Question: How many monks died from the disease?

[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?

 

[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.