Who is the liar in this logic puzzle?

[Eyesaw]

Originally posted by elibol
yes, and he would commit suicide the first night.


what? what was the point in pointing this out?
this has already been apparent thru out our argument.

it is obvious that the amount of marked monks determines the day of suicide.



wrong. why would it take them 7 days to notice that 7 were marked?

it states in the puzzle:
"At the second day's prayer meeting the monks all look at each other, but again that evening nothing unusual occurs."

why would they bother to examine each other if all of them had the markings? why would they sit around and wait 7 days? i cannot bother with this anymore. you are missing the fact that there is no logic in waiting 7 days when they all have markings.



what? explain this if you care to. this doesn't make logical sense that they would all wait 7 days.



yes but you are still assuming a given amount to begin with. you are saying 7... bah! you arent making any sense...

im sorry but until you cease to repeat yourself i will nolonger continue to argue with you. you are not being the least bit rational.

Read the problem again. It says that the monks could not tell each
other who has marks on their heads. So again, look at the simplest scenario where there are only two monks, both with marks on their heads. The first monk sees a mark on the second monks head but he cannot tell whether his own head is marked or not. From the second monk's point of view, it looks the same. So even though both monks are marked and there are only two of them, they cannot know on the first day whether they both have marks.

The only way for one of the monks to know he himself is marked is by the fact that the other monk doesn't commit suicide on the first evening.

If there are 3 possessed monks, it would take them 3 days to figure out they all have marks on their heads. And if there are 4, it would take 4 days.. et al.

 

[elibol]

i was the first to explain the answer man...

it isn't like i don't understand something. its more like you keep adding these weird things to your explanations.

like some crap about 7 monks having marks when the dude that could speak came and spoke.

why you keep dodging what you said before?

i am done.

 

[Eyesaw]

Originally posted by elibol
i was the first to explain the answer man...

it isn't like i don't understand something. its more like you keep adding these weird things to your explanations.

like some crap about 7 monks having marks when the dude that could speak came and spoke.

why you keep dodging what you said before?

i am done.

Nowhere in the original problem does it state that monks are being possessed every passing day, you made up that assumption. But even if a million monks were being possessed every passing day, the
only relevant point is how many monks were marked on the first
day. It takes 7 monks 7 days to figure out they all have marks on their heads.

 

[Eyesaw]

I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide.

 

[elibol]

umm, yep. you sure do.

there is no solution without assumption in this puzzle... assuming there are any number of monks anywhere in your answer is an assumption. what the hell are you talking about? all i hear is bla bla bla...

my assumptions make logical sense, in that the conclusion of the puzzle is solved.

all i hear you say is 7, 7, 7... lucky number 7.
you stand corrected?

you arent the least bit rational.
if you are flawed (trust me, everyone is flawed at one point in their life or another), how will you ever know this?
your ignorance is impossible to deal with.
obviously it is an impossible task to try and convince you to be rational.

you will never be taken seriously if you are not rational.

you can stand corrected, in your room, wherever you are, by yourself. you are the only one that agrees that you are correct. it seems your in your own little world trying to prove people wrong. does it feed your selfesteem to prove others wrong? is your life so depressing that you feed off of the frustration of others inpatience to deal with your ignorance? it probably satisfies you that people give up against your ignorance. it is an obvious paradox, but from your perspective this is not obvious. you just love the feeling and don't really care to take a look at yourself for one second and realize your way out of line.

is it so hard for you to prove people wrong in a logical sense that instead of being rational you have to be ignorant enough to keep writing on and on? until the person you argue with gives up?

arguing over the internet is like running the special olympics.
whether you win or not, you are still RETARDED FOR ARGUING.

state your opinion, and sit down. let others decide whether you "stand correct".

 

[Eyesaw]

Originally posted by elibol
umm, yep. you sure do.

there is no solution without assumption in this puzzle... assuming there are any number of monks anywhere in your answer is an assumption. what the hell are you talking about? all i hear is bla bla bla...

my assumptions make logical sense, in that the conclusion of the puzzle is solved.

all i hear you say is 7, 7, 7... lucky number 7.
you stand corrected?

you arent the least bit rational.
if you are flawed (trust me, everyone is flawed at one point in their life or another), how will you ever know this?
your ignorance is impossible to deal with.
obviously it is an impossible task to try and convince you to be rational.

you will never be taken seriously if you are not rational.

you can stand corrected, in your room, wherever you are, by yourself. you are the only one that agrees that you are correct. it seems your in your own little world trying to prove people wrong. does it feed your selfesteem to prove others wrong? is your life so depressing that you feed off of the frustration of others inpatience to deal with your ignorance? it probably satisfies you that people give up against your ignorance. it is an obvious paradox, but from your perspective this is not obvious. you just love the feeling and don't really care to take a look at yourself for one second and realize your way out of line.

is it so hard for you to prove people wrong in a logical sense that instead of being rational you have to be ignorant enough to keep writing on and on? until the person you argue with gives up?

arguing over the internet is like running the special olympics.
whether you win or not, you are still RETARDED FOR ARGUING.

state your opinion, and sit down. let others decide whether you "stand correct".

I think you are overreacting a little here. All I did was
gave my solution to gnome's puzzle and you make me out as if
I were the evil demon in the problem. I don't understand what
your objection is to my solution. Where am I making assumptions
that are not already stated in the problem? We know there is
at least one monk there from what the visitor said about the
marks on the monks.

You made assumptions that weren't stated in the original problem,
so your solution isn't even valid. Find where the problem tells
us to assume each passing day, new monks were being possessed?
The assumption is not even relevant to the problem.

You are the one out of line here for calling me ignorant and all sorts of names when you can't even produce a good reason why my solution is wrong.

 

[one_raven]

Eyesaw is correct.
Let me try...

If you are 1 of 2 monks, you will look at the other monk.
If he has no mark, then you will kill yourself that night since you know you are posessed.
If he DOES have a mark, then you may or may not have a mark.
Keep in your mind that he is thinking the EXACT same thing.
The next day, since he did not kill himself, you realize that he was waiting to find out just like you were, so that dictates that you must have the mark as well, that night you both kill yourselves.

See how that works so far?

If you are 1 of 3 monks you will look at the other two monks to see if they are marked.
If neither of them are marked, then you must be, so you kill yourself that night.
If one of them is marked, then you wait till tomorrow since if teh marked one does kill himself then you know the he did not see the mark on the others.
If one of them is marked, and on the second day no one is dead, that means you are marked, so that night you will kill yourself.
That also rings true for the other one since no one was dead on day two and you know that only one of the others are marked.
On day three, the monk that saw that you were both marked is still alive since he did not know if he was marked.
He knew that if he was not marked you would both kill yourselves on day two.
So, if you were not both dead on the morning of day three, then he must be marked as well so all 3 would kill themselves that night.

They all look at each other every day for two reasons.
1. To see who is still alive.
2. To make sure that the markings haven't changed.

Now if you continue that logic to seven days the only way all the monks would be dead is if there were seven of them.
If there were only 6, then they would have all been dead by day 6.
If there were more than 7, then it would have taken more than seven days to figure it out.

Personally, i would have said, "Fucx this vow, am I marked?"

 

[Eyesaw]

Seems gnome's puzzle was a spin off of one called
the "blue-eyed monks".

Here's a solution given from a website:

http://ai.eecs.umich.edu/people/dreeves/brainteasers/archives/

I wasn't nuts after all.

 

[one_raven]

Originally posted by Eyesaw
I wasn't nuts after all.

Now who's making assumptions?

 

[gnome]

Well done, Eyesaw. I knew you could do it -- it didn't require any fractions.

I hope you enjoyed that debate as much as I did reading it. By the way, I didn't get it from the U Mich. site. I guess we both got it from some other source, several times removed. I couldn't find where I originally saw it, so I was writing from memory. I may have embellished it slightly, but my source definitely involved a demon and marks on the forehead.

Oh, and one_raven: nice explanation.

 

[elibol]

sorry eyesaw, but when you said:
"I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide."

that really pissed me off...
i take back my remarks.

also, note: i have taken out a large amount of writing that was once this post, i now realize it is unessecary based on what i have concluded to.

 

[elibol]

Originally posted by gnome
I hope you enjoyed that debate as much as I did reading it.

what does this mean?

 

[leto]

Elibol, Eyesaw didn't explain it very well, but I think you might want to stop and think. You appear to me to be making a fool of yourself much like eyesaw did the other day. Don't worry about the criticism, everyone is wrong from time to time.

I think one_raven did a good job at explaining it, but I'll take a stab in my own terms since the concept wasn't easy for me to grasp either.

If there were three monks that all had a marking on their heads, each would see two other monks with a marking on their head on day one. At this point, none of the three monks would be certain there was a mark on his head. On the second day, the monks waited for the two they observed to have markings to kill themsevles; the two monks would realize they had a marking on their head if the observing monk did not. When all three monks came back on the 3rd day, all three monks would realize they had a marking on their heads.

 

[elibol]

i already know this.

this makes perfect sense man...

its much deeper than this though... read my posting on page 7 please. everyone read it, if you care to look into this puzzle further. don't turn your heads the other way, because without what i wrote on page 7, the means of getting to the solution still has holes even though the logic the monks use makes sense...

 

[elibol]

ok, i was wrong, and i was able to figure this out with the blue-eyed monks version's follow up question.

it is essential to accually understanding the purpose of the announcement, and for me, to understanding how the entire problem makes any sense.

it states:
"It's clear that the visitor was necessary in the
proof in order to establish the base case. However, the only information
that the visitor provides (there exist blue eyes on this island) is
something that (assuming more than 1 blue-eyed monk) every single monk
already knew. The question: What was different after the visitor's
announcement? Ie, as a monk on the island, what do you know after the
visitor's announcement that you didn't know before?"

this is what i am trying to say, what does the monk know now that he didnt know before the announcement?

i will answer this after i have addressed my former flaw:

lets take now the 7 marked monks, i will explain it now the way i wanted it to be explained for me to understand it. i knew there was something i was missing, but the flaw in gnomes version (the pointlessness of the announcement) caused me to come to some alternate possible case scenario with an assumed aspect (the daily incrementation).

if i had read formerly the blue-eyed monks version i wouldn't have had this problem.

there are 7 marked monks. each one see's 6 other marked monks, and each one assumes the other monks only see 5 marked monks based on the assumption that he is not marked. now he must look from the perspective of the 5 monks, to know whether they are marked or not they must assume first that they are not, thus seeing only 4 marked monks. he must now look from the perspective of the 4 monks, to know whether they are marked or not they must assume first that they are not, thus seeing only 3 marked monks. he must now look from the 3 monks point of view, to know whether they themselves is marked they must first assume that they are not, thus seeing only 2 marked monks. he must now look from the two marked monks perspective, to know whether he himself is marked, he must first assume that he is not, thus seeing only one marked monk. when this monk does not commit suicide on the first day, he now knows he is marked. thus the addition to his knowledge base on how many monks know that themselves are marked becomes two.

i think now that the assumption they all have of not being marked causes this logical solution to be possible.

i believe the announcement marked a starting point in which all of them could take advantage of this logical method to figure out whether they were blue-eyed or not.

allowing them to begin their evaluation of the situation at the same time.

an attempt to figure out the color of their eyes anytime before this would be impossible, since they have no way of telling when they can start the countdown to get an accurate count on how many monks knew the color of their own eyes.

it is now obvious that it would be illogical for a monk to just begin counting down days with no starting point to base his information on.

 

[leto]

I just finally got the nerve to attempt to explain it with 7, and you go and burst my bubble.

Edit: Anyway, my bad for not reading what you wrote before I replied to you, but I thought the analogy with 3 made the concept easier to understand. Not much changes between 3 monks/3 days, and 7 monks/7 days. It also takes much less explaining.

 

[gnome]

Elibol, I really don't know what I did to make you sick. I certainly had no malicious intent in posting this puzzle. I don't see that there is anything wrong with enjoying a debate. If we all agreed all the time, what would be the point in talking?

And I do not agree that my version of the puzzle is flawed, either. In fact, some of your own objections highlight the fact that the U.Mich. version is flawed. That is, blue eyes are obviously a permanent feature (anybody with blue eyes has always had blue eyes), so your point about what information was provided by the visitor is well taken. Also, the U.Mich. version says they will commit suicide "by sunset", thus allowing the possibility of someone committing suicide immediately at prayer meeting, which can screw up the sequentiality of it. My version specifies that suicides only occur in private in the evening, so the information about the suicides is only available the next day. For these reasons, I think that the U.Mich. version is ambiguous.

My version, on the other hand, allows the interpretation that the marks first appeared on "day 1" when the visitor arrived. Yes, you are correct in observing that if 6 out of 7 monks were marked that day, they would all see marks on either 5 or 6 of their fellows. But the puzzle doesn't tell you how many there are, or how many are marked. So the visitor is certainly necessary for the benefit of the puzzle. IF there were only two or three monks, and IF only one monk (out of any total) was marked, the visitor is necessary for the monks as well. As far as the question of what knowledge is provided by the visitor in the case where there are 7 monks, all marked, I haven't heard or thought about that question before now, and I'll admit I don't know the answer. But if the number of monks is unknown, and the number of marked monks is unknown, and you don't know if any of the monks know if any of the monks are marked, there is no puzzle.

Anyway, while the last few posts were going up, I've been busily working on this explanation which I see is now redundant, but since I spent all this time on it, I'll throw it up here anyway. Maybe it will help someone else.

I still believe that a total of 7 monks, all marked beginning at the moment of the visitor's announcement, is the correct answer. I probably can't explain it any better than the others have done, but I'll try. The key to it is the dimensionality. Monk A is trying to deduce what monk B (and every other monk) is thinking, knowing that monk B is trying to deduce what monk C (and every other monk) is thinking, and so on. So it becomes one of those picture in a picture in a picture in a ... situations. Monk A thinks that Monk B thinks that Monk C thinks that ... etc. I can't see 4 dimensions, let alone 7.

1 of course is trivial. If there is only 1 monk, & the visitor says "at least 1 is marked", the monk commits suicide that evening & the game is over.

2 is not much harder. Monk A tells himself, "only 1 of us is marked, so if I am unmarked, B will see that, deduce that he IS marked, and he will commit suicide on evening 1. When Monk B appears at prayers on day 2, Monk A knows that B sees the mark on A, so A commits suicide on evening 2. You can see that my choice of who is A and who is B is completely arbitrary, so swap the letters, & now BOTH commit suicide on evening 2. Note that this is true only if IN FACT BOTH ARE MARKED. If only Monk B is marked, he sees that A is unmarked and commits suicide on evening 1. This confirms in A's mind that he is unmarked so he does not commit suicide. Period. Either way, the condition that "all of the monks commit suicide on evening 7" does not occur, so 2 monks is not a valid answer.


Let's look at 3.

IF there are 3 and all 3 are marked, on day 1, Monk A sees that his two brethren are marked. Monk A wants to believe that he is unmarked. He knows that Monk B wants to believe that HE is unmarked. Monk A thinks, "If I am unmarked and Monk B believes that he is unmarked, he (B) must assume that C is the only one marked. He (B) would then expect C to commit suicide on evening 1, and therefore he (B) would expect that C would not show up at prayers on day 2. Since C was still alive on day 2 (A reasons), B would have to conclude (on day 2) that C does NOT see two unmarked monks. So, if C is still alive on day 2, he must see at least 1 marked monk. If I (A) am unmarked, B sees that, and therefore he will know on day 2 that HE (B) is that marked monk, so he will commit suicide on evening 2. If B shows up at prayers on day 3, then he must see that I (A) am marked, so on day 3 I commit suicide.

If all of the monks are marked, they all follow the same reasoning, and all commit suicide on evening 3. If only 1 was marked, he would commit suicide on evening 1, and the other two would know that they are unmarked. If 2 were marked, following the same reasoning they would both commit suicide on evening 2, saving the 3rd.

If there are 4...
I don't have time to take this to 4. I'm not sure that it can be described adequately verbally. And even if I did, you'd say "prove it for 5". I believe the inductive proof already described on the U. Mich. site is valid for my scenario, even more than it is for theirs.

If you don't agree, I guess none of us is going to convince you. I hope you - all of you - enjoyed puzzling over it anyway.

 

[elibol]

Originally posted by leto
I just finally got the nerve to attempt to explain it with 7, and you go and burst my bubble.

thanks.

i admit i was wrong. but i don't think i ever would have figured it out or understood it as fully without the blue-eyed version's follow up question...

it might sound a bit pathetic to some of you, but the process it took for me to get to the solution to this problem will be a pretty big asset to my problem solving skills...

know that I am amateur to these problems, and if I am wrong I am humbly requesting that if i ask of someone to explain the problem the way i need it to be explained i would greatly appreciate it. otherwise it seems i keep going and going and... yall saw what happened =[

 

[elibol]

well, i guess the time it took you to post that i had already posted that we have a mutual agreement on how the monks logic work, and fixed a lot of what i said before... you don't make me sick, i have corrected this posting as well.

i was utterly frustrated, and i admit, a bit childish.

thank you for the puzzle, it was very enlightening.

 

[one_raven]

gnome,
I have a suggestion.
The first time I read this I thought, "That visitor must be a prick! Why wouldn't he just tell which monks are marked, knowing that none of them can communicate with each other?"

Maybe next time you can avoid such confusion (about whether the curse spreads and why the visitor's announcement is important) by changing one small tidbit. The visitor is the one who curses them. Maybe it is a demon or devil of some sort. He visits teh island and says, "I will place a curse on at least one of teh monks on this island. I will not tell you which ones, because I do not want you to know, but every one else will know because there will be a mark on your forehead. You will live the rest of your life not knowing whether or not you are cursed." Or something like that. Then he disappears. Seven days later, all the monks are dead of suicide.

I think that makes it a seamless puzzle.
What do you think?

 

[Eyesaw]

Originally posted by one_raven
gnome,
I have a suggestion.
The first time I read this I thought, "That visitor must be a prick! Why wouldn't he just tell which monks are marked, knowing that none of them can communicate with each other?"

Maybe next time you can avoid such confusion (about whether the curse spreads and why the visitor's announcement is important) by changing one small tidbit. The visitor is the one who curses them. Maybe it is a demon or devil of some sort. He visits teh island and says, "I will place a curse on at least one of teh monks on this island. I will not tell you which ones, because I do not want you to know, but every one else will know because there will be a mark on your forehead. You will live the rest of your life not knowing whether or not you are cursed." Or something like that. Then he disappears. Seven days later, all the monks are dead of suicide.

I think that makes it a seamless puzzle.
What do you think?

Well, agreed- that visitor was a prick. But I think gnome wrote a clever version of that blue-eyed monk puzzle actually. And I don't think there are any loose ends. The key clue in the puzzle was that all the monks committed suicide on the same day. With this knowledge and also knowing the day of the suicide, it's kind of implicit that the puzzle wasn't talking about monks getting possessed every passing day but about how many monks were marked at the time
of the visitor's announcement.

Let's try to solve the problem the other way to see why it is
not possible. Let's say on the first day, two monks are marked but one isn't. Let's also assume that on the second evening the demon will come to possesses the third monk. Well, since the two monks marked on the first day have already concluded by the second day that they both are marked, they will kill themselves that evening, so they won't even live long enough to see the mark on the third monk. And once they are dead, the third monk would have no way of knowing if he is marked, so he will live on.

Even if the third monk was marked by the morning of the second day, it still won't prevent the first two monks from committing suicide
on the second evening since their deductions of marks on their heads only depended the marks they saw on the first day.

So you see, if the monks were getting possessed each passing day,
there will always be at least one monk left alive. So the only way all 7 would be dead if they were all marked by the first day.

BTW, I had never saw this puzzle before and I solved it correctly
the first time so it must have been written appropriately. I only bothered to look for second source for the solution to convince a doubting elibol.

 

[gnome]

I agree with Eyesaw (of course, since he agrees with me ) but I'll take this opportunity to be a prick & say, Why should I tell the reader that all the monks become infected simultaneously? Part of the puzzle is to figure that out for yourself.

Eyesaw, you get a gold star for being such a discerning judge of puzzles.

Oh, and elibol: no sweat. We all get aggravated at times, as I did the other day when Eyesaw was enjoying hassling me about* the birthday puzzle.


*[my correct solution to]

 

[Eyesaw]

Originally posted by elibol
sorry eyesaw, but when you said:
"I stand corrected. If elibol was one of the monks,
it would take him quite a few years to commit suicide."

that really pissed me off...
i take back my remarks.

also, note: i have taken out a large amount of writing that was once this post, i now realize it is unessecary based on what i have concluded to.

I thought it was a good joke, my bad. From your first explanation
of the puzzle, I don't doubt you would've solved the problem correctly if the assumptions were made more explicit. I probably just got lucky because the first thought I had when I read the line " all monks committed suicide on day 7" was that they must've
all been marked on the same day. And I also had a strong hunch that 7 was going to be a key number- it just seemed like a great way to complete the puzzle.

 

[Eyesaw]

Originally posted by gnome
I agree with Eyesaw (of course, since he agrees with me ) but I'll take this opportunity to be a prick & say, Why should I tell the reader that all the monks become infected simultaneously? Part of the puzzle is to figure that out for yourself.

Eyesaw, you get a gold star for being such a discerning judge of puzzles.

Oh, and elibol: no sweat. We all get aggravated at times, as I did the other day when Eyesaw was enjoying hassling me about* the birthday puzzle.


*[my correct solution to]

I found the lazy way to solve that problem.

 

[elibol]

Originally posted by Eyesaw
I only bothered to look for second source for the solution to convince a doubting elibol.

hehe, my last name is accually elibol. so for you to put it that way it makes a lot of sense!

:)

 

[marqq]

please stop quoting all posts...please do not post off topic...please post some new puzzles

 

[marqq]

A general is capturing 3 ennemies...being a puzzle fan,he proposes them a possibility to gain their lives
He puts the one behind another , face on the wall(the 1st was seing just the wall, the 2nd was seing the 1st etc)and he brings a bag with 5 hats(3 black and 2 white)
He puts a hat on each head (not the small heads )and asks the colour of the hat that they are wearing.
The 3rd says that he dosen't know...and he dies
The 2rd says that he dosen't know...and he dies
The 1st says that his hat is black and he's free
Explain his logic

 

[elibol]

third guys logic-> he see's the second and first person either is wearing a black hat or a white hat. he would have been able to figure out which color he was wearing if both had a white hat on, so one of them must have had atleast one black hat on. he dies.

second guys logic-> since the guy behind him is dead, he knows one of them must have a black hat on or else the guy behind him would have gone free. if he see's a white hat on the guy infront of him he would guess that he had a black hat, and he would have gone free. this isn't the case, so he is dead.

first guys logic -> since he is logical enough to understand the logic the third and second guy went thru, he knows for sure he has a black hat on. he lives.

 

[Eyesaw]

I think the puzzle with the monks was an offspring
of this hat puzzle. The monk one is harder.

 

[marqq]

We have a long cable with 10 wires inside...we have a battery and a light ball...we have to find the ends of each wire, but in the mean time we can move only 2 times between the sides of the cable
enjoy

 

[gnome]

Pick up end of cable, drag it to the other end, test wires, drag it back.

 

[marqq]

:)

u r joking, right [?] ...



DO NOT ERASE MY POSTS please

 

[gnome]

OK, you don't like that approach, how about this. I'll try to upload a diagram, but in case that doesn't work, maybe you can follow this.

Starting at the left end of the cable, connect wires together in a group of 4, a group of 3, and a group of 2. (Leaving 1 wire unattached to anything.)

Going to the right side, you can use the battery and the bulb (I assume that your "light ball" is actually a light bulb) to test and identify the wires that are in the 4-group, the 3-group, the 2-group and the single one. Label them a - j as I show in the diagram. Now connect b to e, c to h, d to j, and f to i. (a and g are left unconnected.)

Now go back to the left side with the battery and bulb. Label the wires in each group so you will remember which ones they were, disconnect them from each other, and test to find:
- a wire from the group of 4 that makes no circuit with any other wire - that one is a.
- a wire from the group of 4 that makes a circuit with a wire from the group of 3 - those are b and e.
- a wire from the group of 4 that makes a circuit with one from the group of 2 - those are c and h.
- a wire from the group of 4 that makes a circuit with the one that was originally unconnected - those are d and j.
- a wire from the group of 3 that makes a circuit with one from the group of 2 - those are f and i.
- a wire from the group of 3 that makes no circuit - that is g.

 

[marqq]

good job

I SAID "GOOD WORK" DUDE! (do not erase my postreplys please )
i found 2 solutions on this problem but you kinda used them both in your answer

 

[marqq]

:)

so...we have 12 balls.Only one is different(lighter or heavyer)
We also have a balance...and only 3 trys to find the different ball
Enjoy