-1

A little history - I read about a blue-spotted monk version of this problem on another site. A "better" solution came to me, which I posted on my blog (theSilentKnight.info). A couple of people responded and made reference to this site. It turns out my first answer was flawed, but the commenters gave me the insight to find a better answer. In my new answer (Plan B), I find a way so that the monks don't have to go all the way back to zero to start their logic count-up by using the concepts of modular arithmetic. Specifically, whatever number of blue spots you see, start counting not at 0, but at the last multiple 6. So if you saw 10 blue-spotted monks, you would start counting at 6 on Day 1. The possible person seeing 9 blue spots would start at 6, and the possible monks seeing 11 monks would start at six. All the other potential sightings aren't possible. Since everybody started counting at 6, the logic count-up may proceed as usual, and at the appropriate time, all spotted monks will leave the island. Other details are in the comments that follow. Since I have been wrong before, I wouldn't mind if somebody could again point out any error in my ways. What is wrong with Plan B? Thank you.

Silent Knight
  • 7
  • 4
  • 2
    Please include relevant hyperlinks here and/or state your new answer here. – the4seasons Jul 23 '16 at 09:25
  • 1
    Could you possibly link what you're refering to or post your question as a complete here? This does not appear to be a real question and such will be soon flagged. – Radhato Jul 23 '16 at 09:26
  • 5
    Welcome to Puzzling! It's very unclear what question you're trying to ask - could you tell us what exactly you think the "plan B" is so that we can explain why it won't work? – Deusovi Jul 23 '16 at 09:37
  • Post your solution here, this forum is not a place to (merely) advertise your blog. post and include a link is ok. – Jasen Jul 23 '16 at 23:10
  • Sorry, I thought the explanation was adequate to show this was not just a shameless promotion. I think I've come up with a solution, but have been wrong before. I was hoping you could find my errors. The explanation took more than 600 characters (which is all you give me here). In this plan, if you see 10 spotted monks, start counting at the last multiple of 6 instead of 0. In this case, that would be 6. The possible guy who saw nine spots would also start at 6. The possible monk who saw 11 would also start at 6. Nobody else matters. I give more details at theSilentKnight.info. – Silent Knight Jul 24 '16 at 03:53
  • 1
    It appears that your strategy depends on someone being able to go to the exit point, see whether other people are there, and then decide whether to leave or not based on that. This acts as an unauthorized method of communication and changes the problem entirely if it is allowed. – f'' Jul 24 '16 at 17:39
  • 1
    An even faster strategy for your modified problem would be for everyone who sees an even number of dots to go to the exit point on the first day. This immediately separates everyone into a group with dots and a group without dots, and then everyone knows their own status. – f'' Jul 24 '16 at 17:41
  • Yes, in the case that someone sees an exact multiple of 6 spotted monks, it does require them to go to the exit point at a specific time even before they know whether or not they are infected and make a decision based on what they see. I debated the communication aspect, but ruled it out because observing other monks is exactly what they have been doing at every morning meeting. I decided that if that was considered communication, then the whole problem is messed up. At least that's the way I saw it. – Silent Knight Jul 25 '16 at 03:38
  • As far as the even-odd idea, I guess you could be on to something. If the evens went and saw no spots, they would all be ruled out. The odds would still need an exit plan. If the evens went and saw spots, it seems to me they would be in the same boat they were in before going. In my blog, I mentioned that there is most likely a plan that is faster than mine. – Silent Knight Jul 25 '16 at 03:44
  • Hmm, I see that "On Hold" sign is still there. I thought I addressed all concerns in subsequent comments. Oh, do I have to do all of that in the original question? – Silent Knight Jul 28 '16 at 05:57
  • (Didn't see your comments until now.) My point is that you have answered a different interpretation of the problem than the generally accepted interpretation, in which the only information the monks exchange is whether they have already left. – f'' Jul 28 '16 at 07:04
  • For your problem (which, I repeat, is not the same as the most commonly-discussed version), the even-odd strategy works in two days (or even one day if they can choose two different places to meet). The evens meet at the exit point on the first day and the odds meet there on the second day. Whichever group has spots will be able to deduce that and leave on their specified day. – f'' Jul 28 '16 at 07:15
  • OK, duh, I get it now - disregard my comment four comments up about some still needing an exit plan (I mentioned before I still sometimes have trouble wrapping my mind around the notion that we don't need to worry about all the non-existent possibilities). In fact, would the odds have to meet at all? If the evens met on the first day and didn't see spots, the odds would know the second day before any meeting, yes (maybe that's just a technicality. They would still meet on the way out)? – Silent Knight Jul 29 '16 at 03:08
  • About the different interpretation, I guess that's one way of looking at it. Is that bad? Maybe one could argue that the original solvers, due to a lack of imagination, or an assumption that monks were lazy, or to simplify the problem, or something, unduly restricted the scope of the solution. Maybe they did that because otherwise the problem starts to look trivial? I guess the important question would be "What would 'perfect logicians' think?" As I've already mentioned elsewhere, I'm not qualified to answer that question. – Silent Knight Jul 29 '16 at 03:21
  • Closed for being unclear? Really? Did I stutter? Some people seem to understand the question. If my English isn't strong enough, ask them to translate. – Silent Knight Jul 29 '16 at 03:39
  • At most one could argue that the question was not worded as precisely as it should have been. The fact remains that you have found a solution to a different, easier, problem, and the established answer is not wrong. This puzzle is meant to be entirely about logic, not about finding loopholes in the wording. – f'' Jul 30 '16 at 06:03
  • As for the closure, I voted to reopen the question but other users have decided not to. Please also note that I don't get any notification about you replying to my comments unless you use the @[name] syntax (this comment below uses it). – f'' Jul 30 '16 at 06:09
  • @f" OK, sorry about the @[name] thing. I still have a lot to learn. And thanks for the real explanation and consideration. Although I'm tempted to question the "easier problem" part, I probably have a couple other questions (as discussed with @McFry below) to ponder before commenting much further. – Silent Knight Jul 30 '16 at 06:44

1 Answers1

3

Instead of trying to find a better solution, why not just attempt to find a proof that there is no better solution?

The information given on Day 0 is "There is a monk with a blue spot". Suppose there is a monk that sees no blue spots. Then he knows he has a blue spot, and he can act. However, if a monk sees $n > 0$ blue spots, he needs additional information. How can he get this additional information? By waiting until something happens (or not) that actually depends on the blue spot he has.

If there are $n$ blue spots in total, nothing will happen before day $n$.

Proof by induction over $n$:

$n = 1$: The single spotted monk immediately leaves on the first day.

$n-1 \Rightarrow n$: There are $n > 1$ monks with spots. A spotted monk sees $n-1$ spots. He knows that if he does not have a spot, nothing will happen until day $n-1$, by induction hypothesis, and he also knows that if he has a spot, nothing at all will happen before he himself leaves, since all other spotted monks are in the exact same situation as him.

He can therefore act on day $n$ at earliest, since before day $n-1$ he receives no new information whatsoever.

An unspotted monk won't act anyway, in particular not before day $n$. $~~~~\square$

Anon
  • 2,684
  • 1
  • 12
  • 18
  • A single solution would invalidate any proof that there was no solution. I proposed such a solution, Plan B at theSilentKnight.info, and invited all of you to find its flaws (like some of you have already done once before). I gave the link not to promote the blog (which is usually about current events and such, not logic), but because I thought a good explanation would take more than 600 characters. Here's the short version: Start counting not at 0, but at the last multiple of 6. If you see an exact multiple of 6 spotted monks, start heading out today. There may be even better plans. – Silent Knight Jul 24 '16 at 03:40
  • Actually, I agree with this argument, except that more information may be available if you know where to look. My plan complies with your statement about "waiting until something happens (or not) that actually depends on the blue spot he has". That is, if I didn't make another mistake (and didn't cheat). Please let me know which is the case. – Silent Knight Jul 24 '16 at 04:01
  • 1
    So, what happens according to your strategy if there are exactly 6 spotted monks? – Anon Jul 24 '16 at 12:06
  • If there are exactly 6 spotted monks, then six monks will see five, and everyone else will see six. Those who see six (an exact multiple of 6) head to the exit point on Day 1. They see a large number of other monks waiting around with no spots. They go back home and go about their business. That same day, the ones who see five spotted monks start their count at zero, but don't pay attention until Day 5, when they wait to see if those seeing 4 spots leave. On Day 6, they all wake up and realize it's their turn. By this time, those who saw six have known the for days but didn't tell. – Silent Knight Jul 25 '16 at 04:00
  • 2
    @SilentKnight Plan B is effectively trying to circumvent the no communication rule of the puzzle, yes it would work if all the monks knew to use it. However the same will work if all those seeing an even number start at $1$ and all those seeing an odd number start at $0$ or vice versa (and it will work other ways too). Unfortunately the monks cannot communicate their chosen method, so they are back to relying on the fully inductive method that they know all the others, also being perfect logicians, will realise they all should utilise. – Jonathan Allan Jul 25 '16 at 05:14
  • Are you sure that even-odd rule will work? (I haven't yet been able to get it to work for me).Although I start my blog post (theSilentKnight.info) with a story of a possible communication violation, Plan B as discussed above, requires no communication - just the same observational skills required of the monks on a daily basis. – Silent Knight Jul 26 '16 at 03:22
  • [Oops, I guess I hit the wrong key. I wanted to add the following:] In the last paragraph of the blog (titled "Calling All Logicians") I say that there is probably a better plan out there and say what I think it would take (a single plan that is logically better than all others) for the monks to buy in. To preserve the status quo, your mission may be to prove that optimal plan doesn't exist. That proof may or may not be harder than the original proof. Looking for the optimal plan may help you find the proof that it doesn't exist. In the blog I offer a simple spreadsheet, for what its worth – Silent Knight Jul 26 '16 at 04:01
  • Dude, the proof is done. It's right there, in my answer. First thing I posted. Waiting around and letting yourself be seen at the exit point does technically violate the no-communication rule. In the Wikipedia formulation of the question, the six-monks case would not work, since the unspotted monks technically aren't allowed to wake up before dawn (after all, they don't actually know whether they're spotted, they're just acting like they know). – Anon Jul 26 '16 at 09:07
  • Hmm, that Wikipedia version could be a problem. It may sound silly, but on reading that version I didn't take the waking up after dawn as a mandate, just a convenient local custom. But I guess it would be reasonable to ask "then why did they even mention it". Even so, those that see 6 spots can't assume they have no spots, and can't give themselves the luxury of sleeping in. They get up early and pack their bags with full intent on leaving. It is only when they arrive at the exit point that they discover that they broke the law. In that case, when the 6 who really do have the spots . . . – Silent Knight Jul 27 '16 at 04:55
  • . . . wake up, they see they are alone, because every other monk on the island has turned themselves in at the jail. Those 6 would no longer have to count those extra days and would all leave the island the next morning. Eventually the non-spotted monks would be reformed, released, and go on to be productive citizens vowing never to make that mistake again. In your first answer you say "he needs additional information (which he gets) By waiting until something happens...that actually depends on the blue spot he has." The assumption that he must get that info passively could be questioned. – Silent Knight Jul 27 '16 at 05:13
  • Look, anything you're doing just circumvents the abstract puzzle by finding flaws in its analogical visualization. Do you at least agree that my proof holds in the abstract case? Perfectly intelligent abstract agents, who have two abilities: 1) to leave (or not) at each of a fixed discrete common-knowledge set of times, 2) to see at any other time who else is still there and who has left. – Anon Jul 28 '16 at 13:07
  • Although the size of your words might be approaching my upper limit, I think I get it, and I think that's a great question (we may have just started to touch this question above). Perhaps a related question might be "how abstract do we need to get to be fair to the intent of the problem (or maybe not)? To answer these questions, I'm going to need a little sleep and/or "a little help from my friends". Don't go away. – Silent Knight Jul 29 '16 at 03:51
  • I've always thought your proof made perfect sense. Just moments ago a doubt crept into my mind as to whether your proof is rigorous enough to prove what you think it does, but I'm not ready to go there yet. So let's say the proof is still good. Plan B does not discredit that logic. On the contrary, that reasoning is the foundation of Plan B. Plan B added landmarks, but from landmark to landmark, it relies on your proof. To me, the odd/even rule you and one other person mentioned would be the logical end of the landmark path and have the same foundation (although it's not as easy to see). – Silent Knight Jul 30 '16 at 05:21
  • (In fact, based on that, I'm going out on another limb to suggest that as a generalization of the 0 case, the monks would logically pick the evens-go-first option over the equally fast odd-first variation, so it wouldn't be the flip-a-coin type indecision that would paralyze the plan.) But let's get to the important question - did I cheat? Contrary to some, the communication rule was not broken. Having given it more thought, I see no reason for the wake-at-noon rule. Being perfect logicians, I don't see what difference it makes whether a monk watched his fellow monk leave or not. – Silent Knight Jul 30 '16 at 05:43
  • The jail was not an integral requirement of the plan (I look at it as one possible physical implementation of an abstract idea). But your question still remains. Oh shucks, the question about the proof that I'm not ready to answer yet fits right here, which makes the following comments just conjecture. Does the proof really sets n days as the minimum or the maximum? Does it prove that this is one solution or guarantee it is the only solution? Plan B (and successors) suggests different answers to those questions than previously considered. Would the monks, after finding one solution, . . . – Silent Knight Jul 30 '16 at 06:01
  • just give up and move on without looking for optimization? I hate moving forward without fact support, so I'll shut up now. Consider this only food for thought and possibly the direction of my future investigations, or the parts you need to explain better to help me see my error. I didn't set out looking for technicalities to exploit; I wasn't trying to cheat. I do find it useful to change perspective. Like the monks, once I discover (or you explain how) I have cheated, I will turn myself into the local constable (where I expect to see some puzzled expressions). I appreciate your help. – Silent Knight Jul 30 '16 at 06:21
  • 1
    In the six-spotted-monks case, the non-spotted monks prepare to leave on the first day, right? Even though they don't actually know that they have spots, so they shouldn't have reason to actually leave. Acting as if you would leave is exactly the same as shouting "Hey, I see a number of monks divisible by six". By standing at the exit, the monks communicate, to everyone nearby, something about their knowledge that is not just the fact that they know that they have a spot. – Anon Jul 30 '16 at 09:54
  • @Anon You would only be shouting "I see a number of monks divisible by six" to other monks who went to a specific place because they saw a number of spots divisible by six, so this would not be news. You would be communicating the same thing under the standard answer. That you do or do not have a spot on your head IS information useful to the others (which under the standard solution they would have already figured out), but since you don't know that information, you can't be convicted of communicating. – Silent Knight Jun 17 '17 at 04:03
  • @Anon If you meet the requirement (seeing six (or an even number) of spots), this is your only opportunity to leave and avoid violating the rules, so your motives are to obey the law, not gather information. It is the "logical" scheme that allowed for this opportunity to gather information. The question is "would perfect logicians have devised such a scheme?" – Silent Knight Jun 17 '17 at 04:09
  • @f" The inductive proof shown above assumes there is no other way to gather information. Without that (unreasonable) assumption, the proof only gives the upper limit on the time needed. I imagine that faced with, say 40 visible spotted monks, perfect logicians would have plenty of time to question that assumption. I am convinced your "even on Day 1" rule is the logical solution they would all gravitate toward, and if they were perfect logicians, they would arrive at that conclusion before sundown. – Silent Knight Jun 17 '17 at 04:20
  • if the original problem were rewritten to require the monks to commit hara-kiri quietly in their rooms, the new plan wouldn't work. Would that really be a better problem? To say the new answer is more trivial suggests that everybody would have considered it right away and only discarded it because it didn't meet the requirements. Is that really the case? – Silent Knight Jun 17 '17 at 04:26