Uzquiano (Analysis 70:39–44, 2010 ) showed that the Hardest Logic Puzzle Ever ( HLPE ) [in its amended form due to Rabern and Rabern (Analysis 68:105–112, 2008 )] has a solution in only two questions. Uzquiano concludes his paper by noting that his solution strategy naturally suggests a harder variation of the puzzle which, as he remarks, he does not know how to solve in two questions. Wheeler and Barahona (J Philos Logic, to appear, 2011 ) formulated a three question (...) solution to Uzquiano’s puzzle and gave an information theoretic argument to establish that a two question solution for Uzquiano’s puzzle does not exist. However, their argument crucially relies on a certain conception of what it means to answer self-referential yes–no questions truly and falsely . We propose an alternative such conception which, as we show, allows one to solve Uzquiano’s puzzle in two questions. The solution strategy adopted suggests an even harder variation of Uzquiano’s puzzle which, as we will show, can also be solved in two questions. Just as all previous solutions to versions of HLPE , our solution is presented informally. The second part of the paper investigates the prospects of formally representing solutions to HLPE by exploiting theories of truth. (shrink)

In this paper, we present a framework in which we analyze three riddles about truth that are all (originally) due to Smullyan. We start with the riddle of the yes-no brothers and then the somewhat more complicated riddle of the da-ja brothers is studied. Finally, we study the Hardest Logic Puzzle Ever (HLPE). We present the respective riddles as sets of sentences of quotational languages , which are interpreted by sentence-structures. Using a revision-process the consistency of these sets is established. (...) In our formal framework we observe some interesting dissimilarities between HLPE’ s available solutions that were hidden due to their previous formulation in natural language. Finally, we discuss more recent solutions to HLPE which, by means of self-referential questions , reduce the number of questions that have to be asked in order to solve HLPE . Although the essence of the paper is to introduce a framework that allows us to formalize riddles about truth that do not involve self-reference, we will also shed some formal light on the self-referential solutions to HLPE. (shrink)

Rabern and Rabern (Analysis 68:105–112 2 ) and Uzquiano (Analysis 70:39–44 4 ) have each presented increasingly harder versions of ‘the hardest logic puzzle ever’ (Boolos The Harvard Review of Philosophy 6:62–65 1 ), and each has provided a two-question solution to his predecessor’s puzzle. But Uzquiano’s puzzle is different from the original and different from Rabern and Rabern’s in at least one important respect: it cannot be solved in less than three questions. In this paper we solve Uzquiano’s puzzle (...) in three questions and show why there is no solution in two. Finally, to cement a tradition, we introduce a puzzle of our own. (shrink)

Rabern and Rabern (2008) have noted the need to modify `the hardest logic puzzle ever’ as presented in Boolos 1996 in order to avoid trivialization. Their paper ends with a two-question solution to the original puzzle, which does not carry over to the amended puzzle. The purpose of this note is to offer a two-question solution to the latter puzzle, which is, after all, the one with a claim to being the hardest logic puzzle ever.

"The Hardest Logic Puzzle Ever" was first described by the late George Boolos in the Spring 1996 issue of the Harvard Review of Philosophy. Although not dissimilar in appearance from many other simpler puzzles involving gods (or tribesmen) who always tell the truth or always lie, this puzzle has several features that make the solution far from trivial. This paper examines the puzzle and describes a simpler solution than that originally proposed by Boolos.

We present the simplest solution ever to 'the hardest logic puzzle ever'. We then modify the puzzle to make it even harder and give a simple solution to the modified puzzle. The final sections investigate exploding god-heads and a two-question solution to the original puzzle.

In this paper, we first propose a simple formal language to specify types of agents in terms of necessary conditions for their announcements. Based on this language, types of agents are treated as ‘first-class citizens’ and studied extensively in various dynamic epistemic frameworks which are suitable for reasoning about knowledge and agent types via announcements and questions. To demonstrate our approach, we discuss various versions of Smullyan’s Knights and Knaves puzzles, including the Hardest Logic Puzzle Ever (HLPE) proposed by Boolos (...) (in Harv Rev Philos 6:62–65, 1996). In particular, we formalize HLPE and verify a classic solution to it. Moreover, we propose a spectrum of new puzzles based on HLPE by considering subjective (knowledge-based) agent types and relaxing the implicit epistemic assumptions in the original puzzle. The new puzzles are harder than the previously proposed ones in the literature, in the sense that they require deeper epistemic reasoning. Surprisingly, we also show that a version of HLPE in which the agents do not know the others’ types does not have a solution at all. Our formalism paves the way for studying these new puzzles using automatic model checking techniques. (shrink)

If you think that a proposition can have more than one contradictory, or can have none, then you need read no further. What I will show is that if then It is not obvious that this must be so. If p1 and p2 are distinct but logically equivalent, it might appear that their contradictories, q1 and q2, should also be distinct though logically equivalent. In traditional logic, a pair of propositions which satisfy (3)(i) alone are called contraries, and a pair (...) which satisfy (3)(ii) alone are called sub-contraries.2 The relation of contradictoriness defined by (3) is clearly symmetrical: Let p1 and p2 be two propositions which are logically equivalent, and assume (1). Then p1 has a contradictory, q. We first show that Suppose p2 and q were both true. Then, since p1 and p2 are logically equivalent, p1 and q would both be true. But p1 and q are contradictories, so, by (3)(i) p1 and q cannot both be true. We next show that By (3)(ii) at least one of p1 and q must be true. So, since p1 and p2 are logically equivalent, at least one of p2 and q must be true. From (5) and (6) it follows that p2 and q are contradictories, so by (4) From (7) and (1) we have that p1 and p2 are identical; and since all that was assumed about them was that they are logically equivalent, (2) follows. (shrink)