Who is the liar in this logic puzzle?

[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

 

[y3nis]

I got something nice too.
I am a Gabber with a bold head.
Am I lying or telling the truth? [?]


Kindest regard Dj You_nis

 

[marqq]

yes you are

 

[y3nis]

wrong, i aint a skinnhead I am the devil :P

 

[marqq]

Tell me y3nis where did i said that you were a skinhead ?

 

[y3nis]

you didn't. let's make this really philosophy. What exectly does skinhead mean? Someone that is bold i'd say. Else my previous post weren't correct

 

[marqq]

Did I really said smth?
Pls. quote me if did

 

[y3nis]

Originally posted by marqq
Did I really said smth?
Pls. quote me if did

Save My Tits Here=smth?

Well i'll tell you the truth. I aint a gabber and i aint bold.
Did i really tell the truth?

 

[marqq]

Originally posted by y3nis
Save My Tits Here=smth?

something

Btw...try to puzzle me by offering an corect answer to my 12 balls problem

 

[marqq]

Originally posted by 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

you didn't forget about that didn't you?

 

[marqq]

where r the signatures?

 

[tribdog]

next problem

You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?

 

[leto]

I just took the time to answer that horribly long balance puzzle, and my post didn't work. I'm not certain I am right, I am tired and can't think anymore. There is more to it than this, but this is all I had in my clipboard and there is no way in hell I'm typing anymore:

I would take 8 of the 12 balls and weigh them with 4 on each side of the balance.

If they were even, then I would know the odd ball was one of the 4 I haven't weighed. If the balance tipped then I would label the balls on the tipped side as l1 l2 l3 and l4. The balls on the other h1 h2 h3 and h4.

I would then take l1, h4, h3 and weigh against h1, h2 and l2. If the balance was even, then I would know the odd ball was either l3 or l4. if the balance tipped to the first side, then i would know the odd ball was either h4, h3 or l2. if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2.

If it was l3 or l4 then I would use my last balance to measure l3 again any ball other than l4. If the scale was even then l4 is the odd ball. If the scale dips to l3 then l3 is the odd ball.

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.

If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against each other. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.

 

[marqq]

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

If it was h1, h2, or l2, then I would do the same as directly above and find the odd ball.

If my initial weigh was balanced, then I would take 3 of the 4 remaining balls and weigh them against 3 other balls. If the balance stayed level, then I would know the odd ball was the one I didn't weigh. If it did not, then I would know whether the 3 balls were lighter or heavier. I would then take 2 of the 3 remaining balls and weigh them against each other. If the balance was level then the odd ball is the one I didn't weigh. Otherwise, it is the heavier/lighter one, depending on whether the 3 balls were lighter or heavier.

isn't it cute? BUT IS WRONG! remember that u don't know if the ball is lighter or heavyer...

It it was h4, h3, or l2 then I would use my last balance with h4 vs h3. If the balance dips to h4 then h4 is the odd ball. If the balance dips to h3 then h3 is the odd ball. if the scale is even then l2 is the odd ball.

why?
but you had a good start!

 

[marqq]

You are the big winner of a television game show. The host says "Grand prize is behind door #1, 2 or 3. pick one." After you make your choice (lets say you pick #1) the host says "You may be right. It's either #1 or #2 but its not number 3. Now I'm going to give you one last chance do you want to stick with door #1 or would you like to pick again?" What should you do?

whatever you want...you have a 33% chance to find the prize...you pick #1...and you find that the door #3 is a fake...Your chances increase to 50%...so either you take #2 or you rest on #1 you heave the same chances to win the prise :)

 

[leto]

I had one typo, but the fact that I don't know whether it is heavier or lighter was taken into account. This puzzle would be simple if I knew if it was lighter or heavier.

"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l2."

should of been:
"if the scale tipped to the second side, then I would know the odd ball was either h1, h2 or l1."


I can connects the dots if you still think I am wrong, but I'm too lazy to do that until then :P