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On December 13 2012 05:21 TanGeng wrote:Show nested quote +On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? + Show Spoiler [answer] +It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy. ![[image loading]](http://i.imgur.com/iXGLi.png) Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible.
The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions. However if you adjust the swim start position clockwise a little or the wolf start counter clockwise a little to get him going on your red path I think it could work.
I'm trying to think about the calculations but they seem really complicated. You need his run distance to be 1.44*pi*r at least.
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On December 13 2012 08:25 ZapRoffo wrote:Show nested quote +On December 13 2012 05:21 TanGeng wrote:On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? + Show Spoiler [answer] +It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy. Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible. The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions.
Not true.
+ Show Spoiler +At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to.
That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so.
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On December 13 2012 08:55 xxpack09 wrote:Show nested quote +On December 13 2012 08:25 ZapRoffo wrote:On December 13 2012 05:21 TanGeng wrote:On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? + Show Spoiler [answer] +It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy. Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible. The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions. Not true. + Show Spoiler +At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to. That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so.
I understand now, + Show Spoiler +that point is the point where it switches over and the wolf is indifferent, so in reality, the wolf start has to be an infinitesimally small distance further counterclockwise to guarantee he goes counterclockwise.
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On December 13 2012 09:34 ZapRoffo wrote:Show nested quote +On December 13 2012 08:55 xxpack09 wrote:On December 13 2012 08:25 ZapRoffo wrote:On December 13 2012 05:21 TanGeng wrote:On December 12 2012 09:12 TanGeng wrote:On December 11 2012 21:39 eluv wrote:
Not sure if its been posted, but here's one that's less lateral/logical thinking, and more geometry. Suppose you are at the center of a perfectly circular lake. On the edge of the lake is a wolf, who can run four times faster than you can swim. However, you can run on land faster than the wolf, so that if you can reach the bank at a point where the wolf is not already located, you will get away. He will always run towards the point on the shore closest to your current position, with instantaneous (i.e. continuous) reactions to your movement. Assuming you can both change direction in no time, how do you reach the bank without being eaten? There is a harder challenge: The wolf runs 4.5 times faster than you can swim with everything else being the same. How do you reach the shore? + Show Spoiler [answer] +It takes a bit of thought to arrive at the starting point. Then after that it is provable that the tangent swim-out of the inner circle is the optimal strategy. Plugging in the all the numbers it's possible to show that velocity ratios larger than 4 and as high as 4.6 are possible. The way you draw that the wolf is going to run clockwise at the wolfstart/swimstart positions. Not true. + Show Spoiler +At the instant you change paths from circular to tangent, the wolf is running counterclockwise (since you had been running counterclockwise) targeted at the easternmost point of the lake. If we assume you spiral out correctly, at the point where you change paths, your angular velocity is equal to that of the wolf's. After a short time-step, your radius is increasing, but your linear velocity remains constant, so your angular velocity is decreasing. The wolf, however, runs at constant angular velocity--so he will be less than pi radians away from the point on the shoreline you're closest to. That explanation was probably shitty and hard to understand so if someone can make a better one by all means do so. I understand now, + Show Spoiler +that point is the point where it switches over and the wolf is indifferent, so in reality, the wolf start has to be an infinitesimally small distance further counterclockwise to guarantee he goes counterclockwise.
+ Show Spoiler +Does it matter? If the wolf goes clockwise can't you just swim south and get the same result?
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On April 19 2012 03:58 flamewheel wrote:Show nested quote +On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this.
sorry for bringing up this old one, i just found it, thought about it and i really dont see why there wouldnt be a much more simpler solution, i m not sure if i have to spoiler it since, it's not the actual solution:
+ Show Spoiler +Since they have perfect logic, and they see how everyone has either blue or green eyes, they will simply pair people who have identical eye colors. In other words, you dont know ur own eye-colour but u know all the rest, so you can help them and in return they will help you. It's not communication, u just grab two guys with same eye-color and pair them. Even if they wouldnt guess what u are doing, when they see u paired others (or grouped them up for that matter) they could undertand that their partner(s) have the same eye color as they do, since they can see pairs or groups with identical eyes. You will be paired with someone by another person, who sees your eyecolor.
Ok, u dont actually know if u have one of the two eye-colors, u could have purple eyes, but that's the worst case scenario, and still 199 will be saved. I know it's not as fancy, but i think it's valid, since it's not mentioned that they canot interact, only that they canot communicate.
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On December 23 2012 21:45 Geo.Rion wrote:Show nested quote +On April 19 2012 03:58 flamewheel wrote:On April 19 2012 03:24 Bahamuth wrote:
Okay, I have a really good one. Credit goes to the xkcd forums. I'll post the answer tomorrow. The solution is extremely counterintuitive.
There are 200 people on an abandoned island. 100 have green eyes, 100 have blue eyes. The only thing these people can do, is look each other in the eyes. There is no other form of communication. Therefore, they have no way to know what the colour of their own eyes is.
Every night, a boat comes to the island. If you can tell the captain with certainty what colour eyes you have, you can leave the island.
On day 1, a message is given to all inhabitants on the island: "There is at least one person that has green eyes."
The question is: Who can leave the island, and after how long?
Important notes:
- They are all perfect logicians
- Everyone knows the eyecolour of every OTHER inhabitant at all times. The only thing they don't know, is their own eyecolour.
- This can be solved with pure logic, not by coming up with workarounds like reflection in the water or communication with the captain. This was fun :3 + Show Spoiler [solution] +Each green sees:
99g 100b
Each blue sees:
100g 99b
Day 1: There is at least one person that has green eyes.
Start from base case--
1g Xb
The person with g will see only B. Thus at end of first day g will leave.
2g Xb
Each g will see 1g and Xb. Each g will think "if that person is the only g, he will leave at night."
Result: Neither person leaves. Both g will then deduce that the other person was not the only g, and since everybody else is b both g will leave at end of day 2.
3g Xb
Same story: each g will think that the other two g are the only 2 g. Following the [2g Xb] case each g will think "if the two g have left at end of day 2, then they were the only 2 g."
Result: None leave. All three g will deduce that the other two g are not the only ones, and since everybody else is b all three g will leave at end of day 3.
==
Doesn't matter how many blues there are. Inducting upon 1g, 2g, 3g... we arrive at the conclusion that all 100 green-eyed people will leave at the end of the 100th day.
Sucks to have blue eyes. Though not sure if correct. Edit: hmm I feel slightly better after reading other posts in this thread about this. sorry for bringing up this old one, i just found it, thought about it and i really dont see why there wouldnt be a much more simpler solution, i m not sure if i have to spoiler it since, it's not the actual solution: + Show Spoiler +Since they have perfect logic, and they see how everyone has either blue or green eyes, they will simply pair people who have identical eye colors. In other words, you dont know ur own eye-colour but u know all the rest, so you can help them and in return they will help you. It's not communication, u just grab two guys with same eye-color and pair them. Even if they wouldnt guess what u are doing, when they see u paired others (or grouped them up for that matter) they could undertand that their partner(s) have the same eye color as they do, since they can see pairs or groups with identical eyes. You will be paired with someone by another person, who sees your eyecolor.
Ok, u dont actually know if u have one of the two eye-colors, u could have purple eyes, but that's the worst case scenario, and still 199 will be saved. I know it's not as fancy, but i think it's valid, since it's not mentioned that they canot interact, only that they canot communicate.
That is a form of communication. Communication does not entail the verbal use of conventional signs. Providing others information in a goal directed manner suffices.
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This one (mainly the first one, I don't care as much about the rest) has me completely stumped and I can't find anything in google:
http://www.teamliquid.net/forum/viewmessage.php?topic_id=390196
The only thing I could find is someone saying + Show Spoiler + is the answer, but no matter how many times I try to work it out I am not reaching the same number.
edit:
I keep getting + Show Spoiler + when I work it out, which is the reverse of what it needs to be unless I am messing up. Can anyone help me with this? I use the - to denote between rule applications
+ Show Spoiler +474536-14745362
by 1x2 = x rule,
474536-474536
by 6x = 1x rule,
7453-1474536
by 3x = xx rule,
745-14745361474536
by 5yx = x rule,
74-4745361474536
by 4x = x reversed rule,
7-6354741635474
by 7x = 2x rule,
26354741635474
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I also seem someone said they found a number and the number being
+ Show Spoiler +
I am way to tired to even think of trying it to see if they are correct but I will check tomorrow
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So here s this thing:
http://kmdszbonuszfeladat.appspot.com/10523483189552331441321.html
do you guys have any clue? Feel free to PM me, ill know the answer eventually and post it if nobody else figures it out
Solution, well not that cool, but it was pretty hard to figure out. Actually really really hard>
+ Show Spoiler +The numbers in the link were the first 14 fibonacci numbers, mixed up, and the latin quote has 14 letters. To solve this puzzle you had to take the proper order of the Fibonacci numbers and identify each number with one letter. So D-0, I-1, V-1, I-2 , D-3, E-5 etc, and re-arrange the letters so the order is the mixed up order of the numbers in the link. The solution was in Hungarian, "Idei-medve-parti" which translates raughly to "This year's bear party". Pretty sick one, and this is only the 2nd from a 10-20 long bonus-task list.
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sry but, i don't even know what im looking at, can you maybe explain it a bit?
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On April 23 2013 14:51 TerranosaurusWrecks wrote:sry but, i don't even know what im looking at, can you maybe explain it a bit?
there is no more explanation. It has something to do with the Fibonacci numbers, and the quote is in latin, it translates to "divide and concqure" i think
EDIT
yes, that's the picture, but not the solution.
From some googling i found there is a Divide and Concqure algorithm, though im kinda stuck there
http://en.wikipedia.org/wiki/Divide_and_conquer_algorithm
I suppouse the last part of the original link, the numbers, factor in, though i m still trying to figure out how
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Australia8532 Posts
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Spain64 Posts
I can't figure this riddle out.
"Everyone loses us as a child, and neglecting us makes us holy. What are we?"
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On June 05 2013 18:21 XpNc wrote:
I can't figure this riddle out.
"Everyone loses us as a child, and neglecting us makes us holy. What are we?"
+ Show Spoiler +
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On June 05 2013 18:28 refmac_cys.cys wrote:Show nested quote +On June 05 2013 18:21 XpNc wrote:
I can't figure this riddle out.
"Everyone loses us as a child, and neglecting us makes us holy. What are we?" + Show Spoiler +
Can someone explain, for non native speakers, why neglecting+ Show Spoiler + makes them holy? Is that a euphemism used in religious countries for + Show Spoiler +?
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Aotearoa39261 Posts
On June 05 2013 18:43 HaRuHi wrote:Show nested quote +On June 05 2013 18:28 refmac_cys.cys wrote:On June 05 2013 18:21 XpNc wrote:
I can't figure this riddle out.
"Everyone loses us as a child, and neglecting us makes us holy. What are we?" + Show Spoiler + Can someone explain, for non native speakers, why neglecting + Show Spoiler + makes them holy? Is that a euphemism used in religious countries for + Show Spoiler +?
+ Show Spoiler +It makes them full of holes 
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EPT
