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I haven't done a simulation, but I would guess not much more than 10,000. Almost every visit of the counter would decrement the number of prisoners who haven't flipped yet by one.


Good guess. 100 runs: Average: 10163 Lowest: 7762 Highest 12391

If I knew how to paste code in here, I would show you my simple python script to test it.

Edit: 1000 runs: Average: 10109 Lowest: 7389 Highest 13461


http://news.ycombinator.com/item?id=32766

It would probably be helpful if that information was readily available via a link near the "add comment" and "reply" buttons


Thanks.

 import random
 def run():
 	prisoners = {}
 	for x in range(100):
 	 prisoners[x]=0
 	days=0; counter=0; bulb=0
 	while counter<100:
 	 prisoner = random.randint(0, 99) 
 	 days += 1
 	 if prisoner == 0:
 	  if bulb == 1:
 	    bulb = 0; counter += 1
 	 else:
 	   if prisoners[prisoner] == 0 and bulb == 0:
 	     bulb=1
 	return days


Along with all other relevant info for working with text in comments.


There's a solution that takes around 3500 days.


Curse you, amichael! No! No, I am not going to try to solve this problem tonight. I'm going to do ourdoings.com coding on the train ride home. I am not going to think about prisoners, light bulbs and distributed systems. I'll check in tomorrow and see if someone else has posted the answer, but I won't try to come up with it myself. You tempt me, but I resist. :-)


I'm working on a solution using the current day (modulo the population) as a kind of covert channel to keep track of the number of prisoners who already visited the room (the basic idea is that, if the light is on on day 64, at least 64 people have visited the room).

I still have a bug in my algorithm and I'm not even sure it will really work.


can you run mine through and see how many days you get!


First 99 days:

if bulb on: do not touch

if bulb off:

if you haven't been visited before, consider yourself counted, and do nothing.

if you have been in here before, turn it on. you are now the counter. look at the day number (1-based). if it's day 36, then 35 people (including yourself) have visited the room.

day 100:

if bulb off, and you haven't been there before, declare victory (super lucky!)

if off, but you have been there before, you're the counter and the count is 99.

if bulb on: turn it off.

after day 100:

use the original solution with a counter.

speedup:

you get about 12.2 people counted in the first 100 days. normally it's somewhere around one person counted per 100 days. so total time is around 100 + 87.8*100. a bit better. not near 3500 though.


i wrote code. 1000 trials. naive mean: 10389.832 my strategy's mean: 9392.652

i think the improvement was a little less good than i guessed because i jumpstart getting the first dozen or so people counted, and they are a little easier to count than the last people.


Here's a paper on the subject:

http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBul...

It includes my answer :)




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