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

It doesn't(well obviously as it's flawed) and I can post more later if the other poster wanted to work through it otherwise it is a bit pointless - if he feels the shortcut still applies at at lower number then working through/explaining it becomes easier IMO. Though there is also the contradiction approach too.
 
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I'm not sure you understand what logic means...

as I said the answer on the site is not logical, it's all assumption based, therefore it can't be right :rolleyes:

and you're missing a fine detail, the same as the original site did. They have to find out that the individual dragon has green eyes, a very specific detail. The human did not say ALL of them have green eyes therefore by logic it means that 1 or more may not have green eyes. Because logic is based on facts rather than theories they would never be able to confirm said details about their own eyes because like I said they don't talk about it and they have no mirrors (they clearly ignored water/glass etc....). They've known for ages that 99 others have green eyes but nothing has happened because no one has told the others they specifically have green eyes meaning each dragon does not know that they have green eyes and must give up being a dragon.

If and I mean if the dragons took a leap of faith, which they wouldn't do as they're logical, then they could change on the first night. The logic would be that statistically speaking if 99 out 100 have green eyes then the odds are that all of them have green eyes and they'd all change on the first night. However they wouldn't do this because of the facts presented to them.

But then there is one side of the argument that everyone ignores...logically none of them would want to become a sparrow and give up their dragon 'powers'. They have to relinquish the power, it's not taken from them, meaning they have to give it up IF they ever find out they have green eyes, so by adding in the original rules this would never happen, even with the human comment. They would be doing everything they can not to find out.
 
The way which helped me understand it, was asking questions to myself, for example using 4 dragons:

I am a dragon who doesn't know their eye colour, I can see 3 other dragons have green eyes.

I have been told that at least one of us has green eyes, I already know this as I can see them, but because of this surely one of us should change tonight?

Hmm, at midnight nobody changed as they didn't realise they had green eyes, however it can't be me as they're 2 others that have green eyes?

3 days have passed now, the other 3 dragons haven't changed yet, we were told at least one of us has green eyes meaning one of us should have changed however after 3 days. Logically that must mean I have green eyes?!

All 4 of them change on the 4th night following their rules of if they realise they have green eyes, they change.
Now replace those numbers with any other and you'll get the same result.

You have to remember the dragon cannot change until he/she know's they have green eyes, they can't assume they have green eyes and then change!
I think this links into the flawed logic above.
 
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I've not read through the entire thread so not sure if I'm just repeating what others have said (I probably am). The original question asks what exactly is the new information given to the dragons, and that is that all the dragons now know that all the other dragons know that all the dragons know there is at least one dragon with green eyes.

Suppose there are 3 dragons, each with green eyes, A, B, and C
C knows that A knows that there is at least one dragon with green eyes.
C knows that B knows that there is at least one dragon with green eyes.
BUT, he doesn't know that A knows that B knows that there is at least one dragon with green eyes. C doesn't know that A has as much knowledge about what the other dragons know, as C himself does.

For A to know that B knows there is at least one dragon with green eyes, A has to know that C has green eyes, but C doesn't have that information, and therefore C cannot know that A knows that B knows there is at least one dragon with green eyes.

The man gives all the dragons the knowledge that they all know there is at least one dragon with green eyes, and so C now knows that A knows that B knows that there is at least one dragon with green eyes.

In other words, it wasn't the knowledge itself that was new information, it was the fact that all the dragons know about it, and know that each other knows this information. From that point, they can deduce that if no other dragon has come to the logical conclusion that they have green eyes by day 100, then they must all have green eyes.

It's ridiculously hard to explain clearly...

Although they all know that at least X dragons have green eyes, and that all the dragons know about this, they don't know that all the dragons know what they know about each other until the man tells them so.

(That's no better!!)
 
It's sunday again meaning answers and a new question. Here is the answer supplied by io9 for the 3 Question one, its a biggie ...

The solution to last week's puzzle is rather long, mainly because Boolos' puzzle is actually several puzzles bundled into one, and each of the constituent puzzles requires a fair bit of explaining. In his original presentation of "The Hardest Logic Puzzle Ever," Boolos "set out and solved three related, but much easier, puzzles," before tackling the big one. If you'd like to work through those puzzles, and read his solution, click here.


Can You Solve The World's (Other) Hardest Logic Puzzle?
Last week, we asked you to solve 'The Hardest Logic Puzzle In The World." This week,…
Read more
Below are two additional solutions to the riddle, helpfully provided by philosopher Brian Rabern, who helped me moderate comments on the Sunday Puzzle last week. The solutions he presents here are based off the ones provided in his 2008 publication, "A simple solution to the hardest logic puzzle ever."

Brian Rabern's Solutions

Most of the difficulty with this puzzle is in concocting the first question. In the worst case scenario, we address Random with our first question, and Random's answers are useless. How can we make any progress? We'd like to be able to figure out that a certain god is not Random, so that we can address our next questions to him. As Boolos states: "Your first move is to find a god that you can be certain is not Random."

There are different ways to do this, but they all rely on reasoning of the following sort (to make things easier let's simplify the problem by assuming for now that the gods answer in English, and that none of the gods is a liar. We will add these challenges shortly.):

Let's address god B and ask "Is A Random?" B either answer "yes" or "no."

If he answers "yes," then either he is Random (giving us a useless answer) or he is not Random. If B is not Random, then the answer indicates that A is Random. If B is Random, then C is not Random; If B is a truth-teller, then A is Random, and thus C is not Random. So either way we know that C is not Random.

If instead he answers "no," then either he is Random (giving us a useless answer) or he is not Random. If B is not Random, then the answer indicates that A is not random. If B is Random, then A is not Random; If B is a truth-teller, then A is not Random. So either way we know that A is not Random.

So if B answers "yes," C is not Random; and if B answers "no," A is not Random. Thus, after this question we have found a god that we are certain is not Random. From here on out, we will only talk to him.

Ok. But all of our reasoning up to this point has been carried out under two simplifying assumptions:

That the gods speak English, and
That there is no lying god.
First let's deal with the fact that there is a liar in our midst, and then deal with the language barrier.

There is a well-known trick – popularized by Raymond Smullyan's "Knights and Knaves" puzzles, as well as the movie Labyrinth – that transforms liars into truth-tellers while preserving the honesty of the truth-tellers. What we can take advantage of is the fact that even liars are sensitive to the truth of a proposition, it is just that they track the truth-values in the opposite way from truth-tellers. David Lewis wisely stated that "[Liars] are as truthful in their own way as we are in ours. But they are truthful in Liarese... and Liarese is a language like [ours] but with all the truth values reversed." (Lewis 1980, "Index, context, and content", 80)

Consider asking False the following question:

"If I asked you, 'Is 2 even?' would you say 'yes'?"

Since False is a liar, if we were to ask him a question whose correct answer is "yes" – such as "Is 2 an even number?" – he would in fact say "no." So when we ask him about how he would respond to this question, he will lie and answer "yes." A double-negative makes a positive. But True will also answer "yes" to this question. He would tell us the truth about 2's evenness, and he would also tell us the truth about his truth-telling. A double-positive remains positive.

The upshot: Whether you adress the truth-teller of the liar, posing questions of the form "If I asked you Q, would you say 'yes'?" yields "yes" just in case the correct answer to Q is affirmative. Otherwise, it yields "no." (This is a very useful device on Smullyan's island of Knights and Knaves. A similar device, used by Boolos in his solution, is biconditionals of this form: "Are you are a truth-teller if and only if 2 is even?")

Still, we are faced with the problem that the gods answer in a foreign language that we do not understand. However, we do know something about the language: We know that the words for "yes" and "no" are "da" and "ja," in some order. So if "da" means yes, then "ja" means no, and if "ja" means yes, then "da" means no.

First notice that the following question – with "no" in place of "yes" – would be just as useful in overcoming the liar:

"If I asked you, 'Is 2 even?' would you say 'no'?"

True will answer "no," and so will False, since he would in fact say "no" to "Is 2 even?," so he will lie and say "no" to the question about how he would respond. So we can also conclude that asking either the truth-teller or the liar questions of the form "If I asked you Q, would you say 'no'?" yields "no," just in case the correct answer to Q is affirmative (and yields "no" otherwise).

But now we can see that it doesn't matter what "yes" or "no" mean. It only matters that one of them means yes and one of them means no. So, in fact, the embedded questions overcome both the liar/truth-teller problem and the language barrier all at once. This is what Rabern & Rabern (2008) call the embedded question lemma – when either True or False are asked "If I asked you Q, would you say 'ja'?", a response of "ja" indicates that the correct answer to Q is affirmative and a response of "da" indicates that the correct answer to Q is negative.

Now that we can handle a liar in our midst and we can overcome the language barrier we can return to the first question without the simplifying assumption that the gods speak English and that there is no lying god. We simply have to embed our question "Is A Random?"; the disjunctive reasoning is the same.

Question 1: Ask god B, "If I asked you 'Is A Random?', would you say 'ja'?". If B answers "ja," either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is indeed Random. Either way, C is not Random. If B answers "da," either B is Random (and is answering randomly), or B is not Random and the answer indicates that A is not Random. Either way, you know that a particular god is not Random.

Now we know that we can talk to a non-random god. And given the embedding device it doesn't matter whether we address a liar or truth-teller.

Question 2: Go to the god who was identified as not being Random by the previous question (either A or C), and ask him: "If I asked you 'Are you False?', would you say 'ja'?". Since he is not Random, an answer of "da" indicates that he is True and an answer of "ja" indicates that he is False. (Note that other questions would also work here, e.g. "Does Da mean yes?"; if "da," then he is True, if "ja," then he is False.)

Question 3: Ask the same god the question: "If I asked you 'Is B Random?', would you say 'ja'?". If the answer is "ja," B is Random; if the answer is "da," the god you have not yet spoken to is Random. The remaining god can, then, be identified by elimination.

A natural thought that many people have when confronted with the puzzle is this: Are there questions that the gods can't answer? If so does that hold the key to a solution?

This is made especially salient when you start to wonder about how the truth-telling gods would answer questions about Random's future behavior. As was mentioned in the comments to last week's post, there are some interesting things here about how omniscient gods answer questions about random events.

One might reason as follows: Clearly an all-knowing god knows the results of all future coin flips, so why should there be a problem with True answering questions about how Random would answer things? That seems reasonable enough. But the answers are supposed to be completely "random," and this seems to imply that they could not be known or predicted in advance. How do we reconcile this problem? That depends on how we understand "random," and on more general issues relating to the foreknowledge of chancy events. If the gods are limited in this way, then indeed we can make use of this to solve the puzzle. ( See Uzquiano 2010 for a solution that makes us of these ideas.)

Besides, even if we think that the gods can answer any question about Random's behavior, there is another limitation – and this one is insurmountable.

Let me defend a bold claim: god does not exist. Well, at least, a god who always tells the truth does not exist. And we can easily prove it. To be precise, we can prove that a god with the following property does not and cannot exist; being such that one answers all yes-no questions truthfully with either "yes" or "no." We'll call such a god an "absolute truth-telling god."

Proof. Assume (to reach a contradiction) that an absolute truth-telling god exists. Ask her the following yes-no question: 'Are you going to answer 'no' to this question?' If she responds 'yes', then she affirms that she answers "no" but she did not answer "no." Thus, she did not tell the truth. If instead she responds "no," then she denies that she answers "no" but she did answer "no." Either way, we get a contradiction. Thus, an absolute truth-telling god is impossible.

NB: This is essentially Tarski's theorem, i.e. there is no truth predicate for a language, which has the expressive resources to talk about it's own sentences (while employing classical logic), that satisfies the T-schema where for a predicate to satisfy the T-schema is for a valid schema to result when this predicate is substituted for 'Tr' in 'Tr(ψ) <–> s' where instances of this schema are obtained by substituting sentences for 's' and substituting names of the corresponding sentences for ψ).

This shows that the truth-teller (and liar) will be unable to answer "yes" or "no" to some questions. Thus, there are not two but three possible reactions that a truth-teller might have upon being posed a yes-no question: "yes," "no," and head-explode. If so, then three possibilities can be distinguished with one yes-no question. Thus, using such questions allows for a two-question solution to the puzzle. (For the details see Rabern & Rabern 2008 and Uzquiano 2010).

This is why I introduced the new rule: If you are asking questions that, for whatever reason, gods can't answer (e.g. questions about Random's future behavior or paradoxical head-exploding questions), then you have to solve it with only two questions. Allowing yourself a third question is lazy and wasteful. You don't need it.

And here is this weeks question

Today's puzzle will be posed in two halves. The first half is a classic riddle – in fact, I suspect many of you will have heard it before. The second half, however, is an extension of the riddle that reveals its most common solution be be insufficient.

Sunday Puzzle #3: The Bear Hunter (Part 1)

The classic riddle goes as follows:

A hunter walks a mile due south, turns and walks a mile due east, turns again and walks a mile due north, only to find herself back where she started. The hunter draws a bead on a bear and shoots it dead. What color is the bear? Why?

If you've solved the riddle, already know the answer, or have already given up (for shame), proceed to Part 2. If you'd like a hint, click here. If you'd like an even bigger hint, click here.

Sunday Puzzle #3: The Bear Hunter (Part 2)

The bear you shoot will always be a polar bear, but the explanation for why this is – that the hunter must have started at the North Pole – is insufficient. While this is one possibility, the North Pole is not the only point of origin on Earth that satisfies the conditions presented in the problem. Can you think of any other point (or points) on the globe from which the hunter could begin her journey and find herself back at her original location?
 
Hmm that new one doesn't specify whether the journey has to satisfy that outlined in the original question or not :S
 
I like the 'head explosion' in reference to some of the potential questions having to result in an answer that is null or undefined :D

As far as that new problem - I've thought of one right away:

you need a circle 1 mile in circumference with its centre at the south pole - any point 1 mile north of that circle would then satisfy the problem... walking one mile south would hit the circle, walking one mile east would just complete a lap of the circle leaving you at the point you hit it and then walking one mile north would bring you back to your starting point

(so essentially any point 1/2π + 1 miles north of the south pole)
 
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That second question doesn't really mention shooting bar to qualify the North Pole answer therefore it would be at the South Pole location. You would need a location where if went you went south you would come to a point which would have a longtitudinal distance of eg exactly 1 mile, or 1/2 mile or 1/3 mile etc then you could go south one mile to that point then either once around, twice or three times around the globe and of course one mile back north. I am taking this from solely "satisfies the conditions" that is all the second question seems to ask.
 
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There are no polar bears at the south pole, so how can the bear you always shoot be a polar bear if you are at the south pole?

otherwise sound.

It's just asking for the same journey at some other point on the globe.

Hunter could be stalking penguins if you like though :)
 
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To be fair they were always going to struggle to come up with 100 questions with that level of difficulty.

I think they probably played the god question too early.

I would have expected a few more like this latest one to start things off.

I'm sure there will be others that lead to a bit more discussion.
 
If we need to shoot a bear at two points then could also be a bit pedantic and say you could start at magnetic north... then go south, east etc.. relative to magnetic north just as you could with true north.
 
If we need to shoot a bear at two points then could also be a bit pedantic and say you could start at magnetic north... then go south, east etc.. relative to magnetic north just as you could with true north.

Not for long! I hear there is an impending magnetic field flip...
 
Here's Sundays new one :
http://io9.com/youll-need-all-3-clues-to-solve-this-puzzle-1650957105/+Fahey

Sunday Puzzle #4: The Logician's Children

Two former college roommates, both logicians, meet at a conference after many years without contact. While catching up, the two eventually get around to discussing their children. The first logician asks the second how many children he has, and what their ages are. The second replies that he has 3 children, but (ever the logician) he will only reveal clues about their ages. The first logician must deduce for himself the ages of the second's children.

"First," says the logician, "the product of my children's ages is 36."

"Second, the sum of their ages is the same as our apartment number in college."

"Third, my oldest child has red hair."

Upon hearing the third clue, the first logician replies at once with the ages of his friend's children. What are they? How do you know?

and last weeks answer

Last week, I asked you to think of points besides the North Pole from which someone could walk a mile south, a mile east, and a mile north and find herself back at her point of origin.

To do this, one must travel to the very opposite side of the earth – or rather, just shy of it. Many of you came up with the correct solution (or close to it) in the comments, but an exchange between Peter Ravn Rasmussen and whoa, I believe, was the first to provide a complete answer. Rasmussen wrote:

Any point that is 1 + 1/(2 π) miles north of the South Pole will satisfy the conditions of this riddle's setup.
The reasoning is as follows: There is a line of latitude near the South Pole with a circumference of one mile. If one begins at a point 1+1/(2 π) miles north of the South Pole and walks one mile south, she will find herself on the mile-round line of latitude; walking one mile east will therefore bring her in a complete circle around the pole, such that when she turns and walks a mile north, she finds herself right back where she started.

But wait! As whoa points out, this answer is almost right, but not quite:

To [travel a mile south and find one's self on] a circle of circumference 1/n miles, whose radius r is 1/(2πn), one must travel a portion of the Earth's circumference (call the circumference 2πR) given by the angle one will reach off the Earth's axis. This angle is arcsin(r/R), and so the distance to any point on the circle from the South Pole is R(arcsin(1/2πnR)). That's very close to 1/(2πn) for all values of n.
The upshot is that one must begin from any point on a circle drawn at a distance slightly more than 1+1/(2π) miles from the South Pole. Alternatively, one could start at various distances closer to the pole, such that one's walk east brings her not once around the Pole, but two times, three times, four times, ..., or n times.

In other words, there are truly an infinite number of starting points to consider!
 
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I quite like the extra detail on last week's one. Hadn't thought of that as further possibilities.

For this one:

We know xyz = 36 and x+y+z=n (where n is known to both logicians)

We get possible options:

1*1*36 = 36 / 1+1+36 = 38
1*2*18 = 36 / 1+2+18 = 21
1*3*12 = 36 / 1+3+12 = 16
1*4*9 = 36 / 1+4+9 = 14
1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13
2*3*6 = 36 / 2+3+6 = 11
3*3*4 = 36 / 3+3+4 = 10

This means n is either 10,11,13,13,14,16,21,38.

If we assume that both logicians knew their apartment number (and so knew n), we can assume n must be 13 (as this is the only number that has more than one solution). For any other value of n, the logician would would have been able to solve the ages of the children with just those first two pieces of information.

If we now take:

1*6*6 = 36 / 1+6+6 = 13
2*2*9 = 36 / 2+2+9 = 13

The third piece of information only tells us that there is an 'oldest' child. This must mean that the children are 2,2 and 9.
 
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