The riddle thread - Page 8

Which road would your twin tell me to take. Then take the opposite.

It is yes when n < (3^k - 3)/2.

Basically, it is yes when the possible number of outcomes for the weighing is less than the number of balls weighed. Each weighing has 3 possibilities (Left heavier, Right heavier, or equal.). With k weighings, that means that there are 3^k outcomes.

3^k

Because we are constructing our system in such a way that all Left heaver or all Right heavier or all equal are not viable outcomes, those 3 cases are subtracted from our possible outcomes.

3^k - 3

Since we do not know if the odd ball is heavier or lighter, we have half as many outcomes to work with.

(3^k - 3)/2

For 3 weighings, this gives us (3^3 - 3)/2 = (27 - 3)/2 = 24/2 = 12 outcomes, so we can determine the odd ball out of a maximum of 12 balls.

There is a sure way to get this right, but it takes a long time.
The prisoners create a process of selecting a leader. The other prisoners will be ordinary prisoners. The leader has the job of declaring all 100 prisoners have been to the room.

The first time an ordinary prisoner enters the room and finds the light bulb turned off, the prisoner turns the light bulb on. If the light bulb is on, the prisoner does nothing. If the prisoner enters the room and finds the light bulb turned off a second time, the prisoner does nothing.

Whenever the leader enters the room, the leader turns the light bulb off if it is on. If the leader does this, the leader adds 1 to the tally. When the tally reaches 99, the leader declares that all prisoners have been inside the room.

Optimally, the leader is the prisoner that enters the room on the second day. The process takes 20+ years on average.

The alternate solution is to use statistical probability such that the probability of getting all 100 prisoners into the room is so high that it might as well be certainty. Something like a 1000 days is enough time to wait. Each prisoner expects to enter the room 10 times during that long a duration. The probability of each prisoner never going to the room is near zero, but over 100 prisoners, it still adds up. Yet there is still more than 99% probability.

So spend 3 years in jail and take your chances. Easy enough right?

Your solution has one "leader" to tally up counts.
This usually takes around 10,000+ days, or ~28 years. What's interesting is that we can optimize even more. You chose the leader to be the person who enters the room on the second day. If we choose the leader to be the first person to enter the room twice. That is, if the light is still on, then only unique prisoners have entered on everyday. This means that if I enter the room on the 40th day (and it's my second time) and the light is still on, then I know for a fact that 39 unique prisoners have entered the room already. This does a somewhat significant slimming on the solution time (~9000 days instead of ~10,000).

Now with this track of mind, we have been using only one leader to count. But...

What happens if you use more than one? How much does this reduce the average time?
Lastly, what do you need to add or change to the prisoner's planning in order to make it work with the new plan?

For those who think this is too easy, can you think of a better solution if the room now has two lightbulbs instead of 1?

What if there are n prisoners and k light bulbs?
(I really love this riddle, at first glance many find it impossible, and upon hearing one solution people fail to realize how deep the rabbit hole can go).

Nice job. Though LLL/RRR is a usable outcome, so you're one off.

The velocity of the snail is the summ of 2 terms: the velocity of the snail relatively elastic and the velocity of elastic (each point of elastic is moving while it stretching).

where

with new variables

this equation will take simple form

and the solution (after returning to the original variables) will be

Now from the equation

we can find time T required to catch up the bicycle and this time will be

Ask twin A "What would twin B say which way I should pick not to die?" If the answer is left, you go right.

the solution was something retarded with ice bullets, except they're pretty much impossible to make so the solution could be whatever the hell you think it is. i say the bullets were made of fairy dust and disappear upon contact.

Ask one of the twins if the other twin would say that a(choose either) passage is the safe one to go. If its the correct passage the liar would say it isn't. Even if it was the truth teller he'd say it isn't also. If its the unsafe passage the liar would say its safe and the truth teller would say it isnt safe. This way the passage that is always "unsafe" is the correct one.

It looks like a mutli-tiered counting approach would work faster. The fastest is to tier by counting to as close to e as possible. That is 3, and so that gives 4 tiers, and all 100 prisoners have to be involved in one way or another, but the goal is to minimize their involvement.

(strange 4 tiers of 3 = 12, but 6 tiers of 2 = 12 - except with more tiers - don't know what this means - perhaps I want to use 2 instead?)

So that looks like that will be
66->22->8->3->1
which gives a total of exactly 100. nice.

But there's no way to coordinate when the first tier is done - or when any tier is done. Only the last 1 can know when it's down. Hmmm, that means that the prisoners have to guess by calendar date, and there will need to be a way to return to any or all of the tiers if necessary if they are unlucky.

Ahh yes, they have to set calendar dates ahead of time in order to coordinate all the prisoners, and they'll also have to work out calendar dates for when they redo the tiers.

On the first pass, prisoners will want to be lightweight and progress through each cycle with a reasonable chance of 100% completion. If they are too conservative, then they are just wasting time, since if they miss 100% completion, they can just go back through a redo cycle.

On the redos they need at least 100 day per tiers to make meaningful progress. A redo cycle has to be effective at matching up 1 countee and 1 counter. The first redo cycle maybe might have to be more conservative. And there needs to be some kind of tier transition mechanism. I'll have to work that out first. A programming exercise is probably in order.

Conservatively, I would have to think that it would take 4 tier * 3 counts * 3 probability estimate * 100 men = 3600 days or 10 years. Ha!

Cut the rope into 2 pieces, one 500m and one 250m.
Make a noose at the end of the 250m piece
Slip the 500m piece through. You now have a rope that is 500m.
Slide down to the rock and stop. Pull the rope through the noose
Slide down the the bottom