Brian Rabern and Landon Rabern (2008) and Gabriel Uzquiano (2010) have each presented increasingly harder versions of “the hardest logic puzzle ever” (Boolos 1996), and each has provided a two-question solution to his predecessor’s puzzle. In a paper forthcoming in JPL, Pedro Barahona and I solve Uzquiano’s hardened version of the puzzle, and show why any solution strategy must include at least three questions. Finally, we introduce a puzzle of our own:
Three gods, A, B, and C are called in some order, True, Random, and Devious. True always speaks truly, and whether Random speaks truly or falsely or whether Random speaks at all is a completely random manner. Devious always speaks falsely, if he is certain he can; but if he is unable to lie with certainty, he responds like Random. 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 in their own language, in which the words for ‘yes’ and ‘no’ are ‘da’ and ‘ja’, in some order. You don’t know which word means which.
Here are the guidelines:
- It could be that some god gets asked more than one question.
- What the second question is, and to which god it is put, may depend on the answer to the first question.
- Whether Random answers ‘da’ or ‘ja’ or whether Random answers at all should be thought of as depending on the toss of a fair three-sided dice hidden in his brain: if the dice comes down 1, he doesn’t answer at all; if the dice comes down 2, he answers ‘da’; if 3, ‘ja’.
- When Devious is able to lie he does so; but if Devious cannot be sure of telling a lie, then rather than remain silent, he responds randomly like Random, i.e., there is a fair three-sided dice in Devious’ brain that is tossed when he is not certain to lie. If the dice comes down 1, he doesn’t answer at all; if the dice comes down 2, he answers ‘da’; if 3, ‘ja’.
Updated May 19th: A solution is available here.