If you are not aware of the Blue-spotted Monk problem, or haven’t seen my previous post (in which I made an error in logic), please visit The Problem Of Blue-spotted Monks.
Did you ever wonder what would happen if, as a cruel joke, it had been reported to the monks that one of them had the dreaded blue spot disease when in fact nobody was afflicted? (Am I really the only one here with a slight sinister streak?) Since none of the monks saw a spot elsewhere, they would each assume it was them that was infected and the very first day all the monks would have started to gather at the exit station. One of them, as he approached, would notice that none of the others had a blue dot and would start to chuckle to himself, thinking “these clowns can’t even count”. The others, hearing him laugh, would look around for the first time (until then they were so sure of their logical talents that they hadn’t even bothered to check), and by the time he stopped laughing the rest would have disappeared into the forest. Then it would occur to him that maybe he didn’t have a spot either. When word got back to the Guru, the laugher would be expelled for violating the “no communication” rule, but it wouldn’t affect the solution because the other monks only counted the peers that had spots on their forehead and he was never missed.
Plan B (I named this scheme in honor of B, who first put things in perspective for me. If this plan turns out to be flawed, like the last, I take full responsibility.)
When answering nature’s call in the middle of the night a few days after accepting B’s explanation about why the monks had to start at zero, it struck me that since a monk was really only concerned with three possibilities in the number of blue spots anybody has seen, modulo arithmetic might give a way to synchronize everybody’s universe or get everybody on the same page, so to speak. Plan B calls for everyone to start counting not at zero, but at the last multiple of six. For example, the person who saw ten blue dots knows that there are either:
- ten afflicted, not including himself, which means there are ten people who see only nine spots. Everybody else sees ten.
- On the other hand, if he is blue-spotted, there are ten others like him who see ten spots and everyone else sees eleven.
Those who see nine spots know that those who see eleven don’t really exist, and those that see eleven know that the nine-seeers don’t exist. The monk that sees ten must consider each of the other two cases, but not both at once.
Under Plan B he would start counting at six, as would the possible people who saw nine, and those potential people who saw eleven. If everyone knew to start at six, the nonexistent people who saw only six spots would be gone before the second day (we’ll discuss them again shortly), and since that won’t happen, the nonexistent people seeing seven spots will leave on Day 2, etc. Our guy knows he can sleep in until Day 4, when the really possible nine-spot sighters would be scheduled to leave. If he saw spots on Day 5, he would turn himself in and everybody else would live happily ever after.
Starting the inductive thinking process, the first six possibilities start counting at zero, just like the old days. We’ll call that their landmark. As the logic countdown continues past the next landmark (which would be six in this scheme), the hard thing for me was knowing who would start using it first, or even if it was possible to make the switch if that person needed to wait for the person ahead of him (who is still using the old landmark) to make his move. I had the hardest time reconciling the notion that you needed to wait for the people ahead of you with the notion that those people don’t exist. It turns out the solution was easier than I imagined.
In Plan B, if the number of spots you see happens to be six (or any exact multiple thereof), your dilemma is that the possible person who sees five spots started counting at zero, and won’t budge until Day 6. The guy who sees seven spots will start counting at six, and is depending on you to leave now (Day 1) or else he’s lost. Go to the exit point on Day 1. If you are not infected, there will be a very large crowd heading for the exit with no spots on their head while six people with a spot sit comfortably at home. When you spot your peers, you will avoid the crowd and head home knowing that those six afflicted people will all leave on Day 6 as scheduled, and all is right with the world. If you do have the spot, there will be six others with spots heading out that day with you while everyone else waits patiently at home. When those who originally saw seven spots wake the next morning, you will be gone and the problem will be solved.
Lately this problem has turned into an on-again-off-again obsession for me. I first saw a different version of the problem when I was much, much younger; I didn’t figure it out for myself but the answer made perfect sense. Just a few weeks ago, I ran across the spotted monk version I discussed in my last post and again, when reading the answer, I was completely satisfied (although I felt for the commenters who couldn’t accept that it would take 100 days for 100 blue-spotted monks to turn themselves in). Days later, while on a walk contemplating even more difficult (but unrelated) relationship issues, an answer just came to me out of the blue, which is the answer I last posted. It was flawed, and when B. gave his “parallel universe” explanation for why we needed to start at zero, I thought it made even more sense than the conventional answer did, and I was again happy. Several days after that, the modular arithmetic idea just came to me (as discussed). Then it was only my day job and other commitments that slowed me from working toward this solution. But we are not actually finished yet.
Calling All Logicians
All I’ve done so far (if I got it right this time) is to show that starting at zero is not strictly necessary. Plan B is not the only possible plan, however. In fact, I doubt it is the best or fastest plan (I suspect that using a smaller modulus might be helpful). For the monks to abandon zero as their global landmark there would have to be an understanding that there was a single, logically optimal plan to replace it. Based on my difficulties wrapping my mind around the issues so far, I don’t feel I qualify for the “perfect logicians” requirementA of this monastery. It’s just as well; as an aging curmudgeon, that vow of silence probably wouldn’t have worked for me for very much longer anyway. I’m hoping that the real experts will take it from here and God will finally allow me to let go of this problem. If you do see a flaw in this scheme (a check of the references will show it would not be my first mistake on this problem), let me know. I may not be able to fix it, but in keeping with journalistic standards, I am willing to admit and advertise the error. Thank you.
2 thoughts on “The Blue-spotted Monks Revisited”
UPDATE: There has been some discussion about this at puzzling.stackexchange.com/questions/. Here’s a quick summary:
In that discussion, one person presents a “proof” by induction that a faster solution is not possible. When time permits, we may discuss what an inductive proof is and see if his proof really proves what he thinks it does (of course we will check other references first to be sure he didn’t just misstate the proof or that I didn’t just misunderstand it). But in this political climate, there are other matters that need to be addressed. We need to move on for now, but I will keep you posted of any further developments. If you’ve been motivated to pursue this on your own, possibly as a distraction to current political rhetoric, please let me know what you learn.