

The Hardest Logic Puzzle Ever - zatkin
http://qntm.org/gods

======
justinpombrio
Hardly the hardest. Here's a harder:

[http://justinpombrio.net/tell/prisoner-
lightbulb.html](http://justinpombrio.net/tell/prisoner-lightbulb.html)

EDIT: Also, for anyone stuck on the three gods puzzle, the three sisters
puzzle is a good stepping stone:

[http://mathpuzzlewiki.com/index.php?title=Three_princesses](http://mathpuzzlewiki.com/index.php?title=Three_princesses)

~~~
hurin
Is there a link for a solution to your variant (of the prisoners and lightbulb
problem)?

~~~
zatkin
The original problem is on a website that also contains the answer (in a red,
dash bordered box) [1].

[1] - [http://www.cut-the-
knot.org/Probability/LightBulbs.shtml](http://www.cut-the-
knot.org/Probability/LightBulbs.shtml)

~~~
hurin
I know the original problem. I'm curious how he obtains a deterministic
solution for the version where the warden can switch the light-bulb state at
random (excerpt below).

 _“I reserve the right to flip the light on or off between interrogations up
to 12 times in total.”_

Maybe it's poorly worded?

~~~
justinpombrio
You mean the Easy++ problem? The solution isn't very different from the
solution to Easy (i.e. the original problem); it just takes longer to run.

For the Hard problem, you're on your own.

The wording is very careful :-)

~~~
hurin
>> “I reserve the right to flip the light on or off between interrogations up
to 12 times in total.”

> The wording is very careful :-)

Do you mean he can flip the switch 12 times in total between _all
interrogations_? Or that between _every interrogation_ he can flip the switch
up to 12 times.

That's what I mean by poorly worded.

~~~
justinpombrio
The former, hence the phrase "in total", and the number 12 instead of 1.

~~~
hurin
> The former, hence the phrase "in total", and the number 12 instead of 1.

It wasn't very clear.

>> You mean the Easy++ problem? The solution isn't very different from the
solution to Easy (i.e. the original problem); it just takes longer to run.

Btw what's the run-time on your easy++ solution? - I think the run-time on the
solution to the regular problem with a single counter is around ~26 years, so
all your prisoners will die before they can get released unless you're doing
something more clever than the basic solution.

Also I'm not sure I understand the difference between your Easy and Hard
problems? I assume the prisoner who is not interrogated with the other inmates
just estimates the run-time for their solution and waits (which I think is
part of the original problem anyway ? )

------
wmt
[http://www.smbc-comics.com/?id=2886](http://www.smbc-comics.com/?id=2886)

------
SEMW
Another interesting one in the style of the Singaporean puzzle that's been
making the rounds recently: [http://jdh.hamkins.org/transfinite-epistemic-
logic-puzzle-ch...](http://jdh.hamkins.org/transfinite-epistemic-logic-puzzle-
challenge/) (knowledge of transfinite ordinals is useful)

------
im3w1l
Hints: [http://pastebin.com/ypPA5Uab](http://pastebin.com/ypPA5Uab)

------
nhumrich
Do the true/ false gods know how the random god would answer? I don't see how
they could since only he know the randomness. But if they don't know, they
would have no way to answer the question, "what would guy x say ja meant"
where x happens to be the random one.

~~~
qntm
This is a question of implementation, and answering it was part of the reason
why I implemented this problem in the form of a computer program in the first
place. In this implementation, the True and False gods simply pose the
question to Random and take Random's answer - which is random. If you then ask
the same question of Random again, of course you may not get the same answer.

------
SatvikBeri
You can make this question even harder by removing foreknowledge of the words!
So you know the gods have words for yes and no, but don't know what those
words are. Here's an article on the solution:
[http://www.technologyreview.com/view/428189/the-hardest-
logi...](http://www.technologyreview.com/view/428189/the-hardest-logic-puzzle-
ever-made-even-harder/)

~~~
IshKebab
Isn't that the same as stated in the link?

~~~
SatvikBeri
The qntm link tells you the words are da and ja, which makes the problem
easier. In the hardest version you have no idea what the words for yes and no
are.

~~~
Blaine0002
If you actually read the next sentence..

"in which the words for yes and no are da and ja, in some order. You do not
know which word means which."

~~~
SatvikBeri
That's the original version. The harder version is discussed below. In the
harder version, you don't know the words are "da" and "ja". They might be
"potato" and "walrus" or something completely different.

~~~
arfar
I don't think this makes the problem any different, does it? If it were
possible that there was a list of "yes" words and a list of "no" words that
they could prick from randomly or something, then yes, that would be much more
difficult.

But in both examples, there's one word for "yes" and one word for "no" and in
both examples, the asker doesn't know which one is which.

~~~
SatvikBeri
There's a solution that works if you know the words are "da" and "ja", but
doesn't work if you don't know what the words are: “If I asked you if B is
random, would you say ja?”

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jdiomede
It seems like asking about randomness is not a valid question. I'm not sure
how you would implement that (beyond checking for the assignment of random
behavior), you can't really find out until you observe a difference in
behavior over a larger sampling.

~~~
qntm
If you're puzzled about the implementation, I invite you to inspect the
JavaScript code! And if you disagree with my implementation, I'd be interested
to know what you would do differently.

~~~
jdiomede
My statement was conceptual. Boolos himself suggests that we ask if X is
random to get to a solution (though in a more complex iff statement). But
being able to ask if a source is random seems unreasonable, though without
that capability I can't say if there is a solution.

------
NAFV_P
Be very careful what you ask the gods, you might run into:

[http://en.wikipedia.org/wiki/Epimenides_paradox](http://en.wikipedia.org/wiki/Epimenides_paradox)

------
amelius
You can ask multiple questions by compounding them into one. For example:
"what is the outcome of 'is A true' converted to string and concatenated by
the outcome of 'is B true'"?

~~~
ThatOtherPerson
You can only ask yes/no questions.

~~~
amelius
In that case, I'd try enumerating all possible questions to tackle this
problem (breadth first on question size).

------
Zecc
> each question must be put to exactly one god.

> It could be that some god gets asked more than one question (and hence that
> some god is not asked any question at all).

Make up your mind.

~~~
IkmoIkmo
Basically he's trying to say

"Each of the 3 questions must be unique and only asked to a single god, and
each god can receive 0, 1, 2 or all 3 questions out of the 3 unique
questions".

In other words, don't create 1 question and ask it to multiple gods. But if
you create 2 or 3 _different_ questions, it's fine to ask them to the same god
(forgoing to opportunity to ask them to another god).

