# Faster Solution for Blue-eyed Monks [closed]

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.

## closed as unclear what you're asking by Deusovi♦, Gamow, Fabich, Alconja, AlexisJul 23 '16 at 14:32

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• 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 9:26
• 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 9:37
• 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
• 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

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 $n$. $~~~~\square$

• 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 3:40
• 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 4:01
• So, what happens according to your strategy if there are exactly 6 spotted monks? – Anon Jul 24 '16 at 12:06
• @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 5:14
• 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 9:54