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** STICKY ** League of Reason? PROVE YOURSELF!
- Thread starter Fictionarious
- Start date
Re: League of Reason? PROVE YOURSELF!
I'm just out of sixth form college and I've played computergames since I was about nine, so I shouldn't be able to dress mysefl <.<
And about the weighing of the balls, I was halfway done with a modell on how to solve it when I realized I was weighing three balls at both sides of the scale half of the time
Marcus said:
I'm just out of sixth form college and I've played computergames since I was about nine, so I shouldn't be able to dress mysefl <.<
And about the weighing of the balls, I was halfway done with a modell on how to solve it when I realized I was weighing three balls at both sides of the scale half of the time
The Apathetic Despot
New Member
Re: League of Reason? PROVE YOURSELF!
Well done. If I'm following you right that's the very same method I came up with, but I'd be very interested to hear if anyone sees a different way to go about it. That method also works beautifully for the harder 2d version of the problem, though I won't bother asking that one since conveying it without diagrams would be a nightmare.
Here's one in a similar vein that I haven't actually sat down and solved yet myself. You and your friends (n people in total) stand in a circle, fix one person as the "starting point", and beginning from him you each take turns (going clockwise) counting up from one. The statring person says 1, the next person says 2 and so on. When someone says an even number, he immediately leaves the circle, and the last un-eliminated person wins.
The problem is to figure out what "position" in the circle you want to be in to win the game. Obviously doing this one iteratively is boring, since you're basically just reiterating the problem in numbers, so what it needs is a closed form function of n that gives you the winning position. Again, I haven't actually done this problem myself but I'm assuming such a solution exists because otherwise it's not much of a puzzle.
Marcus said:
Well done. If I'm following you right that's the very same method I came up with, but I'd be very interested to hear if anyone sees a different way to go about it. That method also works beautifully for the harder 2d version of the problem, though I won't bother asking that one since conveying it without diagrams would be a nightmare.
Here's one in a similar vein that I haven't actually sat down and solved yet myself. You and your friends (n people in total) stand in a circle, fix one person as the "starting point", and beginning from him you each take turns (going clockwise) counting up from one. The statring person says 1, the next person says 2 and so on. When someone says an even number, he immediately leaves the circle, and the last un-eliminated person wins.
The problem is to figure out what "position" in the circle you want to be in to win the game. Obviously doing this one iteratively is boring, since you're basically just reiterating the problem in numbers, so what it needs is a closed form function of n that gives you the winning position. Again, I haven't actually done this problem myself but I'm assuming such a solution exists because otherwise it's not much of a puzzle.
Fictionarious
New Member
Re: League of Reason? PROVE YOURSELF!
Okay, so I fail at this n logician soot thing.
If there are two of them, they will observe the other laughing and conclude with certainty that they have soot, so they'll simultaneously wipe it off.
If there are three however, any one will observe the other two, see them laughing, and not know.
Because it could be (to them) that their head is clean and the two others are laughing at one another, but not at him.
...But at that point, he will observe that neither of the others is bothering to wipe, and must conclude that he has soot on his forehead, because if he hadn't, the other two would have taken action. And if there be four... the same logic hold.
Ah fuck it, just explain it to me.
Okay, so I fail at this n logician soot thing.
If there are two of them, they will observe the other laughing and conclude with certainty that they have soot, so they'll simultaneously wipe it off.
If there are three however, any one will observe the other two, see them laughing, and not know.
Because it could be (to them) that their head is clean and the two others are laughing at one another, but not at him.
...But at that point, he will observe that neither of the others is bothering to wipe, and must conclude that he has soot on his forehead, because if he hadn't, the other two would have taken action. And if there be four... the same logic hold.
Ah fuck it, just explain it to me.
TwoPoint7182818
New Member
Re: League of Reason? PROVE YOURSELF!
But doesn't that lead to a paradox? Because if there is a way for one to determine whether or not the soot is on their head, they will all wipe their heads. So therefor they must determine that their heads have soot on them from the fact that everyone else is wiping, but if everyone needs to wait for everyone else to wipe before they can precede then no one can move because no one can move first. But if they are determining from the fact that no one is wiping, then every one will require that no one wipe in order to wipe, but since it's symmetrical ARGHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
The only way it can work is if some of the logicians are brighter than others, and everyone knows the relative speed at which everyone else thinks?
But doesn't that lead to a paradox? Because if there is a way for one to determine whether or not the soot is on their head, they will all wipe their heads. So therefor they must determine that their heads have soot on them from the fact that everyone else is wiping, but if everyone needs to wait for everyone else to wipe before they can precede then no one can move because no one can move first. But if they are determining from the fact that no one is wiping, then every one will require that no one wipe in order to wipe, but since it's symmetrical ARGHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
The only way it can work is if some of the logicians are brighter than others, and everyone knows the relative speed at which everyone else thinks?
Fictionarious
New Member
Re: League of Reason? PROVE YOURSELF!
Marcus-
Oh, cool.
I guess we could reason that it will take an increasingly larger length of time for all the logicians to wipe the soot from their heads as each of them waits for the others to take actions they fail to take... there might be some mathematical relationship between the # of logicians and 'moments' of simultaneous cognition.
TAD-
2 people, you want to be 1st
3 people, you want to be 3rd
4 people, you want to be 1st
5 people, you want to be 3rd
6 people, you want to be 5th
7 people, you want to be 7th
8 people, you want to be 1st.
shit. that's a good one.
Marcus-
Oh, cool.
I guess we could reason that it will take an increasingly larger length of time for all the logicians to wipe the soot from their heads as each of them waits for the others to take actions they fail to take... there might be some mathematical relationship between the # of logicians and 'moments' of simultaneous cognition.
TAD-
2 people, you want to be 1st
3 people, you want to be 3rd
4 people, you want to be 1st
5 people, you want to be 3rd
6 people, you want to be 5th
7 people, you want to be 7th
8 people, you want to be 1st.
shit. that's a good one.
TwoPoint7182818
New Member
Re: League of Reason? PROVE YOURSELF!
But how will the first one know?
Either everyone knows at the same time or no one knows. Unless some logicians are smarter than others.
But how will the first one know?
Either everyone knows at the same time or no one knows. Unless some logicians are smarter than others.
Re: League of Reason? PROVE YOURSELF!
The first one knows because no one is wiping. If everybody reacts at the same time and they see everybody else wearing soot, they will all remove it simulateously.
If somebody reacts quicker, then it will result in a chain reaction.
The first one knows because no one is wiping. If everybody reacts at the same time and they see everybody else wearing soot, they will all remove it simulateously.
If somebody reacts quicker, then it will result in a chain reaction.
TwoPoint7182818
New Member
Re: League of Reason? PROVE YOURSELF!
But if everyone can figure out from the fact that no one is wiping that they should be wiping, then everyone will be wiping....
But if everyone can figure out from the fact that no one is wiping that they should be wiping, then everyone will be wiping....
Fictionarious
New Member
Re: League of Reason? PROVE YOURSELF!
If they all know the speed at which they each calculate the logic, then the first on in a group of four will know after the three complete the thinking process narrowing it down to two and then themselves, so my hypothesis is that they'd all wipe at once, but at a progressively distant moment after exiting the tunnel the more there are.
If they all know the speed at which they each calculate the logic, then the first on in a group of four will know after the three complete the thinking process narrowing it down to two and then themselves, so my hypothesis is that they'd all wipe at once, but at a progressively distant moment after exiting the tunnel the more there are.
Re: League of Reason? PROVE YOURSELF!
Yes. Well, if you could play it in rounds it would be alot easier.
For instance if n=4:
All logicians look around and start laughing, realizing that everybody else has soot on their forehead. Nobody wipes because there are two or more others with soot on their forehead.
Round 2:
The logicians realize that because nobody wiped their forehead during the first round the count of sooted foreheads must be >= 3, nobody wipes because they see three or more sooted foreheads.
Round 3: The logicians realize that because nobody wiped their forehead during the second round the count of sooted foreheads must be >=4 and everybody wipes their foreheads, because if the count is four or over then their forehead has to be sooted.
This can go on ad-infinitum.
Yes. Well, if you could play it in rounds it would be alot easier.
For instance if n=4:
All logicians look around and start laughing, realizing that everybody else has soot on their forehead. Nobody wipes because there are two or more others with soot on their forehead.
Round 2:
The logicians realize that because nobody wiped their forehead during the first round the count of sooted foreheads must be >= 3, nobody wipes because they see three or more sooted foreheads.
Round 3: The logicians realize that because nobody wiped their forehead during the second round the count of sooted foreheads must be >=4 and everybody wipes their foreheads, because if the count is four or over then their forehead has to be sooted.
This can go on ad-infinitum.
Marcus
New Member
Re: League of Reason? PROVE YOURSELF!
Basically, you can do it by induction. Firstly, think of the n=3 case. All start laughing, but each figures that he can't have a clean forehead, since otherwise the others would have seen the other dirty headed logician laughing and surmised that he had a dirty head.
The induction works similarly. Everyone knows how the others think. Imagine a situation where one logician postulates that has a clean forehead. He then puts himself in the position of one of his colleagues. That person can see soot on the other n-2 logicians, so as far as he's concerned the clean-headed logician could be laughing at them. This means that if the logician had a clean head, the others would be thinking exactly as they would if the clean headed logician were absent - i.e. it would be the n-1 case.
So, since 3 logicians will eventually realise all have sooty heads, so will 4, 5, 6, ...
Basically, you can do it by induction. Firstly, think of the n=3 case. All start laughing, but each figures that he can't have a clean forehead, since otherwise the others would have seen the other dirty headed logician laughing and surmised that he had a dirty head.
The induction works similarly. Everyone knows how the others think. Imagine a situation where one logician postulates that has a clean forehead. He then puts himself in the position of one of his colleagues. That person can see soot on the other n-2 logicians, so as far as he's concerned the clean-headed logician could be laughing at them. This means that if the logician had a clean head, the others would be thinking exactly as they would if the clean headed logician were absent - i.e. it would be the n-1 case.
So, since 3 logicians will eventually realise all have sooty heads, so will 4, 5, 6, ...
The Apathetic Despot
New Member
Re: League of Reason? PROVE YOURSELF!
Ah nuts, I got half way through a post saying that before I got myself tied back up in logical knots. Next time I'll be sure to just throw ideas at the wall until one sticks :lol: .Marcus said:
Fictionarious
New Member
Re: League of Reason? PROVE YOURSELF!
TAD -
for a circle with a number of people which is a power of two, the pattern seems to be that you'd want the spot which is n - (n - 1).
I'm not certain, someone would have to check that at 16 and then more...
In the meantime, I'll throw this one out -
What is it that the man who makes it does not need, the man who buys it does not use himself, and the person who uses it does so without knowing?
Might be too easy...
TAD -
for a circle with a number of people which is a power of two, the pattern seems to be that you'd want the spot which is n - (n - 1).
I'm not certain, someone would have to check that at 16 and then more...
In the meantime, I'll throw this one out -
What is it that the man who makes it does not need, the man who buys it does not use himself, and the person who uses it does so without knowing?
Might be too easy...
The Apathetic Despot
New Member
Re: League of Reason? PROVE YOURSELF!
A little bit, yes. It's a grave. Also, n-(n-1) just equals 1.
Edit: now that I've thought about it (I have a calculus exam tuesday so I'm a bit distracted), that does make a lot of sense, since you'll eliminate half each time you go through an evan number of people, and half a power of two is also a power of two, which in turn is even, so yes for any n wich is a power of two you want to be first.
Fictionarious said:
A little bit, yes. It's a grave. Also, n-(n-1) just equals 1.
Edit: now that I've thought about it (I have a calculus exam tuesday so I'm a bit distracted), that does make a lot of sense, since you'll eliminate half each time you go through an evan number of people, and half a power of two is also a power of two, which in turn is even, so yes for any n wich is a power of two you want to be first.
Fictionarious
New Member
Re: League of Reason? PROVE YOURSELF!
Oh. Duh. :roll:
Oh. Duh. :roll:
Re: League of Reason? PROVE YOURSELF!
OK, new puzzle:
There's a hall with 1000 doors, and there are 1000 people.
The first person opens all 1000 doors.
Then, the second person goes to every second door, and closes it again.
Then, the third person goes to every third door. If it's open, he closes it; if it's closed, he opens it.
The fourth person does this to every fourth door, and so on.
After the 1000th person has done his job, how many doors are open?
Too bad you missed the 100 Reasoning Leaguers riddle, you would've loved that one.Marcus said:
OK, new puzzle:
There's a hall with 1000 doors, and there are 1000 people.
The first person opens all 1000 doors.
Then, the second person goes to every second door, and closes it again.
Then, the third person goes to every third door. If it's open, he closes it; if it's closed, he opens it.
The fourth person does this to every fourth door, and so on.
After the 1000th person has done his job, how many doors are open?
The Apathetic Despot
New Member
Re: League of Reason? PROVE YOURSELF!
The doors which will be open at the end are those whose index has an odd number of divisors, and if memory serves the numbers with an odd number of divisors are the perfect squares. There are 31 perfect squares below 1000, so 31 doors.
Pulsar said:
The doors which will be open at the end are those whose index has an odd number of divisors, and if memory serves the numbers with an odd number of divisors are the perfect squares. There are 31 perfect squares below 1000, so 31 doors.