a logic puzzle

I think what you're saying is that instead of asking the question that the subject may not be able to answer, you can ask an equivalent alternative question that can always be answered.

But the reason Wanderer's solution works is that involves a question that has three responses - yes, no, and no answer, each of which provides distinct information. An alternative question that can always be answered will only have two responses, and you're back to the problem that the total number of combinations of responses from two questions is less than the number of possible arrangements of the people.

Sylvia.

Ah, I see. You meant to indicate three possible answers, not

I read your post too literally. I stand corrected.

Ed

However:

Nope. Say P1 is the Knight, but does not know who the other two are, so he truthfully answers the question about whether he likes eggs, with a yes or no. Or he might answer with "I am the knight, I like eggs, and I don't know who the others are." Or some other answer that is equally true and yet not helpful. And the Knave could answer the question the same way (either yes or no or with the longer answer).

But it may just be the answer to whether the Knight likes eggs, and imply nothing about identity.

Ed

I think it's an established premise in these kinds of puzzles that the subjects do know all the relevant facts. Thus the knight and knave know exactly who everyone is (so does the randomizer probably, but nothing turns on that). They also know how each other would respond to questions, because their responses are predictable. They don't know how the randomizer would respond because that's not predictable.

He isn't even presented with that option, which is given only to the randomizer.

Sylvia.

another inhabitant)

Nope. Any question asked to the randomizer will yield either a yes or no answer. If you mistakenly ask either question of the randomizer, you'll get a yes or a no.

So the possible set of possible yes/no answers is yes/yes, yes/no, or no/no. Using any one of those pairs, you cannot identify who is who.

We can't assume that the knave or knight will go beyond a yes or no answer, even if they are allowed to.

Ed

another inhabitant)

You might as well group them in a circle facing inwards and say, "Mr. Knight, is the guy to your left a knave?" You'll get all 3 of their identities in one answer.

Just ask each of them a question you all know the answer to. Two answers will be the same and one will be different. You now have identified the one who is different, and can use him to reliably distinguish the other twi.

--
It\'s times like these which make me glad my bank is Dial-a-Mattress

It doesn't work. It presupposes that inhabitants know who the other inhabitants are.

Ed

is that involves a question

Your statement that you think this type of puzzle has an established premise interrupted the reply, making it seem he isn't presented with that option. I stipulated that he did not know who the others are, and with that stipulated, he is inescapably presented with the option.

In a nutshell, your compound questions each start with a question: "If you are the ....".

Only the Knight is guaranteed to answer that honestly. And his answer is helpful only if we assume he knows who the others are. Even then it only implies some information, per your description. And not knowing who is answering hurts, big time. Any one of the three could answer yes or no.

Long description below, if you care to read it.

Ed

Here's how the question fails when rejecting the premise that they all know who each other is.

Without knowing who the others are, the Knight listens to your question: " Q1 to P1 - If you are the knave, then is P2 the randomizer, else if you are the knight, then is P3 the randomizer, else do you like eggs? "

He is not the knave, and knows he is not the knave, so he gets to the part that says "if you are the knight then is P3 the randomizer" and he does not know, so he takes the else to "do you like eggs"? Therefore we can't tell if the answer applies to the eggs question or the randomizer question.

If the knave _does not_ know who the other inhabitants are, he can choose to answer any of the three parts in your compound question, because he is lying when he identifies either P2 or P3. If he is so compulsive in lying that he has to ensure that his answer is wrong, he gets to the "do you like eggs" part and lies. That way he can't inadvertently give a correct answer as to identification of the others.

Now assume we accept the premise that each one knows who the others are.

Examine the knave, answering the same question, and assuming he _does_ know who the others are: "If you are the knave," he (silently) lies about that so he skips the "then is P2 the randomizer" part. He gets to the "else if you are the knight" part and he can decide to answer as if he is the knight or as if he is the randomizer, both of which satisfy his compulsion to lie. That gives him the choice of answering either the "is P3 the randomizer" or the "do you like eggs"? We can't tell which of those questions he is answering, and we can't predict whether his answer will be yes or no.

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The very first statement in the puzzle is "You know the puzzles involving Dichotomy Island". This clearly indicates that the puzzle is a derived class from base classes of puzzles in the namespace "I know". I can therefore embellish the puzzle with any information I have in those base classes of puzzles I know.

Right now, outside my window, the sky is pretty gray...

Using this general idea, and *only* assuming the inhabitants know their own tribe (they *don't* have to know the others' tribes), here are the two questions:

Q1: Mr Knight, is that [pointing to one of them, now known as #1] you? Q2: Mr Knave is that [pointing to a different one of them, now known as #2] you?

#1 is chosen at random; the method for assigning numbers to the others is given below.

Ask Q1. There are two possibilities: #1 says yes, or someone else says no.

Case I: #1 is the knight. Choose #2 at random from the other two, and ask Q2. Whoever answers is the knave; whoever is left is the randomizer.

Case II: Call the "someone else" #3; he's the knight. Ask Q2. Whoever answers is the knave; whoever is left is the randomizer.

Note that the inhabitants needing to know their own tribe is in fact an assumption. It's otherwise conceivable that they're hypnotized into doing their thing without any self-knowledge.

--
Neal Plotkin
my first name dot my last name at nyu dot edu

I would classify that as a condition, rather than a question.

Otherwise even questions such as "would the knave answer yes if asked X?" become an exercise in deciding how many sub questions are involved. Does that question involve the subject asking himself who is the knave, and then possibly lying about it?

There is a bit of an ambiguity about how the "else" binds, but it was intended to bind with "if you are the knight."

Construing the question the other way requires that it be treated as ungrammatical - that is, that it contains a question of the form "is statement, else question."

They're details anyway. If there were a point, then I've no doubt the question could be phrased unambiguously.

There isn't a point, beyond the one I was making about information, because a question of this form eliminates 2 out of the 6 possibilities, leaving only two questions to select amongst four, which is not going to work while the randomizer may still answer one of them.

Sylvia.

B is a randomizer who speaks a dozen times a day. C is a knight, who speaks once a week.

What will A do then?

Easily fixed: "Does B tell the truth more often per question than C?", or equivalently "Is it true that either B is the knight or C is the knave?"

another inhabitant)

** Sorry... that ASS-u-ME-s that one knows who is the Knight and who is the Knave...because one must direct the question to a specific person, not to them as a general group.

False assumption that "one will be different"...think about it.

Robert Baer wrote: ) Matthew Russotto wrote: )> In article , )> David Breton wrote: )> )>>I think this is unsolvable even with an arbitrary number of questions

)> Just ask each of them a question you all know the answer to. Two )> answers will be the same and one will be different. You now have )> identified the one who is different, and can use him to reliably )> distinguish the other twi. ) False assumption that "one will be different"...think about it.

I thought about it. One will be different. The Knaves answer will always be different than the Knights answer. The Randomizer will match the answer of either the Knight or the Knave, but not both. So the person that answered differently than the other two is either the Knight or the Knave.

Thus rendering it perfectly solvable with an arbitrary number of questions.

The only way to make the problem unsolvable would be to assume, counter to the base standard for this type of logic puzzle, that the three people do not know the identities of the others. *AND ALSO* to assume that only answerable yes/no questions are allowed. Which goes against the 'randomizer will answer any question' rule.

(Otherwise you simply ask "What is two plus two" of each of the three, and the randomizer is required by the rules to answer 'yes' or 'no'.)

Oh, and even with the above assumptions, it *is* solvable, you just can't put an upper bound on the number of questions required. Simply repeatedly ask a simple yes/no question to the two remaining unknowns until they don't give the same answer, after which you're done.

The only way to render it unsolvable is to assume that the randomizer will maliciously choose the answer so as to give you the least amount of information. Which goes against the 50-50 rule.

So, the only way to make the puzzle unsolvable with an arbitrary number of questions is to make not one, but two assumptions that go against the stated rules.

SaSW, Willem

--
Disclaimer: I am in no way responsible for any of the statements
            made in the above text. For all I know I might be
            drugged or something..
            No I\'m not paranoid. You all think I\'m paranoid, don\'t you !
#EOT

how about this solution if they do not know each other's behavior

ask all three of them

what would you say if I asked you X (where X is the question you want answered) (since the knave is constrained to answer the question dihonestly he will lie about his potential answer to X and in doing so give the truth.)

thus the majority answer will be the truth. if the first two agree (and they will 2/3 ofd the time) there's no need to ask the third.

Let's say you ask a question of all three, a question to which you know the correct answer, to which each of the three knows the correct answer, which each of the three MUST answer, and which they must ONLY answer with one response selected from "Yes" or "No".

The knight and the knave will never give the same answer - where one answers "Yes", the other will always answer "No". The randomiser must answer either "Yes" or "No", which is bound to be the same as one of the other answers, making the other other answer(!) different. And that different answer will always be given by either the knight or the knave.

As you said...think about it.

--
Richard Heathfield 
Email: -http://www. +rjh@
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