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[QUOTE]On March 15 2009 08:45 imBLIND wrote:
An evil king wishes to kill his general because he thinks smarter than the king himself. He cannot just kill him because then his empire will rebel against him; he must create an excuse to kill him. He knows he cannot prove anything in front of a jury, so he decides to give him an impossible riddle to answer right before his execution. The king goes up to his house, kidnaps him, and throws him in prison garments with no possessions.
The next day, they are on a stage where everyone can see both the king, the general, and the two bags that the king is holding. The king says:
"In my hands, i am holding two bags. One has a black stone and the other has a white stone. If you choose the one with the black stone, you will die. If you choose one with the white stone, i will let you live."
The general knows that this is a rigged trick; both of the bags have black stones in them.
The general takes a stone out of the bag and walks away with his life.
How does he do it?
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[QUOTE]
+ Show Spoiler + he declares that he will choose the black stone, since it's 50/50, the other stone "must be white". he takes out a black stone, so the other stone is white by process of elimination, unless the king rigged it (which he did, but the public doesnt know that). so the general lives
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On March 16 2009 14:27 l10f wrote:
The MBC Game Hero team decided to go on a trip to the USA. Light decided to bring his laptop to practice with him. All the other players were jealous that he could practice while they couldn't. One day, his laptop went missing while he slept.
1. Jaehoon claims that he was busy hooking up with girls last night.
2. Shark says that he never even saw the laptop.
3. Pusan says he saw Sea take the laptop.
4. Sea says he was out last night on a one night stand with a girl he met, and that Pusan is lying
5. Saint says Shark was in Light's room at 3AM last night.
Any of the players could be telling a lie, or could be telling the truth. Who stole Light's laptop?
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I think it's impossible to figure this out due to lack of information.
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On March 16 2009 13:09 Tensai176 wrote:
A man is facedown in the middle of the dessert dead. If he had opened his knapsack he would have survived. What was in his backpack?
+ Show Spoiler +
Guys, the key here is 'dessert'.
He was unknowingly about to devour a strawberry cheesecake, but he has a deadly allergy to strawberry.
If he had opened his knapsack, he would have taken the chocolate pudding in there which just so happens to be his favorite food, and would have eaten that instead thus saving his life.
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Korea (South)922 Posts
Kinda old but
1. How do you put a giraffe into a refrigerator?
+ Show Spoiler +Open the refrigerator, put in the giraffe, and close the door.
This question tests whether you tend to do simple things in an overly complicated way.
2. How do you put an elephant into a refrigerator?
+ Show Spoiler +Did you say, "Open the refrigerator, put in the elephant, and close the refrigerator?" (Wrong Answer)
Correct Answer: Open the refrigerator, take out the giraffe, put in the elephant and close the door.
This tests your ability to think through the repercussions of your previous actions.
3. The King of the Forest is hosting an animal conference. All the animals attend except one. Which animal does not attend?
+ Show Spoiler +Correct Answer: The Elephant. The elephant is in the refrigerator. You just put him in there. This tests your memory.
OK, even if you did not answer the first three questions correctly, you still have one more chance to show your true abilities.
4. There is a river you must cross but it is inhabited by crocodiles. How do you manage it?
+ Show Spoiler +Correct Answer: You swim across. All the crocodiles are attending the animal conference.
This tests whether you learn quickly from your mistakes.
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On March 16 2009 03:55 Z-BosoN wrote:
lost in the desert, a travelr sees a box in the sand. In the lid, there is a hole of the size of a needle head and four colored buttons, one blue, one yellwo, one green, and one red. He tries to look throught the hole but cannot see anything. Next to the box he sees the following letter. What should he do?
"Inside this box there is a small fan, well lubrificated, silent, powerred by batteries and does not vibrate when on.
The buttons on the top of the box, is pressed, stay on the "on" position, and when pressed again, they go back to "off". Only one of them is connected to the fan. Keeping the box closed, you can press buttons however you want and put whatever you want through the hole in the lid. When you open it, you have to figure out immediately which is the button that turns on the fan. A right answer will give you access to a secret oasis. A mistake will encarcerate you to the hienas."
+ Show Spoiler +This probably isn't the right solution but I would just stick a stick or needle or string into the hole and press the buttons one at a time. Eventually the fan will turn on and hit the stick.
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Netherlands6142 Posts
On March 16 2009 07:40 BanZu wrote:
Every year, a monk makes his way to the top of a mountain for a week of intense meditation. In preparation for the annual trip he decides to leave at 6 in the morning so as to make the 12 hour trip to the top before the sun sets. Because the trail is long and steep the monk takes many breaks and travels at a varying pace. Finally, he makes it to the top and spends his week in peace and harmony with nature. At the end of the week of meditating, the monk leaves the mountaintop at 9 in the morning and makes his way down the trail, also at a varying pace and with many breaks. This time, however, he makes the trip in only 9 hours.
Prove that the monk was at the exact same point on the trail at the exact same time of day on his way up and down the trail.
+ Show Spoiler +
I don't see how this is possible. At 6 in the morning on the first day he is at the foot of the mountain whereas at 6 am the 7th day he is on top of the mountain. I've tried reasoning with him going up down up again and with his starting point being the top also but I don't think it's possible that way either.
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On March 16 2009 06:40 ZerG~LegenD wrote:You have 12 balls. They are all coloured differently to help you tell them apart. 11 of these have the same weight, 1 is either slightly heavier or slightly lighter than the rest. Can you, with the help of 3 weightings with a basic balance scale, find out which ball that has a unique weight and whether it's heavier or lighter than the rest? + Show Spoiler +Oh no, it took me many hours to figure out. You're not getting it that easy.
Here's a follow up question of a more mathematical nature. Given n balls and k weightings, when is it the answer to the riddle 'yes'?
On March 16 2009 09:14 Z-BosoN wrote:
Sorry I forgot about the spoilers... fixed it
Zerg-Legend you have the right idea, I'm not gonna say yours doesn't work, because it is logical, but there is one that does not rely on sand cones (would be kinda hard to tell if there is a sand cone on top of flat layer of sand or not)
+ Show Spoiler +I've got another solution based on another doubtful premise; am I allowed to press down some buttons, open the box and then let go of the buttons as the box is opened and observe what happens? If I am, then this is a solution:
First press red and blue, then press blue and green and hold them down. Open the box and let go of blue and green.
If the fan was on when I opened the box and it stayed on when I let go of the buttons it is the red one that is connected to the fan.
If the fan was on but turned off it means it is the blue one.
If it was off and turned on it is the green one.
If it was off and stayed off it is the yellow one.
Though this would mean that the hole in the box is a red herring.
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On March 16 2009 07:40 BanZu wrote:
Every year, a monk makes his way to the top of a mountain for a week of intense meditation. In preparation for the annual trip he decides to leave at 6 in the morning so as to make the 12 hour trip to the top before the sun sets. Because the trail is long and steep the monk takes many breaks and travels at a varying pace. Finally, he makes it to the top and spends his week in peace and harmony with nature. At the end of the week of meditating, the monk leaves the mountaintop at 9 in the morning and makes his way down the trail, also at a varying pace and with many breaks. This time, however, he makes the trip in only 9 hours.
Prove that the monk was at the exact same point on the trail at the exact same time of day on his way up and down the trail.
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+ Show Spoiler +On March 17 2009 00:32 fanatacist wrote: Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p)
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On March 17 2009 00:51 gondolin wrote:+ Show Spoiler +On March 17 2009 00:32 fanatacist wrote: Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p)
What the FUCK are you talking about??
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Netherlands6142 Posts
I get the riddle now, I think fanatacist got it. Is that all there is to it?
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On March 16 2009 17:23 AltaiR_ wrote:
3. The King of the Forest is hosting an animal conference. All the animals attend except one. Which animal does not attend?
+ Show Spoiler +thought of a ephemera... either it didnt hear anything about it yet, or its alrdy dead.
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On March 17 2009 01:18 fanatacist wrote:Show nested quote +On March 17 2009 00:51 gondolin wrote:+ Show Spoiler +On March 17 2009 00:32 fanatacist wrote: Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p) What the FUCK are you talking about??
You obviously don't understand what he is saying but gondolin is actually right.
And to contribute to this wonderful thread I will add to Altair_'s wonderful riddles:
How can you know that there has been an elephant in the fridge?
+ Show Spoiler +
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On March 17 2009 04:41 Sean.G wrote:Show nested quote +On March 17 2009 01:18 fanatacist wrote:On March 17 2009 00:51 gondolin wrote:+ Show Spoiler +On March 17 2009 00:32 fanatacist wrote: Yeah but you need the intermediate value theorem for that so you need to be in classical logic. In intuitive logic this fail, even if it is not obvious. ;-) I think you can find counter-examples in the topos of sheaves over [0,1], but i suck at intuitive logic. (but "topos of sheaves" looks good so i can say it to be a smart-ass :p) What the FUCK are you talking about?? You obviously don't understand what he is saying but gondolin is actually right. And to contribute to this wonderful thread I will add to Altair_'s wonderful riddles: How can you know that there has been an elephant in the fridge? + Show Spoiler +
I wouldn't know if that's true or not because like you said, I don't understand what the fuck he is talking about. I know intermediate value theorem, but I think the graph is self-explanatory, IVT is almost like an axiom to this, an unwritten assumption.
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Take an elastic that can stretch to infinity, an infinitely long straight track, and a bicycler and snail who never tire. The biker is standing still on the track, with the elastic on the end of his bike, the elastic is further fixed to a point at the beginning of the track. The snail is crawling at the fixed point of the elastic at distance 0. The biker starts biking at distance > 0 and at a constant speed stretching the elastic. The snail crawls at a constant speed towards the biker, but it crawls slower than the biker bikes. Does the snail ever reach the biker?
(I do not know the answer, but I was once told what the answer was. I don't get it X-| perhaps some smart people here want to crack this one)
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Yes because eventually the force of the the biker going forwards is equal to the force of the rubber band pulling him backwards so he has therefore 0 acceleration at that point and no velocity. The snail moving at a constant rate would eventually overtake him as he is standing still.
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On March 16 2009 06:40 ZerG~LegenD wrote:You have 12 balls. They are all coloured differently to help you tell them apart. 11 of these have the same weight, 1 is either slightly heavier or slightly lighter than the rest. Can you, with the help of 3 weightings with a basic balance scale, find out which ball that has a unique weight and whether it's heavier or lighter than the rest? + Show Spoiler +Oh no, it took me many hours to figure out. You're not getting it that easy.
This is actually seems quite easy, and no, I did not look it up
+ Show Spoiler +1. Divide the balls into two groups of 6 and put each group on the balance. The ball that is lighter must be in the group of 6 that is lighter.
2. Divide that group of six into two groups of 3 and repeat this process, the ball that is lighter must be in the lighter group of 3.
3. Take any two of the the group of three and weigh them on the balance. If one is lighter, there you have it. If they are in fact, balanced, then the one you didn't weigh from that group of three must be the lighter ball.
edit: hmmm, just realized you said it could be lighter or heavier, must rethink.
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Ok, here's my much more involved new answer:
+ Show Spoiler +It is impossible to go by the elimination method I outlined earlier because you do not know if the ball is heavier or lighter, so you would need an extra weighing to know which group to pick each time. So instead I thought that you could assign a 3 letter code for each ball that would be the result of the three weighings. If the first weighing makes the left side go down, then the first letter in the code is "L", an "R" if the right side goes down, or an "e" if it is even. There are then 14 unique pairs of codes that correspond to each ball being an incorrect ball, as follows:
LLL - RRR
LLe - RRe
LLR - RRL
LRL - RLR
LRe - RLe
LRR - RLL
LeL - ReR
LeR - ReL
Lee - Ree
eLL - eRR
eLR - eRL
eLe - eRe
eeL - eeR
eee - eee
So if ball one's code is eLR and it's heavier, then that corresponds to ball one not being on the scale for the first weighing, then being on the left side for the second then on the right side for the third. If the ball is lighter, then it's code would be eRL, and hence the pairing system.
Our problem now is to assign each ball a unique code pair, and then assign the weighings that will result in that code pair if that ball is the odd ball. Since there are 14 paris and 12 balls, I throw out LLL and eee as uninteresting, though it might be possible to solve it with any combination of the 14... that would require more work to prove, which I won't do here.
Taking my first weighing as
I. 1 2 3 4 vs. 5 6 7 8
I'll assign codes just in order (pretty arbitrary, I think you can make multiple ways work). (Note: 5-8 will use reciprocal codes because they start on the right)
1 - LLe 2 - LLR 3 - LRL 4 - LRe
5 - RLL 6 - ReR 7 - ReL 8 - Ree
So now we have some information about our second two weighings. We must have
II. 1 2 5 _ vs. 3 4 _ _
III. 3 5 7 _ vs. 2 6 _ _
for the codes to work. We also have 4 pairs left to assign, which again, I'll just do in order
9 -> eLL/eRR 10->eLR/eRL 11->eLe/eRe 12->eeL/eeR
again, there are a couple ways to pick which pair you want to use, but the bolded ones are the ones I picked.
Now our three weighings are:
I. 1 2 3 4 vs. 5 6 7 8
II. 1 2 5 11 vs. 3 4 9 10
III. 3 5 7 10 vs. 2 6 9 12
Do these weighings, and match what code the experiment gives you with the correct ball for that code, and viola, you are done, because each ball has a unique code. (If it's a reciprocal code, that simply means the ball is lighter, not heavier, as the code corresponds to which side goes down.)
phew, you were right, that was a bit of work!
edit: a few minor typos upon reading
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On March 16 2009 17:23 AltaiR_ wrote:Kinda old but 1. How do you put a giraffe into a refrigerator? + Show Spoiler +Open the refrigerator, put in the giraffe, and close the door.
This question tests whether you tend to do simple things in an overly complicated way. 2. How do you put an elephant into a refrigerator? + Show Spoiler +Did you say, "Open the refrigerator, put in the elephant, and close the refrigerator?" (Wrong Answer)
Correct Answer: Open the refrigerator, take out the giraffe, put in the elephant and close the door.
This tests your ability to think through the repercussions of your previous actions. 3. The King of the Forest is hosting an animal conference. All the animals attend except one. Which animal does not attend? + Show Spoiler +Correct Answer: The Elephant. The elephant is in the refrigerator. You just put him in there. This tests your memory.
OK, even if you did not answer the first three questions correctly, you still have one more chance to show your true abilities. 4. There is a river you must cross but it is inhabited by crocodiles. How do you manage it? + Show Spoiler +Correct Answer: You swim across. All the crocodiles are attending the animal conference.
This tests whether you learn quickly from your mistakes.
Yours was actually amusing compared to all these boring math ones that they are arguing about. Nice job. 
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On March 17 2009 07:33 LaughingTulkas wrote:Ok, here's my much more involved new answer: + Show Spoiler +It is impossible to go by the elimination method I outlined earlier because you do not know if the ball is heavier or lighter, so you would need an extra weighing to know which group to pick each time. So instead I thought that you could assign a 3 letter code for each ball that would be the result of the three weighings. If the first weighing makes the left side go down, then the first letter in the code is "L", an "R" if the right side goes down, or an "e" if it is even. There are then 14 unique pairs of codes that correspond to each ball being an incorrect ball, as follows:
LLL - RRR
LLe - RRe
LLR - RRL
LRL - RLR
LRe - RLe
LRR - RLL
LeL - ReR
LeR - ReL
Lee - Ree
eLL - eRR
eLR - eRL
eLe - eRe
eeL - eeR
eee - eee
So if ball one's code is eLR and it's heavier, then that corresponds to ball one not being on the scale for the first weighing, then being on the left side for the second then on the right side for the third. If the ball is lighter, then it's code would be eRL, and hence the pairing system.
Our problem now is to assign each ball a unique code pair, and then assign the weighings that will result in that code pair if that ball is the odd ball. Since there are 14 paris and 12 balls, I throw out LLL and eee as uninteresting, though it might be possible to solve it with any combination of the 14... that would require more work to prove, which I won't do here.
Taking my first weighing as
I. 1 2 3 4 vs. 5 6 7 8
I'll assign codes just in order (pretty arbitrary, I think you can make multiple ways work). (Note: 5-8 will use reciprocal codes because they start on the right)
1 - LLe 2 - LLR 3 - LRL 4 - LRe
5 - RLL 6 - ReR 7 - ReL 8 - Ree
So now we have some information about our second two weighings. We must have
II. 1 2 5 _ vs. 3 4 _ _
III. 3 5 7 _ vs. 2 6 _ _
for the codes to work. We also have 4 pairs left to assign, which again, I'll just do in order
9 -> eLL/eRR 10->eLR/eRL 11->eLe/eRe 12->eeL/eeR
again, there are a couple ways to pick which pair you want to use, but the bolded ones are the ones I picked.
Now our three weighings are:
I. 1 2 3 4 vs. 5 6 7 8
II. 1 2 5 11 vs. 3 4 9 10
III. 3 5 7 10 vs. 2 6 9 12
Do these weighings, and match what code the experiment gives you with the correct ball for that code, and viola, you are done, because each ball has a unique code. (If it's a reciprocal code, that simply means the ball is lighter, not heavier, as the code corresponds to which side goes down.)
phew, you were right, that was a bit of work!
edit: a few minor typos upon reading
+ Show Spoiler +The letter codes leads straight into my follow-up question. Would you mind going for that one? There are actually elimination methods that can be used too. Here's one: + Show Spoiler +A ball proven to be among the 11 will be referred to as a ’neutral’, a ball whose weight is unknown will be referred to as a ’candidate’. “The light one” will refer to the odd ball after it has been proven to be lighter than the rest. Similarly “the heavy one” will refer to the odd ball after it has been proven to be heavier than the rest.
First, pick 8 balls and distribute them evenly on the weighting scale. Now this may produce 3 different outcomes.
(a) Side A is heavier => 4 neutrals.
(b) Side B is heavier => 4 neutrals.
(c) Equality => 8 neutrals.
Let’s assume (a). Now, pick 2 candidates and replace them with 2 neutrals. Then pick 1 neutral and one candidate from the same side and switch them with two candidates from the other side. Now this may result in 3 new outcomes:
(aa) Nothing happens.
(ab) The heavier side goes lighter and the lighter side goes heavier.
(ac) Equality.
I’ll deal with (aa) first. This means that the ball is among the 3 balls which we didn’t move. Two of them will be on the same side, let’s assume this is the heavy side. Weight those two balls against each other, equality means the third ball is “the light one”, inequality means the heavier ball is “the heavy one”. The same principle is useable even if the two balls on the same side happen to be on the light side.
(ab) is similar to (aa) except that the odd ball is among the 3 which you switched sides on. 2 of these will be on the same side, and you will know whether that side is the heavier side or the lighter side. Apply the same principle as in (aa).
(ac) means the odd ball is among the 2 you removed from the weighting scale. These two balls came from different sides and from the first weighting you will know which of these two is heavier than the other. Weight the heavier one against a neutral, if it’s heavier it’s “the heavy one”. Else the other ball is “the light one”.
(b) can be treated like (a).
(c) is different, you have to deal with 4 balls you know nothing about in 2 weightings. First weight 2 candidates against 1 candidate and 1 neutral. Equality means that the odd ball is the lone candidate still not on the weighing scale, weight it against a neutral to determinate whether it’s “the light one” or “the heavy one”.
Inequality means that the odd ball is among the 3 candidates on the weighting scale. 2 among these will be on the same side which you know to be either heavier or lighter than the other side. Proceed as in (aa).
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EPT![[image loading]](http://i43.tinypic.com/2cckuh5.png)
