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   100 prisoners & a light bulb
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guido
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Re: 100 prisoners & a light bulb  
« Reply #525 on: Jan 28th, 2007, 11:35am »

Hello all,
  i arrived on this forum a couple of years ago, looking for a few riddles to spend some time. At that time, i saw this one, but didn't really pay attention.. It remained in a corner of my mind, and yesterday i thought to myself "what was it exactly?" after a little search, i could locate again the page, found the riddle and, since i didn't want to think about it  Grin just came to look for the solution..  Lips Sealed
 
Well, none of those that i read in the 21 pages seemed satisfying to me Cool because:
 
1) the riddle make no mention of counters, badges, and so on. Assuming that a general rule for solving a well done riddle is that all that is needed is given in the question, the answer shouldn't  add anything (this invalidate also the idea of keyholes or anything else that would allow to see the light)
 
2) when you think about leaders and drones, you assume that there are "actives" and "passives" prisoners. Now, i think (but it is just my idea) that where the riddle says  
 
", all prisoners are set free and inducted into MENSA"
 
We are getting a clue that they all need to be active and use their little brains.. Cheesy
 
So i arrived to this solution, wich didn't seem to have been posted and seems to work. I am not giving an average time because, as you will see, it could be 100 days as well as 100 years. But really, once a prisoner think that all have been in, they have (or else i made a big mess  Grin)
 
so: we shall split the timeline in 100 days segments, that we'll name "cycles" (c). each cycle takes a value c = (c+1) at the end of the cycle.
 
In this solution, we have to choose a 100 days cycle because we have 100 prisoners.  
 
beginning:
 
light is off.
 
Each prisoner must remember the number of times he entered the room.
 
Each time a prisoner has been (c) times or less in the common room during the cycle, he leaves the swich as it is. If the light is off, he counts the number of days past  (let us say "DP")since the beginning of the cycle and divided them for the number of the cycle: DP/c rounding down if needed. This way, he will know the minimal number of people that have been in the room at least once before him (minpax). If he has been more than (c) times in the room, and the light is off, he switches it on and count as said before. If the light is on, he does nothing but remember the highest (minpax) he ever came around (if any).
 
At the 100th day (end of (c)):  
 
-if a prisoner has been in the room for 100 days  
(that is, if he is the only one, for the whole cycle) he switches the light off.  
 
-If a prisoner finds the light off, he declares that all have been in.  
 
-If a prisoner finds the light on, and he hasn't been in the room 100 days continuously, he switches it off AND MAKES NO DECLARATION (important  Wink not a joke).
 
At the beginning of 2nd cycle - 101 day: if the person that enters find the light on: means he is the second ever to enter the room.
 
 If it is off, and he never entered it before, he knows that at least 3 people have been in.  Else, he counts that he is the second entering.
 
[If he has been in the room 100 days during the fist cycle, enters, finds the light off and think he is the second entering, he is a complete idiot and deserves to be executed. Wink]
 
From now on, each prisoner must take into account the POSSIBILITY to realize a passage of all of them within the end of the cycle.  
 
That is: if a prisoner has spent 2 days in the room, and knows that 1)there have been at least 2 oher people in the room, and 2) there are yet 99 days to the end of the cycle, then he will consider that the condition is possible and turn the light off (or leave it off if it was already).  
 
Else, he will light the bulb.  
 
In both cases, he will remember the highest (minpax) he came across and discard the other (so, if a prisoner came in the 35th day of the 1st cycle and found the light off, and then comes in the 4th day of the 2nd cycle and finds the light off, he will keep 35 as minpax, and discard 4).
 
In the case of a prisoner that has never been in the room before, he will take his minpax if the bulb is off, and will do nothing if it is on.
 
Things are so far quite easy, but they start to get complicated when several prisoners spent more than (c) in the room: in this case, to define things a bit better, we can say:
 
assume that  
 
(on) is the lightbulb turned on
 
(off) is the lightbulb turned off
 
(c) is the cycle number
 
(minpax) is the minimal number of people that have surely entered the room
 
(DP) is the number of days past from the beginning of the cycle
 
(PG) the number of times a prisoner came in the room
 
and both (prisoners) and (cycle) have a value of 100  
 
then
 
When  a prisoner finds (off) he will count :
 
if [(prisoners) - (minpax)] < [(cycle) - (DP)]  then (off)  
 
("<" is lower, isn' it ? Shocked I'm bad at mats..)
 
If [(prisoners) - (minpax)] > [(cycle) - (DP)] then
If  (PG) < or = (c) then (off) Else (on)
 
 
In all cases, he will check for (minpax) to ensure he has the greatest value.
 
If a prisoner found (off) but didn't have (minpax) because at the previous passages he always foud the light on, he will directly use  
 
If  (PG) < or = (c) then (off) Else (on)
 
 
When a prisoner finds (on), he will do nothing.
 
This way, when a prisoner (no matter who) arrives at the end of a cycle (the 100th day) with the light off, he can surely declare that all have passed in the room at least once.
 
As said, i am no good at all in maths, so i reasoned it out with those strange symbols..  
 
Sorry.  
 
I would like to know what people think of this solution..
 
(and about the spelling, i am no english, but surely everybody already noticed.. Grin Grin )
 
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Re: 100 prisoners & a light bulb  
« Reply #526 on: Jan 28th, 2007, 12:00pm »

on Jan 28th, 2007, 11:35am, guido wrote:
1) the riddle make no mention of counters, badges, and so on.
Those are just conceptual objects though, only existing in the minds of the prisoners.
 
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Re: 100 prisoners & a light bulb  
« Reply #527 on: Jan 28th, 2007, 2:04pm »

Yes - the tokens and badges are not real items added to the puzzle. They are just concepts used to make the schemes easier to follow. These schemes could be (and originally were) explained in terms of rules for when you could turn on the light, and maintaining a count of how many prisoners have visited based on which stage the counters found the light on and turned it off.
 
For instance, in Paul Hammond's original staged scheme (as amended by Alex H.), days were divided into 2 stages: Stage 1 would last a set amount of time, then Stage 2 would last for it's set amount, then stage 1 comes again, and then stage 2 again, and so on.
 
10 prisoners are designated secondary counters, and 1 of those 10 is also chosen to be the master counter. During stage 1, all non-counters can turn on the light exactly 1 time (the first time they come in and find it off). Each counter has a count which he starts at 1 (for himself). If he enters the room and finds the light on, and his count has not reached 10, then he turns off the light and adds 1 to his count. Otherwise, he leaves the light alone. He himself never turns the light on in stage 1.
 
In stage 2, non-counters leave the light alone. secondary counters turn it on if their count has reach ed 10. Only the master counter turns it off. In addition to his stage 1 count which he maintains as a secondary counter, he also keeps the master count of how many times he turns off the light in stage 2. When his stage 1 count has reached 10, and his stage 2 count has reached 9, he knows that everyone has visited and declares it to the warden.
 
 
 
As you can see, there is nothing involved here except the light, and information that people keep in their own heads. Expressing the idea in terms of badges and tokens doesn't mean we've added anything to the puzzle that wasn't already there.
 
 
I've not fully worked out how your scheme works yet, but based on what I have seen, and experience with previous schemes, I doubt it will prove any faster that what is already out there. I can tell you right now that the average time for your scheme is much greater than 100 days. It takes a few years before it is even likely that everyone has visited once, and no successful scheme can give an answer before everyone has.  
 
The current best scheme is about 9 years average time until release.
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Re: 100 prisoners & a light bulb  
« Reply #528 on: Jan 28th, 2007, 3:34pm »

hi,
   really sorry, in fact i DID make a mess.. Embarassed
 
It doesn't even works on 2 cycles under certains conditions.. Cry
 
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Re: 100 prisoners & a light bulb  
« Reply #529 on: Jan 30th, 2007, 8:28am »

Hi, I have told ... after a week I will post more information about "improved" algorithm. I have spend a week trying to optimize counter methods in general and for general number of prisoners.
I have written 7 pages of "article skelleton", but more questions appear than answersSad.  
 
(Small example ... when recovery phases on k levels are equaly long and all nonbinary, the optimal level order is not 0,1,2,...,k-1,0,1,2,...,k-1,... its beneficial sometimes to choose i and skip levels 0,1,2,...i.
This does not mean the levels would be skipped, it rather mean, the phase lengths will differ)
 
This unsuccessful attempt to solve problem more in general consumed most of the time so I didn't fine tune the observers method with snowroll and dynamic counter initial distribution for nonbinary 0 level yet.
 
I will work on it with smaller priority....
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Re: 100 prisoners & a light bulb  
« Reply #530 on: Jan 30th, 2007, 7:32pm »

Two more souls lost to this beguiling puzzle! Why, oh why did William have to say "prove your solution is optimal"! Cry Cry Cry
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Re: 100 prisoners & a light bulb  
« Reply #531 on: Feb 7th, 2007, 1:44pm »

Ok, ive read a bunch of probability theories and stuff about this riddle ( most of which i didnt understand) and heres my solution.
 
The most common forms of light bulbs are, 100 hour bulbs and 1000 hour bulbs, so...
 
Every day a prisoner is taken into the room with the light. he looks at the writing on it to see how many hours it will shine for.  Take that number and divide if by 100.   --for example-- a 100 hour light bulb divided by 100 is 1.  if it is his first time in the room he turn the light on for 1 hour then turns it off.  If he has been in the room he doesnt turn it on.  After  all 100 prisoners have been in the room and turned the light on for an hour the 100 hour light bulb will burn out; showing the prisoner that is in there at the time that all 100 prisoners have been in the room and turned on the light for 1 hour.
 
if its a 1000 hour light then the prisoners leave the light on for 10 hours, so on and so forth.
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Re: 100 prisoners & a light bulb  
« Reply #532 on: Feb 7th, 2007, 2:02pm »

A 100 hour light bulb won't burn for exactly 100 hours and then then burn out. It might last 80 hours, or 120.
There is still one of Edison's original lightbulbs burning somewhere, I think.
You can't really depend on the estimated time a lightbulb will burn. If it lasts too long or too short, they'll never get out.
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Re: 100 prisoners & a light bulb  
« Reply #533 on: Feb 7th, 2007, 2:41pm »

yeah your right.    
 
there are a bunch of counter examples for my answer.
 
- the light was used before the prisoners were first put in the room; therefore, it would burn out too early and they all get shot
 
-i guess if it burned for an extra amount of hours ( considering it wouldnt be too many extra --like years-- ) it would still burn out--however if all the prisoners were in there and they stop turngin on the light after 100 hours and it didnt burn out then it wouldnt burn out.
 
i spend like 3 hours at school during class on this riddle, i cant stand the fact that its all probability and math with no non-technical answer !!!!
 
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Re: 100 prisoners & a light bulb  
« Reply #534 on: Feb 7th, 2007, 2:56pm »

on Feb 7th, 2007, 2:41pm, Joel wrote:
i spend like 3 hours at school during class on this riddle, i cant stand the fact that its all probability and math with no non-technical answer !!!!
Well, the probability is just to give an estimate of when they get out.
The important first step is that some day they will get out, and that's purely technical.
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Re: 100 prisoners & a light bulb  
« Reply #535 on: Feb 7th, 2007, 3:06pm »

There is another thread which deal with just 5 prisoners, you might like that better.
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Re: 100 prisoners & a light bulb  
« Reply #536 on: Feb 12th, 2007, 3:22pm »

rmsgrey introduced improvement of snowroll prephase in 5 prisoners forum. It seems to me I didn't seen it in this 100 prisoners thread.
 
3rd day exception ... if the prisoner swithes light off on second day as it is repeated visit, he does not become counter. This is delayed one day.  
If the light is off after 3rd visit ... counter was choosen.
It is initialised to 1 with probability 1/n^2, it is initialised to 2 with probability 3*(n-1)/n^2 and snowroll continues with probability (n-1)(n-2)/n^2
(for n prisoners).  
 
For c counters and n prisoners its improvement roughly (n-1)/n^2*(n/c+n/(n-2c)). But as counter initialised with snowroll is expected to have much higher goal then other counters, the risk that it remains the only active counter is high and the improvement tends to (n-1)/n^2*(n/1+n/(n-2c))
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Re: 100 prisoners & a light bulb  
« Reply #537 on: Mar 22nd, 2007, 12:40pm »

Each prisoner should take a dump on the floor (being careful not to mess up each others' piles) in *one spot only* the first time he's in the living room. Not able to "perform"? Then it's your duty to bite off a body part and lay it in a nice, tidy spot on the floor. First one to spot 100 piles of dung or body parts (or 99 if it's his first visit) wins the grand prize.
 
Self - evidently, it's optimal in terms of efficiency. It will indeed get the prisoners into MENSA as soon as 100 prisoners have actually been in the living room (there is no way to make it faster than that without swaying the choosing of which prisoners get to visit the room when), though I'm not sure how well a reputation they would get within the MENSA ranks.. Undecided
 
If they had a tiny bit of wit in their heads, they would have gone for toilet paper (most prisoners today get toilet paper), a piece of cloth or even a lump of food.
 
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Re: 100 prisoners & a light bulb  
« Reply #538 on: Mar 22nd, 2007, 2:14pm »

So what happens when the guards decide to clean up the room? Or make it messier?
 
The agreement only stated that the guards would leave the light bulb in the on/off state it is left in. It says nothing about the state of the room itself...
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Re: 100 prisoners & a light bulb  
« Reply #539 on: Mar 22nd, 2007, 4:07pm »

on Mar 22nd, 2007, 12:40pm, BobStew wrote:
BobStew - humble advocate of thinking outside the box

 
...who has posted yet again an answer of what is by far the most common variety, repeated endlessly in this thread. Indeed, if you read the thread, you will discover that your exact variant has been posted before - more than once, I believe. Though I do admit none of your predecessors bothered to ask what if you couldn't perform, much less suggested self-mutilation as the answer.
 
And yet once again, I point out that the "outside the box" answer is as uninteresting as it is simple. What is challenging here is to figure out the best way to do this "inside the box".
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Re: 100 prisoners & a light bulb  
« Reply #540 on: Mar 23rd, 2007, 2:02am »

on Mar 22nd, 2007, 12:40pm, BobStew wrote:
First one to spot 100 piles of dung or body parts (or 99 if it's his first visit) wins the grand prize.

If has to be his first visit.  Or someone screwed up earlier.
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Re: 100 prisoners & a light bulb  
« Reply #541 on: Apr 15th, 2007, 6:54pm »

i do believe that it is obvious that any specific prisoner visiting the room should have a count of how many of the 100 prisoners have visited the room on the day of his specific visit(i.e. if each prisoner leaves an article of clothing behind on their first visit, weather it be a shirt or a string left in the far right corner of the room(although each person must leave the same piece of clothing as the next); also, on a regular 60 watt light bulb, the metal of the treads mark easily on paint in which each prisoner leaves a tally mark on the wall or floor on their first visit( but what if the walls are made of cinder blocks, right.)  
 
well, to solve it probablistically, i resorted to the law of union of independent events and it goes like this:
to determine the probabilty of independent happenings you must subtract the probabily of an event not to occur from 1. For a second factor you take the probabilty of the first not to occur multipy it by the probabilty of the second not to occur  and subtract that from 1 and so forth.
to determine any prisoners probability being chosen at random on the first day would be: P=1-(99/100)=0.01 ok  
for the second day the the probability for the each individual prisoner is:
 P=1-[(99/100)x(99/100)=0.0199
 
so the probability of all prisoners visiting the room by day 450 would be:  
P=1-[(99/100)^450]=0.9891398...or just 98.91%
 
if you continue, the probability that all 100 prisoners visited by day 500 is:
P=1-[(99/100)^500]=0.993429 or 99.34%  
 
and on the 600th day, the probabilty would be:
 99.99759%  
which to me are pretty good odds that all 100prisoners visited the room. Six-hundred days is still less than 2 years. you can not find it with 100 percent probability because there is still that uncertainty factor in which one prisoner, or even a few prisoners, may never be chosen; but the odds of that happening decrease with each passing day.
\
if you are asking yourself why i went from a single prisoner's chances to everyones chances kind of segue, the answer is because everyone should have the same chances, but the chance that everyone visits the room by day 1 and 2 are 0; so therefor i chose to leave it at that.
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Re: 100 prisoners & a light bulb  
« Reply #542 on: Apr 16th, 2007, 3:23am »

Welcome to the Wu::forums board, sammos.
 
Since it is a long thread, I will give you a short answer: both ideas (leaving mark/closing/piles/whatever behind and using probability) were suggested before.
The thread's has evolved to looking for the best-achievable (not using optimal, as that may be impossible to prove) time, while staying "inside the box" and guarantying 100% success.
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Re: 100 prisoners & a light bulb  
« Reply #543 on: Apr 16th, 2007, 5:13am »

sorry bout that, im new to this so i didnt read every response to this puzzle in the forum; but thanks BNC for setting me on the right path.
so i figure that dealing with the light switch is the accepted "must" factor in solving this riddle.  
 
probablistically speaking, if i were in that situation, i would take the gamble at any time im selected after a year and a half!
 
well, i really dont know, and i guess the 9 years average time is best although i still dont know by what method the prisoners used to achieve that solution. can somebody fill me in?
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Re: 100 prisoners & a light bulb  
« Reply #544 on: Apr 16th, 2007, 6:34am »

Very roughly, the current best strategy works like this:  every prisoner starts out pretending they're holding a "token".  They can "drop" this pretend token in the common room by turning the light on if they find it off, and can "pick up" a token from the common room by turning the light off if they find it on (meaning someone else has dropped a token.)
 
The strategy starts out with a set of days (a "round") with some fancy rules that allow more than one token to be dropped and picked up at once (because this is more efficient at the start), and which also assign other pretend objects ("badges" and a "crown") to prisoners who pick up the dropped tokens.  From then on, only a prisoner with a badge can pick up a token.
 
After that comes a long round where the prisoners with pretend badges each try to collect a certain number of pretend tokens, "filling" their badges.  This round lasts long enough to make it very likely that all the tokens will have been picked up and so all the badges will have been filled.
 
Finally there comes a shorter round where the one prisoner with the pretend crown tries to collect all the filled badges.  (Badges are dropped and picked up just like tokens, by turning the light on and off.)  If he collects them all, he knows everyone has been in the common room (because they each dropped a token which was picked up by someone with a badge) and can declare victory.  If, at the end of the round, the one with the crown still doesn't have all the badges, it goes back to the previous round to try again.  Eventually, the one with the pretend crown has all the pretend badges (and so all the pretend tokens) and declares victory.
 
For a quick history of the solutions we've considered, see this post by Icarus.  For a more detailed summary, see this series of posts by Icarus.  The details of just the best solution so far (average time of 3460 days) can be found in the links from this post of mine.
 
Hope that helps!
 
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Re: 100 prisoners & a light bulb  
« Reply #545 on: Apr 16th, 2007, 6:42am »

The basic idea behind the solution is pretty simple:
 
Each prisoner turns the light on once, and there's one prisoner who is the only one who can turn the light off. The guy turning the light off keeps track of how many times he does it, and when he's done it 100 times (once for each prisoner, including flipping the switch on then off once for himself) he knows everyone else has been in. This gives a time of around 30 years.
 
To get it down to 9 years, there are two major refinements:
 
First, rather than just having one guy doing all the counting, have 10 guys each counting to 10 (plus themselves) or 9 counting to 11 (with one of them counting to 12 instead) or some similar division for some number of days, then have a second stage where each of them that has completed their count turns the light on once and only one of them turns the light off - then have a repeat of the first stage to mop up any prisoners who missed being counted, then a repeat of the second to pass on the remaining completed counts, etc...
 
The second refinement lies in the process used to select the people counting - the simplest form is to spend the first hundred days with each prisoner leaving the light on unless someone comes in for the second time - at which point they become the leader, know how many people have been in so far, then, after the hundred days are over, anyone who didn't leave the light on earlier turns it on once at the first opportunity. This sort of "snowball" actually works best ending after a shorter time - around 30 days - once a leader has been selected, the rest of the agreed duration is wasted time...
 
Combining those two refinements, along with some minor tweaks like allowing counters to start with a negative count in order to minimise the chances of any given person having to make a double count, gives an initial 30 day stage, at the end of which 10 (not necessarily distinct) prisoners each have responsibility for accumulating a ten-count, and one of them has the additional responsibility of collecting those completed counts, over two alternating stages.
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Re: 100 prisoners & a light bulb  
« Reply #546 on: Apr 16th, 2007, 9:06am »

I know I wouldn't expect someone to read 545 comments before posting.  So, could a moderator just add the summary to the first post of the thread?
 
The longer this thread gets, the less likely new posters will read it, making it more likely they'll post the same old things over and over again, making the thread even longer, etc. etc.
 
[I don't know if this would work (or how well), but if Willy posted from the future, would it always show up at the end?]
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Re: 100 prisoners & a light bulb  
« Reply #547 on: Apr 16th, 2007, 10:01am »

Maybe we should just lock this thread and ask people to read the summary first and make a new thread if what they are thinking is new.
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Re: 100 prisoners & a light bulb  
« Reply #548 on: Apr 16th, 2007, 10:47am »

on Apr 16th, 2007, 10:01am, Aryabhatta wrote:
Maybe we should just lock this thread and ask people to read the summary first and make a new thread if what they are thinking is new.

 
I think adding a note about daily cleaning of the room to the problem statement will get rid of most "out of the box" solutions. Deleting all comments that describe such solutions and the ensuing discusions will shrink the thread at least twice.
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Re: 100 prisoners & a light bulb  
« Reply #549 on: Apr 16th, 2007, 7:48pm »

There used to be a header, before the YaBB changeover, that did this. I'm always a little leery of changing people's posts, so I haven't attempted to do modify the first post for the same purpose. Perhaps I should, but I don't have the time to spend on it right now.
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