Not as easy as 1 2 3
Moderators: jestingrabbit, Moderators General, Prelates

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Re: Not as easy as 1 2 3
If someone asked you whether 2+2=4 if you chose 1, whether 2+2 = 5 if you picked 2 and whether the rein man hypothesis is true if you picked 3, what would you saw.
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 jestingrabbit
 Factoids are just Datas that haven't grown up yet
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Re: Not as easy as 1 2 3
Riemann hypothesis. And you could be talking to someone with a proof/counterexample.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Not as easy as 1 2 3
If we equally distribute that many fans between two sports teams and in a mach between them a team with at least one fan wins, will there be exactly one sad little fan?
PS. Why do you bother putting spoilers on your answers, when it's clear that the point to this is to come up with some wacky answer?
PS. Why do you bother putting spoilers on your answers, when it's clear that the point to this is to come up with some wacky answer?
It's only when you look at an ant through a magnifying glass on a sunny day that you realize how often they burst into flames
Re: Not as easy as 1 2 3
Spoiler:
No complicated maths, something anyone could answer. Also, the "maybe" works as "need further information" rather than "no one knows yet (which the person who you ask might know if he is some sort of oracle)"
EDIT: jestingrabbit: I guess I was unclear with the question. fixed it above.
Last edited by Ankit1010 on Wed Jul 20, 2011 10:17 am UTC, edited 3 times in total.
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
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Re: Not as easy as 1 2 3
@Ankit: Its what you posted on the previous page, and its still not really right. The way I read, the answer for 3 is no, not maybe. If you made it clear that you were fixing some unknown odd number, rather than asking about what happens for all odd numbers, it would be right.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Not as easy as 1 2 3
Didn't go through every spoiler so it may have been posted, but for the original question I'd do it this way:
Spoiler:
 GrueTodayBleenTomorrow
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Re: Not as easy as 1 2 3
Sort of skimmed so I'm not sure if mine has been posted yet:
This question can be used for both the 123 case and the 123456789 case (when asked twice in a row).
The question is kind of a mouthful because I had to make it applicable for both cases (and for two different askings in the second case!).
This answer generalizes. That is, for any n such that your interlocutor is thinking of one of the first 3^{n} natural numbers, you can determine which number it is by asking my question n times.
This question can be used for both the 123 case and the 123456789 case (when asked twice in a row).
Spoiler:
The question is kind of a mouthful because I had to make it applicable for both cases (and for two different askings in the second case!).
This answer generalizes. That is, for any n such that your interlocutor is thinking of one of the first 3^{n} natural numbers, you can determine which number it is by asking my question n times.
“Either the wave function, as given by the Schrödinger equation, is not everything, or it is not right” ~ J.S. Bell
Re: Not as easy as 1 2 3
The way i see this puzzle is that you can always divide the asked group into 3 parts because you can get 3 answers. So for N amount guessables (not sure if english) you will always want to divide the group as evenly to 3 as possible and then come up question that marks each answer group with a group of quessables.
Same can be done with N amount of guessables and it will take x answers if there is N amount of guessables and N=3^x+y where y e {0,1,2}. x= log_3 (NY) There must be a better solution to the number of answers
Spoiler:
Same can be done with N amount of guessables and it will take x answers if there is N amount of guessables and N=3^x+y where y e {0,1,2}. x= log_3 (NY) There must be a better solution to the number of answers

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Re: Not as easy as 1 2 3
I think this is correct, and requires less mathematical analysis on the part of the person guessing the number.
Spoiler:
Re: Not as easy as 1 2 3
The neat idea behind the original problem is that you can get more than 1 bit of information from an ostensibly yes/no question, because there is an additional possible answer (`I don't know'). We can try to take this further, making use of the fact that there are other potential answers, such as `That question makes no sense' (for questions like `Do colourless green ideas sleep furiously?'). In theory, then, we can use a single yes/no question to distinguish between 4 numbers the guard might be thinking of. Here's the best I can do on that front so far:However, we could potentially distinguish between even more options by making use of even more potential answers, like `I know a paradox when I see it, smartypants' for questions such as `Are you going to answer this question by saying `No'?'. So the question arises  how many options can we potentially distinguish with a straightforward yes/no question?
Spoiler:
Re: Not as easy as 1 2 3
sfwc wrote:The neat idea behind the original problem is that you can get more than 1 bit of information from an ostensibly yes/no question, because there is an additional possible answer (`I don't know'). We can try to take this further, making use of the fact that there are other potential answers, such as `That question makes no sense' (for questions like `Do colourless green ideas sleep furiously?'). In theory, then, we can use a single yes/no question to distinguish between 4 numbers the guard might be thinking of. Here's the best I can do on that front so far:However, we could potentially distinguish between even more options by making use of even more potential answers, like `I know a paradox when I see it, smartypants' for questions such as `Are you going to answer this question by saying `No'?'. So the question arises  how many options can we potentially distinguish with a straightforward yes/no question?Spoiler:
Indeed. The idea is the very basis of quantum computing. Quantum bits can be a 1, a 0 or anything in between.
Spoiler:
Re: Not as easy as 1 2 3
zemerick wrote:Indeed. The idea is the very basis of quantum computing. Quantum bits can be a 1, a 0 or anything in between.
That's a slightly simplified way of putting it  anyone interested in the question of what values a quantum bit can take should look here, though I guess it is subtle enough that you need at least a slight understanding of quantum mechanics to understand the explanation there.
zemerick wrote:your example question does not take the individual into account.
I agree. It was the best I could think of off the cuff, but it would be good to have something more clearly nonsensical in the nonsense case and less ambiguous in the yes case.
Re: Not as easy as 1 2 3
A new solution:
Spoiler:
Re: Not as easy as 1 2 3
New nonmath answer that con be used to solve either the 3 by one question or 9 by two question:
Spoiler:
Disclaimer: I am not a scientist.
 TheDancingFox
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Re: Not as easy as 1 2 3
So, I've got a challenge for all you folks then. The 2 questions, 9 numbers form has been answered a dozen times, and pretty much entirely by the same recursive setup.
Could you figure out the same puzzle: You get 2 questions with answers yes/no/I don't know to guess which of 9 questions the man has, however, you must ask both questions together. Which is to say, you must have already given the second question before you receive the answer to the first question.
I would figure there's some way which is simple enough to put it together out of the same recursive 3 and 3 and 3 setup, just by using formulas instead of specifics for the second question, but I'm not positive.
For an extra layer of difficulty, can you do it such that the second question is not influenced by the answer to the first question, ie, they could be answered in either order and you would still be able to produce the correct number.
Which means no
Could you figure out the same puzzle: You get 2 questions with answers yes/no/I don't know to guess which of 9 questions the man has, however, you must ask both questions together. Which is to say, you must have already given the second question before you receive the answer to the first question.
I would figure there's some way which is simple enough to put it together out of the same recursive 3 and 3 and 3 setup, just by using formulas instead of specifics for the second question, but I'm not positive.
For an extra layer of difficulty, can you do it such that the second question is not influenced by the answer to the first question, ie, they could be answered in either order and you would still be able to produce the correct number.
Which means no
Spoiler:
 phlip
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Re: Not as easy as 1 2 3
I'm pretty sure there's been some solutions given that can already solve that variation... but anyway:
Spoiler:
Code: Select all
enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};
void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}
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