a logic puzzle

"If i were a knave, would i tell a lie?"

Rec.puzzling at its finest!

Phil

I think they have to be yes/no questions.

-- Daryl McCullough Ithaca, NY

Then 2 questions give 4 possible outcomes. There's 6 arrangements of the 3 inhabitants, making the deduction impossible.

Herc

Yes, that's what I said: 2 questions are not enough.

-- Daryl McCullough Ithaca, NY

from the OP "who will answer any yes/no question" I see.

If you were to ask me, "if I were to ask you 'do you think the problem is impossible?' what would you say", I would say "yes".

Herc

OK, if we allow the "I don't know/cannot answer" response, and assume that they all know who they all are, then I think the following works. Call them A,B,C.

Ask A: "If I asked you and B the same question, would you both give the same answer?"

If A is unable to answer, then B must be the randomizer, and we can sort out A and C out just by asking C a question to which we know the answer, like "Is B the randomizer?".

Otherwise B is definitely not the randomizer. Also, if A says yes, then A is randomizer or knave, and if A says no then A is randomizer or knight.

In either case, ask B: "If I asked you and C the same question, would you both give the same answer?".

No answer means C is randomizer, and A's answer completes the picture.

A yes/no answer implies that C is not the randomizer, so A must be the randomizer, and B's answer tells us the rest.

Derek Holt.

Oops, mis-read the problem. Damn!

Mark-T wrote: > ... randomizers - > they answer any question yes or no, 50-50. ...

You get NO information from a randomizer. Worse, the presence of the randomizer pollutes the information that you could get from either a knight or knave. I.e., any question that would give you information if answered by a knight or knave is worthless, given that a randomizer might be the one answering it.

I think it's unsolvable.

Bob

Yeah so they do know each other?

As I said, I assumed that they do! If not, I have no idea!

Derek Holt.

Note that this doesn't explicitly say "you can only ask yes/no questions", it just says "they will answer yes/no questions this way".

That said, if the randomizer avoids paradoxes, then I think one of Smullyan's books suggested a question along the lines of:

"If I asked you 'Are you the knight?' twice, would you answer 'yes' exactly once?"

The knight will say no, the knave will say yes, and the randomizer will freeze up.

BZZZZT!!

Disallowed in the problem spec.

Hope This Helps! Rich

Also vaguely Dilbertian. ;-)

Cheers! Rich

Doesn't that kind conflict with the idea of "random"? Just because I toss a head once doesn't mean the next toss is constrained to be a tail, to maintain the 50% average!

Cheers! Rich

Norman ... Coordinate ...

Cheers! Rich

I cannot see anything in the problem specs as given that disallows that. We have to try and answer the problem posed and not some other similar problem. The Smullyan-type solution in which the randomizer freezes up definitely doesn't work, because we are told that the randomizer answers every question randomly yes/no.

Maybe the OP will eventually tell us what solution he had in mind!

Derek Holt.

I understand what you are saying, but the problem specifications don't allow true randomness IMHO. If they do, then there is virtually no way to ferret out the randomizer should he either answer with the correct answer twice, or the incorrect answer twice. In both those cases, he'll appear as if he were one of the others.

OK, as an appology for not reading the original problme properly, I'll post one of my own:

In San Francisco CA, two women were sharing the same umbrella. They walked over a block, outdoors, carefully sidesteping a puddle in teh sidewalk, no less, all the while working hard to control the umbrella in the stiff breeze.

Yet, when they arrived at their destination, they were both quite dry! How come?

Disagree: the randomizer will answer either yes or no, based only on a random chance of 50/50. After all the problem states: "the randomizers

- they answer any question yes or no, 50-50" (and that is 'any question'...)