The World's Hardest Logic Puzzle

[MerkW]

I stumbled across a riddle, titled by logician/philosopher George Boolos and conceived by Raymond Smullyan (mathematician/philosopher/logician/professional magician/pianist), which is, allegedly, one of the (if not the # 1) hardest logic puzzles to have existed. If you have heard of it, do not spoil the answer. Try to solve it on your own. Please do refrain from using external sources to find the answer.

Here it is:

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 all questions 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.

* It could be that some god gets asked more than one question (and hence that some god is not asked any question at all).

* What the second question is, and to which god it is put, may depend on the answer to the first question. (And of course similarly for the third question.)
* Whether Random speaks truly or not should be thought of as depending on the flip of a coin hidden in his brain: if the coin comes down heads, he speaks truly; if tails, falsely.
* Random will answer 'da' or 'ja' when asked any yes-no question.

Go for it.

[Economica]

Originally Posted by Merkw

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

Irrelevant nitpick: Did they have to choose two words that mean yes in other languages (Russian and German among others)?

Edit: Do the gods know which of the other two gods is which?

[MerkW]

Originally Posted by Economica

Irrelevant nitpick: Did they have to choose two words that mean yes in other languages (Russian and German among others)?

I'm relatively sure that such was unintentional, yet I did actually notice that myself.

[Athenian200]

Okay, this is one of the few posts I've seen I have no clue on. But here's an idea (or at least a line of thinking) anyway...

If I ask all of the gods a question to which the answer is known to be true, two of them will have the same answer, and I'll know one of the two was Random. However, this wouldn't work because I wouldn't know which of the gods Random had agreed with... True or False. The same would apply for a question to which the answer was known to be false.

If I ask two of the gods a question to which the answer is known to be true, and they both give the same answer, I know that one of them is Random. But if they give different answers, It could be that they are True and False, or that one of them is Random.

But suppose I ask one of the gods all three questions, two to which the answer is true, and one to which it is false. Then any god who gives the same answers for the first two and changes on the last one would most likely be either True or False, although I would be unable to tell which was which. If the God doesn't give the same answer for the first two, or gives the same answer for the last one, I'll know it was Random. But then Random could luck out here too... darn.

If I ask one of the gods two questions, the first known to be true and the second known to be false, and the replies are opposite... then I'm most likely dealing with True or False. If I then pose a question with a true answer to another God, and they give the same answer as the first one did to the true one, I'll know that was Random. If they give the opposite answer... it could be either the other god or Random. Ack!

I can't make this work... I'm missing something. Perhaps I have to ask a specific question that I probably don't even know about because I'm not strongly mathematical or philosophical that would tear this apart?



I'm probably going to feel so ridiculous when someone comes up with the answer now. That's how it always is...

[scantilyclad]

this REALLY makes my head hurt.

[Wandering]

... head... hurts

[Jennifer]

Kill each one of them with The Subtle Knife, then check their IDs in their wallets. (See? Simple!)

[Athenian200]

I'm still thinking about it, unfortunately. I thought I could just let it go, but I can't seem to help but get my mind caught on these until I figure them out...

I think I have to ask a specific question that some how causes the answer to yield more information than a question should be able to yield. Like maybe asking one of them who the other is or something. I can't figure out how to make that work, because I don't see a pattern that would show one of the answers to be more true definitely, but I know I've got to ask a question phrased in such a way that it yields more information than I think it should now.

[disregard]

What Is:
True, False, and Random.
Three questions to ask to any of them.
Da or ja will be their only response.

Ask A, "Is B Truth?".
Ask him once more.
If his answer changes, he is Random.
If his answer doesn't change, ask B, "Is A Random?"

That's all I've got.

[Kiddo]

Originally Posted by Dana

What Is:
True, False, and Random.
Three questions to ask to any of them.
Da or ja will be their only response.

Ask A, "Is B Truth?".
Ask him once more.
If his answer changes, he is Random.
If his answer doesn't change, ask B, "Is A Random?"

That's all I've got.

I thought you had to ask each God only one question?

My proposed solution:

Ask God A if "da" means yes.
Ask God B if God A said (insert God A's answer) Which is a question you know is yes.
Ask God C if God A said (insert God A's answer) Which is a question you know is yes.

I'm too lazy to work it through, but I think if you knew all the possible outcomes for those three questions in advance then the contradictions would yeild the answers to the identities and to the answer of what "ja" and "da" mean. Of course I could be wrong, but there are 8 possible outcomes no matter how you ask it, assuming you are required to ask each God one question.

1. da, da, da
2. da, da, ja
3. da, ja, da
4. da, ja, ja
5. ja, ja, ja
6. ja, ja, da
7. ja, da, ja
8. ja, da, da

Of course, I could be way off, but I challenge anyone to prove it.