Boolos' Hardest Logic Puzzle Ever can be solved in no less than three admissible questions: axiomatic framework and rigorous proof

TL;DR

This paper introduces a formal axiomatic framework for Boolos' Hardest Logic Puzzle Ever, proving it requires at least three admissible questions for a deterministic solution and analyzing related probabilities.

Contribution

It establishes a rigorous axiomatic foundation and proves that the puzzle cannot be solved in fewer than three admissible questions, providing a formal proof of its complexity.

Findings

Proved that the puzzle requires at least three admissible questions for a deterministic solution.

Introduced the concept of admissible questions within an axiomatic framework.

Calculated probabilities related to solving the puzzle by chance.

Abstract

A formal axiomatic mathematical framework for Boolos' Hardest Logic Puzzle Ever is presented and two theorems about its solvability are proved. By strictly following Boolos' instructions (in particular, the requirement that all gods are always obliged to answer), the novel concept of \textit{admissible questions} for the puzzle is introduced. It is then rigorously proved that Boolos' original puzzle can be solved, in an absolute deterministic way, in no less than three yes-no admissible questions. However, this does not mean that one could solve it in less than three admissible questions by just pure \emph{chance}. Hence, such probabilities are computed here as well.