The hardest logic puzzle ever… The following puzzle was titled "L'indovinello più difficile del mondo", which means "The hardest logic puzzl puzzle e ever ever"" in Itali Italian. an. An Amer American ican philosopher philosopher and logic logician ian named George Boolos gave this Raymond Smullyan logic puzzle the name. It was published in an Italian newspaper called La Repubblica. Think you can solve it?
The Puzzle
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
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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.
Clarifications
Boolos gave the following clarifications regarding this puzzle: p uzzle: It could be that some god gets asked as ked more than one question (and hence that some god is not asked any question at all)
What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
Random will answer 'da' or 'ja' when asked any yes-no question.
P.S. from the author, but I agree with him: I'm posting this logic puzzle here for the enjoyment of puzzle lovers. I actually am actually pretty bad at logic puzzles.
The solution’s 2
in here somewher e…don’t even think about cheating, unless 3
you’ve given up. If so, hint: what’s the quad root 4 of 42 ? 4
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The Solution Boolos states that the first move is to find either the god True or False, not Random. One possible strategy is to use logical connectives in your questions: Boolos' question was: Does 'da' mean yes if and only if you are True if and only if B is Random? Or: Are an odd number of the following statements true: you are False, 'ja' means yes, B is Random?
The puzzle's solution can be simplified by using counterfactuals. The key to this solution is that, for any yes/no question Q, asking either True or False the question: * If I asked you Q, would you say 'ja'? This results in the answer 'ja' if the truthful answer to Q is yes, and the answer 'da' if the truthful answer to Q is no. The reason this works can be seen by looking at the eight possible cases, explained below. * Assume that 'ja' means yes and 'da' means no.
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(i) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'ja', which means yes. (ii) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'da', which means m eans no. (iii) False is asked and responds with 'ja'. Since he is lying l ying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means yes. (iv) False is asked and responds with 'da'. Since he is lying it follows foll ows that if you asked him Q he would in fact answer 'ja'. He would be lying, l ying, so the truthful answer to Q is 'da', which means no. * Assume 'ja' means no and 'da' means yes. (v) True is asked and responds with 'ja'. Since he is telling the truth the truthful answer to Q is 'da', which means m eans yes. (vi) True is asked and responds with 'da'. Since he is telling the truth the truthful answer to Q is 'ja', which means no. (vii) False is asked and responds with 'ja'. Since he is lying it follows that if you asked him Q he would in fact answer 'ja'. He would be lying, l ying, so the truthful answer to Q 'da', which means yes. (viii) False is asked and responds with 'da'. Since he is lying it follows that if you asked him Q he would instead answer 'da'. He would be lying, so the truthful answer to Q is 'ja', which means no.
Now that we know the eight cases, we can proceed:
Ask god B, "If I asked you 'Is A Random?', would you say 'ja'?"
If B answers 'ja', then either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If I f B answers 'da', then either B is Random (and is answering a nswering randomly), or B is not Random and the answer indicates that A is not Random. Either way, A is not Random.
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Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you True?', would you say 'ja'?"
Since he is not Random, an answer of 'ja' indicates that he is True and an answer of 'da' indicates that he is False.
Ask the same god the question: "If I asked you 'Is B Random?', would you say 'ja'?"
If the answer is 'ja' then B is Random; if the answer is 'da' then the god you have not yet spoken to is Random. The remaining god can be identified by elimination.
There you have it - the word's hardest logic puzzle! It doesn't seem that bad, but it can get really confusing. As for me - I can't even begin to wrap my head around it. But hopefully you've found this puzzle is be a tricky treat for your grey matter.
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You’re getting colder…. 63
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Go back, before it’s too late!.. 73
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Earth orbit…
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