|
Germany1297 Posts
Inspired by the thread with this really old math riddle, which is not really a riddle I wanted to post another riddle that really took me a while to solve:
3 men discuss which of them is the smartest, so they walk all up a mountain to a very old and very wise man. They state their problem and he pulls out a sack and sais: "In this sack there are 2 black hats and 3 white hats." Then he tells them to close their eyes and puts a hat on each of them. They cannot see their own hat but the other two. Then he sais: "the first to tell me which color his hat has and why is the smartest of you". (You as the person to solve this know that one guy has a black hat on his head and two guys have a white hat on their heads.) After some minutes one of the guys jumps up and sais: "i know that my hat is .."
How did he know?
Facts in short: Sack with 5 hats, 2 black, 3 white 3 guys, one with a black hat, two with a white they can only see the others hats, not their own how can they know what color their hat has?
|
oh, this type of riddles where people wait around to see if the others react are funny. But maybe like 10 or 20 seconds would be more than enough assuming they are as smart as they claim. Will not spoil answer just yet. gl.
|
+ Show Spoiler +one of the guys (A) with the white hat sees that the other guys have 1 black (B) and 1 white (C) hat each. (A) waits dutifully for a few seconds to see if anybody else gives an answer. since nobody does in that time, he figures he (A) can't have a black hat, because if he did guy (C) would've seen 2 black hats and thus known he (C) had a white hat. so (A) then announces he has a white hat imba for (B).
|
Germany1297 Posts
Feel free to post your own riddles, I'm waiting (:
|
This is a fun one, it was posted a while back in this thread. My answer's there too, if anyone wants to check.
Alice at the logicians convention
Alice was walking among the paths of Never Land when from nearby she heared some voices. Being a curious person she climbed a tree and has been witness to a scene...
Around gigantic table there gathered 31 men. Before them there was the Speaker, funny, wearing scarlet tunic professor with short, white beard. He gestured for everyone to be silent and has given a strangest speech Alice has ever heared.
- Brother logicians. We, the most disciplined minds of the Never Land, have gathered here today, on our 125th annual convention. We will hear the amazing tales of logic, we will be thinking about things unthinkable for common mortal, we will pass the Mountains of Endless Querries and the most demanding Paths of Intelect. But first we must make sure that no intruder is hiding within our circle.
After that the professor started to walk around the table, at each passed logician he paused and sticked a coloured dot on their foreheads. After his return to his spot at the top of the table, he began to explain the rules of this odd experiment.
- Each one of you see the dots on the foreheads of your colegues, but I was careful, so no one saw the colour of his own. Your task is to find out the colour your forehead was marked with.
- There is only one rule and it is simple. Each minute this bell will make a sound. If at this very moment some of you will know the colour of the dot you're wearing, they must stand from the table and join me at the nearby clearing where our convention will be continued. If you still won't know the colour of your dot, you are to remain at the table.
- Someone who will remain at the table, when he should stand up, or someone who will stand up when he should remain at the table, can't of course title himself logician and will be removed from the convention without a right to ever return.
Professor was about to leave when his attention was drawn by the troubled look of the smartest of novices. He cleared his doubts with this words:
- Do not be affraid youngster. It is possible to solve this task. But, of course, you may not contact each other in any way.
The novice smiled because the Speaker of the Most Disciplined Minds of the Never Land may not make false statements.
Before the eyes of completly confused now Alice, professor has left the convention and the experiment has begun.
At the first sound of the bell four persons have left the table. At the second sound, all with red dots stood up and left together. At the third sound no one moved, but at the fourth at least one person has reacted. Novice, who was mentioned before and his sister, with dots of different colours, have left shortly after, but each one of them, before the last sound of the bell.
Tired with long words Alice has fallen into deep slumber before the test was finished. Can you tell her how many times the bell rang before the table was left empty?
|
There exists a king and his n prisoners. The king has a circle-shaped dungeon in his castle, and has n cell doors around the perimeter, each leading to a separate, sound proof room. When within the cells, the prisoners have absolutely NO means of communicating with each other.
The king sits in the central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
In an effort to make it more difficult, the king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice or trying to sneak something around.
The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
|
On July 23 2007 03:58 devour wrote: There exists a king and his n prisoners. The king has a circle-shaped dungeon in his castle, and has n cell doors around the perimeter, each leading to a separate, sound proof room. When within the cells, the prisoners have absolutely NO means of communicating with each other.
The king sits in the central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
In an effort to make it more difficult, the king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice or trying to sneak something around.
The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
my brain exploded from just reading the question.
|
|
Nice one devour. Do the prisoners know k?
|
On July 23 2007 03:58 devour wrote: There exists a king and his n prisoners. The king has a circle-shaped dungeon in his castle, and has n cell doors around the perimeter, each leading to a separate, sound proof room. When within the cells, the prisoners have absolutely NO means of communicating with each other.
The king sits in the central room and the n prisoners are all locked in their sound proof cells. In the king's central chamber is a table with a single chalice sitting atop it. Now, the king opens up a door to one of the prisoners' rooms and lets him into the room, but always only one prisoner at a time! So he lets in just one of the prisoners, any one he chooses, and then asks him a question, "Since I first locked you and the other prisoners into your rooms, have all of you been in this room yet?" The prisoner only has two possible answers. "Yes," or, "I'm not sure." If any prisoner answers "yes" but is wrong, they all will be beheaded. If a prisoner answers "yes," however, and is correct, all prisoners are granted full pardons and freed. After being asked that question and answering, the prisoner is then given an opportunity to turn the chalice upside down or right side up. If when he enters the room it is right side up, he can choose to leave it right side up or to turn it upside down, it's his choice. The same thing goes for if it is upside down when he enters the room. He can either choose to turn it upright or to leave it upside down. After the prisoner manipulates the chalice (or not, by his choice), he is sent back to his own cell and securely locked in.
The king will call the prisoners in any order he pleases, and he can call and recall each prisoner as many times as he wants, as many times in a row as he wants. The only rule the king has to obey is that eventually he has to call every prisoner in an arbitrary number of times. So maybe he will call the first prisoner in a million times before ever calling in the second prisoner twice, we just don't know. But eventually we may be certain that each prisoner will be called in ten times, or twenty times, or any number you choose.
In an effort to make it more difficult, the king is allowed to manipulate the cup himself, k times, out of the view of any of the prisoners. That means the king may turn an upright cup upside down or vice versa up to k times, as he chooses, without the prisoners knowing about it. This does not mean the king must manipulate the cup any number of times at all, only that he may.
Assume that both the king and the prisoners have a complete understanding of the game as I have just explained it to you, and that the prisoners were given time beforehand to come up with a strategy. The king was able to hear the prisoners discuss, however, so also assume that if there is a way to foil a strategy, the king will know it and exploit the weakness. The prisoners must utilize a strategy that works in absolutely every single possible case.
Now you must figure out not only how to keep the prisoners alive, but how to also ensure their eventual freedom. When can any one of them be certain they've all been in the central chamber of the dungeon at least once? And how? Don't try to imagine any trickery like scratching messages in the soft gold of the chalice or trying to sneak something around.
The problem is as simple as it sounds. The prisoners have absolutely no way of communicating with each other except through the two orientations of the chalice. If any of them attempts any trickery at all they will all be beheaded. All the prisoners can do is turn the chalice upside down or right side up, as they choose, whenever they are called into the chamber.
W T F
|
On July 23 2007 04:32 Orome wrote: Nice one devour. Do the prisoners know k?
I guess they have to know it, otherwise they could never be sure of anything.
|
btw can someone post a clever/witty/cute riddle and not one that will make my brain explode. im pretty sure if i told anyone the riddle about the king and his prisoners, id get slapped halfway thru
|
Yes, the prisoners know k. Remember they got to dicuss their "strategy" the night before.
|
On July 23 2007 04:55 zizou21 wrote: btw can someone post a clever/witty/cute riddle and not one that will make my brain explode. im pretty sure if i told anyone the riddle about the king and his prisoners, id get slapped halfway thru
Try this one:
Given the following information, what is 10 + 10?
1+1=0; 2+2=0; 3+3=0; 4+4=2; 5+5=0; 6+6=2; 7+7=0; 8+8=4; 9+9=2; 10+10=?
Note: This was taken from a Japanese kindergarten application
|
A solution to devours riddle: (EDIT: just want to brag: the solution was made up on the spot, not something I've read somewhere. This what you get from years and years of university training. ) EDIT2: corrected a flaw. Also redefined n to not write (n-1) al the time.
+ Show Spoiler + Prisoners choose one guy to be the leader and use the following rules. 1) The leader always leaves the chalice right side up. 2) The n other prisoners always leaves it upside down, but will TURN the chalice only 2k+2 times.
When the leader has found the chalice upside down n(2k + 2) - k times he known that everyone has been in the room.
why? Here is the explanation: There can be three things that can make the leader find the cup upside down a) it may have started upside down. This can happen only once. b) the king may have turned it. This can happen only k times. c) Some of his friends may have turned it. This can happen at most k+2 times for each prisoner.
So after n(2k+2) - k times, all except k +1 have been c) happening. So he knows that at least n(2k + 2) - (k+1) times has a prisoner turned the chalice. Since each prisoner turns it max 2k+2 times, everyone must have turned it at least once by then, even. So he can be sure everyone has been in the camber. Also, whatever the starting position of the chalice, and whatever the king does with his k turns, this will always happen eventually since even if the king turns it back to normal when the prisoners has turned it upside down k times. In that case the chalice will be turned exactly n(2k+2) - k times (2k+2 times for each prisoner, minus the k times the king cancels the turn).
Orome: so we know nothing about how many colors or how many of each color? Hmm, tricky one.
|
On July 23 2007 05:12 devour wrote:Show nested quote +On July 23 2007 04:55 zizou21 wrote: btw can someone post a clever/witty/cute riddle and not one that will make my brain explode. im pretty sure if i told anyone the riddle about the king and his prisoners, id get slapped halfway thru Try this one: Given the following information, what is 10 + 10? 1+1=0; 2+2=0; 3+3=0; 4+4=2; 5+5=0; 6+6=2; 7+7=0; 8+8=4; 9+9=2; 10+10=? Note: This was taken from a Japanese kindergarten application + Show Spoiler +ok, numbers are equal to 2 if they are divisible by a perfect square not equal to 1, 0 if they are not, and 4 if they are divisible by a perfect square (not equal to one) squared. 10+10 is only divisible by the perfect square 4 so I am guessing 10+10=2 edit: Lol, my expanation fails for 2+2=0
|
On July 23 2007 05:21 pat777 wrote:Show nested quote +On July 23 2007 05:12 devour wrote:On July 23 2007 04:55 zizou21 wrote: btw can someone post a clever/witty/cute riddle and not one that will make my brain explode. im pretty sure if i told anyone the riddle about the king and his prisoners, id get slapped halfway thru Try this one: Given the following information, what is 10 + 10? 1+1=0; 2+2=0; 3+3=0; 4+4=2; 5+5=0; 6+6=2; 7+7=0; 8+8=4; 9+9=2; 10+10=? Note: This was taken from a Japanese kindergarten application + Show Spoiler +ok, numbers are equal to 2 if they are divisible by a perfect square not equal to 1, 0 if they are not, and 4 if they are divisible by a perfect square squared. 10+10 is only divisible by the perfect square 4 so I am guessing 10+10=2
haha, I agree with the answer but for a reason that is slightly more accesible for kindergarten level.
+ Show Spoiler +The number is just the number of circles, or loops, on the left hand side.
|
On July 23 2007 05:25 Cascade wrote:Show nested quote +On July 23 2007 05:21 pat777 wrote:On July 23 2007 05:12 devour wrote:On July 23 2007 04:55 zizou21 wrote: btw can someone post a clever/witty/cute riddle and not one that will make my brain explode. im pretty sure if i told anyone the riddle about the king and his prisoners, id get slapped halfway thru Try this one: Given the following information, what is 10 + 10? 1+1=0; 2+2=0; 3+3=0; 4+4=2; 5+5=0; 6+6=2; 7+7=0; 8+8=4; 9+9=2; 10+10=? Note: This was taken from a Japanese kindergarten application + Show Spoiler +ok, numbers are equal to 2 if they are divisible by a perfect square not equal to 1, 0 if they are not, and 4 if they are divisible by a perfect square squared. 10+10 is only divisible by the perfect square 4 so I am guessing 10+10=2 haha, I agree with the answer but for a reason that is slightly more accesible for kindergarten level. + Show Spoiler +The number is just the number of circles, or loops, on the left hand side. Lmao, I just found a flaw in my explanation. Your explanation works better. I am guessing you learned about the blue eyed riddle from university experience.
|
Cascade is correct
|
I have a solution to the king/prisoners/chalice riddle
Hint: + Show Spoiler + First try to solve the riddle without the king being allowed to interfere; by the way, I was already familiar with the riddle in this form, which really helped.
Solution: + Show Spoiler +
One of the prisoners takes a special role to count the number of times he/she has seen the chalice up-side-down, and always resetting it to right-side-up. When the count reaches A, this prisoner announces "yes". otherwise he/she announces "not sure".
The rest of the prisoners always say "not sure". They will flip the chalice to up-side-down if they see it right-side-up, but will do so no more than B times (after doing this B times, they will no longer manipulate the chalice at all).
the question is now how to select A and B so that: - 1st prisoner's counter cannot reach A until all prisoners were inside the room at least once - 1st prisoner's counter is ensured to reach A in finite time
Because of the first criteria, A must be greater than (n-2)*B + k + 1. This is the worst case - The chalice was up-side-down to begin with (1), the king interfered k times (k), and all but one of the other prisoners made all of their B chalice flips ((n-2)*B) - but the last prisoner never went into the room. there is no way to exceed this amount by any other means short of allowing the last prisoner into the room, thus any value of A greater than (n-2)*B + k + 1 ensures that all prisoners visited the room at least once.
As for the second criteria, the maximum guarantied count is (n-1)*B - k. The prisoners make (n-1)*B chalice flips, but up to k of them may not be counted because of the king's interference.
therefore the following inequalities must be solvable: (n-2)*B + k + 1 < A <= (n-1)*B - k
Taking A = (n-1)*B - k is a possible solution, but only if the following inequality holds: (n-2)*B + k + 1 < (n-1)*B - k
k + 1 < B - k
2k + 1 < B
Clearly, taking B = 2k + 2 and A = (n-1)*B - k satisfies the necessary conditions to make this a sound strategy for the prisoners.
|
|
|
|