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Old 09-November-2008, 09:16 PM
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Default The hardest logic puzzle ever

sorry for the corny title but there's something in this.

http://en.wikipedia.org/wiki/The_har...ic_puzzle_ever

Quote:
Three gods A, B, and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
can the BAUT brains tackle this?
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Old 10-November-2008, 09:20 AM
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Random - He can just get away with the answer because you cannot tell whether he is a Liar or telling the truth .
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Old 10-November-2008, 10:15 AM
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I've loved this solution to these types of problems since I first saw this strip.

Minor language warning. Order of the Stick.

As for a real solution, I've got something in mind, but I can't seem to make it work. I think the first step would be to isolate Random, but that has to be done in a way that also makes it clear which word is yes, and which is no. That's where I'm stuck.

Once you get it down to True and False, with a known yes or no, it will only take one question to work them out.
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Old 10-November-2008, 03:53 PM
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I'd guess you could start with something along the lines of asking god A "Would god B and God C give the same answer to the question 'are you a liar'?". Haven't got time to work it out but I'd have thought you might even do it in less than three questions.
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Old 13-November-2008, 10:21 AM
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Quote:
Originally Posted by Sp1ke View Post
I'd guess you could start with something along the lines of asking god A "Would god B and God C give the same answer to the question 'are you a liar'?".
The word "liar" is imprecise. We can describe as a liar someone who speaks falsely on occasion, not necessarily in every statement. Probably you meant the indirect question to be "do you always speak falsehood?".

Putting aside what the indirect question is for the moment, if one of B and C was the random one, then A couldn't know whether B and C would give the same answer to a given question on any given occasion. I suspect that asking a question to which the answer could be undeterminable is bending the rules of the challenge, as you are only allowed to ask yes/no questions.

But also you seem to be asking a paradoxical question ("do you always speak falsehood?")which someone who always speaks falsehood would find impossible to answer without ruining their reputation. I think that asking such a question, even in an indirect way, must also be against the rules.

The challenge as worded does not tell us whether the gods are aware of the propensities of the others. However I see that the proposed answer to the challenge is based on the assumption that they do know each other's propensities.
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Old 13-November-2008, 04:21 PM
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Is the answer 42 by any chance
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Old 13-November-2008, 05:18 PM
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Let's see if this works:
The gods are A, B and C. A(T) means A is the one who always tells the truth, A(L) means A always lies, A(R) means A answers randomly true or false.

Start with some true assertion, such as "Water is wet".

Ask this question of B: "If I asked A if water is wet, would A answer 'yes'?"

Here are the options:
  1. B(T) and answers 'yes'. Then A must be A(T) therefore C(R)
  2. B(T) and answers 'no'. Then A must be A(L) therefore C(R)
  3. B(L) and answers 'yes'. This would mean A(L) also so could not happen (**)
  4. B(L) and answers 'no'. Then A is either A(T) or A(R)
  5. B(R) but this time answers truthfully and answers 'yes'. Then A must be A(T) therefore C(L)
  6. B(R) but this time answers truthfully and answers 'no'. Then A must be A(L) therefore C(T)
  7. B(R) but this time lies and answers 'yes'. Then A must be A(L) therefore C(T)
  8. B(R) but this time lies and answers 'no'. Then A is A(T) therefore C(L)
** Oops, spotted at least one mistake. The assertions about A missed out the complication of A(R). Based on the assumption that A is not A(R), then the rest holds and only statement 4 doesn't identify all cases. Thus the second question just needs to establish whether A is answering randomly and also resolve A's identify for question 4. So not quite there yet...
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Last edited by Sp1ke; 13-November-2008 at 05:19 PM.. Reason: Messed up subscripts in changing from x(F) to x(L)
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Old 13-November-2008, 05:35 PM
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I think the real nature of the challenge is to determine exactly why this can't be done in the number of questions it allows, and determine the minimum number of questions it would take.
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Old 13-November-2008, 06:36 PM
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I recall discussing a much easier (but still very thorny) version of this problem in a class. Ultimately it came down to figuring out whether a particular god was Liar or Random. In desperation, one of the students suggested "Ask him Are you a god?" On the grounds that a god who answers truthfully even some of the time, would never humiliate himself by answering "No".

I forgot the precise phrasing of that particular version.
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