Two more puzzles
- Thread starter Simon F
- Start date
Pirate problem:
If x is the amount of money a pirate is willing to pay for the joy of killing a greedy brethren, then pirate 5 should propose at least 2*x to pirate 1, and at least 1*x to pirate 3.
[Edit]
Hmm. If the pirates are blodthirsty enough to kill a friend without any monetary gain (as long as it is "by the rules"), the offer of 2*x could actually just as well go to pirate 2.
And MfA's comment below means that x=1.
If x is the amount of money a pirate is willing to pay for the joy of killing a greedy brethren, then pirate 5 should propose at least 2*x to pirate 1, and at least 1*x to pirate 3.
[Edit]
Hmm. If the pirates are blodthirsty enough to kill a friend without any monetary gain (as long as it is "by the rules"), the offer of 2*x could actually just as well go to pirate 2.
And MfA's comment below means that x=1.
Pirates: I liked it - great puzzle....
1 pirate...
P1 : all of it
2 pirates: P2 suggests P1 takes all the money. A 50:50 vote is a loss and P1 won't agree to anything unless he gets all the money. Of course P1 may still disagree because he is bloodthirsty and wants P2 to die anyway.
P1 all
P2 zero
3 Pirates: P3 knows that if he is eliminated, P2 won't get anything, so he suggests he and P2 "share" the cash (999 to P3 and 1 to P2). P2 agrees because at least he has some cash and is not dead.
P1 zero
P2 1
P3 the rest
4 Pirates: P4 knows that if he is eliminated that P1 gets nothing, P2 gets 1 and P3 gets 999. He needs 3 voters to win and so improves the outcomes...
P1 gets 1
P2 gets 2
P3 NOTHING
P4 the rest
The majority are better off so agree.
5 Pirates: Same process. P5 still needs 3 voters so suggests
P1 gets 2,
P2 gets 3,
P3 gets 1
P4 nothingP5 the rest
No one's answered the ring puzzle yet...
1 pirate...
P1 : all of it
2 pirates: P2 suggests P1 takes all the money. A 50:50 vote is a loss and P1 won't agree to anything unless he gets all the money. Of course P1 may still disagree because he is bloodthirsty and wants P2 to die anyway.
P1 all
P2 zero
3 Pirates: P3 knows that if he is eliminated, P2 won't get anything, so he suggests he and P2 "share" the cash (999 to P3 and 1 to P2). P2 agrees because at least he has some cash and is not dead.
P1 zero
P2 1
P3 the rest
4 Pirates: P4 knows that if he is eliminated that P1 gets nothing, P2 gets 1 and P3 gets 999. He needs 3 voters to win and so improves the outcomes...
P1 gets 1
P2 gets 2
P3 NOTHING
P4 the rest
The majority are better off so agree.
5 Pirates: Same process. P5 still needs 3 voters so suggests
P1 gets 2,
P2 gets 3,
P3 gets 1
P4 nothingP5 the rest
No one's answered the ring puzzle yet...
I thought about it some more, now I can see why your answer is correct. If they rejected 5's proposal, 4 would offer 1 and 2 one coin each. And they both would have to accept because 3's proposal would mean no coins at all for themXmas said:
btw, If the pirates were infinitely smart, they'd never agree on such a stupid idea
Simon F said:
Yes of course.
Sometimes I am soo confused, that I forget what it is like to live in two dimensions .Anyway the string is infinite...
I am not so sure if a 2d-person living on a 2d world, actually could see the difference.... I think since he has no knowledge about the 3rd dimension, the surface would seem flat to him. But I am not sure on that...
The solution I know is more simple...
A somewhat similar approach, he takes 100 steps in one direction, turns 90°, repeats that another two times, and then measures the distance to the origin of his walk. Then he repeats the procedure with some slight rotation angle to account for the case that this square might be axis-aligned with the torus.hupfinsgack said:
Another possibility: He gives the end of the string to his wife and tells her to walk straight ahead in one direction, and he walks off in another direction. He now measures how much wool is released from the disc. If he lives on a plane, the distance between them should grow linearly, on a torus it does not.
The 2-D physicist doesn't have any way to measure the angles - he's only got some wool and a nagging wife.
What he can do is start walking in a straight line trailing the wool behind him, if his universe is toroidal(sp?) then he will cross his path of wool eventually, if it's flat he'll never cross it. Of course, all this assumes he's capable of walking in a perfectly straight line.
Simon, how about this for an answer to your ring problem. The reaction force from the ground will be equal to the combined masses of the ring and weights * gravity at the top. When the weights are at the widest part of the ring, the reaction force will only equal the mass of the ring * gravity. At the bottom the reaction force will again be the total of ring and weights. This should vary sinusoidally.
Super imposed onto that is the effect of the centripetal acceleration on the weights, which will reduce the reaction force while the weights are above the centre of the disk and increase it when they are below the centre (I can't be bothered figuring out how much)
As was mentioned earlier, there will be a corriolis(sp?) effect as well. At either of the poles, the ring will start rotating (relative to us) when the weights are above the centre, then slow when the weights are below the centre. A E-W orientated ring at the equator will roll to one side, while a N-S one will just fall over. Everywhere else will be a complicated mixture of the above effects.
Edit: Fixed some typos and stuff.
Why wouldn't a 2d physicist have a a means to measure angles?
Nathan said:The 2-D physicist doesn't have any way to measure the angles - he's only got some wool and a nagging wife.
What he can do is start walking in a straight line trailing the wool behind him, if his universe is toroidal(sp?) then he will cross his path of wool eventually, if it's flat he'll never cross it. Of course, all this assumes he's capable of walking in a perfectly straight line.
That's correct. And if he's lucky enough he'll never see his wife again and lives happily ever after
I haven't been answering these, b/c I have seen a lot of them before, and know the answers.
Btw, for the 2d physicist, the torus world and the plane world are NOT the only unique choices of simple 'topologies'. There are even more if you consider worlds that are not simply connected
I'll leave it to you guys to think of ways of distinguishing between such latter 'topologies', the answers given don't suffice. There are even some choices that might win you a fields medal, so there is monetary incentive =)
Btw, for the 2d physicist, the torus world and the plane world are NOT the only unique choices of simple 'topologies'. There are even more if you consider worlds that are not simply connected
I'll leave it to you guys to think of ways of distinguishing between such latter 'topologies', the answers given don't suffice. There are even some choices that might win you a fields medal, so there is monetary incentive =)
Basic:
In order to find out if he made a circle, he could start from his line and start walking in a perpendicular direction. If he is able to return to his original position by making only one crossing with the original string (near his starting point) then he must be on the surface of a torus, since two overlapping circles in a plane must have two intersections.
In order to find out if he made a circle, he could start from his line and start walking in a perpendicular direction. If he is able to return to his original position by making only one crossing with the original string (near his starting point) then he must be on the surface of a torus, since two overlapping circles in a plane must have two intersections.
Akira posted a good reply to question 2 on the first page - unfortunately its absolutely not guaranteed to work given the pre-conditions of the burn rate is irregular. It seems logical that is should - but its not guaranteed without other cited boundary conditions that we are not given.
For instance assume that all but 30 cm in the middle will burn practically instaneously. So regardless of which end you light it still takes and hour.
But does lighting it at both ends guarantee a 30 minute burn? Well no - its logical to hope it might but its not guaranteed. A proof by induction will show you the error of this thinking.
If the wick is chain of molecules one molecule long it probably doesn't matter which end you light - it will burn at the same rate. If the wick was two atoms wide and the first took 59 minuts to burn and the second took one minute to burn lighting it at both ends doesn't guarantee a 30 minute burn. It might happend - but its not guaranteed! Now assume the wick being N atoms takes a time different from 30 minutes to burn even when lit at both ends - add one atom to make the wick N+1 atoms long - does this guarantee the wick must burn at precisely 30 minutes? Well no - not as far as I can see.
Or
Imagine if the molecule chain burns instanteously except for one atom which takes exactly 60 minutes to combust. Lighting this molecule from both sides or even from 1,000 points on its molecular chain is only guarantee to have an irregular reaction - not to perfectly half the burn rate.
You need to add another pre-condition to your puzzle to guarantee it is solveable with the logical answer. Can you see what this pre-condition must be?
For instance assume that all but 30 cm in the middle will burn practically instaneously. So regardless of which end you light it still takes and hour.
But does lighting it at both ends guarantee a 30 minute burn? Well no - its logical to hope it might but its not guaranteed. A proof by induction will show you the error of this thinking.
If the wick is chain of molecules one molecule long it probably doesn't matter which end you light - it will burn at the same rate. If the wick was two atoms wide and the first took 59 minuts to burn and the second took one minute to burn lighting it at both ends doesn't guarantee a 30 minute burn. It might happend - but its not guaranteed! Now assume the wick being N atoms takes a time different from 30 minutes to burn even when lit at both ends - add one atom to make the wick N+1 atoms long - does this guarantee the wick must burn at precisely 30 minutes? Well no - not as far as I can see.
Or
Imagine if the molecule chain burns instanteously except for one atom which takes exactly 60 minutes to combust. Lighting this molecule from both sides or even from 1,000 points on its molecular chain is only guarantee to have an irregular reaction - not to perfectly half the burn rate.
You need to add another pre-condition to your puzzle to guarantee it is solveable with the logical answer. Can you see what this pre-condition must be?
'It seems we have a new challenge. Can anyone post a problem that Fred can't answer? '
Democoders problem from a few weeks ago with the Faro plane had me completely stumped. I was trying to do things with elliptic geometry, but it wasn't quite working. Easily one of the hardest, 'doable' mindtwisters I've ever seen.
A lot of these though tend to get repetitive, and if you've seen a few solutions you kinda 'know' the right way to think about it.
I assume most everyone has seen the following 'classic' problem (but if you haven't its quite fun the first time).
A gameshow host asks a contestant to pick exactly one of three doors. Behind one of the three, is a prize, the other two possess nothing.
Now, when the contestant picks a door, the gameshow host opens one of the remaining two doors and reveals that there is nothing inside.
Now, he gives the contestant the following option: Do you wish to stay with the door you had previously picked or do you wish to change your gamble to the remaining door.
Which is the better choice, and why (or does it matter) ?
Democoders problem from a few weeks ago with the Faro plane had me completely stumped. I was trying to do things with elliptic geometry, but it wasn't quite working. Easily one of the hardest, 'doable' mindtwisters I've ever seen.
A lot of these though tend to get repetitive, and if you've seen a few solutions you kinda 'know' the right way to think about it.
I assume most everyone has seen the following 'classic' problem (but if you haven't its quite fun the first time).
A gameshow host asks a contestant to pick exactly one of three doors. Behind one of the three, is a prize, the other two possess nothing.
Now, when the contestant picks a door, the gameshow host opens one of the remaining two doors and reveals that there is nothing inside.
Now, he gives the contestant the following option: Do you wish to stay with the door you had previously picked or do you wish to change your gamble to the remaining door.
Which is the better choice, and why (or does it matter) ?
