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The Hardest Puzzle of All Time.

 
 
markr
 
  1  
Reply Sat 14 Oct, 2006 08:44 pm
stuh505:

I identified a non-random person with one question.
I determined the meaning of ja/da with one question.

gravenewworld:

I'm quite familiar with iff. Iff is true when both are true or both are false. It is not false when both are false.
0 Replies
 
gravenewworld
 
  1  
Reply Sat 14 Oct, 2006 09:55 pm
Quote:
'm quite familiar with iff. Iff is true when both are true or both are false. It is not false when both are false.


Correct. I missed typed.
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gravenewworld
 
  1  
Reply Fri 20 Oct, 2006 06:15 pm
0 Replies
 
Tryagain
 
  1  
Reply Sat 21 Oct, 2006 12:41 pm
I had a feeling I would not like the answer, or even the basis on which the answer is based. Take the first:


PUZZLE 1:

Noting their locations, I place two aces and a jack face down on a table, in a row; you do not see which card is placed where. Your problem is to point to one of the three cards and then ask me a single yes-no question, from the answer to which you can, with certainty, identify one of the three cards as an ace. If you have pointed to one of the aces, I will answer your question truthfully. However, if you have pointed to the jack, I will answer your question yes or no, completely at random.


SOLUTION TO PUZZLE 1:

POINT TO THE middle card and ask, "Is the left card an ace?" If I answer yes, choose the left card; if I answer no, choose the right card. Whether the middle card is an ace or not, you are certain to find an ace by choosing the left card if you hear me say yes and choosing the right card if you hear no. The reason is that if the middle card is an ace, my answer is truthful, and so the left card is an ace if I say yes, and the right card is an ace if I say no. But if the middle card is the Jack, then both of the other cards are aces, and so again the left card is an ace if I say yes (so is the right card but that is now irrelevant), and the right card is an ace if I say no (as is the left card, again irrelevantly).



Now, put it into the context of T/L/R + Da/Ja and come up with a single question to produce the required result. In my humble opinion, a total impossibility.

I wait to be amazed.
0 Replies
 
Tryagain
 
  1  
Reply Mon 23 Oct, 2006 10:54 am
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darkgirl
 
  1  
Reply Tue 24 Oct, 2006 10:03 am
there is another way to do it.

ask a simple question to god a such as, "is 2+2=5?" let's say he answers "ja". go to god b and ask him " does ja mean ja or da?" if he says ja then you no god b must be the one that tell the truth or the one that some times tells the truth. or if he said da then you know he is the one that lies or some times lies. based on the answer he gives you would be able to determine wuther it ja means yes or no and if da means yes or no.considering that it is obvious that that if the first one says ja then it would mean no because the first one is false. any way the question you would have to ask the last one is " are you the one that tells lies sometimes?" if he says yes then he would be telling the truth. so it would not be the one lies and infact the one who lies sometimes. there your problem is solved.

bye bye
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gravenewworld
 
  1  
Reply Tue 24 Oct, 2006 07:52 pm
only some of what you wrote makes sense to me. I still don't see how you can determine da/ja to be yes/no from the questions you ask. from the first two questions you are proposing to ask, i see no way how you can determine da/ja to be yes or no
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gravenewworld
 
  1  
Reply Tue 7 Nov, 2006 07:47 pm
Well here is what you all have been waiting for, the solution to the hardest logic puzzle of all time- http://people.ucsc.edu/~jburke/three_gods.pdf


When I said the puzzle was hard I wasn't kidding.
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markr
 
  1  
Reply Tue 7 Nov, 2006 08:30 pm
Looks like I was on the right track with the first two questions. I guess I should have stuck with it.
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