Concerning the graveyard logic puzzle (spoilers below if you haven’t clicked on the link):

Some few rabid DN addicts might recall that some days ago I posted a brief comment on this puzzle, but I realized pretty quickly that I got sidetracked into making remarks that weren’t exactly clarifying. (My deep gratitude to Justin for granting me the favor to have them quickly deleted. God knows we read enough useless stuff anyway and I certainly didn’t want to contribute to that.)

Those remarks, however, were motivated by my own struggle to solve the problem, which I eventually did but believing that, even as I realized how it must be solved, the steps involved (there are many) could be shortened. Now I think I know why, though earlier I thought it had to do with the framing of the problem. It doesn’t, if one takes a very strict assessment of that very framing. Ok, once more–spoilers follow.

It’s about two prisoners, ignorant of what each other sees looking out on a graveyard, the first (1) seeing 12 grave statues, the second (2) 8, but they may be freed only if one answers whether they see collectively 18 or 20 statues given a daily question sequentially to (1) then (2). They may pass or answer, but only a correct answer first given frees them, and any other answer either immediately dooms them or they are doomed in another way by a regress of passes.

The answer given in the linked piece assumes that both prisoners are something like perfect logical thinkers. This assumption is crucial for my argument after the answer provided in the link, which follows.

The provided solution requires days of passes between (1) and (2), involving strict deduction by each. On Day 1 (1) passes, which (2) knows by being asked for the correct number. But (2) passes, which (1) knows by being asked again on Day 2 (etc.). The deductions involve elimination of logical possibilities beginning with (1)’s first pass, not seeing 20 or 19 statues (where the answer in that case given it’s 18 or 20 must be 20), which then (2) deduces and passes because she sees 8, not 18 (and she knows now it’s not 20 or 19 that (1) sees). Day 2 (1) deduces from (2)’s pass that she doesn’t see 18 but thus sees at least 2. Successive passes and eliminations ((1) knows every next day that (2) sees at least 2 more statues from her pass) on successive days makes us arrive at a crucial day, Day 5, when the deductions due to passes allow (1) to conclude that (2) sees at least 8 statues, and because (1) sees 12, the total must be 20, not 18.

What bothered me from somewhere deep down was that, given that (1) and (2) are something like perfect logical thinkers, 5 days should not be necessary. I lost a lot of sleep that night I first posted my later deleted obtuse remarks until I realized why.

Clearly (1) and (2) can deduce from the get-go two possibilities for each other–(1) knows that (2) sees either 6 or 8 statues, and (2) knows that (1) sees 10 or 12. They thus deduce these constraints for one another as mutual knowledge.

And both are something like perfect logical thinkers! Thus, they each know the elimination scheme based on the parameters of it, and they both know that they both know this. This is shared metaknowledge that can be used to shorten the deductive process, along with their mutual knowledge about the two different disjunctive possibilities about what they may see. This was what deep down was bothering me all along.

Therefore, each by grasping the elimination game deduces before any questions on Day 1 the crucial day for knowing the upper boundary for when (1) would know the least number of statues (2) sees if (1) sees 12 (instead of 10)–the end of Day 4 when (2) passes (or, the beginning of Day 5 if one specifies when the crucial deduction for (1) actually occurs). Observe that both conclude that Days 1-3 of the elimination game must end with the Day 4 beginning assumption that (2) sees at least 6 statues, since that is the also the minimum number (2) can see consistently with the 18/20 final answer. So on Day 1, given all this, (1) passes on the question as if it were Day 4, and (2), knowing this (because that’s the only way (1) might make an upper-limit deduction of her seeing 8 given (1) seeing 12 statues), passes, thus allowing (1) on Day 2 to know that (2) sees at least 8 statues. So (1) at the beginning of Day 2, seeing 12 statues, gives the right answer–20.

Note that the puzzle only requires that you arrive at which of 18 or 20 statues are seen by (1) and (2)–not how many there actually are. For example, if two statues at their convergent points of view are seen in common by (1) and (2) (they could not know that), then the answer is still 20 despite the fact that there are only 18 statues actually in the graveyard. My worry early on about this distracted me into thinking that something was amiss about the framing of the puzzle. However, the puzzle is sufficiently set up to provide the 5-Day answer or, as I argue above, the 2-Day answer as well.

One small defense of my seemingly strong assumption that (1) and (2) would know that the other would share knowledge of the elimination game, especially from (1)’s epistemic perspective on Day 2: it’s the only explanation by the original one why (1)’s answer on Day 5 is certain rather than a lucky guess (so taken from (1)’s epistemic perspective on day 5). Since all previous days were just constituted by passes, unless one assumes that (1) is epistemically confident that (2) is undergoing the same rigorous deductive schemes, then the happy result of the start of Day 5 just devolves to 50/50 chance (given that (1) has then subjectively operated on the erroneous assumption that the elimination game was going on). In my scenario that same confidence would be behind (1)’s pass on Day 1 to know that (2) would take it as Day 4.