a logic puzzle

Mark-T wrote: ) You know the puzzles involving Dichotomy Island, inhabited ) by knights and knaves, who will answer any yes/no question; ) knights are honest, knaves are compulsive liars. ) ) Now, let's welcome onto the island, the randomizers - ) they answer any question yes or no, 50-50. ) ) You are marooned on Trichotomy Island. You meet 3 ) inhabitants. You are given that one is a knight, another ) a knave, the other a randomizer. You are permitted 2 ) questions, each question addressed to a single (not ) both the same) person, and must determine their ) identities.

Well, given that there are six ways to distribute the identities, and only four possible answers to two yes/no questions, this would seem impossible. The only way would be if there is either a third possible answer (I don't know, I can't answer that, something like that), or if time somehow factors in to the answer (if it is only possible for someone to answer at a certain time).

As soon as you start allowing that, the possibilities are endless.

(Also, must we assume that the three inhabitants know each others' identities ?)

And then there are the real lateral-thinking ones:

"Did you know that the knave slept with the knights wife ?" Result: knight kills knave. All identities are known. ^__^

SaSW, Willem

You've got it backwards. In American, the knights are chronic liars and the knaves are honest.

Who told you that, knight, knave or knowknot?

How generous. In American the knave is not to ask or reason why, but only to pay and die.

The question "which of you is the randomiser" has three possible answers.

Nick

Or maybe 4? The knight and knave respond by pointing to a person, but we are told that the randomizer answers any question yes/no 50/50., so presumably he would rather inconsequently answer yes or no.

There are some questions which could not be answered - I suspect the intended solution makes use of that somehow.

Suppose you ask one of them "If I asked the randomizer 'Is 2+2=3D4?" what would she say?" Then the knight can only answer "I don't know". But it is difficult to predict how the knave would respond.

Derek Holt.

Nick Wedd wrote: ) In message , Willem ) writes )>Mark-T wrote: )>) You know the puzzles involving Dichotomy Island, inhabited )>) by knights and knaves, who will answer any yes/no question; ^^^^^^ )>) knights are honest, knaves are compulsive liars. )>) )>) Now, let's welcome onto the island, the randomizers - )>) they answer any question yes or no, 50-50. ^^^^^^^^^ )>) )>) You are marooned on Trichotomy Island. You meet 3 )>) inhabitants. You are given that one is a knight, another )>) a knave, the other a randomizer. You are permitted 2 )>) questions, each question addressed to a single (not )>) both the same) person, and must determine their )>) identities. )>

)>Well, given that there are six ways to distribute the identities, )>and only four possible answers to two yes/no questions, this would ^^^^^^ )>seem impossible. The only way would be if there is either a third )>possible answer (I don't know, I can't answer that, something like )> that), or if time somehow factors in to the answer (if it is only )> possible for someone to answer at a certain time). ) ) The question "which of you is the randomiser" has three possible ) answers.

Lateral thinking means finding a loophole in the given rules, not blatantly ignoring them.

SaSW, Willem

What about meta-questions? "If I ask him which he is, would his answer be "liar?""

Or something like that.

Cheers! Rich

Vaguely related:

Do the three inhabitants know each other?

Did you miss the part wher he specified "...who will answer any yes/no question..."? The question "which of you is the randomiser" is not a yes/no question.

I think Haley Starshine's answer still works, though you may need more arrows.

1. Ask each the question:

"Is red the same color as blue"

1. Ask each the question.

"Is 2+3=4"

Now,

The knight will always tell the truth: no they (the colors) are not the same, and the addition is incorrect. The knave will always lie: they (the colors) are always the same, and the addition is also correct. The randomizer will answer one way the first question, and because he must maintain a 50% rate of wrong to right answers the second question his answer must be different from the first!

And it doesn't matter whether they know each other or not.

That's six questions (three people, two questions each).

Simple - the riddle is to identify the questions. Just remove the non-restrictive clause yielding "You are permitted 2 questions," ... "and must determine their identities."

Identify the questions by the order they are asked: First and Second.

Ed

I suspect someone has already solved this, but I still want to try.

Just to clarify:

The randomizer will basically flip a coin and say "Yes" or "No?" He has no problems with paradoxes?

Tha randomizer can't be forced to alter his strategy with tricky phrases like "If I asked if you were a truth-teller, and you were to answer that question with the same degree of honesty that you're answering this question..." or anything like that? He'll still just mentally flip a coin and deliver an answer?

I assume we can only ask yes-no questions?

The goal is to identify EACH of the three guys (not just find a truth-teller, or find the better of two possible paths)?

Thanks.

Without yet hearing any answers to these, here are my thoughts on the puzzle as it appears to be stated:

It's impossible. There are 6 possible combinations and you only get 4 "bits" of info. Even if you ask Guy #1 your first question, and then base your second question in some way on how Guy #1 answers, that's still only 4 bits.

Let's say that you'll ask your second question of Guy #2 if Guy #1 answers Yes to whatever you ask him. Otherwise, you'll ask your second question of Guy #3. That still only provides 4 outcomes:

That could well be enough to find a truth-teller, or maybe even figure out which path you should take, but I don't think it's enough to figure out all 3 guys.

I think I'll turn my attention to an alternate version of the problem

-- probably one of these:

"How do you identify 3 guys with 3 questions?" or "How do you find the guy that you want as your guide?" (presumably a truth-teller, although a cute twist might be that a known liar could also be an acceptable "guide").

And it tells you a heck of a lot, assuming that the Randomizer is only allowed to say "Yes" or "No," while the others will actually point to one of the group.

Also, we don't know what the Randomizer just did five minutes ago. Maybe he just told two lies and now he has to tell the truth twice to meet his quota.

(although I think it's more likely he flips a coin every single time he answers, or else his name wouldn't make too much sense)

I don't believe that it is possible with 2 questions. You *can*, however, figure out the identity of one person with two questions, but (in the worst case) you don't know who the other two are.

The trick here is that you can force a knave to tell the truth by replacing any question Q by the question "If I asked you question Q, would you answer 'yes'?" For this convoluted question, both knights and knaves will answer correctly.

To see this, suppose that Q is the question "Is 2+2 equal to 4?".

1. The correct answer is "yes".
2. If you asked the question to a knave, he would answer "no".
3. If you asked a knave how he would answer, he would lie and answer "yes".

Now suppose Q is the question "Is 2+2 equal to 5?".

1. The correct answer is "no".
2. If you asked the question to a knave, he would answer "yes".
3. If you asked a knave how he would answer, he lie, and answer "no".

So if you ask a knave "If I asked you question Q, would you answer 'yes'?" he will give the the correct answer to question Q.

Now, to figure out which one is the knight:

Line the three people up. Ask the first one the following (convoluted) question: "If I asked you if the second person is the randomizer, would you answer 'yes'?"

Now, there are the following possibilities:

1. If the first person answers "yes", then either he is a knight or a knave, in which case the second person is a randomizer, or he is a randomizer. In either case, we know that the third person is *not* a randomizer.

1. If the first person answers "no", then either he is a knight or a knave, in which case the second person is not a randomizer, or he is a randomizer. In either case, we know that the second person is not a randomizer.

So after one question, we know one person who is definitely *not* a randomizer. Now, using the same convoluted question, we ask: "If I asked you if you are a knight, would you answer 'yes'?"

If he answers 'yes', then he's a knight. Otherwise, he is a knave.

So with two questions, you can find out who one person is. But it seems to me that it takes three questions to find out who all three are.

-- Daryl McCullough Ithaca, NY

Ask 2 of the 3 inhabitants, "If I was to ask you to identify all 3 of you, what would you say?".

The knight will identify all 3, the knave should identify all 3, using an extension of your logic, and if you pick the randomiser he will say yes or no. Either way all 3 will be identified.

Herc