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Acclaimed Hardest Logic Puzzle

06/12/2007 4:05 AM

Guys,

Hey, I just saw this on the net...have fun...

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.

Heres the Homepage http://philosophy.hku.hk/think/

Thanks

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#1

Re: Acclaimed Hardest Logic Puzzle

06/12/2007 6:54 AM

I'll have a go -

1. Are you a god?

2. Will both the other gods always tell me the truth?

3. Will both the other gods always tell me a lie?

The lier will answer No, yes, yes. The reliable one will answer yes, no, no. If the random one is always random, there is a good chance of no repetition.

Best can come up with in coffee break time!

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#2
In reply to #1

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 3:27 AM

It says EACH Q must be put to exactly 1 God.

So I take that to mean one question each.

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#3
In reply to #2

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 3:35 AM

I wondered about that, but I interpreted it the other way.

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#4
In reply to #3

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 5:06 AM

I'm certain there's no solution if you are only allowed to speak to one God (what if you chose Random). However on reflection I suppose the wording does allow for you to choose any one of the three depending on previous answers.

One thing I'm sure of is that the correct solution will never reveal which word means yes and which means no.

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#5
In reply to #4

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 5:18 AM

That was the problem I had............

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#6
In reply to #4

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 8:03 AM

It is 3 questions, one to each god.

This little riddle might keep me from getting anything done today, thanks guys! :)

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#30
In reply to #4

Re: Acclaimed Hardest Logic Puzzle

06/27/2007 4:38 PM

I think you are right about there not being a solution to the harder version that can always reveal which is yes and which is no. But you can frame the questions so that you can ask any of the gods any of the questions, and still get the same answer.

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#7

Re: Acclaimed Hardest Logic Puzzle

06/13/2007 9:19 AM

Puzzle aside the website is very informative and I will be reading it further.

Kind of a must for persons who have a boss thats hard to communicate with.

Thanks for the link!

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#8

Re: Acclaimed Hardest Logic Puzzle

06/14/2007 10:11 AM

Let " T " be the god that tells the Truth.

Let " F " be the god that answers Falsely.

Let " R " be the god that gives a Random answer.

Let " Y " = Yes. (ja?)

Let " N " = No. (da?)

Question to god one: Are you god R.

Question to god two: Are you god R.

Question to god three: Did god R answer question one truthfully.

Answer combinations of the first two question to god one and two are YY, NN, YN, NY.

If the answer is YY, only one can be false. And as T would have answered N to both question, only gods one and two can be F or R, and by deduction god three is T.

If the answer is NN, only one can be true. And as F would have answered Y to both questions , only gods one and two can be T or R, and by deduction god three is F.

If answer is YN, god one can't be T, therefore god one is F or R, god two is therefore T or R.

If answer is NY, god two can't be T, therefore god one is T or R, god two is therefore F or R.

By deduction god three is therefore T or F or R.

*****************************************

Answer combining the first two answer with the answer from god three, where god three is known, are:

Group one YY+Y, YY+N, NN+N, NN+Y.

Therefore by deduction YYY = R and F and T.

Therefore by deduction YYN = F and R and T.

Therefore by deduction NNY = T and R and F.

Therefore by deduction NNN = R and T and F.

*****************************************

Group two where god three is unknown.

YN+Y and YN+N = F or R + T or R + T or F or R,

*****************************************

Group two where god three is unknown.

YN+Y and YN+N = F or R + T or R + T or F or R,

By elimination, given YN + Y = F or R + T or R + T or F or R.

If god three is T then answer one is true, and god one = R, god two must be T or R, as god one and two can't be R and god two and three can't both be true, then god three is not T.

If god three is F then answer one is false, as F would not answer that R answered truthfully, god one = T,

god two must be T or R, as god one and two cannot be T, god two must be R.

Therefore by deduction YNY = T and R and F.

By elimination, given YN + N, = F or R + T or R + T or F or R.

If god three is T then answer one is false, and god one = F, god two must be T or R , and as god two and three can't be T, then god two must be R,.= F R T

If god three is F then answer one is true, god one = R. god two = T or R and as god one and two can't be R god two must be T. = R T F

As god three being T or F don't agree, god three must be R, therefore only the answer of god two is true.

Therefore by deduction YNN = F and T and R.

*****************************************

Group three where god three is unknown.

NY+Y and NY+N = T or R + F or R + T or F or R.

By elimination, given NY + Y = T or R + F or R + T or F or R.

If god three is T then answer one is true, god one = T . As god one and three can't be T, god three is not T.

If god three is F then answer one is false, god one = R. As god two is F or R , god one and two can't be R, and god two and god three can't be F. then god three must be R. therefore only the answer of god one was true.

Therefore by deduction NYY = T and F and R.

By elimination, given NY + N = T or R + F or R + T or F or R.

If god three is T then answer one is false, god one = R, As god one and two can't be R, god two must be F.

Therefore by deduction NYN = R and F and T.

*****************************************

wouldn't be the first time I was wrong?

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#9

Re: Acclaimed Hardest Logic Puzzle

06/14/2007 7:08 PM

Giving more though to Ja and da.

The gods that answer truthfully and falsely just answers the question.

The god that answers Randomly gives a doubtfully answer.

Therefore ja and da is the way the gods answer, not yes or no?

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#10

Re: Acclaimed Hardest Logic Puzzle

06/15/2007 12:08 AM

What do know group two is wrong, T would not have answered Y?

getting confused in my old age. Perhaps some one can correct it?

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#11

Re: Acclaimed Hardest Logic Puzzle

06/15/2007 10:03 AM

A philosophical question - Would the truthful and mendacious gods answer if I asked a question that they couldn't answer with either a truthful or false answer - e.g. if I ask the first of them what the the second would answer if I asked the second what the third would answer to the question "Do you ever lie?". Probably doesn't help with the solution, though.

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#12

Re: Acclaimed Hardest Logic Puzzle

06/15/2007 4:56 PM

I haven't looked at the site - yet...

If we can use inabilty to answer as suggested in #11, then we just need a way of identifying the meaning of an answer, and ensuring an equivalently truthful answer from either of T or F.

To identify the meaning of an answer, we conclude our question with: "will you answer me with da"

To ensure the same answer regardless of whether you are asking T or F, we start our question with "I intend to ask you the following question" before the question.

The question in the middle is "if I ask X a question, will he answer me truthfully"

The question is addressed to each ogd in turn, with X being the next ogd in line.

No answer identifies the random ogd as the next in line, so we ignore what he says; the other answer tells us everything else we need to know.

The post that says we don't find out which of ya or da is yes or know was quite right. If you really want to know, two times out of three you will have a question to spare (though you might have more important questions you could ask)...

[IF the above is an allowable answer, it's what I call a "one-loo-session toughie" - but by no means the hardest I've ever seen]

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#15
In reply to #12

Re: Acclaimed Hardest Logic Puzzle

06/16/2007 10:32 AM

The site seemed to be an advertisement for a book, and didn't give any answer...

But I realised that the above method (#12) only ever needs two questions... And I came up with solutions for two versions that were progressively more constrained. I think the question is poorly framed. My version follows - but is probably clumsy and may still be ambiguous, so suggestions for improvements would be welcomed.

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 god will answer only one question. But be careful - if you ask a question whose answer is undeterminable, the gods will strike you dead on the spot.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've already used all the logic techniques that are required, although the wording of my questions does need to be tidied up. If you need a further clue, feel free to e-mail me.

Fyz

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#16
In reply to #15

Re: Acclaimed Hardest Logic Puzzle

06/17/2007 4:59 AM

I see they would strike you dead with a dart or javelin, should duck that question.

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#13

Re: Acclaimed Hardest Logic Puzzle

06/15/2007 8:35 PM

"asking three yes-no questions; each question must be put to exactly one god. "

Looking at the interpretation of the uderlined statement, does this mean that the same wording or meaning can't be repeated? Like my post ( are you god Random ) is asked twice, does this mean I did not give the question to exactly one god only? Therefore does changing X in above constitute a different question. I like the answer, but interpreting the above is giving me problems, and I can do with out more problems.

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#17
In reply to #13

Re: Acclaimed Hardest Logic Puzzle

06/17/2007 9:39 AM

"you like the answer" - where did you find it?. But there is no need to repeat the question - if you make the core of the first question "is god X unreliable", the core of the second question can be "is god Y reliable". The third core could be "is god Z truthful". (X, Y and Z are not necessarily distinct). So you could have one question for each god, and all questions different. It is perfectly possible also to frame the questions so they are all the same - if you chose to make that an additional requirement - but that would make them rather long-winded.

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#18
In reply to #17

Re: Acclaimed Hardest Logic Puzzle

06/17/2007 11:02 PM

I was just being agreeable that it only takes two question to establish who is who, My first two questions where aimed at establishing if R was one of the first two gods, it then follows a YY or NN answer establishes that he was, the temptation then is to conclude a YN or NY answer that he was not, a YNY and a NYN solved by elimination is less conclusive, I resolved it by concluding because T and F don't agree on which is R, and as their answers are not random then one gives into temptation and conclude that god three was in fact R.

Because this appears far two easy "only a one loo stop", has led me to examine whether the logic is in the correct interpretation of the question. Is it a logical question or a misleading one.

I also agree that the web sight is about selling the book? Thou I do find it interesting and will look further into what it offers. Optimusprime is very quiet perhaps he or she may have the answer.

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#19
In reply to #18

Re: Acclaimed Hardest Logic Puzzle

06/18/2007 5:20 AM

I think I already presented the answer to the question as asked - two questions to two different gods - the questions are different because we have changed the name of the god. If you want additional changes, you can swap ya for da in the appropriate place in the second question. You could also swap "falsely" for truthfully in the second question.

However, the question as I believe it should have been posed is more interesting (post #15) - and can be answered using the same techniques - but it does require one additional 'insight'. I will post a hint next week, detail of the 'insight' the following week, and my version of the solution one week later (unless someone else does these first); in the meantime, I am happy to give hints via the CR4 email.

Fyz

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#20
In reply to #19

Re: Acclaimed Hardest Logic Puzzle

06/18/2007 8:58 PM

- two questions to two different gods -

Looks like I'll wait for more insight? more posts like #11 would be helpful.

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#14

Re: Acclaimed Hardest Logic Puzzle

06/16/2007 12:49 AM

"in which the words for yes and no are "da" and "ja", in some order."

Looking at the proposition of an unanswerable question. I see from the web site that the puzzle was first thought up by the philosopher and logician George Boolos, and looking at the underlined statement ( in some order), the gods answers are possibly Boolean logic. Then getting back to the unanswerable question eg: Will the next god be god Random? T answers Y and N truthfully giving you two possibilities , R answers Y or N giving you a random choice, F answer would be in the singular? and it would therefore be doubtful (da) that it can be used in a justifiable (ja) logical conclusion. So R and T would answer ja, and F da?

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#21

Re: Acclaimed Hardest Logic Puzzle

06/20/2007 10:06 PM

Hum??? Wouldn't you just ask 2 of them if they are true. Eiether or any that says no ask that or either if the other one was true? Then it is logic! am

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#22
In reply to #21

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 6:18 AM

Perhaps you are omitting a few stages? But... assuming that you can understand the answer, the only one who ever says no to that question (as you pose it) is old unreliable - and that one sometimes says yes. So you could ask the question as often as you wish without any guarantee of getting any usable information.

Even if you imbed the question to get an answer from T and from F that you can interpret correctly, I don't see how your choice of answerer for the second question works, as presumably he could still be R.

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#23

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 7:30 AM

Right, got it! You only need one question, because what you do, see, is preface it with 'you see this sword at your neck, right, the magic one guaranteed to kill 99% of all known gods, right, well, if you're going to lie, you'd better be good at it see!'

Sorted!

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#24
In reply to #23

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 7:57 AM

Are you sure you aren't one of George W's advisers?

The god might understand English and give a truthful answer, but as it is completely immobile and its vocal system is only capable of "da" and "ya", you still need two or more questions however well you frame them. In any case, it can almost certainly strike you dead between your making the threat and completing the first question - at which point your sword will be of little use.

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#25
In reply to #24

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 9:11 AM

Nnnnnnnno, seems perfectly logical to me!

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#26
In reply to #25

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 11:23 AM

Sorry, didn't realise you were one of the gods. Not certain yet whether you are F or R.

Ya-da-da di-dum-dum-dum

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#27
In reply to #26

Re: Acclaimed Hardest Logic Puzzle

06/21/2007 11:39 AM

Well. it didn't say it had to be nice..

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#28

Re: Acclaimed Hardest Logic Puzzle

06/22/2007 8:28 PM

Looking back at the question, as they are gods and we are only mortals we can not lay hands on them, and they don't answers questions with an undignified yes or no, they presumally sit in judgement? Where they dont agree (da) or judge that they are in agreement (ja), then da or ja is Yes or No depends on the question? And it does not neccesary need to be a direct question, such as are you R? but could be; is 1 and 1 equal to 2, Like a politicion we once had, give me the answer and I'll tell you if your right by agreeing or disagreeing. If the first two question are not right you end up back where you started. This is hindsight not as good as insight, but thats where I keep ending up?

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#29

Re: Acclaimed Hardest Logic Puzzle

06/23/2007 12:22 PM

Promised hint for the version where the gods kill you if you ask a question that they can't answer precisely with da or ya: the wording of the second sentence of Optimusprime's version of the question is significant.

Hopefully, no further explanation will be needed - but I'll post it in about five days if no-one develops it earlier.

Fyz

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#31

Re: Acclaimed Hardest Logic Puzzle

06/27/2007 7:53 PM

Having problems with the unaswerable question, or same answer? I presume that R can answer any question put to him, and a question put to T or F can be answered the same by both, or put in such a way that neither can answer. The problem I then have is, if R is the last asked, how do you determine who is T and who is F of the first two questioned? I think one question each of one god exactly means that a question cannot be addressed to more that one god at the same time simultaneously, requiring one or either god to answer? for example; is one of you god T, requiring either of the gods to answer. Still waiting for insperation.

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#32
In reply to #31

Final clue to original, and even harder version

06/28/2007 4:41 AM

Here is the threatened explicit version of the clue:

Part of the difficulty in this question is in realising what is the literal interpretation of the question.

Although R is named random, the question says that on each occasion his answers are either true or false; that means that they remain related to the question. Therefore, if you can find a wording gives the same answer for both true and false occasions, you have a method of obtaining accurate information from R.

So all you have to do is extend the method you use to get identical answers from T and from F so that it applies to R.


Now to the fully difficult version*: the gods are very touchy. If any one of them considers that the question they are asked has become inessential in the solution of the problem, it will strike you dead before you have all the information you need. And they are somewhat inappropriately pedantic about this: they will strike you dead if you could deduce the ordering from any two of the questions - even if you needed to the other question to decide what was best to ask.
*My invention, I think - and there are still solutions, I promise.

Enjoy

Fyz

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#33
In reply to #32

Re: Final clue to original, and even harder version

06/28/2007 5:53 AM

Sort of in the spirit of the problem "I think I disagree". I do not believe it is possible to gain any useful information from Random [unless you ask him the same question twice and (you happen to) get different answers (which would confirm him as Random)].

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#34
In reply to #33

Re: Final clue to original, and even harder version

06/28/2007 8:01 AM

A general remark: there are six possibilities, and only three questions. Two questions with only yes/no answers could at most sort out four possibilities reliably. Therefore, each question has to provide data, so you can't "waste" any questions.

The wording has to be very careful - but I'm as certain as I know how that I've got a wording that works.

I think you will get there if you think about how you might word a direct substitution. Otherwise, you can pull my answer to pieces in about five days time...

Fyz

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#35
In reply to #34

Re: Final clue to original, and even harder version

06/29/2007 2:40 AM

Not giving any useful information turns out not to be quite as useless as it first seems. It looks as though we have differently interpreted the question, so, I'm going to give my answer, which I only worked out fully when you gave me the extra nudge the other day. First I'm going to explain the principle: partly because I think it makes it easier to understand, and, partly in case I've made a mistake in the detailed questions which I can't be bothered to work all the way through.

The first question to the first god has to force both True and False to correctly indict the second god as Random. Notice (and I think this is the clever bit) that if the first god happens to be Random it doesn't matter because we now know that the third is either True or False.

The second question to the third god simply forces that god to reveal if he is True or False. (The question I'm going to use here seems over complicated, I'm sure there are better ones, but I worked it out earlier when I would have asked it twice of the first two gods.)

For the third question, and, I'm returning to my belief that you can ask the same god two questions (otherwise there is no solution), you simply get the third god to tell you which way round the other two are.

Question 1: to god A "If someone asked you if B was Random would you answer 'da'?".

If the answer is 'da' then C is True of False, if 'ja' then B is True or False.


Question 2 to B or C as determined by Q1: "If other visitors asked the other two gods if they were gods would they both always answer 'da'?".

True must always answer 'ja'; False must always answer 'da'.


Question 3 to B or C as determined by Q1: "If someone asked you if A was Random would you answer 'da'?".

If the answer is 'da' then A is Random.

Now that I've written this out I notice that you could use the first question twice, but, substituting False for Random and B for A. If you did this it would allow you to ask the third question to the last god, but, only if he was revealed as not being Random.

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#36
In reply to #35

Re: Final clue to original, and even harder version

06/29/2007 6:00 AM

!!! I think you have a valid solution to the problem as originally posed.

BTW, once you have this sorted, I think there are multiple choices for Q2 and Q3.
For example, Q2 could be equivalent to Q1, except that the target is god A. Then you know which god is R, and Q3 straightforward (but still might have to be a repeat).

I can't see how you can be certain to ask a each god a question. I suspect it to be impossible - unless you know different?

Now for the modified "really difficult" one? I think that needs another inventive step.

Congrats, and regards

Fyz

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#37
In reply to #35

Re: Final clue to original, and even harder version

06/29/2007 9:31 PM

The wording of Q2, "- would they both always answer 'da'?" (problem with the use of the word always?)

True must always answer 'ja'.

This implies that 'da' is not refering to a yes or no answer but to an understanding between themselves, do you agree?

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#38
In reply to #37

Re: Final clue to original, and even harder version

06/30/2007 9:28 PM

Sorry about the above it's nonsence, Ive had a night sleep since then. First I would like to establish the meaning of A,B and C, as they have two meanings, First they could mean the order of seating arrangement of the gods, A on the left, B in the centre and C on the right; Second A is the first god questioned, B is the second god questioned and C is the last. In the first scenario when B is refered to, the others know who I'm refering to, in the second B is a random choice, In the following A,B and C is Left,Centre and Right.

Q1 If A was R then the answer would correctly indicate whether B or C was T and F, but if A was T or F then the answer would tell you who A is?

Q2 Acting on the assumption that A is R then asking the god indicated by his answer, you can once again ask a yes-no question, If he is T or F the answer tells you who he is.

From this point the combinations are TT or FF or TF or FT. As the first two gods can't be the same, the combination TT and FF establishes that random is not the last god, the temptation now is, if R is not established as one of the first two questioned then he must be the last god, now R might have been a smart B and masqueraded as T or F depending on the answer given by A? but if he was a smart A then he may have answered knowing that the next god would not be able to reveal him as A? So who would you ask next to reveal R?

Asking the gods a second question on the surface appears not to be in the spirit of things, but looking at the problem again prompted by your answer, and by changing the wording to reveal more enlightment (insight) you may be right? Based on Yes-no being hyphenated and the wording, each question to one god exactly. This then gives more insight into the answer, they do not answer, yes or no, to each question directly, but will give a single answer, indicating the yes and no in some order, this may be what Physyist may be getting at in the wording of his questions. Fyz is that correct?

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#39
In reply to #38

Re: Final clue to original, and even harder version

07/02/2007 4:51 AM

No - I just meant one question addressed to each god, which they would answer specifically. But they would get upset if there was any way that they could interpret the question as meaning that their contribution became unnecessary.

However, it so happens that my proposed solution means that they could wait until you had asked all three questions before any of them gave an answer, so you could reword the question to require that if you want. (But we do need to know which god answers which question - we would need five gods to make that unnecessary).

I hope that somewhere in there I've answered the correct question!

Regards

Fyz

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#40
In reply to #39

Re: Final clue to original, and even harder version

07/02/2007 5:50 AM

Thank you for that extra bit of insight I will take in on board, I had not thought of that alternative. I must admit to pages of questions and answers with out a solution, hence the speculations. I am enjoying this and repect people like yourself and Randall who are prepared to put up and be counted. I must now admit to a weakness, I've ordered the book and should recieve it in the next few weeks, Before the end of the months I hope as I'm off to the UK for a holiday. Will not post the book answer unless requested, and if I have enough will power I will not look at it till them???

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#41
In reply to #40

Re: Final clue to original, and even harder version

07/02/2007 6:43 AM

First, the book might give a more precisely worded version of the question than appeared above (I suspect the internet postings might have been reworded for copyright reasons). Second, whatever wording you come up with, there will be multiple solutions.

Hopefully, the book will include discussion and different sets of constraints for different solutions, which will put you in a good position to amend or add to whatever Randall and/or I can come up with - so I can't see any reason to wait before reading it - but there is a chance that it will be unnecessarily hard work due to the use of supposedly mathematical syntax*. Randall's version of the first question was very elegant and spare - I'd be interested to know whether Boolos manages anything nearly as good.

Fyz

*In his "miscellany", Littlewood makes some remarks on that topic which I think are rather apt)..

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#42

Re: Acclaimed Hardest Logic Puzzle

07/02/2007 9:02 PM

Yes I liked Randall's first question and summation and have thought along simular lines, but not simular wording, Yes I think the book version would be copyright and would possible restrict me and postings. So expanding on what Randall and you have said, I would like to put up the following argument. Looking at some of the ground rules if one accepts that more than one question can be directed to the same god, or that they are not required to answer untill all questions have been put, then the following logic could be put?

'Not giving any useful information turns out not to be as useless as it first seem.'

There is a unique question that can only be put to A, if the above rules did not apply, and normally could be said not to reveal any useful info. (Q1 to A) I intend to ask three questions, will my next question to be answered truthfully?

T must answer truthfully or not at all, he cannot answer the above, as I have a random choice as to whom I ask the next question, and as that might be to F or R then he can not answer truthfully? The same applies to F as the next question could be to T or R he cannot answer Falsely. And if A is not R then it would appear that I have no information at all, and if I'm only allowed to ask one question of each god, the second god knows who I've asked and who I'm going to ask next, and it's lost its uniqueness.

Now if the rules are changed, and if I can put the same question to the next god still retaining my choice of a random selection as to whom I put the next question to, I have all the info I need?

===============================================================

(Q1 to A) I intend to ask three questions, will my next question to be answered truthfully?

If A is T of F then there is no answer. But if A is random he will be very upset at being revealed, and being compelled to answer, and will tell you that you have a dead argument (da) and it will leads you no wear.

(Is that in the spirit of things Fyz?)

(Q2 to B if A is not equal to R) Q1 is again put this time to B.

We have now isolated R, if he answered Q1 he is A, if Q2 he is B, else he is C.

(Q2 to B if A = R) Was it possible for A to answer my question truthfully?

If B is T then he will answer that I have a justified argument (ja) and should continue.

If B if F then he will answer that I have a dead argument(da) and it will lead no wear.

Then if the answer is ja, B is T and C is F, if da then B is F and C is T.

(Q3 to A if R is B or C) Was it possible for (B or C, select which ever is R) to answer my question truthfully.

The logical deductions are the same as Q2 above.

This is my second attempt how does it rate, it all depends on Fyz and Randall being right?

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#43
In reply to #42

Re: Acclaimed Hardest Logic Puzzle

07/02/2007 11:38 PM

Q2. Is it possible for X to answer a question truthfully?

And substitute nowhere for no wear, as I doubt he'll give you the bear facts.

JD.

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#44

Getting a consistent answer from "Random"

07/04/2007 9:02 AM

Here, as threatened, is a possible format for a question that should elicit the same answer from any of the three gods - based on the precise wording of the originally posed puzzle. I'm certain it is "functional", but the present form seems rather inelegant to me - so I'm begging for improved wordings.
Once you have this, it becomes straightforward to adapt Randall's solution so that you can ask each god one question. So it's on to the really difficult amended version - if anyone has the stomach for it...

If I were to ask you the following question, and you were to answer it as truthfully as you will answer this question, would you answer da? <Question of choice>

Fyz

P.S. regarding whether this is necessary for the question as posed, the literal interpretation of "each question must be put to exactly one god" is that you may not ask the same question of two different gods. This itself is ambiguous, however. For example: "Are you truthful" addressed to gods B then C is as different logically as asking "Is C truthful" of B and "Is B truthful" of C. If we assume pedantic gods, none of these would be allowable; similarly "Is either of B or C truthful" is the same as "Is A truthful" (even though the answer is inverted). Maybe answering within these constraints and being allowed to ask the same god two questions is challenge enough?

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#45
In reply to #44

Re: Getting a consistent answer from "Random"

07/04/2007 9:33 AM

If you're constraining Random to be either truthful or honest for the whole "transaction" then:

"If someone asked you <question of choice> would you answer 'da'?" does it for all gods.

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#47
In reply to #45

Re: Getting a consistent answer from "Random"

07/04/2007 10:03 AM

The constraint is that he answers the one question either truthfully or falsely.

Your question has to provide a constraint that means that only the same answer as the others would give is true (or false).
If in doubt, try to follow my version through by pretending to be random - decide to answer the question as I pose it truthfully (then falsely), and see what answer you have to give (you may find a hole, which would show that improved wording is needed)

Unfortunately, that simplified version does not provide this constraint.

BTW, my first thought was explicit substitution "If instead of this I were to ask..."; unfortunately, this doesn't work because the true/false nature of the answer can be question-dependent without invalidating randomness of state (random1 x anything_independent_of_random1 is always random).

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#53
In reply to #47

Re: Getting a consistent answer from "Random"

07/05/2007 4:41 AM

I've got a couple of major problems here. The first is in the use of the phrase "the one question".

Second lets forget the da/ja bit for a moment, and try your question on the False god.

If I were to ask you the following question, and you were to answer it as truthfully as you will answer this question, would you answer yes? Do 2 and 2 make 4?

There's an implicit question in "answer it as truthfully".

Say to the false god "Are you honest?", and, he will answer yes.

Now say to him "Ask yourself this question: "Am I honest?" "

Which answer does he now have in his head? This seems pretty tricky for False let alone Random.

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#54
In reply to #53

Re: Getting a consistent answer from "Random"

07/05/2007 7:10 AM

Some neat ideas here - however, I am puzzled whether some of them are relevant to the problem.

Regarding the implicit "second" question: there is no problem to rephrase the internals as conditional statements - I've only used a form that includes an implied question (or questions) because it is appreciably less hard to follow. The form would be: present a set of conditional statements (these would have to include all necessary double negatives), and then ask "Is the final statement true".

The idea that there can "unanswerable" questions is interesting - but not of itself relevant unless you allow the use of confusion to add information (N.B. that I don't think my original version of this would necessarily work, because random does not mean unknowable - in any case, there is no reason that Random's proposed answer should not be determined at the point you specify the question to the other god).

Given that all the gods conform to their individual defined consistent behaviour, they are not setting out to mislead - so I would say they are all honest. So I'll rephrase your question as I understand it (addressed to false, or to Random in false mode):

"If, maintaining the same degree of truthfulness with which you will answer this question, you ask yourself whether the answer you give me answer to this question is true, would you answer da"?

I'm not certain whether the question is unanswerable or not - if the true answer is that the god could not answer at all, then the true answer is "no", and the lying god will reply whatever means "yes".

I've got some other thoughts on the constraints of the puzzle - I'll put these in a note to JD as he is currently on that tack.

Fyz

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#46

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 9:34 AM

Hi, sorry if I've missed something here, but I still don't see how you know which way around Ja and Da are for yes and no, hence how can you identify the gods?

If anyone has the patience please explain and type slowly 'cos I can't read very fast.

Thanks.

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#48
In reply to #46

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 10:34 AM

The whole point is that you never get to find out which way round ja and da are:

Suppose you are True, and, I ask you "If someone asked you if 2 and 2 make 4 would you answer 'da'?". If 'da' means yes you answer 'da'; if 'da' means no you answer 'da'.

Now suppose you're False, and, I ask you "If someone asked you if 2 and 2 make 4 would you answer 'da'?". If 'da' means yes you answer 'da'; if 'da' means no you answer 'da'.

Now do it all again for <if 2 and 2 make 5>.

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#49
In reply to #46

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 11:02 AM

I'll try: the clue is that you don't need to know which is true and which is false. What you do is ensure the answer gives the information you need.

Here follows the "slow" version.

Question (not useful except as a demonstrator):
If I ask you a question whose true answer is yes, would you answer da?

Posed to truthful, with da=yes:
He would answer yes=da to the internal question, so he will answer yes=da to this.
Posed to truthful, with da=no:
He would answer no=ya to the internal question, so he will answer no=da to this.
Therefore, truthful will answer da to a question whose truthful answer is yes - independent of the meaning of da.

Posed to false, with da=yes:
He would answer falsely=no=ya to the internal question, so the false answer to this is yes=da.
Posed to false, with da=no:
He would answer falsely=yes=da to the internal question, so the false answer to this is no=da .
Therefore, false will answer da to a question whose truthful answer is yes - independent of whether the god speaks true or false, and of the meaning of da.

Hopefully, this shows that this form of question will result in the answer "da" to any question whose truthful answer is yes. Following the same logic, you would get "ya" if the truthful answer is no.

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Fyz

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#51
In reply to #49

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 11:28 AM

Corrections for mistypes in descriptions:

Posed to truthful, with da=no:
He would answer yes=ya to the internal question, so he will answer no=da to this.

Posed to false, with da=no:
He would answer falsely=no=da to the internal question, so the false answer to this is no=da .

Apologies

Fyz

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#50

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 11:13 AM

Thanks guys, its starting to make some sense now.

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#52

Re: Acclaimed Hardest Logic Puzzle

07/04/2007 11:08 PM

Yes that's a lot better for us slow readers, as I have to spell it out blow by blow, others it appears can jump over the obvious to get at the nitty gritty. Yes I think my version two brakes to many rules. And through the wording in the problem is ambiguous I think it's intentional, and put in such a way that one has to analyse it as well as find a solution. So if I can recap on the question, as rules are getting tighter, and state my understanding, as it appears they are going to get even tighter?

' Your task is to determine the identities of A , B , and C by asking three yes-no questions; '

This has at least three meanings; One, a specific question requiring a yes or no answer; Two, as yes-no is hyphenated it could be that the question should be worded to contain a yes-no statement, or Three, simular to two but with onus of yes-no being put on the gods and not the questioner, meaning the question has both a yes and no answer.

' each question must be put to exactly one god. '

This to has numerous meanings; "each" can be seen as singular or plural, all questions must be put to just the one god, or only one question to each god. Then again it may mean that a question must be address to a single god, and not to two requiring either to answer, now this last point is interesting, because it opens up the possibility raised by Randall that you can ask more that one question to a god, as it only excludes you from address the gods with a question without specifying who. But if it means we can't cross reference, and must be specific to that god only makes it harder.

' 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. '

Again is open to interpretation, is ja and da specifically yes and no, or are they question related, or if we have put a question with a yes-no statement in it, is the yes-no asked answered in some order.

======================================================

As you have written more slowly, so we mere mortals can understand, and with the new insight, I'll look more closely, and hopefully understand will come to me.

Regards and respect JD.

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#55
In reply to #52

Re: Acclaimed Hardest Logic Puzzle

07/05/2007 9:46 AM

Generally, I think we should assume that the gods are working to the constraints literally as they have been presented to you. Other than that, they are totally free. To my mind, that means we must frame our questions so that any godly behaviour that conforms to these constraints will allow us to extract a valid answer. To that end, this e-mail goes through the constraints as I understand them - and also looks at the implications of some of the behaviours that are left unspecified. There is some new material, although I will rephrase some stuff that exists elsewhere - including some stating of the obvious.

"Three gods A,B, and C". Presumably "gods" implies supreme logicians, and also that they know everything relevant to the exercise. Also, that you can use A, B, and C as identifiers.

"are called, in some order, True, False, and Random" - also usable as identifiers, but probably need to be preceded by "god" or something to avoid ambiguity.

"True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter".
Clearly, for a given answer, Random speaks either truly or falsely. Therefore, if a question implies the same answer whether he speaks true or false, he will give that answer.
N.B.1 that all of the gods' internal thought processes have to be assumed to be logical, to allow them to achieve truth uniquely under all conditions. (This is to clarify what happens when a god asks himself questions as suggested by Randall).
N.B.2 "random matter" does not mean unknowable. Nor, to a god, necessarily unpredictable, as a god may not exist in time in the same way that we do. Merely that there is no rhyme or reason as to which decision he will take at any particular time or for any particular question (both may affect the outcome, provided that there is also a random process). (This invalidates my original two-question method, even if godly confusion were allowed as a method of extending the range of answers)

"Your task is to determine the identities of A, B, and C by asking three yes-no questions;"
Questions with English answers other than yes or no are not allowed (don't ask me what happens if you try - they may strike you dead, or at best simply ignore you. (Given our total lack of understanding of the gods' language, the answer to any other sort of question would be wasted on us. That means that this constraint is implicit in our knowledge if we need to use the questions to discriminate between the gods)

"each question must be put to exactly one god." Semantically**, that means that you may not repeat a question to a different god - although you can ask the same question twice of the same god (not that l can see any benefit from doing that). This leaves an ambiguity - what will the gods consider "the same question". I think we need to be cautious here, as the gods' behaviour is only constrained provided we stick to the rules as they have interpreted them. To be safe, the following must therefore be treated as disallowed (examples):
"Are you god Random?" addressed to two different gods.
"Is C god Random?" addressed to two different gods, therefore also:
"Are you god Random?" to god C, and "Is god C Random" to one of the others, and similarly:
"Are you god Random" to god C, and "Is B god random" to god A.
"Is either of A or B god Random" to any god, and "Are you random" to god C.
This last one is on the basis that two questions that give identical information are the same, even if the wordings are different and the answer inverted. But if this last pair is identical and so is the first pair, then:
"Is either of A or B god Random" to any god, and "Are you random" to any god would also be disallowed.
This is a case where "blind" application of logic leads to a surprising result. IT may not be sensible, but the gods are within their rights to come to this conclusion - so we must account for it.
**Each as an adjective implies the individuals have this characteristic - it is not the same as every, though in this case the constraint in the other half of the sentence ( = "precisely one") means the result would be the same even if it were.

I think this is the best point to cover one aspect that is not defined: when the god will answer. For all you know, the gods may wait until all three questions have been asked before answering. If you ask two questions of one god, there is no reason to suppose that he will answer them in the order you ask them. (Of course, you could with some effort frame the questions to include the order of response in the outcome, and so overcome this issue. But it seems unnecessary.)

"will answer in their own language, in which the words for yes and no are "da" and "ja", in some order." I think we should assume that their language is a language in the sense we understand, so that "da" and "ya" each have consistent meanings, which are the same (yes or no) for all questions and for all gods. Actually, I do not think it is not necessary to know this, but is one less thing we need to think about.

In conclusion, the implications of the above (that have not yet been satisfied in any of our posted answers) are:
We need to ask exactly one question of each god.
No two questions should either appear to elicit the same piece of information or have a wording that would allow it to elicit the same piece of information if addressed elsewhere.
We cannot afford to use the answer to one question to determine another.

I have yet to work out a way of doing this that does not require a properly interpretable answer from god Random - so I assume we need to word at least one of the questions so it will elicit meaningful answers regardless of which god is addressed.

Fyz

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#56

Re: Acclaimed Hardest Logic Puzzle

07/05/2007 7:47 PM

Yes your remarks about Random are interesting. If taken in a literal sense it would mean a answer based on the toss of a coin, but if these gods think internally to express a logical answer, then Random may have a human weakness and his answer may be one of self interest, and what would appear, to us, to be a random answer, would in fact be a calculated answer designed to hide his identity. It is simple to ask a question where both god True and False give the same answer, ( Do you answer truthfully? god True = yes, god False = yes), and god Random would also answer yes to hide his identity. I find it difficult to word a question where god Random is compelled to answer otherwise, as he can internally consider the implications, and answer accordingly. Then looking at the problem, if god Random is the first questioned, then you have a 50/50 chance of exposing him, but if he's the last to put a question to, he's not going to reveal anything. I still like post #11. A question that can be answered with both a yes and no, that neither god True or False can give an answer to, and, still remain in character? (da) don't answer, (ja) judged answerable?

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#57
In reply to #56

Re: Acclaimed Hardest Logic Puzzle

07/06/2007 7:31 AM

No, the literal sense is that the answer would either be be true or false, with truth or falsehood selected randomly. For most questions, that would be the same as being randomly ya or da. But it is possible to frame a question that gives a consistent answer. I'll give an example** that always gives the answer "Da" to illustrate why the answer need not be not random. Obviously, this gives no information - it's just to set you on your way by illustrating that you can control the answer:

"Proposition 1: both da means yes and the question is answered truthfully.
Proposition 2: both da means no and the question is answered falsely.
The question: is either of proposition 1 is or proposition 2 true?"

**Deliberately unhelpful for this problem, and worded to satisfy Randall's objection to embedded questions.

BTW, unless you make false assumptions about which god is which, there is nothing that Random can do to mislead you - so it doesn't really matter if its state of veracity is derived truly randomly or out of malice.

Also, I have yet to come up with a question that is both unanswerable by True and by False, that is also helpful under worst-case assumptions. As it's not necessary, I would regard it as an unprofitable diversion - except as a matter of interest once you have a solution for the original puzzle, of course.

Fyz

P.S. I wonder what happened to Optimusprime (who posted the problem)?

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#58

Re: Acclaimed Hardest Logic Puzzle

07/07/2007 9:38 PM

Well what do you know looking for some logical statement I come across the following. Question and Answer - The hardest logic puzzle ever.

Anyone what to buy a book?

http://en.wikipedia.org/wiki/The_hardest_logic_puzzle_ever

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#59
In reply to #58

?Safe Answer to "Acclaimed Hardest Logic Puzzle", and a harder version

07/09/2007 6:59 AM

Given jdretired's "spoiler", here is what I believe to be a valid answer. Critique welcomed. (I'll look at the wikipedia answer when I have a few minutes)

First - there are many possible answers.
What follows is the simplest that I have found that satisfies Randall's requirement for "no subsidiary questions". I have also chosen a wording that should not require specialist knowledge of mathematical interpretation. The questions can be asked of any god in any order (so long as no god is asked two questions - the reason this is necessary is given in post #55).

Q1. Is an odd number of the following three statements that form part of this question true?
Statement 1: Your answer to this question is truthful.
Statement 2: Either A is False, or both A is Random and B is False
Statement 3: Ya means Yes
That completes this question
[Note: the answer to this question divides the six possibilities into two groups of three, and in each group B can still be any of the three possibilities]

Q2. Is an odd number of the following three statements that form part of this question true?
Statement 1: Your answer to this question is truthful.
Statement 2: B is Random
Statement 3: Ya means Yes
That completes this question

Q3. Is an odd number of the following three statements that form part of this question true?
Statement 1: Your answer to this question is truthful.
Statement 2: B is True
Statement 3: Ya means Yes
That completes this question

Now to my 'more difficult' version.

Three gods A, B, and C are called, in some order, True, False, and Random. True only speaks truly, False only speaks falsely, but whether Random speaks truly or falsely is a completely random matter. The gods will answer up to a total of three yes-no questions; each question must be put to exactly one god, but the gods will not answer any questions if the answers to any two questions make the answer to the third question unnecessary. 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.

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#60
In reply to #59

Re: ?Safe Answer to "Acclaimed Hardest Logic Puzzle", and a harder version

07/09/2007 8:27 AM

Yes it is a spoiler, Im sorry about that I should have been more patient, and should have given fair warning. I will study the next one and see what I come up with.

Sorry. JD.

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#61
In reply to #58

Re: Acclaimed Hardest Logic Puzzle

07/09/2007 11:41 AM

If the unamended wikipedia article does indeed represent George Boolos' answer, there is no way whatsoever that I could recommend that book. It trivialises what could be a reasonable, although not exceptionally difficult, puzzle. (The wikipedia article even advocates the "confusion" protocols that are explicitly excluded by the "one question per god").

I have amended the wikipedia article. I look forward to seeing how my comments are amended/deleted or otherwise handled.

Fyz

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#62

Re: Acclaimed Hardest Logic Puzzle

07/11/2007 6:20 AM

Yes I see your editing on the web page. Interesting.

I recieved the book today, and for those that have been followed this puzzle, the same wording that is in the book is at the following web sight, I will revise the above to make sure I correctly understand it, and get back to it.

http://people.ucsc.edu/~jburke/three_gods.pdf

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#63
In reply to #62

Re: Acclaimed Hardest Logic Puzzle

07/11/2007 9:11 AM

Firts - I could see no differences between the version posted by Optimusprime and the one in Boolos article - have I missed something?

With regard to the literature - I find this most curious. As a semi-bystander, I would have expected the "professional experts" to have solved the problem while making the minimum of assumptions (i.e. no need for an early answer). I would also expect that they would ensure that the questions survive the worst-case interpretation of "each question must be put to exactly one god". The Boolos article seems to fail on both counts. Other contributions fail on the strict interpretation of yes-no questions as well as on the immediate reply assumption.

At the other end, so far as I know, the only assumption that I have made (other than that the gods accept the constraints and operate precisely as described in the question) is the unavoidable one - that all three answers will be given in time for me to deduce the identities of the gods.

I was a bit concerned about the wikipedia page, as my comments appeared to vanish for a while - but it's clearly there now. I'm not certain whether this would be because I was accessing via a proxy server that had not been updated, or it was removed while someone assessed it and was then replaced.

Regards

Fyz

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#64

One-question answer to Boolos' "clarified" version, and answer to Fyz' extension

07/23/2007 8:12 AM

I believe that the question as "clarified" by Boolos (that the gods will answer immediately the question has been asked) can be solved with a single question - for example:

"This question is for the god called False:
Is exactly one of the following two statements true?
Statement1: Either A is called True or you are A and B is called True.
Statement2: Ya means Yes."

Can anyone see a flaw?


Here follows my proposed answer to the question as originally presented here, but with the addition that the gods regard it as ill-mannered to answer any question if there is any possibility that any answers would make another answer redundant is:

Q1:
Is an odd number of the following three statements true?
Statement1: When you answer this question, you will speak truly
Statement2: Either A is called True, or A is called False and B is called True
Statement3: Ya means yes

Q2:
Is an odd number of the following three statements true?
Statement1: When you answer this question, you speak truly
Statement2: Either B is called True, or B is called Random and Ya means yes
Statement3: Ya means yes

Q3:
Is an odd number of the following three statements true?
Statement1: When you answer this question, you speak truly
Statement2: Either B is called False, or B is called Random and Ya means yes
Statement3: Ya means yes

These questions may be asked of any god, provided that each question is asked of a different god* (to remove confusion should they be asked out of order).

Landon Rabern apparently considers the modification to be trivial, and the questions that solve it in some sense inferior. My view is that both sets only guarantee an answer to the puzzle after the third question is answered, and the modified version additionally gives a one-in-three chance that you will also know the meanings of "da" and "ya" at the end of the process.

*If it is required that two questions be asked of the same god, Q2 and Q3 can be modified to allow this. However, that would remove the 1/3 possibility that we could find out the meanings of Da/Ya

Fyz

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#65

Re: Acclaimed Hardest Logic Puzzle

05/23/2008 5:32 AM

Hi Guys!

Whew took time to finish all those reply of yours...

Even I dont know the answer, I just shared it...

soory for th LATE reply =)

Wikipedia, and some other sites already have the answer(?)..

I'll just share again their thouths,,,thanks

http://en.wikipedia.org/wiki/The_Hardest_Logic_Puzzle_Ever

http://www.hcs.harvard.edu/~hrp/issues/1996/Boolos.pdf

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#66
In reply to #65

Re: Acclaimed Hardest Logic Puzzle

06/19/2008 7:35 AM

Hi,

Here's a generalized version of the famous Smullyan/Boolos Hardest Logic Puzzle. I don't know of a solution yet, I even don't know if it is indeed harder although it looks like (but I have a conjecture based on my work on the original one). Here it is, for what it is worth: Suppose we are confronted with n+1 gods, numbered G0 ... Gn. G0 always tells the full truth, i.e. G0(p)=p all p. G1 deviates randomly from G0 on some statements and/or some occasions, i.e. G1(p)<>G0(p)= p for some p. G2 deviates still more from G0 but in such a manner that deviations by G1 are retained. Etc. Gn will deviate from G0 on all statements, so he is the real and only full liar. So there will be a hierarchy of ever more untrustfull Oracles in between one God and one evil Devil (G-O-D). All Gods know their own identity as well as the identity of the other Gods (or at least their immediate precursors/successors in the hierarchy?) Is there a simple way to reveal each creature's identity by asking just n+1 questions (although there are clearly (n+1)! different possible orders? Note: If n=2, we have the original puzzle. If you are interested you may contact me at p.h.vossen@googlemail.com.

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#67
In reply to #66

Re: Acclaimed Hardest Logic Puzzle

06/19/2008 4:44 PM

This puzzle is equivalent to framing a question so that it has self-verifying multi-value (>3 possible) answers. It is achievable if and only if fully-false means that if there is more than one false answer, at least two will be given. An example of a sequence of questions that would serve in that case is:
"Show me the god for whom there are exactly r gods who answer truly more often than he does"
You will need to ask this question for n different values of r [i.e., you only need n questions]

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