THE ULTIMATE RIDDLE PART TWO.

Pamela Anderson's Left Nipple wrote:

Someone feeling slighted? It's too bad that the best you could come up with is still only a guess and not 100% certainty.

I know that you soooooooo badly want to solve this problem, but there is no solution. Chew on this "gold standard":

DUE TO RANDOM SELECTION, IT IS POSSIBLE FOR THE WARDEN TO PICK THE SAME PRISONER, EVERY SINGLE DAY, UNTIL ONE OR ALL DIE. IT IS POSSIBLE THAT NO OTHER PRISONERS WILL EVER SEE THE INSIDE OF THAT ROOM.

As long as that is possible, then there is NEVER an instance where a prisoner can assert with 100% certainty that every prisoner has been inside the room.

Please, cease being so dense.

Nip,

Your logic is flawed. Just because the possibility exists that all 100 prisoners would have a chosen turn, that does not mean it cannot happen. There is a possible solution, tough not one that is guarateed.

MM

Pamela Anderson's Left Nipple wrote:

Nope. If a prisoner enters the room for his first visit and the light is already off, how does he turn it off again?

by first turning it on. it's not rocket science.

Mensa called, they want their card back.

Form the orignal post: "Thus, the assertion should only be made if the prisoner is 100% certain of its validity."

From you: There is a possible solution, though not one that is guarateed (I gave you the H).

To quote from one of my favorite movies, "Fast Times at Ridgemont Times", "What part of 100% don't you understand... Brad?".

If I remember from my graph theory class, you can't.

KnowItAll wrote:

A riddle I got in grade school that I never found an answer for.

There are 3 buildings that provide utility services. Gas, Electrical, Water. There are 3 houses. You can not cross gas, electrical or water lines. How do you get all 3 services to all 3 houses.

This is more of a graphical riddle, so may not be easy to propose solutions on here, but I've never been able to solve the b!tch.

Pamela Anderson's Left Nipple wrote:

Mensa called, they want their card back.

Form the orignal post: "Thus, the assertion should only be made if the prisoner is 100% certain of its validity."

This does not change my previous point. Read the sentence carefully, then think before posting. If you are as bright as you think you are, you will understand my point.

Pamela Anderson's Left Nipple wrote:

Nope. If a prisoner enters the room for his first visit and the light is already off, how does he turn it off again?

He doesn't. He waits until the first time he visits the room and the light is on. Then he turns it off. And he never turns it off again in future visits. Like NK said, each prisoner only turns it off once. When the counter sees that the light has been turned off 99 times, it means 99 DIFFERENT people have turned it off, and he knows everyone has been to the room. I don't see why this won't work.

Yes, it will take a very long time before the counter visits the room 99(or more) times, but like you said, it's a mathematical problem, not a practical problem.

you know there is going to be one jackass prisoner with a death wish and the first chance he gets he is going to call the warden and take everyone out with him

it's a moot topic they are all going to die, face it

The first few times I read this option I didn't really understand it, but after reading this it does make sense. That would take a long time, and I would probably rather take my chances on probability...

dukerdog wrote:

He doesn't. He waits until the first time he visits the room and the light is on. Then he turns it off. And he never turns it off again in future visits. Like NK said, each prisoner only turns it off once. When the counter sees that the light has been turned off 99 times, it means 99 DIFFERENT people have turned it off, and he knows everyone has been to the room. I don't see why this won't work.

Yes, it will take a very long time before the counter visits the room 99(or more) times, but like you said, it's a mathematical problem, not a practical problem.

Pamela Anderson's Left Nipple wrote:

Someone feeling slighted? It's too bad that the best you could come up with is still only a guess and not 100% certainty.

I know that you soooooooo badly want to solve this problem, but there is no solution. Chew on this "gold standard":

DUE TO RANDOM SELECTION, IT IS POSSIBLE FOR THE WARDEN TO PICK THE SAME PRISONER, EVERY SINGLE DAY, UNTIL ONE OR ALL DIE. IT IS POSSIBLE THAT NO OTHER PRISONERS WILL EVER SEE THE INSIDE OF THAT ROOM.

As long as that is possible, then there is NEVER an instance where a prisoner can assert with 100% certainty that every prisoner has been inside the room.

Please, cease being so dense.

Err, yes there is. The broken lightbulb model is the best, by far. Not only does is provide certainty (when all pieces have moved, then all have been in - regardless of repetetive or random selections), but it also accomplishes the goal in the shortest amount of time. No on-off switches, 100 visits, nothing. The broken bulb is genius.

Mensa Mann wrote:

This does not change my previous point. Read the sentence carefully, then think before posting. If you are as bright as you think you are, you will understand my point.

Should I go to dictionary.com and look up the definition of certainty, too?

I do understand your point. However, your point does not meet the requirements dictated by the problem, which asks for a prisoner to make an assertion of 100% validity. Try as you like 99 followed by an infinite number of nines will never equal 100.

the real NK and deeprunXC have the best answer (even though deeprunXC phrased it a little badly). I'll try to make it more clear.

1) We choose one prisoner, call him Bob. Bob is the only person who will turn the light off and he will do it whenever he enters the room (assuming the light is on).

2) Everyone else will turn the light on the first time that they enter the room and find the light off. Otherwise they do nothing.

Essentially what happens is one prisoner turns the light on and eventually Bob turns it off and records that one prisoner has been in the room. Then another prisoner turns it on and Bob eventually records this. It continues untill Bob has recorded 99 prisoners (other than himself) entering the room. He will then know with 100% certainty that everyone has been in the room. On average this will take about 30 or so years.

There is a much more efficient method explained at the bottom of the website in the 4th post, but it is too complicated to explain here. It basically is the same thing, except everyone is involved in counting and they can switch the light to not only mark that they have been in the room, but they have counted another 2^k people have been in the room. It involves setting up a formula beforehand and in it's most efficient form may reduce the time to about 12 years on average.

These are the best ways to solve the problem theoretically (as it was meant to be done).

In this were to somehow actually occur in the real world, clearly it would be best to just wait a few years because the odds would be so small that someone has not been in the room, but any statistics based answer does not really solve the problem because the question asked how he can make the assertion with 100% certainty.

Pamela Anderson's Left Nipple wrote:

Should I go to dictionary.com and look up the definition of certainty, too?

No, but you may want to look up "if" for your own edification.

clearly the broken bulb is bullshit because the problem said all they are allowed to do with the bulb is turn it on and off.

It's not genius. Once the bulb is broken it will be too dark to avoid stepping on the glass. That alone would be unfortunate, but then there would be the problem many more shards of glass strewn about. At least a few of the prisoners would slit their own wrists, don't ya think?

Hell, after reading this, I feel like offing myself.

Just Wondering wrote:

Err, yes there is. The broken lightbulb model is the best, by far. Not only does is provide certainty (when all pieces have moved, then all have been in - regardless of repetetive or random selections), but it also accomplishes the goal in the shortest amount of time. No on-off switches, 100 visits, nothing. The broken bulb is genius.

Another know-it-all... when it comes to knowing crap. Read all of the posts, and then chime in.

What would you break the light bulb with? A stick? What stick? Who said anything about a stick?

Is it within reach? If not would you get on a chair? What chair? Who said anything about a chair?

What if all 100 prisoners never get to go into the room? Waht if only 99 get in? If it's equally random selection, then there is no guarantee that more than one prisoner will ever see the inside of the room. Think random... the solution requires a minimum of 100 days, but the maximum is a number that can not be quantified.

Try this very simple experiment. Put 100 pennies into a jar, pull one out, put a mark on it, and then put it back into the jar. Pull another one out (since you put the first one back in, you could pull it out again), put a mark on it, and then put it back into the jar. Repeat until you've marked every penny.

Does equally random make sense to you yet?

my guess is that the reason the answer is not posted on the internet is because it is such a simple problem. First off, any decent computer programmer would have realized the answer immediately. It's a fun problem, but not exactly MENSA worthy.

To those who believe the problem is unsolvable (hehe):

Using one prisoner as the counter works - it is a sure way that the counter can declare that all of the prisoners have been in the room (if all the prisoners eventually enter the room) - which as part of the problem, it is assumed that they eventually will. Remeber the question? "What is the optimal plan that they can agree on, so that eventually, someone will make a correct assertion?"

Say things like "oh, but what if someone dies of malaria first..." or "oh, what if a bomb blows up the building..." or "oh, what if they die of starvation first..." is just avoiding the whole point of the riddle. You could mimick this exact same logic puzzle using a computer program with sokoban cubes and then you wouldn't have to worry about the imposed mortality of inmates. The problem just uses people as props in order to make it more interesting, so get over the "what ifs" and concentrate on the logic. Obviously... the assumption that they all will eventually enter the room has to be made, and that the problem will go on until this can be determined with absolute assurance by the person doing the counting... otherwise the problem would be pointless.

Pamela Anderson's Left Nipple wrote:

What would you break the light bulb with? A stick? What stick? Who said anything about a stick?

Yikes! You don't seriously think you couldn't break the bulb without some implement, do you?

You've got some serious anger problems, don't you?

Mensa Mann wrote:

Nip,

Your logic is flawed. Just because the possibility exists that all 100 prisoners would have a chosen turn, that does not mean it cannot happen. There is a possible solution, tough not one that is guarateed.

MM

Just enjoying this ditty of a sentence:

Just because the possibility exists that all 100 prisoners would have a chosen turn, that does not mean it cannot happen.

Analyze your sentence, and I'll lgo ook up "if".

Since we're looking for certainty, and your solution is possible, but not guaranteed, I think our discussion is over.

(Answer to your sentence problem: substitute "cannot" with "will" and the sentence will make sense.)

Look I might have ended up being manhandled by the likes of Scott Baio, Tommy Lee, Brett Michaels and Kid Rock, but I'm a lot brighter than you. Just accept it.