Can You Solve 'The Hardest Logic Puzzle In The World'?

AJK said:
The logic does work. Sorry, I've tried. I give up.
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The problem is not the maths but the psychology behind them thinking "if a dragon can see 99 blue eyes" . They know this not to be the case so the whole induction method does not work.
 
Jono8 said:
The problem is not the maths but the psychology behind them thinking "if a dragon can see 99 blue eyes" . They know this not to be the case so the whole induction method does not work.
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Like I said, I give up. You've been offered multiple solutions and explanations from academic sources. The problem is not the puzzle, it's your refusal or inability to understand logical proof (please, forget the word "psychology", it's logic and maths, nothing more.)

Read these, or don't. I really don't mind :)

https://xkcd.com/solution.html
http://io9.com/can-you-solve-the-worlds-other-hardest-logic-puzzle-1645422530
https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol2.pdf
https://cowles.econ.yale.edu/~gean/art/p0882.pdf
 
AJK said:
Like I said, I give up. You've been offered multiple solutions and explanations from academic sources. The problem is not the puzzle, it's your refusal or inability to understand logical proof (please, forget the word "psychology", it's logic and maths, nothing more.)

Read these, or don't. I really don't mind :)

https://xkcd.com/solution.html
http://io9.com/can-you-solve-the-worlds-other-hardest-logic-puzzle-1645422530
https://www.physics.harvard.edu/uploads/files/undergrad/probweek/sol2.pdf
https://cowles.econ.yale.edu/~gean/art/p0882.pdf
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I agree with this chap. He puts it much better than I could.

"You are absolutely right. The flaw here is that the idea of induction doesn't work.

You cannot extrapolate from the 2 sets of blue eyes example. In that example, the other blue-eyed person sees EXACTLY one other set of blue eyes, and correctly concludes that if they did not have blue eyes, the other person would leave.

EXACTLY one set is important... critical to the solution. You cannot get that in the 100 person example. The idea that one of the 99 blue-eyed people you can see "didn't look out and see no blue eyes, and there didn't leave" is disingenuous -- the islander who sees 99 blue-eyed people KNOWS that each of the other blue-eyed people sees AT LEAST 98 other blue-eyed people. The construct that they also don't see ZERO other blue-eyed people doesn't add information, nor does it get the process of induction from n = 2 up to n = 100 going. Because what made the n = 2 example work was that the other blue eyed person looked and saw EXACTLY one blue-eyed person.

The induction that actually works here is what the guru has to say to add information. In the n = 2 example, what the guru is really saying is: "I see AT LEAST ONE blue-eyed person," meaning, "I am reporting one less blue-eyed person than you would see if you were not one of the blue-eyed people." What induction tells you then is that, if there are n blue-eyed people, and the guru says, "I see AT LEAST N-1 blue-eyed people," then a blue-eyed person who also sees N - 1 other blue-eyed people has been given information -- that if none of the other N - 1 blue-eyed people vanish tonight, then he/she is blue-eyed. Everybody else sees N blue-eyed people and can gather no information.

The reason why AT LEAST N-1 counts as new information isn't because "you" can also see N - 1, it's because "you" don't know for sure that the other blue-eyed people see N - 1. The only thing you know for sure is that the other people see at least N - 2! If you're not blue-eyed, and you see N - 1, then the other blue eyes will actually see N - 2. If they do, they'll use the information given to them by the guru, and leave that night! The fact that they don't leave means you're blue-eyed. If they leave, then you don't have blue eyes (and have no idea what color your eyes might be).

Think about it: if the guru adds no information, what is he/she for? You already knew someone had blue eyes (99 or 100, depending on whether you have blue eyes). If you could figure out if you had blue eyes from that information, you'd have done it way before the guru had opened his/her mouth! But you can't -- it's not new information.

By the way, the guy that posted the 100 days of possible execution analogy also erred. The King could perfectly well set the execution day as day 100. There's no rule that says the King couldn't corner himself on day 100. Just because the guy would know it after day 99 proves nothing.... he just won a high stakes game of chicken with the odds.



25th April 2011"

https://www.englishforums.com/English/BlueEyesLogicPuzzle/zmpqw/post.htm
 
^ I've read that twice, and can't make any sense out of it at all. It just doesn't make any sense, or is written terribly. Or both.

Besides, you are not extrapolating from two dragons, you use two dragons as a base for three dragons, then use induction from there.

The dragon one is really quite simple. I can't wrap my head around the problems being posted about the solution.
Remember that all dragons are perfectly logical, and would all choose the quickest method of determining for sure who has the green eyes.
 
Jono8 said:
Think about it: if the guru adds no information, what is he/she for? You already knew someone had blue eyes (99 or 100, depending on whether you have blue eyes). If you could figure out if you had blue eyes from that information, you'd have done it way before the guru had opened his/her mouth! But you can't -- it's not new information
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It does change something though, it adds the expectation of change. Saying at least one dragon has green eyes sets a point in time from which at least one dragon has to change. It is this expectation, and failed expectation, that means come day 100 they all know that they themselves have green eyes.

A simply way to view it is before the humans each dragon thinks "I don't know and I cannot know what my eye colour is but I do know I see 99 pairs of green eyes." After the human leaves this becomes "I don't know what colour eyes I have but I might have green eyes though I do know I see 99 pairs of green eyes"
 
Jono8 said:
Why would any of them suddenly think they have green eyes?
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They don't, it was human way of looking at it rather than with pure infallible logic of the puzzle. Obviously my attempt failed so go back to my other point, it doesn't give them any new information as they each already know 99 dragons have green eyes it simply means that for the humans statement to be true at least one dragon must change. It then is a countdown of how many night does it take to work out what eye colour each dragon has themselves.

Edit: A better way to think of it is their thought process is "I don't know what eye colour I have but I know all other dragons have green eyes" then becomes "I don't know what eye colour I have but I know all other dragons have green eyes and because of what the human said at least one should change tonight". I think the hard part to imagine or work out from is that all 100 dragons are thinking the same thing at the same time about the other 99 dragons.
 
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Greboth said:
They don't, it was human way of looking at it rather than with pure infallible logic of the puzzle. Obviously my attempt failed so go back to my other point, it doesn't give them any new information as they each already know 99 dragons have green eyes it simply means that for the humans statement to be true at least one dragon must change. It then is a countdown of how many night does it take to work out what eye colour each dragon has themselves.

Edit: A better way to think of it is their thought process is "I don't know what eye colour I have but I know all other dragons have green eyes" then becomes "I don't know what eye colour I have but I know all other dragons have green eyes and because of what the human said at least one should change tonight". I think the hard part to imagine or work out from is that all 100 dragons are thinking the same thing at the same time about the other 99 dragons.
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But the dragons know none of them will change on the first night :confused:
 
The whole thing is contradictory. They can't use induction with 100 of them because the first part of it is not valid. They cant ask "if one of us sees all non green eyes then they will know that they have green eyes and turn into a sparrow at midnight" because they all know that this absolutely cannot be the case.

Yes they are infallibly logical, but they also know that all the others can also see at least 98 green eyed dragons therefore asking the first question required for induction is impossible.
 
They don't have to - the question they're asking themselves is whether 99 or 100 of them do.

You can still solve it by induction - what if there was one with green eyes and so on...

If you're struggling then try with two, three, four dragons and so on.
 
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dowie said:
They don't have to - the question they're asking themselves is whether 99 or 100 of them do.

You can still solve it by induction - what if there was one with green eyes and so on...

If you're struggling then try with two, three, four dragons and so on.
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I know how induction works but i do not believe it applies to these 100 green eyed dragons.

1 dragon - yes, two dragons, yes. More than two - no because they are asking themselves a question that doesn't exist in their world.
 
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then you don't really appreciate how it works and it isn't about belief

why not three dragons - supposing there are three dragons - one with blue eyes and two with green - try solving that....

then try three dragons all with green eyes...

then try 4

think it through a bit more - it really isn't a matter for opinion, there is a right answer it's just you're failing to understand it so far
 
Jono - I thought you might get it with my attempt above to explain why the human's statement is significant even though it doesn't give any new facts. Thought I'd done an OK job :( :).

The key is not just what is said (cos they all know that anyway) - it's the fact all dragons are present to simultaneously hear it. The inductive solving process requires all dragons to be engaged in the process starting on the same day because each dragon can only reach a conclusion if he knows he can rely on the conclusions and actions of the other dragons. It's only when the human delivers the statement in the presence of all the dragons simultaneously that each dragon is certain that all the others have started the process. It's that crucial 'Gentlemen, synchronise watches' moment.

So you can see it works for 1 or two dragons - good. Surely you can also see it works for three dragons all with green eyes? That's easy enough to follow - here goes...

Upon hearing the statement, any dragon 'X' can reason to himself...

'IF my eyes are not green, then the other two dragons can only see one green eyed dragon'.

Dragon X can continue to reason that the other two dragons will in that case both reason to themselves...

'If my eyes are also not green, that other dragon can see no green eyed dragons, and will therefore conclude his eyes are green and will leave on the first night'

Dragon X further reasons...

'If nobody leaves on the first night, the other two dragons will conclude they both have green eyes and will leave on the second night'

(aside - dragon X actually knows nobody will leave on the first night as he can already see the other two both have green eyes - he is only really waiting to see what happens on the second night)

On the second night, if the other two dragons up and leave, dragon X knows his original IF statement was correct - he does not have green eyes. However, if on the second night nobody leaves, then he knows he was wrong and he infact has green eyes like the rest, so they all leave on the third night. And that's what happens.


So you can see it also works for three green eyed dragons, right? And to summarise what the logical reasoning was for three dragons, it was just the same reasoning for the two-dragon version plus one more level (one more day). I.e. for three dragons, any dragon X reasons...

IF my eyes are not green THEN the other two dragons will figure out their eyes are green with the two-dragon method and leave in 2 days.
IF that fails THEN my eyes are also green and off we all go on day 3

So can you see the process for any number of dragons is just the same process for 1 less dragon plus another layer? So for N dragons, any dragon X can reason...

IF my eyes are not green THEN the other N-1 dragons will figure out their eyes are green with the N-1-dragon method and leave in N-1 days.
IF that fails THEN my eyes are also green and off we all go on day N

Easy right?
 
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Liampope said:
Jono - I thought you might get it with my attempt above to explain why the human's statement is significant even though it doesn't give any new facts. Thought I'd done an OK job :( :).

The key is not just what is said (cos they all know that anyway) - it's the fact all dragons are present to simultaneously hear it. The inductive solving process requires all dragons to be engaged in the process starting on the same day because each dragon can only reach a conclusion if he knows he can rely on the conclusions and actions of the other dragons. It's only when the human delivers the statement in the presence of all the dragons simultaneously that each dragon is certain that all the others have started the process. It's that crucial 'Gentlemen, synchronise watches' moment.

So you can see it works for 1 or two dragons - good. Surely you can also see it works for three dragons all with green eyes? That's easy enough to follow - here goes...

Upon hearing the statement, any dragon 'X' can reason to himself...

'IF my eyes are not green, then the other two dragons can only see one green eyed dragon'.

Dragon X can continue to reason that the other two dragons will in that case both reason to themselves...

'If my eyes are also not green, that other dragon can see no green eyed dragons, and will therefore conclude his eyes are green and will leave on the first night'

Dragon X further reasons...

'If nobody leaves on the first night, the other two dragons will conclude they both have green eyes and will leave on the second night'

(aside - dragon X actually knows nobody will leave on the first night as he can already see the other two both have green eyes - he is only really waiting to see what happens on the second night)

On the second night, if the other two dragons up and leave, dragon X knows his original IF statement was correct - he does not have green eyes. However, if on the second night nobody leaves, then he knows he was wrong and he infact has green eyes like the rest, so they all leave on the third night. And that's what happens.


So you can see it also works for three green eyed dragons, right? And to summarise what the logical reasoning was for three dragons, it was just the same reasoning for the two-dragon version plus one more level (one more day). I.e. for three dragons, any dragon X reasons...

IF my eyes are not green THEN the other two dragons will figure out their eyes are green with the two-dragon method and leave in 2 days.
IF that fails THEN my eyes are also green and off we all go on day 3

So can you see the process for any number of dragons is just the same process for 1 less dragon plus another layer? So for N dragons, any dragon X can reason...

IF my eyes are not green THEN the other N-1 dragons will figure out their eyes are green with the N-1-dragon method and leave in N-1 days.
IF that fails THEN my eyes are also green and off we all go on day N

Easy right?
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I totally understand what you are saying and the reasoning behind the official answer.

My problem is that no dragon in this situation would ever say "'If my eyes are also not green, that other dragon can see no green eyed dragons, and will therefore conclude his eyes are green and will leave on the first night"

All the dragons know, from observation, that this will never, ever be the case. They know that all the others can see other green eyed dragons. They are questioning and waiting for an outcome that they know will never happen and basing their entire final conclusion on it.

For induction to work all the dragons, on the first day, are "waiting to see if any one of the other dragons see no green eyed dragons", but they know this cannot be the case.

To put it another way, each dragon would think "If one of us can see only blue eyed dragons then he will realise that he has green eyes and leave on the first night. However I know that everyone here will not see all blue eyed dragons and will see at least 98 green eyed dragons. I therefore know that no one will be leaving on the first night because no one can determine what their own eye colour is. I therefore cannot conclude anything from the first night or any night after that"

For the official theory to work, all the dragons have to fabricate a situation in their mind that they know does not exist. They would be coming to a conclusion based upon fabricated evidence.
 
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Ok, so I'm hoping to redeem my inability to read the dragon puzzle by offering some thoughts on the god puzzle.

One of the questions, probably the first question, must relate to something that you know to be factually true. For example, asking God A "does 2 +2 = 4?", would produce Ja or Da. A second question to the same god, God A, could clarify the answer to the first question e.g. "does the answer to Q.1 mean 'yes' in English?"

So this would be the results using the above answers:

"Ja" "Ja" - that god is either the truth god or the maybe god. This means you know factually that one of the other gods MUST be the false god.

"Ja" "Da" - that god is either the false god or the maybe god. This means you know factually that one of the other gods MUST be the truth god.

I'm stumbling at the last hurdle, but I think the last question must relate to the fact that the maybe god will not always give a consistent answer TOGETHER with the truth revealed by the previous two questions (see above). So something along the lines of asking God B 'will God A and God C always give a consistent answer?', or perhaps asking God B 'Will God A always agree with God C?'
What do you guys think? Hopefully with a bit of combined brainpower we can get there!
 
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