At 17:42 11/10/04 +1000, Eric Cavalcanti wrote:
> > And here is another puzzle, which is not entirely
> > unrelated with both the KK puzzles and the current probability
> > discussion: I put three cards, two aces and a jack, on their
> > face in a row. By using only one yes-no question and pointing
> > one of the card, you must with certainty find one of the aces.
> > I know where the cards are, and if you point on a ace, I will
> > answer truthfully (like a knight), but if you point on the jack, I
> > will answer completely randomly!
> > How will you proceed? (The puzzle is one invented by Boolos,
> > as a subpuzzle of a harder one by Smullyan & McCarthy.
> > Cf Boolos' book "Logic, logic, and logic". According to
> > Boolos, it illustrates something nice about the practical
> > importance of the excluded middle principle. And this is
> > a hint, perhaps.)
>
>Bruno,
>
>That is a nice one. It seems on a first thought that
>whatever question I decide to ask, if I point on the jack the
>answer will be random, so that I can't gain perfect information.
I did that reasoning too!
From which I conclude that by the case of pointing on the jack
we should get an symmetrical answer, and from this I derive the
jack should be placed at the middle.
Now, you will refute me.
>I came up with a way out, but by making an assumption that
>is not explicit in your statement - that I can make a question
>which does not have an answer.
>
>I could ask you: "Is the answer to this question no?"
>
>In case I point to an ace, you cannot answer it. In case I point
>to a jack, you answer randomly.
I would accept that solution ... and then reformulate the question.
>One other solution would be if I could ask a question about
>other cards.
This has never been forbidden. Any yes-no question is accepted. You
were right that we should have said "any *answerable* yes-no
question" to avoid your preceding correct but "too easy" solution ;-)
>I could then point to the first one and ask:
>is any of the first two a jack? If you answer 'no' I know that
>card 2 is an ace. If you answer yes I know that card 3 is an
>ace.
Nice! It is an original solution, which refutes my point above.
The solution in the book did not! It is "point on the middle cart
and ask if the card on the left is an ace". Then if you got the
"yes" answer you know the left card (card 1) is an ace, if you
got "no" the right card (card 3) is an ace. (It did confirm my
too quick and wrong reasoning!).
Your solution points on the need of only *logical* symmetry!
--- --- ---
As a Price, I give you the (known?) Smullyan McCarthy
puzzle. You are in front of three Gods: the God of Knights, the
God of Knaves, and the God of Knives. The God of Knight always
tells the truth. The God of Knaves always lies, and the God of Knives
always answers by "yes" or "no" randomly.
You must find which is which, through some questions.
You can ask no more than three yes-no (answerable) questions.
(Each question must be asked to one God, but you can ask
more than one question to a God; only then there will be a
God you can no more ask a question).
And (added McCarthy) I let you know that all the Gods, although
they understand English, will answer the yes-know question by
either "JA" or "DA", and you are not supposed to know which
means "yes" and which means "no".
Bruno
http://iridia.ulb.ac.be/~marchal/
Received on Mon Oct 11 2004 - 08:44:58 PDT