Not as easy as 1 2 3

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blademan9999
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Re: Not as easy as 1 2 3

Postby blademan9999 » Tue Jul 12, 2011 6:59 am UTC

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
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Re: Not as easy as 1 2 3

Postby jestingrabbit » Tue Jul 12, 2011 9:12 am UTC

Riemann hypothesis. And you could be talking to someone with a proof/counterexample.
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Re: Not as easy as 1 2 3

Postby wraith » Tue Jul 12, 2011 9:36 am UTC

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?
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Ankit1010
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Re: Not as easy as 1 2 3

Postby Ankit1010 » Tue Jul 12, 2011 9:58 am UTC

Spoiler:
"Does the number divide a random odd number?"
1 - Yes
2 - No
3 - Maybe

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.

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jestingrabbit
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Re: Not as easy as 1 2 3

Postby jestingrabbit » Tue Jul 12, 2011 11:03 am UTC

@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.
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Tunga
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Re: Not as easy as 1 2 3

Postby Tunga » Wed Jul 20, 2011 8:33 am UTC

Didn't go through every spoiler so it may have been posted, but for the original question I'd do it this way:

Spoiler:
If we call your number x, is 1/(x-2) positive?

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GrueTodayBleenTomorrow
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Re: Not as easy as 1 2 3

Postby GrueTodayBleenTomorrow » Thu Jul 21, 2011 5:01 am UTC

Sort of skimmed so I'm not sure if mine has been posted yet:

This question can be used for both the 1-2-3 case and the 1-2-3-4-5-6-7-8-9 case (when asked twice in a row).
Spoiler:
Take all the numbers that I have yet to determine that you are not thinking of. List the numbers in order from least to greatest. Let us partition the list into three equal parts by drawing a line one third of the way down the list and another two-thirds of the way down the list. Name the upper most sub-list "1" the one in the middle "2" and the one at the bottom "3". Behind my back, I am holding up a positive number of my fingers. Is the number of the sub-list containing your number neither divisible by two nor greater than this number?


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 3n natural numbers, you can determine which number it is by asking my question n times.
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kalakuja
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Re: Not as easy as 1 2 3

Postby kalakuja » Thu Jul 28, 2011 6:29 pm UTC

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.
Spoiler:
For me the easiest way to mark the groups was to use probability type question. For 1,2,3 you will put them into a bag and then ask if the drawed number is larger than numbers i have drawn which are 1.1 or 2.1 and thus get the answer.
So for the 1-9 you can divide it to 3 groups and then into 3 again from the answer. Like i put them in the bag and i got 2 numbers i have drawn 4.1 and 6.1...

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 (N-Y) There must be a better solution to the number of answers :P

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Re: Not as easy as 1 2 3

Postby rocketjock » Fri Jul 29, 2011 6:51 am UTC

I think this is correct, and requires less mathematical analysis on the part of the person guessing the number.

Spoiler:
If I randomly place your number (1, 2, or 3) of points on the circumference of a circle , is it possible to draw a single straight line connecting all of your points and the center of the circle?

Yes would be 1, No would be 3, and I don't know would be 2

DavidSpencer
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Re: Not as easy as 1 2 3

Postby DavidSpencer » Wed Aug 03, 2011 11:37 am UTC

Spoiler:
If someone asked the question "Did Italy win World War <n>" 100 years from today, would the answer be "yes"?

sfwc
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Re: Not as easy as 1 2 3

Postby sfwc » Wed Aug 03, 2011 4:26 pm UTC

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:
Spoiler:
Question: Suppose I started with a bag with your number of balls in it and took out either 2 or 3 balls. Would I be left with an empty bag?

Answers:
If the number is 1 - That question makes no sense
If the number is 2 - Yes
If the number is 3 - I don't know
If the number is 4 - No
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?

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zemerick
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Re: Not as easy as 1 2 3

Postby zemerick » Wed Aug 03, 2011 6:09 pm UTC

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:
Spoiler:
Question: Suppose I started with a bag with your number of balls in it and took out either 2 or 3 balls. Would I be left with an empty bag?

Answers:
If the number is 1 - That question makes no sense
If the number is 2 - Yes
If the number is 3 - I don't know
If the number is 4 - No
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?


Indeed. The idea is the very basis of quantum computing. Quantum bits can be a 1, a 0 or anything in between.
Spoiler:
Current quantum computers mostly just have 3 bits, just the same as the initial answers only have 3. It's much easier to have a yes, no, maybe or 0,1, unknown/both than to add more in there, so they tend to come first.

An example though of the complexity is that your example question does not take the individual into account. Personally, if I didn't spend too much time thinking about it, I would just answer:

1 - Yes ( There was one ball, and you tried to take 2 or 3 out )
2- yes
3 - maybe
4 - no

while someone else might have answered:

1 - doesn't make sense
2 - might not make sense ( what happens if you try to remove 3 of the 2 balls? )
3 - maybe
4 - no

In a way, you have yet another possible answer...but it's more likely to come out as either "doesn't make sense" or "maybe" rather than a usable answer. You could probably add a series of qualifiers onto your initial statement of the question to explain the rules to answering and possible answers so that you get only usable answers. It would probably end up reading like 'legalese', but the precision will be needed for greater than 3 answers to a yes/no question.

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Re: Not as easy as 1 2 3

Postby sfwc » Wed Aug 03, 2011 6:58 pm UTC

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.

mward
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Re: Not as easy as 1 2 3

Postby mward » Tue Aug 09, 2011 1:43 pm UTC

A new solution:
Spoiler:
I am going to toss three coins. My score will be 1.5 plus 0.5 per head. Will my score beat your number?

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Re: Not as easy as 1 2 3

Postby wannabe » Tue Aug 09, 2011 7:08 pm UTC

New non-math answer that con be used to solve either the 3 by one question or 9 by two question:

Spoiler:
set up the question this way...

1=blue
2=red
3=red or blue

is your number blue?

yes = 1
no = 2
IDK = 3

the same can be used by breaking the group of nine into three groups and assigning a color to each group, then follow the above.
Disclaimer: I am not a scientist.

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TheDancingFox
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Re: Not as easy as 1 2 3

Postby TheDancingFox » Sat Aug 13, 2011 8:57 am UTC

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
Spoiler:
"I'm thinking of 1 or 2, is your number higher than mine, and if you answered I don't know on the first question add three, and if you answered no on the first question add six" second question.

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phlip
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Re: Not as easy as 1 2 3

Postby phlip » Sat Aug 13, 2011 11:02 am UTC

I'm pretty sure there's been some solutions given that can already solve that variation... but anyway:
Spoiler:
I have written down two lists of numbers. List A includes the numbers 1, 2 and 3, and may or may not include the numbers 4, 5 and 6 (but it definitely does not include the numbers 7, 8 or 9). List B includes the numbers 1, 4 and 7, and may or may not include the numbers 2, 5 and 8 (but it definitely does not include the numbers 3, 6 or 9).
(1) Is your number on list A?
(2) Is your number on list B?

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