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GendoIkari

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Re: Logic problems
« Reply #50 on: January 11, 2013, 11:31:14 am »
0

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.


Almost. Nicely done; except... You know know if switch 2 started up or down. So if it started up, then Master would count 1 even when there hasn't been a maiden yet. So you could add 1 extra to that (which your count of 50 does), but then if switch 2 happens to start down instad, you would never reach 50, it would stay at 49 forever.
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jonts26

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Re: Logic problems
« Reply #51 on: January 11, 2013, 11:36:33 am »
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100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

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DStu

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Re: Logic problems
« Reply #52 on: January 11, 2013, 11:38:07 am »
0

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.


Almost. Nicely done; except... You know know if switch 2 started up or down. So if it started up, then Master would count 1 even when there hasn't been a maiden yet. So you could add 1 extra to that (which your count of 50 does), but then if switch 2 happens to start down instad, you would never reach 50, it would stay at 49 forever.


I did add 1, as there are only 49 other maidens, the mastermaiden obviously has been in this room, as she is at the moment and has been there 49 times before...

Edit: OK, I did add 1, but I didn't finish reading you're post...
Edit2: On the other hand, we already waited N^2log N, and it only take N log N for everyone to enter the room, so probably we're fine with 49...
« Last Edit: January 11, 2013, 11:43:52 am by DStu »
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DStu

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Re: Logic problems
« Reply #53 on: January 11, 2013, 11:39:22 am »
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100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?


The answer is incredible large...
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GendoIkari

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Re: Logic problems
« Reply #54 on: January 11, 2013, 01:11:07 pm »
0

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.


Almost. Nicely done; except... You know know if switch 2 started up or down. So if it started up, then Master would count 1 even when there hasn't been a maiden yet. So you could add 1 extra to that (which your count of 50 does), but then if switch 2 happens to start down instad, you would never reach 50, it would stay at 49 forever.


I did add 1, as there are only 49 other maidens, the mastermaiden obviously has been in this room, as she is at the moment and has been there 49 times before...

Edit: OK, I did add 1, but I didn't finish reading you're post...
Edit2: On the other hand, we already waited N^2log N, and it only take N log N for everyone to enter the room, so probably we're fine with 49...

Granted, once you reach 49 you're pretty likely to be safe... but there is a way to be sure.It's still your basic answer; just with a minor change.
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eHalcyon

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Re: Logic problems
« Reply #55 on: January 11, 2013, 02:00:28 pm »
+1

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.


Almost. Nicely done; except... You know know if switch 2 started up or down. So if it started up, then Master would count 1 even when there hasn't been a maiden yet. So you could add 1 extra to that (which your count of 50 does), but then if switch 2 happens to start down instad, you would never reach 50, it would stay at 49 forever.


I did add 1, as there are only 49 other maidens, the mastermaiden obviously has been in this room, as she is at the moment and has been there 49 times before...

Edit: OK, I did add 1, but I didn't finish reading you're post...
Edit2: On the other hand, we already waited N^2log N, and it only take N log N for everyone to enter the room, so probably we're fine with 49...

Granted, once you reach 49 you're pretty likely to be safe... but there is a way to be sure.It's still your basic answer; just with a minor change.
Have each regular maiden signal twice instead of just once. Master maiden counts to 99 instead of 49.
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GendoIkari

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Re: Logic problems
« Reply #56 on: January 11, 2013, 03:01:09 pm »
0

What plan do the maidens come up with that night?

Probably not optimal:

You could elect some master maiden.  Use switch 1 as dummyswitch which does not carry any information.  If usual maiden enters the room and switch 2 is "down", and she hasn't switched it already, switch it
if switch 2 is "up", or this maiden has already switched switch 2, switch switch 1.
if master maiden enters room and switch 2 is "up", switch it to "down"
otherwise use switch 1

Now if the master maiden has turned the switch2 down for 50 times, all other maiden have entered the room.


Estimated maiden in rooms ~N^2 log N I think.


Almost. Nicely done; except... You know know if switch 2 started up or down. So if it started up, then Master would count 1 even when there hasn't been a maiden yet. So you could add 1 extra to that (which your count of 50 does), but then if switch 2 happens to start down instad, you would never reach 50, it would stay at 49 forever.


I did add 1, as there are only 49 other maidens, the mastermaiden obviously has been in this room, as she is at the moment and has been there 49 times before...

Edit: OK, I did add 1, but I didn't finish reading you're post...
Edit2: On the other hand, we already waited N^2log N, and it only take N log N for everyone to enter the room, so probably we're fine with 49...

Granted, once you reach 49 you're pretty likely to be safe... but there is a way to be sure.It's still your basic answer; just with a minor change.
Have each regular maiden signal twice instead of just once. Master maiden counts to 99 instead of 49.

Right.
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Jimmmmm

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Re: Logic problems
« Reply #57 on: January 11, 2013, 07:20:15 pm »
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My take on the blue eyes problem:

If there was exactly one person with blue eyes, they would leave the very next morning (morning 1). If there were exactly two, both would realise this after the other did not leave on morning 1, and leave on morning 2. If there were exactly 3, they would all see that the other two did not leave on morning 2, and then all leave on morning 3. And so on and so forth. So each morning that no one leaves increases the minimum number of people with blue eyes. On morning 99 when the other 99 people with blue eyes don't leave, everyone realises there must be at least 100 people with blue eyes, and all the people with blue eyes realise their eye colour and leave on morning 100, which makes everyone else realise they don't have blue eyes, since if they did all the blue-eyed people would hang around until the next morning.

In fact, if there are only two colours then you know already the exact morning you will leave, morning x + 1, where x is the number of people you can see with blue eyes.
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WanderingWinder

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Re: Logic problems
« Reply #58 on: January 11, 2013, 07:40:33 pm »
0

My take on the blue eyes problem:

If there was exactly one person with blue eyes, they would leave the very next morning (morning 1). If there were exactly two, both would realise this after the other did not leave on morning 1, and leave on morning 2. If there were exactly 3, they would all see that the other two did not leave on morning 2, and then all leave on morning 3. And so on and so forth. So each morning that no one leaves increases the minimum number of people with blue eyes. On morning 99 when the other 99 people with blue eyes don't leave, everyone realises there must be at least 100 people with blue eyes, and all the people with blue eyes realise their eye colour and leave on morning 100, which makes everyone else realise they don't have blue eyes, since if they did all the blue-eyed people would hang around until the next morning.

In fact, if there are only two colours then you know already the exact morning you will leave, morning x + 1, where x is the number of people you can see with blue eyes.

This (last point anyway) can't be right, because you can make the same argument for the number of people you can see with non-blue eyes, and x for the two cases need not be equal.

Edit: I was assuming you were talking about the no new person showing up case, which I now realize isn't clear. The person showing up and saying blue breaks the symmetry, making my point not applicable
« Last Edit: January 11, 2013, 07:43:00 pm by WanderingWinder »
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pacovf

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Re: Logic problems
« Reply #59 on: January 12, 2013, 06:04:43 am »
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My favorite puzzles have already been posted (blue eyes, 100 prisoners), so I'll just post a somewhat easy one:

7 pirates have just ended plundering the Caribbean, so they get together to discuss how they will share the 100 gold coin booty. Now, these aren't your normal kind of pirates, they are blood-thirsty, unlimitedly greedy, perfectly rational pirates, and they are ranked by fighting prowess.

Thus, the captain has to propose how to share the booty, knowing that if more than half of his crew starts a mutiny, he will get killed, and the 2nd in command will take his place (who will in turn have to propose how to share the booty, eventually being replaced by the third in command, etc.).

How much gold will each pirate receive?
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Drab Emordnilap

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Re: Logic problems
« Reply #60 on: January 12, 2013, 08:27:49 am »
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My favorite puzzles have already been posted (blue eyes, 100 prisoners), so I'll just post a somewhat easy one:

7 pirates have just ended plundering the Caribbean, so they get together to discuss how they will share the 100 gold coin booty. Now, these aren't your normal kind of pirates, they are blood-thirsty, unlimitedly greedy, perfectly rational pirates, and they are ranked by fighting prowess.

Thus, the captain has to propose how to share the booty, knowing that if more than half of his crew starts a mutiny, he will get killed, and the 2nd in command will take his place (who will in turn have to propose how to share the booty, eventually being replaced by the third in command, etc.).

How much gold will each pirate receive?

Does the captain get a vote? I forget.
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Jimmmmm

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Re: Logic problems
« Reply #61 on: January 12, 2013, 08:45:08 am »
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100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

My first thought is this: Start from the back of the line. If you're at an odd position, say the colour of the hat directly in front of you. There's a 50% chance of this being successful. If you're at an even position, simply say the colour that the person behind you said. This will save at least 50 of them, and on average 75. There may well be a way to do better, especially since using this strategy, the first person to speak will know exactly who will live and who will die.
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WanderingWinder

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Re: Logic problems
« Reply #62 on: January 12, 2013, 09:00:59 am »
0

100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

My first thought is this: Start from the back of the line. If you're at an odd position, say the colour of the hat directly in front of you. There's a 50% chance of this being successful. If you're at an even position, simply say the colour that the person behind you said. This will save at least 50 of them, and on average 75. There may well be a way to do better, especially since using this strategy, the first person to speak will know exactly who will live and who will die.
If you're forced to start at the back of the line and move up, you can save everyone except the dude at the back with 100% certainty, and that guy has a 50-50 shot. Not sure if being able to do better than this by going in some other order, but I don't think so.

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Re: Logic problems
« Reply #63 on: January 12, 2013, 09:05:48 am »
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100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

My first thought is this: Start from the back of the line. If you're at an odd position, say the colour of the hat directly in front of you. There's a 50% chance of this being successful. If you're at an even position, simply say the colour that the person behind you said. This will save at least 50 of them, and on average 75. There may well be a way to do better, especially since using this strategy, the first person to speak will know exactly who will live and who will die.
If you're forced to start at the back of the line and move up, you can save everyone except the dude at the back with 100% certainty, and that guy has a 50-50 shot. Not sure if being able to do better than this by going in some other order, but I don't think so.

I had a feeling there was a way to guarantee either 99 or 100 (I may have heard of it before), but I have no idea how. Like, the only possible information each person can get is from what the very first person said.

Edit: I'm assuming you can't give across any information by the timing etc that you say your colour.
« Last Edit: January 12, 2013, 09:08:09 am by Jimmmmm »
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WanderingWinder

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Re: Logic problems
« Reply #64 on: January 12, 2013, 09:35:05 am »
+1

100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

My first thought is this: Start from the back of the line. If you're at an odd position, say the colour of the hat directly in front of you. There's a 50% chance of this being successful. If you're at an even position, simply say the colour that the person behind you said. This will save at least 50 of them, and on average 75. There may well be a way to do better, especially since using this strategy, the first person to speak will know exactly who will live and who will die.
If you're forced to start at the back of the line and move up, you can save everyone except the dude at the back with 100% certainty, and that guy has a 50-50 shot. Not sure if being able to do better than this by going in some other order, but I don't think so.

I had a feeling there was a way to guarantee either 99 or 100 (I may have heard of it before), but I have no idea how. Like, the only possible information each person can get is from what the very first person said.

Edit: I'm assuming you can't give across any information by the timing etc that you say your colour.

Simple version. Treat say white hats as 1s, or really just count them. Guy in back says white if there's an odd number of white hats in front of him, black if there's an even number. He gets a 50-50 shot - there's no way for him to tell what his own hat is, nobody can see it. Next guy in front of him counts the number of white hats. If guy behind him said white, then he know guy behind him saw an odd number of white hats. If this is the case, he can count number of white/black hats in front of him, and if it's also odd, he knows he's wearing black, or if it's even, he knows he's wearing white. If guy in back said black, he knows that there were an even number of white hats, and if he also sees an even number, he's wearing black, and if he sees an odd number, he must be wearing black. Then everyone else just keeps this even-odd counter in their head. Every time somebody says white, because everyone before them and after the first guy was right, they flip the counter in their head. Then they count the number of whites in front of them and compare, is that number even or odd. If it matches their counter, they must be wearing black. If it differs, then they must be wearing white. Ergo, everyone except the last guy can be sure.[/quote]

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Re: Logic problems
« Reply #65 on: January 12, 2013, 09:40:27 am »
+1

Ah, very nice.
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Grujah

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Re: Logic problems
« Reply #66 on: January 12, 2013, 09:40:29 am »
0

I looove the maiden one. count 50 unique visits with only 2 bits, it's awesome :)
I did not solve it.

100 prisoners are about to be executed. The king, in his mercy, decides to give them a chance for pardon. Each prisoner will be given either a white or a black hat at random and are then arranged in a line such that each prisoner can see every other prisoner in front of him, but none behind him. The prisoners are given some time to formulate a strategy, but once they are lined up, each prisoner can only speak a single word, either black or white. If the prisoner correctly identifies his own hat color, he goes free. If not, he dies.

What is the best strategy for allowing the most number of prisoners to go free?

I think this is it:
The one at the far back counts the number of white hats. He says white if it is even, black if it is odd. He is 50-50.
Guy in front of him now knows if he is black or white (if he sees the same oddity, he is black, otherwise he is white).
Guy in front of that guy now eliminates the last guy from calculations, and does the same.
And so on..

Last guy in line is 50-50, rest are 100-0.


Right?

Edit: Ninja'd by WW, but I didn't know this beforehand ( I did know blue eyes from xkcd :P)
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Re: Logic problems
« Reply #67 on: January 12, 2013, 09:44:20 am »
+1

Ah, very nice.

Love the spoiler tags on this one. :)
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Jimmmmm

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Re: Logic problems
« Reply #68 on: January 12, 2013, 09:46:02 am »
0

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Grujah

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Re: Logic problems
« Reply #69 on: January 12, 2013, 10:20:39 am »
0

My favorite puzzles have already been posted (blue eyes, 100 prisoners), so I'll just post a somewhat easy one:

7 pirates have just ended plundering the Caribbean, so they get together to discuss how they will share the 100 gold coin booty. Now, these aren't your normal kind of pirates, they are blood-thirsty, unlimitedly greedy, perfectly rational pirates, and they are ranked by fighting prowess.

Thus, the captain has to propose how to share the booty, knowing that if more than half of his crew starts a mutiny, he will get killed, and the 2nd in command will take his place (who will in turn have to propose how to share the booty, eventually being replaced by the third in command, etc.).

How much gold will each pirate receive?

Does the captain get a vote? I forget.

Figured this one out. Yeah, cap gets a vote too, checked wiki for exact setup. Also, they want as much money as they can get (greedy), and if they can kill some other pirate while still receiving the same amount of money, the will do so (bloodthirsty).
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Jimmmmm

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Re: Logic problems
« Reply #70 on: January 12, 2013, 10:30:59 am »
0

My favorite puzzles have already been posted (blue eyes, 100 prisoners), so I'll just post a somewhat easy one:

7 pirates have just ended plundering the Caribbean, so they get together to discuss how they will share the 100 gold coin booty. Now, these aren't your normal kind of pirates, they are blood-thirsty, unlimitedly greedy, perfectly rational pirates, and they are ranked by fighting prowess.

Thus, the captain has to propose how to share the booty, knowing that if more than half of his crew starts a mutiny, he will get killed, and the 2nd in command will take his place (who will in turn have to propose how to share the booty, eventually being replaced by the third in command, etc.).

How much gold will each pirate receive?

Captain keeps 97 for himself and gives 1 to each of 3, 5 and 7?
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pacovf

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Re: Logic problems
« Reply #71 on: January 12, 2013, 10:35:05 am »
0

Captain keeps 97 for himself and gives 1 to each of 3, 5 and 7?

Yep! Obviously, if the pirates could leave the ship, I'm sure that wouldn't be the solution :)

Does the captain get a vote? I forget.

Figured this one out. Yeah, cap gets a vote too, checked wiki for exact setup. Also, they want as much money as they can get (greedy), and if they can kill some other pirate while still receiving the same amount of money, the will do so (bloodthirsty).

Ah, you beat me :)

You can solve the problem too if the captain doesn't get a vote; it's slightly harder, and gets you a different result, but the way you get the solution is the same.
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Re: Logic problems
« Reply #72 on: January 12, 2013, 11:01:23 am »
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I'll just leave this here

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

The simpler version that I know of that is:

There are two computers, and two doors. Behind one door is a lion that will eat you, and behind the other is freedom and riches that will leave you happy for life.

One computer is programmed to answer any query made of it truthfully.
The other is programmed to answer any query made of it falsely.

You do not know which computer is which.

You may input one question only into one of the two computers. What question do you ask, and which computer do you ask it of?

I solved this one as a kid, asked in a bit different form, so this might not be a valid query to put in:
Are the truth-telling computer and door with riches on booth the same side? (i.e. left)
If the answer is yes, you go to the door on the same side as the computer that you asked at, otherwise on the other door.)


If not, you need to artificially pair the computers and the doors in some fashion, and than ask a more complicated query, i.e.
"If I name you the "green" computer, and the other one "blue" computer, and I name door 1. "green" and door number 2 "blue", are both truth telling computer and door to the riches of the same "color"?
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Galzria

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Re: Logic problems
« Reply #73 on: January 12, 2013, 11:10:14 am »
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I'll just leave this here

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

The simpler version that I know of that is:

There are two computers, and two doors. Behind one door is a lion that will eat you, and behind the other is freedom and riches that will leave you happy for life.

One computer is programmed to answer any query made of it truthfully.
The other is programmed to answer any query made of it falsely.

You do not know which computer is which.

You may input one question only into one of the two computers. What question do you ask, and which computer do you ask it of?

I solved this one as a kid, asked in a bit different form, so this might not be a valid query to put in:
Are the truth-telling computer and door with riches on booth the same side? (i.e. left)
If the answer is yes, you go to the door on the same side as the computer that you asked at, otherwise on the other door.)


If not, you need to artificially pair the computers and the doors in some fashion, and than ask a more complicated query, i.e.
"If I name you the "green" computer, and the other one "blue" computer, and I name door 1. "green" and door number 2 "blue", are both truth telling computer and door to the riches of the same "color"?


The simpler answer, without needing to define anything at all (and probably derives from the question Jonts linked), is "What door would the other computer tell me contains the riches?"

If you asked the Truth computer, he would answer by directing you to the door with the Lion.
If you asked the Lying computer, he would answer by directing you to the door with the Lion.

In either case, you choose the opposite door.
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Mafia Record:
TOWN Wins: M3, M5, M6, M11, M17, M28, M32, M105, M108, M114, M118, M120, M122, DM1, DoM1, OZ2, RM45, RM47, RM48, RM49, RM55
TOWN Losses: M4, M7, M8, M9, M13, M14, M18, M31, M110, M111, M113, M117, M125, RM3, RM4, RM54
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SCUM Losses: M15 (SK), M102 (Tr), OZ1, RM55

Total Wins: 30
Total Losses: 20

WanderingWinder

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Re: Logic problems
« Reply #74 on: January 12, 2013, 11:36:39 am »
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Another hat problem (the first one I ever heard):
Four buys are captured, yadda yadda,  and put, buried in sand up to their necks, into the following arrangement:
O O O | O
Where O represents a guy's head and | represents a wall that can't be seen through. All of them are facing the direction that corresponds to right in the above diagram. They are told that two of them will have white hats on their heads, the other two black hats. The first thing any of them says must be their own hat color; if they do this, all are freed; if not, they are all executed.
The hats are put on them in the order Black/White/Black|White. Who is able to free them all (none of them, one, or multiple people; if more than 0, specify which)?
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