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# logical deduction – A Possible Solution to George Boolos’ “The Hardest Logic Puzzle Ever”

The full rationalization of the puzzle, by George Boolos, might be discovered right here.

Basically, it goes as follows:

Three gods A, B, and C are known as, in no explicit order, True, False, and Random. True at all times speaks actually, False at all times speaks falsely, however whether or not Random speaks actually or falsely is a very random matter. Your job is to find out the identities of A, B, and C by asking three yes-no questions; every query should be put to precisely one god. The gods perceive English, however will reply all questions in their very own language, wherein the phrases for sure and no are da and ja, in some order. You have no idea which phrase means which.

More clarifications might be discovered by the hyperlink.

I suggest the next:

You ask every god two questions: (1) Will you at all times reply in truth? and (2) does 2 + 2 = 4?

Question 1: “Will you always tell the truth?”
Question 2: “Does 2 + 2 = 4?”

(1): Y
(2): Y

(1): Y
(2): N

Random would, if requested sufficient occasions, reply:
(1): N
(2): Doesn’t matter, since solely Random can reply N to (1).

If a god solutions “ja” for each, then “ja” should be “sure” and that god should be True. Same if he solutions “da” for each. Whichever god is False should reply otherwise for each. As for Random, I assume he might be handled by repeatedly asking the identical query till he contradicts himself.

In different phrases, True will repeat his reply, False won’t, and Random will reply randomly.

This answer just isn’t proposed within the Wikipedia article, so I hoped somebody would look it over. Does it work?

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