# A Possible Solution to George Boolos' "The Hardest Logic Puzzle Ever"

The full explanation of the puzzle, by George Boolos, can be found here.

Basically, it goes as follows:

Three gods A, B, and C are called, in no particular 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.

More clarifications can be found through the link.

I propose the following:

You ask each god two questions: (1) Will you always answer truthfully? and (2) does 2 + 2 = 4?

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

True would answer: (1): Y (2): Y

False would answer: (1): Y (2): N

Random would, if asked enough times, answer: (1): N (2): Doesn’t matter, since only Random can answer N to (1).

If a god answers "ja" for both, then "ja" must be "yes" and that god must be True. Same if he answers "da" for both. Whichever god is False must answer differently for both. As for Random, I assume he can be dealt with by repeatedly asking the same question until he contradicts himself.

In other words, True will repeat his answer, False will not, and Random will answer randomly.

This solution is not proposed in the Wikipedia article, so I hoped someone would look it over. Does it work?

• Welcome to PSE! We have a question about this puzzle here and some tips on solving here. Nov 16, 2022 at 14:32