Two more puzzles

[Fred]

The assumption thats needed is that the cord needs to be sufficiently smooth in terms of its differentiability (or that it can be infinitely spliced which is a stronger condition). If you start quantizing it in molecules, you run into nasty little problems that can get arbitrarily complicated, however.. For all but the pathelogical cases, the corrections will be tiny and the result will be very close to the desired time interval.

 

[Vince]

Cool thread, just started reading it.

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) ?

I feel like I'm probably missing something, but I'd guess that the presence of an observer 'modifies' the probability from 1/3rd to 1/2 - so he might as well stay.. or something indeterminate like flip a coin or ask a blond?!?

 

[Basic]

I assume most everyone has seen the following 'classic' problem (but if you haven't its quite fun the first time).

Closer than you think.
http://www.beyond3d.com/forum/viewtopic.php?t=8577&start=30


akira 888:
Yes, that's it.
And related to that, have anyone here played Populous3: The Beginning? It's a cool game. But the world is quite disorienting. Each planet (level) is a torus. But instead of using fog to cut off distant parts, they cut out a piece of this torus and map it onto a sphere. So instead of disappearing in fog, stuff falls off behind the horizon. This mapping is good enough that you don't notice any error locally. But as you walk around in the world, you could get seriously lost.

And just imagine how irritating it is when you managed to build a good defence line as a closed loop, you see that your enemy is on the other side of the defence, but he can still get to you without any fight.

 

[akira888]

New puzzle:

12 guys are taken to a jail. Upon arrival, they hear this from the warden. "There are two cards on a table in a special room, each red on one side and blue on the other. Both cards are now showing red. Here is your escape method: Everyday I will pick one of you to enter the room. There, you will have to turn over exactly one of the cards, not two or none but exactly one. *The only information you can extract from the cards is their color; I will straighten them if they are tilted.* During this time you will not speak to any other inmate nor will you know how many inmates went ahead of you. I will release all inmates when one inmate informs me that he knows all inmates have entered the room at least once. If an inmate says so falsely you all will be shot. However, I will make one concession: I will give you one day to talk over a strategy." All of the prisoners were later released (EDIT:alive and unshot). How?

*EDIT

 

[StefanS]

New puzzle:

12 guys are taken to a jail. Upon arrival, they hear this from the warden. "There are two cards on a table in a special room, each red on one side and blue on the other. Both cards are now showing red. Here is your escape method: Everyday I will pick one of you to enter the room. There, you will have to turn over exactly one of the cards, not two or none but exactly one. During this time you will not speak to any other inmate nor will you know how many inmates went ahead of you. I will release all inmates when one inmate informs me that he knows all inmates have entered the room at least once. If an inmate says so falsely you all will be shot. However, I will make one concession: I will give you one day to talk over a strategy." All of the prisoners were later released (EDIT:alive and unshot). How?

I have a solution, though it might not be the exspected one
a card can be placed in 4 different easily distinguishable positions (-, /, I, \) times 2 values (red, blue) *time 2 cards that gives 16 possibilities.
So first prisoner takes card 1, turns it that way (I) and turns it over. Second prisoner takes card 2, turns it I and turns it over. 3rd prisoner takes card I, turns it / and turns it over, ...
They rotate the cards clockwise...
This way they could free 4 more inmates...

 

[akira888]

Drat hupfinsack, you got me. That's one way to work it. But I'm going to edit the above post to make the puzzle harder.

 

[Simon F]

Surely you mean a disc of wool? I'm almost loathe to ask this but, how long is the piece of string?

..the string is infinite

Well why didn't you say so in the first place! Next you'll be telling me he has a wheelbarrow and a holocaust cloak.

Simon, how about this for an answer to your ring problem. <SNIP>

It's much simpler than that... maybe I haven't described it clearly enough!

I'll drop some hints at lunchtime today.

 

[Simon F]

New puzzle:

12 guys are taken to a jail. Upon arrival, they hear this from the warden...

There may be a stupidly simple answer to this but I just want to check one aspect: Does he pick the guys completely at random, i.e. everyday there is a 1 in 12 choice therefore being a chance that guy A is chosen 2 days in a row?

 

[akira888]

New puzzle:

12 guys are taken to a jail. Upon arrival, they hear this from the warden...

There may be a stupidly simple answer to this but I just want to check one aspect: Does he pick the guys completely at random, i.e. everyday there is a 1 in 12 choice therefore being a chance that guy A is chosen 2 days in a row?

You are exactly correct.

 

[Simon F]

OK: Here is my solution to the 12 prisoners puzzle:

This assumes that the guard picks a person completely at random each day.

They nominate one person (at random) as "the counter" and they agree the following for the cards:
Red, Red = 0
Red, Blue= 1
Blue,Blue= 2
Blue,Red = 0

When a person is picked, he obviously knows if he has already visited. Each also has to remember another value, X.

The trick is...

When a person enters the room, he calculates
X := X + Cards.

If this is his very first visit, he also does

X:= X+1

If he is NOT "the counter", he sets the cards to the biggest reachable value that is less than or equal to X, and substracts that value off his X.


If he is "the counter", he first checks to see if he's reached 12 or not. If not, he sets the cards to a zero and remembers his "X".

Eventually, assuming a random distribution, all the values will make their way to "the counter".

 

[Simon F]

As promised, a hint for the ring + weights puzzle: "For every action there is an equal and opposite reaction" and "orbit"

 

[Nathan]

OK: Here is my solution to the 12 prisoners puzzle: <snip>

Nice work.

Orbit? How big is this ring? :?

 

[Basic]

Ring problem:
Is it that if you try to balance the weights next to each other, and then let go, it's most likely that both weights fall on the same side, straight down?
The first weight to come off-center will push away the ring so that the other weight will become off-center to the same side. And since the weights are much heavier than the ring, they will just push the ring to the side as they do an (almost) free fall.

 

[Simon F]

OK: Here is my solution to the 12 prisoners puzzle: <snip>

Nice work.

Actually, there might be a slight flaw in that it could take a very long time - there's a simple improvement: After a certain number of days (say 4*12), you change the encoding of the "2" to also mean "1". That way you can always pass at least a one to the other inmates.

Orbit? How big is this ring? :?

It doesn't really matter but it is resting on the ground - you don't have to take relativistic effects into account!

 

[akira888]

Here is my answer to the prisoners problem (it really quite simple and [perhaps] optimal)-

Highlight the quotebox through tripleclick or hold-and-drag. (I'm sick of copying microtext into kate or notepad)

Appoint one prisoner the "leader." If a prisoner enters the room for the first time and sees that the first card is red, then turn it over to blue. If the card is blue, have the prisoner flip the second card.

However, if the leader enters the room and sees the first card is blue, he knows a new prisoner has entered and he flips the first card back to red. If the first card is red when the leader visits, the leader will know no new prisoners have entered the room since his last visit, so he just flips the second card over. Once the leader has flipped eleven blue first cards over back to red, he knows he everyone has visited the room at least once.

 

[Simon F]

Solution to the ring problem:
----- Use Akira's select trick to read ---

Before the weights fall half the height of the ring, the ring lifts off the ground...

When the two weights are at the top of the ring gravity is pulling them down but the ring contrains them to move in a circular direction - one down each side of the ring. At the top, the ring is thus exerting forces on the initially stationary weights that are almost (but not quite) equal and opposite to gravity. By Newton's laws of motion, therefore, the weights are exerting a force equal and opposite force downwards upon the ring.

Now imagine that the weights have slid to the about "2 oclock" and "10 oclock" positions (i.e. 30 degrees). They will now have accelerated due to the influence of gravity.

(In fact we know that if they've dropped a distance of "h" then their energy is mgh which must also be the same as their kinetic energy, 1/2 mv^2. IE we have v^2 = 2gh and so v^2 is proportional to the height dropped.

The weights, however, aren't just falling, they are constrained to travel in a circle (i.e. orbit) and so a centripetal force, directed to toward the centre of the ring, must be acting upon them to keep them going in a circle.

Note that this force is proportional to the v^2 so as the weights fall further, the centripetal force gets stronger.

Now some component of this force can be provided by gravity but (a) that force is constant and (b) the angle changes so it soon has less and less effect (and reaches zero if we get to the half-way mark).

We thus get the situation (before the weights reach half-way) where the forces exerted by the ring are dominant - the left and right components cancel out and all you are left with is a overall downward force being applied to the weights by the ring . This means that the weights are applying an equal and opposite upward force to the much lighter ring and, therefore, it lifts off the ground.

Once the weights go past the halfway mark, though, the ring will be pulled down very hard!

 

[Jabbah]

Thats mad!

Immediate thoughts are, NO dont be silly, that couldnt happen...

guess i better do this so as not to give anything away to antone who is still working on the problem :?

Anyway, I can see and understand the logic to it now, and my brain hurts!

Another way of looking at it - If the weights are not connected to the ring but just placed on the outside of it then when they are released, they will at some point leave the edge of the ring before the centre height due to the horizontal component of the velocity gained from being forced to travel around the outside of the ring, i.e. they wont follow the edge of the ring to the centre height and then drop straight down. So it stands to reason that if the weights are connected to the ring then at the point they would have left the edge of the ring, it will start to lift to try and allow the weights to travel the path they would have if they were not connected.

Nice one Simon, evil, but nice

 

[Simon F]

OK, here's a simpler one (that you can actually easily test for yourself).

You have a small helium balloon in your car. (For safety, tied with long piece of string, say, to the front passenger's seat. Assume that it's at about head height).

You're initially driving along at high speed and then slam on the brakes. What happens to the balloon and why?

 

[Fred]

Btw, the ring problem is driving me absolutely nuts, i'm trying to derive the exact equations of motion atm, and it 'aint' simple to put it bluntly.

There's something a little fishy too with the explanation above. The second you say 'the mass of the ring is much lighter than the heavy weights' you open a huge can of worms since you are explicitly altering the orbital equations with a varying contact force that might a priori depend on the geometry of the ring, however since its so symmetric things should cancel out. But I can't see how they cancel out, and its bugging me to death.

The proper way to treat this is with a Lagrangian btw, if anyone is trying to do the force method.

 

[Jabbah]

Much easier

The balloon will go towards the back of the car, opposite to the direction of travel, as the air inside the car will try to move forwards and the balloon is boyant so will be forced backwards.