mean when you define as communication. if i one vizier tells the others + Show Spoiler +if there is a odd or even number of red hats , that is as much communication as eye conntact.
1. Nevermind, I missed the bit where he closed his eyes.
2. ... I'm not going to list every possible method of communication or define it in textbook terminologies. It is self-evident that communication means the transfer of information, and making eye contact or choosing not to do so is transferring information. Also, what? They are allowed to strategize BEFORE the test, which is when the viziers plan that there will be + Show Spoiler +an even or odd amount of a certain color hats
|
in your solution they are also transfering information. i thought the aim was to let the sultan think that there is no communication, and if you make only subtile eyecontact it looks like the viziers don't communicate. in my oppinion its the same for your solution, it looks like they dont communicate, but they do. but i think the problem is not accurate enough in its description (thats the reason why i normally don't like such problems), so if by "They are not allowed to communicate with each other in any way" is meant they they still can communicate by saying "red" or "blue", the my solution is of course wrong
|
... No offense man but you're sounding pretty dumb. Red and blue is the only thing they are allowed to say, that is said from the beginning - to realize that it can be used as a method is the strategy itself. Let me repeat myself - it's the only thing they are allowed to do once the game starts.
|
i think you don't understand me. it is said: "hey are not allowed to communicate with each other in any way". saying red or blue is a method of communicating, there are no exceptions made for this case. if you solution is right, imao mine should also be right. but let us say you are right and i am wrong, i am tired of arguing and your blog should not be about two people arguing if the solution to a problem is right or wrong.
|
This is the problem with putting riddles in context - DaasEuGen's answer is indeed some sort of solution, though a lot weaker one than the 'correct solution'. The 'correct solution' is better, though, because it doesn't rely on the viziers having to make eye-contact, only on hearing each other's answers.
|
On February 02 2008 21:24 DaasEuGen wrote:
i think you don't understand me. it is said: "hey are not allowed to communicate with each other in any way". saying red or blue is a method of communicating, there are no exceptions made for this case. if you solution is right, imao mine should also be right. but let us say you are right and i am wrong, i am tired of arguing and your blog should not be about two people arguing if the solution to a problem is right or wrong.
I understood you perfectly, but if there was no possible form of communication at all, there would be no point to this problem. The proper solutions works within the limits of the problem - no communication, except the ability to say "red" or "blue". The fact that a person has to come to the conclusion that using this as a form of communication is in itself the answer to the riddle. Trying to use some sort of trick form of communication in order to get to the same result and saying that red and blue are communication too is faulty, seeing as they are explicitly allowed to say red and blue, as it is essential to both the solution and the set-up of the problem. Why not just make them Morse code the color of each other's hats to each other, if we use your method? Then they can have any color hat on their head and they will know!
|
Should make that part of the problem description: they not only know color of every other vizier's hat, they know every previous vizier's guess.
It's possible for a problem to be constructed without even that, where no one knows what anyone else picked.
Also, the given solutions only work if there is a finite number of viziers. I think.
|
On February 03 2008 03:05 BottleAbuser wrote:
Should make that part of the problem description: they not only know color of every other vizier's hat, they know every previous vizier's guess.
It's possible for a problem to be constructed without even that, where no one knows what anyone else picked.
Also, the given solutions only work if there is a finite number of viziers. I think.
Uh, that's pretty obvious, since they say it aloud? And wouldn't that give away the answer?
Once again, I'm not going to include a list of things you can't do or the exact subtleties of what is allowed - the solution and problem are what they are.
And yea, how are you going to have an infinite amount of viziers?
|
Eh. See this problem, here. We're not constrained by real logistics, right?
|
On February 03 2008 10:59 BottleAbuser wrote:Eh. See this problem, here. We're not constrained by real logistics, right?
The mental leap to realizing that they know the hat colors of all those before them, and thus using that information as communication, is essential to this problem, is what I'm trying to say. Give away a part of it in the problem description and that makes the solution that much easier to arrive at.
|
|
|
|
|
|
EPT
