

100 Prisoners and a Lamp - C--
http://theccontinuum.wordpress.com/2013/11/24/100-prisoners-and-a-lamp/

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popularopinion
This puzzle has been on the XKCD puzzles wiki for several years. They have
some very fun puzzles there. It's a huge timewaster, so only click if you have
nothing pressing to do.
[http://wiki.xkcd.com/irc/puzzles#Prisoners](http://wiki.xkcd.com/irc/puzzles#Prisoners)

If you need help on any of the puzzles, hints and answers are on the talk
page. For this specific puzzle, the talk page links to a paper by William Wu
that answers this exact question with asymptotic analysis.

[http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBul...](http://www.ocf.berkeley.edu/~wwu/papers/100prisonersLightBulb.pdf)

~~~
andlima
This is a really nice puzzle! I didn't know William Wu wrote a paper about it.
I saw it many years ago on William Wu's riddles site[1] (highly recommended).
It's one of the most popular problems there. The original thread[2] in the
forum has 25 pages, covering lots of approaches. There is also another
thread[3] summarizing the discussion.

[1] [http://wuriddles.com/](http://wuriddles.com/)

[2] [http://www.ocf.berkeley.edu/~wwu/cgi-
bin/yabb/YaBB.cgi?board...](http://www.ocf.berkeley.edu/~wwu/cgi-
bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1027805293)

[3] [http://www.ocf.berkeley.edu/~wwu/cgi-
bin/yabb/YaBB.cgi?board...](http://www.ocf.berkeley.edu/~wwu/cgi-
bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1236693480)

~~~
jd007
reminds of the good times i had with wuriddles in high school. definitely
wasted many hours talking about the 100 prisoner problem and its variants with
others back then. another one i liked was the sink the sub one

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throwaway0094
Give everyone a unique number during the planning phase. Have everyone scratch
their number into the wall. Leave the light on.

When all numbers are on the wall, you are done.

~~~
shootTheHostage
This is a clever and practical idea, in the sense of having Indiana Jones just
shoot the guy with his pistol, instead of engaging in an elaborate sword
fight, but it violates the spirit of the problem. If we solved all of our
problems like this, we would kill the patients instead of curing the diseases.

The premise of the problem is very clear: How do you tally the presence of 100
unique events, some of which may repeat themselves, when you may use one and
only one bit to register the state of the system?

~~~
throwaway0094
It depends on the disease, doesn't it? At this point (for some diseases) it's
much more cost-effective to kill the patient that would have gone towards
prolonging a state of suffering. Then send the funds towards Against Malaria
or something else. ;-)

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brainburn
The problem description could be a bit clearer. What exactly are the rules?

~~~
spacehome
This is a pretty poor description of the problem if you've never heard of it
before. The salient information is that there is a single binary switch in the
room that can be turned to an 'on' or 'off' position. The prisoners entering
the room can observe what the state of the switch is when they enter, and they
have the option to flip the switch or not.

Also, one of the prisoners is not a 100-bit accumulator. He or she would be a
7-bit accumulator (128 > 100), but this is a very tortured way to say that
this one prisoner can count to 100.

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adam-f
This reminds me of a problem I learned in UCSC (and was accused of cheating by
the teacher when I figured it out in five minutes).

100 Prisoners are told they will be given white or black hats, but they don't
get to see the hat they're wearing, they will be lined up facing the same
direction, and they gun-to-the-head, say "black" or "white" and if they guess
the color of their hat, they get to live.

They get to speak back to front, i.e., the rearmost prisoner sees all the hats
ahead.

One strategy for optimizing the number left is to speak the color of the hat
directly in front of you, in which case the prisoner ahead gets to live by
repeating that color. This saves 50%, but of course there's a better solution.

~~~
mikeash
Spoiler ahead! Potential answer!

My solution is that the first prisoner states the color of the hat in front of
him, the second prisoner states the color he just heard, and so forth with odd
and even numbers until you get to the end. This saves 75% (even numbers are
guaranteed to live, odd numbers live with 50% odds).

Is this optimal?

~~~
ssfermoy
If they decide the guy at the back indicates an odd number of black hats by
saying white or an even number of black hats by saying black. Now the
remaining prisoners all live and the back guy has a 50/50 shot.

~~~
mikeash
Very nice. It took me a while to see why that works, so I wonder how long it
would have taken me to come up with it.

Now that I get it, it seems almost obvious. It's basically the same as my
solution, except I'm essentially using this technique on 50 separate lines of
2 people each.

~~~
ssfermoy
Glad you like it. Heard this as an interview question at a trading company.
Gave something similar to your solution and was asked for a better one,
eventually work my way to the solution above but was given no indication if it
was what they wanted. It wasn't a pleasant experience.

