Question: How many monks died from the disease?

[davee123]

One question, does the solution given solve the problem given my criteria? And not your assumptions?

Wait, what solution? My solution does, I don't see how yours does.

For starters, why do you say that 2 monks die on the first day? And why does another one die on the 2nd day? And after which meals do they die? What is the reasoning of the first two monks who kill themselves, and that of the 3rd monk that dies? How do they successfully deduce that they each have the disease, and how many monks are infected at each stage, given that you seem to be claiming that that state changes?

I guess-- here's one thing I'd like to see: Make me a little table that shows how many monks you're saying are infected at each mealtime. Something like this:

Meal 1: Monks infected: 2 Monks who learn they are infected: 0
Meal 2: Monks infected: 2 Monks who learn they are infected: 2
2 monks die.
Meal 3: Monks infected: 1 Monks who learn they are infected: 0
Meal 4: Monks infected: 1 Monks who learn they are infected: 1
1 monk dies.
Meal 5: Monks infected: 0 Monks who learn they are infected: 0
Meal 6: Monks infected: 3 Monks who learn they are infected: 0
Meal 7: Monks infected: 3 Monks who learn they are infected: 0
Meal 8: Monks infected: 4 Monks who learn they are infected: 4
4 monks die.
etc.

Again I am reminded of the last one where I presented a solution and you felt the need to go on for pages and pages about x and y that aren't specified?

The truth is I want us both to be on the same page here. From what I can see, you posted a solution that was incorrect, and I'm not sure where you went astray. I'm interested in hearing you explain your position so that I can help clarify the point. And if I AM making a mistake in my assumptions, I'd like to know how and why. If that takes pages and pages, I'm fine with that. Understanding is worth taking time.

Again your unavoidably assuming that given the criteria the original solution fails because you say it does, and because you've made up new ideas?

I'm saying there are multiple ways of stating the same problem, and some are better than others. In this instance, I very strongly believe that there's a particular logical method which is attempting to be demonstrated. I can explain that point in very abstract terms, if desired, but it's boring. So usually we like to dress these up as word problems. Hypothetical situations which are interesting to us because there are people and things involved we can relate to. But as soon as you make that leap from purely logical to "realistic", assumptions of all kinds jump out and haunt you.

For example: In your original problem, you did not explictly state that the disease does not infect anyone else. So, I'm *forced* to assume something. I can assume that the disease does not infect anyone else, or I can assume that it MIGHT infect someone else. Either way, it's an assumption. I utterly HAVE to assume something. And it's pretty unclear which way I should assume things! However, the clue about the disease not being contagious tells me that the author *probably* intended to clarify my assumption by stating that nobody else is infected.

If you'd like to dispute which assumptions are acceptable and which aren't, that's fine, and we can even talk about what other solutions might present themselves if we make other assumptions. But it gets really sticky when we *don't know* what the other person's assumptions are.

Please quote my post and the original solution and show how it is wrong,

Heh, alright, I'll try. But keep in mind that 90% of my problem is that I don't understand what your solution is! You may do better not to even read this, but here goes:

===================================================
BEGIN POST SNIPPIT OF SOLUTION
===================================================

Assuming that the only time the monks can be sure that they have it is if only one monk has it

This is wrong, because, as I showed, if only two monks are infected, they are capable of discovering at the 2nd meal that they are infected. Each monk knows that there is at least one monk with the disease, and each monk has seen one monk with the disease. If they had NOT seen any monks with the disease, they would know they were infected. If that had been the case, the monk they were looking at would have died. Quite clearly, that has not happened, and so they realize that it was NOT the case that ONLY the other monk was infected. Therefore, they know that the only other possibility is that THEY are infected.

ergo he dies on the second meal. We then say one more has it

Ok, clarification please. "One more has it"? One more has it when? After the 2nd meal? On the very first day? Has the other monk you just mentioned died, or not?

and one monk is dead and so on until the 11th day after the messenger arrived.

You appear to be suggesting that the only way for monks to die is if they are, at any time, the only monk to have the disease. Hence, you appear to be stating that in between EACH meal, one monk dies, and one new monk becomes infected. This is clearly wrong, because after the first monk(s) die, the remaining monks are not assured again that at least one monk is still infected.

However if you have 2 monks then they find out after 2 meals, if 3 after 3 meals and so on with subsequent deaths leaving potentially no monks infected,

This is actually correct!

however the disease may continue to show up after the first day if it has an incubation period of 1 to x days,which of course all diseases do.

While possible if assuming that the disease can infect new people, the remaining monks have lost their valueable piece of knowledge that at least one monk is infected, and therefore will be unable to make any conclusions after the number of monks infected begins to change, or after any infected monks die.

Since we cannot be sure when the disease presents the symptoms, the only way we can be sure that at least one monk dies on the last day is if one monk dies after every meal with one infected.

Again, this is not possible, because as soon as a single monk dies who was infected, the remaining monks are deprived of the information they had. Part of their information derives from the fact that known infected monks are somehow not killing themselves, and the REASON they're not killing themselves is due to what they can observe. If they can no longer relay that information, the remaining monks' information source is drained.

It is possible that 33 are infected but then no one else must present with the disease, 33 would then die on the last day but then the answer is the same anyway not to mention this is unlikely.

Ok, so, you just sort of solved the problem, but dismissed it due to its being unlikely?

The total is therefore the number of meals:-

2 die on the first day
3 for the following 10 days
1 on the morning of the 11th day after the messenger arrives.

Total 33.

Again, this solution is unclear. Days are rather irrelevant-- only meals are particularly relevant. Which meals for which days do you mean?

Also, this is incorrect. If one dies on the morning of the 11th day, that would have to be *AFTER* breakfast (because if they died from knowing, it would have had to have been after DINNER, which would be in the evening of the 10th day), and the question quite clearly stipulates that the monks do not appear for breakfast on the 11th day, so they are already dead, making the total be 32 instead of 33.

You have two bits of information here that you should adhere to, at least one monk has it,at least one monk dies on the last day, if 3 monks have it then it will take 3 meals before they realize that they have it,

Ok, that's correct, but ONLY if no other monks have died yet, and no new monks have been infected. If new monks have been infected, or if other monks have died, they CANNOT realize they have it.

and you have to assume no one else gets it. Otherwise the system is screwed

Wait, why is it screwed at this point, instead of earlier?

===================================================
END POST SNIPPIT OF SOLUTION
===================================================

The onus is on you to prove your solution correct given my criteria and the solution I present as wrong, I aint going to go round searching for anything until you do it. Think about it.

Done.

But I fail to see why the onus isn't on whoever wants to do the understanding. If you want to understand where the disagreement is, the onus is on you to try and understand it-- ask questions, make points, whatever. If you don't care about understanding, then you can do whatever you want. But if you geninuely want to understand, you'll try and put some effort in.

DaveE

 

[Schrodinger's Dog]

Your assuming the monks who cannot possibly know don't figure out the pattern 1 on 1, anyway, I haven't the time to address your questions. anon.

I'll try and think about it on the morrow. You still haven't answered how you can with no ability to predetermine a causal pattern in the disease, or otherwise having no prognosis determine your answer is right.

Suffice to say I think we're getting somewhere here, maybe the aholes who presented the problem are indeed wrong, and by default so am I, frankly I'd be happier with not being right than being right. Not my problem but no time to consider it, hope we're all having fun.

 

[Hurkyl]

For example, suppose only one monk displayed symptoms at the beginning. He dies two hours after the first meal.

You then assert:

We then say one more has it and one monk is dead​

but you give no reason why we can say another one has become infected, nor have you given a reason why that monk can determine he is infected.

Put yourself in the monk's place: you saw one red spot at the first meal. At the next meal, you found out he has died, and you see no other red spots. What can you possibly glean from this information to distinguish between the case that nobody has the disease during the second meal and the case that you have the disease during the second meal?



The classical solution, which still works with a spreading disease, is a simple inductive argument on the number of meals.

At the N-th meal, if M people displayed symptoms at the beginning, exactly one of the following is true:
(1) M < N, and all of the people who displayed symptoms at the beginning have died, and nobody else has died.
(2) M = N, nobody has died yet, but all of the people who displayed symptoms at the beginning will die in two hours.
(3) M > N, nobody has died, and nobody will die before the next meal.

The base case is obvious. The inductive step is straightforward: each monk (who saw that k other monks had a red spot at the beginning) considers whether the following hypotheses are consistent with his observation:

(1) I am infected
(2) I am not infected

And it's easy to see from the inductive hypothesis that the only time that (1) is consistent with observation and (2) is inconsistent with observation is if k = N-1, and nobody has died yet.

 

[Schrodinger's Dog]

For example, suppose only one monk displayed symptoms at the beginning. He dies two hours after the first meal.

You then assert:

We then say one more has it and one monk is dead​

but you give no reason why we can say another one has become infected, nor have you given a reason why that monk can determine he is infected.

Put yourself in the monk's place: you saw one red spot at the first meal. At the next meal, you found out he has died, and you see no other red spots. What can you possibly glean from this information to distinguish between the case that nobody has the disease during the second meal and the case that you have the disease during the second meal?



The classical solution, which still works with a spreading disease, is a simple inductive argument on the number of meals.

At the N-th meal, if M people displayed symptoms at the beginning, exactly one of the following is true:
(1) M < N, and all of the people who displayed symptoms at the beginning have died, and nobody else has died.
(2) M = N, nobody has died yet, but all of the people who displayed symptoms at the beginning will die in two hours.
(3) M > N, nobody has died, and nobody will die before the next meal.

The base case is obvious. The inductive step is straightforward: each monk (who saw that k other monks had a red spot at the beginning) considers whether the following hypotheses are consistent with his observation:

(1) I am infected
(2) I am not infected

And it's easy to see from the inductive hypothesis that the only time that (1) is consistent with observation and (2) is inconsistent with observation is if k = N-1, and nobody has died yet.

To be honest I've kind of given up on this, it's obvious that the problem although not mine is deeply falwed and the answer given by the particular website is completely wrong. Let's just leave it at that, I really don't know why they have specified that as the answer.

I have suposed this is because given there are 1-34 number of monks present with systems on the first day and then x for the next few days, ie there is no reason to assume they all have the symptoms on the first day, if this is the case then the only solution I can see is if the monks use the counting method, but this doesn't gel with the answer. So I guess either the problem is deeply falwed, or there is something missing from the presented problem, either way there's obviously something wrong here so I suggest we just forget about it, I was misnformed by some dubious web site.

 

[davee123]

To be honest I've kind of given up on this, it's obvious that the problem although not mine is deeply falwed and the answer given by the particular website is completely wrong. Let's just leave it at that, I really don't know why they have specified that as the answer.

Could you perhaps post a link to the place where you found it? As it is, the problem isn't deeply flawed, it's just badly worded. And it's not without a solution-- it just kind of sounds like your interpretation of the answer isn't quite right.

DaveE

 

[Schrodinger's Dog]

Could you perhaps post a link to the place where you found it? As it is, the problem isn't deeply flawed, it's just badly worded. And it's not without a solution-- it just kind of sounds like your interpretation of the answer isn't quite right.

DaveE

Someone else got it from some web site, so I'd have to contact them, I'll have a go anyway. That is the answer given and I pretty much cut and pasted the problem, or at least directly copied it.

 

[davee123]

Someone else got it from some web site, so I'd have to contact them, I'll have a go anyway. That is the answer given and I pretty much cut and pasted the problem, or at least directly copied it.

Here's a few variations of the problem that I found:

http://ai.eecs.umich.edu/people/dreeves/brainteasers/archives/
https://www.physicsforums.com/showthread.php?t=6845&page=4&highlight=monk
http://www.everything2.com/index.pl?node_id=882867
http://forums.warriorsworld.net/other/msgs/111923.phtml

DaveE

 

[davee123]

there's a particular logical method which is attempting to be demonstrated. I can explain that point in very abstract terms, if desired, but it's boring. So usually we like to dress these up as word problems. Hypothetical situations which are interesting to us because there are people and things involved we can relate to.

To expound the point, this is the same problem in a more "pure" format:

-------------------------

A group of N entities exists. A given entity must be in exactly one of 3 states, A, B, or C. Initially, all N entities are in states A or B, none are in state C. These states do not change from A to B or B to A. All entities observe all of each other exactly once in every given period. Entities observing other entities know the state of the other entity. Entities are not capable of observing or learning their own state directly. They are only able to determine their state through logical deduction. If any entity that is in state B has enough information to deduce that they are in state B, they will change to state C at the immediate end of the current period. No entity will change from state A to state C, and no entity will change from state C to state A or from state C to state B.

All entities are informed at the immediate start of period P that at least one
entity is in state B. At the end of the P+Kth period, some number of entities
change to state C. How many entities changed to state C?

----------------------

So, this puzzle has fewer holes, because it doesn't reflect reality. But it's boring as all get-out. Replace the entities with monks, state C with death, or state B as a "disease" and suddenly we've got all kinds of crazy "reality" problems we have to deal with, which aren't really supposed to be part of the riddle.

DaveE

 

[Schrodinger's Dog]

To expound the point, this is the same problem in a more "pure" format:

-------------------------

A group of N entities exists. A given entity must be in exactly one of 3 states, A, B, or C. Initially, all N entities are in states A or B, none are in state C. These states do not change from A to B or B to A. All entities observe all of each other exactly once in every given period. Entities observing other entities know the state of the other entity. Entities are not capable of observing or learning their own state directly. They are only able to determine their state through logical deduction. If any entity that is in state B has enough information to deduce that they are in state B, they will change to state C at the immediate end of the current period. No entity will change from state A to state C, and no entity will change from state C to state A or from state C to state B.

All entities are informed at the immediate start of period P that at least one
entity is in state B. At the end of the P+Kth period, some number of entities
change to state C. How many entities changed to state C?

----------------------

So, this puzzle has fewer holes, because it doesn't reflect reality. But it's boring as all get-out. Replace the entities with monks, state C with death, or state B as a "disease" and suddenly we've got all kinds of crazy "reality" problems we have to deal with, which aren't really supposed to be part of the riddle.

DaveE

I think that's one of the problems, the reality got lost in translation maybe between the website and another forum? Anyway, I'll find out what the missing part is once I can get in contact with the guy who set the problem, if there is one...

 

[Wizardsblade]

I think I see one problem in logic used:

If x>1 monks are infected then each infected monk will see x-1 infected monks. The infected monks will also know that all the other infected monks see x-1 dots as well, so after the second meal when none die all infected monk will die. Realizing that if the x-1 did not die they must be infected. By the 2nd meal all infected monks die everytime.(of course only 1 meal if only 1 monk is infected.

I could not follow some of the other logic used to solve this problem, and this problem seems to have self sustaining problems.

 

[Hurkyl]

I think I see one problem in logic used:

If x>1 monks are infected then each infected monk will see x-1 infected monks. The infected monks will also know that all the other infected monks see x-1 dots as well, so after the second meal when none die all infected monk will die. Realizing that if the x-1 did not die they must be infected. By the 2nd meal all infected monks die everytime.(of course only 1 meal if only 1 monk is infected.

I could not follow some of the other logic used to solve this problem, and this problem seems to have self sustaining problems.

You are a monk. You see 9 other dots. When you retire to your chambers after the second meal, how do you figure out whether x = 9 or x = 10[/color]?

 

[Wizardsblade]

Yea sorry it hit me in the middle of the say today that I was wrong =/. Brain must have been off last night. =)

But if I was a monk I would not be trying to figure it out ;), is not that really the logical thing to do?

 

[Gelsamel Epsilon]

http://www.xkcd.com/blue_eyes.html
Edit:

http://xkcd.com/solution.html

That is the solution to the blue eye's riddle.

DO NOT CLICK THIS IF YOU STILL WANT TO SOLVE THIS MONK PROBLEM, as they are essentially the same.