Conundrums, Puzzles, Logic Problems (1 Viewer)

[The Man from Laramie]

How can 4 be half of 5?

because 7 ate 9?

 

[Peyton Marino]

The Solution

I could give the solution for three dragons only, by breaking the situation down into four cases and handling each case; but I will do it a better, more elegant way. I will generalize this to any number of dragons, some with red eyes and some with blue.

Let the island have n > 0 dragons with red eyes, and m dragons with blue eyes. I claim that all n dragons with red eyes will die on the n-th night after the wizard makes the statement "At least one of you has red eyes." (So for the original problem, the answer is the 3rd night). I prove this using mathematical induction on n.

If n = 1, then the lone dragon with red eyes will look around the island and see that all other dragons have blue eyes. He concludes that he is the only one with red eyes, and thus dies on the first night.

Suppose that for n = k-1, k > 1, that all k-1 red-eyed dragons die on the (k-1)-th night. Then for n = k, this is what happens: each of the k dragons will look around and see that there are k-1 dragons with red eyes on the island. Each dragon thinks, "I don't want to die, so let's assume that I have blue eyes." Then each dragon should expect the other k-1 dragons to die in the (k-1)-th night; so they wait. On the k-th day, each dragon sees that the other k-1 dragons are still alive, and thus their initial assumption that "I have blue eyes" must be false. Thus each of the k dragons realize that they have red eyes, and die on the k-th night.

why would they expect this?

 

[The Man from Laramie]

Way to expose the spoiler.

 

[DTD]

Why don't we have any of these anymore?

I am beginning to understand....

 

[Peyton Marino]

Way to expose the spoiler.

I doubt someone reads "the solution" at the top and then proceeds to read 4 more paragraphs before he realizes that he's reading the spoiler.

 

[(HULK)]

How can 4 be half of 5?

because 7 ate 9?

mytagid = Math.floor( Math.random() * 100 );document.write("Nope.

IV is half of fIVe

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

 

[(HULK)]

There is a common English word that is nine letters long. Each time you remove a letter from it, it still remains an English word - from nine letters right down to a single letter. What is the original word, and what are the words that it becomes after removing one letter at a time?

Spoiler

The base word is Startling - starting - staring - string - sting - sing - sin - in - I

 

[(HULK)]

This is an unusual paragraph. I'm curious how quickly you can find out what is so unusual about it? It looks so plain you would think nothing was wrong with it! In fact, nothing is wrong with it! It is unusual though. Study it, and think about it, but you still may not find anything odd. But if you work at it a bit, you might find out! Try to do so without any coaching!

Spoiler

The letter "e", which is the most common letter in the English language, does not appear once in the long paragraph.

 

[NRV77]

there's a magical island that is populated by 99 dragons. all of these dragons have red eyes, but they do not know this. if at any time it is possible for a dragon to deduce his eye color, the dragon will die the next day at noon. dragons, being the intelligent creatures they are, never discuss eye colors and they avoid looking in water where they might accidentally see their reflection. dragons are also extremely observant, so each dragon knows that the other 98 have red eyes. one day, an evil logician appears on the island and gathers the dragons together. he yells to them, "there is at least one dragon with red eyes on this island!" and suddenly vanishes. the dragons freak out as eye colors are not supposed to be discussed on the island, but eventually they settle down upon seeing many other dragons with red eyes. 99 days later, 99 dragons are dead at noon. why?

mytagid = Math.floor( Math.random() * 100 );document.write("Sounds good to me. If after 1 day, 1 dragon saw 98 pairs of, say, blue eyes, he'd know he was the one with red eyes, and die. This didn't happen. So the dragons know that there are at least 2 dragons with red eyes. On day 2, if a dragon saw 97 blue eyed dragons and 1 red eyed dragon, he'd know he was the other red eyed dragon and die. This didn't happen, so they know there must be at least 3 dragons with red eyes. Repeat.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

mytagid = Math.floor( Math.random() * 100 );document.write("actually, the dragons know that there are at least 98 other dragons on the island with red eyes. they can see them. look at it as if there are two dragons on the island. they both have red eyes. when the logician appears and says "at least one of you MFers has red eyes" each dragon assumes it's the other dragon that has red eyes and assume that they themselves have blue eyes. if it is true that one of those dragons had blue eyes, then the other dragon would see those blue eyes and know he has red eyes, then he'd be dead the next day...

but, both dragons have red eyes, and they can see the other has red eyes. therefore, when neither of them dies on the first day, it is impossible that either of the dragons have blue eyes, because neither of them died. so they both die.

now, just extrapolate that out with 99 dragons.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

This doesn't make any sense at all.For starters, where was it stated that a dragon must have either red or blue eyes? Why couldn't his eyes be any other color?

they all have red eyes. that was stated in the second or third sentence of the original problem. and yes, i've already conceded that my answer doesn't work. welcome.

Are you sure the logician didn't say that there is at least 1 dragon with BLUE eyes?

 

[Buckychudd]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

 

[TheIronSheik]

How can 4 be half of 5?

4 5

 

[NRV77]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

 

[(HULK)]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

Honestly, the more I think about the answer I pulled off the internet, the more I think thats wrong anyways.

 

[shuke]

I applaud (Hulk)'s proof by induction. Full points for nerditude.

Not mine. 100% culled from the internet.I felt the need to google this as I realized the question wasn't correct in the riddle. The initial riddle had no logical explaination.

the guy's question is the same as mine except in his riddle there's only 3 dragons and you have to find N and explain how, other than already being given N and explaining how.and, I know i'm going to catch a ton of flack from all the math wizards here, but when I was trying to answer this a while ago, i too found this guy's blog and it didn't help me. I honestly can't make two ####s out of what he's saying.

The fact that dragon's eyes can only be blue or red makes a difference, you didn't include it in your riddle.

I don't think that makes a difference. They're either red or not red. One specific color doesn't need to be given.

 

[Buckychudd]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

I'm too afraid to look stupid to question your BS.

 

[shuke]

Why don't we have any of these anymore?

I am beginning to understand....

 

[(HULK)]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

That doesn't follow though.At least 1 of them has red eyes. They all already knew that 98 of them had red eyes at least.

 

[Buckychudd]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

Honestly, the more I think about the answer I pulled off the internet, the more I think thats wrong anyways.

Ummm, you think?

 

[ren hoek]

You must use each digit from 0-9 only once, solve:

* *x   *-----  * *+ * *-----  * *

0-9 is 10 unique digits. The problem only uses 9.

 

[Peyton Marino]

Why don't we have any of these anymore?

I am beginning to understand....

the dragons are all different shapes, therefore the puzzle is flawed.

 

[Peyton Marino]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

the information you have in post 109 was given in the original question. "on the 99th day after the logician appeared, 99 dragons are dead at noon." that means they all died at once at noon, not one day at a time. sorry if this wasn't clear.

 

[NRV77]

So if I understand this correctly....the solution to the Dragon puzzle is that you make up a half-way intelligent sounding answer and hope everyone is too afraid to look stupid to question the BS?

see post 109

the information you have in post 109 was given in the original question. "on the 99th day after the logician appeared, 99 dragons are dead at noon." that means they all died at once at noon, not one day at a time. sorry if this wasn't clear.

yeah. it made perfect sense when I typed it. it doesn't anymore though. now my brain hurts.

 

[Aabye]

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

I agree that I don't see it.

mytagid = Math.floor( Math.random() * 100 );document.write("It does work for 3 dragons.

Imagine that there are 3 dragons (A, B, and C). A looks and see 2 dragons with red eyes. At this point, A doesn't know if his eyes are red or not.

Suppose A's eyes are not red. Now B and C will see one dragon with red eyes and one dragon with non-red eyes. At this point, neither of those dragons know if their eyes are red or not.

So one night will pass and no dragon will know the color of its own eyes.

When nobody dies, the dragons get an additional piece of information. They know that no dragon saw two dragons with non-red eyes. But B and C saw one dragon with red eyes and one dragon w/ non-red eyes. So B and C will deduce that they must have red eyes, since if B had non-red eyes, then C would have deduced that his own eyes were red and would have died after the first day and if C had non-red eyes, then B would have deduced that his own eyes were red and would have died after the first day.

So after the second night, B and C will die.

So if A's eyes are not red, B and C will die after the second night. Now, If nobody dies after the second night and A sees two sets of red eyes, he can deduce that his own eyes must also be red. So he (and B and C) will all die following the third night.

Make sense?

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

 

[parasaurolophus]

there's a magical island that is populated by 99 dragons. all of these dragons have red eyes, but they do not know this. if at any time it is possible for a dragon to deduce his eye color, the dragon will die the next day at noon. dragons, being the intelligent creatures they are, never discuss eye colors and they avoid looking in water where they might accidentally see their reflection. dragons are also extremely observant, so each dragon knows that the other 98 have red eyes. one day, an evil logician appears on the island and gathers the dragons together. he yells to them, "there is at least one dragon with red eyes on this island!" and suddenly vanishes. the dragons freak out as eye colors are not supposed to be discussed on the island, but eventually they settle down upon seeing many other dragons with red eyes. 99 days later, 99 dragons are dead at noon. why?

mytagid = Math.floor( Math.random() * 100 );document.write("Sounds good to me. If after 1 day, 1 dragon saw 98 pairs of, say, blue eyes, he'd know he was the one with red eyes, and die. This didn't happen. So the dragons know that there are at least 2 dragons with red eyes. On day 2, if a dragon saw 97 blue eyed dragons and 1 red eyed dragon, he'd know he was the other red eyed dragon and die. This didn't happen, so they know there must be at least 3 dragons with red eyes. Repeat.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

mytagid = Math.floor( Math.random() * 100 );document.write("actually, the dragons know that there are at least 98 other dragons on the island with red eyes. they can see them. look at it as if there are two dragons on the island. they both have red eyes. when the logician appears and says "at least one of you MFers has red eyes" each dragon assumes it's the other dragon that has red eyes and assume that they themselves have blue eyes. if it is true that one of those dragons had blue eyes, then the other dragon would see those blue eyes and know he has red eyes, then he'd be dead the next day...

but, both dragons have red eyes, and they can see the other has red eyes. therefore, when neither of them dies on the first day, it is impossible that either of the dragons have blue eyes, because neither of them died. so they both die.

now, just extrapolate that out with 99 dragons.*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

the more I think about this i dont get it. Wouldnt all of the dragons just say to themselves thanx a lot you dumb wizard I can see for myself that at least one of these dragons have red eyes. In fact 98 of them do smart guy. Why would there be further deduction after that first day? Each dragon would say to themselves well there are plenty with red eyes so this stupid wizard can get the f outta here and they would never think about it again.

edited to add i just reread and saw it was a logician not a wizard, but same crap. this riddle still sucks

 

[parasaurolophus]

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

I agree that I don't see it.

mytagid = Math.floor( Math.random() * 100 );document.write("It does work for 3 dragons.

Imagine that there are 3 dragons (A, B, and C). A looks and see 2 dragons with red eyes. At this point, A doesn't know if his eyes are red or not.

Suppose A's eyes are not red. Now B and C will see one dragon with red eyes and one dragon with non-red eyes. At this point, neither of those dragons know if their eyes are red or not.

So one night will pass and no dragon will know the color of its own eyes.

When nobody dies, the dragons get an additional piece of information. They know that no dragon saw two dragons with non-red eyes. But B and C saw one dragon with red eyes and one dragon w/ non-red eyes. So B and C will deduce that they must have red eyes, since if B had non-red eyes, then C would have deduced that his own eyes were red and would have died after the first day and if C had non-red eyes, then B would have deduced that his own eyes were red and would have died after the first day.

So after the second night, B and C will die.

So if A's eyes are not red, B and C will die after the second night. Now, If nobody dies after the second night and A sees two sets of red eyes, he can deduce that his own eyes must also be red. So he (and B and C) will all die following the third night.

Make sense?

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

i dont get why they get an additional piece of info when nobody dies? They would already know that NO dragon saw one with two non red eyes because they would each see two red eyes. By default Obviously they already know that nobody is seeing two non red eyes. They dont need to wait for nobody to die. Make sense?

 

[Aabye]

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

I agree that I don't see it.

mytagid = Math.floor( Math.random() * 100 );document.write("It does work for 3 dragons.

Imagine that there are 3 dragons (A, B, and C). A looks and see 2 dragons with red eyes. At this point, A doesn't know if his eyes are red or not.

Suppose A's eyes are not red. Now B and C will see one dragon with red eyes and one dragon with non-red eyes. At this point, neither of those dragons know if their eyes are red or not.

So one night will pass and no dragon will know the color of its own eyes.

When nobody dies, the dragons get an additional piece of information. They know that no dragon saw two dragons with non-red eyes. But B and C saw one dragon with red eyes and one dragon w/ non-red eyes. So B and C will deduce that they must have red eyes, since if B had non-red eyes, then C would have deduced that his own eyes were red and would have died after the first day and if C had non-red eyes, then B would have deduced that his own eyes were red and would have died after the first day.

So after the second night, B and C will die.

So if A's eyes are not red, B and C will die after the second night. Now, If nobody dies after the second night and A sees two sets of red eyes, he can deduce that his own eyes must also be red. So he (and B and C) will all die following the third night.

Make sense?

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

i dont get why they get an additional piece of info when nobody dies? They would already know that NO dragon saw one with two non red eyes because they would each see two red eyes. By default Obviously they already know that nobody is seeing two non red eyes. They dont need to wait for nobody to die. Make sense?

OK, think about it this way.If one dragon sees two sets of non-red eyes, he'll know the color of his own eyes. That's obvious.

Now, if one dragon (call him A) sees a dragon with red eyes (call him B) and another dragon with non-red eyes (call him C), what can he figure out about his own eyes?

Nothing directly. But he will know that if his eyes are not red, then B will learn that his own eyes are red (since B would see two sets of non-red eyes and so discover that his own eyes are red).

So if A's eyes are not red, B will die the following day.

So if B doesn't die, A knows that his own eyes are red.

Is this much clear?

 

[parasaurolophus]

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

I agree that I don't see it.

mytagid = Math.floor( Math.random() * 100 );document.write("It does work for 3 dragons.

Imagine that there are 3 dragons (A, B, and C). A looks and see 2 dragons with red eyes. At this point, A doesn't know if his eyes are red or not.

Suppose A's eyes are not red. Now B and C will see one dragon with red eyes and one dragon with non-red eyes. At this point, neither of those dragons know if their eyes are red or not.

So one night will pass and no dragon will know the color of its own eyes.

When nobody dies, the dragons get an additional piece of information. They know that no dragon saw two dragons with non-red eyes. But B and C saw one dragon with red eyes and one dragon w/ non-red eyes. So B and C will deduce that they must have red eyes, since if B had non-red eyes, then C would have deduced that his own eyes were red and would have died after the first day and if C had non-red eyes, then B would have deduced that his own eyes were red and would have died after the first day.

So after the second night, B and C will die.

So if A's eyes are not red, B and C will die after the second night. Now, If nobody dies after the second night and A sees two sets of red eyes, he can deduce that his own eyes must also be red. So he (and B and C) will all die following the third night.

Make sense?

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

i dont get why they get an additional piece of info when nobody dies? They would already know that NO dragon saw one with two non red eyes because they would each see two red eyes. By default Obviously they already know that nobody is seeing two non red eyes. They dont need to wait for nobody to die. Make sense?

OK, think about it this way.If one dragon sees two sets of non-red eyes, he'll know the color of his own eyes. That's obvious.

Now, if one dragon (call him A) sees a dragon with red eyes (call him B) and another dragon with non-red eyes (call him C), what can he figure out about his own eyes?

Nothing directly. But he will know that if his eyes are not red, then B will learn that his own eyes are red (since B would see two sets of non-red eyes and so discover that his own eyes are red).

So if A's eyes are not red, B will die the following day.

So if B doesn't die, A knows that his own eyes are red.

Is this much clear?

not clear at all.not a one of them will see a dragon with any non red eyes. They will all see a two red.

dragon A will see two red. If he thinks his eyes could be blue he knows dragon B will still see one set of red eyes because he knows Dragon C has red eyes. He Knows dragon C will still see one set of red eyes because he knows Dragon B has red eyes. So the prediction of "at least one with red eyes" is legit because every dragon knows that every dragon will see at least one set of red eyes. There would be no further deuction possible.

wouldnt it have to be a prediction of "at least two are red"

 

[Mungo Burrows]

While I see how this would work with 2 dragons, I don't see how it would work with 99 dragons.

I agree that I don't see it.

mytagid = Math.floor( Math.random() * 100 );document.write("It does work for 3 dragons.

Imagine that there are 3 dragons (A, B, and C). A looks and see 2 dragons with red eyes. At this point, A doesn't know if his eyes are red or not.

Suppose A's eyes are not red. Now B and C will see one dragon with red eyes and one dragon with non-red eyes. At this point, neither of those dragons know if their eyes are red or not.

So one night will pass and no dragon will know the color of its own eyes.

When nobody dies, the dragons get an additional piece of information. They know that no dragon saw two dragons with non-red eyes. But B and C saw one dragon with red eyes and one dragon w/ non-red eyes. So B and C will deduce that they must have red eyes, since if B had non-red eyes, then C would have deduced that his own eyes were red and would have died after the first day and if C had non-red eyes, then B would have deduced that his own eyes were red and would have died after the first day.

So after the second night, B and C will die.

So if A's eyes are not red, B and C will die after the second night. Now, If nobody dies after the second night and A sees two sets of red eyes, he can deduce that his own eyes must also be red. So he (and B and C) will all die following the third night.

Make sense?

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***

");document.close();

i dont get why they get an additional piece of info when nobody dies? They would already know that NO dragon saw one with two non red eyes because they would each see two red eyes. By default Obviously they already know that nobody is seeing two non red eyes. They dont need to wait for nobody to die. Make sense?

OK, think about it this way.If one dragon sees two sets of non-red eyes, he'll know the color of his own eyes. That's obvious.

Now, if one dragon (call him A) sees a dragon with red eyes (call him B) and another dragon with non-red eyes (call him C), what can he figure out about his own eyes?

Nothing directly. But he will know that if his eyes are not red, then B will learn that his own eyes are red (since B would see two sets of non-red eyes and so discover that his own eyes are red).

So if A's eyes are not red, B will die the following day.

So if B doesn't die, A knows that his own eyes are red.

Is this much clear?

not clear at all.not a one of them will see a dragon with any non red eyes. They will all see a two red.

dragon A will see two red. If he thinks his eyes could be blue he knows dragon B will still see one set of red eyes because he knows Dragon C has red eyes. He Knows dragon C will still see one set of red eyes because he knows Dragon B has red eyes. So the prediction of "at least one with red eyes" is legit because every dragon knows that every dragon will see at least one set of red eyes. There would be no further deuction possible.

wouldnt it have to be a prediction of "at least two are red"

I still don't get it either. The statement by the logician/wizard gives them no information they didn't already know. If knowledge of at least one other red eye leads to this deduction, they should already be dead because they all know there's at least one (98 or 99, in fact) with red eyes.

 

[FragileFred]

Let me try to explain this one. The solution requires each dragon to believe the other dragons have perfect logic.

Let's start with 3 dragons since some of you seem to think this is where the riddle breaks down. Dragon A sees that Dragon B and Dragon C have red eyes. He first considers what would happen if his eyes were some other color (let's say blue). In this scenario, Dragon B would see one dragon with red eyes and one dragon with blue eyes. Seeing this, Dragon B would consider what would happen if his eyes were non-red (again, let's say blue). In this scenario, Dragon C would see two dragons with blue eyes and would thus conclude that he has red eyes and would die the next day. But he doesn't die, so in a world where Dragon A has blue eyes, Dragon B would conclude on the 1st day after the announcement that his own eyes must be red. Following the same train of thought, Dragon C would conclude the same thing when Dragon B doesn't die. So, Dragon A concludes that if his eyes are indeed blue and the other dragons have perfect logic, both Dragon B and Dragon C will die on the 2nd day. But they don't die, so Dragon C concludes that he must also have red eyes. And of course, Dragon B and C would conclude the same thing about themselves by the same logic, and thus all three would die on day 3.

Clear?

 

[The Man from Laramie]

Let me try to explain this one. The solution requires each dragon to believe the other dragons have perfect logic.Let's start with 3 dragons since some of you seem to think this is where the riddle breaks down. Dragon A sees that Dragon B and Dragon C have red eyes. He first considers what would happen if his eyes were some other color (let's say blue). In this scenario, Dragon B would see one dragon with red eyes and one dragon with blue eyes. Seeing this, Dragon B would consider what would happen if his eyes were non-red (again, let's say blue). In this scenario, Dragon C would see two dragons with blue eyes and would thus conclude that he has red eyes and would die the next day. But he doesn't die, so in a world where Dragon A has blue eyes, Dragon B would conclude on the 2nd day that his own eyes must be red. Following the same train of thought, Dragon C would conclude the same thing when Dragon B doesn't die. So, Dragon A concludes that if his eyes are indeed blue and the other dragons have perfect logic, both Dragon B and Dragon C will die on the 3rd day. But they don't die, so Dragon C concludes that he must also have red eyes. And of course, Dragon B and C would conclude the same thing about themselves by the same logic, and thus all three would die on day 4.Clear?

Your nomenclature is unclear. What you call the 4th day is referred to in the problem as the 3rd day after the announcement. As it turns out all 99 of those dragons die on the 99th day AFTER the fact is revealed.

 

[Majorum]

quick clarification for those who don't understand what extra information the wizard is giving the dragons, using 2 dragons (A and B) to start

BEFORE THE REVELATION

-A knows B has red eyes (therefore A knows at least 1 dragon has red eyes)

-B knows A has red eyes (therefore B knows at least 1 dragon has red eyes)

AFTER THE REVELATION (call this day 0)

-A knows that B knows at least one of A or B has red eyes

-B knows that A knows at least one of A or B has red eyes

This new information is what matters. Now A can deduce that if he(A) doesn't have red eyes, B would know he(B) has red eyes, and die on day 1. Since A does have red eyes, B doesn't die on day 1, and therefore A now knows he(A) has red eyes and dies on day 2. Since the problem is symmetric, same thing goes for B who also dies on day 2.

Moving on to 3 dragons :

BEFORE THE REVELATION

-A knows B & C have red eyes (therefore A knows at least 1 dragon has red eyes)

-B knows A & C have red eyes (therefore B knows at least 1 dragon has red eyes)

-C knows A & B have red eyes (therefore C knows at least 1 dragon has red eyes)

AFTER THE REVELATION (call this day 0)

-A knows that B & C know at least one of A, B or C has red eyes

-B knows that A & C know at least one of A, B or C has red eyes

-C knows that A & B know at least one of A, B or C has red eyes

(actually, this was already known, the additional information that A has is that B knows that C knows (sry for the 2nd edit, this is complicated), this becomes quite convoluted)

Now take dragon C. He© can deduce that if he doesn't have red eyes, the other two dragons are left with the following knowledge:

-A knows B has red eyes

-B knows A has red eyes

-A knows that B knows at least one of A or B has red eyes

-B knows that A knows at least one of A or B has red eyes

This is the problem for 2 dragons solved above, in which both dragons die on day 2. But C has red eyes, so A & B do not die on day 2. Dragon C now knows he has red eyes because of this, and dies on day 3. Same goes for A & B by symmetry.

Add a dragon, have him assume he doesn't have red eyes, and the problem reduces to N-1 dragons, with 1 extra day of waiting time

hope that helps

Majorum

 

[The Man from Laramie]

well done.

 

[FragileFred]

Let me try to explain this one. The solution requires each dragon to believe the other dragons have perfect logic.Let's start with 3 dragons since some of you seem to think this is where the riddle breaks down. Dragon A sees that Dragon B and Dragon C have red eyes. He first considers what would happen if his eyes were some other color (let's say blue). In this scenario, Dragon B would see one dragon with red eyes and one dragon with blue eyes. Seeing this, Dragon B would consider what would happen if his eyes were non-red (again, let's say blue). In this scenario, Dragon C would see two dragons with blue eyes and would thus conclude that he has red eyes and would die the next day. But he doesn't die, so in a world where Dragon A has blue eyes, Dragon B would conclude on the 2nd day that his own eyes must be red. Following the same train of thought, Dragon C would conclude the same thing when Dragon B doesn't die. So, Dragon A concludes that if his eyes are indeed blue and the other dragons have perfect logic, both Dragon B and Dragon C will die on the 3rd day. But they don't die, so Dragon C concludes that he must also have red eyes. And of course, Dragon B and C would conclude the same thing about themselves by the same logic, and thus all three would die on day 4.Clear?

Your nomenclature is unclear. What you call the 4th day is referred to in the problem as the 3rd day after the announcement. As it turns out all 99 of those dragons die on the 99th day AFTER the fact is revealed.

Yeah, you're right. Sorry about that. Original post is now corrected.

 

[Kilgore Trout]

Add a dragon, have him assume he doesn't have red eyes, and the problem reduces to N-1 dragons, with 1 extra day of waiting time

I think this breaks down when applied to 4 or more dragons the way the puzzle is worded. Every dragon sees 98 other dragons with red eyes. His eyes could be red or not red once you got to 4 or more dragons, because it wouldn't matter what he assumed about his own eyes, as he will always know that their is at least one set of red eys, as the logician said.

 

[Aabye]

No, it works with 4 as well. Here's the reasoning.

mytagid = Math.floor( Math.random() * 100 );document.write("

Call the 4 dragons A, B, C, and D.

It easiest, I think, to first suppose that A has blue eyes and then start with how B would reason

B sees one set of blue eyes and two sets of red eyes. So B supposes that his eyes are blue.

If this were the case, then C would see two sets of blue eyes and one set of red eyes. So C supposes that his eyes are blue.

If this were the case, then D would see three sets of blue eyes and conclude that his own eyes must be red.

So if A, B, and C have blue eyes and D has red eyes, D will die on the first day ("first" meaning the first day after the logician appears).

When D does not die on the first day, C realizes that his eyes cannot be blue. So C realizes that his own eyes must also be red. So C will die on the second day.

So if A and B have blue eyes and C and D have red eyes, C and D will die on the second day.

When C and D do not die on the second day, B realizes that his own eyes must also be red. So B will die on the third day.

So if A has blue eyes and B, C, and D all have red eyes, then B, C, and D will die on the third day.

Now, if no dragons die on the third day, A realizes that his own eyes must also be red. So A, B, C, and D will die on the fourth day.

Now look at all those cases. They cover all of the possible distributions of eye color.

If one dragon has red eyes, he will die on day 1.

If two dragons have red eyes, they will die on day 2.

If three dragons have red eyes, they will die on day 3.

If four dragons have red eyes, they will die on day 4.

So if no dragons die on day 3, each dragon realizes that the only possible distribution is 4 sets of red eyes, and so all dragons die on day 4.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

It should be clear that this process can be repeated for any number of dragons.

 

[Majorum]

Add a dragon, have him assume he doesn't have red eyes, and the problem reduces to N-1 dragons, with 1 extra day of waiting time

I think this breaks down when applied to 4 or more dragons the way the puzzle is worded. Every dragon sees 98 other dragons with red eyes. His eyes could be red or not red once you got to 4 or more dragons, because it wouldn't matter what he assumed about his own eyes, as he will always know that their is at least one set of red eys, as the logician said.

I'll try to demonstrate the extra information given by the logician for 4 dragons :A knows the following :B,C,D have red eyesB,C,D know at least 1 has red eyesB,C,D know at least 2 have red eyesnow A does NOT know that B knows he(A) has red eyes, so A has the following information regarding B's knowledge:A knows that B knows :C,D have red eyesC,D know at least 1 has red eyessimilarly, going down the chainA knows that B knows that C knows has red eyesAfter the revelation, the extra information is that :A knows that B knows that C knows THAT D KNOWS AT LEAST 1 HAS RED EYESAfter re-reading this, I don't think this will help as it becomes way too abstract. The new knowledge obtained is not about what each dragon know by itself, but what dragons know about the knowledge of other dragons. This is a nasty problem.Majorum

 

[Kilgore Trout]

No, it works with 4 as well. Here's the reasoning.

mytagid = Math.floor( Math.random() * 100 );document.write("

Call the 4 dragons A, B, C, and D.

It easiest, I think, to first suppose that A has blue eyes and then start with how B would reason

B sees one set of blue eyes and two sets of red eyes. So B supposes that his eyes are blue.

If this were the case, then C would see two sets of blue eyes and one set of red eyes. So C supposes that his eyes are blue.

If this were the case, then D would see three sets of blue eyes and conclude that his own eyes must be red.

So if A, B, and C have blue eyes and D has red eyes, D will die on the first day ("first" meaning the first day after the logician appears).

When D does not die on the first day, C realizes that his eyes cannot be blue. So C realizes that his own eyes must also be red. So C will die on the second day.

So if A and B have blue eyes and C and D have red eyes, C and D will die on the second day.

When C and D do not die on the second day, B realizes that his own eyes must also be red. So B will die on the third day.

So if A has blue eyes and B, C, and D all have red eyes, then B, C, and D will die on the third day.

Now, if no dragons die on the third day, A realizes that his own eyes must also be red. So A, B, C, and D will die on the fourth day.

Now look at all those cases. They cover all of the possible distributions of eye color.

If one dragon has red eyes, he will die on day 1.

If two dragons have red eyes, they will die on day 2.

If three dragons have red eyes, they will die on day 3.

If four dragons have red eyes, they will die on day 4.

So if no dragons die on day 3, each dragon realizes that the only possible distribution is 4 sets of red eyes, and so all dragons die on day 4.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

It should be clear that this process can be repeated for any number of dragons.

I see. Dragons shouldn't think so hard about what color their eyes are though.

 

[Aabye]

No, it works with 4 as well. Here's the reasoning.

mytagid = Math.floor( Math.random() * 100 );document.write("

Call the 4 dragons A, B, C, and D.

It easiest, I think, to first suppose that A has blue eyes and then start with how B would reason

B sees one set of blue eyes and two sets of red eyes. So B supposes that his eyes are blue.

If this were the case, then C would see two sets of blue eyes and one set of red eyes. So C supposes that his eyes are blue.

If this were the case, then D would see three sets of blue eyes and conclude that his own eyes must be red.

So if A, B, and C have blue eyes and D has red eyes, D will die on the first day ("first" meaning the first day after the logician appears).

When D does not die on the first day, C realizes that his eyes cannot be blue. So C realizes that his own eyes must also be red. So C will die on the second day.

So if A and B have blue eyes and C and D have red eyes, C and D will die on the second day.

When C and D do not die on the second day, B realizes that his own eyes must also be red. So B will die on the third day.

So if A has blue eyes and B, C, and D all have red eyes, then B, C, and D will die on the third day.

Now, if no dragons die on the third day, A realizes that his own eyes must also be red. So A, B, C, and D will die on the fourth day.

Now look at all those cases. They cover all of the possible distributions of eye color.

If one dragon has red eyes, he will die on day 1.

If two dragons have red eyes, they will die on day 2.

If three dragons have red eyes, they will die on day 3.

If four dragons have red eyes, they will die on day 4.

So if no dragons die on day 3, each dragon realizes that the only possible distribution is 4 sets of red eyes, and so all dragons die on day 4.

*** SPOILER ALERT! Click this link to display the potential spoiler text in this box. ***");document.close();

It should be clear that this process can be repeated for any number of dragons.

I see. Dragons shouldn't think so hard about what color their eyes are though.

The saddest part is that if they weren't so rational, they wouldn't have to die. RIP Smartdragons.

 

[fatguyinalittlecoat]

Nice work FragileFred and Majorum. That's the answer. It's difficult to conceptualize. The first time this was posted I fought with Maurile for like 3 days before I understood.

 

[snorlax]

The dragons with "days" are much easier to conceptualize than the monks with "moments."

 

[Aabye]

Here's a very weird problem that I ran across a while ago. There are two people, call them Al and Bob.

Bob leaves the room and Al writes two different integers of any size on two pieces of paper, puts one in each hand, and puts his hands behind his back.

Bob then picks either the right or left hand. Al reveals the number written on the piece of paper contained in that hand and Bob has to guess whether the number concealed in the other hand is higher or lower.

Can Bob arrive at a strategy that allows him to guess the right answer with a frequency of better than 50%?

 

[snorlax]

I would guess lower if the first number in the integer is 6 or above and higher if it is 5 or below. Also, I would think that people would more often put the higher number in their right hand. I don't know if either of those things has anything to do with the right answer, as Al could be intentionally deceptive in a variety of ways.

My first inclination is no, but I'm interested in the answer.

 

[FragileFred]

I would guess lower if the first number in the integer is 6 or above and higher if it is 5 or below. Also, I would think that people would more often put the higher number in their right hand. I don't know if either of those things has anything to do with the right answer, as Al could be intentionally deceptive in a variety of ways.My first inclination is no, but I'm interested in the answer.

But they can be integers of any value, not just 1-10.

 

[Aabye]

I would guess lower if the first number in the integer is 6 or above and higher if it is 5 or below. Also, I would think that people would more often put the higher number in their right hand. I don't know if either of those things has anything to do with the right answer, as Al could be intentionally deceptive in a variety of ways.My first inclination is no, but I'm interested in the answer.

The puzzle isn't intended to rely on human psychology and I honestly have no idea which hand is more likely to contain the higher number. In any case, Al could defeat this strategy by simply folding the pieces of paper (so that he can't see them), then randomly choosing one for his right hand and one for his left.As far as the integers go, the can be any whole number greater than, equal to, or less than zero. So -195324789544 is a possible number, 4 is a possible number, 97645436714654641 is a possible number, etc. For any given integer, there will be an infinite number of larger integers and and infinite number of smaller integers. So just randomly picking an integer (say 1500, for example) and guessing "larger" if the revealed number is smaller than 1500 and "smaller" if the revealed number is larger than 1500 should yield an expectation of 50% correct for Bob.EDIT: Also, notice that any psychological edge that one might have by getting into Al's head can be negated if Al simply writes down two numbers which are randomly generated (say, for instance, by a computer program).

 

[FragileFred]

For any given integer, there will be an infinite number of larger integers and and infinite number of smaller integers.

But not an infinite number that can fit on a piece of paper of finite size. That doesn't have anything to do with the answer though does it?

 

[The Man from Laramie]

you wanna guess higher for all numbers lower than negative infinity plus 22.

 

[Anonymous Internet User]

Here's a very weird problem that I ran across a while ago. There are two people, call them Al and Bob.Bob leaves the room and Al writes two different integers of any size on two pieces of paper, puts one in each hand, and puts his hands behind his back.Bob then picks either the right or left hand. Al reveals the number written on the piece of paper contained in that hand and Bob has to guess whether the number concealed in the other hand is higher or lower.Can Bob arrive at a strategy that allows him to guess the right answer with a frequency of better than 50%?

not sure i see where strategies can help here. reducing the problem down:you are given a random integer.guess if the next random integer is higher or lower, restricting that it cannot be the same value?seems to me there is no strategy you can develop to help, given that you only have 1 guess, and given that both values are random. the fact that it cannot be the same value adds nothing except for the elimination of one case, and the fact that they are restricted to integers also seems to add nothing.

 

[FragileFred]

Here's one that has been called the the hardest recreational logical puzzle ever created.

Three gods A , B , and C are called, in some order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A , B , and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer in their own language, in which the words for yes and no are “da” and “ja”, in some order. You do not know which word means which.

I don't know the answer and haven't really ever tried to work it out, but it is apparently solvable. Perhaps with our combined powers...

 

[TinHat]

I have 2 coins in my hand that total 30 cents. One is not a nickel. What are they?

for the record, I hate this style of brain teaserIt ruins the whole genre because everytime there's a brain teaser, it forces me toi finely parse it rather than just jumping into the logic.

 

[Aabye]

For any given integer, there will be an infinite number of larger integers and and infinite number of smaller integers.

But not an infinite number that can fit on a piece of paper of finite size. That doesn't have anything to do with the answer though does it?

Yeah, that's a good observation, but one that has to be disregarded for the sake of this puzzle. Assume that Al can write integers of any size on the sheet of paper.The actual solution is really pretty simple to explain and intuitive to understand (of course, it's easy to say that with the answer in hand!) The point is that it won't be a sort of "dirty trick." The answer won't depend on any external factors like Al's psychology or the finite size of the sheets of paper or the finite time that Al has to write a number. All of that practical stuff can be disregarded.