a logic puzzle

I don't see how you can distinguish a randomiser who happens to give the same answer to one or two questions from the knight or knave. As the question is worded, the randomiser answers "yes" or "no" at random, he doesn't tell the truth or lie at random.

In passing, I've bent my brain trying to work out what a knave will say to your Q2. Does he: Evaluate the inner question - to get "no". Negate it - because he always lies - to "yes" Evaluate the outer question (is "yes" = "yes") to get "yes" Negate it - because he always lies - to "no".

OR

Evaluate the inner question - to get "no" Evaluate the outer question (is "no" = "yes") to get "no" Negate it - because he always lies - to "yes".

I think I can come up with ways he works that allow either depending on where the "lie" module is in his wiring.

You can't. Let's look at the first question: "If I asked you whether the second person was the randomizer, would you answer 'yes'?" You ask the first person this question. The possible orders are

1. (randomizer, knight, knave)
2. (randomizer, knave, knight)
3. (knight, randomizer, knave)
4. (knight, knave, randomizer)
5. (knave, randomizer, knight)
6. (knave, knight, randomizer)

If the real situation is 1 or 2, then the first person could answer yes or no.

If the real situation is 3, the first person will answer yes.

If the real situation is 4, the first person will answer no.

If the real situation is 5, then the first person will answer yes. (This is a little complicated. In this case, the correct answer to the question "Is the second person the randomizer?" is yes. Since the knave always lies, he will answer 'no'. So the correct answer to the question "If asked you if the second person is the randomizer, would you answer 'yes'?" is "no". But since the knave always lies, he will answer "yes".

By the same convoluted reasoning, the knave will answer "no" in situation 6.

Okay, we can see that a "yes" answer is only possible in cases 1, 2, 3 and 5. A "no" answer is only possible in cases 1, 2, 4 and 6. So after a "yes" answer, we know that the third person is *not* a randomizer. After a "no" answer, we know that the second person is not a randomizer.

Suppose you ask a question in which the true answer is "yes". Then the knave will answer that question "no". So the question "What would you answer?" has the true answer "no". So the knave will answer "yes".

I don't see the other possibility.

-- Daryl McCullough Ithaca, NY

Nick wrote: ) snipped-for-privacy@yahoo.com (Daryl McCullough) writes: )> 1. "If I asked you whether the second person was the randomizer, )> would you answer 'yes'?" )>

)> 2. "If I asked you whether you are a knight, would you answer 'yes'"? )>

)> 3. "If I asked you whether the first person was the randomizer, )> would you answer 'yes'? ) I don't see how you can distinguish a randomiser who happens to give the ) same answer to one or two questions from the knight or knave. As the ) question is worded, the randomiser answers "yes" or "no" at random, he ) doesn't tell the truth or lie at random.

You don't need to. That's the whole idea.

If the first question is asked of the knave or the knight, then the answer determines which of the second and third persons is not the randomizer.

If it is asked of the randomizer then it doesn't matter, because then neither the second nor the third is the randomizer.

So, in both cases, you know that either the second is not the randomizer, or the third is not the randomizer.

) In passing, I've bent my brain trying to work out what a knave will say ) to your Q2. Does he: ) Evaluate the inner question - to get "no". ) Negate it - because he always lies - to "yes" ) Evaluate the outer question (is "yes" = "yes") to get "yes" ) Negate it - because he always lies - to "no". ) ) OR ) ) Evaluate the inner question - to get "no" ) Evaluate the outer question (is "no" = "yes") to get "no" ) Negate it - because he always lies - to "yes".

The second would imply that he himself believes he would tell the truth. That's a bit far-fetched innit ?

However, here's a different set of question that does not have this problem:

Question 1 (of first person): "Is it true that the second person is the knight, or the third person is the knave ?"

If asked of the knight, the situation is as follows: Either the second person is the knave, in which case the proposition is false, and he would answer 'no'. Or the third person is the knave, in which case the proposition is true, and he would answer 'yes'.

If asked of the knave, the situation is as follows: Either the second person is the knight, in which case the proposition is true, and he would answer 'no'. Or the third person is the knight, in which case the proposition is false, and he would answer 'yes'.

If asked of the randomizer, the situation is as follows: He will answer yes or no. In either case, both the second and third persons are non-randomizers.

Knight: 'no' implies second person is non-randomizer. 'yes' implies third person is non-randomizer.

Knave: 'no' implies second person is non-randomizer. 'yes' implies third person is non-randomizer.

Randomizer: 'no' implies second person is non-randomizer. 'yes' implies third person is non-randomizer.

So, if the answer was 'no', you ask the next two questions of the second person, otherwise you ask it of the third person.

Second and third questions (of either second or third person): "Is it true that two plus two equals four ?" "Is the first person the Randomizer ?"

These trivially determine the identities of all persons.

Note: Perhaps the OP misquoted the puzzle, and the actual puzzle involves two paths, one leading to certain death and the other to safety. You have two questions to determine which path to take.

SaSW, Willem

That's good for finding out which of two roads you should take with only 2 or 3 questions. But it won't tell you much about the identities of the people you've just spoken to.

I wonder if there's a way to find out which road you should take with exactly two questions. (I feel like there is. Probably a solution where you find one non-randomizer with the first question, then ask about the roads with your second question.)

SOLUTION

Solution for my own improvised puzzle above:

Question 1. Ask A "Does B tell the truth more often (per question) than C?"

Liar A will point to the randomizer as being more honest. Knight A will point to the randomizer as being more honest.

In any case, you then go with whichever guy DIDN'T get pointed to and ask...

Question 2. "If I were to ask you whether I should take the left road, what would you say?"

Assuming that he can parse that question correctly, you'll now get pointed to the correct road.

If you've got an arbitrary number of questions, and you're asking "Does 2+2=4" over and over, that's analagous to testing which of 3 coins is two-headed, two-tailed, and fair by flipping them over and over. You could bet pretty good money that you'll figure it out in less than ten flips.

That's not a Yes/No answer.

Neither is that. Pointing ain't allowed, only Yes/No

And the point is that only Yes/No questions, no pointing was allowed in the original problme.

The riddle as posed does not tell us how the inhabitants will respond to "conditions".

If you're allowed to invent "conditions" which must be obeyed then it's not a riddle. You can simply have them point to one another. "If you are the randomizer, point to the knave but do not answer the question, else if you are the knight answer while kneeling and pointing to the randomizer, else if you are the knave, answer while standing and point to the knight: do you live here?

If you consider the "if you are the" phrase a "condition" that need not be obeyed, then it is meaningless within your compound question.

You are just taking the riddle out of the realm of logic and moving into trickery.

Ed

No you can't.

None are valid yes/no questions.

There are no actions that need to be obeyed, unlike in your examples.

All questions which are an echelon of conditional sub-questions, such that only one sub-question is ever active can be reworded into something which doesn't contain any conditions, but instead contains tawdry arithmetic operations. It's usually possible to then map that arithmetic statement back into something which looks more like natural language and less like mathematics again. However, such a rewording is usually ugly and harder to understand (as the entire expression must be evaluated), it's far better to just let human language do what it's evolved to be good at instead, and keep the conditional wording.

Random example: If you are a knight, is the person to your left a knave, but otherwise, are you a randomiser? could become: Is at least one of the statement that you are a knight and also that the person to your left is a knave, or the statement that you are not a knight and also that you are a randomiser, true?

I disagree. I presume all those who have contributed convoluted compound conditional questions also disagree.

Phil

Puzzles of this nature do NOT allow you to insert whatever the heck you want into the posted conditions. That would cancel them as being logic puzzles.

If we know that there are truth tellers who always tell the truth and liars who always tell lies, and information beyond yes/no in an answer can be useful to expand what we know. So if a truth teller says someting about another tribe, that extra information is fact. But we can't introduce facts, conditions whatever from the last puzzle.

By the way: Believing that there is a class of puzzles involving "Dichotomy Island" is erroneous. A Google search on "Dichotomy Island" (with the quote marks) yields 111 hits. When you click on next on page 1 to go to page 2 you see that there are a total of 15 unique hits and you get the "repeat the search with the omitted results included."

When you expand it to include omitted results and review the 111 (I reviewed only 5 pages) you find that most of them refer to the present postings. When you review the 15 hits prior to expanding it there is only 1 "Dichotomy Island" puzzle found prior to the present posting - hardly a "class". See

Bottom line: it's a damn logic puzzle and should be solved, or declared unsolvable - by logic, not by facts, premises, conditions, whatever, that are not found in the puzzle as posted.

Ed

I didn't recognise the expression 'Dichotomy Island', but I certainly recognised knights and knaves. In puzzles concerning them, unless the puzzle expressly says otherwise in relation to some specific fact(s), knights and knaves always know all the relevant facts, and can accurately reason with them. They just have their particular way of giving answers.

That's introducing non yes/no questions, which is quite a different matter, particularly as the OP indicated by implication that the knights and knaves cannot be relied on to give answers to such questions.

Sylvia.

You don't read well? Quote: "If you are allowed to invent conditions which must be obeyed" etc.

You don't read well? Quote: "do you live here?"

Yes there are. Her question directs the respondent to answer some particular clause in her compound question.

That's better than ok - it's what makes horseraces. :-)

Ed

Bull. As an example, the question "do you live here" may elicit the response, "yes, and so does he". So it is NOT introducing non yes/no questions, as you assert. You are twisting things out of shape, making it impossible to have a logical discussion. No point in continuing.

which is quite a different

As Phil pointed out, conditional questions can be rephrased:

I proposed

"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?"

but if you don't like conditional questions then it can be expressed as

"Is it true that you are the knave and P2 is the randomizer, or that you are the knight and P3 is the randomizer?"

Sylvia.

It was clearly implicit that yes/no questions will be answered "yes" or "no".

Sylvia.

Are you the same Sylvia Else who hangs around aus.politics with some very nice analyses using market driven models?

If so, we must be some of the very few Australians interested in both formal logic and economics. Send me an email if you would like to correspond off-line - I'm not a crank or anything, you just sound interesting.

I don't usually post directly to aus.politics, but respond to postings that are cross posted to other groups. Still, it seems likely that I'm the Sylvia Else you're referring to.

Still, I prefer to keep the conversations in the NGs - it's always nice to get input from others.

Sylvia.

I read perfectly, it's your writing that contains the nonsense.

The immediately prior sentence was "You can simply have them point to one another." which of course, is complete nonsense. Which is why I disagreed with it.

Again, the problem is with what you wrote. That wasn't the full request - you were ordering them to perform an action _as well as_ answer a question. Simply not allowed.

An instruction on how to interpret the question is not an order to perform an additional action.

Phil

But with this implicit assumption the problem is unsolvable (2 bits of information to distinguish 6 configurations). Besides, how should a knight answer the question "will the number that I will tell you after getting your answer be even?"?

Lars